There are two questions one has to answer when asking ‘will $\ce{X}$ react with $\ce{Y}$?’

Thermodynamics

This basically asks ‘will the products of a hypothetical reaction be more stable than the reactants?’, i.e. whether the entire process will release energy. A quick web search tells me that $\Delta_\mathrm{f} H^0 (\ce{TiN}) = - 337.65~\mathrm{\frac{kJ}{mol}}$ and therefore the reaction is thermodynamically possible.
Kinetics

This is asking ‘can I overcome the activation energy required for said hypothetical reaction to occur?’ And this is where it gets complicated.

The process of titanium burning can be split into the following sub-equations:

$$\ce{2 Ti + N2 -> 2 TiN} \\
~\\
\ce{Ti(s) -> Ti(g)}\\
\ce{Ti(g) -> Ti^3+ (g) + 3 e-}\\
\ce{N2(g) -> 2 N(g)}\\
\ce{N(g) + 3 e- -> N^3- (g)}\\
\ce{Ti^3+ (g) + N^3- (g) -> TiN(g)}\\
\ce{TiN(g) -> TiN(s)}$$

Broken down in this way, we should immediately realise:

The first step is generally okay-ish but may require some activation. (We don’t need to make the entire material gaseous in one step; an atom a time is enough.)
The second step requires energy (no surprises there).
The third step is extremely hard to do, since the $\ce{N#N}$ is so strong.
The fourth step also requires energy since nitrogen has an endothermic electron affinity.
The fifth and sixth steps are the ones where huge amounts of energy are liberated; especially the sixth one where lattice enthalpy is liberated.

Now if we consider a block of titanium, as DavePhD mentioned, the the first step is already rather hard to do. The block’s crystal lattice is rather stable as is so we cannot easily extract atoms from there. However, the larger the surface to volume ratio gets the easier this can happen, especially since we have multiple ‘attacking’ points at the same time. Thus, solid metal blocks are usually hard to burn, while metal powder often burns just by putting it in pure oxygen. (Iron wool doesn’t even need to be a powder, the surface to volume ratio is small enough already to oxidise in pure oxygen at room temperature.)

As Jon Custer mentioned in a comment and as I mentioned earlier, the cleavage of dinitrogen molecules is also extremely difficult. It will take a lot of activation energy to do so. This means, that a lot of energy must be gained in the formation of the ion lattice so that at least the dinitrogen cleavage can be overcome and combustion can continue.

Thus, the kinetics part of the question can be answered with: ‘it can, but.’

The question remains how to supply activation energy. Since we are most likely dealing with powder, it could be too slow to ignite titanium in air and then immediately immerse in nitrogen. Since nitrogen cleaving is so hard, many other ignition ideas that rely on, say, a match’s flame won’t work (the match is too weak). Maybe the activation energy is ‘low’ enough that heating to a few hundred degres will suffice. Unfortunately, I do not know the experimental details.


Compressibility factor (Z) of a real gas

From the given data, the molar volume $V_\mathrm m$ is $$V_\mathrm m=\frac{150\ \mathrm{cm^3}}{3\ \mathrm{mol}}=50\ \mathrm{cm^3/mol}=0.05\ \mathrm{L/mol}$$

In terms of molar volume, the van der Waals equation reads:
$$\left(p+\frac{a}{V_\mathrm m^2}\right)(V_\mathrm m-b)=RT$$

So, solving for the pressure gives:
$$p=\frac{RT}{V_\mathrm m-b}-\frac{a}{V_\mathrm m^2}$$

So, to get the pressure you don't need to know the compressibility factor.  Which of these two terms dominates numerically in determining the pressure?


The quantity atomic mass (quantity symbol: $m_\mathrm{a}$) is defined as rest mass of a neutral atom in the ground state.
The dimension of the atomic mass is
$$\dim m_\mathrm{a} = \mathsf{M}$$
The coherent SI unit for atomic mass is ‘kilogram’ (unit symbol: $\mathrm{kg}$).

The quantity relative atomic mass (quantity symbol: $A_\mathrm{r}$) is defined as the ratio of the average mass per atom of an element to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$; i.e. the relative atomic mass is the ratio of the mass of an atom to the unified atomic mass constant:
$$A_\mathrm{r} = m_\mathrm{a}/m_\mathrm{u}$$
where $m_\mathrm{a}$ is the atomic mass and $m_\mathrm{u}$ is the unified atomic mass constant.
The relative atomic mass is a quantity of dimension one (for historical reasons, a quantity of dimension one is often called dimensionless):
$$\dim A_\mathrm{r} = 1$$
The coherent SI unit for relative atomic mass is the unit one (symbol: $1$).
For historical reasons, the IUPAC accepts the use of the special name ‘atomic weight’ for the quantity relative atomic mass. The use of this traditional name is deprecated.

The value of the unified atomic mass constant (symbol: $m_\mathrm{u}$) is defined as 1/12 of the mass of a neutral atom of the nuclide $\ce{^12C}$ in the ground state at rest.
$$m_\mathrm{u} = \frac{m_\mathrm{a}(\ce{^12C})}{12}$$
The recommended value is
$$m_\mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$ (source)
Note that the value in SI units is obtained experimentally.

The unit dalton (unit symbol: $\mathrm{Da}$) and the unified atomic
mass unit (unit symbol: $\mathrm{u}$) are non-SI units that are accepted for use with the SI. Actually, ‘dalton’ and ‘unified atomic mass unit’ are alternative names for the same unit, equal to 1/12 times the mass of a free carbon-12 atom, at rest and in its ground state, i.e.

$$1\ \mathrm{Da} = 1\ \mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$

Thus, the value of the unified atomic mass constant is, by definition, equal to one dalton or one unified atomic mass unit:

$$m_\mathrm{u} = 1\ \mathrm{Da} = 1\ \mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$

The old relative atomic masses (unfortunately called ‘atomic weights’) and the corresponding atomic mass unit (amu) were originally referred to the relative atomic mass of oxygen, which was taken as 16. However, physicists used the atomic mass of the nuclide $\ce{^16O}$ whereas chemists used the average atomic mass of natural oxygen. This unit became obsolete when IUPAP (1960), IUPAC (1961), ISO, CIPM (1967) and CGPM (1971) agreed to assign the value 12 to the relative atomic mass of the nuclide $\ce{^12C}$.


Why conductivity of electrolyte decreases on dilution, but molar conductivity increases?

I've just been doing an experiment on this, and my understanding's not that great, but as I understand it the conductivity decreases because you have a lower concentration of charge carriers in the solution. The molar conductivity increases however, because as the charged ions get further apart they interact and slow each other down less. Essentially each ion becomes more able to move through the solution, but the increase in conductivity from this effect is outweighed by the decrease due to there being fewer ions in a given volume.

Personally I find it easier to think about the other way around. As you increase the concentration of the solution you get more charge carrying ions, so the conductivity increases. But because the ions get closer together they begin to interact and slow each other down so the conductivity per mole of the ions is reduced.

I hope you still find this useful and that I managed to explain what I meant


At what temperature can molecules no longer form?

The bond will break if you can put more than its dissociation energy into it; $\ce{CO}$ has a huge dissociation energy of $\pu{1080 kJ/mol}$, $\ce{I2}$ less than most and is only $\pu{151 kJ/mol}$. This energy can be easily obtained using a laser of the correct wavelength. (You will need to convert units to electron volts to get wavelength needed. $\pu{1 kJ/mol} = \pu{1.036E-2 eV}$, $\pu{1eV} = \pu{8065 cm^-1}$).

However, when we excite using a laser we only use only a small range of energy, using heat at a given temperature is different as temperature is an average measure of a distribution of energies given by the Boltzmann expression $n_i \approx \exp(-E_i/k_\mathrm{B}T)$ where $n_i$ is the number of molecules with energy $E_i$ at temperature $T$, $k_\mathrm{B}$ is the Boltzmann constant. If  dealing with molar units replace $k_\mathrm{B}$ by $R$, the gas constant.

Thus, even at room temperature ($\pu{300 K}$),  there are, in principle, some molecules of $\ce{CO}$ that will be decomposed  even though the dissociation energy is so large; the chance of being decomposed is $\exp(-1080000/(300\times8.314) \approx 10^{-190}$. It is utterly minute even for $6\cdot 10^{23}$ molecules in a mole, but for iodine $\approx 10^{-27}$, still minute so these molecules are stable at room temperature. At $\pu{1000 K}$, 1 in $\approx 10^8$ $\ce{I2}$ molecule decompose and $10^{-4}$ at $\pu{200 K}$ so this is appreciable over a period of time.

(Strictly we should integrate from $T$ to infinity to get the exact answer but the exponential decays so quickly just one value gives us a good idea)


Is Chemical Engineering a good pathway to Biomedical Engineering for graduate school?

If you want to get into a biomedical engineering program in graduate school, chemical engineering is not a bad option. You would of course probably have better luck if you majored in biomedical as an undergrad as well, since your coursework would more directly prepare you for those programs.

A couple of things to note though:

There is a (big) difference between bio-medical engineering and bio-chemical engineering. Biomedical engineering is more closely related to pre-med and materials science than to biochemistry. That said, bio chem courses probably wouldn't hurt you, but if you have the option, taking real biomedical engineering courses as electives (or as a minor) would be better. For bio-chemical engineering, these days regular chem E programs almost always have at least an elective component in biochem engineering, and often times departments are called "Chemical and Biological Engineering" due to the similarities. In other words, chemical engineering undergrad degrees are very well suited (and possibly over-prepare) you for biochemical engineering graduate work.

The other thing to know is: the real secret to getting into a good graduate program is to get research experience as an undergrad in the field that you want to work in. This demonstrates to the acceptance committee that you meet the two main requirements PIs have for grad students:

You are interested in the subject.
You have the aptitude to actually accomplish something.

A chemical engineering degree demonstrates aptitude on its own, and as @rch mentioned, is a pretty good "fall back" plan in case you change your mind later (although chem E is a tough major if it's just a fall back plan - I would not recommend it unless you really like chemistry and physics and could imagine working in a chemical plant, since that is likely what you will do if you "fall back" on it).

The problem is going to be that lots of other people will be able to demonstrate aptitude, with degrees in biomedical engineering, materials engineering, chemical engineering (lots of them get into biomedical later), pre-med, and other stuff that is just as well suited. In addition to aptitude, a lot of them (if not most of them) will also have research experience through undergrad projects or internships. To be competitive with them, you will want some as well.

The best easiest way to get research experience as an undergrad is to find a professor who is working on something you are interested in, and volunteer. Many times they will have paid positions specifically for undergrads, but if you can put in volunteer hours and are reasonably good at communicating, managing your time, and getting things done, most professors would be happy to put you to work for free. If you can manage to get a publication before you finish your undergrad degree, you will have an excellent shot at getting into the school of your choice, provided your GRE scores and grades are good (another reason to carefully consider chem E - not an easy major).

Another option is internships, although a research internship carries more weight in grad school circles than a regular old "gets the coffee and fills out spreadsheets" internship. They are much harder to get, but usually pay pretty well, so it is definitely worth trying for one.

The main point I am trying to make here is - major in (almost) whatever you want, but if you want to go to grad school in a particular field, get research experience in that field as an undergrad. Go for paid work first, but be prepared to volunteer if you have to, at least for a semester. You can quickly turn a volunteer position into a paid one if you do good work. Work experience also goes a long way toward getting you a job if you don't end up going to grad school. I'd put your odds of getting a good job with a biomedical engineering undergrad degree with experience at higher than getting a job with a chemical engineering degree without experience, especially these days.

Also, just so you don't think I'm trying to talk you out of a Chem E degree, Chem E was my major (BS, MS, and PhD) and I really liked it. If I didn't like chemistry and physics so much though, I probably wouldn't have made it through.


How is the polyethylene being attacked?

Polyethylene is attacked by microbes that contain oxygenase or oxidase enzymes.  These enzymes can attack C-H bonds even at moderate temperatures.

  The only agents that I have heard of that can degrade a fully-saturated hydrocarbon at room temperature are fluorine (and related nasties), other halogens (with UV light), magic acid (which is extremely reactive with almost anything).

One prominent example of a hydrocarbon degrading enzyme is methane monooxygenase, which is capable of catalyzing methane oxidation to methanol even at temperatures as low as 0 °C.  The C-H bonds of polyethylene are about as reactive as methane's.  (Actually, they're slightly more reactive.)  Thus enzymes can do the job.  They don't need elemental fluorine or magic acid.

However, your point about halogens reacting with hydrocarbons is very interesting.  Several research groups have reported that adding halide salts to aqueous (and aerobic) suspensions of bacteria accelerates the rate of oxidation.

It's possible that this finding can be explained by the involvement of haloperoxidase enzymes in the degradation.  However, the mechanism of microbial polyethylene degradation is a very new, and very active, area of ongoing research.


Can we recycle garbage with the principles of mass spectrometry?

Yes it is possible, but is very expensive and would be orders of magnitude more costly than what people are willing to pay for recycled materials.  Let me give two data points to explain why it is expensive.

Some commercial material is manufactured by mass spectrometry.  The starting material isn't garbage, but reasonably chemically purified metals of various types.  The product(s) are isotopically separated metals.  Iron 57 ($\ce{^{57}Fe}$) is an isotope that is naturally present in terrestrial iron supplies at an abundance of about 2%.  It is useful to researchers that study iron materials using various spectroscopic techniques.  Pure iron 57 costs hundreds of dollars per milligram.
Mass spectrometry requires very high voltage, a very high vacuum, and large amounts of purified gases; additionally, the throughput of material separated by mass is very low.  I think that for commercial isotope purification, magnetic sector-type instruments are used.  I'm not too familiar with them, so let me go through some rough numbers for a different type of mass spectrometry: time-of-flight: Several thousand volts are applied to packets of ions which leads to their separation. Total ion currents are as high as say $10^9$ particles per second, which corresponds to about 0.1 nanoamperes or $10^{-15}$ moles per second.  So to purify one mole of iron would require $10^{15}$ seconds, or several million years of continuous operation of very expensive machinery!  Even if other types of mass spectrometry do better than time-of-flight by three orders of magnitude, it will still be cost-prohibitive.$%edit$


The relative acidities of these molecules in the aqueous solution$^{[1][2]}$ is:

$$\text{methanol > water > ethanol > isopropanol > t-butanol}$$

The relative acidities of these molecules in DMSO$^{[3]}$ is:

$$\text{methanol > ethanol > isopropanol > t-butanol > water}$$

The problem with only considering the acidities of these molecules in aqueous solution is that it assumes only one explanation for acidity. The inversion of the order of acidity in DMSO indicates that there are multiple factors at play.

The acidity trend in DMSO is explained by polarizability and the anionic hyperonjugation effect$^{[4]}$ of the conjugate bases of the molecules. The larger size of the alkyl substituents in the conjugate bases allows for increased distribution of negative charge over a larger volume. This reduces the charge density, and thereby the Coulombic repulsion between the conjugate base and the solvent. In DMSO, water is much less acidic than even methanol, which is also consistent with the relative polarizability of a methyl group and a hydrogen atom, but fails to explain the relative acidity of t-butanol and methanol.

The trend in alcohol acidity is explained by anionic hyperconjugation, which is weakened in larger alkyl substituents.

  For methanol and the methoxide anion, the primary interaction occurs between a filled lone-pair ($n$) orbital on oxygen and an unoccupied $\pi^*_\ce{Me}$ orbital of the methyl group, resulting in a two-electron stabilizing effect.

  $\hspace{3.8cm}$

  Deprotonation of the alcohol leads to an increase in the energy of the n orbitals on oxygen, and a decreased energy separation between the interacting orbitals $n$ and $\pi^*_\ce{Me}$, and hence a greater stabilizing interaction$^{[5]}$.

Although DMSO does solvate the solute to an extent, because is an aprotic solvent, it is incapable of hydrogen bonding with the solute. This allows DMSO acidities better reflect the intrinsic acidities of the molecules.

In aqueous solution, however, the conjugate bases are stabilized by hydrogen bonding. Because smaller ions are better solvated by water, water's acidity is increased more than methanol's, but because methanol is intrinsically more acidic due to its polarizability, it is still slightly more acidic than water, even in aqueous solution. As the size of the alkyl substituent increases, the alcohol becomes weaker due to the decreasing ability of water to solvate the conjugate base. This has a greater affect on the alcohol's acidity than the dispersion of the negative charge, and overrides the the intrinsic acidity, inverting the order of acidity.

$^{[1]}$ Evans $\mathrm{p}K_\mathrm{a}$ Table, Alcohols
$^{[2]}$ Wikipedia, Ethanol
$^{[3]}$ Bordwell $\mathrm{p}K_\mathrm{a}$ Table, Alcohols
$^{[4]}$ ACS, Acidities of Water and Methanol in Aqueous Solution and DMSO
$^{[5]}$ ACS, Does a Methyl Substituent Stabilize or Destabilize Anions?


Does liquid ammonia react with hydrogen gas?

No, it does not. I do not know what they described but I am pretty sure liq. ammonia doe snot react with hydrogen gas. I believe they use hydrogen gas as a carrier but I found that to be questionable as well, since hydrogen in residential setting can be pretty dangerous.


If the vapor (i.e. gas) phase above the solution can be treated as an ideal gas, then yes, if two species have the same partial pressure, then they have the same mole fraction.

For an ideal gas mixture, $p_i = y_i P$, where $p_i$ is the partial pressure of species $i$ and $y_i$ is the mole fraction of $i$ in the vapor phase, and $P$ is the total pressure.

The ideal gas assumption will be very accurate in many cases commonly encountered in chemistry, but not always.  Binary mixtures of $\ce{CO2}$ and another compound at high pressure would be one example of vapor phases that are not ideal.  Concentrated vapor of aliphatic carboxylic acids is also not very ideal.


Replacement for optical calcite (Glan Thompson prism)?

I have a pair of Glan Thompson polarizers (Karl Lambrecht) that cost a pretty penny many years ago. I also have several other (less expensive) pairs of calcite polarizers. Nothing to date can match high quality calcite in terms of low scattering, low fluorescence and low extinction ratio. In some cases, with the highest quality calcite devices, it is possible to find 'sweet spots' having $10^{-10}$ extinction ratio. (The best I ever got, after hours in a dark room, was about $8 x 10^{-9}$). In contrast, the plastic sheet polarizers are nice and convenient, but nowhere near good enough for serious polarimetry work. Sodium nitrate crystals might have excellent optical properties (I do not know anything about them), but would they need to be protected from water vapor in the air? After all, potassium nitrate is used in gunpowder and I suspect one reason is that it 'keeps the powder dry'!


Is this amide compound electrophilic enough to be attacked by a cyclopentadiene in base?

Although I have no prove for this, I am almost completely certain, that this will not work.

If you look closely at the cyclopentadienyl anion, what you would like to utilise as the nucleophile, you will notice that it is and aromatic system. (It even completely obeys the Hückel rules.) There are no electrons in an orbital of suitable symmetry, that would allow attacking the π* C-O bond.
In the other molecule, as you have noted, the electrons of the nitrogen will be interacting with the π carbonyl bond. Moreover, this part of the π system is also overlapping with the aromatic system of the phenyl moiety. That will certainly remove quite a bit of charge density from the carbonyl carbon. A nucleophilic attack at that position seems highly unlikely.


How can I find out the concentration of chlorine in an acid bath?

The addition of chlorine to the bath will produce hydrochloric acid.  If the chloride ion concentration in the bath sample could be determined, then you could back-calculate to the amount of hydrochloric acid added.  The standard method to determine the amount of chloride ion in an aqueous sample is to precipitate it by adding silver nitrate solution followed by collecting the precipitate by filtration and then drying and weighing it.  You could use that approach here because silver nitrate and silver fluoride (which would be created from silver nitrate and the hydrofluoric acid) are soluble* in water, so only the silver chloride would precipitate if silver nitrate were added to your acid bath sample.  Here's a link to an actual gravimetric procedure for determining chloride ion content.

Note: silver fluoride is soluble to the extent of 1.8 kg/L, therefor silver fluoride should not precipitate upon addition of silver nitrate to your chlorine-contaminated sample because the hydrofluoric acid content of the bath was only 4-5%.


How to properly dispose of sodium...or store it properly

If you handle sodium, the first thing you need is a way to extinguish it if it catches on fire. You should have a bucket of sand ready to use, and additionally a class D fire extinguisher, if available. I'm assuming you know this, but don't let the sodium come into contact with water.

You always want the sodium stored in an inert liquid, usually mineral oil is used for that. Organic solvents like n-hexane are also used for short time periods. You should refill the mineral oil immediately.

The usual way to neutralize sodium is to put small pieces in an organic solvent like n-hexane and slowly add either isopropyl alcohol or ethanol to it. Isopropyl alcohol reacts slower, and is therefore a bit safer. When all the sodium is dissolved, you can add water to make sure there is no elementary sodium left.

How dangerous this is depends a lot on the scale, it's easy to do if you are only getting rid of a bit of sodium you haven't used in your reaction. I have no experience with neutralizing large amounts of sodium, this method should still work but it might take a long time if you are careful as you'll have to do it in small batches.

To be honest, the fact that you asked this question here on the internet indicates to me that you should at least let someone more experienced supervise the procedure.


The multiplying affixes are given by Table 11 of the Blue Book, referenced in R-0.1.4 and R-4.1:
Number  Numerical term | Number  Numerical term
1       mono-          | 90      nonaconta-
2       di-            | 100     hecta-
3       tri-           | 200     dicta-
4       tetra-         | 300     tricta-
5       penta-         | 400     tetracta-
6       hexa-          | 500     pentacta-
7       hepta-         | 600     hexacta-
8       octa-          | 700     heptacta-
9       nona-          | 800     octacta-
10      deca-          | 900     nonacta-
11      undeca-        | 1000    kilia-
12      dodeca-        | 2000    dilia-
20      icosa-         | 3000    trilia-
30      triaconta-     | 4000    tetralia-
40      tetraconta-    | 5000    pentalia-
50      pentaconta-    | 6000    hexalia-
60      hexaconta-     | 7000    heptalia-
70      heptaconta-    | 8000    octalia-
80      octaconta-     | 9000    nonalia-

Therefore, the answer to your question is 2,4,6,8-tetraene.

Extra information
Note that there is no elision of vowels, specified by R-0.1.7:

R-0.1.7.1 Vowels are systematically elided as follows:
(a) the terminal "e" in names of parent hydrides when followed by a suffix beginning with "a", "i", "o", "u", or "y";
(b) in the Hantzsch-Widman system, the final "a" of an element prefix when followed by a vowel;
(c) the terminal "a" in the names of numerical multiplicative affixes when followed by a suffix beginning with "a" or "o", or a Hantzsch-Widman prefix or stem beginning with a vowel;
(d) the terminal "a" of an element prefix in "ababa" repeating unit names (see R-2.3.3.2) and the terminal "o" of a replacement infix when followed by a vowel;

The relevant part is (c), which states that the final "a" in the above affixes are elided when the next letter is "a" or "o", but in this case the next letter is "e" so it is not elided.
The following are given examples of elision:

benzenehexaol > benzenehexol
[1,1'-binaphthalene]-3,3',4,4'-tetraamine > [1,1'-binaphthalene]-3,3',4,4'-tetramine
1,3,5,7-tetraoxaocane > 1,3,5,7-tetraoxocane


Can molecule have just h-bonds between its parts

There are some grey areas in the normal usage of the word 'molecule'
The normal usage of the term molecule is accurately described by your quote from the Khan Academy. The majority of molecules chemists talk about are groups of atoms joined with covalent bonds. But, like many things in chemistry, there are some grey areas and the normal usage gets bent a little.
One of the major areas where the terminology gets fuzzy is in polymers. While on a strict interpretation of the normal definition a single polymer strand (be it DNA or a single strand of polyethylene) is a molecule the terminology is not particularly helpful as the key things that matter in bulk polymer properties are not easily reduced to the properties of individual 'molecules' so the very idea of molecules is less relevant. In polyethylene (a very simple polymer) there will not be a mix of identical molecules but, rather, a wide range of very similar molecules (even for linear polyethylene there will be a wide range of chain lengths (say from 1,000 to 10,000 carbons). So polymer chemists tend not to talk about molecules much.
DNA is a polymer but, by its nature, one where the structure of a given strand is constant (and can be precisely replicated). We do tend to say DNA molecule at least in casual conversation despite the fact that the two strands are connected only by hydrogen bonds. But there are good reasons why this is not an abuse of the normal definition of molecule. One is that many hydrogen bonds along a polymeric chain add up to a strong overall bond even though the individual bonds are merely hydrogen bonds. But the relative weakness of each bond is a key part of the process of replicating DNA as the two strands can be easily unzipped by the machinery in the cell nucleus. You could argue that a DNA strand consists of two molecules using the normal definition but this doesn't help much and, in normal discussion, we talk about both a DNA molecule with two hydrogen-bonded strands and a DNA strand as a molecule. This mild abuse of a strict definition of "molecule" doesn't matter much as it doesn't interfere with any understanding of how DNA works.
Mostly, though, chemists don't consider hydrogen-bonded units as "molecules" even when the bonds have significant effects on the bulk properties of the substance. The physical properties of, for example, acetic acid are heavily influenced by hydrogen bonding. In solution molecules often form pairs (joined by two hydrogen bonds) and this leads to a big elevation in their expected boiling point compared to a similar hypothetical molecule of the same size that couldn't do hydrogen bonds. But there is no benefit in describing this a "molecule" and nobody does this.
But the main point is that many properties in chemistry have an almost continuous range with no clear point where a black and white distinction can be made. The definition of what counts as a molecule is like this. Describing hydrogen bonded DNA as a molecule isn't a terrible use of the terminology; describing an acetic acid hydrogen-bonded pair as a molecule probably is. The term molecule can be useful in some circumstances but not others. The utility of the term is what matters not a simple black or white definition.


I think your last sentence answers your own question very well:

  It is because it forms strong hydrogen bonds in solution and thus forms
  stable solvation spheres (solubility), but its high electronegativity
  prevents $\ce{H+}$ atoms from leaving its vicinity (dissociation).

Dissociation is one mechanism of dissolution, but not the only one. Also, however, the $\ce{H-F}$ bond strength is not solely determined by electronegativity.


Is glycerol/glycerin soluble in ethanol?

As LDC3 indicated above here it states that "Glycerol is completely soluble in water and alcohol." So, this is correct, as also indicated here.

Now, for the reason this happens it is most likely because of its 3 hydroxyl groups ($-OH$) and the fact that it is a polar molecule.


What's wrong with using tin in medicinal chemistry?

Organotin compounds are rather toxic.  They are also persistent in the environment and have a long biological half-life.

The problem is that trialkyltin byproducts from your reaction are difficult to separate from the product.  In the lab, this is painful and (usually) involves multiple columns, but when you want to get a drug past the FDA into the clinic you have to meet purity standards for heavy metal contamination.  To a medicinal chemist, a transformation that won't make it into pilot plant and production is useless.


A don't see major problems in storing it in a PVC bottle,   PVC is not soluble in ethanol, neither in isopropanol. However, in this site they rate isopropyl alcohol as A while ethanol as C (with possible swelling in the long term).


Alloy-component weight based on desired molecule

Yes, you are correct.

7.692 g / (20 g/mol) = 0.3846 mol X (or 2.316 * 10^23 atoms X)

2.308 g / (3 g/mol) = 0.7693 mol Y (or 4.633 * 10^23 atoms Y)

Moles are just a count of atoms, so if you have two moles of Y per mole of X, you have two Y atoms per X atom.


Why is the ionization energy of sulfur anion more than that of oxygen anion?

The ionisation energy of an anion is the same as the electron affinity of the element. $$\text{IE}(\ce{A-}) = \text{EA}(\ce{A})$$

Your reasoning is not wrong and one might expect exactly this behaviour. Another point for this reasoning is the substantial lower electronegativity of sulfur ($2.6$) versus oxygen ($3.4$).  However, experimental findings contradict this reasoning.
$$\begin{array}{lll}
&&/\mathrm{kJ\cdot{}mol^{-1}}\\\hline
\text{IE}\ce{(O)}  &        &=1314\\
\text{IE}\ce{(S)}  &        &=1000\\
\text{IE}\ce{(S− )} &=\text{EA}\ce{(S)} &=200\\
\text{IE}\ce{(O− )} &=\text{EA}\ce{(O)} &=141\\\hline
\end{array}$$

Sources: Wikipedia articles on ionisation energies, electron affinities

Basically it all boils down to electron-electron repulsion. Oxygen ($152~\mathrm{pm}$) is much smaller than sulfur ($180~\mathrm{pm}$) (van der Waals radius, source). This goes hand in hand with the smaller orbitals (lower main quantum number for the valence shell)  The electrons can therefore only occupy lesser space and repulse each other more. This is mainly the reason, why sulfur has a greater electron affinity than oxygen.

The same reasoning may be applied to fluorine and chlorine.


Help understanding the synthesis of HOCl

Chlorine bleach is commonly produced via the chloralkali process as a highly alkaline mixture of sodium hypochlorite (NaOCl) and hydroxide (NaOH) along with some unreacted NaCl as well.
It's usually dangerous to lower the pH of chlorine bleach, as you may inadvertently generate chlorine gas
$$ \ce{2H+(aq) + OCl- (aq) + Cl- (aq) \rightleftharpoons Cl2(g) + H2O (\ell)} $$
$$ \ce{4H+(aq) + 4OCl- (aq) \rightleftharpoons  2Cl2(g) + O2(g) + 2H2O (\ell)} $$
And that is why it is kept at a high pH, to help favor the left side of the above equilibria.


When a system is in thermodynamic equilibrium, there is no tendency for spontaneous change. In other words everything stays the same, the composition, temperature, pressure do not change over time, etc.
Therefore:
Tsurr is constant
and
Tsys is constant.
If you make an infinitesimal change, the system will tend to revert to the equilibrium.
So there is no issue.
When you are talking about a reversible process in thermodynamic terms, you're talking about a system which can be taken from its initial state to another state, and then back to the initial state without any net change to either the system or the surroundings.
The idea being that you make an infinitesimal change to some aspect of the surroundings, the system changes to a new state which is again in equilibrium with the surroundings, lather, rinse, repeat. Of course it would take infinite time for the process to be completed.


Why is the Gaussian Window Function used to enhance resolution in 1D NMR spectra?

Under the conditions you measure a plain 1D proton spectrum, you usually never have any roblems with truncation of the FID. You have a relatively long relaxation delay, you don't gain anything by making your acquisition time much shorter than it can be. So a default setup might measure 64k or even 128k points, which is far, far larger than the values you use in 2D NMR.

In two-dimensional spectra there are several aspects that severely limit the number of points you measure:

Measuring more points in the indirect dimension takes time, if you want twice the points you have to measure twice as long. For that reason you only measure as much points in the indirect dimension as you have to, no more.
Processing can become a bit slow and take quite a bit of space when you increase the number of points in the direct dimension in a 2D spectrum. It's not a big problem with current hardware, but with a 3D (or higher) spectrum this gets much, much worse.
If you need to decouple during acquisition time, the acquisition time is severely limited by the probe hardware and heating effects of the decoupling.

So a typical 1D spectrum will have a lot more points measured, beyond the point where the FID has decayed below the noise. In such a case you won't get any truncation artifacts, as the FID was never truncated to begin with. You only need to worry about truncation artifacts if you actually truncate the FID.

In the following image you can see

lila: QSINE (SSB 4) 10k points in FID
green: GM (LB -1, GB 0.5) 10k points in FID
red:  QSINE (SSB 4) 2k points in FID
blue: GM (LB -1, GB 0.5) 2k points in FID

You can clearly see the truncation artifacts in the gaussian-processed spectra with 2k points, but already at 10k points they are not visible anymore. The quadratic cosine of the truncated FID doesn't look all that nice, but it suppresses the truncation artifacts much better than the gaussian window function.

The SSB 4 I used for this example is also a far more aggressive setting than I'd usually use. I chose it because it gives a relatively similar resolution as the gaussian I compared it to. Usually I'd use SSB 2 for most 2D spectra, which emphasizes signal/noise over resolution.


Why are alpha carbonyl halides most reactive towards SN2 reactions?

I have asked a question previously which concerns the same electrophilic system with the exception that nucleophilic attack happens on the carbonyl carbon and not on the α-carbon.

From the organic chemistry textbook by Clayden, Warren, Wothers and Greeves on pp. 890-891:

The $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$ orbitals add together to form a new, lower-energy molecular orbital, more susceptible to nucleophilic attack. But, if $\ce{X}$ is not a leaving group, attack on this orbital will result not in nucleophilic substitution but in addition to the carbonyl group. Again, this effect will operate only when the $\ce{C-X}$ and $\ce{C=O}$ bonds are perpendicular so that the orbitals align correctly.

This interaction when considered in the context of an attack on the carbonyl is the principle of the polar Felkin-Anh model used to describe a certain diastereoselectivity. However it can likewise be used to explain the lowering of the $\unicode{x3c3}^*(\ce{C-Br})$ orbital which is attacked in a nucleophilic substitution reaction. It may seem as though these statements are contradictory, but the orbital interaction will create a three-centre π-type double-antibonding orbital with only the contribution on oxygen being neglegible due to oxygen’s electronegativity being highest and thus its contribution to high-energy orbitals being lowest.
It will depend on your nucleophile what will actually happen. It can attack both on the carbonyl carbon and on the α-carbon. If the attack on the carbonyl carbon is reversible, only the attack on the α carbon leading to an observed $\mathrm{S_N2}$ process will be observed after the reaction — even if the carbonyl attack is faster. Only if the attack on the carbonyl is not reversible as is the case e.g. with Grignard reagents, a mixture of two products will be observed.
Naturally, if $\ce{X}$ is a very poor leaving group — e.g. $\ce{OTBS}$ — the attack on the carbonyl will dominate this way or that.

To sum up:
The reaction rate is increased because the $\unicode{x3c3}^*(\ce{C-X})$ orbital and the $\unicode{x3c0}^*(\ce{C=O})$ orbital linear combine to give a lower-energy orbital. Since this is the LUMO, the LUMO energy is lowered and nucleophilic attacks are facilitated.


TL;DR
The neutral point is not the same as the equivalence point.
A neutral aqueous solution at room temperature, $25~^\circ\mathrm{C}$ and standard pressure $1~\mathrm{atm}$, has always $\ce{pH}=7$.

If you look for credible sources of definitions of neutral solutions on the internet, most likely you will find something along the lines of this:

  Neutral solution
  (Science) Has a $\ce{pH}$ level of $7$: a solution in which the concentration of hydrogen ions and hydroxide ions are equal (biology-online.org)

Or even worse:

  Neutral Solution Definition:
  An aqueous solution with a $\ce{pH}$ of $7.0$ $(\ce{[H^{+}]} = 1.0 \times 10^{-7}~\mathrm{M})$. (chemistry.about.com)

The second one especially is negligent of a lot of features.
Unfortunately the IUPAC does not provide an actual definition of a neutral solution. There are two possible definitions, which come to the same conclusion (compare the above), given the same external conditions.

The lazy definition:
The concentrations of hydronium and hydroxide ions are identical. $$c(\ce{OH-})= c(\ce{H3O+})$$
In terms of $\ce{pH}, \ce{pOH}, \mathrm{p}K_w$ one arrives at a more complete side.
Pure water is a neutral solution. It's autoprotolysis provides
\begin{align}
\ce{2H2O &~<=>~ OH- + H3O+}.\tag{1}\\
\end{align}
Now we can formulate the equilibrium constant and furthermore assume that autoprotolysis is small compared to the overall concentration of water.
\begin{align}
  K_c &= \frac{c(\ce{OH-})\cdot c(\ce{H3O+})}{c^2(\ce{H2O})}\\
  K_w &= c(\ce{OH-})\cdot c(\ce{H3O+})\tag{2}\\
\end{align}
Equation $(1)$ also provides the lazy definition. $$c(\ce{OH-})= c(\ce{H3O+})\tag3$$
We plug that into $(2)$ and we arrive at
$$c(\ce{H3O+})=\sqrt{K_w}\tag{4}$$
Consider the actual definition of
$$\ce{pH} = −\lg a(\ce{H+}) = −\lg\left( m(\ce{H+}) \gamma_m(\ce{H+}) / m^\circ\right)\tag{5}$$ rewritten using concentrations
$$\ce{pH} = −\lg a(\ce{H+}) = −\lg\left( c(\ce{H+}) \gamma_c(\ce{H+}) / c^\circ\right)\tag{5a}$$
The identities used here are $a(\ce{H+})$ for the activity of a proton in aqueous solution, $\ce{H+ (aq)}$ and we will consider $\ce{H+ (aq) = H3O+}$ to be the same. Now we are going to assume, that the activity coefficient is $\gamma_c(\ce{H+})\approx 1$ for very diluted systems. Using the standard concentration $c^\circ$ we do not have to care about units.
$$\ce{pH} = −\lg\left( \frac{c(\ce{H+})}{c^\circ}\right)\tag{5b}$$
Now we can substitute in $(4)$ and we will have a nice definition of the $\ce{pH}$ of a neutral solution.
$$\ce{pH} = −\lg\left( \frac{\sqrt{K_w}}{c^\circ}\right)\tag{6}$$

The second formulation includes a very crucial point. The autoprotolysis $(1)$ is temperature, $T$, dependent, which is obvious following the definition of the standard equilibrium constant
$$ K^\circ = \exp\left\{\frac{-\Delta G^\circ}{\mathcal{R}T}\right\}\approx K_c.$$
Therefore also the ion product of water is temperature dependent, $K_w(T)$.
Wikipedia actually has a couple of values for the $\mathrm{p}K_w$. Reaction $(1)$ is endothermic. A higher temperature means providing energy, means a higher $\mathrm{p}K_w$. For example, in the human body, $37~^\circ\mathrm{C}$, this value is slightly higher than for the one we usually assume as the "neutral point" $\mathrm{p}K_w(25~^\circ\mathrm{C})=14$.

Since $(6)$ is more correct in the following form,
$$\ce{pH}(T) = −\lg\left( \frac{\sqrt{K_w(T)}}{c^\circ}\right)\tag{6a},$$
the neutral $\ce{pH}(T)$ therefore changes with temperature.

Another note is that it is also pressure dependent, but that might take it a little too far.

The equivalence point is usually defined for an arbitrary chemical reaction between an acid and a base, and it refers to the situation where there are stoichiometric quantities of acid and base are present.
$$\ce{AH + B <=>[\ce{H2O}] A^- + HB+}$$
This means, that at the equivalence point the following statement holds:
$$c_\text{initial}(\ce{HA}) = c_\text{final}(\ce{HB+}).$$
Therefore the $\ce{pH}$ at the equivalence point will only be governed by the reactions of $\ce{HB+}$ and might not be neutral.

The issue of solvent dependency is addressed by Fred Senese quite concise on antoine.frostburg.edu:

  pH is often used to compare solution acidities. For example, a solution of pH 1 is said to be 10 times as acidic as a solution of pH 2, because the hydrogen ion concentration at pH 1 is ten times the hydrogen ion concentration at pH 2. This is correct as long as the solutions being compared both use the same solvent. You can't use pH to compare the acidities in different solvents because the neutral pH is different for each solvent. For example, the concentration of hydrogen ions in pure ethanol is about 1.58 × 10-10 M, so ethanol is neutral at pH 9.8. A solution with a pH of 8 would be considered acidic in ethanol, but basic in water!


Is sodium hypochlorite solution an ionic solution - does it conduct current?

Yes, it is a solution of an ionic salt. a table lists electrical conductivity at various concentrations.


Why is the discovery of chiral molecules in space a big accomplishment?

The story is a big deal because it might help explain why molecules in living creatures have a single chirality

One of the great mysteries of living systems is that, when they use chiral molecules, they only use molecules of one chirality. There is no obvious mechanism to kick-start this preference and, while there are many esoteric theories, there is very little evidence that explains why one chirality was chosen over another.

Proteins are, for example, made from amino acids which are almost all chiral. And they are all left-handed (technically they are all levorotatory, see wikipedia). Most sugars used in biological systems are right-handed. There are simply no convincing theories that explain the origin of this preference.

One modern theory suggests that there is a subtle energetic preference for one chirality caused by the interaction of molecules with the electrons emitted in certain types of radioactive decay (the weak nuclear force involved in radioactive decay is one of the few fundamental forces to have a mirror-image preference). But other experts think the preference is arbitrary.

Resolving this would be big news for biology.

The reason why the observation of chiral molecules in space (the news story only reports the observation of a molecule that is chiral not that we can observe the chiral preference) is that it creates hope than we can observe directly whether the chiral preferences of life are inherent in the universe or are arbitrary.

We don't yet know whether propylene oxide (the molecule in the news story) exists with more of one chirality than the other, but now it has been observed, we can expect to find out as instruments improve. So the big story is that we now have some hope that we can test some of the theories about why life has a handedness preference.


Yellow-amber debris above water filter: What is it?

It looks like ion exchange resin, it is used to specifically capture ions from water and replace them by others (either $\ce{H^+}$ or alkali metal ions.

As such, it is probably also part of the filter.


Irreversible chemical reaction in a battery

The original poster has a serious misconception by equating a reversible chemical reaction with a reaction that is in equilibrium. The electrochemical reaction that allows the battery to deliver energy to the load (such as the copper/zinc reaction mentioned) is always reversible in theory by reversing the current flow. However in addition to the electrochemical reaction, there are other chemical reactions going on inside the battery which are not reversible as well as geometrical changes that may not allow the reactants to be close enough for the electrochemical reaction to run in reverse. In batteries designed to be recharged, these geometrical and irreversible chemical changes are minimized to allow as many recharge cycles as possible. Batteries not designed to be recharged therefore usually can't be recharged fully and have very few recharge cycles not to mention a high probability of gaseous or corrosive liquid leakage.

At equilibrium, both forward and reverse reactions are going at the same rate, but this is only discernable microscopically. At a macroscopic level you see nothing happening. So a battery is never at equilibrium while you are using it since at that point the battery is completely dead. Then (or preferably sooner) it is time to recharge the battery or throw it away.


Can 1 mole of H2 react partially with one mole of O2?

As you have said, you are studying stoichiometry at High School Level. From this I can guess, that you have probably not studied the concept of Limiting Reagent yet.
What is Limiting Reagent?

In a chemical reaction, the limiting reagent, also known as the "limiting reactant", is the substance which is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent since the reaction cannot proceed further without it. The other reagents may be present in excess of the quantities required to react with the limiting reagent.

As per your question, $\ce{H2}$ is limiting reagent. Hence it will get totally consumed when reaction happens and 1/2 Moles of $\ce{O2}$ will be left.

1 mole of H2 and 1 mole of O2 won't make 1 mole of H2O and 1/2 mole O2.

1 mole of $\ce{H2}$ and 1 mole of $\ce{O2}$ will make 1 mole of $\ce{H2O}$ and 1/2 mole $\ce{O2}$ will be left.
$\ce{H2 + O2 -> H2O + 1/2O2 (Left) }$
And here is an advice for you. If your teacher is saying that this will not happen at all, tell him to be sure that reactant in reaction are single molecules, not 1 mole.
Read this to be more clear on concept of limiting reagent.


The pH is determined experimentally with a glass electrode. A glass electrode is a small glass tube limited by a thin membrane and filled with an aqueous KCl solution with a silver electrode inside. There is also a similar silver electrode out of the tube, in a compartment in contact with the outer solution. When you dip a glass electrode into an aqueous solution, the experiment shows that the tension E read between the two electrodes is proportional to the logarithm of the H+ concentration of the external solution. It is :  E  = Constant + 0.059 pH. The Constant depends of the particular glass membrane.
And the experiment shows that pH is -log[H+] at low acidic concentrations, and is measured as

7 in pure water,
3 in 0.001 M aqueous solutions
2 in 0.01 M aqueous HCl solutions,
11 in 0.001 M aqueous NaOH solutions
12 in 0.01 M aqueous NaOH solutions.

These values are in agreement with the usual expression : pH = -log[H+], or : pH = 14 - log[OH-]. At higher concentration, the measured pH values are not exactly equal to the logarithm of [H+]. The measured pH is said to be the logarithm of the "activity" of H+.

The pH does not go only from zero to 14. It may be smaller than zero in concentrated solutions of HCl and of H2SO4.

In non-conducting solutions, the glass electrode does not give any stable values. The measured tension is fluctuating without any reason and can be ± 3, 5, 10 or 0, and this changes constantly.


Are graphite intercalation compounds stable in water?

According to "Graphite sulphate – a precursor to graphene", Siegfried Eigler, Chem. Commun., 2015,51, 3162-3165:

  Graphite sulphate is used as a precursor to graphene for the first
  time. The positively charged graphene layers react with water to yield
  a processable graphene derivative.

This would seem to imply that it would indeed de-intercalate in a beaker of distilled water.


Effect of inductive effect on boiling point,melting point and dipole movement

Not quite sure about melting and boiling points but things like polarity ,dipole moment and inductive effect are kind of interconnected.Inductive effect is a consequence of electronegativity differences.More inductive effect leads to more dipole moment but that also depends from one case to another.For example each of the $\ce{C-Cl}$ bonds in $\ce{CCl4}$ is highly polar but net dipole moment is 0 due to the fact that Dipole moments are added through vector addition and $\ce{CCl4}$ is tetrahedral.


I think the professor is right. I would solve it like any other acid-base problem. Glycine (G, neutral as zwitterion) can either protonate to HG+; or deprototonate to G-. The mass balance and charge balance equations are:

C$_G$=1.0=[HG+]+[G]+[G-] (Mass balance for glycine)

[H+]+[HG+]=[OH-]+[G-]    (Charge balance)

Given these two equations, and the three for $K_{a1}$, $K_{a2}$ and $K_{w}$, I could solve exactly the system of five equations with five unknowns. But that would be overkill, since we can do some approximations:

Since G is both a (not so) weak acid and a weak base, we can assume that it doesn't dissociate that much to HG+ or G-. So, the predominant form would be G, which (given the $pK_a$'s) is predominant in the 3.5 to 8.5 pH range (aprox, it depends on the errors that we tolerate).
In this pH range, and given the high glycine concentration and $pK_a$ values, [HG+]>>[H+] and [G-]>>[OH-]; so that, from the charge balance:

[HG+]=[G-]

This equality solves the problem since, combining the equations for $K_{a1}$, $K_{a2}$, we get:

$K_{a1}K_{a2}=\frac{[G-][H+]^2}{[HG+]}=[H+]^2$ so $[H+]=\sqrt{K_{a1}K_{a2}}$

$pH=\frac{1}{2}log(pK_{a1}+pK_{a2})=6.0$

You can check that at this pH, [G]=1,[HG+]=$10^{-3.2}$,[G-]=$10^{-3.6}$, so that the approximations [HG+]>>[H+] and [G-]>>[OH-] are valid.

(For more details see 6.7.5 at https://chem.libretexts.org/Textbook_Maps/Analytical_Chemistry_Textbook_Maps/Map%3A_Analytical_Chemistry_2.0_(Harvey)/06_Equilibrium_Chemistry/6.07%3A_Solving_Equilibrium_Problems)


Advantages of Travelling wave IMS and Drift tube IMS

In a nutshell, since other sources probably would give much more accurate and complete answers:

Advantages of DT-IMS over TWIMS:

Direct access to absolute collision cross sections. This is a major advantage considering the difficulty to provide accurate CCS calibration scales for TWIMS measurement as can be seen in the recent litterature.
Separative resolution is usually reported as higher, although recent work by Dick Smith's group at PNNL seems to show that it is mostly related to separation device lengths which are different.

Advantages of TWIMS over DT-IMS:

Reduced ion losses and increased sensitivity. The stacked-ion ring basis leads to reduced drift towards the sides. This advantage has been reduced by the addition in several DT-IMS setups of ion funnel electrodes which focus the ion trajectories upon reaching the end of the drift tube.
No need for large potentials and thus large voltage generators and all the related problems. To reach a given electric field, the potential difference does not need to be applied to the whole drift tube.


Partial pressure and chemical equilibirum

No, at new equilibrium $K_c$ would actually is the same value. This constant stands for reaction between these two compounds no matter the pressure, it is only affected by temperature.

As ammonium hydrosulfide is a solid, the reaction is only relevant at the surface of the crystals, and the "concentration" of ammonium hydrosulfide at the relevant areas will remain constant. Therefore same partial pressure of both ammonia and hydrogen sulfide is needed to maintain the equilibrium. Hence, at equilibrium, the partial pressure of both ammonia and hydrogen sulfide should be the same as the initial partial pressure.


$A $ is the cross-sectional area in solution. The  "bulk" is the relatively homogeneous "mess" of charged particles and solvent -- $A $ is not the area of the electrode.

What this equation tells you in terms of $A $, is that, the larger the area through which some number of charges are passing in a given moment, the greater the possible current, because more area means more room for more charges that can pass through in a moment. If you squeeze the area you start to make a bottleneck wherein less charged particles can pass in a given time. (This is why copper wires with larger diameter (or, smaller US gauge) have higher current capacity).

This makes sense, right? Current is measured as charge per unit time. The implication there, is that it is motion of charge through some point (1D) or cross sectional area (2D, $A $ here) per unit time.


Yes, I agree with your teacher's comment that upon increasing the total pressure of a system, there are select reactions "more dependent on collisions", and such reactions have been a matter of study.
For example, per this 2008 work reported in 'The Journal of Physical Chemistry',  The Temperature and Pressure Dependence of the Reactions H + O2 (+M) → HO2 (+M) and H + OH (+M) → H2O (+M), where '+M' here denotes the catalytic presence of a surface, as can be provided by another gas molecule.
To quote from the abstract:

The reactions $\ce{H + O2 (+M) → HO2 (+M)}$ and $\ce{H + OH (+M) → H2O (+M)}$ have been studied using high-level quantum chemistry methods. On the basis of potential energy hypersurfaces obtained at the CASPT2/aug-cc-pVTZ level of theory, high-pressure limiting rate coefficients have been calculated using variable reaction coordinate transition state theory. Over the temperature range 300−3000 K, the following expressions were obtained...Available experimental data on the pressure dependence of the reactions were analyzed using a two-dimensional master equation.

Note, reactions are being studied via surface collisions involving the addition of  Ar and N2 gases. The selection of the gas relates to chemical inertness and the total collisional cross-section of the particular gas particles, all of which are components of more complex theories surrounding collision frequency and collision theory (see, for example, discussion here).


Is it possible to have accumulation of electron density in the inter-nuclear region even if the overlap integral is zero?

If you define overlap as given by the overlap integral,

$
S_{ij} = \langle \psi_{i} | \psi_{j} \rangle
$

than you can have an accumulation of electron density in a specific part in space even though $S_{ij} = 0$.

Of course, if the complete wave functions of the two AOs are completely non-overlapping in all space, for instance because the centers of the two AOs are seperated by a long distance (say more than 5 A), then there can be no accumulation of electron density anywhere. This is of course logical. There is hardly any accumulation of electron density between two atoms that are quite far apart.

An interesting scenario exists where two wave functions have opposite sign and an overlap of these wave functions actually gives a decrease of electron density. This is termed deconstructive interference opposite to constructive interference when there is an accumulation of electron density.


Yes, that is completely correct. Let me summarize the approach.

If you look at the chemical reaction:
$$\ce{N2 + 3H2 \rightarrow 2 NH3}$$

You can see that you can, at best, make 2 mole of $\ce{NH3}$ per mole of $\ce{N2}$ (for the case here where $\ce{H2}$ is abundant). Given that you have $4.46\; moles$ of $\ce{N2}$ you could at best make $8.92\; moles$ of $\ce{NH3}$ which would be $151.9\; g$. Since you have made only $36.5\;g$ that means that your efficiency is $36.5/151.9=24\%$


Metazene 99% or other Alpha Aromatics high heat compounds.

What is the mal-odor caused by, and the complete application?


Does the probability of fluorescence increase when energy of incident X-rays match the electron's binding energy?

Short answer, yes.

As you note, the fluorescence signal is strongest when the exciting photon has an energy close to the excitation energy for the element absorbing the photon.  There are actually two factors coming into play:

1) The closer the resonance of the incident photon with the electronic orbital it is exciting, the higher the cross-section to absorb the photon.  (See here.)

2) The higher-Z the absorbing atom, the more likely it is that once excited, the atom will emit a fluorescence photon (as opposed to an Auger electron).  That is, it has a higher fluorescence yield.  (See here.)

In your case, the Z of the atoms are pretty close, but also in portion of the periodic table where the fluorescence yield is rapidly changing.  The chance of a fluorescence photon being emitted once excited is about 20% greater for Co than Mn (and you can look it up in the literature).

The cross-section for absorption as a function of energy for these three elements looks like this (code for the plot here):

I also drew a vertical line for the exciting Cu-K$\alpha$ photon at 8048 eV, and you can see that the Co curve has the highest cross-section at about 316 cm$^2$/g, and Mn has the lowest at about 270 cm$^2$/g, which means the Co cross-section is about 17% higher than the Mn.  The primary reason for the reduction in the sensitivity for the Mn edge is that the absorption cross-section drops off as $\approx\Delta E^3$.  These data were plotted using the data files described by B. L. Henke, E. M. Gullikson, and J. C. Davis, Atomic Data and Nuclear Data Tables Vol. 54 No. 2 (July 1993)

So both these effects are colluding to increase your fluorescence for Co vs Mn.

As a note, if you want to reduce this effect in your X-ray diffraction patterns, you can change your source target material with a heavier element such as Zr which has it's K$\alpha$ up at 15.7 keV, yet the L lines are still around 2 keV (below the transition metal K-edges) or you can use a lighter target, such as Mn.  Mn, of course, would not be able to excite its K-edge using its own K$\alpha$ fluorescence photon.


What is the possible side product when mixing citric acid, NaCl, Cu, Al, H2O2, and an unknown photoresist?

1) You should not use acetic acid (or any organic acid) in this setup. It potentially may form some (even if minuscule) amount of chloracetic acid which is really hazardous. Furthermore, copper acetate (and other carboxilates) has low solubility. Go for either genuine $\ce{HCl}$ or for $\ce{H2SO4}$ used in lead accumulators.

2) Most likely it is aluminium hydroxide. If the substance is bluish, it may be also copper hydroxide.


I read the question as asking whether a stereocenter bearing a C-Mg bond will keep its configuration. The answer is yes, but not indefinitely. Unfunctionalized (alkane) Grignards of this type racemize/epimerize with a half life on the order of hours. Specific compounds will have more or less stability. By analogy to organolithium compounds, I would expect having Lewis basic atoms in the proximity to stabilize the configuration. I'm not sure if this is still an active field of research, but it certainly was for a long time. Try searching "configurational stability of Grignard reagents" or similar search terms.

There's an additional complication with these chiral Grignard reagents: they can't be prepared by the usual method of oxidative addition of Mg(0) to an alkyl halide. The mechanism of that reaction goes by a single electron transfer, through a radical intermediate, destroying any stereocenter. Chiral Grignards have to be synthesized using less direct methods.

Ref: Side Reactions in Organic Synthesis: A Guide to Successful Synthesis Design, pp. 200-202.


etymology of phosphatidates and phosphatidic acid

Apparently, "phosphatidate" is from "phosphate" + "-ide" + "-ate".

The word "phosphatidic acid" is from "phosphate" + "-ide" + "-ic acid".

Phosphate ($\ce{PO4^3-}$) is the anion of phosphoric acid ($\ce{H3PO4}$).

The suffix "-ide" is used in biochemistry to just about anything. It just denotes an element of a group of related compounds.

Now, "-ate" and "-ic acid" denotes the acidic nature of the compounds.

And no, it is not related to the English word "fat".


Estimating for atomic number for an "unknown element"

Well I found it out. It is seaborgium atomic number 106, mass number 271.

To be found in period B6, group 7.


Reactions between enantiomers in a racemic mixture

If a chiral molecule (an enantiomer) can react with itself, dimerize for example, to form a new compound, then formation of this new compound would contaminate the racemic mixture.

(image source)
Perhaps more importantly is interaction of the racemic mixture with itself in the solid state.  Just as the interaction between a right hand and left hand is different than the interaction between two right hands, the interaction between the R form and S form of the enantiomer will be different than the interaction between two R (or two S) forms.
In liquids and gases, molecular encounters are fleeting and cannot be preserved or "locked in" for significant periods of time as they can be in a solid. In other words, any favorable, lower energy molecular encounters are quickly broken apart by the energetically larger effects of thermal motion.  However, when the racemic mixture is in a solid phase, if the interaction between R and R (or S and S) is lower in energy then the interaction between R and S, then the R and S enantiomers will tend to form separate crystals and a mixture of pure R and pure S crystals will result.  Such mixtures are often termed "conglomerates".  On the other hand if the interaction between R and S is more favorable (lower in energy), then a racemic compound or racemate is formed where each crystal contains R-S pairs.
A conglomerate, being a simple mixture, will show a melting point depression compared to the pure enantiomer.  A racemic compound may have a melting point higher or lower than the pure enantiomer.
Further reading on racemic conglomerates and racemic compounds can be found here and here.


Average bond energy of H−F and H−O hydrogen bonds

So fluorine does have the ability to hydrogen bond. However, they are usually transient and depend significantly on the species that is donating the H-bond along with some other factors (i.e. what phase the molecules are in, temperature, etc...)
Fluorine is so electronegative that it holds its electrons much "tighter" then other atoms, making it a poor H-bond acceptor. Oxygen, being less electronegative, is a better H-bond acceptor since the electrons aren't held onto as tightly, allowing them to interact more with other atoms.
Since bond energy is "the average value of the gas-phase bond-dissociation energy for all bonds of the same type within the same chemical species. The larger the average bond energy, per electron-pair bond, of a molecule, the more stable and lower-energy the molecule" (IUPAC WEBSITE), I would suggest that the H-bond with oxygen has a higher bond energy, as it is the stronger interaction of the two.
EDIT
The OP requested some sources, in which I failed to do initially, so I apologize for that but here they are.

J. Am. Chem. Soc. 1983, 105, 10, 3206–3214

Tetrahedron, Vol. 52, No. 38, pp. 12613-12622, 1996

J. Am. Chem. Soc. 1962, 84, 16, 3221–3222

Here is a quote from reference 2.

Theoretical calculations variously estimate 5 the strength of a F...H bond to be between 2 to 3.2 kcal mo1-1.
This can be compared 6 to an O...H hydrogen bond which is typically between 5 - 10 kcal mo1-1. Consistent
with this the electrostatic influence of fluorine is approximately half that of oxygen 7. Thus the greater
electronegativity and lower polarisability of fluorine over oxygen, suppresses its electrostatic influence and
renders it a poorer hydrogen bond acceptor.

The last reference discusses hydrogen bonding in halides, not too sure how much it compares them to oxygen bonding but it is certainty worth taking a quick look.


Why does protic solvent suggest E1 mechanism rather than E2?

First of all, a protic solvent does not mean a highly acidic solvent; it mainly means a solvent whose molecules contain protons that can reasonably be expected to act as an acid. Take for example methanol: its $\mathrm pK_\mathrm a$ is somewhere around $15$, so calling the hydroxy proton acidic is a stretch by most definitions but there are sufficiently basic compounds that can abstract said proton.

Secondly, the question of how a protic solvent affects the E2 versus E1 ratio has less to do with the intermediate cation and more to do with the leaving group. In an E1 reaction, the leaving group dissociates from the main molecule first taking the bonding pair with it. That typically means this leaving group is now anionic or has a somewhat nucleophilic lone pair on it. The protons of the solvent can ‘block’ that lone pair from reattacking the intermediate cation, preventing the backwards reaction and increasing the odds of proceeding. The resulting cation can then be stabilised by the polar bit of a polar protic solvent: the more negatively polarised part of the solvent molecules will interact favourably with the cation.

In an E2 reaction, the base must interact with the starting material for both the starting material’s proton and the leaving group to be displaced in one step. In this case, having a protic solvent which is capable of protonating the base (reducing its basicity) is harmful meaning the reaction will proceed slower. In E1 reactions, weaker bases can be used that are not able to deprotonate the solvent molecule so this pathway is not much of a problem.


Which are the major truths of paradigms on which the whole body of chemistry is built?

Well the most important driving force I see in chemistry is for atoms and molecules to be most stable when the atoms have a noble gas electronic configeration. Many reactions will occur to try to reduce the energy of the atoms and molecules by gaining / losing or rearranging electrons to bring as many atoms as close as possible to a noble gas configeration.

Another important driving force is to reduce or avoid the mechanical strain in a molecule (Steric effects)

There are effects in pi systems such as aromaticness (4n + 2 pi electrons in rings)

There are entropy effects, by increasing the entropy of a system then an endothermic reaction can be made to occur. For example the dissolution of ammonium nitrate in water occurs with ease but this is an endothermic process.


Adjusting PBS with HCl: pH drift after dilution of phosphate buffer

The differences between PBS and the buffer obtained by mixing $\ce{Na2HPO4}$ and
 $\ce{NaH2PO4}$ are in ionic strength ( because of the presence of $\ce{KCl}$ d $\ce{NaCl}$ in PBS) and in the ratio of [acid]/[conjugated base]. We have to signal that the buffer capacity (buffer effectiveness) is maximum when $\ce{pH=pKa}$.
The best way to dilute 10x PBS into 1PBS (let's assume we want to prepare 1L of 1xPBS from 10xPBs) is to begin by diluting 100 mL of 10xPBS solution with 800mL water, then adjusting the pH and finally completig the volume to 1000mL. In doing so, there will be no drift in pH.


Why is $\ce{C2O4}$ unstable?

Have you tried drawing it?

I get four different kinds of structures for it:

$\ce{(O=)O^{.}C-CO^{.}(=O)}$
$\ce{(O^{.})2C-C(O^{.})2}$
$\ce{O#C-O-CO^{.}(=O)}$
$\ce{O#C-O-O-C#O}$

They are unstable for the following reasons:

Biradical
Tetraradical (This won't really exist anywhere near anything else.)
Radical
Peroxo bond, very easy to break homolytically.

All of these compounds can easily decompose to give (the thermodynamical Mariana trench) carbon dioxide:
$$\ce{C2O4 -> 2CO2}$$

That is why it is so unstable.

Reduce 'em!

  And would adding hydrogens to a carbon oxide make the carbon oxide even more stable?

Yes. If you start reducing the compounds above, you will end up with ketones and water (because geminal hydroxy groups tend to give water and ketones), or even alcohols (if you continue reducing).


How can one reclaim isotopic cadmium from a solution of acid, i.e. H2SO4 or HCl etc.?

Thank you everyone for the discussion.  Sorry, I have been away for a few days, so was unable to follow up.
Reading the ideas, it is a toss up on a procedure.  Scraping the Cd off of an Aluminum electrode might work, but Al contamination would be very bad for the laser oscillation.  It might work for awhile, but would die from contaminants in short order.
The Al oxide, though only a molecular layer, is extremely tough and can withstand a great deal of heat and abuse.  I make Aluminum edge-wound voice coils for my speaker company and we depend on the oxide layer to act as an insulator from winding to winding.  Our coils can withstand 600 degrees and will melt before they fail.
Stripping the oxide gives you only a few seconds before it reforms in the normal atmosphere.  Any other thoughts?
Perhaps reduction of the salt by heating?
Thanks
Dale


According to this MSDS, Astlett Rubber, Inc.'s BUTYL RUBBER BK-1675N product contains multiple stabilizers and antiaglomerates, each of which is described as a white powder. I suspect it is the leaching or other means of migration of these stabilizer's to the rubber surface that causes the white spots. The following is an excerpt from the above MSDS:

  Butyl rubber contains moisture, stabilizer (Agidol-2 or Irganox-1010)
  and antiaglomerate (calcium stearate or zinc stearate). Stabilizer and
  antiaglomerate are not discharged into the atmosphere during
  processing butyl rubber into finished goods.

  Stabilizer Agidol-2 is a white crystalline combustible powder. Flash
  point in the open crucible is 172°C, ignition point is 208°C, the
  standard temperature of spontaneous ignition is 352°C. Agidol-2 is
  permitted for use as a stabilizer of rubber that is used in food
  industry, in water  pipelines, and children toys production.

  Stabilizer Irganox-1010 is white crystalline combustible powder.
  Ignition point is 288°C. Meets all requirements for use as a
  stabilizer in rubber th at is used in medicine.

  Zinc Stearate is white crystalline powder or paste. The standard
  temperature of spontaneous ignition is 900°C. Maximum permissible
  concentration in work area is 4 mg/m3.

  Calcium Stearate is a white powder. The standard temperature of
  spontaneous ignition is 560°C. Maximum permissible concentration in
  work area  is 10mg/m3.


How do I find Rf values & length size of paper in chromatography?

$$\Delta R_f = 0.08$$
Let the separation distance be $\Delta r$ = 1 cm and let $l$ be the length of the paper in cm.

$$\Delta R_f = \frac{\Delta r\, \mathrm{cm}}{l\, \mathrm{cm}} $$

Solve for $l$, obtain $l$ = 12.5 cm.


Its because of the resonance. buta-1,3-diene is resonance stabilized. The Heat of combustion actually will be less than calculated theoretically because some of energy released will be used to destabilize the molecule.


The image gives a thousand words, it is difficult to say how many words the colorized image gives, but might help.


Why is the lanthanide element Thulium malleable?

All the lanthanides are soft metals; there is nothing special about thulium in that regard, AFAIK. For that matter, so are the alkali metals, and gold can be scratched with a fingernail.

Most pure metallic elements are malleable. "This is because of the ability of the atoms to roll over each other into new positions without breaking the metallic bond." Brittleness (the opposite of malleability and ductility) often occurs because of impurities such as carbon, or a metal alloy which separates on cooling. These impurities "pin" dislocations in the metal, preventing them from moving smoothly. It makes alloys harder, as well as more brittle. The early technologists discovered that the alloy bronze was harder than pure copper.


Chemistry of Walter Mitty's negative developer/stop bath/fixer?

basics here: http://science.csustan.edu/nhuy/chem1002/photoexp.htm

In depth: there are many good books about the chemistry of photography, starting with this 120-year old classic:
https://archive.org/details/chemistryphotog02meldgoog

Here is another that's fully available online and recent enough to cover every major aspect of the subject:
https://openlibrary.org/books/OL5953553M/Photographic_chemistry_in_black-and-white_and_color_photography

You can buy more recent books, especially for developments in color.


I am not sure how to elaborate much beyond saying that yes, there are bound states of $\ce{HHe^{2+}}$. I cite three papers below which give numerical calculations of various states, ref. 1 gives calculations for the lowest 20 states of the system.

It appears to be that the 1s$\sigma$ is not a bound state in this system, but the 2p$\sigma$ state is bound. I would expect there are some other bound states as well.

References:

[1] Winter, T. G., Duncan, M. D., & Lane, N. F. (1977). Exact eigenvalues, electronic wavefunctions and their derivatives with respect to the internuclear separation for the lowest 20 states of the HeH2+ molecule. Journal of Physics B: Atomic and Molecular Physics, 10(2), 285.

[2] Laaksonen, L., Pyykkö, P., & Sundholm, D. (1983). Two‐dimensional fully numerical solutions of molecular Schrödinger equations. I. One‐electron molecules. International Journal of Quantum Chemistry, 23(1), 309-317.

[3] Buehler, E. J., Gooch, E. E., Dial, J. L., & Knudson, S. K. (2000). Semiclassical energies of low-lying states of one-electron diatomics. Chemical Physics, 253(2-3), 219-230.


Determination of major product in crossed aldol condensation

There are two main steps in any aldol condensation:-
1. Formation of enolate ion
2. Using the enolate ion as a nucleophile.

The 1st step is a reversible step. If you think that the kinetically favoured enolate ion will be formed fast then you are wrong as it also disappears faster. So, in any case the one which is more thermally stable will form in more amount except in the cases where the formation of stable enolate is too slow or the backward reaction of kinetic favoured enolate is blocked or not enough time has been given to let the reaction reach equilibrium.

Now, I will tell you about above strategies to block reverse reaction or make formation of stable enolate too slow:-

1. Use a strong electron withdrawing group at alpha position to the newly formed CH2(-)COR so that the electron can spend most of time away from the original carbon.
-This blocks reverse reaction

2. Use a bulky base or somehow hinder the alpha carbon to make the extraction of proton hard which will in turn make the reaction slow.

3. Using Non-Protic Solvent stops the protonation of the enolate ion which in turn blocks the reverse reaction.

Now, the answer for your question, the conjugate base of acetaldehyde is more stable when compared to the former but the attack of conjugate base of acetophenone is more easier as compared to latter because of steric reasons and the more electrophilic character of aldehyde and when you make cannonical structures, the number of structures which has negative charge on oxygen is greater than being away which makes it a good nucleophile when you take weighted average of all. So, the major product should be 1-phenyl-but2en1one.


Inductive effect in this sequence in the reason of the increased basicity only in the case of N-methyl piperidine compared to piperidine. Increased basicity in the sequence of azetidine < pyrrolidine < piperidine is connected with the ring strain. Upon the decrease in the C-N-C angle, sp3-hybridized N atom "rehybridize" to increase p-character of orbitals involved in the C-N bonds formation, leading to the increased s-character of orbital involved in the N-H bond and the lone pair of electrons. Since s-electrons are closer to nucleus, this rehybridization leads to the increased stability of the N-H forming orbital and the lone pair in more strained cyclic amines making the lone pair being 'less exposed' to protonation (less basic).


Concentration of product at a time t in a reaction

Your method should work as you know the boundary conditions for your scheme, but it is not the easiest method. Instead, it is easier in this case, as you are given the initial amount of A, to let $x$ be the amount of B produced at time $t$, then $a_0 - a = b = x$ where to save writing brackets, $a,\;b$ are the amounts of A and B at time $t$. The the rate equation is

$$\frac{\mathrm{d}x}{\mathrm{d}t} =k_\mathrm{A}a -k_\mathrm{B}b$$

and substituting gives

$$ \frac{\mathrm{d}x}{\mathrm{d}t}= k_\mathrm{A}a_0-(k_\mathrm{A}+k_\mathrm{B})x$$

This can now be integrated with limits $(0,x),\; (0,t)$ to give

$$\ln \left( \frac{k_\mathrm{A}a_0-(k_\mathrm{A}+k_\mathrm{B})x}{k_\mathrm{A}a_0}   \right)= -(k_\mathrm{A}+k_\mathrm{B})t$$

At equilibrium, the rate of reaction is zero, then $k_\mathrm{A}a_e=k_\mathrm{B}b_e=k_\mathrm{B}x_e$ where subscript $e$ means that it is the equilibrium value. Also, we know that $\displaystyle K_\mathrm{e}=\frac{k_\mathrm{A}}{k_\mathrm{B}}=\frac{x_e}{a_0-x_e}$, where $a_0-x_e$ is the amount of A remaining at equilibrium.

Substituting and rearranging gives the amount of $x$, which is B as a function of time:

$$x=x_e -x_ee^{-(k_\mathrm{A}+k_\mathrm{B})t}$$

Both this and the previous equation show that the equilibrium is approached with a rate constant that is the sum of the two rate constants.


The internal energy $U$ is the sum of all the energies stored in the chemical bonds. If a chemical reaction happens, the bonds are losing or getting energy. Heat is getting in or out of the system, and this heat $\Delta Q$ can be measured. $\Delta U$ =$\Delta Q$, if the volume is constant.  But if the transformation is carried out in contact with the atmosphere, at constant pressure, the volume may change. A fraction of this energy is converted into work $\Delta w$ to repel the atmosphere, because the volume of the system may change. Usually this volume change is small (except if a gas is produced or consumed). And this volume change may be difficult to know with precision. But it must be taken into account to calculate $\Delta U$. So if this small amount of work is neglected, a correction must be added to the internal energy $U$, that depends on pressure and volume. Neglecting this volume correction gives a sort of “apparent internal energy”, which is called enthalpy $H$, with $H = U + PV$. And then, in all transformations made at ordinary pressure, the following expression is valid : $\Delta H$ = $\Delta Q_p $.
Enthalpy is a sort of "apparent internal energy". There is a similar difference between the weight and the "apparent weight", which is measured in air. The weight of an object should be measured in a vacuum. The weight, as measured with a dynamometer in the usual atmosphere, is a little bit smaller than in a vacuum, because of Archimedes.


Let's first imagine that one of these complexes is placed in neutral (pH 7) water. If the resulting solution possesses a pH of > 7, we can conclude that the anions show an "alkaline character" as you describe. If the resulting pH is < 7, we'll conclude that the material is contributing an "acidic character."

The way one might go about answering this question without dunking one of these salts in water is by starting with the a more well know complex such as the hexa-aqua complex of each metal and observing the behavior.

$\ce{AlCl3 + 6H2O -> Al^3+.6H2O + 3Cl^-}$ Chloride becomes a spectator anion

This $\ce{Al}$ water complex ($\ce{Al^3+.6H2O}$) possesses acidic protons, just like any polyprotic acid (eg. phosphoric acid.) Thus, each proton also has an acid dissociation constant or pKa. The first pKa for $\ce{Al^3+.6H2O}$ is approximately 5 and would thus acidify the solution. The pKa's of each of the subsequent protons donated to solution is higher than the last and eventually, no more protons would be donated to the solution.

With this basic understanding out of the way, let's look at the approximate pKa's of each of the neutral hydroxide salts of each of these metals.

For $\ce{Al(OH)3}$ the pKa is > 7. This means that by adding another hydroxide to the complex to make it a tetrakis hydroxide complex and would result in an even higher pKa and a more alkaline solution.

$\ce{Pb(OH)2}$ and $\ce{Zn(OH)2}$ are both weakly basic (pKa > 7) and possessing low solubilites. Adding more hydroxide ligands will make each of them more alkaline and should also make each of them more soluble. The zincate anion $\ce{Zn(OH)4^2-}$ is often formed by dissolving Zinc metal in a strongly alkaline solution.$\ce{^1}$

I hope this helps a little.

ref 1. http://depts.washington.edu/chem/facilserv/lecturedemo/BrassPenny-UWDept.ofChemistry.html


For starting purposes, you should write the half cells, either with their oxidation/reduction potentials.
Now if you want to compare the oxidizing species then you have to consider the oxidizing potentials of all the species and the one with highest value will be the spontaneous half cell and the reverse of the other one will occur.
Remember you should consider only one type of potential at a time(i.e. either oxidation or reduction) for comparison.


Which has higher rate of hydrogenation - methyl propene or trans-2-butene?

We should be able to determine which molecule will hydrogenate faster by comparing the activation energies for hydrogenation of these two molecules.  A catalyst does not alter the reaction mechanism, it merely reduces the activation energy for the process.

The heat of formation of 2-methyl-propene is ca. 1.3 kcal/mol higher than that of trans-2-butene, presumably because of the adverse steric interaction between the geminal methyl groups in 2-methyl-propene.  The transition state for this reaction will start to resemble the corresponding alkane in its eclipsed conformation (due to the cis-hydrogen delivery).  If we only consider methyl-methyl steric interactions, then for 2-methyl-propene as we approach the transition state the two geminal methyl groups, initially separated by a 120 degree ($\ce{sp^2}$) angle, will displace towards a 109 degree ($\ce{sp^3}$) angular separation.  Therefore, the methyl-methyl steric interaction should increase in 2-methyl-propene as we approach the transition state.  2-methyl-propene's  transition state will be more destabilized than the ground state, consequently increasing the activation energy required to reach it.  With trans-2-butene, there are no methyl-methyl steric interactions in the ground state or transition state; its activation energy, the energy separating the ground state and transition state, should not be increased or decreased.

Based on this analysis, the ground state - transition state energy separation would be expected to increase (slightly) for 2-methyl-propene as compared to trans-2-butene.  Consequently, faster hydrogenation of trans-2-butene would be predicted.  However the effect is expected to be very small.


Effect of effective nuclear charge increase on the stabilization of 2s and 2p orbitals

2s orbitals are stabilized more than 2p orbitals by the effective nuclear charge because of better penetration.

Everything you wrote above is correct except for,

  2p orbitals are more stabilized because they penetrate less in the
  space of the 1s orbital

It is correct that they penetrate less, but they are not stabilized more.  Because they penetrate less, they are stabilized less by the effective nuclear charge.  Electron penetration correlates with electron stabilization.

Edit: Response to OP's first comment

  explain then why 2nd period non-metals can't easily form higher
  oxidation states while 3p elements can?

First off, second period non-metals can form higher oxidation states, for example $\ce{NO3^{-}}$ where the nitrogen is in the +5 oxidation state.  But I understand what you're getting at, and the answer comes back to the difference in effective nuclear charge that 2p and 3p electrons feel.  3p electrons are further from the nucleus and screened by all of the 2s and 2p electrons.  3p electrons are therefore held more loosely than 2p electrons.  This leads to lower electronegativities,

image source

lower ionization potentials

image source

and higher oxidation states being more common for third period non-metals compared to second period non-metals.

Edit: Response to OP's second comment

  3p electrons penetrate well into the space near the nucleus (link)
  Therefore it
  should be easier for 2p elements to form higher oxidation states than
  3p elements, but it's not, how?

Your link (Rose-Hulman) to the 2p-3p radial electron distribution is good, it shows that the 3p electrons are "on average" further from the nucleus than 2p electrons - which is the key point.  Here is a 3-dimensional representation of the same thing.  Look how much further away the 3p electrons are from the nucleus than the 2p electrons.  Given that the 3p electrons are 1) better screened (more interior electrons) and 2) much further away from the nucleus than the 2p electrons, it will much easier to remove more 3p electrons than 2p electrons and achieve higher oxidation states.  This is consistent with the electronegativity and ionization potential arguments presented in my first edit.

Edit: Response to OP's third comment

  the effective nuclear charge has bigger increase for 2p than 2s
  (except N-O) according to this chart. How so?

The key here is that the s orbital is spatially symmetric, it has a spherical pattern.  An electron in an s orbital can screen outer electrons from the nucleus equally in all directions.  On the other hand, a p orbital is directional, it is not spherically symmetric.  An electron in the 2px orbital will not screen outer electrons from the nucleus equally in all directions.

As we proceed from boron to carbon, that first 2s electron is effective in screening the second 2s electron because of its spherical orbit (2s effective nuclear charge increase=0.641).  However, the first 2p electron is not as effective in screening the nucleus because of its non-spherical shape.  So when we add the second 2p electron, it "feels" the nucleus more strongly than the second 2s electron did so the change in effective nuclear charge is greater for the second p electron than it was for the second 2s electron (2p effective nuclear charge increase=0.715).

You noticed the "anomaly" for the 2p electrons as we go from nitrogen to oxygen.  This is because with 3 2p electrons, all of the nitrogen 2p orbitals (px, py, pz) have an electron and produce a spherical electron distribution pattern.  The nucleus is better-screened by the 3 2p electrons in nitrogen, so when we add one more p electron to yield oxygen, the effective nuclear charge increase is noticeably smaller (N -> O 2p effective nuclear charge increase=0.619).


How is silver chloride more soluble in brine than water?

I mean won't the common ion effect operate?

The common ion effect always applies for pure substances. With the added sodium chloride, you have changed the effective solvent and changed the ions.

  Or may be this is due to some kind of complex formation as NaCl solution is highly concentrated?

Exactly, before your ionic equation was:
$$\ce{Ag+ (aq) + Cl- (aq) <=> AgCl (s)}$$
but with brine, the ionic equation is:
$$\ce{AgCl (s) + Cl- (aq) <=> [AgCl2]- (aq)}$$
Which has a different solubility constant, and where as before more chloride made the $\ce{AgCl}$ less soluble, if you note that you have switched regimes and $\ce{Cl-}$ has switched from being a reactant to a product, which means now with additional chloride silver chloride becomes more soluble.


Can two lone pairs in different orbitals make a chiral centre?

Chirality refers only to the positions of the nuclei. So, if we have the situation you described where one Carbon is bonded to two other atoms (e.g. CX$_2$ or CXY) it can never be chiral. (i.e. a three atom molecule can always be superimposed on its mirror image by a single rotation)

Now, why don't we consider the atomic and hybrid orbitals  populations when assigning chirality? This comes down to the fact that the atomic and hybrid orbitals are just a helpful tool for rationalizing/predicting the structure and reactivity, but they are not representations of the actual wavefunction. The true wavefunction is defined by a set of ''molecular orbitals.'' These orbitals always have the same symmetry group as the nuclear coordinates. In this case, there is a reflection through the plane defined by the three atoms (some jargon: this is called the C$_{s}$ symmetry point group). So if the nuclear positions are non-chiral, so is the wavefunction.

(More on chirality and symmetry: http://symmetry.otterbein.edu/tutorial/applications.html)


Constructing potential energy surface of a periodic system, how to deal with atoms moving out of the supercell?

I think that it is best to consider the calculation of the potential energy (respecting the periodic boundary conditions) as a separate matter from keeping the atoms within the supercell. For simplicity I'm going to assume a cuboidal box with sides $(L_x,L_y,L_z)$, with periodicity in all three Cartesian directions.

Some people like to put the atoms back into the primary cell whenever they move out. This is done by replacing the atomic position vector $(x,y,z)$ by $(x,y,z)+(n_xL_x,n_yL_y,n_zL_z)$, where $(n_x,n_y,n_z)$ are integers, chosen to make the new coordinate values lie in the desired range, often $(\pm\frac{1}{2}L_x,\pm\frac{1}{2}L_y,\pm\frac{1}{2}L_z)$. If you do this every MD step, the integers will usually just need to take the values $0,\pm1$. Other people prefer not to apply this correction, and just keep track of the copy of an atom that diffuses into the periodic-image supercells.

Even if you do this correction to the atom positions, you still need to apply a similar correction every time you calculate a vector between atoms, for the purposes of computing two-body, three-body, potentials and forces. So, a properly written calculator for these quantities will not rely on the atoms lying within the same supercell. If the atomic coordinates are allowed to evolve freely, without periodic boundary corrections, then the integers $(n_x,n_y,n_z)$ will take whatever values are needed to bring the corrected interatomic vector into "minimum image" form, with components in the range $(\pm\frac{1}{2}L_x,\pm\frac{1}{2}L_y,\pm\frac{1}{2}L_z)$. There are standard prescriptions for doing this.

Often, the energy/force calculation is speeded up by a neighbour list calculation, which involves assigning each atom to a subcell of the original supercell. On the face of it, this seems to require that we put the atoms back into the supercell, before doing the assignment. But actually, a properly written function to identify nearest neighbours will have this mapping built in to it, and may just work with copies of the original coordinates. So, again, there should be no need to put the atoms back into the original supercell "by hand", if the neighbour identification is done this way, and if the vectors between neighbours are corrected as just described.

My impression, from a cursory look at ase, is that this mapping is indeed built in to the functions used to identify neighbours "within range". There is a simple Lennard-Jones calculator, for instance, which seems to rely on the neighbour list function for this purpose. The relevant lines are

        neighbors, offsets = self.nl.get_neighbors(a1)
        cells = np.dot(offsets, cell)
        d = positions[neighbors] + cells - positions[a1]

where I think that offsets contains the integers $(n_x,n_y,n_z)$ and I think that cell contains the cell lengths $(L_x,L_y,L_z)$ in the form of a $3\times3$ matrix (just the diagonal elements would be nonzero in this case).

If you are not using such a function to locate neighbours "within range" and pre-calculate the necessary minimum image corrections to the interatomic vectors, you will need to put it into your calculation of pair and three-body energies and forces, somewhere, yourself. As I mentioned, this is not the same as putting the atoms back into the supercell, when they move out.

Caveat! I am not an expert in ase, and if you want specific advice on that package, you should look at the package documentation, and at simple examples of calculators provided within it. Generally speaking, I believe that advice on debugging programs, and on specific packages, is considered off-topic for this site, so I think that's as far as I should go in any case.


Except pure alloys, are there any compounds with more metal elements proportion of atoms than nonmetal elements in proportion of atoms?

There are some rare but very interesting cases in which two metals react when mixed instead of forming an alloy, creating a non-metallic compound made only with metal atoms. Examples are caesium auride ($\ce{CsAu}$), caesium platinide ($\ce{Cs2Pt}$) and several barium platinides ($\ce{BaPt}$, $\ce{Ba2Pt}$, $\ce{Ba3Pt2}$, and possibly others), all of which best described as ionic salts rather than alloys. These are a remarkable consequence of relativistic effects in heavy atoms.


Why super glue does not stick inside their own bottle?

From explainthatstuff.com:

  Why doesn't glue stick to the tube?

  Adhesives are designed to work when they leave the tube—and not
  before. Different adhesives achieve this in different ways. Some are
  dissolved in chemicals called solvents that keep them stable and
  non-sticky in the tube.

  When you squeeze them out, the solvents
  quickly evaporate in the air or get absorbed by the surfaces you're
  sticking to, freeing the adhesives themselves to do their job.

  Plastic modeling glue works like this. It contains molecules of
  polystyrene in
  an acetone solvent. When you squeeze the tube, the glue spurts out and
  you can usually smell the very strong acetone as it evaporates.

  Once it's gone, the polystyrene molecules lock together to make strong
  chemical bonds.
  Glue doesn't smell when it's dry because all the
  solvent has vanished into the air.

  Some glues (such as synthetic,
  epoxy resins) have to be mixed together before they work. They come in
  two different tubes, one containing the synthetic resin and the other
  containing a chemical that makes the resin harden. The two chemicals
  are useless by themselves but, mixed together, form a tough, permanent
  adhesive.

From sciencefocus.com:

  How super glue works

  The main ingredient in super glue is a chemical called cyanoacrylate, an acrylic resin that forms a bond (cures) almost
  instantly;
  The only trigger it requires are the hydroxide ions found in water;
  Traces of water are found on almost everything and in the surrounding air, which causes the glue to become sticky;
  It works best in moist conditions, although it is not advised to start gluing on a soaking wet surface.

Therefore the lack of water in the tube, or the chance of solvents evaporation, doesn't allow the glue to stick to the tube.

In other words: that strong smell that comes out when we open the tube's cap is due to the solvents evaporation, so once they're gone the glue sticks. If a superglue tube is left open for a long time (exposed to air, allowing the water to come in contact with the glue), it's impossible (or almost) to squeeze it out since it got stuck to the tube.


ATP chemical structure: phosphate groups with Oxygen anions or with OH groups?

in vivo : ATP with oxygen anions.
in vitro : Depends on pH of solution
IN VIVO :
DNA = Deoxy Ribonucleic Acid.
DNA is acidic , because its monomers contain three phosphates bound together and this so-called triphosphoric acid is extremely acidic.
In living organisms, DNA is always accompanied by histones(positively charged proteins) or sometimes by Mg+2 complexes. Therefore the dNMPs ( deoxyribo Nucleoside Mono Phosphates) that constitute the genetic material , are negatively charged .
For rNMPs (ribo Nucleoside Mono Phosphates) the negative charged form is also more preferable.
Generally speaking in living creatures , ATP = -4 , ADP = -3 , AMP = -2 charged.
(It is worth mentioning that Pi which is a by-product in cellular energetics is HPO4(-2). )
You can see in Molecular Cell Biology Lodish et al 8th ed section 12.1 that the overall reaction of cellular respiration is:

Notice the charges.
IMPORTANT : Sometimes during the metabolism of these compounds Proton(H+) may be added to oxygens of phosphate groups only because of the mechanism of the reaction which is temporary. But usually the charges are as I mentioned before.
IN VITRO :
If you purify ATP in a solution isolated from the cell, the number of acidic hydrogens attached to the molecule is related to the pH of solution. you can predict the molar concentration of each chemical species the same way you predict the [H3PO4] , [H2PO4 -] , [HPO4 2-] , [PO4 3-] , [H+] in a solution of H3PO4 and H2O (a typical analytical chemistry question) Either by a complex utilization of Henderson–Hasselbalch equation or by Charge and Mass equations.
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