The concept of oxidation state simply works on concept of electronegativity.The more electronegative atom acquire a negative charge while less electronegative atom acquire a positive charge.
Depending on this hypothesis oxygen have 5 oxidation states.

In all the oxides,oxygen has an oxidation state of $-2$. Eg. $\ce{CO2,CO}$
In all peroxides (oxygen-oxygen linkage), oxygen has an oxidation state of $-1$. For example, consider $\ce{H2O2}$, here $\ce{H}$ is less electronegative so it will acquire a charge of $+1$ and to balance the $2$ positive charge of 2 H-atoms,each oxygen atom will acquire a charge of $-1$.
In all superoxides ($\ce{KO2,CsO2,RbO2}$), oxygen has an oxidation state of $-\frac{1}{2}$,this is because $\ce{K,Cs,Rb}$, being elements of the first group and less electronegative than oxygen acquire a charge of $+1$, to balance it, each oxygen atom acquires a charge of  $-\frac{1}{2}$.
In one of the exceptions $\ce{OF2}$, the fluorine being more electronegative acquires a charge of $-1$ and to balance the $-2$ charge of 2 fluorine atoms oxygen acquires a charge of $+2$.
As last, there is $\ce{O2F2}$, similarly here to balance the $-2$ charge on 2 $\ce{F}$-atoms each oxygen atom acquire a charge of $+1$.


The potentials depend on the form of the compounds.
It is true that in solution

$\ce{Pb^{2+}  (aq) + 2e^- -> Pb (s)} $

is -0.126 V.

But in the case of a battery we have:

$\ce{PbSO4 (s) + 2e^- -> Pb (s) + SO4^{2-} (aq)}$

And in this case the $\ce{Pb^{2+}}$ is in solid form and the potential is -0.356 V.

In a battery the sulphate is insoluble and it is required that it sticks to the electrode, otherwise the reverse reaction can not occur.

A table of potentials can be found here


How do we know that one mole of an ideal gas occupies (exactly) 22.4 litres?

It doesn't.

The current recommended value for the molar volume of an ideal gas at a temperature of $T=273.15\ \mathrm K$ and a pressure of $101.325\ \mathrm{kPa}$ is $V_\mathrm m=22.413962(13)\times10^{-3}\ \mathrm{m^3\ mol^{-1}}$ (source). Note the given uncertainty; i.e. it is an experimental value  that is not exact.

Also note that these conditions do not correspond to STP. According to current IUPAC recommendations, STP corresponds to a temperature of $T=273.15\ \mathrm K$ and a pressure of $100\ \mathrm{kPa}=1\ \mathrm{bar}$. At this state, the molar volume of an ideal gas actually is $V_\mathrm m=22.710947(13)\times10^{-3}\ \mathrm{m^3\ mol^{-1}}$ (source).


Introduction

Let's define buffer capacity quantitatively as

$$\beta=\cfrac{\mathrm{d}c_\mathrm{b}}{\mathrm{d}(\mathrm{pH})}=-\cfrac{\mathrm{d}c_\mathrm{a}}{\mathrm{d}(\mathrm{pH})}$$

that is, the relationship between concentration (in equivalents) of strong base ($c_\mathrm{b}$) or acid ($c_\mathrm{a}$) added to a solution and its change in $\mathrm{pH}$. From now on I'll assume we're adding a monoprotic base or acid, so "equivalents" and "moles" (and their concentrations) can be used interchangeably.

Pure water

Pure water has a very low buffering capacity - its $\mathrm{pH}$ is very sensitive to the addition of acids or bases. For instance, imagine we add a concentration $c_\mathrm{b}$ of base to pure water:

$\ce{H2O + B- <=> OH- + BH}$

The self-ionisation equilibrium of water will be displaced:

$\ce{[OH-]} \approx c_\mathrm{b}$

and therefore

$ c_b \approx \ce{[OH-]} = \cfrac{K_\mathrm{w}}{\ce{[H3O+]}} = 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}}$

so, taking the derivative,

$\beta_{\ce{OH-}} = \cfrac{\mathrm{d}c_\mathrm{b}}{\mathrm{d}(\mathrm{pH})} = \cfrac{\mathrm{d}}{\mathrm{d}(\mathrm{pH})} 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} = 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} \ln{10}$

it's easy to show that, similarly, for the addition of acids,

$\beta_{\ce{H+}} = -\cfrac{\mathrm{d}c_\mathrm{a}}{\mathrm{d}(\mathrm{pH})} = 10^{-\mathrm{pH}} \ln{10}$

and, combining the buffer effect of both semi-systems, we get the total buffer capacity of water:

$\beta_{\ce{H2O}} = \left( 10^{-\mathrm{pH}} + 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} \right) \ln{10}$

Weak acid/base pair

1:1 solutions of weak acids/bases are a typical buffer system around $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$ for the buffer. To understand why, let's imagine we add a weak acid/base pair, $\ce{HA/KA}$, with total concentration $C_\mathrm{A}$, to which we later add a certain concentration of base, $c_\mathrm{b}$. This weak acid/base pair will be described by the equilibrium

$K_A = \cfrac{\ce{[H3O+] [A-]}}{\ce{[HA]}}$

which, taking into account that $C_\mathrm{A} = \ce{[HA]} + \ce{[A-]}$, implies

$\ce{[A-]} = C_\mathrm{A} \cfrac{K_\mathrm{A}}{\ce{[H3O+]} + K_\mathrm{A}}$

From charge balance:

$\ce{[H3O+]} + \ce{[K+]} = \ce{[OH-]} + \ce{[A-]}$

Taking into account that $\ce{[K+]}$ is equal to the formal concentration of $\ce{KA}$ we added, so $c_\mathrm{b} = \ce{[K+]}$, and the previous expression for $\ce{[A-]}$:

$c_b = \cfrac{K_\mathrm{w}}{\ce{[H3O+]}} - \ce{[H3O+]} + C_\mathrm{A} \cfrac{K_A}{\ce{[H3O+]} + K_\mathrm{A}} = 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} - 10^{-\mathrm{pH}} + C_\mathrm{A} \cfrac{10^{-pK_A}}{10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{A}}}$

So the buffer capacity of the solution will be:

$\beta = \cfrac{\mathrm{d}c_\mathrm{b}}{\mathrm{d}(\mathrm{pH})}= \left(10^{\mathrm{pH}-\mathrm{p}K_w}  + 10^{-\mathrm{pH}} + C_A \cfrac{10^{-\mathrm{pH}-\mathrm{p}K_\mathrm{A}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{A}} \right)^2} \right) \ln{10}$

Note that the first two terms are the buffer capacity of water, so the contribution of the acid/base pair is

$\beta_{\ce{HA/A-}} = C_\mathrm{A} \cfrac{10^{-\mathrm{pH}-\mathrm{p}K_\mathrm{A}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{A}} \right)^2} \ln{10}$

Additive buffer capacity contributions can be calculated this way.

For instance, for a typical acetate buffer ($\ce{HAc} \ 0.2 \mathrm{M} $, $ \ce{NaAc} \ 0.2 \mathrm{M} $; $\mathrm{p}K_\mathrm{a} = 4.76$, $C_\mathrm{A} = 0.4\mathrm{M}$), the buffering capacity looks like this:

Ionic strength

Since buffers are moderately concentrated electrolyte solutions, the assumption that activity coefficients in them are approx. 1 shouldn't be automatic. How does this affect the calculations of buffer capacity we've performed until now?

Let's define an effective equilibrium pseudo-constant, $K'$, operating on concentrations, that is mathematically equivalent to the actual equilibrium constant, $K$, operating on activities:

$K_\mathrm{A} = \cfrac{a_{\ce{H3O+}} a_{\ce{A-}}}{a_{\ce{HA}}} \iff K'_\mathrm{A} = K_\mathrm{A} \cfrac{\gamma_{\ce{H3O+}} \gamma_{\ce{A-}}}{\gamma_{\ce{HA}}} = \cfrac{\ce{[H3O+]} \ce{[A-]}}{\ce{[HA]}}$

So reflecting the impact of ionic strength on buffer capacity becomes a question of evaluating $K'_\mathrm{A}$ and using it to replace $K_\mathrm{A}$ in our previous expressions. For instance, we can use the Debye-Hückel relationship:

$\mathrm{p}K'_\mathrm{A} = \mathrm{p}K_\mathrm{A} + A \left( 2 z_{\ce{HA}} - 1 \right) \left( \cfrac{\sqrt{I}}{1 + \sqrt{I}} - 0.1 I\right)$

where $A$ is a constant ($\approx 0.51$ at room temperature), $z_{\ce{HA}}$ is the charge of the conjugate acid and $I=\frac{1}{2}\sum c_i z^2_i$ is the ionic strength of the solution. Note that $\mathrm{p}K'_\mathrm{A}$ has to be solved recursively, however, as it depends on $I$, which depends on the concentration of the different ions, which depend on $\mathrm{p}K'_\mathrm{A}$.

A similar correction can be applied to $K_\mathrm{w}$ to obtain $K'_\mathrm{w}$.

Applying this ionic strength correction to our acetic/acetate $\pu{0.4M}$ buffer above, we can see the impact of ionic strength:

The $\ce{HCl/KCl}$ buffer

So what about the $\ce{HCl/KCl}$ buffer? Unlike the example of acetic/acetate, $\ce{HCl}$ is a strong electrolyte - so it dissociates completely at all values of $\mathrm{pH}$. So we can treat it in two ways: as simply a water buffer with a different starting $\mathrm{pH}$, or as an acid/base conjugate pair buffer with a $\mathrm{p}K_\mathrm{A} < 0$. Both are equivalent, as the acid/base contribution is negligible compared to the water contribution due to the $\mathrm{p}K_\mathrm{A}$ value.

For instance, for a typical $\ce{HCl}\ 0.2\mathrm{M}$, $\ce{KCl}\ 0.2\mathrm{M}$ buffer ($\mathrm{p}K_\mathrm{A} = -6.3$, $\mathrm{pH} \approx 0.7$, $I \approx 0.8$), the contributions to buffer capacity are as follows:

$\beta_{\ce{H2O}} = \left( 10^{\mathrm{pH}-\mathrm{p}K'_\mathrm{w}} + 10^{-\mathrm{pH}} \right) \ln{10} \approx  10^{-\mathrm{pH}} \ln{10} = 0.4594$

$\beta_{\ce{HCl/Cl-}} = C_A \cfrac{10^{-\mathrm{pH}-\mathrm{p}K'_\mathrm{A}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K'_\mathrm{A}} \right)^2} \ln{10} \approx C_\mathrm{A} 10^{\mathrm{p}K'_\mathrm{A}-\mathrm{pH}} \ln{10} = 5.865 · 10^{-8}$

So, for all effects and purposes we have a buffer that behaves just like water. It is, however, a buffer that behaves just like water in a region of the $\mathrm{pH}$ scale which has a relatively large buffer capacity by virtue of being close to $\mathrm{pH}=0$. For instance, this is what buffer capacity looks like around our typical $\ce{HCl}\ 0.2\mathrm{M}$, $\ce{KCl}\ 0.2\mathrm{M}$ buffer (the red dashed line marks the initial $\mathrm{pH}$):

Note that buffer capacity is highly asymmetrical around this point - adding acid increases the buffer capacity of this system, requiring increasingly more acid to decrease $\mathrm{pH}$, while adding base decreases the buffer capacity, making the system's $\mathrm{pH}$ more sensitive to further base additions. This can be easily visualised if we realise that the area under the $\beta$ curve represents the concentration that needs to be added to move in the $\mathrm{pH}$ scale.

Note also that, unlike in weak acid/base pairs, we cannot simply increase concentration to increase buffer capacity while retaining the same $\mathrm{pH}$ - as the concentration-dependent term, corresponding to the $\ce{HCl/KCl}$ base, is negligible. Ionic strength is irrelevant for $\beta$, as well, as $K'_\mathrm{w}$ and $K'_\mathrm{A}$ is not involved - $\beta_{\ce{H3O+}}$ is the only relevant contribution to $\beta$ in this very acidic region, and, by definition, it is not affected by activity concerns (unlike $\beta_{\ce{OH-}}$ or $\beta_{\ce{HCl/Cl-}}$).

Does that mean that $\ce{HCl/KCl}$ buffers with different concentrations all have the same buffer capacity? Not at all: but their different buffer capacity is mediated by $\mathrm{pH}$. In other words, different concentrations produce different $\mathrm{pH}$ which, in turn, produces a different $\beta$ - which is unresponsive to changes in $I$ that do not change $\mathrm{pH}$.

Two final plots to illustrate this. Imagine we prepare solutions of $\ce{HCl/KCl}$ (1:1) at different total concentrations $C_{\ce{Cl-}}=c_{\ce{HCl}}+c_{\ce{KCl}}$. If we compare the two contributions to $\beta$, it's clear that $\ce{HCl/KCl}$ increases buffer capacity almost exclusively through $\mathrm{pH}$; $\beta_{\ce{H3O+}}$ is proportional to $\ce{[H3O+]}$ which increases linearly with $c_{\ce{HCl}}$ as $\ce{HCl}$ is a strong electrolyte:

Conversely, what if we control for $\mathrm{pH}$ and change only $I$? Let's imagine a series of solutions containing a fixed concentration of $\ce{HCl}$ and varying amounts of a non-reacting, strong electrolyte without common ions - such as $\ce{NaBr}$ - so we can change the ionic strength of the solution without affecting neither the $\mathrm{pH}$ nor the $\ce{HCl/Cl-}$ equilibrium. The impact of $I$ on $\beta$ would look like this (note that the minimum $I$ corresponds to a solution with $\ce{HCl}$ and no $\ce{NaBr}$):

This is because although the $\beta_{\ce{HCl/Cl-}}$ and $\beta_{\ce{OH-}}$ terms do depend on $I$, these terms are many, many orders of magnitude lower than the $\beta_{\ce{H3O+}}$ term, which doesn't.

So, after this discussion, we can now directly respond to your questions.

TL;DR

1) Why does the increased ionic strength of the solution improve water's buffer capacity?

It doesn't. The buffer capacity of solutions of strong acids are dominated by the $\beta_{\ce{H3O+}}$ term, which (by definition of $\mathrm{pH}$) is not affected by ionic strength.

There are, however, other reasons to desire an ionic strength in that range - in biological systems, many molecules, particularly proteins, have a privileged ionic strength stability gap - above or below which they tend to denaturalise or precipitate, and outside of which their biological activity can be inhibited or poisoned. Although different systems will require different ionic strengths, buffers are typically prepared with ionic strengths in the range $0.1-1\mathrm{M}$ for that reason.

2) Isn't a higher equilibrium constant required to make water a better buffer? Then isn't it bad that the ionic strength decreases $K$?

The $\beta_{\ce{H3O+}}$ term, by construction, does not depend on $K'_\mathrm{w}$, so it is unaffected by ionic strength. The $\beta_{\ce{OH-}}$ term does, and as you mention, it is decreased at moderate ionic strength - although it is increased at higher $I$. In general, the impact of ionic strength on buffers is as follows: increasing ionic strength begins by lowering $\beta$, then, as $I$ keeps increasing, $\beta$ raises again.

3) Could any other salt have been used to increase the ionic strength, or is there some specific reason $\ce{HCl-KCl}$ is used?

Yes, any other salt could have been used to increase the ionic strength. Adding $\ce{KCl}$ affects the acid-base pair contribution to $\beta$, but as we've seen, that contribution is completely negligible - as will be the case with strong acids, which operate mainly through the $\ce{H3O+}$ contribution.


How many stereoisomers are possible for hepta-1,6-diene?

In the case of your original drawing with the terminal double bonds, there are no E or Z (trans or cis) isomers!

E and Z isomers are only possible if both ends of the $\ce{C=C}$ double bond each carry a substituent other than a hydrogen atom.

As far as the other structure is concerned, E and Z isomers look as follows:


What is the fire point of sucrose with an average particle diameter of 60 micrometres?

I managed to find the answer in the MSDS of sugar given by the US Department of Labour, after some searching.

In the MSDS found here there is an ignition temperature (which I imagine to be the same as the fire point) of 350 C, with a minimum explosion concentration of 0.045 g/l in air. There also seems to be a possible ignition from a dramatic increase in pressure (mentioned in the MSDS to be 5,000 psi/sec), though no further dangers are expressed.


The usual titration of glucose with Fehling's Reagent is to place the Fehling's Reagent in a flask and titrate the boiling solution with glucose. At the end point, all the $\ce {Cu^{2+}}$ ions are converted to $\ce {Cu^{+}}$; this makes the solution colorless. That is difficult to see since the red copper(I) oxide is floating in the solution.

With methylene blue, after the copper has been reduced, the glucose can then reduce the methylene blue (in a basic solution) to a colorless solution. This makes the end point clearer since the methylene blue is much darker than the $\ce {Cu^{2+}}$ ions. The pH of the solution does not change much in this reaction.


There exist plenty of basic salts. For example,

Copper produces a "simple" copper carbonate $\ce{CuCO3}$ as high $\ce{CO2}$ pressure. At lower pressures, the following basic copper carbonates are produced : $\ce{Cu3(CO3)2(OH)2}$ or $\ce{Cu2CO3(OH)2}$ (malachite), depending on the partial $\ce{CO2}$ pressure. Other copper salts produce basic copper salts like : $\ce{Cu2NO3(OH)3}$ , $\ce{Cu4(OH)6Cl2}$, $\ce{Cu4(OH)6SO4}$ .
Zinc produces the two following basic zinc carbonates, according to the chosen pH : $\ce{Zn5(OH)6(CO3)2}$ or $\ce{Zn2CO3(OH)2}$
Iron(III) produces also basic salts like $\ce{FeSO4(OH)}$ which is produced by air oxidation of iron(II) sulfate. Dilution of a ferric acetate solution produces a precipitate of basic acetate $\ce{Fe(CH3COO)(OH)2}$
Aluminum solutions are often hydrolyzed producing basic salts. In the case of acetate, it gives a precipitate of basic acetate $\ce{Al(CH3COO)(OH)2}$
Cobalt chloride solutions treated by cold NaOH solutions give a precipitate of basic chloride $\ce{CoClOH}$ When treated by carbonate solutions they produce a great variety of basic carbonates.
This list is not exhaustive.


Intermediate is a middle step in the chemical process: reactants -> intermediate(s) -> products. Benzyne is formed from benzene and is highly reactive. Thus, it is never the first or last step of the process. To my knowledge benzyne is rarely observed as a product, because its reacting fast with other molecules because its triple bond.

What makes a molecule an intermediate? It is a highly reactive "middle product" of a reaction.
See here and there.


how to measure the efficiency of carbon filter [or of activated carbon/charcoal]?

1/ The iodine method:

I found the answer on this video (from Cody's lab youtube channel). Cody's show how to make activated carbon/charcoal and how to test its efficiency.

It's quite an easy task. The test below measures the absorption's efficiency of your charcoal filter (activated or not).

● Take a gram of your active charcoal (from your filter)
● Put it in a test tube
● Add 2ml hydrochloric acid (to make sure the charcoal is acidified, because otherwise, the alkaline ash in the charcoal will react with the iodine)
● just warm it a little to help the reaction.
● Add 25ml of iodine (iodine with alcohol).
● Wait for 24h, look at the color of your solution: the lighter the color the more effective is the charcoal (the lighter, the more it has absorbed iodine which is dark). You could compare it with some active carbon you bought, or normal charcoal

To measure more precisely the efficiency of your charcoal:
● Take 10ml of the solution you made above.
● Add some thiosulfate to it (drop after drop) until your solution becomes clear.
● Measure the quantity of thiosulfate that was needed.

See Cody's video for more details.

2/ The Formaldehyde (or VOIC) method:

If you want to know if your activated carbon filters works, you can also test it using Thomas Talhelm method (he is the most famous air pollution DIY expert). But this technique require a solution of formaldehyde, and a device to mesure the quantity of formaldehyde in the air.

Buy a formaldehyde solution.
warm it (eg. using a rice cooker) in a closed room.
measure the level of formaldehyde
turn ON your activated carbon filters, and keep measuring.

If your filter is working you should see that:

  With a fan only (red line), formaldehyde levels stayed high. But with
  a carbon filter on the fan (blue line), formaldehyde levels went down.


How to carry out basis set fitting and extrapolation

If I well understand your question you have a set of experimental points $(E_1,X_1), (E_2,X_2), ... , (E_n,X_n)$. You want to fit to this set of points the function :
$$
E(X) = E_\infty + Ae^{-\alpha X}
$$
So, you want to approximately determine the parameters $E_\infty$ , $A$ and $\alpha$.
This requires a non-linear regression calculus. For example see https://mathworld.wolfram.com/NonlinearLeastSquaresFitting.html
This calculus is iterative and requires to set some "guessed" values of the parameters to start the iterative process.
Another method (much simpler but for approximative result only) is not iterative and doesn't require "guessed" values is explained in : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales
You could try it in folowing the numerical example below with $\quad y(x)=E(X)$ , $a=E_\infty$ , $b=A$ , $c=-\alpha$

Note : This simplified method is not reliable if the number of points is too low.


do Couto et al have published an article [1] characterizing the density of states (band structure) of water using DFT theory combined with MC sampling. Extrapolating from clusters to bulk water, the HOMO-LUMO gap is 8.55 eV (Table 3).

From the abstract:

  By extrapolating the results for larger clusters the threshold energy
  for photoelectron emission is $9.6±0.15 eV$ free clusters and
  $10.58±0.10 eV$ embedded clusters. Our results  for  the  electron
  affinity $V_0=−0.17±0.05 eV$ and  adiabatic  band  gap
  $E_{G,Ad}=6.83±0.05 eV$ of liquid water are in excellent agreement
  with recent information from theoretical and experimental works.

From the introduction:

  It is generally accepted that water canbe  described  as  a  very
  large  band-gap  amorphous semiconductor.11,17 However, it is not
  obvious that the band-gap of liquid water can be uniquely associated
  with an optical vertical excitation process, where the gap is
  defined simply as the highest occupied molecular orbital-lowest
  unoccupied molecular orbital HOMO-LUMO energy difference. As recently pointed out
  by Coeet  al.,3 the reactive nature of electronically excited water
  molecules,18 the reorganization of water molecules around charged
  species in liquid phase, and the observed photophysics of anionic
  defects known as the anion problem 4 strongly indicate that an
  adiabatic route can be defined for accessing the conduction-band edge
  in liquid phase. Moreover, given that the time scale ofthe solvent
  relaxation is much larger than the vertical process, the adiabatic
  band gap of liquid water cannot be determined vertically.

Reference
[1] Cabral do Couto, Estácio, and Costa Cabral J. Chem. Phys.123, 0545102005


Cyclic voltammetry (CV) is often performed in an enclosed vessel with the purge gas (argon/nitrogen) constantly bubbling through to prevent the solution from aeration and unnecessary side processes, mainly oxidation by the atmospheric oxygen. You'd rarely see alligator clamps in any CV apparatus also because they are not suited for holding moderately thick round objects (such as electrodes) in place, they can damage the inlet and outlet piping (e.g. the one for the purge gas, or chemicals if one studies kinetics), plus they make the experiments less reproducible since the position of the electrodes is changed each time you change the beaker.

Typically a teflon lid with 4 (or more) holes is used, where all 3 electrodes and the purging line are adjusted in place with rubber rings or plugs so that they are all firmly attached to the lid. This is very convenient since you don't have to manually adjust the apparatus every time the solution needs to be changed. All that's required is take the lid off, rinse the vessel and electrodes, fill the new portion in and put the lid back in place.

Alternatively, one can use a 3-neck round bottom flask with the electrodes attached through the rubber stoppers in the side necks. The downside is problematic cleaning, so this setup is more appropriate for the continuous measurements when one doesn't have to switch solutions too often. This is the simplest setup; a more sophisticated one can as well look like this:

  Image source

Either way, I would discourage from using alligator clips for CV. Even though they might work for simple measurements, in my opinion there are too many downsides.


Neutralization of alkali solution by a mix of strong acids

I think you over-complicated things a bit by focusing on the formation of sodium salts, when in reality all you need to take into account is the net neutralization reaction:

$$\ce{H3O+ + OH- -> 2 H2O}$$

From the equivalence point condition of any neutralization reaction one can find volume of the base $V_\mathrm{b}$ expressed via the volume of acid(s) $V_\mathrm{a}$:

$$V_\mathrm{b} = \frac{C(\ce{H3O+})}{C(\ce{OH-})} \cdot V_\mathrm{a} \tag{1}$$

Since there are two moderately diluted strong acids and a base, one can conclude that the dissociation of each acid and a base is independent and complete:

\begin{align}
\ce{HCl + H2O &<=>> H3O+ + Cl-} \\
\ce{H2SO4 + 2H2O &<=>> 2H3O+ + SO4^2-} \\
& \\
\ce{NaOH &<=>> Na+ + OH-}
\end{align}

Note that concentration of protons is doubled in case of sulfuric acid. Total concentration of protons in the solution is

$$C(\ce{H3O+}) = \frac{n(\ce{H3O+})}{V(\ce{H3O+})} = \frac{n(\ce{HCl}) + 2 \cdot n(\ce{H2SO4})}{V(\ce{HCl}) + V(\ce{H2SO4})} \tag{2}$$

Taking into account that $n = CV$ and that two identical volumes of acid were mixed ($V(\ce{HCl}) = V(\ce{H2SO4}) = V(\text{acid})$) (2) can be transferred into:

$$
\require{cancel}
\begin{align}
C(\ce{H3O+}) &= \frac{C(\ce{HCl}) \cdot V(\ce{HCl}) + 2 \cdot C(\ce{H2SO4}) \cdot V(\ce{H2SO4})}{V(\ce{HCl}) + V(\ce{H2SO4})} \\
&= \frac{\cancel{V(\text{acid})} \cdot (C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4}))}{2 \cdot \cancel{V(\text{acid})}} \\
&= \frac{C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4})}{2} \tag{3}
\end{align}
$$

Now let's plug (3) in (1) and finally find the volume of alkali:

$$V_\mathrm{b} = \frac{C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4})}{2 \cdot C(\ce{OH-})}  \cdot V_\mathrm{a} = \frac{\pu{0.1 M} + 2 \cdot \pu{0.2 M}}{2 \cdot \pu{0.1 M}} \cdot \pu{20 mL} = \pu{50 mL}$$


Thought Experiment: Applying a high frequency high voltage to methane

To decompose methane molecule you have to apply energy about several electron-volts directly to molecule. Since microwave radiation wavelength measured in millimeters, to really decompose methane you'll need ridiculous voltage, much more them mere kilovolts needed to ignite electrical discharge.

And btw, electrical discharge is the easiest way to supply the energy needed for molecule to break.


What is the correct way to notate conformation preferences of polysaccharides?

If I understood the question correctly, your first notation looks correct and less ambiguous.

Common polysaccharides (glycogen, starch, and cellulose are composed of pyranoside subunits (having six-membered rings)). Such molecules can be represented as a series of rigid pyranose rings connected by an oxygen atom bridging two carbon atoms (the glycosidic bond).

  There is, in principle, free rotation about both $\ce{C–O}$ bonds
  linking the residues , but rotation about each bond is limited by
  steric hindrance by substituents. The three dimensional structures of
  these molecules can be described in terms of the dihedral angles,ϕ and
  Ψ, about the glycosidic bond.

The bulkiness of the pyranose ring and its substituents, and electronic effects at the anomeric carbon, place constraints on the angles ϕ and Ψ thus certain conformations are much more stable than others, as can be shown on a map of energy as a function of  ϕ and Ψ:

(a)

(a) When the relative energy (Ʃ) is plotted for each value of  ϕ and Ψ with isoenergy (“same energy”) contours drawn at intervals of 1 kcal/mol above the minimum energy state, the result is a map of preferred conformations.

A map of favored conformations for oligosaccharides and polysaccharides.

The torsion angles  ϕ and Ψ which define the spatial relationship between adjacent rings, can in principle have any value from 0 to 360 degrees.
In fact, some of the torsion angles would give conformations that are sterically hindered, whereas others give conformations that maximize hydrogen bonding.

(b)

(b) Two energetic extremes for the disaccharide Gal(β-3)Gal; these values fall on the energy diagram (a) as shown by the red and blue dots. The red dot indicates the least favored conformation, the blue dot the most favored conformation.

Conformation at the glycosidic bonds of cellulose, amylose, and dextran.

The polymers are depicted as rigid pyranose rings joined by glycosidic bonds, with free rotation about these bonds. Note that in dextran there is also free rotation about the bond between C-5 and C-6 (torsion angle ѡ (omega)).

So the notation considering the glycosidic linkage will be $\ce{Glu(ϕ)-(β1-4)-Glu(Ψ)}$ or another detailed way of illustrating this is $\ce{C1(ϕ)-O4(Ψ)-C4}$ for cellulose.

There are other ways of illustrating polyasacharides notations, and their side groups e.g hydroxyl group (in newman projection)

For the precise specification of the orientation of a polyatomic ring substituent, it is necessary to specify the torsion angle about the exocyclic bond.

  The reference atom in the ring is the carbon atom with the number one
  lower than that of the substituted carbon, unless substitution is on
  the anomeric carbon, when the ring oxygen is the reference atom. The
  reference atom in the exocyclic –CH2X group is X. The exocyclic
  torsion angle is denoted by χ(chi), followed by the atoms to which it
  refers e.g χ(C1-C2-O2-H).

References

Lehninger Principles of Biochemistry 5th ed: Polysaccharides
Symbols for Specifying the Conformation of Polysaccharide Chains


How to replicate compositions of discontinued paints?

I'm wondering if adding 1 ounce of Elmer's glue to 100% acrylic paint would duplicate the old formula?

No.

Paints are a specific mixture of a number of components designed to fulfill their role well in this specific combination. If you go and add a random additional ingredient, which in itself is not a pure substance but a nondisclosed mixture, you cannot predict what will happen. At best, the additional ingredient does no good and no harm. At worst, you can throw away the entire can of paint.

Paint companies also do not disclose key data about their mixtures for the obvious reason that they don’t want to be copied. So even though it may say X, Y and Z on the back of the can, that doesn’t mean you can infer much about the mixture. Even if they give percentages, you don’t know what all the rest is. Therefore, you cannot duplicate the old formula unless you have access to the full blow of analytical methods and a sample of the old paint.

Finally, I’m not even sure if changing vinyl acrylic on the packaging to just acrylic has any implication on the content of the can. It could just be a new choice of description.


The notation $\ce{H^.}$ refers to the hydrogen radical (a hydrogen atom). The dot represents the unpaired valence electron.

An equivalent notation for the reaction would be

$$\ce{e- + H2O -> H + OH-}$$

However, adding the dot emphasizes the fate of the electron on the reactant side.


Why is this PH3 considered to be an unhybridized molecule?

I'll try to explain this by comparing this with the case of NH3.
N is a better electron withdrawing group than P. So in the case of NH3 the electron cloud is more closer to N. And therefore the repulsion in the bonds causes the 3 H atoms to be farther away from each other resulting in a bond angle of about 109°. But in the case of PH3 the electron cloud is farther away from the central atom(P) than in the case of NH3. So the bond pair - bond pair repulsion is comparatively lesser, causing the 3 H atoms to move closer together to an angle of almost 90°, resembling the px, py, and pz orbitals, as a consequence it's lone pairs are present in almost pure s - orbital. So it is is said that PH3 can be viewed as unhybridized.


Ammonium nitrate is very soluble in water, so watering the soil will remove it, though, depending on the amount and dilution, it could then pollute streams, lakes and other bodies of water, possibly causing an algae bloom.

Your best choice might be to experiment with nitrate-tolerant plants, such as lettuce, ornamental cabbage and rhubarb, or perhaps bell peppers.

Citric acid will not help in any way, and would make the soil less likely to support life.

In nature, denitrifying bacteria will slowly reduce the $\ce{NH4NO3}$. In an anaerobic environment, it might be converted to $\ce{NH4+}$.


Are a meso compound and its mirror image isomers?

meso-compound
A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.[1, 2]
isomer
One of several species (or molecular entities ) that have the same atomic composition (molecular formula) but different line formulae or different stereochemical formulae and hence different physical and/or chemical properties.[1, 3, 4]

Your notion is correct in so far as the meso-compound and its mirror image are not isomers of each other since they are actually the same compound.
For example, meso-tartaric acid (preferred IUPAC name: (2R,3S)-2,3-dihydroxybutanedioic acid) can be superimposed on its mirror image “(2S,3R)-2,3-dihydroxybutanedioic acid”; i.e. both structures actually describe the same compound. Thus, they are not isomers of each other. (And hence, “(2S,3R)-2,3-dihydroxybutanedioic acid” is not a correct IUPAC name.)

However, two further stereochemical configurations are possible, namely (2R,3R)-2,3-dihydroxybutanedioic acid and (2S,3S)-2,3-dihydroxybutanedioic acid.

These structures are non-superimposable mirror images of each other, i.e. they are enantiomers, and the above-mentioned meso-tartaric acid is a diastereomer of both of them.
Therefore, there are a total of three isomers of tartaric acid:

(2R,3R)-2,3-dihydroxybutanedioic acid,
(2S,3S)-2,3-dihydroxybutanedioic acid, and
the meso-compound (2R,3S)-2,3-dihydroxybutanedioic acid.

References
[1] Basic terminology of stereochemistry (IUPAC Recommendations 1996). Pure Appl. Chem. 1996, 68, 2193
[2] meso-compound. In IUPAC Compendium of Chemical Terminology, (Gold Book)
[3] Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994). Pure Appl. Chem. 1994, 66, 1077
[4] isomer. In IUPAC Compendium of Chemical Terminology (Gold Book)


Computing accurate vibrational and rotational contributions to the free energy of transition states and loosely bound complexes

Given the accuracy it sounds like you're looking for, and the systems you're studying, this indeed appears to be a Very Hard Problem. I think your intuitions are well founded.

Briefly, to your questions:

1) How to handle the low-frequency modes?

A lot of manual exploration of the PES, doing things like running normal mode scans and optimizing intermediate geometries along suspected internal rotations. This gives you an idea how the various energetic barriers to non-harmonic motions compare to a Boltzmann distribution of states.

If the barriers are high, the harmonic approximation is probably decent and you can just use Gaussian's results
If barriers are low, then you have ~free internal rotation.
If intermediate, then you're facing hindered internal rotation.

There's literature on evaluating enthalpy/entropy contributions of free and hindered internal rotations.  Some can be found in my answer to a different question, here.  Other potentially relevant resources include:

Czaszar (WIREs Comput Molec Sci 2 273, 2012)
Overview of anharmonic force fields and their computation.
Bloino (J Chem Theor Comput 8(3) 1015, 2012)
Overview of the VPT2 anharmonic computational system implemented in recent versions of Gaussian.
Carbonniere (Chem Phys Lett 392 365, 2004)
Description of the vibrational-rotational (Coriolis) coupling anharmonic correction and its potential significance for some systems. More recent references are likely available.
Meal (J Chem Phys 24(6) 1119, 1956)
Meal (J Chem Phys 24(6) 1126, 1956)
Classic discussions of Coriolis coupling.

You should be able to turn up quite a bit more information with some quick further searches, if need be.

2) Do I need to be worried about contributions from centrifugal distortion?

Very likely. Unfortunately, I don't know of very many examples where people have attempted corrections such as this. Certainly you're looking at manually developing a suitable analysis; I can't imagine any software packages out there at present are capable of automatically chugging through this for you.


Wikipedia has the correct definition:

  A more rigorous statement is that with respect to exchange of two identical particles the total wave function is antisymmetric for fermions, and symmetric for bosons. This means that if the space and spin co-ordinates of two identical particles are interchanged, then the wave function changes its sign for fermions and does not change for bosons.

The textbook shorthand you quote, 'No two electron can have same set of quantum numbers', only makes sense when you are using the one-electron approximation. As you move beyond that approximation, you also have to move beyond this shorthand definition.


How are bubbles formed when heating water?

Saying that the liquid is converted to bubbles is not correct. The liquid simply becomes a gas (so inside the bubble there is the gas phase of your liquid) this is a phase transition and depends on the pressure and temperature of the portion of liquid close to the bottom of your container.

The points where the bubbles appear depends on the local variations of temperature and pressure.


How to quantify caffeine content in coffee?

According to a paper published named: "Chemical Test for Caffeine" for the "Hong Kong Chemistry Olympiad for Secondary Schools (2013-14)" (Hyperlink provided below). You can titrate caffeine, with the help of iodine and potassium iodide, and sulfuric acid as the acid. You will have to add an excess of iodine, which means it will be an iodometric backtitration. The iodine will react in two times the number of moles compared to your caffeine.
The only problem I have with this method, is that I cannot found any elaboration on how this reaction is possible on the internet (I haven't tested it myself). If there is someone who knows more about how this reaction if possible, please provide me with more information.

If someone knows how or why the reaction can take place, please keep me notified about it. I will provide the reaction below:
Reaction:
Reactants --- Caffeine + 2 Iodine + Potassiumiodide + Sulfuric Acid
Products ---- Potassium Bisulflate + Caffeine (in a peculiar complex with the remaining substances, notated like a hydrate. No water present.)

$$\begin{align}\ce{C8H10N4O2 + 2 I2 + KI + H2SO4 &-> C8H10N4O2 . HI . I4 + KHSO4}\\
\ce{\color{red}{I2} + 2 Na2S2O3 &-> 2 NaI + Na2S4O6}\end{align}$$

EDIT 1: This URL should lead to the paper I mentioned (if it works).
Chemical Test for Caffeine
EDIT 2: Another URL provided, this is the source used in the actual paper. The site is in Chinese (if I'm not mistaken), but can be translated via google. For anyone interested: Source used in mentioned paper, from "www.med66.com"


Shouldn't all of group 16 form double bonds?

I notice that out of all of group 16 oxygen is the only element that can form a double bond.

This premise is not true. Sulfur in fact forms double bonds in many organic compounds.  Selenium and tellurium both form double bonds with triphenylphosphine as triphenylphosphineselenide and triphenylphosphinetelluride.

The larger chalcogenides do have difficulty in forming double bonds in organic molecules due to lower bond strength, higher reactivity, preference of molecules for oxygen, and steric bulk.  They can however indeed form double bonds; it's just not as common.


Is it reasonable to use natural orbitals of an unrestricted HF calculation as a basis to start a CASSCF calculation for a radical

The best starting orbitals for a CASSCF calculation are optimised orbitals of another CASSCF (or RASSCF) calculation. That sounds a bit ridiculous, but this is probably the best way to figure out the active space.

At first you are probably choosing something quite small, like CAS(2,2) to CAS(4,4). For these calculations it barely matters what kind of starting orbitals you are choosing. The change will be negligible and probably depending heavily on the system you are using.

The advantage of calculating the cations by dismissing the radicals is often a good choice since you can run a very fast RHF calculation and the orbitals are symmetric, which is what you need for CAS.
Natural orbitals of an UHF calculation are also a quite sensible choice, because they already allow for fractional occupation and should also be symmetric. The only downside I can think of is, that this kind of start orbital generation takes a lot longer.
Similarly you can choose to run a ROHF calculation, if you are uncomfortable with too much cationic charge.
As you have said you can choose also an UHF calculation by throwing away one set. I would advise caution here. Sometimes - probably especially when CAS is necessary - spin contamination in UHF is significant. Therefore one set might not give you anything close and you could end up choosing the wrong CAS.

I am not aware of any major pitfalls except for the just mentioned choosing of a completely wrong active space.

In any case it is a good idea to give the CASSCF calculation a few cycles and see how the active space is developing. Sometimes you need to adjust the active space long before the calculation converges and then you will have already a much better guess than with your initial starting orbitals.

I personally use the following routine

R(O)HF [x+] -> CAS(x, ⌈½x⌉ +1)
            -> RAS(x+2n, ⌈½x⌉ +n +1)
            -> CAS(x+2n, ⌈½x⌉ +n +1)
            -> RAS(...)
            -> CAS(...) -> ...

to narrow down the active space and include all orbitals I want and need. In most cases I am not converging the intermediate calculations, except for the minimal CAS.
As you can see the procedure that follows after your start orbitals is much™ more time consuming, so it really does not matter.


Does "covalent" AgCl (aq) exist and if so, how to quantify its concentration?

Some amount of $[\ce{AgCl(H2O)}]$ particles should exist in solution from pure thermodynamic point of view. See, $\ce{Ag+}$ ion can coordinate two ligands, and in solution both $\ce{Cl}$ and $\ce{H2O}$ are present. However, it should be really hard to detect trace amounts of this particle, as $\ce{AgCl}$ is almost insoluble. In case one can get concentrated salt solutions, similar particles may exist and hints for their existence may be got from conductivity measurements and osmotic pressure. Sometimes a direct study using NMR is possible, which can give some structural information if used properly (see 2D-NMR methods). As I remember, $\ce{Cr(CH_3COO)2}$ hydrate is slightly soluble in water but undergoes little to no dissociation.


Does diamagnetic anisotropy only apply to a proton bonded to triple bonded carbon?

In principle, diamagnetic anisotropy can be invoked for any bond or any functional group; however, there's only a few situations where it's commonly used to understand patterns of shielding/deshielding.  A decent discussion of this can be found in Silverstein's textbook (Spectrometric Identification of Organic Compounds), but I seem to recall that a couple of his figures are drawn in a confusing manner...

In short, this picture relies on the fact that pairs of electrons are diamagnetic, i.e., when immersed in a magnetic field, they get magnetized against the direction of the magnetic field:

$M=\chi B_0$

(M is the induced magnetization, B0 is the applied magnetic field, and $\chi$ is the magnetic susc., which is negative for diamagnetic objects).  Thus, any electron density in a stable molecule (ignoring radicals) will become magnetized against the magnetic field (this is why we speak in terms of shielding).

We can picture molecular orbitals or valence bond orbitals of interest to us simply as a cloud of electron density that becomes magnetized in a magnetic field.  Anisotropy means that effect depends on an angle: we're familiar with the picture of a magnetic dipole having magnetic field lines that curve around from the N end back to the S end – so something below or above the electron cloud will be in the (so called) shielding cone whereas things off to the side will be in the deshielding plane.  Note, this is true for anything that is diamagnetic - there's no need for mystical ring currents or anything like that; any electron cloud placed in a magnetic field will be magnetized and, because it's diamagnetic, will reduce the local magnetic field above and below the electron cloud, but increase it off to the sides.

However, the molecules are tumbling rapidly in solution, so most of this effect simply averages away to zero...  We need special situations where a proton of interest is held in a fixed position relative to the electron cloud of interest to get an noticable effect.  Common example:

Flat conjugated systems like benzene, but also larger annulenes and related molecules.  Most chemists are familiar with the argument that hydrogens outside the ring are deshielded by the "ring current" – while some may debate the existence of the ring current, the pi system is, by its nature, diamagnetic, so it's not surprising that any protons held in the outer edge are deshielded while methylenes held in a position above or below the ring are significantly shielded.  There are beautiful examples of annulenes with protons inside the large aromatic ring that have negative chemical shifts due to the large shielding anisotropy of the pi system.
Triple bonds have two pi bonds that are 90º to each other.  It's a little harder to "see", but you can look at various possible orientations in a magnetic field and see that you can rationalize some shielding of the carbons in the triple bond.
Cyclohexane derivatives.  While not as dramatic as the benzene example (and confusingly illustrated in the Silverstein book), you can picture the sigma skeleton of the cyclohexane as a large cloud of electron density, so it should have a shielding cone above/below the ring and a deshielding plane off to the sides of the ring.  This turns out to be true!  For a system that has a relatively fixed chair conformation and a diastereotopic methylene, you'll always find the equatorial hydrogen deshielded relative to the axial hydrogen.

I'm sure there are more examples, but those are the only examples that I have seen work consistently enough to be useful.


How does addition of AgCl affect concentration of iodide in KI solution?

I think the experts here have muddled the problem by stating "no equilibriums," so I'll solve it.
Given the reaction:
$$\ce{KI(aq) + AgCl(s) <=>[excess AgCl(s)] KCl(aq) + AgI(s)}$$
there are equilibriums. There are always equilibriums in chemistry. In this case it is much better to think of the reaction as proceeding quantitatively. That is to say that if the initial concentration of $\ce{I-}$ is $\pu{0.10 M}$, then the final concentration of $\ce{Cl-}$ will be  $\pu{0.10 M}$. (This is a significant figure simplification. The $\ce{[I-]}$ isn't given to 10 decimal places.)
Now Wikipedia gives the $K_\mathrm{sp}$ of silver chloride as $1.77 \times 10^{-10}$ and for silver iodide as $8.52 \times 10^{−17}$.
Since the final concentration of $\ce{Cl-} = \pu{0.10 M}$, the final concentration of silver is given by:
$$\ce{[Ag+]} = \dfrac{K_\mathrm{sp, AgCl}}{\ce{[Cl-]}} = \dfrac{1.77 \times 10^{-10}}{0.10} = \pu{1.77\times 10^{-9} M}$$
So the final concentration of $\ce{I-}$ is given by:
$$\ce{[I-]} = \dfrac{K_\mathrm{sp, AgI}}{\ce{[Ag+]}} = \dfrac{8.52 \times 10^{−17}}{1.77 \times 10^{-9}} = 4.8\times10^{-8}$$
Since the $\ce{[I-]}$ starts out as $\pu{0.10 M}$ and ends up as $\pu{4.8 \times 10^{-8} M}$, it is better to say that the reaction proceeds quantitatively rather than completely.


Acid-Base Titration, why does for a weak acid indicator, [HA] = [A-]

In an Acid titration, we want the end-point to be as close to the equivalence point as possible.

Different indicators have different end-points (they will change colour at different pH values).

When an end-point is reached, the colour of the indicator will change with respect to the [$H^+$] concentration of the solution it is measuring.

As [$H^+$] increases, there will be a point when [HA] = [$A^-$] and depending on the indicator used by the person doing the titration, this would be close to the calculated equivalence-point.

Note, the difference in volume between the end-point and equivalence-point is very very small, hence we can approximate the end-point as the same as the equivalence-point in any experiment.


When does phenolphthalein containing solution change colour?

I am not absolutely sure I understood the question completely, but I will try to shed some light on the colour change of an indicator.

As a rule of thumb the human eye can make out a change in colour when there is a 10:1 ratio of the components. Since phenolphtalein, $\ce{HIn}$, is without colour, and its deprotonated form $\ce{In-}$ is pink, it means that a solution with

$\displaystyle \frac{n(\ce{HIn})}{n(\ce{In-})} = \frac{10}{1}$ will appear clear, while
$\displaystyle \frac{n(\ce{HIn})}{n(\ce{In-})} = \frac{1}{10}$ will appear pink.

Since phenolphtalein is a weak acid, you will see, that the colour change will theoretically happen around equilibrium constant, i.e. $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$. (Which is also a simplification, because phenolphtalein has multiple acidic protons.) At that point you will probably not notice significant changes. Though phenolphtalein is a bit of an exception here, because it has only one colour. The human eye can probably see a smaller surplus of the colourful component. (It is much more difficult for an indicator like methyl red.)

If you substitute the above fractions into the Henderson-Hasselbalch equation, you can approximate the range in which the colour change will happen:
\begin{align}
\mathrm{pH}
  &= \mathrm{p}K_\mathrm{a} - \lg\left(\frac{n(\ce{HIn})}{n(\ce{In-})}\right)\\
  &=%
  \begin{cases}
     \mathrm{p}K_\mathrm{a} - \lg\left(\frac{10}{1}\right); & \text{clear}\\
     \mathrm{p}K_\mathrm{a} - \lg\left(\frac{1}{10}\right); & \text{pink}  \end{cases}\\
\mathrm{pH}
  &= \mathrm{p}K_\mathrm{a} \pm 1
\end{align}

When you are estimating the amount of carbon dioxide dissolved in water, then the following equilibrium will be present:
$$\ce{CO2 + H2O <=> HCO3- + H+ <=> H2CO3}$$

Adding hydroxide ions you are shifting this equilibrium to the right, essentially deprotonating the acid. Once you have fully (in first approximation) converted it to carbonate, you start deprotonating the indicator until you notice the colour change.


Making sense of J coupling data for Hydronium

I read this as indicating 2JH-H = 9.6Hz and 1JO-H = -163.8Hz. That's a large O-H coupling.

For water, experimental values have been reported as 2JH-H = -7.3Hz and 1JO-H = -80.6Hz 1, and calculated values vary, but within an order of magnitude. Gas phase studies for 1JO-H also provide values around -80Hz. I can't explain the large discrepancy here; perhaps there is a significant change in bond angle for hydronium? Or perhaps a poorly refined calculation set?


How do I dissolve sodium oleate in distilled water

A product information by Sigma-Aldrich, a chemical supplier, cites Merck Index regarding the solubility of sodium oleate.  And states "It is
also soluble in water (approximately 100 mg/ml)", in other words, 10 mass%.  Since this value is larger than the 2 mass% requested by you, and provided sodium oleate is a surfactant anyway, just go ahead.  A gentle warming, like 30 to 40 Celsius, may accelerate the process, though.


In cases like this calculate the reaction quotient, Q (https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Equilibria/Chemical_Equilibria/The_Reaction_Quotient) using the initial concentrations, and compare the value with the equilibrium constant, K:

Q=K implies that the system is at equilibrium
Q > K means that there is an excess of products so the system will evolve towards the equilibrium consuming products (which means that you should subtract x from the products)
Q < K just the other way around, the system will shift to the right to reach equilibrium (subtract x from the reactants)


Does purified water corrode copper pipes?

Although it seems pretty clear that removing potentially corrosive compounds from drinking water will result in a net of no effect or possibly decreased corrosion, there is at least one consideration worthy of mention.

While metallic copper has a solubility that is well below most regulatory standards for drinking water (i.e. 2 mg/L in the UK), enough copper can still potentially dissolve to result in a significant loss of copper mass, such that pinholes and destructive oxidative processes can take place.

However, according to the UK's National Physics Laboratory, Good Practice Guide No. 120, "Avoidance of corrosion in plumbing systems", September 2011:

  "...the rate of corrosion and the release of copper ions slow rapidly in the first few weeks of operation as protective films form on the bore of the copper pipes."

Further:

  The surface films (often referred to as a patina) are composed of thin brown or black coloured copper oxide ($\ce{CuO or Cu2O}$) overlaid with a thicker green or turquoise basic copper carbonate layer. These layers, although protective, are not true passive layers (like those which form on stainless steels or aluminium and which are much thinner and invisible to the naked eye). It is when these films on copper do not form properly in the first place or become disrupted later that problems of pitting or blue water can occur.

According to this related document from the UK's "Foundation for Water Research", the low alkalinity and buffering capacity of purified water can result in conditions of slight acidity that can prevent the formation of the desired patina, and recommends the addition of 50 mg/L bicarbonate alkalinity in order to promote the development and maintenance of the protective copper oxide and carbonate layers.

To reiterate, my answer to the question is a net "No" with an interesting caveat.


There is no chemical difference between table salt and kosher salt - both are sodium chloride. Kosher salt just has bigger crystals. However, some table salt brands are iodized, meaning that iodide salts have been added to help us maintain a healthy dietary intake of the essential nutrient iodine. Kosher salt is never iodized.

Sea salt is a different matter. Since sea salt is obtained from sea water, it may have other minerals in it, which perhaps alters the taste.

Kosher salt is used in some recipes as a matter of taste - the bigger salt granules take longer do dissolve, and so there may still be little pockets of salty crunchiness. Kosher salt is also used when this property is necessary, for example in the curing of some meats. For more on the differences in culinary use between the various types of salt, you should consult Seasoned Advice.


How does hyperconjugation lead to the directing properties of alkyl group?

Why methyl group is 2,4-directing? This question describes in detail why the methyl group is o,p directing.

However, I think the major point of your misunderstanding is in the interpretation of resonance structures. Resonance structures are not stable structures in their own right.
A set of resonance structures can be used to represent bonding that cannot be represented using classic Lewis structures. The structures do not exist on their own but the true structure can be viewed as an 'average' of all the contributing resonance structures.

The electrons do not jump between different atoms but are instead delocalized over all of the atoms with a slightly higher chance of being found at the ortho and para positions in the case of toluene and a slightly lower chance of being found at on the methyl carbon. This gives rise to the increased electron density of the ortho and para positions and therefore the increased negative charge and similarly the decreased electron density on the methyl carbon and the positive charge.


How could scientists produce accurate chemical structures before they could capture images of molecules?

They ran chemical reactions on the molecule and identified the fragments.  Then like a jigsaw puzzle, using additional information derived from other chemical reactions, they pieced the molecule back together.  Some Nobel prizes were awarded for such structure determinations.


Which singlet dihalocarbene has the strongest carbon-halogen bond?

Singlet carbene is zwitteronic, meaning the $\mathrm{sp^2}$ orbital is doubly occupied resulting in a negative $\mathrm{sp^2}$ orbital and a slightly positive empty $\mathrm{p}$ orbital. Therefore it attracts electron on one orbital and repulses them around another orbital. The huge reactivity of the singlet carbene comes from this zwitterionic character.

Below is the picture electrostatic potential of on the left a singlet carbene with doubly occupied sp2 orbital and the right triplet carbene.

It can be argued in many ways how a halogen stabilizes the singlet state of carbene.

Two explanation usually used are:

Electronegative species increase the $\mathrm{s}$ character of σ orbital (Bent's Law).
Electron donation from the electron pairs into the empty carbene $\mathrm{p}$ orbital.

Though the second one is usually used, I suggest you follow the first (more physical). Adding electron negative substituents increases the s character of the doubly occupied sigma in-plane orbital of carbene stabilizing it, significantly decreasing the energy of the system. Following this concept you can see that flourine should provide the maximum stabilization and so on.

$$\ce{F} < \ce{Cl} < \ce{Br} \ldots$$


What type of test can I do to show that water is one of the products in a neutralisation reaction?

As was written in comments, it is nigh impossible to detect the formation of more water in an aquaeous solution. Unless you coloured the oxygen atoms in $\ce{NaOH}$ pink or purple you would have no clue which water molecule is new and which isn’t.

Say you had a $1~\mathrm{M}$ solution of both acid and base and used $25~\mathrm{ml}$ each; you would create $25~\mathrm{mmol}~\ce{H2O}$ in the reaction or approximately $0.45~\mathrm{g}$. That’s just about $1~\%$ of the total water added, so you may be able to detect a slight increase in weight if you distilled away the water.

Do not try to get your hopes high, though.


What chemical should I use to thin UV cure epoxy?

Adding solvents that need to evaporate, to your epoxy, can cause pinholing. You need a integratable solvent. Styrene monomer works well. You can get it from a fibreglass supply or by pyrolizing poly-styrene.

The monomer will combine with the epoxy so it won't need to evaporate out of the epoxy.


Quantum Chemistry: Small imaginary frequencies

It's probably okay, but an indication you don't have a true minimum in the geometry.

It's important to calculate frequencies to know if you have a true minimum of the potential energy surface. Presence of imaginary frequencies are indeed indications of saddle points.

That's simple theory. Imaginary frequencies indicate problems in the curvature with respect to that degree of freedom - it's simply not a minimum.

If you want a true minimum, you'll need to improve convergence and numerical accuracy.

Now the practical side...

I'm not familiar enough with ORCA to know if the translational and rotational degrees of freedom are removed from the frequency list. Some programs list these, and some do not. Obviously these include 5 or 6 degrees of freedom which are inherently low energy and due to numerical instability or incomplete geometric convergence can sometimes be imaginary frequencies. Your list includes three very small imaginary frequencies, so they could be translations or rotations.

(I also mention this, because ORCA indicates the first vibration is at ~14 $\mathrm{cm}^{-1}$ and the code is likely removing the lowest frequencies as translations and vibrations.)

For even moderate-sized molecules, and particularly for large molecules, it can be very, very hard to converge geometries to true minima.

Consider your list - with multiple vibrational frequencies of 14-30 $\mathrm{cm}^{-1}$. Those are incredibly low energy breathing modes. Gradients and energy changes along those dimensions will be small and a great number of optimization steps will be required.

This is one reason that geometry optimization methods "give up" if the gradients get very small and displacements get very small. It may not be a true minimum, but in all likelihood, it's very, very close and would require a large computational effort to get the true minimum.

Note: Don't read this and think "oh, I don't need to bother." Your needs may vary. If you're doing accurate work, you absolutely need to push hard on geometry convergence.


Assignment of the 13C NMR spectrum of 4'-bromoacetophenone

Maybe this isn't the answer you are looking for, but $\ce{^13C}$ NMR shifts are often not amenable towards analysis via electron-withdrawing / electron-donating effects—in my experience this is especially so for aromatic rings. Any such analysis is quite unreliable and can lead to wrong conclusions if you try to use it for assigning peaks.
The numbers that ChemDraw predicts are based on empirical tables of chemical shift corrections. These are experimentally observed chemical shift changes that occur at each position of the benzene ring upon substitution. For a ring with multiple substituents you can approximately add these up. See e.g. Hans Reich's website for such tables.
See also long's answer to "Assigning the 13C NMR spectrum of bromobenzene and 4-bromobenzophenone": although most of the discussion is on the ipso carbon (C6 in your labelling), the general point is the same: it's often not trivial to rationalise or predict chemical shift changes in aromatic rings.
[There might be more sophisticated theoretical analyses out there based on diamagnetic and paramagnetic shielding arguments, but I'm unfortunately not aware of these.]


How might ammonia be created in this mechanochemical reaction?

Fine hexagonal boron nitride powder is hygroscopic, so this implies there might be some water in your system. According to Reaction of Hexagonal Boron Nitride Nano-crystals under Mild
Hydrothermal Conditions "The reaction between water and hBN nano-crystals starts at very low pressure and temperature, i.e. 220 C and 1.0 MPa [~10 atmospheres]."

Those conditions are likely to be met in a ball-mill or vibratory tumbler, at least momentarily at the points of impact (in a similar way, cavitation can heat water to luminescence!).

A simple test would be to preheat the tumbler and its contents to 150 or 200 C for an hour or two, then seal it airtight before running the tumbling cycle. If there's no ammonia odor, then it indicates that adsorbed water in the hBN had been reacting with it, perhaps catalyzed by the copper bearings, as well as by impact heating and pressure.


Is the d-block ever part of the valence electrons?

The answer is merely a matter of definition of "valence electrons".

Many texts define "valence electrons" in a way that explicitly includes d electrons of unfilled (but not filled) d subshells.

For example: Chemistry structure and dynamics by Spencer et al. at page 125:[1]

  We can define valence electrons as electrons on an atom that are not present in the previous rare gas, ignoring filled d or f subshells.

Many books published in the last 10 years use this definition.

That d electrons may be valence electrons is also supported by the 18-Electron rule (at least to the extent that there is such a rule).

[1] James N. Spencer, George M. Bodner, Lyman H. Rickard: Chemistry: Structure and Dynamics, 5th Edition. John Wiley & Sons: 2010. ISBN: 978-0-470-58711-9


What does the negative superscript in the term symbol for ground state Oxygen stand for?

You're right that it has to do with a mirror plane. The negative sign means that the overall electronic wavefunction changes sign when reflected in a mirror plane that includes the internuclear axis.

For ground-state oxygen, reflecting in a mirror plane will change the sign of one unpaired $\pi$ electron (the one that sticks out of the mirror plane) but not the other, resulting in an overall change in sign. (This is slightly simplified, a more rigorous explanation would take into account the complex-valued orbitals rather than their real-valued linear combinations.)


Why do you drop the charge in the crossover method?

Let's figure out the molecular formula for calcium chloride.

Calcium has a +2 charge. Chloride has a -1 charge.

So with the crossover rule we get

$$\ce{Ca_{-1}Cl_{+2}}$$

But for molecular formulas a -1 is impossible. The coefficients must be positive integers (usually...). Also have a 1 is superfluous, so it is just $\ce{Ca}$ and not $\ce{Ca_1}$.

Now for the chlorine the + is superfluous, so it is just $\ce{Cl_2}$ and not $\ce{Cl_{+2}}$.

I noted that the coefficients must be positive integers. A case where that is not true is when there is a double salt. Say a mixture of NaCl and KCl. The cations and the anions must balance so the formula can be written as

$$\ce{Na}_x \ce{K}_{1-x}\ce{Cl}$$

thus $(x) + (1-x) = 1$ and the charges balance.

Frankly I find this whole rule thing much to do about nothing. The net charge must be 0. Calcium has a +2 charge and Chlorine has a -1 charge. So you need two -1 charges to balance the +2 charge. Thus $\ce{CaCl2}$. Note that the coefficients must be the lowest common multiple, so it is $\ce{CaCl2}$ and not $\ce{Ca2Cl4}$ which also has a net 0 charge.

So to me the only difficult part is figuring out the charges on the ions. Once you know that, the rest is easy. Making a special rule just seems to be overkill.


How chemists find out the electronic properties of elements (atomic numbers)

One could do it in the following way.

The atomic number $Z$ can be determined by examining the x-ray spectrum of an atom.
The frequency $f$ of the so-called K-alpha line is related to the atomic number $Z$ by Moseley's law
$$
    \sqrt f = k_1 \cdot \left(Z - k_2\right) \, .
$$
The atom is electrically neutral, so the number of electrons is obviously calculated from the atomic number.
And the number of neutrons is also simply to choose: the only requirement at this stage would be the correct atomic weight for the element in question.


For an ideal gas mixture the total pressure is the sum of the partial pressures of all the gases in the mixture. To make intuitive sense of this, consider that since it is a gas mixture, all the gases essentially have access to the whole container, and so will homogenously fill it. Each gas wil the exert a certain pressure on the container walls. The sum the pressures from all the contributing gases will then equal the total pressure the container walls experience.

We also have the relationship $$p_\text{x}/p_\text{total}=n_\text{x}/n_\text{total}$$

The potential yield for the rxn is 2 mol ($\ce{SO2}$ limiting reactant). To get a 80% yield, we therefore have 1.6 mol $\ce{SO3}$ at equilibrium. Therefore there will be 0.4 mol $\ce{SO2}$ and 1.2 mol $\ce{O2}$ left at equilibrium. $n_\text{total}=3.2$

$n_\ce{SO2}/n_\text{total}=0.4/3.2=0.125$

$n_\ce{SO3}/n_\text{total}=1.6/3.2=0.5$

$n_\ce{O2}/n_\text{total}=1.2/3.2=0.375$

To find the partial pressures of the individual gases, we combine the ratios above with relationship 1

$p_\ce{SO2}=0.125\times p_\text{total}$
$p_\ce{SO3}=0.5\times p_\text{total}$
$p_\ce{O2}=0.375\times p_\text{total}$

As you know, the equilibrium can be expressed with partial pressures, and is constant for all changes in pressure. $K_\mathrm{p}=p_\ce{SO_3}^2/p_\ce{SO_2}^2p_\ce{O_2}=0.13$  at 830 Celsius

Plugging in the partial pressures and solving for p(total):

$(0.5x)^2/((0.125x)^2\times0.375x)=0.13$
$\implies x=328.205$ (The unit can be given by the units of the eq. constant.)


How did the temperature in the solution with a lower concentration of CuCl2 increased in a faster rate?

The reaction $\ce{Cu^2+ + 4 Cl- <=> CuCl4^2-}$ may lead to slowing down reaction with aluminium, decreasing the equilibrium concentration of $\ce{Cu^2+}$ in $\pu{0.8 M}$ $\ce{CuCl2}$. That all would lead to lower peak temperature due bigger heat dissipation before reaching maximal temperature.

Contributing factors may be

decreasing of chemical activity of water for the electrode surface reaction $\ce{Al(s) + 6 H2O -> [Al(H2O)6]^3+(aq) + 3 e-}$. These  electrons, charging the electrode, then reduce copper: $\ce{[Cu(H2O)n]^2+(aq) + 2 e- -> Cu(s) + n H2O}$
increasing of viscosity, therefore decreasing of diffusion coefficients and another decreasing of the reaction rate.


Calculating change in enthalpy of combustion of 1.12g of hexane

I can't figure out exactly what you are trying to calculate, but if you want to know how much energy is created when burning 1.15 g of hexane by measuring an temperature increase of water, then this is the right way:

1.) 0.2 kg Water was icreased by 24 °C:
U = 0.2 kg * 4187 J/(kgK) * 24 K = 20,100 J = 20.100 kJ

2.) n(Hexane) = 1.15 g / (86.18 g/mol) = 0.013 mol

3.) 1.15 g caused the temperature increase of 24 °C:
20.100 kJ / 0.013 mol = 1,510 kJ/mol

So you will get 1.510 MJ out of burning 1 mol (86.18 g) hexane.


Can Hückel's rule be applied to inorganic compounds?

Huckel's rule can be applied to "inorganic" cyclic molecules but only if they have a delocalised pi system. They also have to have flat (or close to flat) rings.

Recently some N5 rings which 6 pi electrons have been made, these are "inorganic" aromatic rings.

Also borazine has been the subject of considerable debate, it has been discussed on stack exchange before. It is the subject of a interesting paper: Islas, R.; Chamorro, E.; Robles, J.; Heine, T.; Santos, J. C.; Merino, G. Borazine: to be or not to be aromatic. Struct Chem 2007, 18 (6), 833–839. DOI: 10.1007/s11224-007-9229-z.


Empirical Evidence and Assignment of Orbital Energies (Electron Filling)

As far as I know, it was a long and complex undertaking. You can't look at a single atomic species by itself. The understanding of level assignments to series slowly grew starting from the simplest cases.

Of course the first was hydrogen, where - at least initially - no series are discernible but one - there is almost perfect degeneration on l
. But that helped, as it allowed the first step, the one from wavelengths λ (or wavenumbers k=1/λ) to terms, later identified with energy levels. It was Ritz' combination principle:
k=Tm−Tn.
Already at this introductory level, a difficulty was to be overcome: the difference between emission and absorption spectra. The former is much richer. This was explained as a consequence of atoms in absorption initially being in the ground state alone, whereas in emission photons are emitted by excited atoms jumping to a lower state, not necessarily the ground one. For hydrogen this is especially notable, as there are no absorption lines in the visible region - the Balmer series has n=2

.

Then alkaline spectra were explored. In absorption spectra only lines Tn−T1
are observable, where T1 is what later would become the ground state energy (divided by hc). Note that here n=1, meaning the ground state, but this is not the principal quantum number of hydrogen-like classification, which is 2 for Li, 3 for Na, etc...

The main difference between hydrogen and alkaline spectra is that in the latter l degeneracy is broken. This required to separate terms in several series. Surely you know the origin of symbols S, P, D,... related to a different appearance of lines starting from different series towards the same final term. An instance is transitions S-P and D-P (in emission), In absorption only S-P transition is visible, giving rise to the principal series (therefrom P-terms).

But a second feature appears in alkaline spectra: the so-called "fine structure". It exists in hydrogen lines too, but is much less prominent, whereas the famous sodium doublet requires a modest resolving power to be seen. Its interpretation required the discovery of electron spin and of L-S coupling. The effect was a doubling of columns for all series, S excepted.

Well, this is a rough history, as I'm able to follow. Maybe a book could be of help: G. Herzberg, "Atomic Spectra and Atomic Structure". It's outdated as far as theory is concerned, but thanks to its publication date (first edition 1936) is nearer to the times when the studies were carried out.


How does the ionic strength formula work? (simple version)

Firstly; why is there a factor of a half out the front?

To avoid double-counting. For a salt such as $\ce{NaCl}$ comprising a monovalent cation and a monovalent anion, the ionic strength is defined such that it should be equal to the molar concentration of the salt, as opposed to the sum of both the anion and cation concentrations.  Therefore, since such a salt has two constituents, each constituent contributes half of the total ionic strength.

  Secondly; why do we square the charge number of the ion?

For this there's not much choice but to turn to the Poisson-Boltzman equation, which provides a differential expression for the electric field inside the electric double-layer. (Sorry, 5-year-old version of Jekowl!) Below is the version I have in my graduate transport book, Eq. (15.4-6):

$$
\nabla^2\Phi = -{F\over \epsilon} \sum_i{z_iC_{i\infty}\exp{\left(-{z_iF\over RT}\Phi\right)}}
$$

$\Phi$ is the electric potential, $\mathrm{J} \over \mathrm{C}$ or $\mathrm{V}$
$F$ is Faraday's constant, $96,485\ {C \over \mathrm{mol}\ q}$
$\epsilon$ is the electrical permittivity, $\mathrm{C}\over \mathrm{V\ m}$
$z_i$ is the valence of ion $i$, $\mathrm{mol}\ q \over \mathrm{mol\ ion}$
$C_{i\infty}$ is the concentration of ion $i$ far from the surface of interest, $\mathrm{mol} \over \mathrm{m^3}$
$R$ is the ideal gas constant, $8.314\ {\mathrm{J} \over \mathrm{mol\ K}}$
$T$ is temperature, $\mathrm{K}$
$q$ is the elementary charge

If the magnitude of the electric field is small such that $\left|z_iF\Phi/RT\right|\ll 1$, the exponential can be expanded as a Taylor series:

$$
\nabla^2\Phi = -{F\over \epsilon} \sum_i{z_iC_{i\infty}\left(1-{z_iF\over RT}\Phi+\dots\right)}
$$

If you then pull everything inside the parentheses, ignoring signs the second term becomes:

$$
C_{i\infty}z_i^2{F^2\over \epsilon RT}\Phi
$$

The second-order dependence of $z_i$ is readily visible in the above.


Why does aluminium chloride react with water in 2 different ways?

Yes, adding a small amount of water to anhydrous AlCl3 is different from an aqueous reaction. Here is an excellent educational institution's answer in the first case, depicting the action of water on anhydrous aluminium chloride, to quote:

If a small amount of water - a few drops - is added to anhydrous aluminium chloride there is a hissing sound and hydrogen chloride gas is given off as steamy acidic fumes. The reaction that occurs will depend on the amount of water added, but is probably
$\ce{AlCl3(s) + H2O(l) → AlCl2OH(s) + HCl(g)}$

In the second case, aluminium chloride in aqueous solution appears to undergo a two-stage reaction, starting with the formation of the hexaaquaaluminium ions, to quote again:

If aluminium chloride is dissolved in a large amount of water the solution is acidic, but this has nothing to do with formation of hydrochloric acid. The solution contains hydrated aluminium ions and chloride ions:
$\ce{AlCl3(s) + aq → [Al(H2O)6]^{3+}(aq) + 3 Cl -(aq) }$

This is followed by an acidic hydrolysis paralleling transition metals:

The hexaqua complex ion behaves exactly like ions of similar type formed from transition metals; the small, highly charged metal ion polarises (withdraws electron density from) the water molecules that are attached to the aluminium ion through dative covalent bonds. This makes the hydrogen atoms d+ and susceptible to attack from solvent water, which is acting as a base. The complex ion is deprotonated, causing the solution to be acidic from the formation of hydroxonium ions H3O+:
$\ce{[Al(H2O)6]^{3+}(aq)  + H2O(l) →  [Al(H2O)5OH]^{2+}(aq) + H3O+(aq)}$


Why does bond angle decrease in the order H2O, H2S, H2Se?

Here are the $\ce{H-X-H}$ bond angles and the $\ce{H-X}$ bond lengths:
\begin{array}{lcc}
\text{molecule} & \text{bond angle}/^\circ & \text{bond length}/\pu{pm}\\
\hline
\ce{H2O} & 104.5 &  96 \\
\ce{H2S} &  92.3 & 134 \\
\ce{H2Se}&  91.0 & 146 \\
\hline
\end{array}

The traditional textbook explanation would argue that the orbitals in the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions, thereby forcing the $\ce{H-X-H}$ angle to contract slightly. So instead of the $\ce{H-O-H}$ angle being the perfect tetrahedral angle ($109.5^\circ$) it is slightly reduced to $104.5^\circ$.  On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ have no orbital hybridization.  That is, The $\ce{S-H}$ and $\ce{Se-H}$ bonds use pure $\ce{p}$-orbitals from sulfur and selenium respectively.  Two $\ce{p}$-orbitals are used, one for each of the two $\ce{X-H}$ bonds; this leaves another $\ce{p}$-orbital and an $\ce{s}$-orbital to hold the two lone pairs of electrons.  If the $\ce{S-H}$ and $\ce{Se-H}$ bonds used pure $\ce{p}$-orbitals we would expect an $\ce{H-X-H}$ interorbital angle of $90^\circ$.  We see from the above table that we are very close to the measured values.  We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two $\ce{X-H}$ bonds the angle opens up a bit wider.  This explanation would be consistent with the $\ce{H-S-H}$ angle being slightly larger than the corresponding $\ce{H-Se-H}$ angle. Since the $\ce{H-Se}$ bond is longer then the $\ce{H-S}$ bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.

The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum.  The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the orbitals in the two $\ce{O-H}$ bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital.


Why do you use acetic acid instead of water to create your standard riboflavin solutions?

From Sigma product information:

  The flavins are light sensitive and unstable in alkaline solutions.  The dry solid is stable to diffuse light, but is highly photolabile in solution, especially in alkaline solutions. Neutral and acidic solutions are stable in the dark (3% decomposition per month at 27 °C at  pH 6.0).  Acidic solutions may be sterilized by autoclaving.

So making a slightly acidic solution probably helps to prevent decomposition. If you dissolve it in pure water without a buffer, the pH is not well defined.


So I could have sworn I've cleaned stuff off with saliva before, but I wanted to make sure, so I ran a quick impromptu experiment.

I slathered the underside of my left arm with some maple syrup I had lying around. I waited about 10 minutes for it to dry, then I mentally sectioned off my arm into three parts and cleaned each part off in different ways:

The section closest to my hand was cleaned by putting my other hand under running water and wiping the syrup off (basically, getting my right hand wet and making wiping motions across the arm, then returning the hand to the water, until I felt like I was no longer making any progress).

The middle section was left untouched.

The section closest to my elbow was cleaned by licking my hand, then using it to wipe that section down (essentially the same as the water, except I used my mouth as the source of liquid).

After the cleaning, I let everything dry again for 3 minutes. Then I placed some toilet paper over my entire arm and attempted to remove it. Here are the results:

(Sorry for the horrid image quality)

Unfortunately, I didn't have any skin marking tools on me at the time (I use pretty much only water-soluble inks), so I had to add the lines afterwards.

As we can see, the section of my arm that was cleaned with water and with saliva are pretty much TP free, aside from a small section near my elbow that I think was just me missing some spots. The section cleaned with water was similarly TP-free. The middle section is pretty heavily TP-ed. What you're seeing is what got left behind when I attempted to rip the TP free.

So it seems like saliva actually can clean off stickiness pretty well! This is backed up by the comment that Michiel made, which points out that saliva is 99.5% water.

That being said, running this experiment gave me a possible explanation for why it might seem like it doesn't: it takes a LOT of saliva to do this!

With the water, I was easily able to wash off all of the syrup off of my hand and get a new coating of water on my fingers. Unfortunately, I can't generate that much saliva at once, and so I couldn't wash off the sugar that was already on my fingers that effectively. That made the saliva cleaning take a LOT longer than washing with water, both in terms of how many wipes it took to clean off most of the stuff and how much time it took overall (sadly, I cannot drool at the same rate a bathroom sink outputs water).

There's another possible explanation that springs to mind: when you have syrupy stuff stuck to you, you probably just ate some of the syrupy stuff. That means your mouth probably still has a good bit of sugar in it, which would also make it harder to clean off additional sugar.

Still, based off of this experiment, I feel fairly confident in saying that saliva cleans off sugar stickiness almost as well as regular old water, as long as you have enough of it lying around.

NOTE: I initially tried this with a flavored whiskey, hoping the alcohol would help it dry faster. Terrible mistake! Not only was there not enough sugar in the whiskey to get any sort of real stickiness, most of it basically ran off of my arm right away, so I ended up losing a lot of it. Save your flavored sprits for drinking, use a syrup!


Why does the dipole moment of molecules go from positive to negative?

You have assessed the problem correctly. Physicists and chemists use just the opposite directions. Keep in mind that both are arbitrary choices, however the physicist's convention is consistent with electrostatics. All you have to do is to be aware of this direction in the chemistry literature. I have rarely seen any student ever thinking about it which is rather sad.

If you want more details read this article Misconceptions in Sign Conventions: Flipping the Electric Dipole Moment

  "Most chemists draw the direction of an electric dipole backwards. This inconsistency in sign convention is more than a trivial oversight or matter of convenience. We are introducing an incorrect convention that is central to our student’s understanding of fundamental thermodynamics; which way is up? As a student matures from general chemistry to organic through analytical, physical, inorganic, and biochemistry they are reintroduced to similar, powerful concepts that hold our discipline together. However, this reconnection of chemical principles comes with each subdiscipline’s own idiosyncrasies and a highly contextual framework. Reexamination of a central concept from the perspective of a new subdiscipline should not introduce misconceptions about that concept. When misconceptions introduced through chemical language can be avoided, we should change the way we speak."


First of all, there are no such thing as a "perfect" crystal.

Defects occur from both human error and the experimental setup.

Defects from human error are primarily due to handling issues, which can never be perfected, especially not if the crystals are solvated.

Experimental defects can occur from either the diffraction experiment or the crystal growth experiment. Impurities can affect crystal growth by displacing the molecule of interest in the lattice and unless you are attempting neutron diffraction, the crystal will be significantly damaged during the data collection.

In my field of protein x-ray crystallography, the typical crystals can have diffraction ranging from nothing at all to ~0.8 Angstrom range (best cases) and as such you can have beautiful crystals that have no diffraction and ugly crystals that diffract very well.

Crystal defects can be perfectly acceptable and are a so common that we usually don't think about them too much.


Yes, one would try to find points or lines of symmetry in general to check for equivalent ring atoms. It immensely helps to have a model and a lot of practice.

In your specific example, notice that you have two lines of symmetry in your (idealized) diagram: one passing through the sp3 carbons, and one crossing the two double bonds. You can thus "fold" the diagram into "quarters", leaving you with two non-equivalent atoms.


Aqueous copper and how it reacts with NH3

[Fe(H2O)6]2+ is a high-spin d6 complex, while [Fe(H2O)2(NH3)4]2+ is a low-spin d6 complex. The ligand substitution reaction is not favourable in terms of spin, so aqueous Fe2+ does not react with excess ammonia.

However, both [Cu(H2O)6]2+ and [Cu(H2O)2(NH3)4]2+ are d9 complexes. Their spins are the same, and being a stronger field ligand, NH3 increases the crystal field stabilization energy, so ligand substitution is favourable.


Does the translational part of internal energy mean the kinetic energy of the gas?

Thanks to @IvanNeretin for pointing me in the right direction.

Imagine a box full of an ideal gas. Let the box and gas together be of a mass $m$.

Now imagine that the box is moving with a velocity $v$. Thus the system will have a kinetic energy $\frac{1}{2}mv^2$. Now this energy is of the system as a whole and is therefore not counted as internal energy.

Similarly imagine the box placed at height $h$. It will have a potential energy $mgh$. Again this will not be counted as internal energy.

When talking about diatomic or triatomic gases, we will say that that the total kinetic energy is the sum of all the possible ways in which the particle can move in space. This includes motion like translational motion, rotational motion and vibrational motion. The number of dimensions of such motion are called the degrees of freedom($f$) of the molecule and the total kinetic energy of the system is given by

$$U = \frac{f}{2}nRT + \phi$$

where the first term represents the total kinetic energy and the last one represents all the other energies associated with the gas molecules

For example for a diatomic gas, $f=5$ because it can move in $3$ dimensions and rotate in $2$. So $$U = \frac{5}{2}nRT + \phi$$

For gases in high temperature, the degree of freedom would increase by due to the vibrational motion of the gas particles.

Wikipedia


According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical units that are linked by di- or polyvalent groups can be named in two ways:

by ordinary substitutive nomenclature, in which one of the parent structures is chosen as the senior parent structure and the remainder of the structure is expressed by substituent prefixes
by multiplicative nomenclature, in which two or more parent structures are connected by symmetrical or unsymmetrical single or concatenated substituent groups

Both methods yield correct names, which may be used in general nomenclature. However, the IUPAC recommendations define a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations.

P-45.1 MULTIPLICATION OF IDENTICAL SENIOR PARENT STRUCTURES
P-45.1.1 Multiplicative nomenclature is senior to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical senior parent structures, other than alkanes, in the name of the parent structure (see P-51.3.1). In most cases multiplicative names are shorter than regular substitutive names. A preferred IUPAC name is generated by multiplicative nomenclature when the following criteria for its use are met (see P-51.2.3).
(1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical; and
(2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
(3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.
When these conditions are not met, substitutive nomenclature generates preferred IUPAC names.

The given structure fulfils these requirements; therefore, the preferred IUPAC name is generated by multiplicative nomenclature as follows.

P-15.3.1.3 Multiplicative name formation
Multiplicative names are formed in accordance with the number of occurrences of identical structural units as defined in P-15.3.1.1 and the relationship of the linking multiplicative substituent group to the identical structural units.
When a compound contains identical structural units as defined in P-15.3.1.1 linked by a symmetrical simple, compound, complex, or concatenated multiplicative group (a di- or polyvalent substituent group), it is named by stating successively:
(a) the locants for the positions of substitution of the linking multiplicative substituent atom or group to the identical parent structural unit (the locant 1 is omitted when alone in the name of a mononuclear parent hydride);

Here: 3,3'

(b) the name of the linking multiplicative substituent atom or group;

Here: oxybis(ethane-2,1-diyloxy) for $\ce{-O-CH2-CH2-O-CH2-CH2-O-{}}$

(c) the numerical prefix ‘di’, ‘tri’, etc.; and/or ‘bis-‘, ‘tris-‘, etc., with no elision of the final vowel before the name of the identical parent structural unit;

Here: di

(d) the name of one of the identical structural units including the principal characteristic group and substituents, if any, enclosed in appropriate enclosing marks (see P-16.5).

Here: propanoic acid

The numbering of the identical parent structural unit is retained and, when there is a choice, the locants of the point of substitution by the linking multiplicative substituent groups on the identical parent structure are as low as possible.

Here: 3,3'-(…)dipropanoic acid
Therefore, the preferred IUPAC name is 3,3'-[oxybis(ethane-2,1-diyloxy)]dipropanoic acid.

By way of comparison, the Blue Book contains the similar example 2,2'-[oxybis(ethane-2,1-diyloxy)]diacetic acid.


How to calculate he equilibrium constant change for acid-base reactions in a different solvent?

For all solvents that can dissociate into ions, a dissociation constant can indeed be measured. Expect them to be unpredictable in each direction.

Furthermore, the $\mathrm pK_\mathrm a$ values you know are typically those measured in water, although some organic chemists rely more heavily on the DMSO tables e.g. the one by Evans. Just a quick look at the first column shows you that there is no way you can predict the $\mathrm pK_\mathrm a$ value in one solvent from the one in a different solvent:

$$\begin{array}{lcc}\hline
\text{compound} & \mathrm pK_\mathrm a(\ce{H2O}) & \mathrm pK_\mathrm a(\ce{DMSO}) \\ \hline
\ce{HF} & 3.17 & 15\phantom{.0}\\
\ce{HCN} & 9.4\phantom{0} & 12.9\\ \hline\end{array}$$

While $\ce{HF}$ is the stronger acid in water, $\ce{HCN}$ is the stronger acid in $\ce{DMSO}$.

Couple this with a different dissociation constant means that you can calculate nothing ab initio.


Nucleophilic addition of grignard reagent

The attacking of the Grignard reagent in these kinds of $\alpha$-$\beta$ unsaturated carbonyl compounds depends on two factors mainly. $1$) Hardness of the acid centre $2$) Steric repulsion between the surrounding groups and incoming nucleophile.When the attacking nucleophiles are Grignard reagents,as per your equation, they are generally hard bases.According to Pearson's HSAB Theory, Hard bases prefer to bind to the Hard acid centre and soft bases prefer to bind to the soft acid centre.  In your compound i.e in $\alpha$-$\beta$ unsaturated carbonyl compounds, the carbonyl carbon centre is the Harder acid centre, whereas the $\ce{C=C}$ double bond centre is the softer acid centre.So, Grignard reagents try to first bind with the carbonyl carbons unless there is no other destabilising factor.The second governing factor is the steric crowding, which can sometimes make the attacking nucleophile attach to the least sterically crowded region even though it is violating HSAB theory. So, the addition compound actually depends on these two competititive factors.  Now, in your given compound the attacking nucleophile is $\ce{PhMgBr}$.It is undoubtedly Hard Base. But if it were to bind to the carbonyl carbon i.e. $\ce{C=O}$ double bond, there will be huge repulsion between the lone pair of oxygen and $\pi$-electron cloud of Phenyl ring which actually makes the previously stated second factor to dominate. So, due to this huge steric repulsion, The attacking nucleophile has to attack the less sterically crowded(Note: The repulsion is also there at the double bonded carbon centre , but repulsion between methyl groups and phenyl rings are only due to inefficient spatial orientation which is definitely smaller than inter electronic strong columbic repulsion forces in close vicinity in the case of $\ce{C=O}$) softer basic $\ce{C=C}$ centre which infact gives the product.


Why are water droplets shaped like that?

From the Wikipedia article on surface tension:

  Surface tension is responsible for the shape of liquid droplets. Although easily deformed, droplets of water tend to be pulled into a spherical shape by the cohesive forces of the surface layer. In the absence of other forces, including gravity, drops of virtually all liquids would be approximately spherical. The spherical shape minimizes the necessary "wall tension" of the surface layer according to Laplace's law.

In short, the more surface tension is, the rounder shapes of water you get. And the opposite goes for gravitational potential energy: The lesser gravitational acceleration results in more spherical droplets of water.

The symbol for surface tension is $\gamma$.

  $\gamma (\ce{H2O}) = 72.8~\mathrm{dyn~cm^{-1}}$ (at $20~\mathrm{^\circ C}$)
   $\gamma (\text{mercury}) = 465~\mathrm{dyn~cm^{-1}}$ (at $20~\mathrm{^\circ C}$)[1]

That's the reason you hardly ever see mercury drops out of their spherical shape.

In short

  The spherical shape minimizes then necessary "wall tension" of the surface layer according to Laplace's law.[2]

Oh and I almost forgot: This great article - Antonin Marchand, Joost H. Weijs, Jacco H. Snoeijer, and Bruno Andreott, Why is surface tension a force parallel to the interface?. Physique et Mecanique des Milieux Heterogenes, 2012 - is very nice in case you wanted to do additional study in the case.

References:

R Nave. Surface Tension. URL (accessed April 25th, 2018)
R Nave. Surface Tension and Droplets. URL (accessed April 25th, 2018)


This is somewhat an elementary discussion, but viscosity is typically measured in some sort of viscometer. A typical one is the Oswald viscometer shown below.

Fluid is poured into the apparatus through tube at $f$ to fill neck and bulb at the bottom. The fluid is then sucked up above line $c$. The fluid is then released and drains down the tube to line $c$ at which time the timing for the flow rate starts. When the fluid falls to the line at $d$ is reached then a flow rate can be measured. By using samples of known viscosity, the flow time can be calibrated to the viscosity of the sample.

Now such measurements are in units of pascal-seconds. For reference a short list will be provided.

     Substance           mPa-s
     acetone                0.0306
     water                  0.0894
     sulfuric acid          0.242
     olive oil             81.
     motor oil SAE 40     319.
     glycerol           1,200
     corn syrup         1,380.6
     pitch              2.3×10^8

Wikipedia also lists tables for Viscosity of selected substances

Now let's defined three ranges of viscosity:

Low Viscosity -    0.1 mPa-s >= sample
Medium Viscosity - 0.1 mPa-s <= sample <=1.0 mPa-s
High Viscosity - sample > 1.0 mPa-s

Of course the "exact" ranges are arbitrary. So a viscous liquid is a High Viscosity liquid on our scale.


Why is ethoxyethane the preferred IUPAC name for diethyl ether rather than 1,1'-oxydiethane?

The deciding criterion is in the preamble to P-51.3.1, not in the numbered criteria: “other than alkanes.” The identical parent structure in diethyl ether is a straight-chain alkane, so multiplicative nomenclature does not apply.


The issue is that the rate equation is determined not by the reaction equation, but rather by the actual mechanism that lies beneath it. It turns out that reactions that appear fairly simple, such as the burning of methane in oxygen in the flame of a Bunsen burner, do not necessarily take place in a simple manner, I understand that the even the reaction for the Bunsen flame can in real life contain a very large number of steps. Depending on the reaction, different steps will require different reactants, and will take place at different speeds. Maybe for the reaction $\ce{A + B->AB}$, some step that takes a long time involves $\ce{A}$ decomposing on heating, the products of which go on to quickly react further to eventually form $\ce{AB}$. This step that takes a long time is called the rate-determining step, and decides the form of the rate equation. It may be that this causes only $[\ce{A}]$ to feature in the rate equation.

In this case however, the rate equation is given to you by the comments to the right of the equation. Just because we write it as $\ce{2A}$ or $\ce{A+A}$ doesn't make a difference, it's just how someone decided to write the equation, which does not necessarily reflect the steps of the mechanism. In fact, if a mechanism involved a slow step $\ce{A + A -> B}$, the reaction would be second order with respect to $\ce{A}$.

Hopefully that helps.