Why must polymers have a repeating unit?

Entropy would favour a random orientation of each monomer unit when building up the chain, however whenever you have two different substituents at the ethylene unit, then electrostatics and sterical hindrance will make one side be preferred to make the connection to the growing chain end.

Polymers that do grow via chain growth therefore have a regular structure, but there can be defects. The defect density depends on how strong the preference for one side is during the polymerisation reaction. You can have as little as one in ten thousand. That's actually a bit of a problem, because the regularity often allows polymers to crystallise, which is important for the technical application. You want the number of defects always to be the same, because too high crystalinity makes your material brittle, too low makes it soft.

Polymers that make step growth (any two molecules randomly connect with their end groups, until you have a long chain) could be different, but also there often a regular structure occurs, because of A-B + A-B , either the A+B like to react, or A+A and B+B.


For a typical diprotic base buffer system you can use sodium bicarbonate. Adding hydrochloric acid to sodium carbonate solution

The overall equation for the reaction between sodium carbonate solution and dilute hydrochloric acid is:

$$\ce{Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + CO2(g) + H2O(l)}$$

If you had the two solutions of the same concentration, you would have to use twice the volume of hydrochloric acid to reach the equivalence point - because of the 1 : 2 ratio in the equation

Take a look at he sample titration curve:

The sense of diprotism of a base, is revealed through its ability to accept the two protons from hydrochloric acid to form water.

Referencing and diagrams from

pH (TITRATION) CURVES


Home made dispersion/solution with detectable >0.1μm particles

Try to get some clay and shake it vigorously with water. Let it settle for a several hours. Test your filter with the supernatant water and collect the filtrate in a very clean glass tumbler. Colloids have an interesting property of scattering light. In a dark room, try to shine light (ordinary flashlight might work or perhaps an ordinary pointer used in presentations/offices). Be careful, never ever look at the pointer directly. Look at the filtrate at $90^o$ (right angles to the light beam), if you see a light beam travelling in water, it means that colloidal particles are seeping through the filtrate and the filter is damaged.

Read more about the Tyndall Effect https://www.sciencesource.com/archive/Tyndall-Effect-SS2423500.html or on Wikipedia before doing anything.


1) Do NOT heat the solution, if you want chlorine out. Electrolysis of hot $\ce{NaCl}$ solution produces $\ce{NaClO3}$, then $\ce{NaClO4}$, not chlorine. You apparantly don't want this.

2) traditionally, in such setups some barrier between cathode and anode spaces is used. The easiest one would be unglazed ceramic pot (with no holes and no direct contact of anode and cathode solutions), though I'm not sure if it will work with chlorine ions. It is recommended to perform electrolisys of $\ce{MgSO4}$ solution first. In this case chlorine ions would migrate from cathode space. With barrier introduced you may expect chlorine to evolve from anode space and sodium hydroxide to accumulate in cathode space.

3) do NOT blow air atop sodium hydroxide solution. It reacts with $\ce{CO2}$ from air.


Can substituted acetone undergo Cannizzaro reaction?

As far as I'm aware, the Cannizzaro reaction can't occur with ketones, only aldehydes. The reaction mechanism involves the transfer of hydride from the former carbonyl carbon (of the tetrahedral intermediate resulting from hydroxide's nucleophilic addition to the carbonyl carbon), during which step the carbonyl reforms, yielding a carboxylic acid. Since a ketone, by definition, has no hydrogens bonded to the carbonyl carbon, it obviously can't undergo such a reaction. I would expect ketones with no $\alpha$-hydrogens to be essentially unreactive with hydroxide, since aldol reactions and keto-enol tautomerism are impossible with no $\alpha$-hydrogens, and most hydrates are unstable.


Ether nomenclature: Oxygens on alkyls and alkyls on alkyls

Names for ethers can be formed by five different methods. Furthermore, various traditional names are retained for use in general nomenclature.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), preferred IUPAC names of ethers $\ce{R-O-R'}$, when $\ce{R}$ and $\ce{R'}$ are both aliphatic groups, are formed by substitutive nomenclature. In substitutive nomenclature, when $\ce{R}$ is different from $\ce{R'}$, $\ce{RH}$ is chosen as parent hydride and $\ce{R'-O-{}}$ is cited as a substituent to it.

  P-63.2.2.1.1 Substituent prefix names for $\ce{R'-O-{}}$ groups are formed by concatenation, i.e., by adding the prefix ‘oxy’ to the substituent prefix name for the group $\ce{R'}$. (…)

For the first example in the question, the name of the parent hydride $\ce{RH}$ is ‘decane’.
The substituent prefix name for the group $\ce{R'-{}}$ is ‘2-ethylheptyl’.
Thus, the substituent prefix names for the $\ce{R'-O-{}}$ group is ‘(2-ethylheptyl)oxy’.
Therefore, the preferred IUPAC name of the compound is ‘4-[(2-ethylheptyl)oxy]decane’.

For the second example in the question, the name of the parent hydride $\ce{RH}$ is ‘decane’.
The substituent prefix name for the group $\ce{R'-{}}$ is ‘3-propyloctan-2-yl’.
Thus, the substituent prefix names for the $\ce{R'-O-{}}$ group is ‘(3-propyloctan-2-yl)oxy’.
Therefore, the preferred IUPAC name of the compound is ‘4-[(3-propyloctan-2-yl)oxy]decane’.


Is there a known relation between the diffusion of a molecule and the diffusion of the reactants?

More a set of queries and comments than an answer.

I don't understand why you make A and B vary with time irrespective of your reaction scheme.

You should really analyse using Fick's diffusion equations with the reaction scheme added and do so for each species. This will have to be a numerical calculation in space and time. (There are well established methods to do this. e.g. Crank-Nicholson). By the symmetry of your experiment you may be able to reduce this to one dimension and time.

You should be able to 'guestimate' the diffusion coefficient from kinetic theory at least their ratio which should be quite accurate.

Your second equation ('new formula') is that for a reaction $\ce{A + B ->[k_1] AB ->[k_1] }$ where both rate constants are the same. In your case there seems to be no reason why the two rate constants are the same, only you know enough about your experiment to explain this :)


How to make a CCl3F Fluorine NMR standard?

You need the concentration of your reference peak to be the same order of magnitude as your solute. Accuracy of your integration depends on something called dynamic range, which refers to the relative intensities of the peaks being integrated. The greater the dynamic range; the greater the difference in peak intensities, the bigger the error.

You might want consider that for most medium-high field NMR applications, CDCl3 is used with TMS at concentrations of about 0.03%v/v. Low field and bench top applications use TMS at 0.5-1%v/v concentration.


How to measure concentration of arsenite and arsenate in a water sample

Speciation analysis is very challenging because you would like to separate the species first before detection. In order times electrochemical detection was done by now ion chromatography is the way to go for arsenic speciation. You will need ion chromatography system with inductively coupled plasma with mass spectrometry. Have a look at this paper: Determination of Anionic, Neutral, and Cationic Species of Arsenic by Ion Chromatography with ICPMS Detection in Environmental Samples, Anal. Chem., 1998, 70 (17), pp 3649–3655 The good thing about Analytical Chemistry papers is that most of the methods are reliable.


Reduction half-reaction for the reduction of elemental oxygen

O2 + 2HO+ 4e- —> 4OH-
That is the correct answer


What exactly does the "per mole" unit of heat of atomization mean?

According to Wikipedia,

  Standard enthalpy of atomization is the enthalpy change when $1$ mole of gaseous atoms is formed from its element in its defined physical state under standard conditions $(\pu{298.15K}, \pu{1 bar})$.

$\Delta H_{\text{at}}^{\circ}$ focuses on the production of $1$ mole of atoms from the element. Hence it's defined for the reaction:

$\ce{\dfrac{1}{8}S8 ->S}$

but not for the reaction:

$\ce{S8 -> 8S}$

So whenever you see the term "per mole", it is, by definition, implicitly referring to the product and not to the reactant.


Why do many hypervalant chemicals have non-constant bond angles?

The $\ce{IF7}$ and $\ce{PCl5}$ molecules have 7 and 5 ligands, and pentagonal and trigonal bipyramidal geometries, respectively. These geometries happen

  simply because
  there is no geometrical arrangement which can result in five equally
  sized bond angles in three dimensions.

(Wikipedia)

There is no arrangement in which all the angles would be the same, except in a plane, where the repulsion would obviously be a lot higher than in the pyramids.


What is the oxidation state of Boron in Ammonia Borane?

As per this source, the idea and definition of oxidation state is based on the following principle:

  The oxidation number of an atom in a molecule is based
  on a formalism that forces a covalent compound to possess
  complete ionic character and may be defined as the charge
  that an atom would have if all bonds to it are broken such
  that the ligands retain a closed-shell configuration ;
  an exception, however, refers to homonuclear bonds, in which
  case the bond is broken homolytically and a single electron
  is transferred to each atom.

  The oxidation number may
  thus simply be expressed as
  Oxidation number = charge on compound - charge on ligands

So,evidently, for the presented molecule $\ce{NH3BH3}$, we see that $\ce{NH3}$ is a ligand for the $\ce{BH3}$ moiety. So,you can heterolytically cleave the dative bond between $\ce{N}$ and $\ce{B}$ towards $\ce{N}$(as per the electronegativity trends),leaving no charge on the $\ce{B}$ as of now,as after this action,it's valence shell contains 3 electrons.Now,for finding oxidation state on $\ce{B}$,an interesting thing crops up due to the presence of the three $\ce{B-H}$ bonds.
The article further enumerates:

  In many cases, the charges assigned to simple monoatomic
  ligands do not vary from compound to compound, as illus-
  trated by $\ce{F-}$, $\ce{Cl-}$, and $\ce{O^2-}$. However, a notable exception is
  provided by hydrogen for which both $\ce{H+}$ and $\ce{H-}$ have per-
  missible closed-shell configurations ($\ce{1s^0}$ and $\ce{1s^2}$
  , respectively).
  In this case, the charge assigned to hydrogen is determined by
  the relative electronegativity of the atom to which it is attached.

So again, due to sligthly higher electronegativity value of $\ce{H}$ than $\ce{B}$,the $\ce{H}$ atom becomes the ligand for the $\ce{B-H}$ bond. Hence,all the $\ce{B-H}$ heterolytically cleave towards $\ce{H}$,each cleavage leading to a +1 charge on $\ce{B}$ and -1 on $\ce{H}$. In totality,$\ce{B}$ ends up with +3 oxidation state, as it has lost all three electrons from it's valence shell.

Note: I highly recommend everybody to read that paper cited above. It's really insightful into the difference between valence,oxidation number and coordination number,which are often used interchangeably


The hindered rotation is due to the hydrogens at the naphthyl moiety. The following shows the rotation calculated at the DF-B97D3/def2-SVP level of theory. To better visualise this, I have chosen a mode with large atoms (not actually using the van-der-Waals radii, because that looked very strange; click here for a ball and stick version):

We can observe three minima and three transition states corresponding to that rotation, the depiction you have posted is slightly inaccurate as it shows one local minimum and a transition state.
From the three states that should be observable, the global minimum is achiral (1), while the other two are enantiomers (2, 3). The latter are connected via a low lying transition state (TS-2-3); probably because of the hydrogen bonding interaction guiding it. The global minimum on the other hand is separated via a very high energy barrier. These transition states (TS-1-2, TS-3-1) are again enantiomeric to each other.

I have calculated the associated energy barriers in $\pu{kJ/mol}$ at the DF-B97D3/def2-TZVPP level of theory.
\begin{array}{c*{6}{r}}\hline
\text{State} &
\Delta E_\mathrm{o} &
\Delta G^{\pu{50 K}}&
\Delta G^{\pu{75 K}}&
\Delta G^{\pu{100 K}}&
\Delta G^{\pu{298.15 K}}&
\Delta G^{\pu{500 K}}\\\hline
%
\mathbf{(1)}     &
    0.0 &  0.0 &  0.0 &  0.0   &  0.0  &  0.0\\
\mathbf{\text{(TS-1-2)}}     &
   32.9 & 32.9 & 32.9 & 33.1   & 35.7  & 39.7\\
\mathbf{(2)}     &
    0.7 &  0.7 &  0.7 &  0.8   &  1.4  &  2.3\\
\mathbf{\text{(TS-2-3)}}     &
   12.0 & 12.1 & 12.1 & 12.3   & 14.9  & 18.9\\
\mathbf{(3)}     &
    0.8 &  0.8 &  0.8 &  0.8   &  1.5  &  2.3\\
\mathbf{\text{(TS-3-1)}}     &
   32.9 & 32.9 & 33.0 & 33.1   & 35.7  & 39.8\\\hline
\end{array}

From these values we can estimate rate constants for the energy barrier with the Eyring equation, which employs the transition state theory:
$$
k = \kappa \frac{k_\mathrm{B}T}{h}\exp\left\{\frac{-\Delta G}{RT}\right\}$$

I use $\kappa = 1$, i.e. I am ignoring it, to get a guesstimate at which temperature we will be able to distinguish between all, or some of the states.

\begin{array}{c*{5}{l}}\hline
k/\pu{s-1};\ T= &
\pu{50 K}&
\pu{75 K}&
\pu{100 K}&
\pu{298.15 K}&
\pu{500 K}\\\hline
%
\mathbf{(1) \to (2)}     &
   \pu{4.9E-23} & \pu{1.8E-11} & \pu{1.1E-5} & \pu{3.5E6}  & \pu{7.5E8}\\
\mathbf{(2) \to (3)}     &
   \pu{2.6E-1}  & \pu{5.4E3}   & \pu{7.8E5}  & \pu{1.5E10} & \pu{1.1E11}\\
\mathbf{(3) \to (1)}     &
   \pu{2.8E-22} & \pu{5.9E-11} & \pu{2.8E-5} & \pu{6.1E6}  & \pu{1.3E9}\\
\hline
\end{array}

From this we can see, that at room temperature the rotation is pretty much free with a couple million transitions per second. At around $\pu{100 K}$ the is hindered; from the global minimum (1) to the adjacent minima the methyl groups simply cannot pass. Without haven calculated the values, I'd guess that this rotation starts to become hindered at around $\pu{150 K}$.
The other barrier still is easily manageable. That essentially means, that the enantiomers still readily interconvert. You will spectroscopically observe a superposition, which gives rise to the schematic structures that were used in the book. Even at $\pu{75 K}$ this barrier is taken a couple of thousand times per second
Only around $\pu{50 K}$, the interconversion between the enantiomers will also be hindered, and you should be able to distinguish all three states.

TL;DR:
You will start seeing atropisomerism at about $\pu{150 K}$, but at that temperature you'll only see the superposition of 2 and 3 (probably what is depicted in the book, i.e. left in your question) and the achiral, global minimum (on the right).

Side note:
It should be fairly obvious, that non covalent interactions also play an important role in this molecule. The barrier between the enantiomers 2 and 3 is probably significantly lower because of the hydrogen bond. However, one shall not forget, that this interaction is also present in the local minima to a not insignificant amount. The following pictures show the non covalent interactions analysed with NCIPLOT-3.0 and plotted with VMD; green areas are weakly attractive, blue areas are stronly attractive (hydrogen bonds), red areas are repulsive.


It depends on the ability of the solvent to solvate anions. If you draw structures of $\ce{DMF}$ with its resonance structure and check the side of the molecule that should interact with anions, you will see, that methyl groups prevent the interaction pretty much. Same for $\ce{DMSO}$. Thus these two solvents are commonly used to enhance the reactivity of nucleophiles. They solvate cations well but anions poorly. Maybe a bit strange, but it works well and it's pretty reliable.


Calculating pH when weak base is added to an strong acid

The question
This is the question, corrected according to Poutnik's comment, and adding (i) and (ii) to distinguish the two scenarios:

$\ce{NH3}$ solution of $\pu{0.1 mol dm-3}$ is being added to a $\pu{25.0 cm3}$ of $\pu{0.1 mol dm-3}$ $\ce{HCl}$ solution. Calculate the pH of the solution when volume of added $\ce{NH3}$ solution is
(i) $\pu{25.0 cm3}$ and
(ii) $\pu{26.0 cm3}$.
The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH4+}$ is ${9.25}$.

The chemical reactions
The chemical equation (1) in the question is incorrect and missing something. It should reflect that in aqueous solution, hydrochloric acid is completely dissociated (strong acid), and ammonia is partially dissociated (weak base). One way of writing this is:
$$\ce{HCl(aq) + NH3(aq) -> NH4+(aq) + Cl-(aq)}\tag{A1}$$
$$\ce{NH4+(aq) <=> NH3(aq) + H+(aq)}\tag{A2}$$
Equation A1 shows that the reaction of a strong acid with a weak base goes to completion (no HCl will be present, chloride does not get protonated). Equation A2 shows that ammonium is a weak acid.
If you find it confusing that ammonia first picks up a proton, and then partially loses it again, you could also write it like this:
$$\ce{HCl(aq) -> H+(aq) + Cl-(aq)}\tag{B1}$$
$$\ce{NH3 + H+ <=> NH4+(aq)}\tag{B2}$$
If you learned not to write $\ce{H+}$ but instead  $\ce{H3O+}$, you could write the B-equations slightly differently:
$$\ce{HCl(aq) + H2O(l)-> H3O+(aq) + Cl-(aq)}\tag{B1'}$$
$$\ce{NH3 + H3O+(aq)  <=> NH4+(aq) + H2O(l)}\tag{B2'}$$
In the end, each pair of equations says the same thing: Hydrochloric acid is as a strong acid, ammonia is a weak base. If ammonia is a weak base, we also know that the ammonium cation is a weak acid (with ammonia its conjugate base).
The calculation
There are two steps. You missed one (the one that involves the equilibrium). Let's work with equations A1 and A2 to work through the calculations.
You had the correct intuition about what happens with the hydrochloric acid: When adding 25 mL ammonia solution to the hydrochloric acid solution, the reactants are present at stoichiometric ratio, and the only species that is left to influence the pH is ammonium cation. When adding 26 mL ammonia solution, however, the ammonia is in excess, so some of it turns into the ammonium cation, and some remains. Scenario (i) is called a weak acid in solution, scenario (ii) is called a buffer solution. We can ignore the chloride ion, it is a spectator ion.
The amount of hydrochloric acid present initially can be determined from initial concentration and volume:
$$n_{\mathrm{\ce{HCl},initial}} = c_{\mathrm{\ce{HCl},initial}} \cdot V_{\mathrm{\ce{initial}}}$$
$$\ \ \ =0.10\  \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.025\  \mathrm{L}$$
$$\ \ \ =2.5\times 10^{-3}\  \mathrm{mol}$$

Scenario (i): weak acid

The amount of ammonia present initially can be determined from initial concentration and volume:
$$n_{\mathrm{\ce{NH3},initial}} = c_{\mathrm{\ce{NH3},initial}} \cdot V_{\mathrm{\ce{initial}}}$$
$$\ \ \ =0.10\  \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.025\  \mathrm{L}$$
$$\ \ \ =2.5\times 10^{-3}\  \mathrm{mol}$$
You calculate the concentration of species by first figuring out the amount, and then dividing by the volume of the solution, which is 50 mL. In scenario (i), all of the ammonia was turned into ammonium, so you know:
$$n_{\mathrm{\ce{NH4+}}}=2.5\times 10^{-3}\  \mathrm{mol}$$
$$c_{\mathrm{\ce{NH4+}}}= \frac{n_{\mathrm{\ce{NH4+}}}}{V_\text{solution}} = \frac{2.5\times 10^{-3}\  \mathrm{mol}}{\pu{50 mL}} = \pu{0.05 mol/L}$$
This concentration is lower by a factor two compared to the initial concentration of ammonia because the volume doubled when we added hydrochloric acid.
Now you can calculate the pH (pH of a weak acid with a pKa of 9.25 and a concentration of 0.05 mol/L:
$$ \text{pH} = \frac{1}{2} 9.25 - \frac{1}{2} \log 0.05 = 5.28 $$

Scenario (ii): buffer

$$n_{\mathrm{\ce{NH3},initial}} = c_{\mathrm{\ce{NH3},initial}} \cdot V_{\mathrm{\ce{initial}}}$$
$$\ \ \ =0.10\  \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.026\  \mathrm{L}$$
$$\ \ \ =2.6\times 10^{-3}\  \mathrm{mol}$$
HCl is limiting in reaction A1, so that determines the amount of product:
$$ n_{\mathrm{\ce{NH4+}}} = n_{\mathrm{\ce{HCl},initial}} = \pu{2.5e-3 mol}$$
Some of the ammonia is still present:
$$ n_{\mathrm{\ce{NH3}}} = n_{\mathrm{\ce{NH3},initial}} -  n_{\mathrm{\ce{NH4+}}} = \pu{2.6e-3 mol} - \pu{2.5e-3 mol} = \pu{0.1e-3 mol} $$
You can plug these concentrations into the Henderson-Hasselbalch equation (buffer equation) to get the pH:
$$ \text{pH} = 9.25 + \log \frac{0.1}{2.5} = 7.85$$


Why is gaining hydrogen called reduction when gaining electrons is called reduction? Aren't they opposites

As a blanket statement the gain of a hydrogen atom cannot be considered a reduction. The IUPAC gold book defines it as follows:

  reduction

  The complete transfer of one or more electrons to a molecular entity (also called 'electronation'), and, more generally, the reverse of the processes described under oxidation (2) and (3).

  oxidation

  The complete, net removal of one or more electrons from a molecular entity (also called 'de-electronation').
  An increase in the oxidation number of any atom within any substrate.
  Gain of oxygen and/or loss of hydrogen of an organic substrate.

  All oxidations meet criteria 1 and 2, and many meet criterion 3, but this is not always easy to demonstrate. Alternatively, an oxidation can be described as a transformation of an organic substrate that can be rationally dissected into steps or primitive changes. The latter consist in removal of one or several electrons from the substrate followed or preceded by gain or loss of water and/or hydrons or hydroxide ions, or by nucleophilic substitution by water or its reverse and/or by an intramolecular molecular rearrangement. This formal definition allows the original idea of oxidation (combination with oxygen), together with its extension to removal of hydrogen, as well as processes closely akin to this type of transformation (and generally regarded in current usage of the term in organic chemistry to be oxidations and to be effected by 'oxidizing agents') to be descriptively related to definition 1. For example the oxidation of methane to chloromethane may be considered as follows:

As you can see, the reverse of (3) is your first statement with one significant addition: organic substrate.

To understand this one only has to look at the electronegativities. With an electronegativity of 2.20 (Pauling) for hydrogen, it is less electronegative than most other non-metals, or in general, elements that organic substrates are made out of.[1] If you follow the electronegativity scheme of assigning oxidation states, then adding a hydrogen atom (one proton and one electron) results in an decrease of the oxidation number of the element that the hydrogen atom was added to.[2]

On the other hand, adding oxygen would increase the oxidation number of the element (except fluorine) it was added to, since its electronegativity is the second highest, i.e. 3.44 (Pauling).[3]

For example, adding dihydrogen to ethene, the carbons are reduced, while hydrogen is oxidised.
$$\ce{\overset{\color{orange}{-2}}{C}_2\overset{+1}{H}_4
  + \overset{0}{H}_2
  -> \overset{\color{orange}{-3}}{C}_2\overset{+1}{H}_6}$$
Adding dioxygen to ethane to form ethane-1,2-diol, the carbons are oxidised, while oxygen is reduced.
$$\ce{\overset{\color{orange}{-3}}{C}_2\overset{+1}{H}_6
  + \overset{0}{O}_2
  -> \overset{\color{orange}{-1}}{C}_2\overset{+1}{H}_6\overset{-2}{O}_2}$$
You can go as far as looking at the addition of water to ethane to form ethanol. One carbon will be oxidised and one will be reduced.
$$\ce{\overset{\color{orange}{-2}}{C}_2\overset{+1}{H}_4
  + \overset{+1}{H}_2\overset{-2}{O}
  -> (\overset{+1}{H}\overset{-2}{O})H2\overset{\color{orange}{-1}}{C}-\overset{\color{orange}{-3}}{C}\overset{+1}{H}_3}$$

Just keep in mind that your first statement is only true when hydrogen is added to to more electronegative elements. The reversal is the case when adding it to a less electronegative element, like a metal.
A second point to keep in mind is that oxidation states are bookkeeping tools only. The world of bonding is not strictly ionic or covalent; it is often somewhere in between. Therefore electrons are almost never completely transferred and the actual charge distribution might be quite different.

References:

Hydrogen: electronegativity. WebElements [http://www.webelements.com/]. [Online]; Dated 30th Nov. 2016; https://www.webelements.com/hydrogen/electronegativity.html
Electronegativity Considerations in Assigning Oxidation States
Oxygen: electronegativity. WebElements [http://www.webelements.com/]. [Online]; Dated 30th Nov. 2016; https://www.webelements.com/oxygen/electronegativity.html


How can the stereochemistry of the product of an E1cb reaction be determined?

Since your example compound has a chirality center, let's start by picking one enantiomer. We'll do the other enantiomer later. In the presence of a suitable base, this compound can be deprotonated to form the enolate anion. The enolate group itself can be E or Z, but for this purpose it does not matter.

In the second step of the mechanism, the electrons in the new pi bond from the enolate pi electrons shift toward the beta carbon. The new pi bond will be formed from the enolate p orbital on the alpha carbon and the C-Cl sigma-antibonding orbital on the beta carbon.

For this to work, the C-Cl must be oriented perpendicular to the plane of the pi bond. We can draw Newman projections for our specific enantiomer of the enolate anion. There are two conformations that put the chlorine in the correct orientation. The second one below has less steric hindrance with the larger group (ethyl) near the H atom and the smaller group (methyl) oriented toward the enolate carbon.

The major product forms from this conformation, with the methyl group cis to the enolate/carbonyl carbon.

Now, the other enantiomer forms the same major product, but I would like you to convince yourself of that.


Quick Laboratory method:

$$\ce {CoCl2(aq) + 2NaOH(aq)->Co(OH)2(s) + 2NaCl(aq)}$$
$$\ce {Co(OH)2(s) + 2CH3CO2H(aq)->Co(CH3CO2)2(aq) + 2H2O(l)}$$
$$\ce {Co(CH3CO2)2(aq) + \Delta->Co(CH3CO2)2\cdot 4H2O (s)}$$

You should separate $\ce{Co(OH)2}$ by filteration and washed with water ($\pu {K_{sp} = 1.0 \times 10^{-15}}$) to remove all $\ce{NaCl}$ before subjected to react with $\ce{CH3CO2H}$.

This might work in industrial scale, if be careful. $\ce{Co(OH)2}$ is unstable to heat but no need to dry before react with $\ce{CH3CO2H}$.


MCPBA preferentially reacts at the more electron-rich alkene, in this case the more substituted alkene according to this entry on chemistryportal.net here and this review RSC Advances


TL;DR: $\ce{He^2+}$ is the only preferred notation.
Notations $\ce{He^{++}},$ $\ce{He^{+2}}$ or $\ce{He^{1+}}$ are obsolete and should be avoided.

From IUPAC “Green Book” [1, p. 49], section 2.10.1 Other symbols and conventions in chemistry, subsection (i) The symbols for the chemical elements:

  The ionic charge number is denoted by a right superscript, by the sign alone when the charge number is equal to plus one or minus one.

  $$
\begin{array}{lll}
\textit{Examples} & \ce{Na+} & \text{a sodium positive ion (cation)} \\
  & \ce{^{79}Br−} & \text{a bromine-79 negative ion (anion, bromide ion)} \\
  & \ce{Al^3+}~\text{or}~\ce{Al^{+3}} & \text{aluminium triply positive ion} \\
  & \ce{3 S^2-}~\text{or}~\ce{3 S^{-2}} & \text{three sulfur doubly negative ions (sulﬁde ions)}
\end{array}
$$

  $\ce{Al^3+}$ is commonly used in chemistry and recommended by [74]. The forms $\ce{Al^{+3}}$ and $\ce{S^{−2}},$ although widely used, are obsolete [74], as well as the old notation $\ce{Al^{+++}},$ $\ce{S^=},$ and $\ce{S^{−−}}.$

For the historical perspective, see the corresponding article by Jensen [2]:

  In contrast, the German chemist, Walther Nernst, in his equally inﬂuential 1893 textbook of theoretical chemistry, chose to place an appropriate number of superscripted $+$ or $−$ signs directly above the ion’s atomic symbol (4), a practice which was soon modified by placing them instead to the immediate right of the symbol, as in $\ce{Ba^{++}}$ and $\ce{PO4^{---}}$(5).

  The IUPAC guide to Quantities, Units and Symbols claims that yet a third “algebraic” method of indicating ionic charges was also used in the past in which the charge preceded the numerical value, as in $\ce{Ba^{+2}}$ and
  $\ce{PO4^{-3}},$ even though this particular sequence of symbols was originally intended to represent the inherent sign of a number or exponent and not the number of signs (6). However, inspection of nearly three dozen general, inorganic, and analytical textbooks, spanning the period 1909–1975, revealed that the vast majority employed the modified Nernst notation, with a smaller number — mostly of European or Russian origin — using the Ostwald notation instead. Rather surprisingly, very few examples of texts using the algebraic notation could be found, all of them post-1970 (7).

  Since at least the 1950s IUPAC has ruled that ionic charges or “charge numbers,” as they are now officially called, should be written instead with the number preceding the charge sign, as in $\ce{Ba^2+}$ and $\ce{PO4^3-}$
  (6, 8, 9). There are several reasons for this decision. It is more concise than the typographically inelegant Nernst approach and more physically meaningful than the Ostwald notation. Unlike the algebraic notation, it
  avoids confusion with the conventional symbolism for inherently positive and negative numbers and maintains consistency in how we count physical entities.

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7. (PDF)
Jensen, W. B. The Proper Writing of Ionic Charges. J. Chem. Educ. 2012, 89 (8), 1084–1085. DOI: 10.1021/ed2001335.


I think what confuses you  is that 'salt' has two meanings: In the every day sense it means table salt (which has the formula NaCl) but in the chemical sense it means a certain way for particles to hold together and form a tangible substance.
I'd define the latter as any periodic arrangement/lattice (see e.g. here ) of particles held together by ionic bonds.
The positive ion is almost always a metal, the negative one can be anything (so usually either a halogen (those lack only one valence electron to have a full shell, see e.g. Cl) or a (smallish) molecule-ion like SO4^2−)

  base and acid radicals

I really don't see much connection between salts and either bases/acids and radicals. Do consider that not all ions are bases/acids and all radicals are highly reactive. It would rather be an edge case to have a salt with a radical in it; most salts have ions with energetically favourable number of electrons (as a rough rule this means all "electron shells" are either full or empty.)
NaCl is a good example.

Because different sized salt crystals still have extremely similar behaviour, the salt formulas express only the ratio of anions and cations. Realize that on the whole salts are neutral.

  Why do they call the acid and basic radical a salt?

A single kind of particle is never called a salt, salt means an arrangement of two ions of different polarity. I guess some lab 'recipes' have sloppy language if something is provided as a salt, but only one component is important for the reaction.
Perhaps I can give a better answer I understood what you mean by radical in this context.


How to compare stability of following two alkenes

The second alkene is more stable since it is symmetric. Achiral compounds are always more stable than chiral compounds due to the Gauche Effect. The explanation for why the Gauche Effect takes place is here.


Thermodynamics of photochemical reactions

$h\nu$ is not a thermal energy so thermodynamic considerations do not apply. For example, you cannot predict the probability of being in an electronic excited state from equilibrium thermodynamics.  The main reason is that the excitation is not a result of (classical) thermal energy transfer, but rather the adsorption of a photon which changes the quantum state.

With a constant light source

$$\ce{CO2} + h\nu \ce{-> CO + C}$$

will reach a steady state with constant concentrations that will be different from the thermal equilibrium concentrations.  However, you can not extract a standard free energy difference by
$$ \Delta G^\circ=-RT\ln \left( \frac{p_{\ce{CO}}p_{\ce{O}}}{p_{\ce{CO2}}} \right) $$
because the products and reactant are not in thermal equilibrium.


Does granite become soft as modelling clay at 2000 °C?

Mixtures of different compounds do not usually work like that.  They generally melt at lower temperatures than the constituent pure compounds.  So I would expect your granite to be not like modeling clay, but molten even before getting to the melting point of silica.

Relevant:  https://en.wikipedia.org/wiki/Eutectic_system.

If you want modeling clay, try 1300 °C where the Earth's lithosphere gives way to the asthenosphere.  See the Characteristics section of https://en.wikipedia.org/wiki/Asthenosphere.


The situation you described is not a vacuum because it has a single molecule in it, so what you're asking about is how the concept of temperature applies to an isolated molecule.

First, the idea that you can start at absolute zero even with a single molecule is not consistent because there will always be fluctuations/movement possible in the ground state of a quantum system (the zero point energy or ZPE). The cleanest example of this is the quantum harmonic oscillator: a particle in the ground state has an average position of zero ($\langle x \rangle = 0$), but the fluctuations are greater than zero ($\langle x^2 \rangle > 0$). Let's think about this in terms of the uncertainty principle. The uncertainty principle is given by $\Delta p\Delta x \geq \hbar /2$ where $\Delta x  = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}$ and $\Delta p  = \sqrt{\langle p^2 \rangle - \langle p \rangle^2}$. Because $\langle x^2 \rangle >0$ and $\langle x \rangle = 0$, we can see that the uncertainty in the momentum is also non-zero. As we know, the kinetic energy is $p^2 / 2m$, so our energy uncertainty is non-zero as well.

Second, and onto the main point of your question, you can certainly define a concept of temperature with only one molecule. Consider that even a single molecule has many different energy levels (i.e. electronic, vibrational, and rotational degrees of freedom are all available). The partition function for our system is still defined as $Z = \sum_i \exp{\left(-\frac{E_i}{k_bT}\right)}$ where $E_i$ is the energy of the $i$th state.(See Note) So, if we start very near absolute zero we will see only contributions from the ground state, but as we increase the temperature of the system we will observe non-zero probabilities of populating excited states. Nothing is broken about the concept of temperature in this situation, the energy scales are just considerably smaller than those in an ideal gas.

Note:

Your initial reaction to applying thermodynamic principles to a single molecule might be to think that it doesn't apply because we only have one molecule and not an ensemble. However, within the density matrix formulation of statistical mechanics, the quantum uncertainty of a system is entirely indistinguishable from the statistical uncertainty of a system.


What is chemically different about tea for hard water?

Hard water tea

By definition hard water is defined as water that has more than 0,6mg/l of dissolved CaCO3. Therefore soft water that doesn't have that much minerals has a greater dissolubility and can dissolve more tea. It can also dissolve the tea quicker. So the tea companies need to find what tea blend dissolves quickly and efficiently in hard water.

  A hard water blend needs lots of quality Assams and East African teas, with plenty of “guts” and strength. It needs flavour too, of course, but this flavour will be the rich, malty type that punches through the hard water. -https://allabouttea.co.uk/tea-news/hard-water-tea/ (as Bdrs commented)


Is there open source software for SEM EDX data analysis?

You could use HyperSpy library for Python.

Some examples of what you can do with it are available here.


Why does the electron configuration for some elements not follow the diagonal rule?

The reason of these exceptions (from the diagonal rule) is that some elements are more stable with fewer electrons in some subshells and more electrons in others:

Electronic configuration of molybdenum: $[\ce{Kr}] \ce{4d^{5} 5s^1}$, instead of  $[\ce{Kr}] \ce{5s^24d^{4}}$ according to the diagonal rule, because a half-full $\ce{4d}$ subshell and a half full $\ce{5s}$ subshell are more stable than $\ce{4d}$ filled with four electrons and a full $\ce{5s}$ subshell.
Electronic configuration of Gold: $[\ce{Xe}] \ce{4f^14 5d^{10} 6s^1}$, instead of  $[\ce{Xe}] \ce{4f^14 5d^{9} 6s^2}$ according to the diagonal rule, because a full $\ce{5d}$ and half full $\ce{6s}$ subshell is more stable than $\ce{5d}$ filled with 9 electrons and a full $\ce{6s}$ subshell.
Electronic configuration of Palladium: $[\ce{Kr}] \ce{4d^{10}}$, instead of  $[\ce{Kr}]\ce{ 5s^24d^{8}}$ according to the diagonal rule, because a full $\ce{4d}$ orbital is more stable than $\ce{4d}$ filled with eight electrons.
Electronic configuration of Curium: $[\ce{Rn}] \ce{7s^2 5f^7 6d^{1}}$, instead of  $[\ce{Rn}] \ce{7s^2 5f^8} $ according to the diagonal rule, because a half full $\ce{5f}$ orbital is more stable than $\ce{5f}$ filled with eight electrons.


Molecular orbitals form when you apply the "Linear Combination of atomic orbitals" (LCAO) model to atoms bonding. Where you take a linear combination of the relevant atomic orbitals to create an MO. In each case of combination a bonding and an anti bonding (ab) combination forms. The ab MO form as a result of applying the Huckel model which results in a square root in the expression for the energy, thus giving a negative and a positive contribution to the energy. One of the expressions E1 will be lower in energy and will form the bonding and the second E2 will form the higher energy MO - anti bonding.
Sometimes a non-bonding orbital forms, which doesn't contribute to the total bond energy, where the squared root expression equals zero.

To understand it more thoroughly one needs to understand the hydrogen atom quantum model and it's solution which give the orbitals we know as: s,p,d,f etc. In the case of MO we apply say that each atom "brings" it's set of occupied orbitals from it's valence electronic shell, combining these gives MO diagrams.
It's an approximation but the results it gives describe to some extent what can be measured in the lab.


No doubt about it: it is due to the ionic radii and thus the distance between ions. The Rb-Cl bond is not as strong because of the distance between them. Na has an E.N. value of 0.9 Paulings while Rb is at 0.8; thus no significant difference in the magnetic forces of attraction.

LiCl has an even lower melting point due to excessive polarization (this is when Farjan's rules come into play), and the result, along with electronegativity again, is still an ionic compound, but with more covalent character.


Why is the ester bond stable under Reformatsky reaction conditions?

Yes, technically you have a zinc alcoholate which could display the basicity and nucleophilicity of any other alcoholate and you know that you can transesterificate with e.g. sodium methanolate.

Aside from the oxyanion here being rather closely associated to the zinc cation which has a larger positive charge than e.g. sodium and thus exercises a greater attractive force, there is also the fact that the alcohol in question is tertiary and tertiary alcohols are bad nucleophiles. In most of the cases, $\ce{R3}$ will be primary or at most secondary, thus the transesterification reverse reaction will be more rapid than the forward reaction.

If, however, $\ce{R3}$ is tertiary, too, you would be dealing with a huge steric bulk on that carbonyl carbon, attempting to have two tertiary $\ce{OR}$ groups on the tetrahedric intermediate. That is unlikely to happen for steric reasons.

If you are talking about the absolute final step, i.e. the aquaeous workup, then you will likely be using ammonium chloride solution or something else mildly acidic. In mildly acidic or otherwise buffered solutions, there is not a high enough concentration of hydroxide ions to attack the ester carbonyl sufficiently for an observable reaction to occur.


Brown color at end point during titration with phenolphthalein as indicator

Assuming there are no major color-influencing side reactions, the subtractive color mixing as follows:

#FFA8D9 (Pink, phenolphtaleine) + #DAA520 (Goldenrod, white wine) = #D2691E (Chocolate, result)

actually already gives brown(ish) color. It's hard to pick the right color codes, especially for the white whine, but OP can probably try to perform the same operation with the color he remembered/picked from the photo.


Regarding the free energy requirements for reacting gases

I think what you are fundamentally saying is you are trying to relate classical thermodynamics to the molecular properties. It is best to have some firm familiarity with classical thermodynamics before attempting this. Classical thermodynamics does not require a molecular interpretation at all! (It helps a lot but is by no means necessary)!. I personally feel the answer I am about to write will help you more than trying to wade through terms and ideas/principles.... I hope so at least. If you don't like it then maybe someone else can provide a more quantitative solution.

1) The whole point of statistical thermodynamics is to provide a molecular interpretation. To do this we have to divide the system up into smaller sub-systems as I described earlier. The key to it is that there can be many "configurations" that give rise to the same macroscopic properties. The aim is to find the most probable one ... which is where your distributions come in.

It is the energy of the microstates that is what you put in so that you can calculate the various macroscopic properties you quote above. This is the best way to think about it. When you get to advanced theories this definition may change slightly. For now just accept that it is the energy of the micro state you are putting in.

In my honest opinion I don't know why courses teach the Maxwell-Boltzmann velocity distribution in a statistical thermodynamics course. I feel while a nice introduction to dealing with ensembles ... it leads to confusion (when I was learning this I had very similar questions with regard to what can be related to what, it only comes with time and exposure to the subject. So I know exactly where you are coming from!).

2) Whenever you see kinetic energy you should be thinking temperature rather than trying to get the individual molecular speeds... what does knowing the instantaneous position/velocity of a mole of gas particles at one instant in time really lead to? (not much)... I haven't really seen anyone combine the thermodynamic potentials with something like velocity. If anyone reading this has then I would love to hear what you have to say!! When I was first learning thermodynamics I clung on to what I was familiar with ... potential / kinetic energies ... now I don't really think about them when you say thermodynamics at all ... I don't think about $G$ as being a "part" of $U$ ... I just treat them as if they were separate (hmm how to phrase this) "properties" of a given system under certain conditions.

In fact I would go further to say that I view them as mathematical tools ... functions of variables of the system that I can manipulate to express molecular properties in terms of measurable quantities ....

3) Yes. You can indeed As is done by most theoretical chemists use Gibbs energies to calculate the thermodynamic activation barrier in the starting material and transition state.... The problem, which admittedly I may have confused you earlier about, is that this is done via the Arrhenius equation A kinetics equation. When in early undergraduate years you don't mix kinetics with thermodynamics. Later on it becomes (bar a factor of $2\pi$ or so ) acceptable to do so ... but you should not attempt this yet ....

A crucial point is that the activation energy graph's x-axis is the reaction coordinate. The reaction coordinate is a generalised coordinate which can be used (whatever it is) to describe the progression of the reaction.

Normally we deal very nicely in cartesian $x,y,z$ coordinates ... the idea of a generalised coordinate is that it can be used to reduce the dimensionality of the mathematical problem with some nifty rethinking of how we choose to describe the system .... for instance when talking about a simple harmonic diatomic molecule .. instead of choosing the location of both atoms in space (requiring 6 position coordinates) to fully describe the system as a function of time (as it oscillates back and forth ) i could use the relative separation (the internuclear distance) relate to the equilibrium bond length.... in this way it becomes much easier to describe the system ....

That was just one example of a generalised coordinate.... something that makes the description of the problem al little easier.... (especially if the system is subject to constraints, the generalised coordinates remove the constraint equation... anyway where were we.)...

Bottom line: ** there is a big difference between reaction coordinate and the Maxwell-boltzmann graph you linked.

4) This question in itself has hit on a very big topic. This is the realm of classical thermodynamics. What you need to understand is the difference between the standard Gibbs energy and the Gibbs energy .... For the following just imagine that $A$→$B$.

In a very brief overview it all boils down to this:

The Gibbs free energy is a mathematical function... don't try to pin down WHAT it is... its just another energy we can describe the system in .... It is a function of $P$, $T$ and $N$. Lets make it easier and say that the number of particles does not change. i.e $dN=0$ ...

Since the Gibbs energy is a function of $T$ and $P$ if we change either of these variables ... $G$ will change too .... The starting materials will have a certain $G$ and the products will have a certain $G$ ... The difference between these under standard conditions is fixed by definition that we are controlling the variables. This is like the thermodynamic drive for the process if it were to occur at our specified standard conditions .... Over the course of the reaction this value does not change since as I said it is the difference between the Starting Material and the Products under given conditions. .... This is also for pure values of the starting material and the product...

So far so good right?

As your reaction proceeds the composition varies i.e $dN\neq 0$. As I pointed out in another answer from the definition of the Gibbs energy this means that the value Gibbs function now changes... if you want a molecular picture it is due to the entropy of mixing adding in another factor to the "disorder" that has so far not been accounted for...

For an illustration please follow this link:

http://www.public.asu.edu/~laserweb/woodbury/classes/chm341/lecture_set5/lecture_set5.html

There are therefore two measures of the Gibbs energy. The one under standard conditions... unchanging for a given reaction... it is this one that has the $\Delta G^{\small o}$.. it is this one you get your electrode potentials off... it is this one that leads to the the equilibrium constant....

The other one is ever changing as the reaction proceeds since the composition of the reaction vessel is ever changing as $A$ goes to $B$. At equilibrium the net rate of change of A to B and B to A is now zero right... this in fact defines an equilibrium point.... shifting it slightly in one way or the other will cause the system to respond to re-establish equilibrium .... Most importantly we define equilibrium as the point where $dN=0$ .... for both types (in this case one would go up as the other went down ... ) but at equilibrium $dN=0$ .... This means that $dG=0$ too from the mathematics I showed you before ....

Therefore equilibrium is the point where the change in the Gibbs energies of two substances in a mixture is zero.... The reaction has no more drive if you like .... You can equivalently say their chemical potentials are equal too ... but leave that for now.....

So thats it ... Im sure if you ask question 4 to someone else they will give you a more complete answer.... but since its getting on for 2:00 am here I think I'm done. Hope that helps and again if I have made any errors ... please edit away :)

edit

What you are describing on the end of question 4 is know as the steric factor... you basically compare the frequency of successful collisions in the transition state geometry "linear" for example with respect to a full sphere of reactivity .... this gives you your factor :)

Edit
So the problem is your definition of the energy. I dot know if you have seen this before but perhaps this will help,...

Yes you can indeed use statistical mechanics to obtain a value of $G$ ... The best way to do it is to get an expression for the Helmholtz free energy and then just Legendre transform it to $G$ ... $G=A+pv$.

Whilst yes at a simple level the internal energy does indeed "contain" the kinetic energy ... when you learn a bit more about quantum mechanics .. we never talk about kinetic energy ... I think it would help to actually leave the classical results behind and focus on the actual version of events...

With reference to the uncertainty relation .. how do you think kinetic energy is something we can talk about at a molecular level? The answer is you don't. Once QM comes in it is not useful to talk about velocity.

That being said how do we describe motion ? after all the Schrödinger equation is just an equation of motion. All the Schrödinger equation describes is the time evolution of a vector state in a certain vector space - Hilbert space, Which is the space in which we describe quantum mechanics...

With this in mind ... all motion of the system is quantised ... can only have discrete values .... non-continuous values ... etc etc...

Translational motion is quantised... so is vibrational ... so is rotational ..... each degree of freedom has its own quantised set of discrete energy levels ....

Equipartition theorem uses classical ideas and distributions to distribute all the available energy across these states ... roughly it says that each has energy of $\frac 1 2 KT$....

To say $U$ includes kinetic energy is a primitive way of saying the internal energy is a function that describes the energy of all the modes in the system... (you would need a distribution to see how they are populated mind) ....

Equipartition theorem doesn't get it quite right when you consider the low temperature theoretical and experimental heat capacity of a substance..... Even at low temperature you are still expected to populate all modes with $\frac 12 KT$ ... but this doesn't happen since $KT$ does not have enough energy to populate said modes ... i.e it doesn't reach the quantised energy level requirement .... therefore fewer modes are available at low temperature...

Bottom line is: just try and separate the different preconceived competing ideas you have when it comes to statistical mechanics .... have you had a lecture course on this yet?

it is certainly not about having "enough Gibbs energy to react"... The Gibbs energy between a reactant and transition state $\Delta G ^{\ddagger}$ is not a variable thing as you are saying ... (obviously for a given set of chemicals),.... If it changed all the time then why would we write them down to compare them? The Gibbs energy in your picture is not the same Gibbs energy that goes to zero at equilibrium ...
\begin{equation}
\bigg(\frac{dG}{dq}\bigg)_{T,P}=\Delta _r G^{\small o}+RT\ln Q
\end{equation}
at equilibrium the left hand side .... the gradient of the graph in that link i sent you before ... is zero ...

The rate of change of the Gibbs energy with respect to the extent of the reaction is by definition zero at eqm.

When equilibrium is reached $Q=K$ that is to say the reaction quotient becomes the equilibrium constant ....


What is the difference between absolute and relative configuration in stereochemistry?

Let's say that we have synthesized a new compound, C(1)(2)(3)(4), where (1)-(4) represent 4 different groups attached to the central carbon, so the molecule is chiral.
Let's say that we find a method to separate the two enantiomers present in the new compound and that we placed the enantiomers into separate test tubes.
Next we might take each enantiomer by itself and place it in a polarimeter.  We would find that the material in one test tube rotates light in the (+) direction and the material in the other test tube rotates light in an equal, but opposite direction (-).
At this point, all we know is that we have two enantiomers that, as expected, rotate light in equal and opposite directions.  We don't know which test tube contains the "R" enantiomer or which contains the "S" enantiomer.  We just know their relative configuration with respect to each other, they are mirror images of each other.
To find out which is "R" and which is "S" we would either need to 1) put a crystal of one of enantiomers in an X-ray diffractometer and get a three-dimensional picture of the structure of the molecule from which we could see whether it is the "R" or "S" enantiomer or 2) carry out a series of chemical transformations where the stereochemical consequences of each step are known (e.g. does each step proceed with inversion or retention of configuration at the chiral center) and convert it to a compound of known absolute configuration
Let's say we pick one of our test tubes and are able to convert that enantiomer to a chiral compound C(A)(B)(C)(D) where we know that R-C(A)(B)(C)(D) rotates light in the (+) direction.
We find that the C(A)(B)(C)(D) we created from our C(1)(2)(3)(4) enantiomer rotates light in the (-) direction; therefore we have produced the "S" enantiomer of C(A)(B)(C)(D).
For simplicity, let's say that all of the reactions we used in our conversion of C(1)(2)(3)(4) into C(A)(B)(C)(D) are known to proceed with retention of configuration.  Then we know that the test tube of C(1)(2)(3)(4) used in our reaction contains the S-C(1)(2)(3)(4) enantiomer. Now we know the absolute configuration of our sample.

EDIT: d, l and D,L

The lower case "d" and "l" can be used interchangeably with "+" and "-" respectively.  The letters stand for dextrorotatory and levorotatory.  They simply indicate how the plane of polarized light will be rotated when a sample of the material is placed in a polarimeter.  So l-2-butanol is the same as (-)-2-butanol.  Use of "d" and "l" is discouraged because it can create confusion with "D" and "L".

Upper case "D" and "L" are somewhat archaic.  Way back when, the two enantiomers of glyceraldehyde were separated, purified and placed in a polarimeter.  One enantiomer rotated the light in a dextrorotatory manner; this enantiomer was called D-glyceraldehyde.  The other enantiomer rotated light in the levorotatory direction and was referred to as L-glyceraldehyde.  If you take what ever chiral molecule you are studying and convert the chiral carbon to glyceraldehyde, and it rotates light in the "+" direction, then you've made D-glyceraldehyde, and if all of your reaction steps involved retention of configuration (see above) then you would have the "D" enantiomer of your material. The terms "D" and "L" relate whatever chiral center you are studying to the chiral carbon in glyceraldehyde.


Where's my experimental error coming from?

I would go with a full fledged error propagation analysis on this one, because without systematic approach you might be guessing forever. So if I look at the calculation process you have the following equation to get from known quantities to the answer:

$$X=\left(\frac{W_0}{W_e}-1\right)\left(\frac{M_{CuSO_4}}{M_{H_2O}}\right) \tag{1}$$

where $W_0$ is the start weight, $W_e$ the end weight and $M_i$ is the respective molecular mass. $X$ is the desired answer.

To find the expected error in the answer, $\Delta X$, assuming no error in the molecular masses, we calculate:
$$\Delta X=\sqrt{\left(\frac{\partial X}{\partial W_0}\Delta W_0\right)^2+\left(\frac{\partial X}{\partial W_e}\Delta W_e\right)^2} \tag{2}$$
which works out to:
$$\Delta X=\frac{\sqrt{(\Delta W_0)^2 W_e^2+(\Delta W_e)^2 W_0^2}}{W_e^2}\left(\frac{M_{CuSO_4}}{M_{H_2O}}\right)  \tag{3}$$

So now the final thing left to do is estimate how big $\Delta W_0$ and $\Delta W_e$ are. For $\Delta W_0$ I think it is safe to say that this is only the inaccuracy of the scale, so perhaps $0.01$ g. For $\Delta W_e$ I would go for the worst case scenario: that all of the 0.6% impurity in the original copper sulfate has evaporated, meaning that the weight will be off by 0.6% times $W_0$ i.e. $0.08$ g.

Now plugging in the numbers in Eq.$(3)$ I find $\Delta X=0.14$ mol water per mol copper sulphate. Which is very close to the actual error you find.

This error is almost completely determined by the error in $W_e$. If, like @brinnb suggests, an additional reduction in $W_e$ could be expected due to adhered water say that the error in $W_e$ is $0.1$ g you already get to $\Delta X=0.17$ mol water per mol copper sulphate.

Conclusion

The error you find falls within the worst-case expected error due to impurity and potentially adhered water.


Reading CGTF basis from Basis Set Exchange site

Yes it is correct that for $p$-type basis functions that you have all three $p$-functions, i.e. $p_x$, $p_y$, $p_z$.

For basis functions of larger angular momentum, you are also correct that there is two "different" conventions. The cartisean coordinate and the spherical coordinate convention.

In many electronic-structure programs both conventions are used, since the integrals are calculated in cartisean coordinates and then afterwards transformed to spherical coordinates, i.e. as you stated for $d$-type functions going from 6 basis functions to 5 basis functions.

In regard to make you own electronic-structure code, I would recommend keeping everything in cartisean coordinates, because this will make your life simpler. (Keep in mind that if you compare you energy with from a Hartree-Fock calculation with other programs you will have a slightly lower energy, because you have 1 more $d$-type basis function).

For a simple electronic-structure program a very nice guide for molecular integrals can be found here, http://joshuagoings.com/2017/04/28/integrals/


Why does NaOH catalyze H2O2 decomposition?

According to THE HOMOGENEOUS BASE-CATALYZED DECOMPOSITION OF HYDROGEN PEROXIDE J. Phys. Chem., 1961, vol. 65 , pages 304–306:

$\ce{H2O2 + OH- <=> HO2- + H2O}$

$\ce{H2O2 + HO2- ->}$ [6-membered ring with 2 "H"s bridging 2 "O-O"s, with another H bonded to one of the "O"s] $\ce{-> H2O + O2 + OH-}$

  It is clear that the rate will go through a maximum
  when the peroxide is 50% ionized, that is
  $[\ce{H2O2}] = [\ce{HO2-}]$.

So OH- catalyzes decomposition of peroxide, but a maximum is reacted when the amount of hydroxide added is half the amount of peroxide.  If more hydroxide than peroxide is added, the catalytic effect is lost.


How to measure saturated salinity and mineral levels for flaked sea salt?

Why not use salt (NaCl) bought from a store, or even better at a defined purity grade from a chemical vendor?

When you use sea salt, you get lots of other ions but more importantly you get colloidal particles, organic substances, and other compounds that will affect the crystallization. Chunky results sounds to me like a colloid particle that started crystallization in all directions at once.

If you want perfect crystals, do it perfectly, i.e. with a pure solution, in a clean and new glassware or polished stainless container. If you want perfect crystals you have to reduce the new-age-ness of it ;)

As for understanding, nobody fully understands crystallization, but a lot of things are known. There are books on this topic, check amazon and search for "crystallization". As for instruments, x-ray fluoroscopy and diffraction wouldn't be amiss but you will get very far with an optical microscope, I know I can watch crystals through a microscope for hours and still find something interesting. Barring XRF, XRD and microscope, and within a household budget there are salinity measurement probes and conductance measurement probes which will be able to tell you something of the brine such as it is.


Intramolecular Reaction with carbocation rearrangement

Your approach is feasible. Thermal cyclizations of appropriate dienes, enones, enynes, and other related unsaturated systems are known in literature. For example, there are extensive examples of intramolecular ene-reactions in Organic Synthesis and some of them are reviewed (Ref.1 and Ref.2). To justify your first step, I'd include the abstract of reference 1, which states that:

  Thermal cyclizations of appropriate dienes, enynes and related unsaturated systems, some of them carried out on an industrial scale, demonstrate increasingly the preparative power of the intramolecular ene reaction. A variety of substituted, fused and bridged ring systems, including natural products, are thus easily accessible in a regio‐ and stereo‐selective manner. Numerous examples are discussed systematically illustrating the possibilities, limitations, and common features of this cyclization reaction and its reverse ring‐opening process.

Nonetheless, see following example given in reference 1 for thermal cyclization:

Although those are different than the problem in hand, the intramolecular cyclization has possible clear path. For example, various types of intramolecular cyclizations has been discussed in Ref.1. Now, I modified your mechanism using arrows to make it clear:

The electrophile here is the carbonyl carbon of 1, which would get more electron deficient by carbonyl oxygen’s complexation to a Lewis acid such as $\ce{AlCl3}$, or  $\ce{AlEtCl2}$, or even $\ce{AlEt2Cl}$ (Ref.2). This is evident in the intermediate I. The neucleophile, the alkene group of the molecule is in close proximity to make a bond with elecrophilic carbonyl carbon (intermediate II). The driving force here is making a less strained 5-membered ring and relatively stable $3^\circ$-cabocation. Keep in mind that all of these intermediates are reversible. The reversible hydride transfer is possible because energy different between resulting $3^\circ$-cabocations are minimal. Yet, this transfer facilitate pinacol-pinacolone type rearrangement of methide shift to give final stable product 2.

References:

Wolfgang Oppolzer, Victor Snieckus, “Intramolecular Ene Reactions in Organic Synthesis,” Angew. Chem., Int. Ed. Engl. 1978, 17(7), 476-486 (https://doi.org/10.1002/anie.197804761).
Barry B. Snider, “Lewis-acid catalyzed ene reactions,” Acc. Chem. Res. 1980, 13(11), 426-432 (https://doi.org/10.1021/ar50155a007).


By "rapid switching" they technically mean Ligand-to-Metal Charge Transfer (LMCT). A more modern framework is Ligand Field Theory. I would have to teach a class to fully explain it in those terms, but I'll try to explain it in terms of this hybridization you brought up.

A chemical bond implies a higher probability of finding electrons between two bound nuclei. Atomic orbitals describe the electron density for individual nuclei. A bond between two nuclei necessitates some physical overlap of the relevant atomic orbitals.

It is in this way that we say that the linear combination of atomic orbitals, say two consisting of a $2s$ and $2p$ orbital, results in a molecular orbital that describes the molecule. We would write the wave function for the complex as:

$$\Psi = C_1 \psi(2s) \pm C_2 \psi(2p)$$

The amount of "mixing" is just a matter of adjusting the coefficients, $C_n$. And so we might say that the bond in this made-up molecule is $sp$ hybridized. $sp^3$ hybridized would mean the following kind of wave function:
$$\Psi = C_1 \psi(2s) \pm C_2 \psi(2p) \pm C_3 \psi(2p) \pm C_4 \psi(2p)$$

Similarly, $sd^3$ could mean a wave function of the following nature:

$$\Psi = C_1 \psi( (n+1) s) \pm C_2 \psi(n d) \pm C_3 \psi(nd) \pm C_4 \psi(nd) $$ where $n=3$ in the case of Mn or Cr.

If one performs the proper molecular orbital calculations on the valence $3d$, $4s$, and $4p$ for manganese or chromium and the $2s$ and $2p$ for oxygen in tetrahedral symmetry then you can draw the molecular orbital energy diagram for your complex once you match your eigenvalues to the correct trace of the appropriate transformation matrix... and you can hear a bunch more jargon on it, or I can explain the main point of our models: why we think MnO$_4$ $^-$ is purple.

The bonds in, specifically permanganate  (I have never actually done the calculations on dichromate, although they should be similar in principle) are between Mn$^{+7}$ and four O$^{-2}$ in the geometry shown here. The electronegativity of oxygen clearly dictates most of the electron density and, as a result, has them sitting on ligands (oxygen) instead of the metal. An electronically excited state can be reached with light absorption in the range of 500-600 nm light due to the relative weakness of the interactions between the ligand and metal. If they were stronger interactions then it would take more energy to promote electrons into a higher energy state and the color would be shifted into the UV. A colorwheel tells you that light absorption in the 500-600nm range should be approximately violet, which is what we see for the color of this particular complex. Anyway, this excited state means some electrons temporarily move from the ligand to the metal, resulting in a LMCT band.


What is the mechanism of addition of glycine into 5-chloro-N-methylisatoic anhydride?

I propose the following mechanism. Contrary to the answer provided by DHMO, I do not believe that the hemicarbamate intermediate is stable. Instead, I assume that it breaks down into carbon dioxide and an amido anion immediately upon formation or very shortly afterwards. This gives us a nucleophilic amido residue which can attack glycine’s carboxylate function. If need be, we can supply catalytic protons to the carboxylic group before the amido nitrogen’s attack but I have omitted that in the scheme.

Scheme 1: Proposed reaction pathway.

I consider the amino group of glycine to be the first attacker since it is most nucleophilic, and I consider it attacking the ester group since that should be more susceptible than a substituted hemicarbamate. Upon breakdown of the tetrahedric intermediate to give the first amide group, the free hemicarbamate anion is liberated which should immediately liberate carbon dioxide as stated above. The formation of the final amide linkage is thermodynamically favourable, even in spite of the unfavourable ring size.


Why are most hydrocarbons less dense than water?

There is no truly simple explanation for bulk properties of liquids. But the balance of two factors probably goes a long way to explaining the difference between water and simple hydrocarbons.

The density of a liquid is determined by the mass of the molecules and the efficiency of how they pack in the liquid.

Molecules made of heavier atoms will tend to be denser (e.g. chloroform where the heavy chlorines give the liquid a density about 50% higher than water). Water is made of moderately heavier atoms than a hydrocarbon (oxygen is about 30% heavier than carbon).

The other factor is how well the molecules pack in the liquid. This depends on their shape and the molecular interactions among them. Water molecules have very strong hydrogen-bond interactions causing them to, on average, stay closer than molecules of equivalent size that don't hydrogen bond. Hydrocarbons have only weak van der waals interactions so don't, on average, stay so close.

The interaction of the molecular mass and the forces determine the overall density. Water wins on both compared to hydrocarbons.


Where does computational chemistry overlap with organic chemistry?

The first thing that comes to mind: read the backlist of Henry Rzepa's and (especially) Steven Bachrach's blogs.  Depending exactly on what resources you have available, there are lots of things they do on the computational side of organic chemistry that you could at first imitate and then use as a springboard to innovate.

In particular, I've found Bachrach's posts on reaction dynamics to be especially fascinating.  This is the phenomenon where the topography of the potential energy surface near a transition state is such that the reactions don't always follow the minimum-energy path due, effectively, to the inertia of the nuclei as they cross the TS.

I also have a personal interest in Bader's Quantum Theory of Atoms in Molecules, which uses the topology of the electron density to divide molecules up into atoms, where each atomic volume is enclosed by a 'zero-flux surface' in the electron density.  (There's some ongoing controversy over the theory, but it makes good sense to me, for whatever that's worth.)  Bader's book is a good place to start; or, you can search the journal literature for Richard Bader and a whole bunch of examples will surely turn up. I engaged with Dr. Rzepa last year in the comment thread of a post of his about electrides that touched on QTAIM, the electron localization function (ELF), non-covalent interactions (NCI), etc.

Finally, I've been very curious lately about 'charge-shift bonding', a relatively newly described class of bond where the nuclear framework tries to cram so many electrons into a small space that in the end Pauli repulsion drives them apart, causing what might otherwise appear to be a covalent bond to instead take the form of a resonance between two ionic structures.  (The electronic structure of $\ce{F_2},$ for example, is better described as a balanced resonance between $\ce{F- \!- F+}$ and $\ce{F+ \!- F-}$ overlaid atop a very weak covalent interaction, than as a purely covalent $\ce{F\!-F}$ species.) Search the literature for Sason Shaik's work for more information and examples.

Other things that I've come across that have seemed interesting include carbenes, cyclic alkynes, unusual Diels-Alder reactions, and fullerene systems.

I would also subscribe to, e.g., Computational Chemistry Highlights, CCL, and Jan Jensen's Computational Chemistry Daily -- the posts and/or questions that come across there might give you some inspiration.


Concentration of Amino acid calculation?

To clarify:
$\mathrm{p}K_\mathrm{a1}: \ce{NH3^{+}-R-COOH <=> NH3^{+}-R-COO- + H+}$
$\mathrm{p}K_\mathrm{a2}: \ce{NH3^{+}-R-COO- <=> NH2-R-COO- + H+}$
For aminoacids, the the pH calculation is the same as for acidic salts of diprotic weak acids $\ce{MHA}$.
Unfortunately for you, $\mathrm{pH}=\dfrac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$ as the very good approximation, so it is almost independent on the concentration, so you cannot determine the concentration from $\mathrm{pH}$.
More exactly. you could, but by a much more complex formula as you listed above, as it would be very inacurate because of high $\dfrac{\mathrm{d}c}{\mathrm{d(pH)}}$
By other words, aminoacid solutions act as low capacity pH buffers ( because of quite a big difference of their $\mathrm{p}K_\mathrm{a1}$ and $\mathrm{p}K_\mathrm{a2}$.


A lot of questions have been posed on electrolysis and electrochemistry on this forum recently, with many recurring themes. I will explain electrolysis in depth in order for a true understanding to come out of this; if you're in a hurry then skip to the last three paragraphs.

One of the most important things about learning electrochemistry is the abstract property of voltage. Voltage can be considered a driving force, which denotes which direction charges are going to flow in a system. A high positive voltage means that charges most definitely wish to flow in that direction, and a high negative voltage means that the charges wish to flow strongly in the opposite direction. If we really want charges to flow in the direction of negative voltage, we must supply energy to the system in order for that to happen—in other words, we're supplying energy to move a charged particle against an electric field (work must therefore be done to do this, as with mass in a gravitational field).

And so it is with $\ce{H2O}$! The reaction corresponding to our notion of electrolysis is:

$$\ce{2H2O -> O2 + 2H2}$$

At least two of the intermediate species in electrolysis are going to be $\ce{OH-}$ and $\ce{H+}$, and so the energy requirement for this process takes into account the energy required to move an electron from $\ce{OH-}$ to the anode, and also to move an electron from the cathode to $\ce{H+}$.The voltage requirement for this process is  $\pu{-1.23V}$. In reality the voltage requirement is closer to  $\pu{-2V}$ taking into account overpotential (the electrochemical equivalent to activation energy).

So this negative voltage tells us that this process will not happen spontaneously, and we will need to supply a voltage of  $\pu{2V}$ in order for this to happen. So far so good: we never observe electrolysis in a standard beaker of water just sitting in the lab!

Your question asked what happened to the $\ce{KMnO4}$ during electrolysis. $\ce{Mn}$ is quite happy in a $\ce{Mn^7+}$ and $\ce{Mn^4+}$ oxidation state. Therefore in solution, the following reaction can happen:

$$\ce{Mn^7+ + 3e- -> Mn^4+}$$

Note that the above is a simplification, $\ce{Mn^7+}$ and $\ce{Mn^4+}$ don't occur simply as isolated ions in solution. In our case, these oxidation states of manganese are tied up in molecules:

$$\ce{MnO4^- -> MnO2}$$

But the above isn't balanced, so we ask ourselves what other likely molecules are going to be produced in order to balance our oxygens. It's common for these redox systems to produce $\ce{H2O}$, so we add $\ce{H2O}$ on the right to balance the oxygens, and add $\ce{H+}$ on the left so we have enough hydrogens. Thus:

$$\ce{MnO4^- + 4 H+ -> MnO2 + 2 H2O}$$

The voltage requirement for this reaction is $\pu{+1.68 V}$, so this means that the electron transfer happens spontaneously. In other words, this reaction is more likely to occur than the electrolysis of water at a less than neutral pH. If you start your electrolysis of water using distilled water (neutral pH implied) then the reduction of $\ce{Mn^7+}$ doesn't happen initially. It's only when the electrolysis starts and $\ce{H+}$ is generated that the reduction of $\ce{Mn^7+}$ can happen; this is then favoured over electrolysis on thermodynamic grounds.

As a last point, it's unlikely your copper electrode was burning, but these can get very hot if the voltage is high enough and hence melt. Turn the voltage down as only $\pu{2V}$ is needed for water electrolysis.

Experimental hint: try using $\ce{NaOH}$ and $\ce{KMnO4}$ in combination and see if there is any change in your observations. Then try increasing the amount of $\ce{NaOH}$ relative to $\ce{KMnO4}$ to see if you observe any changes which would suggest my theory. Note that other manganese reactions (pdf via The Internet Archive) can still happen at lower voltages, which does complicate things.

To answer your question directly, nothing additional is observed using $\ce{KMnO4}$ as it reacts instead of behaving as a charge carrier for electrolysis. A quality of a good electrolyte is that it does not interfere with your reaction of interest.


How do gelatin and aluminium sulfate strengthen paper?

The process of adding gelatin and such to paper (cellulose) is called sizing. The idea behind using gelatin is that it has both a hydrophilic end and a hydrophobic end to the molecule. The aluminum sulfate (alum) is added as a hydrogen bonding linker between the cellulose and the hydrophilic end of gelatin (or some other similar molecule). Mostly this is just to prevent the paper from getting wet in the first place.

Your second paragraph is spot on. Typically other reagents are used to cross link ( the cellulose in order to retain strength upon getting wet (wet strength). Those reagents are a bit more aggressive. Like alkyl ketene dimer

or urea-formaldehyde polymer.


Thanks for bringing up this topic, and I would have appreciated it a few years earlier, however!
The important topic I am referring to is the apparent exclusive gas-phase formation of the molecule H2SO3, as correctly noted in Wikipedia on H2SO3, to quote:

There is no evidence that sulfurous acid exists in solution, but the molecule has been detected in the gas phase.[1] The conjugate bases of this elusive acid are, however, common anions, bisulfite (or hydrogen sulfite) and sulfite. Sulfurous acid is an intermediate species in the formation of acid rain from sulfur dioxide.[2]

The implication for acid rain formation has previously been noted, for example, in an MIT article, with cited  Reactions (1) to (3) below:

$\ce{SO2 + .OH + M → .HOSO2 +M (1) }$
$\ce{ .HOSO2 +O2 → SO3 + .HO2 (2) }$
$\ce{SO3 + H2O → H2SO4 (3) }$

However, in this recent 2019 work: A New Mechanism of Acid Rain Generation from HOSO at the Air–Water Interface, some important chemistry:

The photochemistry of SO₂ at the air–water interface of water droplets leads to the formation of HOSO radicals. Using first-principles simulations, we show that HOSO displays an unforeseen strong acidity (pKₐ = −1) comparable with that of nitric acid and is fully dissociated at the air–water interface. Accordingly, this radical might play an important role in acid rain formation.

where the net photolysis of gaseous sulfurous acid (in addition to SO2) likely proceeds as follows:
$\ce{H2SO3 (g) + hv -> .OH (g) + .HOSO (g) }$
Supporting source: See Page S6,Table S2, Eq (1), Eq (2), Eq (5) and Eq (12) in this available supplement. Also, related results for the photolysis of nitric acid, to quote:

Here we present both field and laboratory results to demonstrate that HNO3 deposited on ground and vegetation surfaces may undergo effective photolysis to form HONO and NOx, 1–2 orders of magnitude faster than in the gas phase and aqueous phase. With this enhanced rate, HNO3 photolysis on surfaces may significantly impact the chemistry of the overlying atmospheric boundary layer in remote low‐NOx regions via the emission of HONO as a radical precursor and the recycling of HNO3 deposited on ground surfaces back to NOx.

Note, there is thus an implied similar possible acceleration in the rate of photolysis of gas-phase SO2 after becoming H2SO3 at the air–water interface.
The implications of the above chemistry is that in addition to the cited Reaction (1) above (which is a sink for the removal of the hydroxyl radical, that otherwise could be involved in an ozone depletion cycle), the UV photo-induced decomposition of also gaseous H2SO3 likely leads to more problematic radicals cited in the acid rain formation and even ozone depletion.


How do Cryocoolers that use Helium as Coolant work?

A helium based cryocooler works just like an air conditioner, but helium is the gas rather than a halocarbon compound. The boiling point of helium is −268.928 °C, so such a cryocooler could get really cold.

Helium is a rare gas and relatively expensive. So to cool a small experimental apparatus it is much cheaper to buy one helium based cryocooler rather than dewars and dewars of liquid helium.


Maybe your electrode is incorrectly calibrated. But it may be another effect : you should know that at high acidic concentration, the pH values is not obtained by taking the logarithm of the concentration of $\ce{H3O+}$. At high acidic concentration, the concentration should be replaced by the activity, which could be rather different from the concentration. Just for the fun try this experimentation : Take your $\ce{HCl}$ solution with an electrode showing  pH $2.1$. Add an equal volume of saturated $\ce{NaCl}$ solution (which is neither acidic nor basic). You will not believe your eyes by reading your electrode : The pH values goes down to $1$, as if diluting the initial solution produces an increase of concentration !
The activity is sometimes described as the amount of $\ce{H3O+}$ ions divided by the volume of the so-called "free water" (and not by the volume of the solution). In concentrated solutions the "free water" is this small amount of water that is free to move, and not attracted and fixed around the ions. One Liter $1 M$ $\ce{HCl}$, should contain much less that $1$ Liter free water, maybe about $0.1$ Liter free water, or less.


Why is the difference between atomic radii of Na and K greater than that of Rb and Cs?

Electronic orbitals do not come in fixed sizes.  A given orbital will shrink due to greater nuclear charge as more protons are addded to the nucleus.  Going from $\ce{Rb}$ to $\ce{Cs}$ you are adding more orbitals than when you go from $\ce{Na}$ to $\ce{K}$.  But you are also adding more orbital-shrinking protons to the nucleus.

The shrinkage effect on each orbital with more protons gets stronger as we go to heavier atoms, when relativistic effects kick in.  Put in very rough terms: when you solve the Schroedinger equation for electronic wavefunctions the size of the orbital would decrease if the mass of the electron were increased.  In nonrelativistic quantum mechanics this does not occur, but in relativistic mechanics the energy needed to squeeze the wavefunctions (which in this case comes from electrostatic attraction) acts like mass, and thus allows for even more shrinkage.  For alkaline earth metals we end up with almost no change at all going from $\ce{Ba}$ to $\ce{Ra}$.  See https://periodictablegroups.wordpress.com/alkaline-earth-metals-2/.


How to move forward in balancing this equation?

The reason you cannot balance the reaction is because you did not consider everything that happens. In the Prilezhaev epoxidation, mCPBA is a stoichiometric reactant but only one single atom of mCPBA is transferred to the product. Therefore, these compounds are often written on the reaction arrow rather than as reactants and products:

$$\ce{C14H12 ->[C7H15ClO3] C14H12O}\tag{1}$$

However, that still does not tell you what exactly happened. If you take a close look at the mechanism, you will see that mCPBA remains essentially unchanged except that it loses its peroxy function; the second product is mCBA or meta-chlorobenzoic acid. Thus, a full reaction equation would be:

$$\ce{C14H12 + C7H15ClO3 -> C14H12O + C7H15ClO2}\tag{2}$$

Notice how this balances automatically.


What defines matter's phase at room temperature?

It's all in the intermolecular forces (and the air pressure). Intermolecular forces are the forces of attraction and repulsion between molecules to neighboring particles. To understand how they determine physical properties like enthalpies of vaporization, fusion, etc..., you need to think about what states of matter are and how they affect the molecules. When you heat up water to a temperature (average kinetic energy) of 90 °C, the energy within the molecules isn't strong enough to overcome the intermolecular bonds (and the atmospheric pressure above the sample) and become a gas. However, it does have enough energy to overcome the bonds to a certain extent and thus can remain in a liquid state instead of in a solid state. This principle applies to any substance. In order for there to be a change in state of matter, the energy within the molecules has to overcome the intermolecular bonds and atmospheric pressure.

As mentioned several times above, another factor is the atmospheric pressure. A common experiment is to put water in a vacuum and watch it boil, something many first-time-experimenters wouldn't expect. The reason water is able to vaporize at room temperature is because it doesn't have the atmosphere to prevent it from vaporizing and therefore, in order to reach an equilibrium, the water is able to vaporize with much less energy than normally required at STP (standard temperature/pressure).

As a direct response to your question about neon, the reason that the noble gases are usually gases unless at extreme temperatures has to do with the weak intermolecular forces they have (van der Waals forces, or more specifically, london dispersion forces). Because the noble gases don't usually bond intramolecularly with anything else, including themselves, and have full energy levels, they have extremely weak intermolecular forces allowing them to almost always be in the gas phase.

I've listed some helpful links below if you want to take your research to the next step.

http://en.wikipedia.org/wiki/Intermolecular_force
http://wps.prenhall.com/esm_hillpetrucci_genchem_4/16/4219/1080299.cw/index.html
http://www.youtube.com/watch?v=A30lguyG0WM
http://chemed.chem.purdue.edu/genchem/topicreview/bp/intermol/intermol.html
http://en.wikipedia.org/wiki/Van_der_Waals_force


Old samples of magnesium oxide become contaminated with magnesium carbonate. Suggest how this contamination takes place

Carbonate contamination in oxides compounds often happens after a certain time and air exposure because of the $\ce {CO2}$ in the air.

For your magnesium oxide it follows the reaction

$\ce {MgO + CO2 -> MgCO3}$

with $\ce {MgCO3}$ being the magnesium carbonate.

A solution to prevent this kind of polution is to keep the oxide in a dessicator or to treat them in a furnace at a temperature around to $500 °C$ wich is the decomposition temperature of most of the carbonate compounds.


I have calculated pressure in container one and that is p1 = 342817.92 Pa, and in container two p2 = 326979.2275 Pa. Then I have substracted this two values which gives me 15838.6925 Pa. After that I have divided this value with two, to figure out how much do I have to decrease pressure in container 1, and that gives me 7919.34625 Pa.

You cannot generally just average both pressures. As $pV=nRT$, then ( for $\mathrm{T}$ and $\mathrm{V}$ constant ) :
$V \cdot \mathrm{d}p=\mathrm{d}n \cdot RT$
$\frac{\mathrm{d}p}{\mathrm{d}n}=\frac{RT}V$.
So pressure averaging would work only in the special case if $\frac{T_1}{V_1}=\frac{T_2}{V_2}$. Therefore, a weighted average must be used.

I suppose the scenario is vaguely defined. It can be the polytropic  expansion, or it may be the "locally isothermal" expansion. But for the former, it has like infinite number of possible solutions, depending on the rate of the transfer and of heat exchange.
I consider below the latter, i.e. isothermal conditions in each of containers.

We can consider the ideal behaviour, where
$$p=\frac{nRT}V$$
For equality of pressures, if $\Delta n$ is the moved molar amount of xenon from the container 2 to the container 1, the following equality must be true:
$$p=\frac{(n_1 + \Delta n)RT_1}{V_1}=\frac{(n_2 - \Delta n)RT_2}{V_2}$$
From that, we can easily calculate $\Delta n$:
$$(n_1 + \Delta n)RT_1V_2=(n_2 - \Delta n)RT_2V_1$$
$$\Delta n \cdot R(T_1V_2 + T_2V_1)=n_2RT_2V_1 - n_1RT_1V_2$$
$$\Delta n=\frac{n_2T_2V_1 - n_1T_1V_2}{T_1V_2 + T_2V_1}$$
Now you plug in the numbers here in the final formula, instead of working with literal numbers all the way:

In two containers we have Xenon. The volume of the first container is 2780 L, temperature is 300 degree Celsius, and n = 200 mol. Volume of the second container is 4000 L, temperature is 1300 degrees Celsius and n = 100 mol. Temperature in both of the containers is constant. How much Xe do we have to transfer from one container to another so that the pressure in both containers is the same?

$$\Delta n=\frac{(\pu{100 mol})(\pu{1573.15 K})(\pu{2.780 m3}) - (\pu{200 mol})(\pu{573.15 K})(\pu{4.000 m3})}{(\pu{573.15 K})(\pu{4.000 m3}) + (\pu{1573.15 K})(\pu{2.780 m3})}$$
In the sense below, if to be precise, we should round the temperature values for presented formula, because temperature was not known in 2 decimals accuracy ( otherwise there would be $\pu{300.00 ^{\circ}C}$, resp. $\pu{1300.00 ^{\circ}C}$), internally still computing with non rounded values.
$$\Delta n=\frac{(\pu{100 mol})(\pu{1573 K})(\pu{2.780 m3}) - (\pu{200 mol})(\pu{573 K})(\pu{4.000 m3})}{(\pu{573 K})(\pu{4.000 m3}) + (\pu{1573 K})(\pu{2.780 m3})} \approx \pu{-3.18 mol}$$
The negative value means the $\pu{3.18 mol}$ of xenon is to be moved from the first container to the second one, as we have formally assumed the opposite direction.

Internal full calculator precision of computation still makes good sense, to avoid artefacts of rounding errors.  But for presentation of final or intermediate values of quantities from the real world, the values have to reflect error propagation rules. As input data and constants have limited accuracy.


Is real water made up of heavy water and light water?

Water has formula H2O. Oxygen has 3 stable isotopes (99.76% 16O, 0.039% 17O, 0.201% 18O), and hydrogen has two (99.985% 1H, 0.015% 2H). Thus, there are 9 natural isotopic configurations for water: 3 possibilities for oxygen, multiplied by 3 possibilities for 2 hydrogens with 2 possible isotopes.

Out of those 9 possible configurations, only 4 have a natural abundance larger than 1 ppm:

So, pure water is a mixture of 9 different isotopic configurations. You will note that D2O, which is what is usually called heavy water, is not actually among the 4 more common isotopic configurations given by natural abundance.

Also of note: water that is not “heavy water” is sometimes called “light water”, but never “soft water”. “Soft water” is a term used for a chemical process of natural (non pure) water that has been depleted from certain metal cations. (It is used in contrast to “hard water”.)


How were elements and molecular formulæ determined in the old days?

First thing, they couldn't tell elements and compounds apart. See Dalton's list of elements with their symbols.

Lime is right there--and it's not actually an element. Neither is Baryte(s). Soda and Potash probably refer to the oxides of Sodium and potassium, though terminology may have been different back then.

In those days, elements were things that could not be synthesized by addition reactions. If you have a reaction $\ce{A + B ->C}$, then $\ce{C}$ is a compound (can't say much about the other two). Alternatively, if the substance can be decomposed (via anaerobic heating/etc), then it's a compound.

Most probably they would stuff things into the list "innocent until proven guilty" (I don't know about this).

Molecular mass can be determined by osmotic pressure, ebullioscopy, Cryoscopy, etc. (Usually stuff delendant on colligative properties)

Determination of molecular formula is much harder.

For organic compounds, it can be done by burning in a controlled environment. The ratios of oxides produced lets us know the relative quantities of all elements but oxygen. The quantity of oxygen can be determined by finding the decrease in Oxygen content of the chamber, and subtracting it from the Oxygen content of the oxides. The molecular formula can be found if you use the emperical formula and determine molecular mass.

For inorganic compounds, they probably had various tests for ions/etc (like the silver nitrate test for halides). They also had flame tests (and the borax bead test), which could help identify certain ions. I'm not too sure here, though.


For this reaction

$$\ce{A -> B}$$

a $\Delta G$ value of $1\ \mathrm{kcal\,mol^{-1}}$ will result in an equilibrium value

$$K = e^{-\frac{\Delta G}{RT}}$$

of approximately 0.18 favoring reactant. Every additional increment or decrement by this value will change the relative ratio by roughly 5.


Carius method for quantitative analysis of halogens

1) $\ce{HNO3}$ is a more oxidizing acid than $\ce{H2SO4}$.

2) The silver nitrate would not dissolve well in the organic solution.

3) Sulfide is an example of something you might worry about interfering as it forms a very insoluble compound with silver. But silver sulfide and most anything else that might interfere will decompose in the fuming nitric acid.


HPLC Columns For Compounds Not Soluble In Non- Or Low Polar Solvents

Try searching Google Scholar and write HPLC separation and the names of your compounds. You will find plenty of publications and of course column names. If your compounds are new, you should test the solubility in MeOH, ACN, and 50:50 MeOH/H2O and 50:50 ACN/H2O. Do they dissolve in such solvents systems? If they do, you would use C-18 RPLC columns. There are literally hundreds of columns with similar C-18 chemistry.

If your compound is highly polar (carboxylates, amines, phosphates etc.), even bare silica column can do the separation. This technique is called HILIC. I have written several articles on HILIC. Try Google Scholar for reviews on HILIC.


Looking at the formal charges on the atoms, each oxygen has a charge of -2. With the added electron, the iodine has a formal charge of +5. This leaves 2 electrons in the valance shell for the lone pair. So there are 4 items (3 oxygens and a lone pair) to place around the iodine atom, giving it a tetrahedral shape. Since each oxygen has the same charge of -2, there must be 2 bonds to each oxygen. The extra electron is placed on the molecule, not the iodine atom. Since you have a tetrahedral shape with a different corner (lone pair), it has a dipole (like ammonia).


It is possible to make an homogeneous solution of water and oil?

Generally polar and non polar liquids do not dissolve in one another to any great extent. Practically, you would have to find a temperature composition phase diagram for the oil you want to use. This will tell you if there is a region in which the water and oil are mixed as a homogeneous solution. Alternatively if you add surfactant then a ternary phase diagram (triangle plot) is what you need to look for since than micellar or liquid crystal phases may form.

As you mention the free energy I have tried to summarise what happens on dissolving a non-polar molecule in water.
The enthalpy $\Delta H$ change on placing, alkane, into water is small, slightly positive or negative depending on the compound, -9kJ/mol for ethane, +1.6 for toluene for example.  This can be thought of as the cost  making a void in the water to hold the molecule.

Water is unusual in having a strong and highly directional hydrogen bonding nature. Thus adding a non-polar molecule into water is determined by entropic effects. This generally means that $T\Delta S$ is negative and so $\Delta G$ is positive, for cyclohexane into water $\Delta G$ = +28 kJ/mol.

The entropy of water decreases on contact with non-polar molecules since these molecules cannot form hydrogen bonds. The water has to bridge around the non-polar molecules, as this is a lowering of entropy compared to not bridging. This bridging partly restricts the water molecules compared to bulk water. This option is preferable as the alternative of breaking hydrogen bonds, and leaving H atoms pointing at the non-polar molecules,is too energetically costly. Bridging makes it possible to form some H bonds at the expense of entropy loss. The figure shows this (from Southall et al. J. Phys. Chem. p521, 106, 2002 )

As the temperature increases the enthalpy increases as does T$\Delta S$, but $\Delta G$ remains largely constant and remains positive, see sketch below. This suggest that  water and oil will not easily mix.