text:
Why do disproportionation reactions take place?
I want to know what are the conditions when disproportionation reaction is feasible. Is it just because of stability factors or it has other dimensions to it. As an example:
$$\ce{3HNO2 ->  HNO3 + 2NO + H2O}$$
Nitrous acid disproportionates to nitric acid and nitrogen monoxide. Here stability of +5 in nitric acid is understandable but how +2 in the monoxide is stable if the disproportionation takes place to gain stability.
Am I ignoring some other factors like pH or hydration enthalpy. I would be very thankful if someone gives me some general ideas on these reactions.

Related: Disproportionation with aqueous NaOH

"Stability" is not really a driving force of a reaction. That would be the change in Gibbs Free Energy or ∆G. That takes into account both entropy and enthalpy, as well as concentrations of reactants and products, and temperature.
If you haven't yet discussed these in class, you should be able to read about these concepts by looking up "Gibbs Free Energy" in your textbook's index.


text:
Why are higher acids solid while lower acids are liquid?
I was just going through acids on Wikipedia in my free time, and noticed this neat trend:

Formic acid, Ethanoic acid, uptil Nonanoic acid, are all liquids. The first few are colourless, while the latter are yellowish and also "oily" (more viscous).
Decanoic acid and all its elder brothers are white crystals (or "powders")
Aromatic acids like benzoic acid or picric acid are solids.

I am unsure if there could be a reasonable logic for the colours. But, I am sure there could be a good logic for the physical state of these acids at room temperature.
To me, all these acids have hydrogen bonding common, so we can rule out that factor. Apparently, the physical state seems to be related to the "size" of the acid. The heavier acids  are solids, while the lower ones are liquid. My question is: why is it so?

PS: List of all acids at one place is at the bottom of this Wikipedia page.

Van der waals interactions

Come to think of it, higher everything is solid (alkanes, alkenes, alcohols, etc.)

On a cool day, you probably would not be asking this question. Formic acid: mp. 8 deg. C (47 deg. F); acetic acid: 17 deg. C (62 deg. F).

@user55119 haha, sharp observation! ;) Though on a more serious note I have always hoped (but have never seen or read about) that chemical laboratories of high standard have temperature control builtin, so they maintain temperature at around 25degC throughout their room.

Not when it isn't working!

Acids with a larger size have greater intermolecular forces than 'smaller' acids, meaning it would take more heat energy to break those bonds in larger acids than in smaller ones. Hence at room temperature the heavier acids would be solid.

Really, that's it? This was so straightforward :/ Thanks though! :)

@GaurangTandon Well, this question was rather trivial and I'm pretty sure similar ones were asked here many times... :(

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Abnormal ionisation energy trend: Group 13 and 14
I was going through some ionisation energy data, where I came across the following:
Ionisation energy order for Group 13 and 14:
B > Al ≈ Ga > In < Tl
C > Si > Ge > Sn < Pb
What could be the possible reason for reversal in trend after fourth period?
I'm not sure how accurately it can be attributed to screening effect.
An explanation would be great.

d-orbital contraction.

@Cyclohexanol. You meant the lanthanide contraction I guess?

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Adding more solid to a solid/gas equilibrium
$$\ce{CaCO3(s)<=> CO2(g) + CaO(s)}$$
In this equation, what happens if you increase the amount of one of the solids while there are existing quantities of everything else?
According to the calculation for the equilibrium constant, equilibrium is only dependent on the concentration of $\ce{CO2}$, so it technically should not change anything, right? However, if you decrease the pressure of the $\ce{CO2}$ enough, then shouldn't there be a point where the equilibrium constant can no longer be met because there aren't enough moles of $\ce{CO2}$?
Additionally, at this point, shouldn't adding more $\ce{CaCO3}$ change the point of equilibrium?

Related questions:

How does equilibrium change when the products are solid and gas?
How do equilibrium shifts affect solids?
Le Chatelier's principle with pure solids and liquids
Adding pure solid/liquid in heterogeneous equilibrium

The concentrations of the solids change only negligibly with temperature or other reaction conditions and so are involved in the equilibrium only as constants. The amount of solids present does not change the concentration of each solid in its crystal structure. Therefore the equilibrium of the reaction is written as:
$\mathrm{K = constant}\times \ce{[CO2]}$
If one of the reactants ($\ce{CaCO3, CaO, CO2}$) is not present, there is no equilibrium.


text:
Why is the suffix carbaldehyde valid here?
The compound in question, and the name generated by ChemSketch:

I think this compound should be named as "cyclopentylmethanal" because the principal functional group (carbonyl) lies in the methane chain and not in the ring. So, the ring should be treated as a substituent for methanal, but apparently this is not the right answer.
I also noted that the following compounds validate my logic (ring is being treated as a substituent):
So, what is the exceptional rule with the first compound?


text:
Determination of pKa by absorbance and pH of buffer solutions
I've been tasked (as part of a report) to calculate the $\mathrm pK_\mathrm a$ of 4-nitrophenol by measuring the pH and absorbance of six 'buffer solutions' containing the 4-nitrophenol:
$$\begin{array}{cccc}\hline
\text{Solution} & \text{Volume of base / }\pu{cm^3} & \text{Volume of acid / }\pu{cm^3} & \text{Volume of phenol / }\pu{cm^3} \\ \hline
1 & 2.80 & 7.20 & 1.00 \\
2 & 5.00 & 5.00 & 1.00 \\
3 & 6.20 & 3.80 & 1.00 \\
4 & 7.20 & 2.80 & 1.00 \\
5 & 8.70 & 1.30 & 1.00 \\
6 & 10.00 & 0.00 & 1.00 \\ \hline
\end{array}$$
We have to then plot a graph of $\mathrm{pH}$ versus $\log_{10}[A/(A_f-A)]$ and use that graph to calculate $\mathrm pK_\mathrm a$. $\text{A}_{f}$ is the absorbance of Solution 6 - we assume that the weak acid in the buffer is present only as its conjugate base - and A is the absorbance of the buffer solution.
My two questions about this are:
1) Where does the expression $\log_{10}[A/(A_f-A)]$ come from? I thought the derivation would be as follows, which clearly gives a different result ($\text{A}_{f}$ on the numerator rather than $\text{A}$):

Using the Beer-Lambert Law,
$$
\begin{align}
A &= \epsilon cl \\
c_\text{base} &= \frac{A_{f}}{\epsilon l} \\
\end{align}
$$
And substituting into the Henderson–Hasselbalch equation:
$$
\begin{align}
\mathrm{pH} &= \mathrm pK_\mathrm a + \log_{10}\left(\frac{[\text{conj. base}]}{[\text{acid}]}\right) \\
\mathrm{pH} &= \mathrm pK_\mathrm a + \log_{10}\left(\frac{A_{f}}{A_{f}-A}\right) \\
\end{align}
$$

2) Once I have the graph, how do I calculate $\mathrm pK_\mathrm a$ from it? Assumed it would be using the gradient of the line of best fit, but that won't work since $\mathrm pK_\mathrm a$ is given by the difference of $\mathrm{pH}$ and $\log_{10}[A/(A_f-A)]$ rather than the quotient.

Make the plot $10^{\mathrm{pH}}(A_{\ce{L-}} - A)$ versus $A$,
the adjusted equation will be of the form
$$
10^{\mathrm{pH}}(A_{\ce{L-}} - A)
  = 10^{\mathrm{p}K_\mathrm{a}}A
  - 10^{\mathrm{p}K_\mathrm{a}}A_{\ce{HL}},
$$
the slope is $10^{\mathrm{p}K_\mathrm{a}}$, that is, $\mathrm{p}K_\mathrm{a} = \log_{10}(\text{slope})$.


text:
Is the reaction of SiC and H2O a redox reaction?
In the reaction: $$\ce{SiC + 2H2O <=> SiO2 + CH4}$$ The oxidation state of Si is +4 in both SiC and SiO2, while the oxidation state of carbon is -4 in both SiC and CH4 (confirmed by this).
Here (page 69, middle) it says that silicon carbide is oxidized in humid air.
So actual oxidation states of C and Si do not change, but this is considered a redox reaction? What a I missing?

I'd appreciate it if downvoters could let me know what is wrong with the question. I thought it was legitimate when I posted it...

Silicon carbide is very inert chemically, so this is more of a terminology discussion than discussion of a chemical mechanism. There are two ways of describing the reaction:
1) water oxygenates the silicon and hydrogenates the carbon. Or
2) water oxidizes the silicon and reduces the carbon.
The question predisposes an answer like #2, but maybe #1 is more descriptive. Or, perhaps better:
3) water oxidizes the silicon and hydrogenates the carbon. Or
4) water reacts with SiC to give xyz (CH4, CO2, CO, SiO2). Avoid "redox" entirely.
You can't compare this high temperature reaction with redox reactions at room temperature where voltage is applied to electrodes in solutions.


text:
Isocyanides undergo addition reactions, but cyanides do not. Why?
My book mentions addition reactions of isocyanides with halogens, ozone, sulphur, etc. but it doesn't mention any such reaction for cyanides. Instead, it bluntly states: "Cyanides do not undergo addition reactions". My question is: why is this so?
 I understand the structures of the ions, but am unable to relate the exact correct logic in this situation.

Here are the addition reactions of isocyanides I am talking about:
$$
\begin{align}
\ce{R-NC + O3 &->RNCO + O2}\text{(methyl isocyanate)}\\
\ce{R-NC + 2HgO &-> RNCO + Hg2O}\text{  (methyl isocyanate)}\\
\ce{R-NC + S &-> RNCS} \text{  (methyl isothiocyante)}\\
\ce{R-NC + Cl2 &-> RNC\cdot Cl2} \text{ (methyl imino carbonyl chloride)}\\
\end{align}
$$

I would think that the nitrogen in cyanide is much less nucleophilic than the carbon in isocyanide, from the look at their resonance structures and how much they might contribute to the real structure. Maybe some reason along those lines?

@ralk912 Well I wish I knew the reason myself :P if you think you're in the right direction, please consider posting an answer. Thanks!

Well, with ozone is likely to be through a 1,3-dipolarcycloaddition, but I can't figure out how the ozonolide would collapse. I can't really find many references either, so I'm not confident on the answer. Interesting question for sure.

Isocyanide can undergo addition reaction due to the fact that it is capable to attach both the electrophile and nucleophile in the same carbon atom while it is not achieved by cyanide. The negative charge on carbon atom accepts the electrophile to form an intermediate species in which the positive charge on nitrogen atom gets neutralized by the addition of the nucleophile to the carbon atoms and the nucleophile gets attached in the same carbon atom. Cyanide is incapable of undergoing such mechanism because of less nucleophilicity of nitrogen atom of cyanide and thus don't undergo addition reaction. The mechanism is as follows:

Nilay, might want to omit the first link. Chrome flares up with a 'not safe' warning when I used it.

@HarryHolmes Thanks for informing. I removed the links.

The Google books one was okay! It was the other one.

@HarryHolmes Google links may sometimes be problematic.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Alkyl chain substituent has higher preference than benzyl substituent?
The compound in question, and the name generated by ChemSketch:

I have looked at relevant (outdated) documentation for naming of substituted amines. I understand tertiary amines are named as: $\text{N-alkyl}_1\text{N-alkyl}_2\text{alkyl}_3\text{amine}$.
Now, I am confused at the first point: when we have to select the parent carbon chain. For this compound, I would have selected "benzylamine" as the parent chain, as it obviously has more number of carbon atoms and more unsaturation than both methyl and propyl. Thus, my preferred IUPAC name would have been "N-methyl-N-propylbenzylamine".
But, for some reason "propane" is the main carbon chain here instead. So, what is the logical reasoning for the naming these tertiary amines?

Notice propane not -yl here? Can you make such change in benzyl?

@Mithoron No, I can't... But still, that small detail shouldn't stop benzyl from being the main chain. There must be another way to name benzyl as the main chain I hope?

The longest carbon chain is the base name. Benzyl has a "one carbon chain". Compare the two structures below. In the top example, the butane chain, even though it bears a phenyl group, is the main chain. On the bottom example (yours), the phenyl is attached to a "one carbon" chain. therefore, propane is the main chain.

"Benzyl has a "one carbon chain"" That's interesting, but then, what about all the unsaturation it has? Is it just being ignored?

You look for the longest chain, just like acyclic alkanes. The aromatic ring in the top example is just a substituent. In your example, aromatic ring is part of the benzyl group.

After checking out phenylacetic acid and 1-phenylbutan-2-ol, I think I get your point. Apparently benzyl can't be the part of the main chain as long as it doesn't have the functional groups directly attached to it (like phenol or aniline)


text:
What are the ways to abiogenically synthetesize Uric acid?
The only link I could find was, but I couldn't find it anywhere:
Horbaczewski, Johann (1882). "Synthese der Harnsäure" [Synthesis of uric acid]. Monatshefte für Chemie und verwandte Teile anderer Wissenschaften. 3: 796–797.
This is not a homework question, as I don't have a chemistry background, but I do understand thing about biology. I am trying to work the chain of events in the opposite order: https://upload.wikimedia.org/wikipedia/commons/b/b8/Nuc_acid_deg.png
If you look for synthesis of urea, I only find biological sources.


text:
Relation between R and k
$R$ is the universal gas constant equaling $8.314\ \mathrm{J/(mol\ K)}$.
$k$ is the Boltzman constant equaling $1.38\times10^{-23}\ \mathrm{J/K}$.
$$R=nk$$
$n$ is no. of moles.
$n$ can change.
$R$ is a constant, $k$ is a constant,  but $n$ changes. Then how is $k$ a constant?
Is the equation correct?

R is for one mole. One mole is a constant.

The form of the equation is correct, but one of the variables is wrong. It should be:
$$\mathrm{R} = N_{\mathrm A}k_{\mathrm B}$$
The Boltzmann constant ($k_{\mathrm B}$) is for one molecule (or atom); the universal gas constant ($R$) is for one mole of molecules (or atoms).
Therefore, in this relationship - and which is clear from dimensional analysis ($R$ is per mole) - $N_{\mathrm A}$ is the Avogadro constant, and not number of moles: in your argument, $n$ is a dimensionless integer, and thus the units do not work out. You need to use $N_{\mathrm A}$ instead of $n$. Note that the Avogadro constant can be denoted by either $L$ or $N_{\mathrm A}$.
The proper relationship is:
$$R = (\pu{6.022*10^{23} mol-1}) \cdot (\pu{1.380648*10^{-23} J*K-1}) = \pu{8.314 J*K-1*mol-1}$$
Note that I have included more significant digits for the Boltzmann constant here to get your value for $R$.


text:
Order of filling of molecular orbitals in heteronuclear diatomic molecules such as CO
Currently studying for possibly my last Chemistry exam in my life, can't understand why physics undergrads have to learn chemistry as high level as this, but oh well.
So I'm currently dealing with Molecular Orbital theory to predict the strength of bonds between various atoms and whether the resulting molecule will be paramagnetic or diamagnetic.
Now, I have discovered that for homonuclear diatomic molecules, there are two different molecular orbital filling orders, as in, for Li$_2$ all the way to N$_2$, the two $\pi_{2p}$ orbitals are filled before the $\sigma_{2p}$ orbital is filled.
From O$_2$ onwards, the $\sigma_{2p}$ orbital is less energetic than the two $\pi_{2p}$ orbitals, and thus the $\sigma_{2p}$ orbital is filled before.
For certain heteronuclear diatomic molecules, such as CN$^-$, the order of filling is the same as in N$_2$, and in molecules like OF$^+$, it appears to be the same as in O$_2$.
However, in "mixed" molecules such as CO or CO$^+$, how does one predict the order of filling of the molecular orbitals? It appears to be $\sigma_{2p}$ first, then the two $\pi_{2p}$ orbitals, following the fashion of O$_2$.
Is there a rule of thumb on how to predict which molecular orbital will be more energetic in molecules where the two atoms follow different orbital filling orders in their corresponding diatomic molecules, or is there not?
We are only asked to be able to do this for atoms lighter that Ne, it seems, so we only care about $2p$ orbitals.

Good question. Just a point of clarification, however, you talk about 'orbital filling' - orbitals are filled bottom up, always. I think what you're trying to get at is the ordering of the orbitals themselves (i.e. does s-p mixing take place).

@NotEvans. Yes, that is the question. Pretty much which orbital is more energetic in which case.


text:
Deciding if a molecule is polar and non-polar in cyclic carbon compounds
While discussing about dipole moment of the compounds, our chemistry teacher told that the dipole moment will be 0 for 1,4-Dichlorobenzen, and for those compounds where same atoms or group of atoms are attached in para position if the hybridization of the atom of the group which is directly attached to the carbon of the benzene ring is sp or sp2 but if it is not the case i.e if OH is attached at para positions then the dipole moment wont be 0. I want to know if there is any reason or are they just unexplained examples. Are there more such exceptions?
Please check the picture for examples.

Yes there is a reason, symmetry.

No but if the copound is oh instead of no2 the hybridisation will be sp3 and dipole moment wont be 0

Exactly, because it won't be symmetrical. Is the OH going to be linear?

Ok then why true for no2. It has lone pairs more towards doubly bonded oxygen

Resonance, both the oxygens are the same

Oh yes. But in OH attached to C why the dipole moment wont be 0 why the O and H and C are in the same plane and lone pairs are invisible

@avnish can you tell that instead of OH if CH3 was on either side , what would have been the dipole moment?

Both sides or one

On Both the sides CH3 there on opposite vertices. My teacher said that the dipole moment will not be 0 I can't understand why. Do you have the same opinion?

The dipole moment is 0 see this

The actual reason behind sp3 atoms if its not CH3 type structure meaning all visible 4 'equivalent'bonds having dipole moment not equal to 0 is the exsistance of infinite conformers for the same molecule out of which only one is 0 and rest are not. The reason why only sp3 shows such variation is tgat sp2 hybridized or single atom or sp hybridized atoms would be equally planar on both tge sided so as to cancel dipole moment of each other in all possible ways meaning for every conformer there is a counter conformer so the final dipole moment is 0.


text:
Why Mohr's salt and not FeSO4 in titration?
During titration of $\ce{KMnO4}$, we use Mohr's salt instead of ferrous sulphate salt to prepare the standard salt solution. Is it done to prevent oxidation of $\ce{Fe^2+}$ to $\ce{Fe^3+}$ before the titration itself? But to prevent that sulphuric acid is used. So why we do not use ferrous sulphate salt instead?

https://en.wikipedia.org/wiki/Ammonium_iron(II)_sulfate#Applications

To put an answer, and prevent link rot, quoting Wikipedia with my emphasis, "Mohr's salt is the preferred source of ferrous ions as the solid has a long shelf life, being resistant to oxidation. This stability extends somewhat to solutions reflecting the effect of pH on the ferrous/ferric redox couple. This oxidation occurs more readily at high pH. The ammonium ions make solutions of Mohr's salt slightly acidic, which slows this oxidation process"


text:
Calculating Triplet-Triplet Spin-Orbit Coupling
According to the Q-Chem manual [1], the singlet-triplet spin-orbit couplings between a singlet excited state $I$ and a triplet excited state state $J$ is:
$$\langle\Phi_\text{singlet}^I|\hat{\mathrm H}_{SO}|\Phi_\text{triplet}^J\rangle = \sqrt{\sum_{m_s=0,\pm1}|\langle\Phi_\text{singlet}^I|\hat{\mathrm H}_{SO}|\Phi_\text{triplet}^{J,m_s}\rangle|^2}~.$$
The further elaborate on the triplet-triplet spin-orbit coupling to be:
$$\langle\Phi_\text{triplet}^I|\hat{\mathrm H}_{SO}|\Phi_\text{triplet}^J\rangle = \sqrt{\sum_{m_s=0,\pm1}|\langle\Phi_\text{triplet}^{I,m_s}|\hat{\mathrm H}_{SO}|\Phi_\text{triplet}^{J,m_s}\rangle|^2}~.$$
But shouldn't the latter actually be a double sum like:
$$\langle\Phi_\text{triplet}^I|\hat{\mathrm H}_{SO}|\Phi_\text{triplet}^J\rangle = \sqrt{\sum_{m_{s,i}=0,\pm1}\sum_{m_{s,j}=0,\pm1}|\langle\Phi_\text{triplet}^{I,m_{s,i}}|\hat{\mathrm H}_{SO}|\Phi_\text{triplet}^{J,m_{s,j}}\rangle|^2}~?$$
As of porphyrin's comment, I'd like to add, why I think so.
The triplet-triplet SOC matrix looks like the following:
\begin{array}{c|ccc}
\hline m_{s,i}\text{\\} m_{s,j} & 0 & -1 & 1 \\\hline
  0 & 0 & a & a \\
  -1 & a & b & 0\\
  1 & a & 0 & b\\
  \hline
\end{array}
If the rules from the "single sum" would apply, only $b$ would count into the sum and $a$ would be neglected, because the summation only accounts for the diagonal elements.

[1] http://www.q-chem.com/qchem-website/manual/qchem50_manual/sect-tddft.html

I would agree if the total from all $m$ is needed then this would seem logical.


text:
How does an acid prevents or discourges oxidation of iron?
In the titration of $\ce{KMnO4}$ with Mohr's salt, I have learned that sulphuric acid prevents oxidation of $\ce{Fe^2+}$ to $\ce{Fe^3+}$ in the solution before the actual titration. How does the acid hampers oxidation?

This is largely due to the extreme insolubility of iron (III) hydroxide, i.e. $\ce{Fe(OH)3}$.
One can write the oxidation reaction, somewhat simplified, as:
$$\ce{Fe^2+(aq) + 6 H2O(l) <=> Fe(OH)3(s) + 3 H3O+(aq)} + e^-$$
The $K_{sp}$ (solubility product) value for $\ce{Fe(OH)3}$ is very small: $2.79 \times 10^{-39}$ (Wikipedia).
From the equilibrium reaction it's easy to understand that acidic conditions (high $[\ce{H3O+}]$) push the equilibrium to the left, thus preventing the oxidation.
It also explains why Mohr's Salt resists oxidation much better than simple ferrous sulphate, as the ammonium ion provides some acidity.
In alkaline conditions the oxidation of ferrous ions to ferric ions is very fast.

Yes, I understood you used the Le Chatelier's Principle on the reaction. That answers my questions. But why does extreme insolubility has a role to play in this.

Well, precisely because of Le Chatelier. As precipitation removes $\ce{Fe^3+}$, the equilibrium shifts toward more $\ce{Fe(OH)3}$. Lower solubility obviously promotes precipitation.

I'll put a bit of 'mathematical meat' on this answer tomorrow.

Tomorrow never came...

Would you still like me to add that part?


text:
How to compare energy levels in hybridized sp orbitals?
How to compare energy levels in $\ce{sp, sp^2, sp^3}$ orbitals?
Since a higher energy level implies lower stability, an $\ce{sp-sp}$ bond must have the lowest energy level, since it is formed by the overlap of one sigma and two pi bonds in total, more than those in $\ce{sp^2-sp^2}$ or $\ce{sp^3-sp^3}$. Hence, $\text{energy level(}\ce{sp < sp^2 < sp^3}\text{)}$.
Is my calculated trend correct? If so, then is this the intended reasoning? If not, then why?

How to compare energy levels in $\ce{sp, sp^2, sp^3}$ orbitals?

Background

Electrons in an atomic $\ce{2s}$ orbital will be lower in energy than electrons in an atomic $\ce{2p}$ orbital since the s orbital is closer to the nucleus.
When we describe a hybrid orbital as $\ce{sp^n}$, this is really a shorthand notation for $\ce{s^1p^n}$, which means that the orbital is made up from 1 part s orbital and n parts p orbital.

Analysis
As the "n" in $\ce{sp^n}$ grows larger we will have more p character and less s character in the hybrid orbital and, as explained above, it will consequently be higher in energy. So an $\ce{sp}$ hybrid orbital that has 50% s character will be lower in energy than an $\ce{sp^3}$ orbital which only contains 25% s character.
The energy ordering you proposed in your question is correct: $\ce{sp < sp^2 < sp^3}$.


text:
Understanding the CIP configuration for cyclic substituents with hetero-oxygen atoms

As per CIP sequence rules, the topmost $\ce{C}$ on the left gets 1st priority. Moving on to the right side of the double bond, we have:
Top $\ce{C}$: ($\ce{C}$,$\ce{O}$,$\ce{O}$,$\ce{H}$)
Bottom $\ce{C}$: ($\ce{C}$,$\ce{O}$,$\ce{O}$,$\ce{H}$,$\ce{C}$...)
Thus bottom $\ce{C}$ gets priority one on the right side.
Now, since 1st priorities are opposite, it's an E isomer. But the solution manual states that it's a Z isomer. Why is my method incorrect?

CIP rules for cyclic substituents are different. See
this question for the rules. (The rules are given in the question itself).
This will be the hierarchical digraph for the right side of the molecule.

The atoms encircled are the ghost atoms. From the digraph one can see why the compound is Z.


text:
What is a suitable acid to remove limewash from masonry brick
Diluted Hydrochloric acid is often touted as the only thing to remove limewash from masonry but the process is dangerous and can adversely affect the bricks, color changes can readily occur.
Is there a viable alternative? I noticed that a kettle descaler product used formic acid and wondered if that could be used instead.

Glycolic acid (hydroxyacetic, HOCH2CO2H) is a mild organic acid, not very expensive, and the calcium salt is quite soluble. It is essentially a non-smelly vinegar! It has been used for brick washing.
One issue is that whereas HCl will dissolve CaCO3 until the HCl is all gone, the pH of glycolic acid solution increases as it dissolves CaCO3, so the dissolution reaction slows down and finally stops at maybe 50% completion.
Citric acid and tartaric acid are very soluble, but their calcium salts are not. Calcium citrate and tartrate may come down as fine crystals in the pores of the brick and be difficult to remove. Calcium formate is moderately soluble and so would be useful, but formic acid is a vesicant, so it must be used with as much or more care than HCl.
Muriatic acid is readily available at 32% and 20% concentrations. Diluting by a factor of 50 would give a pH about 2, which is safer for people and brick, but that's a lot of water to clean a brick wall.

I've found that while Muriatic is very strong, the tasks is still difficult and time consuming even with 32% strength acid. I also discovered that at 32% and also at a 50% dilution it would turn London yellow 'stock' bricks (around 100 years old) a slightly different colour, more golden, or browner. I tried some phosphoric acid to see if that changed some of these 'stains' afterwards, but with no effect whatsoever.  Would I be correct in saying that Glycolic, Citric and Tartaric are much less active compared to Muriatic?

Glycolic and the other organic acids are no match for muriatic for speed, but glycolic has the advantage of less (or no) discoloration on many bricks.

An issue with limewash is that it is absorbed into the pores of the bricks, where it absorbs CO2 and does a good job of plugging up the pore. Dissolving the contents out of the pore channel is a long term process.

Another issue, relative to discoloration, is possible additives to the limewash. In particular, addition of portland cement, which is used for a harder coating, hydrates to silicates which are more difficult to dissolve. If silica or cement is present, restoration cleaners which contain hydrofluoric acid might help - but they are even more hazardous than muriatic acid. (The problem is that these cleaners don't hurt much if you spill them on yourself or your clothing, but if you leave the fluoride on for an extended time, plan to visit the hospital emergency room.

The limewash is pure white to look at, but seems pretty hard with minimal dusting. As an alternative to chemical removal I tried spinning the faces on a few bricks to see how that process would compare. On those spun bricks the colour just under the surface seems normal. Is the discolouration via additives that you mentioned something that would happen on original application or as a reaction to the muriatic?

I'm not familiar with additives to limewash other than what's in Wikipedia - there are many! With regular brick (especially with lighter colors), it seems that strong acids like muriatic (i.e., low pH) can dissolve metals that are light colored in the original baked brick, then deposit them as darker salts on the brick, giving discoloration. Glycolic acid is milder, so does not do this (as much). Additives to the limewash is a whole new possibility.

Is the potential to stain the brick linked primarily to the ph rather than the acid ? In other words, if both glycolic and muriatic were diluted to ph of 3, would they have the same stain outcome?

It would seem so, but I have no comparative data, no guarantees. At least I have not heard of any cases where glycolic stains worse.  The problem is that pH 3 HCl is very dilute (0.36 grams per liter), and would become used up fast (to a higher pH) and therefore ineffective. Glycolic may be slower but more satisfactory overall if discoloration is an issue.


text:
SN1, SN2 or acid-base reaction

I am asked to give the type of reaction (SN1, SN2 or acid-base reaction) for the following reactions. I know that both $\ce{MeOH}$ and $\ce{EtOH}$ polar protic solvents, so because the first is an SN2 reaction shouldn't the second reaction be that as well? It says that it is an acid-base reaction as $\ce{MeO-}$ reacts with the $\ce{EtOH}$ to from 1-bromopropane, $\ce{MeOH}$ and $\ce{EtO-}$ and then goes on to from a SN2 reaction.
I would appreciate if someone could explain why the second reaction would be an acid-base reaction at first instead of a SN2 reaction.

EtO- would be more basic than MeO-...the proton xfer is fast

Note: $\ce{MeO-}$ is a strong base while, $\ce{EtOH}$ is a weak base. My question: how can you perform an acid-base neutralisation reaction then? + acid-base neutralistion also produces water as a product!

In both cases the alkoxide will be in an equilibrium with the alcohol. This will be occurring on a faster timescale than the substitution but in the first case it's irrelevant as  the same species are present on either side of the equilibrium.

Methanol is more acidic than ethanol, however, so the reaction lies in favour of methoxide and ethanol.

But EtOH is in vast excess so you are going to end up with EtO-


text:
How to rationalize the colour change of the thermochromic bis(diethyl-ammonium) tetrachloridocuprate(II) complex
The colour of the compound changes from deep green/green-blue to yellow upon raising the temperature. I'm trying to rationalise this by using the d-orbital splitting, but I've got myself in a logic bind.

At low temperatures, CuCl42- is square planar or distorted square planar.
At high temperatures, CuCl42- is tetrahedral.
Δt is ~0.44 Δo, but Δsp is ~1.4Δo, therefore Δt < Δsp.
This therefore implies that at low temperatures (square planar geometry), the photon absorbed to give a colour is of higher energy, and so lower wavelength.
Green corresponds to an absorption of 650 nm (in the red), whereas yellow corresponds to an absorption of 450 nm (violet absorption).

Therefore, the colours contradict my predictions. What is the cause of this?

The colour you observe by eye is not the same as the wavelength that the species absorb:  for example Chlorophyll is green but absorbs strongly in the red , 680-700 nm, ( as well as in near uv 400nm).  (Also you mean 650 nm and 450 nm not wavenumbers)

Thanks, I've corrected the units.
Also, for the colours - green corresponds to an absorption of 650 nm i.e. absorption of red light, and yellow to 450 nm (violet light absorbed). Therefore I believe that I have already taken your comment into account?

See the last part of this answer

Please correct me if I'm wrong, but in that answer, it is stated that the colour changes to yellow due to a transition from square planar to tetrahedral, and gives a MO diagram. However, I don't believe it explains why the wavelength of light absorbed falls during the transition? This corresponds to an increase in splitting which is typically not the case from square planar to tetrahedral - transitions to the b1g dx2-y2 should require more energy than transitions to the t2 dxy, dxz, dyz?


text:
What is responsible for the odor of gasoline?
Gasoline as we know it is a blend of combustible hydrocarbons and has a characteristic odor. The natural gas we use to heat our homes and (for some of us) cook our food has its own odor, except...
Natural gas as piped to our homes consists of methane, which is odorless, plus an odorant so that we can be aware of its presence for safety reasons. So the odor we think of as "gas" isn't the fuel at all, but an odorant.
So this leads to my question: is the familiar odor we attribute to gasoline that of one or more of the components which are the fuel itself (heptane, octane, etc.) or an additive for the purpose of giving it an odor, again for safety reasons?

@Jacob That video comes as a total surprise. I have always found the odor of gasoline to be rather noxious, which is why I wondered about the presence of an odorant.

A roomful of Amoco  PHD chemists and chem engineers did not know which compounds caused the odor of  gasoline in the '70s.  They started a very expensive research project to determine the causes with the intention of improving /eliminating the odor.The Arab oil embargo came and it turned out customers did not care what gasoline smelled like, they only cared about the price.  So the project was cancelled so I don't have an answer.

@blacksmith37 so... short answer, not an odorant but one or more of the fuel components

Yes. Very likely "more".

@Jacob That video is pretty good proof that youtube is a very bad source for high quality scientific evidence.

Many of the components of gasoline (petrol) have notable odours
A simple starting point for understanding the odour of gasoline is the typical composition of the product. That composition varies a lot and always has done as different refineries use different processing techniques and different crude oil sources have different components to start with. And regulators have changed the rules over time (for example to reduce the amount of benzene, a carcinogen). All that matters, in practice, is the octane rating which is what engines care about and this is achievable in a variety of ways by varying the composition and the additives that modify the combustion characteristics.
This (old) paper gives a crude overview of a typical composition.

The typical composition of gasoline hydrocarbons (% volume) is as follows: 4-8% alkanes; 2-5% alkenes; 25-40% isoalkanes; 3-7% cycloalkanes; l-4% cycloalkenes; and 20-50% total aromatics (0.5-2.5% benzene).

Modern blends will now have far less benzene (but will still have a lot of other aromatics). Without getting too specific on the individual chemicals we can already tell some of the sources of the distinctive odour of the mixture. It is pretty meaningless to try to pin it to a single component.
Aromatics as a class are actually named because of their distinctive smell. even if we substitute other aromatics like toluene or xylene for the benzene, there will be a distinctive "aromatic" component in the product. But plenty of the other hydrocarbons also have notable odours. Cycloakenes (maybe 7% of the mixture) often have sharp smells (cycopentene is even said to have "a petrol-like odour"). Even linear and branched alkanes have some smell (though it is far less intense than the unsaturated components of gasoline).
In short, trying to pin down the typical smell of gasoline to a single compound is a little silly when there are often scores of components with well-known and strong odours and the specific composition varies a lot. aromatics are a big part with other unsaturated compounds providing some other odiferous components with a smaller but notable contribution even from unsaturated hydrocarbons.
Also, gasoline is not "odourised" as it doesn't need to be. But it sometimes has specific ingredients added to make tracing the source easier. But this is done with complex instrumentation not the human nose.

LPG has its smell because of a smelling agent called ethyl mercaptan (ethanethiol). It is added for safety purpose around the household or industrial area

Natural gas is also orderized . Methyl mercaptan may also be used.

Knowing what makes LPG smelly doesn't help understand what makes gasoline smelly.

Gasoline has it's distinctive odor because of benzene, which is found naturally in crude oil, which gasoline is derived from. Benzene is a carcinogenic, highly flammable liquid with a high octane number, but because of it's toxicity is not really used for non industrial uses.
Source: https://en.wikipedia.org/wiki/Benzene

The United States Environmental Protection Agency introduced new regulations in 2011 that lowered the benzene content in gasoline to 0.62%. (Wikipedia)

Since gasoline has hundreds of components, assigning the smell to one doesn't sound like a high quality answer.

@matt_black, if it makes you happy, benzene is responsible for the main odor of gasoline. Benzene is extremely pungent, and sticks out from among the other compounds. If you ever smell benzene (and you probably shouldn't) you'll notice its odor smells distinctly like gasoline. I'm not editing my answer to reflect this, because it should be pretty self-evident that that was my intent.

@Rafael having worked with benzene, toluene, xylene and a wide variety of other compounds, I disagree. Not least because the higher volume aromatics are also pungent.

@matt_black, I agree that other aromatics are pungent; However gasoline's distinct odor is from benzene.

@Rafael I disagree. If you want to demonstrate I'm wrong you will need to find far better sources of evidence than the one you seem attached to which shows no sources to justify the idea that benzene is uniquely pungent compared to the other aromatics present in gasoline in much larger concentrations.

Benzene has an odor threshold similar to that of toluene, but toluene is present in higher concentrations in gasoline as it's less toxic. So between the two you already expect benzene to be less important, and that's ignoring all the other aromatics that matt black notes.

A bit late though this link seems to suspect benzene too https://www.discovermagazine.com/health/why-some-people-love-the-smell-of-gasoline


text:
Eggshell experiment
I have recently done the eggshell experiment, exposing several eggshells to different beverages. What surprised me most is that, after 3 days of immersion in lemonade, the eggshell turned corrosive, this means, had lots of bumps on it. In fact, it gained weight...
It would be very gratifying if you could answer why did the eggshell in lemonade turn corrosive and why did it gain weight.
Thanks!

I believe eggshells are made of calcium carbonate, a weak base.
Upon addition of any acid to a carbonate, CO2 and water are released and a salt of the acid and base is formed.
It may be possible that the salt has deposited itself onto the the eggshell. It might be possible to dissolve that salt by adding surplus water.

Oh thank you very much James! So do you think that because of the salt the eggshell gained weight?

It could be phosphate

can you further develop your thoughts? Why do you think this?


text:
Fatty Acid synthesis - enzymes that vertebrates cannot make
In the fatty acid elongation and desaturation reactions, how are enzymes that insert double bonds after the 9th carbon different from those that insert  at the 9th or before? Vertebrates (at least mammals) don't have these enzymes.


text:
Why is bond angle of H2O greater than that of PH3?
Central atom of water has two lone pairs and that of phosphine has one lone pair, then why isn't bond angle of water smaller?

If you added another hydrogen on water, the bond angle would likely decrease...

related https://chemistry.stackexchange.com/questions/14791/hybridization-in-ph3


text:
Why will a strong base neutralize as much acid as a weak base?
I don't understand how the volume of 0.200 M NaOH needed to neutralize 250.0 mL of 0.010 M HCl (0.0125 mL) is the same volume of 0.200 M NH3 that would be needed to neutralize 250.0 mL of 0.010 M HCl.

Well, depends what you mean by neutralize. Solution of NH4Cl  is still weakly acidic.

Related: https://chemistry.stackexchange.com/questions/60407/why-will-a-strong-acid-neutralize-as-much-base-as-a-weak-acid/60409#60409

@IvanNeretin Thanks for providing that link. Can you tell me if I understand the concept correctly?: So from what I understand, since the HCl ionizes completely there is an abundance of H+ ions that can neutralize the small amount of OH- ions that are ionizing from the weak base. As a OH- ion is neutralized the equilibrium shifts to replace the neutralized OH- ion which again gets neutralized by the H+ ion present in solution. This cycle continues until the strong acid is completely neutralized by the weak base. In terms of the strong base NaOH, since it ionizes completely, there are alrea-

dy the needed amount of OH- ions present in solution to neutralize the H+ ions in solution. Is this all I need to understand for a Chemistry 12 course for this concept?

Well it's because the strength of an acid or base is a measure of it's dissociation in water. But for the the reaction of an acid with a base the equilibrium lies completely to the right as the products (salt and water) are much more stable than the acid and base.
You're trying to compare the dissociation of a species in water with the equilibrium constant of an actual reaction. I understand why, but think of them as two different situations.

Thanks for the answer. So while the weak base ionizes only slightly, because the products are more stable than the acid and base, the weak base will continue to neutralize the H+ ions and continue to ionize until the products have been formed or in other words until the HCl is neutralized completely?

Yes, the weak base will ionize only slightly. The extent of that determines the stability of the forward reaction. A slight ionization means the forward reaction is not preferred/ 'products' not as stable.

Yes, that's the idea.
For dissociation:
Yes, the weak base (ammonia) will ionize only slightly in water. Water is already a stable molecule to start, so to break the O-H bonds in water and form ions requires a good deal of energy.
For neutralization:
All reactions are a measure of enthalpy (more important) and entropy (less important). If we use this and compare the energy level of the acid and base molecules versus the energy of the salt and water formed, we see that the products are at a lower energy and thus preferred.


text:
What happens to the H electron during acid reaction?
In this generic acid-base reaction:
$\ce{HX + B <=> X- + HB+}$
It's explained that the acid HX donates a proton H+ to its conjugate base HB+.  What happens to the electron that was originally in the acid's H atom?

"What happens to the electron that was originally in the acid's H atom?" It's stuck as an extra electron with the $X$ atom, and that explains why $\ce{X}$ ends up with a uninegative charge.

The electron belonging to elemental hydrogen forms part of a covalent bond with $\ce{X}$. In this acid-base reaction the $\ce{HX}$ bond breaks heterolytically, meaning that both electrons constituting the bond go to $\ce{X-}$, hence the negative charge.

The base then donates electrons to the proton, forming the electron deficient species $\ce{HB+}$.

Thanks!!  why can't they just say that in the textbook??

I'm not sure what level you're working at or what textbook you're using but this page has some diagrams that you might find helpful. Otherwise, if you feel I've answered your question, please consider accepting my answer!


text:
Can you make deuterium-depleted water through a freezing method?
Deuterium-depleted water (DDW) is sold commercially by some companies. It is apparently manufactured by fractional distillation, which leverages differences in boiling points of heavy and light water.
My question is this: can DDW be manufactured through a freezing method? Heavy water has a freezing point of 38.8 °F and normal water of 32 °F. If a jar of water were kept in a water bath of say a constant 35 °F, wouldn't the heavy water freeze and sink to the bottom of jar, allowing the DDW to be siphoned off of the top? I am surprised there appears to be no discussion of the merits or demerits of this idea anywhere on the internet.

related https://chemistry.stackexchange.com/questions/2476/what-is-deuterium-depleted-water-actually-used-for

Here is a link to a clinical trial: https://www.ncbi.nlm.nih.gov/m/pubmed/23441611

No, slightly deuterium enriched ice will not sink. Tap water is around 150 part per million (1,000,000) or 99.985 %  H2O.  At these concentrations deuteriated water will most likely be of the form DHO vs D2O.  DHO has a freeze temp between D2O and H2O and it has a very high chance its going to freeze to other H2O  molecules forming a water crystal that has a lower density than liquid water, thus it  will float.

Not easily. For all practical purposes, $\ce{D2O}$, $\ce{DHO}$ and $\ce{H2O}$ behave the same -- including complete miscibility. Though some small fraction of the liquid that freezes first would be enriched in $\ce{D2O}$, and to a smaller extent in $\ce{DHO}$, it would still be composed largely of $\ce{H2O}$.
According to Wikipedia, the Girdler process is the most energy efficient means of separation.

To clarify, the objective is not to obtain heavy water. It is to obtain light water. So it is not an issue if the ice contains only some deuterium, but mostly light water, as the point is that the liquid water now has less deuterium than it did before.
I actually just got a reply from a DDW manufacturer who said the following:
"Deuterated ("heavy") water has a higher freezing point than ordinary water, there is indeed a fractionation slightly above zero Celsius, but it does not mean that all heavy water will be frozen.
By holding the temperature at let’s say 1°C, you can get some ice that contains D2O or DHO in a higher concentration than in liquid because the freezing point for H2O is lower.
This way you can reduce the D-concentration of the water with 8-10 ppm in one step. If you wish to achieve further decrease you have to freeze this water in further steps.
According to our knowledge, fractional distillation is the best way to produce DDW in large scale."
This same manufacturer says that home water distillers can at best reduce deuterium concentration by 1 or 2 PPM per pass, which is much less efficient than commercial evaporative methods. But an 8 to 10 drop in PPM from a freezing method (per pass) is considerable (if true), as the early clinical trials in this field suggest that even drops of say 20ppm are potentially clinically significant.

"Clinical trials"? This would seem to be pseudoscience aimed at marketing.

@DrMoishe Pippik, for most, that statement would mean 150 ppm average tap water to 130 ppm. But I think deuterium depleted water used in various studies used 120 ppm to as low as 20ppm. Anyhow the intent is to off set food preparation enrichment of deuterium. Have those clinical result been used to advance and market deterium depleted water, sure. How important are the clinical results, and their meaning, that is really another discussion.

I would say. Diffusion rates play a role. The closer the surface of ice formation is to the freezing point of deuterium water. The purer the deuterium ice will be. The process will be slowed because of this but a purer form of deuterium depleted water would result. Since I have no experience this is speculation on my part. With the formation of ice some heat would be liberated and a natural form of convection would occur.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Overshooting titration effect on K

a. Explain how overshooting the titration of the aqueous layer (ie. using too large a value for the volume of thiosulfate to react with the iodine and of triiodide ion) would affect the concentrations of iodide ion, iodine and of triiodide ion, and therefore of the calculated $K$.

My thoughts regarding this question
overshooting the titration will increase the concentrations of the iodide ion, iodine and triiodide ion
- According to me  a change in $K$ comes only when the temperature is changed.

b. Explain how using too large a value for the minimum volume of thiosulfate to react with the iodine (organic) in the mineral oil layer would affect the concentrations of iodide ion, iodine and triiodide ion, and therefore of the calculated $K$

I am unsure of the answer for this one too.
My reasoning
Using too large a value for the minimum volume of thiosulfate will increase the concentrations of the iodide ion, iodine and triiodide ion and nothing will happen to $K$ again.

Is my reasoning correct?

Your reasoning is correct that $K$ is independent of concentration. But I think you are misreading the question: it wants you to tell how you will miscalculate K with respect to the correct value if you put in the wrong concentrations in the formula, that is, if there was an overshoot in your titration.


text:
Racemization of ketones in presence of catalytic acid
Question:
Which of the following ketones racemise in aqueous solution containing acidic or basic impurities?

My answer:
A and B should racemise in aqueous solution, for the carbonyl groups in A and B contain alpha-hydrogens, leading to the possibility of enolisation, hence racemization. Compound C can't form an enol, hence shouldn't racemise.
It's very clear that compound C doesn't racemise, my confusion is about A and B.
My logic (stability of carbanion) says that A and B should both racemise, however, the answer I have is only B.
I agree that the carbanion formed by A would be less stable than that formed by B, but not so much so to ignore the possibility of enolisation and racemization.
Could someone please help? Is there anything obvious that I'm missing? or is it related to data? A good explanation would help. Thanks a lot.

To answer you need to look for a chiral centre that has a labile proton.

For recimisation of carbonyl compound alpha carbon must be chiral. Also it must contain alpha hydrogen atom.As in A alpha carbon is not chiral it is not recimise.

You are right that the racemization occurs through the enol form.
The chiral center is on the carbon attached to the chlorine atom. In B, that carbon participates in the enol form, but in A not. Therefore, in A the enol is formed but there is no racemization.


text:
Fock operator invariance under unitary transformation
I know how to show that the Coulomb operator of the Fock operator is invariant under a unitary transformation of the orbitals, as on page 121 of Szabo and Ostlund, but the indices in my proof for the exchange operator are just not working.  What might I be missing?
I have
\begin{align}
\sum_i\hat{K}_i'(1) &=\sum_i\int dr_2\chi_i'^*(2)\chi_j'(2)/r_{12} \\
 &=\sum_i\int dr_2\sum_kU_{ki}^*\chi_k^*(2)\sum_lU_{lj}\chi_l(2)/r_{12}\\
 &=\sum_{ikl}\int dr_2U_{ki}^*U_{lj}\chi_k^*(2)\chi_l(2)/r_{12}\\
 &=\sum_{i}\int dr_2\chi_i^*(2)\chi_i(2)/r_{12}\\
 &\neq\sum_{i}\int dr_2\chi_i^*(2)\chi_j(2)/r_{12}.
\end{align}

How are you contracting $U_{ki}^* U_{lj}$?

The exchange operator is not a operator by itself, it is only defined with the orbital it is working on:
$%
\newcommand{\ll}{\left\langle}\newcommand{\rr}{\right\rangle}
\newcommand{\lb}{\left|}\newcommand{\rb}{\right|}
\newcommand{\op}[1]{\mathbf{#1}}$
$$\begin{align}
  && \op{F}_i &= \op{H}^\mathrm{c} + \sum_j (\op{J}_j - \op{K}_j),\\
\text{with}&&
     \op{J}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr,\\
\text{and}&&
     \op{K}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr.
\end{align}$$
While in the Coulomb operator case the $\lb \phi_i\rr$ doesn't do anything, so you can take it through the proof as a constant:*
\begin{align}
  \sum_j\op{J}_j\lb \phi_i\rr
  &= \sum_j
     \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
  &= \sum_k \sum_l \sum_j U_{kj}^*U_{lj}
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_l(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
  &= \sum_k \sum_l \delta_{kl}
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_l(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
  &= \sum_k
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_k(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
  &= \sum_k \op{J}_k \lb \phi_i\rr
\end{align}
That is not the case in the Exchange operator, as it switches the orbitals:
\begin{align}
  \sum_j\op{K}_j\lb \phi_i\rr
  &= \sum_j
     \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr\\
  &= \sum_k \sum_l \sum_j U_{kj}^*U_{lj}
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_i(\op{x}_1) \rr \lb \phi_l(\op{x}_2) \rr\\
  &= \sum_k \sum_l \delta_{kl}
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_i(\op{x}_1) \rr \lb \phi_l(\op{x}_2) \rr\\
  &= \sum_k
     \ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
     \lb \phi_i(\op{x}_1) \rr \lb \phi_k(\op{x}_2) \rr\\
  &= \sum_k \op{K}_k \lb \phi_i\rr
\end{align}

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


text:
Is it possible to calculate the concentration percentage using chromatography paper?
I'm trying to calculate the concentrations of specific compounds using paper chromatography. Would I be able to calculate a rough estimate by comparing the size of each concentrated spot? If not is there another way to calculate this?
Thanks!

Yes it is possible. You must be thorough. To have a better resolution, place your mixture in two small points very close "oo" instead of one big "o". It is best to calibrate the TLC system using the compounds you are looking for (if available). Then measure not only the area of spots, but also their optical density.

Aweomse, thank you. I hadn't thought about using multiple spots for better consistency. Unfortunately, I don't have a spectrophotometer as these tests will simply be at my office instead of a lab. That does give me a good start and something to hopefully buy down the road. I'll most likely be using this method more and more often if all goes well.

Yes, it can be done.
I have done it on silica plates to compare the content of a natural product in different samples of plants.
It is , obviously, better to use HPLC, but in that case the sample preparation was very tedious so we decided to do it through TLC.
The content in the different samples was between 0 and 2%. Standards equivalent to a content of 0.5, 1, 1.5 and 2% we're prepared.
What you need to have succes is to be very accurate in the quantity that you apply on the plate and to have a good developer.
The quantities applied must be all the same, preferably not too much because then the spots are very big and it is more difficult to compare them.
Our product was developed with $\ce{KMnO4}$ yielding an intense yellow spot on a purple background.

Very cool, thank you for the answer. I think that the case I will be using this for will be a relatively unknown concentration of each compound. Hopefully, with several tests, I'll get an estimate anyway. Thanks again for your help!


text:
Simple/ Primitive Tetragonal Bravais Lattice
Are there any elements which exhibit the Simple(Primitive) Tetragonal Bravais Lattice?

base-centered cubic lattice can be redrawn as a primitive tetragonal lattice, therefore we do not include it in the list of Bravais lattices.    Is from here

How is there a base-centered cubic lattice?  Cubic lattices are supposed to be primitive, fcc or bcc.

Possible duplicate of Why can a face-centered cubic lattice not be redrawn as a body-centered tetragonal lattice?

Colleagues, please consider reopening this question. Just how is it "too broad"? It asks one specific thing in no uncertain terms, and has a clear answer, even if my own answer could have made it seem otherwise.

Chances are there are no such elements. This link seems to suggest so, but I wouldn't put too much trust in it. (To begin with, it claims only one crystal structure for each element, ignoring any polymorphs.) So what? This is not a fact of any consequence. It is about as (un)important as the knowledge that only one of the element names in English starts with "K", and none start with "J". Besides, both may change over time. New high-pressure crystal modifications of elements are discovered every now and then, and will be for a while, because no matter how far you reach, there is always a higher pressure.
Come to think of it, there are millions of different crystal structures out there. Elemental compounds are just a very tiny minority. Surely there are examples of all Bravais lattices (not that it matters much).
So it goes.


text:
What's the actual or official definition of activation energy?
In some sources I have seen that activation energy is the minimum energy required to cause a reaction. This implies that activation energy doesn't change with temperature (not considering very high changes).
In other sources it is the difference between the energy required to cause the reaction & the energy of the reactants. This implies that activation energy is temperature dependent since the energy of the reactants change with temperature. So which is correct?
Also another source (khan academy's guardian chemist) says that "The official definition of activation energy is a bit complicated and involves some calculus."

Activation Energy is the minimum amount of energy needed extra than the average potential energy of the reactants to cause a reaction.And it does not change with the change of temperature, and for clarification the energy of the reactant molecules don't also change with the reaction. It is just the rate which changes with the reaction according to the arrhenius equation k=Ae^{-\frac{Ea}{RT}}So,the fraction of molecules having energy higher than activation energy changes with change in temperature. I think the easier definition can be stated like this

The free energy of activation depends on temperature. Free energy has enthalpy and entropy components, and the entropy component is multiplied by the temperature.

The FREE ENERGY  ($\Delta$ G) and ACTIVATION ENERGY both are different.The FREE ENERGY  change is of total reaction but ACTIVATION ENERGY is only regarding min energy which is independent of temperature.  Don't combine them without any logic.. Think twice before adding comments and downvoting.

Actually both exist: there is the free energy of reaction ($\Delta G$) and also the free energy of activation ($\Delta G ^{\ddagger}$).

Free Energy (e.g. Gibbs) is different from Energy. The Arrhenius equation considers Energy and not Free Energy. In that regard, what you wrote is correct, Energy does not depend on temperature. However, free energy depends on temperature.


text:
how to control relative of Fe2+ and Fe3+
hi   how can I control relative of Fe2+  andFe3+  when  we  adding  raw iron to acid solution  To get the ratio 2:1

Most acids aren't strong oxidisers and really only yield $\ce{Fe^2+}$ when dissolving iron. The most common oxidising acid is nitric acid, $\ce{HNO3}$, which does fully oxidise the iron to $\ce{Fe^3+}$.
If you don't have access to nitric acid (now a restricted material in most European countries), you can dissolve the iron in hydrochloric acid, $\ce{HCl}(aq)$, then oxidise the $\ce{Fe^2+}$ subsequently with hydrogen peroxide $\ce{H2O2}(aq)$ solution. Then boil that solution for a bit to get rid of any excess peroxide.
You can also buy ferric chloride solution, $\ce{FeCl3}(aq)$, as etching solution.
Now how do you create a solution with a known $\frac{\ce{Fe^2+}}{\ce{Fe^3+}}$ ratio?
Add a calculated amount of steel wool to a known quantity of the ferric solution. The following reaction takes place:
$$\ce{Fe^3+ + Fe -> 2Fe^2+}$$
Control the concentration and quantity of ferric solution and the amount of steel wool used and you can obtain any $\frac{\ce{Fe^2+}}{\ce{Fe^3+}}$ ratio. For example, if you set aside precisely half of your ferric solution, then add an excess of steel wool to the other part, filter of the unreacted iron (leave it to react overnight, e.g.) and mix both parts then your ratio will be $\frac21$.
This how many instructables for ferrofluid are written. I've done this myself and it works well.


text:
Is H2PO4- stronger than HCN?
Wouldn't $\ce{HCN}$ be stronger, since $\ce{H2PO4-}$ predominantly acts as a base in aqueous solution?

I don't know why you're saying that it predominantly acts as a base. The pKa values clearly disagree with your premise: http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

Looks like I had made a mistake in my calculations. Thanks for the input

@Zhe your comment seems to be the answer.  Please convert.

@OscarLanzi It's kind of a link only answer, and I would probably down vote myself... :/ If you would like to elaborate into a suitable answer, I would not mind at all.

As Zhe points out, dihydrogen phosphate ion is not predominantly basic in water.
But the larger answer is that one thing can be both a stronger acid and a stronger base than another.  For example, zinc hydroxide reacts with both acids and bases in  water whereas carbon monoxide reacts with neither (at least under normal laboratory conditions).  So zinc hydroxide is both a stronger acid and a stronger base than carbon monoxide.
Similarly any basicity in the dihydrogen phosphate ion does not rule out the possibility that it could also be an acid.  But there is a limit to this, things that are both acidic and basic, called amphoteric, are generally weak in both directions.


text:
How to use the calculate velocity correlation from VanHove correlation?
I am trying to derive an expression for the velocity correlation function $C(t) = \langle \mathbf{V}(t) \mathbf{V}(0)\rangle$ from the VanHove correlation:
$$
G(\mathbf{r},t) = \frac{1}{N}\left( \sum_{i,j} \langle  \delta(\mathbf{r}- \mathbf{r}_i(t) +\mathbf{r}_j(0)   \rangle \right)
$$
However I am not sure how to get the time dependency and the $\dot{\mathbf{r}}(t)$ from the VanHove correlation. The average $\langle\rangle$ is over the ensemble.
I tired to figure out how to derive a term for velocity autocorrelation from the original paper, but without any luck.

Probably by now you have found the answer elsewhere, but for the benefit of others who may come across this question, here is my answer.
The link to the original paper seems to be broken.
However the problem you describe is fairly well known in the field of liquid state physics,
and the lecture notes that you refer to give a good introduction to the relevant correlation functions,
including the van Hove function. Certainly this is explained in
Theory of Simple Liquids by JP Hansen and IR McDonald (Academic Press, 4th edition).
That is a comprehensive graduate-level text which doesn't pull its punches:
the liquids may be simple, but the theory isn't!
There are two problems in making this connection.
Firstly, as you have noted, the van Hove function relates to the positions of the atoms,
not their velocities.
Therefore, we must anticipate that it will need to be time-differentiated (twice, in fact).
The second problem is that the velocity correlation function
involves the correlation of the velocity of a particle $i$ at time $0$
with the velocity of the same particle $i$ at time $t$.
However, the van Hove function involves cross correlations
of the positions of different atoms $i\neq j$
as well as positions of the same atom $i=j$.
These are
respectively the distinct part and the self part of $G(\mathbf{r},t)$,
as defined in the notes that you cite.
Formally, the velocity correlation function may be linked to the self part.
If you are attempting to connect it to the full $G(\mathbf{r},t)$,
you need to develop a justification for neglecting the cross correlations.
The self part of the van Hove function is defined
$$
G_s(\mathbf{r},t) = \frac{1}{N}\left\langle \sum_{i=1}^N
\delta[\mathbf{r}+\mathbf{r}_i(0)-\mathbf{r}_i(t) ]
\right\rangle
=
\left\langle
\delta[\mathbf{r}+\mathbf{r}_i(0)-\mathbf{r}_i(t) ]
\right\rangle
$$
where I have chosen one arbitrary atom $i$ since they are all equivalent.
It is convenient to Fourier transform this in space,
to give the self-part of the intermediate scattering function
(so-called because of its importance in inelastic neutron scattering experiments):
$$
F_s(\mathbf{k},t) = \int G_s(\mathbf{r},t) \exp(-i\mathbf{k}\cdot\mathbf{r})  d\mathbf{r}
=
\left\langle
\exp(i\mathbf{k}\cdot\mathbf{r}_i(t)) \, \exp(-i\mathbf{k}\cdot\mathbf{r}_i(0))
\right\rangle .
$$
For simplicity, we will define $\mathbf{k}$ to lie along the $z$-axis:
$$
F_s(k,t)
=
\left\langle
\exp(ikz_i(t)) \, \exp(-ikz_i(0))
\right\rangle .
$$
Then we differentiate with respect to time, twice.
Let $v_i$ be the $z$-component of the velocity of particle $i$.
$$
\frac{d^2}{dt^2}F_s(k,t)
= -k^2\left\langle
 v_i(t) \exp(i k z_i(t)) \, v_i(0) \exp(-i k z_i(0))
\right\rangle .
$$
A subtle trick has been used on the right hand side: the so-called dot-shifting trick.
Because of stationarity $\langle \dot{A}(t) B(0)\rangle = -\langle A(t) \dot{B}(0)\rangle$
where $A$ and $B$ are any variables, and the dot denotes time differentiation.
This is how we are able to bring down $v_i(t)$ from the first exponential,
and $v_i(0)$ from the second one.
The derivation is completed by dividing both sides by $-k^2$ and taking the limit $k\rightarrow 0$:
$$
-\lim_{k\rightarrow 0} \frac{1}{k^2}\frac{d^2}{dt^2}F_s(k,t)
= \left\langle
 v_i(t) v_i(0)
\right\rangle .
$$
This is the answer.
In some circumstances one can make the Gaussian approximation,
namely that the self part of the van Hove function is a Gaussian function of $\mathbf{r}$ at all times.
This is true at short times (free-particle limit) and at long times (diffusion limit),
but has been shown to be inaccurate (by about 10 percent) in typical liquids at intermediate times.
If we nonetheless assume that it is true,
then the intermediate scattering function is a simple function of the mean-squared displacement
$$
F_s(k,t) = \exp\left[-\frac{1}{6}k^2 \langle |\mathbf{r}_i(t)-\mathbf{r}_i(0)|^2\rangle\right]
$$
where now $k$ represents the magnitude of the $\mathbf{k}$-vector.
This can be related straightforwardly to the velocity correlation function.

Thank you! This is great. Any chance you add a comment if it's applicable in the theory of liquid crystal?

In a liquid crystal the van Hove function, and the velocity correlation function, will both lose their spherical symmetry. For example, in the nematic phase, with the director along the $z$-axis, they will both have cylindrical symmetry about that axis: motion in the $x$ and $y$ directions will be equivalent, but motion in the $z$ direction will be different. However, I don't think that this will affect this particular relationship. It should go through for each of the $x$, $y$ and $z$ components separately, pretty much unchanged.


text:
Comparison of basic strength of benzylamine and ammonia
According to me, ammonia should be the stronger base for two reasons:

$\ce{-C_6H_5}$ group shows $-I$ effect, due to which electron density on $\ce{N}$ atom of benzylamine decreases
steric hindrance in benzylamine is more than that in ammonia

But their $\mathrm{pK_b}$ values are: benzylamine: $4.70$ and ammonia : $4.75$. High $\mathrm{pK_b}$ value of ammonia shows that it is actually less basic than benzylamine, contrary to my reasoning. Why is it so?

See nitrogen inversion $\ce{->}$Wikipedia

Due to ring inversion tendency in NH3, IT IS lesser basic wrt benzylamine. The ring inversion tendency is lesser, in benzylamine

Adding methyl amine pKb = 3.4 into the discussion might help to see what a phenyl group can do. Why quibble over a 0.05 difference in pKb when there is a whole factor of ten to worry about.

Everything you said is correct, except that it is more relevant to consider the inductive effect of methyl on nitrogen: electron density is donated to nitrogen.
The inductive effect of phenyl is on the methyl, decreasing its ability to donate electron density to nitrogen, but apparently not completely.
I suppose that if you increased the negative inductive effect of the phenyl group with a couple of nitro groups, the basicity of the compound would decrease.

Benzylamine is $\ce{C6H5CH2NH2}.$ Therefore, $\ce{NH2}$ group is bonded to $\ce{C6H5-CH2}$ unit which has greater +I effect. Therefore, it increases electron density on $\ce{N}$ and makes benzylamine more basic.


text:
Can we determine the presence of carbon by simply heating or charring the substance?
Char is made of Carbon, but just because a substance char, does it immediately mean it contains carbon?
Also,
Why is charring not a good test for detecting carbon in organic compounds?
I searched that because the test can't differentiate an elemental carbon from an organic carbon. Can someone elaborate or give an example?

Other compounds also char, not just carbon. Also, some carbon compounds won't char. That is why it is is a poor test.

If you can fix CO2, than definitely your sample contains C in a form or another. For the rest see answer.  If in concomitance with CO2 you see water, this point to organics. By the way, I won't check for carbon in organic compounds. I would rather check if they are organic and flame as well as conc. Sulphuric acid do quite well to this end.

Not all carbon-containing chemical char. For example, sodium carbonate, $\ce{Na2CO3}$ just produces $\ce{CO2}$. Sodium carbonate might be viewed as inorganic, though.
An organic chemical that behaves in a similar way is hexamethylenetetramine, or hexamine, $\ce{(CH2)6N4}$. On heating, it decomposes primarily into formaldehyde, $\ce{H-CHO}$, and ammonia, $\ce{NH3}$, though there are traces of other gases such as $\ce{HCN}$ and $\ce{CO}$. For that reason, hexamine is used as a smokeless fuel (no char left behind, as the saying goes).


text:
Difference between Codeine Phosphate and Codeine Sulphate
I am concerned about Codeine medication. What are the differences between the two medications Codeine Phosphate and Codeine Sulphate and why are these two different types available as medication? Are they both equally water soluble?

Both have high water solubility of greater than 100mg/ml.
According to this ref. the sulfate has enhanced stability over the phosphate   https://doi.org/10.1002/jps.2600750917


text:
Can the choice of which reference electrode to use in a basic solution significantly affect the results?
I am performing a combined chronoamperometry/chronopotentiometry experiment in a solution of 0.1 M NaOH and with Ag/AgCl reference electrodes. I was initially told by someone that typically Ag/AgCl reference electrodes are used with most applications because they work well enough and are inexpensive. However, I haven't been able to get consistent or even normal-looking results yet. After doing some more searching, I've found some information that Ag/AgCl electrodes are not good for basic solutions. They recommend Hg/HgO electrodes instead.
I trust this is true, but could my choice of electrode be enough to cause such sporadic, inconsistent and unexpected data? Because Hg/HgO electrodes are much more expensive, I'm hesitant to just label that as the root of my problem if it wouldn't change the results a whole lot. Also, I would expect the main effects on my results to increase with time (i.e. the more tests I run, the more the electrode is affected by the solution), but my results have been consistently inconsistent from the beginning, despite trying to mimic other's parameters of similar tests.

Could you give more information about the Ag/AgCl electrode you are using? What is the frit on the tip? Porous glass frits like Vycor are known to get damaged very quickly in basic solutions.


text:
State solid synthesis of LiMn2O4 mechanism
I have difficulties finding the mechanism of the state solid synthesis of $\ce{ LiMn2O4}$ in literacy.
Starting from lithium acetate and manganese acetate, I've found the general equation easily:
$$\ce{Li(OAc) + 2 Mn(OAc)2 + 43/4 O2 -> LiMn2O4 + 10 CO2 + 15/2 H2O}$$
What I can't find is a mechanism of this reaction. I know that $\ce{Mn}$ is oxidized and $\ce{Li}$ keeps its oxidation state. Is $\ce{Mn3O4}$ a precursor in this reaction? Is $\ce{Mn2O3}$ a by-product?

$\ce{LiMn2O4}$ has been made by rotary kilning of $\ce{Li2CO3}$ (or $\ce{LiOH}$ or $\ce{LiOH.H2O}$) + $\ce{MnO2}$. I made hundreds of grams on a small scale and 1000 pounds on a larger scale about 25 years ago.
The fine powders are added to a rotary kiln (similar to making Portland cement) and heated to about 800 degrees C. The kiln rotates, stirring the mix; the lithium compound melts and reacts with the $\ce{MnO2}$ as the powdery mass gets mushed around and around. The product was used as the cathode material in a lithium battery.
The oxidation state of the manganese drops from +4 to +3.5. The significant process is that oxygen gets lost, so to balance the equation, take one lithium, two manganese and four oxygens on the product side to make $\ce{LiMn2O4}$; everything else from the raw material side goes away.
Perhaps $\ce{LiMn2O4}$ can be made in water from dissolved salts, but gaseous $\ce{O2}$ would not be a convenient reagent.


text:
Most stable chair conformation

I am asked to draw the most stable chair conformation of (1S,2R,4S)-1-chloro-2-methyl-4-(propan-2-yl)cyclohexane 1.

After doing so I am a little uncertain which of the conformations is the most stable. The most stable conformation is the one where the most bulky group is positioned equatorial. However, do I prioritize Cl over the methyl- and isopropyl-group or are the two groups more prioritized due to them being bonded to Hydrogens which take up more space than a Cl-atom. From what I have read, I would say that the chair conformation on the top is the most stable, prioritizing the methyl- and isopropyl-group. Is that correct?

Check your hand-drawn structures, I believe them to be wrong.

Neither of those structures correspond to compound 1...

Could you please explain what is wrong with them. Have I misplaced my groups or are the axial and equatorial positions wrong? @Zhe

Oh, okay I see what I did. So the isopropyl-group should be in axial-down position in the first conformation and equatorial-down position after the ring flip. Correct? But then which is the most stable? @Zhe

The biggest groups tend to be equatorial. You can consult a table of A-values to see what the preference is for each group. t-Bu is quite large. i-Pr is smaller but still quite large.

The lowest energy conformer for a cyclohexane ring with multiple substituents will be the one in which the largest group occupies an equatorial position.
Steric parameters are tabulated in many organic chemistry textbooks, the most common of which is the A-value. The A-value represents the difference in energy (ΔG) between a group being axial/equatorial on a cyclohexane ring, thus, as a group gets bigger, its A-value increases, reflecting the increasingly high energy needed to flip the ring.
Bulky substituents such as tert-butyl are often said to "lock the conformation", that is, with a tert-butyl group on a ring, the energy to flip is sufficiently high that ring is "locked" in the conformation with the tert-butyl group equatorial.
In your example, building a simple model with a kit will show you that having the iso-propyl group axial isn't terribly favourable (interestingly the iso-propyl group has a similar A-value to ethyl, due to the ability of the group to rotate, avoiding having a methyl group pointing into the ring).
I ran a few simple calculations (molecular mechanics followed by DFT optimisation - gas phase) on the axial and equatorial conformations, which showed that the two conformations are separated by around 3.4 kcal/mol.  The minimum energy conformation of your example is shown below:


text:
How to use a modified Peng-Robinson equation to calculate a density?
I'm trying to use the Peng-Robinson-Stryjek-Vera 2 (PRSV2) Equation of State to calculate the density of a compound, in this example: methanol. At ambient temperature and pressure, shouldn't I be able to solve the cubic equation of state for the molar volume and calculate a density corresponding reasonably well to the density of the liquid given in the literature at those conditions of temperature and pressure? The values are considerably off, at least farther off than I would expect from such a well tested equation of state. For instance, for methanol, I obtain a density of 0.672 g/mL rather than the literature value of 0.791 g/mL. For acetonitrile, it's even worse: 0.480 g/mL rather than 0.786 g/mL. Does anyone have experience with such cubic equations of state? Are they simply not that accurate? Am I doing something wrong? I've checked my implementation numerous times for typos, but cannot find any.
This implementation in MATLAB comes from the following two papers:
Stryjek, R. and J. H. Vera, (1986). "PRSV: An Improved Peng-Robinson Equation of State for Pure Compounds and Mixtures", Can. J. Chem. Eng., 64, 323-333.
Stryjek, R. and J. H. Vera, (1986). "PRSV2: A Cubic Equation of State for Accurate Vapor-Liquid Equilibria Calculations", Can. J. Chem. Eng., 64, 820-826.
T = 298; % Temperature, in K
P = 1; % Pressure, in atm

% Values for methanol
M = 32.04; % Molar mass, in g mol^-1
T_c = 512.58; % Critical temperature, in K
P_c = 8095.79; % Critical pressure, in kPa
omega = 0.56533; % Acentric factor, from PRSV2
kappa1 = -0.16816; % From PRSV
kappa2 = -1.3400; % From PRSV2
kappa3 = 0.588; % From PRSV2

R = 8.3144598; % Ideal gas constant, in Pa m^3 mol^-1 K^-1
R = R*1000; % Ideal gas constant, in kPa cm^3 mol^-1 K^-1
T_R = T/T_c; % Reduced Temperature

kappa0 = 0.378893 + 1.4897153*omega - 0.17131848*omega^2 + ...
    0.0196544*omega^3;

kappa = kappa0 + (kappa1 + kappa2*(kappa3 - T_R)*(1 - (T_R)^0.5))*...
    (1 + (T_R)^0.5)*(0.7 - T_R);

alpha = (1 + kappa*(1 - (T_R)^0.5))^2;

a = (0.457235 * R^2 * (T_c)^2 / P_c)*alpha;
b = 0.077796*R*(T_c)/(P_c);

P = P*101.325; % Pressure, in kPa

% Using the PRSV EOS cubic expansion for molar volume
C3 = 1;
C2 = b - R*T/P;
C1 = a/P - 2*R*T*b/P - 3*b^2;
C0 = (b^2 + R*T*b/P - a/P)*b;

cubic  = [C3 C2 C1 C0];
v = roots(cubic); % Molar volumes, in cm^3 mol^-1

rho1 = M/v(1) % Density, in g cm^-3
rho2 = M/v(2) % Density, in g cm^-3
rho3 = M/v(3) % Density, in g cm^-3

I am not sure he made a mistake. The Peng Robinson equation is widely used for gas behavior, but I wouldn't guess it's just as suitable to predict liquid density.  Plus his values are not completely off, they are quite good estimates depending on what you intend to do.

According to this source:

The popular EoSs such as Soave-Redlich-Kwong (SRK) [1] and Peng-Robinson (PR) [2] predict liquid density with an average absolute error of about 8%, much higher than several good density correlations. This large magnitude of error is not acceptable by industry; therefore, they are not used for this purpose.

Which is what I first thought when I thought of your question. Liquid density is a hard issue, specially in compounds with hydrogen bonding like ethanol. I wouldn't predict a gas-oriented (as in its primary use) EoS to perform just as well for liquids.
The article further suggests

In order to overcome this deficiency, volume translated methods have been developed.

So that's what I'd recommend you to look into.


text:
Reaction of Halogens with water
The reaction of fluorine with water is very different from that of chlorine with water.
$$\ce{2F2 + 2(H2O)->O2 + 2HF}\tag1$$
$$\ce{Cl2 + H2O ->HCl + HOCl}\tag2$$
I want to understand two main things:

Why hypochlorous acid is formed in second reaction why doesn't it proceed like the first one?
Has it got to anything with reduction potentials? chlorine and oxygen are very close and is that reason?

I found  this where a related question was discussed here I expect the answer to describe what happens in reaction of any halogen with water and how does its reduction potential play a role in it also are $\ce{HOI}$ and $\ce{HOBr}$ also formed?

Why hypochlorous acid is formed in second reaction why doesn't it proceed like the first one?
Has it got to anything with reduction potentials? Cl and O are very close and is that reason?

Your intuition in 2. is a good guess and could answer the question mathematically but would be an indirect explanation. If you look closer at reaction 1 you will see that oxygen is oxidized by fluoride where as in reaction 2 oxygen is not oxidized. From a re-dox perspective in the first reaction your equation 1 looks something like:
$$\ce{H2O -> 2H+ + 1/2O2 + 2e-}\tag A\\
\text{and}$$
$$\ce{F2 + 2e- -> 2F-}\tag B$$
Here the fluorine molecule takes 2 electrons from the oxygen atom in an irreversible process. But for equation 2 the redox reaction is more like:
$$\ce{Cl2 <=> Cl- + Cl+}\tag C$$
Here the chlorine is oxidize by the other chloride molecule and reacts with the water to form $\ce{HCl}$ and $\ce{HOCl}$
$$\ce{Cl+ + OH- <=> HClO} \qquad\text{and}\qquad \ce{Cl- + H+ <=> HCl}\tag D$$

side note: $\ce{HOCl}$ would more properly be presented as $\ce{HClO}$ since chlorine has a $+1$ formal charge but not $\ce{ClHO}$ since $\ce{ClO-}$ is the conjugate base.

Note that the reactions for clorine are revresable now whereas fluorine was not. Combining equations C and D give your original equation 2
$$\ce{Cl+ + Cl- + OH- +H+ \implies Cl2 + H2O <=> HCl + HOCl}\tag2$$
Now for the why. The answer comes from the electronegativities of the elements. $(X_\ce{F} = 4.0 > X_\ce{O} = 3.5 > X_\ce{Cl} = 3.2)$ as you can see fluorine's electronegativity is greater than oxygen's which allows flourine to oxidize oxygen and oxygen's electronegativity is greater than chlorine's which mean chlorine cannot oxidize it. Further Hydrogen is already oxidized and so the only thin for chlorine to oxidize is itself.
Since the other halogens also cannot oxidize oxygen, this equation can be generalized for all of the halogens that are not fluorine as well:
$$\ce{X2 + H2O -> HX^{(I-)} + HX^{(I+)}O}\tag{X = Cl, Br, I}$$

In addition to A.K's answer there is a more subtle effect, having to do with the fact that $\ce{HF}$ unlike the other hydrogen halides is only a weak acid.  Thereby, hydrogen can readily bond to fluorine forming $\ce{HF}$ molecules, facilitating displacement of oxygen as the element.  We cannot form $\ce{HCl}$ and heavier congener molecules so easily in water, so we don't break all the original bonds between hydrogen and oxygen and thus we get products with hydroxyl groups like $\ce{HOCl}$.


text:
Why were elements discovered "out of order" after 1950?
I'm referring to the fact that there is one element that corresponds to one proton in an atom, a second that corresponds to two protons, up to, I believe, an element with 115 protons. These elements are listed in the periodic table discovered by Mendeleev in the 19th century, or modern expansions thereof.
By "out of order," I'm referring to a situation where elements were not discovered consecutively by number of atoms, e.g. where an element with 103 protons was discovered before an element with 102 protons, and again with 109 protons before 108, according to this table. I use the qualification "after 1950", that is, the "modern era," because this would have given scientists time to sort out Mendeleev's Table after it gave us a road map to atoms. Perhaps more to the the point, the scientific process is a lot more organized than it was in Mendeleyeev's time.
What were the circumstances of these out-of-order discoveries? Isn't it exponentially harder to discover elements with 103 protons than 102 or 101, etc.? Or wouldn't scientists concentrate on creating the atom with the "next" number of protons, e.g. 116 using advanced technology nowadays? Because this is a process that is different from "randomly" coming across atoms commonly found in nature.

You can find a table of the elements which you can sort by year at the Wikipedia article Timeline of chemical element discoveries

@MaxW: I rephrased (and expanded) the question. Thanks for your help.

Elements weren't discovered in order before 1950. Why would you expect that they would be afterwards?

Could you reword your title? The word “been” is confusing me.

@Zhe Because, in the olden days, new elements were "discovered" in the sense of being found, in whatever order they were found in. But, today, new elements aren't discovered but created, which implies that the scientist chooses which element they're going to create.

Why would scientists concentrate on finding the next one? If everybody tries to find the same thing, there'll be a lot of wasted effort.

@DavidRicherby: Key phrase from the accepted answer: The big issue is that nuclear stability isn't linear with atomic number." That was the answer I was looking for, except that my command of chemistry wasn't good enough to frame the question well.

The major reason is arguably nothing to do with chemistry (unless you count nuclear physics as nuclear chemistry).
The big issue is that nuclear stability isn't linear with atomic number. Some configurations of protons and neutrons are more stable than others (for reasons that would require quite a lot of nuclear physics to explain). Nuclei with even numbers of both protons and neutrons tend to be more stable, for example. There are particular magic numbers of both that impart some extra stability. And Nuclei containing odd numbers of protons and even numbers of neutrons are less stable than nuclei containing even numbers of protons and odd numbers of neutrons.
The basic takeaway of these observations is that the stability doesn't change in a simple fashion with atomic number (the number of protons). Nuclei with odd numbers tend to be less stable, for example. This means that the easiest ones to make are sometimes not the next element but the next but one element. Or something beyond that. There is even a theory that says there is an island of stability where we will discover some unusually stable heavy elements somewhere slightly beyond the list we currently have.
So there is not reason to expect that we should discover new elements in the order of atomic number.
More detail here.

It's also a lot easier to bombard nuclei with alpha particles than with just protons.

@OrangeDog But simple bombardment often doesn't work. Most larger transuranic were created by using much larger "bullets" than alpha particles.

I know, but reading your answer people would assume the early discoveries went in 2s because they were more stable, rather than because people were using alpha particles to create them.

@OrangeDog Could it be a bit of both? Alpha particles were convenient things to bombard with, and a jump of two protons was more likely to be stable-enough to spot.

Accepted for "The big issue is that nuclear stability isn't linear with atomic number."

@OrangeDog Early discoveries used both protons and alpha particles. And the dominant reason seems to be stability not the choice of bullet. And the bullet doesn't determine an even or odd end point atomic number since you can start with even or odd atomic numbers.

@TomAu except "The big issue is that nuclear stability isn't linear with atomic number."  isn't the right answer. The actual reason concerns how science is conducted and how new elements are synthesized. It's levineds' answer that is correct.

@OrangeDog Nope. Too simplistic. The product of the reactions, for example, are not always obtained by adding the protons in the bullet to the protons in the target. Pu (94 protons) was first made by bombarding U (92 protons) with deuterons (1 proton) or neutrons (no protons).

@OrangeDog Even if all heavy elements were made by alpha bombardment (which as noted elsewhere is incorrect), if nothing else was involved you'd just have an 'odd target' and an 'even target' and could alternate between them going up 1 at a time.

The synthetic trans-uranic elements (the "modern era" elements as you call them) are synthesized by bombarding a certain isotope of one element with a certain isotope of another element with a lot of energy in order to get nuclear fusion. The reason they are "out of order" is that the building blocks of these elements have to be very specific isotopes (extra rich in neutrons for example, somewhat abundant for another). This limits the possible sources considerably and so the heavy elements synthesized (and hence discovered) earlier were those where there was a convenient pair of smaller elements to combine to form the larger ones. You don't just bombard each successive element with protons to make the next one.
A lot of the early discoveries were made by bombarding sources with alpha particles, so there are several discoveries that are increases by 2 protons (e.g. Cm->Cf, Am->Bk). Indeed, the table you linked gives more examples of the pairs of elements used to make the heavy actinides and beyond.

Or bombard with neutrons (e.g. in a nuclear reactor) and wait for beta-decay to turn them into protons.


text:
Vapour-Liquid Equilibrium
I have been studying flash distillation, and I am reading the VLE tables my textbook has provided.
I am looking at these tables and I am not understanding the data, I think.
Why is the vapour mole percent ($y_{benzene}$) and liquid mole percent of benzene ($x_{benzene}$) zero at 110.62 $^0 C$? Does that mean benzene just disappear from the flash distiller at that temperature?

Absolutely not. It means that for the mixture to be boiling at 110°C, it must not contain any benzene.


text:
Why does helium have such a low van der Waals coefficient but not as low a Lennard-Jones coefficient?
Helium has the lowest polarizability of any atom, and therefore ought to have the smallest London dispersion force. Indeed, if you look at the van der Waals constant of helium, you find that it has the lowest value of $a$ by a lot. $a$ roughly corresponds to the amount attraction between particles.
The long-distance force between neutral particles can be roughly modelled by the Lennard-Jones potential. Even though helium should have the weakest $r^{-6}$ term of any atom, if you look up the experimental values of the Lennard-Jones coefficients, you find that they're the same for Helium as they are for the rest of the noble gases.
What gives? Why is its Lennard-Jones potential roughly the same when its London dispersion force is so weak? (Perhaps my source is just wrong?)

Can you post your source? When I look up the values for $\sigma$ for the LJ potential, I find helium has the smallest of everything in the table, and is always smaller than the other noble gases. Keep in mind that that a small change in $\sigma$ can have a large change in the attraction because the attraction goes as $\sigma^6$ and repulsion $\sigma^{12}$.

Even your linked source shows that helium has a smaller value of $\sigma$.

Lennard-Jones Parameters
To quote the original source of those Lennard-Jones parameter values (Gordon and Kim),

For all the systems involving atoms larger than helium, the predictions appear quite reliable... Our approach thus provides the first successful prediction of the intermolecular potentials for the rare gases (except helium)

So I would not put much faith in the He-He LJ parameters in the table you linked. The analysis was good in 1972, but even still wasn't great for helium! The authors go on to state that

Polarizing (induction) forces are not included

which, arguably, is the strongest evidence not to trust the LJ parameters too much. The Lennard-Jones (6-12) potential's strongest theoretical justification is probably that dispersive forces can be shown to approximately follow $r^{-6}$ dependence. If dispersion's not even included, then that's a big red flag since the $r^{-12}$ part has no theoretical justification. It seems they just tried to fit the LJ potential to their calculations, probably for reasons of comparison with previous results. The experimental value that they compare their calculations to is $16.5 \times 10^{-16}$ ergs, which would be 1.0 in the table instead of 3.9, and thus would continue the expected trend.
van der Waals parameters
I will just be brief and say that the quantum gases, including helium, are not amenable to the use of the classical van der Waals equation. Helium's critical temperature is 5.2 kelvin, and there is a bunch of quantum stuff going on in that regime. You just can't compare the quantum gas parameters with those of fluids that behave classically; it's quantum apples and classical oranges.


text:
Why does the isoelectric range decrease as an Alanine chain grows longer?
The pK1 and 2 values for alanine are 2.3 and 9.7. In the dipeptide Ala-Ala, they are 3.1 and 8.3 and in tri-peptide Ala-Ala-Ala, they are 3.3 and 8.0. Why does this trend in pKa values exist? Why does the isoelectric range decrease?
The reason I can think of is the extra Alanine dumps electrons towards the C terminal and so makes the O-H bond less polar and thus less acidic while it also pulls more electrons from the N terminus to make that more polar and thus more acidic.
Thanks!

I would not say that the extra alanine unit dumps or pulls electrons, rather it "insulates" the inductive effect of charged functional groups that we can see in the monomer.
The postitively charged alanine monomer has the structure
$\ce{NH_3^+}-\ce{C(H)(CH_3)}-\ce{COOH}$
The dimer has this:
$\ce{NH_3^+}-\ce{C(H)(CH_3)}-\ce{C(O)NH}-\ce{C(H)(CH_3)}-\ce{C(H)(CH_3)}-\ce{COOH}$
When the carboxyl group ($\ce{COOH}$) is deprotonated the monomer gets some stabilization of the negative charge from the positively charged, electron withdrawing function $\ce{NH_3^+}$ two positions away on the chain.  But the dimer has a neutral, less electron drawing peptide ($\ce{C(O)NH}$) function at the same position, making the deprotonation less favorable.
After the first deprotonation the molecules, now zwitterionic, are as follows:
$\ce{NH_3^+}-\ce{C(H)(CH_3)}-\ce{COO^-}$
$\ce{NH_3^+}-\ce{C(H)(CH_3)}-\ce{C(O)NH}-\ce{C(H)(CH_3)}-\ce{C(H)(CH_3)}-\ce{COO^-}$
So we move on to the second deprotonation which neutralizes the ammonium function.  The double deprotonated monomer then has an amino group, $\ce{NH_2}$, two places from the carboxylate anion.  The diner has a peptide function, $\ce{C(O)NH}$, at that location and this is more electron withdrawing, thus more stabilizing to the negative charge, than the monomer's amino group.  So in this second stage the dimeric anion is more stable and forms under less strongly basic conditions.
Thus we see inductive effects working in opposite ways in the two amino acid functions.  It's an example of how the two ends have opposing characteristics.

It is because of the inter- and intramolecular associations that occur.
In Ala, the first loss of a proton forms a two point intermolecular association which is very favourable.
Since the equilibrium tends to the right, more $\ce{H^+}$ are liberated and the p$K_a$ is lower.

For the second step, this stable dimeric association must be destroyed, so now the equilibrium tends to the left and the p$K_a$ is higher then it would be without this association.

In the case of Ala-Ala-Ala, this type of dimeric association can not occur because the distance is too long.
There is intermolecular association but with different molecules, not in pairs.
Here the first p$K_a$ is higher than in the case of Ala and the second lower because the equilibrium is not forced so much to the associated forms.
Finally, in the case of Ala-Ala, besides the intermolecular association between different molecules like in Ala-Ala-Ala,
there is also an intramolecular association.
This one is less stable than the association in pairs of the individual amino acids, but more than the random intermolecular ones.
Therefore, the two p$K_a$  are in between of the ones for Ala and Ala-Ala-Ala.


text:
Cubic zirconia additives and toxicity
What are cubic zirconia (CZ) gemstones usually colored with?
According to wikipedia the following dopants are being used:

Cerium
Chromium
Neodymium
Erbium
Titanium.

Chromium is quite colorful and can be used for green CZ. However I'm worried about Cr(VI) which is carcinogenic being used, or other equally toxic substances. (manufacturing can take place in countries with less strict regulations than EU and US).

On the other hand, when it comes to pottery used for cooking food if it is properly fired practically no lead migrates to food.
So, if pottery used for boiling acidic food is safe when "properly fired", I'm assuming that the CZ (which only comes in contact with the skin) should be safe regardless of the dopants used since it can reach temperatures of 1400 °C.
Questions:

Is my above assumption correct?
Is there any research on CZ toxicity or allergies.
Could any other toxic dopants be used?

It seems very unlikely that anyone has done research on this as 1) dopants are used in very small concentrations 2) the rate of migration is likely to be so low it is unmeasurable at room temperature.