Water color additive that will maintain color when water is a thin film

Use fluorescein or other fluorescent dye, along with an effective UV filter (AKA haze or skylight filter) on the lens and an ultraviolet source, such as a xenon strobe (speed light) or UV flashlight. A UV-transmission filter on the strobe can increase contrast, and can be made cheaply, but might not be needed. Because fluorescence makes the fluid become a source of light, it is quite obvious against a dark background. You can even use most laundry detergents, which contain blue-fluorescing dyes -- though foaming might be a nuisance.

See this image of fluorescent soap bubbles or this one. Soap bubble film may be just a few hundred nm thick, yet show clearly with fluorescence. Also watch a demo of strobe illumination of water droplets using sodium fluorescein for contrast. In the video, notice, in particular, that the stream tends to break into pairs of droplets, one perhaps a millimeter in diameter, and the sibling, just a few microns across, but all clearly visible!

Choose a near-UV light source and fluorescent dye for safety, since shorter wavelengths are more damaging to eyes and skin. A search for UV flashlight returned items from US$10 and up, some of which might be suitable. Most were in the 395 to 365 nm spectrum, relatively innocuous.


Do 2-hexyne and hexa-2-yne mean exactly the same thing?

For 2-hexyne, the extra naming possibility isn't really a help.

However, consider a more complex molecule, say, a hexynol.  It has both an alkyne and an alcohol.  A name like 2,4-hexynol would be very confusing.  Which group is at carbon 2 and which is at carbon 4?  That is why IUPAC doesn't provide for that possibility.  Instead, Writing it as hex-4-yn-2-ol is actually clearer and more understandable.


No both are right.

If you add the latter equations, then you will get the former.

If sodium hydroxide solution was highly concentrated like you said, then both equations are wrong since sodium hydroxide solution will strip away the second hydronium as well forming carbonates.

All state notations with (aq) implies that it is dissolved in water. If it was pure NaOH, it would be in a solid state under room temperatures. It is also noteworthy to know that NaOH must be contained in a dry environment as it will spontaneously absorb water from the air.


Why does AsF3 boil at 60.4 Celsius and AsF5 at -52.8?

This has to do with the dipole moment of those compounds. As the difference in electronegativity between As (2.18) and F (3.98) is very large, those bonds are highly polarized.

Using the VSEPR theory, $\ce{AsF3}$ is an $\ce{AX3E}$ molecule. As such, it has a pyramidal structure. Therefore, there is a separation of charge between $\ce{F^\delta-}$ and $\ce{As^\delta+}$ in the molecule, leading to a dipole moment. This implies dipole-dipole interactions between those molecules, which greatly increases the boiling temperature.

However, $\ce{AsF5}$ is an $\ce{AX5}$ molecule and has a trigonal bipyramidal structure. The centers of charges coincide and the net dipole moment is null. There is no dipole-dipole interaction between the molecules and the boiling point is much lower.


Which of the following can be a source of error in Spectrophotometry? Mark the wrong answer

It is a very open ended question, but first three choices will cause errors. Regarding your pH issue, you are right, it is multiplicative error. Take the example of phenolphthalein, as you say that its color (=absorption spectrum) is different in acidic or basic medium. So pH change can shift the lambda max of analytes, hence the cause of error.
Phenolphthalein still absorbs light in the acidic medium. I cannot find phenolphthalein spectra, but let us take phenol red indicator, you can see drastic changes in absorbances at the lambda max as a function of pH.
If by some random chance, you were measuring at the isosbestic point, then pH will not affect absorbance but in all other cases pH will change the A values drastically.


What is the hybridisation of the carbons in the allyl radical?

Radicals are electron-deficient compounds. You should not treat a radical as a lone pair, as they are pretty different.

A good rule of thumb is to assume an orbital that contains the highest possible p-contribution for the single electron. The more s-contribution is mixed in, the ‘closer’ the single electron gets to the nucleus which is less favourable than having an electron pair in a similar orbital. A pure p-orbital fulfils the condition perfectly by having no s-contribution whatsoever. Thus, you should assume radicals to be in p-orbitals whenever possible.

This is similar to carbocations which have an orbital fully devoid of electrons. However, this empty orbital must always be a p-type orbital for electronic reasons. Radicals have the possibility of undergoing hybridisation and e.g. ending up in an $\mathrm{sp}^n$ orbital (examples: radicals on $\mathrm{sp^3}$ hybridised bridgehead atoms; phenyl radical).

Since we just established that the radical will occupy a p-type orbital and since there is a π bond nearby, we can delocalise. In resonance structure terms, this can be visualised in the following way:

$$\ce{H2C=CH-CH2^. <-> H2C^.-CH=CH2}$$

The radical is delocalised across two carbon atoms making the overall system more stable than if it weren’t. In molecular orbital terms, this is basically a stabilisation of the double bond at the expense of the radical-containing orbital which benefits the overall system. The delocalisation can, of course, only occur if there is a π bond nearby; the n-propyl radical does not feature one.


Why does buffering capacity increase with the pKₐ of the acid component?

Buffer capacity depends on the $K_\mathrm{a}$ of an acid. I will use the explanation that can be found in David Sheehan: Physical biochemistry: principles and applications. John Wiley & Sons, 2013.

Any aqueous solution containing both $\ce{A-}$ and $\ce{AH}$ is, in principle, capable of resisting change in $\mathrm{pH}$ according to
$$\ce {A^- + H^+ <=>> AH}.$$
If alkali is generated in the solution (that would tend to remove protons) we have
$$\ce{AH + OH- <=> A- + H_2O}.$$

Most buffers consist of mixtures either of a weak acid and its salt or of a weak base and its salt. The ability to resist change in $\mathrm{pH}$ is finite, especially if the number of protons involve is especially large. This limit is represented by the buffering capacity of the buffer $\beta$.
This is defined as the amount of substance in moles of $\ce{[H^+]}$ which must be added to a liter of the buffer to decrease the $\mathrm{pH}$ by one unit. It can be mathematically calculated by
$$\beta = \frac{2.3 K_\mathrm{a}[\ce{H+}][C]}{(K_\mathrm{a}+[\ce{H+}])^2},$$
where $[C]$ is the sum of conentrations of $\ce{A-}$ amd $\ce{AH}$. This relationship means that buffering capacity increases with the buffer concentration. Buffers work best at $\mathrm{pH}$ values around their $\mathrm{p}K_\mathrm{a}$  most pH are effective one $\mathrm{pH}$ unit above an one below their $\mathrm{p}K_\mathrm{a}$.
I hope that the formula previously reported help in clarifying how the buffer capacity is related to $\mathrm{p}K_\mathrm{a}$ (since they are indeed related).

Example (based on ZUMDAHL, Steven. World of chemistry. Cengage Learning, 2012.)

A chemist needs a solution buffered at $\mathrm{pH}~4.30$. choosing from ethanoic,  propanoic, butanoic acid. We can calculate the ratio  $\ce{[HA]/[A^-]}$ required for each system. Considering that a $\mathrm{pH}~4.30$ corresponds to $\ce 5\times 10^{-5} M$ using the equation
$$[\ce{H+}]=K_a \frac{[\ce{HA}]}{[\ce{A^-}]}$$
we substitute the required $[\ce{H+}]$ and $\ce{K_a}$ for each acid to calculate the ration $[\ce{H+}]/[\ce{A-}]$ needed in each case is:

\begin{array}{llr}
\text{Acid} &
[\ce{H+}] = K_\mathrm{a} \frac{[\ce{HA}]}{[\ce{A^-}]} &
\frac{[\ce{HA}]}{[\ce{A^-}]} \\  \hline
\text{Ethanoic} &
5.0 \times 10^{-5} = 10^{-4.76} \frac{[\ce{HA}]}{[\ce{A^-}]} &
2.88 \\[1ex]
\text{Propanoic} &
5.0 \times 10^{-5} = 10^{-4.87} \frac{[\ce{HA}]}{[\ce{A^-}]}  &
3.749\\[1ex]
\text{Butanoic} &
5.0 \times 10^{-5} = 10^{-4.9} \frac{[\ce{HA}]}{[\ce{A^-}]}  &
3.98  \\ \hline
\end{array}

In this case the choice would be ethanoic acid since it has the ratio closes to 1.

If we need a buffer near $\mathrm{pH}~5.5$ the best choice is butanoic acid:

\begin{array}{llr}
\text{Acid} &
[\ce{H+}] = K_\mathrm{a} \frac{[\ce{HA}]}{[\ce{A^-}]} &
\frac{[\ce{HA}]}{[\ce{A^-}]} \\  \hline
\text{Ethanoic} &
3.2 \times 10^{-6} = 10^{-4.76} \frac{[\ce{HA}]}{[\ce{A^-}]} &
0.18 \\[1ex]
\text{Propanoic} &
3.2 \times 10^{-6} = 10^{-4.87} \frac{[\ce{HA}]}{[\ce{A^-}]}  &
0.23 \\[1ex]
\text{Butanoic} &
3.2 \times 10^{-6} = 10^{-4.9} \frac{[\ce{HA}]}{[\ce{A^-}]}  &
0.25 \\ \hline
\end{array}


Standard Hydrogen Electrode

A SHE consists of four parts: a platinum lead wire, a platinum foil contact, an acid solution, and a hydrogen gas flow. The word standard refers to the fact that all concentrations and pressures are at standard state, thus the acid is 1M and the gas flow is 1 atm. At these conditions, the electrode has a potential of 0 V (this is how potential is defined). If there are any deviations from these values, the electrode is no longer considered a SHE. The potential can then be calculated using the Nernst equation.

The electrode is one half of an electrochemical circuit. The other half cell determines whether the SHE will behave as the cathode or the anode. If it is the anode, the electron flow comes from the external system to the platinum contact. 2 electrons reduce 2 acidic hydrogens from the solution to form one molecule of hydrogen, as in $2H^+ + 2e^- → H_2$. The molecular hydrogen then bubbles up above the solution. If the electrode is cathodic, hydrogen gas is absorbed on the platinum contact plate and oxidized to two acidic hydrogens which remain in solution. The electrons then flow to the external circuit. Electrons do not flow from hydrogen gas to the acid, or vice versa.

Reaction Quotient

The reaction quotient is indeed unitless. While it is common to use concentrations and pressures to calculate Q, it is more properly calculated by using activity of each species, which are also unitless. The activity is the apparent concentration or pressure of a species (can be viewed as a correction to concentration or pressure). Because in most chemical applications activity is very close numerically to concentration, using concentrations or pressures directly for calculations is a close approximation.


Comparing rates of hydrolysis in cyclohexane ring (involving neighbouring group participation)

The third reaction infact occurs via neighbouring group
  participation.

(https://en.wikipedia.org/wiki/Neighbouring_group_participation)

In the third compound $\ce{S}$ atom attacks carbon containing $\ce{Cl}$ by $\ce{SN_{NGP}}$ forming sulphonium ion. An attack by $\ce{H2O}$ on the least hindered carbon of the three member ring occurs. In this case both the carbopns in three member ring will be similar though. So the third compound has highest rate of reaction.

(I'd like to add "This can only happen in the trans configuration, however it requires ring flip into the higher-energy trans-diaxial conformation." as pointed out by - @orthocresol in the comments to the orginal question.)

  Adding as a proof that trans reacts faster than cis form during NGP in cyclohexane ring.
  http://meskvmcollege.org/upload/study_material/30275.pdf (See page 10
  of the pdf)

The first compound has second highest rate. Anchimeric assistance cannot occur in cis form. Negative inductive effect by $\ce{Cl}$ enhances the rate. Reaction mechanism is $\ce{S_N1}$ and water itself acts as nucleophile.
The central compound has the lowest rate among the three. $\ce{S_N1}$ mechanism occurs and $\ce{Cl}$ is replaced by $\ce{OH}$ where water itself acts as nucleophile.

  So summing up: $$\ce{rate_3>rate_1>rate_2}$$


Part of the discrepancy appears to be due to the need for two $\ce{O-H}$ dissociation energy terms to compute the $\ce{OH}$ bond energy for water: one for the $\ce{OH}$ bond in water (i.e. $\ce{H-OH}$), and another for the $\ce{OH}$ bond in the $\ce{OH}$ radical (i.e. $\ce{^.O-H}$). Values for these are reported by Ruscic et al. [1]. Using the average value of $\ce{D_0}$ for $\ce{H2O}$ (the $\ce{OH}$ bond energy for water), namely $\pu{459 kJ/mol}$, gives

$$\ce{2H2 + O2 -> 2H2O} \quad \Delta H = \pu{-466 kJ/mol}$$

This point is addressed in the Wikipedia page for "bond dissociation energy", the reference cited therein providing a slightly different bond energy for water of $\pu{461.5 kJ/mol}$.

Given various uncertainties common to the measurement of bond dissociation energies, and the potential effect of temperature on the bond energy, the final discrepancy seems reasonable.

References

Ruscic, B.; Wagner, A. F.; Harding, L. B.; Asher, R. L.; Feller, D.; Dixon, D. A.; Peterson, K. A.; Song, Y.; Qian, X.; Ng, C.-Y.; et al. On the Enthalpy of Formation of Hydroxyl Radical and Gas-Phase Bond Dissociation Energies of Water and Hydroxyl. The Journal of Physical Chemistry A 2002, 106 (11), 2727–2747. https://doi.org/10.1021/jp013909s.


How to use gas laws to calculate the partial pressure of a gas that has partly condensed from a mixture?

You would need the initial temperature to be able to solve this.

You could use the combined gas law (or Gay-Lussac's Law) to determine the total pressure of the gas at -100°C, assuming none had condensed.  This ideal gas pressure would differ from the 82.5kPa by the partial pressure of the condensed gas at -100°C.

The question doesn't specify at which conditions it wants to know the partial pressure of the missing gas.  If it's -100°C, you'd be done at that point.  If it's at the original temperature, you would have to apply the same gas law in reverse to get back to your original conditions.

The boiling point information you added should allow you to determine which of the three gases condensed.  Only one would be a liquid or solid at -100°C.


You can do your experiment. The active ingredient in most antacids is bicarbonate, and bicarbonates are amphoteric: they can be both acids and bases. In most cases, bicarbonate is a weak base, so despite your bromothymol blue outcome, bicarbonate is a much weaker acid than $\ce{HCl}$, and you should be able to do your titration.


Why is trans-1,2-dimethylcyclohexane an enantiomer?

No, they can't.

A model from any chemical modeling kit will convince you right away.  Or maybe a cheaper alternative made out of chewing gum and matchsticks will suffice.

Or better yet, you may do it in your head. Place the molecule in a horizontal plane (or rather, as close to a plane as possible, since the thing is not planar). Now place your eyes in the same plane, so as to look at the molecule from the side. Walk around until you are facing the bond that connects two methylated carbons in the ring. Look at that bond. Look also at the two methyl groups to the right and to the left from it. You either have the "left one up, right one down" isomer of your molecule, or vice versa. You might think "what the heck, I'll turn it around." No way. You turn each enantiomer around, and it still ends up looking like itself, and not like the other one.

Another way to think of it is via symmetry elements. For a molecule, not being chiral implies having certain kinds of symmetry. Earlier revisions of your question suggest that you were looking for some symmetry plane. Just where it might be? Think of it. And if you don't find that plane after a while, consider the possibility that it might be simply not there (which is true).

Ring inversion is irrelevant altogether. It can be done, but it will switch all equatorial substituents to axial and vice versa, so you won't be able to recognize the same molecule until you invert it again to the original form.

So it goes.


Should the measurement uncertainty be included in the random uncertainty?

Yes - to calculate the total uncertainty when adding or subtracting measurements, it's the room-sum-squared of all the uncertainties.

$$
e_{tot} = \sqrt{(e_{meas})^2 + (e_{rand})^2}
$$
The more generalized form of this will be the root-sum-squared of all sources of error and uncertainty.
$$
e_{tot} = \sqrt{\sum{(e_n)^2}}
$$

To get the quantity "$e_{meas}$" you apply the root-sum-squared formula of all the uncertainties of your measurement devices used to transfer liquids (transfer beaker, graduated cylinder, burette, etc...).

$$
e_{meas} = \sqrt{(e_{beaker})^2 + (e_{burette})^2+(e_{grad.cyl})^2+(e_{burette})^2...etc}
$$

=========

To answer whether you apply the burette uncertainty to your measurement, the answer is also yes. Assuming that the burette has an uncertainty of $±0.01 cm^3$, the actual reading of your measurements will be: $33.3cm^3 ± 0.01 mL$ and $33.4cm^3 ± 0.01 mL$. The two uncertainties in the measurement will be applied to the $e_{tot}$ equation above.


Meaning of the term "vertex" in the topic of skeletal formulae

The term vertex (plural vertices) comes from the nomenclature used in graph theory, a branch of mathematics which describes how points in a given space are connected to each other. This turns out to be quite fundamental to chemical nomenclature, as the vertices of a graph can (generally) be chemically interpreted as an atom, and the lines coming out of a vertex can (generally) be interpreted as covalent bonds. Two graphs which contain the same vertices, but where the connectivity between the vertices is different, corresponds in general to different chemical compounds.

To be clear, the term vertex applies to any point in a skeletal formula which is not "in the middle" of a connecting line. This means that every explicitly labelled atom, every "kink" in a chain (which implicitly represents carbon atoms) and every termination (also implicitly a carbon atom) is a vertex. Hydrogen atoms are often not vertices in a skeletal formula, though this can vary.

Regarding to your example, with no other context, chemists would call that a skeletal drawing of methylcyclopentane, which contains six vertices in its skeletal formula, one for each non-hydrogen atom. For it to be understood as a methylcyclopentyl radical/substituent, either the context should clearly state it, or preferably we would use other standardised graphical notation, shown below:


Mass chromatogram : Total Ion Chromatogram (TIC)

Does this mean it is a (mixture) sample injected into a MS without the preceding chromatographic separation (as is typical in, say, a GC-MS or an LC-MS)?

No. A chromatogram is the measurement of something after separation by chromatography. So you got chromatography beforehand. In case of direct injection into MS the term is usually used too, if you record over time like you would do using chromatography.

  What's the point in doing this?

Ok so let's imagine I'm running a sample of an unknown mixture and I'm recording m/z 100-1000. Now I don't know at all what to expect, so selecting randomly a specific mass for my chromatogram would, most likely, lead to nothing. What we can do is just integrate over the whole mass range and take the total number of all ions, so we see a peak this means there are more ions than for baseline noise. That should correspond to a substance reaching the detector.

  Wouldn't the resultant spectrum be very noisy since it's essentially the sum over all components of a mixture of unknown proportions.

Depending on the method: yes that might be a problem. And if the noise is very high then even a strong signal of a single m/z wouldn't really contribute to the total number of ions, so we won't see a peak in the TIC.

A way to circumvent this problem is using a base peak chromatogram. Base peak is the highest peak in your spectrum, and you are plotting the intensity of the highest peak over time. Now for noise there might be a large amount of total ions, but the intensity of each m/z is usually small. If you got a specific signal from a substance you will get much higher signals there, so the base peak chromatogram will show a peak.

  What's the alternative?

Collecting data in TIC mode is an alternative to collecting data in SIM mode. SIM stands for "single ion monitoring" or "selected ion monitoring". Let's say that you know your sample contains acetophenone, and the four largest peaks for acetophenone are at 120 (molecular ion), 105 (base), 77, and 51 m/z. Instead of integrating over all ions, you could monitor only those masses (or only one of them). You would know where the acetophenone eluted and could quantify it relative to a standard, but you would know little about the rest of the sample.

By analogy, consider a LC system with a UV/Vis photodiode array detector. The detector can monitor single wavelengths or it can capture a full absorbance spectrum at each time interval. A mass spectrometer collecting data in TIC mode is like a PDA detector collecting a full spectrum. A mass spectrometer collecting data in SIM mode is like a PDA monitoring individual wavelengths.


Ethanol's molecular mass $M(\ce{C2H6O})=46\ \mathrm{g/mol}$

$$n=\frac mM,$$
so
$$\frac{460\,000\ \mathrm g}{46\ \mathrm{g/mol}} = 10\,000\ \mathrm{mol}$$

So having 10 000 moles we can multiply that by the molecular mass of carbon dioxide and multiply again by two because there are two carbon dioxides.

$$10\,000\ \mathrm{mol}\times44\ \mathrm{g\ mol^{-1}}\times2=880\,000\ \mathrm g=880\ \mathrm{kg}$$


Why are acetic acid and sodium bicarbonate mixed to unblock a drain?

Quoting text from this source under the section "Why could baking soda and vinegar clean clogged drains?" :

Bicarb soda and vinegar react because of the acid-base reaction. Bicarb soda is bicarbonate $(\ce{NaHCO3})$ and vinegar is acetic acid $(\ce{CH3COOH})$.

When bicarb soda and vinegar react, they fizzle and sizzle and they expand. This is why they are able to remove clogs from drains. The pressure from the expanding product shoves the clog down as it moves along. The vinegar solution (vinegar plus water) can remove soap residue that clings to the walls of the pipes.

So it seems like your conjecture is justified, that it is not actually the reactivity of these two substances with the waste that is responsible for majority of the cleansing, but simply the mechanical action of the expansion of the carbon dioxide gas and the forming product that pushes the waste down the drain, and leads to unclogging.
The relevant chemical equation is:
$$\ce{CH3COOH (aq) + NaHCO3 (s) -> CH3COONa (aq) + CO2 (g) + H2O (l)}$$
This is similar to the mechanism by which baking powder makes a cake fluffy on addition of a little vinegar as explained here.


When and why do hot material take heat from cold material?

The statement is wrong and/or poorly formulated. Heat cannot spontaneously flow from cold areas to hot areas, there are no exceptions known. It contradicts Clausius' statement of the second law and hence also the more general observation/statement that the entropy of an isolated system is increasing.

Of course you can drain heat from cold regions and transport it to hotter regions, but this should always go paired with some other (larger) increase in entropy somewhere else, e.g. by doing work like in a Linde refrigeration. An other example is that some salts have a very high entropy of solvation and can cool water to ~ -15 °C in a lab at room temperature. These salts can be/are used in practical chemistry for cooling a reaction mixture.


Basic and nucleophilic character of NaOH (aq)

Even if, The reagents you are using are the same for both substitution or elimination - the halogenoalkane and either sodium or potassium hydroxide solution. In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factor.

Nucleophilic substitution reaction has $\ce{SN2}$ reaction mechanism.

For $\ce{SN2}$ favors the reaction with primary substituted alkyl halide more than secondary than tertiary, because its effected by steric hinderance. The more crowded it is, the more difficult it is for a nucleophile to attack.

So, whatever you do with tertiary halogenoalkanes, you will tend to get mainly the elimination reaction, whereas with primary ones you will tend to get mainly substitution.

For a given halogenoalkane, to favor elimination rather than substitution, use:

higher temperatures
a concentrated solution of sodium or potassium hydroxide
pure ethanol as the solvent

To favor substitution rather than elimination, use:

lower temperatures
more dilute solutions of sodium or potassium hydroxide
more water in the solvent mixture


Significance & applications of polarizability for nucleophiles

The examples the site you mentioned gave of uncharged nucleophiles were: the hydrides of Group V and VI. They say that the bigger atoms can have a more effective overlap of atomic orbitals and thus, the bigger the uncharged nucleophile, the stronger it is.

I would like to dispute this reasoning and answer your question regarding this trend.

How we should see a nucleophile should always primarily be in terms of electron donation because a nucleophile donates electron density to electrophiles to form bonds. This is a very important consideration.

For bigger atoms, the valence electron pair, which is being donated to the electrophile, is further away from the nucleus and is more shielded from the nuclear charge. Thus, it is easier for donation to take place. This should be the reason why uncharged nucleophiles with bigger atoms are more nucleophilic; that is, they are more capable of donating the electron pair of theirs. My explanation of the trend has nothing to do with polarisability.

The reasoning stated is incorrect because although bond formation is easier for bigger atoms, it does not mean that the overlap of atomic orbitals is more effective. In fact, it is likely less effective because the atomic orbitals used are more diffused (Recall that as the principle quantum number n increases, the orbitals become more diffuse in terms of electron density). Thus, their reasoning is nonsensical.

Hope I have clarified everything you were not sure of.


Dangers of using activated carbon past its lifespan

Every adsorbent has a saturation capacity depending on its active surface area. The idea to replace carbon filter is not only due to the saturation capacity but imagine fungi, bacteria may be adsorbed as well. It is worth replacing.


Why does the equilibrium constant change on scaling the stoichiometric coefficients of a reaction?

Have a look at this answer of mine.
There you can find the derivation of the formula that defines the equilibrium constant

\begin{equation}
\log \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \log K = -\frac{\Delta G^{0}}{RT}
\end{equation}

of a general reaction

\begin{equation}
 \ce{\nu_{1}A + \nu_{2}B + ... + \nu_{$x$}C <=> \nu_{$x+1$}D + \nu_{$x+2$}E + ... +\nu_{$n$}F}
\end{equation}

and it is also shown how this definition relates to the maybe more familiar definition

\begin{equation}
 K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .
\end{equation}

Now, the reason why the equilibrium constants of the general reaction above and the same reaction multiplied with a constant factor $\alpha$:

\begin{equation}
 \ce{\alpha\nu_{1}A + \alpha\nu_{2}B + ... + \alpha\nu_{$x$}C <=> \alpha\nu_{$x+1$}D + \alpha\nu_{$x+2$}E + ... + \alpha\nu_{$n$}F}
\end{equation}

are not the same but are related to each other via:

\begin{equation}
 K_{\alpha} = K_{1}^{\alpha}
\end{equation}

comes from the definition of $K$.

That there must be a difference is also plausible: The defining formula for $K$ relates it to the standard Gibbs free energy of reaction $\Delta G^{0}$ which in turn is defined to be the amount of Gibbs free energy that is produced per formula conversion (meaning that it is defined for an extend of reaction $\xi = 1 \, \text{mol}$).
So, it is the amount of Gibbs free energy produced by the reaction of $\nu_{1}\, \text{mol}$ of $\text{A}$ and $\nu_{2}\, \text{mol}$ of $\text{B}$ etc. to $\nu_{x+1} \, \text{mol}$ of $\text{D}$ and $\nu_{x+2}\, \text{mol}$ $\text{E}$ etc.
But if you multiply the reaction by $\alpha$ the standard Gibbs free energy of reaction is the Gibbs free energy produced by the reaction of $\alpha\nu_{1}\, \text{mol}$ of $\text{A}$ and $\alpha\nu_{2}\, \text{mol}$ of $\text{B}$ etc. to $\alpha\nu_{x+1} \, \text{mol}$ of $\text{D}$ and $\alpha\nu_{x+2}\, \text{mol}$ $\text{E}$ etc. and so there should be $\alpha$ times as much Gibbs free energy as before:

\begin{equation}
 \Delta G^{0}_{\alpha} = \alpha \Delta G^{0}_{1} \ .
\end{equation}

And this is indeed the case when you have $K_{\alpha} = K_{1}^{\alpha}$ because according to the logarithm laws

\begin{equation}
RT \log K_{1}^{\alpha} = RT \alpha \log K_{1} \ .
\end{equation}


How to separate a 5 atom thick layer of Cu from Au

In most cases I've seen, ALD is performed on the intended substrate - that is, there's no need to separate it. In other cases, you use a sacrificial substrate that you can remove easily.

If you wanted to separate the film from the substrate, there are various techniques, but nothing comes to mind for separating Au from Cu. I find it incredibly unlikely in the specific case of a 5-atom thick layer of Cu, which would likely not survive typical techniques (vs. bending, mushing, forming a larger packing, etc.)

Some typical techniques for separating two films:

Adhesive tape (as mentioned in a comment) - works for many films, e.g., graphene, other layered materials, where you physically pull apart layers
Swelling (e.g., separating Au from mica) - including various intercalation methods used in graphenes and graphene oxides and other layered materials, so that the layers separate more easily
Dissolving, etching, or eroding the substrate (e.g., back-etching a Si wafer to expose an $\ce{SiO2}$ or thin metal film deposited on the wafer)

In my group, for example, we make "template-stripped gold" where Au is deposited on cleaved mica substrates. Then the thin gold layer is glued to glass substrates and the mica is separated from the Au by swelling / degredation, leaving a very smooth Au surface.

This technique can work with fairly thin layers of gold (~10-30 nm) but there's a large difference between Au and mica, making separation possible. Not true with Au/Cu.


Indeed, much of the chemistry that takes place in the universe involves the breaking of covalent bonds and the formation of new covalent bonds. The key is that the energy of the products needs to be in a lower state than that of the reactants. This is the thermodynamic requirement anyway. Lets consider one of your examples:

$$\ce{2H2(g) + O2(g) -> 2H2O(g) + \pu{572kJ}}$$

Although we had to break the covalent bonds of hydrogen and oxygen, the new bonds that were formed are even stronger, thus the reaction is thermodynamically favored.

The complicating point of the above example is the fact that there is a large kinetic barrier to the reaction. This basically means that there is an intermediate product in the steps of this reaction that is energetically very high (unfavorable), so some catalyst is required (say, a spark) for this reaction to get over that energetic barrier and proceed on any reasonable time scale.

Summary, TL;DR:
The breaking and creating of covalent bonds plays a major role in chemistry. The criteria for covalently bound compounds to participate in a chemical reaction is that the products have a lower energy state than than the reactants. If this is not the case, the reaction will not happen. Even if it is the case (that the products are lower energy / more stable than the reactants), there may be kinetic limitations requiring a catalyst in order for the reaction to proceed. But, no catalyst can make a reaction proceed if the above criteria is not met, that the products be more stable than the reactants.


Using a gas stove instead of Bunsen burner for a flame test?

Yes, you can use a common stove to test for cations.
But a stove is designed to minimize the risk of incomplete combustion (which could lead to production of carbon monoxide), hence its flame always appears as an intense blue flame.
Such color contamination could be problematic when testing for cations.
In contrast, the combustion (and the color of the flame) can be regulated in a Bunsen burner.

If you want to experiment at home, an old (but gold) method is that of placing a wick through the metal lid of a closed glass vial, thus creating a burner. You can put alcohol in it, and light the wick.

You can adjust the flame by moving the wick.

Plus, as reported in
[1],
a solution of the salt in alcohol produces a long lasting colored flame in that setting.

References

Dragojlovic, V.; Richard F. Jones. Flame Tests Using Improvised Alcohol Burners. J. Chem. Educ. 1999, 76 (7), 929. https://doi.org/10.1021/ed076p929.


Can superglue (cyanoacrylite) be thinned with acetone and still harden?

As you can see from the reaction of cyanoacrylate from Wikipedia, the carbonyl ($\ce{C=O}$) groups do not react with the cyano acrylate which means that acetone should not react with your super glue.

That does not mean though that any water present in the acetone won't. So you should make sure your acetone is dry. But even if the super glue does cure, it is dissolved in acetone and will still act as a binder for the Iron oxide.


What is the correct order of chemical reactivity in terms of oxidising property for the following elements: F, Cl, O, and N

See, F is the strongest oxidising agent in the entire periodic table. Also, according to periodic table trends, oxidising character decreases down the group and increases across a period. You should check the reactivity of any element according to the question given, here, according to the oxidising character. If nothing is given, usually IE and Electronegativity is considered.
Therefore, I think the order should be F>O>Cl>N. Cl>N because Cl is more oxidising than N.
I must add that a simple Google search would have yielded the same result.


Deprotection of carboxybenzyl (Cbz) in the presence of Methyl ester?

Just leave out the last step, i.e., LiOH/MeOH and you should be just fine, the Cbz group gets removed under reductive conditions which leave the methyl ester unaffected. If you lose the methyl ester, you could just add it again afterwards.


Why does addition polymerisation of high density poly(ethene) use low temperature and pressure?

In fact, when polymerizing ethylene using a radical initiator at high temperature and high pressure, we favor the formation of low-density polyethylene. This is because the radical at the end of the growing chain can be transferred to any carbon atom of the chain by radical rearrangement (favored at high temperature) and due to close distances among molecules (high pressure) a branch chain can grow on the principal chain, and we get a branched polyethylene.  The chains of a branched polymer cannot align together, and eventually the density of the resultant polymer is low. We have got LDPE.

On the other hand, using a specific catalyst, known as Ziegler-Natta catalyst and under mild conditions of temperature and pressure, we get linear chains of polyethylene that can be aligned together and the density of the resultant polymer is high. We have got HDPE.


How does acidity affect electrode potentials?

There are numerous reactions that involve iron and many involve either $\ce{H+}$ and $\ce{OH-}$. Take a look at this table and search for "$\ce{Fe}$". The non-standard potential for each one depends on the concentration of the species involved, and for the ones with hydrogen or hydroxide ions this means it includes the pH. This is described by the Nernst Equation. The potential varies as the natural log of the concentration, so you would add or subtract the pH or pOH, depending on the half reaction in question.


How to calculate the variation of enthalpy in a reaction that does not occur in standard conditions?

For a pure substance you can write generally that the enthalpy depends on pressure as follows:

$$\left(\frac{\partial H}{\partial p}\right)_T = V - T \left(\frac{\partial V}{\partial T}\right)_p \tag{1}$$

If you evaluate this for an ideal gas you find that  $\left(\frac{\partial H}{\partial p}\right)_T = 0$. If the substance is present in a condensed phase with a density (or molar volume) that is weakly dependent on the temperature, then roughly

$$\left(\frac{\partial H}{\partial p}\right)_T = V$$

If you speak of standard enthalpies then by definition there is no pressure dependence, as they are defined at a particular pressure.

If your reaction involves gases which can be assumed ideal, then you can use the standard enthalpy for all reagents and products, since according to my answer the enthalpy of an ideal gas is independent of pressure. On the other hand if it involves condensed phases you need to evaluate the molar volume and perhaps also the thermal expansion coefficient $\alpha$, since Eq. (1) can be rewritten as

$$\left(\frac{\partial H}{\partial p}\right)_T = V (1- T\alpha)$$

For the enthalpy of reaction you might write

$$\left(\frac{\partial \Delta_r H}{\partial p}\right)_T = \sum_i \nu_i V_{m,i} (1- T\alpha_i)$$

where $\nu_i$ are stoichiometric coefficients. In the summation in the rhs account should be taken for sign changes depending on whether species are reactants or products.

If you want to evaluate the enthalpy of reaction at one pressure given the value is known at another pressure, and the molar volumes are known, you can integrate the previous equation, giving (at constant T)

$$\Delta_r H(p)  =  \Delta_r H(p^\circ) + \int_{p^\circ}^{p} \sum \nu_i V_{m ,i}(1- T\alpha_i) dp$$

where $p^\circ$ is the reference pressure at which the enthalpy is already known. This last expression may be simplified in various ways depending on the state of the species and the pressure integration range. If the molar volumes and expansion coefficients are approximately constant in the pressure range then

$$\Delta_r H(p)  =  \Delta_r H(p^\circ) + (p -p^\circ) \sum \nu_i V_{m,i} (1- T\alpha_i)$$

If the species have negligible thermal expansion coefficients then

$$\Delta_r H(p)  =  \Delta_r H(p^\circ) + (p -p^\circ)\sum \nu_i V_{m,i}$$

and so on. For instance, for a reaction converting a solid between two forms $\text{react}$ and $\text{prod}$ having constant molar volume and negligible thermal expansion coefficients in the pressure range $p^\circ$ to $p$,

$$\ce{\text{react}<=>\text{prod}}$$

$$\Delta_r H(p)  =  \Delta_r H(p^\circ) + (p -p^\circ)(V_m(\text{prod})-V_m(\text{react}))$$


You essentially answer this question yourself. The chemical shifts of transition metal hydrides are often largely negative because they are highly shielded. That's what chemical shifts are a measure of - the amount of shielding from the external magnetic field. A very small hydride nucleus buried in a great big fluffy cloud of d-shell electrons of a transition metal is incredibly highly shielded from the base magnetic field.

The negative number of the chemical shift is arbitrary; what is relevant is that chemical shifts to the right are more shielded, and those to the left are less shielded.

Read the information about chemical shifts on this great online resource.


What is the difference between temperature and thermodynamic temperature?

Temperature is more of a relative notion--heat flows from higher temperature to lower temperature, but that's about it. The zero in a temperature scale is meaningless; it is an arbitrary convention. Usually, we measure temperature in Celsius, Fahrenheit, Rankine, etc (note that Kelvin measures temperature as well--but it also measures thermodynamic temperature)

Thermodynamic temperature is such that zero temperature is absolute zero (unlike Celsius, where -273.15°C is absolute zero). It is related to the energy (specifically the kinetic energy) of the molecules of a substance-- for example, in a gas, kinetic energy per molecule is $\frac12k_BT$ (rotational energy, etc can be obtained if you know the type of molecule--but temperature is principally a measure of kinetic energy). It also has strong ties with entropy.

We generally don't use "temperature" in thermodynamics, thermodynamic temperature is more convenient. Temperature is only useful in equations involving temperature difference--whereas equations involving the absolute value of temperature (Arrhenius equation in kinetics, entropy, Maxwell distribution equation, etc) demand thermodynamic temperature.


Is this a counterexample to ΔU = 0 J for isothermal processes?

The situation you cite calls for the use of the more general form of the first law of thermodynamics (which I assume your textbook provides), given by $$\Delta U+\Delta (KE)+\Delta (PE)=q+w$$where KE is the organized kinetic energy of the system and PE is the potential energy.  Applying this to the problem at hand gives $$\Delta (PE)=w$$and the change in internal energy is zero.  So the thing that was wrong with your approach was not using the proper form of the first law.


Using Szabo and Ostlund's notation (page 85):

\begin{align}
J_{ij} &= (\psi_i\psi_i|\psi_j\psi_j)&
\text{and}&&
K_{ij} &= (\psi_i\psi_j|\psi_i\psi_j)\tag1\label{integrals}
\end{align}

For minimal basis set $\ce{H2}$ (page 162):

\begin{align}
\psi_1 &= [2(1+S_{12})]^{-1/2}(\phi_1+\phi_2)&
\text{and}&&
\psi_2 &= [2(1-S_{12})]^{-1/2}(\phi_1-\phi_2)\tag2\label{minimal}
\end{align}

Substitute equation $\eqref{minimal}$ into $\eqref{integrals}$, set $S_{12}$ and any integral involving both $\phi_1$ and $\phi_2$ to zero (large $R$) and you see that

$$J_{11} = K_{12} = \frac{1}{2}(\phi_1\phi_1|\phi_1\phi_1)$$

Remember that $(\phi_1\phi_1|\phi_1\phi_1)=(\phi_2\phi_2|\phi_2\phi_2)$ since you are dealing with a homonuclear diatomic.


Determining the half-reaction equations for extraordinary redox reaction?

You have rightfully assumed the reaction takes place in either basic or neutral medium, otherwise $\ce{Cr(OH)3}$ wouldn't be the product.
This automatically suggests $\ce{H+}$ are not used for balancing half reactions.
Your solution is fine for the provided reaction given that chromium(III) hydroxide stays precipitated.
Here I've basically formatted your solution in a form I'm more familiar with:

\begin{align}
\ce{\overset{+6}{Cr}O4^2- + 4 H2O + 3 e- &→ \overset{+3}{Cr}(OH)3 + 5 OH-} &|\cdot 2 \tag{red1}\\
\ce{\overset{+4}{S}O3^2- + 2 OH- &→ \overset{+6}{S}O4^2- + H2O + 2 e-} &|\cdot 3 \tag{ox1}\\
\hline
\ce{2 CrO4^2- + 3 SO3^2- + 5 H2O &→ 2 Cr(OH)3 + 3 SO4^2- + 4 OH-} \tag{redox1}
\end{align}

Resulting full redox equation:

$$\ce{2 Na2CrO4 + 3 Na2SO3 + 5 H2O → 2 Cr(OH)3 + 3 Na2SO4 + 4 NaOH} \tag{R1}$$

This implies that reaction begins in neutral medium; however there is remaining sodium hydroxide ($\ce{4 NaOH}$ on the right), so the pH will increase over time.
This will result in further dissolving of $\ce{Cr(OH)3}$ in $\ce{NaOH}$ yielding $\ce{Na3[Cr(OH)6]}$ complex, which is likely the final product.
Alternatively, addition of a base to this mix would also accelerate formation of chromium(III) hydroxocomplex.
With this in mind, I'd say the more appropriate way to write the reaction is

\begin{align}
\ce{\overset{+6}{Cr}O4^2- + 4 H2O + 3 e- &→ [\overset{+3}{Cr}(OH)6]^3- + 2 OH-} &|\cdot 2 \tag{red2}\\
\ce{\overset{+4}{S}O3^2- + 2 OH- &→ \overset{+6}{S}O4^2- + H2O + 2 e-} &|\cdot 3 \tag{ox2}\\
\hline
\ce{2 CrO4^2- + 3 SO3^2- + 2 OH- + 5 H2O &→ 2 [Cr(OH)6]^3- + 3 SO4^2-} \tag{redox2}
\end{align}

Resulting full redox equation:

$$\ce{2 Na2CrO4 + 3 Na2SO3 + 2 NaOH + 5 H2O → 2 Na3[Cr(OH)6] + 3Na2SO4} \tag{R2}$$


Why are Salts Removed before Ion-Exchange Chromatography?

Think about what the ion-exchange material does, and the mechanism of separation by this method. The ions of interest interact with the ion exchange sites to a different degree than do the other ions or other species in the solution, thus you get separation.  What if your solution is loaded down with ammonium sulfate?  The ion exchange resins have a limited capacity to accommodate ions and all of the sites would be swamped in ammonium ions and all of the much lower concentration ions of interest thus would not be able to be separated.


Difference between spin-orbit coupling and the Russell-Saunders Effect?

I'm not aware of the Russell–Saunders effect, but the Russell–Saunders coupling scheme is definitely a thing. As you noted, the Wikipedia page on "spin-orbit interaction" doesn't talk about it, but a different Wikipedia page does, and basically tells you the same thing as I will.
The answer is... yes and no.
The word "coupling" refers to the coupling of several sources of angular momentum, namely the spin component and orbital component. Now, the issue is that in an atom you typically have many electrons, and every electron has its own orbital angular momentum $\vec{l}$ and spin angular momentum $\vec{s}$,* so you have many, many sources of angular momentum.
The challenge is to bring all of these together in a way that allows us to describe the electronic state of an atom using its angular momentum properties (for example, this is what a term symbol does). There are two approaches towards coupling all of these angular momenta together:

(Russell–Saunders or LS-coupling) Couple all the individual $\vec{l}$'s together to form one gigantic orbital angular momentum $\vec{L}$, and couple all the individual $\vec{s}$'s together to form one gigantic spin angular momentum $\vec{S}$. Then couple these two together to form the total angular momentum $\vec{J}$.
If you have studied term symbols before it is probably using the Russell–Saunders scheme, where you calculate $L$, $S$, and $J$, then write the term symbol $^{2S+1}L_J$.

(jj-coupling) For each individual electron, couple $\vec{l}$ and $\vec{s}$ together to form the total angular momentum $\vec{j}$ for that one particular electron. Then bring all the electrons' total angular momentum together to form $\vec{J}$.
Note that we haven't ever mentioned $L$ and $S$ here, so the term symbols under this coupling scheme are different. Instead, you'd label the term symbols with the individual values of $j$ for each electron. For an example see e.g. Atkins Molecular Quantum Mechanics.

Now, which you use depends on whether electron-electron repulsions or the spin-orbit coupling is a "bigger" term. If spin-orbit coupling is very significant, then it means that the spin and orbital angular momenta on their own (i.e. $\vec{L}$ and $\vec{S}$) are not very useful quantities,† since the interaction between them is large. In this scenario, jj-coupling is a more appropriate way of describing the effects of spin-orbit coupling on the electronic state.
On the other hand, if the spin-orbit coupling is relatively small, then $\vec{L}$ and $\vec{S}$ are useful quantities which are still applicable to the atom's electronic state, so the Russell–Saunders scheme is appropriate.
And of course, sometimes we get stuck in the middle ground where neither scheme is fully appropriate.
TL;DR The Russell–Saunders and jj schemes are both methods which can be used to describe the effects of spin–orbit coupling, but they are not the same thing as spin–orbit coupling.‡

* Well, sort of, anyway: the electrons are indistinguishable, so it's more accurate to say that the $n$ electrons in the atom have $n$ orbital angular momenta $\{\vec{l}_1, \vec{l}_2, \cdots, \vec{l}_n\}$ and $n$ spin angular momenta $\{\vec{s}_1, \vec{s}_2, \cdots, \vec{s}_n\}$.
† To be precise, $L$ and $S$ are not "good quantum numbers" because the operators $\hat{L}$ and $\hat{S}$ don't (approximately) commute with the total Hamiltonian $\hat{H}_0 + \hat{H}_\text{so}$, where $\hat{H}_\text{so}$ is the spin-orbit coupling Hamiltonian and $\hat{H}_0$ is the rest of the Hamiltonian (which does commute with $\hat{L}$ and $\hat{S}$).
‡ If you read the previous footnote, then the spin–orbit coupling itself is represented by the Hamiltonian $\hat{H}_\text{so}$. The two coupling schemes can be thought of as ways of dealing with this term as a perturbation to $\hat{H}_0$. In Russell–Saunders the perturbation is small, and consequently, the "good" quantum numbers are similar to those of $\hat{H}_0$. In jj the perturbation is large.


Reflect means to bounce back, as in a ball bouncing off a wall. The average density of an atom is very low, so the observed reflection was startling... like throwing many billiard balls at a mass of fluffy cotton  candy and every now and then a ball comes bouncing back!
The explanation offered was that there must be some "hard", dense object inside that fluff... the nucleus. Electrons are effectively spread out in their orbitals, and tend to act individually (bound by electromagnetic forces). An alpha particle, massing $\pu{4 \times 10^{9} eV}$, could not be reflected by an electron massing ~$\pu{5 \times 10^{5} eV}$ -- that would be like bouncing a baseball off a ping-pong ball.
Since both alpha particle ($\ce{He}$ nucleus) and gold nucleus are positively charged, and the gold nucleus is held together by powerful nuclear forces (so all its particles' masses are effectively combined to ~$\pu{2 \times 10^{11} eV}$), it explained the reflection. Since reflection occurred only rarely, it implied that the nucleus was a small, dense target.


(Q) is the least acidic compound because the conjugate base formed is anti aromatic and hence unstable.

(R) is more acidic than (Q) because the conjugate base is stabilized by the electron withdrawing inductive effect of two sulphur atoms.

Conjugate bases of both (P) and (S) are stable due to resonance. Among the two, conjugate base of (S) is more stable because a negative charge on Oxygen atom is more stable than a negative charge on carbon atom. Here are the resonance structures:

This is verified by @Ivan Neretin's comment on the OP.

  Let's look at some data. pKa of cyclopentadiene is about 16, while that of acetylacetone is about 9.


Fock operator invariance under unitary transformation

The exchange operator is not a operator by itself, it is only defined with the orbital it is working on:
$%
\newcommand{\ll}{\left\langle}\newcommand{\rr}{\right\rangle}
\newcommand{\lb}{\left|}\newcommand{\rb}{\right|}
\newcommand{\op}[1]{\mathbf{#1}}$
$$\begin{align}
  && \op{F}_i &= \op{H}^\mathrm{c} + \sum_j (\op{J}_j - \op{K}_j),\\
\text{with}&&
     \op{J}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr,\\
\text{and}&&
     \op{K}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr.
\end{align}$$

While in the Coulomb operator case the $\lb \phi_i\rr$ doesn't do anything, so you can take it through the proof as a constant:*
\begin{align}
  \sum_j\op{J}_j\lb \phi_i\rr
  &= \sum_j
     \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
  &= \sum_k \sum_l \sum_j U_{kj}^*U_{lj}
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_l(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
  &= \sum_k \sum_l \delta_{kl}
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_l(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
  &= \sum_k
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_k(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
  &= \sum_k \op{J}_k \lb \phi_i\rr
\end{align}

That is not the case in the Exchange operator, as it switches the orbitals:
\begin{align}
  \sum_j\op{K}_j\lb \phi_i\rr
  &= \sum_j
     \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr\\
  &= \sum_k \sum_l \sum_j U_{kj}^*U_{lj}
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_i(\op{x}_1) \rr \lb \phi_l(\op{x}_2) \rr\\
  &= \sum_k \sum_l \delta_{kl}
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_i(\op{x}_1) \rr \lb \phi_l(\op{x}_2) \rr\\
  &= \sum_k
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_i(\op{x}_1) \rr \lb \phi_k(\op{x}_2) \rr\\
  &= \sum_k \op{K}_k \lb \phi_i\rr
\end{align}

* Let
$$
\mathbb{U}=\left(\begin{matrix}
  U_{11} & U_{12} & \dots \\
  U_{21} & U_{22} & \dots \\
  \vdots & \vdots & \ddots\\
  \end{matrix}\right);
\mathbb{U}^\dagger\mathbb{U}=\mathbb{E}
\implies U_{ij}^* U_{kl}= \delta_{ij} = \begin{cases}
   1; & i = j = k = l \\ 0; & i \neq j \dots\\ \end{cases}
.$$


Molecular orbitals symmetry - states notation

The symbol $B$ is explained in greater detail here: Letter codes in molecular term symbols. In this context, it indicates that it is the second excited state of the same multiplicity as the ground state. The ground state, labelled with $X$, is a doublet; the first doublet excited state is labelled with $A$, and the second doublet excited state is labelled with $B$.

The superscript $2$ denotes the multiplicity, i.e. this is a doublet state, with total spin $S = 1/2$ (technically the spin quantum number) such that $2S + 1 = 2$. The dinitrogen cation $\ce{N2+}$ will have one unpaired electron in its low-lying electronic states, so it's not too surprising that the total spin is $1/2$.

Finally, there is a bit more to the symbol $\Sigma$ than just being a substitute or a generalisation of $\mathrm{A}$. In the linear point groups, this denotes the total orbital angular momentum about the internuclear axis. In this state, there is no orbital angular momentum about this axis, and the relevant quantum number is $\Lambda = 0$, which is denoted by the letter $\Sigma$. ($\Lambda = 1$ is a $\Pi$ state, $\Lambda = 2$ a $\Delta$ state, and so on.)

If you think about the ground state electronic configuration of $\ce{N2+}$, it is essentially the same as $\ce{N2}$, but with one unpaired electron in the $\sigma_\mathrm{g}$ orbital. In order to generate any kind of orbital angular momentum, you need to have an asymmetric population of $\pi$-type MOs. Since this isn't the case, the ground state of $\ce{N2+}$ has no orbital angular momentum, and is consequently a $\Sigma$ state. A similar argument can be applied to the second excited state, which seems to arise from a $\sigma\to\sigma$ excitation.


What is causing the buffer region in a weak acid - strong base titration?

The linked question's answers were too long for me, so I'll attempt to summarize quickly.

The equation that tells you everything you need to know is the Henderson-Hasselbalch equation:

$$\mathrm{pH} = \mathrm{p}K_{\mathrm a} - \log\frac{\ce{[A-]}}{\ce{[HA]}}$$

Here $\ce{HA}$ is a weak acid, and it's conjugate base is also weak. But we can imagine also a weak base with a weak conjugate acid. The magic happens when you've half neutralized a weak acid or base, creating a mixture of a weak acid and a weak base. The fraction on the right is fairly close to 1 (within an order of magnitude, let's say).

As you add more acid, the weak base can neutralize it. As you add more base, the weak acid can neutralize it. Assuming these amounts are reasonably small, the logarithm of the fraction on the right doesn't change by much, so the overall pH of the solution does not change by much. For example, changing the fraction on the right by a factor of 10 only affects the pH value by $\pm 1$.

Of course, if you add enough acid or base to neutralize (or close to neutralize) one of the buffer components, then the buffer stops working. This is also when the fraction gets very large or goes to zero and even the logarithm can't change that.


Will sulphuric acid react with water produced in an esterification process?

Sulfuric acid acts as catalyst meaning it is regenerated at the end of the reaction. In the presence of water or other bases, sulfuric acid will primarily be deprotonated. $$\ce{H2O + H2SO4->H3O+ + HSO4-}$$

It is occasionally more correct to consider the water acting as a catalyst. Any time that you have a strong acid acting as a catalyst, because its deprotonation is much more likely than the organic reaction, it is nearly completely deprotonated at a given time. Thus the protonated $\ce{H3O+}$ will act as your catalyst. protonating the carboxylic acid and deprotonating the ester.


Collidinium chlorochromate and gamma-picolinium chlorochromate are known oxidising agents. e.g. Synth Comm. 31 2001 1253

I see no reason why other methyl pyridines should not also work.


The charge on a centre in Lewis structures is not of special relevance, except that it helps to provide the information which is useful: the number of electrons on that centre.

A positive charge of $X$ means $X$ electrons were lost, while a negative charge, $-Y$ means $Y$ electrons gained.

From this you can either simply remove/add the corresponding number of dots from/to the diagram, or calculate how many valence electrons are now present, and then draw the diagram ab initio from this information.

A "normal" lithium centre has 1 valence electron, so the positive ion just loses it (leaving no dots) while the negative ion gains one more (and thus has two dots on the diagram).


Wavelength of iPhone 4S camera light for visible spectroscopy calibration

According to the site, wavelength calibration is meant to be done with a compact fluorescent lamp (for the mercury lines), but at the bottom there's a tool for calibrating using any two known wavelengths. If the LED on the phone has two distinct lines that you can calibrate against, it may work, but something like a pair of monochromatic LEDs or laser pointers (say one blue and one red) or something will probably work better.


Why Use Lewis Structures to Determine Radicals

Lewis structures are excellent for showing radical character for an odd number of electrons.

The reason they fail for molecular oxygen is that they Lewis representation carries no intrinsic information about the molecular spin, and as such it cannot distinguish the unpaired electrons in molecular oxygen's antibonding orbitals as spin matched (triplet) or spin opposed (singlet).

However, when there are an odd number of electrons, it is clear there will be one unpaired spin, and it is not relevant whether what the spin orientation is for these cases.

That being said, and somewhat outside the scope of this question, Lewis structures cannot distinguish between doublet and quartet states.


How long did the most difficult compound to synthesize last?

There is no such thing as the most difficult compound to produce. Instead, there are some hundreds of thousands of "stable" compounds of which we know pretty much everything - melting point, boiling point, heat capacity over a wide range of temperatures, refractive index, speed of sound, magnetic susceptibility, electrical conductivity, crystal structure, and many more. Around that field is a much wider "grey zone" of the not-so-stable compounds (millions of them), which miss some of these properties because they can't be measured.

As we reach out to the far outskirts of that zone, we meet more and more compounds with less and less known properties in each one, because they are increasingly frail and short-lived, to the point that the very words "produce" or "synthesize" (or even "exist", for that matter) do not quite apply to them anymore. Some of them last but a fraction of second; some live longer, but require very peculiar conditions. There is absolutely no meaningful way to arrange them in order of increasing "difficulty".

Also, welcome to Chem.SE.


Calculating molecular weight of an acid based on given values of mass, volume, pH and pKa

Ok here's what you're doing wrong.

When you write $$C = \frac{n}{V}$$

$C$ is meant to be the concentration of the monoprotic acid dissolved initially, not at equilibrium. Plugging in the equilibrium concentration is disasterous, as it is bound to be different from the actual ratio of amount of substance in moles of $\ce{HA}$ to the volume of solution in liters.

This approach will better suit the problem:

$$\ce{HA \ <=> \ H+ \ + \ A-}$$
$[\ce{HA}]_{\mathrm{initial}}=\frac{m}{M}.\frac{1}{V}$

Assuming $x$ moles/liter to be consumed at equilibrium,

$[\ce{HA}]_{\mathrm{eq}}=\frac{m}{M}.\frac{1}{V}-x$

Now $x$ moles/liter each of $\ce{H+}$ and $\ce{A-}$ would have appeared.

$[\ce{A-}]_{\mathrm{eq}}=[\ce{H+}]_{eq}=x$

$x$ happens to be $10^{-5}$, as $\mathrm{pH}$ is given to be 5.

$K_a$ would be $10^{-6}$, as you have already mentioned.

$$K_a = \ce{ \frac{[H^+]_{eq}[A^-]_{eq}}{[HA]_{eq}}=10^{-6}}$$

Plugging in, I get

$$10^{-6}=\frac{10^{-5} \times 10^{-5}}{\frac{0.011}{M \times 0.5}-10^{-5}}$$

Simplifying a little, I get

$$M=\pu{200 g/mol}$$


There exists related anions called monofluorophosphate ion ($\ce{PO3F^2-}$) and difluorophosphate ion, ($\ce{PO2F2-}$). The hypothetical neutral compound, $\ce{PO2F}$ is called phosphenic fluoride. It is also observed that irradiating potassium difluorophosphate with gamma rays create free radicals $\ce{PO2F^{.−}, PO3F^{.−} and PO2F2^.}$.
Moreover, there also exist a polymer $\ce{(PO2F)_n}$ which is a decomposition product of a white solid of composition $\ce{P7O10F15}$. It was apparently named polymeric phosphorus oxyfluoride species. Its structure has been proposed by Wannagat and Rademachers1:

It is suggested that the parent ions of Group I result from the simple
ionization of a linear polymer with the general formula, $\ce{(PO2F)_n}$,
where each of the terminal phosphorus atoms of the chain is bonded to
two oxygen atoms, one of which is a bridging atom, and to two fluorine
atoms.

Conclusion: There is no such compound called POOF, but there certainly is a radical ion and a polymer of exact formula.
Notes and References

High molecular weight phosphorus oxyfluorides by D.W.Muenow, O.M.Uy, J.L.Margrave, Journal of Inorganic and Nuclear Chemistry, Volume 31, Issue 11, November 1969, Pages 3411-3415, DOI: 10.1016/0022-1902(69)80324-6
Other decomposition products were $\ce{PF5, POF3}$ and $\ce{P2O3F4}$: Technique of Inorganic Chemistry, Hans Boegh Jonassen, Arnold Weissberger Volume 1
Interscience Publishers, 1963


Can the oxygen in an ester participate in resonance donation?

Sure, structure III in the following diagram is a valid resonance structure.

  However, resonance withdrawal by the ring via the oxygen doesn't seem
  possible

No, that's possible too.  It might not be a major contributor, but structure V is a valid resonance structure.

  would that make benzoic acid the stronger acid?

I think you're asking which is the stronger acid, benzoic acid or the half-ester of carbonic acid. Resonance structures like III and VII will stabilize both the acid and the corresponding carboxylate anions.  One molecule doesn't appear to have a more resonance stabilized starting acid or a more resonance stabilized carboxylate anion than the other.  So resonance arguments don't seem to suggest that one molecule should be more acidic than the other.  This conclusion, that resonance factors do not control the acidity in these two compounds, can also be arrived at by noting that the starting acids and particularly the carboxylate anions are analogous to the allyl MO system which has a node on the central carbon atom in the HOMO ($\ce{\Psi 2}$). The node effectively  disconnects the substituent (-OR or phenyl) from the MO, meaning that the substituent cannot interact via resonance with the MO.

Inductively one can argue that the electron withdrawing oxygen in the half-ester would destabilize the already positively polarized carbonyl carbon in the starting acid and then also provide some inductive stabilization of the electron-rich carboxylate anion, more so than a phenyl group would.  So perhaps, based mainly on inductive arguments, one could argue that the carbonic acid half-ester would be more acidic than benzoic acid.

I couldn't find any pKa values for carbonate half-esters, but we can compare the pKa of

carbonic acid 3.6

to

benzoic acid 4.2

The values are close, as expected - even in the right direction, but that's probably just luck.


If by "bond dipole moment" you mean "electric dipole moment", then the electric dipole moment is the first term in a multipole expansion of the molecular charge distribution. In that respect, an electric dipole moment is an approximation.

For a static charge distribution, we are interested in the resulting electric fields or potentials. For a general charge distribution, this does not have a closed form expression, but can be increasingly approximated by higher order terms in a multipole expansion. The dipole approximation is good enough for many cases. Depending on the experimental setup, the only surviving term might be the dipole term. So in that sense it maps to an experimental value and can be considered "real".


What is the relative size of the (M+2) peak?

Let’s assume your compound is $\ce{C_nH_xO_y}$. Thankfully, both hydrogen and oxygen are elements that only have one significant naturally occuring isotope. Therefore, we can treat the entire contribution of $\ce{H_xO_y}$ as a constant $c$. All we need to answer is how large the $M+1$ and $M+2$ peaks are given the number of carbons, $n$. (This treatment is not entirely correct as highlighted by orthocresol, but it is close enough for my purposes.)

Assuming you do not have any isotope enrichments, each carbon will independently have a chance of either being $\ce{^12C}$ or $\ce{^13C}$ (again, we will ignore all other isotopes such as $\ce{^14C}$). The independence is the big key word here. We can use general principles of stochastics to calculate the result.

The probability of all carbon atoms of one molecule of your compound are $\ce{^12C}$ is:
$$P(\ce{^12C_nH_xO_y}) = 0.989^n$$
The $M+1$ peak is represented by one single atom being $\ce{^13C}$. Again, the principles of stochastics apply:
$$P(\ce{^12C_{n-1}^13C1H_xO_y}) = \left ( n\atop 1\right) \times 0.989^{n-1} \times 0.011$$
And finally for the $M+2$ peak:
$$P(\ce{^12C_{n-2}^13C2H_xO_y}) = \left ( n \atop 2\right) \times 0.989^{n-2} \times 0.011^2$$

Using this, we can plot what the probability of each peak is – and use that as their relative heights.

We see that from $n=90$ the $M+1$ peak is actually larger than the $M$ peak. From $n=128$, even the $M+2$ peak will be larger than the $M$ peak. And from $n=181$ the $M+2$ peak becomes the largest of these three.

Of course, further peaks will also start appearing, meaning that the mass spectra of very large molecules will become difficult to analyse.

Now the numbers from which $M$ is no longer te principal peak are quite alrge. Unless you are synthesising maitotoxin, you will probably not need to resort to any analysis that does not only involve $M+1$.


The potential as it is given there is vs. standard hydrogen electrode(SHE) which has an arbitrary assigned value of 0.0000V. Through the choice of another reference, it is very easy to make all of those numbers positive or negative. The absolute value of the half cell potential can only be useful in comparison to others. So you could make arguments like Cu has a more positive potential compared to Ni(by 0.59V based on the above numbers), but that is it.


Why V can't have the valence 7 by taking an electron from the 3p orbital and putting it into 3d?

Vanadium has a stable valency of 5.
Electrons cannot be transferred from $3p$ orbital to $3d$ orbital as the $3p$ orbital is completely filled and has attained stability.
It requires high energy so as to transfer electrons from $3p$ orbital to $3d$ orbital.
Hence,Vanadium has a valency of 5 only and not of 7 or 6.


sp2 hybridisation of alkyl radicals causing formation of racemic mixture

In the first step, due to the selectivity of bromine in a free radical reaction, the tertiary carbon which has a $\ce{-CH3}$ group attached to it forms the radical.
Now, since the radical is a $\ce{^{.}CR3}$ radical, the carbon radical in the intermediate becomes $\ce{sp^2}$. This means that $\ce{^{.}Br}$ can attack from both the top and the bottom since this lone electron in the $\ce{^{.}CR3}$ radical is in a p-orbital. This is because the electron is equally likely to be found in both lobes of the orbital.
Now after the attack of the $\ce{^{.}Br}$ radical, you get two products. The first one shows the product formed when $\ce{^{.}Br}$ attacks from the bottom and the second one shows the product formed after the $\ce{^{.}Br}$ attacks the intermediate from the top.
You may have noticed that both products are enantiomers of each other and hence the answer would be (c).
Reference
Chem Libretexts


Approach:
The approach taken in this proposed synthesis of chloramphenicol is largely based on Greene's synthesis of paclitaxel, using dihydroxylation methodology proposed by Sharpless.
Whilst not particularly elegant (I was hoping for some kind of asymmetric hydrogenation to set the stereocentres), it does get to the product in a respectable 7 steps (including setting both stereocentres, as required), and avoids having to introduce protecting/directing groups at any stage.
Forward synthesis:

Step 1: Horner Wadsworth Emmons olefination

The first step in this proposed synthesis is a Horner Wadsworth Emmons olefination to introduce the (E)-alkene required for the planned dihydroxylation. Without the nitro-group, the exact reaction has been done many times and with good selectivity, so no reason to assume that this HWE would be particularly challenging (the nitro group clearly makes the aldehyde electron poor).

Step 2/3: Sharpless dihydroxylation, epoxidation sequence

With the (E)-alkene formed, a Sharpless asymmetric dihydroxylation can take place using AD-MIX-B (taking the aryl group as 'large' and the ester as 'medium'). This step (without the nitro) is present in the Greene paclitaxel synthesis (above) and initially gives 82% e.e, though a single recrystallisation of the product brings this up to >95%.
Once recrystallised, treatment with mesyl chloride and subsequent heating (same pot) affords the necessary epoxide.

Step 4/5: Epoxide opening and reduction

Now comes the slightly dubious step....
In the paclitaxel synthesis, the epoxide is opened with sodium azide at the benzylic position (expected, the aromatic ring can stabilise the developing partial positive charge here), however in our case, we have an incredibly electron withdrawing nitro group in the 4-position, completely shutting down any stabilisation at the benzylic position.
Based on this lack of benzylic stabilisation, we could reasonable assume that it will take place with the opposite regioselectivity to the paclitaxel case. There is some evidence for this in the literature, (including where there is an ester present on the other side, but sadly not using azide as the nucleophile), and a cursory flick through an organic textbook will tell you that SN2 adjacent to carbonyls is generally fast.
Assuming it works fine, a Staudinger reduction converts the azide to the secondary amine with retention of stereochemistry.

Step 6: Ester reduction

Lithium borohydride is a mild and selective reagent, able to reduce esters in the present of various other functionality, including nitro groups. One draw-back of this is that lithium borohydride is significantly slower than its counterparts (not to mention more expensive), but, using LAH or other highly reactive hydride species would result in side reactions involving the nitro group.

Step 7: Amide bond formation

The required acyl chloride is commercially available, and should do the required reaction in place of the possibly competitive ester formations (nitrogen LP higher in energy).
Conclusion:
7 steps overall starting from the cheap, commercially available 4-nitrobenzaldehyde. The majority of steps have good precedence on similar systems (ignoring the epoxide opening, which may/may not be fatal), and all of the proposed reactions can (and have been) demonstrated to be possible on large scale (again, ignoring the epoxide opening, but TMS-azide is often substituted in the case of process syntheses involving azide).


Why is water wet?
Chemically speaking saying that water is wet has not much sense.We could, however, say that "Wetting process" is caused by the presence of a thin liquid particles layer over a material. In our specific case, this depends on the chemistry of water and the chemistry of skin. So the main causes of the fact that water is wet, is its wetting properties determined by a force balance between adhesive and cohesive forces over the skin. Mercury, although is a liquid, will not "wet" you because it has a strong cohesive force that doesn't permit the molecules to get in touch with the sensory cells.  When you touch water or a water wetted material part of the water molecules are transfer to your hands (a hydrophilic material) and so to the receptor cells giving the sensation of wet, this sensation can be caused however in some case of parasthesias even if there are no liquid particles. In fact, there is no "wet receptor" on your skin but the sensation is due to the elaboration of different stimuli from tactile, pressure and temperature receptor present in your skin.
Try to play with plasticine after a while handling it your hands will become hydrophobic you will no more fell the wet sensation when you wash them even if water is still "wet"!
What causes fire to be hot as an inherent property?
Fire is an exothermic process. So energy stored in a chemical bond is released as kinetic energy and electromagnetic wave. The fire has a high temperature because the molecules have great kinetic energies when you get in touch with fire (of course wormed air could be a medium as well for short distance) this kinetic energy is transfer to your skin and so to your thermoreceptors . And you feel what human call hot. In fact, you can feel the sensation of hot even without touching the fire because it even irradiates energy as electromagnetic waves that can be absorbed by your skin and so be feel for his receptors.


You can predict this with the help of electrochemical series.

On going up the series reducing power increases, for you case it can be seen that Fe is a better reducing agent than bromide ion as Fe is above Br2 so bromide can't reduce ferric ion.


Significant figures in pH calculations: does pH 0.30 count as 2 or 3 significant figures?

Short Answer:

  The number of significant figures in the mantissa of a value expressed in scientific notation equals the number of significant figures to the right of the decimal in the logged value. So in your example, assuming the likely error is the last digit, I'd use two.

Now back to your question, you have $[\ce{HCl}]=0.500$ assuming, the error is in the last digit, I'd recommend using 2 significant figures, $\pm$ error in the last.

Additionally, just because pH is a scale doesn't mean you can treat $0.30$ as a number with three significant figures, it still has two.

The general principles are given below.
Concepts like this are best illustrated with an example.

Consider, a random number $0.0000273$ Let us write this in scientific notation for the sake of clarity, $0.0000273 = 2.73×10^{–5}$ We have three significant figures.

Now take the log on both sides

$\log(0.0000273)= \log(2.73×10^{–5}) = \log(2.73) + \log(10^{–5})$

Now let us examine the two terms separately,

First, $\log(2.73) = 0.436$: the answer has three significant figures, reflecting the possible error in the last digit of $2.73$

$\log(10^{–5}) = -5$  This is an exact number and has infinite significant figures, i.e there is no error.

Going back, $\log(0.0000273)= \log(2.73) + \log(10^{–5}) = 0.436 -5.0000...$

In this situation the rules for significant digits for sums, dictate that we can only add to the same decimal place as the value with the least significance. In this case, that is the third decimal place from 0.436, so

$\log(0.0000273)= \log(2.73) + \log(10^{–5}) = 0.436 -5.0000... = -4.564$

$\log(0.0000\overline{273}) = -4.\overline{564}$  (the overline indicates the 3 significant figures)

End of example. Now back to your question, you have $[\ce{HCl}]=0.500$ assuming, the error is in the last digit, I'd recommend using 3 digits after the decimal, $\pm$ error in the last. So, $0.301 \pm 0.00\epsilon$


What does negative pressure mean in the context of this manometer?

The manometer shown doesn't measure absolute pressure, just relative pressure. In this car the pressure of the gas in the closed tube versus the pressure of the gas in the bulb.

As you have drawn it, the pressure of the fas in the tube is higher. If you measure pressure in the bulb relative to the pressure in the tube the answer is negative. If you measured absolute pressure (which this manometer can't do because of the lack of any external reference in the closed system) both would be positive but one would be higher than the other.


Evaluation of a synthesis from aspirin to paracetamol from this Nile Red video

You are absolutely correct that $6~\%$ yield is not acceptable for a synthesis — especially if it is a reaction as simple as the nitration of phenol. A quick search in SciFinder gives numerous reaction conditions that allow for up to $98~\%$ yield of the para-substituted product, depending on which catalysts and sets of conditions you want to use. One of the simplest sets of reagents is probably Ducry’s and Roberge’s Angewandte paper which, however, uses a microreactor. While their best selectivity is $1.2:1$, their yields are way above $12~\%$ (which would be the corresponding yield for $6~\%$ of the para-product in a $1:1$ mixture).[1]

Having an acetate group rather than a hydroxy group greatly modifies the properties of that particular oxygen atom. I synthesised a phenol and the corresponding acetate and the chemical shift of the ortho carbon changes by $7.5~\mathrm{ppm}$ upon acetylation while the meta carbon remains essentially unshifted. Therefore, I wish to challenge your assertion that acetate be ortho and para-directing — please back this with appropriate references. However, I will acknowledge that an acetate group is less meta-directing that a carboxylic acid; therefore the preferred reaction site should still be ortho/para to the acetate.

Whatever the selectivity, I do not agree that nitration of acetylsalicylic acid or salicylic acid would be more efficient than nitration of phenol. Remember that the reaction you are trying to perform is an electrophilic aromatic substitution $\mathrm{S_EAr}$. For an aromatic species to be attacked electrophilicly, it should have a high electron density. The carboxyl and acetate groups are both electron-withdrawing and therefore hinder the reaction. It should be harder and require harsher conditions to nitrate a benzoic acid than a phenol. The choice of steps is therefore better in the originally proposed way, in my opinion.

I will agree that decarboxylation of a meta-nitrobenzoic acid should be easier than of salicylic acid. Since the yield of carboxylation is already rather acceptable already, I do not think that switching the order would provide any significant advantage here.

Finally, it should also be rather easy to increase the yield of the final-step acetylation. $18~\%$ needs a lot of work.

Reference:

[1]: L. Ducry, D. M. Roberge, Angew. Chem. Int. Ed. 2005, 44, 7972. DOI: 10.1002/anie.200502387.


Is it possible to displace cerium from cerium sulphate using aluminium to get pure cerium metal?

No
The std. electrode potential for Ce3+/Ce is -2.33V, the std. electrode potential for Al3+/Al is -1.66V. This shows that Al is not reactive enough to reduce Ce(III), it is reactive enough to take Ce(IV) to Ce(III) so I think you would get a mixed salt of Ce/Al sulfate.
This reference here states that calcium metal is used with cerium fluoride as a preparative method.
Note: you can't do this in aqueous solution. Your reactive metal will react with the water preferentially.


Electron density definition of single, double, triple bonds?

Below are plots of the electron density for staggered ethane, ethene, and ethyne, computed from electronic structure theory.  For ethene, which is planar, imagine that the hydogens are jutting forward and backward into the paper, so that the contour plots should contain any putative $\pi$ bonds. It is the orthogonal plane that still contains the two carbon atoms.

A couple of points that I wish to make:

You are correct (in this case for sure) that there is some association between stronger bonds and higher electron density at the C-C bond critical point $(\star)$.  This was noted by Bader and co-workers.
There is no evidence (to my eyes) that there are any side-wise $p$ orbital interactions forming $\pi$ bonds. These look like single $\sigma$ bonds!

My question is, where do we draw the dividing line in the correlation between electron density at the bond critical point, and what constitutes single, double, and triple bonds?

Responding to @michaelm: I might have expected this, if for example, ethene were to have a "double bond" between the carbons:


Sulfur trioxide is planar with 120 degree O-S-O bond angles.  If the molecule were unhybridized and we used the 3 orthogonal p-orbitals for our sigma bonds, then we would expect the O-S-O angle to be 90 degrees.  Further, in the unhybridized scheme, there are not enough s orbitals to form 3 sigma bonds that are pure s. Just considering s and p orbital involvement, we could say that the sulfur is $\ce{sp^2}$ hybridized.  This would produce 3 $\ce{sp^2}$ orbitals for sigma bonds from sulfur to oxygen and leave one p orbital for one pi bond.  We can draw 3 resonance structures for this arrangement, so we would expect all bond lengths to be equivalent.


Reaction of alkyl halides with strong unhindered bases

I agree with you, I'd expect some $\mathrm{S_N2}$ product too. I'm not sure about the relative quantities of the $\mathrm{E2}$ and $\mathrm{S_N2}$ products though.

I wouldn't flat out say that you'd expect more $\mathrm{S_N2}$ product than $\mathrm{E2}$ product, because although the substrate is unhindered, the solvent is ethanol, which is protic. The solvent interacts with the nucleophile through hydrogen bonding.

This hinders the nucleophile. This slows down the rate of the $\mathrm{S_N2}$ reaction because now the nucleophile is dragging along lots of solvent molecules, and it also increases the bulk of the nucleophile.


Physical Meaning of the Difference between MO and VB Wave Function of H2?

I'll give a somewhat simplistic answer. VB assumes that all bonding is between only two atoms. MO bonding considers bonding as a molecular phenomenon.

In $\ce{H2}$ there isn't really much difference. However consider benzene, $\ce{C6}\ce{H6}$. VB theory would have three double carbon bonds and three single carbon bonds which isn't really true. So VB bonding then hybridizes over the six carbon atom ring to predict the aromatic character of the molecule. However in MO theory the aromatic character is fundamental to the model.

From wikipedia

  "In particular, valence bond theory correctly predicts the dissociation of homonuclear diatomic molecules into separate atoms, while simple molecular orbital theory predicts dissociation into a mixture of atoms and ions. For example the molecular orbital function for dihydrogen is an equal mixture of the covalent and ionic valence bond structures and so predicts incorrectly that the molecule would dissociate into an equal mixture of hydrogen atoms and hydrogen positive and negative ions."


When measuring avidity should you use the ligand concentration or the nanoparticle concentration?

In the case where there is not cooperativity between binding events, the Scatchard Equation is the generalized solution to multiple ligand binding to a macromolecule. The Scatchard plot uses the free ligand concentration.

There are multiple models used to address cooperative binding. The Hill Equation is the most frequently encountered in biochemistry literature:
$$ \theta = \frac{[L]^n}{K_d+[L]^n} $$
Where $[L]$ is the concentration of free ligand, $\theta$ is the fraction of bound ligand-receptor complex, $K_d$ is the apparent dissociation constant, and $n$ is the Hill coefficient, describing cooperativity.


Can an aqueous solution conduct electricity forever?

Since nobody has posted a counter example, I will go ahead and say that my claim is true for direct current (which is what I originally had in mind). The current drives a chemical reaction and, at some point, the reactants will run out and the solution will no longer have enough ions to effectively conduct electricity.

Hopefully someone will post a counter-example and make things more interesting :)

In the case of alternating current, it seems possible to plate material onto one electrode and remove it from the other. When the current switches direction, this process would reverse and the electrodes would remain the same mass. In theory, this could go forever, but in reality I wonder if the electrodes would deteriorate and break due to the many cycles of deposition and erosion.


Calculating the volume of 2 M solution needed to make a 0.1 M solution

If you have a simple solution of say $\pu{0.1 M}$ sodium chloride, you have $\pu{0.1M}\, \ce{NaCl}$ in $\pu{100 ml}$.
Which is equivalent to "$\pu{100 mL}$ of $\pu{0.1 mol}\,\ce{NaCl}$ per litre", or,
$$n_\ce{NaCl} =\frac{100}{1000} \times 0.1 = \pu{0.01 mol}$$
$\pu{5 mL}$ of $\pu{2 M}\ \ce{NaCl}$ is equivalent to "$\pu{5 mL}$ of $\pu{2 mol}$ $\ce{NaCl}$ per litre", or,
$$n_\ce{NaCl} = \frac{5}{1000} \times 2 = \pu{0.01 mol}$$
So congratulations, you did it right. If ever you aren't sure, just write out all the steps and follow it through.
You could also look at it like this - $\pu{2 M}$ is $20\times$ more concentrated than $0.1\times\frac{2}{0.1}$, so you expect to need $20\times$ less of the $\pu{2 M}$ than the $\pu{0.1 M}$.
$\pu{5 ml}$ is $20\times$ less than $\pu{100 mL}$.


Balancing reduction equations for galvanic cell

To construct reduction equations there are a set of steps you can follow.

For a balanced equation, the number of atoms of each element, and the total charge, must be the same on each side of the equation.

Firstly identify the main reactants and products which you have correctly done in this case as: $$\ce{[Sn(OH)4]^{2-}(aq) -> Sn(s)}$$

Under acidic conditions the only things we can add to balance the reaction are $\ce{H=}$, $\ce{H2O}$ and $\ce{e-}$. Under alkaline conditions we can add $\ce{OH-}$ instead of $\ce{H+}$. Usually, when dealing with alkaline reactions the best way of dealing with them is to treat them as acidic and then neutralize the acid later. However, in this instance, since the only reactant involves hydroxide ions, we can immediately balance the hydrogens and oxygens just by adding four hydroxide ions to the products: $$\ce{[Sn(OH)4]^{2-}(aq) -> Sn(s) + 4OH- (aq)}$$

Then balance the charges to get the overall equation: $$\ce{[Sn(OH)4]^{2-}(aq) + 2e- -> Sn(s) + 4OH- (aq)}$$

The alternative method which works for any alkaline reactions, which are sometimes less obvious than the example above, is to balance the reaction under acidic conditions and then neutralize the acid.

Since we only have one option for balancing the oxygens we add $\ce{H2O}$ first: $$\ce{[Sn(OH)4]^{2-}(aq) -> Sn(s) + 4H2O(l)}$$

Now balance the hydrogens: $$\ce{[Sn(OH)4]^{2-}(aq) + 4H+(aq) -> Sn(s) + 4H2O(l)}$$

Now the charges: $$\ce{[Sn(OH)4]^{2-}(aq) + 4H+(aq) + 2e- -> Sn(s) + 4H2O(l)}$$

If this were in acidic solution then this would be the final balanced equation. However, since we are in alkaline solution, we add $\ce{OH-}$ ions to cancel out the $\ce{H+}$, forming $\ce{H2O}$. $$\ce{[Sn(OH)4]^{2-}(aq) + 4H2O(l) + 2e- -> Sn(s) + 4H2O(l) + 4OH- (aq)}$$

Finally we can cancel the excess $\ce{H2O}$ to give: $$\ce{[Sn(OH)4]^{2-}(aq) + 2e- -> Sn(s) + 4OH- (aq)}$$

This is the same result as we got earlier by the quicker method.


I think you can come at this approximation in two ways. Using more advance methods, the approximation is obtained as a truncation of the Laurent series of $\sqrt{x(x+2)}$ about $x=\infty$.

This is possible, but I think needlessly complex in this case. Using just algebra, we can note $$\sqrt{n(n+2)}=\sqrt{n^2+2n}\approx\sqrt{n^2+2n+1}=\sqrt{(n+1)^2}=n+1$$
By looking at a plot, we can see this approximation is very good, giving essentially the exact result at $n=10$.


Equilibrium, temperature change, and prediction in constant volume

For this particular scenario, the system in equilibrium satisfies (assuming ideal gases):

$$K_{\text{p}}^{(T)} = \frac{p_\text{C} p_\text{D}}{p_{\text{A}}^3 p_\text{B}} = \frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(T)\,2}}}$$

The equilibrium constant at temperature $2T$ can be calculated as:

$$K_\text{p}^{(2T)} = K_\text{p}^{(T)} \exp \left(\frac{\Delta H}{2RT}\right)$$

After increasing the temperature at constant volume, a new total pressure, $p_\text{tot}^{(2T)}$ is attained. Since the molar fractions are not changed at the beginning, the reaction quotient is:

$$Q_{\text{p}}^{(2T)} = \frac{p_\text{C} p_\text{D}}{p_{\text{A}}^3 p_\text{B}} = \frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(2T)\,2}}}$$

So, for the reaction to proceed forward, we want $Q_{\text{p}}^{(2T)} < K_\text{p}^{(2T)}$, that is:

$$\frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(2T)\,2}}} < K_\text{p}^{(T)} \exp \left(\frac{\Delta H}{2RT}\right) = \frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(T)\,2}}} \exp \left(\frac{\Delta H}{2RT}\right)$$

Simplifying:

$$\frac{1}{{p_\text{tot}^{(2T)\,2}}} < \frac{1}{{p_\text{tot}^{(T)\,2}}} \exp \left(\frac{\Delta H}{2RT}\right)$$

But $\displaystyle p_\text{tot}^{(T)} = \frac{n_\text{tot}RT}{V}$ and $\displaystyle p_\text{tot}^{(2T)} = \frac{2n_\text{tot}RT}{V}$, so:

$$\frac{1}{4} < \exp \left(\frac{\Delta H}{2RT}\right)$$
$$\Delta H > - 2RT \ln{4}$$

As long as $\Delta H$ satisfies this relationship (which will always imply a range of exothermic reactions since $T$ is always positive), this reaction will proceed when increasing the temperature at constant volume.

Notice that if this was a reaction where $\Delta n = 0$, the $p_\text{tot}$ cancels out and we get:

$$1 < \exp \left(\frac{\Delta H}{2RT}\right)$$
$$\Delta H > 0$$

Which implies that, when change in number of moles is not a factor to consider, the equilibrium will only shift to the right upon temperature increase at constant volume for reactions with $\Delta H > 0$, i.e. endothermic reactions.


cyclopropane's reaction with lithium aluminium hydride

Cyclopropanes do not react with LiAlH4. Substituents of cyclopropanes rings may be reduced with std LiAlH4 conditions with the cyclopropane ring being untouched

example in this patent


Help me understand electrochemistry and batteries

You can't "connect" with Zn/Cu? You'll need to be more clear, I don't understand what you mean. (You seem to be implying that you understand Al and steel in a more comprehensive way than zinc or copper, but that doesn't seem plausible to me.)
The zinc half cell reaction is Zn°(s) + H2O + H+ → Zn++(aq) + 2e- + H2O + H+ (pH << 7 ) or eliminating the "by-standers" Zn°(s)→ Zn++(aq) + 2e-. The chemical reaction at the copper electrode (and it would be the same reaction for many electrically conductive and inert electrodes, such as Pt (platinum), C (graphite), Au (gold), Aq (silver), and many more) is 2H+(aq) + 2e- → H2↑ (pH << 7) where I've 'canceled' the by-standers again. There are two chemical reactions going on, one at each electrode. The simple analogy is two cups with a tube between them. If the cups and the tube are all filled with water, then water will flow from one to the other if one is higher (and of course one lower) than the other. The water flows because the gravitational potential energy is greater for the water in the higher cup. Well, energy flows if the half-cell reaction (the reaction on one electrode) is more energetic than the other. Please refer to http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm or http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/electpot.html or https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page) for tables of just a few of the possible redox reactions (note that most of the half-reactions listed are inorganic, with a couple of exceptions. There is an enormous number of organic redox reactions which aren't generally tabulated.) The hyperphysics page links to a further discussion. Aluminum is always (at least, on this planet's surface) covered by a insulating layer of Al2O3. I don't know what would happen if you connected a piece of Al to a piece of "steel". First because I don't know what your electrolyte is, second because commercially available aluminum is, just like steel, an alloy of several elements. In other words, neither is a pure (or nearly pure) element, unlike the copper or zinc you can get. But, there's no reason why you couldn't substitute a piece of stainless steel for the Cu electrode, or even carefully open up a zinc-carbon battery's carbon electrode (if you can find one).=============== The important facts are that each half-cell reaction has a "potential" and the larger the difference is between the two half-reactions, the greater the current flow and voltage will be between electrodes. Since many of the half-cell reactions won't work in the same electrolyte, salt bridges (see Wikipedia) can be used to connect the fluids. (BTW, there's another reason why the use of steel can be a problem, there can be various reactions on the surface of the steel (corrosion) in which one part of the surface acts as an anode, and another part acts like a cathode. If an electrode is "too reactive" in a particular electrolyte, these side reactions can decrease the amount of electricity flowing to the other electrode. Cu/Zn is popular because all you need is a copper wire, a lemon (or even some vinegar (and salt)), and something that's either made of zinc, or has been galvanized (like a galvanized nail, which is coated with a layer of (mostly) zinc). While it doesn't seem like the hydrogen half-reaction involves the copper, actually the voltage (potential) of that reaction depends on the nature of the surface of the metal, sanding the surface until it is 'bright' will increase the amount of H2 produced. So, to sum it up, first you need two different oxidation/reduction half-cell reaction components, then you need to connect them in an electrolytic system so that the circuit is complete. An important fact, but sometimes overlooked in beginning discussions of half-reactions, is that the reaction should be considered reversible with the arrow → reversed ← as well as changing the sign of the potential (multiplying it by -1, so if it is negative it becomes positive, and vice versa). Also note that many tables assume an aqueous environment, and often an acidic or basic pH. The tendency of any "randomly chosen" electrodes to work, depends on the system they create (including your choice of electrolyte and pH). In a sense, half-reactions are a clever way of separating chemical reactions, like putting a paddle wheel into a stream of flowing water, to use some of the energy being release by the chemical reaction. If you didn't keep the two reactions separate, all that would happen is the chemical reactions would generate heat instead of electricity (and heat). In other words, putting zinc metal into acid will generate hydrogen gas (as long as a surface, maybe even the zinc) exists for the gas to form on), and won't generate any electricity (because the electrons aren't being forced through the circuit). This answer is a little disorganized, sorry.


Miller Indices of Goldnanoparticles from Turkevich and Frens method

Assuming you refer to a record like this:

which I found in Kimling et al. in «Turkevich Method for Gold Nanoparticle Synthesis Revisited» (2006JPhysChemB15700) when reporting X-ray diffraction.


In Thomson's atomic model, why did he not assume that there are positively charged particles embedded in a negatively charged sphere?

J. J. Thomson is recognized with blowing apart a long-held belief that atoms were indivisible through his discovery of electrons while studying the deflection of discharges from cathode ray tubes.  He postulated that the particles formed in this way were small with a very high charge to mass ratio.  At this point he had two pieces of information: (1) an atom has an overall neutral charge and (2) part of the atom consisted of these small negatively charged entities which we now call electrons.  He therefore postulated that the rest of the atom consists of positive charge to balance the negative charge of electrons.  It is believed that the plum pudding analogy stems from his proficiency as an instructor and a desire to relate the atomic model to something his students could understand.


Bubble formation when aluminium reacts with copper(II) sulfate solution

As Ivan Neretin noted in a comment, your supposition is absolutely correct.  The aluminium metal is reactive enough with water that it forms hydrogen gas and aluminum hydroxide (or other aluminum oxyhydroxide species) where it's exposed to the solution:

$$
\begin{align}
\ce{6H2O + 6e-} & \ce{->~ 6OH- + 3H2\uparrow} \\
\ce{2Al(s) + 6H2O } & \ce{->~2 Al(OH)3 + 6H+ + 6e-}
\end{align}
$$

Overall reaction:

$$
\ce{2Al(s) + 6H2O -> 2Al(OH)3 + 3H2\uparrow}
$$

If the $\ce{CuSO4}$ solution is acidic enough, the $\ce{Al(OH)3}$ rapidly dissolves, exposing a fresh aluminium surface, ready for further reaction.

On the other hand, the redox reaction of $\ce{Fe(s)}$ and water is not favorable enough for hydrogen evolution to occur, which explains the lack of bubbles you observed.  The single-displacement reaction between $\ce{Cu^2+}$ and $\ce{Fe(s)}$ is quite favorable, however, leading to paired reactions like the following:

$$
\begin{align}
\ce{Fe(s)} & \ce{->~ Fe^2+ + 2e-} \\
\ce{Cu^2+ + 2e-} & \ce{->~ Cu(s)}
\end{align}
$$

Overall reaction:

$$
\ce{Fe(s) + Cu^2+ -> Cu(s) + Fe^2+}
$$

This results in the rapid exchange of metallic Cu for metallic Fe on the iron surface, leading to a film of plated copper on the surface.


Calculation of volume required for titration with dibasic acid

We first observe the balanced equation. $$\ce {H2SO4 + 2NaOH <=> Na2SO4 + 2H2O}$$ we see the ratio of $\ce {NaOH}$ and $\ce {H2SO4}$ is 2:1. Given $ 0.050~\mathrm{L}$ of $\ce {1M~NaOH}$ we find moles of $\ce {NaOH}$ and use stoichiometry to find how many moles of $\ce {H2SO4}$ is needed. We find the volume to be $25~\mathrm{ml}$.


$\ce{H2O}$ is a both a neutral molecule and an amphoteric molecule — the two are not mutually exclusive terms. An amphoteric molecule is simply one that can act as either an acid or a base, while a neutral molecule is one in which the total number of protons is equal to the number of electrons, such that there is zero net charge. Water can act as a Brønsted acid (i.e., an $\ce{H+}$ donor), as per the following reaction:

$\ce{H2O + NH2- -> OH- + NH3}$

However it can also act as a Brønsted base (i.e., an $\ce{H+}$ acceptor):

$\ce{H2O + HCl -> H3O+ + Cl-}$

When proton transfer is involved, as in the Brønsted definition of acids and bases, the more precise term is amphiprotic. However, water can also act as both a Lewis acid (an electron pair acceptor) and Lewis base (an electron pair donor). Notably, as a Lewis base, it is not limited merely to proton abstraction, but can act as a nucleophile with any number of electrophilic molecules.