Is the order of orientation of electron box diagrams meaningful or arbitrary?

Raise hell with the grader. There's no reason whatsoever for this answer to be rejected. A Google book search can substantiate your argument, and the first example from a textbook is actually in your order:

I think the guidance “allow single arrow in either direction” is not exclusive with “allow double arrows in either order”, and maybe only single arrows were explicitly mentionned because it was obvious to the writer that order in double arrows doesn't matter.


Which is a better nucleophile: hydroxide anion or amide anion?

I was not able to find data on this nucleophilicity of $\ce {NH2^-}$ and $\ce {OH^-}$ in dimethylformamide, $\ce {Me2NCHO}$. However, I have a plausible explanation as to why the answer may not be what you expected.

Firstly, it is important to remind ourselves that nucleophilicity is always relative to the electrophile that we are concerned with. However, it is also possible to discuss nucleophilicity generally, i.e. not with respect to any particular electrophile. This is the case now.

The fact that the solvent is polar and aprotic suggest that hydrogen bonds cannot be made with the $\ce {N}$ and the $\ce {O}$ of the nucleophiles. However, it does not restrict us to form hydrogen bonds between the lone pair on the carbonyl oxygen with the partially positively-charged $\ce {H}$ atoms on the ions. This is not at all unlikely considering that the strong delocalisation of the lone pair on the $\ce {N}$ in amides, gives a significant resonance structure with this carbonyl oxygen bearing a formal negative charge.

$\ce {NH2^-}$ has two such $\ce {H}$s and $\ce {OH^-}$ has only one such hydrogen. Thus, perhaps, we may still expect greater extent of solvation for the former. Relative to $\ce {OH^-}$, the greater amount of solvation of $\ce {NH2^-}$ increases the activation energy required to initiate the subsitution reaction since solvation can be said to bring down the energy of the reactant in the reaction energy profile. Thus, the rate of substitution with $\ce {OH^-}$ would be faster than $\ce {NH2^-}$ in this polar aprotic solvent.

Alternatively, the answer provided could very well be wrong. That is a possibility that cannot be discounted...


Different solutions in cyclic voltammetry

Recording a CV assess both so-called half cell reactions.  If balanced individually, some of them only need only the transfer of an electron, e.g. for
$$\ce{Fe^{2+} -> Fe^{3+} + e-}$$
Other half-reactions include either $\ce{H+}$ or $\ce{OH-}$, and their consumption / generation alters the pH value of the solution, e.g.
$$\ce{MnO4^- + H2O + 3 e- -> MnO2 + 4 OH-}$$
or
$$\ce{MnO4^- + 4 H3O+ + 2 e- -> Mn^{2+} + 6 H2O}$$
Since each of them are characterized by a standard electrode potential, altering the pH value of the solution thus may alter the reaction pathway accessible; the reaction of permanganate to manganese dioxide is typical for a reaction in a basic aqueous reaction, the one to yield $\ce{Mn^{2+}}$ typical for an acidic one. Indeed pH electrode is an application of this pH dependence which may be quantified by the Nernst equation.
While the examples shown deal with small inorganic ions, the same principle is valid for recording a CV for organic molecules, too.  For the couple quinone / hydroquinone, the variation of the pH value thus may yield a plot like the following:

(reference)
The changes for the redox couple, recorded in different solutions of different pH value equally were plot with the cell potential in function of said variation of the pH value of the analyte:

Note: compare the half reactions with each other.  If both the number of electrons, as well as the number of protons exchanged are equal, the potential of the cell will not be affected by a change of the pH value.
reference: pH Dependent Redox Couple: An Illustration of the Nernst Equation, Walczak et al., J. Chem. Educ. 1997, 74, 1195-1197; doi 10.1021/ed074p1195


How is Electrode Potential Developed on Inert Electrode?

So, let's take a solution of copper sulfate in a beaker and insert an active electrode: a copper wire. OK so far? We connect this wire to a voltmeter.
Then lets connect another wire (we are quite well off, even rich, so we will use a shiny platinum wire) to the other connection of the voltmeter. Platinum is nice, inert, always shiny. How shall we connect to the solution? Just stick the inert wire into the solution! The wire is inert chemically in that it does not replace copper from copper sulfate or hydrogen from water, but it is not an insulator - it does do something: it carries electrons when they are pushed or dragged.
If we had inserted an inert electrode (Pt) into the solution first so we had two inert Pt electrodes, well, nothing would happen. Nothing that we could measure. Those little ions and molecules are hopping all over the place, but to no net effect.
So the copper atoms near the solution can donate electrons which then float thru the voltmeter to the platinum which can accept those electrons, handing them off to water, forming OH$^-$ and H$^.$ (two of which will form H$_2$). This occurs more easily than the other way around. (It's even a bit more complicated: copper doesn't do this so easily, but if O$_2$ is available, it will consume the hydrogen atoms produced at the Pt wire and drive the equilibrium toward copper dissolution. I just didn't want to start off the discussion with an active metal like zinc, and zinc wires, etc., because it's more difficult to paint a picture with unfamiliar objects. I probably could have used aluminum wires...)
The philosophical question then is: Is the inert electrode really inert? At the copper electrode, a copper ion just plops down on the metal (yeah, some activity), while at the platinum electrode, electrons come in, go out into H$_2$O, split it, form OH$^-$ and H$^.$, help that form H$_2$, then bubbles - it's a crazy mess compared to the copper electrode.


Arrhenius equation in exothermic reactions

The Arrhenius equation is absolutely applicable to exothermic reactions and it will give results in accordance to Le Chatelier's principle.
Applying the Arrhenius equation for the forward reaction yields:
$$k_\mathrm{f} = A \exp\left(\frac{-E_\mathrm{f}}{RT}\right),$$
applying the Arrhenius equation to backward reaction yields
$$k_\mathrm{b} = A \exp\left(\frac{-E_\mathrm{b}}{RT}\right).$$

We know that $E_\mathrm{f} -E_\mathrm{b} = \Delta H$. Dividing $k_\mathrm{f}$ by $k_\mathrm{b}$ we obtain:
 $$\frac{k_\mathrm{f}}{k_b} = A \exp\left(\frac{-\Delta H}{RT}\right).$$
For an exothermic reaction $\Delta H < 0$. So $-\Delta H/(RT)$ is a positive number. As $T$ increases, the exponent decreases and thus $k_\mathrm{f}/k_\mathrm{b}$ decreases, which is nothing but the decrement of the equilibrium constant.
Thus the result is in perfect accordance with Le Chatelier's principle.

Also, you can apply the van't Hoff's equation which states:
$$\ln\left(\frac{k_2}{k_1}\right) = \frac{\Delta H}{R}\cdot\left(\frac 1{T_1} - \frac 1{T_2}\right),$$
which also gives the decrease of rate of reaction with increase in temperature as $\Delta H < 0$.


Why does technetium not exist in nature?

Technetium does occur in nature. From the Los Alamos page on technetium:

Technetium was the first element to be produced artificially. Since its discovery, searches for the element in terrestrial material have been made. Finally in 1962, technetium-99 was isolated and identified in African pitchblende (a uranium rich ore) in extremely minute quantities as a spontaneous fission product of uranium-238 by B.T. Kenna and P.K. Kuroda.
Technetium has been found in the spectrum of S-, M-, and N-type stars, and its presence in stellar matter is leading to new theories of the production of heavy elements in the stars.

The journal article describing the 1962 discovery is Technetium in nature. B. T. Kenna and P. K. Kuroda.  Journal of Inorganic and Nuclear Chemistry 26(4) 493-499 (1964).


Will the temperature of a faucet actually be lower than the countertop or will it only feel cooler due to specific heat?

The short answer:

A metal object as the faucet drains heat from fingers due its high thermal conductivity faster than the countertop.

The fingers get therefore colder than if touching the countertop.

The faucet perceived temperature is therefore colder, even if the faucet and countertop temperature are the same.

If it is really colder due cold water, the perceived coldness is even stronger.

The longer answer:

Metals have high thermal conductivity.
That causes higher thermal flow at the same temperature difference than for most non metal materials of the same temperature. ( Rare exception are diamonds, as their thermal conductivity is superior to all metals.).

That leads to perception like if metals were colder ( if being colder then body temperature ) or warmer ( if being warmer ) than they really are.

OTOH, metal (mass-)specific heat capacity  is mostly lower than for non metals, due heavy metal atoms, that in fact somewhat decreases the effect of high thermal conductivity.

For your specific case, it depends also on the real metal temperature, e.g if it is partially sharing temperature with cold or warm tap water.

Similarly, water feels like colder than air of the same temperature, as thermal flow from human body is higher. But at this time, the higher volume specific thermal capacity plays major role.

2 examples how our senses can be deceiving:

If you had not too much hot herbal tea in an aluminium cup, you could still be sipping the tea itself without burning your tongue, but the cup would feel like burning your lips. If you did the same with the stainless steel cup, it would not be feel like burning as much, but still hot. If we used thermometer sensors, we would realize the metal surface not in direct contact with tea is very slightly colder than tea. But in both cases you would have impression the metal is warmer than the tea itself.

This time our senses monitor real thing ( temperature ), but rather their own temperature, instead of the temperature  of the sensed object.

Let have 3 basins with cold water, lukewarm water and warm water. You measure their temperature with thermometers, so it is a real thing. Now you put your left hand to cold water and your right hand to warm water for a minute. Then you put both hands to lukewarm water. The left hand will give you perception it is warm water, while the right hand will give you perception it is cold water.

This time our senses process not the temperature itself, but rather the temperature spacial and temporal gradient.


Can a spark ignite hydrogen? Also, what about the tip of a cigarette?

Yes, if the $\ce{H2(g)}$ concentration in oxygen or air is greater than about 4% not only will it burn, but most likely explode. You should avoid all heat sources when dealing with it, including direct sunlight. Just so you know, hydrogen safety datasheets include words such as "extremely flammable" and "DANGER" in red capital letters.
Besides its high flammability, another thing that makes it so dangerous is that you can barely see its flames when there's a high oxygen concentration, as seen below. It also reacts violently (A.K.A. explodes) with most oxidizing compounds, so you should be careful where you are storing it.
A rocket engine at full thrust. There's a barely visible bluish flame below it, that's the hydrogen-oxygen mix used as fuel.


Is there a simple and inexpensive way to generate nitrogen gas?

Per Wikipedia, we have this laboratory method:

$$\ce{NH4Cl(aq) + NaNO2(aq) → N2(g) + NaCl(aq) + 2 H2O (l)}$$

Also per Wikipedia, this method generates small quantities of impurities of $\ce{NO}$ and $\ce{HNO3}$, which can be eliminated by passing the gas through potassium dichromate $\ce{(K2Cr2O7)}$ dissolved in sulfuric acid $\ce{(H2SO4)}$

Both ammonium chloride and sodium nitrite are reasonably available chemicals, and the other products (table salt and water) are undeniably safe. Sulfuric acid might a bit tougher to get depending on where you live, and potassium dichromate will be the most difficult given that the use of hexavalent chromium is discouraged (and you should research procedures to reduce the $\ce{Cr2O7^{-2}}$ to $\ce{Cr2O3}$). But otherwise, this looks like the best approach.


What is the activation energy of creating molecular hydrogen from atomic hydrogen?

As porphyrin pointed out, two hydrogen atoms coming together from infinite distance will tend to bond, if the energy of the bond can be dissipated. This is unlike the situation where two magnets click together and stay together (because that interaction hits an absolute minimum, rather than a roller-coaster upswing as in the diagram. Ref 1). Two hydrogen atoms coming together from infinity would be attracted, gain kinetic energy from the potential energy of their separation, come too close, then repel each other - to infinity again. Over and over!

Perhaps instead of thinking of an activation energy, it is more relevant to consider a deactivation energy, which is all the energy that the bond needs to dissipate. You could ask whether it is possible for the two atoms to come together and emit a photon of the exact energy needed to quell the oscillation. Or even a lower energy photon, just to keep the molecule bonded, even if still in a high energy state. Alas! Selection rules seem to prevent this or at least minimize its likelihood, and recombination of hydrogen atoms is reported to occur thru three-body interactions. Ref 2
Can you do this any other way? Perhaps instead of observing them in free space, you could attach both atoms to a metal surface and let them find each other in small steps. Some metals, like zinc, evolve H$_2$ in acidic solution. Very very pure zinc reacts only very slowly with acids. Ref 3 My interpretation of that situation is that when the first evolution of hydrogen occurs, the hydrogen is in atomic form, and is attached (not strongly bonded) to the surface, whether that is actually metallic zinc or a nanothin layer of ZnO, perhaps as a ZnO  -  H moiety. A more stable state would be for two of these loosely attached hydrogen atoms to bind together to form H$_2$, but movement is inhibited (by a high activation energy). In less pure zinc, the presence of cathodic sites (like a little copper) provides a better site for H$_2$ evolution, lowers the activation energy and the process spreads over the rest of the zinc to allow rapid corrosion. Less pure zinc could be coated with a thin film of mercury to provide another surface for hydrogen atoms to be relatively immobilized on, and this also gives a high overpotential (a high activation energy)for hydrogen evolution. Ref 4
Some metals, noble metals which have no oxide layer, like Pt, Au and Pd, enable rapid migration of H$_2$ across their surface (thus not only a low, near zero activation energy, but also connection with a massive metal substrate that can accept some vibrational energy when two hydrogen atoms get together and dance for a while before settling down. They are good surfaces for evolving H$_2$.
A number of organic reactions were done with atomic hydrogen, using high activation energy to prevent  formation of molecular hydrogen; this was classified as due to nascent hydrogen, a term now in disfavor. Ref 5
Ref 1. http://www1.biologie.uni-hamburg.de/b-online/library/newton/Chy251_253/Lectures/LewisStructures/Dihydrogen.html
Ref 2. I. Amdur,  J. Am. Chem. Soc. 1938, 60, 10, 2347–2355
Ref 3. https://en.wikipedia.org/wiki/Zinc
Ref 4. https://en.wikipedia.org/wiki/Overpotential
Ref 5.https://en.wikipedia.org/wiki/Nascent_hydrogen


Determining Fluorescence Quantum Yield of C60 in Toluene

Toluene instead of benzene can be a poor choice when H-atom abstraction by the excited state is possible. Here, you're safe.

Try to get a copy of Fluorescence spectra and quantum yields of buckminsterfullerene (C60) in room-temperature solutions. No excitation wavelength dependence.

The authors apparently used 9-cyanoanthracene in toluene as a standard to determine $\Phi_f$ of C60.

You might also want to have a look for review articles by Dirk Guldi from the time around 2000; I remember him giving a number of talks on C60 around that time.


With hydroxyl groups, the oxygen is donating electron density into the ring (resonance). Additionally, the overlap between oxygen and carbon pi orbitals is sufficient enough to make up for the electronegativity of oxygen.
Halogens also donate electron density, however, the orbitals do not overlap as well with halogens and carbon, causing less electron donation then oxygen. Thus, electronegativity trumps resonance.


Why don't proteins form branched polymers?

Note that branching is not unheard of, for instance glutathione is an abundant short branched peptide formed by condensation of the carboxylic group on a glutamic acid side chain to the main chain amino group of a cysteine aminoacid. This requires a dedicated enzyme both for synthesis (glutamate cysteine ligase) and degradation of the gamma peptide bond. Such gamma peptide bonds could serve as the basis for branching but afaik occur only in special cases.
An important example of branching also occurs during ubiquitylation, in which ubiquitin is attached to a protein's C-terminus via lysine side chains to target the protein for degradation. A variety of polyubiquitin chain arrangements (including multiple branching) have been identified.
I suppose one can only speculate from its absence (or non-ubiquity) that branching either (1) does not generally confer a significant evolutionary advantage or that (2) there are evolutionary barriers that have impeded development of advantageous biochemical machinery for generation and regulation of branching.


Direction of gas molecules in kinetic model of gases?

You have pointed out the problem of how to handle the infinitesimal in probability theory. So yes a molecule could traveling in the $yz$ plane and have 0 $x$ velocity. However the probability of the motion being in exactly the $yz$ plane can be made as small as you like.

Let's make the problem one dimensional and choose a random angle between 0 and 359.999... degrees. So the question is what is the probability that the chosen angle will be for a vector on the x-axis?

To a degree if I chose 0 or 180 then the vector will be on the x-axis. Odds 2/360
To a tenth of a degree if I chose 0.0 or 180.0 then the vector will be on the x-axis. Odds 2/3600
To a hundredth of a degree if I chose 0.00 or 180.00 then the vector will be on the x-axis. Odds 2/36000

and so on...

So what probability should be assigned to a randomly chosen angle being exactly along the x-axis?


Is there a terminology contradiction about whether the conjugate of a strong acid is a "weak base"?

(Would rather have made this a comment than an answer but still new. So please don't hang me.)

Through high-school, I was also bugged by this. No, you are not going mad, there are many different views on this matter when you google it. The problem is that like many other scales humans create, the way we define 'strong' or 'weak' is a rather relative concept, especially when it gets to the mid-ranges. It is only a way for us to explain what we observe and get a handle on things.

One thing all chemists would agree on is that you should rather look at how strong an acid or base is relative to another (based on Ka or Kb values). In the end, these are the things you would actually use. However, as you need an answer in a school context I would suggest the following:

At school (at least where I come from) we mainly dealt with Bronsted-Lowry acids and bases. Here the strength of an acid depends on how easily it donates ("accepts" for bases) protons. If your students already know a bit about Ka values then anything with a Ka>1 can be considered (reasonably) strong: on a school level, this is usually limited to HI, HBr, HCl, HNO3, H2SO4 and sometimes HClO3 and HClO4. You might want to look at the relative strengths of these 'strong acids' (I seem to recall an exam question about why HI is a stronger acid than HCl (how stable are the ions) and why HF isn't considered a strong acid - the answer is H-bonding). Anything with Ka<1 can be seen as weak.

As to conjugates. Your statement in 2 about 'The stronger the Acid the weaker the conjugate base' and its inverse applies. Here you also work comparatively rather than absolute. You could try explaining it by saying if something gives away a proton easily (acid strength) why would it want to take that proton back later on? Mathematically this can be expressed by relating the Ka value to the Kb value of its conjugate Ka = Kw/Kb (Kw = 1e-14). You would notice that very few 'weak bases' (Kb<1) will give you a Ka>1, from there the confusion about "the conjugate of a weak base is a weak acid". eg. H2PO4^- (Ka = 6.23e-8) is neither a particularly strong base nor acid, but if it reacts as a base and forms H3PO4 this conjugate still only has a Ka = 7.52e-3 - which is still very weak compared to HI.


Determining the amount of chlorine in a swimming pool

If you need a no-frills solution, get some quick-test paper, such as this.

A slightly more complex solution would be a test kit like this, which makes use of the formation of Bandrowski's base from 1,4-phenylenediamine in the presence of oxidants, such as chlorine.


Breathing pure oxygen is potentially toxic. At reduced pressures, pure oxygen can be inhaled for days, or weeks. It is used to treat individuals suffering from stroke, heart attack, etc. and can be comfortably used for up to 24 hours. The body can be overwhelmed by oxygen during long periods or higher pressures, free radicals in the oxygen cause damage through oxidative stress which can result in tissue damage and death.


What are the states of HX when reacting with alcohol and alkenes respectively, and Br2 when reacting with benzene?

The first reaction ($\ce{HX}$+Ethanol) happens in solution.  High concentration of HX mixed with ethanol will produce the product. However, for $\ce{X}$ other than $\ce{Cl}$, more often one put concentrated phosphoric or sulfuric acid with alcohol and a salt of X (for example, $\ce{KBr}$ or $\ce{KI}$).  Concentrated phosphoric/sulfuric acid are easy to obtain and store in high concentration, as opposed to $\ce{HX}$. It is, therefore, a liquid phase reaction.

For ethene + $\ce{HX}$, it can be done in gas-phase, in an aqueous environment, or you can also add $\ce{HX}$ gas to an ethene solution in a solvent such as $\ce{CCl4}$. So, it can be either gas phase or liquid phase.  It is almost always more preferable to do a reaction in solution in a laboratory set up. (So much that we call the experimental part of the chemistry wet chemistry.)  This constraint is, however, lifted in the industrial world, with the availability of expansive specialized reactors.

And for benzene + $\ce{Br2}$, both reactants are liquid and $\ce{Br2}$ is also soluble in benzene, so the reaction is in also the liquid phase.


Good question! For a solid dissolved in liquid, obviously the solid behaves as a liquid rather than as a solid, since its particles are separated and moving freely.

A gas, though, is defined as free to fill its container, unlike a liquid with a definite volume. Since a dissolved gas (as opposed to one chemically bound to a liquid) is free to disperse, e.g. $\ce{CO2}$ bubbling out of seltzer, one might still consider it gaseous, rather than liquid, just loosely bound to the liquid.

That said, in a closed container in equilibrium, or over a short time, dissolved gases do behave as liquids. One can pour $\ce{NH4OH}$ ($\ce{NH3}$ aq.) into $\ce{H2SO4}$ ( $\ce{SO3}$ aq.), and the majority of the dissolved gases stay in solution -- though some $\ce{NH3}$ gets out, leaving one tearing and coughing, if not done under a hood.

It appears to be a matter of definition, then if dissolved gases are considered as gas or liquid.

BTW, gases dissolved in solids, such as methane clathrate, are more tightly locked in place and may behave as solids until warmed.


Should you work in metres-cubed or litres in chemistry?

Values for the gas constant are tabulated for various different units as a matter of convenience, since when carrying out calculations, the units of the gas constant need to match out with the other values being used (which in turn depends on the expression being used in the calculation) and its often easier to look up a value of the gas constant than to convert everything else into a specific set of units.

Generally, chemists aim for SI units, however there is a great deal of discrepancy between, for example, chemists in the USA vs. chemists in Europe. This is especially apparent in physical chemistry.

For example, the SI unit for energy (+work, +heat) is joules, such that the 'correct' way would be to express energy is terms of kJ/mol or J/mol, however, for historical reasons, many american chemists still work in calories (ending up with kcal/mol etc).

In terms of the volume question, we have equally confusion.

For small quantities, volumes are measured in units like millilitres, micro litres etc, though cubic centimetres (CC) do occasionally get used. For calculations involving moles/concentrations however, we default to decimetres and metres cubed. (you should note that 1 decimetre cubed == 1 litre)

In conclusion, in an ideal world we'd all use SI units, but various quirks have meant that we don't. If in doubt, look at what other people in the literature are doing, and, whatever units you choose, make sure that the units in your calculation match up.


What is a "ynoate" in the context of alkynes?

ynoate refers to a derivative of an alkyne acid. e.g. ethyl-3-butynoate  structure here

The simplest of these is propiolic acid  https://en.wikipedia.org/wiki/Propiolic_acid


You are certainly thinking in the right direction.  Fluorine strongly prefers the axial position (for reasons described here), so at low temperature, the isomer that you've drawn on the right (2 axial and 2 equatorial fluorines) is the only isomer present.  As a result we have an $\ce{A2B2X}$ spectrum. The 2 equivalent equatorial fluorines (spin = 1/2) split the phosphorous signal into a triplet.  Each line of this triplet is further split into a triplet by the 2 equivalent axial fluorine's. BTW, the same reasoning that explains fluorines axial preference, also explains why generally $\ce{J(P-F)_{equatorial}}$ > $\ce{J(P-F)_{axial}}$

At room temperature, pseudorotation is rapid (more thermal energy is available and the barrier to pseudorotation can be surmounted) and all fluorines become equivalent on the nmr timescale. Therefore, at room temperature and above a quintet is observed in the $\ce{^{31}P}$ nmr spectrum.


Why is a buffer solution most effective when concentrations of salt and acid are the same?

Using the Henderson-Hasselbalch-Equation:
$$\mathrm{p}\ce{H}=\mathrm{p}{K_a}+\log\frac{\left[\ce{salt}\right]}{\left[\ce{acid}\right]}$$
and buffer capacity:
$$\phi=\frac{\text{number of moles of acid/base added to 1L solution}}{\Delta\mathrm{p}\ce{H}}$$
where:
$$\Delta \mathrm{p}\ce{H} = \left|\log\frac{\left[\ce{salt}\right]}{\left[\ce{acid}\right]}\right|$$
it is minimum when $$[\ce{salt}]=[\ce{acid}]$$


Calculating Vmax (or Kcat) given specific activity of an enzyme over length of time incubated with an activator

The enzyme catalysis reaction following MM kinetics can be depicted as:

$$\Large\ce{E + S<=>[k_f][k_r]ES->[k_{cat}]E + P }$$

Note that $V_{max} = k_{cat}.[E_0]$ i.e. conversion rate constant (turnover number) times the total amount of enzyme.

You should note that the enzyme here is p38 and substrate is MSK1 (which in turn is an enzyme). When MSK1 gets phosphorylated, it assumes an active form. MSK1 activity (and specific activity) would reflect the concentration of phosphorylated MSK1 (MSK1-P) i.e. the product. Note that the $k^{^{MSK1-P}}_{cat}$ is its intrinsic property and it does not change; only the $[E^{^{MSK1-P}}_0]$ changes.

I hope your doubt is clear.


How to prepare solutions of different concentrations of potassium from stock solutions with different concentartions?

If you mix two solutions of different concentrations with each other you can easily determine the resulting concentration from the general formula for it:
$$c = \frac{n}{V}\tag1\label{concentration-general}$$

You have one solution with $c_1 = \pu{100 mM}$ and one with $c_2 = \pu{5 mM}$, therefore with $\eqref{concentration-general}$ you can determine the amount of substance for each, and the total amount of substance:
$$n_\mathrm{tot} = c_1\cdot V_1 + c_2\cdot V_2\tag2\label{amount-of-substance-total}$$

Assuming there is not volume contraction, we can simply add them to know the total volume:
$$V_\mathrm{total} = V_1 + V_2 \tag3\label{volume-total}$$

The total concentration therefore is:
$$c_\mathrm{total} = \frac{n_\mathrm{total}}{V_\mathrm{total}}
  = \frac{c_1 V_1 + c_2 V_2}{V_1 + V_2}\tag4\label{concentration-total}$$

For your dilution series you will need to modify it, a $\eqref{volume-total}$ is fixed.
$$V_\mathrm{total} = V_1 + V_2 = \text{const.}$$

Therefore we modify $\eqref{concentration-total}$:
\begin{align}
  c_\mathrm{total} &=
    \frac{c_1 V_1 + c_2 (V_\mathrm{total} - V_1)}{V_\mathrm{total}}
    \tag{4a}\label{concentration-total-mod-a}\\
  c_\mathrm{total} &=
    \frac{c_1 (V_\mathrm{total} - V_2) + c_2 V_2}{V_\mathrm{total}}
    \tag{4b}\label{concentration-total-mod-b}\\
\end{align}

Now you have one remaining unknown, solve for it:
\begin{align}
  V_1 &= \frac{c_\mathrm{total} - c_2}{c_1 -c_2} V_\mathrm{total}\\
  V_2 &= \frac{c_\mathrm{total} - c_1}{c_2 -c_1} V_\mathrm{total}\\
\end{align}

Unless you have very accurate equipment, however, I do suggest using larger quantities than $\pu{4 mL}$.
\begin{array}{rrrrrr}
c_\mathrm{total}/\pu{mM} &
V_\mathrm{total}/\pu{mL} &
c_1/\pu{mM} &
c_2/\pu{mM} &
V_1/\pu{mL} &
V_2/\pu{mL} \\\hline
10 & 4 & 100 & 5 & 0.21 & 3.79 \\
20 & 4 & 100 & 5 & 0.63 & 3.37 \\
30 & 4 & 100 & 5 & 1.05 & 2.95 \\
40 & 4 & 100 & 5 & 1.47 & 2.53 \\
50 & 4 & 100 & 5 & 1.89 & 2.11 \\\hline
\end{array}


Your proton-charged water will not last long enough to react selectivy with any electropositive metal or a base that you add.  Electrostatic forces are powerful.  The excess protons will grab electrons, generating ions that themselves grab other electrons, until eventually the beta particle electrons that were emitted to begin with get consumed to rebalance the charges.  You might get some unpredicted chemical reactions but no lingering excess charge.


Why is imidazole considered an electron deficient heterocycle?

As indicated in the comment below, a rule of thumb for these particular compounds, is that aromatic nitrogens with double bonds (e.g., not pyrrole) are "electron deficient."

The term is a bit ill-defined. It does not mean the aromatic system (which usually still has 4n+2 electrons) actually has a different electron count.

More broadly, there can be several ways of declaring "electron rich" or "electron deficient," for organic compounds, but in the context of a "push-pull" conjugated or aromatic system (in the paper you link), these refer to rings or functional groups that are electron donors or electron acceptors.

So:

Electron deficient - tends to gain negative charge as an electron accepting group or ring
Electron rich - tends to gain positive charge as a donor group or ring

The charge transfer need not be a full electron - it is almost always much less (e.g., a small fraction).

The second part of your question is how you can determine this.

Conjugated systems are typically imagined as a particle-in-a-box.. that the electrons are highly delocalized. The "electron donor" is a region with higher potential energy and the electron acceptor is a region with lower potential energy. Thus, the electrons have slightly greater probability on the electron acceptor.

Therefore, you determine "electron rich" or "electron deficient" in comparison with the rest of the conjugated system.

Typically, electron deficient rings have more negative HOMO and LUMO energies - making them more difficult to oxidize (i.e., they are "deficient" in electrons and it's hard to give up another) and easier to reduce (i.e., adding an electron makes them less "deficient").

Electron rich rings have more positive HOMO and LUMO energies - making them easier to oxidize and give up an electron.

Again, this is a rule of thumb and it's almost always partial charge transfer.


Electronegativity of heavier elements of Group 15

electronegativity value decrease down the group but amongst the heavier elements difference is not that much pronounced.

The reason for this is that in heavier elements much more filling of inner d and f orbitals starts. For example as we move down from Sb to Bi, just 6 (3+3) more electrons fill p sub-shell, and 2 electrons enter to s sub-shell. Whereas on the other hand, 24 (10+14) electrons make their way to d and f sub-shell. This results in much poor shielding of electrons in outermost shell and effective nuclear charge thus remains approximately same in heavier elements. So, they have nearly same value of electronegativity on electronegativity scales.


Can the human body withstand the heat from setting plaster?

The thermal effect depends on the amount of water used. If you mix plaster of Paris with 0.6 times its amount of water (in weight), the temperature will not go higher than 37°C.


That quote had false statements all over it.

Type I ultra pure water (Milli-Q water, or others) are fairly common among all labs.  They are not corrosive to stainless steel.

As @chipbuster said, purified water is never distributed through stainless steel partly because concern of contamination.  Metal reacts not only with oxygen, but also many pollutants in the air.  Overtime on metal surface will build up a layer that has all kind of compounds in them.  When water flows through, ions get picked up.

However, the most major reason would be that glass is better in all respect.  The reason glasswares are commonly used in the lab have little to do with pure water.   Glasswares are relatively stable across wide range of pH and very hard to be oxidized or reduced.   Plus, glasswares are cheap, extremely easy to manufacture, and transparent, so you can see through.  You will really only need metal if extreme pressure or temperature is required.

As a matter of fact, it is also unwise to use glassware to distribute ultrapure water--you will get ion contaminations from glass as well.  Purified water are actually almost always stored in plastic containers and distributed through plastic tubes.


Have the shapes of synthesized anthropomorphic molecules (nanoputians) been verified experimentally?

I think to solve the question "seeing one of the nanoscholars, nanotexan, etc." in this particular conformation shown requests (at least) two steps:

The deposit of a single molecule on a clean, flat, and cold surface.  From the outsider's perception of work done @IBM Zurich and other groups, preparing such a pad accommodating an organic molecule is something understood well; perhaps some adjustments in how the molecules are then deposited are needed.  Are nanoputains as substance well sublimable, would be OMBE feasible?
The cold surface tightens the contact between molecule and substrate, lowers thermal vibrations of the molecule deposited; contributes that the then collected AFM is crisp.  Provided the earlier mentioned conformational flexibility of the material (not necessary flat molecule, either), how do you -- if necessary -- decoil the members of a nanoputain?  To move small molecules entirely, optical tweezers come to my mind, but here it were moving just a portion of a molecule representing a leg, redressing bow tie and hat that I assume as even more delicate.

So my speculation is the second part is the more challenging one.  Maybe it was already addressed as the original work was published in 2003, a citation analysis may shed some light on this aspect.


What causes that foul taste with combination of toothpaste and orange juice?

Why do juice taste awful?

Awful taste is due to sodium laureth sulfate, also known as sodium lauryl ether sulfate (SLES), or sodium lauryl sulfate (SLS)- depending on which toothpaste you use.

SLES and SLS are surfactants(wetting agent). Both chemicals are added in toothpaste to create foam and make the paste easier to spread around your mouth.

Both chemicals they suppress the receptors on our taste buds that perceive sweetness, inhibiting our ability to pick up the sweet notes of food and drink. Also break up the phospholipids on our tongue.  These fatty molecules inhibit our receptors for bitterness and keep bitter tastes from overwhelming us, but when they're broken down by the surfactants in toothpaste, bitter tastes get enhanced.

Thus, anything you eat will be less sweeter and more bitter!

Source: mentalfloss

  How to prevent awful taste after brushing?

After a bit research I found few ways to avoid awful taste:

Avoid brushing. (not recommended)
Wait for about 30 minutes.
Look for SLS-free toothpaste.
eat a little bit of cheese first.(suggested by repurposer)
hot water will do the trick.


In regards to the addition of hydrogen to alkynes, does the amount of moles of Hydrogen matter?

Your question is not entirely clear, but I think that you ask whether or not an excess of hydrogen gas would interfere with the reduction of an alkyne to an alkane.

This is not the case. A lot of the time hydrogen is just drawn into the reaction flask directly from a large reservoir (20L $\ce{H2}$ gas cylinder under pressure or a simple balloon). When the pressure in the reservoir does not change anymore (in the case of the balloon when it does not shrink anymore) it means that the reaction is finished, i.e. no more hydrogen can be added.

In general over-reduction cannot be prevented by limiting the amount of hydrogen present. It will just result in an awful mixture of products (alkanes/alkenes/alkynes). A poisoned catalyst is required to prevent this (e.g. Lindlar's catalyst).


How can I separate sodium ions from brine or molten NaCl?

You can't simply produce isolated sodium ions, you also need to have an equivalent amount of counter ions of the opposite charge. So, in this case, you are going to get sodium chloride unless you introduce some other negatively charged species.

So you could just evaporate water which will give an increasingly concentrated solution of sodium chloride and, eventually, solid sodium chloride (which will contain all your sodium ions alongside an equal number of chloride ions.

This sort of separation is commonly used for a different reason: to produce pure water. Sea water is too salty to drink but there is far more of it than fresh water. So distillation (retaining the fresh water) or reverse osmosis (which produces more concentrates salt on one side and fresh water on the other). These techniques produce both fresh water and more concentrated salty water (which is one way of getting more sodium ions in your product).


Does a smell imply particles or chemicals in the air?

Smell works because your nose detects substances in the air.

Everything you can smell is the result of a molecule carried by air interacting with complex proteins in the nose/mouth (smell and taste are related but taste copes with solids and liquids as well as volatile airborne stuff). We broadly understand how smell works but the exact details are often a complete mystery.

The key is that smell-sensing nerves contain some proteins with very specific sensor sites that can detect particular groups of substances when they "fit" into the site. Exactly how the sites translate the fit into nerve signals is often a mystery. But the point that matters for this question is that the stuff being detected has to be carried into the nose via the air (which means it has to be a gas or, possibly, a very small particle that is easily carried by air).

So, yes, smell implies there are chemicals or particles in the air. However, the nose is extremely sensitive to some chemicals and being able to detect the smell does not imply a problem (even nasties like hydrogen sulphide are detectable by smell at below parts per billion concentration, hundreds of times below the level causing harm). Your blinds probably are emitting something but it is most likely both harmless and will probably fade with time like the smell characteristic of a new car.


The buffer capacity of a weak acid-conjugate base buffer is defined as the number of moles of strong acid needed to change the $\ce{pH}$ by 1 unit.
$$\beta = \frac{\mathrm{d}[A]}{\mathrm{dpH}} $$
and the acid is present as $$[A]= \frac{K_\mathrm{w}}{[\ce{H+}]}-[\ce{H+}] +\frac{C_\mathrm{B}K_\mathrm{a}}{[\ce{H+}]+K_\mathrm{a}}$$ where $K_\mathrm{w}$ is the water ionization equilibrium constant, $10^{-14} $, $K_\mathrm{a}$ is the acid dissociation constant, and $C_\mathrm{B}$ is the total concentration of buffer.
You assume that your protein is the weak acid. Change $\ce{pH}$ to $\ce{pH}=-\log_{10}([\ce{H+}])$ to differentiate or use the product rule. You should then find that $$\beta = 2.303\left[ \frac{k_\mathrm{w}}{[\ce{H+}]}+[\ce{H+}] +\frac{C_\mathrm{B}K_\mathrm{a}[\ce{H+}]}{([\ce{H+}]+\ce{K_\mathrm{a}})^2}\right]$$
you can then plot $\beta$ vs $\ce{pH}$ and from this you should be able to find what you want.


Formulas for the number of spectral lines

To my understanding, it is simply single atom versus many number of atoms. For example, suppose one atom with an electron at energy level 7 ($n_2=7$). That electron can "de-excite" from $n_2=7$ to $n_1=6, 5, 4, 3, 2,$ or $1$. All those transitions give one spectral line for each. Thus, total of $1 \times 6 = n_1(n_2-n_1)$ (foot note 1) spectral lines would be present in the spectrum.

Similarly, when there were more than one atom in the sample, excited electrons ($n_2$) would be in different states $(n_2=2, 3, 4, 5, 6,....,\infty)$. For example, suppose we have aom population having electrons in all levels up to energy level 8 ($n_2=8, 7, 6,...$). Suppose those electrons "de-excite" to energy level 2 ($n_1=2$). Thus, electrons in $n_2=8$ can "de-excite" to energy levels $7, 6, 5, 4, 3,$ and $2$ meaning total of 6 spectral lines $(8-2=n_2-n_1)$. Some atoms with electrons in energy level $n_2-1=7$ can also "de-excite" to energy levels $6, 5, 4, 3,$ and $2$ meaning total of 5 spectral lines $(7-2=n_2-1-n_1)$, etc., etc. Thus, total numbers of spectral lines ($s$) in this case would be:
\begin{align}
s&=6+5+4+3+2+1=21=\frac{42}{2}=\frac{7\times6}{2}\\
&=\frac{(8-2+1)(8-2)}{2}\\
&=\frac{(n_2+1-n_1)(n_2-n_1)}{2}
\end{align}

Foot note 1: Total number of spectral lines for single atom where $n_2=7$ should be: $1 \times 6 = (n_2-n_1)$ in the spectrum, not $n_1(n_2-n_1)$ as I originally suggested (Thanks @porphyrin for careful reading).


Generating geometries for substituted molecules

For most of the geometry creation/manipulation I do, I use the Atomic Simulation Environment (ASE) - all in python.

It's pretty straightforward to use and has a gui so you can make sure you're doing the right thing as you go along. In particular, I'd recommend playing with it in a jupyter notebook.

Checkout https://wiki.fysik.dtu.dk/ase/tutorials/manipulating_atoms.html for some examples.

The ASE also has a dataset of existing molecules that you can import and add to your structure in various places.


Obsidian

This is a volcanic glass, formed when rhyolitic lavas cool too quickly for crystals to form.  From a chemistry point of view, it's a mixture of silicon dioxide, aluminium oxide, with sodium ,potassium, calcium and iron oxides in various quantities. These lavas will also contain significant volatiles (water and $\ce{CO2}$) held in solution by pressure, although they should be lost at the pressures required for glass formation (higher pressures means slower cooling due to more insulation).

Melting

Note that when molten, rhyolite lavas can have a very high viscosity. So it's possible that you have melted your sample, but just didn't give it enough time to flow. Indeed, with substances like this you will see partial melting - some components will melt at a lower temperature than others. If you are getting bubbles, this would indicate that your glass formed under enough pressure to retain some $\ce{CO2}$ and/or $\ce{H2O}$ in its lattice, and you are now allowing this to be released by heating at surface pressure.  This is all a bit speculative, without a closer inspection.


Why can't oxygen in furan be sp-hybridized?

Is this only because the overall structure will not be consistent, because sp-hybridization will make the COC line straight, and thus the OCC angles would have to be lesser than the carbons' sp2-hybridization would allow?

Yes, this is the correct answer. A very common misconception is that hybridisation can be used to predict the geometry, or that hybridisation somehow involves an energy cost associated with 'promoting' electrons into the hybrid orbitals. This is entirely wrong.

Hybridisation is always determined by geometry. You can only assign hybridisation states to an atom if you already know its geometry, based on some experimental or theoretical evidence. The geometry of the oxygen in furan is trigonal planar and therefore the hybridisation must be $sp^2$.


I do not know the exact process in industry, but these are the steps that I thought of (someone likely thought of this before me, it is a very simple synthesis).  The main synthesis of the ring is done in Fischer indole conditions.  Therefore I broke up the molecule into a compound known as 4-acetamidobutanal and p-hydrazinoanisole.  The first compound is an amide that is available from the reaction of 4-aminobutanal and acetyl chloride or acetic anhydride in the presence of a base.  The latter compound is available from the reaction of p-anisidine with sodium nitrite and hydrochloric acid to form the diazonium chloride, and then reduction with sodium sulfite and sodium hydroxide.  This is just an example of a synthesis.  The overall synthesis shouldn't be more than 5 or 6 steps.

I found an article highlighting the syntheses, you can take a look.  Apparently the synthesis that I described above is very similar to the most effective synthesis, a one-pot synthesis done in very similar conditions and having similar starting materials.


The color of resorcinol-formaldehyde resin

The polymerization of resorcinol creates multiple (spaghetti-like) chains containing the aromatic rings of resorcinol.  This is a simultaneous growth in multiple directions. Eventually, the chains trap each other in the solid state. In the analogy used, it is like stirring a pot of spaghetti and then freezing the content.
It may happen that an arene ring is just in proximity of an arene ring of an other chain.  Given a growing polymer chain may be flexible, it may happen that there are intramolecular "close couples", too.  If these arenes are close enough to each other, their $\pi$-electron systems may interact with each other as if they were coupled as one.  This extension of the $\pi$-electron system may move the observable absorption band toward longer wavelengths. Then,

If there are just enough of these "close encounters" per unit of volume with the same arene-arene distance, the perception of color may be intensified (e.g., pale yellow -> bright yellow).

With variation of the arene-arene distance, the intermolecular interaction of these $\pi$-electron systems equally varies. This may yield the absorption of light of different wavelengths / the subtraction of different colors from the white light by these arenes.  Thus it may happen that the color of the polymer is not uniform. If the range of light absorption is wide enough, the polymer may appear just black.


Are there any examples of nuclear spin isomers having consequences for chemical reactivity?

Yes!

Papers in the field of preparation or selection of nuclear spin isomers have introductions that justify the work by potential applications and some of them include chemical reactions, either under that name or under the more descriptive terminology of “reactive collisions”. While no specific cases are cited as examples, the potential exist but only in limited situations such as bimolecular reaction in molecular beams.

From what I have seen, the canonical reference is “Detailed symmetry selection rules for reactive collisions”, Mol. Phys. 1977, 34, 477–504. This is the theoretical paper laying out the foundations of the effect of nuclear spin isomerism on reactions. I don't have access to the full text, but here is the abstract:

It confirms what I suggested in my earlier comment: the different nuclear spin isomers have different rovibrational states, and accounting for effect changes the selection rules indicating which collisions are reactive or not.

One specific example which I found is “Observation of Ortho-Para H3+ Selection Rules in Plasma Chemistry”, Phys. Rev. Lett. 1997, 78, 3844–3847:

  Large dependences of the intensity ratios on the parent gases have been observed, indicating the spin memory of protons even after chemical reactions. The results clearly demonstrate the existence of selection rules on nuclear spin modifications in chemical reactions theoretically anticipated by M. Quack.

Their results are obtained on the formation of H3+ ions from a H2 plasma.


If there are two longest chains possible in an organic compound and both have the same number of substituents, how do we decide the parent chain?

If two longest chain are possible with same number of side chain in both of them, then we consider sum of locants of the side chain. The longest chain in which the sum of locants of the side chain is least is taken into consideration.


Percent Dissociated from Titration Curve

I'm not sure I follow your logic.
For a monobasic acid S $(\ce{HA})$ dissociation degree $α$ is

$$α = \frac{[\ce{H+}]}{c_\mathrm{a}},$$

where $c_\mathrm{a}$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_\mathrm{b}$:

$$c_\mathrm{a}V_\mathrm{a} = c_\mathrm{b}V_\mathrm{b} \implies c_\mathrm{a} = \frac{c_\mathrm{b}V_\mathrm{b}}{V_\mathrm{a}}$$

Finally, taking $V_\mathrm{a} = V_\mathrm{b} = \pu{50 mL}$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[\ce{H+}] \approx \pu{1e-4 mol L-1}$:

$$α = \frac{[\ce{H+}]V_\mathrm{a}}{c_\mathrm{b}V_\mathrm{b}} = \frac{\pu{1e-4 mol L-1}\cdot\pu{50 mL}}{\pu{0.1 mol L-1}\cdot\pu{50 mL}} = \pu{1e-3}~\text{or}~0.1\%$$


Why does water evaporate spontaneously at room temperature despite ΔG > 0?

Standard Gibbs free energy of formation of liquid water at 298 K is −237.17 kJ/mol and that of water vapour is −228.57 kJ/mol. Therefore, $$\ce{H2O(l)->H2O(g)}~~\Delta G=8.43~\mathrm{kJ/mol}$$
Since $\Delta G>0$, it should not be a spontaneous process but from common observation, water does turn into vapour from liquid over time without any apparent interference.

Your math is correct but you left out a very important symbol from your equations.  There is a big difference between $\Delta G$ and $\Delta G^\circ$.  Only $\Delta G^\circ$ means the Gibbs energy change under standard conditions, and as you noted in the question, the free energy values you quoted are the standard gibbs free energy of water and water vapor.
Whether or not something is spontaneous under standard conditions is determined by $\Delta G^\circ$.  Whether something is spontaneous under other conditions is determined by $\Delta G$.  To find $\Delta G$ for real conditions, we need to know how they differ from standard conditions.
Usually "standard" conditions for gases correspond to one bar of partial pressure for that gas.  But the partial pressure of water in our atmosphere is usually much lower than this.  Assuming water vapor is an ideal gas, then the free energy change as a function of partial pressure is given by $G = G^\circ + RT \ln{\frac{p}{p^\circ}}$.  If the atmosphere were perfectly 100% dry, then the water vapor partial pressure would be 0, so $\ln{\frac{p}{p^\circ}}$ would be negative infinity.  That would translate to an infinitely negative -- i.e. highly spontaneous -- $\Delta G$ for the water evaporation reaction.
Small but not-quite zero dryness in the atmosphere would still lead to the $\Delta G$ of water vapor that is more negative than liquid water.  So water evaporation is still spontaneous.
Extra credit: given the standard formation energies you found, and assuming water is an ideal gas, you could calculate the partial pressure of water vapor at which $\Delta G = 0$ for water evaporation.  And the answer had better be the vapor pressure of water, or else there is a thermodynamic inconsistency in your data set!


http://antoine.frostburg.edu/chem/senese/101/acidbase/indicators.shtml
This is a link to a webpage that talks all about pH indicators and the different pH ranges if you're looking for more info.

When it comes to your question, the color should go from yellow to green because in your breaker is where you add the indicator fluid and HCl acid and you titrate that with the NaOH base. If you titrate further than your green color you have now made your solution more basic.


Efficiency of various Ewald-type schemes for the summation of electrostatic interactions

I also unsuccessfully tried to look for such a paper in the past but below are some useful resources I've found. The only definitive answer I can give you is that traditional Ewald summation will be slower than a particle-mesh technique for any relevant system sizes (this is backed up in several of the references below).

Chapter 4.9 of the GROMACS user manual has some very nice discussion on their various implemenations of long-range electrostatics and appropriate references.

The LAMMPS documentation also has quite a few references and some bits of helpful discussion.

You can also try digging through the AMBER manual which uses a PME implementation but I haven't read any of this myself.

Also here you can find implementation details for the PPPM method in HOOMD (helpful if you're interested in GPGPU implementations).

I think the most relevant bit of discussion is probably this excerpt in the GROMACS manual:

  The Particle-Particle Particle-Mesh methods of Hockney & Eastwood can
  also be applied in GROMACS for the treatment of long range
  electrostatic interactions [99]. Although the P3M method was the ﬁrst
  efﬁcient long-range electrostatics method for molecular simulation,
  the smooth PME (SPME) method has largely replaced P3M as the method of
  choice in atomistic simulations. One performance disadvantage of the
  original P3M method was that it required 3 3D-FFT back transforms to
  obtain the forces on the particles. But this is not required for P3M
  and the forces can be derived through analytical differentiation of
  the potential, as done in PME. The resulting method is termed P3M-AD.
  The only remaining difference between P3M-AD and PME is the
  optimization of the lattice Green inﬂuence function for error
  minimization that P3M uses. However, in 2012 it has been shown that
  the SPME inﬂuence function can be modiﬁed to obtain P3M [100]. This
  means that the advantage of error minimization in P3M-AD can be used
  at the same computational cost and with the same code as PME, just by
  adding a few lines to modify the inﬂuence function. However, at
  optimal parameter setting the effect of error minimization in P3M-AD
  is less than 10%. P3M-AD does show large accuracy gains with
  interlaced (also known as staggered) grids, but that is not supported
  in GROMACS (yet).

If you're really looking to dive in, you can look at the source code for these packages (I believe you need a license for AMBER) to see the actual implementations.


How Does The Mordant, Ferric Acetate, Interact With Tannins When Ebonizing Wood?

A mordant is a substance used to set dyes on fabrics or tissue
  sections by forming a coordination complex with the dye which then
  attaches to the fabric or tissue (source).

Tannin is a biomolecule consisting of many carbohydrate and polyphenol residues with plenty of hydroxyl and carboxyl groups. Tannic acid is a common type of tannin:

$\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ form colored complexes with these oxygen-containing functional groups. These iron-tannin complexes change the color of the wood. The color intensity increases with tannin content of the wood and can be increased by the addition of tannin-rich tea. Structures with a varying coordination sphere of iron are formed, some of which are shown here (sources: 1 and 2):


How to determine the value for n in the Gibbs free energy and redox potential equation?

The accepted answer is not correct, the number $n$ is not equal to 2 simply because 2 is the lowest common multiple of 2 and 1. Furthermore $n$ is not the "number of moles of electrons", which would have units of $\pu{mol}$; $n$ is in fact dimensionless.

In fact the number $n$ can be anything you like, and it depends on how you combine the two half-equations.

To use a different (not very realistic, but I don't care) example consider the half-reactions (data from Wikipedia)

$$\begin{align}
\ce{Na+(aq) + e- &-> Na(s)} & E_1 &= \pu{-2.71 V} \tag{1} \\
\ce{Br2 (l) + 2e- &-> 2Br-(aq)} & E_2 &= \pu{+1.07 V} \tag{2} \end{align}$$

We have $E_\mathrm{cell} = E_2 - E_1 = \pu{+3.78 V}$. To construct a full reaction for this cell the "usual" procedure taught is to take (1), multiply it by 2, and subtract it from (2).

$$\ce{Br2(l) + 2Na(s) -> 2Na+(aq) + 2Br-(aq)} \tag{3}$$

In this case, we have $n = 2$ as there are two electrons being transferred in this reaction, from bromine to sodium. Equivalently, you can think of (3) as being the sum of two half-reactions, each featuring the gain or loss of two electrons:

$$\begin{align}
\ce{2Na(s) &-> 2Na+(aq) + 2e-} \tag{4} \\
\ce{Br2(l) + 2e- &-> 2Br-(aq)} \tag{5}
\end{align}$$

Note that the stoichiometric coefficient of $\ce{e-}$ - the number in front of the electrons - is 2, for both half-equations.

However, you could easily construct a different full reaction, simply differing from (3) by a multiplicative factor. There's no reason why you cannot, for example, divide (2) by 2, then subtract (1) to get:

$$\ce{1/2 Br2(l) + Na(s) -> Na+(aq) + Br-(aq)} \tag{3*}$$

and here you should only have $n = 1$, because there is only one electron being transferred here, from bromine to sodium. Again you can decompose this into its half-reactions:

$$\begin{align}
\ce{Na(s) &-> Na+(aq) + e-} \tag{4*} \\
\ce{1/2Br2(l) + e- &-> Br-(aq)} \tag{5*}
\end{align}$$

and the stoichiometric coefficient of $\ce{e-}$ here is simply 1.

The bottom line is that

$E_\mathrm{cell}$ doesn't change with the stoichiometric coefficients.
$n$ does, and hence $\Delta G = -nFE_\mathrm{cell}$ does. Your choice of the full cell reaction determines the value of $n$ you work with.
The best definition of $n$ is the stoichiometric coefficient of the electrons in the half-reactions into which the cell reaction can be divided (Atkins & de Paula, Physical Chemistry).


Why does dichloridobisethylenediaminecobalt(III) have only three optical isomers?

In simplistic terms, we can see $\ce{[Co(en)2Cl2]^+}$ as an $\ce{[MA4B2]}$ pseudooctahedral complex. We can initially consider the $\ce{A\bond{->}M}$ and $\ce{B\bond{->}M}$ bonds identical. Trying this out on a model octahedron or molecular modelling kit will show you that only two possible different configurations exist: cis and trans.

Here, however, we have one of the ligand types actually being a briding ligand. Because the bridge is too short to allow for a trans configuration, the two coordination sites of these bridging ligands will always be cis to each other.

For the complex in which the two chlorido ligands are trans to each other, only one possible arrangement of the $\ce{en}$ ligands is possible. At first approximation, the entire complex should have $C_\mathrm{2v}$ symmetry if not more: a $C_2$ axis with two perpendicular $\sigma_\mathrm{v}$ planes; one bisecting the $\ce{en}$ ligands through the $\ce{C-C}$ bond, the other being orthogonal and mapping each $\ce{en}$ onto the other. Due to the presence of planes of symmetry, this complex cannot be chiral.

In the complex in which the two chlorido ligands are cis to each other, one could think of up to three potential arrangements for the bridges; however, as I pointed out, it is not possible to have the two coordinating atoms of one $\ce{en}$ ligand be trans to each other, reducing the total down to two. These two possibilities do not differ in whether one of the chlorides is in an ‘axial’ position (remember, we are assuming equal bond lengths initially so there is no defined ‘axial’ position). Rather, in one case you have a leftwards-turning screw and in the other case a rightwards-turning screw if you assume the $\ce{en}$ ligand to be a plane and wanted to screw the entire thing into your paper.

Neither of these two structures have a plane of symmetry (or higher symmetry) and both are therefore optically active. Careful analysis will reveal, that they are in fact mirror images and thus enantiomers of one another. They are called the Δ and Λ isomers, respectively. All three isomers are attached in the image below.

Figure 1: isomers of $\ce{[Co(en)2Cl2]^+}$. From left to right: trans, cis-Λ, cis-Δ.

Thus, there are three isomers of the compound, two of which are optical isomers of each other (the third is a geometrical isomer).

Finally on the topic of whether axial bonds are actually longer, i.e. whether there is a Jahn-Teller type distortion or not. The answer is no. $\ce{Co^{III}}$ is a $\mathrm{d^6}$ complex. It could be assumed to be high or low spin. In case it accidentally ends up to be low spin, the $\mathrm{t_{2g}}$ orbitals are fully populated while the $\mathrm{e_g}$ orbitals are entirely unpopulated; there is no need for Jahn-Teller distortion.

In the potentially more likely high-spin case, one of the $\mathrm{t_{2g}}$ orbitals is fully populated while all other orbitals are half-populated. We might expect a Jahn-Teller distortion to remove the degeneracy of the $\mathrm{t_{2g}}$ orbitals. However, this distortion would be minor at best: the $\mathrm{t_{2g}}$ orbitals do not interact well with the ligands since they are oriented to point in-between the ligand orbitals. (On a group theory basis, the $\mathrm{t_{2g}}$ orbitals are nonbonding.) Therefore, any symmetry reducing effect will be very minor at best and not lead to a noticeable distortion. Thus, it is reasonable to assume near-identical bond lengths and no principal (‘axial’) direction.


Your answer of $2500$ and the book's answer of $2523$ both have four significant figures.

If you want to keep your answer to the same degree of accuracy as the given data with the least number of significant figures, i.e. $2.8$ kg, it is best to express your answer as $2.5\times10^{3}$.


Does the delocalization of p-orbitals ensure conductivity?

Delocalized $\pi$ orbitals do allow for mobile charges. The catch there is "allow." Just because these systems can have mobile charges does not make them conductive.

Conductivity can be defined as:
$\sigma = n e \mu$
where $e$ is the charge on an electron, $\mu$ is the charge mobility, and $n$ is the number of charges.

In the case of $doped$ conjugated polymers like you indicate above, the halogen vapors or other dopants usually oxidize the polymer and create more charges (i.e., $n$) and the conductivity increases. This does happen with pretty much every conjugated system I've seen: oxidize or reduce the neutral system and you'll have more charge carriers = greater conductivity. (NB: most conjugated organics are easier to oxidize than reduce, so you get + charges (holes) as the charge carrier, rather than electrons.)

That's the quick explanation as to why adding halogens increases the conductivity -- you get more charge carriers.

Incidentally, there are other ways to increase the charge carriers. Chemical or electrochemical doping usually makes a conductive system, but you can also use electrical bias (e.g., a field-effect transistor) or light (i.e., a photovoltaic) which will also increase carrier concentrations.

As to why trans-polyacetylene isn't a metal in the first place?

If we use a simple particle-in-a-box model, the difference in energy levels goes to zero as the box becomes very large. This would suggest trans-polyacetylene should be a metal (i.e., no band gap).

Consider also the Hückel model for polyacetylene. We can build up the nearly-infinite system by solving for ethylene, butadiene, etc.

Notice that the energy gap between levels becomes smaller and smaller. We can plot the HOMO-LUMO gap as a function of chain length:

The gap becomes smaller and smaller!

Wikipedia even gives the general formulas:

$$E_k = \alpha + 2\beta \cos \frac{k\pi}{(n+1)}$$
$$\Delta E = -4\beta \sin \frac{\pi}{2(n+1)}$$

So both particle in a box and Hückel predict no band gap when $n$ gets very large. Theory predicts a metal!

Why doesn't the real polyacetylene behave as a metal. For any one-dimensional system, there will be a Peierls distortion which lowers the total energy by distorting the geometry (long-short bond alternation) and increasing the band gap.1

In short, a truly one-dimensional metal is impossible, but doped polymers are certainly very good electrical conductors. Record films of trans-polyacetlyene have conductivity close to conventional metals.2

1 The long-short bond alternation is the Peierls distortion. It means some electrons are localized (shorter double bonds) and thus lower in energy. This creates a band gap (i.e., lower E for valence band).

2 The film isn't a 1D material, but the trans-polyacetylene chains are each 1D.


Some people define a "closed system" as one that can exchange neither heat nor work not mass with the broader surroundings.  Others, like us engineers, define a "closed system" as on that can exchange both heat and work with its surroundings, but not mass.  We engineers call a system that can exchange neither heat nor work with the broader surroundings an "isolated system."  Which definition of a closed system are you subscribing to?


Destabilization of antibonding MO vs stabilization of bonding MO

TL;DR

  "The destabilisation of an antibonding MO is equal to the stabilisation of a bonding MO, relative to the constituent AOs."

Several key assumptions are inherent in this statement.

Born–Oppenheimer approximation (not discussed further)
"Orbital approximation": wavefunction of $\ce{H2, He2, ...}$ can be expressed as an antisymmetrised product of one-electron wavefunctions (called MOs)
Said MOs take the form of the stationary states of $\ce{H2+}$, without modification
LCAO approximation: stationary states of $\ce{H2+}$ can be expressed as a linear combination of 1s orbitals on both hydrogens
The overlap integral $\langle \mathrm{1s_A} | \mathrm{1s_B} \rangle$ is neglected

If we make all five assumptions, then the statement above is correct.

If we only make assumptions 1 through 4, then MOs will still have the same form as above (i.e. everything else about the MO diagram is correct), but the statement above will be incorrect: the antibonding orbital will be more destabilised than the bonding orbital is stabilised.

The question linked in the comments does give a qualitative explanation, so depending on what you are looking for, it may be worth checking out.

Assumptions

The MOs that you see are one-electron wavefunctions that are obtained by looking at the single-electron species $\ce{H2+}$. Finding the exact stationary states of this system (by analytical solution of the Schrödinger equation) is a bit complicated, so the strategy is to find a set of approximate stationary states for the electron in $\ce{H2+}$, which we can label as $\{|i\rangle\} (i = 1, 2, 3 \cdots)$. These states are called molecular orbitals.

For multi-electron molecules, like $\ce{H2}$, a further approximation is usually invoked, in expressing the electronic wavefunction of $\ce{H2}$ as a properly antisymmetrised product of these states $\{|i\rangle\}$. Therefore, the MOs, which were only "rigorously" derived for $\ce{H2+}$, may be extended to molecules such as $\ce{H2}$. Very loosely speaking, we may assign one stationary state $|i\rangle$ to one electron, which corresponds to placing an electron in an orbital. Because of the Pauli exclusion principle, each stationary state may only be assigned to a maximum of two electrons.

These are obviously crude approximations and it is a surprise that it even works decently well. In particular the issue of electron-electron repulsions in multi-electron systems is glossed over; in this very simple model, the MOs, being derived from a one-electron system, are entirely unaffected by them, which doesn't really make sense. As the comments on the question have alluded to, there are much more sophisticated ways of obtaining MOs nowadays, but since the claim in the question is derived using this model, we have to use it.

MOs of hydrogen molecule ion

Once we have assumed that the stationary states (i.e. MOs) are of the form

$$\psi = c_1\phi_1 + c_2\phi_2$$

where $\psi$ is an MO and $\phi_1, \phi_2$ 1s atomic orbitals on the two hydrogens, then one can use the variational principle to optimise the two coefficients $c_1, c_2$. The idea is that one wants to minimise the energy of the wavefunction $\psi$:

$$\mathcal{E} = \frac{\langle \psi | H | \psi \rangle}{\langle \psi | \psi \rangle}$$

with respect to both $c_1$ and $c_2$. To this end, both the following partial derivatives must be equal to zero:

$$\left(\frac{\partial \mathcal{E}}{\partial c_1}\right)_{\!c_2} = \left(\frac{\partial \mathcal{E}}{\partial c_2}\right)_{\!c_1} = 0$$

The maths will be omitted and if one is interested, one may find it in a variety of textbooks. For example it is covered in Atkins & Friedman's Molecular Quantum Mechanics 5th ed. (search for Rayleigh–Ritz method), or in Coulson's Valence 3rd ed. It is also described thoroughly in this handout from QMUL. The key result is the secular equations, which are a set of generalised eigenvector equations:

$$\mathbf{Hc} - \mathcal{E}\mathbf{Sc} = \mathbf{0}$$

where the $2 \times 1$ vector $\mathbf{c}$ simply contains the coefficients $c_1$ and $c_2$, the elements of the Hamiltonian matrix are $\mathbf{H}_{ab} = \langle a | H | b \rangle$ and the elements of the overlap matrix are $\mathbf{S}_{ab} = \langle a | b \rangle$. The basis set used is $\{|\phi_1\rangle, |\phi_2\rangle\}$. For a non-trivial solution (i.e. $\mathbf{c} \neq \mathbf{0}$) we require the secular determinant to vanish:

$$|\mathbf{H} - \mathcal{E}\mathbf{S}| = 0$$

Now we can use symbols to fill in these matrix elements:

$H_{11} = H_{22} = \alpha$ (this must be so, as both hydrogen atoms are equivalent by symmetry)
$H_{12} = H_{21} = \beta$ (assume that $\beta$ is real, so by the hermiticity of $H$, we must have $H_{12} = H_{21}$)
$S_{11} = S_{22} = 1$ (the atomic orbitals are normalised)
$S_{12} = S_{21}$ (again, assume that this is real. Assumption 5 is that this goes to zero, but for now we will not neglect it; later we can set it to zero to see what happens.)

The exact form of the Hamiltonian, as well as more explicit expressions for these quantities, are given in Atkins MQM. For now they need not concern us. Solving the determinant we obtain two values of $\mathcal{E}$ and two associated vectors $\mathbf{c}$, which can simply be written as the MOs $\psi = c_1\phi_1 + c_2\phi_2$. The MOs (i.e. approximate stationary states) are now quoted without proof, with their eigenvalues (i.e. their approximate energies):

$$\begin{align}
\psi_1 &= \frac{1}{\sqrt{2 + 2S}}(\phi_1 + \phi_2) & & & \left\langle \psi_1 \middle| H \middle| \psi_1 \right\rangle &= \mathcal{E}_1 = \frac{\alpha + \beta}{1 + S_{12}} \tag{1} \\
\psi_2 &= \frac{1}{\sqrt{2 - 2S}}(\phi_1 - \phi_2) & & & \left\langle \psi_2 \middle| H \middle| \psi_2 \right\rangle &= \mathcal{E}_2 = \frac{\alpha - \beta}{1 - S_{12}} \tag{2}
\end{align}$$

$\mathcal{E}_1$ is the ground state energy (i.e. $\mathcal{E}_1 < \mathcal{E}_2$) since $\alpha$ and $\beta$ are both negative quantities (again MQM has more info).

Stabilisation and destabilisation energies

Now we have actually reached the part where the original question can be answered. What is the stabilisation of the bonding MO? The energy of the constituent AOs is $H_{11} = H_{22} = \alpha$. So, the stabilisation must be:

$$\begin{align}
E_{\mathrm{stab}} &= \alpha - \mathcal{E}_1 \\
&= \alpha - \frac{\alpha + \beta}{1 + S_{12}} \\
&= \frac{\alpha S_{12} - \beta}{1 + S_{12}}
\end{align}$$

Similarly, the destabilisation of the antibonding MO is:

$$\begin{align}
E_{\mathrm{destab}} &= \mathcal{E}_2 - \alpha \\
&= \frac{\alpha S_{12} - \beta}{1 - S_{12}}
\end{align}$$

The quantity $S_{12}$ is positive (again it should be in MQM), so $E_{\mathrm{destab}} > E_{\mathrm{stab}}$.

The "incorrect" piece of information comes from a further simplifying assumption - namely, that $S_{12} \ll 1$. That is not the greatest assumption, since the values of overlap integrals can sometimes be quite non-negligible (for a $\ce{H-H}$ bond length of $74~\mathrm{pm}$, $S_{12} = 0.753$!), but it is a useful way to predict approximate energies and the general forms of the MOs.

If $S_{12} \ll 1$, then $1 \pm S_{12} \approx 1$ and equations $(1)$ and $(2)$ become:

$$\begin{align}
\psi_1 &= \frac{1}{\sqrt{2}}(\phi_1 + \phi_2) & & & \left\langle \psi_1 \middle| H \middle| \psi_1 \right\rangle &= \mathcal{E}_1 = \alpha + \beta \tag{3} \\
\psi_2 &= \frac{1}{\sqrt{2}}(\phi_1 - \phi_2) & & & \left\langle \psi_2 \middle| H \middle| \psi_2 \right\rangle &= \mathcal{E}_2 = \alpha - \beta \tag{4}
\end{align}$$

Now it turns out that, with the additional neglect of overlap approximation, $E_{\mathrm{stab}} = E_{\mathrm{destab}} = -\beta$. That's where the "destabilisation of antibonding MO = stabilisation of bonding MO" idea comes from.

The arguments above can be extended to the case where you have constituent AOs which are different. The main difference is that $H_{11} \neq H_{22}$, so you have to write $H_{11} = \alpha_1$ and $H_{22} = \alpha_2$. The result you obtain will be rather similar: the destabilisation of the antibonding MO is greater than the stabilisation of the bonding MO, but in the limit of $S_{12} \to 0$, they become equal.


Chlorination of Water using sodium hypochlorite

The reaction as shown goes only a little to the right. The $\ce{HClO}$ shown is poorly ionized; on the other hand, the $\ce{NaOH}$ is completely ionized. How far the reaction goes to the right will be indicated by the pH, which is probably about 10, considering that sodium carbonate is added to commercial bleach to raise the pH to 11 for greater shelf life.

When chlorinating a swimming pool, long shelf life (in the pool) is not desired. You want some more activity, even if at the expense of having some of the $\ce{NaClO}$ divert to other products.

There are two variables in the swimming pool:

the chlorination amount
the activity, which will be controlled by the pH.

So you put in the chlorine, then dial up the best pH. As far as moving the equilibrium, perhaps it is better to show the reaction as,
$$\ce{ClO- + H2O -> HClO + HO-}$$

The original sodium hypochlorite will be totally ionized; the hydrolysis reaction is the $\ce{ClO-}$ ion attracting a proton from water to form somewhat nonionized HClO and some $\ce{OH-}$ ions. Then the prospect of adding more $\ce{OH-}$ ions is easily seen to move the equilibrium back toward the left, generating more $\ce{ClO-}$ ions. Likewise, adding acid generates more $\ce{HClO}$ in the solution.


How to create the total reaction out of the two steps behind denitrification?

After balancing carbon and nitrogen, I'd recommend you go for balancing the charge in the equation:

$$ \ce{6NO3- + CH3OH -> 3N2 + CO2 + H2O + 6OH-} $$

Now, we've to balance oxygen and hydrogen without disturbing the already balanced species. It would be interesting to note that here the stoichiometric coefficients of methanol and carbon dioxide should be the same or the number of carbon atoms wouldn't be equal on both sides of the equation.

So, let's assume the stoichiometric coefficient for both these two species to be $x$. Also, let's assume the stoichiometric coefficient of water on the right hand side to be $y$. The equation now looks like this:

$$ \ce{6NO3- + xCH3OH -> 3N2 + xCO2 + yH2O + 6OH-} $$

We now balance the number of oxygen atoms:

$$ 18 + x = 2x + y + 6 $$

And the number of hydrogen atoms:

$$ 4x = 2y + 6 $$

Solving these two equations, you get $x=5$ and $y=7$. And therefore your overall equation finally is:

$$ \ce{6NO3- + 5CH3OH -> 3N2 + 5CO2 + 7H2O + 6OH-} $$


Temperature is directly proportional to diffusion rate.

Diffusion, being the dispersion of molecules throughout a space, is affected by the speed of molecules and the density of the space through which the molecules are being dispersed.

An increase in temperature increases the speed at which molecules move at. This is because the average kinetic energy of the gas molecules is proportional to the absolute temperature of the gas. Increased speed will yield faster and more even diffusion.

Average Kinetic Energy of gas molecules = $\frac{3}{2}RT$.

Therefore, decrease in temperature will cause molecules to slow down (due to reduction in average kinetic energy). Hence rate of diffusion decreases.

I hope this helps!


Fires need oxygen to burn properly. By that I mean if there isn't sufficient oxygen supplied to the fire, the material that is burning does not completely burn up. They come to the middle of the combustion reaction but they run out of oxygen. These bits of incomplete combustion are called soot or smoke. $\ce{CO2}$ is another by-product of combustion. There are particular woods that give off certain smells and flavours like hickory, but smokey smells are probably only caused by carbon.


Are halogens a further classification of non-metals, or are they another group themselves?

Halogens are merely just a subset of non-metals. Consider this periodic table:

Clearly halogens do come under the fluoroscent colored elements, which signifies non-metals. There are several other groups, which have their own unique name. For instance, the oxygen group members are called chalcogens, and the nitrogen group members are called pnictogens.

However, there is something special about halogens though. They are infact the most non-metallic elements known, considering their great tendency to gain electrons. This also makes the halogen group the only group other than inert gases to have only non metals in them (ignoring Ununseptium of course).

In short, halogens are a subset of non-metals and not an entire class by themselves. But they are, the most important subset of non metals.


Name for analog of salt hydrate with non-water solvents?

The corresponding rule in the Red Book (‘Nomenclature of Inorganic Chemistry – IUPAC recommendations 2005’) reads as follows:

  The names of the individual components of such a generalized addition compound are each constructed by using an appropriate nomenclature system, whether compositional, substitutive or additive. The overall name of the compound is then formed by connecting the names of the components by ‘em’ dashes; the proportions of the components are indicated after the name by a stoichiometric descriptor consisting of arabic numerals separated by a solidus or solidi. The descriptor, in parentheses, is separated from the compound name by a space.

$\ce{CaCl2.8NH3}$ calcium chloride—ammonia (1/8)

$\ce{AlCl3.4EtOH}$ aluminium chloride—ethanol (1/4)

$\ce{BiCl3.3PCl5}$ bismuth(III) chloride—phosphorus(V) chloride (1/3)

$\ce{2Na2CO3.3H2O2}$ sodium carbonate—hydrogen peroxide (2/3)


Galvanic cell involving a hydrogen electrode and copper electrode in sodium hydroxide

In this case, "saturated" means that there is enough dissolved copper (II) hydroxide such that no more will dissolve, i.e. the equilibrium represented by the $K_\mathrm{sp}$, $\ce{Cu(OH)2_{\ (s)}<=>Cu^2+_{(aq)} + 2OH-_{(aq)}}$ holds true. So first, let's figure out the concentration of copper (II) hydroxide that will dissolve in 0.1 M NaOH (making the assumption, as mentioned in the comments, that $[\ce{OH^-}] = \mathrm{C_\ce{NaOH}})$:

The $K_\mathrm{sp}$ is really really small, so we can actually ignore
its contribution to the $[\ce{OH^-}]$, but for fun, you can do the
whole ICE table thing and end up with:
$$\begin{align}
     [\ce{OH^-}] &= .1 + 2x \\
     [\ce{Cu^2+}] &= x \\
     K_\mathrm{sp} &= [\ce{OH^-}]^2[\ce{Cu^2+}] \\
     1.6\times 10^{-19}  &= (.1 + 2x)^2(x) \\
     x = [\ce{Cu^2+}] &= 1.6 \times 10^{-17}\ \mathrm{M}
     \end{align}
   $$

Next, using the Nernst equation, we can calculate the copper half-cell reduction potential (Assuming T = 298.15 K):

$$\begin{align}
    E &= E^\circ - \frac{RT}{zF}\ln\frac{1}{[\ce{Cu^2+}]} \\
      &= 0.34\ \mathrm{V} - \frac{RT}{2F}\ln\frac{1}{1.6 \times 10^{-17}} \\
      &= -0.16\ \mathrm{V}
    \end{align}
    $$

Finally, the whole cell potential:

First, negate the calculated reduction potential to get the oxidation potential:
$$\begin{align}
    E_\mathrm{ox} &= -E_\mathrm{red} \\
      &= 0.16\ \mathrm{V}
    \end{align}
    $$
Then, because the SHE's reduction potential is defined as 0 V, the whole cell potential is the same as the oxidation half-cell potential:

$$\begin{align}
    E_\mathrm{ox} &= E_{(\ce{H})\mathrm{red}} + E_{(\ce{Cu})\mathrm{ox}} \\
                  &= 0\ \mathrm{V} + 0.16\ \mathrm{V} \\
                  &= 0.16\ \mathrm{V}
    \end{align}
    $$


Do endothermic reaction have higher energy products than reactants?

The simple answer is yes - as energy is required to be absorbed for the reaction, the products have more energy than the reactants.


How do I determine which liquids are immiscible with one another?

If you want high accuracy for solubility, UNIFAC is a good way to go. The basic idea behind it is to break the molecules down into groups then generate correlations for how each of the groups interact. The weighted sum of all the group contributions gives you the total molecular interactions. From there you can get solubility.

The Dortmund Data Bank maintains a comprehensive list of all the group coefficients and offers a program to get the various properties, like solubility. Unfortunately, the free version is pretty limited. You an only access a handful of fluids.

If you want to put in a bit more work, someone has put an Excel version on the internet. It doesn't explicitly do solubility, but basically if the activity coefficients are very far from 1, you can assume they are insoluble.

Furthermore, I'm not sure what the ferrofluids will do. I would guess that the concentrations would be low enough that you could just calculate solubility using the surfactant. Otherwise, you might have to search the literature for how to handle magnetic nanoparticles with UNIFAC.


Can you give examples for a halide ion reacting with interhalogen compound to give polyhalides?

TLDR: There is no exact generalized reaction. The reaction depends on type of interhalogen compound(value of n) and the nature of metal halides(ionic or covalent)

$\ce{XX^{'}}$ form

Let us take iodine monochloride($\ce{ICl}$). If it is reacted with ionic chloride like $\ce{KCl}$, it will form $\ce{ICl2-}$ ion.

$$\ce{MCl + ICl -> M+ICl2-, (M=K, Rb)}$$

If a different counter-ion is taken like bromide, it will form $\ce{ClBrI-}$ ion

$$\ce{ICl + KBr -> K+[ClBrI]-}$$

But if we add pure diatomic halogen, it will form a additive product

$$\ce{Cl2(excess) + ICl -> ICl3}$$

Covalent halides like $\ce{AlCl3}$ however gives different product:

$$\ce{ICl + AlCl3 -> I+ + AlCl4-}$$

$\ce{XX^{'}3}$ form

Let us take iodine trichloride. It will form $\ce{ICl4-}$ ions on reacting with ionic chloride:

$$\ce{MCl + ICl3  -> M+ICl4-}$$

If we take any other counter ion like fluorine, it will form $\ce{MICl3F}$ which is again kind of an additive product.

$\ce{XX^{'}5}$ form

Taking chlorine pentafluoride as examples, reacting with covalent fluorides like $\ce{AsF5}$ and $\ce{SbF5}$ will give adducts: $\ce{ClF5.AsF5}$ and $\ce{ClF5.SbF5}$

Iodine pentafluoride gives iodine heptafluoride on reacting with fluorine at 250 C

So, there is exact no generalized reaction. It all depend upon type of inter-halogen compound, reaction conditions and nature of metal halide. The close generalization I can make is that interhalogen compound of form $\ce{XX^{'}_n}$ if reacted with ionic metal halide will form metal polyhalide of form $\ce{MXX^{'}_{n+1}}$ or $\ce{MXX^{'}_{n}X^{''}}$(if a different counter-ion is taken). Additive product is formed when inter-halogen compound is reacted with pure diatomic halogens provided the second halogen is big enough to occupy electrons like bromine or iodine. In some case, covalent halides forms adducts with interhalogen compounds.

Thus the generalized reactions are(of course, there are certain exceptions):

$$\ce{X^{'-} + XX^{'}_n -> XX^{'-}_{n+1} (ionic)}$$

$$\ce{X^{'}_2 + XX^{'}_n -> XX^{'}_{n+2}}$$

Reference

http://www.vpscience.org/materials/Study%20Material%20on%20d%20Inter-halogen%20%20Compounds%20By%20Dr.%20D.%20M.%20Patel.pdf (reaction sources)


Why is there a factor of half in the definition of bond order?

Because there are two electrons in a covalent bond! Say all the electrons you have is two bonding electrons... Would you call that a bond order of two? Or is it just a good old single bond?


When does copper leach to water? Conflicting faqs

The second answer would be correct in an analytical context as the auto-ionization of water decreases in pH as temperature increases. And new copper does not have the surface of reacted material to shield the metal underneath, as such, scenario 2 always occurs, scenario one occurs faster, however, it's conditional statement that the water be acidic and the copper be corroded are invalid.
Cheers,


Amount of substance produced in chemical reaction

In the reaction:

$$\ce{N_2 + 3H_2 \to 2NH_3}$$

If 15 moles of $\ce{N_2}$ is used along with 30 moles of $\ce{H_2}$, then $\ce{H_2}$ will be the limiting reagent (Reactant which is consumed completely). This is because 15 moles of $\ce{N_2}$ will need 45 moles of $\ce{H_2}$ ($15 \times 3$, by the balanced reaction) for reacting completely, but as there is only 30 moles available hence it will react in a fixed proportion, hence you will have to assume the reacting moles of $\ce{N_2}$ as $x$. By this the equation goes,

$$\frac{\ce{N_2}}{\ce{H_2}}=\frac{1}{3}=\frac{x}{30}$$
From this
$$x=\frac{30}{3}=10$$

Hence from 15 moles of $\ce{N_2}$ only 10 moles will react, hence the amount of $\ce{NH_3}$ formed will be 20 moles ($10 \times 2$, by the balanced reaction).


What does ingredients being "biologically active" mean?

Chemists usually marks as biologically active compounds, that have clear effect on living being in small amounts. For example, methane is not biologically active, but potassium cyanide clearly is. Of course, the borderline is quite blurry and can move if context changes.


Suitable coating material for protection of sensors

Various DIY capacitive moisture meters use a PCB for the sensor:

While not directly related to your question I thought I'd mention that you can buy these sensors for several dollars:

Purchasing a readymade kit and relying on the discount of mass production (along with a manufacturer's warranty) is probably easier than designing your own, and etching your own PCB.

I think how you plan to make the probe plays a part in how you plan on coating it; for example you could fit the parts in a test tube, that would meet all the requirements in your list. Such a capacitive sensor would operate similar to the touchscreen of a cellphone.


Energies of d orbitals relative to the barycentre

The passage you quoted is perfectly correct and contains one word which you may have overlooked:

  the weighted mean of these two sets of perturbed orbitals is taken as zero [...]

In this case the weight corresponds to the degeneracy of the set. The lower-energy $\mathrm{t_{2g}}$ set contains three orbitals and the higher-energy $\mathrm{e_g}$ set contains two. If we place the $\mathrm{t_{2g}}$ set at $x \Delta_\mathrm{O}$ and the $\mathrm{e_g}$ set at $y\Delta_\mathrm{O}$, then

$$\begin{align}
y - x &= 1 & &\text{(difference must be equal to }\Delta_\mathrm{O}) \\
\frac{3x + 2y}{3+2} &= 0 & &(\textit{weighted}\text{ mean is equal to zero)}
\end{align}$$

and $(x,y) = (-0.4, 0.6)$.


Finding structure of a compound given its reactions

I have taken the liberty of identifying your unknown compound.  If you'd like to work through the problem yourself, then carefully scroll down to the part of the answer that deals with the iodoform test.  Good luck!

What we know

Just looking at the molecular formula we see that there are too many hydrogens for the 6 carbons to be in a benzene ring.
does not change the colour of acidic dichromate solution - acidic dichromate readily oxidizes primary and secondary alcohols to aldehydes and ketones respectively; tertiary alcohols are not oxidized. The aldehyde produced from a primary alcohol can be further oxidized to a carboxylic acid, this results in a solution color change from orange to green. But there was no color change, therefore there is no primary $\ce{-OH}$ or aldehyde group in compound A.
Compound A on treatment with H2SO4 produces alkene - this is a common reaction that causes deydration (elimination of water) and converts an alcohol into an olefin.  Therefore compound A is (as you suggested) an alcohol (but not a primary alcohol).
oxidative ozonolysis gives a molecule $\ce{C6H10O3}$ - ozonolysis will cleave the double bond in the compound produced upon dehydration.  The key fact here is that the resulting product still contains 6 carbon atoms.  If the double bond involved carbons outside of a ring we would wind up with an ozonolysis product containing fewer than 6 carbons.

(image source)

Only if the double bond is contained within a ring will we retain all 6 carbon atoms.  Therefore our double bond must be in a ring.

positive iodoform test - This is our final clue, the iodoform test (or more generally the haloform reaction) is a test that detects the presence of methyl ketones and generates iodoform ($\ce{HCI3}$) in the process.

Therefore we know that there is a methyl ketone in our ozonolysis product.

Combining all of this information (non-aromatic, ring, non-primary alcohol, methyl group) leads to the following structure

(mouse over to see the answer)

which is called

   1-methylcyclopentanol.

Note: some methylated-cyclobutanols might also be considered, but when these dehydrate and produce an alkene, the double bond is likely to be placed external to the ring (or [advanced topic] external to a cyclopropane ring formed upon rearrangement) and we would not wind up with 6 carbons in our ozonolysis product.


How will metallic lithium react with atomic oxygen?

As explained in Long-term ozone decline and its effect on night airglow
intensity of Li 6708 ˚A at Varanasi (25◦N, 83◦E)
and Halley Bay (76◦S, 27◦W) Journal of Earth System Science 120, No. 2, April 2011, pp. 291–300:

$\ce{Li + O -> LiO}$

$\ce{Li + O2 -> LiO2}$

$\ce{Li + O3 -> LiO + O2}$

$\ce{LiO2 + O -> LiO + O2}$

The article gives rate coefficients for each of these reactions.

The article also gives $\ce{Li}$, $\ce{O}$ and $\ce{O3}$ concentrations as a function of altitude for the range 80-99km (85km being considered the lower boundary of the thermosphere)


What is the difference between enthalpy (H) and heat (q)?

Enthalpy and heat are entirely different things.

Enthalpy is a function of state. If you know the state of a system, you know its enthalpy. If you know the starting and ending states of a process, you can find the enthalpy change.

Heat, on the other hand, is an inexact differential. Knowing the initial and final states of a process is not enough information to tell you the heat transfer. Instead, the heat transfer depends on the particular path taken between the states.

For a simple example of why this is important, consider a heat engine. The process is a cycle, meaning the initial and final states are the same, so a cycle of a heat engine has zero enthalpy change. However, the entire point is that the heat engine converts heat into work, so the heat exchange during a cycle is not zero.

That is why the heat exchanged is equal to the enthalpy change only under constant pressure. We need to give extra information (such as that the pressure is constant) to know the path taken during the process in order to say anything meaningful about heat exchange.


Thermal decomposition of group 2 permanganates

It is observed that true composition of alkaline earth metal manganate is not known. It is proposed to be a mixture of manganite-manganate and has a general formula, $\ce{R3Mn2O8.H2O}$. However the composition/formula varies according to given method of preparation. It is found on some instance that heating the permangantes of alkaline earth metal gives manganates while in other case it gives manganites.

Mangesium permanagate is prone to decomposition and it explode
violently in aqueous solution giving oxygen, ozone and the
manganate?(the reaction product is not defined properly?)
Calcium manganate is formed by decomposing calcium
permanganate.
Strontium permanaganate is also prone to decomposition and
forms either manganite or manganate depending on reaction condition
but either way, strontium manganate is formed by heating
strontium manganite at 1600 °C.
Barium permanganate also decomposes at 320 °C to form barium
manganite but if the permanganate is mixed with barium hydroxide,
barium manganate of composition $\ce{Ba3Mn2O8.H2O}$ is formed.
The reaction of beryllium permanganate decomposition is unknown to
the best of my knowledge mainly due to poor study on the compound.


Realistic chemical spill accident in high school chemistry class

Collecting together the responses from the comments and Ben Norris's answer and adding one of my own, individual solutions in water of the following chemicals are realistic options for a liquid spill in a high school chemistry lab, leading to destroyed clothes and time under a safety shower:

Potassium permanganate $(\ce{KMnO4})$
Sulfuric acid $(\ce{H2SO4})$
Hydrochloric acid $(\ce{HCl})$
Nitric acid $(\ce{HNO3})$
Sodium hydroxide $(\ce{NaOH})$

As for an experiment leading to a spill, one possible example would be in the course of filling a buret in preparation for a titration. Burets are usually quite tall and narrow (click image for larger version):

      (Marketing image from Microtech Scientific)

A slip while filling one could easily result in spilling a few hundred milliliters or more of liquid on oneself.


What will be the order of acidity in the following compounds?

I think you are correct. The $\mathrm{p}K_\mathrm{a}$ of 4-methylphenylacetic acid is quoted as 4.37 here. The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is quoted as 4.19 by the same source here.


What's the chemical formula of "Everitt's salt"?

According to various sources Everitt's Salt has the formula $\ce{K2Fe[Fe(CN)6]}$, where both $\ce{Fe}$ are in the $+2$ Oxidation State. Everitt described it as a yellow precipitate, resulting from the reaction:

$$\ce{2 K4Fe(CN)6(aq) + 6 H2SO4(aq) \to 6 HCN(g) + 6 KHSO4(aq) + K2Fe[Fe(CN)6](s)}$$

Acc. Everitt, the wet salt can be oxidised to Prussian Blue by the action of air (oxygen).

Source.


Will aluminum continue to corrode after its surface is oxidized?

No. Aluminium stops corroding once a strong oxide layer is formed.

Aluminium is a fairly reactive metal. The reason why it usually survives in air and water under normal conditions is because the aluminium oxide layer on the surface is very strong and normally protects the metal from further attack. Even if the layer is disrupted, it reforms quickly preventing any further reaction. Aluminium oxide is a very strong compound and impervious to water and oxygen. In bulk form it is known as corundum and appears in nature as ruby and sapphire, just about the hardest non-diamond minerals.

But there are substances that can disrupt that layer and expose aluminium to attack from air and water. Mercury and its salts and gallium metal both do this. Once they penetrate the surface layer, aluminium will be destroyed fairly rapidly and will crumble to a weak bulky mess. See examples on youtube from NurdRage and NileRed.


A chemical compound with these conditions?

I will satisfy 4 of the five conditions, but you will have to research the fifth. Lead(II) iodide has a layered structure, is slightly soluble in water, is ionic (but only weakly). I hope that this helps.


How to determine whether a compound is gas or liquid or solid?

As the molecular weight increases (more halogens or carbon atoms), so does the electron density and consequently the van-der-waals forces(dispersion forces) between the molecules rises. These forces increase intermolecular attraction and therefore increase the boiling point. T^his is valid in case of most other organic compounds' homologous series, not just alkyl halides.