About the use of pascal or atmosphere in chemistry

Generally, the use of quantity equations is preferred and is strongly recommended. They have the advantage of being independent of the choice of units.

For example, instead of “speed in kilometre per hour is 3.6 times the quotient of distance in metres by time in seconds”, which corresponds to the numerical value equation

$$v/(\mathrm{km/h})=3.6\frac {l/\mathrm m}{t/\mathrm s}$$ or $$\left\{v\right\}_{\mathrm{km/h}}=3.6\frac{\left\{l\right\}_{\mathrm m}}{\left\{t\right\}_{\mathrm s}}$$

you should write “speed $v$ is the quotient of distance $l$ by time $t$”, preferably in form of the quantity equation

$$\displaystyle v=\frac lt$$

Generally, you can work with any units you like; however, this requires the use of conversion factors. The use of conversion factors is never required when only coherent SI units are used; therefore, the use of coherent SI units (the SI base units and the SI coherent derived units, including those with special names) are preferred.

In order to avoid large or small numerical values, decimal multiples and submultiples of units may be formed with the SI prefixes. However, when prefixes are used with SI units, the resulting units are no longer coherent, because a prefix effectively introduces a numerical factor in the expression for the unit in terms of the base units.

Furthermore, various non-SI units are accepted for use.

In the International System of Quantities, pressure $p$ is defined as
$$p=\frac{\mathrm dF}{\mathrm dA}$$
Where $\mathrm dF$ is the force component perpendicular to the surface element of area $\mathrm dA$.

The coherent derived units in the SI for the pressure has the special name pascal and the symbol $\mathrm{Pa}$. According to the definition of pressure, the pascal can be expressed in terms of other SI units as $1\ \mathrm{Pa}=1\ \mathrm{N/m^2}$. The pascal can also be expressed in terms of SI base units as $1\ \mathrm{Pa}=1\ \mathrm{m^{-1}\ kg\ s^{-2}}$. The last form is especially useful to check whether units in equations and calculations are converted correctly.

The SI prefixes may be used with any of the names and symbols, but when this is done the resulting unit will no longer be coherent.

The bar (symbol $\mathrm{bar}$) is included in the list of non-SI units that are used by special interest groups for a variety of different reasons. Although the use of SI units is to be preferred for reasons already emphasized, authors who see a particular advantage in using these non-SI units should have the freedom to use the units that they consider to be best suited to their purpose. In particular, the bar is included because since 1982 one bar has been used as the standard pressure for tabulating all thermodynamic data. (Prior to 1982 the standard pressure used to be the standard atmosphere, equal to $1.01325\ \mathrm{bar}$, or $101\,325\ \mathrm{Pa}$.)

The following other non-SI units can be used, which requires the use of conversion factors. However, the use of these units is deprecated.

$$\begin{array}{lll}
\hline
\text{Unit} & \text{Unit} & \text{Conversion} \\
\text{name} & \text{symbol} & \text{factor} \\
\hline
\text{standard atmosphere} & \mathrm{atm} & 1\ \mathrm{atm} := 101\,325\ \mathrm{Pa}\\
\text{kilogram-force per square metre} & \mathrm{kgf/m^2} & 1\ \mathrm{kgf/m^2} = 9.80665\ \mathrm{Pa}\\
\text{technical atmosphere} & \mathrm{at} & 1\ \mathrm{at}=1\ \mathrm{kgf/cm^2} = 98\,066.5\ \mathrm{Pa}\\
\text{conventional millimetre of water} & \mathrm{mmH_2O} & 1\ \mathrm{mmH_2O}:=10^{-4}\ \mathrm{at} = 9.806 65\ \mathrm{Pa}\\
\text{conventional millimetre of mercury} & \mathrm{mmHg} & 1\ \mathrm{mmHg} \approx 13.5951\ \mathrm{mmH_2O} \approx 133.3224\ \mathrm{Pa} \\
\text{torr} & \text{Torr} & 1\ \mathrm{Torr}:=\frac1{760}\ \mathrm{atm} \approx 1\ \mathrm{mmHg} \approx 133.3224\ \mathrm{Pa} \\
\text{pound-force per square inch} & \text{psi}, \mathrm{lbf/in^2} & 1\ \mathrm{lbf/in^2}\approx6\,894.757\ \mathrm{Pa} \\
\hline
\end{array}$$

Note that expressions for units shall contain nothing else than unit symbols and mathematical symbols. Any attachment to a unit symbol as a means of giving information about the special nature of the quantity or context of measurement under consideration is not permitted.

For example, write “the gauge pressure is $p_\mathrm e=0.5\ \text{bar}$”, not “$p=0.5\ \text{bar(g)}$”.

(The symbol $p_\mathrm e$ is recommended for gauge pressure, defined $p-p_\mathrm{amb}$, where $p_\mathrm{amb}$ is the ambient pressure. Thus the gauge pressure is positive or negative depending on whether $p$ is larger or smaller than $p_\mathrm{amb}$.)


Nucleophilic addition reaction of ketone to form alcohol - no acid treatment?

The nucleophile is the H-, a hydride. In your SP3 ketone, the tiny Hydride will have no issue producing a high yield of the alkoxide.

The heat will simply make the reaction faster and drive the equilibrium further.

Ethanol + water is a mild acid, so no issue with protons to complete your alcohol, and the heat makes that mix slightly more active as a proton donor.

You end up with a new equilibrium of your 2 alcohols competing for the proton, however with plenty of methanol and some careful washing with hexane or toluene you should be able to get your alcohol out of that mixture...


How is Negative Temperature Hotter than Infinite Temperature?

The other answer given is correct, but for completeness, I will give the thermodynamic, rather than statistical mechanics, explanation.

Thermodynamics gives us that $$\frac{\partial E}{\partial S}=T$$

This means that  when entropy--as a function of energy--increases, the temperature must also increase.

On a macroscopic level, this makes perfect sense because you're never going to think about a system reaching a maximum temperature because energy and entropy are always going to increase together.

Of course, it must be remembered that this definition of temperature was being given long before people were thinking about quantum mechanics. Even more, they were only concerned with temperatures because they realized that they could use a temperature difference to get work out of a system. A good example of this is the Carnot heat engine. The reason I reference the Carnot heat engine is that it demonstrates that when temperature was being defined mathematically, which was around this time, they were concerned with heat flow. We know that heat flows from a hot body to a cold body, so this means, almost exclusively, that heat flows from high energy to low energy.

For instance, in a system with only two energy states containing $N$ particles, the state of maximum entropy is when $\frac N2$ particles are in both states 1 and 2. The state of maximum energy, however, is when all $N$ particles are in energy level 2.

That means, if we push beyond the state of maximum entropy, energy will increase as entropy decreases, hence, $\frac{\partial E}{\partial S} < 0$.

So, negative kelvin can be "hotter" than infinite temperature because our definition of temperature has nothing to do with "hotness" as we feel on the macroscopic level, it is merely an exchange between energy and entropy.

Furthermore, negative kelvin is hotter than infinite temperature, not because it would feel hotter, but because heat flows from the negative temperature object to the positive temperature object because we can demonstrate mathematically that the negative temperature object is in a higher energy state. It makes sense to think about it that way because 90% of chemistry is just looking for the low-energy state.

In regards to the issue of our temperature scale going in a loop, there is a solution to this using the thermodynamic beta, $$\beta=\frac{1}{kT}$$Consider this graph which shows that temperature as a function of energy is asymptotic and thus discontinuous, while $\beta$ is continuous as a function of energy.

Also worth pointing out is that $\beta$ has units of energy, and is thus more physically relevant than temperature because energy is fundamental to all matter, while temperature is basically nothing more than a mathematical construct to describe "hot" and "cold".


Calculating the pH of a 'buffer' solution?

Question: Determine the $\mathrm{pH}$ of the solution resulting when $\pu{100 cm^3}$ of $\pu{0.50 mol dm-3}$ $\ce{CH2ClCOOH}$ is mixed with $\pu{200 cm^3}$ of $\pu{0.10 mol dm-3}$ $\ce{NaOH}$.

I'm not sure what level of chemistry is OP's in, but the given solution for the question is for the chemistry students with appreciable knowledge of stoichiometry of chemical reactions and buffers. That's probably why OP seems so confused of the solution. The answer elsewhere explains to certain level but has not explore the total picture. Hence, I thought it would be beneficial to OP if I explain it a little further:
$\ce{CH2ClCOOH}$ is a weak acid (about 100 times acidic than acetic acid) with $K_\mathrm{a}$ of $1.3 \times 10^{-3}$ (which is $1.8 \times 10^{-5}$ for acetic acid). On the other hand, $\ce{NaOH}$ is a strong base, which reacts with strong or weak acids irreversibly. When add $\ce{CH2ClCOOH}$ and $\ce{NaOH}$ together, following reaction would take place:
$$\ce{CH2ClCOOH (aq) + NaOH (aq) -> CH2ClCOONa (aq) + H2O (l)} \tag1$$
The net reaction is:
$$\ce{CH2ClCOOH (aq) + OH- (aq) -> CH2ClCOO- (aq) + H2O (l)} \tag2$$
The acid to base mole ratio of the reaction is 1:1. However, added amounts of acid and base are:
Weak acid, $\ce{CH2ClCOOH}$: $\pu{0.50 mol dm-3} \times \pu{0.100 dm3} = \pu{0.050 mol}$
Strong base, $\ce{NaOH}$: $\pu{0.10 mol dm-3} \times \pu{0.200 dm3} = \pu{0.020 mol}$
That makes $\ce{NaOH}$ the limiting reagent (recall your general chemistry knowledge). Thus, $\pu{0.020 mol}$ of $\ce{NaOH}$ reacts with $\pu{0.020 mol}$ of $\ce{\ce{CH2ClCOOH}}$ to give $\pu{0.020 mol}$ of $\ce{\ce{CH2ClCOONa}}$ or $\ce{\ce{CH2ClCOO- + Na+}}$, if you look at the net reaction.
The unreacted amount of $\ce{\ce{CH2ClCOOH}}$ in the mixture ($\pu{0.300 dm3}$ in total volume) is: $\pu{0.050 mol} - \pu{0.020 mol} = \pu{0.030 mol}$.
Therefore, the concentration of acid, $\ce{\ce{CH2ClCOOH}}$: $\frac{\pu{0.030 mol}}{\pu{0.300 dm3}} = \pu{0.10 mol dm-3}$
Simillarly, the concentration of salt, $\ce{\ce{CH2ClCOONa}}$ or $\ce{\ce{CH2ClCOO-}}$: $\frac{\pu{0.020 mol}}{\pu{0.300 dm3}} = \pu{0.067 mol dm-3}$
Since, $\ce{\ce{CH2ClCOOH}}$ is a weak acid, it would ionize in aqueous solution as follows:
$$\ce{CH2ClCOOH (aq) + H2O (aq) -> CH2ClCOO- (aq) + H3O+ (aq)} \tag3$$
Suppose, $\alpha$ amount (in $\pu{mol dm-3}$) of $\ce{\ce{CH2ClCOOH}}$ ionizes at $\ce{\ce{CH2ClCOOH}}$, the equilibrium concentrations of $\ce{\ce{CH2ClCOOH}}$, $\ce{\ce{CH2ClCOO-}}$, and $\ce{\ce{H3O+}}$ are $0.10 - \alpha$, $0.067 + \alpha$, and $\alpha$, respectively.
$$K_\mathrm{a} = \frac{\ce{[H3O+]}\ce{[A-]}}{\ce{[HA]}} $$
$$\therefore \ \ce{[H+]} = \alpha = \frac{K_\mathrm{a} \times \ce{[HA]}}{\ce{[A-]}} = \frac{1.3 \times 10^{-3} \times (0.10  - \alpha)}{(0.067  + \alpha)} \approx \pu{1.9 \times 10^{-3} mol dm-3}$$
$$\mathrm{pH} = -\log\alpha = -\log (1.9 \times 10^{-3}) = 2.72$$
On the other hand this can be easily solved using the Henderson-Hasselbalch equation as directed in the other answer elsewhere:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac {\ce{[A-]}}{\ce{[HA]}}\right) = 2.89 + \log \left(\frac {0.067}{0,10}\right) = 2.72 $$


How to use C1V1=C2V2 How do the units work?

Concentration has units mass over volume; or particles (moles by convention) over volume.

It's fine to use $g/mL$ for concentration in a basic dilution equation.

$C$ is never a mass.

So your second equation has proper units.

If you want mass of aspirin in a given solution with concentration $C$, the mass is given by $m = C\times V$


Can you find 5 mistakes in this organic molecule?

The five mistakes are:

Oxygen in the 5-member ring needs a positive charge [mentioned]
Carbon at the bottom should be a methylene, not methyl [mentioned]
Alkyne Group is not drawn in a linear fashion [mentioned in comments]
Carbon with $3^o$ alcohol has two substituents going out of the plane [mentioned in comments]
Chiral Carbon to the left of the oxygen in the 5-member ring has no stereochemistry indicated.

Now, 5 is not technically wrong but since the rest of the molecule is labelled for stereochemistry it is odd that this chiral center is not marked.


Determining the mechanism followed when alcohols undergo a reaction with sodium bromide and sulphuric acid

yeah so basically no reaction is purely SN1 or purely Sn2 its always a mixture of both and the overall order which we see is nothing but what that majority of reaction follows so now going by ur question, we do have to check about the substrate to inspect SN2 mechanism but in this case, since the reaction conditions are made in such a way that the overall reaction follows SN1 (as it is acid-catalyzed) hence we would compare the rate of reaction in terms of RDS (FORMATION OF CARBOCATION) and hence the more stable the carbonation is the faster the reaction proceeds.


Why is there only one bond length for sulfur hexafluoride?

This is an old question and is largely answered in the comments, but here's a full answer for the sake of completeness.

There are really two questions here:
1) is the given bonding orbital description correct for SF6?
2) why is SF6 octahedral?

The first question is easier: no, your description is incomplete as you've described it, but perhaps correct as intended by the author (I don't have the textbook to see the exact wording). It is correct if you explain further that there are three possible representations with two $sp$ bonding orbitals and two 3c/4e bonds. The three different representations look identical except that the $sp$ orbitals are aligned on different Cartesian axes (ie one for $x$, one for $y$ and one for $z$). The true structure is a composite of these three, meaning that each S-F bond is 1/3 $sp$ and 2/3 3c/4e (which is essentially 1/2 of a $p$ orbital) which adds up to a total of 2/3 of a bond with a contribution of 1/6 from $s$ and 1/2 from $p$.

I suspect that the textbook uses this representation because it connects readily to earlier discussions of integer hybridization, i.e. "$sp$" orbitals are familiar and easy to visualize. And presumably the concept of a 3c/4e bond has also been previously introduced. Making a composite of familiar concepts is perhaps easier than introducing the idea of delocalized molecular orbitals, but in this case they are simple enough that it's worth going through them to show that the result is the same.

First, the $s$ orbital is spherically symmetric, so it necessarily interacts the same with all six identical ligands. The lowest energy molecular orbital is thus represented by the $s$ orbital on sulfur and the six fluoride ligand orbitals all in the same phase (images below). That's where the 1/6 $s$ comes from. The $p$ orbitals each are able to interact with two of the ligands, one on each end of orbital. The other four ligands do not contribute at all, because they are exactly on the nodal plane of the $p$ orbital. Since each $p$ interacts with two ligand orbitals, each S-F bond gets 1/2 $p$ to go with the 1/6 $s$, exactly the result we got from the composite of hybrids. Below are drawings of the four bonding molecular orbitals:

Now on to the second question: why is SF6 octahedral if it only makes a total of 4 bonds?

Given that there is a lot of negative charge on the ligands and positive charge on S, a simplistic representation is to make the atoms solid charged spheres. A simple electrostatic potential energy minimization would then lead to an octahedral arrangement of the ligand spheres around the sulfur sphere. That arrangement minimizes the repulsive interactions between the ligands and maximizes the attractive interaction between ligand and sulfur.

Without a lot of empirical data about atomic interactions, though, we cannot rule out the possibility that there could be favorable bonding interactions with a strong distance dependence that would make it more favorable to, for example, push four of the ligands even closer to the sulfur and push the other two farther away. However, it turns out that the distance dependence of the bonding interaction is similar enough to the distance dependence of the electrostatic interaction that the geometric shape of the minimum energy conformation is the same - all six ligands at equal distance from the sulfur and spaced evenly around it.


Which one is the stronger Lewis base: diethyl ether or tetrahydrofuran?

Which is "better"? In some specific reaction, with all other circumstances identified. Ether is a little too volatile for my preferences, but THF is not bad. I like the smell of ether, but the odor of THF is OK. Little problem with oxygen (fire and peroxides) with both of them, so you must be careful with both.
The difference in enthalpy of formation (11.6 kJ/mol) is small compared with activation energies of typical organic reactions (~100 kJ/mol), so "better" might be related to human preferences more than than chemical performance.
A chemist might already have a feel for the molecular configurations of these molecules, but a physicist could benefit from making the models or looking at them:

The ability of THF to poke the BF3 molecule better than ether can is apparent. Whether this is the controlling factor in which is "better" remains to be seen. I'll go out on a limb and say that either one will work, although the results (convenience, yield, expense, etc.) could easily be +/- 20-50%, depending on the situation. In a lab, one might work better; in industry, the other might be preferable.
Either solvent could probably be optimized for best performance under slightly different conditions, and these optimum conditions will not be the condition you choose for the first experiment. Probably a dozen experiments will suggest which solvent could be the better choice. The literature should suggest the better (or more typical) choice for your needs.
Ref. https://en.wikipedia.org/wiki/Activation_energy


The rusting of steel may be promoted by brushing with HCl followed by brushing with hydrogen peroxide solution. What is the chemistry involved?

I understand the point of your question, here is a discussion of the associated science.
The process remains a classic electrochemical process where here H2O2 is a replacement for oxygen and the require $\ce{H+}$ is provided by the $\ce{HCl}$ which is also a source of chloride ions noted to promote corrosion as well as forming salts which are excellent electrolytes.
Some references, starting with a good depiction of the underlying electrochemistry, to quote:

Iron (or steel) rusting is an example of corrosion -- an electrochemical process involving an anode (a piece of metal that readily gives up electrons), an electrolyte (a liquid that helps electrons move) and a cathode (a piece of metal that readily accepts electrons). When a piece of metal corrodes, the electrolyte helps provide oxygen to the anode. As oxygen combines with the metal, electrons are liberated. When they flow through the electrolyte to the cathode, the metal of the anode disappears, swept away by the electrical flow or converted into metal cations in a form such as rust.

And some comments from ScienceDirect on the referenced chloride properties:

The chloride ions present in the water promote local corrosion such as crevice corrosion and pitting, and the higher the chloride content, the more corrosive the water.

I hope this helps.


Why is Electrostatic Potential Energy positive when the charges are like?

There is no way to know the actual potential energy of a system, we can only determine it relative to some predefined zero value.  In the case of electrostatic potential, zero potential energy is defined as the energy between charges when their separation is infinite.

The charge of the electron doesn't change in this scenario; it is always $1.6\times 10^{-19}\ C$.

When oppositely charged particles near one another, they lose potential energy (their energy becomes more negative.)  When particles with the same charge near one another, they gain potential energy.  The attractive force between oppositely charged particles does work as their separation decreases, and the repulsive force between similarly charged particles requires work to be done on them in order to move closer to one another (so potential energy goes up.)


What is the range of photon energies can be effectively used to split a diatomic molecule like hydrogen?

If the photon's $E=\hbar\omega=h\nu=\frac{hc}{\lambda} $ energy is greater than the energy of the bond which you want to break up it will break it up. It's always best to use higher energy photons but not too high because it may break up bonds which were unwanted to break up in your reaction . So the best for you is to use blue or higher energy UV light for your experiment.You need to deal with the rate of the reformation of bromine molecules, there will be an equlibrium between the production and decomposition of bromine molecules depending on the intensity, not the energy of the light. If the photons has higher energies than a given bond they will always break it up, this is why gamma radiation is so dangerous. The absorbtion spectrum doesn't have a lot connections to this phenomena. You can use the Maxwell-Boltzmann distribution
$$f_{v}(v_{i})=\sqrt{\frac{m}{2\pi k_{B}T}}exp\left[-\frac{mv_{i}^{2}}{2k_{B}T}\right]$$
to determine the Doppler-effect's impact on the absorption spectrum depending on the temperature and to construct an absorbtion function explicitly. The doppler effect will affect the percieved frequency and energy of the light by gas particles so if you want to be as accurate as possible, you need to calculate it.The Maxwell-Boltzmann distribution is only suitable in low temperatures (below several thousand Kelvins).


Does any chirality-dependent inorganic chemical reaction exist?

A crystalline form of silicon dioxide known as α-quartz is optically active and can spontaneously crystallize in one of the two optically active form. However, which one of the two you get is defined purely by chance.

This, however, is different from "asymmetry" in particle physics, see here for details.


Titration calculations - working out cation

You must have forgotten a factor $2$ somewhere.
You are right is thinking that $24.10$ mL is needed for the final titration. But now let's redo the calculations.
$24.10$ mL $\ce{NaOH} ~0.25$ M contains $24.10·0.25 = 6.025$ millimole $\ce{NaOH}$. Obviously this $\ce{NaOH}$ has neutralized the same amount $6.025$ millimole $\ce{HCl}$. As a consequence, there was $10$ times more $\ce{HCl}$ in the $250$ mL flask. This is $0.06025$ mol $\ce{HCl}$, and was the residual amount of $\ce{HCl}$ remaining after reaction of HCl plus the unknown carbonate. It also means that after adding $0.100$ mole $\ce{HCl}$ in the initial flask, the difference $0.100 - 0.06025 = 0.03975$ mol $\ce{HCl}$ had been used to destroy the carbonate.
But whatever the formula of the initial carbonate, $2~\ce{H+}$ are needed to react with  $1~ \ce{CO3^{2-}}$. So the amount of unknown $\ce{XCO3}$ or $\ce{Y2CO3}$ carbonate in the original powder is half the amount of $\ce{HCl}$ and it is : $0.03975/2$ mol = $0.019875$ mol $\ce{XCO3}$.
As the mass of this carbonate is $2.50$ g, its molar mass is: M = m/n = $2.50/0.019875 = 125.8$ g/mol. As the molar mass of $\ce{CO3}$ is $60$ g/mol, it means that the molar mass of $\ce{X}$ is $125.8 - 60 = 65.8$ g/mol.
This final value may be twice the atomic mass of a monovalent cation, or the atomic mass of a bivalent cation. As there are no monovalent cation with an atomic mass of about $32.9$, the unknown element must be a bivalent cation. And the only element with such an atomic mass is Zinc : A(Zn) = $65.39$
So the unknown carbonate is $\ce{ZnCO3}$


Is there a relationship between London dispersion forces and hydrophobic interactions?

To summarize the excellent but rather lengthy and detailed Stack Exchange answer referred to by Andy above, hydrophobic interactions do not require London forces between the hydrocarbon chains. They are a result of the very strong attraction of water for itself.

I would further quote the following comment to that entry (by https://chemistry.stackexchange.com/users/5633/thomij), which I think sums it up nicely:

"This is a good answer, but the last part about van-der waals forces being required to pull the hydrophobic solvent together is incorrect. If you had particles with no attractive forces between them in a solvent that has attractive forces, as long as the enthalpy drop for phase separation is larger than the entropic penalty, they will separate. Van der waals forces help to make the enthalpy drop larger, but they aren't necessary. –  thomij Jun 23 '14 at 19:02"


What are the border definitions in the Ramachandran plot?

I wish I could answer your question outright, but this is outside my area of expertise. The best I can give you is a place to start. I expect any recent book on proteomics would cover this topic. For example Introduction to Proteomics: Principles and Applications (Methods of Biochemical Analysis) by Mishra and Blobel, Wiley 2010.

I found some promising papers with a little searching with SciFinder. Neither I nor my institution subscribe to Protein Science, but the abstract looks promising. I have skimmed the article in PNAS. Hopefully one of these articles has your answer.

Zhou, et al, "Revisiting the Ramachandran plot from a new angle", Protein Science 2011 1166-1171.

Porter and Rose, "Redrawing the Ramachandran plot after inclusion of hydrogen-bonding constraints", PNAS, *2011, 108, 1, 109-112


Can an organic compound undergo ring contraction to give more stable carbocation?

Your idea of ring contraction to give more stable carbocation is correct and similar contraction has observed before (see example 4 below):

Courtesy of Department of Chemistry, Michigan State University, East Lancing, MI: link to webpage


Is Linus Pauling's General Chemistry still readable and up-to-date today?

I currently have the Dover copy of Pauling's book on my shelf.  It indeed is a nice read but it is in no way 'up-to-date' as it was written many years ago.  That being said, a lot of conceptual information presented in the book still apply just as much today as they did 6 decades ago (I would stray away from using the book as a source for benchmark measurements and such).  It is a book that should find its way onto any chemist's bookshelf, and it makes for a fascinating read into many areas of general chemistry.


Defining the Turbomole named BP86 functional in Gaussian

From the Gaussian 09 manual:

  User-Defined Models. Gaussian 09 can use any model of the general form:
      P2EXHF + P1(P4EXSlater + P3ΔExnon-local) + P6EClocal + P5ΔECnon-local

  The only available local exchange method is Slater (S), which should be used when only local exchange is desired. Any combinable non-local exchange functional and combinable correlation functional may be used (as listed previously).

  The values of the six parameters are specified with various non-standard options to the program:

  IOp(3/76=mmmmmnnnnn) sets P1 to mmmmm/10000 and P2 to nnnnn/10000. P1 is usually set to either 1.0 or 0.0, depending on whether an exchange functional is desired or not, and any scaling is accomplished using P3 and P4.
  IOp(3/77=mmmmmnnnnn) sets P3 to mmmmm/10000 and P4 to nnnnn/10000.
  IOp(3/78=mmmmmnnnnn) sets P5 to mmmmm/10000 and P6 to nnnnn/10000.

  For example, IOp(3/76=1000005000) sets P1 to 1.0 and P2 to 0.5. Note that all values must be expressed using five digits, adding any necessary leading zeros.

  Here is a route section specifying the functional corresponding to the B3LYP keyword:

  #P BLYP IOp(3/76=1000002000) IOp(3/77=0720008000) IOp(3/78=0810010000)

  The output file displays the values that are in use:

IExCor=  402 DFT=T Ex=B+HF Corr=LYP ExCW=0 ScaHFX=  0.200000
ScaDFX=  0.800000  0.720000  1.000000  0.810000

  where the value of ScaHFX is P2, and the sequence of values given for ScaDFX are P4, P3, P6 and P5.

For BP86 specified normally, this looks like:

IExCor=  404 DFT=T Ex=B Corr=P86 ExCW=0 ScaHFX=  0.000000
ScaDFX=  1.000000  1.000000  1.000000  1.000000

Those 3 IOps don't completely define the functional, however, IOp(3/74) is also needed. See the IExCor field, which corresponds to Becke 1988 exchange (400) combined with Perdew 81 + Perdew 86 correlation (4).


Possible E2 products of S-2-bromo-2-methyl-1-phenylbutane

The major products will be (E)-2-methyl-1-phenyl-1-butene and (Z)-2-methyl-1-phenyl-1-butene, formed in a 1:1 ratio. There is are reasons both for why elimination occurs specifically across the first and second positions, and as for why the product is formed in a 1:1 ratio.

The hydrogen on position 1 are called benzylic hydrogen, and are much more acidic than aliphatic hydrogen due to the resonance stabilization of the benzyl anion. This will cause these hydrogen to be removed much more quickly than the others, resulting in a majority of 1,2-unsaturated products. This is of course not to say that the other products are not formed, but they are to a lesser extent.

The products are formed in a 1:1 ratio because there are two ways for the $\ce{H}$ and $\ce{Br}$ to be anti-periplanar. This is easily seen when constructing the Newman projections of the starting material:

$\hspace{3.6cm}$

Upon extraction of the hydrogen, the electron pair forms a $\pi$-bond with the adjacent carbon, ejecting the $\ce{Br-}$, and locking in the stereochemistry.


Which is the better leaving group in the following Cannizzaro reaction?

If you'd like to read more about this mechanism, you should out this paper:

Swain, C. G.; Powell, A. L.; Sheppard, W. A.; Morgan, C. R. Mechanism of the Cinazzaro Reaction, 1979, Am. Chem. Soc. 3576

They did a kinetic study with deuterium oxide. They gave two equivalent mechanisms:

and

I think that this mechanism, which is essentially what you have shown except with intramolecular hydride shift (it cannot shift through solution) is the way I would represent it. It makes sense that this can happen if you think of the transition state. In my mind, it makes sense that this hydride transfer can occur given the very electrophilic nature of the aldehyde.

This is the transition state:


Deviation angle for 1st order diffraction

The grating equation you've used is correct, and in order to find the first order diffraction angle ($n = 1$) you need to use arcsine trigonometric function and carefully place data with respective units:

\begin{align}
n \lambda &= d \sin{\Theta} \tag{1}\\
d &= \frac{\pu{1 mm}}{N}~[\pu{mm}] \tag{2}
\end{align}

where $N$ is number of lines per unit length. Combining (1) and (2) and converting the wavelength from nm to mm:

$$\Theta = \arcsin{\left(\frac{n \lambda}{d}\right)} = \arcsin{\left(\frac{n \lambda N}{\pu{1 mm}}\right)} = \arcsin{\left(\frac{1 \cdot \pu{5e-4 mm} \cdot 560}{\pu{1 mm}}\right)} = 16.26^\circ \tag{3}$$


Deriving a reduction potential from two other reduction potentials

That is because you cannot simply add the electrode potentials algebraically. In both cases, the electrode potentials have to be multiplied by $n$. What you can do instead is use Gibbs free energy change for each reaction which then can be added algebraically.

$$\Delta G^\circ = -nFE^\circ$$

Using this you can get the $\Delta G^\circ$ for each reaction which can be then added algebraically in the same way you add the reactions. The resultant $\Delta G^\circ_\mathrm{net}$, on equating with $-nFE^\circ_\mathrm{net}$, we can get the value of $E^\circ_\mathrm{net}$, which should be your answer.

In this approach, since the value of $n$ for both the reactions are different, the relative weightage of the electrode potentials of each reaction in determining the resultant $E^\circ$ will differ and therefore the final $E^\circ$ will be different than what you get by just adding up the electrode potentials.

For your case:

$$\begin{align}
\ce{Co^3+ + e- &-> Co^2+} & E^\circ_1 &= \pu{+1.82 V} & \Delta G_1^\circ &= \pu{-175.6 kJ mol-1} \tag{1} \\
\ce{Co^2+ + 2e- &-> Co} & E^\circ_2 &= \pu{-0.28 V} & \Delta G_2^\circ &= \pu{+54.0 kJ mol-1} \tag{2} \\
\end{align}$$

$$\begin{align}
\Delta G^\circ_\mathrm{net} &= \Delta G_1^\circ + \Delta G_2^\circ \\
&= \pu{-121.6 kJ mol-1} \\[3pt]
E^\circ_\mathrm{net} &= -\frac{\pu{-121.6 kJ mol-1}}{3F} \\
&= \pu{+0.42 V}
\end{align}$$


Adjusting the ph of salicylic acid using sodium hydroxide

The Henderson-Hasselbalch equation is the correct equation, but ask yourself what you are putting in there:
$$pH=pK_a + \log\left(\frac{c(A^-)}{c(AH)}\right)$$

Since $A^-$ and $HA$ occupy the same volume, you can rewrite:
$$pH=pK_a + \log\left(\frac{n(A^-)}{n(AH)}\right)$$

When adding Sodium Hydroxide consider the following equation
$$\ce{AH + NaOH <=> A- + H2O + Na+}$$

Given that you will find
$$ n(A^-) = n(\ce{NaOH})\\
   n(HA) = n_0(HA)-n(\ce{NaOH})$$

Then you can substitute and solve for $n(\ce{NaOH})$:
$$pH=pK_a + \log\left(\frac{n(\ce{NaOH})}{n_0(HA)-n(\ce{NaOH})}\right)$$

You then divide by the concentration, yielding your volume.

Intermediate results:

Initial amount of $HA$:

       $n_0(HA) = 0.5825\ \text{mmol}$

Amount of $\ce{NaOH}$:

   $n(\ce{NaOH}) \approx 0.4993\ \text{mmol}$

Final Volume:

   $V(\ce{NaOH}) = 3.22\ \text{mL}\ \text{(3sf)}$


What is the difference between TIC and TICC?

Ok, I had to read the actual text on pages 11 and 12 of the textbook link to understand.

TIC - In the olden days of mass spectrometers there was a way to measure the ion current from the sample (all masses). This total current is the current of all masses that is being injected into the mass spectrometer. Since the current in detector was also being measured, the spectra is plotted as the ratio of currents vs the mass (scan time).

TICC - The current from the sample is ignored and the spectra is just plotted as current in the detector vs mass (scan time).


Why is water taken in calculating K, equilibrium constant

The question they gave you is somewhat flawed, in that there are four compounds and you don't know the range of their mutual solubility.

To really answer the question, you would need information such as from Solubility, liquid–liquid equilibrium and critical states for the quaternary system acetic acid–ethanol–ethyl acetate–water at 293.15 K Fluid Phase Equilibria vol. 313, pages 46–51.

The reactants, ethanol and acetic acid, are completely miscible; however, ethyl acetate only has limited solubility in water.

When you evaluate the equilibrium constant, all the reacting compounds, including water, which are in a single phase, must be included.


Removing trace amounts of water from agricultural grade anhydrous ammonia

Calcium oxide can get water down to 4ppm in anhydrous ammonia according to http://www.google.com/patents/US4075306.


Best Way to Store Copper Acetate Solution?

Electroless copper plating requires a copper salt solution, of course (https://en.wikipedia.org/wiki/Electroless_copper_plating), but there are many other considerations, like the reducing agent and surface preparation.
I think that storage of the homemade copper acetate solution is a minor issue in your project and a plastic container with an air-tight lid should be perfectly fine, without refrigeration, for a few years.
However, for the actual electroplating, your copper solution may not be adequately characterized. There may be excess acetic acid, other stuff from the vinegar (which could be perfectly safe to consume, but not for sensitive chemical reactions). Commercial copper salts are available inexpensively and with high purity, if the project you are working on has significant value.


Absorption spectrum of Hydrogen molecule in UV region 200 nm - 300 nm (not Atomic Hydrogen)

Being a homonuclear diatomic molecule, molecular hydrogen does not posses a pure rotational of vibrational (electric dipole) spectrum. The lowest excited singlet state is the B state (for historical reasons there is no A state in H$_2$) and transitions from this state to the X ground state are known as the Lyman band which starts around 120 nm, thus outside the range you are interested in. In principle, electric quadrupole transitions may be observed but these are several orders of magnitude weaker than electric dipole transitions. In addition, the dissociation energy of molecular hydrogen lies around 36000 cm$^{-1}$, so that only radiation longer than about 277 nm can result in bound-to-bound transitions. However, I expect that the Franck-Condon factors to these highly excited vibrational states are very poor.

So there reason that you cannot find a spectrum is that there is none.


$\ce{MgSO4}$, otherwise known as epsom salt, is soluble. $\ce{CaSO4}$ ($K_\mathrm{sp}=7.1 \times 10^{-5}$), is slightly soluble. It is extremely likely that unless there are some other unknown metal ions in this solution, the white precipitate you describe in the solution is $\ce{CaSO4}$.


Calculating weight of soluble salt in a solution

Ok... It is hard to read you post since the details are scattered around.

Problem Statement

Given: 50 ml of a 0.250 molar solution of barium nitrate with a density of 1.035 g/ml  and that barium phosphate has a $K_\mathrm{sp} = 4.03 \times 10^{-25}$.

Question: How much $\ce{Na3PO4}$ must be added to reduce the concentration of barium to less than 7.00 ppb.

Complications with problem

Initial density

  The density of the initial solution, 1.035 g/ml, is useless.

Final density

  We need a density for the final solution to convert 7.00 ppb $\ce{Ba^{2+}}$ to a concentration of grams per liter. There is no density for the final solution, so the assumption will be that the density is 1.000 g/ml which means that a liter of the final solution will weigh 1000. grams. Since each $\ce{Ba^{2+}}$ cation will be replaced by 2 $\ce{Na^+}$ cations, the final solution will be at least 0.50 molar in $\ce{NaNO3}$. Thus the assumed density of 1.000 g/ml will be too low.

A pragmatic point here about the experimental procedure.

  You typically wouldn't want to add solid $\ce{Na3PO4}$ to a solution containing $\ce{Ba^{2+}}$ since $\ce{Ba3(PO4)2}$ would precipitate on the particles of $\ce{Na3PO4}$ sealing them from the solution and thus preventing complete dissolving of the $\ce{Na3PO4}$ particles. So experimentally you first need to dissolve the $\ce{Na3PO4}$ in a small amount of water and then add the solution of $\ce{Na3PO4}$ to the solution of $\ce{Ba(NO3)2})$. Thus the initial volume would really be over 50 ml.

Rounding errors

  The calculations will be carried to 5 significant figures to try to avoid rounding errors due to intermediate results. The final result will be properly rounded to 3 significant figures

Calculations

Moles of barium in the initial solution

In the initial solution $\ce{[Ba]} = 0.250\ \mathrm{m}$ so 50 ml contains

$\mathrm{m}\ \ce{Ba^{2+}} = 0.250\ \mathrm{m/l}\times 0.050\ \mathrm{l} = 0.0125\ \ \mathrm{m}$

Moles of barium in the final solution

In the final solution for Ba:

$\ce{Ba^{2+}} = 7.00\ \mathrm{ppb}$
$\quad\quad = (7.00\times 10^{-9})\times 1000\ \mathrm{g/l}=7.00\times 10^{-6}\ \mathrm{g/l}$
$\quad\quad  = \dfrac{7.00\times 10^{-6}\ \mathrm{g/l}}{137.33\ \mathrm{ g/m}} = 5.0972\times 10^{-8}\ \mathrm{m/l} $

For 50 ml of solution this is:
$\mathrm{m}\ \ce{Ba^{2+}} = 5.0972\times 10^{-8}\ \mathrm{m/l}\times 0.050\
\mathrm{l} = 2.5486\times10^{-9}\ \ \mathrm{m}$

$2.5486\times10^{-9}\ \ \mathrm{m} \ll 0.0125\ \ \mathrm{m}$ so the residual amount of barium in solution can be ignored.

    (1) *In determining the amount of barium phosphate that will precipitate, the amount of barium left in solution is so small that it doesn't matter if the solution has a density of 1.000 g/ml or 1.05 g/ml. So the density of the sodium nitrate solution doesn't really matter for that part of the problem.

    (2) $2.5486\times10^{-9}\ \ \mathrm{m} \ll 0.0125\ \ \mathrm{m}$ This is a significant figures argument. In other words for intermediate calculations I'm using only 5 significant figures.
    $$\begin{align}
0&.012\,500\,000\,000\,0 \\
-0&.000\,000\,002\,548\,6\\
\hline
0&.012\,499\,998\,551\,4
\end{align}$$ Now when trimmed to 5 significant figures the intermediate result is 0.012500. But the 0.0125 is exact, so I don't need the two extra zeros at the end.

Residual phosphate in the final solution

For the residual phosphate, in excess of that needed to precipitate the stoichometric amount of barium is:
$K_\mathrm{sp} = \ce{[Ba^{2+}]^3[PO4^{3-}]^2}$ or $\ce{[PO4^{3-}]} = \sqrt{\dfrac{K_\mathrm{sp}}
{[Ba^{2+}]^3}}$

$\ce{[PO4^{3-}]} = \sqrt{\dfrac{4.03\times10^{-25}}{(5.0972\times 10^{-8})^3}} = 5.5164\times10^{-2}\ \mathrm{m/l}$

For 50 ml this would be:

$\mathrm{g}\ \ce{PO4^{-3}}\ \mathrm{as }\ \ce{Na3PO4} = (5.5164\times10^{-2}\ \mathrm{m/l})\times 0.050\ \mathrm{l}\times 163.94\ \mathrm{g/m} = 0.45218\ \mathrm{g}$

  This is too great of an amount of residual phosphate in solution to be ignored. Thus we need to add this to how many grams of trisodium phosphate are needed to precipitate 0.0125 moles of $\ce{Ba^{2+}}$ as $\ce{Ba3(PO4)2}$.

Stoichiometric amount of phosphate to precipitate all the barium

Each mole of $\ce{Ba3(PO4)2}$ requires 3 moles of $\ce{Ba^{2+}}$. So for 0.0125 moles of $\ce{Ba^{2+}}$ there would be

$\mathrm{moles}\ \ce{Ba3(PO4)2} = \dfrac{\mathrm{moles}\ \ce{Ba^2+}}{3}= \dfrac{0.0125}{3} = 4.1667\times10^{-3}$

But each mole of $\ce{Ba3(PO4)2}$ requires two moles of $\ce{PO4^{3-}}$

$\mathrm{moles}\ \ce{PO4^{3-}} = 2\times(\mathrm{moles}\ \ce{Ba3(PO4)2})= 2\times( 4.1667\times10^{-3}) = 8.3333\times10^{-3}$

$\ce{Na_3PO4} = \mathrm{MW}\ 163.94\ \mathrm{g/m}\\
\quad\quad\quad\quad= 163.94\times8.3333\times10^{-3} = 1.3662\ \mathrm{g}$

Total amount of trisodium phosphate needed

$\mathrm{Total}\ \ce{Na_3PO4} = 1.3662 + 0.4522 = 1.8184\ \mathrm{g}$

Now rounding the final result to 3 significant figures we get 1.82 grams.

Now the other pertinent fact is how much $\ce{Na3PO4}$ can be dissolved in 50 ml at 25 °C? Wikipedia shows 14.5 g/100 mL (25 °C), so 7.25 g/50 mL (25 °C). Thus it should be possible to dissolve the $\ce{Na3PO4}$.


Why are polysaccharides non-reducing sugars?

For polysaccharides made with only glucose (starch, cellulose, glycogen, etc), only 1 unit can be reduced from hundreds, thousands or tens of thousands of units. With the same mass of dextrose and starch, the amount reduced for starch is 1/100, 1/1000 or 1/10000 the amount for dextrose. This is usually below the detection limit for most ways of detecting the reaction.


What is difference between density of the equilibrium and vapour density of the mixture

Let's start from $m = 0.925 g˙\ce{SO3}$ that will be heated at $900 K$. This is the amount of $\ce{SO3}$ that will occupy a volume of $1.00 L$ after heating it at $900 K$ at $1 atm$. and after partial decomposition. In moles it is also  $n_o$ moles $\ce{SO3}$,  $$\ce{ $n_o = \frac{m}{80 g/mol} = \frac{0.925 g}{80 g/mol} = 0.01156 mol$}$$ At $900 K$,  a fraction $\alpha$ of these moles are decomposed into $\alpha$ moles $\ce{SO2}$ plus $\alpha/2 $ moles $\ce{O2}$. Then the total number of moles at equilibrium will be : $$\ce{$n$ = $n_o$˙($1$ - $\alpha$ + $\alpha$ +    \alpha/2}) = n_o(1 + \alpha/2) = 0.01156 mol (1 + \alpha /2)$$ The total volume is : $$\ce{$V = nRT/p= 0.01156 mol$(1 + \alpha/2)$RT/p = 0.001 m^3$}$$  Simple rearrangement of this formula gives : $$\ce{1 + \alpha /2  = \frac{0.001 $m^3·p$}{0.01156 $mol·RT$} = \frac{0.001 $m^3$·101325 $Pa$}{0.01156 $mol$·8.314 $J mol^{-1}K^{-1}·900 K$} = 1.172 }$$
The degree of dissociation $\alpha$ is equal to $\alpha = 2 ·0.172 = 34.4$ %


Why is the molecular structure of water bent?

I mean, there is a time and place for VSEPR, and this is probably as good a time as any, because all beginning chemistry students go through it. The actual model has already been explained multiple times, so I will only briefly say that according to this theory, there are four pairs of electrons around the central oxygen. In order to minimise electron-electron repulsions, these pairs adopt a tetrahedral arrangement around the oxygen. It does not matter which two are lone pairs and which two are connected to hydrogen atoms; the resulting shape is always bent.

What's worth bearing in mind (and hasn't been explained very carefully so far) is that VSEPR is a model that chemists use to predict the shape of a molecule. The truth is that there is no real way to predict the shape of a molecule, apart from solving the Schrodinger equation, which is not analytically possible for water. Everything else is an approximation to the truth. Some of these approximations are pretty accurate, such as the use of density functional theory. Some of them are extremely crude, and VSEPR falls into this category: it essentially treats electrons as classical point charges, and seeks to minimise the electrostatic repulsion between these point charges. As a physics student you should know better than to do this. Thus, while it predicts the correct result in this case, it is more in spite of the model rather than because of the model. And you should not be surprised to hear that in some slightly more complicated cases, VSEPR can predict entirely wrong outcomes.

As you learn more chemistry you will find that there are increasingly sophisticated ways of explaining molecular geometry. Most revolve around molecular orbital theory. For a qualitative method, you have Walsh diagrams which have been explained at Why does bond angle decrease in the order H2O, H2S, H2Se?. For a more rigorous method you would likely have to run some quantum chemical computations, e.g. Are the lone pairs in water equivalent?. Of course, the drawback of this is that it becomes more and more difficult to extract true chemical understanding from the numbers. Although it should also be said that you cannot extract any true chemical understanding from the VSEPR model.

What interests me more is the followup question:

  Also, wouldn't the Schrödinger equation provide an equally plausible structure for water with the lone pairs on the opposite side of the oxygen from what we assume (imaging the electrons on the top or on the bottom of the oxygen in the Lewis structure)?

Because the Hamiltonian of the water molecule is invariant upon rotation, this means that indeed, any orientation of the water molecule is equally likely. However, this only refers to the orientation of the water molecule as a whole. It does not say anything about the internal degrees of freedom, such as the bond angle.

In the absence of any external force, the molecule is free to bend in whichever direction it likes, and most water molecules indeed do do this as they float through space or swim in a lake. But it will always be bent.

  If that were true, then there would be a resonance structure between the two states and we would get a linear geometry.

If you were to think of a single particle in a double-well potential, say something with

$$V = \begin{cases}
\infty & x < -b \\
0      & -b \le x \le -a \\
\infty & -a < x < a \\
0      & a \le x \le b \\
\infty & x > b
\end{cases}$$

then because of the symmetry of your system, in every eigenstate of your system, the expectation value of $x$ would be $\langle x \rangle = 0$. This is quite similar to your argument. In the case of water, let's set the oxygen nucleus to be at the origin. Because it can point either up or down, the expectation value of the hydrogen nucleus position along the up-down axis would be exactly level with the oxygen atom, i.e. 0. In fact, don't stop there: it can point to the left or the right, and to the front or the back. So the hydrogen nucleus has a position expectation value of exactly $(0, 0, 0)$, i.e. right inside the oxygen nucleus.

Does that mean it's actually there, though? In our contrived double-well system, it's patently impossible for the particle to be at $x = 0$, because $V = \infty$ there. If you were to measure its position, you would never find it at $x = 0$; you would only find it in the left-hand side $[-b, -a]$, or the right-hand side $[a,b]$. Just because the particle has an expectation value of $\langle x \rangle = 0$ does not mean that it is physically there, or that $x = 0$ is somehow its equilibrium state. You're confusing an expectation value with a genuine eigenstate (which is what a resonance structure is).

In exactly the same way, if you ever were to measure the properties of water (and bear in mind that practically every interaction with a water molecule is, in effect, a measurement), we would find that it is indeed always bent.


Is magnesium sulfate basic, neutral or acidic?

Looking at the following references, it seems that most agree that a water solution of $\ce{MgSO4}$ is slightly acidic.

"The pH of an aqueous magnesium sulfate solution is related to the molarity of the MgSO4. Typically, the pH is between 5.5 and 6.5 due to magnesium's affinity for hydroxide ion (OH-). As the sulfate goes into solution, hydroxide anions associate with the magnesium, increasing the relative ratio of H+ to OH-. This shift results in more acidic solutions."

"The pH of a pure magnesium sulfate solution is approximately 6.4"

"The pH of [magnesium sulfate] hydrates is average 6.0 (5.5 to 6.5)."

"pH: Neutral to litmus"

Why not perform the experiment in class, using an accurate pH meter and/or indicators (see the pH indicator chart in Wikipedia for suggestions).


The negative charge is due to the unpaired electron in the oxygen atom, remained after all bonds(molecular orbitals) are filled. Oxygen has 2 unpaired electron in its valence shell and is singly bonded to carbon in this molecule. Hence being un-bonded, the other electron exhibits the charge.

Check this link, hover the cursor on the $\ce{O}$-atom, you will get a graphical representation of the same.


how to separate a mixture of sodium chloride and sodium hydroxide

One solution would be to reduce the amount of solvent (under vacuum for instance), then filter out the precipitated NaCl at high temperature (at higher temperatures NaOH has as much as >10x the water solubility of NaCl).
The common ion effect will play in your favour, as the high water solubility of NaOH at higher temperatures will help you reach NaOH concentrations that strongly disfavours NaCl solubility.
The way to go may be to first reduce the solvent mass under vacuum at acceptably safe temperatures, and once the desired mass is reached redissolving most of the precipitated NaOH by pushing the temperature up before filtration to a level at which most NaOH has redissolved.
Use appropriate protection equipment and contact material to handle very hot NaOH.


It looks like you are describing continuous wave (CW) and not pulsed NMR. The latter is generally considered a more modern way of doing NMR as it was developed after the CW technique. The pulsed technique was also accompanied by the development of FT processing which resulted in a Nobel price being awarded to Richard Ernst.
In CW NMR you usually "sweep" the field - increase the external field $B_0$ in small increments, as you explain - while keeping constant the frequency generated by a "read" RF coil. Coincidence between the read frequency and the Larmor frequency $\omega = \gamma \left(1-\sigma \right)B_0$ - the resonance condition - results in a signal registered with a pickup coil. Here $\sigma$ is the chemical shielding which reduces the effective magnitude of the applied field $B_0$  at the nucleus.

by increasing the external magnetic field, won't we increase the energy differences between the spins and cause the resonance frequency to be higher which conflicts with the idea that we have a lower resonance frequency?

This is exactly the point of increasing the field. The Larmor frequency is related to the splitting in the energy levels, and you need a greater field to bring about a splitting in a more shielded nucleus, so that the resonance condition is satisfied.
The resonance frequency in such a CW experiment is always $\omega = -\gamma B_1$, but as the field at the shielded nucleus is $\left(1-\sigma \right)B_0$ and the shielding opposes the applied field, it is necessary to apply a larger field ("go upfield") to excite a more shielded nucleus.


As far as I know, the short answer is no; in general, it is not possible to know the reaction mechanism a priori.

In general, the approach is:

Postulate a reaction mechanism/pathway
Use this to calculate the relevant reaction rate(s)
Compare these rate expressions with the results of experiments

If your proposed reaction mechanism matches the empirical results, it is likely (but you can never be sure!) that the mechanism is correct. If not, a new mechanism should be postulated and tested.

That being said, in many examples it may be possible to make an educated guess of a plausible reaction mechanism - based on your own experience or the experience of others (i.e. from articles etc.) - and then investigate whether the proposed mechanism matches the empirical facts of relevant experiments. Often, the proposed mechanism is more or less correct.

In addition, modern advances in computational chemistry may give useful information regarding likely reaction pathways based on the structure and interaction of the involved molecules. To my knowledge, it is not yet possible to use this in a standalone fashion to figure out the detailed mechanism, but it may help out in terms of making a better educated guess.


The Trend Between Boiling Point and Solubility in Organic Chemistry

As andselsik pointed out, definetley no in the general sense. Just think of some practical examples like PTFE (Teflon) - which melts at about 330 °C but is definetly not water soluble.

However there is one connection I'd like to point out that allows to relate the enthalpy of melting to the solubility:

Consider phase equilibrium between a soluble compound C and water. At equilibrium, we have equality of chemical potentials:

$$\mu^{solid}_c(p,T,x_s)=\mu^{liquid}_c(p,T,x_l)$$

Assume that $x_s$ at equilibrium is 1 (there would be a tiny fraction of water in the solid, but its fair to neglect that). Then the LHS is just the chemical potential of the pure solid:

$$\mu^{*,\:solid}_c(p,T)=\mu^{liquid}_c(p,T,x_l)$$

Now using the solvent convention for the chemical potential of the liquid mixture:

$$\mu^{*,\:solid}_c(p,T)=\mu^{*,\: liquid}_c(p,T)+RT\mathrm{ln}(x_c^l \: \gamma_c)$$

Rearranging this and recognising that a chemical potential difference of a pure substance across a phase boundary is the gibbs energy of that phase transition gives

$$\Delta g_{c;\: melt} = \Delta g_{c;\: S \rightarrow L}=\Delta \mu_{c;\: S \rightarrow L} = RT\mathrm{ln}(x_c^{l} \gamma_c)$$

and

$$\Delta g = \Delta h + T \Delta s$$

using the limiting behaviour of the activity we find that for $x_c , \gamma_c \rightarrow  1$, the RHS tends to zero; we can reexpress the entropy of the phase change by setting $\Delta g$ in the above equation zero, and substitute the result into the equation obtained from equality of chemical potential:

$$\Delta h_c^{*,\:melt} \Big(\frac{T}{T^{melt}}-1\Big) = RT\mathrm{ln}(x_c^l \gamma_c)$$
...
$$x_c^l\gamma_c=\mathrm{exp} \Bigg( \frac{\Delta h_c^{*,melt}}{R} \Big(\frac{1}{T^{*,melt}}-\frac{1}{T}\Big)\Bigg)$$

The activity coefficient itself is obviously also a function of the composition, so this does not give us a straightforward way to calculate the solubility of compound C (it does however provide us with a route to $\gamma$, which is normally harder to measure than a solubility) but it shows that phase change properties have all sorts of implications!


How to predict disproportionation products from Frost diagram?

Interesting question. First of all understand what is on the x-axis and on the y-axis. You y-axis is Gibbs free energy/Faraday constant. Choose three consecutive oxidation states, the middle oxidation state, if it lies above the two, is likely to disproportionate.Read more here on Frost and Latimer digram

In your case look at the example of $Mn^{3+}$, if you connect $Mn^{2+}$ and $MnO_2$, $Mn^{3+}$ lies above the two, so 3+ oxidation of Mn is unstable with to disproportionation.

In short, just by using Frost diagram you cannot predict disproportionation of Mn (VII) because if it were to disproportionate, you will need Mn(VIII), which is not shown, so Frost diagram analysis only works for three points and you cannot say anything about the boundary values.


Why does hydrogen so easily take the place of oxygen in organic molecules during reduction reactions?

But, why does it generally follow that hydrogen takes the place of an oxygen molecule during a hydrogenation/reduction reaction [...]

Ok, if at all, it's an oxygen atom ;)

I don't see that general trend towards complete deoxygenation, neither in catalytic hydrogenations, nor in hydride transfer reactions.

It usually takes

rather drastical conditions
quite some special reagents
a particular type of derivatisation

or a combination of any of these to convert oxygen-containing compounds to alkanes.

Let's have a look at some examples:

$\ce{R1-CO-R2 -> R1-CH2-R2}$

Ketones and aldehydes may be directly reduced to alkanes in the Clemmensen reduction. The "mildest" variants use anhydrous $\ce{HCl}$ and amalgamated $\ce{Zn}$ in organic solvents.
The clasical Wolff-Kishner reduction, using hydrazine and sodium in diglyme, involves the in situ functionalization of the carbonyl and isn't a prototype for a mild reaction either. Admittedly, milder variants exist, using N-tert-butyldimethylsilylhydrazone and catalytical amounts of a Lewis acid, e.g. $\ce{Sc(OTf)3}$
Tosylhydrazones may be converted to alkanes using $\ce{NaCNBH3}$, $\ce{Na(CH3COO)3BH}$, or boranes.
Catalytic hydrogenation over Raney nickel is possible after converting the carbonyl to a dithiane; so it's actually a desulfurization.

$\ce{R1-CH2-OH -> R1-CH3}$

I'm not aware of any transformation that does not involve functionalization.

Alkyl tosylates can be deoxygenated using $\ce{LiEt3BH}$.
Acylation of alkohols with thiocarbonyl chlorides furnish thiocarbonates, which may be deoxygenated under using either tin hydrides ($\ce{R3SnH}$, Barton deoxygenation) or $\ce{B(CH3)3}$ in water/benzene.
Alkanes may also be obtained from alkohols under Mitsunobu conditions ($\ce{Ph3P}$, $\ce{DEAD}$, o-nitrobenzenesulfonylhydrazine).

$\ce{R1-CH2-COOH -> R1-CH3}$

Direct transformations of carboxylic acids to alkanes typically involve decarboxylation, i.e. the loss of $\ce{CO2}$. This may be achieved either by the Barton decarboxylation, or via photoinduced electron transfer reactions with electronically excited (electron) acceptors. In the course of the latter reactions, a carboxylate is oxidized to an acyloxy radical, which then undergoes decarboxylation.

To sum it up: It's not that easy ;)


Pressure of carbon dioxide due to two simultaneous equilibriums

It is a conceptual subtlety. It is needed all the species in the hypothetical reaction to require that the equilibrium constant equation be satisfied. This is normally non understood in most chemistry textbooks. The sentence Two equilibria are established is unhappy.

Briefly you can not use this two equations system till you be sure that all the species exist in equilibrium state.

Initially, both solids will react to form the products, once a pressure of 3 atm is reached the only the second reaction will progress. But, as it will increase pressure above of 3 atm the reaction 1 will go in reverse. This situation will continue while CaCO$_3$ is present. Notice that until now there is no requirement that equilibrium constant equation be satisfied due to the states in which the system was were non equilibrium state.
This situation will continue till all BaO disappears. So, the pressure will increase to 5 atm and the first equilibrium won't take place.

Edit: "what happens if CaCO$_3$ is exhausted? then wont CaO react back to reduce the pressure of CO$_2$?"

If CaCO$_3$ is exhausted before reach 5 atm (after that it is not possible), let say at 4 atm, CaO could react back, but as a soon as a it happens (that a little piece of CaCO$_3$ enough big to have the intensive properties of a macroscopic crystal is formed), there will be more tendency to go forward again.

In such case the final pressure will be 4 atm, different from both equilibrium pressure. It will be another equilibrium state. Nothing will change for BaCO$_3$ in this analysis.

Above I exposed two possibilities: The first one, before the edition, that the vessel is small enough to allow the CO$_2$ pressure reach 5 atm. This is the supposed "right answer".
The later where final pressure of CO$_2$ greater than 3 atm and smaller than 5 atm.
If the vessel is very big, it can happen that reactives are exhausted before reach 3 atm, so other equilibrium state will take place.


How are odor and electronic structure related to aromaticity?

I can address the part of your query pertaining to odor and nomenclature (more accurately, I can link the UC Davis ChemWiki page that does so and quote it here):

Benzene derived products are well known to be pleasantly fragrant. For this reason, organic compounds containing benzene rings were classified as being "aromatic" (sweet smelling) amongst scientists in the early 19th century when a relation was established between benzene derived compounds and sweet/spicy fragrances. There is a misconception amongst the scientific community, however, that all aromatics are sweet smelling and that all sweet smelling compounds would have a benzene ring in its structure. This is false, since non-aromatic compounds, such as camphor, extracted from the camphor laurel tree, release a strong, minty aroma, yet it lacks the benzene ring in its structure (See figure 1). On the other hand, benzene itself gives off a rather strong and unpleasant smell that would otherwise invalidate the definition of an aromatic (sweet-smelling) compound. Despite this inconsistency, however, the term aromatic continues to be used today in order to designate molecules with benzene-like rings in their structures. For a modern, chemical definition of aromaticity, refer to sections Aromaticity and Hückel's Rule.

You will find (in the linked section on Aromaticity) that presently:

The three general requirements for a compound to be aromatic are:

The compound must be cyclic

Each element within the ring must have a p-orbital that is perpendicular to the ring, hence the molecule is planar.

The compound must follow Hückel's Rule (the ring has to contain 4n+2 p-orbital electrons).

Putting it all together: The historical naming convention for molecules that smelled good is responsible for the term aromatic. It turns out that many of the molecules that fall into the aromatic category happen to be rather stable/unreactive. These molecules happen to have certain physical characteristics which, taken together, are indicative of a molecule being classified as aromatic or not (based on present-day definitions not relating to odor/smell).


What Type of Gas Would Vinegar and Epsom Salt Create Under Electrolysis?

You're most likely forming $\ce{H2}$ gas at the part.

If you're plating anywhere above about $2~\mathrm V$, that is probably sufficient overpotential to cause part of the passed current to go toward electrolysis of water.

There is negligible toxicity hazard to inhaling the small amounts of hydrogen that would be generated from an electrolytic cell on this scale.

There's a small ignition hazard if anything sparks near electrolyte of the plating cell as it's operating, but as long as you're not running it inside a tightly closed, small space you should have nothing to worry about with how small the cell is.


Reference request on formate potash process

The reference is correct, from [1, p. 888]:

Potassium formate: $\ce{HCOOK},$ colorless, deliquescent, rhombic crystals; $M_\mathrm{r}~84.12,$ density $1.91$, m.p. $\pu{167.5 °C}.$ P. is formed by neutralization of aqueous formic acid with potassium hydroxide or potassium carbonate. It is an industrial intermediate in the formate-potash process (see Potassium carbonate).

However, formate-potash process doesn't yield in elemental potassium, rather potassium carbonate [1, p. 886–887]:

Potassium carbonate, potash:
[…]
P. is also produced by the formate-potash process, in which potassium sulfate and lime are treated under pressure with carbon monoxide:
$$\ce{K2SO4 + Ca(OH)2 + 2 CO -> 2 HCOOK + CaSO4}.$$
The resulting potassium formate is oxidized to P. by heating it in air:
$$\ce{2 HCOOK + O2 -> K2CO3 + CO2 + H2O}.$$

English Wikipedia page claims formate-potash process is for “production of potassium” probably because the author attributed “P.” from the encyclopedia entry to potassium.
Whereas in reality it refers to the entry title, “Potassium carbonate”, which is a standard referencing method in data sources such as a handbook or an encyclopedia.
German Wikipedia page doesn't have this flaw and informs that “Formiat-Pottasche-Verfahren” is a technically insignificant method of producing $\ce{K2CO3}.$
References

Concise Encyclopedia Chemistry, English language ed., 2nd ed.; Scott, T., Eagleson, M., Eds.; de Gruyter: Berlin; New York, 1994. ISBN 978-0-89925-457-9.


How do I calculate the equilibrium constant of a reaction from two related reactions' constants?

Equilibrium constants depend upon the choice of reaction equations. I have noted the extended way to calculate $K''_\mathrm{c}$ below:

$$\begin{align}\ce{2S + 3 O2 &<=> 2 SO3}\tag{2}\\[0.4em]
K''_\mathrm{c} &= \frac{[\ce{SO3}]^2}{[\ce{O2}]^3}\end{align}$$

(Sulfur does not take part in the equilibrium constant because its activity, as a solid, is $a(\ce{S}) = 1$.)

Having this, we can approach the question. First, we need to add particles in an equivalent way (i.e. the same to reactant and product side) to get from equation $(3)$ to something that can be expressed as a linear combination of $(1)$ and $(2)$ ($a(1) + b(2)$):

\begin{align}\ce{2SO2 + O2 &<=> 2SO3}\tag{3}\\[0.4em]
\ce{2SO2 + 3 O2 &<=> 2 SO3 + 2 O2}\tag{$3'$}\\[0.4em]
\ce{2 SO2 + 3 O2 + 2 S &<=> 2 SO3 + 2 O2 + 2 S}\tag{$3''$}\end{align}

   $\displaystyle (3'') = -2 \times (1) + 1 \times (2)$

With that in hand, we can determine what the equilibrium constant of $(3'')$ would look like and how that can be traced back to the other equilibrium constants:

   $$\displaystyle K_\mathrm{c} = \frac{[\ce{SO3}]^2[\ce{O2}]^2}{[\ce{SO2}]^2[\ce{O2}]^3}\tag{4}$$

 $$\displaystyle K_\mathrm{c} = \frac{[\ce{O2}]^2}{[\ce{SO2}]^2} \times \frac{[\ce{SO3}]^2}{[\ce{O2}]^3} = (K'_\mathrm{c})^{-2} \times K''_\mathrm{c} \tag{5}$$

And thus, your final solution is:

   $$(4.2\times10^{52})^{-2} \times (9.8\times10^{128}) = 5.6\times10^{23}\tag{6}$$


Rates of hydrolysis in bicyclic and monocyclic compounds

I’m going to shamelessly steal electronpusher’s drawing for clarification:

Two general hydrolysis mechanisms are possible: $\mathrm{S_N2}$ and $\mathrm{S_N1}$. First, let us consider $\mathrm{S_N2}$ since the analysis is more simple.

Both of the two species are tertiary bromides meaning that $\mathrm{S_N2}$ reactions are already severely inhibited. However, the second compound is even worse: if the nucleophile tries to attack from behind to interact with the $\sigma^*_{\ce{C-Br}}$ orbital, it would need to come from inside the bicyclo[2.2.2]octane system. This is completely impossible, not only due to sterics. Thus, while compound 1 is very slow in $\mathrm{S_N2}$ reactions, it is not possible for 2 to react in an $\mathrm{S_N2}$ manner.

Conclusion: 1 should react faster by $\mathrm{S_N2}$ — albeit still unmeasurably slow.

To undergo an $\mathrm{S_N1}$ reaction, first the halide must dissociate to give a carbocation. Both species are, as noted above, tertiary, meaning that carbocations are well stabilised by triple hyperconjugation. Initially, we expect both $\mathrm{S_N1}$ mechanisms to be fast.

A carbocation when formed must practically always adopt a planar, $\mathrm{sp^2}$ hybridised configuration. Thereby, the unoccupied orbital has the highest energy and all bonding electrons are as low as possible in energy. Any distortion away from the perfect $\mathrm{sp^2}$ geometry requires the empty orbital to be stabilised at the cost of populated orbitals — very unfavourable.

For bridgehead carbons as in 2, especially those that bridge small rings (which is the case here), the general rule is that they do not undergo reactions that require planar transition states for exactly that reason: they are too strained to adopt an $\mathrm{sp^2}$ configuration. (This configuration would not only require the bridgehead to ‘move inwards’, but would also strain the bonds of the three neighbouring carbons, whose angles would have to be decreased.)

Conclusion: while 1 reacts rapidly via $\mathrm{S_N1}$, compound 2 does not react via $\mathrm{S_N1}$.

Complete conclusion: 2 is unreactive towards hydrolysis. 1 therefore reacts much faster.


Why do transition metals form particularly stable complexes?

To illustrate my answer, I am going to use a modified version of the picture of a simplified molecular orbitals scheme Professor Klüfers (LMU Munich) uses in his internet scriptum to the coordination chemistry I course.

On the left you can see the metal’s orbitals, on the right the group orbitals of six $\unicode[Times]{x3C3}$ donors (assuming no $\unicode[Times]{x3c0}$ interactions of any kind). $\mathrm{e_{g}}$ and $\mathrm{t_{2g}}$ correspond to the metal’s d-orbitals, $\mathrm{a_{1g}}$ is the 4s orbital and $\mathrm{t_{1u}}$ are the 4p-orbitals.

You can clearly see that the 3d-orbitals are similar in energy to the ligand group orbitals, and that they have the correct symmetry to mix. The greatest stabilisation comes from the $\mathrm{e_g}$ orbitals.

Empty d-orbitals can thus be thought to help with coordination by stabilising the coordination compound. Completely filled d-orbitals (e.g. $\ce{Zn^2+}$) are contraproductive and therefore zinc’s coordination chemistry is much like sodium’s.


If electrolysis splits water, why does only either hydrogen or oxygen (but not both) bubble up on one pole?

Water, as you may know, has a dualist nature between covalent and ionic bonding; oxygen is the second most electronegative element in the periodic table, while hydrogen's simplistic construction makes it very zen about how it forms bonds (it defines the center point of most electronegativity scales). While the bond between the hydrogens and oxygen of a water molecule are termed covalent, the electrophilic nature of the oxygen makes the bonds border on the ionic, and so water is just as easily thought of as $\ce{2H^+ + O^2-}$, or in more stable terms as $\ce{H^+ + OH^-}$.

In electrolysis, this ionic character of water is exploited in a series of "half-reactions"; at each node, a gas is formed by adding or removing electrons from water to liberate one of its two components, and when that gas precipitates out, what's left remains dissolved in the water as charged ions, which are repelled by the node that created it and attracted to the other node.

When electrons are added to water at the cathode, they are attracted to the partial positive charge of the "hydrogen side" of water, and can then form an orbit around a hydrogen. That liberates the hydrogen; it no longer needs the electron it's sharing with the oxygen to balance its charge, but it still needs a second electron to stabilize its 1s shell, and the best bet is to pair with another hydrogen to form hydrogen gas, which precipitates out. That leaves soluble hydroxide ions in the water, which are repelled by the cathode (and each other, so these ions wouldn't form any compounds) and move toward the anode. It's highly unlikely that both hydrogen atoms would be liberated from a single oxygen, and even if it did, the oxygen, with extra electrons, couldn't bond with another negative-charge oxygen to form gas; it would be much more likely to deprotonate another water molecule forming two hydroxides.

$$\ce{4H2O + 4e^- -> 2H2 (g) (^) + 4OH^- (aq)}$$

At the anode, the positive potential causes water to give up electrons, liberating the hydrogens again, this time as naked protons, which are attracted to water's partial negative pole to form hydronium ($\ce{H_3O^+}$ or simply $\ce{H^+ (aq)}$). The now chargeless but bond-seeking oxygen is left to pair up as oxygen gas, which precipitates out. Hydrogen peroxide is a possible intermediate, but very short-lived, as this peroxide will either find another peroxide and pair up the extra oxygens, or it will give up its hydrogens as protons under the positive voltage potential the same way water would, and become oxygen gas.

$$\ce{2H_2O - 4e^- \to O_2(g)\uparrow + 4H^+ (aq)}$$

... Then, somewhere in the middle, the hydronium and hydroxide meet, and as these proton donors and proton acceptors will do, they reform water:

$$\ce{4H^+(aq) + 4OH^- (aq) \to 4H_2O }$$

You'll notice that only one gas precipitates out at each node; the other atoms are trapped as soluble ions, either protons/hydronium or hydroxide, which cannot form other compounds because their like charges repel them from each other. Instead, they remain dissolved as local acidic and basic solutions. If the potential charge in the nodes is high enough, dipping a litmus strip at each end of the water bath (or dropping in a few mL of pH indicator dye) will yield an acidic reading at the anode and a basic reading at the cathode.


How to meausre ion interferences in solution?

What you can do is to add a reagent which forms a highly coloured complex with one of the metals, this will allow you to make a measurement which is good for one of the metals. You may need to do three different experiments on three seporate aliquats from the same solution.

For example for the iron, I would suggest that you use 2,2'-bipyridine, or 2,2'-bipy as it is sometimes known. This forms a very deep red complex with ferrous iron. This would allow you to measure the iron.

You would need a series of calibration samples with different amounts of iron, please do be aware that the matrix can sometimes interfere with this measurement. I have found that oxalic acid does inhibit the formation of the iron tris bipy complex.

Maybe you should read this paper which is about 2,9-dimethyl-4,7-diphenyl phenanthroline. This is a reagent which has been reported to be good for copper determination.


Double Replacement Reactions form Precipitates

Precipitation of silver chloride is a common example of a double replacement reaction. e.g. $$\ce{AgNO3~_{(aq)} + NaCl_{(aq)} -> AgCl_{(s)} + NaNO3~_{(aq)}}$$

Initially all four ions are present in solution. When the solutions are mixed the silver and chloride ions bond together and precipitate, leaving the sodium and nitrate ions remaining in solution along with a small number of silver and chloride ions because the precipitation is an equilibrium so a small number of dissolved ions will still be present (thanks to @Martin for pointing this out). The reaction happens because the precipitation of silver chloride removes these ions from the solution equilibrium, driving the reaction to completion.

Silver chloride is insoluble because the $\ce{Ag-Cl}$ bond has a high degree of covalent character due to the high polarizing ability of the silver cation and the high polarizability of the chloride anion and the similar size of both ions which allows a shorter bond length and so a stronger bond requiring more energy to break apart in order to dissolve the solid. See this question for more detail on this point.


60 tons on $\ce{CaCO3}$ per hour means half a million tons per year. That's quite some calcium carbonate.

Hens need $\ce{CaCO3}$ to produce eggs. So the most natural idea to recycle it is to produce feed lime.

Or consider producing agricultural lime. The annual consumption of agricultural lime is about 2.5 million tons in Germany, and about 25 million tons in Brazil. Your production is in this scale. The presence of 3% proteins is probably of little concern if not beneficial.


The best way to demonstrate this kink is with pictures. Usually we have some motivation to try to represent the molecule on a single line. So for oleic acid (cis-octadec-9-en-1-oic acid), we often get pictures like the following:

Elaidic acid, the trans-isomer is also sometimes depicted unhelpfully to save space:

The second structure for elaidic acid is not as misleading as the corresponding structure for oleic acid. A better bond-line structure for oleic acid is:

Better yet, here are some space-filling structures from Wikipedia based on crystal structures of both molecules.

Oleic

Elaidic

And here's $\gamma$-linolenic acid, which has three cis alkenes in it:


Which material to use for athletic accessories?

For plastics, my first instinct for the first two properties would be a polycarbonate plastic. Polycarbonates are known to be very strong plastics (tensile strength up to ~70 MPa) and deform without breaking under many conditions, which is why they're commonly used in things like safety glasses and bullet-resistant glass. It also looks very interesting as it's quite optically clear. PC is 3D printable, I've seen a few people using it in hobby printers (http://www.protoparadigm.com/blog/2011/12/printing-polycarbonate/) and in very high end StrataSys printers, but it's not a super common material so it might take some experimenting.

Another option is a nylon. There are a number of different types available and many have similar properties to polycarbonates (minus the optical clarity). Nylons seem to be a bit more popular for 3D printing and I've heard good things about the Taulman filaments.

Of course, you're not limited to a single material and there are many things that could potentially be used in this kind of application. Often, the combination of multiple materials makes sense (e.g. a metal insert in a plastic part).

Something like carbon fibre-reinforced epoxy construction might be appropriate. It's not 3D printable, but can be constructed in layers around a lightweight (expanded polystyrene or the like) form. This approach is often used for parts that need to be very light while having high tensile strength (think bicycle frames, ski poles, etc.) I could see a carbon fibre-wrapped plastic insert making a very strong and light shoe heel.

Ultimately, these things are balancing games between strength, mass, fabrication cost, aesthetics, etc. but these are a few potential materials that come to mind.


Cis and trans substituents on rings different stereodescriptors

Edit: strongly related to your previous question How to determine the configuration of C atom bounded to two groups with same configuration? and Yolo123's question that was suggested to you Number of diastereomer pairs of 1,3-dichloro-1,2,3-triphenylpropane where the nomenclature was explained in detail by Jerrepierre. The same situation applies here.

Although the cis/trans nomenclature might seem 'outdated', it is still by far the easiest way to describe disubstituted ring systems, and is commonplace in both conversation between organic chemists, and in the organic literature when such relationships are introduced into a molecule.

As these ring systems possess some symmetry, the (R) and (S) descriptions commonly used are not applicable, but (r) and (s) may be assigned to carbons ('pseudo asymmetric'), based on the IUPAC 2013 recommendations for describing achiral cyclic compounds.

I've included the assignments for the two molecules you presented in the image below. In essence (taken from March's Organic Chemistry):

  The small letters used for the psuedoasymmetric center are assigned using established rules. [...] The descriptors “r” and “s” are used to denote such centers; they are assigned in accordance with Sequence Rule 5, taking into consideration that “(R)” has precedence over “(S)” in the order of priority.

This additional example, taken directly from the IUPAC blue book 2013 may clarify the reasoning behind assignment of (r) and (s):

Its worth noting that this nomenclature only applies when you can find symmetry, in many cases this is non-existent, and normal (R) and (S) nomenclature can be applied.


Does H3CO3+ ion exist and what is its structure?

A protonated carbonic acid (trihydroxymethyl cation) $\ce{H3CO3+}$ does exist, and is in fact incredibly stable in the solutions of superacids [1] up to $\pu{0 °C}.$ One way of obtaining $\ce{H3CO3+}$ is dissolving inorganic carbonates and hydrogen carbonates in magic acid at $\pu{-80 °C}$ [2]:

  $$\ce{CO3^2- or HCO3- ->[FSO3H-SbF5/SO2][\pu{-80 °C}] H3CO3+ ->[\pu{-10 - 0 °C}] CO2 + H3O+}$$

As for the structure, the $C_\mathrm{3h}$-symmetrical molecule (triskelion shape, structure 7 below) was found to be the global minimum [2] (reference numbers updated):

  We also carried out ab initio and IGLO calculations […] on bicarbonate as well as free and mono-, and diprotonated carbonic acids to rationalize the observed experimental results. Initially the possible geometries were optimized at the HF/6-31G* level. The most stable isomers were subsequently determined from
  relative energies at the MP2/6-31G*//HF/6-31G* level. Finally, the lowest energy structures were further optimized at the MP2/6-31G* level. MP2/6-31G* optimized structures were used to calculate $\ce{^{13}C}$ NMR chemical shifts.

  […]

  Protonated Carbonic Acid. The symmetrical $C_\mathrm{3h}$ structure 7
  was found to be the global minimum. This structure was suggested for the observed species in superacid solution by Olah et al. [1] on the basis of $\ce{^1H}$ and $\ce{^{13}C}$ NMR spectroscopy. The ion at $\pu{-80 °C}$ shows single sharp peaks at $δ(\ce{^1H})~12.05$ and $δ(\ce{^{13}C})~165.4$ in the proton and $\ce{^{13}C}$ NMR spectra, respectively.

  Protonation of carbonic acid on the hydroxyl oxygen atom (8) was shown to be unfavorable by $\pu{23.4 kcal/mol}$ over protonation on the carbonyl oxygen (7). Frequency calculation at the HF/6-31G*//HF/6-31G* level showed that the structure 8 is not a minimum, as it contains two imaginary frequencies.

$\ce{H3CO3+}$ shares structural similarities with its triaza-analog, the guanidinium ion, as both are possessing resonance stabilization via their onium forms [3, p. 60].

References

Olah, G. A.; White, A. M. Stable Carbonium Ions. LXIV. Protonated Carbonic Acid (Trihydroxycarbonium Ion) and Protonated Alkyl (Aryl) Carbonates and Hydrogen Carbonates, and Their Cleavage to Protonated Carbonic Acid and Carbonium Ions. The Possible Role of Protonated Carbonic Acid in Biological Carboxylation Processes. J. Am. Chem. Soc. 1968, 90 (7), 1884–1889. https://doi.org/10.1021/ja01009a036.
Rasul, G.; Reddy, V. P.; Zdunek, L. Z.; Prakash, G. K. S.; Olah, G. A. Chemistry in Superacids. 12. Carbonic Acid and Its Mono- and Diprotonation: NMR, Ab Initio, and IGLO Investigation. J. Am. Chem. Soc. 1993, 115 (6), 2236–2238. https://doi.org/10.1021/ja00059a020.
Dewar, M. J. S., Hafner, K., Heilbronner, E., Itô, S., Lehn, J.-M., Niedenzu, K., Rees, C. W., Schäfer, K., Wittig, G., Boschke, F. L., Series Eds.; Topics in Current Chemistry; Springer Berlin Heidelberg: Berlin, Heidelberg, 1979; Vol. 80.


How to derive the total number of molecular orbitals based on the given number of atomic orbitals?

The question itself deals with molecular orbital theory and by extension with the approximation of combining atomic orbitals to form these molecular orbitals; in quantum chemistry this approach is called LCAO (Linear combination of atomic orbitals).

Let us look at one of the simple cases, the dihydrogen molecule. You have one 1s orbital at the first hydrogen, i.e. $\chi_A$, and one 1s orbital at the second hydrogen, i.e. $\chi_B$. You can now combine these orbitals to form a bonding $\eqref{eq:bonding}$ and an antibonding $\eqref{eq:antibonding}$ combination to create the molecular orbitals $\phi_{A+B}$ and $\phi_{A-B}$.
\begin{align}
\phi_{A+B} \equiv 1\sigma_\mathrm{g} &= \chi_A + \chi_B
  \tag1\label{eq:bonding}\\
\phi_{A-B} \equiv 1\sigma_\mathrm{u} &= \chi_A - \chi_B
  \tag2\label{eq:antibonding}
\end{align}
A scheme of this can be found on Wikipedia created by the user CCoil.

This does in principle mean that for every bonding configuration of a molecular orbital you will always have an antibonding configuration. In other words the number of atomic orbitals is the same as the number of molecular orbitals.

In your case, when there are four orbitals on one atom and four orbitals on the other atom, you will end up with eight molecular orbitals. A quite complicated case of bonding that shows this is the $\ce{C2}$ molecule, which has previously been discussed here: Bonding in diatomic C2, a carbon-carbon quadruple bond?


Is [7]сirculene aromatic or anti-aromatic?

TL;DR: We don't count π electrons here, rather the π conjugated circuits [1]. [7]-сirculene has all conjugated circles of size $4n + 2,$ and none of $2n,$ what makes it an aromatic compound.
Strictly speaking, the aromaticity $4n + 2$ and antiaromaticity $4n$ rules in their original interpretation are very limited and cannot be universally applied.
According to the original Hückel $4n + 2$ rule, if a monocyclic, planar molecule has $4n + 2$ π electrons, it is considered aromatic.
But the count of π electrons is not what matters for polycyclic compounds, therefore a generalized $4n + 2$ rule exists: if a conjugated molecule has only $4n + 2$ π conjugated circuits, it is considered aromatic.
Non-benzenoid compounds can have both $4n + 2$ as well as $4n$ number of conjugated circuits counted by taking into account all Kekule structures. Among non-benzenoid non-alternant polycyclic hydrocarbons [5]-сirculene and [7]-сirculene both have only $4n + 2$ $(6, 10, 14, \dots)$ numbers of circuits.
Further reading: Is there any special rules for checking the aromaticity of polycyclic compounds?
Reference

Gutman, I.; Cyvin, S. J. Conjugated Circuits in Benzenoid Hydrocarbons. J. Mol. Struct. THEOCHEM 1989, 184 (1), 159–163. DOI: 10.1016/0166-1280(89)85141-3.


so wouldn't it be that you
  would have an even more positive carbon and 2 partially negative
  oxygens

Yes, your analysis is correct to this point.  A chemist would say that the bonds in $\ce{CO2}$ are polar (or polarized) and therefor each $\ce{C=O}$ bond has a bond dipole moment.  However the molecule itself is linear and the two bond dipole moments are oriented 180 degrees with respect to one another and cancel each other out, so overall the molecule does not have a dipole moment and is non-polar.

EDIT:  There are a couple of reasons why $\ce{CO2}$ is more soluble in water than $\ce{O2}$.  Because the two $\ce{C=O}$ bonds in $\ce{CO2}$ are polarized (whereas in $\ce{O2}$ the bond is not polarized) it makes it easier for the polar water molecule to solvate it and to form hydrogen bonds.  Both of these factors will stabilize a $\ce{CO2}$ molecule more than an $\ce{O2}$ molecule in water; stabilization translates into greater solubility.  Another factor enhancing the solubility of $\ce{CO2}$ in water is the fact that $\ce{CO2}$ reacts with water to set up an equilibrium with carbonic acid.
$$\ce{CO2(aq) + H2O <=> H2CO3(aq)}$$
This reaction will also enhance $\ce{CO2}$'s solubility in water compared to oxygen which does not react with water.


Order of second iozination energies of lithium,berillium,neon,carbon and boron?

To reach the second ionization potential we remove a second electron from an already ionized element
$$\ce{M^{+1} -> M^{+2}}$$
Here are the electron configurations for the singly charged ions in your question, along with their second ionization potentials (kJ/mol)
\begin{aligned}
\ce{& Li^{+1} & 1s^2~~ &~~ 7297}\\
\ce{& Be^{+1} & 1s^2~ 2s^1~~&~~ 1757}\\
\ce{& B^{+1} & 1s^2~ 2s^2~~&~~ 2426}\\
\end{aligned}
We see that both $\ce{Li^{+1}}$ and $\ce{B^{+1}}$ have completely filled shells, the stable "inert gas" configuration.  We would expect it to be difficult to remove one electron from these two ions and destroy this stable configuration.  Further, we would expect removal of one electron from $\ce{Li^{+1}}$ to be more difficult than from $\ce{B^{+1}}$ because in the former case we are removing a 1s electron which is closer to the nucleus than the 2s electron in the case of $\ce{B^{+1}}$.  On the other hand, if $\ce{Be^{+1}}$ loses one electron it will have a $\ce{1s^2}$ electron configuration and achieve a stable "inert gas" configuration.  This should make it much easier to remove one electron from $\ce{Be^{+1}}$ than from the other two ions.


Why do aromatic compounds have an upfield shift upon coordination to metals?

Chemical shift data dug up for a number of iron compounds in the website of Hans Reich reveals that the shift is not unique to aromatic compounds, but rather appears due to Fe(2+):

The base shift for a vinyl proton (C=C-H) is 5.25 ppm.

The following is a series of spectra from magritek, illustrating the effect of Fe(2+) complexation on the shift of the cyclopentadienyl ligand.

Sandwich compounds based on magnesium do not show such large upfield shifts:

This seems to eliminate your conjecture 3 as an explanation, since cyclopentadienyl is aromatic and addition of Fe(2+) causes an upfield shift, whereas Mg(2+) does not cause such a large one.

Now Fe(2+) complexes may be paramagnetic or diamagnetic depending on the the ligand (low/high field). Paramagnetic metals have unpaired electrons that can induce strong dipolar magnetic fields. In porphin, for instance, paramagnetic Fe(3+) results in strong shielding in the aromatic ligand, with one proton shifted upfield ~20 ppm. A diamagnetic Zn(porphin) complex does not show such strong shielding. Ferrocene is diamagnetic.

The quick and sloppy answer therefore appears to be that the shift observed in ferrocene is mainly due to Fe(2+), even though the compound as a whole is diamagnetic.

Regarding the bonding arrangement, the wikipedia has this to say:

  In terms of bonding, the iron center in ferrocene is usually assigned
  to the +2 oxidation state, consistent with measurements using
  Mössbauer spectroscopy. Each cyclopentadienyl (Cp) ring is then
  allocated a single negative charge, bringing the number of π-electrons
  on each ring to six, and thus making them aromatic. These twelve
  electrons (six from each ring) are then shared with the metal via
  covalent bonding. When combined with the six d-electrons on Fe2+, the
  complex attains an 18-electron configuration.

To add a little more detail, Mulay and Fox hypothesize on the nature of metal-ligand bonding in ferrocene based on the magnetic anisotropy:

  The anisotropy between K3 and the average of K1 and K2 was found to be
  $\pu{49.5×10—6 cgs units}$, and is evidence in favor of a single dπ—pπ bond between each of the rings and the iron atom. Pauling's "resonance
  structures" also seem plausible; the studies do not support Fischer's
  "complete π‐electron donation" structure. Theoretical calculations
  based on Langevin's formula for diamagnetism show a close agreement
  with the experimental value.

A more detailed account of bonding in ferrocene can be found in Miessler and Tarr's textbook.

[6]: Magnetic Anisotropy of Ferrocene and Chemical Bonding, J. Chem. Phys. 38, 760 (1963); https://doi.org/10.1063/1.1733734, L. N. Mulay and S.N.D. Mary Eleanor Fox


What is the Solubility of Common Noble Gases in Concentrated Phosphoric Acid, 85% w/v?

Value seems to be $k_H^{\ominus}=4.3\times10^{-3}\;M/atm$ for Xenon in water.For it in conc. phosporic acid seems not to be available as probably no one has carried the experiment in this solvent since water is most common and used commonly.Maybe you would?

Also, A simple way to describe Henry’s law as a function of temperature is:
$$\LARGE k_H=k_H^{\ominus}\times e^{\left(\frac{-\Delta_{soln}H}R\left(\frac1T-\frac1{T^{\ominus}}\right)\right)}$$
where $\Delta_{soln}H=$enthalpy of solution. Here, the temperature dependence is:
$$\frac{-d\ln k_H}{d(1/T)}=\frac{\Delta_{soln}H}R$$

Useful Links:
Table of constants, The NIST Chemistry WebBook, The Pesticide Properties Database (PPD)


Why is the boiling point of trichlorofluoromethane lower than that of chloroform?

The main reason acting here is Dipole Moment. The higher the dipole moment of a molecule, the greater will be the intermolecular attractive forces and the higher will its boiling point be.
Dipole moment is the displacement of electron density in a molecule and it is a vector quantity. The net dipole moment of a molecule is the vector summation of all the individual dipole moment of 2 bonded atoms created by the electronegativity difference between them.
So, at first, let's get the structure of the molecules. Both of them are sp3 hybridised and have a tetrahedral structure.

For CHCl3, if you consider the bond between carbon and chlorine, its highly polar. Due to the high electronegativity of chlorine, there will be a net negative pole on the bottom part caused by the three Carbon chlorine bonds. And Hydrogen being less electronegative,however, will not attract the electrons to a great extent. Hence, there will be a pretty significant net dipole moment in CHCl3.

But in the case of CFCl3, yes, there will be a charge separation by the chlorine atoms, but on the upper portion lies the F atom which is very electro-negative. As a result, it will also pull its bond pair of electrons towards itself. Since, the net dipole moment is the vector summation of the individual dipoles, the charge separation of C-F, being in the opposite direction of the charge separation by the 3 chlorine atoms, will reduce the net dipole moment of the molecule.

For this reason, the dipole moment of CHCl3 is more than that of CFCl3 and so, CHCl3 has a higher boiling point.


Calculating pH in titration of monoprotic weak acid with strong base

Your calculation is right.  Regarding your question to take into account the conjugate base's contribution to deprotonation of water, I'd say:

If you calculate the equilibrium constant of the neutralization reaction $$\ce{CH3COOH + OH- <=> CH3COO- + H2O}$$
$$K= \frac{K_a}{K_w}=\frac{10^{-4.75}}{10^{-14}}= 10^{9.25}$$

Now, if you write  of reaction of acetate ion with water:$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$
It's the opposite of the neutralization reaction. Its  equilibrium constant  is  $10^{-9.25}$.

You can see clearly that the neutralization reaction is the "predominant reaction". So, the pH is determined by this reaction as if it's the only one happening in the aqueous solution. So, you can neglect the reaction of of acetate ion with water.


Why are chemistry droppers bad at holding bromine?

Since the standard dropper is designed to hold common liquid such as water or ethanol, the capillary size have some particular size. When it comes to bromine, the density is 3 times of water. In other words, the dropper need to hold a super heavy droplet, so you could expect that it won't be as efficient as usual. In addition, your previous question hinted on the density, so it is pretty obvious right?


Why is the chemical formula for ethanoic acid CH3COOH and not C2H4O2?

The first one is preferred since its makes it easy to see that it has a carboxyl group and a methyl group, so its easier to draw the Lewis structure and figure out the molecule's properties. For example, from the first formula one can see that it is an acid. However, its more difficult to figure that out from the second one.


Why does this E2 reaction occur on the less substituted carbon?

E2 elimination is stereospecific for anti elimination. In both variants of compound B that you have drawn, the OTs and the proton on the tertiary centre are syn to each other, meaning that the elimination is disfavoured.

For this reason, the NaOMe deprotonates at the secondary centre, generating the products as drawn, despite the fact that this goes against the Zeitsev rule which favours the most substituted alkene.


How does alkaline hydrolysis of peptides occur?

The unprotonated amino group ($\ce{R-NH2}$) does not protonate $\ce{OH-}$ because it is an extremely weak acid. Primary and secondary amines have a $pK_a$ of around 35, while the $pK_a$ of hydroxide is about 13-14 in aqueous solution. Deprotonating $\ce{NH2}$ groups requires much stronger bases in non-aqueous medium. Therefore, the carboxylic acid group of the amino acid is deprotonated in aqueous alkaline solution, as shown in the scheme from your textbook.