Potential term in the hydrogen atom Hamiltonian Here is how I saw the Hamiltonian being written in one Quantum Mechanics book:

$$\hat{H} = -\frac{\hbar^2}{2m_\mathrm{e}r^2} \left[\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{\sin \theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial}{\partial \theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2} \right] - \frac{e^2}{4\pi\varepsilon_0}\frac{1}{r}$$

Well, that's all very nice and complicated, but looking at the last term, and seeing the use of $r$ in there,

$$-\frac{e^2}{4\pi\varepsilon_0}\frac{1}{r}$$

creates a small doubt in me - what is this $r$ there?

In electrostatics, this was simply the distance between the two charged particles, and that would mean that this was the distance between the proton and the electron, but isn't the position of the electron non-deterministic? Shouldn't that thing be a probability distribution of the distance instead? And why isn't it squared?

What is the $r$ there?

$r$ is the same in the potential energy term of the Hamiltonian as in the kinetic energy term of the Hamiltonian.

$r$ is the distance from electron to the proton.

  But isn't the position of the electron non-deterministic?

In the generally accepted Copenhagen interpretation of quantum mechanics, the position of the electron is not deterministic. However, expressing the kinetic and potential energy of the the electron at each position in space does not contradict this interpretation. It is only saying, for each point, if the electron is at this point, the energy is _.

  Shouldn't that thing be a probability distribution of the distance instead?

The wave function solutions represent probability amplitude in the Copenhagen interpretation.  The square of the absolute value of the wavefunction represents a probability distribution.

  Why isn't it squared?

Force would be inversely proportion to $r^2$, but potential is inversely proportional to $r$.


How do "basic" amino acids (lysine, arginine) act as base under physiological conditions? Why do we consider amino acids with high side chain pKa's basic, yet with a higher pKa, less of the amino acid will exist in basic form at equilibrium compared to acid form at physiological pH?

What will the concentration of the base be given the high pKa of these amino acid side chians. The answer is that the higher the pKa, the lower the concentration of the base, since with 1) with a higher pKa, then 2) you will have a lower Ka (pKa = - logKa) and therefore 3) a lower concentration of base, due to Ka = the conjugate base (times H3O+) / acid, then the numerator, the conjugate base will be  lowered in concentration with a lower Ka. This can also be confirmed with the Henderson-Hasselbach equation, where pH = pKa + log [A-]/[HA] and since pkA is increasing will make pKa > pH, meaning pH - pKa = a negative number = log [A-]/[HA], where [A-]/[HA] would then equal 10^negative number, meaning the concentration of the acid will be greater than the concentration of the base.

My question is, when you look at titration curves, we know that parts of an amino acid will act more as a base, than an acid, at higher pKa's. Meaning, if you have the amine group of an amino acid, with a pkA of 9-10, it doesn't want to give up it's hydrogen, until the pH is pretty high (10-12). Why are amino acids with high overall pKas considered basic, yet, when you look at a molecule's pKa at a high pKa, less of the molecule will exist as the base at equilibrium (meaning it is wrong to say that a molecule with a high pKA will exist primarily it is basic form)?

I will paraphrase the question:

  Why are the amino acids arginine and lysine called basic? Why are the amino acids glutamic acid and aspartic acid called acidic?

This nomenclature makes sense for situations where the amino acids are dissolved in a non-aqueous solvent. In that case, the side chains will be neutral and have nitrogenous base groups and carboxylic acid groups, respectively.

When the amino acids are dissolved in water at neutral or physiological pH, however, the name does not make sense because they have already acted as base (Arg and Lys) or as acid (Glu and Asp), and are now the conjugate acid and conjugate base, respectively. For glutamic and aspartic acid, it is common to refer to them as glutamate and aspartate (the name of the conjugate base). For arginine and lysine, even when protonated, those names are usually used nonetheless.

  [OP in comments:] So, in physiological situations (aka in the human body), how do arginine, lysine, and histidine act as bases, if they exist in their protonated form, considering how they are often used in enzymatic catalysis (same question asked of acidic amino acids in their deprotonated forms)?

Histidine

Histidine is a common acid/base catalyst. Its pKa is close to 7, so an appreciable fraction is protonated and an appreciable fraction is deprotonated. Often, this amino acid will act as acid and as base in different steps of the same mechanism (e.g. RNase).

Lysine

Lysine is not a typical acid base catalyst. It has roles in binding negatively charged ligands (such as ATP), acting as a "positive charge on a stick". It also acts as a nucleophile, forming e.g. Schiff's base with aldehydes and ketones. Lysine's pKa in the context of a folded protein varies considerably (see e.g. https://www.pnas.org/content/108/13/5260), so in some contexts, it will be an efficient acid/base catalyst.

Arginine

Arginine is not a typical acid base catalyst. Like lysine it has roles in binding negatively charged ligands but different from lysine, it can act as a bidendate hydrogen bond donor (e.g. making a nice salt bridge with aspartate or glutamate with two hydrogen bonds). It also has a role in cation-pi interactions, stacking its planar guanidinium group on aromatic rings. This web page has more on the functional role of arginine: http://www.russelllab.org/aas/Arg.html (and there are similar pages for the other amino acids on that site)


How do electron donor groups affect the polarity of an O-H bond in alcohol? My textbook (Chapter Alcohols) , says:

  An electron releasing group ($\ce{-CH3}$, $\ce{-C2H5}$) increases the electron density on oxygen tending to decrease the polarity of $\ce{O-H}$ bond.

Now, my question is that if the electron density increases on oxygen, then shouldn't the polarity of the bond increase? Why does it decrease instead?

Oxygen tries to draw electron density from the two moieties it's bound to, thus making both bonds more polarized than the normal carbon-carbon bond. Some functional groups offer more resistance than others to this polarization depending on their electronegativity. For this reason, if one of the groups "puts up more of a fight" than the other, oxygen will draw more electron density from the weaker one.

In your case, hydrogen is always the same since we're always talking about alcohols. Alkyl substituents donate electron density to the oxygen atom, so it is less eager to draw electron denstity from the hydrogen bond and will make it less polar. Change the alkyl for a primary alkyl chloride for example and the effect will be reversed.


"Drinking water" plastic - can I store beer in this? This is an outdoors & chemistry question.

I have an old and functional water bladder, shown below. On it, there are markings that coffee, beer, wine, hot drinks etc. must NOT be stored inside, only water.

While I realize that wine is acidic, damaging the bladder and becoming poisonous, beer seems to me quite close to water. Furthermore, I indend to test it with weisbeer, which does not contain such a large amount of gases as to explode.

Would be safe for me to store non-fizzy beer for up to a day every day in a "do-not-store-beer-inside" water bottle and then safely drink it?

I doubt that this is a chemistry problem. Food safe plastics which can hold wine and coffee are so abundant and cheep, there is no reason that they went to the trouble of finding one which is not suitable for these, but suitable for water, and made the bladder out of it.

All beverages beside water are a good growing medium for one microorganism or another. And the bladder is hard to get clean, the sucking straw even more so. So, it looks to me that they are warning you to not put in anything beside water because you'd risk starting your own bacterial or fungal colony. Beer is as bad as other drinks - don't store it there.


How much oxygen is formed in this first order reaction? We have a reaction
$$\ce{2N2O5 -> 4NO2 + O2},$$
which has a rate constant $k=6,2\cdot10^{-4}/\mathrm{min}$ at temperature $T=318.15\ \mathrm{K}$. How much oxygen is formed in litres when $1.00\ \mathrm{mol}$ of $\ce{N2O5}$ is at $318.15\ \mathrm{K}$ in $104\ \mathrm{kPa}$ pressure for $20$ hours. Correct answer: $6.67\ \mathrm{l}$

My attempt:

For gases we use partial pressures to measure rates. So using equation
$$\ln P_t=-kt+\ln P_0$$
we get $P_t=49421.76426\ \mathrm{Pa}$. This means that oxygen has a partial pressure of $P_{\ce{O2}}= \frac{104000\ \mathrm{Pa}-49421.46426\ \mathrm{Pa}}{5}=10915.64715\ \mathrm{Pa}$.

Now we say $n_{\ce{O2}}=x$ so according to the reaction $n_{\ce{N2O5}}=1-2x$. If the total volume is $V$ we can type two equations:
$$
\begin{align}
P_t&=\frac{n_{\ce{N2O5}}RT}{V}\\
P_{\ce{O2}}&=\frac{n_{\ce{O2}}RT}{V}
\end{align}
$$
Solving for $x$ and $V$ yields $x=0.153195\ \mathrm{mol}$ and $V=0.0371657\ \mathrm{m^3}$. Now calculating the volume for oxygen at $104\ \mathrm{kPa}$ gives
$$
V_{\ce{O2}}=\frac{0.153195\ \mathrm{mol}\cdot 8.3145\ \mathrm{\frac{Pa\ m^3}{K\ mol}\cdot318.15\ \mathrm{K}}}{104000\ \mathrm{Pa}}=3.900828\cdot 10^{-3}\ \mathrm{m^3}\approx3.9\ \mathrm{l}
$$
My answer is somehow incorrect but what I am doing wrong here?

If we assume that the reaction takes place in a large amount of inert gas (like nitrogen) and that the reaction is first order in $\ce{N2O5}$, we can reach the numerical answer given by neglecting the volume change (the reaction increases the number of particles, but if there is a large amount of inert gas, the volume change will have little effect on partial pressures). The path given by Maurice is fine (no reason to figure out the partial pressure first). If you convert the first-order rate constant into a half-life (18.6 h), you can quickly estimate that a bit more than half of the reactant will be gone after 20 h.

If we assume, like the OP, that you start with pure $\ce{N2O5}$, the rate law still applies, but the integrated rate law does not (because while the reaction goes on, dilution also happens through the volume increase). You could numerically integrate, or try to find the integrated rate law that includes the dilution effect. Overall, you would expect the reaction to go slower, i.e. produce a smaller amount of oxygen. Because the final volume is larger, the partial pressure will be lower (compared to the inert gas case) by even more.

The value calculated by the OP is incorrect for either case because it assumes an integrated rate law that does not apply for the assumed conditions.


A comprehensive list of theoretical approximations that are used in computational chemistry I am having trouble recognizing all the approximations that are used in computational chemistry. I would like to start an answer list (similar as to the list of resources for learning chemistry) that addresses this question.

I will try to answer this question myself. I am hoping that anyone who is knowledgeable in the topic will correct and/or contribute. I am planning to start from general (e.g. Born Oppenheimer, LCAO) to specific (e.g. pseudopotential, functional). I am also planning to include why this approximation is necessary.

The goal of computational chemistry is to obtain the properties of a system. This is done by solving Dirac's Equations.
Treating particles as point particles with mass
In most computational software, particles are treated as points with some mass. Neutrons and protons may be lumped into a nucleus. However, this is not true in all cases as pointed out in this question: Does computational chemistry include neutrons in ground state calculations? .
Electrons, protons, and neutrons are not simply point particles with some mass and charge. The theory and forces can get quite complex. I haven't been able to find journals that address the effect of this approximation. But I will try to post something if I come across it.
Neglecting Relativistic Effects
Wien2k has a nice summary of relativistic effects that need to be considered [8]. The relativistic effects that need to be included are:

Mass-velocity correction
The Darwin term
Spin-orbit coupling
Indirect relativistic effect

However, relativistic effects usually become important for elements further down the fifth row in the periodic table. This is because relativistic effects are dependent on nuclear charge. The velocity of the electron as well as spin orbit coupling increases as the nuclear charge increases.
This does not mean that relativistic effects do not affect "light" atoms. A good example that shows this is the sodium doublet which is a result of spin-orbit coupling. Nevertheless, solving Schrodinger's equation is enough in many cases.
Most computational chemistry software have the option of scalar relativistic calculations. The scalar relativistic technique was developed by Koelling and Harmon. These are calculations that include the Darwin term and the mass velocity correction. According to Koelling and Harmon's paper:

We present a technique for the reduction of the Dirac
equation which initially omits the spin-orbit interaction (thus keeping spin as a good quantum number), but retains all other relativistic kinematic effects such as mass-velocity, Darwin,and higher order terms.[9]

Some offer fully relativistic calculations, which include all (or most) relativistic effects. But these are rarer to find, and are only available for certain cases.
So why are relativistic effects neglected? Well, relativistic effects are small for light molecules and relativistic calculations are expensive

This is so because relativistic calculations need self-consistent solutions of about twice as many states as non-relativistic ones. [10]

Thus, scalar relativistic calculations offer a nice middle ground between efficiency and accuracy.
The Born-Oppenheimer Approximation
I am not sure what motivated Born and Oppenheimer to use this approximation. The seminal paper is in German.[1] It appears that the motivation was to simplify the Hamiltonian. According to the introductory course at MIT,[2]

it allows one to compute the electronic structure of a molecule
without saying anything about the quantum mechanics of the nuclei

And according to Wikipedia,[3] it reduces the amount of computations:

For example, the benzene molecule consists of 12 nuclei and 42
electrons. The time independent Schrödinger equation, which must be
solved to obtain the energy and wavefunction of this molecule, is a
partial differential eigenvalue equation in 162 variables—the spatial
coordinates of the electrons and the nuclei.

As many approximations, it does not hold true in all scenarios. Cases where the Born-Oppenheimer approximation fail are:[4]

ion-molecule, charge transfer, and other reactions involving obvious
electronic curve crossings

Qualitatively, the Born-Oppenheimer approximation says that the nuclei are so slow moving that we can assume them to be fixed when describing the behavior of electrons. Mathematically(?), the Born-Oppenheimer approximation allows to treat the electrons and protons independently. This does not imply that the nuclei and electrons are independent of each other. In other words, it does not mean that the nuclei are not influenced by the motion of electrons. The nuclei still feel the motion of the electrons. In addition, the Born-Oppenheimer approximation does not say that the nuclei does not move. It only means that when describing the motion of electrons, we assume that the nuclei are fixed.
No Analytical Solution to Dirac/Schrodinger Equation
Unfortunately there is no analytic solution to the Dirac equation for any atom that has more than one electron even after the Born-Oppenheimer approximation (a list of quantum-mechanical problems that have an analytical solution is available on Wikipedia[5]). Many texts state that the reason as to why the Schrödinger equation is not exactly solvable for more than one electron is due to the Coulomb repulsion between electrons.[6]
However, this is not entirely true. A counterargument is Hooke's atom. The Hamiltonian for Hooke's atom has an Coulomb electron-electron repulsion term. However, it has an exact solution for more than one electron under certain circumstatnces.[7]
The true reason as to why the Schrödinger equation is not solvable for multi-electron atoms is due to the fact that the motion of electrons cannot be decoupled from each other. In other words, the Hamiltonian is not separable for a multi electron system. If we were to get rid of the electron-electron Coulomb repulsion, the motion of the electrons can be decoupled. This may be the reason as to why the electron-electron Coulomb repulsion (a.k.a. electron correlation) is used as the reason why the Schrödinger equation is not exactly solvable.
From non-interacting to the real thing
Since the Dirac equation cannot be solved analytically, we must make  models that are solvable and add approximations to it. These approximations are further refined to get more accurate results.
The most simple model (and the foundation for computational chemistry) is the system of non-interacting electrons. As the name suggests, the electrons do not interact with other electrons. This allows us to write the Hamiltonian for all electrons as the sum of one-electron Hamiltonians.
The one electron Hamiltonian consists of a kinetic energy term and a potential energy term.
$$\mathcal{H}=\sum^N_i \left(\frac{\hbar^2}{2m_i\nabla^2 }+V_i\right)$$
The solutions for the non-interacting system of electrons are analytical. And they give a starting point for other calculations. Hartree-Fock, DFT, and solid state physics use this simplified model.
Note: it appears that there are several terms that say something similar to this. There is the independent electron approximation and the central field approximation. I'll dig into literature to see what are the differences between these three terms.
Variations of the non-interacting system of electrons
There are several variations of non-interacting electron systems. In all these, the electrons do not interact with each other. What sets them apart from each other is the potential energy term of the Hamiltonian $V_i$. This potential energy term is often called the effective potential.
Free electron model
This model is used as the starting point in solid state physics. The free electron model describes the behavior of valence electrons in a metal or semiconductor. Here, the potential $V_i$ is equal to 0 for all electrons. It does not feel the effect of other electrons or nuclei. The wavefunctions of this model are plane waves.
Nearly free electron model
The nearly free electron includes a weak periodic potential. In other words $$V_i(r)=V_i(r+k)$$ It is used to describe periodic systems such as ideal crystals. The wavefunctions of the nearly free electron model are Bloch waves. Bloch waves are plane waves that are multiplied by a periodic function.
Hartree method
In the Hartree method, the potential considers repulsion between two electrons and attraction to the nucleus. In this model,

Hartree assumed that each electron moves in the averaged potential of
the electrostatic interactions with surrounding electrons [11]

Hartree replaced the electron-electron interaction with an effective potential that only depended on the coordinates of the $i$th electron. This effective potential describes an electron interacting with an electron cloud.
From one-electron orbitals to the multielectron orbital
As mentioned above, the total Hamiltonian can be approximated as the sum of one-electron Hamiltonians for non-interacting electrons. However, the wavefunction has to satisfy antisymmetry.
Exact exchange
Electron correlation
Some define electron correlation as everything that the Hartree-Fock method leaves out.
Electron correlation: DFT edition
Electron correlation: Post-Hartree-Fock edition
Periodic systems and pseudopotentials

References

M. Born and R. Oppenheimer, Ann. Phys. 1927, 389, 457–484.
doi: 10.1002/andp.19273892002
Born Oppenheimer Approximation. Open Courseware MIT:Introductory Quantum Mechanics. Fall 2005. Section 12 Lecture. (https://ocw.mit.edu/courses/chemistry/5-73-introductory-quantum-mechanics-i-fall-2005/lecture-notes/sec12.pdf)
(https://ocw.mit.edu/courses/chemistry/5-73-introductory-quantum-mechanics-i-fall-2005/)
Born-Oppenheimer approximation. Wikipedia (https://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation)
L. J. Butler, Annu. Rev. Phys. Chem. 1998, 49, 125-71.
PMID: 15012427
doi: 10.1146/annurev.physchem.49.1.125
List of quantum-mechanical systems with analytical solutions. Wikipedia
LibreTexts: 9.1: The Schrödinger Equation For Multi-Electron Atoms
Hooke's atom. Wikipedia
Summary of relativistic effects. Wien2k  (http://www.wien2k.at/reg_user/textbooks/WIEN2k_lecture-notes_2013/Relativity-NCM.pdf)
A technique for relativistic spin-polarised calculations. Journal of Physics C: Solid State Physics, Volume 10, Number 16
(http://iopscience.iop.org/article/10.1088/0022-3719/10/16/019/meta)
The Scalar Relativistic Approximation. Takeda, T. Z Physik B (1978) 32: 43. doi:10.1007/BF01322185 (http://link.springer.com/article/10.1007/BF01322185)
Density Functional Theory in Quantum Chemistry. Tsuneda, T. 2014. ISBN: 978-4-431-54824-9. Page 36.


Is soap the natural form of Sodium Lauryl Sulfate? I was told that "Sodium Tallowate, sodium cocoate and sodium palm kernelate are the 'natural' forms of SLS".  Is this true?  I thought "natural" SLS would be SLS derived from a natural product such as coconut oil.

Conventional soaps are the sodium salts of long chain carboxylic acids (aka fatty acids). These are formed by the hydrolysis of triglycerides, a process known as saponification. This process breaks the triglyceride at its ester linkages and yields three fatty acid molecules and glycerol molecule.

As the aforementioned wikipedia article indicates, sodium lauryl sulfate is formed by refunctionalising the fatty acid lauric acid. In this case, the $\ce{R-OOH}$ (carboxylic acid) head group has been replaced with an $\ce{R-OH}$ (alcohol) group, which is in turn replaced by $\ce{R-OSO3^{-}}$ (sulfate). Here I am using $\ce{R}$ to refer to the long alkyl chain of the molecule.

I'm not sure whether sodium lauryl sufate occurs naturally, however the other molecules you mention are different in the sense that they have carboxylate ($\ce{R-COO^{-}}$) head groups whereas sodium lauryl sulfate has the sulfate head group ($\ce{R-OSO3^{-}}$). Both classes of molecule will act as surfactants as they have polar head groups which are hydrophilic, and nonpolar alkyl tails which have an affinity for fats and oils.

Lauric acid derivatives spotter's guide

(These were drawn using the nifty ChemFig package for (La)TeX.)

The last two (sodium laurate and sodium lauryl sulfate) are sodium salts. Both carboxylic acids ($\ce{R-COOH}$) and hydrogen alkyl sulfates ($\ce{R-OSO3H}$) are fairly willing to ionise to give the corresponding carboxylate ($\ce{R-COO^{-}}$) or sulfate ($\ce{R-OSO3^{-}}$). Sodium ions happily associate with these anions, balancing the negative charge.


What are overlap integrals? I came across something called overlap integrals. I could understand that it is something about the overlapping of orbitals. What is it actually?

Reduce the problem to one dimension first. The overlap integral between two functions, $f = f(x)$ and $g = g(x)$, is defined by

$$S_{fg} = \int_{-\infty}^\infty [f(x)]^*g(x)\,\mathrm{d}x$$

where $^*$ denotes complex conjugation. $S$ is the common symbol for the overlap integral; the subscript $fg$ indicates that it is the overlap between $f$ and $g$.

So, for example, let's try a simple case: let

$$f(x) = g(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \exp\left(-\frac{m\omega x^2}{2\hbar}\right)$$

(the ground state of the quantum harmonic oscillator). This function is real, so the complex conjugation doesn't do anything. We have

$$\begin{align}
S_{fg} &= \int_{-\infty}^\infty \left(\frac{m\omega}{\pi \hbar}\right)^{1/2} \exp\left(-\frac{m\omega x^2}{\hbar}\right)\,\mathrm{d}x \\
&= \sqrt{\frac{m\omega}{\pi\hbar}} \int_{-\infty}^\infty  \exp\left(-\frac{m\omega x^2}{\hbar}\right)\,\mathrm{d}x \\
&= \sqrt{\frac{m\omega}{\pi\hbar}} \sqrt{\frac{\pi\hbar}{m\omega}} \\
&= 1
\end{align}$$

(I used a standard integral, $\int_{-\infty}^\infty \exp{(-ax^2)}\,\mathrm{d}x = \sqrt{\pi/a}$). So this tells you that the overlap of this function with itself is equal to 1. That's not very interesting, because if you substitute $g(x) = f(x)$ into the original definition, you'll find that that is simply the normalisation condition for a wavefunction. I just wanted to use it as an illustration.

In general, the overlap between two normalised states ranges from 0 (when they are orthogonal) to 1 (when they are identical). This applies to three dimensions as well.

Of course, the notion of overlap is usually talked about in a three-dimensional context. So, all you have to do is to generalise the definition to three dimensions. Let's say we have two functions $\psi_1(x,y,z)$ and $\psi_2(x,y,z)$. Then we simply have

$$S_{12} = \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty \psi_1^*\psi_2 \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$

or in more concise notation

$$S_{12} = \int \psi_1^*\psi_2 \,\mathrm{d}\tau$$

where $\mathrm{d}\tau$ is the volume element (note that Wikipedia denotes it as $\mathrm{d}V$) and the limits of integration are assumed to be over all space.

This won't get you anywhere with atomic orbitals, though. Orbitals are frequently expressed in terms of spherical coordinates. So, you need to know what the volume element in spherical coordinates is, and that is:

$$\mathrm{d}\tau = r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$$

(note again that Wikipedia can be confusing sometimes, because the symbols used for spherical coordinates are different: $\theta$ and $\phi$ in chemistry have exactly the opposite meaning from $\theta$ and $\phi$ in mathematics.)

The limits of the integral are still "over all space", but in spherical coordinates this is expressed differently. $r$ takes values from zero to infinity, $\theta$ from $0$ to $\pi$, and $\phi$ from $0$ to $2\pi$. So:

$$S_{12} = \int\limits_0^{2\pi} \int\limits_0^\pi \int\limits_0^\infty \psi_1^*\psi_2 r^2 \sin\theta \,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$$

All you need to do now, is to find the form of the function, plug it into the integral, and evaluate the integral. Warning: That's best done by a computer.


Why is water's IR spectra indistinguishable from the IR spectra of a solution of glucose in water? Anhydrous glucose has an absorption spectra (glucose IR spectra) that is quite different from the spectra of water (Liq. water absorption spectra). However, when a solution of glucose is prepared in water, the absorption spectra is almost indistinguishable from the absorption spectra of water alone. Is this because glucose forms additional bonds when in water? Pointers to links that explain this in detail is also appreciated. Most links I searched on Google just stated this as an observation without explaining the underlying reason.

I see the spectra in the following way:

The sample and measurement: glucose in water and IR absorption measurement

Considering free glucose molecule, we should expect the vibrations from $C-H$, $O-H$, $C-O$ and $C-C$ bonds and other associated vibrations.
The approximate vibrational frequencies are :
a) $C-H$ = ~2800 - 3300 $cm^{-1}$ (stretching)
b) $O-H$ = ~3800 - 3200 $cm^{-1}$ (stretching)
c) $C-O$ = ~1200 - 1030 $cm^{-1}$ (stretching)
d) $C-C$ =  usually not good for interpretation
e) $C-H$, $CH_{2}$ and $O-H$ bend(in plane) = 1500-1200 $cm^{-1}$
and now considering water,
the $O-H$ stretching band  spectrally overlaps with the bands of glucose (~3750 - 3200). The combination band (~2300) and scissor mode (~1600) are also relatively broad and strong. Due to this spectral overlap and also larger infrared  absorption cross-section of water, glucose is not clearly identifiable in solution.

Instrument and spectral quality : For obtaining a high resolution spectra with good s/n ratio the instrument should be good (FT-IR) and long exposure is needed.
The spectra shown by @user1155386 has the $x-axis$ as nanometers (not wavenumbers  ! ), indicating that the spectral resolution is low. It is noisy showing that exposure time can be improved (the exposure time shown there is only 10 milliseconds).

However, still it would be difficult (but not impossible) to differentiate glucose solution and water. The main hurdle would be get a good spectra and the reason 1. mentioned above.
Additionally, I should mention Raman spectroscopy is better for aqueous solution measurements.


What is the mechanism for oxidation of ketones by the Popoff's rule? Ketones are relatively much harder to oxidise, but they do undergo oxidation reactions at extreme temperatures. Our teacher also told us about Popoff's rule which states that, on oxidation of unsymmetrical ketones, the $\ce{>C=O}$ group remains with the smaller alkyl group. Could you please give me a hint as to how this happens, and the probable reaction mechanism for this?

Also, please give the complete balanced equation when butan-2-one is oxidized to a carboxylic acid.

I am uncertain about the "Popoff's" rule you mention. There are two reactions that can oxidize ketones, and one seems to follow the behavior you are suggesting, but does not form a carboxylic acid. The other reaction does form carboxylic acids, but is more complex.

The Baeyer-Villager oxidation is an oxidation of ketones to esters using a peracid in the presence of a mild base:

$$\ce{CH3COCH3 + RCO3H->[\ce{Na2HPO4}] CH3CO-OCH3 + RCOOH}$$

The mechanism involves the fragmentation of one of the $\ce{C-CO}$ bonds. Unsymmetrical ketones fragment in a predictable pattern, but not always that the carbonyl remains with the smaller group. The fragment that would be a more stable carbocation (even though carbocations are not formed in this reaction) is the one to move. The migratory aptitude is tertiary alkyl > cyclohexyl > secondary alkyl, aryl > H > primary alkyl > methyl . For exampl, with 2-methyl-3-octanone, the isopropyl group moves because it is attached via a $2^\circ$ carbon:

$$\ce{(CH3)2CHCOCH2CH2CH2CH2CH3 + RCO-O-OH ->[\ce{Na2HPO4}] (CH3)2CH{\bf O}COCH2CH2CH2CH2CH3 + RCOOH}$$

Nitric acid chews ketones apart into two carboxylic acids. Concentrated $\ce{KMnO4}$ in acid also does this. These reactions are a little bit harder to find information on, since they tend to be considered uncontrollable.

The reaction goes through a series of oxidations from ketone to $\alpha$-hydroxyketone to cleaved acids through several enol intermediates.

$$\ce{RCH2COCH2R <=>[\ce{HNO3}] RCH=C(OH)CH2R ->[\ce{HNO3}] RCH(OH)COCH2R}$$
$$\ce{RCH(OH)COCH2R <=>[\ce{HNO3}] RC(OH)=C(OH)CH2R ->[\ce{HNO3}] RCO}$$

If the ketone is unsymmetrical, there is no guarantee that is will cleave predictably. For example, 2-butanone could cleave into propanoic acid and carbon dioxide or two equivalents of acetic acid.

$$\ce{CH3CH2COCH3 ->[\ce{HNO3}] n(CH3CH2COOH + CO2) + (1-n)(2CH3COOH)}$$

A balanced equation for the formation of acetic acid would look like:
$$\ce{CH3CH2COCH3 + 3NO3^- +  -> 2 CH3CO2H + 3NO2^-}$$

A balanced equation for the formation of propanoic acid and carbon dioxide would look like:
$$\ce{CH3CH2COCH3 + 4NO3^- +  -> CH3CH2COOH + CO2 + 4NO2^- + H2O}$$

The permanganate reactions are tougher to balance, since permanganate is a three electron oxidant.

Update:

When I wrote this answer, I had never heard of Popoff's rule. No textbook I own mentions this rule, and a Google search about it brings up this question as the top hit (and similar questions at other sites like ask.yahoo.com as the other hits). I now know that Aleksandr Popov published a paper in Liebigs Annalen in 1872 describing a variation of the reaction that would become the Baeyer-Village oxidation. This article is behind a paywall for me, and the first page preview confirms I would not be able to make much of it. I only know a small amount of German and the PDF sadly looks like a low quality copy of a copy of a copy.

However, from the title "Die Oxydation der Ketone als Mittel zur Bestimmung der Constitution der fetten Säuren und der Alkohole" I can parse what the paper was about. Roughly this paper was about "The oxidization of ketones - a means for determining the constitution of the fatty acids and alcohols". Thus, his method involved the oxidation to the ester and hydrolysis of said ester in one step. I don't know what reagents Popov used, but I'm pretty sure it was not a peroxyacid (since this reagent will not hydrolyze the ester). I have previously answered a question about migratory aptitude in Baeyer-Villager reactions. Since I cannot read Popov's paper, nor can I find any authoritative resource on his rule (on- or offline), I have to assume that the known migratory aptitude for the Baeyer-Village reaction is the same as Popov's rule.


Calculating Delta H for Non Ideal Gasses We have to calculate delta H for a gas that follows the Van der Waals equation of state:

$$\left(p + \frac{a}{V_\mathrm m^2}\right)\left(V_\mathrm m-b\right)=RT$$

I know that $H = U + pV$ but I don't know how to approach this question.

Since I also know that $\Delta H = \Delta U + \Delta (pV)$, does that mean that I can calculate this simply by calculating $\Delta U$ (which I know how to do) and then $\Delta (pV) = p_2V_2 - p_1V_1$? Is that an appropriate method?

Because previously in other questions I have also seen it done as
$\mathrm{d}H = \mathrm{d}U + V\mathrm{d}p + p\mathrm{d}V + \mathrm{d}V\mathrm{d}p$ so I am wondering when is which equation applicable?

Are you familiar with the following equation?:  $$dH=C_pdT+\left(V_m-T\left(\frac{\partial V_m}{\partial T}\right)_P\right)dP$$


Could Calcium Oxide be made from CaCO3 in a bonfire? I have some chalk and I need to turn it into CaO. Would a big bonfire be hot enough to make it? If anyone has personal experience with this reaction please share, whether succesful or not

According to wikipedia, CaCO3 will turn into Calcium if heated above 825 °C (1,517 °F).

As for your fuel

Wood burns in several phases.

In the first phase, which can be referred to as dehydration, the heat produced by the wood's combustion drives of water from it in the form of vapour. It takes a lot of energy to do that which limits the fire's temperature to ~200-250°C.

The second phase, which may be referred to as pyrolysis, kicks in as the wood gradually becomes dehydrated. In that phase the fire's temperature will go from 250°C to topical peaks up to ~800°C.

As the wood further breaks down, you will encounter a third phase which is equivalent to the burning of charcoal. The temperature at the heart of your ambers will be at a minimum of ~700°C and could reach as much as 1300°C if you bring enough oxygen into it (by injecting compressed air into the heart of your ambers through a metal pipe for instance). This temperature would be fully sufficient to produce CaO from CaCO3 in a reasonably short amount of time.

At this point you could also add regular coal into the ambers, which requires a temperature above ~700°C to start burning. In ideal conditions, coal can burn up to ~ 2100°C which would significantly push the limits of what is possible for you.

If you would like to push these limits further, you could add coke on top of your regular coal, which would push the theoretical maximum temperature your fire could reach up to ~ 2500°C.

More practically

As Francis mentioned, there is a wide gap between theory in ideal conditions and practical reality.

The first aspect I already mentioned that will affect the practical temperature your chalk will be exposed to, is the amount of oxidising molecules that is brought to the heart of your fire. Injecting air through a pipe may be the safest option (though it is already significantly less safe than just making a regular bonfire, as it will significantly increase the combustion temperature and is at risk of propulsing fuel around if not adequately set up).
Theoretically, you could achieve a yet better result (at the cost of a lot of danger) by injecting pure compressed O2 into the heart of your ambers (which, for safety reasons, you should NOT do).

The second and very important aspect is insulation. The theoretical maximum temperature that is reached within your fire is of little practical importance. Insulating your fuel combustion zone without compromising the supply of oxidisers into it is key not only to increasing the maximum temperature that is locally reached inside of your fire, but more importantly to just keep calories inside of the combustion zone, which will significantly increase the average temperature inside of it (and tremendously lessen the required fuel expense for achieving any given amount of CaCO3 calcination). You can insulate your fire in numerous ways. Coal itself can act as a (burning) insulator if you place your CaCO3 at the heart of your combustion zone and cover it in a pile of coal/charcoal. Or you could build a structure around your fire: SiO2, Al2O3, CaO and MgO are particularly suitable materials for their resistance to high temperatures and for their refractive properties. You can look up their properties on wikipedia (MgO for instance melts at 2,852 °C, has a thermal Conductivity 45–60 W·m−1·K−1 and a Refractive Index of 1.7355).

Chemistry

My last point is that there might be the possibility for you to make things easier by adding some additional chemistry into the process (if you are inclined to do so). It would seem, for instance that Ca(OH)2 might turn into CaO starting around 512°C, which seems to be lower than CaCO3. So you might be able to lower your requirements on the combustion side by turning your CaCO3 into Ca(OH)2 prior to calcination. There are many ways to do this but one way would be to react CaCO3 with HCl to yield CaCl2 , which you'd subsequently react with NaOH (the very low solubility of Ca(OH)2 would drive the reaction forwards).


Valency and chemical formula If an element valency corresponds to the group number in the periodic table, oxygen would be 6 right?

The question states; Species: Li, O
I assume Li is 1 and O is 6, so my chemical formula is $\ce{Li6O}$ but the correct answer is $\ce{Li2O}$
Why is the valency of Oxygen 2?

similarly Pb(II) and Iodine, which iodine is in group 7 but just like the answer it said the valency is 1.

 If I drew the electron shells for oxygen I realize there are 2 electron spaces, if I did the same for Iodine there are 7 spaces. so this did not answer my question

then I saw number of bonds = (full valence shell) - (n. of valence electron)
for oxygen it returned 2, through: $8 - 6$
then it was the same for Iodine: $8- 7 $

This did not answer my question either, as Lithium and Sodium correspond with the group number. Am I missing something OR doing something wrong OR is there some sort of principle among this periodic table?

Then it asked, $\ce{Ag}$, $\ce{OH-}$. because Ag is a transition metal, so how do I find its valency? OH is a polyatomic ion, is there a way to find out the valency for polyatomic ion or am I suppose to memorize it for exams?

 The question in my book mostly gives 2 species then asked for the chemical formulas. I'm just very confused at the valency part.

Max's comment sums it up. Beware, I'm gonna have to use many simplifications in order to explain this.
A bit of introduction
The species you mentioned are ionic compounds. For your level it's useful to take ionic bonds as bonds between atoms that have such different electronegativities (ChemGuide typically provides useful introductions, I recommend clicking on that link and reading it carefully) that we can imagine them as different ions. Hence, this bond is the attraction between charged particles. What does that mean? It means that one of the species will have a positive charge and the other a negative charge.
Electronegativity increases in diagonally, as we move from bottom left to top right.
$\hspace{5ex}$
$\hspace{30ex}$ Electronegativity (Periodic Trend) – Source
For now just don't apply the electronegativity concept to noble gases. If you really want to know how their electronegativities are determined, take a look at interesting posts like this.
The more electronegative the atom, the less the chance that it'll bond by losing its electrons. Thus, we can conclude two important rules that occasionally become useful, while they're not highly accurate at higher levels:

Metals in general, and especially alkali and alkali earth metals, will bond ionically by losing their valence electrons.
Non-metals in general, and especially halogens, oxygen and nitrogen, will bond ionically by gaining electrons.

I repeat: The two bullet points above are just rules of thumb. Don't expect them to work all the time. But they're good starting points, so I included them in the answer. Don't hesitate to resort back to the first sentence whenever you ran into cases where you don't get a definitive answer by them.

Lithium oxide's case
So let's apply what we learnt to the $\ce{Li2O}$'s case. Lithium will lose its electrons in order to gain the configuration of a noble gases, in which case it would be helium.
$$\ce{Li(s)->Li(g)}^{~1}$$
$$\ce{Li(g)->Li+(g)}$$
Oxygen, however, doesn't favor losing its electrons at all. It will, instead, gain electrons and its electronic configuration would arbitrarily look like that of neon, a noble gas.
$$\ce{O2(g)->2O^2- (g)}$$
Finally, the two ions will form a lattice. Billions of billions of these ions form $\ce{Li2O}$.
$$\ce{Li+ (g) + O^2- (g) -> Li2O(s)}$$
Hence the overall reaction is
$$\ce{2Li (s) + O2 (g) -> 2Li2O(s)}$$
So let's review what we understood: In lithium oxide, oxygen will gain electrons while lithium will lose electrons, since their electronegativities differ greatly. So we should see how many electrons it will lose in order to get to the configuration of the previous noble gas in the periodic table, while oxygen will gain electrons to reach the configuration of the next noble gas in the periodic table. See this Q/A to gain more insight on how to figure this out using their electronic configuration or group number if you don't already know how.
Note that the formula for ionic compounds is the smallest relation between its ions. By $\ce{Li2O}$ we indicate that for every two lithium ions, one oxygen ion is present in the lattice, crudely, and assuming the crystal is without defect. Think of it like this: There was a need for two lithium ions to provide a single oxygen ion with the electrons it needed to fulfill its stable electronic configuration.
Silver hydroxide's case
Let's solve the case of silver and hydroxide.
The hydroxide ion has a charge of $-1$, hence the trouble of determining the charge it will have while bonded ionicallly with silver is gone! Unfortunately, for now, the best way to find the charge out is memorizing it. You may get some fuzzy speculations by drawing Lewis structures, but poly-atomic ions are sometimes exotic chemical species and unpredictable, since many effects are at play to determine the charge of the ion. You can find a list of them here, or more just with a simple Google search.
You have no choice but to memorize the charge of transition metals though, since they tend to assume abnormal electronic configurations. Over time, you'll face it so much you'll learn by heart that $\ce{Ag+}$ is silver's most stable ion.
Now try to apply the logic and reasoning we used back in $\ce{Li2O}$'s case. What would the formula be?

 $\ce{AgOH}$

Useful tips:

Don't forget, that if the poly-atomic ions are more than one in the ionic formula, you should indicate their number with them in parenthesis.
$\Longrightarrow \rm\color{red}{Wrong:} \quad \ce{(NH4)F,~Ca3PO42}$
$\Longrightarrow \rm\color{green}{Correct:} \quad \ce{NH4F,~Ca3(PO4)2}$
Beware of poly-atomic ions that only consist of one atom. These make for the biggest traps and confusions of the students. Good examples are peroxide, superoxide, and azide. Hence $\ce{Li2O2}$ and $\ce{LiO2}$ are correct! Simply because, the ions in them are not the oxide ion, $\ce{O^2-}$, but $\ce{O2^2-}$ and $\ce{O2^-}$, respectively. 2
The oxidation states of the elements, if mentioned, give clue as to what charge their corresponding ion will have. $\ce{Pb(II) \Longrightarrow Pb^2+ ,~Fe(II) \Longrightarrow Fe^2+ ,~Fe(III) \Longrightarrow Fe^3+}$ and so on.

1: It's simplified if I recall correctly, but we take the lattice as formed by ions in the gas state. It helps in defining lattice enthalpy, and I haven't read anywhere that it's not what's really happening.
2: Actually, you can't necessarily make any ionic compounds with any ions you want. Some of them are too unstable to reasonably exist, like lithium superoxide. But it seems the examiners want you to be a magician anyway.


Is omega 3 a specific fatty acid? Is eicosapentaenoic acid (EPA) omega 3 or that's just a acid that comes with a omega 3? Same question for docosahexaenoic acid (DHA).

Just want to know are both of those two acid omega 3 despite being different acids.

There exists a range of ω-3 acids.

Fatty acids are made of a chain of carbon atoms (generally linear), with a carboxylic acid at one end. Usually the carbon atoms are linked with a single-bond, but in unsaturated acids one or more of these are double-bonds.
In order to describe the positioning of these double bonds, the carbon-atoms are numbered, starting either from the carboxylic acid end, also called the α-end,  or from the methyl-end (the other end), which is called the ω-end. Therefore, an ω-3 acid is a fatty acid with a double-bond from carbon number 3 to 4, counted from the methyl/ω-end.

Fatty acids can have several double-bonds in its chain, making it a polyunsaturated acid. An example is Eicosapentaenoic acid, which has 5 double bonds. Edit: As pointed out in comments, it is the position of the first double-bond that determines if the acid is called an ω-3 acid. Likewise: if the first double bond is from carbon number 6 to 7, the acid is an ω-6 acid.

Wikipedia has some very informative articles about both ω-3 acids and fatty acids  in general.


Why does stannous chloride occur despite the octet rule? Shouldn't reaching an octet be any atom's "goal"?

However, I've recently learned about cases that are either expanding octets, or have lesser than "enough" electrons for an octet abiding. e.g.:

S in sulfur hexafluoride (Expanding octet into $\text{3d}$)

Source

B in boron trifluoride (It's "hextet", instead of octet)

Source

However, both $\ce{SnCl2}$ and $\ce{SnCl4}$ are existent. The latter is explainable with octet, but not the former. Surprisingly, $\ce{SnCl2}$ is more stable!

How?
Why is the phenomenon happening?

Dave pointed out that the octet nonabidingness is happening in the vapor phase. So I redirect the question to ask about the vapor phase; since that's what I'm looking for.

According to Theoretical and experimental studies of the electronic structure
of SnCl2 by photoionization with synchrotron radiation AIP Conf. Proc. 258, 60 (1992)

$\ce{SnCl2}$ has a molecular orbital structure as below (HOMO at the top)

Sigma-antibonding: 36% Sn s, 22% Sn p, 41% Cl s

Nonbonding 99% Cl p

Nonbonding 99% Cl p

Nonbonding 7% Sn p , 92% Cl p

Nonbonding 3% Sn s, 6% Sn p , 90% Cl p

Sigma bonding 14% Sn p, 83% Cl p

Sigma bonding 48% Sn s, 6% Cl s 45% Cl p

Nonbonding 3% Sn p, 96% Cl s

Nonbonding 5% Sn s, 2% Sn p, 92% Cl s

So overall it seems:

There is a total bond order of only 1.

Much of the Sn p and some Sn s electron density has been donated to Cl p orbitals.

There is little mixing of Sn s and p orbitals expect in the antibonding orbital.

additional references:

A Symmetry Rule for Predicting Molecular Structures (This reference explains how, based upon second order Jahn Teller effect, the symmetry of molecular orbitals predicts geometry, for example bent vs linear, and says that the symmetry predicts bent for SnCl2)

and Metal oxidation state effect in photoionization of gas‐phase metal halides (which further discusses the molecular orbitals of SnCl2 discussed above)


Why grain boundary is reactive to acid? I have read about grain boundary etching. They use acid to make grain boundary visible for light microscope. I know that grain boundary is reactive to acid, but why? Thanks.

I think this is due to the disordered atomic arrangement at the grain boundary which implies higher  internal energy of the constituting atoms and consequently more reactivity (attack by acid, corrosion,...).


Are these three molecules optical isomers? In the structures below (2) vs (3) seem obvious optical isomers to me.

But what about (1)? Is that just a stuctural isomer (e.g. Butane vs Isobutane) or is there a more specific way to describe the relation of (1) vs (2) / (3)?

Alternative sketch of (2) & (3) based on the comment by @DGS below:

There is no exocyclic methylene in structure 1.  Is it not an optical isomer of 2 and 3.

Likely there is a pericyclic rearrangement between 1 and 2/3 which would formally render them tautomers.

2 and 3 are diastereomers.


How to interpret orbital transition in TDDFT? Consider the following TDDFT run with GAMESS:

!   File created by the GAMESS Input Deck Generator Plugin for Avogadro
 $BASIS GBASIS=N311 NGAUSS=6 $END
 $CONTRL SCFTYP=RHF RUNTYP=ENERGY TDDFT=EXCITE  DFTTYP=B3LYP $END
 $CONTRL ICHARG=0 MULT=1 $END
 $TDDFT NSTATE=9 $END
 $STATPT OPTTOL=0.0005 NSTEP=99 METHOD=RFO UPHESS=MSP HSSEND=.T. $END
 $SYSTEM MWORDS=1000 PARALL=.TRUE. $END
 $SCF DIRSCF=.T. DIIS=.T. DAMP=.T. $END

 $DATA
Title
C1
O     8.0     0.00000    -1.27900     0.00300
C     6.0    -0.00000    -0.05800     0.00100
C     6.0     1.29700     0.69100    -0.00000
C     6.0    -1.29800     0.69000    -0.00000
H     1.0     1.35900     1.32900    -0.90600
H     1.0     1.35900     1.33200     0.90300
H     1.0     2.15700    -0.01300     0.00100
H     1.0    -2.15700    -0.01400     0.00100
H     1.0    -1.35900     1.32900    -0.90600
H     1.0    -1.35900     1.33200     0.90300
 $END

From this we obtain the following part of the output. (The complete calculation can be found here.) How can we interpret the orbital transitions, in terms of for example $\mathrm{n}\to\sigma^*$ or $\pi\to\pi^*$?

To visualise the orbitals of your calculation, use a program of your choice. For Gamess, there are a few options availabile. I use ChemCraft and Molden, and they work quite well. Here is a compilation of some with the former mentioned:

You can further use the summary to identify the most interesting, strongest transitions:

                          SUMMARY OF TDDFT RESULTS

   STATE             ENERGY     EXCITATION  TRANSITION DIPOLE, A.U.  OSCILLATOR
                    HARTREE          EV         X       Y       Z     STRENGTH
   0  A         -193.0290234748    0.000
   1  A         -192.8724089055    4.262     0.0001  0.0000  0.0001    0.000
   2  A         -192.7831335626    6.691     0.4051 -0.0004  0.0000    0.027
   3  A         -192.7317175333    8.090     0.0018 -0.0976  0.0001    0.002
   4  A         -192.7220472153    8.353    -0.4522 -0.0001  0.0000    0.042
   5  A         -192.7210047911    8.382    -0.0121  0.0000 -0.0006    0.000
   6  A         -192.7176816741    8.472    -0.0001 -0.0013  0.0724    0.001
   7  A         -192.7167365427    8.498     0.0006 -0.0002  0.0075    0.000
   8  A         -192.6964881601    9.049     0.0029 -0.3152  0.0008    0.022
   9  A         -192.6862850361    9.326    -0.0022  1.1274 -0.0093    0.290

For the calculated 9 states that is the last one, with an oscillator strength of 0.290. Now skip back one section to where you find:

      -------------------
      SINGLET EXCITATIONS
      -------------------

Look for excited state number 9:

 STATE #   9  ENERGY =    9.326387 EV
 OSCILLATOR STRENGTH =    0.290423
 LAMBDA DIAGNOSTIC   =    0.561 (RYDBERG/CHARGE TRANSFER CHARACTER)
 SYMMETRY OF STATE   =    A
                 EXCITATION  DE-EXCITATION
     OCC     VIR  AMPLITUDE      AMPLITUDE
      I       A     X(I->A)        Y(A->I)
     ---     ---   --------       --------
      9      17    0.137515       0.018404
     15      17    0.694034      -0.078753
     13      18   -0.085768      -0.006093
     14      19   -0.059549      -0.004510
     16      19   -0.289828       0.018031
     15      20    0.051487       0.010566
     11      21    0.053380       0.009416
     13      21   -0.060120      -0.009392
     16      22   -0.607830       0.005676
     11      24    0.052546       0.012226
     13      25   -0.039117      -0.008987
     16      26    0.112895       0.016712
     16      30   -0.041659      -0.014274
     16      33    0.052880       0.019973
     16      36   -0.031783      -0.011699
     13      40   -0.030352      -0.013004

There you find the orbital transitions. You can also have a look at the amplitudes to identify the most dominant one. In this case it is probably 15 to 17 and therefore corresponds to $\pi\to\pi^*$.

The more complicated the molecules get, the more confusing will this process be. Also the higher the level of theory, the more transitions you will need to consider. In this case Natural Transition Orbitals will certainly become very helpful, see Richard L. Martin, J. Chem. Phys., 2003, 118, 4775-4777.


What is the purpose of halogen in oxidation of thiols? Why do you need molecules like $\ce{Br2, I2}$ as one of the reagents in oxidation of thiols? I am saying this because when you look at the mechanism of reaction, it seems that you can have same product of disulfide by simply joining two protonated thiols (by strong base like $\ce{OH^{-}}$) together. Thanks.

Try drawing a mechanism without halogens present, starting with your base DE-protonating one thiol. See how far you get. The problem is that you'd have to somehow kick out H- (hydride), which is a terrible leaving group (enzymes can accomplish this, but standard lab reactions cannot do this easily). The fact is, combing two thiols to give the disulfide is an oxidation reaction, and there is an electron pair being lost from the reactants. It has to go somewhere, i.e. to reduce a diatomic halogen (playing the role of an oxidizing agent).

The thiol-disulfide redox equilibrium you are asking about has great biological significance; see disulfide bridges (for protein structure) and glutathione (an antioxidant).


Calculating charge of amino acids (am I doing it right?) Please consider this picture of a peptide chain:

I've done some examining on the different amino acids and concluded that the chain is as follows:

Lys-Leu-Gly-Ser-Citrullin (variant of arginine).

I would now like to calculate the charge of this peptide at neutral pH.

I'm using this table to orient myself: http://www.sigmaaldrich.com/content/dam/sigma-aldrich/life-science/biochemicals/migrationbiochemicals1/common-amino-acids-table.jpg

So here's my train of thought:

As we're at neutral pH the +1 and -1 of the mandatory amino and acid groups will cancel out for all amino acids.

Lysine has a pKx of barely 11 which is some higher than 7 so it is basic in our solution. It means that it will accept a proton so Lysine has a charge of +1 in our respective solution.

Leucine, Glycine and Serine have no charge on their x group so their charge is 0 and does not contribute anything to the total charge.

Citrullin, I read, is uncharged at neutral pH, opposed to Arginine which is, so it does not contribute either.

So my total answer is +1. Is this correct? I am in no way 100% sure of my method but it would be great to get a confirmation that I am doing it correctly or to be corrected on my possibly faulty method.

Thank you!

I second Lighthart's "Looks good to me"

  Let's break out the Henderson-Hasselbalch equation to be sure.

The Henderson-Hasselbalch equation is usually used to determine the pH of a buffer, but we can work backwards to get the ratio of an acid to its conjugate base at any pH.

$$\ce{pH}=\ce{p}K_a +\log{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}$$
$$\log{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}=\ce{pH}-\ce{p}K_a$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}=e^{\ce{pH}-\ce{p}K_a}$$

Let's take the three possible cases:

  $\ce{pH}>\ce{p}K_a$

If the pH is greater than the pKa, then

$$\ce{pH}-\ce{p}K_a>0$$

$$e^{\ce{pH}-\ce{p}K_a}>1$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}>1$$
$$[\ce{A-}]>[\ce{HA}]$$

For OH and SH functional groups in proteins, when $\ce{pH}>\ce{p}K_a$, then the group is deprotonated and has a negative charge.

For amino groups, $\ce{HA}$ is $\ce{RNH3+}$, so for amino groups at $\ce{pH}>\ce{p}K_a$, they are not protonated and neutral.

  $\ce{pH}<\ce{p}K_a$

If the pH is less than the pKa, then

$$\ce{pH}-\ce{p}K_a<0$$

$$e^{\ce{pH}-\ce{p}K_a}<1$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}<1$$
$$[\ce{A-}]<[\ce{HA}]$$

For OH and SH functional groups in proteins, when $\ce{pH}<\ce{p}K_a$, then the group is protonated and neutral.

For amino groups at $\ce{pH}<\ce{p}K_a$, they are protonated and positively charged.

  $\ce{pH}=\ce{p}K_a$

In this case you will find that:
$$[\ce{A-}]=[\ce{HA}]$$

Half of the groups are protonated and the other half are not.


Reaction of aldoxime and a ketoxime with acetic anhydride In mt book I saw these two reactions:

I don't know what is the mechanism for these reactions & that's why I am unable to understand how these products are formed & why they are so different. Please explain the mechanism in detail.

This might sound obvious, but since the reagent and the reaction conditions are identical, the reason for the different outcome of the reactions must be related to the starting materials.

In the case of the aldoxime, the carbon atom of the former $\ce{CHO}$ group still bears a hydrogen atom.

   Together with the $\ce{OH}$ group on the adjacent nitrogen atom, this allows (formally) for the elimination of water from the aldoxime under suitable conditions. High temperatures and the presence of a dehydrating agent facilitate exactly that.

The formation of a nitile from an aldoxime in boiling acetic anhydride is (formally) a dehydration reaction.

For obvious reacons, this pathway is not possible in the case of the ketoxime.

What is the role of acetic anhydride here?

I mentioned that the formation of nitriles from aldoximes is formally a dehydration. In reality, however, it is more likely that acetic anhydride actually reacts with both oximes.

Both the aldoxime and the ketoxime are

   acylated. For the ketoxime (R' = alkyl), the story ends here. In the case of the aldoxime (R' = H), a subsequent elimination (of acetic acid) eventually furnishes the nitrile.


Do noble gasses besides Helium form diatomic molecules at low temperatures? I know that at extremely low temperatures (mK and lower), Helium can form diatomic molecules. Do the other noble gasses also form molecules at extremely low temperatures?

Yeah, dimers of noble gas atoms are stable due to dispersion aka London forces. More generally, neutral, non-polar molecules can still form very weak bound states arising entirely from the induction and interaction of instantaneous dipole moments (that is, fluctuations in the electron density around an atom arising from the motion of electrons). My understanding is that physicists would explain this interaction as a special case of the Casimir-Polder interaction. This interaction means that even noble gases are nonideal.

Correctly predicting the negative interaction energy of a noble gas dimer is a pathological test of dispersion-corrected DFT functionals, as F'x's reference1 and the references therein indicate, and many methods fail qualitatively. I recall that Zhao and Truhlar used noble gas dimers to test the efficacy of their various metahybrid functionals2,3. One of their papers2 makes reference to a table of binding energies and equilibrium distances derived4 by Wang and Ogilvy, giving for instance a $\ce{Ne\cdots{}Ne}$ binding energy of 0.35 kJ mol-1 (that's tiny) and less than half that for $\ce{He\cdots{}Ne}$. The $\ce{Ne\cdots{}Ne}$ equilibrium distance is given as 309 pm, which is extreme. As the paper5 you refer to indicates, however, the interaction energy for the $\ce{^4He\cdots{}^4He}$ dimer is much, much smaller than either of these and has an equilibrium distance of 5.2 nm (!!!), which is something else entirely.

(1) Ruzsinszky, A.; Perdew, J.P.; Csonka, G.I.; Binding Energy Curves from Nonempirical Density Functionals II. van der Waals Bonds in Rare-Gas and Alkaline-Earth Diatomics; J. Phys. Chem. A; 2005, 109, pp. 11015-11021

(2) Zhao, Y.; Truhlar, D.G.; Hybrid Meta Density Functional Theory Methods for Thermochemistry, Thermochemical Kinetics, and Noncovalent Interactions: The MPW1B95 and MPWB1K Models and Comparative Assessments for Hydrogen Bonding and van der Waals Interactions; J. Phys. Chem. A; 2004, 108, 6908-6918

(3) Zhao, Y.; Truhlar, D.G.; Comparative DFT Study of van der Waals Complexes: Rare-Gas Dimers, Alkaline-Earth
Dimers, Zinc Dimer, and Zinc-Rare-Gas Dimers; J. Phys. Chem. A; 2006, 110, pp. 5121-5129

(4) Ogilvie, J.F.; Wang, F.Y.H.; Potential-energy functions of diatomic molecules of the noble gases: II. Unlike nuclear species; J. Mol. Struct.; 1993, 291, pp. 313-322

(5) Grisenti, R.E.; Schöllkopf, W.; Toennies,  J.P.; Hegerfeldt, G.C.; Köhler, T.; Stoll, M.; Determination of the Bond Length and Binding Energy of the Helium Dimer by Diffraction from a Transmission Grating; Phys. Rev. Lett.; 2000, 85, pp. 2284-2287


What is the correct term of a chemical formula for very large molecules if not just "chemical formula"? I know that many molecules, wihch I as a non chemist, might define as "small" if their two dimensional diagram can be sketched on an A4 page, are often formulated;
That is → by one or more processes given a chemical formula;
Maybe after being isolated from a natural deposit or organism.

For example, here is such formula for vitaminB12 of the type hydroxocobalamin:

C62H89CoN13O15P

My problem

Whenever I examined a wiki article about a protein molecule, I understood it as too big to be represented in a "chemical formula", let alone in a two dimensional diagram on an A4 page, such as those I have described.

I though for example:

  As these are isolated molecules after all, they do have a "chemical formula" and should be standardly represented somehow.

  Such large molecules doesn't have to be proteins → maybe proteins and other molecules in similar size have different ways to be represented.

My question

What is the correct term of a chemical formula for very large molecules (such as proteins), if not just "chemical formula"?

From prior research I understand a "protein sequence" is the standard parallel for "chemical formula" in the context of proteins but I don't know about a standard parallel to such large molecules in general.

Chemical formulas of the type you describe are just as valid for large molecules as for small ones. The reason they are not commonly used is that the number of possible molecules described by a given formula goes up exponentially with the number of atoms.

Thus, it is much more common to describe larger molecules with more precise terms. In the case of polymers like proteins or nucleic acids, the primary sequence of the constituents (amino acids or bases) is often used if space allows. Alternatively, a protein or nucleic acid  might be referred to by a common name with which a sequence is associated.

Even for small molecules, IUPAC formal names or common names are used in preference to molecular formula in many cases, because the names provide information about the atom connectivity that is absent in the molecular formula. Other representations that include connectivity information are forms such as SMILES, which is a standardized system for representing connectivity in text.

In the scientific literature, names of molecules are often accompanied by pictorial representations that provide atom connectivity information in an easily understood format. For very large molecules like proteins whose connectivity cannot be easily represented in a small image, pictorial representations are more often used to show secondary and tertiary structural features.


Does this activation of carbonyls usually involve a SN1 or a SN2? The process I have seen in class to activate a carbonyl function is acetalisation.

One of the steps involved is the following :

So I was wondering, is that a $\ce{S_{N}1}$ or a $\ce{S_{N}2}$ ?

(PS : drawing molecules with paint is awful, anyone knows a good free molecules drawer ?)

It is not an $\ce{S_{N}1}$ or an $\ce{S_{N}2}$ reaction.  It is commonly referred to as an "addition-elimination" reaction.  You've drawn the key piece of evidence in your diagram - the tetrahedral intermediate.  Not an $\ce{S_{N}1}$ reaction because there is no carbonium ion formed in this reaction.  This is evidenced by many facts such as the lack of carbonium ion type rearrangements when substituents are present alpha to the carbonyl.  Nor is it an $\ce{S_{N}2}$ reaction, the attacking group does not start to bond and the leaving group does not start to leave and break its bond simultaneously.  Instead there is a real tetrahedral intermediate that exists in equilibrium with the reactant as well as the product.  You "add" a nucleophile to the protonated carbonyl compound to form the tetrahedral intermediate, then you "eliminate" one the substituents in the tetrahedral intermediate and either go back to reactant or forward to product.

Here's another, more specific drawing for the case at hand.


Reasons for dry conditions in substituition reactions with benzene All substitution reactions of benzene must be carried out in dry conditions with a catalyst that produces a powerful electrophile.

This was a statement from my book. My question is, why must it be carried out in dry conditions?

What will occur if moisture is present?

Electrophiles
are reactive cations, they will react with water to form
eg an alcohol (or some weak acid etc)  and protons.
No reaction with benzene.


Computing reaction mechanisms using solvation effects I am about to compute reaction steps for a series of compounds using DFT.
A key point in our discussion is the dependency of the experimental rate of reaction w.r.t. solvent used, so it is very important that we incorporate solvation effects in our computational models.

I am aware of the best practices in choosing functionals, dispersion corrections, basis sets, etc..
By reading a little I decided on using the SMD model for implicit solvation.
I may use either ORCA or NWChem.

My questions are:

a. Should geometries be optimised in solvent, in the gas phase or both?
b. In case I use only gas phase geometries, should I calculate frequencies using implicit solvation (I doubt) or can I use the calculated free energy of solvation (done in a single point calculation) to correct the gas phase free energy?
a. In case I decide on using (one or two) explicit solvent molecules to better model the system, are there best practices on using them together with an implicit solvation model (in terms of calculating free energy of solvation and in deciding when it has converged/is adequate)?
b. What about the points addressed in 1. (gas phase versus condensed optimisation and frequency calculations) for this case?

References from literature are greatly appreciated.
I am looking forward to hear from you!

1) Yes, geometries must be calculated both in the gas phase and in the solvent. To analize your results in the solvent you need something to compare that's why the gas phase calculation is always needed and often is considered "the only source of truth".

Frequencies must be calculated both in the solvent and in the gas phase as well, on the respective converged geometries.

2) I never worked with explicit solvent but I feel that the use cases of two methods are not overlapping. When performing calculations on a multi step reaction usually you don't have enough computational resources to calculate all the solvent molecules needed. This is why people use implicit solvent calculations. If you want to benchmark your solvent calculations, compare it with the gas phase ones and evaluate wether there is some improvement in the replication of experimental data.


Decomposing chemical compounds - Ease of Decomposition Someone asked me a question, that which of the following can be decomposed chemically:

Ammonia
Iron
Neon
Hydrogen
Fluorine

My Analysis:

At first attempt I answered Ammonia but changed it to 'I can't answer'. My analysis of the question is as follows

If Fluorine (the name implies fluorine gas) is decomposed, but it will require to 242kJ/mol (bond dissociation enthalpy). So, its quite difficult that we will be able to decompose it chemically.
If Hydrogen (the name implies Hydrogen gas) is decomposed, but it will require to 435.88kJ/mol (bond dissociation enthalpy). So, its even more difficult that we will be able to decompose it chemically.
If If Ammonia (the name implies Ammonia gas) is decomposed, but it will require to 389kJ/mol (bond dissociation enthalpy). So, its significantly difficult that we will be able to decompose it chemically.
Its useless to say that we will be decomposing Iron or Neon, because they are already in elemental state.

Why I tempted to answer Ammonia: I thought about Haber's Process, and remembered that the enthalpy change of the main reaction is negative (N2+3H2=NH3), about -46.1kJ/mol; this implies that the enthalpy of the just opposite reaction is positive (Enthalpy change of NH3=N2+3H2). This made me to conclude that the reaction is not feasible, thermodynamically and therefore ammonia cannot be decomposed easily. Therefore I changed my answer to 'I can't Answer'

My Question:
I asked my friend my what is the answer, he said Ammonia. I was surprised but still confused. I have written all of my thoughts above, but neither I am able to justify my answer, nor is my friend able to understand my logic. I want your discussion and a final review of the question understanding
'

It really depends on what the definition of "decompose" is in this case. If you take the definition of "decompose" to be:

  ... to break down or cause to break down into component elements or simpler constituents

then the only option would be ammonia because the others are elements whereas ammonia is a compound which can be broken down into elemental Hydrogen and Nitrogen by the Haber Process. The energetics is not of concern in this case.


Why there is hydrogen bonding involved in salt bridge formation between glutamic acid and lysine? I saw this figure on Wikipedia:

After seeing this image I got really confused about the difference between these two, or maybe the similarities. Of course, I understand that a hydrogen bond can be seen as +/- interaction (because of the dipole). However, I don't understand why this Lys + Glu interaction would also make a hydrogen bond. If it would, I would expect the hydrogen bond between the double bounded oxygen and the $\ce{N}$ atom from the $\ce{NH3+}.$
I hope someone can explain me what this figure means in respect to both of the interactions shown.

An electrostatic interaction is one between a centre of positive charge and a centre of negative charge. In your example, instead of a primary ammonium cation ($\ce{H3\overset{+}{N}R}$) you could also consider the fully methylated quarternary ammonium cation $\ce{Me3\overset{+}{N}R}$ which would still be able to build up an electrostatic interaction with the acid anion. The centre of positive charge is somewhere between the three ammonium hydrogens and the centre of negative charge somewhere between the two carboxylic oxygens.

A hydrogen bond, on the other hand, is a directional interaction of the type $\ce{X-H\bond{....}Y}$, where $\ce{X}$ and $\ce{Y}$ are electronegative elements and $\ce{Y}$ must feature a lone pair. Thus, hydrogen bonds always contain three atoms and only those three. One one of the three ammonium hydrogens can form a hydrogen bond with only one of the two oxygen atoms.


Does temperature affect heat capacity? My chemistry  book says that heat capacity varies with temperature. There is no further discussion.
Does the heat capacity of a given amount of substance really vary with temperature? If yes, why is it so?

The heat capacity is the slope of the plot of internal energy U with temperature T. The internal energy is energy that due to the rotation and vibrational energy a molecule possesses and as the temperature increases more rotational and vibrational energy levels becomes excited and so the internal energy increases.

Initially the internal energy is small and changes slowly as energy is added, mainly because there is not enough energy to excite many rotations or vibrations. Thus the slope of U vs. T is small and so is the heat capacity.  Then as temperature increases more of the energy levels become excited & the internal energy rises rapidly and so does the slope of U vs T and so heat capacity increases.

At very high temperatures  many, many energy levels are populated and so the rate of increase becomes constant as the temperature increases and so the heat capacity becomes constant, i.e. it levels out to a maximum value.  The plot shows the change of heat capacity with energy just for vibrational levels, but it has the same general shape when rotational and translational energy are added.

internal energy vs temperature

vibrational heat capacity vs temperature

low temperature region of heat capacity vs temperature.The 'bumpy' plot is due to the discrete values of vibrational energy levels


Why will internal energy decrease in this reaction? 1 mol zinc reacts with 1 mol sulfuric acid in a bomb calorimeter. I know since volume is constant work done will be zero .. but why will the internal energy decrease??
I know that the reaction is not endothermic or exothermic so there is no heat change involved, and as E = w + Q, the internal energy should also have been unaffected. But in an mcq book, the answer clearly states that the internal energy is decreasing.

If the reaction is exothermic, and the system you choose includes both the insulated chamber and its mixture of reactants and products, then the change in internal energy of this system will be zero.  However, if the system you choose includes only the mixture of reactants and products, then heat will be transferred from the mixture to the chamber, and the internal energy of your mixture will decrease.


Ionic Strength of 0.1M of HEPES? HEPES, the buffer, has the chemical formula $\ce{C8H18N2O4S}$.

Using the Ionic Strength formula from Wikipedia.

I have calculated that $0.1M$ of HEPES would have a strength of $1.7M$, would it be possible to confirm whether this is correct or not? (I have little chemistry knowledge).

The Ionic Strength is calculated as:

$$I=\dfrac{1}{2}\cdot\sum{c_iz_i^2},$$ where where $c_i$ is the molar concentration of ion i ($M,\,mol/L$), $z_i$ is the charge number of that ion, and the sum is taken over all ions in the solution.

This is your HEPES molecule ($C_8H_{18}N_2O_4S$):

Only one proton at $SO_3H$ group can dissociate, as far as I see. So in your solution you will have two different types of ions with charge equal to 1 (assuming that molecule will dissociates 100%, and than:

$I=0.5\cdot(0.1M\cdot1^2+0.1M\cdot(-1)^2)=0.1M$

So your calculations are wrong.

But also be aware that this molecule is Zwitterion!
READ THIS Do zwitterions contribute to the ionic strength of a solution?
The resume of the articles:

"Hence, electrostatic effects due to zwitterions are important, even if the zwitterions do
not contribute directly to ionic strength."

My calculations are primitive. They are also far from the truth, but they show how the principle works for you.


When are two orbitals orthogonal? When are two orbitals considered to be orthogonal?

It will be helpful if you can provide an example molecule (or molecular orbitals in a molecule) since I can't really think of a scenario where molecular orbitals can be exactly 90° to each other.

I'm not well-versed in quantum mechanics, so would prefer a simple answer - but all explanations are welcome.

Unfortunately, the sense in which orbitals are orthogonal is more or less impossible to define rigorously without recourse to functions of some kind.  So, I'll give an explanation a shot using some simple, 1-D functions to illustrate the concept, followed by the pictorial orbital example you've asked for.

At a basic level, in order to have any two functions be orthogonal, you need to define a "mixing rule" for them. The most useful mixing rule for orbitals turns out to be integrating the product of the two functions$^\dagger$ over some defined domain.

So, consider two real 1-D functions, $f(x)$ and $g(x)$, defined over an interval $x=[a,b]$. The mixing rule $\Pi\,_a^b\!\left(f,g\right)$ I'll define as follows:

$$
\Pi\,_a^b\!\left(f,g\right) \rightarrow \int_a^b{f(x)g(x)\, dx}
$$

Simply put, two functions $f(x)$ and $g(x)$ are orthogonal under the mixing rule $\Pi\,_a^b$ (or any other mixing rule!) if and only if $\Pi\,_a^b\!\left(f,g\right) = 0$.

A pretty standard example of two orthogonal functions in this framework would be the following:

\begin{align*}
  f(x) &= \cos(2\pi x) \\
  g(x) &= \sin(2\pi x) \\
  a    &= 0 \\
  b    &= 1
\end{align*}

The integral to check from the above is:

$$
\Pi\,_0^1\!\left(f, g\right) \rightarrow \int_0^1{\cos(2\pi x)\, \sin(2\pi x)\, dx}
$$

Since this is Stack Exchange and not a math exam, I will now cheat and just link to Wolfram Alpha to demonstrate that the above integral is, indeed, zero. Thus, on the domain $x=[0,1]$, $\sin(2\pi x)$ and $\cos(2\pi x)$ are orthogonal.

On the other hand, to pick some random examples, $f(x)=-x$ and $g(x)=\ln(x)$ are not orthogonal on the interval $x=[0,1]$.

$$
\Pi\,_0^1\!\left(f,g\right) \rightarrow \int_0^1{\left(-x\right)\ln(x)\, dx} = 0.25
$$

Similarly, the original two functions are not orthogonal on the interval $x=[0.2,0.9]$:

$$
\Pi\,_{0.2}^{0.9}\!\left(f, g\right) \rightarrow \int_{0.2}^{0.9}{\cos(2\pi x)\, \sin(2\pi x)\, dx} = -0.0444852
$$

Ok, enough with the math lesson. How does this help define orthogonal orbitals?

Well, orbitals, both atomic and molecular, are (usually) defined just as as 3-D functions over all space.  If you have two orbitals $\phi\!\left(\vec r\right)$ and $\psi\!\left(\vec r\right)$ and want to know if they're orthogonal, you use a mixing rule somewhat similar to $\Pi$ above and calculate the appropriate integral. If the resulting value is zero, then the orbitals are orthogonal.

For example, consider a $1\mathrm s$ and a $2\mathrm p$ orbital (yes, these are atomic orbitals, but the principle holds for molecular orbitals, too):

Here, the colors of the two lobes of the $2\mathrm p$ orbital are important: one color, say orange, represents positive function values; the other, say blue, represents negative values.  For the $1\mathrm s$ orbital, though, it's all orange, or all positive).

What happens if, per $\Pi\!\left(\phi, \psi\right)$, you center these two orbitals on top of one another, multiply their values together, and integrate over all space? Well, the half of the $\mathrm p$ orbital that is positive will result in positive contributions to the integral of $\Pi$, and the half of the $\mathrm p$ orbital that is negative will provide negative contributions to the $\Pi$ integral.  And, further, due to the spherical symmetry of the $\mathrm s$ orbital and the reflection symmetry of the $\mathrm p$ orbital, those two positive and negative contributions will exactly cancel. Thus, the $1\mathrm s$ and $2\mathrm p$ atomic orbitals are orthogonal.

Epilogue: So, given all of this, why do we also say that vectors that are $90^\circ$ from one another are orthogonal?

Well, it just involves a different mixing rule (the domain is implicit, deriving from the dimensionality of the two vectors, which must be the same):

$$
\Gamma\!\left(\vec v_1, \vec v_2\right) \rightarrow \vec v_1\cdot\vec v_2
$$

Under the (extraordinarily common) mixing rule $\Gamma\!\left(\vec v_1, \vec v_2\right)$, the two vectors $\vec v_1$ and $\vec v_2$ are said to be orthogonal  when $\vec v_1\cdot\vec v_2=0$.

Note: In theory, I could have just linked to this question on Math.SE. However, none of the answers there were really couched in suitably readable terms.

$^\dagger$ Technically, integrating the product of one function with the complex conjugate of the other: $\int_a^b{f^*(x)g(x)\,dx}$


What dangerous gases could electrolysis with water and sodium carbonate produce? My friend was very nervous when he noticed that I'm performing electrolysis in a room that we've just been residing. (it was under open window to let $\ce{H2}$ out) After a sharp argument, he made me shut it down, though I still thought that it was safe.

I'm using $\ce{Na2CO3}$ as electrolyte and as far as I know, it doesn't get electrolysed. I use normal pipe water. Such water surely contains some minerals however in really small amount. My friend pointed out that it also contains chlorine and that's what he's afraid of. He would disregard my notice that he's drinking this water often.

I'm using copper electrodes, producing copper hydroxide and copper carbonate. I failed to find out whether these compounds are hazardous. They're in solid state anyway.

From what I've said you can see that I'm very sure that the reaction is not hazardous. But I'm not chemist. There could be something I didn't think of in which case I've been poisoning myself for quite a while.

Could you make me sure or warn me?

My friend pointed out that it also contains chlorine and that's what he's afraid of.

The amount is laughable at. However, using distilled (deionized) water is recommended for other reasons.

  I failed to find out whether these compounds are hazardous. They're in solid state anyway.

They are reasonably harmless. Of course, eating them is not recommended, but copper compounds have low toxity. Move to graphite or platinum electrodes is still recommended.


Copper Sulfate Pentahydrate experiment, help please? My science project specifies me to make seed crystals of copper sulfate. We must use the powder form to make seeds. There are two types of copper sulfate pentahydrate; one is granular form (small seeds), and the other one is powder form. We must produce crystal seeds from copper sulfate (powder form). We are not allowed to use the granular form, because it forms seeds on its own already.

Steps I have taken:

I saturated 50 ml of distilled water (room temperature) with copper sulfate pentahydrate (powder form).
I added about 15 ml of purified sulfuric acid to the mixture, so it prevents the hydrolysis with water.This purified sulfuric acid I got from boiling battery acid. I boiled about 50 ml of batter acid (sulfuric acid mixed with ionic water). I waited till it reached about 15 ml (water evaporation). The fumes were getting strong at that stage. I'm not sure if purified enough.
I then again mixed more copper sulfate to the mixture, and stirred it very well.
I put the china dish (mixture) on a hot plate, and waited for it to reach crystallization point, while stirring it very well.
5 I know the water has to evaporate for the mixture to produce seed crystals. I waited for the mixture to evaporate about to 20 ml.
I had no results.

I'm not sure what I'm doing wrong. I'm doing my project according to this video which I found on youtube. It is a very short video, and my main problem is the sulfuric acid part.

https://youtu.be/SAU-gptAFe0

Help.

Frankly you're going about this all wrong. First you want to exploit the difference in solubility of the salt in hot water versus cold water.

Copper sulfate's solubility is 32 g/100ml water at 20 C, and 114 g/100ml water at 100 C.

2-3 drops of sulfuric acid is all that is needed. You're just trying to prevent the formation of $\ce{Cu(OH)2}$ or $\ce{CuCO3}$.

  I'm not sure about battery acid. I assume that you're using new battery acid, not actually from an old battery. So the sulfuric acid will be a little less concentrated. Use 5-6 drops instead of 2 or 3.

After you make a saturated solution of the copper sulfate (aim for 100g /100ml water) in hot water, let it cool slowly and don't stir at all. Stirring will promote nucleation of many small crystals.

To get a really large crystal you'd make a saturated solution at room temperature, filter it, and add one of these seed crystals. You'd then let the water evaporate over weeks very slowly.


Properties of starch in packaged pudding I had a mug of packaged pudding. The top layer of the pudding was of typical pudding consistency, while the bottom was more runny. What could have caused this?

Starch makes a glue or a paste with water. Humidity from the air may be sufficient to transform your starch into a glue. Water produces strong hydrogen bridges between starch molecules. This transforms a pile of independent starch molecules into a strong block of interconnected starch molecules.


Does Cute Poison actually work? For those of you that have watched Season 1 of Prison Break (TV Series), in the "Cute Poison" Episode Micheal Scofield combined $\ce{CuSO4}$ (Copper Sulfate) and $\ce{H3PO4}$ (Phosphoric Acid) to weaken the metal. Is it true?

The Wikipedia page on "Cute Poison" episode says that it wasn't anhydrous copper (II) phosphate, but Gypsum.

I haven't seen the episode, and the Wikipedia page's reaction seems to be the backwards reaction of yours. But I imagine yours is the one the pro/antagonist is gonna use to escape the prison; as

Gypsum is commonly found and accessible:

  Gypsum is a soft sulfate mineral composed of calcium sulfate dihydrate, with the chemical formula $\ce{CaSO4·2H2O}$. It can be used as a fertilizer, is the main constituent in many forms of plaster and in blackboard chalk, and is widely mined. A massive fine-grained white or lightly tinted variety of gypsum, called alabaster, has been used for sculpture by many cultures including Ancient Egypt, Mesopotamia, Ancient Rome, Byzantine empire and the Nottingham alabasters of medieval England.

Calcium phosphate's solubility will decrease with temperature increase and it will precipitate, leaving you with a solution of sulfuric acid.

It's possible for a double displacement reaction to occur in aqueous medium (with a spark):
$$\ce{2H3PO4(aq) + 3CaSO4(aq)·2H2O(l) \leftrightharpoons 3H2SO4(aq) + Ca3(PO4)2(aq) + 6H2O(l)}$$

$\ce{H2SO4}$ is sulfuric acid. It's a very strong acid in water, a diprotic acid. Its $p {\rm K_a}$s are −3 and 1.99 according to Wikipedia.

  Its corrosiveness on other materials, like metals, living tissues or even stones, can be mainly ascribed to its strong acidic nature and, if concentrated, strong dehydrating and oxidizing properties.

Later in the same article, you can read how it reacts with different material. It can weaken metal pretty fast and easily, as 3 moles of it will be present in any 7 moles of products, which is high above the dilute concentration (arbitrarily) defined in the Wikipedia page.

Copper sulfate also reacts with phosphoric acid and forms sulfuric acid, but it's hardly accessible in prison, I reckon.


Calculating the equilibrium constants for parallel reactions So, I am attempting to calculate the equilibrium constant for two separate reactions.
$$\ce{A + B<=>C + D}$$
$$\ce{A + B <=> C + E}$$
I have figured out the final pressures at equilibrium for $\ce{A}$, $\ce{B}$, $\ce{C}$, $\ce{D}$, and $\ce{E}$.
I was wondering if the equilibrium constant is calculated as normal using just the final pressures and molar ratios in the equations, or if the final pressures have to be adjusted somehow to compensate for the fact that the second reaction is also consuming the same products and producing one of the same reactants.

You can combine the two chemical reactions into a single global reaction scheme by adding them:
\begin{align}
\ce{A + B&<=>C + D}\tag{1}\\
\ce{A + B&<=>C + E}\tag{2} \\ \hline
\ce{2A + 2B &<=> 2C + D + E}   \tag{Global}
\end{align}
and the equilibrium constant for this overall reaction scheme is equal to the products of the two equilibrium constants:
\begin{align}
K_\mathrm{eq,global} &= K_\mathrm{eq,1} K_\mathrm{eq,2} \\
&= \frac{\ce{[C]^2[D][E]} } {\ce{[A]^2[B]^2}}
\end{align}
(where $\ce{[i]}$ denotes either molar concentration or partial pressure of component i). You do not need to adjust the final total pressure or mole amounts.


What properties of carbon dioxide make it a greenhouse gas? Why is it that heat from the Earth reflects back off carbon dioxide and other greenhouse gases but not gases like nitrogen or oxygen?

According to the Intergovernmental Panel on Climate Change (IPCC):

  "Greenhouse gases are those that absorb and emit infrared radiation in the wavelength range emitted by Earth."

In order for a molecule to absorb and emit in the infrared (IR) region, its chemical bonds must rotate and vibrate in a manner that affects something called the molecule's dipole moment. It turns out that due to the symmetry of diatomic molecules like $\ce{O2}$ and $\ce{N2}$, this process cannot happen, and thus these types of molecules cannot absorb in the infrared, which is the wavelength range in which heat is radiated.

Because $\ce{CO2}$ and other greenhouse gases do not have this kind of symmetry, they can vibrate at specific frequencies within the IR in a manner that affects the molecule's dipole moment, and thus absorb this radiation resulting in the transfer of heat. Much of the heat radiated from the Earth's surface is of the proper wavelength (energy) to be absorbed by these gases, so the criteria given by the IPCC for a greenhouse gas are thus met.

As I'm not certain of your level of understanding of chemistry, I tried to direct my answer somewhere between purely technical and purely non-technical. If I've left anything unclear to you, please don't hesitate to leave a question in the comments.


What is the difference between primitive Hexagonal unit cell and Hexagonal closed packing HCP There are 14 types of Bravais crystal lattices (Primitive + Centered) in 3D. Every solid in this universe is made by the unit cells of these lattices.

With the help of these Bravais lattice can HCP be formed?.
Is there any relation between Hexagonal unit cell and Hexagonal close packing
As we see that Hexagonal unit cell exists only in primitive AAA form then how can HCP ABA be formed from it .

Primitive Hexagonal unit cell

Hexagonal closed packing HCP

If there is a clear distinction then Should HCP also be placed among 14 Bravais lattice.

With the help of these Bravais lattice can HCP be formed?

Yes. One choice of unit cell for HCP is a primitive hexagonal cell with two atoms per unit cell. If you place the first atom on the origin, the second will be at (1/3 2/3 1/2). Here is a 3D view of the situation and some more information: http://lampx.tugraz.at/~hadley/ss1/crystalstructure/structures/hcp/hcp.php

  Should HCP also be placed among 14 Bravais lattice?

No, because it is already in there, just hiding a bit. If you look further into it, the space group is #194, P$6_3$/mmc, see http://img.chem.ucl.ac.uk/sgp/large/194az1.htm


for a polydisperse polymer, why is its weight average molecular weight always bigger than its number average molecular weight? The statement comes from my lecture slides on polymers. I hope someone can show me the maths proof for the inequality.
I will try come up with the proof, too. It is actually a maths question:D Here is the basic definition for the two average weights: https://en.wikipedia.org/wiki/Molar_mass_distribution

Mathematics has many examples of identities and concepts which are extremely powerful due to their simplicity, such as the pigeonhole principle and the triangle inequality. One such identity will guide us through the problem.

We start with the very innocent identity, that for a real number $z$, the following is true:

$$z^2 \geq 0$$

Now let $z = x-y$. We have:
$$(x-y)^2 \geq 0$$
$$x^2 -2xy + y^2 \geq 0$$
$$x^2 + y^2 \geq 2xy $$
$$\frac{x^2 + y^2}{2} \geq xy $$

Now this last inequality has a very powerful interpretation. To show it more clearly, we can replace $x^2$ with $a$ and $y^2$ with $b$, yielding:

$$\frac{a + b}{2} \geq \sqrt{ab} $$

In words, the arithmetic mean of two positive real numbers will always be greater than or equal to their geometric mean. This is called the AM-GM inequality, and it often shows up in the most surprising of places.

Let's get started. We define the number average molecular weight ($M_{n}$) and the weight average molecular weight ($M_{w}$) of a polydisperse polymer mixture as the following, respectively:

$$M_{n}=\frac  {\sum\limits_{i} N_{i}M_{i}}{\sum\limits_{i} N_{i}} \quad , \quad M_{w}=\frac  {\sum\limits_{i} N_{i} M_{i}^2 }{\sum\limits_{i} N_{i}M_{i}}$$

We then define the dispersity index $Đ$ (or $PDI$) as:

$$Đ = \frac{M_w}{M_n} = \frac{\sum\limits_{i} N_{i} M_{i}^2 \sum\limits_{i} N_{i}}{(\sum\limits_{i} N_{i}M_{i})^2}$$

To make things easier, let us start by analyzing the first non-trivial case ($i=2$):

$$Đ = \frac{M_w}{M_n} = \frac{(N_{1} M_{1}^2 + N_{2} M_{2}^2)(N_{1} + N_{2}) }{(N_{1} M_{1} + N_{2} M_{2})^2}$$

We multiply the bracketed terms and expand the binomial to obtain:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{1}N_{2}M_{1}^2 + N_{1}N_{2}M_{2}^2}{N_{1}^2 M_{1}^2 + 2N_{1}N_{2}M_{1}M_{2} + N_{2}^2 M_{2}^2}$$

Rearranging the terms slightly:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}}$$

Notice that the terms in the numerator and denominator match except for the coloured parts.

Here is where the magic happens. From the AM-GM inequality, we must have:

$$x^2 + y^2 \geq 2xy \quad \Rightarrow \quad M_{1}^2 + M_{2}^2 \geq 2M_{1}M_{2}$$

We then replace the term $(M_{1}^2 + M_{2}^2)$ in the numerator with the term $2M_{1}M_{2}$. By doing so, the new numerator is smaller than or equal to the old numerator. Thus:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}} \geq \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}}=1$$

$$Đ = \frac{M_w}{M_n} \geq 1$$

What about the next non-trivial case ($i=3$)? We have:

$$Đ = \frac{M_w}{M_n} = \frac{(N_{1} M_{1}^2 + N_{2} M_{2}^2 + N_{3} M_{3}^2)(N_{1} + N_{2} + N_{3}) }{(N_{1} M_{1} + N_{2} M_{2} + N_{3} M_{3})^2}$$

Multiplying everything out and performing the appropriate arrangements yields:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3} M_{3}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2) + N_{1}N_{3}(M_{1}^2 + M_{3}^2) + N_{2}N_{3}(M_{2}^2 + M_{3}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3}^2M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}}$$

This time, we can use the AM-GM inequality thrice:

$$M_{1}^2 + M_{2}^2 \geq 2M_{1}M_{2} \quad , \quad M_{1}^2 + M_{3}^2 \geq 2M_{1}M_{3} \quad , \quad M_{2}^2 + M_{3}^2 \geq 2M_{2}M_{3}$$

The end result is the same:

$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3} M_{3}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2) + N_{1}N_{3}(M_{1}^2 + M_{3}^2) + N_{2}N_{3}(M_{2}^2 + M_{3}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3}^2M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}} $$
$$\geq \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3} M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3}^2M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}} = 1$$

$$Đ = \frac{M_w}{M_n} \geq 1$$

Now stating the general case:

$$Đ = \frac{M_w}{M_n} = \frac{\sum\limits_{i} N_{i} M_{i}^2 \sum\limits_{i} N_{i}}{(\sum\limits_{i} N_{i}M_{i})^2} = \frac{\sum\limits_{i} N_{i}^2 M_{i}^2  + \color{#ff0000}{\sum\limits_{i,j\ (j>i)} N_{i}N_{j}(M_{i}^2 + M_{j}^2)}}{\sum\limits_{i} N_{i}^2 M_{i}^2 + \color{#0000ff}{2\sum\limits_{i,j\ (j>i)} N_{i}N_{j}M_{i}M_{j}}} \geq \frac{\sum\limits_{i} N_{i}^2 M_{i}^2  + \color{#0000ff}{2\sum\limits_{i,j\ (j>i)} N_{i}N_{j}M_{i}M_{j}}}{\sum\limits_{i} N_{i}^2 M_{i}^2 + \color{#0000ff}{2\sum\limits_{i,j\ (j>i)} N_{i}N_{j}M_{i}M_{j}}}=1$$

$$Đ = \frac{M_w}{M_n} \geq 1$$


How does a redox reaction between ClO- and Br- take place? How does a redox reaction between $\ce{NaClO}$ and $\ce{KBr}$ take place?

will bromide be oxidised to bromine?

$\ce{2 Br- + CIO- +  H2O  ->  Cl- + 2 OH- +  Br2}$

or will bromide be oxidised to $\ce{BrO-}$?

$\ce{Br- + CIO-   ->  BrO- + Cl-}$

In the pH range between 10 and 14, chloride and hypobromide are formed (DOI):

$$\ce{OCl- +Br- ->  Cl- + OBr-}$$


How does the MG811 sensor work? The MG811 is an electrochemical sensor that measures $\ce{CO2}$. I think is something like a redox reaction but i don't understant how it produces $\ce{e-}$ so i can read them and get the $\ce{CO2}$ concentration.

I copy you the reaction that makes the sensor measures

When the sensor exposed to $\ce{CO2}$，the following electrodes reaction occurs：

Cathodic reaction: $\ce{2Li+ + CO2 + 1/2O2 + 2e- -> Li2CO3}$

Anodic reaction：$\ce{2Na+ + 1/2O2 + 2e- -> Na2O}$

Overall chemical reaction：$\ce{Li2CO3 + 2Na+ -> Na2O + 2Li+ + CO2}$

The datasheet says that it workings adopting the electrolyte cell Principle and that it's composed by the following cells:

Air，$\ce{Au}$|NASICON|| carbonate|$\ce{Au}$, air，$\ce{CO2}$

I need you to explain me how actually the sensor is working and with which material are made the electrodes.

The link you gave in the question, actually shows the equation of how this sensor works. You don't need to produce any electrons, you just need to measure the voltage.

Try to write the Nernst equation for the above overall reaction and that will show you what is going on.