How is heat physically released in an exothermic process?

I think that you are on the right track. I try to explain how I see things below.

As a bond forms the reactants must absorb some energy from their surroundings so that the activation barrier between reactants and products may be surmounted. (By the Boltzmann distribution there are always a few collisions with far more energy than the average). At the top of the barrier is the transition state (TS), for this to be crossed the excess energy has to be taken away from the TS more rapidly that it may cross back to reactants.

The transition state is effectively a very hot species; it has far more energy than the surrounding medium. This can be thought of as a species having a large number of vibrational and rotational quanta excited. As far as is known$^*$ a transition state lasts for only a few tens of femtoseconds to a few picoseconds because vibrational/rotational energy is lost very rapidly to the surroundings. This excess energy is carried away also as kinetic energy of the products if, say, two molecules of product are formed. As time progresses this energy excites more and more vibrational and rotational modes in the surrounding medium, which may be a solvent or protein's backbone and residues. Eventually this energy diffuses away from the reaction site (via anharmonic vibrational couplings) and the product eventually has vibrational and rotational modes excited with quanta in accord with the prevailing temperature.

$*$ Although transition states are universal there is a paucity of actual, i.e. time resolved spectroscopic data on their properties.

(A more detailed approach has been given by Kramers (see M. Daune ' Molecular Biophysics for a discussion). In the condensed phase 'friction' or 'drag' on the reaction coordinate slows the reaction (compare to gas phase) and the reactants 'diffuse' towards and across the transition state. 'Friction' here means vibrational and rotational energy affecting the reaction coordinate, by impeding or promoting, by turns, the progress of the reaction. The Kramers approach allows a consistent approach to  gas phase reactions from isolated molecules, then in the presence various pressures of inert gas right through to reactions in viscous solution.)


What are the implications of having different conformations?

That's an interesting question. To start off, what Mith said in the comments is absolutely right. Let's say that two conformations of a molecule have different smells (which is in fact not an absurd proposition by any means). The problem is that, because there is a very low barrier to the interconversion of the conformations, the individual conformations do not really hang around long enough to be individually detected; you can only really detect a weighted average of the conformations.

To give you an example, you can devise di-substituted cyclohexanes that have two predominant chair conformations. These can be investigated by proton NMR spectroscopy, since axial-axial coupling constants are larger than equatorial-axial or equatorial-equatorial couplings. From the integrals of the relevant peaks you can determine the relative proportions of both conformations.

However, if you somehow managed to synthesise a sample of the pure diequatorial conformer (the one in blue) and put it into a NMR tube, some of it would have converted into the diaxial conformer (the one in red) before you could even run your NMR. In fact, there's probably no way you can synthesise a pure conformer to begin with.

Likewise, if you replace the NMR machine with your nose, there's no way you can open a bottle of a perfume X, with conformation X1, and smell something, then open a bottle of the same perfume X with conformation X2 and smell another thing. They would both smell the same because conformers aren't isolable. (Well, not unless you are working in an argon matrix at temperatures near absolute zero or something like that!)

Mith then goes on to say that you can bind to a molecule to manipulate its conformation. That's the more interesting part, in my opinion at least! If you think about it, it's going to be very hard for us to devise some kind of surface that particularly favours one conformation over another. You'd need a very complex binding partner, with a very specific 3D shape, that "fits" with one conformation but not another conformation.

Does that ring a bell? It might, because nature has made those binding partners for us already. That previous paragraph is a perfect description of an enzyme: a protein that catalyses a biological reaction. Lysozyme is one of the most famous enzymes; it was the first enzyme whose structure was successfully solved by X-ray diffraction. It is also a great example of an enzyme that works by manipulating conformations.

Lysozyme binds to a series of carbohydrates in the peptidoglycan cell walls of some bacteria. Together with water, it hydrolyses the glycosidic bonds (a fancy name for an acetal) in these carbohydrates:

(image from Wikipedia)

I'm not going to go into the full mechanism of this, but the striking thing is that upon binding to lysozyme, the "D ring" (which is the NAM residue on the left-hand side of the diagram above) is distorted from its usual chair conformation into a half-chair conformation. This is because it "fits" better into the exact 3D conformation of the enzyme (not shown in the diagrams):

(image modified from Wikipedia)

This makes the ring more unstable and prone to hydrolysis (strictly speaking, more prone to nucleophilic attack by the aspartate-52 residue). Without this manipulation of the conformation, it's likely that the hydrolysis cannot proceed.

So, there you have it! One way in which the conformation of a molecule, upon binding to another molecule, affects its properties.


Is there an actual specific heat for aluminum in its gaseous state?

By "bring aluminum from its liquid state to a gaseous state", I assume you mean to a gaseous state at a temperature above aluminum's boiling point—since to merely convert it from a liquid to a gas, you only need the enthalpy of vaporization.
According to https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node18.html,
most metallic vapors have molar heat capacities very close to those of ideal gases, i.e., $C_v = 3/2 R = 12.47$ J/(mol K) and $C_p = 5/2 R = 20.79$ J/(mol K).  Thus you can obtain aluminum's specific heat capacity by dividing by the atomic mass, giving $C_v = 0.46$ J/(g K)  and $C_p = 0.77$ J/(g K)
The reason for this is that metals don't form metallic bonds in the gaseous state (metallic bonding is a collective phenomenon that requires bulk metal), and are thus monatomic*.  And the heat capacities of monatomic gases are simple, since they have only translational degrees of freedom.
[*As real gases, they may not be exactly monatomic, since they can still transiently associate via van der Waals forces.  But treating them as monatomic is a good approximation.]
The exception to the above would occur when the temperature is sufficiently high to cause the metal to ionize, in which case you would have a higher heat capacity because you would now have two (or more) particles for every ionized aluminum atom. But since aluminum's first ionization energy is 5.985 eV/atom = $577$ kJ/mol, and $RT$ at $3000$ K is $25$ kJ/mol, I don't expect there to be significant ionization close to its $2792$ K boiling point. [Also, some might argue that, once it starts ionizing, it is no longer a gas, but a gas-plasma mixture.]


Conversion  1

Convert ethene to ethanol.
Ethene is mixed with steam and passed over a catalyst consisting of solid silicon dioxide coated with phosphoric(V) acid. The temperature used is 300 °C and the pressure is about 60 to 70 atmospheres (chemguide —
The mechanism for the acid catalysed hydration of ethene).
$$\ce{CH2=CH2 + H2O <=>[H3PO4] CH3CH2OH}$$

React the ethanol with oxalic acid to get diethyl oxalate.
$$\ce{(COOH)2 + 2C2H5OH ->[CCl4,\Delta] (COOC2H5)2 + 2 H2O}$$
The details of this reaction is given on Organic Syntheses — Ethyl Oxalate and in the paper by Jewel and Butts [1].

Conversion 2
Convert ethene to oxalic acid by first converting to ethyne and then making oxalic acid from it. Use the oxalic acid to make diethyl oxalate by reacting it with absolute alcohol.
$$\ce{CH2=CH2 ->[Br2][CCl4] Br-CH2-CH2-Br ->[alc. KOH][\Delta] CH#CH ->[alc. KMnO4] (COOH)2}$$
Reference

Jewel, P. W.; Butts, J. S. A Method for the Preparation of Diethyl Oxalate. J. Am. Chem. Soc. 1931, 53 (9), 3560–3561. DOI: 10.1021/ja01360a052.


Relative humidity above various mixtures of table salt and water

Your information on relative humidity (RH) above saturated $\ce{NaCl}$ is valid within about 0.5% over the range of 0 °C to 80 °C. For other standard humidity solutions, such as $\ce{MgCl2}$ or $\ce{K2CO3}$ the U. S. National Institute of Standards & Technology provides a convenient reference.

However, though it is possible to adjust RH by varying concentration, there are, "special  problems...  because  their  concentrations  must  be  measured  and controlled." If you need only a few accurate RH points, it would be simpler to use saturated solutions of some alternate salts.

BTW, it may take some time for RH to equilibriate because of stratification. It would help to keep the solution on a slow magnetic stirrer, moving not only the liquid but causing eddies in the air above it.


Zinc evaporation/toxicity at room temperature

Zinc will amalgamate with mercury, but whether that reduces the vapor pressure of mercury very much is debatable. At best, the increased volume of zinc amalgam and the ability to amalgamate with fine droplets of mercury would make it easier to clean up the mercury as amalgam.

It might be better to spread some sulfur dust (flowers of sulfur) over the area. The sulfur should react with the mercury and tie it up as sulfide, with very low vapor pressure. Sulfur powder (and possibly the HgS) might also be easier to clean up than zinc dust, because sulfur is more like ordinary dust than a dense metal dust.

BTW, I love mercury thermometers. And I hate it when I break them!


What's the difference between PBE and B3LYP methods?

PBE

The PBE functional${}^{[1]}$ belongs to the class of generalized gradient approximation (GGA) functionals for the exchange-correlation energy $E_{\mathrm{xc}}$. Considering that the dependence $E_{\mathrm{xc}}[\rho]$ may be non-local, i.e. $E_{\mathrm{xc}}$ may depend on the density $\rho$ at a given point (locality), but also on $\rho$ nearby (non-locality) the assumption made by the local spin density approximation${}^{[2,3]}$ (LSDA) that $E_{\mathrm{xc}}[\rho]$ is strictly local leaves a lot room for improvement. But still LSDA is a good starting point since it is a rather simple efficient model that gives good results. The GGA functionals improve on LSDA in a way that's pretty much in the spirit of a Taylor series with LSDA as a starting point: When we are at a point, what happens further off depends not only on $\rho$ at that point, but also the gradient of $\rho$ at the point, etc. (much as in a Taylor series: if you want to go further you may need not only the gradient, but also the Laplacian, etc.). So, these functionals are typically constructed by adding gradient corrections to the LSDA functionals. But that is not a trivial thing to do and doing it "naively" leads to wrong results. Therefore, some GGA functionals are parametrized by fitting experimental data. But there are also some GGA functionals that achieve the inclusion of gradient correction without introducing experimentally fitted parameters which makes them valid for a wide range of systems. One example of such a parameter-free GGA functional is the one developped by Perdew, Burke and Ernzerhof (PBE). It is known for its general applicability and gives rather accurate results for a wide range of systems.

Side note: GGA functionals are frequently termed non-local
functionals in the literature. This is a somewhat misleading and actually sloppy terminology that should be avoided and since my above statements might be misread in that direction I want to make one thing perfectly clear: All GGA functionals are perfectly
local in the mathematical sense: the value of the functional at a point $\vec{r}$ depends only on information about the density $\rho(\vec{r})$, its gradient $\nabla\rho(\vec{r})$, and possibly other information at this very point and is absolutely independent of properties of $\rho(\vec{r}^{\, \prime})$  at points $\vec{r}^{\, \prime} \neq \vec{r}$. Calling these functionals 'non-local' is only motivated by the fact that these functionals go beyond the 'local' density approximation and of course the observation that knowledge of
the gradients is the first step towards accounting for the inhomogeneity of the real density.

B3LYP

The B3LYP approach${}^{[4,5,6]}$ belongs to the hybrid approximations for the exchange–hybrid correlation functional. The approximation is famous, because it gives very good results and, therefore, is extremely popular. The distinguishing feature of such hybrid approximations is that they mix in a certain amount of the exact Hartree–Fock exchange energy into the exchange and correlation obtained from other functionals. There is actually some justification for that: At the lower limit ($\lambda = 0$, where $\lambda$ is the coupling constant) of the so called coupling constant integration the exchange-correlation hole is equal to the exact exchange hole. This observation led Becke${}^{[7, 8]}$ to conclude that a fraction of exact exchange should be mixed with GGA exchange and correlation. The simplest such hybrid functional is

\begin{align}
E^{\mathrm{hyb}}_{\mathrm{xc}} = a E_{\mathrm{x}}^{\mathrm{exact}} + (1 − a) E^{\mathrm{GGA}}_{\mathrm{x}} + E^{\mathrm{GGA}}_{\mathrm{c}}
\end{align}

where the constant $a$ can be fitted empirically or estimated theoretically ${}^{[9, 10, 11]}$
as $a \approx \frac{1}{4}$ for molecules.
So far so good, but it smells like a witch's brew for the B3LYP exchange–correlation potential $E_{\mathrm{xc}}$: take the exchange–correlation energy from the LSDA
method, add a pinch (20%) of the difference between the Hartree–Fock exchange energy
$E^{\mathrm{KS}}_{\mathrm{x}}$ (well, in fact, this is Kohn–Sham exchange energy, because the Slater determinant wave function, used to calculate it, is the Kohn–Sham determinant, not the Hartree–Fock one) and the LSDA $E^{\mathrm{LSDA}}_{\mathrm{x}}$. Then, mix well 72% of Becke exchange potential${}^{[12]}$ $E^{\mathrm{B88}}_{\mathrm{x}}$ which includes the 1988 correction, then strew in 81% of the Lee–Yang–Parr correlation potential${}^{[13]}$ $E^{\mathrm{LYP}}_{\mathrm{c}}$. You will like this homeopathic magic potion most if you conclude by putting in 19% of the Vosko–Wilk–Nusair potential$^{[14]}$ $E^{\mathrm{VWN}}_{\mathrm{c}}$:

\begin{align}
  E^{\mathrm{B3LYP}}_{\mathrm{xc}} = E^{\mathrm{LSDA}}_{\mathrm{xc}} + 0.2 (E^{\mathrm{KS}}_{\mathrm{x}} - E^{\mathrm{LSDA}}_{\mathrm{x}}) + 0.72 E^{\mathrm{B88}}_{\mathrm{x}} + 0.81 E^{\mathrm{LYP}}_{\mathrm{c}} + 0.19 E^{\mathrm{VWN}}_{\mathrm{c}}
\end{align}

So, you can see B3LYP contains lots of empirical parameters but that's fine as long as you only want the results and don't want to know why this is working :)

References

${}^{1}$ J. P. Perdew, K. Burke, M. Ernzerhof, "Generalized Gradient Approximation
Made Simple", Phys. Rev. Lett. 1996, 77, 3865; J. P. Perdew, K. Burke,
M. Ernzerhof, "Erratum to Generalized Gradient Approximation Made Simple", Phys. Rev. Lett. 1997, 78, 1396 (E). (Link)

${}^{2}$ W. Kohn, L. J. Sham, Phys. Rev. 1965, 140, A1133. (Link)

${}^{3}$ U. von Barth, L. Hedin, "A local exchange-correlation potential for the spin polarized case", J. Phys. C: Solid State Phys. 1972, 5, 1629–1642. (Link)

${}^{4}$ A. D. Becke, "Density-functional thermochemistry. III. The role of exact exchange", J. Chem. Phys. 1993, 98, 5648-5652. (Link)

${}^{5}$ P. J. Stephens, F. J. Devlin, C. F. Chabalowski, M. J. Frisch, "Ab Initio Calculation of Vibrational Absorption and Circular Dichroism Spectra Using Density Functional Force Fields", J. Phys. Chem. 1994, 98, 11623–11627. (Link)

${}^{6}$ K. Kim, K. D. Jordan, "Comparison of Density Functional and MP2 Calculations on the Water Monomer and Dimer", J. Phys. Chem. 1994, 98, 10089–10094. (Link)

${}^{7}$ A. D. Becke, "A new mixing of Hartree–Fock and local density‐functional theories ", J. Chem. Phys., 1993, 98, 1372. (Link)

${}^{8}$ A. D. Becke, "Density‐functional thermochemistry. III. The role of exact exchange", J. Chem. Phys., 1993, 98, 5648. (Link)

${}^{9}$ J. P. Perdew, M. Ernzerhof, and K. Burke, "Rationale for mixing exact exchange with density functional approximations", J. Chem. Phys., 1996 105, 9982. (Link)

${}^{10}$ C. Adamo and V. Barone, "Toward reliable density functional methods without adjustable parameters: The PBE0 model", J. Chem. Phys., 1999 110, 6158. (Link)

${}^{11}$ M. Ernzerhof and G. E. Scuseria, "Assessment of the Perdew–Burke–Ernzerhof exchange-correlation functional", J. Chem. Phys., 1999, 110, 5029. (Link)

${}^{12}$ A. D. Becke. "Density-functional exchange-energy approximation with correct asymptotic behaviour", Phys. Rev. A 1988, 38, 3098-3100. (Link)

${}^{13}$ C. Lee, W. Yang, R. G. Parr, "Development of the Colle-Salvetti correlation-energy formula into a functional of the electron density", Phys. Rev. B, 1988, 37, 785-789. (Link)

$^{14}$ S. Vosko, L. Wilk, M. Nusair, "Accurate spin-dependent electron-liquid correlation energies for local spin density calculations: a critical analysis", Can. J. Phys. 1980, 58, 1200-1211. (Link)


Acid titration of sodium acetate in the presence of sodium ethoxide, is it possible?

The pH of 0.1 M NaAc is calculated to be 8.9. https://pubs.acs.org/doi/pdf/10.1021/ed079p29.1
The pH of NaOH will be much higher. The 30-40% of ethanol may have some effects, but in general, you should be able to add about twice as much NaOH as you think will react with all the EtAc, reflux and back titrate the NaOH. As Poutnik comments, there won't be significant NaOEt.
The reaction will look like this:
EtAc + 2 NaOH --> NaAc + EtOH + NaOH
If you can get some thymolphthalein, you could use that as an indicator.

https://en.wikipedia.org/wiki/PH_indicator
Or, you can use pH paper strips if you are satisfied with a cruder analysis.
Possibly a more intricate back titration might work: add about twice as much NaOH as you think will react with all the EtAc in your ethanol, reflux as above, then add the same number of moles of acetic acid as you added moles of NaOH. Then the reaction will look like this:
EtAc + 2 NaOH --> NaAc + EtOH + NaOH;   reflux;   + 2 HAc -->
NaAc + EtOH + NaAc + H$_2$O + HAc
Then back titrate the excess acetic acid with phenolphthalein (common enough - you must have some). If there were no EtAc in your ethanol, the pH of the solution after adding the 2 HAc will be about 8.9; if there is any EtAc in your ethanol, you will have excess acetic acid due to the saponification of the EtAc. And you back titrate that with NaOH. So this is sort of a back to back titration, or a back and forth.


How do I calculate the enthalpy change when a gas is being used to heat water?

You're on the right track.

\begin{eqnarray}
Q & = & m\cdot c\cdot \Delta T\\
\mathbf{unit\ analysis} \qquad \textrm{kJ} & = & \textrm{kg}\cdot \frac{\textrm{kJ}}{\textrm{kg} \cdot \textrm{K}} \cdot \textrm{K}
\end{eqnarray}

For water, the specific heat capacity $c = 4.18 \large{\frac{\textrm{kJ}}{\textrm{kg}\cdot \textrm{K}}}$. You have that right.

The temperature difference $\Delta T = (50.7 - 7.44) = 43.26\ \textrm{K}$. We're talking about differences, so we can use kelvin as the unit. You have that right too.

Now, what is the mass of $724\ \textrm{cm}^3$ water?

Choose any unit you want, but adjust the outcome of your calculation according to the unit analysis above!

Now you know the kJ you need.

From the combustion equation, you know the amount of heat delivered by $n$ moles of butane. Note the $n$!

Now you can calculate how many moles of butane you will need.

From there, the calculation of the mass is not difficult either.


Is it possible for an acid and base to react to form an even stronger base or acid?

$\ce{6HF + Al(OH)3 (s) -> H3[AlF6] + 3H2O}$
$\ce{6HCN + Fe(OH)2 (s) -> H4[Fe(CN)6] + 2 H2O}$
$\ce{TlOH + HOCl + H2O -> Tl(OH)3(s) + HCl}$

May be one of these will work. Even if they don't, you should get the idea where to search.


Does adding slices of lemons to water reduce its chlorine content?

Yes , a quick and cheap way to reduce chlorine residual is to add ascorbic acid
 (VITAMIN C), which if added in an equal molar amount to free chlorine would completely eliminate chlorine residual within 1 min.

I am quoting from ABSTRACT from a paper from, Int. Journal of Engineering Research and Applications , ISSN : 2248-9622, Vol. 3, Issue 5, Sep-Oct 2013, pp.1647-1651.

"Society’s preference for bottled water mainly comes from a dislike of the mildly pungent taste and odor of tap water. Such odor and taste may be caused by chlorine residual (in the form of free chlorine or chloramines), which is maintained in a water distribution system to prevent regrowth of microorganisms and is thus present in tap water.

"Two methods of reducing chlorine residual, boiling tap water and adding ascorbic acid to
tap water, were tested. The results showed that, for free chlorine, a quick and cheap way to reduce chlorine residual is to add ascorbic acid, which if added in an equal molar amount to free chlorine would complete eliminate chlorine residual within 1 min "

Reduction of Chlorine Residual in Tap Water with Temperature

source : https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf

As shown aboce, the chlorine residual decreased faster at higher temperatures.

Reduction of Free Chlorine and Monochloramine in Tap Water through Boiling

source : https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf

As shown in Figure 2, both residuals of free chlorine and monochloramine decreased with boiling time, and the residual of free chlorine decreased slightly faster than that of monochloramine.

The boiling time required to completely remove free chlorine was about 33 min.$\ce{^1}$

Effect of Ascorbic Acid on Free Chlorine and Monochloramine

Ascorbic acid added in a molar ratio of 1:1 (free chlorine to ascorbic acid) completely eliminated free chlorine. The DPD indicator(A DPD solution can also be used for the measurement of chlorine residual) did not produce a red color when added within 1 min,indicating complete elemination chlorine residual within 1 min.$\ce{^2}$

The reduction of free chlorine by ascorbic acid can be expressed as follows:

$\ce{C6H8O6}$ + $\ce{HOCl}$  →$\ce{C6H6O6}$ +$\ce{HCl}$ + $\ce{H2O}$

Ascorbic acid $\ce{(C6H8O6)}$ is oxidized to dehydroascorbic acid $\ce{(C6H6O6)}$ . Simultaneously, free chlorine $\ce{(HOCl)}$ is reduced to chloride ions (HCl).

In the experiment conducted it was determined that 10 mg of ascorbic acid is required to treat 1 L of tap water. and the quickest way to reduce the concentration of free chlorine in tap water is to add powdered vitamin C (ascorbic acid) tablets in less than 1 min.$\ce{^3}$

Thanks to N4v i will be using this cost effective method to remove chlorine from water.(wow).

Reference

https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf
https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf
https://www.ijera.com/papers/Vol3_issue5/JN3516471651.pdf


What is the difference in (alc) and (aq) states in reaction mechanism?

Nucleophilic substitution always competes with Elimination reaction in such cases.

What is the difference between them?

As put there by philipp:

  In alcoholic solution the $\ce{KOH}$ is basic enough ($\mathrm{pK}_{\mathrm{a}}$ 15.74) to deprotonate a small amount of alcohol molecules ($\mathrm{pK}_{\mathrm{a}}$ 16–17)), thus forming alcohlate salts $\ce{ROK}$. The alcoholate anions $\ce{RO-}$ are not only more basic than pure $\ce{OH-}$ but they are also bulkier (how much bulkier depends on the group $R$). The higher bulkiness makes $\ce{RO-}$ a worse nucleophile than $\ce{OH-}$ and the higher basicity makes it better at $\mathrm{E}2$ eliminations.

Mechanisms below:


Does every solute spend at least $t_M$ in the mobile phase?

Dead time ($t_{\bf M}$, also called holdup time) is the time it takes for the mobile phase (eluent) to traverse one length of the column, for a given flow rate. Because moieties in the eluent do not interact chemically with the stationary phase, $t_{\bf M}$ is therefore a lower bound on retention times for the species that do interact (chemically) with the stationary phase.

Simply put, anything that does interact with the immobile phase cannot elute faster that the pure mobile phase: retention times will be greater than $t_{\bf M}$.


Heats of combustion of cycloalkane homologues

Have a look at this table from Organic Chemistry by Wade and Simek [1, p. 134]:

Every time a $\ce{-CH2}$ group is added, the heat of combustion is increased by almost 650 kJ/mol; whereas the maximum relief in ring strain attained by adding that $\ce{-CH2}$ group is not more than 25 kJ/mol.

So there is no competiton between increase in heat of combustion due to more number of carbon atoms and decrease in it caused by ring stability. Everytime you see two hydrocarbons with different number of carbon atoms, be assured that the one with more carbon will realease more heat on combustion.

P.S. If in case you are unfamiliar with how ring strain is quantitatively measured [1, p. 133]:

  The molar heat of combustion of cyclohexane is nearly twice that of cyclopropane, simply because cyclohexane contains twice as many methylene ($\ce{-CH2}$) groups per mole. To compare the relative stabilities of cycloalkanes, we divide the heat of combustion by the number of methylene ($\ce{-CH2}$) groups. The result is the energy per $\ce{-CH2}$ group. These normalized energies allow us to compare the relative amounts of ring strain (per methylene group) in the cycloalkanes. The reference value of 658.6 kJ (157.4 kcal) per mole of $\ce{-CH2}$ groups comes from an unstrained long-chain alkane.

References

Wade, L. G.; Simek, J. W. Organic Chemistry, 9th ed.; Pearson: Glenview, IL, 2017. ISBN 978-0-321-97137-1.


Aluminum does assume negative oxidation states, see the list of elemental oxidation states in Wikipedia. Aluminum assumes a negative oxidation state when forming compounds with more electropositive elements like magnesium.

The magnesium compounds referred to above include $\ce{Mg_{17}Al_{12}}$.  In this compound there is enough magnesium to half-fill the valence $p$ orbital on aluminum and thus attain a $-2$ oxidation state for the aluminum, which is the minimum in Wikipedia's list. (Part of the magnesium would then be below $+2$).  However, this is only a formal designation. Magnesium aluminides are far from ionic in character.


Can F2/HF be separated using acid/base solutions in series?

The answer is no, because as you remark $\ce{F2}$ will react with water and produce $\ce{HF}$.  So it is not the case that $\ce{F2}$ passes through unreacted.

I'm puzzled by this sentence:

  Despite the Method not mentioning F₂, it commonly produces data for both HF and F₂

Can you clarify what you mean by "it produces data"?


My answer to this question is the following. Nodes can happen for highly excited states, whose average distances are typically much larger than the Bohr radius. Consider Na, Cs etc, which have one outermost shell electron. When the outermost shell electron is highly excited, the case is not much different from that of a hydrogen atom (i.e. the indistinguishability of electrons may be ignored when electrons are well separated.) Thus there can be nodes in these highly excited states.

A more non-trivial question is, whether the total electron density can have nodes for the ground state. I tend to think that it's not possible. My following naive, sketchy explanation may be seriously wrong. For a second-order eigenvalue problem, we have the sturm theorem (?; in doubt), which guarantees that the roots of successive states occur in an interlacing manner. That means, where state $i$ has a node at $r_i$, state $i+1$ cannot have a node at $r_i$. Thus the density, which is equal to the sum of square of wave function of all the occupied states, cannot be zero at one point. We have a similar second-order eigenvalue problem for each angular momentum $l$, and so the argument goes.

The above too sketchy, so let's formalise it a bit based on the one-particle approximation. The total ground-state electron density is proportional to $ \sum_{\{n,l, m, s\}; \textrm{occupied}}|\psi_{nlm}(r)|^2 |Y_{lm}(\theta, \phi)|^2$. Here $s$ is for the spin, $n$ labels the energy state (1 for ground, 2 for first excited, etc), and the sum is over all occupied states. Note that in order for the electronic density to be zero at $r_0$, each term has to vanish at $r_0$. But remember that for $l=0$, $m=0$, the $n=1$ radial wave function $\psi_{100}(r)$ can have no node, and $Y_{00}$ is a constant, which means the term $\psi_{100}(r)Y_{00}$ can vanish nowhere in space. This state has to be occupied in the ground state.


Can antimony pentafluoride be made by adding hydrogen fluoride to antimony trifluoride?

It can be prepared using hydrogen fluoride, in a double displacement reaction:

  $\ce{SbCl5 + 5 HF -> SbF5 + 5 HCl}$

However, hydrogen holds onto fluorine more tightly than does antimony in $\ce{SbF5}$.

Consider the reverse of your proposed reaction, which is preferred energetically:

  $\ce{SbF5 + H2 -> SbF3 + 2HF}$


Oxidation of metals/halogens by oxygen gas in acidic aqueous solution

As noted in your question, the standard reduction potentials are given as such in tables.

  $$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{Ag+ + e- &→ Ag(s)} &\quad E^\circ &= \pu{+0.799 V} \\
\ce{Br2(l) + 2 e- &→ 2 Br-(aq)} &\quad E^\circ &= \pu{+1.065 V}
\end{align}
$$

To get the the standard oxidation potentials you reverse the products and reactants and you must also flip the sign of the reaction. So:

  $$
\begin{align}
\ce{Ag(s) &→ Ag+ + e-  } &\quad E^\circ &= \pu{-0.799 V} \\
\ce{2 Br-(aq) &→ Br2(l) + 2 e- } &\quad E^\circ &= \pu{-1.065 V}
\end{align}
$$

Now there are two possible overall reactions (ignoring the Nernst equation):

  $$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{Ag(s) &→ Ag+ + e-  } &\quad E^\circ &= \pu{-0.799 V} \\
\ce{O2(g) + 4 H+(aq) + 4 Ag(s)&→ 4Ag+ + 2 H2O(l)} &\quad E_\text{Total}^\circ &= \pu{+1.23 - 0.799 = +0.43 V}
\end{align}
$$

and the other overall reaction is

  $$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{2 Br-(aq) &→ Br2(l) + 2 e- } &\quad E^\circ &= \pu{-1.065 V} \\
\ce{O2(g) + 4 H+(aq) + 4 Br-(aq) &→ 2Br2(l) + 2 H2O(l)} &\quad E_\text{Total}^\circ &= \pu{+1.23 - 1.065 = +0.16 V}
\end{align}
$$

Since the EMF for both overall reactions is positive, the reactions should be spontaneous. If the overall EMF was negative then the reaction would proceed in the opposite direction (in other words, swap the products and reactants.)

Notes:

(1) I don't know why the silver reaction doesn't happen. The EMF for the cell indicates that silver should oxidize in these conditions.

(2) The EMF's calculated ignore the Nernst equation.

(3) The oxygen half cell reaction is a bit weird. Obviously the oxygen must be dissolved in the aqueous solution to react. I'm assuming that the notation is trying to point out that the dissolved oxygen is in equilibrium with the atmospheric oxygen which has a constant concentration which is inexhaustible.


Remove empty space from structures with R and S groups

Are you sure that the R and S are in the mol file and not part of the xml envelope?

Stereochemical R and S descriptors would not make any sense for a molecule like anthracene!

Is it conceivable that the letters come from Risk and Safety statements?

It would be helpful if you could upload the original mol file somewhere.

EDIT 1

The file that you kindly uploaded is supposed to be a V2000 mol file, but it is not standards-compliant.

(The full specifications can be obtained from Accelrys upon registration.)

The first flaw is already in the header! The header is supposed to have a length of three lines.

The following CTAB should start with the counts line (in line 4 - note that we're not using zero-based indexing here).

With this flawed format, OpenBabel, a standard tool for the conversion of chemical data formats, casts an error.

EDIT 2

The rest of the file, including the stereochemical descriptors, seemed to be fine.

Once I had inserted the missing blank line in the header, a PNG could be generated using

obabel -imol 498268e9.txt -opng -O stuff.png

OpenBabel comes with bindings for Python, Java, C++ and whatnot. You might want to talk to your IT about some refactoring ;)


Confusion about the choice of axis for Molecular Orbital Theory

Although we could not see what is written in OP's "reference books", the mentioned difference most likely is the result of different choices for the bond axis.

Conventionally, and as it is (implicitly?) done in the MIT video lecture, it is chosen to be the $z$-axis, in which case $\pi$ bonds are indeed formed from $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals while $\mathrm{p}_z$ can take part in $\sigma$-bonding. If, however, the bond axis is chosen to be the $x$-axis, then $\pi$ bonds are formed from $\mathrm{p}_y$ and $\mathrm{p}_z$ orbitals while $\mathrm{p}_x$ takes part in $\sigma$-bonding.


What is the n factor of h2o2 undergoing disproportionation reaction liberating oxygen

We can split the reaction
$$\ce{H2O2 -> H2O + O2}$$
into the respective reduction and oxidation half-reactions.
$$\ce{H2O2 -> O2 + 2H+ + 2e-}$$
$$\ce{H2O2 + 2H+ + 2e- -> 2H2O}$$
Since the n-factor of $\ce{H2O2}$ for both these half-reactions is 2, the n-factor is:
$$\frac{1}{n_f} = \frac{1}{2} + \frac{1}{2} = 1$$
$$n_f = 1$$
Based on the comments under your post I believe you have gotten this far, but some sources claim that the n-factor for this reaction is 2. Well, how do you verify what is correct? In such a case, I'd simply write the balanced chemical reaction and verify whether the law of chemical equivalence holds or not.
$$\ce{2H2O2 -> 2H2O + O2}$$
The n-factor for $\ce{O2}$ should be $2$ as can be seen in the first half-reaction, and for $\ce{H2O}$ the n-factor should be $1$, since we have $2$ moles of $\ce{e-}$ in the second half-reaction and $2$ moles of $\ce{H2O}$, and n-factor is defined per-mole of the substance.
From the law of chemical equivalence, the number of equivalents of $\ce{H2O2}$, $\ce{H2O}$ and $\ce{O2}$ should all be equal. If the claim I make above that the n-factor of $\ce{H2O2}$ is $1$ is true, this should hold.
Plugging in the moles from the balanced chemical equation and using the n-factors described above,
Equivalents of $\ce{H2O2} = 2 \times 1$, $\ce{H2O} = 2 \times 1$, $\ce{O2} = 1 \times 2$ (moles $\times$ n-factor)
Thus the n-factor is indeed $1$. The n-factor of $\ce{H2O2}$ for the half-reactions is $2$.
(Note that all the n-factors above are with respect to the reaction $\ce{H2O2 -> H2O + O2}$ and they (most probably) will change for a different reaction)
Update (the above part answered the question before the edit):
From [1]:

The "volume strength" of hydrogen peroxide is defined as the number of times its own volume of oxygen a sample of hydrogen peroxide solution will evolve if decomposed naturally.
$$ \ce{H2O2 -> H2O + O2}$$

As you've mentioned in your question,
$$\pu{Normality = \frac{Volume~strength}{5.6}}$$
and
$$\pu{Molarity = \frac{Volume~strength}{11.2}}$$
This does not match with the reasoning that the n-factor of $\ce{H2O2}$ for the above reaction is $1$, which would imply that the normality and molarity should be equal.
From [2]:

Now, equivalent weight of $\ce{H2O2} = \frac{34}{2} = 17$

A (N) $\ce{H2O2}$ solution contains 17 gms of $\ce{H2O2}$ per litre.

So in fact when we are relating the normality of an $\ce{H2O2}$ solution to it's volume strength, we are not referring to the normality of $\ce{H2O2}$ in the decomposition reaction. You would be right in saying that the normality of $\ce{H2O2}$ in the decomposition reaction is equal to its molarity.
One way of justifying the n-factor of $2$ of an $\ce{H2O2}$ solution (not undergoing any reaction) is due to the $-2$ charge on the $\ce{O2^{2-}}$ peroxide ion.
I can't really explain why this has been done, since it would make just as much sense to consider the n-factor of $\ce{H2O2}$ in the decomposition reaction, it is just a historical fact. Since n-factor is not a constant for a particular compound, it should be really important to specify context about normality when we relate it to volume strength, which is rarely done when the relation is taught which is what leads to this confusion. Both the n-factor's are correct in different contexts, so I would just recommend sticking to the n-factor of $2$ when talking about volume strength since that is what you have been taught.
[1]: Chemical Calculations by King A archive.org
[2]: An Introduction To Chemistry Vol. 1 Ed. 3rd by Das, Ranajit archive.org


What does it mean by 'the residue was taken up in solvent' in synthesis protocol?

You add the DCM to the material left after removing the toluene and then evaporate it off under vacuum. The reason is to ensure that all the trifluoroacetic acid is removed as this would adversely affect the next reaction, and TFA is readily soluble in DCM.


Master equation for catalyst and substrate binding?

The way to approach this is not how you have mentioned in your answer.

The representation of the master equation is correct and the generalized form of the chemical master equation (CME) [1] would be:

$$\frac{\partial P(x,t)}{\partial t}=\displaystyle\sum _{\mu=1}^M a_\mu(x-v_\mu)\ P(x-v_\mu,t) - a_\mu(x)\ P(x,t)$$

Where:

$x$ denotes the state of the chemical system (number of molecules of each species)
$\mu$ denotes a type of reaction
$a_\mu$ denotes the probability of happening of reaction-$\mu$ in a small interval (also called reaction propensity)
$v_\mu$ denotes the change in the state due to the reaction-$\mu$
$M$ denotes the total number of possible reactions

As far as I can understand (also from your question on Biology.SE), you are interested in knowing the distribution of the time interval when no reaction happens. A chemical reaction can be modelled as a Poisson process which is based on 3 main postulates:

$\lim\limits_{h\to0+}\dfrac{P(N_h=1)}h=\lambda$  i.e. the probability of occurrence of one event in a very small interval of time is equal to the macroscopic rate or intensity ($\lambda\,$).
$P(N_h\geqslant2)=o(h)$ i.e. the probability of occurrence of more than one events in an infinitesimal interval is essentially zero.
Consecutive events are independent of each other.

A chemical reaction seems to follow all these postulates. The detailed derivation of the CME can be found in [1]; I am trying to present a short derivation based on Poisson postulates.

  The probability that system is in state $x$ at time $t+\Delta t$ would
  be the probability that system is in state $x-v_\mu$ at time $t$ times
  the probability that reaction $\mu$ happens in the interval $\Delta t$, or the probability that system is in state $x$ at time $t$ times probability that no reaction happens in the interval $\Delta t$.

$$P(x,t+\Delta t)=\sum_\mu P(x-v_\mu,t)\ P(\mu) + P(x,t)\left(1-\sum_\mu P(\mu) \right)$$

when you rearrange this equation, divide by $\Delta t$ and take the limit of $\Delta t \to 0$, you will end up with the CME. Note that, as per the first postulate: $$\lim\limits_{\Delta t\to 0}\frac{P(\mu)}{\Delta t}=a_\mu$$

Now, the probability that $k$ reactions would have happened in a time interval $\tau$ follows the Poisson distribution (see [2] for the derivation).

$$P(N=k)=\frac{(\lambda\tau)^k e^{-\lambda \tau}}{k!}$$

This is a well known property of the Poisson distribution that the time interval between reactions will follow exponential distribution. The probability that no reaction has happened in time interval $\tau$ would be:

$$ P(N=0)=\frac{(\lambda\tau)^0 e^{-\lambda \tau}}{0!}=e^{-\lambda\tau}$$

in this case, $\lambda=\displaystyle\sum _{\mu=1}^M a_\mu$ i.e. probability of any reaction.

Conversely, the probability that some reaction would happen in time $\tau$ would be $P(r\leq \tau) = 1-e^{-\lambda\tau}$. Where $r$ denotes the first reaction time. This is the cumulative distribution function for the exponential distribution.

In other words, the reaction time between intervals would follow an exponential distribution such that the probability that no reaction would happen will exponentially reduce with the size of the interval $\tau$.

In your specific case there are two reactions — association and dissociation, the propensities of whose will be:

$$a_f=\frac{k_f}{V}n_E.n_S\\[1em] a_r=k_r n_{ES}=k_r(n_{E_0}-n_E)$$

Where $k_f$ is the rate constant of association, $k_r$ is the rate constant for dissociation (as per mass action kinetics), $n_i$ denotes the number of molecules of species-$i$ and $V$ is the volume (of the reaction vessel). See [3] for details. Note that the amounts of the species change with time, and so will the reaction propensities. You can calculate the exact reaction time distribution when the system is at the deterministic steady state (using the steady state concentrations).

I think this answers your question on [biology.se] as well.

References:

Gillespie, Daniel T. "A rigorous derivation of the chemical master equation". Physica A: Statistical Mechanics and its Applications, 188 (1992): 404-425.
Hogg, Robert V., and Allen T. Craig. Introduction to mathematical statistics. New York: Macmillan, 1978.
Gillespie, Daniel T. "A general method for numerically simulating the stochastic time evolution of coupled chemical reactions." Journal of computational physics 22.4 (1976): 403-434.


You can displace chlorine from hydrogen chloride.

With sodium chloride, if you did form sodium oxide it would be a strong base.  As such it extracts a $\ce{Cl^+}$ moiety from any potential chlorine molecules.  Sodium chloride is oxidized by oxygen, but the oxidized chlorine ends up as hypochlorite oxyanions such hypochlorite (i.e., oxide coordinated to chlorine).


Calculate the concentration in milimoles per litre (mM) of a solution that is 20% H2SO4 by mass and that has a density of 1.198 g/L

the solution has a density of 1.198 g/L.

This has to be wrong. I assume that the density is 1.198 g/ml.

Thus a liter has a mass of

$$1000\text{ ml/L} \times 1.198\text{ g/ml}= 1198\text{ g/L}$$

20% of the mass is $\ce{H2SO4}$, thus the mass of $\ce{H2SO4}$ in 1.000 L of solutiom is:

$$0.20 \times 1198\text{ g/L} = 239.6 \text{ g/L}$$

$\ce{H2SO4}$ has a molecular mass of $98.079\text{ g/mol}$ so the number of moles is

$$\dfrac{239.6 \text{ g/L}}{98.079\text{ g/mol}} = 2.443\text{ moles/L}$$

Now converting to millimoles

$$2.443\text{ moles/L}\times 1000\text{ mM/mole} = 2443\text{ mM/L} \ce{->[rounding]} 2.4\times 10^3\text{ mM/L} $$


First of all, the point group for $\ce{XeF6}$ is disputed. The latest I saw was $C_\mathrm{3v}$, not $O_\mathrm h$ but the $O_\mathrm h$ is so energetically close to its $C_\mathrm{3v}$ that it is likely in a perpetual state of transition between the two. As for polarity, due to this it is hard to think it would have a permanent dipole. One would assume it would be easy to induce one, though.

$\ce{XeF6}$ is not the typical molecule the different models were made for, so YMMV.


The lone pair of electrons will be in a $\mathrm{sp^2}$ hybridized orbital and not in  a pure p-orbital. Thus, it wil be in the same plane as the molecule. Only the lone pairs which are in an orbital perpendicular to the molecular plane can take part in resonance, so those $\mathrm{sp^2}$ electrons will not be counted in Hückel's rule.


Does grignard reagent react with oxygen?

This paper says it does and notes that while the yield of alcohols from aliphatic Grignards is 60-90%, formation of phenols is much less satisfactory. Other methods exist for the formation of phenols from aryl organometallics with much better yields.


Why doesn't water play a role in reactions in aqueous reactions?

Water is far from a bystander in aqueous solutions. It has a number of features that greatly facilitate chemical reactions, as well as being an important ligand (the aqua ligand) of transition metal complexes.

Here are some (unbalanced) examples:

Water has a rather large dielectric constant and as such is very effective at shielding charges from one another, facilitating the charge separations inherent in the dissolution of ionic compounds. This is why you can easily dissolve sodium chloride in water, but not in hexane. The nature of this shielding is often articulated as several concentric 'solvation shells', where 'shells' of water molecules cluster around an ion with their dipoles approximately aligned towards or away from the ion.

Here's an image from Wikipedia qualitatively depicting the first solvation shell surrounding a sodium ion:

More generally, the large dipole moment of water allows it to partake in hydrogen bonding, where water will weakly associate with other polar molecules or functional groups. This extends to water forming (short lived, highly fluxional) clusters with itself.
Water is also happy to accept (forming hydronium, $\ce{H3O+}$) and donate (forming hydroxide, $\ce{OH-}$) protons in solution. So much so that it will actually do this with itself, in a process called autoionisation. This process makes water molecules significant in a wide range of acid/base and redox equilibria.


Mechanism of Fritsch–Buttenberg–Wiechell rearrangement

As your heading correctly states, this reaction take the path of Fritsch–Buttenberg–Wiechell rearrangement (Wikipedia).   The reaction mechanism of Fritsch–Buttenberg–Wiechell rearrangement (FBW) is described as:

  The strong base deprotonates the vinylic hydrogen, which after alpha-elimination forms a vinyl carbene. A 1,2-aryl migration forms the 1,2-diaryl-alkyne product. The mechanism of the FBW rearrangement was a subject of on-surface studies where the vinyl radical was visualized with sub-atomic resolution (Ref.1).

However, if you carefully read this reference, you'd realize they use special method to generate diradical from gem-dibromo-alkene stepwise. Thus, the proposed mechanism for every FBW-type reactions. Earlier, A. A. Bothner-By (Ref.2) from  Harvard reported that in a diaryl-substituted system of a bromoalkene (e.g., see structures 1 and 4 in the scheme 1), the major migrating group is the aryl moiety located anti to the halogen (the fact, which has been proven by $\ce{^{14}C}$-labeling experiments). Particularly, this migration of an aryl group between multiply-bonded carbons is one of the several unusual features of the rearrangement (Ref.2):

Yet, there are plethora of data to support the 1,2-phenyl group migration through single bonds (e.g., the benzilic acid rearrangement), and at least one example for migration of an aryl group between multiply-bonded carbons (e.g., the Beckmann rearrangement). Thus, based on these facts, I suggests following arrow-pushing mechanism, which would be very reasonable:

References:

N. Pavliček, P. Gawel, D. R. Kohn, Z. Majzik, Y. Xiong, G. Meyer, H. L. Anderson, L. Gross, “Polyyne formation via skeletal rearrangement induced by atomic manipulation,” Nature Chemistry 2018, 10, 853–858 (https://www.nature.com/articles/s41557-018-0067-y).
Aksel A. Bothner-By, “Stereochemistry of the rearrangement involving migration between multiply-bonded carbons,” J. Am. Chem. Soc. 1955, 77(12), 3293–3296 (https://doi.org/10.1021/ja01617a042).


Why does diamond conduct heat better than graphite?

Diamond is one of the best thermal conductors known, in fact diamond is a better thermal conductor than many metals (thermal conductivity (W/m-K): aluminum=237, copper=401, diamond=895).  The carbon atoms in diamond are $\ce{sp^3}$ hybridized and every carbon is bonded to 4 other carbon atoms located at the vertices of a tetrahedron.  Hence the bonding in diamond is a uniform, continuous 3-dimensional network of $\ce{C-C}$ single (sigma) bonds.  Graphite on the other hand is formed from $\ce{sp^2}$ hybridized carbon atoms that form a continuous 2-dimensional sigma and pi bonding network.  This 2-dimensional network forms sheets of graphite, but there is little connection between the sheets, in fact, the sheet-sheet separation is a whopping ~3.4 angstroms.  This might lead us to suspect that heat conduction in the 2-dimensional sheet of graphite would be superior to diamond, but that heat conduction between graphite sheets would be very low.  This is, in fact, an accurate description of thermal conduction in graphite.  Thermal conductivity parallel to the graphite sheets=1950, but thermal conduction perpendicular to the sheet=5.7.  Therefor, when we consider thermal conduction over all possible directions (anisotropic) diamond would be superior to graphite.


How does one calculate the pH of an aqueous solution of a weak acid and a weak base?

Equations and variables

To calculate the proton concentration in a solution as given in the question the following five equations are used for the following five unknown variables: $\ce{[H+ ]}$, $\ce{[HA]}$, $\ce{[A- ]}$, $\ce{[HB]}$ and $\ce{[B- ]}$.

Ion balance:
$$ \ce{[H+ ]} = \ce{[A- ] + [B- ]} + \frac{K_\text{w}}{\ce{[H+ ]}} $$

Mass balances:
$$\begin{align}
\ce{[A]_{tot}} &= \ce{[HA] + [A- ]} \\
\ce{[B]_{tot}} &= \ce{[HB] + [B- ]}
\end{align}$$

Law of mass action:
$$\begin{align}
K_\text{a, 1} &= \frac{\ce{[H+ ][A- ]}}{\ce{[HA]}} \\
K_\text{a, 2} &= \frac{\ce{[H+ ][B- ]}}{\ce{[HB]}}
\end{align}$$

Solution

Combining the above equations we get:
$$
\ce{[H+ ]} = \frac{K_\text{a, 1} \ce{[A]_{tot}} }{\ce{[H+ ]}+K_\text{a, 1}} + \frac{K_\text{a, 2} \ce{[B]_{tot}} }{\ce{[H+ ]}+K_\text{a, 2}} + \frac{K_\text{w}}{\ce{[H+ ]} }
$$

After some reforming we get a fourth-order polynomial for the proton concentration:
$$\begin{align}
0 = & \ce{[H+ ]^4} + \ce{[H+ ]^3}(K_\text{a, 1} + K_\text{a, 2}) \\
  &- \ce{[H+ ]^2}(\ce{[A]_{tot}} K_\text{a, 1} + \ce{[B]_{tot}} K_\text{a, 2} - K_\text{a, 1} K_\text{a, 2} + K_\text{w}) \\
 & -\ce{[H+ ]}\left[ (\ce{[A]_{tot}} + \ce{[B]_{tot}}) K_\text{a, 1} K_\text{a, 2} + K_\text{w}(K_\text{a, 1} K_\text{a, 2}) \right] \\
 & - K_\text{w} K_\text{a, 1} K_\text{a, 2}
\end{align}$$


Why is ethanol written as C2H5OH not C2H6O?

It is better to write $\ce{C_2H_6O}$ as $\ce{C_2H_5OH}$, because there is more than one compound with the same overall formula, and emphasising the hydroxyl group $\ce{-OH}$, which is what characterizes an alcohol makes it ethanol and nothing else.  So when you ask, "What is OH functional group doing here", it is exactly that the product is an alcohol, characterized by a hydroxyl group. And yes, alcohols are acidic in the right situations, given that their hydroxyl groups do produce acidic hydrogen ions, and yes, that is why it's so soluble in water.


How many stereoisomers can exist for 4,5-bis(3-hydroxyoct-1-en-1-yl)cyclopentane-1,3-diol?

The question is tricky, because

the two ends of the wavy bond are both stereogenic units (chiral cyclopentane ring carbon atom, and the double bond), so there is some additional unintentional ambuguity, but it's not a big deal, because the legend at right explains the situation somewhat;
it in fact asks “How many … stereoisomers does this structure represent?” (because of some stereogenic units with unspecified configuration) instead of “how many … stereoisomers does it have?”.

The unspecified-configuration stereogenic units are two chiral carbons and one double bond, which give total 8 stereoisomers.

The first one (top left) has a plane of symmetry, the other 7 stereoisomers are optically active.


From what I've been able to find, it is similar to how graphite acts as a lubricant. Boric acid is flat and it hydrogen bonds to form layers of flat networks. These slide on each other like layers of graphite do, reducing friction.


What reaction could rubidium undergo in a vacuumed chamber with Macor ceramic and glass?

One candidate for reaction with the rubidium is silica, $\ce{SiO2}$.  Reference [1] reports that black powders are formed when alkali metals are contacted with silica.  Such powders are more stable in dry air than ordinary alkali metals.  The reference suggests that electrons are imparted from the alkali metals into the silica forming an electride complex.

Reference

Shatnawi, M.G., Paglia, Gianluca, Dye, James, Cram, Kevin, Lefenfeld, Michael, Billinge, Simon, "Structures of Alkali Metals in Silica Gel Nanopores: New Materials for Chemical Reductions and Hydrogen Production", Journal of the American Chemical Society 129, 1386-92 (2007).  DOI 10.1021/ja067140e


(1) Meaning of the textbook solution

$$\ce{2HCl + Ca(OH)2 = CaCl2 + 2H2O}$$

This reaction clearly tells us that

$$\frac{moles_{acid}}{moles_{base}}=2$$

moles=molarity* volume(in litres)

Therefore, $$\frac{moles_{acid}}{moles_{base}}=\frac{M_{a}V_{a}}{M_{b}V_{b}}$$

(2) What is wrong with your interpretation?

You wrote $2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$.

$2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$ actually implies that more number of moles of base need to be taken. For neutralisation, for every one mole of base, we require two moles of acid since each calcium hydroxide will release 2 hydroxide ions (for which two H+ are required).


Hybridisation of hexaamminecobalt(III) complex

It has d2sp3 configuration since NH3 is a strong field ligand which causes the electron to pair up and use the inner d orbitals(3d) .


Is varnish with a flash point above 40 celsius safe in a microwave oven?

The flash point is for the solvent, i.e. for undried paint. Dried paint will have a considerably higher flash point, well over 100 degrees Celsius.

You should also consider toxicity. I don't think alkyd based varnish is approved for use in items you put in you mouth (like mugs).


Is the 'n' in Delta G = -nFE, ever negative?

No. This result occurs from the Nernst equation.

The "n" is the number of electrons transferred. If no electrochemical reaction occurred, then n = 0.

Otherwise "n" is positive.

Having a negative number of electrons transferred would be impossible.


Why not to consider hydrogen and oxygen moles to determine an empirical formula

The answer is $\ce{C3H4O4}$. It was found by using the mass (percentages) divided by the molar mass of each element.

Exactly! That's the way to do it.

  But they didn't consider the (2moles) of hydrogen and oxygen in calculations. Why? In this case it will be $\ce{CH3O3}$

No, they did not! As you have pointed out yourself, it is about the empirical formula of a compound in question, such as one of the following:

It is completely irrelevant whether an element usually exists as a diatomic molecule in nature. If that would matter, how would you then treat sulfur-containing compounds? Note that natural sulfur typically consists of $\ce{S8}$ molecules! How would you treat elements that only appear in the form of compounds in nature but never as pure elements.

In summary, the "natural appearance" of elements in nature does not matter here, it is only about the relative atomic abundance of an element in the compound in question.


How can we replace volume in ideal gas equation with V - nb?

The actual volume occupied by the gas is less than the measured volume(volume of the container).
In reality gaseous particles are not point objects and don't occupy zero space, which is against one of the postulates of the Kinetic Theory of Gases.
Thus while dealing with real gases we have to account for the volume which has been occupied by the gaseous particles. We do so by assuming the volume of gaseous particles to be $nb$.
Now, the ideal volume(actual volume) occupied by the gas is the measured volume(volume of container) minus the volume occupied by the gaseous particles.
Thus, $V_{ideal} = V_{real} - nb$ and not the other way around.
Note, $V_{ideal} =V_{actual}$ and $V_{real} =V_{measured}$.
The mistake you are making is formula you are using and thus
$P_{ideal}(V_{ideal}-nb) ≠ nRT$, but
$P_{ideal}(V_{real}-nb) = nRT$


How should you name the solid-solution compound ScY(SeO4)3?

This is similar to a double salt, but because of the chemical similarities between scandium and yttrium, $\ce{ScY(SeO4)3}$ is a solid solution in this case. And the answers is you just name the cations in alphabetical order, then the anions like any normal salt. Thus the name is scandium yttrium selenate.

If the ratio between cations had a large difference such as $\ce{Sc_{1.9}Y_{0.1}(SeO4)3}$ It might be called yttrium-doped scandium selenate, but could still correctly be called just scandium yttrium selenate too.


What is non-hückel double bond of Aromaticity?

In the given molecule Pyrene , the π  bond shown in green is in cross conjugation (shown in green in figure 1). This bond( figure 2) is excluded from interaction with rest of conjugated system. Hence , the central $\ce{C=C}$ isn't hosted by the largest p orbital loop, so it's not counted towards Hückel's rule for aromaticity.

The outer periphery (shown in blue in figure 3), is a loop of $\ce{14 π}$ electrons , therefore aromatic. This may be possible hückel double bond you are referring to.

The π bond in cross-conjugation (shown in green) maybe non-hückel double bond.


Stability of furanose vs. pyranose form of glucose?

If you look up the statistical distribution of glucose isomers in aquaeous solution, you will find $36\,\%$ $\unicode[Times]{x3B1}$-pyranose and $64\,\%$ $\unicode[Times]{x3B2}$-pyranose. All the remaining forms lumped together only account for less than $1\,\%$ of glucose in solution; that includes both furanoses, the open-chain form and the (possible, but unlikely) seven- and four-membered rings.

Neither five-membered rings nor six-membered rings contain any relevant amount of angle strain. You can easily verify this buy getting yourself a molecular model kit and attempting to bild these rings from four-bond sp³ carbons: both can be built smoothly, like a charm. So angular strain is not a reason why the pyranose forms are preferred.

Considering reaction mechanisms, the formation of five-membered rings is usually kinetically preferred due to easier reachable transition states. So this also cannot be the reason for preferring six-membered rings.

The actual reason is not so much angular strain, but rather dihedral angle strain: the same strain that prevents ecliptic conformations in alkanes. The pyranose form has the perfect $60^\circ$ dihedral angle between two non-ring atoms which minimises this type of strain. Furanoses must choose between envelope or twist conformations to at least relieve some of this strain.

All this is, however, happening at an equilibrium, which is semi-rapid in adjusting itself. So there will always be a certain percentage of furanose form in solution being formed and breaking apart. If any kind of reaction selectively only happens with furanose anomers, then this will cause the formation of new furanose which again will be captured away. As I wrote before, the attack which creates a furanose is preferred over that which creates a pyranose. (But the breaking apart of pyranose is slower than that of furanose.)


Can hydrogen on platinum reduce carboxylic acids and esters?

This is based off my memory, as far as I can remember what my teacher taught me.

Note, instead of $\ce{Pt}$, you can go for $\ce{Ni}$ (or Raney Nickel) or even $\ce{Pd}$ as a catalyst.
Functional groups such as acid halide, cyanide, nitro, aldehyde, ketone, alkene, and alkyne are easily reduced by $\ce{H2/Ni}$.  Some other groups can be reduced with $\ce{H2/Ni}$ as well, but if someone's asking you a question in an exam, they want you to refer to the list of groups which are easily reduced (more in the reactivity order below).  So the answer in your case would be a no.
Next, here's a reactivity order:
First, the groups which are easily reduced:
$$\begin{align} \text{acid halide} > \ce{R-C#C-R} > \ce{R-C=C-R} > \ce{R-CHO} > \\ &\text{ketone}>  \ce{R-CN} > \ce{R-NO2} > \ldots \end{align}$$
Then the groups that are reduced only with difficulty:
$$ \ldots > \ce{R-X} > \text{benzene} > \ce{R-COOH} > \text{amide} > \text{ester} > \text{acid anhydride}$$
$\ce{R-X}$ would require temperatures around $\pu{50^\circ C}$, benzene would require a temperature of about $\pu{100^\circ C}$ and groups from carboxylic acid to acid anhydride would require temperatures around $200-\pu{250^\circ C}$.


Why is some ionic substance insoluble, such as BaSO4?

The electronegativity is not the main factor. The entropy factor is much more important. When a salt like $\ce{BaSO_4}$  tries to get dissolved, its ions are attracting water molecules that get fixed or attached around each ion. This phenomena creates order, and decreases the entropy of the solution. The same thing happens around sulfate ions $\ce{SO_4^{2-}}$. If there is too much water attached around each ion, the substance does not dissolve. This is the case for $\ce{BaSO_4}$, because the ion $\ce{Ba^{2+}}$ is bigger than $\ce{Mg^{2+}}$, and so creates more order in attracting more water molecules around it. The entropy decreases too much if $\ce{BaSO_4}$ gets dissolved. This salt is practically not soluble. This is not the case for $\ce{MgSO_4}$.


Well, this gets into the concept of basis sets.

LCAO implies that the number of molecular orbitals will be the same as the number of atomic orbitals involved.

In fact, we usually use more basis functions than just the simple number of atomic orbitals. We might add diffuse or polarization functions to better describe electron density away from the nucleus.

Moreover, rather than the physically correct Slater-type orbitals:

$$R_l(r) = A(l,\alpha) r^l e^{-\alpha r}$$

We usually use Gaussian-type orbitals to express the radial component.

$$R_l(r) = B(l,\alpha) r^l e^{-\alpha r^2}$$

Now GTOs don't express the radial part of the wave function well -- they decay too quickly as you get far from the nucleus. So we use a linear combination of multiple Gaussian orbitals to approximate the physically-correct Slater terms. Still, it's much faster to use a set of multiple GTOs.

Beyond STOs and GTOs, people do use other types of orbitals. For example:

Wavelets
Plane waves
Muffin-tin


Is there a word for a compound that has both ionic and covalent bonds?

There may be a strong case to call this a "salt", like other ionic compounds.  Very often, the anion (negative ion) of a salt is an ion comprised of different atoms with covalent bonds.  For example, $\ce{ SO_4^{2-} }$ is the sulfate anion, and the bonds between $\ce O$ and $\ce S$ are covalent to some degree.  So $\ce{ Na_2SO_4 }$ is an ionic compound under this definition.  Thus, any compound with some component that is strongly ionic can be called an ionic compound; i.e. a salt, regardless of the bonding structure of the components.


Why does formic acid give positive Fehling's test?

Here is a short simple available answer provided online by Prachi Sawant in 2016 to quote:

  Formic acid has both aldehydic (-CHO)  and carboxylic (-COOH) functional groups.
  Hence, it gives both Tollen's and Fehling's test positive.

  Reaction of formic acid with Tollen's reagent:

  $\ce{HCOOH + 2[Ag(NH3)2]+ +2OH- → 2Ag + CO2  + 2H2O  + 4NH3 }$

  Reaction of formic acid with Fehling's reagent:

  $\ce{HCOOH + 4OH- + 2Cu^{2+}(in complex)  → CO2 + 3H2O + Cu2O }$

Now, here is my attempt on the possible underlying mechanic, which is likely complex. I start with the above net reaction, which in reality, is probably not completely accurate, but nevertheless, may provide some insight into actual half-cell reactions, beginning with the reduction of the organic cupric complex:

$\ce{Cu^{2+} +  e^-  ->  Cu^+ }$

where the insoluble Cu2O (the test indicator) is formed. The corresponding other half-cell is less clear accept for the liberation of an electron:

$\ce{COOH^- -> .HOCO + e^- }$

which may speculatively may be followed by:

$\ce{.HOCO + .HOCO -> H2O + CO + CO2 }$ (speculation)

And/or, this paper favoring non-conventional ionized species in the context of formic acid as possessing more stability as demonstrated, for example, by:

$\ce{HCOOH -> [.HCOOH]^+ + e^- }$

With its source cited below, followed by:

$\ce{[.HCOOH]^+ -> H2O .CO^+ }$  (a distonic ion, per 2nd cited source)

$\ce{OH^- + H2O .CO^+ -> H2O + COOH^- }$ (speculation)

To quote a cited source above :

  Rate constants for the radical-induced hydrogen abstraction from formic acid, HCOOH, are presented here. Only those reactions leading to the formation of $\ce{.HOCO}$ were investigated.

Per a source: 'The gas phase chemistry of the formic acid radical cation $\ce{.[HCOOH]^+}$ Mechanism for exchange of the hydrogen atoms: a quantum chemical investigation', to quote:

  Earlier labelling experiments had indicated that prior to the loss of H the hydrogen atoms in ionized formic acid, $\ce{.HCOOH^+}$, can become positionally equivalent via an unknown pathway. Using ab initio molecular orbital calculations we have located a minimum energy reaction pathway for hydrogen exchange reactions which takes place close to the dissociation level and proceeds via flexible ion/dipole complexes. Along this path three equilibrium structures of different atom connectivity are found: $\ce{.HCOOH^+}$, 1, ionized formic acid; two forms of the surprisingly stable ion/dipole complex $\ce{.HOHCO^+}$, 2, 3; and the distonic ion $\ce{H2O CO.^+}$, 4. Ion 1 is separated from ion 2 by a barrier of 142 kJ mol−1 and this barrier corresponds to a 2A′ → 2A″ surface crossing. The barrier for the proton transfer 3 → 4 is only 25 kJ mol−1 relative to 3; the transformation $\ce{.HCOOH^+ -> H2O .CO^+}$, formally a 1,2-hydrogen shift, takes place via the ion/dipole complex $\ce{HO· HCO^+}$, $\ce{H2O .C^+}$ is the reacting configuration for decarbonylation and this reaction is calculated to take place at the dissociation limit in full agreement with experiment.

Where ‘decarbonylation’ can also be effected by several transition metal complexes, (source), as in the current context with an organic copper complex.


First, let's assume these reaction happen in aqueous solution and everything is soluble. The two reaction you are interested in are:

$$\ce{H2SO4(aq) + NaOH(aq) -> HSO4-(aq) + H2O(l) + Na+(aq)}\label{rxn:1}\tag{1}$$

$$\ce{H2SO4(aq) + 2NaOH(aq) -> SO4^2-(aq) + 2H2O(l) + 2Na+(aq)}\label{rxn:2}\tag{2}$$

Your saying that reaction \eqref{rxn:2} might happen even when $\ce{NaOH}$ is limiting, and you would end up with some sulfuric acid and some sulfate. In a thought experiment, let's say that is what happened. Now, we can discuss whether sulfuric acid and sulfate, which are present in equimolar amounts, will react with each other according to:

$$\ce{H2SO4(aq) + SO4^2-(aq) <=> 2HSO4-(aq)}\label{rxn:3}\tag{3}$$

Sulfuric acid is a strong acid, and sulfate is a weak (very weak) base. This is like mixing hydronium ($\ce{H3O+}$) and hydroxide ($\ce{OH-}$) ions, which (mostly) form water. So just like water, which occurs mostly undissociated, the equilibrium of reaction \eqref{rxn:3} lies on the side of the products when the reactants are present in equimolar amounts.

Of course, if you add more $\ce{NaOH}$, you would start making a lot of sulfate.

If you wanted to know all of the products and their concentrations, you would have to consider water auto-dissociation as well. The argument I made is not meant to be quantitative, but to show why we don't expect sulfuric acid to be left over. In aqueous solution, there is never much sulfuric acid (even in the absence of $\ce{NaOH}$) because, as a strong acid, it dissociates.


Why are bonded ions more stable than separate ions?

What's meant by stability is whether the compound "wants" to be in current state. If it wants to change the state (separated ions want to bond) then the system is less stable.
In your example you can simply imagine 2 spherical magnets placed close to each other - they would rush towards one another. They had potential energy that turned into kinetic energy. And once they met - they don't have potential energy anymore, thus they are at lower energy state (more stable) and don't want to separate. Then you'd need to apply force to separate them.


Does sodium form complexes like transition metal ions?

Although less common than transition metal complexes, sodium does form complexes with some ligands, particularly oxygen based ligands.

Aqua complexes are formed in aqueous solution, the most common being $\ce{[Na(H2O)6]+}$.

Sodium forms many complexes with crown ethers, cryptands and other related ligands. For example, 15-crown-5:


What is an "attack" in Organic Reactions?

You might have heard the term "attack" in statements like "the nucleophile attacks the electrophile", and terms like frontside and backside attack, and what this basically means is that the nucleophile approaches the electrophile, and begins to share its electrons with the electrophile to form a bond between them (and possibly break a different bond in the molecule too). Frontside and backside attack mean the nucleophile approaches the front or the back of the molecule and forms a bond.

If you've learnt about arrow pushing, you'll know that an arrow has its tail starting at the nucleophile (the electron rich species) and it's head points to the electrophile (the electron poor species) - you can also think of "attack" as the nucleophile "attacking" the electrophile with electrons to react/form a bond, in the direction the arrow is pointing. I suppose we call it attack because it makes it sound more action-packed :-)


There are multiple types of collagen variants, so the answer here is somewhat non-specific.
Pepsin is an endo type of serine protease and it has low specificity. Proteolysis depends on residues upstream (P1’) and downstream (P1) of the recognition site. Hydrophobic residues at P1’ and P1 increase the likelihood of successful proteolysis. An aromatic residue such as Phe, Trp and Tyr at P1’ and a hydrophobic residue Phe, Leu and Met at P1 are associated with highest likelihood of successful proteolysis. Positively charged residues such as His, Lys and Arg at position P1 are detrimental to proteolysis.
A collagen fiber contains multiple fibrils that are made up of tropocollagen molecules, strengthened by cross-links. Tropocollagen is in fact comprised of three left-handed procollagen molecules, forming a right-handed triple helical structure.
Treatment of acid-soluble tropocollagen with pepsin releases multiple peptides whose composition mainly includes Glycine, Tyrosine, Leucine and Phenylalanine amino acids among others.


How can I relate the reactivity series to electronegativity and ionization energy?

With less electronegativity, there is also less ionization energy. This means for metals, it is easier to lose the electrons. When it is easier to lose the electrons, the metal becomes more reactive because it uses less energy to fulfill its goal; to gain a full outer shell.

For nonmetals, the main goal is to gain electrons and not lose them. Thus, an increase in electronegativity and ionization energy means an increase in reactivity because the nonmetals want to hold on to their electrons and gain more vigorously to achieve the full outer shell.


How do point defects in a crystal relate to Entropy?

I would strongly recommend this resource:
http://www.tf.uni-kiel.de/matwis/amat/def_en/
It is a good complete on line introduction to defects theory in crystals.
Some key points listed below.

The concentration of point defects $c_{PD}$ at equilibrium can be expressed by an Arrhenius formula:
$$c_{PD}=A \cdot e^{-\frac{G_F}{kT}}=A \cdot e^{-\frac{S_F}{kT}}\cdot  e^{-\frac{H_F}{kT}}$$
where $G_F$ is the free energy of formation, $H_F$ the enthalpy of formation and $S_F$ the entropy of formation. In solids  $H_F$ can be confused with the energy of formation, $S_F$ is associated the variation of the vibration entropy of the crystal due to the introduction of a point defect.
Formation enthalpies, and thus formation energies can be obtained by:
$$ H_F(D) = E_{tot}(D) - E_{tot}(X)+\sum_i n_i \mu_i +q (E_F -E_V) $$

$E_{tot}(D)$  and $E_{tot}(X)$: total energies of the system with and
without the defect D.
$n_i$: number of atom removed from the system,
a negative value denote the addition of one atom.
$\mu_i$: chemical potential of the atom $i$. This should consider the chemical environment at the equilibrium condition (which reservoir is to take into account).
$E_F -E_V$: Fermi energy relative to the valence band maximum.
$q (E_F -E_V)$: energy change due to the exchange of electrons and holes with the carrier reservoirs.

For the equilibrium concentration in the multiple defects case an additional entropy term comes from the standard configuration entropy which can be calculated from the Boltzmann formula
$$S=k_B \cdot \ln P$$
where P is the number of configurations for the same macrostate.


While doing calculations for Huckel's rule, in polycyclic compound only peripheral pi electrons are considered for aromaticity.
So, 7 double bonds in the periphery i.e. 14 pi electrons.
14=4(3)+2
Huckel's rule satisfied. Aromaticity achieved.
Why are only peripheral pi electrons considered?
The reason is that only these elections are participating in delocalization for conjugation. If multiple bonds occur in the centre like in pyrene these central bonds will not be localized. Thus, only peripheral pi electrons are considered in conjugation.

Therefore, first 2 compounds in this image are aromatic and last one is weakly Anti aromatic or Non aromatic.
We can't commit compound to be aromatic on ONLY Huckels rule with this compound must be

Cyclic
Planar
Complete cyclic conjuction
Huckel's rule

Here in 3rd compound,
Due to crowding of Interior Hydrogen compound loses its planer nature. Thus, not called to be Aromatic.
(I have attached 3d structure of cyclotetradeca-1,3,5,7,9,11,13-heptaene (3rd compound)
Please refer it.)


Is this a violation of Equipartition theorem?

For a non-linear polyatomic molecule like methane, there are a total of $3N$ degrees of freedom, where $N = 4$ is the number of atoms in the molecule. This is broken down as follows:

$$\begin{array}{c|c} \hline
\text{Component} & \text{# Degrees of freedom} \\
\hline
\text{Translational} & 3 \\
\text{Rotational} & 3 \\
\text{Vibrational} & 3N - 6 \\ \hline
\end{array}$$

(Note that for a linear molecule, or for an atom, this breakdown is different.)

Under "normal" temperatures (close to room temperature) the vibrational contribution to the internal energy can be neglected. (The vibrational gap $\Delta E = \hbar\omega \gg k_\mathrm{B}T$ for "normal" $T$, so higher vibrational states are not populated and the equipartition theorem does not hold.) This is probably the reason why your book states that there are six total degrees of freedom.

The key is to realise that the kinetic energy of a molecule arises only from the translational degrees of freedom. From my limited knowledge of physics, it seems to be a common thing to separate translational degrees from the "internal" degrees of freedom, i.e. rotational, vibrational, and electronic. This allows the translational motion of a body to be modelled as the movement of a point mass, located at the centre of mass of the body.

In this paradigm, rotations and vibrations preserve the centre of mass of a molecule, and therefore don't contribute to the kinetic energy of the molecule. Therefore, the correct answer is $(1.5~\mathrm{mol})\cdot RT$.

Another way of looking at it is to consider the definition of the kinetic energy

$$E_k = \frac{1}{2}m\langle v^2\rangle = \frac{1}{2}m(\langle v_x^2\rangle + \langle v_y^2\rangle + \langle v_z^2\rangle)$$

where $v$ is the velocity of the centre of mass, $v_\mathrm{cm}$. For each quadratic term the contribution to the energy is $kT/2$, for a total of $3kT/2$ per molecule. These three terms correspond to precisely the three translational degrees of freedom.


If you look at Eli Lilly patents and other literature, there doesn't seem to be a simple synthesis... I've put together an example of how it could be done. Although there are many ways you could go depending on starting materials, this way is based primarily on literature reactions with the exception of the last 3 steps which are Eli Lilly patents. I stopped after about 16 steps as I think that's more than enough to demonstrate how complex these designer molecule syntheses can be. I also left out reagents/reaction conditions just to show the general pathway, and to save on eye strain.

So really the simplicity (or complexity) will depend on where you are able to start on the synthetic pathway (e.g. what starting materials are available to you).


Note that binding energy of nucleons does not mean contained energy, but the energy needed to break it.

Therefore kernels with lower binding energy per nucleon contain more nuclear energy per nucleon and have higher mass per nucleon.

It means forming kernels with higher binding energy per nucleon releases energy and vice versa.

It is the same as for chemical bonds, just some 6-7 orders higher.

E.g. nitroglycerine has lower binding energy per atom than reaction products formed by its explosion.


The radiation of hydrogen in plasma with low density comes from discrete energy levels, resulting in spectral lines. Those spectral lines can be observed in sunlight, too.

But the sun consists of a plasma with a much higher density. Most photons are the result of a long series of energy exchange inside the sun following the emission of high energy photons and kinetic energy of the nuclear processes. They are part of a continous spectrum as there is not a discrete number of energy levels at the interactions in this high density plasma.

In the photosphere, some of them are absorbed again and result in negative spectral lines (Fraunhofer lines), but the main portion of light remains a continuous spectrum.


Life on Earth happens in pretty tight chemical limits, except for some exotic organisms we rarely encounter.

So, on the human side, the stuff we want to eat is not really very reactive. Carbohydrates, fats and proteins will give you no great fireworks no matter how you mix the three of them. The trace elements we need would be capable of some spectacular reactions, but we don't eat them in their pure form.

Which brings me to the "what is available" side. for almost all of human history, humans ate what they found in nature. And the stuff that reacts violently when mixed with organic matter is not really lying around to be used as an ingredient. Sure, some natural events are capable of producing pure oxygen, fluorine, or whatever, but they don't lie around for a long time waiting to be harvested, they react away. And plants and animals (=our food) tend to evolve in a way such that their bodies use the nitrogen and sulfur to build amino acids, not TNT and sulfuric acid.

There are a few substances in the kitchen which can give you a bit of a bang when mixed. If you want to impress a second grader, such classics as baking soda + vinegar, or mentos + cola will do nicely. But stuff which tends to pack more energy, and likes to free it in a flash, is not really food. You will notice that you need advanced processing methods to get all these ingredients - even though some of them (vinegar fermentation) were known in antique times, they are not something that grows on trees.

You can certainly find some edge cases - I keep pure NaOH on hand for pretzels, it is also used for some styles of noodle - but they are rare and concern ingredients which are not willy-nilly mixed into random dishes.


The designation PBEPBE is an actual terrible artefact from researchers using Gaussian. The actual publication[1, 2] only refers to it as PBE, and most other program packages (I know) implement it as such. Often they make the correlation and exchange parts separately available via PBEC and PBEX, or similar. Unfortunately this designation also made it into the literature.
The confusion further continues since Gaussian uses the PBE1PBE designation for the (standalone) hybrid functional, which is in most other packages known as PBE0.

But it doesn't stop there, for TPSS you may find similar naming/ reference schemes. The BP86 functional keyword requests the VWN(III) version, while many other programs use VWN(V) for this. The latter is available via BVP86, but misses any implementation of dispersion. We could go on with this...

In the end what is important is the correct referencing of the methodology so that the results are reproducible, i.e. look at the literature cited. Gaussian actually does a good job referencing those sources. And when you publish, you'd do us all a favour and not use the keyword designations of Gaussian.

[1] John P. Perdew, Kieron Burke, and Matthias Ernzerhof, Phys. Rev. Lett. 1996, 77, 3865. DOI: 10.1103/PhysRevLett.77.3865. Mirrored at Burke's page: dft.uci.edu/publications.php

[2] (Erratum) John P. Perdew, Kieron Burke, Matthias Ernzerhof, Phys. Rev. Lett. 1997, 78, 1396. DOI: 10.1103/PhysRevLett.78.1396


Doesn't the fact that the lone pair on the nitrogen in amides occupy a p orbital contradict Hund's rules?

two electrons fill the p orbitals while all three $\ce{sp^2}$ orbitals are
  singly occupied

Not really, hybridization occurs in response to bonding, so we can only talk about an atom's hybridization when it is in a molecule.  In amides the nitrogen $\ce{sp^2}$ orbitals form molecular orbitals (MO's) with whatever else they are bound to.  In the case of a simple amide with the nitrogen bound to hydrogen, the MO would involve a nitrogen $\ce{sp^2}$ orbital and a hydrogen 1s orbital and 2 electrons would occupy this orbital (one from the nitrogen $\ce{sp^2}$ orbital and one from the hydrogen 1s orbital)  - so what was the $\ce{sp^2}$ orbital will be filled with 2 shared electrons when a bond is formed.  Therefore the 3 $\ce{sp^2}$ orbitals, now MO's, and the p orbital on nitrogen will each be filled with 2 electrons in the molecule.

  Carbene (CH2) it's $\ce{sp^2}$ hybridized with a lone pair in one of these
  orbitals

Carbenes can exist in either a singlet or triplet state, as shown below.  There are 2 possible triplets depending upon how the carbene is hybridized, a bent $\ce{sp^2}$ hybridized carbene, or a linear $\ce{sp}$ hybridized carbene. In the case of methylene ($\ce{:CH2}$), and actually for most carbenes, the triplet states are lower in energy due to Hund's rule (don't pair electrons unless you must); and the triplet has one electron in the p orbital and one in the $\ce{sp^2}$ orbital.


Determining the formula of zirconium oxide

First, let's solve the problem.
Both relative atomic weight $A_\mathrm{r}$ and relative molecular weight $M_\mathrm{r}$ are a historical dimensionless terms denoting relative atomic mass and molecular mass, respectively:

$$A_\mathrm{r} = \frac{m_\mathrm{a}}{m_\mathrm{u}}$$

$$M_\mathrm{r} = \frac{M_\mathrm{B}}{M_\mathrm{u}}$$

where $m$ corresponds to atomic mass, $M$ – to molecular mass, and index $\mathrm{u}$ refers to unified atomic mass unit ($= m_\mathrm{a}(\ce{^{12}C})/12$).
Since $\ce{Cl}$ in chlorides is monovalent, then formula of unknown zirconium chloride is $\ce{ZrCl_x}$, where $x$ coincides with the valency of $\ce{Zr}$.
From this the relative weight of $\ce{ZrCl_x}$ is

$$M_\mathrm{r}(\ce{\ce{ZrCl_x}}) = A_\mathrm{r}(\ce{Zr}) + xA_\mathrm{r}(\ce{Cl})$$

relative atomic masses of both elements were known ($A_\mathrm{r}(\ce{Zr})\approx 91$, $A_\mathrm{r}(\ce{Cl})\approx 36$), so that the valency of zirconium can be found via $x$:

$$91 + 36x = 233 \implies x \approx 4$$

Since oxygen in oxides has valency of $2$, then the formula of unknown zirconium oxide is $\ce{ZrO2}$.

  By 'volatile chloride', are they referring to zirconium oxide or something else?

No, they refer to zirconium chloride $\ce{ZrCl4}$.
They mention volatility probably to underline how zirconium and hafnium are separated — by differences in sublimation of their volatile compounds.

  Also, by the statement 'assuming that zirconium in the oxide has the same valency' are they saying that zirconium has the same valency as chlorine (i.e. -1)?

No, this means that zirconium in both oxide $\ce{ZrO2}$ and chloride $\ce{ZrCl4}$ has the same valency.


Galvanic corrosion between nickel brass nut and stainless bolt

While I cannot comment in a technical level, I can speak from a background in aerospace experience (in my family) and lifetime of building things that live outdoors.

In time, dissimilar metal corrosion always sets in. Most stainless steels are inhibiting, not a 'way out' of the problem.

On top of that, I think the ambient salts will kills you on this in the marine environment.  If you have a regular maintenance plan, then perhaps you can re-service the bolts yearly, as needed.


How would one express pemoline in terms of phenethylamine?

I would like to know what its chemical name would be in terms of its phenethylamine backbone.

Artificial numbering and naming schemes that only exist in ones own cloud cuckoo land are pointless.

IUPAC nomenclature exists for a reason.

Besides the IUPAC nomenclature, other naming schemes - based on established (trivial) names exists and are widely used - for good reason and with acceptance by the scientific community. Typical examples are found in the field of natural products chemistry (steroids, alkaloids, etc.)

It is however crucial to understand that in these cases, prefixes and/or suffixes denote a variation of the underlying core structure, such as

additional substituents
lack of methyl groups
an elongated side chain, etc.

In the current case, the central structure is a completely different heterocyclus, so markedly different from the original phenethylamine that it would not make any sense to apply a phenethylamine-based name here.

Neither would one try to apply such a naming scheme to the morphine skeleton - despite the fact that one might spot a six-membered ring with $\ce{CH2-CH2-NCH3}$ side chain here.

This does not mean that a member of the phenethylamine family is not among the ancestors of morphine in the biosynthesis! In fact, reticuline, a morphine precursor, most likely is a grandchild of dopamine.

But again, at a certain point structures have become so markedly different that a valid nomenclature (= valid for a particular compound class) no longer is applicable.

Trying to describe similarities between structures is not wrong, but trying to apply an inappropriate naming scheme is not helpful here.

Similarities between molecular entities should be expressed otherwise.
For sequences (DNA, RNA, etc.), Hamming and Levenshtein distances can be used. For the title molecule and the whole dopamine/amphetamine class, I'd consider the comparison of their fingerprints (Tanimoto coefficients).


Evolution of gas during chemical reaction

Likely, any difference you've observed is due to the high solubility of $\ce{CO2}$ in water... it allows accumulation of enough dissolved gas to nucleate and form large bubbles, as in seltzer. BTW, $\ce{Na}$ or $\ce{K}$ etc. do not make tiny bubbles when put in water.


Determine the structure for C6H13Cl using the following data.

Here's a useful hint:  when you have a triplet and a quartet with the area ratio given as above, it indicates the presence of ethyl groups.  Since the total area under those peaks is 10 H, there are two of these  groups equivalent to each other.  Only one isomer with the given formula does that.


Is there a way to turn cellulose into sugar at home?

A method used to break down cellulose into sugars is hydrolysis :

You could use enzymes to hydrolyze the cellulose to produce glucose, but enzymes cost a lot and as you are looking for a cheap way, using enzymes is not an option. A second method is to use an acid (e.g. sulfuric acid) to produce the glucose (just google "cellulose acid hydrolysis"  and you'll find some useful links)


Can I use the hypotenuse of the pythagorean theorem to compute the work?

The work would be area of triangle, which is no way related to the hypotenuse (other than $\small H^2=P_0^2+V_0^2$)$$W=\frac12P_0V_0$$
Heat is :
$$Q_{in:A-C}=(C_V+R/2)\left(3\frac{P_0V_0}{R}\right)+C_V\left(2\frac{P_0V_0}{R}\right)$$
Efficiency is :
$$\eta=\frac{W}{Q_1}=\frac{\frac12P_0V_0}{(C_V+R/2)\left(3\frac{P_0V_0}{nR}\right)+C_V\left(2\frac{P_0V_0}{R}\right)}=\frac12.\frac{R}{3C_V+3R/2+2C_V}$$
$$\eta=\frac{R}{10C_V+3R}=\frac{R}{5fR+3R}=\frac1{5f+3}$$
which for  a monoatomic,diatomic and polyatomic(linear/non-linear) are:
$$\begin{array}{|c|c|}\text{Monoatomic}&1/18\\
\text{Diatomic}&1/28\\
\text{Polyatomic-Linear}&1/28\\
\text{Polyatomic-Non-Linear}&1/33\\
\end{array}$$


The T that is held constant on the LHS of your equation has a different meaning from the T that is allowed to vary on the RHS.

The van 't Hoff equation answers the following:

Suppose I have a reaction that takes place at const T, p at T = T1.  Now I change to a new temperature, T2, and allow that reaction to again take place at const T, p, except now T = T2.  How does the equilibrium constant change?

Now that you understand this, you can see there is no inconsistency between "constant T" and "changing T", since they refer to two different things.

I.e., the fact that the reaction is constrained to take place at constant T in no way prevents me from testing the reaction at a different, but still constant, T.

The analogy I'd give is this.  Suppose we want to determine the effect of altitude on 100 m sprint times.  The 100 m is always run on a level surface (no change in altitude during the race).  Yet this does not in any way prevent us from comparing 100 m run times at different altitudes (e.g,  sea level vs. 10,000 feet).

So your question is like asking:  "How can we determine the effect of changing altitude on 100 m run times, given that the 100 m run is always done at constant altitude?"  Well, we can, of course.


It's hard/impossible to pinpoint the reason for your problems without seeing the  code and knowing which terms you use. If you feel like sharing it, then you should probably head over to Computational Science and make a post there.

A few general things come to mind:

Check that you have $\sigma$ and $\epsilon$ values inserted in the correct order.
Check the correctness of the combination rules you are using. It seems the force field you linked previously contains noncovalent terms from OPLS/AA. Make sure that the rest of the FF is compatible with OPLS/AA.
Check the units you are using, and make sure they are all the same. Probably won't make your simulation blow up, but it's still worth to take a look, especially if you are using parameters from different FFs.


Balancing Redox reaction : with 2 species being oxidized

Your best bet is to consider the entire reactant $\ce{CrI3}$ as the reducing agent.

The chromium (III) iodide starts out as that compound and gets oxidized to a combination of chromium (VI) (in the chromate) and iodine (VII) (in the periodate) with a ratio of chromium to iodine matching the original compound:

$\ce{CrI3 -> Cr^{VI} + 3 I^{VII}}$

Clearly the originally neutral reactant has been oxidized to atoms having a combined oxidation state of $+27$, so we balance this reaction with $27$ electrons:

$\ce{CrI3 -> Cr^{VI} + 3 I^{VII} + 27 e^-}$

Each chlorine molecule takes up two of the electrons so the redox ratio must be $\ce{2 CrI3 : 27 Cl2}$.  Put those in and balance the remaining compounds as you normally would.


Using the mass of calcium carbonate, what volume of carbon dioxide could theoretically be obtained if the reaction progressed to completion?

You are confused about chemical equations. The stoichiometric coefficient (the numbers before a compound) do not indicate the mass of the compound, they indicate the amount of substance. For example
$$\ce{CaCO3 -> CaO + CO2}.$$
This equation means that $\pu{1 mol}$ of calcium carbonate will decompose to give $\pu{1 mol}$ of $\ce{CaO}$ and $\pu{1 mol}$ of $\ce{CO2}$.
It does not mean that if 1 gram calcium carbonate were to decompose, 1 gram of calcium oxide and 1 gram of carbon dioxide will be obtained.
In your question the mass of $\ce{CaCO3}$ is given. From it you can calculate the moles of $\ce{CaCO3}$ involved. Same number of moles of $\ce{CO2}$ will be obtained and you can put that in the ideal gas equation.
Just so you know,
$$n_{\ce{CO_2}} = n_{\ce{CaCO_3}} = \frac{\pu{0.5 g}}{\pu{100.09 g/mol}}.$$
I recommend you study the chapter about the mole concept again.
As a side note: You mentioned in your question that the ratio of products is $1 : 1$. That's actually the ratio of the amounts of substances, not the mass ratio.


Chemical equilibrium for simultaneous dissociation reactions

Short answer (After some trials): It does not matter which dissociation reaction you use, The system of equation will be solved anyway since the reactions are not independent.

However, using the first reaction as a dissociation mechanism would generate a Kp equation that's a lot easier to solve than the others.


Determine reaction rate constant from the reaction time?

Assuming the reaction is a one step reaction, and does not have any elementary reactions, this could potentially work. However, if there are elementary reactions, this would change the rate law, and it would not be Rate = k[KIO3][NaHSO3].
You would also need to know the reaction order for both the potassium iodate and the sodium bisulfate, as well as the coefficients for both in the original reactions, yes you would be able to do this.
Knowing the coefficients, you could figure out the rate using rate -dA/adT. You could plug this rate into the rate law, Rate = k[KIO3]^m[NaHSO3]^n. However, you would need to know the reaction orders for KIO3 and NaHSO3, to plug in for m and n, and these can only be determined experimentally. If you know these, however, this equation would work for you.


Standard electrode potential can only be measured experimentally by comparing the half cells, in your case $\ce{Cr^{VI}/Cr^{III}}$ and $\ce{I2/I-}$ with a standard hydrogen electrode or a calomel electrode. The net standard electrode potential will be obtained by subtracting those of cathode-anode. You can find the standard electrode potential of most half cells in your textbook or online.
See in your reaction, two moles of $\ce{Cr^{VI}}$ turns to $\ce{Cr^{III}}$. So $n = 2 \times (6 - 3) = 6$. You can try out the same thing with the other half cell. The answer will be the same.