text:
What is reduced and oxidized when potassium bromate reacts to oxygen and postassium bromide?
I'm trying to help lead my daughter to a solution and my high school chemistry is almost 30 years ago.

For the following reaction, indicate the species being oxidezed and
  reduced and show the oxidation numbers above these symbols. Lastly,
  indicate the number of electrons that have been lost or gained.
$$\ce{2 KBrO3 -> 2 KBr + 3 O2}$$

What is being oxidized, and what is being reduced? My daughter tells me that since the charge on the two components doesn't change there isn't reduction or oxidation. You could almost say the $\ce{KBr}$ is being 'de-oxidized' because it's losing oxygen, but that would mean the oxygen is being oxidized, which in this case seems nonsensical.
If someone can explain to me which it is and why, so I can walk her through it, or direct me to a place where this is done, I'd appreciate it. This is beyond my knowledge.

One would hope that a concept that fundamental is explained at a high school level in the high school level textbook, which the school is using to teach a high school chemistry class.

She's taking the course online, and has talked about some of the oversights and misplaced information she's encountered. I have the sense this course was not constructed well.

Defining oxidation number (O.N.), Oxidation and Reduction

Oxidation number, also called oxidation state, is an indicator of the degree of oxidation (loss of electrons) of an atom in a chemical compound
Oxidation is the loss of electrons or an increase in oxidation number by a molecule, atom, or ion.
Reduction is the gain of electrons or a decrease in oxidation number by a molecule, atom, or ion.

To calculate the oxidation number of $\ce{Br}$ in $\ce{KBrO3}$:
$\ce{O}$ is the most electronegative element in the compound, assume it gets all the electrons and it therefore has the oxidation number $-2$. $\ce{K}$ has $+1$, so $\ce{Br}$ gets $+5$ according to the following calculation:
\begin{align}
1 \times (\text{O.N. of K}) + 1 \times (\text{O.N. of Br})
+ 3 \times{(\text{O.N. of O})} &= 0\\
1 \times (+1) + 1 \times (\text{O.N. of Br})
+ 3 \times (-2) &= 0\\
\text{O.N. of Br} &= (+5)
\end{align}
To calculate the oxidation number of $\ce{Br}$ in $\ce{KBr}$:
\begin{align}
1 \times (\text{O.N. of K}) + 1 \times (\text{O.N. of Br}) &= 0 \\
1 \times (+1) + 1 \times(\text{O.N. of Br}) &= 0\\
\text{O.N. of Br} &= (-1)
\end{align}
Look at $\ce{O2}$, it gets $0$.

In total that will give you:
$$
\ce{
2\overset{\color{black}{(+1)}}{K}
\overset{\color{red}{(+5)}}{Br}
\overset{\color{green}{(-2)}}{O_3}
->
2\overset{\color{black}{(+1)}}{K}
\overset{\color{red}{(-1)}}{Br}
+3 \overset{\color{green}{(0)}}{O2}}$$
When you look at the reaction: it is an auto-redox reaction, in which

$\ce{Br}$ went from $+5$ to $-1$, it gains electrons, so it is reduced.
$\ce{O}$ went from $-2$ to $0$, it lost electrons, so it is oxidized.

Ah ha. Of course. The reaction does not necessarily involve changing the charge on each molecule, but can rearrange how they are 'on the atoms' of each molecule. Thank you for explaining this so simply, my high school chemistry was long, long ago.

This is not a reaction where ions exchange electrons, but the formal oxidation numbers do change definitively.
Electronegativity of O is 3.5, Br 2.7. That's a rather large difference, so in $\ce{KBrO3}$, that gives O its preferred formal oxidation state -2, K has +1, so Br gets the difference +5. I think you can take it from here.
And yes, the oxygen gets "oxidised", nothing wrong with that! Look at the reverse reaction: Oxygen oxidises KBr to the Bromate, so it obviously itself gets reduced.


text:
How does aniline react with HNO3 in the absence of sulfuric acid?
How does aniline react with $\ce{HNO3}$ in the absence of sulfuric acid? I read that nitration occurs at slow rate, but it didn't explain how nitronium ion is formed.

Autoprotolysis of HNO3?

forms anilinium nitrate

https://chemistry.stackexchange.com/questions/31046/does-anisole-react-with-concentrated-nitric-or-sulfuric-acid

https://chemistry.stackexchange.com/questions/9083/does-the-outcome-of-aromatic-nitrations-depend-on-the-concentrations-of-nitric-a

According to my chemistry teacher, in dilute aqueous nitric acid, the mononitrated product would be obtained while in concentrated acid, the trinitrated product would be obtained.

In the absence of sulfuric acid, aniline does react with nitric acid to form anilinium nitrate. It is observed as an intermediate product during the production of diazobenzene nitrate (benzenediazonium nitrate). But the species is too unstable for mass production and cannot be stored for long hours. Here is the description:

20 g of aniline are placed in a beaker, well cooled, and “boiled-out”
  nitric acid, previously diluted with half its volume of water,
  carefully added till the mixture sets to a thick crystalline paste –
  aniline nitrate.[...]

Caution: The synthesis is not to be carried out due to high explosive nature of diazobenzene nitrate. Explode in dry state.

source: Stack Exchange

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

sort: votes


text:
GAMESS problem set question
I am new to GAMESS and I would like to ask for help. I have a problem set in our class but I am having difficulty on how to proceed with GAMESS. The problem is from Shriver and Atkins,
Perform energy calculations on lithium, sodium, potassium, rubidium and cesium cations using the 3-21G model and display electron density surfaces side by side on screen. Make a copy of the set, and display each cation as a space-filling model underneath the density surface. What is the radius of each of the five cations (you can obtain this from either the surface area or volume)? How do the radii compare with the van der Waals radii? Do the sizes of the cations increase or decrease with increasing atomic number? Rationalize your result.
Can you please help me on how to provide an input file for this? I would really appreaciate it. thanks

Welcome to chem SE! If you haven't yet, take the [tour] and visit the [help]. At the moment, the question is likely to be closed as homework, since it doesn't explain your attempts to solve the problem. What specifically do you have trouble with in terms of making an input for these jobs that isn't in the manual or is leading to errors when you try to run it? For the other part of the problem, what visualization software are you using to view the density surface? @RonaldValdez

First of all, there are three versions of Gammess: UK, US and Firefly. I will assume that you are using Gammess US.
Take a look at the examples at the manual.
Geometry
You have to put the coordinates of the atom. It is very simple in your case because you have one atom, that you can locate at $(0,0,0)$.
Model
You also have to set your basis.
Usually I explicitly write it in the input after taking the basis from Basis Set Exchange but since you have such a simple basis you can use the ones bundled with Gammess US. Furthermore, the basis is not a complete description of the model you are going to use. That should be explained in the book so I suggest you to read it carefully. You make an ansatz about your wavefunction, may it be monodeterminantal, expansion of deteriminants or other you will be using a model like Hartree-Fock, Density Functional Theory, Configuration Interaction, Coupled Cluster, explicit methods, ...
I don´t know which you are going to use but probably it is Hartree-Fock, since it is only to get a grasp. Caution: Li, Na and K all have odd number of electrons so they are not closed shell. Instead their ground state is a doublet in all of them. Knowing this and that you are only interested in the density and not in the spin-density you can do restricted Hartree-Fock (RHF) while notifying that it has one unpaired electron.
Symmetry
Isolated atoms have spherical symmetry but the closest symmetry you can get in Gamess is octahedrical ($O_h$) so use it.
Density
In addition you have to let the program know that you want to have the information of the wavefunction printed at the end of the program execution. Then you have to set the AIMPAC=true variable. Then you can use, lets say, Multiwfn to plot the density isosurface.
Also you will find the following tutorial useful.
The visualization program Avogadro has an input generator but you still have to know what you are doing, it is only a tool to help you to create the file

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


text:
What kind of drugs can be absorbed through skin?
Some drugs such as nicotine can be administered through skin. I thought the layers of skin are designed to prevent in-flow of any chemical/germs. Not all drugs get absorbed in this fashion. So do drugs like nicotine have any specialities(such as a special chemical group) that fool the barriers of skin? A link to an article/site would be very helpful, or even a book recommendation.

The skin is able to prevent some things from passing in and out of a person but it is not a perfect barrier.
The lipophilic things like nicotine, hydrogen flouride, VX, hexane and iodine vapour can pass through undamaged skin. These things are able to dissolve in the fatty parts of skin. Then they diffuse in through the skin into the person. Thus they enter the person.
Some water soluble ionic things can enter / leave a person through the skin, this can occur through the sweat pores.
For example cesium in humans does come out in sweat, in a 1980s case in Brazil where a lot of people were internally contaminated with cesium-137. It was found that cesium kept appearing on the surface of the skin of the contaminated people.


text:
Dependency of rate of diffusion on mean free path and velocity of molecules?
What is the dependency of rate of diffusion ( Graham's law) on mean free path, molecular size , average velocity, collision frequency and pressure? I have searched everywhere on internet but just found that rate of diffusion is inversely proportional to square root of density or molecular mass.

A proper textbook on physical chemistry should explain this in detail. You might have better luck searching in one than on the internet

My reference is Concepts in Thermal Physics, Blundell and Blundell, Ch. 7, 2nd ed. Everything here is from there.
Consider a gas trapped in a rigid box. We begin by evaluating the molecular flux incident on one part of one wall of the box. Let there be $n$ particles per unit volume, so $n$ is the particle density of the gas in the box. Let $A$ be the area of a small patch of the wall, and let $\theta$ be the angle between the particle's direction of approach and the normal to the wall. Let $f(v)$ denote the speed distribution function of the particles, which we assume to be a Maxwell-Boltzmann distribution.
Consider all the molecules with speed between $v$ and $v+\text{d}v$ approaching the wall at an angle between $\theta$ and $\theta+\text{d}\theta$, which we write in shorthand as ($v,\theta$). There are $n$ particles per unit volume, and a fraction $f(v)\,\text{d}v$ of them are moving within the desired speed range. A further fraction $\sin\theta\,\text{d}\theta/2$ of them are moving in the desired direction, so the particle density for ($v$, $\theta$) is $$nf(v)\,\text{d}v\sin\theta\,\text{d}\theta/2.$$ (To derive the fraction $\sin\theta\,\text{d}\theta/2$, we consider a unit sphere with surface area $4\pi$. The surface area of the differential annulus is $2\pi\sin\theta$ (circumference) $\times\;\text{d}\theta$ (differential width), so the fraction of surface area occupied by the annulus is as stated.)
Now consider how many particles traveling at ($v,\theta$) hit the patch of wall in a time $\text{d}t$. These particles must be contained within a volume $$A\,\times\,v\,\text{d}t\cos\theta$$ (distance particle travels normal to the wall in a time $\text{d}t$), for all particles traveling at ($v,\theta$) must impinge on the wall within $\text{d}t$. Actually this area is slightly enlarged by a width $v\sin\theta\,\text{d}t$ uniformly, but this contribution to the area is of order $\text{d}t^2$ and vanishes in the small $\text{d}t$ limit.
All in all, then, the number of molecular collisions in $\text{d}t$ on a patch of wall of area $A$ at ($v,\theta$) is $$nf(v)\,\text{d}v\sin\theta\,\text{d}\theta/2\,\times\,Av\,\text{d}t\cos\theta.$$
The flux $\Phi$ is the rate of molecular collisions per unit area, so we divide by $A$ and $\text{d}t$, and integrate over all $v$ and $\theta$. This yields $$\Phi = \int_0^\infty\text{d}v\int_0^{\pi/2}\text{d}\theta\,nvf(v)\sin\theta/2\cos\theta = \frac{1}{4}n\langle v\rangle = \frac{p}{\sqrt{2\pi mk_BT}},$$ assuming an ideal gas and given the average speed of gases within the Maxwell-Boltzmann distribution.
By definition of the molecular flux, the rate of effusion is simply $\Phi A'$ for $A'$ the cross-sectional area of the hole out of which gas effuses, and the resulting dependencies can be read off from the equation for $\Phi$.

Interesting notes.

Due to the extra factor of $v$ in the integral for the molecular flux, the distribution of speeds of the effusing particles is $v^3e^{-\beta\epsilon}$ , rather than the standard $v^2e^{-\beta\epsilon}$. This extra factor of $v$ comes about because speedier molecules hit the walls of the box faster than do slower ones.
The hole through which particles effuse must be small, and the length scale that governs smallness is the mean free path. Make the hole too large, and the box eventually reaches equilibrium with the outside world. Consider a system of two boxes placed side by side, and a small hole between the neighboring walls of the boxes. For holes small enough that effusive behavior is observed, at equilibrium the molecular fluxes must balance and hence $$\frac{p_1}{\sqrt{T_1}} = \frac{p_2}{\sqrt{T_2}}.$$ On the other hand, if the holes are too large, then we simply have regular thermodynamic equilibrium, with $$\frac{p_1}{T_1} = \frac{p_2}{T_2}.$$


text:
Salt used in Perkin Reaction
The Perkin reaction gives an α,β-unsaturated aromatic acid by the aldol condensation of an aromatic aldehyde and an acid anhydride, in the presence of an alkali salt of the acid. Here is an a diagram showing what the reaction does

The anion of the salt acts as a base for abstracting the β-hydrogen in the acid anhydride. Later it attacks the carbonyl carbon of an intermediate but however leaves in the same step. Since the base is not ultimately involved in the product, is there any kinetic or thermodynamic reason as to why the salt of the acid of the acid anhydride is preferred over any other base?

To prevent side reactions.

Can you give an example? I think since the base is not involved in the product, different bases can only give different side product but the major product will be the same? Is this correct?

I was thinking about nucleophilic attack between the base and acid anhydride. Choosing the base to be the conjugate base corresponding to the acid anhydride negates this issue because then the acid anhydride reforms upon reaction. But this paper suggests other bases may be used, so I will defer to actual organic chemists.

The base, as you pointed out is nearly always the salt of the acid corresponding to the annhydride.
This is because the base which differs in structure will react with the given carbonyl compound and yield some other compound rather that abstracting a proton.
Eg.
Let's say you have tertbutyl salt there instead of what we are having right now.The tert butyl could have directly attacked on the Carbon having the double bonded oxygen as it has a partial positive charge due to polarization.Which will result in to eliminating the other symmetric half of the annhydride and replacing it by tertbutyl.This changed your basic reacting molecule.And hence the
 product you obtained is definitely not the right one. Whereas using the salt of the acid that formed the annhydride will not change the annhydride even if it reacts with the annhydride the way I stated above.
Using the salt of the acid they produced the annhydride does nothing but abstracts the proton or simply regenerate the original carbonyl compound/annhydride if it reacts rather than abstracting a proton.


text:
Fluorination of Carbonyl Compounds using Molybdenum Hexafluoride
What is the mechanism of halogenation of carbonyl compounds using molybdenum hexafluoride? How does it work?
Here's a reaction I came across:

Can we use it for all carbonyl compounds? Are there any limitations to the method?

What makes you think this particular fluorinating agent was so thoroughly researched?

Never seen this used, the standard reagents are DAST or Deoxfluor

Molybdenum(VI) fluoride can be used as a fluorinating agent though not as popular as the selective DAST and Deoxyfluor as @Waylander noted. Molybdenum(VI) fluoride fall under the category of electrophilic fluorinating agent where it acts as a Lewis acid. The reaction is run in an inert solvent like $\ce{CH2Cl2}$ and a catalyst like boron trifluoride is used. The following is the reaction procedure taken from an e-book$\ce{^{[1]}}$. Also yield of various gem-difluorocarbonyl compounds from corresponding carbonyl compounds are listed:

A solution of $\ce{MoF6}$ (102g, 0.49g) in anhydrous $\ce{CH2Cl2}$(500mL) is cooled at 20-30 C. While passing a gentle flow of $\ce{BF3}$ through the solution, the carbonyl compound(0.5g) is added in small portion. The mixture is stirred while warming upto rt. Anhyd. $\ce{NaF}$ and silica gel were added. After filtration, the solvent is distilled in vacuo. The residue is distilled/purified by chromatography.

Further, it is written that:

In general, the yields are not that high due to the sensitivity of the
  reaction towards electronic and steric factors. Yield increase with
  the elctronic withdrawing group of the substituent of the substrate
  and decrease with the steric hindrance of the carbonyl group.
  $\ce{MoF6}$ attacks ether, amino, hydroxy, carboxy as well as
  $\ce{C=C}$ bonds. Ester, amide, cyano, nitro are inert towards the
  reagent. Halo substituent are not exchanged for fluorine.

The general reaction is given by$\ce{^{[2]}}$:

If a cyclic compound is used then R1 and R2 will be fused together.
If fluorination is proceeded at high temperature then catalyst is not needed. Aromatic carboxylic acids are converted into into gem-trifluoro derivatives without any catalyst.
Also, Aryl trifluoromethyl derivatives can also be obtained under less drastic conditions by fluorination of aromatic acid chlorides.
Now, the same reaction can be achieved by using DAST or Deoxylfluor and they are the more popular fluorinating reagent than $\ce{MoF6}$. Their reaction mechanism has been studied extensively. The mechanism is more or less similar than the mechanism achieved by $\ce{MoF6}$. The following is an explanation of the mechanism$\ce{^{[3]}}$(emphasis mine):

The electronic effect of the substituent exerted a significant
  influence on the reactivities of benzophenones. The presence of an
  electron-withdrawing group is believed to facilitate the nucleophilic
  attack of the fluoride anion to the carbonyl group, causing more
  facile cleavage of the carbon–oxygen double bond. Among the
  benzophenone substrates examined, 4-nitrobenzophenone exhibited the
  highest reactivity and the corresponding gem-difluorinated product was
  obtained in 91% yield[...]

Notes and References

Houben-Weyl Methods of Organic Chemistry Vol. E 10b/1, 4th Edition Supplement: "Organo-Fluorine Compounds - Synthesis of Fluorinated Compounds I, Transformations of Fluorinated Compounds" by Bernd Baasner and Georg Thieme Verlag, 14-May-2014 (link)
http://reag.paperplane.io/00002051.htm#2
Ying Chang, Amit Tewari, Avi-Izak Adi, Chul sung Bae "Direct nucleophilic fluorination of carbonyl groups of benzophenones and benzils with Deoxofluor" Volume 64, Issue 42, 13 October 2008, Pages 9837-9842
DOI: https://doi.org/10.1016/j.tet.2008.08.009
https://www.alfa.com/media/docs/FluorinatingAgents.pdf
Direct carbonyl group attack can be avoided and α-fluorination of ketones can be achieved by using highly selective fluorinating agent, NFSI. α-fluorination of cyclohexanone has been observed.

Enantioselective Organocatalytic α-Fluorination of Cyclic Ketones
Piotr Kwiatkowski, Teresa D. Beeson, Jay C. Conrad, and David W. C. MacMillan
Journal of the American Chemical Society 2011 133 (6), 1738-1741
DOI: 10.1021/ja111163u


text:
Napthalene - Friedel Crafts Alkylation and Acylation
Where does Friedel Crafts Alkylation and Acylation occur in Napthalene? Are the reactions kinetic-thermodynamic controlled?
Here's something I came across -

While alkylation occurs at alpha position, acylation occurs at beta position. Why is it so? Are the other possible minor products also formed? Does kinetic-thermodynamic control operate here?
Although I'm worried about Friedel Crafts Alkylation and Acylation reactions at the moment, my question isn't specific to them. In general, given an electrophile for EAS with Napthalene, how do we decide which product is major? (The one formed by attack at alpha, or beta position?)
References would be of great help too.

https://chemistry.stackexchange.com/questions/10736/nitration-of-naphthalene-and-anthracene https://chemistry.stackexchange.com/questions/28754/naproxene-syntheses-electrophilic-aromatic-substitution-on-activated-naphthalen https://chemistry.stackexchange.com/questions/10736/nitration-of-naphthalene-and-anthracene


text:
Mechanism for formation of Quinol from a Phenolic Ether

What's the mechanism for formation of Quinol from the given phenolic substrate? The reaction seems to be an acid catalysed hydrolysis, but I'm not able to reach the final product. I guess the first step should be addition of water to the nucleophilic π bond, but what after that?

The other product is acetaldehyde. So where is the double bond protonated?

Couldn't get to it yet. Could you please post the mechanism as an answer? Would be great help. Thanks.

Here is a mechanism for this enol ether hydrolysis.


text:
Identification for qualitative analysis

The flow chart is indicative of a compound with a silver cation, because $\ce{Ag+}$ reacts with $\ce{I-}$ to form yellow $\ce{AgI}$. Since $\ce{Ag+}$ does not form a hydroxide complex, it forms the brown $\ce{Ag2O}$ precipitate with excess $\ce{NaOH}$ solution. With $\ce{BaCl2}$, $\ce{AgCl}$ white precipitate forms. When excess $\ce{NH3}$ solution is added to that mixture, the soluble complex ion $\ce{Ag(NH3)2+}$ forms.
So, I can choose between answer choices A and B. However, I am not sure which cation is correct. The answer key says the answer is A, but I am not sure why, because both form white precipitate with $\ce{BaCl2}$.

When $\ce{Ag^+}$  reacts with alkali, it do form a complex, but the formula is too complicated to define. So we write $\ce{Ag2O.xH2O}$ for simplicity. 2. I am pretty sure you meant anion($\ce{NO3- , SO4^2-}$) in the 2nd para.

@NilayGhosh Yes, the edits are correct. However, what is the correct anion?

Both the salts satisfy all the conditions. It's the cation($\ce{Ag+}$) that matters not the anion.

@NilayGhosh so there are two answers?

Seems like 2 answers.


text:
Why is the intermolecular hydrogen bond between phenol and aniline stronger than the respective hydrogen bonds?
From NCERT class 12 part 1

The way I see hydrogen bonding is that it occurs due to intermolecular charge interaction. The more electronegative nitrogen/oxygen due to the inductive effect the respective hydrogen will be more electropositive.
Since the compound cointaining the more electronegative atom (oxygen/nitrogen) has greater respective charges.Then the hydrogen bonding will be greater between the molecules of that compound i.e either the hydrogen bonding between  phenol-phenol and aniline-aniline should be greater than phenol-aniline.


text:
Which has a greater atomic radius, Cl- , K+ , S-2 , Ca+2 , Ar?
How can I know that?
I only know that while we going to the right side at the table so the radius is decreasing.
And while we going down, so the radius is bigger.
But how can I check that radiuses in this case ?

Think of it this way:
1)  Count the electrons (all of them) in each atom or ion.
2) What happens to the orbitals, and thus to atoms or ions with a given number of electrons (let us say, 18 electrons), when you increase the atomic number of the element (more protons in the nucleus)?


text:
Stereochemistry of 1,4 disubstituted cyclohexane

How are these cyclohexanes identical?

Take the molecule on the left and ideally flip it by rotating 180 degrees along the vertical axis of the diagram. This inverts the cyclohexane without having to break or modify the molecule in any way


text:
How are the molar mass and molecular mass of any compound numerically the same?
This observation is really annoying me, and the internet isn't providing me with any solid answers. Either their definitions of molar mass completely differ, or they don't stick to their own definition while actually calculating the molar mass of a certain element or compound.
Now, according to Wikipedia,

The molar mass M is a physical property defined as the mass of a given
  substance (chemical element or chemical compound) divided by the
  amount of substance.

and it's unit is

g/mol

My understanding of this is that we're calculating the total mass of all the atoms(or molecules) present in 1 mole of that element(or compund). This means that we would have to take the atomic mass of that element and multiply it by the number of atoms present in 1 mole of that element, which is 6.022 × 10^23 atoms.
So for example, the relative molecular mass of water is 18.02 u, and it's almost the same as the molar mass of water, which is 18.015 g/mol.
So my question is, how is 18.02 u × (6.022 × 10^23) = 18.015 g/mol?

Approximations? Conceptually, they should be identical - considering the same material, i.e. either the relative molecular mass and the relative molar  mass, or both isotopic.

Related: Quick and simple explanation of molar mass, molecular mass and atomic mass

So for example, the relative molecular mass of water is 18.02 u, and it's almost the same as the molar mass of water, which is 18.015 g/mol.
So my question is, how is 18.02 u × (6.022 × 10^23) = 18.015 g/mol?

That is not quite correct.
The relative molecular mass $M_\mathrm r$ of water is a dimensionless quantity, i.e. the unit for relative molecular mass is the unit one and not $\mathrm u$:
$$M_\mathrm r=18.015\,28$$
It's the average mass per molecule $m_\mathrm f$ that has the unit $\mathrm u$:
$$m_\mathrm f=18.015\,28\ \mathrm u$$
Note that
$$1\ \mathrm{u}=1.660\,538\,921(73)\times10^{-27}\ \mathrm{kg}$$
Also note that you are missing the unit of the Avogadro constant:
$$N_\mathrm A=6.022\,140\,857(74)\times10^{23}\ \mathrm{mol^{-1}}$$
Thus
$$\begin{align}
M&=m_\mathrm f\cdot N_\mathrm A\\[6pt]
&=18.015\,28\ \mathrm u\times N_\mathrm A\\[6pt]
&=18.015\,28\times1.660\,538\,921\times10^{-27}\ \mathrm{kg}\times N_\mathrm A\\[6pt]
&=18.015\,28\times1.660\,538\,921\times10^{-27}\ \mathrm{kg}\times6.022\,140\,857\times10^{23}\ \mathrm{mol^{-1}}\\[6pt]
&=0.018\,015\,28\ \mathrm{kg\ mol^{-1}}\\[6pt]
&=18.015\,28\ \mathrm{g\ mol^{-1}}
\end{align}$$

Note that this answer was valid until the redefinition of the SI base units came into force on 20 May 2019.


text:
Does a solution of pH zero contain hydroxide ions?
Are there hydroxide ions in a solution of pH 0?

You mean pH 0? Yes, of course.

Yes, but very few.

Yes. There are $10^{14}$ times as many $\ce{H+}$ as $\ce{OH-}$ in a solution of pH 0. From the formula $$K_w=\ce{[H+][OH-]}$$ you can see there will always be both $\ce{H+}$ and $\ce{OH-}$ in an aqueous solution regardless of the pH.

Yes -- assuming you mean an aqueous solution.
In an aqueous solution there are always some hydroxide ions present, but if the pH is very low, there will be very few. The relationship is:
$\ce{[H^+][OH^-] = 10^{-14}}$
At pH = 0, $\ce{[H+] = 10^{-0} = 1}$ (by definition of pH), so $\ce{[OH^-] = 10^{-14}}$
$10^{-14}$ M is very, very low concentration -- but not none.


text:
What substances are hypergolic with liquid O2?
If something flammable is surrounded just by oxygen, I should see flame, right? The object would be even surrounded by more molecules than in the case of gas oxygen. On the youtube I see that people first light-up the objects and then they add oxygen. That means that the things won't oxidize by themselves. Why? Are there materials that would just burn without prior ignition?

https://en.wikipedia.org/wiki/Hypergolic_propellant

Organic material will burn just fine in liquid oxygen.  The reason people add liquid oxygen to already burning material instead of soaking the material in LOX and then lighting it is that they are not idiots.  If you soak organic material in LOX before lighting it, it becomes a high explosive, called an oxyliquit.

Super cool information! As cool as LOX.


text:
Stability of keto-enol vs. conjugation
I am trying to figure out which tautomer of creatinine:

is more stable.
We see that the enol tautomer is more highly conjugated: the $\pi$ electrons from the alcohol to the double bond to the $\pi$ electrons of the Nitrogen to the imine to the final $\pi$ electrons of the amine.
In the keto form, we have the $\pi$ electrons from the carbonyl to the imine to the two amines.
I know the enol form is more conjugated, but in general I also know that the keto form is more stable. How do I reconcile this?

I feel that you should also look at another possibility, that of having an aromatic pyrrole ring. I guess that's the most stable form. Nevertheless the real molecule will be a hybrid of all these.

The NMR spectrum seems to show that the tautomeric state of creatinine is that of the keto-form (bottom structure).
If the ketone is the only functional group, the keto form is more stable than the enol form. One the other hand, if the carbon-carbon bond is part of a conjugated system, the enol form might be more stable (e.g. acetylacetone or phenol).
You also suggest either the imino or the amino tautomer. There is a Raman study that addresses this question and also raises the dependence on pH.


text:
Reaction of 1,3 Butadiene with HBr in the presence of peroxides
I have learnt that the reaction of 1,3 Butadiene with HBr depends on the temperature as the temperature determines whether the reaction is under kinectic control or thermodynamic control. I believe that this is due to the hybrid nature of the carbocation and relative stabilities of the products. Will the same effects take place in the presence of peroxides? (because the allylic radical is also resonance stabilized)


text:
Conversion of oxalic acid to malonic acid

The reagents I have used here are $\ce{LiAlH_4}$ which reduces the carboxylic groups into alcohol.Used $\ce{HIO_4}$ which breaks up the molecule into methanal. Then  I applied Grignard reagent which consisted ethyl for the alkyl part.Then I tried to oxide the terminal carbon into $\ce{COOH}$  but could not find a suitable reagent.
What is a better approach and a method to solve this question.

Arndt–Eistert, maybe?

https://en.wikipedia.org/wiki/Arndt%E2%80%93Eistert_reaction

I would be wary of using a multistep reaction to try to homologate only one carboxylic acid group of oxalic acid... maybe forming a monoester first would work.

I would fully reduce to the diol and convert to the epoxide. Then, form the halohydrin and oxidize to $\alpha$-chloro acetic acid. Should be pretty straightforward after that. Going through the halohydrin would allow you to desymmetrize the starting material.

The Arndt-Eistert procedure suggested above seems very applicable to this problem. To use it we need ethyl oxalyl chloride. This can be prepared in 2 steps from oxalic acid by first esterification to form diethyl oxalate  procedure here then reaction with PCl5 to form ethyl oxalyl chloride procedure here. This can undergo the Arndt-Eistert mechanism here with excess diazomethane then either irradiation or treatment with silver oxide to give monoethyl malonate. Simple base hydrolysis then gives the required.

Assuming you want oxalic acid as the only source of carbon
$\ce{(COOH)2 ->[\Delta, -CO2] CO ->[Cl2] COCl2  }$
$\ce{(COOH)2 ->[LiAlH4; H3O+] (CH2OH)2 ->[H3O+,\Delta] CH3CHO }$
$\ce{CH3CHO ->[B:-] CHO-CH2- ->[COCl2, huge excess; H3O+] CHO-CH2-COOH ->[Ag2O] CH2(COOH)2   }$
The key here is to split target molecule into $\ce{C1}$ and $\ce{C2}$ fragments and then find a way to find a way to make enolizable $\ce{C2}$ compound from oxalic acid. There is more then one way to obtain phosgene, one I pointed here wouldn't work in a lab. AFAIK phosgene can be obtained by thermal decomposition of oxalyl chloride, but I couldn't find trustworthy looking reference with quick googling.
rearrangement of ethylene glycol might required rather harsh conditions.
.....
That one was fun.

By Arndt–Eistert any luck?

@starunique2016 The common restriction in this type of problems is either to not use outer sources of carbon or not use other organic compounds. Besides, Arnds-Eistert isn't very common. For this reasons I do not consider Arnds-Eistert a valid step for this problem unless OP states otherwise  by quoting exact wording of the problem.

If i want by Arndt-Eistert only?

That ethylene glycol one i have my doubts

@user57048 Rearrangement or synthesis ? Rearrangement is referenced in literature, it is a variant of  pinacol–pinacolone rearrangement.

I specifically wanted Arndt-Eistert  so i didnt accept this answer,but +1 anyways


text:
Nickel-63 and surge protector / voltage regulator
I've read that nickel-63 is used in surge protectors and voltage regulators.
What is it about nickel-63 that makes it particularly suitable for those applications?

It is about radioactivity, I presume.

Nickel-63 is a beta emitter with a half-life of 101 years. That makes it reasonably active, yet long-enough lasting for the application you hint at. Now, what is that application? Gas discharge tubes.
A gas discharge tube (GDT) is used in a variety of applications to divert overvoltage to ground (such as a lightning strike), protecting what is behind the tube. At the simplest, one can consider it two parallel plates with gas in between. One plate is connected to ground, one is on an input line (voltage to a device, from an antenna, etc.). At a high enough voltage, the gas is supposed to break down, usually through an avalanche process, providing a low impedance path to ground and protecting what is downstream of the tube. The problem is, an avalanche process needs a few free electrons around to get it going in a timely fashion. For a high-voltage GDT (1000's of volts or higher), field ionization will likely provide enough electrons quickly enough to have it fire quickly enough to protect the circuit. For lower voltage rated tubes, it is helpful to have a low level of ionization pre-existing in the tube.
A beta emitter inside the tube is one way to consistently have a low concentration of free electrons ready to avalanche in case of enough voltage being present across the device. Nickel is easy to plate on the contacts.
Further, Nickel-63 is readily formed by thermal neutron irradiation of Nickel-62, with a cross-section of nearly 1000 barns (quite large for neutron absorption). So, it is also easy to make by exposing nickel to a reactor environment. (Note one would just expose naturally-occurring nickel, counting on transforming some of the naturally-present Nickel-62 to the 63 isotope).


text:
How is the structure of triiodide ion (I3-) possible?
I have seen the structure of triiodide ion ($\ce{I3-}$) but I cannot understand why this structure is even possible. I have seen in my textbook that

$\ce{I_3^-}$ is formed by combination of $\ce{I2}$ and $\ce{I^-}$ ion in which $\ce{I^-}$ ion acts as a donor and $\ce{I2}$ molecule act as acceptor. It accommodates electrons in empty d orbitals:

But I think $\ce{I-}$ is stable due to its complete octet. Then why it should combine with $\ce{I2}$ molecule to form another new compound? I am confused because the Lewis structure of $\ce{I3-}$ would not make any sense if we consider the octet of central iodine atom.

Lewis structures are not very helpful  in deciding what is possible and what isn't.

(1) There is negligible iodine 5d orbital involvement in the triiodide anion. See e.g. https://chemistry.stackexchange.com/q/60456/16683 (2) Anyway, octet-complete species react all the time, for example: $$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$

So complete octet of any atom does not generally means it is inert.

That's correct, or else we would be in deep trouble, since water and oxygen (amongst others) are both octet-complete. I don't mean to say your question is wrong. It's still valid to ask "what drives the formation of I3(-) from I2 and I-?", but it's just not really correct to say "I don't think it should happen because they all have complete octets".

Actually these confusions arises because of what we are taught in earlier classes and what we are being taught now.

https://chemistry.stackexchange.com/questions/87485/formal-charges-on-polyatomic-ions

First things first: the bonding orbitals of $\ce{I3-}$ do not contain any significant d-orbital contributions. In fact, the $\ce{I-I}$ bond lengths are significantly longer than in $\ce{I2}$, suggesting a lower bond order.
In fact, the triiodide anion should rather be imagined as a 4-electron-3-centre bond (bond order approximately 0.5) which might be easier to understand using this mesomeric depiction:
$$\ce{\overset{-}{I}\bond{...}I-I <-> I-I\bond{...}\overset{-}{I}}$$
What happens orbital-wise in this? Simplistically speaking, a 4-electron-3-centre bond is made up of the following molecular orbitals stemming from $\mathrm p_z$ orbitals of the individual iodine atoms:

an all-bonding lowest level orbital ($\ce{{+}I-\bond{...}-I{+}\bond{...}{+}I-{}}$, where $-$ and $+$ denote the different phases of a p orbital)
A far-reaching bonding orbital to which the central atom does not contribute ($\ce{{+}I-\bond{...}OIO\bond{...}-I{+}}$, where O denotes no orbital lobes; there is no p-type orbital on the central iodine that could contribute. s-type contribution is possible.)
an all-antibonding highest level orbital ($\ce{{+}I-\bond{...}{+}I-\bond{...}{+}I-{}}$) which remains unoccupied.

Why does this happen although the starting compounds iodide and $\ce{I2}$ are stable on their own? Obviously, most reactions happen although both starting materials are full-octet compounds and inherently stable in some way or another. And most happen because the products are more stable. So we just need to determine why it is favourable for $\ce{I- + I2}$ to react to form $\ce{I3-}$.
The simplest answer is distribution of charge. Rather than have a single negative charge on one atom you end up with a structure that has approximately half a negative charge on two different atoms if you take the resonance structures at face value (hint: don’t). Whatever the final picture is, the negative charge will be more distributed over a greater number of atoms than before. This is, very generally speaking, a great stabilisation and we can stop here.

Expanding on Jan's reference to three-center four-electron bonding:
When the three iodine atoms all come together in a line the overlap among all three atoms gives the combination of $p$ orbitals rendered thusly (source: Wikipedia by Arun Sridharan):

Only the lowest orbital gives a bond but, because it involves overlapping all three atoms instead of just two, it's a stronger bond than just the two-center bond in $\ce{I_2}$.  However, the three-center bond is distributed over two linkages instead of one and so each individual linkage is less strongly bonded than an $\ce{I_2}$ molecule.
Which leads to the reverse problem.  Why is this a property of iodine?  Why don't other halogens do the same thing?  In fact the other halogens do form three-atom complexes, as we shall see; but they do so with increasing difficulty as we move to lighter halogens.
What enters here is the fact that you don't just have the trihalide ion.  It must be bound to counterions by ionic bonds or to polar solvent molecules by ion-dipole interactions.  These additional, nonmolecular electrostatistic interactions, are more favorable with a compact spherical ion consisting of one atom instead of an extended three-atom chain.  The nonmolecular interactions with surrounding ions or dipole thus tend to break up the trihalide ions.  With iodine the single atom is still a bit bulky and therefore the nonmolecular interactions have little power to break up the triiodide complex.  With smaller-atom halogens the nonmolecular interactions become stronger and more important.
What is the lightest known trihalide ion, then, given the fact of these evil nonmolecular electrostatic interactions?  The answer may surprise a few people.
Tribromide ion
This ion is actually fairly commonplace if you know where to look, which could be an organic chem lab.  Pyridinium tribromide, $\ce{(C5H6N)^+(Br3)^-}$, is used as a brominating agent for reactive substrates such as phenols; it is easier to handle than liquid bromine.  We therefore move on to a bigger, or smaller, challenge:
Trichloride ion
The $\ce{Cl_3^-}$ ion has been obtained from electrochemical studies in suitable solvents such as acetonitrile, for example see Ref. 1.
Trichloride ion is also obtained as an ionic liquid by adding chlorine to an ordinary chloride salt 2.  These species can be used as chlorinating/oxidizing agents in circumstances where organic solvents would be prone to side reactions with the chlorine.  The reference given here includes a list of trichloride ion bearing materials. All are liquid under ambient conditions, and all involve bulky, irregularly-shaped cations for which the electrostatic preference for a single spherical-atom anion is minimal.
Trifluoride ion
Surely we have reached our wit's end?  This reference [3] reported that trifluoride ion has been observed in a low-temperature noble gas matrix and at low pressure in the gas phase.  The same reference also reports that the main contributing structures for this ion involves radical combinations rather than the ion-molecule model we might ordinarily expect for this type of complex.
So at least under suitable conditions we can go all the way to $\ce{F_3^-}$!
References

M. C. Giordano, V. A. Macagno, and L. E. Serano (1973). "Trichloride ion formation constant in acetonitrile solutions". Anal. Chem. 45, 1, 205–207. https://doi.org/10.1021/ac60323a026.

Xiaohua Li, Arne Van den Bossche, Tom Vander Hoogerstraete and  lKoen Binnemans (2018). "Ionic liquids with trichloride anions for oxidative dissolution of metals and alloys".Alloys. Chemical Communications 55(5), 475-478.

Benoit Braida and Phillipe C. Hubertus(2004). "What makes the F3- ion so special?" A breathing-orbital valence-bond ab initio study. _Journal of the American Chemical Society, 126(45), 14890-14898. https://doi.org/10.1021/ja046443a.

The hyperlink on the word source is not proper attribution. Please make sure that you follow the license when you use external content. Thank you.

I've changed it so you see what I mean. Wikipedia requires in this case (cc-by-sa) that you give attribution, i.e. credit the author directly. In any case, it's better to include a human readable reference for those cases where links rot away. We have a better chance of finding the right resource. I've noticed that you have used a link through Google, you might want to avoid that too, because they obfuscate the real source. I'm on mobile right now, otherwise I'd have changed that, too. Thanks for your continued contribution; I know this is nitpicking, but licenses can be a bit problematic.

Thanks.  I think I know what you are talking about with the Google link.  When a link downloads to my phone instead of being online I struggle to attribute it any other way, I end up copying and pasting the Google search.  Hoping you have a solution.

I don't really have a solution for that. Just hope that someone else finds it and likes it and improves the attribution; or you come back later for yourself at the PCs. If there is, copying the DOI is a good way to reference the publication uniquely. Also we can then use the Userscript of Gaurang to fill it in automatically. Albeit that only works on a computer...


text:
Thermochemistry: What is the difference between the calculation of bond enthalpy and Hess's Law
I know that Hess's law is an indirect way to find the heat of reaction using the known heat of reaction of 2+ thermochemical equations, thus the formula is delta heat of reaction = enthalpy of formation (product)- enthalpy of formation(reactant). But then, bond enthalpy also finds the change in heat of reaction with the formula: change in heat of reaction = bonds broken (reactant) – bonds formed (product). If it's negative, it's exothermic, and if positive, it's endothermic.
I am confused at the relationship between these 2 concepts since they are rather opposing each other.

You are confused as to why we take "reactants minus products" when calculating the enthalpy of reaction based on bond enthalpies.
Consider as reference state all your reactants and products in their atomic states, that is, with all bonds broken. Forming a bond releases energy, so the reactants and products all have a negative enthalpy $$\Delta H = -\sum\Delta H_\text{bonds}$$ with respect to this reference state. The enthalpy of reaction remains $$\begin{align*}\Delta H_\text{rxn}&=\Delta H_\text{prod}-\Delta H_\text{reac}\\ &= -\sum\Delta H_\text{bonds,prod}-\left(-\sum\Delta H_\text{bonds,reac}\right)\\ &= \sum\Delta H_\text{bonds,reac}-\sum\Delta H_\text{bonds,prod},\end{align*}$$
so the inconsistency you've pointed out is just the neglect of a minus sign.

ΔH of a reaction in terms of bond enthalpies = Σ bond enthalpies (products) - Σ bond enthalpies (reactants).
Therefore if the products are at a lower energy than the reactants, the reaction would be exothermic therefore ΔH would be negative.


text:
Regioselectivity of bromination of alkenes
Recently, in my chemistry lesson, we were discussing the mechanism of the bromination of alkenes. The teacher brought up the involvement of the cyclic bromonium intermediate. She also mentioned that the nucleophile attacks the more substituted carbon atom in the bromonium ion. However, this raised many questions among my classmates. Firstly, based on steric reasons, the nucleophile is more likely to attack the less-substituted carbon since there is less hindrance. Hearing this, the teacher then said that the reason for the preference is based on electronic factors. However, this makes little sense as well, considering that the presence of more alkyl groups will decrease the partial positive charge on the more substituted carbon, making it less attractive to the nucleophile (be it $\ce {Br^-}$ or $\ce {H2O}$).
I went to seek clarification by looking at more authoritative texts, such as Advanced Organic Chemistry: Part A: Structure and Mechanisms by Carey and Sundberg, as well as Guidebook to Mechanism in Organic Chemistry by Peter Sykes. However, they were citing very weird reasons that were not very intuitive. Here are some extracts from relevant chapters of the two texts:
From the former,

Unsymmetrical alkenes nevertheless follow the Markovnikov rule because the partial positive charge that develops is located predominantly at the carbon that is better able to accommodate an electron deficiency...

From the latter,

With an unsymmetrical alkene, e.g. 2-methylpropene (32), the more heavily alkyl-substituted carbon will therefore be preferentially attacked by the residual nucleophile, $\ce {Cl^-}$.

Why do the alkyl groups result in greater cationic character when they are in fact, electron releasing?

Clayden has a section on acid-catalysed epoxide opening, I think. The same principles apply here.

@orthocresol Is it possible to rationalise this using MO theory? Has it got something to do with the hyperconjugative interactions of the alkyl groups C-H and C-C sigma MOs with the C-Br sigma * MO?

Hmm... that's not the way it's conventionally taught, but it is possible. You can see some involvement of the CH bonds in the LUMO that Martin provided, consistent with the mixing that you described. I think it's important here to distinguish between discussing the ground state (i.e. the bromonium ion itself) and the transition state (for ring opening). Martin's answer talks about the GS, which is actually very convincing (to me), but the textbooks usually go for an explanation based on the TS. Although perhaps they are just two sides of the same coin...

With bromination, the first step is the formation of an adduct complex, which can be explained that the π-bond polarises the dibromide. The next step is the formation of the bromonium ion, which carries an overall positive charge. A symmetry-changing induction like substituents will always result in asymmetric carbon-bromine bond lengths in the bromonium ion. Since it is a cation, similar rules for the stability/instability as for carbocations apply. Therefore the higher substituted carbon will have the longer $\ce{C-Br}$ bond length, and the higher charge (more positive/less negative). This makes it a better target for the residual nucleophile $\ce{Br-}$ to attack there.
Here is the bromonium ion for the reaction
$$
\ce{\underset{alkene}{H3C-CH=CH2} + Br2 ->
[\underset{bromonium ion}{H3C-CH(Br)CH2}]+ + Br- ->
\underset{$trans-$alkane}{H3C-CHBr-CH2Br}}
$$
calculated at the DF-b97d3/def2-SVP level of theory, charges with NBO6, to illustrate the concept:

Additionally the HOMO has the higher coefficient at the lesser substituted carbon, while the LUMO has the higher coefficient at the higher substituted carbon.

From this, you can see that both quotes from the textbooks are correct.

What could have caused the C-Br bond to the more substituted carbon to be longer? Orbital interactions? Electronic effects?

Why does the calculation show that the carbons are bearing negative charges? Shouldn't they bear a partial positive charge?

@Tan The molecular structure is always by a delicate balance of electronic effects and orbital interactions. The molecule will adopt the structure, which is lowest in energy (or all structures accessible at a certain temperature). I am not a big fan of calling things +/-I or +/-M effects, as that are zeroth order approximations. We know that there will always be hyperconjugation from a methyl group, which will certainly the case here, too. And that much should be clear, that a positive charge will be stabilised by it. It follows basically the same trend as carbocations.

@Tan No they should not. You might be confusing formal charges with partial charges. There is a significant difference in electronegativities for carbon and hydrogen, most of the electron density of the hydrogen is polarised to the carbon. Even such an over-simplified model like oxidation numbers would predict this trend. The values are certainly within expectation.

Are you certain that there will be hyperconjugation? The geometry of bromonium is slightly different from that of the conventional carbocation. Maybe the orbitals (i.e. C-Br sigma * and C-C sigma) may not have the right symmetry/orientation? Could you illustrate more quantitatively the hyperconjugation that is taking place? I am very curious...

@Tan I am certain there is hyperconjugation, as there already was hyperconjugation in a substituted alkene. There probably is always something to which we refer to as hyperconjugation. You are thinking in localised orbitals, which is only one viewpoint. I can't promise that I'll be able to add any more information, I am also unsure whether I will have the time.


text:
How would one disassociate the carboxylic acid on the end of a perfluorooctanoic acid molecule?

This is what a perfluorooctanoic acid molecule looks like. The carboxylate head on the end would normally disassociate in water because carboxylic acids are weak, but because of the fluorocarbon tail, the molecule is very stable. How do you remove the head from the rest of the molecule? Could you use UV radiation to perform homolytic fission for the decarboxyliation?
To provide some context, this molecule is used in flame retardants, and those flame retardants are polluting drinking water reservoirs. The head of this molecule, once the pollutant is inside the body, binds to proteins and behaves like hormones, disrupting normal hormonal function.

A decarboxylation reaction would get rid of it: https://www.organic-chemistry.org/namedreactions/barton-decarboxylation.shtm and decarboxylative cross-couplings are all the rage now e.g. https://pubs.acs.org/doi/10.1021/acs.accounts.8b00209 so you could theoretically replace the CO2H with pretty much whatever you wanted on the end.

"The carboxylate head on the end would normally disassociate in water": this isn't what carboxylic acids do in water. What did you mean by this?


text:
When using a "dilute hydrochloric acid" solution, how diluted should it be?
Quite often when reading through some instructions or documentation or an outline that I've found online (EG: acid-base extractions), theres a step that involves using a "dilute hydrochloric acid solution", but oddly enough, it often doesn't detail exactly how diluted the HCl should be. Heres one example (little past half way into the 3rd paragraph)

Dilute hydrochloric acid is often used in the extraction of basic substances from mixtures or in the removal of basic impurities. The dilute acid converts the base such as ammonia or an organic amine into water soluble chloride salt.

So my question is: How does one know exactly how diluted the HCl should be? Is there a commonly used concentrated amount of HCl that I just don't know about? Sometimes it'll suggest using Muriatic acid instead, which I believe is basically 31.45% concentrated HCl, is that a commonly used concentration?

Muriatic acid is typically loaded with iron, hence its yellow color, so it is not used in chemistry.

@MaxW - Thanks for the comment. I was actually wondering why it was yellow if all it supposedly contained was diluted HCl..

The exact concentration doesn't matter that much. The stock solution that I used to use was 1 M aqueous HCl (~3%). If your compound is acid-sensitive then you might want to try using even more dilute HCl (e.g. 0.01–0.1 M), or skip this step altogether and just purify it using a different method. The important thing is to make sure that you have enough HCl to protonate all the basic stuff you want to remove – a relatively simple stoichiometry exercise.
I would suggest not using concentrated HCl (12 M, or ~37%) as for a typical acid-base extraction it's simply not necessary. You should save that for other circumstances, such as when you want to adjust the pH of a solution without diluting it too much with water.

When would one know that enough HCl has been added and the base is fully (but not overly) protonated? When the pH reaches ~7?

There's nearly never any harm in adding too much HCl, because any extra acid is retained in the aqueous layer, which you are going to remove anyway. For most purposes you therefore only need to do some maths: count how many moles of base you used, and hence the volume of HCl you need to add. If (for whatever reason) you really needed to monitor the pH, though, you could use pH paper to make sure the pH is around 7, as you suggested.

Interesting - Thanks for the input! Also, when you say "or skip this and purify using a different method", what other methods would yield similar results to A/B? (sorry if thats too general of a question)

Chromatography, distillation, or recrystallisation are the most common methods. Often even after an acid-base extraction you will still need to carry out one of these three purification methods. The acid-base extraction just makes it easier because you don’t have so many things left to separate.

Recrystallization is actually what I usually follow up with, actually.


text:
What is the net ionic equation of the following?

Aqueous chlorine is added to a solution of aluminum bromide in a single replacement reaction which forms no precipitate and bromine liquid. Determine the net ionic equation.

Since this is a single replacement reaction, it will take the form of  $$\ce{A + B-C -> A-C + B}$$
Since this reaction forms no precipitate, the new compound will not be a solid.
$$\begin{multline}
\ce{aqueous chlorine + solution of aluminum bromide ->}\\
\ce{ non-solid compound + liquid bromine}
\end{multline}$$
\begin{align}
\ce{Cl2(aq) + Al^{3+}Br3^-(aq) &-> \text{non-solid compound} + Br2(l)}\\
\ce{Cl2(aq) + AlBr3(aq) &-> \text{non-solid compound} + Br2(l)}\\
\ce{Cl2(aq) + AlBr3(aq) &-> Al^{3+}Cl3- + Br2(l)}\\
\ce{Cl2(aq) + AlBr3(aq) &-> AlCl3 + Br2(l)}
\end{align}
Since most chloride salts are soluble (except for silver halides), $\ce{AlCl3}$ is predicted to be aqueous.
\begin{align}
\ce{Cl2(aq) + AlBr3(aq) &-> AlCl3(aq) + Br2(l)}\\
\ce{3Cl2(aq) + 2AlBr3(aq) &-> 2AlCl3(aq) + 3Br2(l)}
\end{align}
Here is the complete ionic equation:
$$\ce{6Cl⁻(aq) + 2Al^{3+}(aq) + 6Br-(aq) -> 2Al^{3+}(aq) + 6Cl-(aq) + 3Br2(l)}$$
Cancelling out the spectator ions, we get
$$\ce{6Br-(aq) -> 3Br2(l)}$$
In lowest terms:
$$\ce{2Br-(aq) -> Br2(l)}$$

You are making a mistake splitting up your reaction equation from \eqref{full} to \eqref{wrong} in order to filter out the spectator ions.
\begin{align}
\ce{3Cl2(aq) + 2AlBr3(aq) &-> 2AlCl3(aq) + 3Br2(l)}\tag1\label{full}\\
\ce{6Cl⁻(aq) + 2Al^{3+}(aq) + 6Br-(aq) &-> 2Al^{3+}(aq) + 6Cl-(aq) + 3Br2(l)}\tag{wrong}\label{wrong}
\end{align}
What you are essentially doing is not conserving the charges. Dichlorine $\ce{Cl2}$ is a neutral molecule, and the following is what your transition implies, but is also wrong:
$$\ce{Cl2 -> 2Cl-}\tag{also wrong}$$
The correct way to transform \eqref{full} is to split (dissociate) only the ionic compounds $\ce{AlBr3}$ and \ce{AlCl3}, and then cancel the spectator ions:
\begin{align}
\ce{3 Cl2 (aq) + 2 AlBr3 (aq) &-> 2 AlCl3 (aq) + 3 Br2 (l)}\tag1\\
\ce{3 Cl2 (aq) + 2 Al^3+ (aq) + 6 Br- (aq) &->
    2 Al^3+ (aq) + 6 Cl- (aq) + 3 Br2 (l)}\tag2\label{split}\\
\end{align}
Next step is to cancel the aluminium ions on each side, and reduce the equation to \eqref{net-ionic}:
\begin{align}
\ce{3 Cl2 (aq) + 6 Br- (aq) &-> 6 Cl- (aq) + 3 Br2 (l)}\tag3\\
\ce{Cl2 (aq) + 2 Br- (aq) &-> 2 Cl- (aq) + Br2 (l)}\tag4\label{net-ionic}
\end{align}
I don't think the reaction will actually produce liquid bromine, but aqueous bromine instead; that doesn't matter to the solution though.

You do not render the chlorine correctly. Keep the chlorine as diatomic molecules on the reactant side and note that because of the charges the chlorine atoms do not cancel out. Your net ionic equation should therefore include both halogens.


text:
Are the 2s and 2p subshells degenerate in the He+ ion?
I know that the 2s and 2p (as well as 3s, 3p, and 3d) electron subshells are degenerate in the hydrogen atom. I was wondering if they are also degenerate in other one-electron species, such as the $\ce{He+}$ atom.

He+ is pretty much the same as H.

I'd point out that there is only one 2s orbital so it can't be degenerate. The various p, d, and f orbitals can be, and they are degenerate in the $\ce{He+}$ atom in the ground state. However if the electron is in a 2p orbital say, then the other two are degenerate, but the one with the electron has a different energy.

@MaxW Sorry, I meant subshells, not orbitals. I know that the 2px, 2py, 2px orbitals are degenerate, but I was wondering if the 2s and 2p subshells are degenerate in all one-electron species.

Yes, because this system is identical to the original if we simply rescale the charge (or, if you'd like, modify the units of our new system). On a more fundamental level, degeneracies are caused by symmetries of the system, and these are affected not by the magnitudes of various constants or parameters, but rather by the form and structure of the relevant interactions.
The degeneracy of the subshells in the hydrogen atom corresponds to the SO(4) symmetry of the hydrogen atom, which leads to the conservation of the quantum-mechanical form of the Laplace-Runge-Lenz vector. This symmetry depends on the inverse-square law nature of the Coulomb force and appears also in celestial mechanics, where the corresponding inverse-square law force is now gravity.
Changing the charge of your hydrogenic atom will not break this symmetry. On the other hand, fine structure corrections to the hydrogen atom will: if you start looking at relativistic and spin-orbit corrections, you change the form of the Hamiltonian, break the symmetry, and hence resolve the degeneracy.

Thanks for your answer. To clarify, all subshells of given energy level in a one-electron atom (like H, or Li2+) are degenerate?

@user8081591, yes.


text:
What does "off-hue" constitute in the context of analyzing a sample using the Pt-Co color scale?
ASTM D1209, the method describing how to perform the Pt-Co color scale test for a sample, recommends reporting material that can't be matched to a particular standard as "off-hue". For color based analyses, how is "hue" measured/quantified and what is the distinction to "color"?

I'm not buying the standard to answer the question so this is an educated guess. I assume that you'd make standards of various concentrations which would have various hues of yellow. You can't make standards of every concentration so the sample is bound to be between standards. Yellow is thus the color, the "deepness" of the yellow color is the hue.

My disconnect is how one differentiates between the color (how to define the difference between yellow and orange - pick some wavelength?) and the "deepness". Should I be looking for distinctions in some spectrometer measurements (absorbance, reflectance, transmission, etc.) or something else all together?

Sorry, did some poking around without being able to find the standard itself. I  had the wrong interpretation of hue. The scale seems to be for yellows and yellows only . Thus if some sample is orange, then it is off hue.

I think of hue meaning the difference between something like sunflower yellow and a faded yellow-I guess I think "hue" and "shade" are synonyms and if I had a sample that was orange I'd think I have some major problem if I am expecting some shade of yellow.


text:
Determining absorption spectrum, magnitude of Δo and spin for Iron Oxalate
Iron Oxalate ions form an emerald green solution when dissolved. If the ions in solution appears emerald green to our eyes, what color of visible light is it mostly absorbing? What does that tell you about the magnitude of Δo (recall that E = hc/ƛ)? Determine if the complex is likely to be high spin or low spin based on this observation.
I know that the crystal field splitting affects the color of the compound, and that compounds with different magnitudes of Δo will appear different colors, corresponding to the wavelength of light that enters our eyes. Since Iron Oxalate ions appear green (490-560nm), I would guess that it is absorbing red light (650-750nm), which has the lowest energy wavelength on the visible light spectrum, which leads me to suspect that the complex is low spin. Is this correct? How might one go about finding the magnitude of Δo?


text:
Nomenclature confusion regarding carboxylate and ester
I suddenly thought about these 2 compounds that seem to have the same name. Aren't both of them methyl propanoate? How do we differentiate them apart name wise?

From context (in spoken word) and/or the presence or absence of a space (in written word). The one on the left is a neutral molecule whereas the one on the right is a charged molecule. In mentioning the given compounds there should be context to indicate a charge. For example, the one on the left is "methyl propanoate", whereas the one on the left would be "2-methylpropanoate" (also called isobutyrate).  The isobutyrate would usually be described as an ion (e.g. "sodium methylpropanoate") which would provide clarity as well.
Now it is possible to see the 2-methylpropanoate written as just "methylpropanoate" since there are no isomers to cause ambiguity. In either case, though notice that names for the left molecule have a space between "methyl" and "propanoate" whereas names for the right molecule do not. That is how you can tell between an ester from an anion.

There is not supposed to be a space between the substituent prefixes (e.g. methyl) and the parent structure (e.g. propanoate), so your second structure is methylpropanoate, or more correctly 2-methylpropanoate.
In the ester there is a space between methyl and propanoate, i.e. methyl propanoate.

Lack of attention to this minor detail is a frequent cause of errors in nomenclature.


text:
Detail from Integral Evaluation in Szabo (Appendix)
This is a step from Appendix A in Szabo and Ostlund's Quantum Chemistry. In short, the question is how to get from \ref{A.20} to \ref{A.21}. We can factor out some constants $K$ in the expressions to get:
\begin{align}
  (A| -Z_C/r_{1C} |B)
  &=
    K \left(\frac{1}{2\pi^2}\right)
    \int \mathrm{d}\mathbf{k} \; e^{-k^2/4p} k^{-2}
    \exp\big(-i\,\mathbf{k}\cdot(\mathbf{R}_P - \mathbf{R}_C)\big)
    \tag{A.20}\label{A.20}\\
  &=
    K \left(\frac{2}{\pi|\mathbf{R}_P-\mathbf{R}_C|}\right)
    \int\limits_0^\infty \mathrm{d}k \; e^{-k^2/4p} k^{-1}
    \sin(k|\mathbf{R}_P-\mathbf{R}_C|)
    \tag{A.21}\label{A.21}
\end{align}
Part of the note between these steps is that (paraphrasing) we can let $\mathbf{R}_P-\mathbf{R}_C$ lie on the $z$-axis and use that $k\cdot (\mathbf{R}_P-\mathbf{R}_C)= k|\mathbf{R}_P-\mathbf{R}_C|\cos\theta$, "then we can easily perform the angular part of the integration over $\boldsymbol{k}$ to obtain [\ref{A.21}]."
What I have done: It seems that Euler's formula must be applied to the second exponent in \ref{A.20} but using the hint we'd have something like $$\cos(k|\mathbf{R}_P-\mathbf{R}_C|\cos\theta) + i\sin \dots \text{etc}. \tag{???}$$
 I guess that the sine in \ref{A.21} follows integration of the (real) cosine in Euler's expansion. In line with this, \ref{A.21} is multiplied by a factor of $2\pi$.
I will keep at this but I think if it puzzles me, then others may have stumbled over it so I am posting.
Hopefully removal of the common factors ($K$) makes this easier to read but the problem is on pages 413-414 of Szabo and Ostlunds book.

Don't have my Szabo at hand (heresy, I know), will look into it today/tomorrow...

There is an integration 'missing' effectively  $\int \exp(-a\cos(\theta) ) d\theta$. Substituting $z=\sin(\theta)$ and some fiddling about makes the integral easier.

Using porphyrin's substitution and Euler's relation I get an answer which is off by a factor of $2\pi ~k^2.$ If a spherical volume was intended by the original vector notation then this seems to be OK. Would still like an authoritative answer...

As commented, do note that the integral (A.20) is taken over all $\mathbf{k}$-space, whereas the integral (A.21) is one-dimensional. The explicit calculation of this integral follows as
$$\begin{align} \int\text{d}\mathbf{k}\,\Psi(k)e^{-i\mathbf{k\cdot r}} &=\int_0^\infty\text{d}k\,k^2\int_0^\pi\text{d}\theta\int_0^{2\pi}\text{d}\varphi\,\sin\theta\Psi(k)e^{-ikr\cos\theta}\\
&= 2\pi\int_0^\infty\text{d}k\,k^2\Psi(k)\int_0^\pi\text{d}\theta\,\sin\theta e^{-ikr\cos\theta}\\
&= 2\pi\int_0^\infty\text{d}k\,k^2\Psi(k)\left[\frac{e^{ikru}}{ikr}\right]_{u=-1}^1\\
&= \frac{4\pi}{r}\int_0^\infty\text{d}k\,k\Psi(k)\sin(kr).\end{align}$$
In the first equality we convert this integral from Cartesian to spherical coordinates, in the second equality we evaluate the integral over $\varphi$, in the third equality we substitute $u = -\cos\theta$, and in the fourth equality we make use of Euler's formula to get the exponential representation of $\sin(kr)$.

This looks right to me...$\Psi(k)$ has absorbed the $k^{-2}$ in A.20, right?

@daniel, yes.$\ $


text:
In Diazonium Ion why does N contain 4 bonds?
Nitrogen's valenc number is 3. Sometimes it can form a co-ordinate bond by donating it's lone pair electrons. In diazonium ion N has 4 bonds, then one of those 4 bonds must be a co-ordinate bond. Actually I can't define the co-ordinate bond in diazonium ion.Again In diazonium structure, it is shown the first N with 4 bonds is positive but why??

Why wouldn't it, really?

I suppose you could consider it to be the product of on of the terminal lone pairs of N2 forming a dative bond to the Ph+ ion

One way to think about this structure, the benzenediazonium cation, is to consider the structure of benzonitrile - C6H5CN - just replace the positive nitrogen with a neutral carbon atom. This has the same basic electronic structure as the diazonium ion, but differs in that the carbon nucleus has one less positive charge than the nitrogen nucleus.
I don't think the concept of "coordinate bonds" is very helpful here. There are many different species with positively charged nitrogen, such as tetramethylammonium salts, that have four full covalent bonds to nitrogen.


text:
Polar compounds that doesn't dissolve in water?
I'm studying for my exam introduction to chemistry. In the course, there's an exercise where we have to explain why something dissolves in water and why other compounds doesn't.
For me, if the substance is polar, than it dissolves in water because H2O is polar to. (and non-polar dissolves in non-polar).
The first example is AluminumOxide. In the course, it's mentioned that this doesn't dissolve in water. They say that the forces are to high? But it's a polar compound? How do you know if a polar compound will dissolve?

Useful read: Enthalpy of solution

You may accidentally have chosen an awkward example because I'm not sure that $\ce{Al2O3}$ is ever properly molecular. It is usually crystalline and may even be in the form of nano-crystals, in which case there will be considerable lattice binding energy to overcome. In its doped form, with the odd Cr$^{3+}$ ion, it is the mineral Ruby which is clearly not soluble in water.
In the general case polarity is not the only consideration for solubility. There is entropy change of mixing which is always favourable when two species are mixed and also enthalpy change when placing molecules of one type into the other. The enthalpy is where the polarity part comes in.
In water in particular there are other  entropy  changes to consider because water is so hydrogen bonded. The H-bonding structure is broken up when a solute is added and water has  then to 'bridge the gap' as it were round the solute. This forms a more structured water arrangement than before adding the solute and so entropy is reduced. This is primarily responsible for insolubility in water.
We measure the process by considering the free energy change as $\Delta G = \Delta H - T\Delta S$ where $\Delta H$ is the change in enthalpy, $\Delta S$ that of the entropy. When the process if favourable $\Delta G$ will be negative.
Surprisingly adding, for example, non-polar  cyclohexane to water has a change in enthalpy ($\Delta H$) that is close to zero, whereas for a polar molecule $\Delta H$ will be negative. The H-bonding bridging with cyclohexane will make the change in entropy negative so making the free energy positive, which is unfavourable.
With a polar molecule in water H bonding bridging is not necessary to such an extent as for a non-polar molecule and the change in entropy is smaller and so with its negative $\Delta H$ the polar molecule has a negative $\Delta G$ and can dissolve.

It depends on how polar is polar; phosphine is technically polar, but its actual dipole moment is ... C'mon man!
Such cases aside, partially chlorinated hydrocarbons such as dichloromethane are good at not dissolving in water despite a respectably large polarity.  Having hydrogens attached only to carbon and chlorine as the electronegative element means dichloromethane and similar compounds can't replace the strong hydrogen bonds that they would have to break upon dissolving in water.


text:
Is there an algorithm that takes atom coordinates and produces VSEPR geometries?
That is, you would give the coordinates of an atom, the coordinates of four (for example) other atoms, and it would return "tetrahedral" or "square planar" or whatever.
I'm currently writing a function that does just that, but (1) I feel like I'm making a lot of arbitrary decisions in the angle tolerance cutoffs I am using, and (2) I'm not sure it's very efficient the way I am doing it.
I have searched but haven't found a prior implementation of this - is there one?

Quantum chemistry software usually exploits molecular symmetries to speed up computations. To do so they need some way to determine the point group for a given set of atoms and their coordinates. So this is basically what you are asking about, just that get something like $D_{4h}$ instead of "square planar", or $T_d$ instead of "tetrahedral".
I am not sure whether this is actually used by the mentioned software, but it is quite simple and efficient if you only want to test for certain point groups:

Each symmetry operation can be represented as a $3\times 3$ matrix, for example mirroring in the $xy$ plane is

\begin{equation}\sigma_{xy} = \begin{pmatrix}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & -1
\end{pmatrix}\end{equation}

We can represent the atom coordinates in a $3\times N$ matrix, where $N$ is the number of atoms. The molecule needs to be properly oriented (the center of mass should be in the origin of the coordinate system and the principle axes of inertia should coincide with the $x$, $y$ and $z$ axes).
For example $\ce{H2}$ (or any other homonuclear diatomic):

\begin{pmatrix}
 0 & 0 \\
 0 & 0 \\
 1 & -1
\end{pmatrix}

To get the transformed coordinates according to a certain symmetry operation, we just apply the matrix of the symmetry operation to the coordinates matrix.

\begin{equation}\begin{pmatrix}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & -1
\end{pmatrix}\times
\begin{pmatrix}
 0 & 0 \\
 0 & 0 \\
 1 & -1
\end{pmatrix}=
\begin{pmatrix}
 0 & 0 \\
 0 & 0 \\
 1 & -1
\end{pmatrix}
\end{equation}

We can then compare the resulting set of coordinates with the original one. Since for $\ce{H2}$ both atoms are the same, we can interchange the 2 columns and see, that we have identical matrices. Thus, $\sigma_{xy}$ is a valid symmetry element of our input coordinates. At this point you may implement a tolerance threshold, by allowing the norm (distance) between two "same" points being larger than 0.
The set of found symmetry operations will then directly give us the point group of that molecule (this information is available in character tables). So we can check for a certain point group by checking whether the symmetry operations of that point group can be found in our input geometry. This however means we need to define which point groups we want to test for.

There are also flow charts (available in books about molecular symmetry or via your favorite search engine), telling you which symmetries elements to check for and then guide you to the point group.
I am not aware of a more general approach. If someone knows more, please extend.


text:
The delocalised nature of hyperconjugation
I have always thought I understood hyperconjugation well until recently, in my post on Regioselectivity of bromination of alkenes, I began to question my understanding of this concept. Hyperconjugation is essentially an MO concept which can be generalised to include any sort of intramolcular orbital interaction between one set of occupied molecular or atomic orbitals with another such set of unoccupied orbitals, belonging to adjacent atoms or groups of atoms in a molecule. A most excellent introduction to this concept can be found here. This is not the IUPAC definition, but anyway, the emphasis of my question is not on the definition so there is no point being pedantic here. This term is most commonly evoked to explain the mechanism by which alkyl groups "donate" electron density to stabilise carbocations.
I am aware that there have already been multiple posts, with good answers, on this concept:

What is hyperconjugation?
Stabilisation of carbocations through hyperconjugation
Inductive effect and hyperconjugation - one elephant, different parts?
Hyperconjugation involving other bonds

In my post which discusses the nucleophilic attack on the cyclic bromonium ion intermediate, I was enlightened by Martin through one of his comments:

I am certain there is hyperconjugation, as there already was hyperconjugation in a substituted alkene. There probably is always something to which we refer to as hyperconjugation. You are thinking in localised orbitals, which is only one viewpoint...

Just like he said, my conception of hyperconjugation had always been how Ron frequently describes it as - a localised phenomena. Specifically, in the context of stabilising carbocations, I see it as an interaction between a $\sigma$ bonding MO of the $\ce {C-H}$ or $\ce {C-C}$ bond with the empty p orbital of the carbocation. However, this comment made me think about the delocalised nature of these interactions. MO theory is supposed to be based heavily on delocalised interactions, yet we are considering rather localised interactions here. Well, it is not as localised as a 2-centre 2-electron bond, but still it is a rather localised picture. I would not like to contend this. But... I would just like to pose the following related questions below:
If hyperconjugation should be more seen as a more delocalised phenomena, would distance and geometry play such important factors? Please take note of my examples below and pay attention to them as you attempt a response at my question.
For example, if I have a tert-butyl group bonded to a carbocationic centre, would there also be (significant enough) hyperconjugative interactions between the $\ce {C-H}$ $\sigma$ bonding MOs with the empty p orbital of the carbon atom. Note that by looking at the more localised picture, such as that given by Ron, we would likely only expect the $\ce {C-C}$ bonds to be significant hyperconjugative donors. Additionally, specifically, in my case of the bromonium ion, would there be significant hyperconjugative interactions between the bonding MOs of the $\ce {C-C}$ and $\ce {C-H}$ bonds of the adjacent alkyl groups with the low-energy $\ce {C-Br}$ $\sigma$* LUMO?
Any help is much appreciated!

This post seems kinda unclear to me.

I'm not sure I understood your question correctly, especially what you mean by delocalized phenomenon. Delocalization occurs between atomic orbitals that have a similar geometry, imo, it does not make sense to talk about delocalization between molecular orbitals. Anyways, distance and geometry does play major role between orbital interactions

@GwendalGrelier Maybe you can try to answer the 2 example-based questions I have highlighted regarding bromonium and t-butyl? I think those questions are quite clear right?

Though your question is written in a slightly confusing fashion and seems to assume a simplistic view of the concept of hyperconjugation, I will try to answer to "If hyperconjugation should be more seen as a more delocalized phenomenon, would distance and geometry play such important factors?".  IT DEPENDS on the specific case. However, from fundamental principles, distance and geometry are the critical factors in all types of interactions involving charge transfer, including hyperconjugation, irrespective of the magnitude of the interaction in a particular case.[1] Delocalization of electrons itself is caused both by distance and geometrical factors. Orbital sizes are defined finitely; thus correct donor-acceptor distances are crucial for orbitals to be able to overlap or hybridize. In most cases, a proper orbital distance (assuming an adequate symmetry, see below) is proportional to the stabilization energy that is achieved, with the concomitant delocalization of the electron or electrons (double hyperconjugation).[2] Of course, geometry (topology, technically) is just as crucial, because if the donor and acceptor orbitals are not allowed to overlap in space by symmetry, then no 'hyperconjugative stabilization' is likely to occur (perhaps other types of factors, stabilizing in some cases, such as exchange repulsions from the periphery, and/or Coulombic factors, could, directly or indirectly, create the illusory picture that the system is somehow stabilized hyperconjugatively, even though no geometrically correct orbital matching was even there. This all, adds to the complexity of dissecting energy contributions accurately, even for 'simple' molecular systems). Going back to the dynamic nature of hyperconjugation, and to your question on the importance of the degree of localization or delocalization in defining the mere concept, it was found that hyperconjugative and not steric reasons are behind the preference of ethane to adopt a staggered conformation over an eclipsed one;[3] it was found explicitly that vicinal hyperconjugation (σCH–σCH*)  and not geminal hyperconjugation (σCC–σCC*)  factors lead to the staggered equilibrium structure of ethane. On the other hand, and as a contrasting example, it is geminal hyperconjugation which is linked to the low ring strain of cyclopropane, and not vicinal.[4] The views expressed in ref. 3 are, however, disputed by Bickelhaupt and Baerends.[5] Based on these two examples, it seems as if tight, or strained hydrocarbon species in space, and hindered small systems, display a more "localized" type of hyperconjugation, semantically speaking, hence favoring geminal hyperconjugation, whereas relaxed systems may prefer vicinal hyperconjugation. These are just two basic examples I could pick out from the literature and which you can use with those you already asked and were answered by Gwendal to increase your intuition about the phenomenon, but eventually, you will realize that these stabilizing interactions are not detectable quantities, and thus it is impossible to define them or generalize cases wholly from basic concepts or from hardcore quantum chemical methods, at the moment. Keep in mind that hyperconjugation is a highly controversial topic, all the way from the days of Mulliken and Dewar.[6–8] A comprehensive review may help out.[9]

References

Solà, M. Why Aromaticity Is a Suspicious Concept? Why? Front. Chem. 2017, 5, 22. DOI: 10.3389/fchem.2017.00022.
Wu, J. I.; Schleyer, P. v. R. Hyperconjugation in hydrocarbons: Not just a “mild sort of conjugation”. Pure Appl. Chem. 2013, 85 (5), 921–940. DOI: 10.1351/PAC-CON-13-01-03.
Pophristic, V.; Goodman, L. Hyperconjugation not steric repulsion leads to the staggered structure of ethane. Nature 2001, 411 (6837), 565–568. DOI: 10.1038/35079036.
Inagaki, S.; Ishitani, Y.; Kakefu, T. Geminal Delocalization of .sigma.-Electrons and Ring Strains. J. Am. Chem. Soc. 1994, 116 (13), 5954–5958. DOI: 10.1021/ja00092a052.
Bickelhaupt, F. M.; Baerends, E. J. The Case for Steric Repulsion Causing the Staggered Conformation of Ethane. Angew. Chem. Int. Ed. 2003, 42 (35), 4183–4188. DOI: 10.1002/anie.200350947.
Mulliken, R. S. Intensities of Electronic Transitions in Molecular Spectra IV. Cyclic Dienes and Hyperconjugation. J. Chem. Phys. 1939, 7 (5), 339–352. DOI: 10.1063/1.1750446.
Mulliken, R. S.; Rieke, C. A.; Brown, W. G. Hyperconjugation. J. Am. Chem. Soc. 1941, 63 (1), 41–56. DOI: 10.1021/ja01846a008.
Dewar, M.; Schmeising, H. A re-evaluation of conjugation and hyperconjugation: The effects of changes in hybridisation on carbon bonds. Tetrahedron 1959, 5 (2-3), 166–178. DOI: 10.1016/0040-4020(59)80102-2.
Alabugin, I. V.; Gilmore, K. M.; Peterson, P. W. Hyperconjugation. Wiley Interdiscip. Rev.: Comput. Mol. Sci. 2011, 1 (1), 109–141. DOI: 10.1002/wcms.6.

Some paragraphing might be helpful here.

I really appreciate your very detailed and well-substantiated answer. It has definitely helped me to better understand this phenomenon.

I'll try to answer your question through the exemples you give us.
Where are the electrons ?
If you look at the hybridized orbitals making the C-H and the C-C bonds, you can see that the more the bond is polarized, the worse electron donor you have. It's quite understandable since if the bond is polarized, the electron density will be closer to the heteroatom but further to the carbocation that needs stabilization.

Usually the donor ability goes in this order increasingly:
C-F < C-O < C-N < C-H < C-C
Hyperconjugation from C-C to carbocation
Following the "rule" stated before, the stabilization of a carbocation will be best with an adjacent C-C bond. And this is in accordence with what we learn in school since the more alkyl groups nearby, the more stable is the carbacation. Not only because of the increased overall electron density (donor inductive effect) but also because of the hyperconjugation. This is also the base of some reactivity (1,2-C-C bond migration)
Hyperconjugation in the bromonium case
We can easely extend the thing we just saw for the exemple of a bromonium intermediate. In fact it is important to dissociate the cases of the bromonium and the transition state for the bromonium openning.
The bromonium has two bonds with a carbon atom. One could expect the opening reaction to occur in competition between these two sites. We can see that the C-Br bond (Scheme 2, left) can be aligned with a C-C bond, so if we take a look at the molecular orbitals, we will see that this filled sigma C-C bond can interact with the empty C-Br sigma*. Thus creating a destabilization of the C-Br bond (filling a anti-bonding molecular orbital destabilizes the bond) trough an hyperconjugation intercation. But if we look at the second C-Br bond (right) we see that there are no such interaction since there are no filled molecular orbital which can interact with the sigma* of the C-Br bond.

In term of reactivity, the destabilization of the left-hand C-Br bond will favor the breaking of this  bond instead of the other one.

Ok thanks for the insights that you have provided. Could I just clarify: do you mean to say that hyperconjugative interactions do not really occur at larger distances since the region of donor electron density would be further away and thus, such interactions would be largely insignificant?

Also, by your electronegativity difference argument, wouldn't C-H be a better donor than C-C? And this seems to contradict what you wrote.

@TanYongBoon Yes the distance is a crucial parameter for orbital overlapping, even for hyperconjugation.
Concerning which is the best donnor between C-H and C-C, the tendency I wrote is just an commonly accepted tendency. I would not be able to give an explanation about it. It might be due to the diffusion difference between H orbital compared to C orbital...


text:
Interatomic distances between hydrogens in allene

If $d_1$ and $d_2$ are the interatomic distances between the pairs of hydrogens shown above, what is the relation between $d_1$ and $d_2$?
I think that $d_1$ and $d_2$ might be equal since the terminal hydrogen atoms are perpendicular to each other.

Yes, that’s right.

Propadiene is an allene which has $D_\mathrm{2d}$ symmetry. This dictates that the distances between the hydrogen atoms at the opposite terminal carbons are the same. In mathematical terms:
$$d(\ce{H^a-H^c}) = d(\ce{H^a-H^d}) = d(\ce{H^b-H^c}) = d(\ce{H^b-H^d})$$
Below shows the resymmetrised ($R\gg0.99$) geometry on the GFN2-xTB semi-empirical level of theory. In faint colours the $C_2$ axis are displayed.


text:
Chemical equation that can be balanced in infinite number of ways?
When I was studying for my chemistry test, I found an equation that illustrates how to get water gas (carbon monoxide + hydrogen) out of methane. It is stated like this:
$$\begin{align}\ce{2CH4 + CO2 + H2O &->[\space \Delta ] 3CO +5H2}\tag{1}\\
\end{align}$$
First, I tried to balance it by myself in order not to memorize it but I ended with this equation:
$$\begin{align}\ce{3CH4 +CO2 +2H2O &->[\space \Delta] 4CO +8H2}\tag{2}\\
\end{align}$$
Then, I noticed that if we make the coefficient Of CO2 one, we can get infinite number of ways to balance the equation by changing the coefficient of CH4
We can make a third trial:
$$\begin{align}\ce{4CH4 +CO2 +3H2O &->[\space \Delta] 5CO +11H2}\tag{3}\\
\end{align}$$
Have you seen the pattern? I can conclude my observations in this equation:
$$\begin{align}\ce{nCH4 +CO2 +(n -1)H2O &->[\space \Delta] (n +1)CO +(3n -1)H2}\tag{4}\\
\end{align}$$
I am confused. Which one is true? If all of them are true, why my textbook mentions only one of them? What really happens in reality? Is equation $2$ just a simplification of more complicated equations? Does this have to do anything with the reaction mechanism?
I need to get more intuition about what is happening and the causes of this special feature.

https://chemistry.stackexchange.com/questions/58069/can-a-chemical-reaction-have-two-balanced-equations

Welcome to Chemistry.SE!  Please take a minute to look over the help center and tour page to better understand our guidelines and question policies.

We first note that your reaction (4) can be broken down into two reactions by splitting up those chemical species that depend on $n$ and those that don't: $$\begin{align}\ce{CH4 + H2O &-> CO + 3H2}\\
\ce{CO2 + H2 &-> CO + H2O},
\end{align}$$
so your reactions (1) through (3) are just statements about the relative rates of these two reactions.

Which one is true? If all of them are true, why my textbook mentions only one of them?

I presume none of them are really true, in the sense that the ratio of the reaction rates isn't something as nice as 1:1 or 1:2 or 1:3, etc., and we haven't accounted for side reactions either. But assuming a 1:1 ratio is certainly a useful simplification.

What really happens in reality?

The ratio of the rates will depend on the temperature and pressure and composition of the feed. There will be side reactions that occur. You can perform experimental measurements of the product composition to determine what happens.

Is equation 2 just a simplification of more complicated equations?

Yes.

Does this have to do anything with the reaction mechanism?

No. The overall reaction does not, in general, tell you anything about the reaction mechanism.

The more basic equations are:
$$\ce{CH4 + H2O <=> CO + 3H2}$$
$$\ce{CH4 + CO2 <=> 2CO + 2H2}$$

After reading your answer, I think the general basic equations should be:
$$\begin{align}\ce{(n-1)CH4 +(n-1)H2 &-> (n-1)CO +(3n-3)H2}\tag{5}\\end{align}$$
$$\begin{align}\ce{CH4 +CO2 &-> 2CO +2H2}\tag{6}\\end{align}$$
If we add those two equations, we get equation 4 and each one of them has only one way to balance it. Thank you for this illustration.

@AmerEsmail Your first one is just $(n - 1)$ times CH_4 + H_2 -> CO + H_2


text:
Why are some chemicals packaged in incompatable containers?
I've noticed that some organic chemicals are sold in HDPE containers despite compatibility charts indicating that such containers are not appropriate acetone.

For example, this acetone listing from Walmart above uses an HDPE container, but when incompatible solvents are packaged in HDPE containers the containers can experience a relaxation in the polymer stress from blow molding causing the bottle to contract as shown below.

How can they use incompatible containers and still safely package solvent products?

This will never lead to the bottle or cannister breaking. Also I strongly suspect you only see the effect so clearly because the cannisters were half empty and temperature changes made the outside air pressure compress them. Also empty PE bottles deform permanently in this case.

@Karl, Break? Porbably not. But I have had full bottles of oleic acid and PEG 400 panel. Oleic acid will even diffuse to the outside and oxidized to form a viscous grease. but paneling has little to do with atmospheric pressure. If you look, the bottle in the left most image has no head space and atomospheric pressure does not swing that drastically in a single location. You can leave bottles closed through all kinds of weather and not get anything that looks like paneling.

The shrinkage is called panelling and bottles that are otherwise incompatible are made compatible by a process called fluorination. In fluorination the HDPE is exposed to fluoride gas with precise control in time, temperature and pressure. This exposure converts the outer layers of polyethylene to a highly fluorinated polymer similar to teflon.
$$\ce{-(C2H2)_n-~+ 4 F2~~ \longrightarrow ~~-(C2F4)_n-~ + 4HF}$$
This fluoropolymer coating greatly reduces diffusion of the solvent into the plastic and thus the solvent is unable to soften the plastic, preventing paneling.


text:
Carbon emission in absence of oxygen
When wood is pyrolysed in absence of air is $\ce{CO2}$ emitted.
If yes why is it not combustible?

Wood contains oxygen on its own.

CO$_{2}$ is already oxidized, if that is why you ask. Or, the wood does not have enough oxygen to fully burn it, but what there is (in various forms) is reacted and driven off as gases.

$\ce{CO2}$ is emitted yes. It is not formed in the usual combustion manner however. There a likely two routes to $\ce{CO2}$ formation during wood pyrolysis. Either through the elimination of carboxyl groups e.g., $$\ce{PhCOOH -> Ph + CO2}\text{,}$$ or by a water gas shift reaction: $$\ce{H2O + CO -> CO2 + H2}\text{.}$$ My opinion is that the latter is more likely as there are many more reactions that would liberate $\ce{CO}$ and $\ce{H2O}$ than would yield $\ce{CO2}$.


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Reaction kinetics - Organic chemistry

On an exam in organic chemistry I was asked why the reaction B proceeds faster than reaction A.
I answered that: in reaction B, I- is a better leaving group than Cl-, the reason being it is a weaker base.
The answer our teacher gave after correcting the exam was:
I- is a good nucleophile and at the same time a good leaving group.
I scored 1,5p/2p for my answer.
Are the two answers very different or is there some incorrect information in mine?

Your answer is simply missing something. Yes, I is a better leaving group, therefore an Sn reaction of an iodide proceeds faster than the one of the chloride. But your starting material in both cases is the chloride. You first need to exchange that chloride for an iodide and that only works well if the iodide is a good nucleophile.


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Spectral database in the UV-Vis/NIR and IR regime
I'm searching for a bunch of spectra (some tens), preferably with intense bands in the Vis spectrum, to play around with (calculating chromaticity, etc.).
They should range from

250 nm to 2500 nm (UV-Vis/NIR)
400 cm-1 to 4000 cm-1 (IR)

and should be available as raw data like jdx format or the like.
I allready know the famous NIST and SDBS, but those have large gaps in the desired spectra, if not one completely missing.
Anybody knows databases where you can search by availability of spectral data in a specific regime?


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Wurtz Reaction with acidic alkyl halides
My chemistry book says we cannot use alkyl halides containing acidic hydrogen for Wurtz reaction. Why is this so?

The intermediate is an alkyl sodium, a highly basic species which will be quenched by any proton source. Acidic hydrogens will also react directly with the sodium metal.

The Wurtz reaction involves the reaction of alkyl halide with sodium in dry ether. In this reaction, two molecules of alkyl halide react to form a alkyl-alkyl chain and sodium halide is also produced -
$$\ce{R-X + 2Na + X-R ->R-R + 2NaX} \tag{$\ce{X}$ = halogen}$$
Now, sodium is a strongly electropositive metal, and is above hydrogen in the electrochemical series. Thus it replaces acidic hydrogen wherever it occurs. For example,
$$\ce{CH3COOH + Na ->CH3COO- Na+ + 1/2 H2}$$
So, sodium reacts with the acidic hydrogen present in the starting alkyl halide.
Also, acid-base reactions are, in general, faster than a substitution reaction. If there was no acidic hydrogen, alkyl sodium intermediate would have formed and an SN2 reaction would have taken place.
However, in this case, if any alkyl sodium intermediate is formed, it would accept the acidic hydrogen and convert into an alkane, stopping the reaction.
$$\ce{R- Na+ + H-M -> R-H + NaM}$$
(R = alkyl group, M represents the functional group or chain attached to the acidic hydrogen)


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Reduction of carbonyl group
A compound contains both alpha,beta unsaturated carbonyl group and a normal carbonyl group. Is there any method to reduce the normal carbonyl group keeping the alpha,beta unsaturated carbonyl group intact?

Most methods I can think of  e.g. Borohydrides, LiAlH4, Zn/HCl, Wolff-Kishner will reduce both unless there are features of the molecule  (eg steric hindrance) that make one less reactive.

Look selective Reduction here.


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Would boiling distilled water in a kettle full of limescale increase the hardness of the water?
If I had a kettle which had limescale on it (due to several episodes of boiling hard water), and I would use it to boil distilled water, would that mean that some of the precipate dissolve into the distilled water, making that water harder than originally? If so, by how much?
I understand CaCO3 has very low solubility in normal conditions and requires CO2 to dissolve in higher amounts, but I'm not sure about how much is "low".

Yes, boiling distilled water in such a kettle would eventually dissolve any calcium carbonate deposit. Wikipedia lists the solubility as 0.013 g/L @25 C.

This still seems like a fairly low number? That would be only 5mg Ca. My tap water has around 130mg/L Ca. What about magnesium?

...same result. Magnesium carbonate would eventually dissolve too.

If it precipitated out of solution, it can go back in to solution (barring a chemical reaction that makes it insoluble).

But if solubility of Ca is only 13mg/L, how come my tap water has 130mgCa/L ?

@CanGencer, you give the anion as Ca++ but what is the cation? Calcium bicarbonate,  Ca(HCO3)2,  is far more soluble, 16.6 g/100 ml. See https://en.wikipedia.org/wiki/Solubility_table

Yes that's probably the solution in the water. However when it precipitates it precipates as CaCO3 if I'm right? So how could it dissolve back as Ca(HCO3)2?


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Can catecholamines be quantified by HPLC-MS/MS at high pH?
It seems that most HPLC-MS/MS methods used for quantitation of catecholamines (epinephrine, norepinephrine and dopamine, specifically) operate at low pH (commonly 3.0), and therefore demand the addition of ion pairing reagents if a C18 column is to be employed. From my understand, these conditions are used to improve catecholamine stability, and avoid formation of quinones by oxidation. Fair.
My question is; why are high pH solvents not used (> pKa + 2.0, so pH ~ 10.5), with the addition of antioxidizers, such as EDTA? I understand that this would require MS operation in negative ion mode, but should circumvent need for ion pairing reagents. Might this be possible? Has this been attempted?

As an update - it seems that oxidation of the catechol group is slow (Bonhomme et al, 1990; https://www.sciencedirect.com/science/article/pii/S0196064405822814?via%3Dihub), so perhaps a reducing agent is unnecessary?
Hoping someone can provide some insight..