text:
Why atropisomers are called conformers?
Changing one atropisomer to another requires bond breaking (in some cases the removal and reattachement of steric groups, according to my understanding) so how it is possible that according to IUPAC atropisomers are classified as conformers (conformational isomers)?
According to IUPAC:

Atropisomers - A subclass of conformers which can be isolated as
separate chemical species and which arise from restricted rotation
about a single bond.
Conformers - One of a set of stereoisomers, each of which is
characterized by a conformation corresponding to a distinct potential
energy minimum.
Conformation - The spatial arrangement of the atoms affording
distinction between stereoisomers, which can be inter-converted by
rotations about formally single bonds. Some authorities extend the
term to include inversion at trigonal pyramidal centres and other
polytopal rearrangements.

This definition of "conformation" seems vague to me, but it is often said that changing one conformation to another involves no bond-breaking and changing one configuration to another requires bond-breaking. This is the difference between them.
IUPAC gives examples of atropisomerism: e.g. (E)-cyclooctene (given in Gold Book "planar chirality" entry). This definitely should involve bond-breaking. Other IUPAC examples of atropisomerism include substituted ortho-biphenyls.

Think of the whole bond-breaking narrative in a different way: the bonds in both isomers are exactly the same.

They're conformers, but with very high energy barrier of rotation.

@IvanNeretin I don't understand? Bonds are the same in (configurational) chiral centres too, but they are oriented differently and require bond breaking as well as (E/Z)-cyclooctene which supposedly is atropoisomeric.

@Mithoron Yes, some of them, like (E)-cyclooctene requires a lot of energy to break the double bond temporarily to turn it in to (Z)-cyclooctene, but this bond-breaking is the problem I have trouble understanding. How come you can call something that involves bond-breaking atropisomerism and conformerism?

@user35715: The transformation of E- to Z-cyclooctene is a configurational change that involves bond breaking be it sigma or pi bonds. But this is not the issue. Think about the E-double bond of cyclooctene "jumping rope with the methylene chain", which interconverts enantiomers. Granted it is difficult for this ring but no bonds need to be broken.

I think your notation of "Changing one atropisomer to another requires bond breaking (in some cases the removal and reattachement of steric groups, according to my understanding)" is not correct. To my knowledge, the free rotation of one single bond of an atropisomer is not restricted by connecting bond(s). I think you get confused by some literature explaining breaking a bond selectively directed to pair of atropisomers (e.g., Ref.1):

According to your post, IUPAC definition of atropisomers:

Atropisomers - A subclass of conformers which can be isolated as separate chemical species and which arise from restricted rotation about a single bond.

A variety of sources confirm this definition. For instance, Ref.2 and 3 described atropisomers as:

Atropisomers are stereoisomers (rotamers) resulting from hindered rotation about single bonds (having an energy barrier to rotation about a single $\sigma$ bond) usually due to steric hindrance. This energy barrier to rotation is high enough to allow the isolation of the conformers (rotamers).

Nonetheless, the name, coined by German biochemist Richard Kuhn in application to a theoretical concept 1n 1933 (Wikipedia) is derived from Greek (a = not and tropos = turn; thus, atropos meaning "without turn"), which supported the definition. However, the atropisomerism was first detected in 6,6'-dinitro-2,2'-diphenic acid by Cristie, et al. in 1922 (Ref.4):

It is also important to know the nomenclature of atropisomers (Wikipedia):

Determining the axial stereochemistry of biaryl atropisomers can be accomplished through the use of a Newman projection along the axis of hindered rotation. The ortho, and in some cases meta substituents are first assigned priority based on Cahn–Ingold–Prelog priority rules (Ref.5 & Ref.6). One scheme of nomenclature in based on envisioning the helicity defined by these groups (Ref.7). Starting with the substituent of highest priority in the closest ring and moving along the shortest path to the substituent of highest priority in the other ring, the absolute configuration is assigned $P$ or $\Delta$ for clockwise and $M$ or $\Lambda$ for counterclockwise. Alternately, all four groups can be ranked by Cahn–Ingold–Prelog priority rules, with overall priority given to the groups on the "front" atom of the Newman projection. The two configurations are termed $R_\mathrm{a}$ and $S_\mathrm{a}$ in analogy to the traditional $R/S$ for a traditional tetrahedral stereocenter (Ref.8}:

Atropisomers are also found in nature. The main pigment of cotton seeds, gossypol is an atropisomer, which exists in both form of $(R_\mathrm{a})$- and $(S_\mathrm{a})$-conformations. Most commercial Upland (Gossypium hirsutum) cottonseeds have an $(R_\mathrm{a})$- to $(S_\mathrm{a})$-gossypol ratio of ~$2:3$, but some Pima (Gossypium barbadense) seeds have an excess of $(R_\mathrm{a})$-gossypol. Between two isomers, $(R_\mathrm{a})$-gossypol is more toxic and exhibits significantly greater anticancer activity than its $(S_\mathrm{a})$-atropisomer:

Two fun facts about atropisomers are (KU.edu):

Atropisomers are detectable by $\mathrm{NMR}$ if half lives exceed $\pu{10^{-2} s}$.
Atropisomers are isolatable if the half-life is above $\pu{10^3 s}$.

References:

Gerhard Bringmann, Thomas Hartung, “Atropo-enantioselective biaryl synthesis by stereocontrolled cleavage of configuratively labile lactone-bridged precursors using chiral H-nucleophiles,” Tetrahedron 1993, 49(36), 7891-7902 (https://doi.org/10.1016/S0040-4020(01)88014-5).
Alan R. Katritzky, Christopher A. Ramsden, John A. Joule, Viktor V. Zhdankin, In Handbook of Heterocyclic Chemistry, Third Edition; Elsevier Limited: Amsterdam, The Netherlands, 2010 (ISBN: 978-0-08-095843-9).
Jonathan Clayden,  “Atropisomerism,” Tetrahedron 2004, 60(20), 4335 (https://doi.org/10.1016/j.tet.2004.03.002).
George Hallatt Christie, James Kenner, "LXXI.—The molecular configurations of polynuclear aromatic compounds. Part I. The resolution of $\gamma$-6 : 6′-dinitro- and 4 : 6 : 4′ : 6′-tetranitro-diphenic acids into optically active components," Journal of the Chemical Society, Transactions 1922, 121, 614–620 (https://doi.org/10.1039/CT9222100614).
R. S. Cahn, Christopher Ingold, V. Prelog, “Specification of Molecular Chirality,”  Angew. Chem. Internat. Ed. Engl. 1966, 5(4), 385-415 (https://doi.org/10.1002/anie.196603851) and Corrigendum: Angew. Chem. Internat. Ed. Engl. 1966, 5(5), 511-511 (https://doi.org/10.1002/anie.196605111).
G. P. Moss, "Basic terminology of stereochemistry: IUPAC Recomendations 1996," Pure and Applied Chemistry 1996, 68(12), 2193-2222 ()(PDF).
http://goldbook.iupac.org/terms/view/H02763: IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.
http://goldbook.iupac.org/terms/view/A00547: IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.

While the user seems to have vanished, its worth pointing out the misconception they are having. Its not that the (E) and (Z) enantiomers of cyclooctene are atropisomers. Rather, (E)-cyclooctene has two atropisomeric conformations, which can be interconverted with out breaking the double bond. Rotation is hindered around this bond, but it can still occur due to rotation of the two surrounding single bonds. A very nice depiction of this can be found on Henry Rzepa's old website
(In case the link breaks, here is the DOI for the main hub of the site)
DOI: http://doi.org/10042/a3uy9


text:
Understanding water models?
I generally work on fluid flow, so some basic concepts in chemistry are new to me.
 Work I'm doing requires use of molecular dynamics (MD) to do classical potential simulations of water–NaCl systems but this question is about water models more than simulation and software details.
The simulations require definition of a "water model."  There's apparently one already implemented in an example/tutorial I'm using (below).  I want to compare the results of simulations based on different water models.  My problem is that I'm having a lot of trouble understanding how to change my water model (which I think is SPC/E) because I don't really understand what it is.
I have found tables that provide parameters for different water models, like this table at Wikipedia:
https://en.wikipedia.org/wiki/Water_model
However, when I look at my software, I can't relate what I'm seeing to the table.
 I can't find an example of where someone has taken information from this sort of table and explained how to implement it in a simulation.  So my first request is, please point me to anything out there!  Here's the part of my software that I think defines my water model:

&FORCEFIELD
  &BEND
    ATOMS H O H
    K 0.
    THETA0 1.8
  &END BEND
  &BEND
    ATOMS O H H
    K 0.
    THETA0 1.8
  &END BEND
  &BOND
    ATOMS O H
    K 0.
    R0 1.8
  &END BOND
  &BOND
    ATOMS H H
    K 0.
    R0 1.8
  &END BOND
  &CHARGE
    ATOM O
    CHARGE -0.8476
  &END CHARGE
  &CHARGE
    ATOM H
    CHARGE 0.4238
  &END CHARGE
  &NONBONDED
    &LENNARD-JONES
      ATOMS O O
      EPSILON 78.198 ! this is K, = 0.155 kcal/mol = 0.650 kJ/mol
      SIGMA 3.166
      RCUT 11.4
    &END LENNARD-JONES
  &LENNARD-JONES
      ATOMS O H
      EPSILON 0.0
      SIGMA 3.6705
      RCUT 11.4
    &END LENNARD-JONES
    &LENNARD-JONES
      ATOMS H H
      EPSILON 0.0
      SIGMA 3.30523
      RCUT 11.4
    &END LENNARD-JONES
  &END NONBONDED
&END FORCEFIELD

I can generally see that everything up to &NONBONDED relates to water somehow because it's about charge, H, and O (I have no idea what "K" is).  But what about the "LENNARD-JONES" stuff?  I know in general what a Lennard Jones model is, but cannot see how that would be related to water models.  If it is, how? Thank you for any advice on how to understand this.

jonsca, sorry for leaving another question, but my reputation doesn't allow me to comment on your comment. I don't see a checkmark icon. I know what it should look like because I had an account here in the past. If you could possibly mark the answer as correct, that would be great. I would like to make sure that the answerer gets credit for his great explanation. Thank you.

I can explain some of this, and actually some of it is explained on the Wikipedia page you referenced.
Most water models represent the oxygen using a Lennard-Jones potential, which encapsulates the repulsive interaction that stops the molecules getting too close (this is physically because the electron shells don't want to overlap each other) and a long-range attraction due to Van der Waals forces. The Lennard-Jones potential can be parameterized in several ways, two of which are:
$$
v_{\text{LJ}}(r)
= 4\epsilon \left[ \left(\frac{\sigma}{r}\right)^{12} -
\left(\frac{\sigma}{r}\right)^{6} \right]
=
\frac{A}{r^{12}} - \frac{B}{r^6}
$$
The first of these appears in your force field specification file.
The $\epsilon$ parameter is actually given in Kelvin, which is $\epsilon/k_{\text{B}}$, where $k_{\text{B}}$ is Boltzmann's constant.
Hence the note "this is K".
This may seem an odd choice, but there are some historical reasons.
However, you'll see on the same line,
in a comment for the user,
that this number is translated into SI units and also into old-time cgs units of kcal/mol.
The $\sigma$ parameter is also given in your file, in old-style Angstroms.
Just to be clear, this is the interaction between oxygen atoms,
hence it is described as the O-O Lennard-Jones term.
The second form appears on the Wikipedia page. The values of $A$ and $B$ appear in the table, in old-time units. If you do the necessary multiplications
you can check that (to within a few significant figures, anyway),
$4\epsilon\sigma^{12}$ and $4\epsilon\sigma^{6}$ do indeed agree with the
tabulated $A$ and $B$, taking $\epsilon$ in kcal/mol and $\sigma$ in Angstroms,
for the SPC/E model.
There are some redundant entries in your file: BEND and BOND. For a molecule like this,
one could allow the OH bonds to vibrate, and the HOH bond angle to vary,
and these terms would need some force constants and equilibrium bond lengths and
angles.
But SPC/E is a rigid model: the bond lengths and angles are fixed,
so it looks like the force constants ($K$) have been set to zero.
One might imagine that the H atoms deserve also to be represented as Lennard-Jones atoms,
but they are quite small, and in this model this aspect is neglected.
Hence we see that $\epsilon$ is zero for the H-O and H-H interactions.
So then, all that is needed to complete the model is the charge distribution,
and it looks like you've identified the three charges,
one on each atom,
characterizing SPC/E.
This isn't the most sophisticated model,
as you can see if you scroll down that page.
For more details concerning the input file,
I would hope that your software package and/or exercise instructions,
give enough information.
Hope this helps at least to understand the correspondence between the
numbers in your file,
and the numbers on the Wikipedia page.

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text:
Is d3-acetic acid more acidic than acetic acid?
The following table is from Solomons, Fryhle and Snyder Organic Chemistry, chapter 3, An Introduction to Organic Reactions and Their Mechanisms: Acids and Bases:

Table 3.8 Isotopic substituents and inductive effect
  $$ \small
\begin{array}{ccc}
\hline
\text{Acid} & \ce{pK_a} & \text{Inductive effect of the substituent} \\
\hline
\ce{CH3-COOH} & 4.75 & \text{+I} \\
\ce{CD3-COOH} & 4.60 & \text{+I} \\
\hline
\end{array}
$$
  +I effect of deuterium is stronger than that of hydrogen. +I ($\ce{-CD3>-CH3}$)

Why is $\ce{CD3-COOH}$ more acidic than $\ce{CH3-COOH}$ despite being less electronegative? The book also said that $\ce{CD3} > \ce{CH3}$ in terms of + Inductive effect.

Which edition you referred to? In the global 12E, there is no table 3.8 and in the Indian 2017 edition, the value for $\ce{CD3COOH}$ is 4.85.

To compare acidity, you must have made the conjugate bases of these acids, that is, $\ce{CH3COO-}$ and $\ce{CD3COO-}$ . Now, the hyperconjugation effect is usually a more dominant electronic effect than the inductive effect, and as the $\ce{C-D}$ bond is stronger than $\ce{C-H}$ bond, so the reduced $\ce{+H}$ effect in the conjugate base of $\ce{CD3COOH}$ makes it more acidic.

But there are zero alpha hydrogens on the carbon atom next to the oxygen anion so how is hyperconjugation possible ?

There are α-hydrogens and α-deuteriums present with respect to the $\ce{COO-}$ group, so to see the hyperconjugative structures, break the $\ce{C-H}$ bond and shift the electron pair so that you get two negatively charged oxygens on one carbon, which can protonate and revert to the original carboxylate ion. Anyway, hyperconjugation is a theoretical effect, and it is not defined clearly for carbanions due to absence of vacant orbitals, so I would advise you not to set much store by the actual mechanism of hyperconjugation here, as the di-negative  carbanaion formed was extremely unstable.

Again, I would like to add here that since hyperconjugation is completely theoretical, so the only concrete thing we can say here is that by whichever path the $\ce{+H}$ maybe happening, it is clear that it will increase electron density on the $\ce{COO-}$ group, so please don't take these structures I just suggested as absolute

The following articles suggest that the data reported from Solomons, Fryhle and Snyder Organic Chemistry is likely incorrect.

Maya Paabo, Roger G. Bates and R. A. Robinson ("Dissociation of Acetic Acid-d3 in Aqueous Solution and Related Isotope Effects from 0 to 50°", J. Phys. Chem. 1966, 70, 2, 540-543) report that the $\mathrm{p}K_\text{a}$ of the deuterioacid is greater (less acidic) than that of the protio analogue.
See Table 1 on p. 134 of C. L. Perrin's chapter ("Advances in Physical Organic Chemistry", Volume 44, edited by John P. Richard) where it is also reported that the $\mathrm{p}K_\text{a}$ of the deuterioacid is greater (less acidic) than that of the protio analogue.
Rose McConnell, Walter Godwin, Brant Stanley, and M.Shane Green ("Acidity Studies of Deuterated Acids and Bases Commonly Used as Buffers in NMR Studies", Journal of the Arkansas Academy of Science, Volume 51 Article 21) also report (see Table 1 on p. 140) similar findings.

These findings indicate that acetic acid-d3 is less acidic than acetic acid-h3.  This would be consistent with the operation of a simple inductive effect where protium is more electronegative than deuterium (as expected).  Hence, the more electron withdrawing $\ce{CH3}$ group is better able to stabilize the electron rich carboxylate anion in the the ionized acid as compared to the less electron withdrawing $\ce{CD3}$ group.

What I have read is that one 'only' needs to see the inductive effect of the chain attached to compare the acidic strength of carboxylic acids. However, at multiple places in Chemistry SE, people have written that hyperconjugation effect and inductive effect of alkyl groups are related, so one should also consider hyperconjugation along with inductive effect. What's your take on it?

@ron Based on this, can we say that migratory aptitude in a carbocation species follows the order: $\ce{CH3 > CD3 > CT3}$? It is due to the fact that if methanide migrates then the other two isotopic methyl group (due to more $\pu{+I}$ effect) will be able to stabilize the positive charge of carbocation better in comparison to other two scenarios?

@Apurvium  The $\ce{CH3}$ group is more electron withdrawing, hence it would be expected to migrate more slowly (stabilize the carbocation less in the TS).

@ron $\ce{CH3}$ is more electron withdrawing, thus, formation of $\ce{CH3-}$ will be easier and in turn, it can migrate fast.

@Apurvium But it's all about the migrating groups ability to stabilize the cationic center.

@ron But what about the stability of new carbocation (after rearrangement) formed?

@Apurvium That has a very small effect on the transition state.

@ron I am confused... So, posted this question

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text:
Recover Solder (Sn-Pb) from wires
I have began to do an "art project" which utilizes solder as main material.
While I know that solder is cheap and easily found in multiple shops, I have set my mind on using recycled solder from out of use PCB's.
I've recovered a somewhat pure amount of material by using solder suckers, but a significant amount is trapped in my DIY desoldering braid which is simply salvaged wires. Those wires might be copper or some grey wires salvaged from a PC power supply.
What I have been able to come with is:

Some video I watched used (IIRC) $\ce{HCl}$ to dissolve solder from PCB's. My main issue is that I can't easily procure high concentration $\ce{HCl}$. My second issue is the acid reacting with the copper and maybe rendering the solder not usable (if this turns out to be a non-issue, I could figure a way to procure the acid).
I have been thinking of pulverizing the solder covered wires using a coffee grinder then heating the powder to the meting point of the solder, mechanically separating the two.

I wanted to ask if any of the two ways above would work without specialized equipment, or if there was a third way.

Trying chemically recover solder from solder braid is a losing battle.

I can't imagine an individual collecting significant solder by these methods. Can you adapt zinc to your project ? Much,much more available and similar physical properties to solder.

@blacksmith37 here the "art" in the project comes in. I am trying to use  recycled materials, and I don't have a source for zinc, that I know of. It was a try I guess.  A futile try, as Max said.  Thanks!

Zinc is very common in autos and consumer products , fourth most common after iron, aluminum and copper.  Also called pot metal and white metal.

Unfortunately the reason solder works is that it partially dissolves copper as shown in Figure 1 to form $\ce{Cu3Sn}$ and $\ce{Cu6Sn5}$ phases (also known as bronze) shown in Figure 2. Because of this you cannot simply melt solder off the solder wick to recover it as the tin will continue to dissolve the copper forming bronze. There is also the issue of needing rosin in your recovered solder to be useable. This 1960's video from NASA is an excellent video for learning more about the soldering process.

Figure 1. Reactions of the Copper-Solder Interface

Figure 2. Copper-Tin Binary Phase Diagram
To address the acid method. Physically, it could work (I would not recomend doing so with leaded solder) as most ROHS compliant electronics are $95\%+$ tin solders with either silver or antimony added as alloying agents to prevent tin whiskers from forming. That said it economically dissolving in acid then precipitating then reducing tin in a way which does not also produce copper would be a tedious and time consuming process. You are far better off just buying solder which does not cost that much and has rosin needed to flux the solder.

First of all, thank you for your detailed answer, I am looking at the NASA video right now. My follow up question would be: would the acid way work? Not rendering the solder unusable after the separation would be something I don't want. Edit: Hit Enter too soon.

There is no way to use acid to leach the copper and leave the solder.

I would have written a very similar answer if I had seen the question first


text:
What studies show that d orbitals don't play an important role in covalent bonding? (> period 2 elements)
Take a look at all the following molecules: $\ce{SO3^{2-}}$, $\ce{SO2}$, $\ce{SO4^2-}$, $\ce{PF5}$ and many more you can add. In all these there is a controversy of one of these two types:

Is formal charge more important than octet rule - as to justify hypervalency- on determining Lewis structure? As a rule I'd say the answer is no. For example in $\ce{SO2}$,$\ce{SO3}$, $\ce{SO4^2-}$, Do calculations show it?
What is the central atom hybridization? For example, in $\ce{PF5}$ (see here). Does it involve d-orbitals?

In any case, both previous statements are related to the d-orbitals energy.
There is -at least- one straightforward way to answer the question:

Calculate* the Molecular Orbitals and see which of previous options fit better (the use of d-orbitals or the non use of them).
*sure some can be found in some papers.

So the question is:
Are there theoretical studies of many concrete examples as to remark that d orbitals don't play any relevant role in covalent bonding?

Sorry, can you clarify your question? What info are you looking for that is not in the related posts?

@Greg look at the title, I think the posts below show $d$ orbitals are not important in covalent bonds. I ask what is the Evidence.

Some related pub.: 1. Gillespie, R. Coord. Chem. Rev. 2002, 233-234, 53–62. DOI: 10.1016/S0010-8545(02)00102-9 2. Gillespie, R. Inorg. Chem. 1995, 34 (4), 978–979. DOI: 10.1021/ic00108a032 3. Smith, D. J. Chem. Educ. 2005, 82 (8), 1202. DOI: 10.1021/ed082p1202 4. Jensen, W. J. Chem. Educ. 2006, 83 (12), 1751. DOI: 10.1021/ed083p1751 5. Halgren, T. J. Am. Chem. Soc. 1977, 99 (21), 6793–6806. DOI: 10.1021/ja00463a002

Is this only meant to be about supposedly hypervalent compounds of main group elements?

Why don't you look for studies that do show a significant d orbital contribution to covalent bonding. Proving nonexistence is hard.

@Karl I have found 2 treating this questions, but couldn't read them completely yet, and both are quite dificult to understand. One from cooper 1994 (https://pubs.acs.org/doi/abs/10.1021/ja00089a033), the other one from magnusson 1990 (https://pubs.acs.org/doi/abs/10.1021/ja00178a014).


text:
How do I identify how many reactions are occurring when precipitating a mixture?
An example of a question is:

After dissolving $x$ grams of a mixture of $\ce{Na2SO4}$ and $\ce{NaCl}$ in water, you add an excess of $\ce{BaCl2}$ solution. You filter off the precipitate which forms, dry it, and find that it weighs $y$ grams. What is the percent by mass of $\ce{Na2SO4}$ in the original mixture?"

If I am correct, the reaction for this would be:
$$\ce{BaCl2 + Na2SO4 + NaCl -> BaSO4 + 3NaCl}$$
However, there are some questions which are of the same format but have two separate reactions:

A $x$ gram mixture of $\ce{CaCl2}$ and $\ce{RbCl}$ is dissolved in water. Addition of $\ce{AgNO3}$ causes the chloride ion to precipitate as $\ce{AgCl}$. The mass of $\ce{AgCl}$ collected is $y$. What is the percent by mass of $\ce{RbCl}$ in the original mixture?

For this one, would the reaction be two equations?
\begin{align}
\ce{CaCl2 + AgNO3 &-> Ca(NO3)2 + 2AgCl}\\
\ce{RbCl + AgNO3 &-> RbNO3 + AgCl}
\end{align}
I know how to do these types of questions past the point of deciding whether or not there are two equations. I just don't understand why I couldn't write that last one as a single equation, in the same way as I did the first example. Could anybody help me? Thanks in advance.

I have updated your post with chemistry markup. If you want to know more, please have a look here and here.

You could: Ag+ + Cl- -> AgCl

You asked two very different questions.

(1) How do I identify how many reactions are occurring when precipitating a mixture?

There is no information in way the reactants are shown in a chemical equation as to what the products will be. That is to say that you have to know the products of the reaction.

(2) For this one, would the reaction be two equations?

Chemists use shorthand all the time. Reactions are typically netted out to the pertinent aspect under consideration. However if I were writing the reactions for the second problem I'd do it much differently that you have presented. I think I'd write:
\begin{align}
x\ce{CaCl2} + y\ce{RbCl &->} x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{Cl-}\\
x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{Cl-} &\ce{->[excess AgNO3]} x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{AgCl v}\\
\end{align}
From an analytical chemistry perspective there are two steps. First the chloride salts are dissolved and then the silver nitrate is added. You wouldn't want to add the salts to a silver nitrate solution since the silver chloride could coat the Ca and Rb salts preventing a complete reaction.
I even might write the second reaction as:
$$ (x+y)\ce{Cl-} \ce{->[excess AgNO3]} (x+y)\ce{AgCl v}$$
I'm not a fan of one line jobs like:
$$x\ce{CaCl2} + y\ce{RbCl ->} x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{Cl-} \ce{->[excess AgNO3]} (x+y)\ce{AgCl v}$$


text:
An equal amount of hydrogen gas is dissolved in a 1L vessel and a 2L vessel, both filled with water. Which vessel has higher entropy?
This question came up in our chemistry class and we could not determine which would have the higher entropy. Clearly, Vessel 2 has a larger volume, but Vessel 1 has a higher concentration of $H_2$. How do we resolve this?

Why does the concentration matter?

Please don't use MathJax in the tile. it make the question less searchable

If a seemingly simple question is hard to answer, then it is likely badly put. ;-) Entropy is an extensive property. The 2l vessel already contains twice the entropy before dissolving hydrogen in it.

Denote by Vessel I the vessel with volume $V$, moles of water $N$, and moles of hydrogen $N'$. Denote by Vessel II the vessel with volume $2V$, moles of water $2N$, and moles of hydrogen $N'$.
Increasing the volume will increase the number of microstates available to the system, and hence also the entropy. Increasing the moles of water will increase the number of microstates available to the system, and hence also the entropy.
Both statements imply that Vessel II has the greater entropy.


text:
Solubility curve for sodium tetraborate in water
What is the solubility curve for sodium tetraborate decahydrate ($\ce{Na2[B4O5(OH)4]·8H2O}$), otherwise known as borax, in water ($\ce{H2O}$)?
Additionally, what is the solubility of the aforementioned substance, borax, at $\ce{100°C}$?

The solubility curve of borax is taken from here.

We can infer that solubility of borax at 100°C is around 520g/L. It is observed that borax in water exhibits supersaturation . When the solution is at 100°C and then you drop at 40°C, then a clear solution is obtained when amount of borax is 60 grams per litre. More than that, there is a tendency of the solution to overflow. See ref.1 for the more detailed explanation of supersaturation and ref.2 for the dissociation reactions happening in water. First the tetraborate ion is formed and it is a weak base in solution and thus reacts with additional water to form boric acid, a weak acid and hydroxide ion.
$$\ce{[Na2B4O5(OH)4].8H2O <=> 2Na+ + B4O5(OH)4^{2-} + 8H2O}$$
$$\ce{B4O5(OH)4^{2-} + 5H2O <=> 4H3BO3 + 2H2O}$$
Reference

https://www.enotes.com/homework-help/how-much-borax-should-dissolve-hot-water-order-721345
https://www.google.co.in/url?sa=t&source=web&rct=j&url=https://www.smc.edu/AcademicPrograms/PhysicalSciences/Documents/Chemistry_12_Experiments/Solubility%2520and%2520Borax.pdf&ved=2ahUKEwjr5b6W1tXdAhWRbisKHb1rBUAQFjAXegQIBBAB&usg=AOvVaw3c9IA0jqEROPS534lSedA8

Does this solubility curve indicate weight of decahydrate or anhydrous borax? In general, when solubility in water is given in grams, would it typically refer to the common crystal form including water of crystallisation or the equivalent amount of anhydrous crystals? Seems often to not be specified.


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Structure of atom
Consider one He+ ion in exited state(n=5).In the answer it is given that no. of spectral lines produced is 4 not 10.can't we use the general equation?

It depends on how the question is worded. There are 4 possible lines directly from the n=5  (5-4, 5-3, 5-2, 5-1). There are 10 possible lines if you include all lines from n=4 (4-3, 4-2, 4-1), n=3 (3-2, 3-1), and n=2 (2-1) as well.

In the solution it's given that multitude of He+ will produce 10 possible lines and single He+ produce max. 4 lines.is it because of the fact that an electron at a time in an atom can occupy only one exited state.


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How are elements used in the human body (for example) encoded for in the genome?
Given that bones require calcium and iron is used in red blood cells, how are these materials gathered by the developing organism? Is it possible that very small quantities used as "references" are somehow stored in the human egg? Or is there some way of encoding for an element in DNA without any amount of the element already available?

The idea that DNA "encodes" something is a stupid antropomorphism (or computeromorphism), perpetuated by journalist writers who try to explain biology to the people without understanding it themselves. Don't fall for such supersimplified nonsense.

I would say that is a very broad use of the term "stupid" -- there is something about the proteins produced that has some relationship to iron, etc. I believe you are also the poster who suggested that urinating on gunpowder made it tougher -- perhaps "stupid" is more applicable to that idea.

The guys who urinate on gunpowder seem tougher. And I said "stupid antropomorphism", not just "stupid".

@Karl: Okay, that's not stupid at all but asking about mechanisms in chemistry in a way that uses software terminology is -- gotcha.

Allright, no problem. ;-)

@Karl I would disagree with your characterization of DNA, I don't think it is terribly far fetched to say that information is encoded in our DNA.

@releseabe A developing organism can't generate calcium or iron, but it can be given some as is the case for developing human babies. In general, organisms find these resources and they are incorporated as needed into existing and newly formed structures. I assume its a somewhat open question how living organisms started to incorporate inorganic materials into their internal workings.

@Tyberius: I did not mean to imply that an organism generates an element. I meant that in theory small quantities could exist in the egg to be compared against although it doe not sound like that is the case.

@Tyberius Of course it's true that DNA contains information. But the statement is also a pointless supersimplification.

@Tyberius  "I assume its a somewhat open question how living organisms started to incorporate inorganic materials into their internal workings." To most chemists, it's rather obvious. ymmv

@Karl we don't really even know how life started, I don't think we can really say we know how organisms "decided" to start using precursors of hemoglobin to transport oxygen. I definitely don't see what makes it obvious chemically. We know there is some sort of evolutionary process, but I don't believe the chemical particulars of those steps are known.

No. DNA is not a blueprint, it is an integral part of the machine. There aren't little people in your cells who can read an instruction set. DNA, together with a living cell, produces (actually is used to produce RNA, which goes into producing) specific proteins, some of which bind to e.g. iron ions and integrate them into the enzyme mechanisms. It's all about making the right protein (e.g. hemoglobin to bind iron) in the right place and time. A living cell is a reactor, and the genes are your catalyst.
What kind of proteins your DNA produces depends on the surroundings. If in a skin cell, it produces proteins that directly or indirectly are necessary to produce skin. If (in an embryo) close to the surface, but not yet a skin cell, it produces proteins that later help turn it into a skin cell. And so on.
If an alien got hold of just our DNA and RNA, he could never ever even find out how a human works or what he looks like.

What is about the proteins that makes them bind to iron or calcium without knowing what iron or calcium are?

It's about making the right protein. One that binds to iron, instead of one that binds to copper or whatever. Chemistry. Nobody told alizarine yellow to have a specific colour. You synthesise the compound, and lo and behold, it's yellow.

There is no knowing. Molecules do not know a thing.

@IvanNeretin: I should put quotes around "know" -- what I mean is, how does a relatively large protein bind to a small atom of one element over a different element? And how are iron atoms even gathered in sufficient quantity to bind to  the protein? Are various atoms routed in some sense to where they can be bound?

Hemoglobin has a very high affinity to iron, and just iron. If there is one around, it will get it. It's the diameter and electronic structure of Fe ions. There possibly are other, less specific transport proteins. You'll have to ask the biologists.

It is like a key and a lock. The lock would recognize the right key among many, but certainly not because there is a copy of that key inside it.

@IvanNeretin A very nice picture, thank you! ;-)

Naked DNA doesn't really code or do anything. DNA inside a living organism provides instructions that enable the organism to build things such as proteins which then do the work. In that context DNA also provides instructions to signal which things get made (some of the "code" tells the organism which things get made in which contexts).
As for how elements used in living things are "encoded" they are not in any meaningful way in the DNA. But proteins and other products made by cells have very, very specific functions. Some of those functions are designed to carry, transport or exploit the properties of elements. In plants, for example, a variety proteins in some cells are designed to assemble chlorophyll, the molecule at the heart of photosynthesis. Chlorophyll contains a haem group designed to bind to a magnesium ion and form the heart of a very complex system embedded in other proteins and plant membranes (all built from other cell machinery). The overall system enables plants to use sunlight to produce sugars from the carbon dioxide in the air.
But the ability of the plant to use magnesium derives from the specific structure of the haem in chlorophyll. This is magnesium specific the same way that much simpler chemical ligands built by chemists can be very element-specific (eg crown ethers and their congeners can be designed to very specifically bind alkali metal ions). But this specificity is serial steps away from the "coding" in DNA. The steps might be DNA codes for a bundle of proteins, which, in turn, build other proteins or other molecules, which in turn make up the machinery to make yet more molecules which are assembled by other proteins and molecular machines to create complex systems like the one used in photosynthesis.
The ability of organisms to exploit specific elements is based on the ability of parts of those final machines to very specifically bind particular elements in particular ways (even chemists know how to do that). But the machinery that achieves those goals is a number of steps removed from DNA which doesn't need to encode anything about those elements.


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Why doesn't cyclopropyl methyl carbocation stabilises itself by ring expansion?
I have been taught that ring expansion stabilises smaller cyclic compounds to a great extent. So why does cyclopropyl methyl carbocation shows such type of resonance

rather than expanding its ring

thereby decreasing it's angle strain, further it will be stabilized by hyperconjugation effect and inductive effect.

perhaps the angles in the 4-ring are bad for hyperconjugation? the intermediate 4-ring is even more badly strained than the 3-ring? the resonance takes a lot of strain from the cyclic structure? (last point is definitively true)

It can't be said that the cyclobutane carbocation does not occur. March's Advanced organic chemistry, 10.C.i Neighboring-group participation by π and σ bonds: nonclassical carbocations, tries to explain the stabilization, but also cites that (chloromethyl)cyclopropane by hydrolysis provides 48% cyclopropylmethanol, 47% cyclobutanol, and 5% "allylmethanol".

I have found a similar question already answered: https://chemistry.stackexchange.com/questions/19090/what-is-the-reason-for-the-exceptional-stability-of-the-cyclopropylmethyl-carboc

I already checked this answer but this doesn't explain why ring expansion is not preferred

Question itself does - it's exceptionally stable (while cyclobutyl isn't)

Unfortunately, the premise of your question is wrong.
There is no cyclopropylcarbinyl or cyclobutyl cation. They're the same thing. Unlike cations which you might be more familiar with, for example, t-butyl cation, the cation you're looking at here, also called homoallyl, is a non-classical cation.
The open form and the two structures you drew in the second line are not distinct. They're just resonance forms of each other.
You ask does this ion undergo ring expansion? Well, yes and no. Yes, in that there is a contribution from the 4-member ring to the resonance structure. No, in that you can't expand something that doesn't exist.


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differentiate between the subscript(s) and coefficient(s) that make up the chemical formula for sulfuric acid, 2H2SO4
Im confuse on whats the question is asking.
differentiate between the subscript(s) and coefficient(s) that make up the chemical formula for sulfuric acid, $\ce{2H2SO4}$.

Welcome to Chem.SE. This might be related :) Should chem.SE use Georgia font?

Do you know in general, what subscripts and coefficients in a chemical formula are? Do you have problems with both terms or only one of them (which one?)?

@mykhal I don't it matters much for someone who clearly doesn't know formula for sulphuric acid...

Subscripts and Coefficients both show the quantity of elements. The difference is that subscripts show the number of atoms in a molecule, while coefficients show the number of molecules there are. Also, subscripts are placed immediately after the atom it describes: $\ce{H2O}$. the 2 is describing the number of H atoms there are. $\ce{3H2O}$ is 3 molecules of $\ce{H2O}$.


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Is the Lowdin Orthogonalization used in diagonalizing the atomic orbitals really a basis change?
This is kind of more on the mathematics side of quantum chemistry, but I can't quite figure out why the Lowdin Orthogonalization is called a basis set change. I get how it works from the perspective of matrices, and that it creates an orthonormal set as long as you have linearly independent input, but I don't get why it works as a change of basis.
Typically to change the basis set of a matrix $M$ you use a non-singular linear transformation $A$, which need not be unitary, as such: $$M'=A^{-1}MA.$$ Quantum mechanics, we are dealing with Hilbert Spaces, which produce an isometry between the Hilbert space and the dual (I think?), and then for post HF methods most methods often use unitary transformations, so that $A^\dagger=A^{-1}$, and so it becomes trivial.
The Lowdin transformation between nonorthogonal AO $|\phi\rangle$ and orthogonal AO $|\phi\rangle_\perp$ involves the overlap matrix $S$ as:
$$ |\phi \rangle_\perp = S^{-\frac{1}{2}}|\phi\rangle  $$
Then, the corresponding bra just has the adjoint, which is also $S^{\frac{-1}{2}}$ because the matrix is self-adjoint: $$ \langle \phi|_\perp = \langle \phi|(S^{-\frac{1}{2}})^\dagger = \langle \phi|S^{-\frac{1}{2}}$$
But, considering the matrix analog, why don't we use the inverse?
For instance, we have the identity: $$ D^{AO} = C  D^{MO}C^T $$ where $D$ is the 1-electron reduced density matrix. In my thinking this has to be a transformation from the MO to the AO on the right of $D^{MO}$, and from the AO to the MO on the left, which gives you $D^{AO}$. However, $C$ is not unitary obviously, because it comes from the Lowdin procedure, and so while $C$ on the left is from the AO to the MO, $C^{-1}$ is not equivalent to $C^T$.
Furthermore to get $D^{MO}$ matrix in terms of the $D^{AO}$ you DO have to apply the inverse relationship with $C^{-1}$. $$D^{MO} = C^{-1}D^{AO}(C^T)^{-1},$$ because $C^{-1}$ and $C$ and related non-trivially through $S$.
Hopefully this is appropriate to ask here. I think the answer involves Hilbert spaces but I can't find a lot on non-orthogonal transformations. Thanks!

I might be misunderstanding your question, but isn't matrix inversion more computationally intensive than taking the adjoint?

It is, but unless the matrix is unitary, then the adjoint and the inverse are not the same, so applying will not yield the same result. Here, the C matrix for instance is not unitary.

Consider a nonsingular, linear transformation $\mathbf A$ of a set of vectors arranged into a matrix $\mathbf V$ $$\mathbf V'=\mathbf {VA}$$
We can say the matrix $\mathbf V'$ is orthogonal (and thus that we have linearly transformed the vectors to form an orthogonal basis) if $$\mathbf {V'^{\dagger}V'}=\mathbf {A^{\dagger}V^{\dagger}VA}=\mathbf {A^{\dagger}SA}=\mathbf{1}$$
In general, this occurs when $\mathbf{A}=\mathbf{S}^{-1/2}\mathbf{B}$, where $\mathbf{B}$ can be any unitary matrix. The Lowdin orthogonalization commonly used is just the case where $\mathbf{B=1}$, so we can see that we are properly changing the basis.
But your issue is specifically with transformations of the density matrix, where the transformation matrix is not unitary in general. The standard change of basis formalism would suggest that to go to AO basis from MO, we would do a transformation something like this:
$$\mathbf{C}\mathbf{D}^{MO}\mathbf{C}^{-1}=\mathbf{D}^{AO}\mathbf{S}$$
Ensuring orthogonality of the MOs requires that $\mathbf{C^{\dagger}\mathbf{S}\mathbf{C}}=\mathbf{1}$, which we can sub in to the equation above to give: $$\mathbf{C}\mathbf{D}^{MO}\mathbf{C^{\dagger}\mathbf{S}\mathbf{C}}\mathbf{C}^{-1}=\mathbf{C}\mathbf{D}^{MO}\mathbf{C}^{\dagger}\mathbf{S} =\mathbf{D}^{AO}\mathbf{S} \to \mathbf{C}\mathbf{D}^{MO}\mathbf{C}^{\dagger}=\mathbf{D}^{AO}$$
So the form in terms $\mathbf C$ and its adjoint is a direct result of the conventional similarity transform combined with the orthonormality condition on the MOs.
This latter derivation was adapted from$^1$ which can be found here.
As to why we avoid using transformations involving the inverse of the transformation matrix, I suspect Zhe is correct that it is a matter of cost, since obtaining the transpose/adjoint of a matrix requires almost no work at all, while an inverse is relatively difficult to obtain.

T. Helgaker et. al Chemical Physics Letters 327 (2000). 397–403

Well, I have no problem with this. It's just the inverse and the transpose are not 1-1 for a non-unitary matrix. So the effect of $CDC^{-1}$ should not be the same as $CDC^{T}$, and it isn't. And generally, the first of those is presented as the 'change of basis' sort of formalism.

@ScottS. I've edited my response; let me know if this better addresses your concern.

Ah, okay this is better, thank you @Tyberius. The paper helps as well. The proper similarity transformation here is actually between $D^{AO}S$ and $D^{MO}$, not $D^{AO}$ and $D^{MO}$. This makes sense, as now the eigenvalues are preserved.....I guess then, my only question is, what is the interpretation of $D^{AO}S$? And so, it seems going from $D^{AO}$ to $D^{MO}$ might not actually be a simple change of basis because you have to change the matrix eigenvalues, i.e. not a similarity transformation?

@ScottS. I'm not sure what it means. If you orthogonalized the AOs, then it becomes a conventional similarity transform, but otherwise I'm not sure how the basis are related. My best guess is that they are in some sense "similar with respect to a metric" for lack of better terminology. I posted a question on Math SE about if there is a clear meaning behind this pseudo-similarity transform.

Thank you @Tyberius, that makes sense, and I think it helped me move in the right direction. I started looking into metric vector spaces and bilinear forms....a particular book$^\dagger$ was helpful.
It seems that there is a congruence relation between the two density matrices $D^{AO}=CD^{MO}C^T$. Then, these would share the same bilinear (or sesquilinear) form just in different bases. Also it is not an isometry, meaning it does not preserve length. If it did, then it would be unitary. I also am not positive about $D^{AO}S$, but it might just be an artifact of the congruence relation?

$\dagger$ Roman, Steven. Advanced Linear Algebra. New York: Springer, 2008. Print.

@ScottS still don't have a clean way of phrasing this, but it seems to be that you can't similarity transform between a matrix with an orthogonal basis (like the MO density matrix) and a matrix with a nonorthogonal basis (like the AO density matrix) and so to make the transformation work, you have to put the AO one in terms of an orthogonal basis which seems to occur by multiplying by the overlap matrix.


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Why can't a quaternary carbon product be obtained by alkane isomerization in presence of AlCl3 and HCl?
This question actually deals with the reaction in another question: Isomerisation of alkanes under Lewis acidic conditions. I have a further doubt, regarding this reaction my book says that a quaternary carbon cannot be generated. Why is this so? Also if a pure alkane is used without alkene traces will the reaction fail to take place?

You're going to have a hard time coming up with a mechanism by which you will generate said quaternary carbon center.

@Zhe using a tertiary alkene?

What is a "tertiary alkene"?

@Zhe I meant something like 2-methylprop-1-ene, though tertiary alkene is not the right term

An alkene can only provide a tertiary cation on protonation. How will you migrate another group to make a quaternary center in a favorable way unless the migrating group is already a quaternary center? Nevermind the steric implications here...

You're basically asking if you can t-butylethylene will undergo rearrangement under, say, hydrochlorination. It can. But notice that you already started with a quaternary center.


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What's the reaction between H2SO4 and NaH2PO4?
I'm guessing it's going to be
$$\ce{H2SO4 + 2 NaH2PO4 -> 2 H3PO4 + Na2SO4}$$
Could anyone please let me know what the actual reaction is?
Thank you

This will do.$,$

Actually you have two reactions, since sulfuric acid can give off two protons:
$\ce{H2SO4 + NaH2PO4 -> NaHSO4 + H3PO4}$
$\ce{NaHSO4 + NaH2PO4 -> Na2SO4 + H3PO4}$
The $\ce{NaHSO4}$ is called sodium bisulfate.  Bisulfate is the thing you form first when sulfuric acid acts as an acid, then you get the sulfate ion if you have a strong enough base to react with the almost but not quite strongly acidic bisulfate ion.  In this case, actually, you can look up some acid dissociation constants and discover that phosphoric acid has almost exactly the same amount of strength as the bisulfate ion acting as the "second stage" of sulfuric acid.  So the second reaction is really reversible:
$\ce{NaHSO4 + NaH2PO4 <=> Na2SO4 + H3PO4}$
For more on what makes acid-base reactions reversible, see here.

Well, it depends on concentrations too, but pKa2 for sulfuric acid is 1.99 and pka1 for phosphoric acid is 2.15. So for equal molarities of sulfuric and sodium dihydrogen phosphate you'd end up with a significant amounts of $\ce{NaHSO4 \text{,}  Na2SO4 \text{,} NaH2PO4}$, and $\ce{H3PO4}$.

That's why it's reversible.

Sorry, stated my point poorly. I'm trying to suggest that there is an equilibrium between the four species, and that all four species will likely have significant concentrations. (I say likely since I am too lazy to actually do the math...)

Ok, it's a two-step reaction Thank you, Oscar Lanzi and MaxW for the information


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Full equation when using vinegar, hydrogen peroxide, and salt to etch copper
What is the full chemical equation when you use the following reactants to etch copper:

vinegar / acetic acid (5%) $\ce{CH3COOH (aq)}$
hydrogen peroxide (3%) $\ce{H2O2 (aq)}$
salt $\ce{NaCl (s)}$
copper $\ce{Cu (s)}$

Yes there are previous questions asking the same thing (one and two). However, the answers to both give only parts of the reaction, not the full equation.
Looking elsewhere on the web, I've come across this Instructables article that says copper(II) chloride $\ce{CuCl2}$ is formed when the copper ions bond with chloride ions from the salt. And therefore the solution is greenish in color instead of the blue it would be if copper(II) acetate $\ce{Cu(CH3COO)2}$ was formed as answered in question two. (When doing this at home, I also got a distinctly greenish color when salt was used, and a distinctly blue color when it was omitted).

Table salt, or $\ce{NaCl}$, brings chloride ions $\ce{Cl-}$ to which $\ce{Cu^2+}$ ions will bond to form cupric chloride or $\ce{CuCl2}$ instead of being left in solution (those ions would endlessly come back and forth from $\ce{Cu}$ to $\ce{Cu^2+)}$. You can see this during the reaction (if you leave the reaction with no salt it will turn blue which is the color of copper(II) acetate and stop, whereas if you put salt it will turn green, the color of cupric chloride, and carry on).

Another article on the web says both copper(II) acetate and copper(II) chloride are produced, but no equation is provided:

Equal parts vinegar and peroxide worked fine, dump in plenty of salt (to the point of having undissolved salt in the bottom of jar) since it is consumed in the reaction. You should see a vigorous fizzing at the copper surface. Add more salt as the reaction slows.
The solution will turn green as the copper is converted to copper acetate and/or copper (II) chloride

Can someone please give the complete chemical equation of the reaction that occurs when acetic acid, hydrogen peroxide, and salt are used to etch copper? I can't seem to find it anywhere on the web.

Context:
I want to dispose the products of the etching process safely. If copper(II) acetate is the product as suggested in answer two, then following the recommended approach for disposal is safe:

You can bypass the problem by collecting the etch solution, and submerge aluminum foil.
$$\ce{3Cu(CH3COO)2 (aq) + 2Al (s) -> 2Al(CH3COO)3 (aq) + 3Cu (s)}$$
React with the aluminum until the bluish color disappears, then collect the solid copper + remaining aluminum to throw in the trash. You can now flush the solution down the drain.

However, if the product is instead copper(II) chloride, the approach is not safe as the $\ce{AlCl3}$ formed is considered a hazardous waste and shouldn't be flushed down the drain.
$$\ce{2Al (s) + 3CuCl2 (aq) -> 3Cu (s) + 2AlCl3 (aq)}$$

Possible duplicate of What reaction occurs when etching copper to create circuit boards?

@Mithoron I quote that specific question (as one) in my question! And I even go on for a bit explaining why I do not find the given answer adequate!

It's exactly the same question and is answered therefore it's a duplicate. That's it.

If you aren't satisfied you can comment under old post or put a bounty on it. Asking again isn't proper solution on this site.

Not to say that " Cu2+(aq)+2OAc− and the second with Cu(CH3COO)2" isn't a good argument, really.

@Mithoron the question was asked three years ago. The OP hasn't been on the site since. Even if I placed a bounty on it, there's no way for me to specify why I find  the current answer inadequate.

@Mithoron "... isn't a good argument" - are the two equivalent? I wouldn't know...

Yeah, pretty equivalent.

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

The acetic Acid is used to dissolve the copper, however it cannot do so directly as the reaction of copper with protons is not spontaneous.
$$\require{cancel} \ce{Cu_{(s)} + 2H+_{(aq)} \cancel{->} Cu^2+_{(aq)} + H2_{(g)}}$$
The hydrogen peroxide is used to oxidize the copper since copper does not spontaneously react with acids directly.
$$\ce{\frac12 H2\color{red}{O2} +e- -> OH-}\\ \ce{Cu_{(s)} -> Cu^2+_{(aq)} + 2e-}$$
However, hydrogen peroxide is unstable in acid so the acetic acid is used because it reacts with hydrogen peroxide to form a peroxyacid which is more stable.
$$\ce{CH3C(O)OH + H2\color{red}{O2} -> CH3C(O)\color{red}{OO}H + H2O}$$
This peroxyacid can react directly with copper to dissolve it.
$$\ce{CH3C(O)\color{red}{OO}H_{(aq)} + Cu_{(s)} -> Cu[CH3C(O)O]OH_{(aq)}} \\ \ce{Cu[CH3C(O)O](OH)_{(aq)} + CH3C(O)OH_{(aq)} -> Cu[CH3C(O)O]2_{(aq)} +H2O}$$
As far as the sodium chloride goes, I'm not sure what that is for. My best guess is that it is catalytic in dissolving the copper similar to how salt corrodes steel
$$\ce{CH3C(O)\color{red}{OO}H_{(aq)} + Cu_{(s)} ->[\text{cat.} Cl-_{(aq)}] Cu[CH3C(O)O]OH_{(aq)}}$$
As far as the final species goes, you probably have a complex from the excess chloride
$$\ce{Cu[AcO]2 (aq) + 2NaCl (aq) -> [Cu(AcO)2Cl2]^2- (aq) + 2Na+ (aq)}$$
which is in dynamic equilibrium
$$\ce{[Cu(AcO)2Cl2]^{2-} (aq) <=>[Cl-][AcO-] [Cu(AcO)Cl3]^{2-} (aq) <=>[Cl-][AcO-] [CuCl4]^{2-} (aq)}$$
The color comes from the excess chloride that acts as a ligand to the copper From Wikipedia:

  Aqueous solutions of copper(II) chloride. Greenish when high in [$\ce{Cl−}$], more blue when lower in [$\ce{Cl−}$].

Thanks for the response! Is the "chloride complex" (not sure if that's the right term) responsible for the green color that appears when salt is added (instead of the blue it would otherwise be)? Does it have a common name?

@JetBlue see edits. Probably not the answer you want , but yes it is likely copper chloride.

Ah I see. So the CuXCl compound should be disposed off as if it were CuCl2?


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Mechanism of Water Potential
The idea of water potential still confuses me. I understand the general idea: water moves from an area of higher water potential to an area of lower water potential. However, I still do not understand the mechanism by which this occurs.
Take the example of putting salt on a slug. Putting salt on a slug lowers the dilute potential which in turn lowers the water potential on the outside of the slug. This lower water potential on the outside then draws water from the higher water potential of the inside of the slug.
I do not understand how this occurs. All the answers I find is that this just “occurs” because “nature wants to find equilibrium” (such as the second response on this Quora question). That makes sense in theory, but how does the cells of the slug “know” that there is now a lower water potential? What has changed?

If I understand you correctly you are asking why osmosis happens the way it does. And obviously you want a more "real-life-approach" to this than the pure mathematically description via potentials.
Think about a cell filled with water that is divided by a permeable membrane that will only let water molecules pass through it. In this initial state the water molecules on both side will occasionally pass through the membrane switching sides. They do so by pure chance and as the particle-concentration on both sides are equal (there are 100% water particles on both sides) the pass through on both sides with an equal rate leaving the overall particle-distribution unchanged.
Now we go ahead and dilute salt in one half of the box. Leaving aside effects like hydration-hulls forming around the ions (which will magnify the effect described below) the effect of this dilution is that we now have a lower water-particle-concentration on that side. Therefore every now and then when a water particle would have been "thrown" at the membrane to pass through it it's not a water particle but an ion instead that can't pass the membrane. Therefore the rate at which the water-particles switch to the non-dilution-side is lower than the (unchanged) rate in which the water-particles flow into the dilution (by pure statistics). This leads to an increase in the overall particle count on the dilution-side.
This increase in particles also increase the pressure and therefore the rate at which random particles (on the dilution side) are being "thrown" at the membrane until this higher rate ultimately compensates the lower water-particle-concentration. At this point both sides are at an equilibrium again. However this equilibrium is in a state in which there are more water-particles on the dilution-side than there have been in the beginning.
And this is what we call osmosis.


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Can a bond form between two atoms that don't have 8 electrons?
Generally, atoms like to have 8 valence electrons. But, I'm wondering if it's possible for two types of atoms to form a bond where neither of them gains 8 electrons. Take Carbon, which has 4 valence electrons; and Hydrogen, which has 1. We know that 4 Hydrogen atoms + 1 carbon atom makes a molecule of CH4. But can carbon make a molecule with just one hydrogen atom?

some atoms can form compounds with up to 12 electrons in their valence shell, like sulfur. i don't have a specific example in my mind though

Lithium Hydride?

works pretty well for $\ce{H2}$

Absolutely! Perhaps the conceptually simplest example to consider is when the prototypical valence-deficient atom, boron, binds to itself.
There are a large number of known, stable diboron compounds containing boron-boron covalent bonds. In the simplest case of a diboron compound with a single covalent bond between boron atoms, we have a diboron(4) compound (the 4 suffix establishes there are four other non-boron atoms bound directly to the boron atoms, implicitly specifying there is a single covalent B-B bond). These are not to be confused with diborane, which is actually an example of a diboron(6) compound and has no boron-boron bond.
There's actually a great open-access review article on the rich chemistry of diboron(4) compounds, which I recommend at least for browsing the chemical structures. Unsurprisingly these compounds often react as you would expect electron-deficient compounds to, readily accepting electron pairs from Lewis bases, for example.
Not all representatives are very stable; alkyl- and hydrogen-substituted diborons are sensitive to water and air and can combust readily. However, they can be stabilised when atoms containing lone electron pairs are next to the boron atoms, as the lone pairs can be partially donated into the empty boron orbital. In these cases it not as clear-cut to say the boron atoms have an unfilled octet.
Diboron(4) compounds are hardly just laboratory curiosities. In particular, tetrahydroxydiboron esters such as bis(pinacolato)diboron are widely used in Miyaura borylation and other borylation reactions, setting things up for Suzuki-Miyaura cross-coupling, one of the most important and widely used C-C bond-forming reactions today.
Also interestingly, it is possible to create double and even triple covalent boron-boron bonds in diborenes and diborynes, though it is certainly rarer. A nice review on this subject can be found here.
Regarding your specific query using carbon, it's rather more difficult to construct something similar. Dicarbon ($\ce{C2}$) is known, though it is very reactive and is found only at high temperatures, such as in flames. It could be said that both atoms in dicarbon have incomplete octets, though pinning down the exact nature of the bonding in this situation can be a delicate matter. I can't think of any simple way to get two valence-deficient carbons to bond.


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Why can't terminal alkynes take part in Birch reduction?
According to my book terminal alkynes are not observed in Birch reduction.Why is this so? The only property I know about them is that they are slightly acidic. Does this have something to do with it? Please help me understand this.

As Waylander has already mentioned, acetylide formation occurs instead: the introduction of this article clearly states this. @Hema
Anyway, I am looking for a textbook with the mention that terminal alkynes are not suited for Birch reduction; would you mind telling me the name of the book you were looking at?

The alkyne proton is acidic enough to react with the dissolving metal to give the anion, giving the metal alkynide as it shown by these procedures sodium acetylide prep,  sodium phenylacetylide prep; this does not reduce.


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Why is the concentration of protons equal to the concentration of a strong acid?
First, I know that strong acids are assumed to dissociate completely in water, so that the concentration of $\ce{H+}$ which they provide equals their concentration.
What I am asking for is not the concentration of $\ce{H+}$ which they provide, but $\ce{[H+]}$ for the solution, since the $\ce{H+}$ ions that they provide will not stay aloof, but rather will react with water and with $\ce{OH-}$ ions that are produced in the self-ionization of water.
Won't the reaction of $\ce{H+}$ ions with water shift the self-ionization of water
$$\ce{H2O <=> H+ + OH-}$$
to the left and won't their reaction with $\ce{OH-}$ shift it to the right according to Le Châtelier’s principle?
So, what will $\ce{[H+]}$ be for the solution?

Autoionization of water (though it certainly does happen) is quite negligible here.

@VicLineal Thanks for polishing this post. For future reference, we prefer to not use MathJax in the title field, see here for details. So while you're editing a question, please remove it.

Is this a case of common ion effect?

The statement $[\ce{H^+}] = [\ce{A^-}]$ is an approximation for not too diluted acids. However you can pretty fast tell that this won't hold if you add a strong acid to water so that the acid has a concentration of let's say $\pu{10^-8 mol//l}$. Via the formula $\mathrm{pH} = \log\left([\ce{H^+}]\right)$ this would result in $\mathrm{pH} = 8$ so if that was true, you just made the solution less acidic by adding an acid.
In fact the total concentration $[\ce{H^+}]$ is given by $[\ce{H^+}] = [\ce{A^-}] + \pu{10^-7 mol//l}$ where the last term stems from the autoionization of water. However in most use-cases $[\ce{A^-}]$ is far greater than $10^{-7}$. In those cases the autoionization of water can be neglected and thus the approximation $[\ce{H^+}] = [\ce{A^-}]$ gives sufficient results.
However, if you are working with really diluted solutions of acids this approximation is no longer applicable (as demonstrated above) which forces you to actually respect the autoionization of water.
If you want to be on the safe side you can always respect the autoionization. This will always give you the correct pH (even for not so diluted solutions) but you may neglect it if the acid-concentration is high enough (this of course depends on the accuracy of the results you need).
EDIT:
As @MaxW pointed out in the comment a general expression would be $[\ce{H+}] = [\ce{A-}] + [\ce{OH-}]$ which for a neutral solution at 25°C (thanks to @Martin for pointing this out) becomes the formula proved above.

Please note, that the value for the auto-ionisation is only true at about room temperature, where $\mathrm{p}K_\mathrm{w}^{\pu{25 ^\circ C}} = 14$

RE: "In fact the total concentration $[\ce{H^+}]$ is given by $[\ce{H^+}] = [\ce{A^-}] + \pu{10^-7 mol//l}$" --- This is not the correct expression. It should be  $\ce{[H+] = [A-] + [OH-]}$

@MaxW Well I was assuming a neutral solution... Though in a general case you are of course correct. I'll add your comment to my answer.

The solution can't be neutral and acidic at the same time. Yes the solution can start out neutral, but after adding the acid $\ce{[A-] > [H+]}$ because some of the $\ce{H+}$ has reacted with the $\ce{OH-}$ from the autoionization of the water.

What about the reaction of some hydrogen ions with some hydroxide ions? Won't this reduce the amount of the hydrogen ions?

I want to know what happens to the water autoionization equilibrium when a new substance is present($A^-$)? Will the solution have two equilibria(The autoionization of water and the equilibrium of the acid with its conjugate base)?

For "strong" acids the usual assumption is that the acid is completely ionized. So rather than being an equilibrium
$$\ce{HA -> H+ + A-}$$
Also this means that the conjugated base does not protonate.
$$\text{NO! }\quad\ce{A- + H+ -> HA}$$
The autoionization of is given by the reaction
$$\ce{H2O <=> H+ + OH-}$$
The addition of $\ce{H+}$ ions from the acid react with $\ce{OH-}$  ions from the water to shift the reaction to the left according to Le Châtelier’s principle.  Most pKa's are only good for two significant figures, so in most cases (eg when pH > 5) you can assume $\ce{[A-] = [H+]}$ for strong acids. But since the charges in the solution must balance the exact expression is
$$\ce{[H+] = [A-] + [OH-]}$$
Nominally the equilibrium for the autoionization of water would be given by
$$K_{eq} = \frac{\ce{[H+][OH-]}}{\ce{[H2O]}}$$
But the assumption is that for aqueous solutions $\ce{[H2O]}$ is always $55.5 \text{ molar}$ so instead the expression used is:
$$K_w = 1.0\times10^{-14} = \ce{[H+][OH-]}$$


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Hydride shift in carbocation

I was practising some problems on hydride shift and got stuck in this particular question. I wanted to know if hydride shift would take place in this particular case and which carbon would get the positive charge. Will a hydrogen from left or right carbon would shift??

The molecule is symmetric?

I strongly suspect that a hydride shift is not observed in this example.
The reason is that it seems much more likely that you will have an alkyl shift instead, accompanied by ring contraction and formation of a carbonyl. This is essentially a pinacol rearrangement.
EDIT:
Adding more details per @OscarLanzi comment...
I would think that this is the expected product after proton transfer.

It looks like the pinacol arrangement of 4,8-decalindiol has been known for quite some time. Hückel, Danneel, Schwartz, Gercke, Ann. 1929, 474, 121. and Meiser. Chem. Ber. 1899, 32, 2055.

Could you render the expected product?  Your description seems to suggest a spiro-type product and that does not go well with my intuition.  Thanks.

@OscarLanzi That's exactly what I'm suggesting. Let me edit with details.

@zhe but why hydride shift would not take place?? I thought that hydride shift would have been easier than an alkyl shift?? And why ring contraction would happen?? Please explain . Thanks

Well, for one thing, a hydride shift still gives a secondary cation, but a alkyl shift provides a carbonyl group and a smaller, not-very-strained ring. I'm not sure if this is a thermodynamic or kinetic effect, but it seems reasonable that this difference can lead to a preference in either.


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What happens when (+) 2-iodobutane is treated with NaI in acetone?
After treating the compound with NaI, we should follow SN2 mechanism and the product will follow an optical inversion, giving us (-)2-iodobutane, but the answer is (±)2-iodobutane.
After going through various articles, I am getting reasons, such as there will be an equilibrium created between the reactants and the products, hence we will get both the products (±)2-iodobutane.
And there is the same question where we are treating (R)-2-bromobutane with NaI in acetone, but the answer is (S)-2-iodobutane.
What could be the reason?
I believe we should simply get one product in the question, and there will be no equilibrium be created, as there shouldn't be enough NaI left to form another compound.

Iodide is attacking and kicking out another iodide, so you cannot ever run out of it.

If iodide can react with the (R)-enantiomer, why not the (S)-enantiomer? As orthocresol states, it goes on and on until racemization occurs. This is classic experiment that shows that racemization is twice the rate of SN2 displacement. Imagine you have two (R)-molecules. You only have to invert one to (S) to achieve racemization. At that point, only one of the two (R)'s has reacted. Think about how radioactive iodide would be useful in this regard.

There is no equilibrium in the second reaction because the sodium bromide formed by the iodide displacement is not soluble in acetone so there is no Br- to displace iodide. In contrast with the first reaction, NaI is soluble in acetone so there is always excess I- around to continue the reaction. This is called the Finkelstein reaction, further information here finkelstein reaction

But in the first reaction, could the final major product be only (-)2-iodobutane?

No, because the iodides keep exchanging so you get the mixtures.


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Show that this mechanism is consistent with the rate law - how to?
The problem is from Principles of general chemistry, 2nd edition by Silverberg, chapter 16 problem no.81:

"Even when a mechanism is consistent with the rate law, later work may show it to be incorrect. For example, the reaction between hydrogen and iodine has this rate law: $rate = k\ce{[H2][I2]}$. The long accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary:
$$\ce{H2_{(g)} + I2_{(g)} -> 2HI_{(g)}}$$
In the 1960s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism:
$$\quad \ce{ I2_{(g)} <=> 2I_{(g)} \quad\quad[fast]}\tag {i }$$
$$\ce{H2_{(g)} + I_{(g)} <=> H2I_{(g)} [fast]\tag{ii}}$$
$$\ce{ H2I_{(g)} + I_{(g)} -> 2HI_{(g)} [slow]}\tag{iii}$$
Show that this mechanism is consistent with the rate law."

I cannot understand what the problem asks. I think, this reaction is consistent with the rate law by the third reaction, $\ce{H2I + I->2HI}$, which is the rate-determining step. Is my guess correct?

Have you learnt about the pseudo-steady-state approximation?

You should express the rate law in terms of reactants if you can. Neither $\ce{H2I}$ nor $\ce{I}$ are reactants in the original reaction equation.

write down the rate of production of product say $k_3[H_2I][I]$ this means that you have to find $[H_2I]$ and $ [I]$. The rate equation for the intermediate $[H_2I]$ is $d[H_2I]/dt= \cdots$ (it three terms) and set this to zero and find $[H_2I]$. All that is left is to find $[I]$ from the first equation.

You have a three step mechanism where only one step is slow and the other steps are fast.
Since the first two steps (which are, in fact equilibria) are fast, it can be assumed that each of the equilibria holds throughout the reaction. Therefore, the expression of the equilibrium constant for each step can be used:
$$\ce{K_1=\frac{[I]^2}{[I2]} \,\,\, and \,\,\, K_2=\frac{[H2I]}{[H2][I]}}$$
So, $\ce{[I]=\sqrt{\ce{K_1[I_2]}}\,\,\,\,and,\,\,[H2I]=K_2[H2][I]=K2[H2]\sqrt{\ce{K1[I2]}}}$
As only the last step is slow, the rate of formation of the product (which is the rate of the overall reaction) only depends on the last step.
$$\ce{rate=\frac{1}{2}\frac{d[HI]}{dt}=k_s[H2I][I]}$$
Here, $\ce{k_s}$ is the rate constant of the last, slow step. Substituting the expressions for $\ce{[H2I]}$ and $\ce{[I]}$ in the rate law should give you an expression of the form $\ce{rate=constant\times[H2][I2]}$, which is what the given rate law is.


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Bromination with NBS on cyclic structures

I found this question and the answer as shown in my book. I didn't understand why the bromine particularly entered the position shown in the answer.
First of all, if a free radical was generated there it would be 2°, and also will have resonance. Is any other product possible or this is the only major product?
Also, is this allylic bromination? Every allylic position is also a vinylic position — I am not getting it.

Are you sure that this a radical bromination? :)

I have a question on that too , I really confused , like there is no allylic position and every allylic position is like a vinylic position. I am not getting this

This is aromatic azulene and it's simple reaction with Br2

Usually, these types of problems can be explained by looking at the intermediate that has resonance structures that don't disrupt the aromaticity of the other ring to where you've attacked.

Can anyone tell me where to learn how to write chemical formulas and reaction is stack exchange ?

As a follow-up to the excellent comments provided thus far, the following thoughts are added. Electrostatic Potential Map of azulene (1a) shows that there is high electron density in the 5-membered ring (red color signifies high electron density). The 5-membered ring is reminiscent of the aromaticity of cyclopentadienide anion and the 7-membered ring bears a positive charge and has aromaticity similar to cycloheptatrienyl cation. This phenomenon is represented in structure 1b. The reaction does not involve bromine atom (a radical) because a bromine atom would be expected to attack the electropositive ring, not the electron-rich ring. The reaction is ionic and likely involves bromine rather than N-bromosuccinimide (NBS, 4) directly. Bromine is liberated from NBS by trace amounts of HBr formed by the slow decomposition of NBS. In the event, bromination of azulene (1c) produces a resonance stabilized cation 2a. Its canonical structure is 2b that is identical to 2b'. Loss of a proton from 2b (or 2b') provides bromoazulene 3.


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Was nitroglycerin invented as an explosive?
Although I have researched this as carefully as the web allows, it is not clear to me if the inventor was planning to make an explosive and if so, why he thought the process he undertook would yield one. Were there similar compounds created with acids that were also explosive? I understand that in the early 1800s, there was a lot of apparently random experimentation.

Comments are not for extended discussion; this conversation has been moved to chat.

Probably ... but he doesn't actually say
$Nitroglycerin^1$ was discovered by Dr. Ascanio Sobrero in 1847. Both Sobrero and Alfred Nobel were students of Prof. Théophile-Jules_Pelouze.
It is rather unfair to describe chemistry of this era as "random". 1847, of course, was mid-1800s rather than early 1800s, and it was some two decades after Wöhler's famous demonstration that organic chemicals could be synthesized from inorganic ones. Organic chemistry of this era was more in its adolescence than its infancy and was being pursued in a systematic scientific manner with publications in peer reviewed journals.
Some of these journals have now been digitised and are searchable, so we have some idea of Pelouze's work, and his students. Much of it involved preparation and study of nitro-esters of simple carbohydrates. They don't seem to say why, but in the context of the era, it is very likely they were looking for practical explosives. Braconnot had discovered nitrostarch in 1832; Pelouze apparently first prepared nitrocellulose in 1838, but didn't recognise it as a new substance. Then in 1846 -- just one year before Sobrero nitrated glycerin -- no fewer than 3 chemists prepared and described nitrocellulose in the same year, with Otto being first to publish.
For a long time, it appeared that nitrocellulose was not a practical material as it is very difficult to purify, and extremely unstable when impure. So it is not surprising that a year later, we find one of the early nitrocellulose investigators now conducting a program of testing nitrates of a range of simple organic compounds, especially carbohydrates. (Cellulose itself is a carbohydrate.) They don't say exactly why in anything I have read. It is possible that they are using nitrations to investigate fundamental chemistry; for example, such a process does help to identify the structure of simple sugars. However, they test them all for explosive properties and refer to them all as "fulminant" (i.e. explosive.)
Sobrero's paper on his discovery of nitroglycerin has been digitised. (Note: article is in French.) He mentions earlier work with sugars and notes that glycerin is not a carbohydrate, but doesn't seem to say what inspired him to try to nitrate it except that it is "analogous" with other materials that his group had nitrated. Incidentally, in this paper he describes purification methods at some length, and also discusses the toxicity:

... for it is sufficient to take a very small quantity (which can be taken
  by slightly wetting the tip of the little finger) on the tongue to
  feel a powerful migraine for many hours. This action on the human body has
  been observed by many people in my laboratory ...

However, he doesn't mention its extreme shock sensitivity. Later, Sobrero came to the conclusion that nitroglycerin was too dangerous to use, and it was, of course, his colleague Alfred Nobel who did the considerably longer and more complicated studies to find a way to render it practicable. (The same situation occurred with nitrocellulose; discovered by Schönbein, Böttger and Otto, but it was Abel who did the real spade work to make a practical industry for it.)

Note that whilst many of the substances in this answer were historically called nitroX where X is the original moiety, none of them are nitro compounds in the modern sense. Rather, all explosives discussed in this answer are nitrate esters. So nitroglycerin, for example, should really be called glyceryl trinitrate.


text:
Second ionization energies of copper(Cu) and silver(Ag)
The ionization energies of copper and silver are

First ionization energy: Cu-745.5 kJ/mol Ag-731.0 kJ/mol
Second ionization energy: Cu-1958 kJ/mol Ag-2073 kJ/mol

Now, looking at the ionization energies, it is quite clear why $\ce{Cu^2+}$ is formed readily but only $\ce{Ag+}$ is usually observed.
The first ionization energies can be rationalized by the fact that Ag is larger in size than Cu.
However, the second ionization energy of Cu should be higher than second ionization energy of Ag because in case of Cu the electron is being taken from the 3d orbital while it is being taken from 4d orbital in case of Ag. 4d orbital should be larger than 3d orbitals and so, the energy required to pull electron from 4d orbital should be lower than that in case of 3d orbital.
Is there any explanation to this anomaly?


text:
How does the Wave-Mechanical model reconcile with the particle definition of an electron?
While just starting chemistry, I have a little pit of particle physics knowledge. I was introduced with the wave-mechanical model of an atom and was slightly confused, because from particle physics, I know an electron is a fundamental particle.
How does this reconcile with the idea of electrons as "standing waves"? Or is the standing wave idea something that explains how electrons never really have a definite location? I have had zero years of calculus so please keep calculus equations and jargon to a minimum.

https://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality

Particles are not particles.

Wave particle duality is a fundamental idea in the quantum mechanics required to understand the behaviour of fundamental particles
One of the key ideas in quantum mechanics is that particles fundamental behaviour cannot be explained if we assume their "real" behaviour is Newtonian (or billiard-ball like). The behaviour of particles like electrons depends on how we measure them and sometimes they appear wave-like sometimes particle like.
This is very important in chemistry as it leads to good explanations of chemical bonding and the shapes of molecules. Electrons in atoms and molecules, for example, were originally thought to behave like a mini solar system where they orbited the nucleus. But this explanation doesn't work well to explain much in chemical bonding or atomic behaviour.
Schrödinger developed a better way to think about their behaviour with his famous wave equation. This describes the shapes of electron clouds around atoms. One way to think of the output of this equation is that it describes clouds of probability describing the possible locations of an electron in an atomic orbital. Those orbitals are the "standing waves" of electrons bound to atoms and they have specific definite shapes (related to spherical harmonics which are the 3D equivalent of the standing waves in a vibrating violin string).
This description of the properties of electrons is the core of our understanding of chemical bonding and the shapes of molecules.
What are electrons? Quantum mechanics doesn't tell us what they really are: it just tells us how they behave when we make certain measurements. And, in chemistry, the measurements that matter mostly give results where the wave-like behaviour is what matters.


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Experimental determination of pH
I am trying to determine the experimental pKa for two weak acids that were titrated against 0.20M NaOH.
I have read elsewhere that you can take the point where the graph becomes steep and divide the value of base added by two the corresponding pH value would then be the pKa, but how do i choose which value since it may not be obvious which point the graph becomes steep.
Below I can see that the pka for acetic acid should be close to the theoretical value calculated of 4.76 and the Tris-HCl pka should be approximately 8.3 but there must be a better way than just guessing from a graph.
My textbook doesn't explain how to experimentally find pH just that it's the point where $[A^-]/ [HA]$. I am hoping someone can give me an equation to work with or guide me in the right direction.
Thank you,

So I think what you heard is about the right idea. The flat region is your buffering region, and isn't super helpful to deduce the pKa. Adding base consumes the weak acid, and because it is weak, you know you are mostly consuming the [HA] form, and equilibrating back to around the pKa. This is why your pH doesn't change much around the pKa.
As an example, if you had 10 mmol of HA to start, then at the pKa point you would have 5 mmol of HA and 5 mmol of A-. Now, if you keep adding base, at some point you will essentially consume the remaining 5 mmol of HA. Then, there will be negligible amount left and it can't buffer anymore - i.e. your pH will change rapidly because you are adding strong base. Here you will have ~0 mmol of HA, and ~10 mmol of A-. Note that you have twice the amount of A- now. The pKa will have been at the point where you had half of this.
Experimentally, I know two simple ways. Using a pH meter and no indicator, you have to measure the sharp region more carefully (i.e., drop by drop), because you want to be able to find the exact point where the pH changes the fastest. By approximating the derivative, i.e.
$$ \frac{d}{d(V_{base})} pH(x_i) \approx \frac{pH(x_{i+1})-pH(x_{i-1})}{x_{i+1}-x_{i-1}}$$
you can find the point of greatest change, or the equivalence point.  Obviously, better data in that region will give you more clarity. I used to do one titration quickly so that I would know the approximate region where I would need to be and then slowly titrate the rest.
The other method, i.e. using an indicator solution and then just very carefully reaching the equivalence point essentially gives you the same information, and I know people who can very quickly do this. But in general, this would not be as accurate.


text:
Why there will be only limited shells in a particular atom?
What is the maximum possible principal quantum number for an atom(any atom not a specific one) ?

There is no hard limit. See for example the hydrogen spectral series..

In fact there are an infinite number of bound solutions to the Hydrogen atom. Rydberg atoms generally are in states (usually mixed) with large quantum numbers.

If it has an infinite number of shells then the energy required to remove an electron from the inner would be infinite..

Imagine a universe that is not expanding, and contains only one large star and one small planet. It is technically possible to put the two objects into orbit no matter how far apart they are. At some very large distance, the gravitational attraction is so small that it is nearly negligible, but if given infinite time you will be able to trace the orbits for the planet-star system. If, however, you introduce more planets and stars close to the planet, then the influence of the star that is very far away will become irrelevant to the motion of the planet.
We can see a similar case when we look at an electron around an atom. Each time you excite it, you move the electron into a higher-level orbital that is more weakly bound. Eventually you get the electron into orbitals that are nearly degenerate and asymptotically approach zero binding energy. If you have a universe that contains only the electron and the atom, and get rid of the vacuum fluctuations, then yes, it is theoretically possible to put an electron into an orbital with arbitrarily high level of n.
In practice, at some point the electron becomes so weakly bound that electromagnetic forces in the environment will dominate and take the electron away from the atom, ionizing it.
It is experimentally possible to isolate an atom from the environment quite well and then excite the electron to a pretty high n without ionizing it. An atom that is excited in this way is called a Rydberg atom.


text:
Uses of Chlorine
Why in the reaction between sodium chlorate and water is chloric acid produced as well as Na+ and OH- ions. Shouldn't both the ions just react to produce NaOH?

What happens if you throw NaOH into water?

It splits into Na+ ions and OH- ions. But why wouldn't they combine to produce NaOH?

You have to make up your mind, because you can't have it split into ions and combine at the same time.

Well, H and OH manage to split and combine all at the same time. Detailed balance and whatnot...

It seems at first that there are three materials involved, but then we subdivide them and sometimes lose track of what's important.
The first material mentioned is sodium chlorate. It is a solid, and if you could break it up into very tiny pieces, at last you would get Na+ ions and ClO3- ions (they could break apart further, but let's not be brutal).
The second material mentioned is water. Break it down to H2O molecules - a few break down further to H+ and OH- automatically, but only about once in 10 million times at pH=7.
The third material is the solution of sodium chlorate in water: sodium ions and chlorate ions dissolve and separate in water molecules (with a very few H+ and OH- ions). Chloric acid is a strong acid; in water the ClO3- doesn't tend to hold on to its H+ or grab H+ from water molecules, so extra OH- ions are not produced.
Overall, nothing much is happening beyond the dispersal of the Na+ and ClO3- ions into water and some association with individual water molecules that makes dissolution possible. If you evaporate the water, you get solid sodium chlorate back; there is no chloric acid produced or detectable.

Na+ is more stable in water than in the form on NaOH.
The effective charge on Na+ is reduced because of the Oxygen in H2O. Therefore why will it prefer a less stable form?


text:
Converting pounds to grams
I wanted to figure out how many grams were to a lb and instead of looking it up I used my textbook and found 1 kg = 1000 g and 1 kg = 2.205 lb. I set up a cross multiplication equation to solve for x. The number I got was close but off by a few hundredths. I got 453.5147 g = 1 lb. I looked to see if I was right and found that the actual number is 453.59 g to lb. What did I do wrong?

I respect the effort in asking and answering your own question, but unit conversions are not within the purview of this site.

I'm voting to close this question as off-topic because it is not about chemistry.

http://chemistry.bd.psu.edu/jircitano/sigfigs.html.

Google says that 1 kg is 2.20462 pounds. Which is 0.2% off of what you wrote. That means that your converted answer will also be off by ~0.2%.

1kg = 2.20462lb. Textbook just rounded.


text:
Keeping the electronic configuration in the ground state is expensive in Ab-initio MD?
In E. Sandré & A. Pasturel's An Introduction to Ab-Initio Molecular Dynamics Schemes

If using small time steps (typically in the order of a few tenth of femto-second), the changes in the electronic configuration is small enough so that convergence of the electronic configuration is easily achieved using a few conjugate gradient steps. However, the computational cost of dynamical simulation appears to be mostly due to keeping the electronic configuration in the ground state.

I don't understand this. DFT and HF are methods that can only give electron ground state. Since it is either DFT or HF that is used in Ab-initio MD simulations, then what does this excited state is referring to?
My understanding:
First of all, the number of electron orbitals calculated by either DFT or HF approach equals the number of electrons in the system, so we get no electronic excited state.
Under Born-Oppenheimer approximation, while varying the atomic configuration, the electrons are expected to stay in their ground state.
Since the validity of BO approximation dictates no abrupt change in electronic energy.
So, the reason of 'Keeping the electronic configuration in the ground state is expensive' is that when the atomic configuration changes in MD simulation, it is hard to make the electronic energy difference small.
Is my understanding correct?

I believe they are saying that when you take a small time step, not much can change so it shouldn't take long to converge the energy. This makes sense. The confusing part is how one can go from that to saying that it is hard to keep the electronic configuration in the ground state. So, I believe what they mean is that it is possible to accidentally get onto an excited surface during the time step. This seems like it would be system-dependent. The point is you have to expend time to verify you are still on the same surface somehow in addition to verifying the orbitals are converged.

They refer to approximate ground state (ie nit exact but close), not a specific excited state. Ie in spite the SCF requires only few steps, it is still the most expensive part of the calculation if we want to system wave function to be calculated with SCF in every single step. A common spread up is to let somehow a looser convergence on the electronic wave function, and calculate the forces without calculating every single time a well converged wavefunction.

Let us first note that the calculation of the first derivative of the total DFT or HF ground state energy of a molecule with respect to its nuclear coordinates (called the gradient) depends on a converged self-consistent field (SCF) solution. The solution is obtained in an iterative fashion and is represented as a collection of molecular orbital (MO) coefficients. In some sense, they are unique for the nuclear coordinates. However, if we move the nuclei only very slightly, repeat the calculation, and compare the new set of MO coefficients with the previous ones, they will be almost identical.
Almost, that's the key here. The gradient will become more of an approximation as you move away from the nuclear coordinates for which the underlying SCF solution was originally calculated (and merely recalculate the gradient at new coordinates using old MO coefficients). One way of dealing with this problem is to take the old SCF solution as a starting point for a new SCF solution, which should converge in just a few steps. But those few SCF iterations can easily dominate the cost, given that a gradient calculation after obtaining the SCF solution typically has a computational cost on the order of 1-3 SCF iterations.$^{1}$
To summarize: I understand the sentence "However, the computational cost of dynamical simulation appears to be mostly due to keeping the electronic configuration in the ground state." in the quoted section of the question to mean: recalculation of the SCF solution is necessary at least every few steps of the molecular dynamics (MD) simulation and that recalculation is the most expensive part.

$^{1}$ A specialized quantum chemistry program, that would keep certain integrals and other intermediates across the different SCF procedures, would help somewhat with the computational cost at the expense of more on-disk storage and more uncertainty, since all those quantities implicitly depend on the nuclear coordinates.


text:
Acid HClF6 and HIF6
I was wondering about $\ce{HClO4}$ which is a very strong acid . I thought it can be made stronger by replacing $\ce O$ with $\ce F$ to make $\ce{HClF6}$ but didn't find its existence . I thought that it might be unstable due to repulsions between $\ce F$ . So than i switched to $\ce{HIF6}$ and there was no existence of this also as well as $\ce{HBrF6}$ . Is there a reason that the halogens cannot form $\ce{HXF6}$ acids similar to $\ce{HSbF6}$?

Related (involving sfur as the centrasl atom):  https://chemistry.stackexchange.com/questions/102290/do-polyfluorosulfuric-acids-like-h2so2f4-exist/102294#102294.  You tend to lose hydrogen halide molecules from your proposed complexes.

ClF6- , BrF6- and IF6- all exist, as does IF8- . However I doubt the parent acids will be stable, but my good chemistry books are a long way away at the moment.

There are a whole bunch of these ions.  But the reference does not document any case where anions can be protonated to acids.

Why can't they protonated to acids.  is there any known reason

Can't they be protnated by superacids

Nothing wrong with protonation as such.  The issue could be they decompose when protonated; compare with a nitrite or a bicarbonate..


text:
Understanding how the definitions of pH and activity are compatible
At $T=\pu{25 ^\circ C}$ and $P=\pu{1bar}$, the density of water is $\pu{0.997 kg L-1}$.
Neglecting nonideality of the mixture $\ce{H2O(l), H+}$ and $\ce{OH-}$ (all activity coefficients are close to $1$), we have the following for $\ce{H+}$:

amount concentration (molarity): $C =\pu{1 \times 10^{-7} mol L-1}$.
mole fraction: $x = 1.807 \times10^{-9}$

An old definition for pH is $\mathrm{pH} = −\log_{10}(C)$, which gives the familiar $pH=7$ for pure water.
A more modern definition is $\mathrm{pH} = -\log_{10}(a)$, where $a$ is activity.
This definition would make sense if the activity of a component in an ideal solution were equal to its molarity. However, isn't the activity defined as
$a = \exp\left(\frac{\mu - \mu_0}{RT}\right)$,
where $R$ is the gas constant, $T$ the thermodynamic temperature, $\mu$ the chemical potential and $\mu_0$ the standard chemical potential of reference, in which case it is equal the mole fraction in an ideal solution?
My impression is that the definition of $\mathrm{pH}$ was changed in order to incorporate the effect of nonideality, in which case I would expect it to be redefined as $\mathrm{pH} = -\log_{10}(\gamma C)$, where $\gamma$ is the activity coefficient. Maybe there is an alternative definition for activity such that $a = \gamma C$.