Why doesn't curie temperature follow a periodic trend in ferromagnetic materials

TL;DR : There's no reason for it to show any periodicity. It doesn't depend in a simple way on the outer electron configuration of atoms. The specifics of crystal structure are important in ferrmoganetic (also in antiferromagnetic, and ferrimagnetic) materials.

This by no means is a complete description of the phenomenon. If you're interested, consult textbooks, the internet etcetera. and come back with more specific questions. The person who originally asked the question seemed like he was a high school student (or lower) so I tried to tailor an explanation that he would understand. I welcome more complete desscriptions

First of all, ferromagnetism is a property that goes beyond the chemistry of a material, and is highly dependant on crystalline structure and microstructure.

Existence of several ferromagnetic metal alloys whose constituents are not themselves ferromagnetic (Heusler alloys), and conversely, the existence non-magnetic alloys, composed almost exclusively of ferromagnetic metals (for example some types of steel) are evidence of this fact.

The OP originally spoke of unpaired electrons, so I am guessing he knew something about electron spin.

Spin, just like electron charge is an intrinsic property, and its behaviour is described using quantum mechanics. Spin and charge on the electron give rise to a magnetic dipole moment (people like to talk about classical charges spinning and generating a magnetic dipole, however, I would like to discourage this analogy).

Due to its quantum nature, the spin of the electron can be in one of only two states; with the magnetic field either pointing "up" or "down" (what I call up and down is somewhat arbitrary; what's important here is that there are two arbitrary, mutually exclusive eigenstates).

The spin of the electrons in atoms is the main source of ferromagnetism, although there is also a contribution from the orbital angular momentum of the electron about the nucleus.

Materials made of atoms with filled electron shells have a total dipole moment of zero, because the electrons all exist in pairs with opposite spin, every electron's magnetic moment is cancelled by the opposite moment of the second electron in the pair.

So of course, a ferromagnetic material needs to have spin unpaired electrons, but that alone isn't sufficient. Hund's rules, posit that the first few electrons in a shell tend to have the same spin, thereby increasing the total dipole moment.

These unpaired dipoles (often called simply "spins" even though they also generally include angular momentum) tend to align in parallel to an external magnetic field, an effect called paramagnetism. Such materials are hence called paramagnetic. Molecules with spin paired electrons are called diamagnetic.

Ferromagnetism involves something extra. In a few substances the dipoles tend to align spontaneously, giving rise to a spontaneous magnetisation, even when there is no applied field. This is ferromagnetism.

Now let's develop some specifics. For instance, how and why do spins align? Additionally, why can we find iron in both magnetised and unmagnetised states, shouldn't it always be magnetised?

When two nearby atoms have unpaired electrons, whether the electron spins are parallel or antiparallel affects whether the electrons can share the same orbit as a result of the quantum mechanical effect called the exchange interaction. The exchange interaction is related to the Pauli exclusion principle, which says that two electrons with the same spin cannot also have the same "position".

When the orbitals of the unpaired outer valence electrons from adjacent atoms overlap, electrostatic repulsions are reduced when spins are parallel to each other. In simple terms, the electrons, which repel one another, can move "further apart" by aligning their spins, so the spins of these electrons tend to line up. This difference in energy is called the exchange energy. In ferromagnetic materials, the exchange interaction is much stronger than the competing dipole-dipole interaction. For example, in iron (Fe) the exchange force is about 1000 times stronger than the dipole interaction.

Therefore, below the Curie temperature virtually all of the dipoles in a ferromagnetic material will be aligned. (we will get to Curie temperature in a minute)

So why are iron and other ferromagnets ever found in an "unmagnetised" state ?

The reason for this is that a bulk piece of ferromagnetic material is divided into tiny regions called magnetic domains. Ferromagnetic materials spontaneously divide into magnetic domains because the exchange interaction is a short-range force, and over long distances of many atoms the tendency of the magnetic dipoles to reduce their energy by orienting in opposite directions wins.

If all the dipoles in a piece of ferromagnetic material are aligned parallel, the space surrounding it contains a lot of magnetostatic energy, which the material wishes to  reduce, and it does so energy by splitting into many domains pointing in different directions, so the magnetic field is confined to small local fields in the material, reducing the volume of the field.

This means that within each domain, the spins are aligned, but (if the bulk material is in its lowest energy configuration, i.e. unmagnetised), the spins of separate domains point in different directions and their magnetic fields cancel out, so the object has no net large scale magnetic field.

The domains are separated by thin domain walls a number of molecules thick, in which the direction of magnetisation of the dipoles rotates smoothly from one domain's direction to the other

Although the exchange interaction keeps spins aligned, it does not align them in a particular direction. Without magnetic anisotropy, the spins in a magnet randomly change direction in response to thermal fluctuations and the magnet is super-paramagnetic. There are several kinds of magnetic anisotropy, however, let's not go into that here

If one now places such a material in a strong enough external magnetic field, the domains will reorient so more of the dipoles are aligned with the external field. Moreover, the domains will remain aligned when the external field is removed, creating a magnetic field of their own extending into the space around the material, thus creating a "permanent" magnet.

The domains do not spontaneously return to the original minimum energy configuration when the field is removed because the domain walls tend to become fixed in place by the crystal structure, and defects in the bulk solid. Thus the magnetisation, and the resulting magnetic field, is baked into to the crystal structure of the material, making it very difficult to demagnetise (it is metastable in this sense).

As the temperature increases, thermal motion, competes with the ferromagnetic tendency for dipoles to align, and when the temperature rises beyond a certain point, (the Curie temperature), the system can no longer maintain a spontaneous magnetisation, as reorientation takes place.

This is an example of a second order phase transition

Thus the material looses its ferromagnetism; unpaired electrons are still present, and thus it is still paramagnetic.

Now clearly, you can see that ferromagnetic properties don't depend in a simple way on unpaired spins, and there is no reason for the curie temperature to follow a periodic trend the way you expect it to.


Lewis Acid / Base reaction for SO2 and H2O

Molecules that contain polar multiple bonds can function as Lewis acid because the central atom is electron deficient with a vacant orbital that can accept an electron pair.

When $\ce{SO2}$ dissolves in water , it forms the weak diprotic acid $\ce{H2SO3}$ (sulfurous acid) from Lewis acid/Lewis base reaction:

$$\ce{O\bond{=}S\bond{=}O + H2O <=> H2SO3}$$

The $\ce{O}$ atom of $\ce{H2O}$ molecule donates a lone pair to the $\ce{S}$ of  $\ce{SO2}$,forming a new $\ce{S\bond{-}O}$ $\sigma$-bond and breaking $\ce{S\bond{=}O}$ $\pi$-bond.


Why is the pressure correction constant 'a' independent of temperature?

The parameter $a$ is independent of temperature because we defined it this way. There is no other reason. Indeed, van der Waals equation is very much not a law of nature. It is just an empirical approximation. There are other approximations, maybe a dozen or more. You may create your own, where $a$ would decrease with temperature. Wait, it's been done already - look at the Berthelot equation!


A rubber balloon does require some force to get filled. A mylar balloon (typically filled with helium) can be purchased empty (or can be emptied) and requires almost no force to fill; then can be squashed manually to eject the gas. They are available in many sizes.

Rubber balloons also tend to go flat or leak, so their storage utility, especially for low m.w. gases, is short-term (a day or so). Aluminized mylar balloons can hold helium or hydrogen for long periods (weeks and even a month or two). For short-term, you could also use a plain polyethylene sandwich bag - after all, many objects are shipped with bags of air in the boxes to help protect them from damage. Use a needle to insert the gas and retrieve it; tape the area around the needle insertion - like an intravenous operation! Or, you can tape a plastic or glass tube into the bag.

Longer storage would require a gas sampling or collection tube, a glass container with a valve at one or both ends. The internet has many references to these.


Is there any graph that depicts the deviation of real gases from Charle's law?

Check out the van der Waals equation wikipedia page for a start ... it gives an explanation of one of the most common modifications of the ideal gas law that allows for modeling of deviations from ideality in the way you describe.  This approach is quite nice in my opinion, because it includes two coefficients (a and b) that account for specific deviations for a given gas ... a scales with the strength of intermolecular interactions, and b accounts for the actual volume of the gas molecules.

You should be able to use that information to do what you want to do regarding Charles Law.  The graph on the page shows isotherms, but you should be able to generate the isobar curves that you are after with a little bit of work.


Does the emission spectrum for a dye depend on the frequency of light used to excite the dye?

Does the emission spectrum depend on exciting the dye at its peak absorbance wavelength $\lambda_{max}$?

No, it is independent of that. The excitation $S_0 \rightarrow S_1$ is vertical, bond length are not changed in the excitation process. Compare the mass of the nuclei with that of the electrons and you know why. You'll just end up in a different vibronic level of the electronically excited state.

However, before the emission takes place, this initial Franck_condon states relaxes to the vibrational ground state of $S_1$ with an equilibrated geometry and emission occurs from that (Kasha's rule).


Why does increasing the pressure cause ice to melt more?

At equilibrium, the change in Gibbs free energy is zero.  So, to maintain equilibrium, if the pressure increases, the temperature must change in tandem in such a way that the change in free energy of the solid is the same as that of the liquid.  We have, for either phase, $$\mathrm{d}G=-S\mathrm{d}T+V\mathrm{d}P$$So, $$\mathrm{d}(\Delta G)=-\Delta S \mathrm{d}T+\Delta V\mathrm{d}P=0$$where the deltas refer to the changes between liquid and solid.  We also have that, at equilibrium, $$\Delta S=\frac{\Delta H}{T}$$So, combining these equations gives $$\mathrm{d}T=\frac{T(V_\mathrm{l}-V_\mathrm{s})}{(H_\mathrm{l}-H_\mathrm{s})}\mathrm{d}P$$Since $V_\mathrm{l}<V_\mathrm{s}$, this equation indicates that the equilibrium temperature for the solid-liquid transition decreases as the pressure increases.

This shows that the melting temperature decreases. So if the pressure is increased, and the actual temperature is not changed, the temperature of the ice will be higher than the melting temperature, and the ice will melt.


As an extension of Mithoron's comment, if you drew the structure as N triple bonded to O, you would fulfil both the octets without placing the free radical electron.

Now, you'd have to extend the octet of either nitrogen or oxygen, but you have no free orbitals to place the extra electron.

Hence, the first one is the correct Lewis structure.


How to write down the composition of eluent mixtures for TLC

Describe what you did. The key is make sure your communication allows your experiment to be reproduced.

  ... Rf 0.34 (10 mL of 9:1 DCM/MeOH + 3 drops of 1.0 M Et3N in MeOH).

If you do all of your TLC with the additional of triethylamine, then you can include a sentence in the general methods about this procedure.

  All TLC plates were developed in a 9:1 mixture of DCM and MeOH that was made alkaline by the addition of three drops of 1.0 M Et3N in MeOH for every 10 mL original solution.

The procedure you describe is common, especially in the elution of basic molecules (since silica is slightly acidic). You might find examples of how to state this sort of procedure in the synthesis of amines and alkaloid natural products. Ultimately, in a publication, you should err on the side of thoroughness for your experimental methods. If it's too long, a reviewer of the editor will tell you.


How to determine from the formula that there is a phenyl group present in the molecule?

It's a matter of convention.

$\ce{C6H5}$ is conventionally used to represent phenyl because it is a common moiety. Your long, highly unsaturated, chain is also $\ce{C6H5}$, but it wouldn't conventionally be written like that without expansion, unless you're setting out to confuse someone.


Why does iodine have different colors in different media?

color is result of electron transfer between orbitals and anything that affect energy of these orbitals can change color of that specie, so iodine in different mediums have different interactions and because of that iodine orbitals have different energies in different medium,so iodine have different color in different mediums.


Calculate limiting molar conductivity of BaCl2

I'll answer the second part of your question first: you have one equation with three unknowns, so you need the value's of two more unknowns, namely the molar conductivities of Ca++ and Cl-.  They are readily available from the CRC Handbook of Chemistry and Physics.  Every chemist and chemistry prof and library has one.  You could also find those values from online resources, just be sure to check with reliable sources.  An internet search will turn up many.  Once you have those two values you just plug them in to your equation and you are done.  You are correct with your V values.


Number of resonance structures of nitrobenzene

We never really need to know the number of resonance structures. The whole concept is qualitative, so we can't draw any quantitative conclusions from it. But if you want to do it anyway, then of course 1 and 5 are the same, while 2 and 4 are not (they have positive charges on different atoms).

Also, both oxygens are in fact equivalent, so you might want to add the structures with their roles reversed.


As @nilay points out, we can make "polychlorides" with "nonclassical" stoichiometries. But for more typical laboratory conditions it is true that heavier halogens are more favorable for forming polyatomic ions.
This is because polyhalide ions involve two competing forces. On the one hand, the delocalized covalent bonding gives more stability to a polyatomic ion than we would have with the separate monoatomic ions and diatomic molecules. On the other hand, surrounding counterions or dipoles are more strongly attracted to compact, spherical monoatomic anions than to bulky polyatomic ones. With lighter halogens forming the most compact monoatomic anions, the electrostatic effect wins out and we need severe conditions to overcome it. Heavier halogens, with larger atoms and more polarizable electrons, make the polyatomic ions more competitive, as in $\ce{C_5H_6N+Br_3−}$ (http://www.organic-chemistry.org/chemicals/oxidations/pyridiniumhydrobromideperbromide.shtm).


Boron trifluoride, as described in Wikipedia is proved to exist the way everything in chemistry is ultimately proved, by experiment.  The octet rule is really only a model that describes the electronic structure of some but not all molecules.  Boron trifluoride is one of those (many, actually) molecules for which the octet rule is not really correct.  The actual electronic structure is not just a single Lewis structure but a combination of multiple contributing structures, some of which do satisfy the octet rule and some of which don't.


What we have, in the Lewis theory, is an acid-base displacement reaction.  The Lewis acid, $\ce{H^+}$, is initially combined with one Lewis base, $\ce{NO2^-}$.  Then another Lewis base, $\ce{HPO4^{2-}}$, takes the $\ce{H^+}$ away displacing the $\ce{NO2^-}$.

Similarly when $\ce{HCl}$ is dissolved in water and dissociates, the water is displacing the chloride ion.  The Lewis acid involved in this displacement is again $\ce{H^+}$.


Why do some materials turn black and others white after combustion?

The objects which turn black are usually carbon based/organic compounds. When burned/combusted they follow one of the following unbalanced reactions:
\begin{align}
\ce C_x(\ce H_y\ce O_z)_\text{(solid/liquid)} + {\ce{O2}}_\text{(gas)} &\ce{->} \ce C_\text{(solid)} + \ce{H_2O}_\text{(liquid)} + {\ce{CO_2}}_\text{(gas)}\\
\ce C_x(\ce H_y\ce O_z)_\text{(solid/liquid)} + {\ce{O2}}_\text{(gas)} &\ce{->}  \ce{H2O}_\text{(liquid)} + {\ce{CO2}}_\text{(gas)}\\
\ce C_x(\ce H_y\ce O_z)_\text{(solid/liquid)} + {\ce{O2}}_\text{(gas)} &\ce{->}  \ce {H2O}_\text{(liquid)} + {\ce{CO2}}_\text{(gas)} + \ce{CO}_\text{(gas)}
\end{align}

You'll notice that some of the reactions will produce solid carbon, which is the black thing you see.

On the other hand, inorganic compounds begin to decompose into their ions and/or release their  "trapped" water molecules. As you'll notice, most ionic compounds are white solid salts.


Classical hybridization theory does not allow for noninteger hybridizations. However, ab initio calculations can be interpreted using a bond order analysis method such as NBO, where the MO coefficients are used to provide the closest analogue possible to a classical hybridization picture.

For example, one of the pure $sp$ orbitals in Pauling's valence bond theory (where hybridization was first introduced) has wavefunction

$$ \phi_{sp} = \frac{1}{\sqrt 2} \phi_{s} + \frac{1}{\sqrt{2}} \phi_{p_x} $$

Taking the square of the coefficients, this orbital is $\frac 1 2$ s character and $\frac 1 2$ p character, i.e. it is an $sp$ orbital.

The basic idea of a bond order analysis method is to reexpress a molecular orbital into a form similar to

$$ \phi = c_1 \phi_{As} + c_2 \phi_{Ap_x} + c_3 \phi_{Ap_y} + c_4 \phi_{Ap_z} + \dots$$

If the other coefficients are very small, the ratio

$$ n = \frac {c_2^2 + c_3^2 + c_4^2}{c_1^2} $$

would yield a number that could be used interpret $\phi$ as a $sp^n$ orbital on atom $A$.

Edit: As for inferring hybridization states from a direct inspection of nuclear geometries, there is in principle no such direct relationship in electronic structure theories that are more sophisticated than VSEPR. The former is an electronic property whereas the spatial arrangement of atomic nuclei are not, and the relationship between the two becomes much more complicated.


Why does diesel have a higher energy density than gasoline?

Diesel and gasoline have roughly the same energy per unit mass (lower heating value), about $41~\pu{MJ/kg}$).

The density of diesel is about $833~\pu{kg/m^3}$ compared to $740~\pu{kg/m^3}$ for gasoline. This gives diesel about 13% higher energy density per volume.


What is the difference between the chemical potential μ and the molar Gibbs free energy Gm?

The chemical potential of a species is the partial derivative of the Gibbs free energy of the mixture of components with respect to the number of moles of that particular species, holding the temperature, pressure, and number of moles of all the other species constant.  For a single component system, this is reduces to the total free energy divided by the number of moles.


What are the key chemical characteristics that determine the octane rating of fuel?

Combustion
 is (usually) chain radical reaction with heavy branching. More
branching means faster combustion and less branching is slower
combustion. And detonation is wave of super-fast combustion.

So, if you can suppress branching or deaden some branches, you can
increase durability of air-fuel mixture. Actually, this means, that some
 radical trap is needed. The oldest trap was tetraethyllead, but it is
currently dropped (lead oxides as fine dust is not a health friendly
substance). Non-saturated hydrocarbons of different types also can trap
active radicals, transforming them into more stable allyl radicals, so
fuel companies tries to keep as much of them in fuel as possible. Some
other compounds, like ferrocene, also can be used to increase octane
rating.


It doesn't matter that the electronegativity difference between nitrogen with itself is 0. It just means the bonds it makes with another nitrogen atom have $100~\%$ covalent character.

A typical chemical bond basically spans the spectrum of bonds with $100~\%$ ionic character, uneven ionic/covalent character, and $100~\%$ covalent character.

$\mathbf{100~\%}$ ionic character: $\ce{A^+B^-}$ or $\ce{A^- B^+}$, with no sharing of electrons at all
Uneven ionic/covalent character (i.e. polar): $\ce{{\stackrel{\delta+}{A}}-{\stackrel{\delta-}{B}}}$ or $\ce{{\stackrel{\delta-}{A}}-{\stackrel{\delta+}{B}}}$, with deviation from perfect sharing of electrons and direction of favored negative/positive end of dipole determined by electronegativity difference
$\mathbf{100~\%}$ covalent character (i.e. $100~\%$ nonpolar): $\ce{A-A}$ or $\ce{B-B}$, with perfect sharing of electrons

It just so happens that $\ce{N2}$ is one example of $100~\%$ covalent character. You can argue the same for $\ce{O2}$, and any similar homonuclear diatomic molecule.

(Granted, there are some exceptions to 100% covalent character, like in iron's metallic bonding, but anyways...)


The energy difference between the orbits gets smaller the higher "up" you go.

For the classical bohr model the energy of a given orbit $n$ can be approximated by:

$E_n=-\frac{Z^2R_E}{n^2}$

where $Z$ is the charge of the nucleus and $R_E$ is the Rydberg energy. Since there is a $n^2$ in the divider the difference $E_{n+1}-E_n$ gets smaller and smaller.

This is explained in more detail here.

As for why this is the case: here we have to look into quantum mechanics and I guess this is outside the scope here. But if you are interested look up explanations of a particle in a 1D box. The spacing of the energy levels there follows the same trend and you can find explanations there on why this is the case.


Heat effect on diluting concentrated NaOH solution

The solution gets warmer by diluting with extra water, but no violent reaction. I suggest to do a test, if the temperature raise is concerning you, e.g. because of the tubing temperature resistance.
What  can also be done is applying solution with graduating concentration, so warming would be spreaded in sevceral steps. But as hydroxide is already quite diluted, I expect temperature change rather few than many degrees.
Another concern is initial dosage would be affected by water, decreasing concentration of the applied volume.


Applications of magnesium sulfate monohydrate MgSO4 · H2O

Magnesium sulfate is frequently used in the laboratory, especially after aqueous work-up.
Aqueous work-up is a common technique in the lab to get rid of residual impurities after completion of a reaction.
For this the organic solution of the reaction is cooled to room temperature and mixed with water. Specific impurities will then diffuse into the aqueous phase and can be separated with a separatory funnel.
Unfortunately some residual water will stay in the organic phase and this can have negative impact for following characterization of the compound.
Therefore magnesium sulfate is added, which is able to catch the residual water in its crystal lattice. After filtration of the solid magnesium sulfate, the water is reduced to a non-significant amount and the compound ready for further investigation.


When it is said 'at constant pressure, at constant pressure ', is it about pressure & temp. of the system or surroundings?

When we say

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

we are referring to $\Delta G^\circ$, $\Delta H^\circ$ and $\Delta S^\circ$ of the system. As such, the variable $T$ in the equation refers to the temperature of the system. The condition for constant pressure $p$ is implicit; the equation above does not directly show why $p$ has to be constant. For more information, refer to this question and answer.

The sentences that you have given are not the best examples to explain this, because they do not relate to Gibbs free energy. I will give you a different example:

  When hydrochloric acid is added to sodium carbonate at atmospheric pressure, the standard enthalpy change $\Delta H^\circ$ and the standard entropy change $\Delta S^\circ$ are measured to be $x~\mathrm{kJ~mol^{-1}}$ and $y~\mathrm{J~K^{-1}~mol^{-1}}$ respectively. Calculate the Gibbs free energy of the reaction and hence determine the equilibrium constant. (The temperature at which the reaction is carried out is $25~\mathrm{^\circ C}$.)

The wording of the question may seem to imply that it is talking about the temperature and pressure of the surroundings, and that is probably not a coincidence. For a simple chemical reaction like the one above, we would probably carry it out in a beaker. We weigh out a certain mass of $\ce{Na2CO3}$, and add a certain volume of $\ce{HCl}$, and pour the acid into the beaker and watch the bubbles.

Under such conditions, we can speak of the system being at mechanical and thermal equilibrium. Without going into too much detail about reversible and irreversible processes (I know there are loopholes in the explanation):

Mechanical equilibrium, in general, means that the net force acting on the system is zero. Correspondingly, this must mean that the pressure of the system and the pressure of the surroundings are equal at all points in time. If, at any point in time, this was not the case, e.g. if we had $p_\mathrm{syst} < p_\mathrm{surr}$, then the surroundings would "push" on the system and cause the system to contract until it reaches the point where $p_\mathrm{syst} = p_\mathrm{surr}$.
Thermal equilibrium means that there is no heat transfer into, or out of, the system. Analogously to above, this must mean that the temperature of the system and the temperature of the surroundings are equal at all points in time. We conventionally assume the surroundings to be a "reservoir" which maintains a constant temperature, and this is pretty true as far as our simple example is concerned. Therefore $T_\mathrm{syst} = T_\mathrm{surr}$.


See Shakashiri's chemical demonstrations. He carefully outlines the steps. As I recall it requires taking sodium nitrate into a molten state and applying a current. The elemental sodium is precipitated on the inside surface of an incandescent bulb.


How to prepare 2-bromobutane from 1-butanol?

The hydroxyl fragment isn't a good leaving group, unless the medium is acidic. Using just HBr, you can convert the hydroxyl group into an oxonium ion (the H2O fragment would be a good leaving group). Upon heterolysis of the C-O bond, a hydride shift occurs simultaneously to create a secondary carbocation. The exogenous bromide can act as a nucleophile towards the secondary carbocation to form 2-bromobutane.


Why can't an enolate anion successfully displace a halogen from a benzene?

explain why substitution reactions with enolate anions don't take place at other sp2 carbon atoms

As explained in this earlier answer, $\ce{S_{N}2}$ reactions do occur at $\ce{sp^2}$ carbons, they're just higher energy and consequently, less common.  So the question becomes, "why is this pathway higher in energy?"

Since the $\ce{S_{N}2}$ reaction involves "backside" approach to the carbon undergoing substitution, steric considerations play a role.  However, the attacking reagent could alter its approach by 15 degrees or so lessening steric interactions (e.g. in benzene the attacking nucleophile could approach at an angle of 15 degrees above the plane of the benzene) and not increasing the activation energy too much.

One way to explain the high energy associated with $\ce{S_{N}2}$ attack at a vinylic carbon is as follows.  Consider the ethylenic double bond as a bent bond system, that is, we have a two-membered ring (instead of a traditional sigma-pi double bond) and each carbon in the ring is $\ce{sp^{4.3}}$ hybridized.  This bent bond description is equivalent to the alternate pi bond description in all regards including electronic (btw, it makes the reason [strain] for the high heat of combustion of olefins more palpable). If possible, build a model to help visualize the situation.  As the attacking nucleophile approaches and begins to interact with one of the ring carbons, that carbon will attempt to rehybridize to an $\ce{sp^2}$ carbon (just like in the traditional case of $\ce{S_{N}2}$ attack on an $\ce{sp^3}$ carbon).  In this case we're forcing the already highly strained bond angle in our two-membered ring to try and open further - towards 120 degrees. Hence, this is a high-energy pathway that is not often observed.

Alternately, this same concept can be rephrased in terms of traditional $\ce{sp^2}$ sigma-pi hybridization, by noting that we start with an $\ce{sp^2}$ hybridized carbon which becomes $\ce{sp}$ hybridized in the transition state.  It is known that $\ce{sp}$ hybridized carbons are roughly 5 kcal/m less stable than $\ce{sp^2}$ hybridized carbons (more electrons in higher energy p orbitals, not offset by the energy lowering of the electrons shifting from $\ce{sp^2}$ bonds to $\ce{sp}$ bonds) thereby increasing the energy required to reach the transition state.

Another point to consider is that since the $\ce{S_{N}2}$ reaction proceeds with inversion of configuration, such a reaction in a ring system would lead to the formation of a trans-double bond.  This would be an extremely high energy process in the case of a small ring system such as benzene.

Finally, $\ce{sp^2}$ hybridized carbon is more electronegative than $\ce{sp^3}$ hybridized carbon due to the increased s-character.  This will increase the electron density around an $\ce{sp^2}$ carbon and, as a nucleophile approaches, the increased electron repulsion will make this a higher energy pathway that nucleophilic attack at a less electron rich $\ce{sp^3}$ carbon.


Calculate specific heat of a substance given other information

Your equation is correct, as are most of your measurements.

The unit for $c$ is $\ce{Jg^{-1}K^{-1}}$ if using grams as mass, ($\ce{Jkg^{-1}K^{-1}}$ if you convert the mass to kilograms).

Using your formula, however, does not give the answer you gave. When calculating using a calculator, consider that a parenthesis needs to be put around the denominator, as such:

$$c = \frac{q}{(m\Delta T)}$$

or else the calculator takes it as being $\ce{979J / 20g \times 45K}$ which gives the value you got.

It needs to be $\ce{979J / (20g \times 45K)}$ which gives an answer that is much more reasonable.


I'm not entirely clear on what you want - I guess a database of a bunch of 3D coordinates for many organic compounds?

By far the largest such database is NIH PubChem with over 30 million compounds - mostly organic. You can search by structure or name or identifier. Most of these have 2D and 3D geometries you can visualize or download and many computed properties.

A similar database is the RSC ChemSpider with over 34 million compounds. Much like PubChem, names, 2D and 3D geometries and other properties are available.

I'll put in a plug for the Pitt Quantum Repository which contains ~64,000 molecules and growing, all with geometries computed using quantum chemical methods. One benefit of PQR is that you can easily access the site using smartphones and tablets and use a QR code or DOI to link to the compound.


Aluminium, Carbon, Aluminium Sulphate battery. What's the reaction?

Very interesting: no gas, no change in the electrode, yet a significant electrical output. This could be the beginning of perpetual power (similar to perpetual motion!).

But no, probably the output power is barely significant, and something is happening somewhere, possibly in the solution. Aluminum sulfate doesn't usually produce a clear solution, so possibly you have missed an increased turbidity.

Possibly oxygen is reacting with the metal under the acidic conditions to produce Al(OH)3 which is removed by the acid solution (without reducing the acidity, because Al(OH)3 is not much of an alkali). The equation could be 4 Al + 3 O2 + 6 H2O --> 4 Al(OH)3. If you restrict oxygen access, the voltage and current should decrease. If you increase it, say with an aquarium pump, you might get a higher output.


Long-lived, non-lethal radioisotope for fiction

Good candidates for your research would be the fallout residues of atomic bombs having exploded in the world between $1945$ and $1964$. These bombs were made by USA and the former USSR (Russia today) during the "cold war". Each time one of these superpowers had developed a "more powerful" bomb, the other one managed to create a still more powerful one. And so on for about $20$ years. This stupid course stopped in $1964$, because of the level of the fallout radioactivity measured all over the world. In $1964$, the radioactivity from the fallout had much increased all over the world. The whole planet had been contaminated by the winds to $10$% of the limit value where all food would have to be decontaminated before being eaten. So both Russians and Americans knew about this danger. They managed to meet in Geneva and decided to stop their nuclear rivalry. All that were made for "in the interest of the peace in the world". It was magnificent.
As a result the radioactivity slowly decreased all over the world. At the end of the century it was back again to the level $1945$.
One of the isotopes present in the fallout is the strontium-$90$, who is assimilated like calcium, and milk contains much calcium. So $\ce{^{90}Sr}$ was found in the teeth of the newborns overall. Its half-life is about $28$ y. It was followed in teeth all over the world from $1964$ to today. It was observed that this isotope disappears from the body with much less than $28$ y. Apparently calcium (and strontium) are naturally and continuously renewed in the body.
Another isotope that can be interesting for you is the radium-$226$ who was deposited on the watch dials before $1964$. Like strontium, radium is assimilated by the body like calcium. Workers that were painting the dials with radium used brushes, and have been contaminated by this isotope. So they are radioactive. They emit constantly gamma rays. They can be detected with a Geiger counter. The level of the radioactivity is not dangerous. But they emit rays. Is it what you want ?


Question on organic chemistry where we are required to find enol content

Thermodynamically, the greater enol content can be associated with greater acidity. Thus, the question can be rephrased as ‘which is the strongest acid of these four?’ We can ignore the interconversion of the methyl ester A into the ethyl ester B — it will happen because we are in ethanolic solution, but extremely slowly as we do not have any strong bases present. (And even if we had strong bases, these would deprotonate A more further reducing the susceptibility of nucleophilic attack on the carboxy group.)

For reference and posteriority, these are the compounds we are talking about:

methyl acetoacetate (A)
ethyl acetoacetate (B)
acetoacetone (C)
3-methylacetoacetone (D)

Simply by looking at the names we can see how highly related they are and how much subtle differences must matter. We can predict all four of them to be rather acidic since if the inter-carbonyl hydrogen is released and an enolate is formed, this enolate will always be in contact with an electron-withdrawing π system on the right-hand side: either another keto-group or an ester group.

That immediately gives us our first hint to compare the four: an ester group is less electron-withdrawing than a ketone, so we predict A and B to be less strong acids than C and D. This is convenient as it means we do not have to concern ourselves with the acidity difference of a methyl versus ethyl ester.

The difference in acidity between C and D is also subtle but more obvious. In both cases, the corresponding enolate anion would be planar. However, the methyl group is much larger than a hydrogen atom meaning that the enolate of D is subject to 1,3-allylic strain — which cannot be lessened without destroying planarity and thereby reducing conjugation. Thus, the enolate of C is more stable than the enolate of D.

This is backed by acidity data from the Bordwell tables — see table.

$$\textbf{Table 1: }\text{acidity data of the four compounds}\\
\begin{array}{lcccc}\hline
\text{solvent} & \mathrm{p}K_\mathrm{a}(\textbf{A}) & \mathrm{p}K_\mathrm{a}(\textbf{B}) & \mathrm{p}K_\mathrm{a}(\textbf{C}) & \mathrm{p}K_\mathrm{a}(\textbf{D}) \\ \hline
\ce{H2O} & -/- & 10.7 & \phantom{0}9.0 & -/-\\
\text{DMSO} & -/- & 14.2 & 13.3 & 15.1\\ \hline\end{array}$$


Calcium oxalate solubility depends on acidity, as the weak oxalic acid is protonated in highly acidic sulphuric acid.

Additionally, oxalic acid would be dehydrated if sulphuric acid was concentrated enough.

If we for convenience mark oxalate anion $\ce{(COO)2^2-}$ as $\ce{Ox^2-}$, then

$$\ce{CaOx <<=> Ca^2+ + Ox^2-}$$

$$\ce{Ox^2- + H+ <=>> HOx^-}$$

$$\ce{ HOx^- + H+ <=>> H2Ox}$$

$$\ce{  H2Ox ->[H2SO4] H2O + CO ^ + CO2 ^}$$

Therefore oxalic acid has no chance to create concentrated sulphuric acid.


Chemically Speaking, What's Bad About Inhaling Plastics?

The primary dangers from plastics themselves come when the monomer, and any compound put in to make the polymer matrix flexible, separate from the polymer and are released (leached) into whatever the plastic is containing. There are several such compounds which are known to be problematic to human biochemistry:

Phthalates - These are aromatic diesters, with a structure incorporating a phenyl ring with two ester linkages, typically to alkanes like ethyl or butyl groups. Phthalates are used in certain plastics to link the monomers, such as in poly-ethyl terephthalate (PET aka polyester), and in others as a plasticizer to make the normally rigid plastic flexible (PVC). Phthalates mimic certain hormones in the human body and so can cause issues with growth and maturation, as well as feminizing characteristics in males (reduced body hair, weight gain, moobs). PET is OK for packaging and storage of cold foods (it is, for instance, the plastic used for soda bottles), but you don't want to heat it because that will degrade the plastic to release the phthalate.
The ones used in PVC are especially problematic because they're not well-bonded; instead of being an integral piece of the polymer chain, they form weak hydrogen bonds between the polymer chains (thus the flexibility, even in relatively thick layers of the plastic), and so these phthalates easily leach out into water or "off-gas" into the air, even without heat. These phthalates are a big part of the "plastic smell" of new plastic products, and of water stored in them; the more flexible, and the longer the plastic has been kept enclosed, the more intense the smell. There was a big stink (NPI) in the early 90s when PVC was being used for water supply lines in home construction, and then they found all these problems with the stuff and a lot of homebuilders and chemical companies went bust paying for all this piping to be replaced with copper.

Bisphenols - These are variations of 2,2-bis-(dihydroxyphenyl)propane (which is the infamous Bisphenol A or BPA). Bisphenol A is the monomer for most of the original formulations of high-impact polycarbonate, such as Lexan. Again, when heated, the polymer breaks down and leaches the water-soluble monomer, which is structurally similar in many ways to phthalate and again has hormone-mimicking, endocrine-disrupting properties, which is why newer polycarbonates intended for food storage are made with other bisphenol monomers (most of which have similar endocrine properties, but because they aren't Bisphenol A they're still legal for now and allow the manufacturer to label the product "BPA-free").

Vinyl Chloride - The monomer of poly-vinyl chloride or PVC, this is typically a hazard in production facilities for PVC and when old, aged PVC begins to break down. The vinyl chloride monomer is a known carcinogen, which is the other reason PVC is no longer used for food packaging or water supply, and is being phased out from wastewater lines in favor of nylon.

As far as answering the question "shouldn't we just be able to pass it out of our system", the same could naively be asked of a whole host of chemicals that are bad for us. The biggest problem with that statement is the First Law of Toxicology; anything, in sufficient amounts, is harmful, even if it's absolutely essential for life in the right amounts. Selenium is an essential trace mineral in human biochemistry, required in certain protein synthesis mechanisms, but the LD50 blood concentration is about 3 mg/kg, and harmful effects such as stunted growth begin to be observed at concentrations within foodstuffs of as low as 4 to 5 ppm. Heck, you can die from drinking too much water, faster than you could from lack of it (it causes a condition called hyponatremia where the added water dilutes electrolytes to the point the body can no longer function).
What causes many chemicals to be harmful is in fact our body's own affinity for them as a beneficial chemical. Either the chemical is beneficial to us, or it looks like one that is, and the chemical is normally so rare in our diet and environment that our bodies grab all of it that they find, without an "off switch" to regulate intake.
Which leads to the second problem with that statement; sometimes our bodies don't efficiently metabolize and dispose of chemicals. Vitamin A, for instance, is an essential nutrient, and our body absorbs all it can, and gets rid of it only very slowly. As a result, it "bioaccumulates", and eventually becomes toxic. That's why beta carotene is the preferred "form" of Vitamin A in supplements; it's in fact an intermediate that our bodies can use to synthesize the vitamin as needed, but being water-soluble unlike actual Vitamin A, any unused excess of beta carotene is simply flushed out via the kidneys.
That process in itself can cause problems; the latest round of drug recalls the lawyers are running commercials on are for drugs that did exactly what they were supposed to, but were then metabolized into carcinogenic compounds that concentrated in the urine and caused various cancers of the excretory system.
And the last problem, tying back into the case in point, is that even if the body selectively absorbs it, efficiently processes it, and it isn't acutely toxic in any form or intermediate along the way, as long as there's an input of the chemical into your system, you'll always have some of it. Plastics are ubiquitous in consumer products, and especially food prep and packaging. As a result, virtually every resident of the industrialized world has a detectable amount of artificial phthalates in their blood, because it is simply impossible not to come into contact with them. Even USDA-certified organic foods have phthalates in them, because the cows eat food that's been stored or served in plastic bins and it shows up in their milk and meat, or the vegetable was fertilized with compost stored in a plastic bin or watered from a PVC irrigation system (even the farmer's market items from someone's backyard were likely grown using water collected in a plastic rain catchment and/or dispensed from a hose or a drip irrigator system that leaches plasticizer). What you store it in or prepare it with once it gets to you is immaterial; the damage is done.


How long does it take to saturate H2O with O3

The solubility of ozone in water as per one website is:

First thing is to establish which curve you need. In 1 m$^3$ gas you have 70 L ozone:

$\frac{70\ L\ \ce{O3}}{1\ m^3}\frac{1\ mol}{0.082\cdot 298\ L}\frac{16\cdot 3\  g\ \ce{O3}}{1\ mol}= 138\ g\ \ce{O3}/m^3$

Let us take the curve of 150 g/m$^3$ which has an ozone solubility of 30 mg/L water.
Then:

$\frac{30\cdot 10^{-3}\ g\ \ce{O3}}{1\ L\ \ce{H2O}}\frac{1\ mol}{16\cdot 3\ g\ \ce{O3}}\frac{0.082\cdot 298\ L}{1\ mol}\frac{1\ min}{5\cdot 0.07\ L\ \ce{O3}}= 0.044\ min$

Which is about 3 seconds.


Routes of formation of esters with highest yield

Option 2 generally gives higher yields.

This is because option 1 (Fischer esterification) usually results in an equilibrium mixture which contains significant amounts of products and reactants, limiting the yield. This is mitigated by using a stoichiometric amount of a concentrated acid catalyst, such as sulfuric acid, which is also a strong dehydrating agent, and therefore removes the water from the equilibrium as it forms, favouring the products. Additionally, the reaction mixture can be heated to evaporate the water, or you can use a large excess of one of the reactants, typically the alcohol because it also acts as the solvent.

Conversely, option 2, more easily gives high yields because the reaction is more or less complete, with very little reactants left at the end. This is because the reaction produces hydrogen chloride gas instead of water, which is lost from the reaction mixture, driving the reaction forward. The downside of this option is that acid chlorides are generally more dangerous than carboxylic acids and are harder to prepare, or more expensive to buy. Aditionally, hydrogen chloride gas is produced which must be disposed of safely.

Another, often used method, is esterification using acid anhydrides. These are intermediate in reactivity between carboxylic acids and acid chlorides and are easier to use than acid chlorides, but still provide excellent yields. A classic example of this is the acetylation of salycylic acid with acetic anhydride to produce Aspirin.


condensation of tellurium on inside of ampoule?

This is a general phenomenon. All substances at any temperature above $0\ K$ will shed a few atoms from the solid or liquid, forming a gas however dilute. The tendency for this to happen varies with temperature, and it can be measured by determining the equilibrium pressure of a substance inside an evacuated chamber. The key concept behind this is vapour pressure. There are several previous Chem.SE questions regarding vapour pressure, so you may find something useful, such as perhaps this.

As a rule of thumb, the higher the boiling/sublimation point of a substance, the lower its vapour pressure at any given temperature (in fact, the definition of the boiling/sublimation point is the temperature where the vapour pressure equals the pressure exerted by the surroundings onto the solid/liquid substance, which is in many cases equal to $1\ bar$). Tellurium is a relatively low boiling-point solid, with a vapour pressure of $1\ bar$ at $988\ °C$. According to this article, the vapour pressure of tellurium at its melting point ($450\ °C = 723\ K$) is approximately $70\ Pa$. This means there is a fair amount of tellurium vapour present inside the ampoule at this temperature. It is entirely conceivable that some of this vapour could deposit in small droplets on the upper surface of the glass ampoule, outside the main body of the liquid. Note that theoretically the deposition of tellurium on the upper surface should also happen at room temperature, though the equilibrium vapour pressure of tellurium at $25 °C$ is only about $5 \times 10^{-4} Pa$, which would make the transfer of a macroscopically visible amount of tellurium an extremely slow process.


What is the chemistry behind cleaning silverware with aluminium foil

Is any one of these reaction more "true" (occurring more often naturally) than the others or is it the case that a little bit of everything is happening?

I think the best answer is "a little bit of everything".  Silver sulfide forms faster but requires exposure of the silver to sulfur-containing materials (like human skin, food, etc.).  Silver that isn't exposed to sulfur would still be oxidized by the air, but at slower rates.  So depending on the life history of the silver object in question, different amounts of the sulfide and oxide could be present.


Can the 1st generation Grubbs Catalyst be used for ring closing metathesis on non-terminal alkenes?

Yes, it appears that that ring closing metathesis of non-terminal alkenes has been performed using the 1st generation Grubbs catalyst several times:

Asymmetric syntheses of enantiopure C(5)-substituted transpentacins
via diastereoselective Ireland–Claisen rearrangements
Stereoselective Synthesis of Deuterium-Labeled (2S)-Cyclohexenyl
Alanines, Biosynthetic Intermediates of Cinnabaramide
Ru-catalyzed metathesis of octadienylether xyloside Synthesis
of (−)-Agelastatin A by [3.3] Sigmatropic Rearrangement of Allyl
Cyanate
A chiral auxiliary cleavable by ring-closing alkene
metathesis — Efficient synthesis of chiral nonracemic
cycloalkenes
Conversion of chiral unsaturated cyanohydrins into
chiral carba- and heterocycles via ring-closing metathesis
Total Synthesis of Pancratistatin Relying on the [3,3]-Sigmatropic
Rearrangement
Stereoselective Chelate-Controlled Addition of Grignard Reagents to Unsaturated Medium-Ring Heterocycles


In helium-2, would each electron experience a single positive charge from the nucleus?

To answer this question, I am going to assume that a $\ce{^2He}$ nucleus actually exists long enough for two electrons to surround it — this is probably not a valid assumption since Webelements.com does not list $\ce{^2He}$ while it does list the isotopes $\ce{^6He}$ and $\ce{^8He}$ (although they admittedly have long half-lifes measured in tenths of seconds). But for our case, let us assume so.

The charge radius of a proton — our closest experimental measurement of its diameter, if ‘diameter’ is even a valid term at subatomic scales — is given on Wikipedia as $\pu{8.75e-16 m}$ or $\pu{0.875fm}$. At this point, take a step back to remember that the Bohr radius, approximately the most probable distance between the electron and the proton in a hydrogen atom, is $\pu{5.29e-11 m}$ or $\pu{52.9pm}$. This is a difference in scale by a factor of $10^5$.

Therefore, a reasonable assumption for the $\ce{^2He}$ nucleus — and indeed for practically any nucleus — is that it can be approximated by a dimensionless point of $+2$ charge and not any sphere, ovaloid, double-sphere or other three-dimensional object. Therefore, if we want to use quantum mechanics to calculate the wave function of $\ce{^2He+}$, we assume a spheric potential caused by a positive charge in the origin of a coordinate system; the value of the positive charge is $2e$ with $e$ being the elementary charge.

A two-electron system cannot, to the best of my knowledge, accurately be calculated. However, we may approximate it as just taking the wave function solutions of $\ce{^2He+}$ and adding another spin-opposite electron. (Then, we slightly adjust the energy values to account for electron-electron repulsion.) This way of approaching the problem shows that the helium atom should really better be considered a central positively charged dot with the outer space occupied by the negatively charged electrons, each of which see a $+2$ charge in the centre and an additional negative charge ‘buzzing around them’ (Note: physically incorrect macroscopic picture).

Replacing $\ce{^2He}$ (as is present in your question) with either of the stable isotopes $\ce{^3He}$ or $\ce{^4He}$ would not change the picture; the size of a neutron (see my note in the second paragraph on subatomic scales) is roughtly that of a proton. If anything, adding one or two neutrons to the two protons will make the nucleus smaller since neutrons increase the intranuclear attractive forces.

In many introductory chemistry books, the core and shell model of an atom and the relevant size differences are exemplified by macroscopic comparisons. My chemistry book offered: if you enlargen an atom so that it would fit the Munich Olympic Tower inside it, the nucleus would inflate to the size of a cherry stone, match head or something of that size. This again shows how good the assumption of poiut-sized charges in the nucleus is.


Fundamental equation of thermodynamics for magnetocaloric materials

The fundamental equation for paramagnetic system is as follows
$$dU=TdS+BdM+\mu dN$$
Where B is the external magnetic field intensity and M is the magnetic moment (Here I have neglected the P-V work of the system). For your chemically pure and non-reactive system the above equation simplifies to
$$dU=TdS+BdM$$
Using Legendre Transformations:
$$y(0)=U(S,M)$$
$$dy(0)=dU=TdS+BdM$$
$$dy(1)=dA(T,M)=-SdT+BdM$$
$$dy(2)=dG(T,B)=-SdT-MdB$$
Now changing the order of the equation, we can find enthalpy,
$$y(0)=U(M,S)$$
$$dy(0)=dU=BdM+TdS$$
$$dy(1)=dH(S,B)=-MdB+TdS$$


Does SN2 favor strong nucleophile and SN1 favor weak nucleophile? Does E2 favor a strong base, and E1 favor a weak base?

That's not the right view.

The $S_{N}2$ mechanism does not favor strong nucleophiles. That statement doesn't make sense at a high level. Mechanisms can't favor anything.

The key here is to look at what controls reactivity in $S_{N}2$ and $S_{N}1$ reactions.

The former is concerted where the bond breaking and bond forming take place at the same time. Therefore, the strength of the nucleophile is important as is the propensity of the leaving group to leave.

In the latter mechanism, there are two steps, bond breaking and bond forming, where bond breaking is the slower and more difficult step. Since the nucleophile is not involved in that first step, the strength of the nucleophile simply doesn't matter.

Similar considerations apply when looking at elimination reactions.


It is not true that heat exchanged at constant pressure is always reversible.
But, if you want to determine the change in entropy from thermodynamic equilibrium state 1 at $(T_1,P)$ to state 2 at $(T_2,P)$, you need to forget entirely about the actual irreversible process path that took you from state 1 to state 2.  It is of no further use.  You instead need to focus exclusively on the two end states.  And, in order to apply the entropy equation, you need to devise a reversible path between these exact same two ends states.  Any reversible path will do, because they will all give the same value for the entropy change.
In the case of your example, you can choose a constant pressure reversible path for convenience. So, for such a path, from the first law we know that $\mathrm{d}H = \mathrm{d}q$. But, now how do we figure out a reversible path between the two states in which the system temperature is changing reversibly along the path?  If we contact the system with at constant temperature reservoir held at the final thermodynamic equilibrium temperature $T_2$, that path would not be reversible because there would be a finite temperature difference between the system and the reservoir over most of the path. But, now, suppose we contact the system, not with a single constant temperature reservoir, but with a sequence of constant temperature reservoirs, each at a slightly different temperature over the range $T_1$ to $T_2$. This would guarantee that, over the entire path, the system would differ only slightly from the reservoir it is currently in contact with. So, in this case, $\mathrm{d}q_\mathrm{rev} = \mathrm{d}H = C_p\,\mathrm{d}T$, where $\mathrm{d}T$ is the change in the system temperature over the present increment of heat exchanged. We use the equation
$$\mathrm{d}H = C_p\,\mathrm{d}T$$
because of the precise definition of constant pressure heat capacity in thermodynamics:
$$C_p\equiv\left(\frac{\partial H}{\partial T}\right)_P$$
So now, for the change in entropy, we have:
$$\Delta S = \int{\frac{\mathrm{d}q_\mathrm{rev}}{T}} = \int_{T_1}^{T_2}\frac{C_p \,\mathrm{d}T}{T}$$
So the key thing to remember here is that, if you want to determine the change in entropy for an irreversible process, you first need to determine the initial and final end states (say, using the first law of thermodynamics) and then devise a reversible path between these same two end states to calculate the integral of $\dfrac{\mathrm{d}q_\mathrm{rev}}{T}$ for that path.


Why is the overall change in entropy of Photosynthesis positive?

To answer this question, we need to be clear whether we are talking about the change in entropy of the system, the surroundings and the Universe.
With regards to the entropy of the Universe, the answer is direct: photosynthesis is a process we observe to happen on a macroscopic scale. Therefore, by the second law of thermodymamics, photosynthesis must cause the entropy of the Universe to increase, period.
With regards to the entropy of just the chemical reagents, if we consider those as our system, then indeed the entropy of formation of glucose is negative, equal to approximately $\mathrm{-40\ cal\ mol^{-1}\ K^{-1}}$.
The way to reconcile these facts is to note that there is a significant increase in entropy which comes from the destruction of high-energy visible photons from the Sun, whose energy is scattered into chemical bonds and multiple low-energy far infrared photons.
In a simplified analysis, if three photons of $\mathrm{700\ nm}$ light are absorbed to generate one molecule of glucose and excess energy is radiated away with black body spectrum at $\mathrm{300\ K}$, then photosynthesis is allowed by the second law of thermodynamics, so long as the photosynthetic efficiency is less than 88%. The observed efficiency of photosynthesis using red light is on the order of 59%, so this requirement is cleared with room to spare.
Depending on the level of detail you want, it gets complicated to work through all the checking and balancing of entropy changes at every individual microscopic step, but the overall results must hold as stated above.
References:
Negative entropy and photosynthesis. Wesley Brittin and George Gamow, Proceedings of the National Academy of Sciences, 1961, 47 (5), 724-727. DOI: 10.1073/pnas.47.5.724
Entropy balance in photosynthesis. Wolfgang Yourgrau and Alwyn Van Der Merwe, Proceedings of the National Academy of Sciences, 1968, 59 (3), 734-737. DOI: 10.1073/pnas.59.3.734
Entropy production and the Second Law in photosynthesis. Robert S. Knox and William W. Parson, Biochimica et Biophysica Acta (BBA) - Bioenergetics, 2007, 1767 (10), 1189-1193. DOI: 10.1016/j.bbabio.2007.07.004


Why does sodium hydroxide react with aluminum oxide?

If you use a "strong acid" to break down the passive oxide you need to be careful which acid you choose.  Sulfuric or nitric acid is oxidizing and thus could maintain the oxide.  Hydrochloric acid would be best, among common mineral acids, for actually getting to the metal while being relatively easy to handle.


Calculating enthalpy per gram mole of oxygen bond in haemoglobin

The heat flows from each Hb molecule during the reaction at constant p and T, so $q_{p, rxn} = \Delta H_{rxn}\cdot\Delta X$.

The energy of the reaction is $\Delta H_{rxn}$ as per mole of reaction.

The reaction itself is Hb + 4O$_{2}\rightarrow$ Hb$\cdot$(O$_{2}$)$_{4}$.

The reaction is exothermic.

The number of moles of Hb present is $$n_{\rm Hb} = {5\over 64,000} = 0.078\;{\rm mmol},$$

all of which reacts: $\Delta X = -\Delta\,n_{\rm Hb} = n_{\rm Hb}$.

That much heat raises the temperature of the water (assuming all the heat is contained in the water):

$$0 = \Delta H_{\rm water} + q_{p, rxn} = C_{p}\Delta T + q_{p, rxn}$$
from which it follows that
$$q_{p, rxn} = -C_{p}\Delta T = -100\times 4.18\times 0.031 = -13\;{\rm J}$$
and
$$\Delta H_{rxn} = {q_{p}\over\Delta X} = {-13\;{\rm J}\over 0.078\;{\rm mmol}}= -167\;{{\rm kJ}\over{\rm mol-rxn}}$$
There are 4 moles of O$_{2}$ used so
$$\nu_{\rm O_{2}} = 4\;{{\rm O_{2}}\over{\rm rxn}}$$
and on an oxygen basis
$$\Delta H_{rxn} = -42\;{\rm kJ}$$


Your reaction product is a double bond isomer of a diazirine.

Halogene-substituted derivatives of the latter are typically prepared via oxidation of amidines with hypochlorites (Graham oxidation).

Heating your hydrazide in an aqueous solution of a mineral acid will only turn it to the corresponding carboxylic acid.


Ammonium is a weak acid but isn't it considered a strong conjugate acid as a salt's ion?

Just because something is a weak base does not mean its conjugate acid has to be a strong acid. Likewise, just because something is a weak acid does not mean that its conjugate base is a strong base.

The only connection between acidity of an acid and basicity of a conjugate base, henceforth labelled $\ce{HA}$ and $\ce{A-}$, is that they are related by the ion product of water, $K_\mathrm{w}$ by the following equations:

$$\begin{align}\ce{HA &<=> H+ + A-}&\quad K_\mathrm{a} &= \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\tag{1}\\[0.8em]
\ce{A- + H2O &<=> HA + OH-}&\quad K_\mathrm{b} &= \frac{[\ce{OH-}][\ce{HA}]}{[\ce{A-}]}\tag{2}\\[0.8em]
K_\mathrm{a} \times K_\mathrm{b} &= \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \times \frac{[\ce{OH-}][\ce{HA}]}{[\ce{A-}]}\\[0.6em]
&= \frac{[\ce{H+}][\ce{A-}][\ce{OH-}][\ce{HA}]}{[\ce{HA}][\ce{A-}]}\\[0.6em]
&= [\ce{H+}][\ce{OH-}] = K_\mathrm{w}\tag{3}\\[1.2em]
\Longrightarrow \mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} &= 14\tag{4}\end{align}$$

Therefore, the sum of the acid and base constants must equal $14$. But if $\mathrm{p}K_\mathrm{a} = 6$, then $\mathrm{p}K_\mathrm{b} = 8$ — meaning the conjugate base of a weak acid is a weak base. This basically mirrors ammonia’s case.


How do manufacturer remove oxygen from metal during melting?

They use flux, that's how. That is, an inert substance which melts easier than the metal and floats on top, preventing it from oxidation in the air. Common fluxes for aluminum seem to be based on eutectic mixtures of various chlorides (say, $\ce{NaCl + KCl}$).

Covering your crucible with a lid does not offer complete protection, and working in inert atmosphere is much more costly (though sometimes they do that, too).


a) Is this a valid rule of thumb: avoid intramolecular proton
  transfers unless the transition state has 5 or 6 members in its ring?

Yes

  b) What's more significant with regard to proton transfer - proton
  tunneling or physical proton transfer? Or is the answer an "it
  depends"?

It depends, at temperatures around absolute zero, tunneling can become significant.  At ambient temperatures or above, physical proton transfer is much more common.

As to the methyl mercaptan anionic example you cite, to the extent that the various carbon-sulfur double bond resonance structures contribute, I would expect a lowering of the barrier around carbon anion planarization - and consequently a lowering of the barrier to inversion at the carbon atom in the anion.  Therefore, it is not clear to me that tunneling of the sulfur bound proton would be preferred to carbon inversion in order to bring about the syn-anti interconversion, especially since any presumed experiments to generate the anion would not be performed near absolute zero.


Boiling points of Butan-2-ol and Butan-1-ol

You have certainly understood the main idea.

In general the boiling point of a substance is determined by the way the molecules come together. The closer they are, the bigger are the intermolecular forces making it harder for one molecule to leave the entire group.

In this particular case the molecules are similar, since both substances have a hydroxyl group, thus being to able to create H-bonds. But in the case of 1-butanol, the molecule is linear while 2-butanol is not. This stereochemical difference affects the way the molecules attract each other. The linear molecules are able to come closer, thus creating stronger bonds or in other words raising the boiling point, because a higher amount of energy has to be applied in order for one molecule to obtain enough, to leave the liquid phase.


This generates an R−O−SOClR−O−SOCl molecule, and the resonance-stabilized O−SOCl−O−SOCl− ion is a good leaving group. This is the substrate that the Grignard can perform an SN2 type reaction on.

I could be wrong about this, because I always doubt everything that I do, but I don't think that's quite right.

Here is a good discussion about thionyl chloride works.  It should turn your alcohol into an alkyl chloride, and now that chloride is your leaving group for the SN2 reaction.

Knowing that, we can work retrosynthetically.

Your product is octane.  You get to the octane by reacting something with ethyl magnesium bromide (a great nucleophile), so that's a strong indication that you did a substitution reaction.  If you add on an ethyl group and get octane, then your substrate was an ethyl group shorter, so it would have contained a hexyl chain.  You got to that hexyl species by treating an alcohol (as it was described in the initial question) with SOCl2, and you know that SOCl2 turns alcohols into chlorides, so your hexyl species that you treated with EtMgBr was 1-chlorohexane, which you obtained by treating hexan-1-ol with SOCl2.


What did Richter mean regarding an acid base reaction?

I am afraid you are mixing all modern and old concepts. Moles and the concept of limiting reagents did not exist in Richter's time.

  It took 615 parts by weight of magnesia (MgO), for example, to neutralize 1000 parts by weight of sulfuric acid

This relatively famous statement has nothing to do with law of multiple proportions but it illustrates the concept of equivalents. All this observation shows is that it always took the same amount of X to react with Y. Hence the ratio of X/Y remained constant.

If we take 1000 parts of sulfuric acid as a standard, then 615 parts by weight of MgO will neutralize it. It will take 672 parts of ammonia to neutralize the same amount of acid.

Note carefully this concept does not even need the knowledge of atomic weights.

Addendum
Also note that Richter's values are incorrect as shown in the comments i.e. if we use moles. Most likely the reason was purity of his magnesia and sulfuric acid. The inaccuracy of his values was realized in the late 1860s but we should respect his idea of constant ratio of reactants. I don't know enough German to check his entire three volumes of books nor have access to them. However a table from 1860s shows the comparison of corrected values (which are still slightly off):

An Introduction to Chemical Philosophy, according to the Modern Theories, by Adolphe Wurtz, 1865.

"It will be noticed that for many bodies Fischer's numbers are widely different from the theoretical figures, and consequently the analyses of Richter from which
they are calculated were not nearly so accurate as those of Wenzel. The inaccuracy of these analyses, and the obscurity of a perplexed explanation, have not been much noticed in giving Richter credit and some authority to
his works."


What would carbon-oxygen polymer be called?

The generic name is oxocarbon. In particular, "Carbon suboxide spontaneously polymerizes at room temperature into a carbon-oxygen polymer, with 3:2 carbon:oxygen atomic ratio." This is not quite the formula you provide, but the polymer has been synthesized.


How do you calculate the total heat capacity of a reaction

Two things are wrong in your calculation. Well, it's really just one thing, but the other one bugs me more.

The total mass of the solution.
Units.

1. Mass of the Solution

To calculate the mass of the total solution, I think it is fair to make the following assumption:
$$ \rho_\ce{HCl} \approx \rho_\ce{NaOH} \approx \rho_\ce{NaCl} \approx \rho_\mathrm{H_2O} \approx 1000~\mathrm{kg\, m^{-3}}$$
$$ V_\text{tot} = V_1 + V_2$$

It says that we assume the densities of all the solutions to be the same, and that the total volume is the same as the sum of the two initial volumes.

Using the above assumptions, we can calculate the total mass of the reaction mixture:
$$ m_\text{tot} = \rho V_\text{tot} = \rho (V_1 + V_2) = 1~\mathrm{g\, cm^{-3}} \times (200 ~\mathrm{cm^{3}}) = 200~\mathrm{g}$$

You have calculated the reaction enthalpy to be $\Delta H = -57.36~\mathrm{kJ\,mol^{-1}}$.

Now we have to calculate the total heat that was released during this reaction, which leads me to

2. Units

Using unit analysis (we have $\mathrm{kJ\,mol^{-1}}$ and need $\mathrm{J}$) we know that we need to multiply the reaction enthalpy with some amount of substance (unit: $\text{mol}$). Now that we know what we have to plug in, we need to know how much of it. Since the reaction enthalpy that you calculated is per $\text{mol}$ of reactant, and we have exactly $0.25~\mathrm{M} \times 0.2~\mathrm{L} = 0.05~\mathrm{mol}$ of reactants, we calculate the total heat as follows:
$$ q_\text{tot} = n_\text{tot} \times -\Delta H_\text{reaction} = 0.05~\mathrm{mol} \times -57.36~\mathrm{kJ\,mol^{-1}} = 2.868~\mathrm{kJ}$$
Why the $-\Delta H$? Well, the heat that goes out from the reaction goes into the mixture. That is why the sign has to change here.

We also need the temperature difference, which was measured to be $\Delta T = 3.38~\mathrm{K}$.

The final result (the heat capacity) is calculated via
$$ q_\text{tot} = m_\text{tot}c_\text{tot}\Delta T \Rightarrow c_\text{tot} = \frac{q_\text{tot}}{m_\text{tot} \Delta T} = \frac{2.868~\mathrm{kJ}}{200~\mathrm{g} \times 3.38~\mathrm{K}} = 4.243~\mathrm{J\, g^{-1}\, K^{-1}
}$$
This is a little bit more than the specific heat capacity of water, so we're good!

I hope my walk-through has demonstrated all crucial points of where you went wrong, and that it clarified things for you.


Determine minimum C-H bond length in ethane derivatives

One way to compare $\ce{C-H}$ bond lengths is to compare their stretching frequencies using IR spectroscopy. However, since ethane and ethylene are gases at ambient conditions, we are going to have some difficulties in accuracy. Yet, it is worth comparing. Since we are comparing $\ce{C-H}$ bond in each case, we can suggest that the stronger the stretching frequency of $\ce{C-H}$ bond is, the shorter the bond should be.

1,2-Dibromoethane is a liquid at room temperature and $\pu{760 torr}$ with boiling point in the range of $\pu{131-132 ^\circ C}$. 1,2-Dibromoethene is from Aldrich is a mixture of cis- and trans-isomers, which boils at $\pu{110 ^\circ C}$ under $\pu{754 torr}$ pressure. IR spectra of both are extracted from NIST.gov webbook and illustrated as follows:

1,2-Dibromoethene: Range of $\ce{C-H}$ stretching is around $\pu{3150-3050 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{3100 cm^{-1}}$):

1,2-Dibromoethane: Range of $\ce{C-H}$ stretching is around $\pu{3010-2950 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{3000 cm^{-1}}$):

Accordingly, we can easily suggest that $\ce{C-H}$ bond in 1,2-dibromoethane is longer than that in 1,2-dibromoethene. However, this is hardly a surprise since we are comparing a $\mathrm{sp^3-s}$ bond versus a $\mathrm{sp^2-s}$ bond. Nonetheless, this is a good exercise because it confirm the validity of the test, in a way. Yet, the $\ce{C-H}$ stretching at around $\pu{3000 cm^{-1}}$ is undoubtedly suggested that  $\ce{C-H}$ bond in 1,2-dibromoethane is shorter than a $\ce{C-H}$ bond in a regular alkane.

Now, let's see two good liquid alkane and alkene examples to represent gasious ethane and ethene, respectively.

2-Methylpent-1-ene (a terminal alkene) is a liquid at room temperature and $\pu{760 torr}$ with boiling point of $\pu{62 ^\circ C}$. n-Pentane (the shortest n-alkene at ambient condition) is a liquid at room temperature and $\pu{760 torr}$ with its boiling point around the range of $\pu{35-36 ^\circ C}$.

2-Methylpent-1-ene: Range of $\ce{C-H}$ stretching is around $\pu{3100-2800 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{3075 cm^{-1}}$):

2-Methylpent-1-ene: Range of $\ce{C-H}$ stretching is around $\pu{2990-2800 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{2900 cm^{-1}}$):

Thus, we can confidently suggest that the $\ce{C-H}$ bond lengths of given compounds are in following order:

ethane > 1,2-dibromoethane >ethene > 1,2-dibromoethene


Addition of hydrogen bromide to 1,3-butadiene: thermodynamic and kinetic control

Thermodynamic and kinetic control

This is a classic example of the concept of thermodynamic versus kinetic control of a reaction. Take a look at this energy profile diagram.1 The horizontal axis is a reaction coordinate, and the vertical axis represents Gibbs free energy.

In this scenario, the starting material $\ce{A}$ can react to form either $\ce{B}$ or $\ce{C}$. The formation of the product $\ce{C}$ has a lower activation energy, which means that it will form faster:

$$E_{\mathrm{a},\ce{AB}} > E_{\mathrm{a},\ce{AC}} \qquad \Longrightarrow \qquad k_\ce{AB} < k_\ce{AC}$$

If we keep the temperature sufficiently low, the molecules of $\ce{C}$ - which are inevitably formed faster - will probably not have enough energy to overcome the reverse activation barrier. The forward reactions $\ce{A->B}$ and $\ce{A->C}$ are, under such conditions, effectively irreversible. Since the formation of $\ce{C}$ is faster, it will predominate, and the major product formed will be $\ce{C}$. This is known as kinetic control; $\ce{C}$ is the kinetic product.

Now, what happens if we heat it up? At higher temperatures, $\ce{C}$ is still going to be the product that is formed faster. However, it also means that all the reactions will be reversible. This means that molecules of $\ce{C}$ can revert back to $\ce{A}$ - and since the system is no longer limited by temperature, it will try its best to minimise its Gibbs free energy, which is the thermodynamic criterion for chemical equilibrium. This means that, as the most thermodynamically stable molecule, $\ce{B}$ will be predominantly formed.2 The reaction is said to be under thermodynamic control, and $\ce{B}$ is the thermodynamic product.

A simple definition is that the kinetic product is the product that is formed faster, and the thermodynamic product is the product that is more stable. This is precisely what is happening here. The kinetic product is 3-bromobut-1-ene, and the thermodynamic product is 1-bromobut-2-ene (specifically, the trans isomer).

A disclaimer

Note that not every reaction has an energy profile diagram like this, and not every reaction has different thermodynamic and kinetic products! If the transition state leading to the formation of $\ce{C}$ were to be higher in energy than that leading to $\ce{B}$, then $\ce{B}$ would simultaneously be both the thermodynamic and kinetic product. There are plenty of reactions in which the more stable product (thermodynamic) is also formed faster (kinetic), so do not assume that every reaction fits this mould!

The reaction mechanism

The first step is the protonation of one of the $\ce{C=C}$ double bonds. In butadiene (1), both double bonds are the same, so it does not matter which one you protonate. The protonation occurs regioselectively to give the more stable carbocation:

The more stable cation is not only secondary, but also allylic, and therefore enjoys stabilisation via resonance (or conjugation). This is depicted in the resonance forms 2a and 2b above.

This allylic carbocation, more properly denoted as the resonance hybrid 2, has two carbons which have significant positive charge, and the bromide ion (here denoted as $\ce{X-}$) can attack either carbon. Attacking the central carbon, adjacent to the site of protonation, leads to the kinetic product 4; attacking the terminal carbon, distant from the site of protonation, leads to the thermodynamic product 4.

There are some people who write that 3 results from attack of $\ce{X-}$ on resonance form 2a, and 4 from attack of $\ce{X-}$ on resonance form 2b. This is not correct! Resonance forms do not separately exist, and they are not distinct species that rapidly interconvert. As such, one cannot speak of one single resonance form undergoing a reaction.

With that out of the way, let's examine why 4 is the thermodynamic product, and why 3 is the kinetic product.

The thermodynamic product: trans-1-bromobut-2-ene

It is perhaps simple enough to see why 4 is more stable than 3. It has an internal, disubstituted double bond, and we know that as a general rule of thumb, the thermodynamic stability of an alkene increases with increasing substitution. So, compared to the terminal, monosubstituted alkene 3, 4 is more stable.

Both the trans isomer 4 as well as the cis isomer 5 can be formed via attack of the nucleophile at the terminal carbon, and both are disubstituted alkenes. However, the trans isomer 4 is more stable than the cis isomer 5, because there is less steric repulsion between the two substituents on the double bond. As such, 4 is the thermodynamic product.

The kinetic product: 3-bromobut-1-ene

There are quite a few explanations on the Internet,3–6 and not all of them are correct.

The worst possible argument, which I have thankfully not seen yet, goes something along the lines of: resonance form 2a, being an allylic secondary carbocation, is more stable than resonance form 2b, which is an allylic primary carbocation. Therefore, resonance form 2a exists in greater proportion, and the nucleophile preferentially reacts with it, leading to the formation of 3.

However, this is patently incorrect, as discussed earlier! Individual resonance forms do not exist, and on top of that, such an argument suggests that we are looking for the more stable intermediate. In fact, we should be looking for the more stable transition state. The carbocation is an intermediate, and not a transition state.

The most common argument that I have seen is actually: since resonance form 2a is more stable than 2b, it contributes more towards the resonance hybrid 2. As such, the positive charge on the internal carbon is greater than the positive charge on the terminal carbon. The nucleophile, being negatively charged, is more strongly attracted to the more positively charged or more electrophilic carbon, and therefore attack there occurs faster (the transition state being stabilised by greater electrostatic interactions).

That's actually a very sensible explanation; with only the data that has been presented so far, we would not be able to disprove it, and it was indeed the accepted answer for quite a while.

The catch

In 1979, Nordlander et al. carried out a similar investigation on the addition of $\ce{DCl}$ to a different substrate, 1,3-pentadiene.7

This experiment was very ingenious, because it was designed to proceed via an almost symmetrical intermediate:

Resonance forms 7a and 7b are both allylic and secondary. There is a very minor difference in their weightages, arising from the different hyperconjugative ability of $\ce{C-D}$ vs $\ce{C-H}$ bonds,8 but in any case, it is not very large. Therefore, if we adopt the explanation in the previous section, one would expect there not to be any major kinetic pathway, and both 1,2- and 1,4-addition products (8 and 9) would theoretically be formed roughly equally.

Instead, it was found that the 1,2-addition product was favoured over the 1,4-addition product. For example, at $-78\ ^\circ\mathrm{C}$ in the absence of solvent, there was a roughly $75:25$ ratio of 1,2- to 1,4-addition products. Clearly, there is a factor that favours 1,2-addition that does not depend on the electrophilicity of the carbon being attacked!

The authors attributed this effect to an ion pair mechanism. This means that, after the double bond is protonated (deuterated in this case), the chloride counterion remains in close proximity to the carbocation generated. Immediately following dissociation of $\ce{DCl}$, the chloride ion is going to be much closer to $\ce{C-2}$ than it is to $\ce{C-4}$, and therefore attack at $\ce{C-2}$ is much faster.

In fact, normal electrophilic addition of $\ce{HX}$ to conjugated alkenes in polar solvents can also proceed via similar ion pair mechanisms.9,10 This is reflected by the greater proportion of syn addition products to such substrates.11

Notes and references

1 Smith, M. B. March's Advanced Organic Chemistry, 7th ed., p 272

2 This does not mean that all of $\ce{A}$ will be converted to $\ce{B}$; the reaction is still an equilibrium, and equilibria always go forward and backward. In general, the minimum system Gibbs free energy ($G_\mathrm{syst}$) will occur at a certain proportion of $\ce{A}$, $\ce{B}$, and $\ce{C}$. However, since $\ce{B}$ has the lowest Gibbs free energy, it will be formed in a greater proportion than $\ce{C}$. See, for example, the colourful graph in this answer of mine.

3 Thermodynamic and Kinetic Control. http://www.masterorganicchemistry.com/2012/02/09/can-opener-economics/ (accessed August 8, 2017).

4 Addition of Hydrogen Halides to Dienes. http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch10/ch10-4-1.html (accessed August 8, 2017).

5 Electrophilic Attack on Conjugated Dienes-Kinetic and Thermodynamic Control. https://chem.libretexts.org/Core/Organic_Chemistry/Conjugation/Electrophilic_Attack_on_Conjugated_Dienes/Electrophilic_Attack_on_Conjugated_Dienes-Kinetic_and_Thermodynamic_Control (accessed August 8, 2017).

6 1,4-Addition. http://www.ochempal.org/index.php/alphabetical/a-b/14-addition/ (accessed August 8, 2017).

7 Nordlander, J. E.; Owour, P. O.; Haky, J. E. Regiochemistry of the addition of hydrochloric acid-d to trans-1,3-pentadiene. J. Am. Chem. Soc. 1979, 101 (5), 1288–1289. DOI: 10.1021/ja00499a045.

8 Because of the larger reduced mass and lower zero-point energy, a $\ce{C-D}$ bond is stronger and therefore less willing to donate electron density into an adjacent empty $\mathrm{p}$ orbital. This is the origin of some secondary kinetic isotope effects; in our case, it means that 7a is marginally less stable than 7b.

9 Fahey, R. C.; McPherson, C. A. Mechanism of the hydrochlorination of tert-butylethylene and styrene in acetic acid. J. Am. Chem. Soc. 1969, 91 (14), 3865–3869. DOI: 10.1021/ja01042a030.

10 Addition of $\ce{HX}$ to butadiene in the gas phase gives approximately a $1:1$ ratio of 1,2- to 1,4-addition product, suggesting that an ion pair mechanism (which would favour the 1,2-addition product) does not operate. See: Mascavage, L. M.; Chi, H.; La, S.; Dalton, D. R. Surface-catalyzed hydrochlorination of alkenes. The reaction of the gases hydrogen chloride and 1,3-butadiene. J. Org. Chem., 1991, 56 (2), 595–601. DOI: 10.1021/jo00002a021.

11 For more details and references to primary literature, see Carey, F. A., Sundberg, R. J. Advanced Organic Chemistry - Part A: Structure & Mechanisms, 5th ed., pp 478–482.


$\ce{H2SO4}$ is one of common strong acids, meaning that $\ce{K_{a(1)}}$ is large and that its dissociation even in moderately concentrated aqueous solutions is almost complete.

Arrhenius dissociation:

$$\ce{H2SO4 <=> H+ + HSO4-}~~~~~~~~~~\ce{K_{a(1)}}=\ce{large}$$

Brønsted-Lowry Dissociation:

$$\ce{H2SO4 + H2O <=> H3O+ + HSO4-}~~~~~~~~~~\ce{K_{a(1)}}=\ce{large}$$

This accounts for the vast majority of protons donated by the acid. However, since it is diprotic, you may want to take into account the second dissociation, which is technically weak but has a larger $\ce{K_a}$ than many weak acids.

Arrhenius 2nd Dissociation:

$$\ce{HSO4- <=> H+ + {SO_4}^2-}~~~~~~~~~~\ce{K_{a(2)}}=1.2\times10^{-2}$$

Brønsted-Lowry 2nd Dissociation:

$$\ce{HSO4- + H2O <=> H3O+ +{SO_4}^2-}~~~~~~~~~~\ce{K_{a(2)}}= 1.2\times10^{-2}$$

This second dissociation may need to be taken into account for some calculations, but it is negligible in concentrated solutions.


How to explain the excited states in the dinitrogen cation?

Excited states make things a bit more complicated. The electron configuration only represent the HF ground state, and changing the occupied orbitals will have an effect on all the orbitals (they are correlated). So this picture becomes more and more inaccurate the higher lying the electronic state. What you do to resolve this is expand the electronic state in many configurations (which all most belong to the same IRREP).

For diatomic molecules there is a lot of spectroscopic data available. So let's check it: $\ce{N2+}$ in the NIST Chemistry WebBook

The states in that table are ordered by energy, with the ground state at the bottom. So scrolling down shows us the 3 lowest lying states: $\rm X ^2\Sigma_g^+$, $\rm A^2\Pi_u$, $\rm B^2\Sigma_u^+$.

Basically, this means you are right about the first state. To explain why we need a bit of symmetry. The first excited state has $\Pi_u$ symmetry, in the $D_{\infty h}$. Since we are working with $D_{2h}$ here, we will need to reduce it. Sparing the details, the result is $\Pi_u$ becomes $B_{2u}\oplus B_{3u}$, which means it will show up once in each IRREP, which is fine because $\Pi$ states are doubly degenerate anyway (just as $\pi$ orbitals are doubly degenerate).

The other state ($\Sigma_u^+$) becomes $B_{1u}$. This one you can for example get by excitation from the anti-bonding $2\sigma_u$ into $3\sigma_g$, with respect to the ground state configuration (see this reference). So there is an electron "jumping over" the $1\pi_u$ orbitals.

Note, that the ordering of the states can change if you start to move away from the equilibrium distance towards dissociation.


Does aniline react with diazonium ions at C or N?

This is a really interesting question and the answer is that the reaction of benzenediazonium chloride with aniline is a bit different to most of the reactions of benzenediazonium salts in that the initial product is compound 1, diazoaminobenzene. It is possible to run the reaction to isolate diazoaminobenzene prep here.

These diazoaminobenzene compounds are unstable with respect to reversion to a diazonium salt + nucleophile, and so many references suggest that compound 2, 4-aminoazobenzene, is directly produced. The procedure for the transformation of diazoaminobenzene 1 to 4-aminoazobenzene 2 is by heating to 50 °C with aniline (prep here).


Following is the list of solvents previously checked for the solubility (From reference 1 below):

$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\textrm{Solvent} & \textrm{Temperature, } \pu{^{\circ}C}  & \textrm{Solubility, } \pu{\%} (w/w) \\ \hline
 \textrm{Water} & 25 & 2.2  \\ \hline
 \textrm{water} & 100  & 66.7  \\ \hline
 \textrm{Ethanol} & 25 & 1.2 \\ \hline
\textrm{ Ethanol} & 60 & 4.5 \\ \hline
\textrm{Ether} & 25 & 0.3 \\ \hline
\textrm{Ethyl Acetete} & 25 & 2.5 \\ \hline
\textrm{Chloroform} & 25 & 18.0 \\ \hline
\textrm{Acetone} & 25 & 2.0 \\ \hline
\textrm{Benzene} & 25 & 1.0 \\ \hline
\end{array}
$$

My suggestion is to use ethanol in $\pu{60 ^{\circ}C}$. However, I'm not sure how ethanol would behave with the presence of tannins. For example, in a separate literature report, $\pu{1.0 mg}$ of tannin was isolated from a $\pu{g}$ of grape skins using 100% ethanol while the amount was increased to about $\pu{5.5 mg}$ when solvent was switched to 100% acetone (Ref.2).

References:

Q. V. Vuong, P. D. Roach, “Caffeine in Green Tea: Its Removal and Isolation,” Separation & Purification Reviews 2014, 43(2), 155-174 (DOI: 10.1080/15422119.2013.771127).
M. O. Downey, R. L. Hanlin, “Comparison of Ethanol and Acetone Mixtures for Extraction of Condensed Tannin from Grape Skin,” South African Journal for Enology and Viticulture 2010, 31(2), 154–159 (DOI: https://doi.org/10.21548/31-2-1412).


I am sorry to give my opinion here but I feel compelled to do so!

Security officers or the people writing the safety data sheet are motivated by only one thing: they don't want to be held liable! For them, everything is deadly and they don't even try to mitigate that (true) fact according to their audience.

As a result, all the risks are presented at the same level and nobody can really adjust his/her practices according to that.

I vividly remember one of my students at Oxford University. The rule there is(was?) that the student should copy the whole safety sheet on a form submitted to the Safety Officer before he/she can even get the chemical. My student had to do that for a ridiculous amount of cyanide and he was so stressed out that he ended up shaking and putting cyanide everywhere! I mean on the balance, around the balance, on the floor etc... I had to be besides him "It's OK, it's no big deal" while making signs to the others to not come around because there was cyanide everywhere.

So @Heather, I really advise you to ask your co-workers about the true risks in your field. You won't die because you had a whiff of chloroform or methanol, even though you should always work under a hood so you don't have a whiff of nasty materials. Also the safety sheets must be adapted to your own risks.


Half cell method of voltage calculation in an electrochemical cell

Think about what state $\ce{Cu^0}$ is in. $\ce{Cu^2+}$ is dissolved in the solution, but is this true of the unionized copper? Try to write out the half-reactions with the physical states included. What activity (effective concentration) do solids have in an equilibrium expression?

Also, though there are some reactions that could happen involving the acid, and I don't know whether they actually contribute to the cell potential in this case, the way the question is written, it looks like the two half-cells you have are the significant ones.


Aromatic Apparent Substituion Reaction Mechanism

The question is slightly misleading. It is usually referred to as a substitution; nucleophilic aromatic substitution to be precise; but it is actually an addition-elimination. What the question is trying to say is that it is not a regular $S_N1$/$S_N2$ substitution (which I think you correctly realised anyway).

The mechanism involves attack at the carbon attached to the bromine to form an intermediate anion which is stabilised by the electron withdrawing acetyl group, para to the site of attack.

I'll let you draw the arrows and come up with a complete mechanism.


How do we find the angular, radial nodes and the quantum numbers when the radial probability distribution curve's equation is given?

For a given orbital, there are two types of modes: radial nodes and angular nodes.

The total number of nodes is $n-1$
Number of angular nodes is $l$
Number of radial nodes is $n-l-1$

In this question: If the given function represents the radial part of the wave function of an atomic orbital, then we get the number of radial nodes: $n-l-1=2$. i.e.  $n-l=3$:

We have several possibilities:

$n=3$, $l=0$: then we have the orbital $\ce{3s}$
$n=4$, $l=1$: then we have the orbital $\ce{4p}$
$n=5$, $l=2$: then we have the orbital $\ce{5d}$

and so on.


Of course, you shouldn’t start from $\ce{H2C+-O-}$ unless you note that it is a mesomeric form of $\ce{H2C=O}$ — the more relevant mesmeric structure.

The mechanism as you wrote it and as it is typically written is given without any counterion, i.e. using just $\ce{OH-}$ and not $\ce{NaOH}$. You could instead write the mechanism using a counterion for $\ce{OH-}$ — $\ce{Na+}$, which would result in the first step being:

$$\ce{H2C=O + NaOH -> H2C(ONa)(OH)}\tag{1}$$

This is essentially the compound you are suggesting. Hence it exists, but it is a minor constituent of the overall solution. It’s not part of the second step, it is the first step. For the second step to happen, you need another molecule of $\ce{H2C=O}$, which has not yet reacted to the geminal diolate $\ce{H2C(ONa)(OH)}$.

See this scheme for the transition state of the Cannizzaro reaction and where $\ce{H2C(ONa)(OH)}$ comes into play:


How to remove dichlorometane from water?

There are two concepts here which should not be mixed:

The solubility of dichloromethane in water is indeed quite low, around 2% as in the question. But this is not the azeotrope, which is formed when there is 1.5% water - 98.5% dichloromethane mixture. That indeed boils at $38^\rm{o} \rm{C}$. See this. It's worth mentioning that you can't possibly bring the azeotrope to $40^\rm{o} \rm{C}$ - because it will boil at a lower temperature by definition, and the vapour will have exactly the same composition as the liquid.

Because what you have isn't the azeotrope, I'd say boiling your mixture (which will most likely happen near the boiling point of pure water) will remove the dichloromethane content.


What are the properties of COₓ where x ≥ 3?

At least the trioxide $\ce{CO3}$ exists, and it has even been detected in at least two isomeric forms.  Wikipedia reports that it forms as a transient species in reactions between carbon dioxide and atomic oxygen or ozone, in irradiation of solid carbon dioxide, and even in oxidation of carbon monoxide by $\ce{O2}$.