Do superhydrophobic/hydrophobic materials generally have a lot of friction or only a little friction

Since you are talking about hydrophobicity I am assuming you are interested in liquid-solid friction. If you are interested in solid-solid friction I can say only one thing: hydrophobicity and solid-solid friction are in general unrelated.

So for the liquid-solid case. I am going to give you an answer that gives some insight, but no concrete rules, because, as far as I know, there aren't any.

Superhydrophobic surfaces typically have some surface structure that allows the liquid to 'float' partly on air and partly on top of the surface structure . Friction is related to the amount of contact area where there is a no-slip boundary so that will decrease therefore friction will be lower. A cool application of this effect can be found in this 2013 PNAS article by Karatay et al.

However, if you have a droplet moving, instead of a full liquid stream, then there is a counter-acting effect which is called contact angle hysteresis, which allows a droplet to sustain a difference in contact angle at the front and back of the droplet. This causes a force counter-acting motion of the order of the surface tension of the liquid $\gamma$ and the difference between the contact angles $\Delta \theta$: $F=\gamma l \Delta \theta$, where $l$ is a lengthscale of the order of the droplet radius. This force basically acts as a static friction on the droplet.

The value of $\Delta \theta$ is a complex function of the amount of surface roughness, being 0 for a atomically smooth surface but also approaching 0 for roughness with a small typical lengthscale. Therefore it is hard to make a general statement whether superhydrophobic surfaces will have a lot or a little friction. It all depends on the details of the surface roughness.


Why mass spectra of 2-methylbutane do not show big peak at 71 coming from tertiary carbocation?

How do you know that losing a CH3 radical gives you a secondary carbocation?
Maybe it will perform a hydride shift and you land at the tertiary carbocation.
Always take this into account, because you can shoot yourself in the foot if you assume carbocations to be too stable. In EI, they rearrange all the time (e.g. benzylium <-> tropylium).
However, seeing that the ion with $m/z~57$ subsequently looses $\mathrm{\text{:}CH_2}$, it seems reasonable to assume that the secondary carbocation is formed here.

The important point is the following: You propose that instead of loosing the methylradical, yielding a secondary carbocation (or a tertiary, considering the argument from above), there should be a loss of a hydrogen radical.
However, the hydrogen radical is far less stable than the methyl radical, and I assume this is the reason for the regioselectivity, here.

Also keep in mind that you supply a lot of energy (70 eV). Hence, the induced fragmentation processes are highly endothermic, and should be under kinetic control.

If you really want to find out what is going on, you should probably have a look at McLafferty's/Turecek's classic textbook "Interpretation of Mass Spectra".


Deciding more electrophilic centre between ester and thioester for SN type reaction

There is also another effect of delocalisation of the electron pair on $\ce{O}$ and $\ce{S}$ atom adjacent to the cabonyl carbon, which is dominant over the electron-withdrawing effect ($\ce{-I }$ effect).

In both the parts i.e. $\ce{OCOCH_3}$ and $\ce{SCOCH_3}$, there are two lone pairs on each of $\ce{O}$ and $\ce{S}$, which can delocalise with the carbonyl carbon. Thus, the bond between $\ce{O}$ and $\ce{C}$, and $\ce{S}$ and $\ce{C}$, will get a partial double bond character. Due to this delocalisation, the electrophilicity of the carbonyl carbon is decreased.

When the delocalisation happens in case of $\ce{O}$, the overlap between  $\ce{O}$ and $\ce{C}$ is $\ce{2p$\pi$ - 2p$\pi$}$ overlap, which is a strong overlap and thus the double bond character between $\ce{O}$ and $\ce{C}$ is more. Thus, electrophilicity is reduced to a great extent. But in case of $\ce{S}$, there is a relatively weaker $\ce{3p$\pi$ -2p$\pi$}$ overlap which introduces very feeble double bond character between them so, the electrophilicity is reduced to very less extent and thus the first electrophilic attack will happen at the $\ce{C}$ adjacent to $\ce{S}$. This is the reason to your question.


Why 2p Orbitals overlap more effectively than 3p orbitals?

As the quantum number increases, the size of the orbital, or more precisely, the area in which electrons are found, increases. The same 1s orbital of H combines with greater overlap with 2p of F,  while the larger 3p of Cl can’t be approached to that extent because of its size.
For the sake of demonstration, consider a child shaking hand with the other child, which is a full overlap of palms, while a child shaking hand with an adult is not!


There are two reasons, depending on the anion of interest:

1) For the flame emission of any species, anion or cation, metal or not, to be observable, it must emit light in the visible wavelength range, and most non-metal anions do not do this very strongly.

2) Even if there is some emission from the non-metal anion, in many cases their flame excitation efficiency is very low.

The answer then is the combination of these two things, and the fact that many metals have the opposite tendencies (e.g. high excitation efficiency, strong emission in the visible region of the spectrum), which overrides any weak visible emission from the non-metal anions.

As noted in a comment, this is not always the case, but it is the explanation for this strong trend.


Comparing C=O bond length in CO2 and CH2O

Drawing the Lewis structure for CO$_2$ and CH$_2$O shows that the C atom in CO$_2$ is sp hybridized and therefore forms a double bond (which is composed of a $\sigma$ and $\pi$ bond) with either O by overlapping one of its sp orbital with one of the sp$^2$ orbitals of O (since O is sp$^2$ hybridized) to form the $\sigma$ bond and one of its p orbitals with the p orbital of O to form the $\pi$ bond.

In CH$_2$O, the C atom is sp$^2$ hybridized, meaning the C$=$O bond is formed by an overlapping sp$^2$ orbital from C and an sp$^2$ orbital from O (since O is sp$^2$ hybridized) for the $\sigma$ bond and an overlapping p orbital from C and a p orbital from O for the $\pi$ bond.

So, the difference in the C$=$O bonds in the two compounds is in the constituent orbitals of their respective $\sigma$ bonds. The $\sigma$ bond of the C$=$O bond is shorter in CO$_2$ because it involves the overlap of an sp orbital with an sp$^2$ orbital. An sp orbital has more "s character" than an sp$^2$ orbital (you can think of an sp orbital as 50% s character, 50% p character and an sp$^2$ orbital as 33% s character, 67% p character), therefore the electron density in an sp orbital is on average closer to the nucleus of the C atom than in an sp$^2$ orbital, leading to a shorter $\sigma$ bond length. Because the $\pi$ bonds are formed from identical (up to orientation) p orbitals in each compound, the shorter $\sigma$ bond in CO$_2$ gives it a shorter C$=$O bond length than CH$_2$O.


In acid/base chemistry, are there amphoteric substances that undergo something that resembles disproportionation in redox chemistry

As I mentioned in the comments, I know of one case where a polyprotic acid has a second proton which is more acidic than the first: the aqueous pervanadyl complex, $\ce{[VO_2(H_2O)_4]^+}$. According to Wikipedia,

$$
\begin{array}{rcl}
\ce{[VO_2(H_2O)_4]^+ &<=>& H3VO4 + H+ + 2 H2O} &  \mathrm{pK_{a_1}} = 4.2\\
\ce{H3VO4\ \  &<=>& H2VO4^- + H+} &  \mathrm{pK_{a_2}} = 2.60\\
\ce{H2VO4- &<=>& HVO4^2- + H+} &  \mathrm{pK_{a_3}} = 7.92\\
\ce{HVO4^2- &<=>& VO4^3- + H+} &  \mathrm{pK_{a_4}} = 13.27
\end{array}
$$

Though it certainly feels quite odd that $\mathrm{pK_{a_2}<pK_{a_1}}$, I suspect this feature is not absurdly uncommon in larger/more complex systems, where structural reorganisation upon deprotonation can be severe. Perhaps what makes the pervanadyl complex unusual is its comparative simplicity.

However, Wikipedia provides no source for those acidity constants. I looked around for an article discussing the speciation of vanadium ions in aqueous solution, and discovered this nice reference:

  Huang, J.-H.; Huang, F.; Evans, L.; Glasauer, S. Vanadium: Global
  (Bio)Geochemistry. Chemical Geology 2015, 417, 68–89.
  DOI:10.1016/j.chemgeo.2015.09.019

The relevant portion in this article is section 4.1. (Aqueous speciation chemistry of vanadium). The authors use thermodynamic data obtained elsewhere to establish the following equilibria:

$$
\begin{array}{rcl}
\ce{[VO_2(H_2O)_4]^+ &<=>& H3VO4 + H+ + 2 H2O} &  \mathrm{pK_{a_1}} = 3.67\\
\ce{H3VO4\ \  &<=>& H2VO4^- + H+} &  \mathrm{pK_{a_2}} = 3.40\\
\ce{H2VO4- &<=>& HVO4^2- + H+} &  \mathrm{pK_{a_3}} = 8.06\\
\ce{HVO4^2- &<=>& VO4^3- + H+} &  \mathrm{pK_{a_4}} = 13.28
\end{array}
$$

As you can see, the first two constants are rather different from the values quoted on Wikipedia, and now they've become difficult to distinguish. However, clearly this is still an unusual case. They take these data and then calculate speciation curves to obtain the following graph:

Evidently, the speciation curves don't really look anything out of the ordinary. It's likely this is partly due to the proximity of the values of $\mathrm{pK_{a_1}}$ and $\mathrm{pK_{a_2}}$ used, and also because the solution is highly dilute. It would be nice to know what happens in more concentrated solutions, but unfortunately we get something like this:

Like some other transition metals, vanadium has a tendency to oligomerise in solution, forming complex mixtures of polyoxometalates. Therefore, it's basically impossible to get a good visualisation of the effects of $\mathrm{pK_{a_2}<pK_{a_1}}$.

Fortunately, we can forget the chemistry and just study how the mathematics behaves. For simplicity, let's consider a diprotic acid $\ce{H2A}$, where we choose the values of $\mathrm{pK_{a_1}}$ and $\mathrm{pK_{a_2}}$. Also for simplicity, let us safely ignore the autodissociation of water by working with relatively concentrated solutions. Therefore, the mole fractions of the species $\ce{H2A}$, $\ce{HA-}$ and $\ce{A^2-}$ will be:

$$X_{H_2A} = \frac{[H^+]^2}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$

$$X_{HA^-} = \frac{K_{a_1}[H^+]}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$

$$X_{A^{2-}} = \frac{K_{a_1}K_{a_2}}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$

Let's keep $\mathrm{pK_{a_1}}=5$ and decrease $\mathrm{pK_{a_2}}$. Here are the individual speciation curves for $\mathrm{pK_{a_2}=10}$ to $\mathrm{pK_{a_2}=-1}$, in decrements of 1 unit of $\mathrm{pK}$ (click on graphs for a larger version).

As you can see, there is no sudden catastrophe as the second proton becomes more acidic than the first. As the second ionisation becomes more favourable, it smoothly and continuously suppresses the formation of $\ce{HA-}$ . Something nice does happen at the crossover point, where $\mathrm{pK_{a_1} = pK_{a_2}}$, namely all species are present in the same concentration when $\mathrm{pH = pK}$. However, this not surprising if you look at the equations above.

For good measure, here are all the plots once again, now with a logarithmic y-axis. It's nice to see that even with the suppression of $\ce{HA-}$, it still exists as a trace in the background, and still behaves smoothly.

And to finish things off, now here are graphs showing the mole fraction of each unique species as $\mathrm{pK_{a_2}}$ varies from 10 to -6, in decrements of 1 unit of $\mathrm{pK}$. I rather like the subtle way the curves for the fully protonated and fully ionised species change, as the intermediate ion stops being a significant contributor in the equilibria.


Entropy Change for mixing water at two different temperatures

Well, your procedure is entirely correct. There's only one problem: the units. Since the number of moles of water is specified, your final answer should not include $\text{mol}^{-1}$. And since there's exactly $1 \text{ mol}$ of both the cold and hot water, the numerical answer should be the same.

I don't know where your error arises from but I suspect it may be because you used the units of $C_{p,m}$ (molar heat capacity at constant pressure) instead of $C_p$ (heat capacity at constant pressure). The former is an intrinsic property, whereas the latter is an extrinsic one, equal to the former multiplied by the amount of substance: $C_p = n\cdot C_{p,m}$. And their relation to the $\Delta S$ of heating at constant pressure:

$$\begin{align}
\Delta S &= C_p\ln{\left(\frac{T_2}{T_1} \right)} & &\text{Units: J K}^{-1} \\
\Delta S_m &= C_{p,m}\ln{\left(\frac{T_2}{T_1} \right)} & &\text{Units: J K}^{-1}\text{ mol}^{-1}
\end{align}$$

I'm entering nitpicking mode again so I shall also pedantically point out that as far as this equation is concerned, entropy comes with a triangle in front of it. Meaning you write $\Delta S$ instead of $S$.


Determining boiling point on a created temperature scale

Unlike an absolute temperature scale like degrees Kelvin, Celsius and your hypothetical °G are relative scales, where the zero points are set to the melting points of water and gallium, respectively. In your first step, you've accounted for the difference in the size of one degree Celsius and one °G, but not the difference in the zero points.

Instead of sodium chloride, think about what the melting point of gallium is on each scale. You've defined it as 0 °G, but you know it's not 0 °C. What do you have to do to correct for this to get a value in Celsius?

Incidentally, this is also why the conversion between Celsius and Fahrenheit is not just multiplying by a constant.


I myself while searching and talking to my dad found the answer.Sounds weird, sorry for posting.

THE NAMES OF ORGANIC COMPOUNDS

Cracking the code

A modern organic name is simply a code. Each part of the name gives you some useful information about the compound.

For example, to understand the name 2-methylpropan-1-ol you need to take the name to pieces.

The prop in the middle tells you how many carbon atoms there are in the longest chain (in this case, 3). The an which follows the "prop" tells you that there aren't any carbon-carbon double bonds.

The other two parts of the name tell you about interesting things which are happening on the first and second carbon atom in the chain. Any name you are likely to come across can be broken up in this same way.

Counting the carbon atoms

You will need to remember the codes for the number of carbon atoms in a chain up to 6 carbons. There is no easy way around this - you have got to learn them. If you don't do this properly, you won't be able to name anything!

code  no of carbons
meth    1
eth 2
prop    3
but 4
pent    5
hex 6
Types of carbon-carbon bonds
Whether or not the compound contains a carbon-carbon double bond is shown by the two letters immediately after the code for the chain length.
code    means
an  only carbon-carbon single bonds
en  contains a carbon-carbon double bond
For example, butane means four carbons in a chain with no double bond.

Propene means three carbons in a chain with a double bond between two of the carbons.

Alkyl groups

Compounds like methane, $\ce{CH4}$, and ethane, $\ce{CH3CH3}$, are members of a family of compounds called alkanes. If you remove a hydrogen atom from one of these you get an alkyl group.

For example:

A methyl group is $\ce{CH3}$.

An ethyl group is $\ce{CH3CH2}$.

These groups must, of course, always be attached to something else.

Types of compounds

The alkanes

Example 1:  Write the structural formula for 2-methylpentane.

Start decoding the name from the bit that counts the number of carbon atoms in the longest chain - pent counts 5 carbons.

Are there any carbon-carbon double bonds? No - an tells you there aren't any.

Now draw this carbon skeleton:

Put a methyl group on the number 2 carbon atom:

Does it matter which end you start counting from? No - if you counted from the other end, you would draw the next structure. That's exactly the same as the first one, except that it has been flipped over.

Finally, all you have to do is to put in the correct number of hydrogen atoms on each carbon so that each carbon is forming four bonds.

If you had to name this yourself:

Count the longest chain of carbons that you can find. Don't assume that you have necessarily drawn that chain horizontally. 5 carbons means pent.

Are there any carbon-carbon double bonds? No - therefore pentane.

There's a methyl group on the number 2 carbon - therefore 2-methylpentane. Why the number 2 as opposed to the number 4 carbon? In other words, why do we choose to number from this particular end? The convention is that you number from the end which produces the lowest numbers in the name - hence 2- rather than 4-.
If you want more detail goto: http://www.chemguide.co.uk/basicorg/conventions/names.html


Is shielding of electrons really best described as shielding or is it really Coulombic repulsion between electrons?

The shielding model works as follows:

 (source)

So it is a mixture of attraction and pushing from protons and non-valence electrons, in accord with the Coulomb's law. The prediction of the effective core charge is not an easy task - you will have to use quantum chemistry methods (e.g. Hartree-Fock) to predict those, like its done in this reference.

I think its best viewed as a pragmatic model to boil down the higher, not very graphic, quantum theories, to one number, the effective nuclear charge:
$$ Z_\text{eff} = Z - S$$


Is there a means to measure pH without indicators?

Have you tried with red or black tea? It turns lighter in acidic media and turns dark in basic media. I tried once adding the pulp juice of one lemon to a cup of tea and it lighted up (mainly for acetic acid), then I added NaOH-based salt and turned very black and opaque.

Maybe that suits for your solution.


Technically speaking, one typically wouldn't write the left hand side unitless, as it is the ratio of the mass of two different components (usually mg of some chemical per gram of the surface).
But, otherwise your intuition is correct and $k$ just needs units that will cancel with $p^{\frac{1}{n}}$ to give the units on the left hand side. So $k$ could have units of $\left(\dfrac{\text{mg chemical}}{\text{g surface}}\right)\left(\mathrm{bar^{\frac{-1}{n}}}\right)$.
We could also reformulate Freundlich's equation in terms of concentration (since we are dealing with an isotherm) as: $$q=KC^{\frac{1}{n}}$$ where $q=m/x$, $C$ is concentration of the chemical and $K$ is a new constant with units of $\left(\dfrac{\text{mg chemical}}{\text{g surface}}\right)\left(\mathrm{{\dfrac{L}{\text{mg chemical}}}}\right)^{\frac{1}{n}}$.
Source


How can I easily identify if a substance is an acid or base based on its chemical formula?

I was going to mark this as too broad, but it's not too broad, because the answer is quite simple: in general, you cannot determine the acid or base properties of a substance from its formula.

As a concrete example, consider the formula $\ce{C4H7NO3}$.

Here are two possible isomers for that formula:

The left compound is an acid. The right compound is a base.


How is phosphate stable disobeying octet rule?

The octet rule is just a rule. Once you leave the 2nd row of the periodic table, the octet rule doesn't apply. Further reading: https://en.wikipedia.org/wiki/Hypervalent_molecule


Does adding NH3 to an HCl solution facilitate the dissolution of aluminum compounds?

Aluminum metal is attacked by $\ce{HCl}$ solutions. But adding $\ce{NH3}$ or $\ce{NH4Cl}$ has no effect on aluminum reactions.
The same thing can be said for aluminum compounds. Adding ammonia or ammonium compounds is not recommended, because it does not help. Aluminum does not make complexes with ammonia or ammonium salts.
But I suspect this addition of ammonia or of ammonium compounds may have another effect. It may help dissolution of other metals if they are deposited on the glass together with aluminum, making complexes containing the metal and ammonia.


Having asymmetric carbon atoms does not in itself make a compound chiral.

A compound is chiral if and only if it is not super(im)posable on its mirror image.

If there's a plane of symmetry, then reflecting in that plane gives the same thing, by definition, and that meets the criterion for super(im)posability - the mirror image is superimposable on the original and therefore the molecule is not chiral.

Only D- and L- chiro-inositol are chiral - all the other stereoisomers have planes of symmetry and are therefore not chiral.


Both are high-spin.

No, square planar complexes are nearly always low-spin. With a $\mathrm{d^8}$ configuration, the lowest four orbitals would be filled. That would make the complex diamagnetic.

For the octahedral $\mathrm{d^8}$ complex, regardless of whether it's high- or low-spin, it will have the same electronic configuration $\mathrm{(t_{2g})^6 (e_g)^2}$. By Hund's first rule, the two electrons in the $\mathrm{e_g}$ orbital will both be unpaired, so the complex will be paramagnetic.

Here's a quick sketch (energy levels not to scale):


Calculating dew point and bubble point of gas mixtures

In the picture above, the C is the critical point. There seems to be some easy methods to calculate the critical temperature, but not so with the critical pressure.

For bubble point it would give me a good indication of where the max range is.

But I realised that the critical temperature point won't help me finding the min and max range for dew point calculation. As you can see in the picture, the point M (max temperature point) would be a good indication of where the max range is for dew point.

Anyway I found a much simpler solution. I consider the bubble point and dew point calculation functions as a black box.

As I'm only seeking the bubble point / dew point for a specified temperature, I step the possible pressure range (1 kPa - 10 MPa), with steps of 1 kPa. Then I calculate the numerical derivate of using current step and previous step.

This function derivate is consistent and negative in the range of the function domain where a bubble or dew point value is valid.

Outside the function domain the derivate is close to zero (+-1E-15).

So this gives me a good starting point for the root finding algorithm.

I also changed from False Position method to Bisection method, as it's less sensitive to derivates.

I hope this will help someone else with similar problem.


It should be like:

The saturated partial pressure of each component of an ideal mixture of liquids is equal to the saturated vapour pressure of the pure component multiplied by its mole fraction in the liquid mixture.

These data are independent on the actual composition of the gaseous phase, that may, or may not be in the equilibrium with the liquid.
OTOH, composition of the liquid phase does depend on the amount and composition of the original liquid mixture put into a container, as evaporation changes liquid composition in favour of less volatile component.
If the pure liquid $\mathrm{A}$ has saturated vapour presure $p_\mathrm{0,A}$
and the pure liquid $\mathrm{B}$ has saturated vapour presure $p_\mathrm{0,B}$
and if the $\mathrm{A}$ in the liquid has molar fraction $x_\mathrm{A,l}$,
then the gaseous molar fraction of $\mathrm{A}$ ( in the saturated gaseous phase, with assumption of ideal behaviour given by the Raoult law ) is:
$$x_\mathrm{A,g} = \frac{x_\mathrm{A,l} \cdot p_\mathrm{0,A}}{ x_\mathrm{A,l} \cdot p_\mathrm{0,A} + (1-x_\mathrm{A,l}) \cdot p_\mathrm{0,B}}$$
For calculation for particular volumes of the container and the initial mixture of volumes, you would need more data, like these volumes and densities, and assumptions how to treat gases, like. e.g. as the ideal gas, but it goes behind the Raoult law. In all the complexity, it is solving a set of equations for set of variables, like molar amounts in both phases and phase volumes.
If the fraction of components in the gaseous volume is negligible, e.g. if we consider the total vapour pressure in a full or almost full container, we can consider total mixture composition is the same as liquid composition.


Why ionic radii of Cu2+ is less than Zn2+?

To answer your question, it is good to see the full picture. I used data from the Wikipedia data page on ionic radii to plot the dependence of the ionic radius of dications by the number of d-electrons. The plot includes 1st d-row data for high-spin and low-spin states (these are terms from crystal field theory and are properly introduced elsewhere). It also includes data for Pd-Cd, Pt and Hg.

As we can see, a rather complex dependence is observed. However, it is clear, that the increase in ionic radius from the d8 to the d10 state is not a random fluctuation, it is observed for all three rows. To understand the nature of this dependence it is critical to understand that the values are for an octahedral ligand field (the values would be difference for tetrahedral and other coordinations). In such a field d-orbitals become unequal — three of them are lower in energy while two are higher. In high-spin complexes the difference is not enough to overcome electron pairing energy, while in low-spin ones it is. As result, for low-spin complexes we see a clear minimum at d6, corresponding to the three lower d-orbitals being fully filled, while for high-spin complexes we see changes in the trend for the 4th (first on higher orbital) 6th (all orbitals half-filled) and 9th (first paired on higher orbital) electrons. Now, let's check out the orbitals mentioned (thanks to chemwiki)

$\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$ are the orbitals with higher energy. As it can be seen, they are located between the metal atom's center and the ligands. As such, electrons on said orbitals have clear effect on metal-ligand distance, so for low-spin state the observed ionic radius monotonously descreases up to 6 electrons (fully occupied lower orbitals) and then monotonously increases. For the high-spin state, the increases are also located on steps when an electron is added to higher orbital, but in this case the electrons are added on d4-d5 and d8-d10 configurations, so the increase happens in two steps. For the 2nd and 3rd rows, only few last steps on the curve are observed as most elements in these rows tend to form metal-metal bonds in low oxidation states and thus their true ionic radii are unknown.

And now, why are these dependences of limited use. Consider the graph of $\ce{M-Cl}$ distances for chlorides of the first d-row below. The data was obtained from the open crystallography database. No correlation with the plot above can be established, despite most of the chlorides except for $\ce{ZnCl2}$  having their metal in sixfold coordination. This is because said chlorides are not isostructural. Layered $\ce{MoS2}$ and $\ce{CdI2}$ structures together with chains of squares and 3D-lattices occur. Furthermore, for iron two different structures (layered and rutile-like ones) were found with, obviously, different $\ce{M-Cl}$ distances (the layered one is plotted. As a fun fact, the longer distance is observed in the higher-pressure rutile modification). An obvious conclusion for the plot is that only for isosctructural fragments we can conclude something based on electron structure of the central metal atom.


Removing potassium chloride crystals from a mineral composed of manganese dioxides

$\ce{MnO2}$ is insoluble in water -- I would expect with a good deal of confidence that the mineral cryptomelane is also insoluble. $\ce{KCl}$ by contrast is highly soluble.

I would just treat the sample with cold water and filter out the dissolved $\ce{KCl}$ (e.g. with a Buchner funnel and filter paper). Then dry with heat.

I would choose a fine-grain filter paper though because, though insoluble, $\ce{MnO2}$ particles could still be small enough to filter through course paper, like coffee filters (from experience).


Even without seeing your molecule, I can say the referee is very likely correct. If you think of doing a scan of the dihedral of hydrogen peroxide, by symmetry the path from $\theta=180\to360$ should just be the reverse of $\theta=0\to180$. In your case, I don't know the symmetry around the dihedral, but at a bare minimum the energy should be the same after a $360$ degree rotation. The fact that it isn't suggests that some scan step (probably that point around $130$) irreparably distorted the molecule geometry so that additional steps didn't bring it back towards the initial geometry.

Generating the scan point by point may help, but you still have to ensure the rotations you are making are physical and aren't distorting the molecule too much. Sharp changes in the energy between steps are a sign that something is going wrong.

You don't mention this detail in your question, but its also important to consider whether what you want is a rigid scan (where you just change this angle) or a relaxed scan (where you change this angle and then allow the rest of the structure to relax).


What are the magnetic quantum numbers for specific d orbitals?

On p. 238 of my copy of McQuarrie (1983; his Table 6-6), he lists the following for the hydrogenlike atomic wave functions (expressed as real functions), describing these as "commonly used":
\begin{align*}
n=3, l = 2, m=0:     & \ \psi_{3d_{z^2}} \ ... \\
n=3, l = 2, m=\pm 1: & \ \psi_{3d_{xz }} \ ... \\
       : & \ \psi_{3d_{yz }} \ ... \\
n=3, l = 2, m=\pm 2: & \ \psi_{3d_{x^2-y^2 }} \ ... \\
       : & \ \psi_{3d_{xy }} \ ... \\
\end{align*}


What are the mechanisms for capture of As(III) and As(V) by magnetite from water?

Unfortunately there are no (or only very hard to find) studies to the mechanism of the actual sorption of arsenic (acids) to magnetite. Most literature does describe it as a simple Ligand exchange, although I find that hard to believe. From what I have read[1-4] I will try to compose some hints on what may happen.

First of all lets have a look at arsenic acid. It is a medium strong acid and a quite well oxidising agent.
\begin{align}\ce{
H3AsO4 &\equiv AsO(OH)3\\
AsO(OH)3 + H2O &<=> H3+O + AsO(O^- )(OH)2\\
AsO(OH)3 + 2H3+O + 2e- &<=> As(OH)3 + 3H2O
}\end{align}

In presence of any electron rich material and acidic environment it will therefore not be unusual that arsenous acid will be formed.[5] We may therefore also focus on this, amphoteric molecule:
\begin{align}\ce{
H3AsO3 &\equiv As(OH)3\\
As(OH)3 + H3+O &<=> As+(OH)2 + 2H2O\\
As(OH)3 + {}^{-}OH &<=> AsO^{-}(OH)2 + H2O
}\end{align}

Depending on the pH of the solution one would expect to find many of different of these  species. Arsenic is also somewhat oxophil, meaning that the $\ce{As-O}$-bond is quite strong.

The magnetite in aqueous solution reacts basic, also forming complex like structures or hydroxides, for example this fully hydrated cluster (ref. [2], stoichiometry omitted):
\begin{aligned}\ce{
 Fe3O4 + $x$\, H2O &<=>  Fe2(OH)2(H2O)8^4+
}\end{aligned}

Of course one has to pay attention, that the used ferric nano particles for sorpting arsen(o)ic acid(s) are far larger than the used model complex. However, we can see, that there are functional groups like $\ce{-OH}$ or $\ce{-OH2+}$ (or maybe even $\ce{-O-}$) covering the surface.

Using this as prerequisites we can think about what might happen next. I generally would suggest to start with physisorption, i.e. the cluster and the acid molecules come in close range. Since we are dealing with contaminated water it is maybe safe to assume, that we can deal with soluble salts of arsenoic acid, hence focusing on $\ce{AsO3^{3-}}$.

When the complexes come into close range hydrogen bonds will form, with
$
\newcommand{\nanoFe}{\bbox[lightgrey,3pt,border:1px solid black]{\ce{Fe}}}
\nanoFe$ being the nano particle:
$$\ce{\nanoFe-OH\cdots{}O-AsO2^{3-}}$$

Since the whole surface is covered with hydroxide groups, several of these hydrogen bonds may form. As we can see, with this we may now enter the realm of chemisorption. Once these complexes have formed, it is from an entropic point of view favoured to release smaller molecules, water in this case. Also the oxygen atoms bound to arsenic are now in close range to iron atoms. Since we are dealing with nano particle, there will be non-coordinated irons in proximity. This allows for a more permanent complet to form:
$$\ce{\nanoFe-O-AsO^3- -O\cdots{}HO-\nanoFe}$$

With this monodentate complex we have an already stable complex of a bound arsenic to an iron. It is furthermore possible, and probably favoured, to form a bidentate complex with a somewhat similar mechanism.

Let us also consider a more basic scenario. In this case the surface of the magnetite will not be entirely covered with hydroxide atoms, but more freely accessible oxygen atoms will be present. Direct bond formation between $\ce{\nanoFe-O\cdots{}AsO3^{3-}}$ may occur.

After writing all this I also found a credible source that somehow proves my theory. In “Understanding Arsenate Reaction Kinetics with Ferric Hydroxides”[6] they used a model system of $$\ce{HAsO4^2- + Fe2O10H15+ -> Fe2O10H15-HAsO4-}$$ to explain the mechanism.

In this they describe that physisorption is barrierless, leading fast to the adduct that is $\Delta G^\circ_\mathrm{r}=-58~\mathrm{kJ/mol}$ more stable in energy. However, chemisorption and developing the equilibrium is a more lengthy process.

When the formed complexes become neutral, the activation barrier to release arsenate is very high, compared to adsorption. I will stop right here, please read the publication for further details.

Notes, references and further reading:

Dimitri A. Sverjensky, Keisuke Fukushi, Geochimica et Cosmochimica Acta 2006, 70 (15), 3778-3802.
Nianliu Zhang , Paul Blowers , and James Farrell, Environ. Sci. Technol. 2005, 39 (13), 4816–4822.
Arnold F Holleman; Egon Wiberg; Nils Wiberg: Lehrbuch der Anorganischen Chemie. 102. Auflage. Walter de Gruyter & Co., Berlin, New York, 2007. (german)
Egon Wiberg, Nils Wiberg: Inorganic Chemistry. Academic Press, 2001. (english)
Thompson, A. & Goyne, K. W. (2012) Introduction to the Sorption of Chemical Constituents in Soils. Nature Education Knowledge 3(6):15
Heavier halogens ($\ce{X=I, Br, Cl}$) should do the job and they will most certainly be present in usual ground water (more or less acidic solution) :
$$\ce{AsO(OH)3 + 2H3+O + 2KX <=> As(OH)3 + 3H2O + X2 ^ + 2K+}$$
James Farrell and Binod K. Chaudhary, Environ. Sci. Technol. 2013, 47 (15), 8342–8347.
Also interesting: (a) Suvasis Dixit and Janet G. Hering, Environ. Sci. Technol. 2003, 37 (18), 4182–4189 (b) Dimitri A. Sverjensky and Keisuke Fukushi, Environ. Sci. Technol. 2006, 40 (1), 263–271.


Schrödinger's equation and spin quantum number

Good question.

This was one indication that there had to be something more than just the Schrodinger equation. Another problem which was perceived quite quickly is that the Schrodinger equation treated time differently than the spatial dimensions which is not in the spirit of special relativity. Thus, Paul Dirac constructed what we now call the Dirac equation. When you solve the Dirac equation in a Coulomb potential, spin, and hence the fine structure of the hydrogen atom energy levels in an electric field, falls out naturally.

This proves to be fairly technical and I won't try to do any of the mathematics for this (I don't completely understand it all), but you can find solutions to the Dirac equation for some common cases here.

As orthocresol points out, we first observed the intrinsic angular momentum of the electron in the Stern-Gerlach experiment, so people guessed what this would look like at first (this is how it was used in the Schrodinger equation) and Dirac came along and explained that people had guessed correctly. This can also be seen from the fact that the Dirac equation reduces to the Schrodinger equation in the non-relativistic, low energy case.


Why do acids ionize in solution if they are covalently bonded?

Because it is energetically favourable ($\Delta{G}<0$) for hydrogen chloride to react with water to form hydronium( $\ce{ H3O+}$) and chloride ions.

Remember that $\ce{ H+}$ does not exist as $\ce{ H+}$ in water, but rather as $\ce{ H3O+}$.


What is the actual reason behind PCl₅ existing as an ionic solid?

In the gas phase ions cannot be stabilized through solvation or lattice forces so ions are destabilized relative to non-ionic entities in the gas phase.  $\ce{NaCl}$ is a case in point, in the gas phase it exists as an equilibrium of monomers and dimers.  Yet in the solid phase it exists as a highly ionic crystal.  This example demonstrates that going from a gas to a solid can dramatically affect the relative stabilities of ionic and covalent structures.

The same is true in the case of $\ce{PCl_5}$.  In the gas phase it exists as a covalent compound with a trigonal bipyramid structure.  However, in the solid phase, it turns out that enthalpic affects favor the ionic tetrahedral/octahedral structure $\ce{[PCl4]^{+}[PCl6]^{-}}$.  Simply said, the ionic structure turns out to be thermodynamically more stable in the solid phase where positive and negative charges can be arranged to stabilize the entire lattice; whereas in the gas phase the positive and negative ions cannot be stabilized so the covalent form is favored since it is now lower in energy than the ionic form.

Another reason why the ionic form is preferred in the solid phase is that packing regular geometric shapes such as tetrahedra and octahedra together allows for a more stable arrangement since they can be packed closer together and in a more ordered arrangement than non-regular (kind of an odd shape) trigonal bipyramids can be arranged.

Edit: Response to OP's comment on ion stabilization

If we generate an ion in the gas phase, it will have a certain energy associated with it. If we put that ion in a solution its energy will be much less.  We say the ion has been stabilized by placing it in solution.  Some solvents will stabilize the ion more than other solvents.  This relates to the dielectric constant of the solvent which is a measure of the solvent's polarity.  The higher the dielectric constant for a solvent, the more polar the solvent and the better its ability to stabilize and lower the energy of an ion dissolved in it.

Here is a picture showing a sodium ion dissolved in water.  Note how the negative ends of the dipoles in the water (the negatively polarized oxygen end of the water molecule) molecules congregate and orient themselves around the positive ion.  This orientation of the negative ends of multiple water molecules around a positive sodium ion is what stabilizes the ion in solution.  There is no such stabilization in the gas phase.  In the gas phase it is just a "naked" ion without any external stabilization, so in the gas phase an ion will have a higher energy.

image source see the "Solvation od Ions" section

Here is another picture showing showing (on the right) sodium chloride in a crystal lattice.  The sodium and chloride ions pack around each other to perform the same function as a solvent.  The opposite charges arrange themselves around each other to minimize electrostatic repulsion between like ions and maximize the stabilizing electrostatic attraction between oppositely charged ions.  Through these types of arrangements, the "lattice" can stabilize ions much like a solvent can stabilize ions.  Consequently, both solvents and crystal lattices stabilize ions and lower their energy compared to the ion's energy in the gas phase where no such stabilization exists.

image source


Why is the dissolution of AgCl considered endothermic?

The typical way of explaining this is using Le Chatelier's Principle. That is, we could consider the reaction,$$\ce{AgCl(s) + heat<=>Ag+(aq) + Cl^-(aq)}$$This then demonstrates that energy in the form of heat is required to have the reaction proceed. Thus, at a higher temerature, the reaction will shift to the right as predicted by Le Chatelier's principle.

There are obviously more thorough explanations possible, but this is a good place to start.


Understanding smoke chemistry of tobacco and/or marijuana

Question 1 is easy to answer: Yes.

The ‘burning’ of whichever plant you’re (or somebody else is) smoking is an incomplete combustion on the ember end of the cigarette/joint/whatever. In any incomplete combustion of nitrogen-containing organic compounds, $\ce{HCN}$ and ammonia can (and likely will) be a side product. Aromatic amines are probably also always generateable as are $\ce{NO_x}$. And PAHs, aldehydes etc. can also result from the incomplete combustion of practically all plant or animal material.

Question 2 is slightly harder. To the best of my knowledge, however, the answer is: We do not know.

While we may be able to say that all the molecules in your question would be independently reacting, the mechanisms are so many and so complex that I very much doubt they have been the target of investigations. I think I recall somebody once studying the reaction mechanism of the complete combustion of methane and coming up with tens if not hundreds of partial reactions and steps. That was complete combustion (i.e. enough oxygen present), so it all boiled down to:

$$\ce{CH4 + 2O2 -> CO2 +  2H2O}$$

In incomplete combustion, we cannot even be fully sure of the reaction equation (and the stoichiometries involved therein), let alone a mechanism.

Side note: the complete combustion of a cigarette in pure oxygen takes about seven seconds ($\pu{7s}$) and oxidises it entirely to $\ce{CO2}$ and $\ce{H2O}$ (plus something from the nitrogens and likely $\ce{SO3}$ from sulphur/$\ce{P4O10}$ from phosphorus)


The first thing to clear up is – Why do you need to store 1000 L of KOH in liquid form? It is as others have pointed out, very concerning that someone who is planning to store this quantity needs to ask on the internet.

My advice is don't store 1000 L of liquid KOH, store 40 × 25 L of liquid KOH. This will reduce the risk considerably, and help ensure that a single leak isn't a catastrophic event, but is something you can deal with. This is a common philosophy, and is used, for example, in trailers of dangerous gases, which are organized into racks of 20 smaller gas bottles instead of one large tank.

Keep your 25 L containers off the floor in a bunded area for spill containment, and have appropriate means to deal with a spill. I'm not going to tell you the best way to handle a spill because I don't know. – You need to find out. Also, make sure the store is properly monitored to detect any problems.

Now, going back to my initial point: Let's accept that, for whatever reason, you need to store 1000 L of KOH (as you haven't shared with us what you are doing.) Solid KOH is much safer than liquid KOH, so why do you even need to mix up the solution all at once?

One of the causes of the Bhopal disaster (the worst industrial disaster ever, which killed many thousands of people due an uncontrolled venting event) was the unnecessary storage of a toxic and volatile compound (methyl isocyanate.) If the pesticide they were making had been synthesized via a different route, or even if the methyl isocyanate had been made only in the quantities needed, the size of the disaster could have been reduced or even eliminated (despite the extremely poor state of maintenance of the plant.)

I suggest you mix up the KOH in small batches as you need it to reduce the risk

  the temperature in the process might touch 500 degrees Celsius

By "process" I assume you mean mixing solid KOH with water, and that you do not actually need to use the KOH at this temperature. If you expect to reach these temperatures, you are doing it wrong. As can be seen from https://koh.olinchloralkali.com/TechnicalInformation/Phase%20Diagram%200-80.pdf KOH solution in all compositions from 0 to 85% is liquid at below 150 °C. Note also that concentrations above 50% are solid at room temperature – so do you really mean 85% conc KOH?

The safe way to do this is to add the dangerous material to the safe material, particularly as water has such a high heat capacity and can lose heat by boiling. You are going to need either a lot of patience, or mechanical screw feeder, to perform the addition at a safe rate. You will also need decent stirring equipment. Also, I would expect most of the heat will be released in mixing the first 50% of the KOH, because after that you are effectively mixing KOH with KOH.

In any case, if you are planning to do this, you should check the thermochemistry yourself and include it in your Risk Assessment, along with any relevant details from industry experience regarding specific hazards of mixing (spattering for example?)

I really hope you make sure what you doing before you do this. Alkalis are a lot worse on human flesh than acids, partly because they turn the oils in your skin to soap, and partly because the spills won't evaporate and therefore persist until something is done about them.

It would be a good idea to dye your KOH so you know where any spills are. I personally have spilt 50L of antifreeze solution in the lab (it was intentionally dyed beforehand in case of spills, and needless to say we used propylene glycol as it is much less toxic than the ethylene glycol used in car radiators.) Whatever you intend, something in your experiment will go wrong.

Please also check the legal aspects in your area. Also, consult with your supplier! The fact that you need to ask on the internet instead of talking to them suggests you are up to something unusual. Would you care to share with us what it is?

EDIT - Now you've clarified you're an engineer, one more very important point. Keep KOH away from aluminium. Alkali will dissolve the oxide layer that protects aluminium from corrosion, and unmask the true (very high) reactivity of the metal. A lot of heat and hydrogen gas will be generated by the reaction $\ce{3 KOH + Al -> K3AlO3 + 2/3 H2}$.

I'm guessing most stainless steels should be fine (but obviously check.) I have used equipment from the catering industry in some experiments, but stay away from stainless steel cookware that has the base thickened with aluminium. Even though it is external, it's not worth the risk.

Also, note that mixtures of acids and bases with water are often most corrosive at a medium concentration, so don't assume that just because a material is OK at 90% concentration, it's going to be OK at 50% concentration.


Equivalence point vs inflection point in the titration curve of a weak acid

Let's suppose we want to titrate a solution containing an unknown monoprotic and weak acid. We use a strong base, such as $\ce{NaOH}$. When the number (and moles) of hydroxide ions is equal to the amount of hydronium ions, here we have the equivalence point.

The equivalence point is, when the molar amount of the spent hydroxide is equal the molar amount equivalent to the originally present weak acid. At this point,  $[\ce{H3O+}]<[\ce{OH-}]$, so $\mathrm{pH} \gt 7$.

  When titrating a strong monoprotic acid the equivalence points coincides with the inflection of the titration curve, and pH is seven.

For the approximative experimental chemical approach, it is true.

But mathematically, even this is not exactly true, as the curve is slightly asymmetric. When there is the same molar amount of hydroxide as was the acid, the solution volume is 3 times bigger. And hydroxide concentration 3 times lower.

It would be symmetric, if we constructed the graph for the constant total volume and complementing strong acid and hydroxide solution volumes, as shown in the formula below.

$$\mathrm{pH}=f\left(\frac{V_\mathrm{acid}}{V_\mathrm{acid} +  V_\mathrm{hydroxide}}\right)$$

And the concentrations of both should be the same.

  In our case, however, the aqueous solution results to be slightly basic at the equivalence point. Therefore, does the inflection point on our curve still coincide with the equivalence point or it is to be found slightly before?

For weak acids, the coincidence is mathematically even less true, as the curve shape before and after equivalence differs.

But for the chemical experimental point of view, the difference is still negligible, as the equivalence/inflection coincidence error is much less then experimental errors.

Both inflection and equivalence points ( approximately identical ) lay in the alkalic region.

  In other words, what does the inflection point indicate, that $\mathrm{pH}$ is seven (so the inflection is distinct but may coincide with the equivalence point, and shows where the rate of increase in $\mathrm{pH}$ is maximum) or the equivalence point itself?

For weak acid/strong base ( and vice versa ) titrations, the neutral $\mathrm{pH}$ does not play any special role, so this point on the curve is insignificant.

Neither is there at the neutral point any inflection point, unless by chance $\mathrm{p}K_\mathrm{a}=7$. In such a case, the inflection point  at the maximal acid/salt buffering capacity at the half equivalence would be at $\mathrm{p}K_\mathrm{a}=7$.

For further reference, see e.g  Titration_of_a_Weak_Acid_with_a_Strong_Base


Difficulties understanding how salt bridge works

Regarding Difficulty 1)  When the zinc bar (unconnected) is first immersed into the solution, some zinc ions immediately dissolve away.  Now the zinc bar has extra electrons and the solution has extra positive zinc cations.  This is entropically acceptable, however, because cations in the solution (both zinc cations and hydronium cations) can align close to the surface of the zinc metal (creating what's called the 'Helmholtz layer').

One might imagine that as soon as the zinc bar was connected, the extra electrons in the zinc would drain to the other side of the cell.  But they don't because the electrostatic pull of the zinc and hydronium cations in the left-beaker solution is holding them back.

There's an analogous situation in the right cell:  the copper metal can't accept any electrons because of the electrostatic repulsion of the sulfate and hydroxide ions in right-beaker solution.

Regarding Difficulty 2) When the cell is running, there is an electrostatic gradient across the membrane where the salt bridge meets the right-side solution (among other places, for more info see note 1).  Cations (such as sodium) are driven out of the salt bridge;  anions (such as sulfate) are drawn into the salt bridge.  Sodium cations are smaller and diffusion more quickly than the larger, slower sulfate anions. Overall, many more sodium ions will leave the salt bridge compared to the number of sulfate anions that will enter the salt bridge.  There is an analogous situation at the membrane where the salt bridge meets the left-beaker solution.

Note 1, Electrostatic Gradients) When the cell is running, the zinc bar has a slightly negative charge and the copper has a slightly positive charge.  If we let ourselves ignore the metal wire for the moment, this creates an electrostatic field throughout  the cell very much like the classic "electrostatic field around adjacent positive and negative charges in a uniform media".  Notice that in the diagram of that situation, there are electrostatic field lines pointing from the positive charge straight-across to the negative charge.  Another way to describe this situation would be to say that there is an electrostatic gradient between the two charges.  If you drop a much smaller, and otherwise free-floating positive charge any where in the neighborhood around those two charges, it will move down electrostatic gradient to the negative charge:  the same way rain-water flows downhill, (in that case due to a gravitational gradient).  Now imagine there was an imaginary plane exactly halfway between the two charges:  you could now also say that there is an electrostatic gradient across that plane, from the one side to the other. If you dropped a positive charge right beside the plane, but on the same side as the bigger positive charge, the electrostatic gradient would drive the positive charge across the plane.  Similarly, if you dropped a negative charge next to the plane,  on the same side as the bigger negative charge, then again the electrostatic gradient would drive the negative charge across the plane.

In the electrochemical cell problem, the electrostatic field lines are going to be more complicated because there are two materials (air and water / salt-bridge) with different permittivity.  Long story short: the electrostatic fields will actually be much stronger in the water than in the air.  The strongest part of the electrostatic field will go from the copper electrode, across the right-side solution, through the salt bridge, and across the left-side solution to the zinc.   So, among other places, when the cell is running, there will be an electrostatic gradient across the membrane where the salt bridge meets the right-side solution.


It is an apparatus for taking a sample of a liquid from a big container. Let's suppose you have at your disposal a big bottle (at least one liter) which has two holes on a level with one another at middle height. Now imagine that a flow of a small river is entering the bottle from the upper neck, at such a rate that the level of the water inside is always on a level with the holes. The water gets out through the lateral holes. Now you put a punched rubber stopper in one of these holes. And the long thin pipe of your glassware is horizontally inserted into the stopper hole. Your glassware can then be turned and placed in such a way that the broader tube is vertical with its neck above. With this equipment, the broad tube is all the time half-filled with water. If the composition of the water slowly changes, a sample of its composition may be taken by turning the broad tube so as to empty it upside down, then putting it vertically again.


How does 1,3-dichloro-1,3,5-triazinane-2,4,6-trione (dichloroisocyanuric acid) generate hypochlorous acid when dissolved in water?

$\ce{N-Cl}$ bond is polarized toward nitrogen: $\ce{N^--Cl+}$ This means, that any chloramine can produce $\ce{HOCl}$ in water by mean of hydrolysis of $\ce{N-Cl}$ bond.

$$\ce{R2NCl + H2O -> R2NH + HOCl}$$

The mechanism is more subtle matter. Basically, it should be one of $S_N$ mechanisms, but it is hard to choose right one here ($S_N1$ or $S_N2$). The central atom in mechanism here should be $\ce{Cl}$, and $\ce{R2N-}$ is leaving group.


A problem regarding the first law of thermodynamics

The question didn't specify which process the given molar heat capacity was specified for. So we will assume that it is molar heat capacity for the first process. The first process is isobaric so $$q=nC_p\,\Delta T$$
You calculated $\Delta T$ correctly. The only mistake you did was assume that the gas was monoatomic. The expression for $\Delta U$ you wrote was $\Delta U=\frac32nR\,\Delta T$. This expression would be correct if this was a monoatomic gas. But we don't know that do we?
The correct expression to use here is $$\Delta U=nC_V\,\Delta T$$
It is given that molar heat capacity for first process (constant pressure) is $C_p=25\ \mathrm{J/(^\circ C\ mol)}$ and  we know that $$C_p=C_V+R$$
So you can calculate the value of $C_V$ and use it to get the value of $\Delta U$ which I calculated to be $$\Delta U=667.2\ \mathrm J$$
Plug this into first law of thermodynamics and you will get the correct answer.
Edit: I used the sign $\Delta U$ for Change in Internal Energy intead of $\Delta E$


Is combustion considered a redox reaction?

In general, yes.  If the reaction involves oxygen going from the oxidation state of zero in $\ce{O_2}$ to an oxidation state of -2, then there is oxidation and the reaction is a redox reaction.


CO poisoning - What I've been taught is a hoax?

Each hemoglobin molecule has 4 subunits (4 hemes each having one iron atom). Each time a molecule (such as oxygen or CO) binds to one subunit, it changes the binding properties of the other subunits. This is referred to as positive "cooperative binding" in the case of hemoglobin, because binding one molecule makes the other sites bind more easily.

Initially, considering hemoglobin with no CO or oxygen bound, CO competes with oxygen for the same binding sites (ligating one of the iron atoms).  CO has a binding constant 300 times lower than oxygen, meaning a factor of 300 lower CO concentration will result in an equal percent of bound protein.

Considering CO has reached a level such that 2 out of the 4 sites can have CO bound, the remaining 2 sites would still provide enough oxygen carrying capacity for a person to live if the sites were function as normal; however, the 2 remaining sites then bind oxygen too tightly and don't release it.

Other iron-containing proteins such as myoglobin can also bind CO and are also inhibited.

See Pharmacokinetics and Mechanisms of Action of Carbon Monoxide for more infomation.


Is this reaction with H+ and OH− a possible reaction?

You can try this in the lab: Take some $\ce{CaO}$, add $\ce{HCl}$ and see what happens. Actually, no don’t. Don’t try this unless you know exactly what you are doing.

A single proton does not exist, especially not in aquaeous solution. Using $\ce{H3O+}$ as a proton-donating species is a much better way to formulate reactions. A professor once said, the smallest actually proven proton-equivalent be $\ce{H9O4+}$, so four water molecules that shift the additional proton between them, but since it was first-year, he didn’t go into details. Also in organic solutions or others, protons will always be bound to something, however weakly.

Now that I got that out of the way, let’s take a look at the $\ce{H3O+}$ particle I just postulated. It has three hydrogens connected to an oxygen and in a very, very, very broad way of arguing, every hydrogen can be considered a potentially acidic proton. Therefore, we can think of hydronium as a triprotic acid with the following conjugate bases:

$$\ce{H3O+ ->C[- H+] H2O ->C[- H+] HO- ->C[- H+] O^2-}$$

Aha! Oxide ions can be considered as fully deprotonated water. Of course, the entire reaction series is reversible, so add ‘protons’ to oxide ions (whatever the source) and find that they get protonated to hydroxide and finally to water. You knew of the second step (hydroxide to water). And here’s the big surprise: The first one is identical, only the equilibrium is much, much further on the hydroxide’s side.

Adding excess acid to oxides would just create water. However, the acid-base neutralisation reaction is exothermic. Very exothermic. Which is precisely why I wrote do not try this unless you know exactly what you are doing at the top. If you have ever neutralised semi-concentrated hydrochloric acid and sodium hydroxide, you know how much heat can be created. Think more than twice that.

In fact, water has what is termed a nivelling effect: Water itself is too strong an acid to allow oxide bases to exist in its vicinity; so $\ce{O^2-}$ cannot be dissolved as is, it will immediately get protonated to hydroxide. Find some $\ce{CaO}$, dissolve it in water, and you would immediately get $\ce{Ca(OH)2}$. Oxides are only able to exist in aquaeous solutions if they are either complexed entirely by cations ($\ce{Al}$ can form oxido-hydroxido multi-core complexes at $\ce{pH 7}$ which include internal oxide ions) or if they are in a bound state, so not technically oxides, such as in the vanadyl cation $\ce{(VO)^2+}$.

Finally, you stated something in your question saying that a strong acid would react with something to a strong base which you think is impossible. That is not the case; if the product of the reaction of something with a strong acid is a strong base, then ‘something’ just needs to be an even stronger base. Which is precisely the case of oxide ions. Just to recall for a second that strong acids and even stronger bases react very, very exothermicly. I feel I have not stressed it enough: Do not try this unless you know exactly what you are doing!


What does A485 mean? And what does the values on the y-axis expresses?

$A_{485}$ means the same thing as the text of the multiple choice question "absorbance of the solution at 485 nm".

"nm" is an abbreviation for nanometers.

In other words, the absorbance of light having a wavelength of 485 nanometers is plotted on the y-axis of the graphs.


Is it the Na+ or the Cl- that gives table salt its taste?

The salt receptors are sodium ionic channels, i.e. peptides which are selectively permeable for Na ions [1].
Ionic current causes receptor activation and neural responce.

Anion plays minor role here, imposing upon salty taste. The result may appear acidic, bitter and even sweet.
For example, I believe that sulphates are overally bitter.
Carbonates or alkali probably just burn the receptors and may have random taste.


How to select the parent chain in mechlorethamine?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), secondary and tertiary amines can be systematically named in various ways similar to the methods for primary amines:

  P-62.2.2 Secondary and tertiary amines

  P-62.2.2.1 Symmetrical and unsymmetrical secondary and tertiary amines are named only by the same methods described in P-62.2.1.2.

  (1) substitutively using the retained name ‘aniline’ or the suffix ‘amine’ and the name of a parent hydride with further N-substitution;

  (2) substitutively, by prefixing, in alphabetical order when required, the name(s) of the substituent group(s) R, R' or R" to the parent hydride name ‘azane’. In order to avoid ambiguity, the second prefix in a secondary amine, and the second and the third prefixes in a tertiary amines must be enclosed in parentheses when these prefixes denote simple substituents.

  (3) substitutively, by prefixing, in alphabetical order when required, the name(s) of the substituent group(s) R, R' or R" to the parent hydride name ‘amine’. In order to avoid ambiguity, the second prefix in a secondary amine, and the second and the third prefixes in a tertiary amines must be enclosed in parentheses when these prefixes denote simple substituents.

  Method (1) generates preferred IUPAC names. Functional parent names like diethylamine and triethylamine are deprecated. The prefixes in names of such secondary and tertiary amines formed by method (2) are set off by parentheses to distinguish them from these deprecated names.

Therefore, the preferred IUPAC name for the tertiary amine given in the question is generated according to method (1) using the suffix ‘amine’, the name of a parent hydride, and further N-substitution. (Other methods may be used in general nomenclature.)

Concerning the choice of the principal chain, the rules for amines state:

  P-62.2.2.2 Selection of the principal chain or senior ring system in secondary and tertiary amines.

  Names of amines formed substitutively by using the retained name aniline or the suffix ‘amine’ are based on a principal chain and a senior ring system (see P-44.1). When a choice for parent hydride is possible between a ring and a chain, the ring is preferred. In names using ‘amine’ as a parent hydride, substituent groups expressed as prefixes are cited in alphanumerical order; the prefix(es) immediately preceding the term ‘amine’ is (are) enclosed in parentheses.

Thus, the rules essentially refer to the usual seniority order for parent structures that is stipulated in P-44.1 and the following subsections. The relevant criteria for the compound given in the question are

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (…) and the seniority of suffixes (…).

and

  P-44.3.2 The principal chain has the greater number of skeletal atoms (…)

In case of the compound given in the question, both methane (methylamine) and ethane (ethylamine) are parent structures that have the maximum number of substitutents corrsponding to the principal characteristic group (i.e. the suffix ‘amine’) since the compound contains only one amine group. The principal chain for the compound given in the question is ethane rather than methane since it has the greater number of skeletal atoms (i.e. the longest chain).

In accordance with the above-mentioned method (1), using the name of the parent hydride ‘ethane’ and the suffix ‘amine’ with the appropriate locant yields ‘ethane-1-amine’. The chloro substitutent is placed at atom ‘2’ of the parent hydride ‘ethane’, which leads to
‘2-chloroethan-1-amine’. Further N-substitution with the substituents ‘methyl’ and ‘2-chloroethyl’ results in the preferred IUPAC name
2-chloro-N-(2-chloroethyl)-N-methylethan-1-amine.

Note that the locant ‘1’ is omitted in ‘ethanamine’; however all locants must be cited in ‘2-chloroethan-1-amine’ and ‘2-chloro-N-(2-chloroethyl)-N-methylethan-1-amine’.

The alternative name bis(2-chloroethyl)(methyl)amine can be generated according to method (3). This is not the preferred IUPAC name, but it is a correct systematic name that may be used in general nomenclature.

Note that, according to method (3), the parent hydride in this case is ‘amine’ (i.e. neither ethane nor methane). The names of the substituent group(s) R, R' or R" are added as prefixes in alphabetical order, for example: ‘ethyl(methyl)(propyl)amine’. The multiplicative prefixes ‘di’ or ‘tri’ are added as usual, for example: ‘diethyl(methyl)amine’. However, the multiplicative prefixes ‘bis’ and ‘tris’ are used for substituted substituents (such as ‘2-chloroethyl’) and other compound or complex entities, which results in the name bis(2-chloroethyl)(methyl)amine for the compound given in the question.

Note the correct use of parentheses according to Subsections P-62.2.2.1 (3) and P-62.2.2.2.

The name N,N-di(2-chloroethyl)methanamine, which has been proposed in the question, is simply wrong. It violates multiple nomenclature rules.


How does the equipartition theorem apply to vibrational modes?

As you probably already know, the equipartition theorem states that each quadratic
degree of freedom will, on average, possess an energy $\frac{1}{2}kT$, where a ‘quadratic degree of freedom’ is one for which the energy depends on the square of some property.
Now lets consider the kinetic and potential energies associated with translational, rotational and vibrational energy:

Translational degrees of freedom:  $~E = \frac{1}{2}mv^2$

Rotational degrees of freedom:  $~E = \frac{1}{2}Iω^2$

Vibrational degrees of freedom: $~E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$

As you can see that for translational and rotational degrees of freedom, there is only one quadratic. Hence each translational and rotational state will contain $\frac{1}{2}kT$ energy.
However for vibrational states, the energy contains two quadratic contributions. Therefore each vibrational state will contain $kT$ energy.
Now lets consider the reasoning behind this. You are right when you say that the vibrational degree of freedom depends on kinetic and potential energy and hence has 2 quadratic contributions. In the equation for the energy of the vibrational state, $\frac{1}{2}mv^2$ represents the kinetic energy and $\frac{1}{2}kx^2$ represents the potential energy.
A molecule in the vibrational state acts like a harmonic oscillator. So let consider $H_2$. You can imagine that the bond between the 2 hydrogen atoms is like a spring which contracts and expands in length.

The average bond length can be thought as the equilibrium length of the bond. When the bond the contracts, the displacement from this equilibrium length is known as 'x'. Now using Hooke's Law, the potential energy of the system is $\frac{1}{2}kx^2$ where k is the spring constant while the kinetic energy is $\frac{1}{2}mv^2$. Hence the total energy of the system is given by: $~E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$


Why does F replace the axial bond in PCl5?

Recently, there has been a lot of discussion of Bent's rule (see for example "What is Bent's rule?") here in SE Chem. Simply stated, the rule suggests that $\mathrm{p}$-character tends to concentrate in orbitals directed at electronegative elements.

Why does $\ce{F}$ replace an axial bond in $\ce{PCl5}$?

In order to answer this question, we need to start by understanding the bonding in $\ce{PCl5}$ and its fluorinated isomers. In introductory courses and texts, it is usually stated that $\ce{AX5}$ type molecules adopt a trigonal bipyramid geometry and are $\mathrm{sp^{3}d}$ hybridized. As the comments by Martin and permeakra point out, and as is learned in more advanced classes, this hybridization scheme is likely incorrect. There are several reasons that argue against $\mathrm{sp^{3}d}$ hybridization including: 1) the fact that $\mathrm{d}$ orbitals are relatively high in energy compared to $\mathrm{s}$ and $\mathrm{p}$ orbitals and therefore it is energetically quite costly to involve them in bonding; and 2) $\mathrm{d}$ orbitals in non-metals are very diffuse leading to poor overlap with other orbitals and any resulting bonds would be very weak.
A reasonable hybridization alternative involves what is termed hypercoordinated bonding; where 3 center, 4 electron bonds are involved. Applying this concept to $\ce{PCl5}$ we would say that the central phosphorus atom is $\mathrm{sp^2}$ hybridized.  Thus, there would be 3 $\mathrm{sp^2}$ orbitals (these will be used to create the equatorial bonds) and a $\mathrm{p}$ orbital (this will be used to create our axial bonds) emanating from the central phosphorus atom.

The $\mathrm{p}$ orbital contains 2 electrons and will form bonds to 2 ligands (chlorine or fluorine in the case at hand). For simplicity, let's say that these ligands also use $\mathrm{p}$ orbitals for bonding (but it could be any type of orbital, $\mathrm{sp^3}$ or whatever is appropriate) and each of these orbitals contains one electron for bonding. This is our 3-center-4-electron bond and its MO diagram is pictured below. Notice how the four electrons are distributed - there are two electrons in the HOMO which is a non-bonding M.O., so the bond order in this bond is reduced.

This reduced bond order in the bond using the phosphorus $\mathrm{p}$ orbital explains why the axial bonds in $\ce{AX5}$ type molecules are longer than the equatorial bonds.

Now that we understand the bonding in $\ce{PCl5}$ we can consider the case of $\ce{PCl4F}$ and how Bent's rule applies to the situation.  First, note that fluorine is more electronegative than chlorine.  As stated above, Bent's rule suggests that more electronegative ligands prefer to form bonds with orbitals that are high in $\mathrm{p}$-character. Why is this? $\mathrm{s}$-Orbitals are lower in energy than $\mathrm{p}$-orbitals. Therefore electrons are more stable (lower energy) when they are in orbitals with more $\mathrm{s}$-character. The two electrons in the $\ce{P-F}$ bond will spend more time around the electronegative fluorine and less time around phosphorus. If that's the case (and it is), why "waste" precious, low-energy, $\mathrm{s}$-orbital character in an orbital that doesn't have much electron density to stabilize. Instead, save that $\mathrm{p}$-character for use in phosphorus hybrid orbitals that do have more electron density around phosphorus (like the $\ce{P-Cl}$ bonds). So, as a consequence of Bent's rule, we would expect phosphorus to use the orbital with lowest $\mathrm{s}$-character, the axial $\mathrm{p}$-orbital, to form the $\ce{P-F}$ bond; and the orbitals with more $\mathrm{s}$-character, the equatorial $\mathrm{sp^2}$ orbitals, to form $\ce{P-Cl}$ bonds.


The difference between Thermodynamic and Kinetic products is the difference in activation energy and the difference in (Gibbs)-Free-Energy.
When the activation energy is low, the reaction proceeds fast, while the product might not have as low a Free-Energy as an other product (with higher activation energy). When the reaction conditions allow however, the thermodynamically more stable product (with lower Free-Energy) will be formed. Even though the kinetic product first forms, most complexes are form in equilibrium reactions. Therefore, the kinetic product reverts to starting materials and then, slowly forms the thermodynamic product. This thermo dynamic product, though it is also in equilibrium, will not revert to starting materials as easily as the kinetic product will, due to the lower Free-Energy.

To conclude: First, the kinetically favored product will form. Then, when the reaction conditions allow it, the equilibrium shifts towards the thermodynamic product.
Note that the equilibrium is essential for the thermodynamically favored product to form.


Time required for a thin layer of ethanol to evaporate?

In others words, it is to know approximately how much time it would take for a small amount (about 5 ml) of pure alcohol to evaporate if poured over a surface such as a table. Would it take about 15 seconds/30 seconds/2 minutes?

It be simpler and faster to do the experiment than to try to predict the time needed.

The kinetic molecular theory explains why liquids evaporate at temperatures lower than their boiling points. At any temperature, the molecules in a liquid have a range of kinetic energies described by a Boltzmann distribution. Some percentage of the molecules have enough kinetic energy to escape into the gaseous phase. These molecules contribute to the vapor pressure of the liquid. As the temperature increases, more molecules are above the gas threshold and the vapor pressure increases. As the vapor pressure increases.

In a closed system, equilibrium would be established with an unchanging ratio of liquid to vapor. Molecules would be exchanged between the liquid and the vapor, but their relative amounts would remain constant. The surface you describe is not closed - the vapor molecules can wander away by diffusion or convection. Equilibrium is disrupted, and Le Châtelier's Principle tells us that the equilibrium shifts to compensate. As more ethanol molecules escape, more vaporize to replace them until there are no molecules remaining in the liquid. In the following equation, $K$ is the equilibrium constant, $p$ is the partial pressure of ethanol vapor (the vapor pressure) and [$\ce{C2H6O}$] is the concentration of ethanol in the liquid.

$$\ce{C2H6O (l) <=> C2H6O (g)} $$
$$K=\frac{p_{\ce{C6H6O}}}{[\ce{C2H6O}]}$$

After establishment of equilibrium (fast), the rate-determining step of the evaporation is probably the diffusion of the gaseous ethanol molecules away. The average kinetic energy of a gas particle can be expressed as a function of mass ($m$, in kg) and root mean square velocity ($v^2_\text{rms}$) and separately as a function of temperature ($T$, in kelvin) times the Boltzmann constant ($k_\text{B} = 1.38 \times 10^{-23} \frac{\text{J}}{\text{K}}$). We can derive a formula for rms velocity.

$$\overline{E_\text{k}}=\frac{1}{2}mv^2_\text{rms}$$
$$\overline{E_\text{k}}=\frac{3}{2}k_\text{B} T$$
$$v^2_\text{rms}=\frac{3k_\text{B} T}{m}$$
$$v_\text{rms} = \sqrt{\frac{3k_\text{B} T}{m}}$$

You could calculate the velocity of ethanol particles escaping. If you set an arbitrary distance (one meter is probably fine), then you can calculate a time it takes one particle to travel that distance (on average). If we know the equilibrium constant, we can determine how much vapor is above the liquid and then calculate the amount of time it takes to move.

But how do we know the equilibrium constant? It varies with temperature! The value $\Delta G^\circ_\text{vap}$ in the equation below is the free energy change of vaporization of ethanol in the standard thermodynamic state. $R$ is the ideal gas constant.

$$K=\mathrm{e}^{-\frac{\Delta G^\circ_\text{vap}}{RT}}$$

The model above ignores complicating factors like mean free path, the fact that evaporation is endothermic (meaning the liquid cools as it evaporates and the vapor pressure decreases with time), the temperature and heat capacity of the surface govern how much kinetic energy is available to the liquid to begin with, and that any amount of air current in the vicinity will move the vapor away significantly faster than diffusion.

A complete model includes consideration of the following:

  Constants

the Boltzmann constant
The ideal gas constant
the free energy change of vaporization of ethanol (not actually constant, but it varies only slightly over the temperature range)
the vapor pressure of ethanol as a function of temperature
the mass of an ethanol molecule
the heat capacity of the surface

  variables

temperature of the air
volume of ethanol
temperature of the surface
atmospheric pressure (needed for mean free path corrections)
velocity of air currents

So, in principle, you could do it. Even so, the best answer cannot be had without some serious calculus. In practice, it would be faster to do the experiment (if often is). We often forget that science is empirical.


How to calculate the composition of a borate buffer with a defined pH using the Henderson-Hasselbalch equation?

Boric acid and sodium borate forms an acidic buffer (A mixture of weak acid and salt of weak acid and strong base). The acid is weak hence the concentration of basic ion from acid is much less compared to the concentration of basic ion obtained from the salt which can ionize completely. Yes. Some of the acid dissociates to form hydronium ion and a basic ion. That's how a buffer works.

Consider a weak acid HA and the salt BA.
Its easy to show that the concentration of $H^+$ due to ionisation of HA is $$[H^+]=k_a\frac{[HA]}{[A^-]}$$
It is reasonable to assume that the concentration of $A^-$ from BA is much larger than that from HA due to complete ionisation of BA. Hence $[A^-]=[BA]$=[salt]
So, $$[H^+]=k_a\frac{[acid]}{[salt]}$$
Taking logarithm, you get the desired equation.


The heat content of the products is more than that of the reactant in an ............. reaction

Don't blame yourself.  This is a badly-worded question. Products and reactants don't have "heat content", they have thermal energy. I.e.,  heat is not a property of substances—they don't 'contain heat.'  Instead, heat is a property of processes; specifically, it is the flow of thermal energy across a boundary. So, for instance, in an exothermic reaction in a system with diathermal walls (walls that allow free flow of heat across them), thermal energy flows out of the system, and thus the heat (q) for that reaction is negative.

Having said that, it seems what they mean by "heat content" is the thermal energy of the system after a reaction if the system is adiabatic (no heat flow in or out, i.e., perfectly insulating walls).  Thus, if you have an exothermic reaction in an insulated container, the products you get will be hotter than the reactants you started with.

A much better way of asking the question would be:  Suppose a reaction takes place at constant pressure in an adiabatic (insulated) container, and the temperature increases.  Is the reaction exothermic or endothermic?  Answer: exothermic.


Is it possible to figure out how many carbon-carbon double bonds are present using just the molecular formula?

Both you and Rapid99 have missed the word "acyclic" in the problem statement.  So you don't have to worry about rings (because they said acyclic).  [Also no triple bonds.]

You also know that it's a carboxylic acid ("COOH group", formula but not structure of carboxylic acid group), so don't worry about something funky with the oxygens.  The formula is then $\ce{C19H31-COOH}$.

If the part of the compound on left was full saturated and sans $\ce{COOH}$ it would be $\ce{C19H_{(2*19+2)}}=\ce{C19H40}$.  Remember any acyclic alkane has formula $\ce{C_nH_{2n+2}}$.

Accounting for the $\ce{COOH}$ group location means the saturated alkane would then be $\ce{C19H39-COOH}$ (take away one $\ce{H}$ "spot" to attach the COOH group).

So we are really 8 hydrogens short: 39-31 (from above) = 8.  Each double bond takes away two hydrogens.  So if there are 8 missing hydrogens, there must be 4 double bonds.  X=8/2.


Why are we still using cis/trans system while the E/Z system is better than it?

That is because the E/Z-system is an extension of the cis/trans-system. So both systems can be applied to a certain amount of structures and the E/Z-system is then able to treat more advanced systems.

From my perception, it seems that Chemistry is a very stubborn … or maybe just very conservative … field where especially new naming conventions get hardly accepted. At least not very fast. So we are still using Angstrom instead of nanometer or picometer and so on. This also means that we stick with the cis/trans-system as long as it fits and we only change in case we can’t apply it.


How does spin flipping of triplet carbenes occur?

In these reactions you are not observing optical transitions between singlet and triplets, but the interconversion of a triplet to singlet state in competition with rotation of part of the molecule. Only in the singlet can a bond be formed as now each electron has a unique set of quantum numbers. As no photons are involved it is necessary for the triplet and singlet levels to become effectively equal in energy before a transition can occur. Thus there has to some kind of perturbation for this to happen and this has to accommodate one unit of angular momentum change, the difference between singlets and triplets. It can be useful to visualise singlet & triplets with a vector model. This is shown below. The arrows show the spin vectors of the two electrons in a triplet and singlet state.

The triplet state corresponds to the three images on the left and the singlet to that on the right. If you look at the triplet with $m_s=0$ you will see that to convert it to a singlet all that is needed is a re-phasing of the spins from parallel to anti-parallel.  In the diradicals you study, the electrons are rather distant from one another thus their interaction is weak, and the difference in energy between singlet and triplet small. This means that only a small perturbation can cause a crossing between levels. This can, for example, be produced by an external magnetic field which raises and lowers different triplet energy sub-levels and so reach the singlet's energy. A nearby solvent molecule can produce this energy perturbation as it also has electrons and it will be diffusing by rotation and translational motion, i.e. producing a time and spatially varying magnetic field. Similarly the diradical is rotating with whole body motion as are segments of it and in doing so presents different orientation of the electrons to the earth's magnetic field which may also induce changes in energy levels and triplet-singlet crossing. These interactions will also have to involve orbital angular momentum.The time scale for rotation will be sub-nanosecond (in most normal mobile solvents) but only occasionally will a perturbation occur at the same time as the groups are proximal, because segmental rotational diffusion will mainly keep them apart, and the low frequency of perturbations of sufficient magnitude and the geometric requirement just mentioned are likely to make the reactions slow.


Why is $1s-1s$ orbital overlapping stronger than $2p_x-2p_x$ orbital overlapping?

Your question is hinged on a false premise. On a deeper level, the premise makes no sense at all and can't even be classified as true or false. If a book says otherwise, then maybe it contains some very specific piece of context that would turn the situation upside down and make everything fall in place. Or maybe the book just says nonsense; such things happen, after all.

To begin with, orbital overlap depends on the internuclear distance more than slightly. Now, $1s$ orbitals are a good deal smaller than $2p$ (not like "you vs Dwayne Johnson", but more like "you vs an elephant"). Naturally, at a distance where $2p_x$ orbitals are in their best overlap, the $1s$ orbitals barely feel each other. Likewise, if the distance is good for $1s$, it is way too short for $2p_x$ and makes them feel pretty uncomfortable. So you see that considering one of these overlaps to be universally stronger than the other is plain untrue.

There is more to it, however. See, the $1s$ and $2p$ orbitals of the same atom are pretty far apart in energy. They never mix. In other words, they can never be important at the same time.

If both your atoms are $\rm H$, then their $1s$ orbitals overlap and make a good strong covalent bond, while the $2p$ orbitals are empty and hence effectively don't exist.
On the other hand, if both your atoms are (say) $\rm O$, then their $2p_x$ participate in bonding (together with other $2p$'s and $2s$), while the $1s$ orbitals are pretty much kept apart. What's more important, both $1s$'s are fully occupied, hence whatever little interaction they have will end up producing a bonding and an antibonding orbital which are also both occupied, thus negating the effect.

At this point you might be wondering what if one of your atoms is $\rm H$ and another $\rm O$, but that's a story for another day.

So it goes.


How is laughing gas produced in the Nef reaction?

The Nef reaction does not produce nitrous oxide directly. It produces nitroxyl as a reactive intermediate, which rapidly decomposes via a bimolecular reaction to nitrous oxide and water$^{[1]}$. You are correct that the base removes the alpha hydrogen; the nitro group is more than capable of accepting this negative charge. The crux of the Nef reaction is this:

The mechanism for the formation of nitrous oxide is as follows:

$[1]$ Wikipedia: Nitroxyl, Reactions


$\ce{Na2CO3 + H2CO3 -> 2NaHCO3}$, yielding sodium "bicarbonate", as an example, and in similar fashion, $\ce{Na2SO4 + H2SO4 -> 2NaHSO4}$, or sodium "bisulfate". The "bi" in this nomenclature meant two different anions, and is confusing, as it is the opposite of "bi" in mercury bichloride, two of the same cations. For that reason, sodium hydrogen sulfate or sodium acid sulfate might be preferred.


Why are square planar coordination compounds with four different ligands optically inactive?

Even though the central atom has 4 different ligands this does not necessarily mean that the compound is optically active. The condition for that rather the following (from Wikipedia):

A molecule is achiral (not chiral) when an improper rotation, that is a combination of a rotation and a reflection in a plane, perpendicular to the axis of rotation, results in the same molecule[...] For tetrahedral molecules, the molecule is chiral if all four substituents are different.

So for 4-coordinated carbon atoms which are tetrahedral having four "ligands" actually establishes optical activity. But in the case of a square planar metal complex you always have a mirror plane, namely the one made up by the central atom and the ligands:

And as a result you don't get optical activity. But it might be possible that some vibrational modes destroy the mirror plane symmetry element of the metal complex and this might lead to some optical activity depending on how well this mode excited.


What about browsing across

the section Laboratory Experiments in the Journal of Chemical Education?  This month (march issue), for example with "Unboiling an Egg: An Introduction to Circular Dichroism and Protein Refolding" by Hoben et al., (DOI:10.1021/acs.jchemed.6b00319), or "Demonstrating the Effect of Surfactant on Water Retention of Waxy Leaf Surfaces" by Chiu et al., (DOI: 10.1021/acs.jchemed.6b00546)
more broad than the journal, visiting educational ressources the American Chemical Society offers
as an example from the other side of the pond, a collection provided by the Royal Society of Chemistry (you may refine your search by age of the audience)
albeit less chemistry is not the barycentre, bring science home in the Scientific American

and then, choosing while taking into account what resources are at your disposition.


Is the correlation energy of a system always negative?

The answer to this question rests on the fact that Hartree-Fock is a variational method.
A variational method is one where you supply a guess wavefunction ($|\Psi\rangle$) with some parameter(s) ($\alpha$) and then use those parameter(s) to minimize the energy. Doing this, you are guaranteed that your resulting energy is an upper bound to the exact ground state energy.
$$
E_{\text{exact}} \leq \frac{\langle\Psi_{\alpha}|\hat{H}|\Psi_{\alpha}\rangle}{\langle\Psi_{\alpha}|\Psi_{\alpha}\rangle}
$$

So, if we are talking about the exact correlation energy (such as one given by full CI) then the answer is that the correlation energy must always be negative. This is  because Hartree-Fock always provides an upper bound to the exact ground state energy. However, if we use a non-variational method to approximate the correlation energy (such as coupled cluster) then we could find cases where we calculate a positive correlation energy, but this is distinct from the exact correlation energy.

(For an authoritative description of this in the context of quantum chemistry with examples and practice problems, I recommend the first chapter of Szabo and Ostlund's Modern Quantum Chemistry)


Regarding the mechanism for biological activity of tyramine, amphetamine and ephedrine

Question: When these foreign amines have been taken up into the vesicles how do they proceed?...Do they re-enter the synaptic cleft by exocytosis...How are they eventually cleared from the synaptic area?
Here is the mode of action of these agents (indirectly acting sympathomimetics):

The most
important drugs in the indirectly acting sympathomimetic amine
category are tyramine, amphetamine and ephedrine, which are
structurally related to noradrenaline.
Amphetamine enters the nerve terminal via the noradrenaline
transporter (NET) and enters synaptic vesicles via the vesicular
monoamine transporter (VMAT), in exchange for noradrenaline
(NA), which accumulates in the cytosol.
Some of the NA is degraded
by monoamine oxidase (MAO) within the nerve terminal and some escapes,
in exchange for amphetamine via the noradrenaline transporter, to act
on postsynaptic receptors. Amphetamine also reduces NA reuptake via
the transporter, so enhancing the action of the released NA.

Exocytosis is not involved in the release process, so their actions do not require the presence of Ca2+. They are not completely specific in their actions, and act partly by a direct effect on adrenoceptors, partly by inhibiting NET (thereby enhancing the effect of the released noradrenaline) and partly by inhibiting MAO.
These agents are cleared by metabolism via oxidative deamination with monoamine oxidase(MAO) enzyme, however  are poor substrates for MAO and, thus, show a prolonged duration of action.
Amphetamine is an example of a drug metabolized through oxidative deamination:

Question: Why do the noradrenaline re-uptake proteins transport noradrenaline out of the axonal terminal under the influence of these foreign amines?
I am not really sure about the “why” part but instead the “how” part AFAICT, this is more of a displacement mechanism:
These drugs, do block reuptake 1(NET) releasing stored transmitter from noradrenergic nerve endings and they displace vesicular norepinephrine into the cytoplasm. The norepinephrine is then transported out of the nerve terminal by the reverse action of reuptake 1 and activates adrenoceptors.
So essentially the drugs increase the availability of noradrenaline to stimulate adrenergic receptors by:
i. releasing or displacing NE (noradrenaline) from sympathetic nerve endings,
ii. blocking the transport of NE into sympathetic neurons or
iii. blocking the metabolizing enzymes, MAO or COMT.
How this helps
References

Rang and Dale’s Pharmacology 8th Edition. Section 2: Chemical Mediators

Medical Pharmacology at a Glance M. JNeal: Drugs acting on the sympathetic system

Lippincott Illustrated Reviews: Pharmacology 6th Edition: Adrenergic Agonists

Goodman and Gilman Manual of Pharmacology and Therapeutics: Agents Acting at the Neuromuscular Junction and Autonomic Ganglia

Principles of Clinical Pharmacology 2nd Ed Atkinson etal
Basic and Clinical Pharmacology B. Katsung et al.


Corrosive/ Acid cabinet gets wet from time to time

The white powdery crystals are probably $\ce{NH4Cl}$, that is formed thanks to leakage form amonia. You can try to measure the pH of the wet layer, should be acidic because of the leakgae from other acids bottles that condensate on inner walls. Most experiment results should be ok, but if you are working in an analytical lab, the results can be shifted.


Why are phosphate-phosphate bonds weaker than phosphate-water bonds?

ATP is an unstable molecule in unbuffered water, in which it hydrolyses to ADP and phosphate. This is because the strength of the bonds between the phosphate groups in ATP is less than the strength of the hydrogen bonds (hydration bonds), between its products (ADP + phosphate), and water acid anhydrides are generally prone to hydrolysis into two acid molecules.

Ignore that description Wikipedia just gave you, it is just blatantly bad from every perspective.

Phosphorous(V) is generally a strong electrophile and will be readily attacked nucleophilicly by water molecules — think of an $\mathrm{S_N2}$ mechanism. On the other side, $\ce{ROPO3^2-}$ is a good leaving group, too.

While the exact details may or may not be like I presented here, all acid anhydrides can easily hydrolysed similarly, be it disulphonic acid ($\ce{H2S2O7}$), organic acid anhydrides ($\ce{R-CO-O-CO-R}$) etc. Technically, the $\ce{P-O-P}$ bond in ATP/ADP is no different from the bonds in $\ce{P4O10}$ which we all know to be very readily hydrolysed.


What gives milk its characteristic white color?

There is no "pigment" that makes milk white.

According to Wikipedia:

  A pigment is a material that changes the color of reflected or transmitted light as the result of wavelength-selective absorption.

Going by that definition of a "pigment", then there is no (white) pigment in milk. In fact (as @Ivan points out in the comments) there is no such thing as a "white pigment", because that would require the "pigment" to absorb no visible radiation (if it absorbed all visible radiation, it'd appear black)

That's because the real reason for the white color of milk is a physical phenomenon called Scattering. (Source: Optics, 4th edition, E. Hecht)

(Diagram's from Wikipedia)

Milk is an emulsion of fat dispersed as these really tiny droplets in water. These fat globules (and some milk proteins) scatter light of all wavelengths in the visible spectrum. This is why milk appears white

This is the same reason (albeit different cause) as to why clouds often appear white; except here, the scattering is due to suspended water droplets and dust particles.


Several answers (full disclosure - mine) are available that shed at least some light. So read them first.

https://physics.stackexchange.com/questions/196686/what-causes-a-eutectic-point

Plot Binary Eutectic from Temperature of Melting and Enthalpy

OK. Now to your confusion (I think). What does a straight line across a plot of free energy vs composition actually mean? Well, mathematically that straight line represents the weighted average of the pure free energies, $x_{A}G_{A} + x_{B}G_{B}$ (perhaps more obvious using $x_{B} = 1-x_{A}$). So, $G_{A}$ is the intercept at $x_{A}=1$ and $G_{B}$ is the intercept at $x_{B} = 1$. If the free energies are a fixed value, the chemical potentials are a fixed value. So, the tie-line that is a common tangent to two free energy curves is the solution where the free energies (chemical potentials) are equal in both phases.

As for why the one-phase region does not chemically separate, the question is why should it? There is no second composition that has a common tangent, so there is not way to make the chemical potentials the same in two phases while in a one-phase region.


You are lucky, because not only the pyrazole motif, but the complete structure drawn by you is part of the Tautobase database

with a preference for the second structure drawn.
By the way, the values of this database are accessible to the public on GitHub, and since January 2020 implemented as (equally [sub]structure) searchable data file within DataWarrior.  Thus, you may train your chemical eye on 1680 tautomers.
Tautobase: An Open Tautomer Database, Wahl, O.; Sanders, T. J. Chem. Inf. Model. 2020, 60,1085-1089; doi 10.1021/acs.jcim.0c00035


If change in free energy (G) is positive, how do those reactions still occur?

$\Delta G^\circ_m$ is the difference in molar Gibbs free energy between the reagents and products in their standard states (in the case of $\ce{AgI(s)}$, the standard state for the reagent is the pure solid and that for the product is the solute in a $\pu{1 molal}$ ideal solution, all at $p^\circ =\pu{ 1 bar}$).

If $\Delta G^\circ_m>0$ you cannot convert the reactants completely into products without doing some accompanying work (for instance, applying an electrical voltage, in the case of an electrochemical cell).

If you want to see where the reaction is headed, for instance, if you start with only reactants (say $\ce{AgI(s)}$), then you should compute the following:

$$\Delta G_m=\Delta G^\circ_m+RT\log(Q)$$

For a reaction $\ce{A<->B}$ this is illustrated in the following plot (generated assuming an equilibrium constant $K_{eq}=10$, and $\Delta G$ in arbitrary units):

Here $\zeta$ represents a progress coordinate ($\zeta=Q/(1+Q)= \chi _B$).

Note that some authors place a "tilde" following $\Delta G_m$ to emphasize that it is a partial molar quantity, that is, it measures the change in free energy with reaction progress.
At the start of the reaction, when there is no solute dissolved, Q=0 and $\Delta G_m=- \infty$. In other words, the driving force for the reaction to occur is theoretically unlimited. At equilibrium that force goes to zero. Past the equilibrium point, the reverse reaction becomes more favorable.

See "What is the difference between ∆G and ∆G°?" for an elaborate answer addressing a similar question.


In a mass spectrum, are there peaks corresponding to a combination of 13C and bromine isotopes?

It does appear. The graphs you're looking at online probably just dumbed it down.

Here's some proof if you like, a compound synthesised by yours truly, with molecular formula $\ce{C7H7BrN2O2}$. I clipped the relevant part of the mass spectrum I got. This was done with ammonia gas CI, so instead of the $\ce{M^{+.}}$ radical cation we expect to see the $\ce{[M + H]+}$ quasi-molecular ion. We expect the following monoisotopic masses (I used this website to calculate these):

$$\begin{array}{ccc} \hline
\text{Molecular ion} & m/z & \text{Predicted relative abundance} \\ \hline
\ce{[C7H7^{79}BrN2O2 + H]+} & 230.9769 & 100.00 \\
\ce{[C6^{13}CH7^{79}BrN2O2 + H]+} & 231.9798 & 8.48 \\
\ce{[C7H7^{81}BrN2O2 + H]+} & 232.9749 & 98.01 \\
\ce{[C6^{13}CH7^{81}BrN2O2 + H]+} & 233.9777 & 8.29 \\ \hline
\end{array}$$