How to correctly prepare a standard solution?

The reason you are using a weighing glass is because it is a lot smaller and it has a much smaller mass than a volumetric flask. These glasses usually are more stable, less affected by temperature, and pressure changes, or drafts in the laboratory. The smaller surface also tends to not accumulate much dirt due to static, your hands, etc.

If you were using method one, then you could skip the part with the weighing glass altogether. This method suffers from getting a stable reading as due to the size it is more affected by the conditions in the laboratory.
Method two is certainly the better method, because the total error tends to be much smaller. When you are using such equipment, then your concentrations are typically smaller than 1 mol/L and you are working with small quantities of solid substances. Continuously rinsing out the weighing vessel will ensure to transfer almost all of your solute.
If you were to produce a saturated solution, then the mass of the solute would not matter in the first place.

If your compound does not readily dissolve then the attempted concentration is too high to be used for analytical chemistry and you should rather start from a saturated solution. All primary standards are easily dissolved. If you are not dissolving a primary standard, then your should check the concentration against one.

Therefore, method one can be used if accuracy is of a minor concern (leave out the weighing glass) because it has to be determined again with a primary standard. Method two should be used when preparing accurately concentrated solutions of primary standards.


The zinc-carbon cell uses a zinc anode and a manganese dioxide cathode. It is manganese dioxide that takes part in the reaction, not carbon. Carbon is added to the cathode to increase conductivity and retain moisture(catalyst). It also contains a layer of ammonium chloride. Following is the overall reaction.
$$ \ce{Zn(s) + 2 MnO2(s) + 2 NH4Cl(aq) → Mn2O3(s) + ZnCl2 (aq) + 2NH3(aq)} $$
This is why the half reactions are as follows,

Anode: $\ce{2Zn -> Zn^{2+} + 2e–}$

Cathode: $\ce{2NH4+ + 2MnO2 + 2e– -> Mn2O3 + H2O + 2NH3}$

An electrolyte is a substance that produces an electrically conducting solution when dissolved in a polar solvent, such as water. Here, ammonium chloride solution acts as the electrolyte immobilized in the form of a paste. Basically its a chemical medium for flow of charge.


Is there any emerged master principle in chemistry, like the theory of evolution in biology?

You may be asking for something quite a bit deeper, and it's rather hard to strictly adhere to the requirements in the question, but after Eric's interesting edit showing some outside-the-box thinking, I went along those lines and came up with some options. I'll let you decide which (if any) make the cut.

Two good candidates for master principles in chemistry may be acid-base theory and reduction-oxidation theory. I think ultimately they're just somewhat coarse measures of the relative positions between HOMOs and LUMOs for different species, but after a couple centuries of research and some tabulation, we've come up with general top-level rules for understanding the chemical behaviour of large classes of compounds, with significant predictive power, without necessarily resorting to complicated calculations.

The periodic table itself could also work; different elements possess certain chemical similarities which allows them to be grouped with one another. Mendeleev famously used this to predict a handful of new elements, such as germanium (though he did also predict some other elements which never came to be).  A related principle which is simple but of great importance is that elements are entirely characterized by the number of protons in their nuclei. Every single atom in the Universe with 79 protons in its nucleus is a gold atom, no matter how many neutrons and electrons are present.

Also, I wonder if you'd count materialism as a master principle which emerged in chemistry; every material in the Universe is made of atoms (not counting dark matter). If two materials are the same, then they must be made of the same types and proportions of atoms in the same configurations. If two materials are different, then either the atoms are different, or they are present in different proportions, or they are present in different configurations.

Finally, I myself would actually have included the octet rule as a master principle. It has many holes and limited applicability, but it is uncanny that such a simple concept works at all. I find it a surprising way to effectively skip tons of gritty quantum mechanical calculations, helping chemistry education to begin in school around ages 14-15.


What product is formed in the Wurtz reaction of 1-bromo-3-chlorocyclobutane?

Br is a much better leaving group than Cl.First of all presuming that Cyclobutane is stabilized by substitutent groups is important since it wil gradually rearrange itself to buta diene.But since it is a theoretical question hence  we can say that it should add to the Br side since Br leaves more efficiently being larger in size than Cl and hence having less efficient orbital overlap with Carbon as well as a large electronic repulsion due to the large number of electrons it has is the reason behind this.And the major product will probably be a bicyclo [1.1.0] butane since it is the kinetic product.
so basically your answer-


In Flame Emission Spectroscopy, what part of the flame is used to measure the concentration of the analyte?

I have read the article but it was never mentioned what part of the flame does the transducer or sensor was placed on.

It was never mentioned because no sensor or transducer is actually placed on or in the flame. What spectroscopist do is that they focus the light (coming from the flame) into a device called the monochromator. The job of the monochromator is to separate the wavelengths of the light from the flame. The desired wavelength is then focussed on the detector. Thus the detector is really far from the actual flame.

Now the key question is which part of the flame is useful. There is no single correct answer. The criterion is we scan the height of the flame and check which part of the flame is showing the maximum intensity of the desired signal. This is done by changing the height of the burner. Most spectrometers do it automatically but old lovely instruments had knobs for manual adjustment.

Normally for most elements one would like to see the hottest part of the flame, where the concentration of the atomic emission is the highest.


Activity of different positions in phenol with respect to electrophilic aromatic substitution

As the figure below illustrates, substituents on aromatic rings that are electron donating like the methyl and t-butyl groups (and also the hydroxy group [e.g. phenol]) increase the reaction rate at all positions (ortho, meta and para) relative to benzene.

On the other hand, electron withdrawing substituents (the chlorine and ester carbonyl examples) decrease the reaction rate at all positions relative to benzene.

So for phenol, or any other electron donating substituent, while the meta position does react more slowly than the ortho and para positions, it still  reacts faster than benzene.  Therefore, it would be incorrect to say that the meta position is deactivated. Only when the rate at a given position is less than the rate for a single position in benzene could one say that the position is deactivated.

Statement "a" in your question is correct.

(image source)

For future reference a note on terminology. The above figure tells us that nitration of toluene proceeds 24.5 times faster than the nitration of benzene; that is to say that the relative rate for nitration of toluene is 24.5 times greater than benzene.  The meta position in toluene reacts 3 times faster than a single position in benzene, this is referred to as the partial rate factor.


Your equation is wrong. In acidic medium, water is the one which gives an oxygen to the nitrite. And that should be an oxidation half-reaction (as nitrite oxidizes to nitrate), but you wrote a reduction where two electrons are spent, not released.


How to separate sulfur and carbon powders?

Two possibilities:

Distillation: Sulfur has the comparatively low boiling point of 718 K (445 °C). Heating in apparatus not affected by sulfur (which leaves out iron) will leave behind the carbon. This is, after all, how sulfur winds up at volcanic vents, where it is harvested. You might even create a retort out of concrete, which would simulate volcanic rock.
Sulfur is somewhat soluble in toluene and in carbon disulfide, but both are toxic and flammable, with $\ce{CS2}$ being particularly easy to ignite. These are not safe to use without proper training and equipment.


Which component acts as the electrophile in the synthesis of the Sudan-1 azo dye?

It was mentioned in the comments but I shall mention it again.

Attack does not have to mean nucleophilic attack. In your case, the attack is electrophilic, the electrophile phenyldiazonium attacking the naphtholate.


Is there a way to "electroplate" wood with copper?

The most common way to apply metals by electrodeposition to nonconductive materials is to apply a "strike" of underlying metal, usually nickel or copper, via a method like electroless plating. In electroless, the electrons for the reduction of the metal ions to the zero-valent state are supplied by a reducing agent in solution:

$$
\begin{align}
\ce{Red &-> Ox + 2 e^-} \\ \\
\ce{Ni^{2+} + 2 e^- &-> Ni^0}
\end{align}
$$

This "strike" coating is then used to provide the electrical connection necessary to apply the metal of interest—here, copper—by typical galvanic electrodeposition methods.


What caused this reaction between chlorine and brake fluid?

Pool Chlorine is calcium hypochlorite and brake fluid is polyethylene glycol. Here is a proposed mechanism for the reaction.


Why don't metals form covalent bonds in bulk metal?

This is due to the low ionization energies of the metals. It's easier for them to release few electrons from the outer shell to obtain a noble gas configuration rather than consuming several ones. However, the difference between an ionic and a polar covalent bond is always fringe.

Again, it's not true that metals don't form covalent bonds at all. I guess you never heard of quadruple bond or δ-bond. There are several examples such as $\ce{K2[Re2Cl8]·2H2O}$ and Chromium(II) acetate hydrate.


How does the Rf value change when changing the eluent composition?

The most important interaction is that of the mobile phase with the stationary phase. If you increase the mobile phase’s polarity, that interaction will increase, decreasing the interaction of the solute with the stationary phase. Therefore, a higher ethyl acetate content in the mobile phase will increase $R_f$ values.


Using sodium peroxide to scrub carbon dioxide emissions

The problem with carbon capture is not so much that we lack the technology to do it, it's more that we lack the technology to do it in a way that is economically feasible.

Take your example - according to the US Energy Information Adminstration, it takes about 1.09 lbs of coal to generate 1 kWh of electricity. A typical coal-fired plant might produce an average of 500 MW throughout each day (this is a rough estimate), which works out to roughly 12,000 MWh, or 12,000,000 kWh. So that's something like 12 million pounds of coal each day.

If we assume that coal is 100% carbon (close but not quite), then it's about 37.8 mole C per pound of coal. That's 453.6 million moles of carbon each day being converted to $\ce{CO2}$, and since the combustion of carbon follows:

$\ce{C + O2 -> CO2}$

that means a single medium sized-coal plant produces roughly 454 million moles  of $\ce{CO2}$ each day. By now you are probably getting a sense of the scale of this problem. Since sodium peroxide and carbon dioxide have a 1:1 ratio in the absorption equation:

$\ce{2Na2O2 + 2CO2 -> 2Na2CO3 + O2}$

you will need an equal number of moles of sodium peroxide. That works out to about 78 million pounds per day.

Alibaba lists sodium peroxide at around \$400-500 per ton, so that means it would cost you (at a minimum) \$15.6 million dollars per day to scrub the $\ce{CO2}$ emissions from one medium-sized power plant. The real costs would be much higher because of the additional equipment and operating expenses.

Electricity currently sells for about \$0.10 to \$0.20 per kWh in the US, and around half of that is usually for production, with the rest being delivery. That means our 500 MW coal plant is going to be bringing in something like (at the high side) \$1.2 M each day. As you can see, the costs for scrubbing with sodium peroxide far exceed the potential revenue for the plant.

Now, there are other options:

One is to reuse the sodium peroxide, although the costs of regenerating it might exceed just buying more, and you would still need to find a place to store the $\ce{CO2}$.

Another is to use a cheaper/more effective absorbent - according to Wikipedia, amine-gas scrubbing is the most promising technology so far, although I am sure it faces the same kinds of cost scaling problems.

Another more popular approach is carbon capture, where carbon dioxide is basically just pumped into underground reservoirs in the hopes that it will stay there. While this is probably going to turn out to be the most economical approach, it has the risk of releasing large amounts of $\ce{CO2}$ accidentally, as well as having unknown effects on the local geological formations.

Yet another option is to use photosynthesis to scrub carbon dioxide. This has the advantage of being very cheap in comparison to other methods, but the downside is that the rate at which plants and algae can consume $\ce{CO2}$ is much slower than the rate at which we produce it.

It's a big problem, and all of the proposed solutions so far have been very expensive to implement - which is the main reason the fossil fuels industry opposes them.


Conversion factor for chemical to physical scale of atomic masses


Osmotic pressure is the additional external pressure that needs to be applied to a solution to compensate its osmotic flow to a second solution, so that there is not net flow between them.

Let's imagine a tube with a piston in each extreme, so we can control the pressure we apply to both sides, with a semipermeable membrane inside it. If the concentration of solutes on both sides is different, there will be an osmotic flow from the hypotonic side to the hypertonic side. To compensate for this osmotic flow, we will need to apply more pressure to the hypertonic side than to the hypotonic side - and in mechanical equilibrium, this additional pressure is equal to the difference in osmotic pressure between the solutions.

Alternatively, if we want to get rid of external pressures, osmotic pressure can be understood not as a pressure exerted on a solution, but by the solution with a certain osmotic pressure on neighbouring solutions with which it can exchange solvent. So, the hypertonic solution exerts more pressure on the hypotonic solution than viceversa, and the result is that solvent moves from the hypotonic solution (which receives more osmotic pressure) to the hypertonic solution (which receives less osmotic pressure) until they become isotonic, just as would happen if it exerted more hydrostatic pressure. This is mathematically equivalent to the original definition.


Tert-butyl hypochlorite, cas no 507-40-4

Reagent for N-chlorination of amines and amides


Now, as I understand it, all processes proceed so as to maximize the "randomness" of its constituent particles. (Oversimplified version of the Second Law of Thermodynamics, yes, I know... just don't chew me out in the comments section...)

Yes. True.

  This Law can easily be observed in, and verified by, natural processes.

Sure. Still with you.

  Now the formation of snow is a natural process, agreed? The way my brain sees it, is that water droplets ought to freeze into random, and by virtue of its "randomness", highly unsymmetrical shapes. But this is not the case here!

Nope. Let's consider a counter argument: chemical bond formation between atoms is a natural process, and natural processes seek to maximise entropy. Thus, all molecules should take on highly unsymmetrical shapes, shouldn't they?  The answer is obviously no. Same applies for snow flakes. First of all, "order" and "symmetry" aren't precluded by thermodynamics. Molecular and crystal structure emerges within the realms of laws of thermodynamics and quantum mechanics.

Like I said in the comments, the precise discussion of how a snow flake's shape merges will be quite involved. There will be a lot of variables that affect the growth of snow crystals: temperature, vapour pressure humidity, wind speed and direction etcetera.

The final shape of the snow flake would depend on the precise path it followed, and the conditions it experienced as it descended, and is impossible to predict/calculate.

Of course, the underlying six-fold symmetry comes from the arrangement of water molecules in the ice crystal lattice.

Also, there are no natural processes coordinating the growth of the arms to be perfectly symmetrical; their growth is governed by simple statistics.

The arms of a snow crystal all grow independently, and grow under randomly fluctuating conditions; however, the effects of these fluctuations are somewhat averaged over the entire snow flake as it tumbles through the sky, and you end up with a somewhat regular shape.

There is in fact, asymmetry in snow flakes. Just take a close, hard look at a photograph

Sure there is some symmetry: the snow flake has 6-major arms, and then each arm has sub-branches, and each sub-branch branches out further. The overall branching pattern is the same, because of random averaging--the snow flake randomly rotates and reorients as it falls, each arm feels (on average) the same conditions. But are each of the sub-branches or the sub-sub-branches identical in shape, number, size etcetera? No.


You will need to start in basic conditions. Dimethyl malonate is a good nucleophile on account of the two ester groups on either side of the central carbon atom. It has a $\mathrm{p}K_\mathrm{a}$ of 12.9, which is extremely acidic for a $\ce{C-H}$ bond. The first step is a Michael addition.

The resultant enolate will equilibrate, enabling it to form a much more stable 6 membered ring. We then have a Claisen condensation.

Lastly, decarboxylation will yield dimedone.


Correlation Diagram Doesn't Match Selection Rules?

No, it's not an error. Take a look at the Tanabe Sugano diagram for $d^2$ (from Wikipedia):

You can see that again, the $^1D$ state is below the $^3P$ state. Why? If we were considering a $p^2$ configuration, the $^3P$ is indeed lower in energy than $^1D$. And again anyone might predict that $^3P$ is lower for $d^2$ too, for the exact reasons you give.

We know this result from spectroscopy. The difference is because of the Racah parameters for the different terms (i.e., the electrostatic repulsion between electrons in the same orbital. The Wikipedia article had a nice link to how you'd compute the difference in this case.

Now of course a $^3F \rightarrow ^1D$ transition is not an allowed transition, which is shown in the textbook on the next page.

The moral of the story is that Hunds' rules and the basics will always get you the lowest energy term symbols for multi-electron atoms, but the details can be tricky for higher-level states.


Will equilibrium ever establish if one of the products starts to disassociate?

Let's make this an abstract problem with species A, B, C, and F:
$$\ce{A <=> B + F}\tag{1}$$
$$\ce{B <=> C + F}\tag{2}$$
We can also write down the sum of the two reactions, which will also be at equilibrium:
$$\ce{A <=> C + 2 F}\tag{3 = 1+2}$$
And let's use a for the concentration of A and so on.
Finding the solution using reactions 1 and 3
The concentration of F is equal to the equilibrium constant for reaction (1), so the concentrations of A and B have to be equal:
$$a = b$$
For every mole of B made, we also get a mole of F. For every mole of C made, we get two moles of F. We know the concentration of F is 2.5 M, so we get:
$$b + 2c = 2.5 M$$
Finally, we started with 4 M of A, which makes equimolar B and makes equimolar C, so the sum of all three has to be 4 M:
$$a + b + c = 4 M$$
Now we have three equations and three unknowns. We can combine the first and third equation to eliminate $a$:
$$ 2b + c = 4M$$
We can multiply the second equation in preparation for solving for $c$:
$$ 2b + 4c = 5 M$$
Subtracting the latter equation gives the solution for $c$:
$$ c = \frac{1}{3} M$$
$a$ and $b$ come out as 1.833 M.
Le Chatelier

Lets say the first reaction reached equilibrium and then $\ce{N_2O_3}$ started decomposing. This means more amount of $\ce{O_2}$ will be produced and the equilibrium for the first reaction will get disturbed

Without the second reaction, the first will attain equilibrium when $a$ is about 1.85 M. The second reaction makes F and uses up B, so the first reaction almost stays at equilibrium, and the concentration of A does not change much.
Will it ever stop?

Due to this the oxygen level will drop and equilibrium for the second reaction will get disturbed and $\ce{N_2O_3}$ will decompose further to form oxygen.
But this process will go on forever. There is definitely wrong with this. Will this really happen or is there some other mechanism at work?

If you look at this step-wise (figure out how much one reaction would go, disregarding the other), you are never done. This is similar to Zeno's paradox. However, the corrections get smaller and smaller, and your approximation gets better and better.
In general, reactions approach equilibrium, they don’t reach it. Nothing special about this system.


Is the molar mass of a molecules lowered by the binding energy of the chemical bonds?

The difference is the best observed for products of nuclear reactions, related to nucleon binding energy due the strong force.

By comparing the mass of $\pu{2 mol}$ of $\ce{^1H2}$ and $\pu{1 mol}$ of  $\ce{^4He}$, the difference is about $0.7\%$.

The chemical bonds have difference more then million times smaller.

E.g. if there is an exothermic reaction producing $\pu{500 kJ/mol}$, the mass difference per mol is $$\Delta m = \Delta E/c^2 = \pu{-5.56 ng }$$

That means, we have hardly ever scales with the needed accuracy.


What is the entropy of mixing a gas into the whole atmosphere?

Performing an actual calculation could be quite complicated, even if we use the ideal gas approximation, because the atmosphere has enormous variations in both temperature and density.

To simplify the problem, let's think of the atmosphere as an ideal gas of uniform temperature and density.   Then we need to determine how the entropy of the argon released into the atmosphere increases as a result of this release.

Since we're doing an ideal gas approximation, the argon doesn't "know" about the other atmospheric gases – the only thing that affects its entropy is the increase in volume.  But, when doing an entropy calculation, we do need to account for the fact that the atmosphere already contains 0.93% Ar.   Effectively, the presence of the existing argon removes a fraction of the configurational possibilities, since if we replace argon with argon, we don't change the entropy.   Hence the entropy increase is calculated from the effective configurational volume that remains for the argon relased from the box.

Specifically, for the purpose of calculating entropy, the effective final volume for the argon released from the box is:

$$V_\mathrm f = V_\text {atmosphere} - V_\text{Ar in atmosphere}+ V_\text {Ar in box}\approx V_\text {atmosphere} - V_\text {Ar in atmosphere} = (1- x_\text {Ar})V_\text {atmosphere},$$
where $x_\text {Ar}$ is the mole fraction of argon in the atmosphere, which is 0.93% = 0.0093.

Proceeding, we start with the following general expression for a pure substance (again, the ideal gas approximation allows us to ignore the non-argon molecules):

$$\left(\frac{\partial S}{\partial V}\right)_{\!T} =\frac{\alpha}{\kappa},$$
where $\alpha$ is the coefficient of thermal expansion, and $\kappa$ is the isothermal compressibility.

For an ideal gas, $\alpha = \frac{1}{T}$ and $\kappa = \frac{1}{p}$, hence:

$$\left(\frac{\partial S}{\partial V}\right)_{\!T} =\frac{p}{T}$$

If we additionaly assume the expansion is isothermal (i.e., T of the box is the same as that of the atmosphere--which we've already assumed is of uniform T) we obtain, for an ideal gas:

$$\mathrm dS = \frac{p}{T}\,\mathrm dV = \frac{n R}{V}\,\mathrm dV \Rightarrow \Delta S = n R \int\limits_{V_\mathrm i}^{V_\mathrm f}\frac{\mathrm dV}{V} = nR \ln\frac{V_\mathrm f}{V_\mathrm i}=nR \ln\frac{0.9907 \,V_\text {atmosphere}}{V_\mathrm i}$$

where $V_\mathrm i$ is the initial volume available to the argon, and $V_\mathrm f$ is the final volume available to the argon.

As you can see, releasing a mole of argon into the atmosphere will result in a significant entropy increase, once the argon is fully dispersed.

Note that, in the case where the atmosphere is 100% Ar:
$$V_f = V_\text {atmosphere} - V_\text{Ar in atmosphere}+ V_\text {Ar in box}=V_{\text{Ar in box}} \Rightarrow \frac{V_f}{V_i}=1\Rightarrow \Delta S = 0, \text{ as expected.}$$

We can also calculate the corresponding free energy change using:

$$\Delta G = \Delta H - T\,\Delta S$$

With the ideal gas approximation, there is no enthalpy of mixing, hence:

$$\Delta G = -T\,\Delta S = -nRT \ln\frac{V_\mathrm f}{V_\mathrm i}.$$

If we assume STP ($273.15\ \mathrm K$, $1\ \mathrm{bar}$), and take the volume of the atmosphere to be the area of the earth's surface times, say, 50 kilometers*, then we have:

$$\Delta S = (1 \ \mathrm{mol})\ R\ \ln\frac{2.53 \times 10^{22}\ \mathrm L}{22.7\ \mathrm L} = 0.403\ \frac{\mathrm{kJ}}{\mathrm K}$$

$$\Delta G = - (1 \ \mathrm{mol})\ R\ (273.15\ \mathrm K) \ln\frac{2.53 \times 10^{22}\ \mathrm L}{22.7\ \mathrm L} = -110\ \mathrm{kJ}$$

*This is all arbitrary, since we're unrealistically assuming constant pressure and temperature; but I needed to pick a number for the calculation, so I chose the top of the stratosphere ($50\ \mathrm{km}$).  $94\ \%$ of the atmosphere's gases are found below this altitude, so about $94\ \%$ of the argon would end up there as well.

Then again, because of the logarithm, the choice of atmospheric volume doesn't significantly affect the calculation, as you can see if you compare my answer with Buck Thorn's. Indeed, even if we allow the argon to expand to fill the observable universe ($4 \times 10^{83} L$), this only increases $\Delta S$ and $\Delta G$ by a factor of four! [$\frac{83}{22}\approx 4$.]

Calculation done in Mathematica (in Mathematica, ln is represented by "Log"):


The energy level diagram which you give and is to be seen in the standard textbooks is assumed to apply to all elements. This is not the case, nor would we expect it to be so. After calcium the 3d drops below the 4s so fills first. Coulombic repulsions tend to push electrons up into the larger 4s orbital where repulsion is less. This means the 4s electrons are the outer electrons which are lost first in chemical reactions and define the atomic radius of the atom.


Confusion regarding mechanism of dehydration of alcohols

I think your mechanism is reasonable.
Coming to rate of dehydration,
In first case Alpha hydroxy ketone will have the least rate of dehydration because a carbocation is formed which is destabilized by Ketone group and even notice there is hydrogen bonding in reactant making reactant more stable and thus rate of dehydration is decreased in first case.
In second case, though carbonation is not stabilized,  here breaking of H will be easier because the H are more acidic being next to electron withdrawing group , so 2 has highest rate .
In third case, since carbocation is not so unstable and even the H is not much acidic. So  3 has moderate rate of dehydration.
Therefore,  rate of dehydration:2>3>1


How can a β,γ-unsaturated ketone be converted to 1-pyrroline?

I suggest two possible approaches. Narasaka, et al. have effected similar cyclizations photochemically. I prefer to start with the allylic alcohol 1 to avoid complications with oxime formation. Formation of the bis acetate 2b is not a problem. However, given the greater acidity of oximes relative to alcohols does not preclude the formation of oxime acetate 2a. In the photolysis the naphthalene serves as a sensitizer while 1,4-cyclohexadiene functions as a hydrogen atom donor. The addition of hydrogen is likely from the convex face as shown. If not, after saponification of acetate 3b and Swern oxidation, base equilibration would likely lead to ketone 4.
Alternatively, a Barton-type radical cyclization (5 --> 3a) might be considered.
ADDENDUM: Direct photolysis of a structure similar to 5 without Bu3SnH has been achieved. See ref 2.

1) M. Kitamura, Y. Mori and K Narasaka, Tetrahedron Letters, 2005, 46, 2373.
2) F. Gagosz and S. Z. Zard, Synlett., 1999, 1978.


Confusion regarding specific conductivity

Let's take a wire inside cable for example. That wire's conductance is just the inverse of the resistance this cable is making when electricity passes through it. This depends on things like the length, what is made of, the maximum/minimum width of certain areas of the wire, etc.

On the other hand, the conductivity of that wire is a direct property of the material it's made of. If the wire is made of copper for example, it'll have (at the same temperature) the same conductivity as any copper wire in the world (because they are all made of copper), but not the same conductance (because they are not physically equal)

I hope this helps, I'm sorry if I was wrong!


Simple way to understand why chromophores often have double bonds?

No. Without going into quantum mechanics and molecular orbital theory, there is no way to understand why chromophores typically have double bonds.

Everything related to colour is related to the excitation of electrons from one orbital to another and therefore any explanation that does not rely on orbitals in at least some way is doomed to fail.

If you decide to accept a little MO theory then you can be helped. Double bonds are formed by an underlying σ bond (or single bond) and a π bond. Each bond is formed by two atomic orbitals (one from each atom) that combine to give a bonding and an antibonding orbital. The energy difference between the bonding and antibonding orbital corresponds to the excitation energy: a photon that carries exactly that energy will be absorbed and the electron excited to the higher level. The greater the overlap between the atomic orbitals, the higher the energy difference between bonding and antibonding.

Now in π orbitals, the atomic p orbitals are oriented perpendicular to the bond axis, so any overlap is doomed to be much smaller than the corresponding single bond’s σ overlap. Therefore, the energy difference from the bonding to the antibonding π orbital is not far into the ultraviolet region; typical values would correspond to a wavelength of $150$ to $200~\mathrm{nm}$. Extending the π network will move those orbitals closer together energetically (again, the complete answer requires much more MO theory) which is why benzene has an energy difference corresponding to a wavelength of $250~\mathrm{nm}$. Thenceforth, it is not a long way until you reach visual wavelengths ($400{-} 700~\mathrm{nm}$).


The $4n+2$ rule applies only for a conjugated ring.  If you assume that the amino group contributes a pair of pendant  $\pi$ electrons into the ring system then perforce you do not have a conjugated ring.  And even if you don't assume such a delocalization from the amino group, the cross-link within the ring system is still a stretch for the $4n+2$ rule.


Is there a name, or any specific jargon for a precipitation equilibrium involving a solid reactant?

AFAIK, such a reaction has no name. However, if one were so inclined, one could call it a replacement reaction, because the $\ce{MgCO3}$ is being converted into $\ce{Mg(OH)2}$.

To calculate the "equilibrium point," which I take to mean the concentration of all the ions at equilibrium, we would need to know the $k_\text{sp}$ of both $\ce{MgCO3}$ and $\ce{Mg(OH)2}$, as well as the $\ce{pH}$ of the solution (which gives us $[\ce{OH^-}]$).

The solubility products of $\ce{MgCO3}$ and $\ce{Mg(OH)2}$ are $10^{-7.8}$, which is approximately $1.58 \cdot 10^{-8}$, and $5.61\cdot 10^{-12}$, respectively.

To calculate how much magnesium hydroxide precipitates, we would first need to find the concentrations of $[\ce{Mg^{2+}}]$ and $[\ce{CO3^{2-}}]$ using the $k_\text{sp}$ of magnesium carbonate. This gives us the solubility of $\ce{MgCO3}$, which is $1.26 \cdot 10^{-4} \,\, \ce{mol lit^{-1}}$. That means that we have $[\ce{Mg^{2+}}] = 1.26 \cdot 10^{-4} \,\, \ce{mol lit^{-1}}$ in the beaker.

Now, to find out how much $\ce{Mg(OH)2}$ will precipitate, we need to find the reaction quotient $Q_c$ of the reaction.
$$Q_c = [\ce{Mg^{2+}}][\ce{OH^-}]^2$$
(I cannot calculate $Q_c$ here because it depends on $[\ce{OH^-}]$, which will have to be given.)

If $Q_c > k_\text{sp}$, then the reaction
$$\ce{Mg(OH)_2(s) <=> Mg^{+2}(aq) + 2OH^-(aq)}$$
will proceed in the backward direction and magnesium hydroxide will precipitate out. This will continue until the concentration of magnesium ions decreases to such a value that $Q_c = k_\text{sp}$, at which point the reaction will be at equilibrium. Here, I've assumed that the concentration is too large to be appreciably changed by it precipitating out of the solution. The exact value of $[\ce{Mg^{2+}}]$ as well as how many moles of $\ce{Mg(OH)2}$ will precipitate can be calculated with this assumption. (However, if we want, we can do it without this assumption too. The calculations would become more messy and involve solving a quadratic.)


This answer is mostly a compilation of the excellent comments by Spontification and Chet Miller.

Differential rate law

The overall reaction order tells you what happens to the rate when you change all concentrations by the same factor at the same time. For example, if you decrease all concentrations by a factor 2 and the overall order is 1, the reaction will be slower by a factor of 2. If the order is 2, slower by a factor of $2^2 = 4$, if the order is 3, slower by a factor of $2^3 = 8$ and so on (overall reaction orders higher than 3 are unusual for elementary reactions). Dilution by a factor 2 would be one way of lowering concentrations; however, not all species will decrease by a factor of 2 (solvent, hydronium concentration in a buffer, any other species buffered by a fast equilibrium).

Integrated rate law

Integrated rate laws describe how the rate changes (decreases) over time. All species concentrations that are part of the differential rate law will change over time unless they are catalysts. If one species concentration is very high compared to others (e.g. the solvent) then its change will be negligible. So the integrated rate law depends on more than the differential rate law.

However, if all of the species in the differential rate law are reactants, and these reactants are present at stoichiometric ratios, the overall reaction order is directly related to the integrated rate law (e.g. overall second order reaction gives second-order integrated rate law). If not, you can get, for example, a pseudo-first order reaction even though the overall order of reaction is two.


[...] is there potential heat energy in the temperature of an object?

The energy associated with temperature is called thermal energy. If you know the heat capacity of an object, you can figure out how much energy is required to change the temperature by a certain amount.

  If heat is just a change in temperature, but temperature isn't energy, doesn't that imply that energy is created when two objects of different temperature interact [...]?

Heat is energy transferred from one object to another. When heat is transferred from a hotter to a colder object, the hotter object loses thermal energy and the colder object gains it. No energy is created.

  Where is the energy from heat and other chemical processes coming from, exactly?

A chemical reaction can give off heat because electrons can be in higher states (weak bonds) and in lower states (strong bonds). Potential energy can turn into thermal energy, and vice versa. It takes energy to melt ice, so melting ice cools down the surroundings. An explosive reaction releases energy, so it increases the temperature of the surroundings by transferring heat to it.


In ionic solids such as NaCl, do the cations fill interstital holes in the cation's cubic close pack structure?

The relative radii of the ions, and the stoichiometric ratio of the ions needs to be considered.

For NaCl, the radius of Na+ is 0.695 that of Cl-.

Na+ ions occupy octahedral holes in the cubic close packed lattice of Cl-.

If the radii are more equal (ratio greater than 0.732) then each type of ion will form a simple cubic lattice, with each ion of one lattice filing a cubic hole of the other lattice. For example, CsCl.

If the radii are very different (ratio less than 0.414) the smaller ions will fill tetrahedral holes in the cubic close packed lattice of the larger ions. For example ZnS.

See http://depts.washington.edu/chemcrs/bulkdisk/chem162A_sum06/handout_Lecture_11.pdf for more information.


Additional Questions Regarding the Auto-Ignition Temperature

To understand the answer to (1), you have to think about what combustion actually is. It isn't a single reaction, with a single set of reactants, and a single set of products. It's actually a whole bunch of tiny steps (thousands) that all occur together. Those reactions happen among a whole bunch of unstable radical species (hundreds). When you put a match to a pool of gasoline, all that energy from the match starts tearing apart bonds in the fuel, leading to the formation of radicals. That's what triggers the initial release of energy, and what creates the cascade of ignition. So theoretically, no, you could never reach combustion products, through the pathway of combustion, because you'd never have that activation energy, the big kick of energy needed to create radicals and keep them alive long enough for them to create a chain reaction.

The answer to (2) is, no, it doesn't matter how much fuel. Remember you're measuring a mass-independent property. The autoginition temperature isn't "5 degrees per gram". It's a single temperature. What matters is the ratio of oxygen to fuel.

As to (3), it's a lot easier to operate an open flask than it is to operate a piston-cylinder, and it doesn't have a substantial impact on the air-fuel ratio if you do it correctly.


Should I plate cast iron with copper first before nickel plating?

The standard technique is to copper plate then nickel ,then chromium, if you want chrome. The copper protects the steel then the nickel provides a silver surface color ,normally significantly thinner than the copper.


Is there a way to make a liquid cold by a chemical reaction?

Ammonium thiocyanate and barium hydroxide octahydrate when mixed will draw a huge quantity of heat. This reaction often used as a chemistry demo of endothermic reactions where the chemicals are mixed in a beaker that's placed on a wet piece of wood. The wood freezes to the beaker.


Are these different neutralization pathways?

Both fast reactions happen in parallel, acting as "first comes, first served".

Concentration of water molecules ( $\pu{55 mol/L}$ in water) is much higher than $\ce{[H3O+]}$  for diluted acids, so oxide anions meet water molecules much more frequently.

If the oxide anion meets water molecule, it will not wait for hydronium.
Additionally, hydronium ions would be rather depleted by reaction with hydroxide ions next to the oxide surface.

Be aware that $\ce{Na+}$ ions are just bystander/spectator ions, so it is more illustrative to write equations as ionic equations:

The main reaction:
$$\ce{O^2− + H2O -> 2 OH−}\tag{1}$$

The secondary alternative reaction:
$$\ce{O^2− + H3O+ -> OH− + H2O}\tag{2}$$

The final reaction:
$$\ce{H3O+ + OH− -> 2 H2O}\tag{3}$$


sodium persulfate microwave heating cause crystallized particles

It looks to me like copper sulphate ($\ce{CuSO4·nH2O}$). If I'm right then you can't use it for etching. For this you need oxidant (persulphate). Try to solve this solid in water. You should get light blue solution. Stick into it a nail (i.e some metal iron source). If it cover itself with a brownish powder like in wiki-link you gave, then it's a copper(+2) salt.


How many particles are necessary to observe a well-defined state of matter?

Computationally:

It depends on the material but in general $10^2 - 10^4$ atoms/molecules is a representative range. I learned this concept as the thermodynamic limit; i.e. the number of particles in a system at which increasing particle number does not change density significantly. For obvious reasons it's also called the macroscopic limit. In Statistical Mechanics we bridge the gap between small "countable" systems and macroscopic "mole-quantity" systems with the thermodynamic limit.

Our lab develops and applies models for small molecules like $\ce{H2, CO2}$, noble gases, hydrocarbons etc. for use in computer simulation. The usual method is to parametrize a model such that phase transitions (i.e. density changes as a function of temperature or pressure) are accurately mapped, and intermolecular potentials are comparable to highly accurate QM calculations (CCS(D/T/Q) methods).

One can reproduce accurate phase diagrams of noble gases using just 40-100 atoms by putting the system in a fluctuating periodic box and changing particle positions until an energy minimum is reached (the computational method is called Monte Carlo) under given $T$ and $p$ -- this is called the isothermal-isobaric ensemble, or $NPT$. You can achieve identical results using the $\mu VT$ (grand-canonical ensemble) wherein chemical potential, volume, and temperature are fixed, but $N$ is allowed to change. For professional model development it is more common to use $10^3 - 10^4$ particles for a good fit.

Below is a snapshot of output from a Monte Carlo program obtaining the density of $\ce{He}$ at STP using 100 atoms (the reported value is 0.1786g/L). You can see that $\ce{He}$ in this model is a near-ideal gas with compressibility factor $Z \approx 1$ :

I think the answer is more appropriately expressed via computational (Statistical-Mechanical) methods because the concept of the thermodynamic limit arose in and is used by SM. In general, the central-limit theorem is where the concept came from, which states that fluctuations to the mean of a quantity (density here, $N/V$) are expressed as $\frac{1}{\sqrt{N}}$, thus for $N=10,000$, the mean density can expectedly vary by $0.01$. For $N=N_A$, the variance would be certainly negligible.


Dosing leaves with Zinc Nitrate for pollution experiment - Need specific concentration

It seems to me that it will be difficult to get a uniform concentration of Zn in a non-uniform material like dried leaves, unless you measure only in bulk (this pile of dried leaves weighs X grams and contains 2000X micrograms of Zn). Things get more confusing if the insects don't eat all the dried-leaf structure (leaving stems, for example).

I'm also not sure how you'll get all the zinc from 1 ml of liquid onto a 0.1g leaf sample. (And, of course, the leaf won't be dry for a while after that!) I think a more concentrated solution would make your task easier. Your math for adjusting the concentration to match the amount of leaf matter appears correct.

As an aside, is your question confusing LD50 (micrograms per gram of insect) with leaf concentration (micrograms per gram of leaf)? I'm not sure why the concentration in the leaves would need to correspond to the LD50, unless there's something significant about the insects eating their own body weight in dried leaves...?


Why do laws of thermodynamics "outweigh" any other law ever formulated?

The laws never outweigh each other, nor do they have to. Newton's laws of motion, Maxwell's equations of electromagnetism, Einstein's theory of relativity are all equally fundamental, in that they've never been overthrown within their respective frameworks of applicability. Sure, these frameworks have their limits: Newton fails at near-light speeds where Einstein takes over, Maxwell and Einstein fail at quantum limit. So what? Thermodynamics has its limits as well. If you don't believe this, then good luck trying to apply it to a single molecule.

If anything, the laws of thermodynamics are less fundamental, in that they are true only statistically, that is, with less precision than the rest. With relatively little effort, you can measure the mass of a body to six-digit precision. Can you do the same to its temperature?


Heterogenous equilibrium nernst equation

You can mix activity with fugacity in the expression of equilibrium coefficient or reaction quotient. In fact that is quite common when the reaction involves both solution and gas phase reactants/products.

The definition of the equilibrium constant actually involves dividing the standard pressure $p^\unicode{x29b5}=1\,\rm atm$ or the standard concentration $c^\unicode{x29b5}=1\,\rm mol/L$. In your case, the equilibrium constant will be

$$K^\unicode{x29b5}=\frac{[a(\ce{H+})/c^\unicode{x29b5}]^2}{[f(\ce{H2})/p^\unicode{x29b5}]}$$

which is dimensionless, and is related to the change of Gibbs free energy by $\Delta G=\Delta G^\unicode{x29b5}-RT\ln K^\unicode{x29b5}$.


Copolymer purity, what does AN / MA / MS mean?

I contacted GoodFellow directly and they gave me the following answer:

93.9% AN: 93.9% AcryloNitrile
5.8% MA: 5.8% Methyl Acrylate
0.3% MS: 0.3% Methyl Sulfonate

I guess different manufacturers may name their products as they see fit. In particular, for polymers that can share the same abbreviations, only the context and a good background knowledge of their synthesis pathway may allow one to decide...
There are, however, IUPAC guidelines for polymer abbreviations available here[1]. In addition several other websites offer their own abbreviation list e.g. here and there.

[1]. J. He et al., Abbreviations of polymer names and guidelines for abbreviating polymer names (IUPAC Recommendations 2014), 2014. DOI: 10.1515/pac-2012-1203


What is behind of the attraction between a burned matchstick with a magnet?

Some matchstick heads contain iron(III) oxide as a colorant.

The yellowish color of the burning match indicates that it has low oxygen, i.e. a reducing flame. It reduces the iron oxide to iron which is attracted by the magnet.

The reduction reactions that occur are probably quite complex. Below are two simplified possible reaction equations. $\ce{(CH2)_n}$ in equation (2) stands for paraffin from the splint's impregnation.

$$\ce{Fe2O3 + 3 CO -> 2 Fe + 3 CO2} \tag{1}$$

$$\ce{Fe2O3 + \frac{1}{n} (CH2)_n -> 2 Fe + CO2 + H2O} \tag{2}$$


Dehydration of primary alcohols without a β-hydrogen

In this case, $\mathrm{E2}$ elimination is impossible regardless of condition used, because of lack of $\beta$-hydrogens. However, in acidic conditions, it is possible to have elimination reaction. Since 2,2-dimethylpropanol is a $1^\circ$-alcohol, initial carbocation formation is difficult. However, this formation of carbocations is accompanied by a structural rearrangement, which is the driving force for the $1^\circ$-carbocation formation, which is otherwise not favored:

Such rearrangements take place by a 1,2-shift of a neighboring alkyl group or aryl group or hydrogen, and are favored when the rearranged carbocation is more stable than the initial cation. In this case, migrating group is methyl group and carbocation changed from $1^\circ$ to $3^\circ$, which is very stable compared to initial carbocation and favored. Once $1^\circ$-carbocation formed, it quickly rearranged to $3^\circ$-carbocation and as a result, the reaction make forward progress. It can also be considered a concomitant elimination of water and methide migration where methide group acting as an intramolecular nucleophile (See the mechanism shown in bottom box of the scheme). Either way, the initial step is very slow because of very high activation energy:


How to calculate which ionic compound has the highest melting point?

The strength of the ionic bond depends on Coulomb's law for the force acting between two charged particles where larger force translates to a stronger ionic bond. The equation is

$$F = \frac{-k\cdot q_1 \cdot q_2}{r^2}$$

$k$ is a constant; $q_1$ and $q_2$ are the charges on the ions and $r$ is the distance between the ions.

So the larger the charge the greater the force, and the smaller the radius the smaller the distance between the ions (smaller $r$ in the equation) again leading to a stronger force.

So, first you should calculate the charge on each ion in a molecule. As can see, the charge on strontium ($\ce{Sr^2+}$) and calcium ($\ce{Ca^2+}$) ions is more than the charge on sodium ions ($\ce{Na+}$). Also, calcium ions are smaller than strontium ions. Fluorine ions are also smaller than Chlorine ions. So $\ce{CaF2}$ will have higher bond strength which means it will have higher boiling point since stronger bonds need more energy to break.


How could uranium be extracted from ore to be melt?

There is another process for extraction of Uranium from the uranium ore in the form of UO2(NO3)2 by dissolving the Uranium ore in nitric acid and extract of uranium  using 20% solution of Tributyl phosphate/dodecane. TBP form strong complex with uranium nitrate and thus you can easily remove uranium from other impurities. Obtained UO2(NO3)2 can easily be converted to (NH4)2U2O7 by reacting with ammonia and finally reduction with hydrogen gave you UO2.
You can't obtained pure Uranium because it is susceptible to oxidation easily.


That is indeed a fascinating reaction!  The following is a part answer, providing links to some resources that may be of use.

The following resources may be of assistance:

The Belousov-Zhabotinsky page from the Kansas Academy for Mathematics and Science Education provides a bit about the background, chemical processes and most relevant to your request, an experimental procedure.

Mary Newton of the Georgia Regent University has provided a reasonably detailed procedure on the webpage Procedure for B-Z reaction.

Slightly different, is the information provided in the Belousov-Zhabotinsky Reaction (Leigh) page from the Flathead Valley Community College takes a different approach, with a focus on the mathematical derivations relevant to the experiment.


Chemoselectivity in synthesis of paracetamol

The reaction is likely with one equivalent of acetic anhydride and under thermodynamic control.

Kinetically, the nucleophilic attack of phenol on acetic anhydride should be a lot faster than that of aniline. Aniline’s lone pair is taking part in resonance across the phenyl ring so it is barely able to attack. Phenol has two lone pairs on oxygen, though, of which only one is able to resonate with the aromatic system. Under kinetic control I would thus expect acylation on oxygen.

However, there is a second process which renders us thermodynamic control: The initial product can react with a second molecule in a transacylation reaction.

$$\ce{RHN-C6H4-OAc + H2N-C6H4-OR <=>> RHN-C6H4-OH + HAcN-C6H4-OR}$$

Both the oxygen of a free phenol and the nitrogen of a free aniline (the latter slower, but still) can attack a corresponding acylated atom. In overall thermodynamic stability, the acetylamine is more stable than the acetate (and is also kinetically more inert) so after equilibration has taken place the acetamide will prevail.


Separating NaClO from its aqueous form in bleach

Sodium hypochlorite, $\ce{NaClO}$, is unstable and can't be separated from the water in which you usually find it without it decomposing.

That's the short answer.  To learn more about it check Wikipedia for sodium hypochlorite.


Fermentation in beer (reaction schemes)?

We can break the entire process into three parts:

breakdown of maltose and maltotriose into glucose
glycolysis to give pyruvate
reductive decarboxylation to give the alcohol.

Writing out full equations for each actual step should hopefully not be required; especially glycolysis is explained in detail in many places across the internet and practically every metabolism textbook. Hence I will restrict myself to generic equations for each part of the process. Furthermore, each set of reactions is catalysed by a specific set of enzymes; I will be ignoring these.

Maltose breakdown

$$\ce{\underset{maltose}{C12H22O11} + H2O -> 2 \underset{glucose}{C6H12O6}}\tag{1}$$
Glycolysis

$$\begin{multline}\ce{\underset{glucose}{C6H12O6} + 2 ADP^3- + 2 HPO4^2- + 2NAD+ -> } \\\ce{2 \underset{pyruvate}{C3H3O3-} + 2 ATP^4- + 2 H2O + 2 NADH + 2 H+}\end{multline}\tag{2}$$
Alcoholic fermentation

$$\begin{align}\ce{\underset{pyruvate}{C3H3O3-} + 2 H+ &-> CO2 + \underset{acetaldehyde}{C2H4O}}\tag{3}\\
\ce{\underset{acetaldehyde}{C2H4O} + NADH + H+ &-> \underset{ethanol}{C2H6O} + 2 NAD+}\tag{4}\end{align}$$


Getting the molality of a solution given molarity and density

If a solution of $\ce{HNO3}$ has a molarity of 16 and a density of
    $1.42\rm{g/mL}$, what is the molality of the solution?

Molarity is an old word for amount concentration and it is defined by the IUPAC Goldbook as

  Amount of a constituent divided by the volume of the mixture. Also called amount-of-substance concentration, substance concentration (in clinical chemistry) and in older literature molarity. For entities B it is often denoted by  B. The common unit is mole per cubic decimetre ($\mathrm{mol\,dm^{−3}}$) or mole per litre ($\mathrm{mol\,L^{−1}}$) sometimes denoted by $\mathrm{M}$.

In mathematical terms $$c(\ce{HNO3(aq)})=\frac{n(\ce{HNO3})}{V(\ce{HNO3(aq)}}.$$

   In this case the concentration is presumably given as $c(\ce{HNO3(aq)})=16~\mathrm{mol\,L^{-1}}$

The molality, $b$, is defined by the IUPAC Goldbook as

  Amount of entities of a solute divided by the mass of the solvent.

In mathematical terms this means
$$b(\ce{HNO3(aq)})=\frac{n(\ce{HNO3})}{m(\ce{H2O})}.$$

The (mass) density, $\rho$, is defined by the IUPAC Goldbook as

  Mass of a sample or body divided by its volume.

In mathematical terms this means
$$\rho(\ce{HNO3(aq)})=\frac{m(\ce{HNO3(aq)})}{V(\ce{HNO3(aq)})}.$$

   In this case the density is given as $\rho(\ce{HNO3(aq)})=1.42~\mathrm{g\,mL^{-1}}.$

Now you can go ahead and derive the formula for conversion all by yourself, which I recommend, but you can also check if someone has already done that. I chose to be lazy today, so I checked Wikipedia:
\begin{align}
c &= \frac{\rho b}{1 + bM} &
b &= \frac{c}{\rho - cM}
\end{align}

Now you only have to figure out the molar mass of nitric acid, and you can calculate the result.

   I assume $M(\ce{HNO3})=63~\mathrm{g\,mol^{-1}}$.

Solution:

   You should find $b(\ce{HNO3(aq)})=38.8~\mathrm{mol\,kg^{-1}}$.

So in principle you have done quite good. If you would have used proper notation and included all the units in every step, you would maybe have noticed the mistake all on your own.

(Hover over the hidden fields to reveal their content.)


Why are lead and zinc commonly found together?

Lead and Zinc have in common two electrons "s" in their outer shell. That is why they may sometimes have a similar behavior. You would object that Lead has also two electrons "p" in its outer shell, which are missing in Zinc. It is true. But these p electrons are "modified" and can be "forgotten" by an unusual phenomena in chemistry, namely Einstein's relativity.

When the number of protons in the nucleus is as high as about $80$, and it is $82$ for $\ce{Pb}$,  the nucleus charge is so high that the electron has to move at a relativistic speed to stay rotating around the nucleus, according to Bohr's model. Of course Bohr's model is wrong and the electron is not rotating and has no speed. But the calculations made with it are still valid, as was shown by Pekka Pyykko, Relativity and the Periodic Table, Accounts of Chemical Research, Vol. 12, No. 8 (1979) p. 276 - 281.

At relativistic speeds, the dimensions of all moving objects decrease. And Pekka Pyykko has shown that this effect is proportional to the sum of the two quantum numbers n and l. Lead has the following electronic configuration : $\ce{[Xe] 4f^{14} 5d^{10} 6s^2 6p^2}$. The sum n+l is equal to $8$ for $\ce{4f, 5d, 6p}$. It is $7$ for $\ce{6s}$. So the electrons $\ce{4f, 5d, 6p}$ are so contracted that they are mixed with inner electrons and "disappear" from the periphery. Only the two electrons $\ce{6s}$ , with n+l = 7, remain in the outer shell of the atom. That is why Lead $\ce{Pb}$ often behaves like an atom of the $2$nd column : The most usual oxidation number of Pb is $2$ (like in $\ce{Pb^{2+}}$), and not $4$ (like the $4$th column), which exists but is not frequent. Also $\ce{PbSO_4}$ is insoluble in water, like $\ce{BaSO_4}$. So $\ce{Ba}$ is made of $\ce{[Xe] 6s^2}$, and Pb may be considered as a sort of [pseudo-$\ce{Xe] 6s^2}$...

To go back to $\ce{Zn}$ and $\ce{Pb}$ configuration, relativity explains why $\ce{Pb}$ looks like a noble gas (Xenon) plus $2$ electrons "s" in its outer shell. By comparaison, Zinc is $\ce{[Ar] 3d^{10} 4s^2}$. Like $\ce{Pb}$, it also looks like a noble gas (Argon) plus $2$ electrons "s" in its outer shell, if the filled shell $\ce{4d^{10}}$ is "forgotten". Mother Earth is probably not able to recognize the existence of this $\ce{4d}$ shell and has often mixed $\ce{Zn}$ and $\ce{Pb}$ minerals.


What are the properties of various solutions at pH <2 that may cause injuries?

Injury is not solely related to how strongly acidic a solution is, as there are several mechanisms by which organic matter is damaged from exposure to acids, not just protonation. It is also highly dependent on the type and length of exposure. Aqueous $\ce{HF}$ is likely considerably corrosive after skin contact even at pH 2, mostly because of $\ce{F^-}$ which is a rather reactive ion with a strong tendency to abstract $\ce{Ca^{2+}}$ (in fact, the fact that  $\ce{HF}$ is a weak acid makes it even more dangerous because the undissociated $\ce{HF}$ is much less polar and the molecules can cross the skin barrier more easily, allowing direct access to muscle and bone which can be severely damaged). Aqueous $\ce{HCN}$ also forms weak acids, but again $\ce{CN^-}$ ions are extremely toxic as they block cell respiration. The risk is compounded by the fact that $\ce{HCN}$ is a very volatile substance and can be released as a gas from concentrated or acidic solutions.

With regards to $\ce{HCl}$, you need to be a bit more specific. What kind of injury? $\ce{HCl}$ itself is a relatively safe acid in dilute solution, even though it's a strong acid, as it does not have strongly oxidising/reducing/dehydrating/toxic properties and is only volatile at rather high concentrations. Remember that $\ce{HCl}$ is even used as an integral part of our digestive system, being the species that allows stomach acids to reach pH 1-2.


How does copper reduce dilute nitric acid to nitric oxide and concentrated nitric acid to nitrogen dioxide?

If we examine the Nernst equation: $E = E^\circ-\frac{RT}{zF}\ln Q$, the logarithmic term is what changes with concentration. The reaction quotient $Q$, gets smaller as the ratio of reactants to products increases, meaning that the log term will decrease (and become negative below $Q = 1$), so both equations will become more favourable as the ratio of reactants to products increases.

The correctly balanced second reaction is $$\ce{Cu + 2NO3- + 4H+ <=> Cu^2+ + 2NO2 + 2H2O}.$$

If we look at just the logarithmic term for each reaction, assuming the products part of the reaction quotient is unity and that the proton activity is equal to the nitrate activity:

\begin{gather}
\frac{25.7~\mathrm{mV}}{6}\ln\frac{1}{[\ce{NO3-}]^{10}}\tag{cyan}\\
\frac{25.7~\mathrm{mV}}{2}\ln\frac{1}{[\ce{NO3-}]^6}=\frac{25.7~\mathrm{mV}}{6}\ln\frac{1}{[\ce{NO3-}]^{18}}\tag{red}
\end{gather}

y axis: log term (mV), x axis: $[\ce{NO3-}]$

As expected, when multiplied to normalize the number of electrons transferred, the reaction that consumes more reagents becomes more favourable more quickly as the reagent concentration is increased and less favourable more quickly as the concentration is decreased.

Conceptually, the reason for this is that the reactions travel down a slope towards the equilibrium point (see figure) and the reaction that consumes more of the reagents to produce a given amount of the products involved in the equilibrium reaches equilibrium with fewer reactions. Because ∆G is simply the partial derivative of the Gibbs energy with respect to reaction extent, the reaction that reaches equilibrium with less reaction extent has a steeper slope.


Chemical Kinetics / Rate Question -- Error in Textbook Question/Answer?

The exercise is absurd. None of the proposed answers can be right, as for any answer A,B,C,D the rate is a function of the concentrations [XH4],[O2] but the table suggests that the rate is constant over time, e.g. 2.0$~$M/min for entry 1. Thus:

$$
R = 2.0~\text{M/min} = \text{constant} \cdot [\mathrm{XH_4}]^0 \cdot [\mathrm{O_2}]^0
$$
and is therefore a zeroth order kinetics.

I assume that the initial rate $R_0$ or a rate at some point $R_{t>0}$ is meant, and the question just heavily confuses things. So from here on, let's assume that $R$ means some rate at some point in time.

But even then, it is not possible to answer the question. This is because entries 3 and 4 have the same domain ([XH4] and [O2]) but a different codomain (different rates R). This means that there is no function mapping from the concentration of the reactants to the rate of the reaction.

However, there obviously is such a functional relation for all answers A,B,C,D.

Therefore, the question is absurd. It is probably meant to be that [O2] is 2.0 in entry 4.


Why does a hydride like NaH act as a base, but a hydride like NaBH4 act as a nucleophile?

First of all, borohydride as a group is not same as hydride. So the mechanisms are very different.

Because a simple $\ce{H-}$ anion is more basic than an enolate $\ce{C=C-O^{-}}$(product of $\alpha$ hydrogen deprotonation of a ketone), it just deprotonate before reduction will happen. Deprotonated ketones (enolates) will no longer be a substrate for reduction.

Borohydride $\ce{BH4-}$ is not as basic as enolate, so deprotonation hardly happen. Thus a reduction will be the predominant reaction.


How do we know the concentration of hydroxide ions in this derivation?

In a solution of acetic ions, say from sodium acetate, some acetate will react with water to produce acetic acid and hydroxide anion (your "hydrolysis" equation"):

$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$

Now, let's say you dissolved 1.00 moles of sodium acetate in 1.00 L of water. For the sake of convenience, you would describe this solution as $\mathrm{1.00\ M\ }\ce{CH3COONa}$. However the actual concentration of acetate ion is less than $\mathrm{1.00\ M}$ because of the reaction with water.

Enter $C$, the "total concentration of acetate". This concentration is the sum of all concentrations of species that are related to or derived from acetate. In this case the relevant species are the acetate ion and acetic acid, so $C = [\ce{CH3COO-}]+[\ce{CH3COOH}]$.

What is $\alpha$? $\alpha$ represents that percent dissociation or percent ionization of the acetate ion. Therefore, $C\alpha$ represents the concentration of acetate that has ionization (reacted with water). In this case, since every acetate that reacts with water produces acetic acid, $C\alpha = [\ce{CH3COOH}]$.

How does this derivation lead to $C\alpha = [\ce{OH-}]$? You need to look back at the reaction equation for the process. For each acetate ion that reacts with water, one acetic acid and one hydroxide ion are produced. Based on stoichiometry, the concentrations of acetatic acid and hydroxide ion must be the same.

How do we determine $[\ce{OH-}]$? We plug the three expressions for the three species into the law of mass action and solve for $\alpha$ (we know $C$):

$$K=\dfrac{K_w}{K_a} = \dfrac{[\ce{CH3COOH}][\ce{OH-}]}{[\ce{CH3COO-}]}=\dfrac{(C\alpha)^2}{C-C\alpha}$$

Now, if there was some additional hydroxide $x$ already present, say because the pH was fixed at 12.00, how do we handle that? The expressions for the concentrations of acetate and acetic acid are still valid. $C$ is still the total concentration of acetate + acetic acid. $C\alpha$ is still the concentration of acetic acid. $C-C\alpha$ is still the concentration of acetate ion. However, the concentration of hydroxide ion is different in this case. There is some concentration from the initial conditions $x$, and there is some increase from the ionization of acetate ion $C\alpha$.

If the initial hydorixde concentration is from the autoionization of water, we can account for that:

$$\begin{aligned}
[\ce{CH3COO-}] &= C-C\alpha \\
[\ce{CH3COOH}] &= C\alpha \\
[\ce{OH-}] &= 1\times 10^{-7} + C\alpha \end{aligned}$$

However, except for the most dilute solutions, the contribution from the autoionization of water is usually orders of magtitude smaller than $C\alpha$ and gets lost in the insignificant figures. Let's compare the same calculation for our $1\ \mathrm{M}$ acetate solution. Note that you get the same outcome regardless if you include the hydroxide concentration form $K_w$.

$$\begin{array}{|c|c|c|} \hline
 & \text{without autoionization} & \text{with autoionization}\\ \hline
C & \mathrm{1.00\ M} & \mathrm{1.00\ M} \\
[\ce{CH3COO-}] & C-C\alpha = 1-\alpha & C-C\alpha = 1-\alpha \\
[\ce{CH3COOH}] & C\alpha = \alpha & C\alpha = \alpha \\
[\ce{OH-}] & C\alpha = \alpha & 10^{-7} + C\alpha = 10^{-7} + \alpha \\
K & \dfrac{K_w}{K_a} = \dfrac{10^{-14}}{1.74\times 10^{-5}} & \dfrac{10^{-14}}{1.74\times 10^{-5}}\\
K & \dfrac{\alpha^2}{1-\alpha} & \dfrac{(10^{-7} + \alpha)(\alpha)}{1-\alpha}\\
\alpha & 2.40 \times 10^{-5} & 2.39 \times 10^{-5} \\
[\ce{CH3COO-}] &  1.00-2.40 \times 10^{-5} \approx 1 & 1.00-2.39 \times 10^{-5} \approx 1 \\
[\ce{CH3COOH}] & 2.40 \times 10^{-5} & 2.39 \times 10^{-5} \\
[\ce{OH-}] & 2.40 \times 10^{-5} & 10^{-7} + 2.39 \times 10^{-5} = 2.40 \times 10^{-5}  \\ \hline
\end{array}$$


What is degree of dissociation for polyprotic acids?

I think this resembles the question Confused about strong/weak dibasic and tribasic acids?
Saying that the degree of dissociation $(\alpha)$ of citric acid is greater because its triprotic is not correct. The degree of dissociation should be calculated for each dissociation separately. The degree of dissociation of citric acid's first dissociation $\ce{C6H8O7 <=> C6H7O7- + H+}$ is greater than that of acetic acid, because its $\mathrm{p}K_\mathrm{a}$ is smaller (3.13 vs 4.76)$^1$, meaning it's a stronger acid.
The second value of $\alpha$ for citric acid, however, will be very to close to acetic acid's, because the second dissociation has approximately the same $\mathrm{p}K_\mathrm{a}$ (4.76)$^1$. The third $\alpha$ will be smaller, since the $\mathrm{p}K_\mathrm{a}$ of the third dissociation is greater (6.40)$^1$.
This can be seen in the common polyprotic acids dissociation graphs, like this one (source):

Note that $\alpha$ is represented on the $y$-axis in a 0-1 scale. For each of the four states of the molecule, there is a different $\alpha$, and different values of $\alpha$ happen simultaneously.
$^1$ $\mathrm{p}K_\mathrm{a}$ values taken from Wikipedia.


There is only one source of hydroxyl ions, ionization of water (Eq. 3 in the OP). Ionized water contributes equal amounts of hydroxyl and hydronium ion. Similarly, dissociation of acid $\ce{HA}$ results in equal amounts of hydronium and of conjugate base $\ce{A-}$. Unless a salt of the conjugate base is explicitly added then the acid is the only source of $\ce{A-}$. In absence of salt the total hydronium concentration in solution is therefore

$$\ce{[H+]}=\ce{[OH-]}+\ce{[A-]}$$

Note that this equation reflects the charge balance condition.

Solving for $\ce{[A-]}$

$$\ce{[A-]} = \ce{[H+]}-\ce{[OH-]}\tag{1}$$

We also know that the original amount of acid equals the sum of undissociated acid and conjugate base:

$$C_\ce{HA}=\ce{[HA]}+\ce{[A-]}$$

Note that this equation reflects the mass balance condition.

Solving for $\ce{[HA]}$

$$\ce{[HA]}=C_\ce{HA}-\ce{[A-]} \tag{2}$$

Substituting Eq. (1) into Eq. (2) we obtain

$$\ce{[HA]}=C_\ce{HA}-(\ce{[H+]}-\ce{[OH-]}) \tag{3}$$

The derivation that lead to this expression is not general. It was assumed that no conjugate salt was added to the solution ($C_\ce{A-}=0$). If a completely dissociating salt $\ce{MA}$ is added at concentration $C_\ce{A-}$ then it is necessary to modify the mass balance expression:

$$C_\ce{HA}+C_\ce{A-}=\ce{[HA]}+\ce{[A-]}$$

and the charge balance expression:

$$\ce{[H+]}+\ce{[M+]}=\ce{[OH-]}+\ce{[A-]}$$

or since $\ce{[M+]}=C_\ce{A-}$

$$\ce{[H+]}+C_\ce{A-} =\ce{[OH-]}+\ce{[A-]}$$

Solving for $\ce{[A-]}$

$$\ce{[A-]} = C_\ce{A-} + \ce{[H+]}-\ce{[OH-]}$$

Combining the two balance expressions leads once again to Eq. (3):

$$\ce{[HA]}=C_\ce{HA}-(\ce{[H+]}-\ce{[OH-]})$$

By the way, the quote you cite should read:

"result because the analytical concentration of HA is diminished by the amount of hydrogen ion produced in Reaction (2), which in turn is the total hydrogen ion concentration minus the hydroxyl ion concentration."

It might be that "typo" ("3" written instead of "2") that led to the confusion.


Does the "Bat Gen" claim to produce more energy than it uses?

I'm not sufficiently familiar with the chemistry involved to state authoritatively that this guy is full of bologna, but there are some pretty fantastic claims made in the second link.  Also, if that's a US patent application, it's a disaster.

The device is able to use any light from 900-200nm, mechanical agitation, sound, or magnetic fields (and unspecified "other") as sources of power.
He has also made all of his cells inside Tupperware containers sealed by silicone caulking, so the 100 year claim is a bit sketchy.
The inventor's testing methodology is pretty dubious as well.  The device was placed in sunlight for 30 minutes and it experienced a change of pH from 7.3 to 7.1, therefore it works.  Also, in his own words:

  It is worth saying that such variation from 0.2 to 1.0 pH unit; and the reverse reaction that occurred once the containers were placed in a dark place, was anticipated by our theoric background, when we performed the experiments, we known already what was going to happen, as the matter of fact, we didn't do many experiments, we only made two or three times, obtaining the results we expected.  {All grammar and spelling mistakes are those of the original author(s).}

That said, the idea of a cell that can catalytically break down water into hydrogen and oxygen using sunlight and then recombine them into water, extracting the energy, is not impossible by any stretch of the imagination.  Designing a cell that is economically viable compared with silicon based devices has yet to be accomplished, to the best of my knowledge.


How is energy released when a chemical bond is formed?

It is easier first to think of what happens when a bond is broken in an exothermic reaction. The energy is initially released into vibrations,  rotations and translational motion of the two fragments. This energy is then lost as these fragments collide with other gas molecules or with solvent molecules and so eventually ends up as heat.

In exothermic bond formation the extra energy vibrationally and rotationally excites the new molecule and this is then lost via collisions with surrounding gas or solvent molecules and again ends up as heat.

(If the new molecule were isolated in space (so no collisions) it would remain 'hot' until the energy is radiated away from vibrational levels and until it (radiatively) comes into equilibrium with the surroundings. This is a  slow process, by many orders of magnitude,  compared to collisions in solution or in a gas at normal atmospheric pressure )

In a very few types of reaction an electronically excited state is produced in the product and then photons radiate (as visible light) some of the energy away. This is called chemiluminescence.


Will an iron nail sunk in a highly concentrated salt solution rust?

Rust is iron oxide, $(\ce{Fe2O3.nH2O})$

It forms from iron(III)hydroxide made when iron reacts with oxygen in the presence of water. The chemical reaction can be summarized as follows:

$\ce{4Fe(s) + 6H2O +3O2 ->4Fe(OH)3(s)}$

Now, water 'naturally' contains some dissolved oxygen (8-14%), depending upon the source of the oxygen (photosynthesis from underwater plants, diffusion from the atmosphere) balanced against oxygen demands (such as fish breathing and decomposition of organic matter). The temperature of the water also affects the solubility of oxygen (gases are less soluble at higher temperatures).

The ocean (which is typically 3.5% salt) typically has 20% less dissolved oxygen in it than freshwater. This is because the water molecules are attracted to the dissolved ions from the salt (solvation), which has the tendency to decrease the weak affinity of non-polar oxygen molecules to water, thereby driving dissolved oxygen out.

Now, the salinity of the Dead Sea is about 290 (that's 29% salt!), which is over 8 times the salinity of the ocean!  At such high levels of salinity, most of the oxygen is 'salted out' of the water, which means that the levels of dissolved oxygen are very low, especially below the surface, so an iron nail totally immersed in water with 29% sodium chloride (salt) will not rust.


Does trans-decalin really have a plane of symmetry?

It's not easy to see from a diagram, because it distorts bonds and angles. I recommend building it with a balls-and-sticks model set. You can also use a molecular viewer to model it; there are a couple of open-source (or at least free) ones out there.

I have calculated the molecule on the DF-BP86/def2-SVP level of theory. The point group of the molecule is C2h. In the following model I have highlighted the plane of symmetry, and the rotational C2-axis. At their intersection is an inversion centre Ci.

(I needed to downscale this a lot, click on the image to get to a high resolution still. Images created with ChemCraft, assembled with GIMP.)


The answer is that it is a sliding scale. There is no such thing as "purely polar" or "purely nonpolar." Water is very polar and hydrocarbons are very nonpolar, but there is no clear dividing line between the two categories.

Further confusion in biological molecules such as amino acids can occur, because different individuals categorize them in different ways, considering either just the side chain or the whole molecule. For example the amino acids have the formula $\ce{H2N-CHR-COOH}$ and in the simplest case, Glycine, R=$\ce{H}$. This is clearly a very non polar side chain, but the molecule as a whole is mostly polar, as the side chain is such an insignificantly small part of the molecule. Thus authors who categorize amino acids by side chain (mainly chemists) put Glycine into the non polar category, while authors who categorize amino acids by the whole molecule (mainly biochemists and biologists) put it into the polar category. When it comes to Alanine, R=$\ce{CH3}$ all authors agree it has a non polar side chain and most seem to categorize it as a non polar molecule.

A simple search of the internet shows that there is very little agreement about the amino acid Tyrosine, R=$\ce{CH2-(C6H4)-OH}$. It is classified as polar, non polar, having a polar side chain and having a non polar side chain by different authors. Clearly it has some characteristics of both, having both a benzene ring and a hydroxyl group. The correct definition would be "moderately polar."

You might hope that it is possible to give molecules a number which describes their polarity. This can be done to some extent. Googling "solvents by polarity" will give you lists for common solvents. This is, however, complicated if the molecule of interest happens to be acidic or basic. An acidic molecule will be polar in acidic solution but will become charged (and therefore much more polar) in basic solution, due to loss of $\ce{H+}$. Similarly, a basic molecule will be polar in basic solution but will become charged (and therefore much more polar) in acidic solution due to gain of $\ce{H+}$.

An amphipathic or amphiphilic molecule has a part that is highly polar and a part that is highly non polar. These terms also generally mean not just a molecule that is moderately polar, but one in which there is some distance between the polar and non polar parts, so the molecular weight would be at least into the hundreds.

The example you give of a fatty acid is a good one. (The deprotonated, negatively charged fatty acid anion in basic solution would be an even better one.) Another example would be many cell membrane proteins, which have a non polar, hydrophobic part (that stays in the cell membrane) and polar, hydrophilic parts (that extend out into the aqueous media inside and outside the cell.)

In summary, different substances should not be pigeonholed as polar or non polar as it would require that you draw an arbitrary line, but in general the polarity of two different substances can be compared fairly objectively. And pH may make a difference.


Is there an isotope effect for diffusion of an ion through a crystal lattice?

$F = ma$, so a less massive isotope should move faster, depending on the forces moving it.


How to identify the ratio of oxidation states in terbium ions in its mixed oxide?

There is certainly a formal way. Let $x$ and $y$ be the number of $\ce{Tb^3+}$ and $\ce{Tb^4+}$ ions respectively. Then:

$$\begin{align}
3x + 4y &= 22 \\
x + y &= 6
\end{align}$$

This is a simple system of two linear equations in two unknowns, and I am sure you know of many ways to solve it. Let's just solve it using matrices for the fun of it:

$$\begin{align}
\begin{pmatrix}
3 & 4 \\
1 & 1
\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix} &=
\begin{pmatrix}
22 \\
6
\end{pmatrix} \\
\begin{pmatrix}
x \\
y
\end{pmatrix} &= \frac{1}{3\cdot 1 - 1\cdot 4}
\begin{pmatrix}
1 & -4 \\
-1 & 3
\end{pmatrix}
\begin{pmatrix}
22 \\
6
\end{pmatrix} \\
&= -
\begin{pmatrix}
1\cdot 22 - 4\cdot 6 \\
-1 \cdot 22 + 3\cdot 6
\end{pmatrix} \\
&= \begin{pmatrix}
2 \\
4
\end{pmatrix}
\end{align}$$

as you found. It seems that trial and error may well be faster!


How to find the molar volume of 1 gas in a 2 gas system using PV_m=ZRT?

If you want to get the partial molar volume of a species in a non-ideal gas mixture, you can treat it as an ideal solution.  In this approximation, the partial molar volume of the species is equal to the molar volume of the pure species at the same temperature and pressure as the solution.  Another method is to evaluate the molar weighted averages of the critical pressures, critical temperatures, and ascentric factors for the various species in the mixture, and then treating the mixture as a pure gas.  You then take the partial derivative of the volume with respect to the number of moles of the species under consideration.  I would try it both ways to see how the results compare.


How to check for geometrical isomerism in cyclic compounds?

I hope the statutes of limitation haven't run out. Structure 1 is a (Z)-alkene. Using Cahn-Ingold-Prelog (CIP) protocol, the rings are deconstructed as shown in diagram 2. The larger red arrow marks the higher priority carbon (C,C,C). The smaller red arrow points to the lower priority[carbon (C,H,H). For the blue arrows, sulfur takes priority over carbon. [Addendum: Strictly speaking, the two blue arrows should point to the methylene groups next to the blue sp2 carbon such that C(S,H,H) > C(C,H,H).] The two larger arrows establish a (Z)-configuration. See: What are the CIP rules for cyclic substituents?


Are sp, sp2, and sp3 hybridisation only relevant to the carbon atom?

Not only $\ce{C}$,but also other elements (e.g., $\ce{B}$, $\ce{N}$, $\ce{O}$, etc.), which have partially filled $\ce {p}$ orbital can undergo $\ce {sp^3}$, $\ce {sp^2}$, or $\ce {sp}$ hybridization.

To understand how it works for molecules other than carbon, which do not show four bonds, let's take a water molecule as an example. Oxygen atom has $\ce {6 e-}$s in its outer most energy level (n = 2), which has electron configuration: $\ce {2s^2 2p^4}$. When water molecule is formed, it was quantum mechanically argued that all these 4 orbital ($\ce{2s, 2p_x, 2p_y, and 2p_z}$) undergo hybridization to form four equally energized $\ce {sp^3}$ orbitals (the name, $\ce {sp^3}$ is given to the newly formed hybridized orbitals because they were made up by $\ce {1\times s + 3\times p}$). Since they are equally energized, they would occupy the space in such a manner that they have equal distances and equal angles to each other. The only way it can arrange to achieve that is in tetrahedral shape. That four orbital can have maximum of eight electrons. When you put six electrons, which are belong to oxygen in its four $\ce {sp^3}$ hybridized orbitals, you end up with two fully filled and two half-filled orbitals, because you are supposed to avoid pairing them whenever possible to avoid pairing energy so that you keep the atom in its lowest possible energy state. Now, two hydrogen atoms can occupy these two half-filled orbitals with one  $\ce {s}$ electron of each, making 2$\sigma$ bonds. We called it $\sigma$ bonds because there are two nuclei involved to make the bond and it was head-to-head overlapping (precisely, $\ce {sp^3-s}$ bond). What about other two $\ce {sp^3}$ orbitals, which are already filled with electrons from oxygen? We called they are lone pairs.

Now $\ce {O}$ of $\ce {H2O}$ molecule is $\ce {sp^3}$ hybridized. Therefore, the theoretical shape of $\ce {H2O}$ molecule is tetrahedral. Yet, actual shape is angular because if you get a snap shot of $\ce {H2O}$ molecule, you can't see the electrons. Only you see is three nuclei, one from oxygen and other two from hydrogen, arranged in angular manner.

Similar to water molecule, oxygen atom in tert-butanol (2-methyl-2-propanol), in addtion to all four carbon atoms, is $\ce {sp^3}$ hybridized. The $\ce {O-H}$ bond is $\ce {sp^3-s}$ bond while the $\ce {O-C}$ bond is a $\ce {sp^3-sp^3}$ bond.

Now, you can look how it work for $\ce {sp^2}$ and $\ce {sp}$ hybridization: Only three $\ce {sp^2}$ hybridized orbitals can be made since only three orbitals used to make them. Only way three can arrange in space with equal energy is with trigonal planar geometry (say they are in $xy$-plane). Thus, the $\ce {p}$ orbital not used for the hybridization occupy as $z$-axis to that $xy$-plane. For example, oxygen and middle carbon it attached to in acetone are $\ce {sp^2}$ hybridized so that the $\ce {O-C}$ bond is a $\ce {sp^2-sp^2}$ bond. Similarly, one $\ce {O-C}$ bond in acetic acid is $\ce {sp^2-sp^2}$ bond while the other $\ce {O-C}$ bond is $\ce {sp^3-sp^2}$ bond (both oxygen attached to $\ce {sp^2}$ hybridized carbon).


Can Oxybuprocaine hydrochloride react with lidocaine hydrochloride?

Looking at the functionality, oxybuprocaine looks more likely to react with itself rather than lidocaine. As it's a salt and also no doubt from numerous medical trials, this doesn't occur,  the anticipation would be that the two compounds would be fine.  The $pK_a$'s of the tertiary amines would also be expected to be similar, so one shouldn't neutralise the other.


Raise the pH of a solution and prevent build up of carbonates

How about $\ce{NH4OH}$? This does not for carbonates as readily, but it does form complexes with many metals. It depends on your intended use.

It might be easier to keep things away from the air:

Use a closed container.
Float an inert liquid (e.g. paraffin oil) atop your solution, or just a layer of pure water, which might take a while to mix.
Use an inert gas over the container to displace air.


On what basis can we say that one nuclide is more stable than another one?

In general it is very difficult to estimate the stability of a nucleus based on the number of protons and neutrons it contains, as it is an interplay of different forces (repulsion between protons and the Strong force). However, a crude estimate can be made based on these empirical rules:

There are no stable nuclei with $Z>83$ (Also no stable nuclei for Technetium ($Z=43$) and Promethium ($Z=61$)).
The stability of a nucleus depends on the ratio $$\eta=\frac{A-Z}{Z}$$
there are no stable isotopes with $\eta<1$, except for $^1$H and $^3$He. Stable nuclei with relatively small atomic number have $\eta \gtrsim 1$, while nuclei with larger $Z$ need more neutrons to stabilize the nuclear charge, for instance $^{206}_{\,\,82}$Pb has $\eta=\tfrac{124}{82}\approx1.51$.
Most stable nuclei have an even number of protons and neutrons, there are only a few stable nuclei that have both an odd number of protons and neutrons.
Nuclei that have a number of protons or neutrons that equals the so-called magic numbers 2, 8, 20, 58, 50, 82, 126 are especially stable. Nuclei that have both a magic number of neutrons and protons (such as $^4_2$He, $^{16}_{\,\,8}$O, $^{40}_{20}$Ca, $^{48}_{20}$Ca, and $^{208}_{\,\,82}$Pb) are exceptionally stable.

In your example, $^{40}_{20}$Ca belongs to this special category.


What is the LUMO of a molecule with an alkene and a carbonyl?

I think here you are missing the point on what exactly a HOMO and a LUMO are. They are descriptions of the energy splitting that occurs when quantum wave functions constructively, or destructively, interfere with eachother when in proximity. Because this system is not conjugated, the alkene and the carbonyl with both INDIVIDUALLY have their own HOMO and their own LUMO. In the case of both of these systems, the HOMO will be the pi bonding orbital and the LUMO will be the pi* anti-bonding orbital. Now, when it comes to which of these LUMOs will be lower in energy, it honestly will depend on the compound. Electronegative atoms, electron withdrawing groups, etc. narrow the energy gap between bonding and antibonding orbitals, making the LUMOs more reactive. So in this particular example, the carbonyl will probably have a larger pi* coefficient and thus be lower in energy. For more complex systems, reactivities need computer algorithms, clever thinking, or just experimentation to determine which energies are lower.


Which reactions are more temperature sensitive: the ones with higher Ea or the ones with lower Ea? And why?

The reaction with the lower activation energy can proceed smoothly without the increase in temperature whereas the the reaction with the higher activation energy cannot. When you increase the temperature, it won't make much difference to the reaction with a lower activation energy since it was already fine with a lower temperature. On the contrary, the reaction with higher activation energy will now gain sufficient energy making the molecules more likely to react and thus, the temperature change affects this reaction much more.


Why can't primary alcohols react with sodium bromide via an SN2 mechanism?

For $\mathrm{S_N2}$ to occur, the $\ce{-OH}$ has to be "helped along" to leave, by converting it to a good leaving group (good leaving groups = weak bases). If you protonate it with a strong acid, $\ce{Br-}$ can attack, and a molecule of water will be kicked out as the leaving group.

So why doesn't it work with $\ce{NaBr}$? A proton is a strong lewis acid, meaning $\ce{HBr}$ can protonate a small fraction of the $\ce{-OH}$ groups. $\ce{Na+}$ is not, on the other hand; it is too electropositive (just as $\ce{Na}$ tends to give up an electron and become $\ce{Na+}$). It simply cannot hold on to the electrons from a formally neutral oxygen-atom at all. Hence, it's like you are still trying to get $\ce{HO-}$ to leave.

  Extra - when oxygen has a formal negative charge

If you go on and study carbonyl chemistry you will find places where $\ce{Li+}$ (which has a better size to interact with $\ce{O-}$ than $\ce{Na+}$) can stabilise the so-called tetrahedral intermediate, and help keep it from collapsing.

This is useful to stop the reduction of, for example, esters at the aldehyde, instead of continuing the reduction to the primary alcohol.

To get the lithium-ion to dissociate, you just add water. As DHMO explained, as soon as you have water, the positively charged group I metal ions are simply solvated so well by the water, that they are virtually inert.


Treatment of copper (0) with sodium hydroxide

$\ce{Cu(0)}$ will not be attacked by $\ce{NaOH}$, and will not decompose in this aqueous environment at room temperature. So, this is not to create some reaction with the copper. Two uses I can think of:

$\ce{NaOH}$ increases the solubility of $\ce{Cu2O}$, so maybe this is a way of getting rid of the last remaining copper oxide.
To bind the thiols with copper, you'd better be in alkaline conditions rather than acidic conditions, right? If so, then maybe this is the way of making sure you don't have any acidic groups or molecules remaining adsorbed on the copper surface, to improve bonding at the next step.


Just because they move faster doesn't mean there's more space between them. Here's why:

We need to look at several important things in this case: how many molecules, what's the temperature and what's the space they are in. For this example it is most easy to understand if we say we got the same space and the same amount of molecules, we just want to change the temperature.

If we got 1000 molecules (to make the math easy) in a liter (=1000 mL) then in average each molecule has free space around it of 1 mL. It doesn't matter how fast the molecules move, there are always 1000 molecules in 1000 mL so on average each molecules does always have 1 mL of space.

And in addition: between molecules is just nothing.


Copper layer doesn't stick after Copper electroplating

Roughen the deposition surface, perhaps by buffing with very coarse sandpaper or an abrasive scrub pad or similar.

If the surface isn't rough enough, the deposited coating has nothing to mechanically "grab onto" and it can be peeled off just as you've observed.  With a sufficiently rough surface, the deposition penetrates into the nooks and crannies of the surface texture and adheres very strongly.


Can redox reactions involve only non metals?

A redox reaction is one in which oxidation state of some atom changes, and such reactions need not involve metals, nor any ionic species. Oxidation state is the idealized charge that a given atom would have if all its bonds are treated as completely ionic. Thus, strictly for the purpose of calculating oxidation state, the electrons of a covalent bond are typically assigned to the more electronegative atom. (I'd note, however, that IUPAC provides a different set of rules for calculating oxidation number.) Hence, in order for a redox reaction to occur, there does not need to be a complete transfer of electrons (in the sense of ionization). Examples of redox reactions involving molecular compounds are numerous.

For example, consider the reaction $\ce{N2 + 3H2 <=> 2NH3}$. This is the Haber process for the production of ammonia. Here nitrogen is reduced from an oxidation state of $0$ to $-3$, while hydrogen is oxidized from $0$ to $+1$. Again, nitrogen being the more electronegative atom, is assigned the electrons of the bonds. This idealization is useful in that it provides some rough idea of electron distribution, and hence has some utility in predicting reactivity.

Edit: Also, the statement "[...] the elements in the molecular compound will not have charges what so ever [...]" is inaccurate. There is a concept of formal charge that applies to neutral molecules as well as ions.


What conditions to use to crystallise Equine Myoglobin?

Does the 1958 Nature paper by Kendrew include any useful details? (Sorry, I'm behind a paywall right now).


What is has a higher priority in naming organic compounds, alkyl groups or halides?

The current IUPAC recommendations (2013) read as follows:

Numbering
When several structural features appear in cyclic and acylic compounds, low locants are assigned to them in the following decreasing order of seniority.
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g.
Furthermore,

The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference (…)

Therefore, the example is named as 3-chloro-2,2-dimethylbutane rather than 2-chloro-3,3-dimethylbutane since the locant set ‘2,2,3’ is lower than ‘2,3,3’.


Can the ions of dissolved salts be physically separated with a strong enough external electric field?

Could an external electric field (e.g. the electrically charged plates in Millikan's oil drop experiment) physically pull apart the dissolved ions in the water if it was strong enough?
What forces would it need to overcome?

Yes, this is the principle behind electrophoresis and electric double layer supercapacitors where charge is separated by an electrostatic field shown below.

Diagram of Double Layer Capacitor charging and Discharging

What would be the outcome if the $\ce{Na+}$ and $\ce{Cl-}$ ions were then physically separated into two different containers?

Assuming you could so this without any interaction with the container, you would have 2 Leyden Jars that were opposite in charged and storing a lot of energy.


How much of Ag has formed on the Cu tile?

In this experiment the tile lose mass due to oxidation and dissolution as Cu dication (Cu++) and acquire mass due to the reduction and deposition of Ag.

This is what is measured by weighting before and after.

This difference in mass correspond to 2n moles of Ag minus n moles of Cu so is just matter of solving a first degree equation.

Quickly, without being consistent in digits and truncation.,
solving it for the current data gave about 0.6 g of Ag.

All others outcomes seems to me to assume an unlike stoichiometry or refuses perpetuating around.

A way to get into an "effective stoichiometry" of 1:1 is to consider formation of Cu+ followed by disproportionation AND assuming that all metal copper finely precipitates without depositing back to the tile from which originally came from. While, partially occurring, this scenario would explain discrepancies found performing a real experiment, it seems unlikely that is intended as to give a fixed result as for it  requires the assumption in italic above. Doing that the amount of silver is now about 1.06 g.


How can I safely store highly outgassing self-pressurizing chemicals?

Bottles for 30% or stronger $\ce{H2O2}$ usually have vented caps to prevent sudden out-gassing, splashing and boil-over that you experienced, as well as to prevent bursting the container. The vent may be as simple as thinned spot, which may be scored or slitted, or it may have a polythelene, PTFE or other sponge or "paper" separator that helps retain liquids while allowing gas to escape.

In either case, I would not suggest storing a mix of $\ce{H2O2}$ and $\ce{HCl}$ for more than a few hours for a few reasons.

As gases evolve, the solution becomes weaker and less effective an etchant.
Stored on a shelf, the gases will corrode the shelving surprisingly quickly. You would not want shelving with caustic or toxic chemicals to collapse.
If gases given off are other than $\ce{O2}$, you probably don't want to breathe them for long.

Ferric chloride is a slow etchant at room temperature, but gentle warming speeds etching greatly (PDF). At 54 °C, $\ce{FeCl3}$ removes ~35 um of copper per minute, so standard PC boards ("0.5 oz" - "1 oz") should be etched in ~1 - 2 minutes.

The biggest nuisance I find using $\ce{FeCl3}$ is that it tends to stain containers, tables and hands, but Iron Out and similar bisulfite or hydrosulfite stain removers can be used on cleanup.


Perfluorobutane is inert and has almost twice the density of sulfur hexafluoride. It is non-toxic enough that it is used in fire extinguishers and injected as a contrast agent for ultrasound. Boiling point: $-1.7\ \mathrm{^\circ C}$.

Perfluoropentane is similar and rarer but somewhat higher density $(\sim13\ \mathrm{kg/m^3})$ in proportion to its higher molecular mass. Its boiling point is $28\ \mathrm{^\circ C}$ (uncomfortably warm, but your thermostat can go that high). This is the densest gas that strictly meets all the criteria in the OP.

If we relax the criteria a bit:

Perfluorohexane is just over the boiling point limit at $56\ \mathrm{^\circ C}$, but it has a molar mass of $338\ \mathrm{g/mol}$ which makes it slightly denser in gas form than tungsten hexafluoride ($\ce{WF6}$).  It's also inert and non-toxic, unlike $\ce{WF6}$.

The ultimate gas density would be uranium hexachloride using depleted uranium and $\ce{Cl-37}$, with a molar mass of $460\ \mathrm{g/mol}$, which makes it over 50 % denser in gas form than tungsten hexafluoride, and 3 times denser than sulfur hexafluoride, but it has a boiling point of $75\ \mathrm{^\circ C}$, decomposes on contact with air, is toxic and slightly radioactive.


Why does AlCl3 dimerise to Al2Cl6 at low temperatures?

Perhaps, we can employ the Gibbs free energy equation for spontaneous reactions...

$$\Delta G=\Delta H-T\cdot\Delta S$$

Since the dimerisation process involves the formation of bonds, $\Delta H$ should be negative, while $\Delta S$ will be negative too, since two $\ce{AlCl3}$ molecules become one $\ce{Al2Cl6}$. As such, when temperature is low, the magnitude of $T\cdot\Delta S$ will be smaller, and $\Delta G$ will be negative or more negative, thus favouring the dimerisation process.