How is this mystery chemical reducing carbon emissions? My chemistry is not good enough to solve this problem, and it's driving me crazy.

At my job, a sales agent approached me to be their representative in my area for their product.  It's a product that when mixed with diesel or gasoline, it reduces carbon emissions in the vehicle to almost 0.

Now this seems impossible to me.  The burning of hydrocarbons necessarily creates $\ce{CO}$ or $\ce{CO2}$, and the $\ce{CO}$ can be converted to $\ce{CO2}$ in the catalytic converter by oxidizing it.

In the test results, $\ce{CO}$ dropped to almost 0, and $\ce{CO2}$ stayed the same or went down (also my own calculations show the change in $\ce{CO2}$ was not statistically significant with the small sample size - 3 before and 3 after tests on 3 cars).

I've seen the test results (yes, they could have been faked, but they filmed the testing process).  My buddy also personally tested it on his car and got the same results.

Anecdotal evidence shows that the car's mileage is the same.

Can anyone point me to somewhere to try and figure out what is going on here? It sounds too good to be true, and as such, there is no way I'm going to recommend my company represent this product until I know what's going on.

Here are the mean test results (ppm).  Some of them have high variance.  The only statistically significant changes are $\ce{CO}$ reduction in all cars, and all changes in the first car.  The $\ce{NOx}$, $\ce{CO2}$, and $\ce{CxHy}$ changes in car2 and car3 are not significant.

        Before                          After
        CO    CO2    CxHy    NOx        CO    CO2    CxHy    NOx
Car1    3491  9.4    0.39    46.7       33.3  7.8    0.07    9.7
Car2    164   11.5   0.13    2.7        12.9  11.4   0.37    0.7
Car3    2868  11.1   0.18    14         123   10.9   0.06    39.3

I'm suspicious this may just be ethanol, but the manufacturer swears it isn't.

The advertised ratio is 1 oz to 12.5 gallons of fuel which is less than 0.1% (I believe they used double that in the tests).  Too low to be ethanol anyway, right?

Follow-up:

I have confirmed that I was originally mistaken in that all measurements are ppm.  I received a new copy of the original results, and $\ce{CxHy}$ and $\ce{CO2}$ are indeed measured in percent, rather than ppm.  It was incorrectly rewritten in the report.  The numbers are the same, though.

With reference to the discussion in the question comments, this product does not appear to appreciably reduce carbon emissions in total. It does however appear to reduce carbon monoxide emissions. As carbon monoxide is a product of incomplete combustion, it is plausible that the additive is helping the fuel burn more completely. Broadly speaking, you would hope that the majority of the carbon in your exhaust gas would come out as carbon dioxide, which is 'merely' a greenhouse gas and not truly nasty like CO or PAHs. Prior to the advent of catalytic converters, carbon monoxide poisoning by vehicle exhaust was a much greater risk.

It should be noted that conservation of mass tells us some powerful things about what to expect from an engine - namely that the carbon that goes in to the engine as a component of fuel must come out as exhaust gasses, particulates, polycyclic aromatic hydrocarbons etc., or else be sequestered inside the engine (though the latter is achievable by putting a banana into the exhaust pipe, it is generally done as a prank).

Reduction of carbon emissions in general is achieved through obtaining more work per mass of fuel burnt, which, when coupled with the mass and other design features of the car confers the fuel efficiency. Unfortunately, the data provided don't tell us anything about that. The more complete burning of the fuel, evinced by the reduction of CO concentration in the exhaust gas does imply some increase in efficiency, however CO is a very small component of exhaust gas, indicating alongside the low concentration of unburnt hydrocarbons that the fuel is already burning virtually to completion.

Without knowing the specific claims made by the manufacturer of the additive, I will not make any judgment of good or bad faith, however my distilled conclusions are:

The product does not dramatically impact total carbon emissions
The product appears to appreciably reduce carbon monoxide emissions


What is the purpose of UV-VIS Spectroscopy? I understand the purpose behind infrared spectroscopy and mass spectroscopy, but UV-VIS doesn't make much sense to me.

Through research, I see that it has to do with electron orbitals, and it has a relation to Beer's Law.

But what exactly can I learn about a molecule from its absorbance/transmittance of UV/Visible light?

The broad absorption bands of the electronic transitions observed in UV/Vis spectroscopy/spectrophotometry, as well as the myriad types of electronic transitions that might lead to absorption at a given wavelength, indeed make the technique much less powerful for identification of unknown compounds as compared to, e.g., IR, NMR, microwave, and mass$^*$ spectroscopies.

Jon Custer is correct in his comment to the question that UV/Vis spectra can provide key information in identifying unknown species.  However, I know of only two approaches to obtaining predictions of a given UV/Vis spectrum:

Experimental spectra of compounds suspected to be structurally related to the unknown species
Quantum chemical computation of the electronic transition spectra of candidate species

The former approach requires the relevant experimental data to exist; the latter is not trivial to perform accurately. Both suffer from significant uncertainties: How similar is a reference experimental spectrum to that of the unknown species? Can I trust that my quantum computation will give realistic band locations and strengths?

In my experience, the strength of UV/Vis lies in its facility for inexpensive, convenient, routine, non-destructive quantitation of known solutes, with known absorption spectra and/or extinction coefficient(s) at given wavelength(s).  The following list of strengths is almost certainly incomplete:

The instrumentation and consumables are relatively inexpensive.
No auxiliary cooling, gas supply, etc. are required.
Preparation and analysis of fixed-volume samples is trivial: pour liquid into cuvette; place cuvette into instrument; measure.
Continuous analysis of dynamic systems (e.g., chromatographic eluates and ongoing reactions) is possible using a flow cell.

A simple example is described here, unfortunately paywalled.
A wide variety of configurations is presented at a commercial vendor here.
(No endorsement intended. There are numerous other suppliers out there; I chose this one because it had a good variety shown in schematic.)

Interference from most solvents is minimal, particularly in the visible range.
Beer's Law in its linear form has a relatively high dynamic range for most UV/Vis-active solutes$-$that is, the concentration where absorbance deviates from linearity is much greater than the minimum concentration detectable:

$$
\frac{C_{\max}}{C_{\min}} \gg 1
$$

Another implication of Beer's Law is that interference(s) from a small number of absorbing co-solutes are relatively straightforward to handle, either by measuring at isobestic points or by measuring multiple wavelengths and solving the linear absorbance equations for the unknown concentrations. For example, if two species $\ce{X}$ and $\ce{Y}$ are present and absorbances are measured at wavelengths $\lambda_1$ and $\lambda_2$, the concentrations can be found by solving the following two expressions:

$$
A_{\lambda_1} = \epsilon_{\ce{X},\lambda_1}\cdot C_\ce{X} + \epsilon_{\ce{Y},\lambda_1}\cdot C_\ce{Y}
$$
$$
A_{\lambda_2} = \epsilon_{\ce{X},\lambda_2}\cdot C_\ce{X} + \epsilon_{\ce{Y},\lambda_2}\cdot C_\ce{Y}
$$

The steady-state nature of the measurements allows for 'deep' examination of a particular sample.

The full UV/Vis spectrum for a given volume of sample can be measured, giving a rich "fingerprint" of the particular solution and the solutes in it
The absorbance at a particular wavelength can be tracked over time, allowing for detailed kinetic analyses

At least in the visible range, the quality of the substance being measured is directly perceptible to the human senses. This is actually quite viscerally satisfying, to see that a sample anticipated to contain significantly more of a solute does indeed produce a noticeably stronger coloration of the solution.

While IR and microwave spectroscopies share some of the above advantages, they are far less suited for analysis of solutes, as the solvent absorbance is often much more likely to confound measurements.  Further, the instrumentation and sample preparation are generally much more involved.

Mass spec is terrifically expensive and complicated, is an intrinsically destructive method, and is less well suited for direct quantitation since there is no way to know a priori what fraction of the injected sample is actually making it to the detector and thus careful calibration is required.

NMR, though non-destructive, also requires quite expensive equipment that is complicated to operate properly, and interpretation of the spectra requires quite a bit of practice.

I know too little about EPR/ESR, EXAFS, XANES, and others to comment on them here. I believe the above strengths hold in comparison to those methods as well, however.

$^*$ Technically, MS is mass spectrometry, not spectroscopy, since the measurement is of discrete ion impacts on a detector, as opposed to absorbance of portions of the electromagnetic spectrum by a detector.


A semi-transparent material that is a solid I am trying to find something I can get the ingredients for and make that starts out a liquid and sets into a solid. My end goal is to create a simple 8bit art figure so I want to pour it into a mold and have it harden. Also I would really like something fully or partly transparent (I was looking at aerogel because of the nice blue translucency but it seems too brittle to work with). So my question is what would be the best compound to try to create this figure out of?

There are other plastics that have better mechanical properties, but probably nothing is easier to find than polyester casting resin. It works really well in silicone moulds and there are various pigments you can buy to change its colour and opacity. It isn't as optically clear as some polyurethane or epoxy resins (you won't be making lenses out of polyester), but it's not bad and you can find it in a lot of art supply stores.

As for something with aerogel's smokey translucency, I don't know. Aerogel's translucency (its colour, as well) is due to scattering off its nanoporous structure. It may be possible to mimic the look by blending silica particles or something into a plastic to cause scattering, but I can't recommend anything specific.


In order to solve for this equation for heat capacity, do I need to treat it as a system of equations? For a lab, we took the temperature of 15ml of tap water, $q_{cool}$, then added 20ml of tap water warmed to 80 degrees Celsius, $q_{warm}$.

We recorded a baseline temperature of 24.8, a temperature of 56 after adding the water, and a temperature of 42.9 100 seconds later.

I have the following equations:

$q=m × 4.184 J/(g*C) × ΔT$
$q_{warm} + q_{cool} + q_{calorimeter} = 0$
$q_{calorimeter} = - (q_{warm} + q_{cool} ) = C_{calorimeter} × ΔT$.

In the first formula $ 4.184 J/(g*C)$ is the specific heat of water.

To the best of my knowledge, specific heat is usually expressed in grams-Kelvin; however, the TA may have provided the info in grams-Celsius (their units are equal, so I presume they're the same thing). However, she also uses $C$ to designate heat capacity as she does in the third equation.

If, in the first equation, $C$ designates heat capacity, then I have three equations, and three unknowns: $q_{calorimeter}, q_{warm}$, and $q_{cool}$, and I'd treat it as a system of three equations and solve for them.

However, if, in the first equation, $C$ designates 'Celsius' then I could just plug in the values for $q_{warm}$ and $q_{cool}$ and solve for $q_{calorimeter}$.

Two related questions:

In the first equation, what does $C$ designate.
If it designates the heat capacity of the system, do I solve for $q$ by solving a system of equations?

In the first equation, what does C designate.

Since the final value is energy in joules, then it should be the the unit for temperature. She should have used °C to be clear.

  If it designates the heat capacity of the system, do I solve for q by solving a system of equations?

No, because you don't have the correct equations. You need the amount of heat released from the hot water minus the amount of heat absorbed by the cold water to give you the heat absorbed by the calorimeter.


Is heat just a change in temperature? According to my chemistry textbook, heat is defined as "Energy that is exchanged because of a difference in temperature or a change in phase." My textbook also says, "Heat is a form of energy. Temperature, on the other hand, is a measure of how quickly the molecules (or atoms) in a substance are moving," implying that temperature is not a form of energy, just a measurement of a physical condition.

This seems all wrong to me. If heat is just a change in temperature, but temperature isn't energy, doesn't that imply that energy is created when two objects of different temperature interact, or is there potential heat energy in the temperature of an object?

Also, the textbook says that the reason we feel hot is because of heat from temperature change, and not temperature itself, but this doesn't explain why chemical reactions occur faster at different temperatures. Nor does it explain why there are different states of matter at different temperatures, or even why our bodies need to stay at a certain temperature at all. Where is the energy from heat and other chemical processes coming from, exactly?

Other sources online also say that heat is a change in temperature and temperature is not energy, only a measurement, so I don't think it's just my textbook. I feel like I'm missing something.

[...] is there potential heat energy in the temperature of an object?

The energy associated with temperature is called thermal energy. If you know the heat capacity of an object, you can figure out how much energy is required to change the temperature by a certain amount.

  If heat is just a change in temperature, but temperature isn't energy, doesn't that imply that energy is created when two objects of different temperature interact [...]?

Heat is energy transferred from one object to another. When heat is transferred from a hotter to a colder object, the hotter object loses thermal energy and the colder object gains it. No energy is created.

  Where is the energy from heat and other chemical processes coming from, exactly?

A chemical reaction can give off heat because electrons can be in higher states (weak bonds) and in lower states (strong bonds). Potential energy can turn into thermal energy, and vice versa. It takes energy to melt ice, so melting ice cools down the surroundings. An explosive reaction releases energy, so it increases the temperature of the surroundings by transferring heat to it.


Why is the cathode reaction required in a fuel cell? So, at the anode of a fuel cell hydrogen is split into a positively charged proton and an electron. Then, there is the membrane which can let only protons through to the cathode, whereas electrons need to follow external circuit (effectively powering it).

Question: why do we need any other reaction on the cathode? Isn't the fact that positively charged ions travel through the membrane to the cathode enough to create the EMF?

Trying to solve it I assume that the reaction is there to actually apply some kind of force to move the ions though the membrane to the cathode, otherwise they'd just stay and reunite with electrons on the anode. What is that force and how does it work?

Why do we need two half reactions, one at the anode and another at the cathode?

The whole thing eventually grinds to a halt without the other reaction. The electrons generated at the anode travel through the circuit to the cathode. What happens then? Something needs to happen to the electron.

Additionally, what happens to that proton? As the concentration of protons increases, what will balance the charge that is produced in the cell, given that the electron is elsewhere?

We can imagine that we can set up a cathode in which the reverse reaction of the anode occurs:

Cathode: $\ce{H2 -> 2H+ + 2e-}$
Anode: $\ce{2H+ +2e- -> H2}$

However, in this case, there is overall no net reaction, which means that $\Delta G =0$ and since $\Delta G = nFE$, then the cell potential $E=0$. We need a different reaction for the cathode to consume electrons.

For example:

$$\ce{O2 + 2H2O +4e- -> 4OH-}$$

Since this reaction produces hydroxide anions, then we have something for the proton to do:

$$\ce{H+ + OH- -> H2O}$$

Now we have the following combination of reactions:

Anode: $\ce{H2 -> 2H+ + 2e-}$
Cathode: $\ce{O2 + 2H2O +4e- -> 4OH-}$
Neutralization $\ce{H+ + OH- -> H2O}$
Net: $\ce{2H2 + O2 -> 2H2O}$

This net reaction is exergonic, and thus produces a potential.


Why do we need to add FeBr3 for bromination of anisole, but not for bromination of phenol? I understand that we need to add $\ce{FeBr3}$ while brominating benzene in order to polarize the bromine molecule. However, we do not need to add such a Lewis Acid in the case of phenol, because it is activated by the $\ce{-OH}$ group, and so polarizes the bromine molecule on its own. Anisole is also highly activated, however we still add $\ce{FeBr3}$ in order to brominate it. Why?
(An answer which compares bromination in phenol and anisole will be highly appreciated.)

I don't think polarization is the issue. Adding $\ce{FeBr3}$ is just to reduce the $\ce{Br-Br}$ bond strength, speeding up the reaction (attack of the aromatic ring on the bromine is part of the rate determining step, which is slow in the bromination of benzene). It functions as a catalyst being recycled when the $\ce{Br-}$ grabs a $\ce{H+}$ to restore aromaticity in the ring. I'm pretty sure — at least I see no reason to doubt — that the catalyst requirements for the bromination of anisole and the bromination of phenol do not differ much, if at all. That is, no catalyst is required for the bromination of anisole.


What is the difference between convection and conduction? I was having a conversation with my dad the other day and he described something as happening due to convection, and it dawned on me that I have no real clue what the difference between and convection and conduction is.

I started to do some reading on this and haven't really found any good clarification.

Here is what I do understand:

Convection:

Involves transfer of heat within a single substance by mass flow
Does not take place in solids due to a solid's inability to behave like a fluid
The heat flow is governed by diffusion of higher energy molecules and thus a distribution of that energy evenly, or by advection which basically seems to be bulk diffusion to me

Conduction:

Transfer of heat within a body by diffusion
This diffusion differs from convection because convection supposedly happens through a material carrier
Can take place in solids

Confusion:

No doubt, some of the points I just put up there are wrong or else I basically see no point in making a distinction between convection and conduction when the processes can be described more easily by diffusion which redistributes energy by principle of maximizing entropy.

So why define these two different terms when they are essentially the same process perhaps just happening in different mediums?

Contradictions:

I was looking at the Wikipedia pages, which I usually find to be reliable, for convection and conduction and was only confused more.

For instance, from the conduction page:

  "In gases and liquids, conduction is due to the collisions and diffusion of molecules during their random motion."

Now, from the convection page:

  "Convective heat and mass transfer take place both by diffusion – the random Brownian motion of individual particles in the fluid – and by advection, in which matter or heat is transported by the larger-scale motion of currents in the fluid."

Those two statements seem to be basically the same. The mechanism for the heat transfer is the diffusion of randomly moving moving particles.

On the other hand, the conduction page explains that conduction takes place in solids through,

  "In solids, conduction is mediated by the combination of vibrations and collisions of molecules, of propagation and collisions of phonons, and of diffusion and collisions of free electrons."

That seems like a perfectly good explanation of what takes place when heat is being distributed towards thermal equilibrium in a solid.

Questions:

What about the above statement that conduction in solids is mediated through collisions of electrons and vibrating atoms makes that process conduction rather than convection?

What of the above points are incorrect?

On a molecular level, (i.e. using typical chemistry ideas) what makes a distinction between convection and conduction?

Finally, what is meant by the statement that convection happens through a material carrier?

Does that imply that conduction does not happen through a material carrier? If so, how?

I hope you can see my confusion between these two terms. They seem very convoluted to me.

I would point out that good and correct answer need not address each of my sub-questions specifically because a good answer may clear up my confusion and hence eliminate one of those sub-questions.

Convection is a mechanism that enhances conductive heat transport, rather than being a separate physical phenomenon.  It operates by bringing hotter parcels of fluid and colder parcels of fluid into closer physical proximity with one another so that conduction can take place more easily and rapidly.  One way this is accomplished is by shearing the fluid near a heated wall in such a way that the colder faster-moving layers further from the wall flow over the hotter slower-moving layers closer to the wall to physically cause the temperature gradient normal to the wall to increase.  This enhances the rate of conductive heat transfer.  Another situation is where the fluid is being mixed by a stirrer to create thin striations of hotter fluid and colder fluid throughout the fluid.  Conduction between these hotter and colder striations increases, again because of the much larger temperature gradients that have been created.  This is typical of how convection works.

What I'm saying is that the deformations experienced by the fluid while flowing are such that they cause the temperature gradients in the fluid to get higher, and this increases the rate of heat transfer.

Imagine a deck of cards split in half, with all the red cards (hot) in one half and all the black cards (cold) in the other half. If you bring these two slabs of hot and cold material together and wait, heat transfer will occur very slowly between the hot and cold regions. However, if you shuffle the deck so that there is a hot (red) card sandwiched in between every pair of cold (black) cards, and a cold (black) card sandwiched in between every pair of hot (red) cards, the heat transfer between the hot and cold layers will be much more rapid. This shuffling effect is what convection does.


Does the 'X' of Grignard's reagent react? Till whatever I have studied in organic chemistry, Grignard's reagent has been the most helpful thing everywhere, whether is it preparation or reaction of organic compounds. But I haven't ever seen the Halogen part of RMgX reacting with any compound. Does the Halogen ever react in any reaction?

Lets look at Grignard's reaction route. Key step is nucleophilic addition of R. And MgX is removed with acid on the next step. Nobody cares about X - if you want MgX2 in solution you just add MgX2. The main reason to use Grignard's reagent is nucleophilic addition of R.


How many basis functions used in STO-3G and 6-31+G** for the water molecule? I am a little confused about how to decide how many basis functions that are used in a particular basis set for a given molecule.

STO-3G is a so-called "minimal basis set", meaning that only one basis function is used for each atomic orbital in the atoms of which the molecule is made from.

The water molecule has two H atoms and one O atom, with 1s1 and 1s2 2s2 2p6 electron configurations, respectively. When STO-3G is used, would that then mean that three 1s type basis functions, one 2s type basis function, and 3 p-type basis functions are used, totalling at 7 basis functions that are used, each being made up from a linear combination (LC) of three simple Gaussian functions?

The 6-31+G** is a split-valence basis set, which includes both diffuse and polarisation functions. The two asterisks tell that polarization functions are used for both the H atom and the O atom, p and d polarization functions respectively. I am not sure "how many" polarization functions are used per atom - just 1, or one for each of the p-orbitals? (Some basis sets are of the form 6-31G(3pd); does that mean that 3 p and 1 d polarization functions are used, or 9 p functions?) The single "+" sign means that just one diffuse function is used for the O atom.
So, I think that the 6-31+G** for the water molecule equates to a total of 16 basis functions:

1s in oxygen: one LC of 6 GTOs
2s in oxygen: one LC of 3 GTOs + 1 GTO
2p in oxygen: one LC of 3 GTOs + 1 GTO (for each of px, py, and pz)
1s in hydrogen 1: one LC of 3 GTOs + 1 GTO
1s in hydrogen 2: one LC of 3 GTOs + 1 GTO
Diffuse: one diffuse basis function for the oxygen atom
Polarization: one polarised basis function for the oxygen atom
Polarization: one polarised basis function for the hydrogen atom
Total: 16

Is this correct? I do not feel confident carrying out this analysis, especially the point concerning how many polarization functions for the p/d-orbitals that should be used.

Notation

** is just a synonym for (d,p), so that 6-31+G** basis is just a different name of 6-31+G(d,p) one.
The meaning of (d,p) is that one set of d-type polarisation functions is added for heavy atoms and one set of p-type polarisation functions is added for hydrogens.
6-31G(3pd) is essentially 6-31G(3p1d), i.e. it is 6-31G plus 3 sets of p-type polarisation functions and 1 set of f-type polarisation  functions on heavy atoms.

How many functions are there in a set?

There are always 3 functions of p-type, but the situation with d- and f-type functions is a bit more involved.
There are either 5 or 6 functions of d-type and either 7 or 10 functions of f-type depending on is a calculation restricted to use only spherical harmonics or not.

Cartesian vs. spherical harmonics

Basis functions can be written in two forms: as spherical harmonics
$$
  N(n, \alpha) Y_m^l r^n e^{-\alpha r^2}
$$
or Cartesian ones
$$
  N(l_x, l_y, l_z, \alpha) x^{l_x} y^{l_y} z^{l_z} e^{-\alpha r^2} \, .
$$
And for $l \geq 2$ the number of Cartesian harmonics is greater than spherical harmonics. For instance, a set of d-type function can have five spherical harmonics ($d_{3z^2-r^2}$, $d_{xz}$, $d_{yz}$, $d_{xy}$, $d_{x^2-y^2}$) or six Cartesian harmonics ($d_{zz}$, $d_{xx}$, $d_{yy}$, $d_{xz}$, $d_{yz}$, $d_{xy}$). Note that these six Cartesian harmonics can be transformed into five above mentioned spherical harmonics of d-type plus one function of s-type that can be neglected. Similarly, there are 10 Cartesian f-type functions and 7 spherical ones, and the later  could be constructed from the former with 3 p-type functions as a by-product.

The choice which harmonics to use can be controlled through the corresponding option in an input file and default choice varies from program to program. If you're specifically interested in Gaussian, then, if I remember correctly, by default Cartesian d-type functions and spherical f-type functions are used for the Pople 6-31G basis set. Thus, there will be 6 functions in a set of d-type functions and 7 functions in a set of f-type functions. This might be the case that Pople basis sets were even originally defined to use Cartesian d-type functions and spherical f-type ones, but I'm not sure.


Phase subscript for supercritical reagent? When writing a chemical reaction, one can use subscripts to denote the phase of a reagent, e.g. $\ce{SiO2 (s)}$, $\ce{H2O (l)}$, $\ce{F2 (g)}$ or $\ce{Na^+ (aq)}$. However, I need to write a reaction involving a reagent that is above its critical point. What is the subscript for a supercritical reagent?

The accepted if not standardised notation is Sc or sc - I've seen both.  Capital "S" followed by a lower case "c", e. g.  ScCO2 is the one I've seen most often in chemistry related papers.
EDIT - apologies, this is a prefix and not a subscript. Hopefully you get the point


why "amu" is used as a unit in relative atomic mass? Unified atomic mass unit (amu) is defined as $1/12$ of the mass of a Carbon-12. I want to know is this amu equal to $\pu{1.99e-23 g}$ and why do we use amu which is $1/12$ the mass of carbon? Why use "amu"?

$$
\pu{1 u} = \pu{1.660539040(20) \cdot 10^{-27} kg}
$$

The u (amu is the old unit name) is $1/12$ of the weight of an $^{12}\ce C$ atom.

The way the u is chosen ensures that all core and atom masses are multiples of 1($\pm $0.1) u. The u is a count of nucleons in an atom.

Until the 60s, chemists used $1/16$ of the mass of a average atom in a oxygen-isotope-mixture and physicist used $1/16$ of the mass of a $^{16}\ce O$ atom.

It's really just about choosing what is the best to measure and have a consistent result with. So you would chose an element with a very consistent isotope ratio (isotope-pure $^{12}\ce C$ e.g.) and one where you can measure weights easily.


Percent concentration of 500 μM of H2O2 What is the percent concentration of hydrogen peroxide which has a concentration of $500~\mu\mathrm{M}$?

Is $500~\mu\mathrm{M}$ $\ce{H2O2}$ toxic to the eyes?

A particular honey has a concentration of $\ce{H2O2}$ of $500~\mu\mathrm{M}$ after a 24-hour incubation time, after it has been diluted. I am trying to figure out what the concentration is in percent (%).

I recalculated and the result is actually $0.00153\%$ for $500~\mu\mathrm{M}$. $0.0012\%$ is the equivalent to $400~\mu\mathrm{M}$, but I need someone to confirm this. Someone on researchgate wrote that $30\%$ of $\ce{H2O2}$ is $9.8~\mathrm{M}$.

I will assume you are asking for percent mass/volume, $\mathrm{g/ml}$, since to calculate percent mass/mass you need the density of the liquid (unless you assume it is water which is $1~\mathrm{g/ml}$ so will give the same answer).

We have $500~\mu\mathrm{M}$ of hydrogen peroxide, which means we have $500~\mu\mathrm{mol}$ of hydrogen peroxide per 1 litre of solution. $500~\mu\mathrm{mol} = 0.0005~\mathrm{mol}$. Now you can work out the molar mass of hydrogen peroxide and multiply it by the number of moles to get the mass of hydrogen peroxide per litre of solution. This gives $0.017~\mathrm{g}$. There are $1000~\mathrm{ml}$ in 1 litre. So you have $0.017~\mathrm{g}/1000~\mathrm{ml} = 0.000017 = 0.0017\%$.

For your question about whether hydrogen peroxide is toxic to the eyes, you can search the internet for the Materials Safety Data Sheet (MSDS) of hydrogen peroxide (Sigma-Aldrich is a good site to search). This contains all the hazards associated with this chemical, and when you do laboratory work you always have to look these up for each chemical before you start a reaction. The MSDS states that hydrogen peroxide can cause severe eye damage, among other hazards.


Can the hydride ion act as a ligand? Hydride ion, as I know is a pretty powerful base, much stronger than hydroxide ion and cannot exist in an aqueous phase.

Can it act as a ligand in coordination compounds? Hydride ion has its electronic configuration as $\mathrm{1s^2}$ in a symmetrical s orbital so the tendency to donate a lone pair into a metal atom would be less. But can its high nucleophilicity allow it to bind to the metal atom despite the symmetry of its orbital?

Yes, there are many examples. For instance

The common reducing agents LithAl, $\ce{LiAlH4}$, and sodium borohydride, $\ce{NaBH4}$ (last one you might argue about)
$\ce{[ReH9]^{2-}}$, see https://en.wikipedia.org/wiki/Potassium_nonahydridorhenate
Metal carbonyl hydrides, see https://en.wikipedia.org/wiki/Metal_carbonyl_hydride
More generally many organometallic compunds, e.g. $\ce{(C5H5)_2ReH}$

A quick google also shows up http://www.ilpi.com/organomet/hydride.html


MO diagram when three orbitals of the same symmetry is allowed to mix How exactly should we combine three orbitals when we have three orbitals that are of correct symmetry to interact?
Take $\ce{H2O}$ for example. $\mathrm{2s},$ $\mathrm{2p}_z$ and the $A_1$ group $(\ce{H2})$ orbitals can interact. It would seem trivial that the lowest bonding orbital is $\mathrm{2s}$ in phase with $A_1,$ but what about the anti binding orbital? Most resources suggest $\mathrm{2p}_z-A_1$ out-of-phase interaction, but is $\mathrm{2s}$ with $A_1$ out of phase not correct?
Could this be due to we try to construct the anti bonding orbital with least overlap possible?

You can estimate what happens by using a Huckel MO approach and, even though it is approximate, it shows the trends. With three similar orbitals on three atoms, such as 3 p orbitals these can be combined in two ways depending on whether the molecule is triangular or linear in shape. The MO's produced always sum to the same energy as that of the three p orbitals.
Without going into details, if we place them in a triangular arrangement each of the p orbitals can interact with each of the other two and the MO orbitals produced have relative energies -2 ,+1, +1 so when three electrons are added there is an overall lowering of energy with two in the lowest energy orbital and one in either of the two degenerate orbitals.
If linear, the three p orbitals will only interact significantly with the one next to it, 1 to 2, 2 to 3 for example. The MO orbitals all have different energy with one lowered a lot and two raised less so, so that the total energy is the same as the initial p orbitals. None are degenerate.
The conclusion is that the energy of the MO orbitals produced depend on the interactions between the p orbitals and this often depends on the geometry with closest p orbitals interacting most.
The Huckel method is described in many places, Wikipedia and most phys. chem. textbooks, if you want to try the calculation for yourself.


How to distinguish between Na2CO3 and NaHCO3 by a chemical test? $\ce{Na2CO3}$ and $\ce{NaHCO3}$ have similar reactions. For example, they can react with acid to produce $\ce{CO2}$.
So my question is:

  Can we use a chemical test to distinguish between $\ce{Na2CO3}$ and $\ce{NaHCO3}$?

I know that $\ce{NaHCO3}$ can be thermal decomposed into $\ce{Na2CO3}$ via this reaction: $2\ce{NaHCO3->} \ce{Na2CO3} + \ce{CO2} + \ce{H2O}$
However, I am not sure if thermal decomposition is a chemical test.

Heating salt is an important test in determining the anion in the salt by checking the gas evolved. This test is called dry heating test and is a important part of qualitative analysis of salt. But this test goes under physical examination of salt (checking color, appearence etc.).

One important chemical test to differentiate carbonate and bicarbonate is $\ce{MgSO4}$  test.

$$\ce{Na2CO3 + Mg^2+ -> MgCO3↓(white) + Na+}$$

$$\ce{NaHCO3 + Mg^2+ -> MgCO3↓(white) + H2O + CO2 ^}$$

When carbonate salt is reacted with $\ce{MgSO4}$, a white salt precipited while bicarbonate salt on reacting with $\ce{MgSO4}$, not only a white salt precipited but also a brisk effervescence occur. The gas  is detected  by passing it through lime water.


What is the correct equation for calculating the electric potential of an electrochemical cell? When I'm studying electrochemistry, we were given an equation that all chemistry students are familiar with, which is:

                    Ecell = Eoxidation - Ereduction

However, when I came to study electroanalytical methods, I came upon this equation:

          Ecell = Eoxidation + Ejunction - Ereduction  - Eelectrode

where Ejuntion is the uncertain potential produced from the salt bridge and the Eelectrode is the potential produced from the reference electrode used in the experiment.

My question is, is this the real equation when calculating problems for an electrochemical cells? I'm thinking that they probably cut Ejunction out of the equation because it is impossible to know the exact value for it. In addition, I think that they cut out the Eelectrode in the equation because most of the experiment data came from a SHE electrode where Eelectrode = 0.00 V.

I am afraid your equations are other way around.

  Ecell = Eoxidation - Ereduction

This is not the accepted convention among electrochemists. Either the reference is outdated or there is a typo. All electrochemists in the world now follow the convention of writing

Ecell= Ereduction-Eoxidation or Ecell = Ecathode-Eanode  ---Eq(1)

The reason is historical and I have written details elsewhere, but briefly the main reason is that Americans used one sign convention and the Europeans used another. It took more than 50 years to come to a consistent sign convention, which was originally used by a German chemist, Ostwald, 100 years ago. Given that all electrode potentials are now written as reduction potentials the above Eq. (1) is valid. If you look up the tables, the signs of the electrode potentials correspond to the actual electrostatic sign (+) or (-) of the electrode.

Now this simplistic equation (1) ignores any "connections" among the half cells. In reality we always need an electrically conducting connection to the half cells to keep them separated yet electrically connected. This "connection" could be a salt bridge, or a porous ceramic frit (used in pH probes and so on).

There is a general interesting electrochemical fact whenever there is an interface (the location where two surfaces meet) one can build up difference of electrical potential. This is a fundamental material property. Recall the simple voltaic cell, which has copper and zinc pieces separated by a piece of wet paper containing some common salt.

Liquid junction potential (Ej) is another type of potential which originates at the interface of two electrolytes with different concentration of ions (=charges). In that case we add the liquid junction potential to this Ecell as follows

Ecell= Ecathode-Eanode+ Ej

I have not seen your other equation.  Ecell = Eoxidation + Ejunction - Ereduction  - Eelectrode. Which book is using this expression and what is its year of publication?


Effect of pH on charge of jasmonic acid It has been such a long time, that I totally forgot how it works.

I am interested in the charge of jasmonic acid at different concentrations.
More specifically, I want to know at what pH my molecule has no charge (or I should rather say jasmonic acid would occur as a zwitter ion).

Only the decaboxylic group can be deprotonated at higher pH.

Is it correct to assume that most of the time, when you have an
acidic compound it would predominantly occur in a nonionic form?
Does the $\log{D}$ say anything about this? does the maximum
$\log{D}$  value corresponds to the pH at the iso electric point?

If we would take aspirin for example

https://disco.chemaxon.com/apps/demos/logd/

at what pH should I expect the uncharged version?

Jasmonic acid has one group that is significantly ionizable, the carboxylic group:

Most molecules in the sample will be negatively charged when the $\mathrm{pH}$ is higher than the $\mathrm{p}K_\mathrm{a}$ of the molecule and an equal number will be neutral or charged when $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$. The same will hold for aspirin, which also has a carboxylic acid group. Below a $\mathrm{pH}$ equal to the $\mathrm{p}K_\mathrm{a}$ it will be mostly uncharged, and above this $\mathrm{pH}$ it will mostly be charged.

If you look up $\mathrm{p}D$, you might encounter the following equation:

$$-\mathrm{p}D = \log D_\mathrm{oct/wat} = \log\left( \frac{[C]^\mathrm{ionized}_\mathrm{octanol} + [C]^\mathrm{unionized}_\mathrm{octanol}}{[C]^\mathrm{ionized}_\mathrm{water} + [C]^\mathrm{unionized}_\mathrm{water}}\right)$$

Since the concentration $[C]$ of the ion is typically higher in water than in octanol, increased ionization at high $\mathrm{pH}$ will increase the solubility in the water phase, and $\log D_\mathrm{oct/wat}$ will decrease (become more negative) at high $\mathrm{pH}$, that is, $\mathrm{p}D$ will increase (become less negative). You can expect the solubility in octanol to continue to decrease as you near and pass the isoelectric point.

For more complex molecules with more ionizable groups you will need to consider individual contributions from different groups. Basic groups will respond in the opposite manner to acidic ones.


Is enthalpy the same thing as potential energy? From what I know in a closed system the potential energy + the kinetic energy is constant since no external work is done on the system. In my textbook it says that enthalpy is equal to the sum of the KE + PE meaning total energy. My question is how can the energy be lost to the environment after the collision of two particles. Since as the particle approach each other KE is turned into PE until activated complex is reached. Once it is reached they break up into smaller particles, repel each other and convert the PE back to KE. In conclusion all energy in the system is constant and no energy is lost. I don't quite understand why we have Enthalpy vs Time graphs. Since the energy in the system was never changed it simply converted between forms. For my homework when I pretended that enthalpy was simply PE and that endothermic = increase in PE and exothermic = decrease in PE I usually got it corrected(Such as question below). Is it not true that if you decrease PE by a amount KE would increase by that amount hence no change?

For instance a question from my homework "After a reaction, the product have less KE than original reactant molecules. Is the reaction endothermic or exothermic?"

Im thinking to myself that if KE has decreased PE must increase hence no change in energy so neither endothermic or exothermic. But the answer is endothermic.

In conclusion my main question is can I ignore the definition the textbook gave me and define enthalpy as total PE and delta enthalpy = change in PE rather than total energy.

In any way, ignoring the definitions given in a textbook is a bad idea. As for the question, the main problem in your thoughts is that you mixed too many things from different fields of science. Your example (two particles, kinetic energy, potential energy etc) looks like classical mechanics. So if you want to explain it, please, use tools offered by classical mechanics. Enthalpy is a measure of total energy in thermodynamics. It is used to explain processes taking place in large systems, which can be characterized with volume, pressure, temperature etc. It is not used to describe one particle.
Maybe you can take it into account and revise your question. Make it clear exactly what you want to understand: change in the kinetic and potential energy in interactions of two particles or change in enthalpy in exothermic and endothermic reactions.


What kind of string to use for the ice fishing experiment for kids? Classic ice fishing experiment for kids.

I used nylon because I saw it in a stationery, and apparently that doesn't work because I guess it's too slippery. I tried thread from my mom's sewing kit, but that doesn't work either.

I'm not sure I can find yarn or kite string in the stationery or nearby so please suggest alternatives expected to be found in the average household, supermarket, grocery store or convenience store.

Also how long should we wait? The above video takes only 10 seconds while this video takes about 2 minutes (off screen). What factors influence waiting time? Amount of salt? Type of string?

This resource has an explaination of the behavior, which relies on the salt-dissolved water wicking into the string. This would mean that you need a type of string which will absorb water. It's surprising to me that your nylon string doesn't work (as long as it's thick/low-density enough to absorb water, probably unlike fishing line), is it possible it has a hydrophobic coating?

I would recommend any cotton-based string or twine. Cooking twine and sewing thread is often cotton, if you have access to that. I've had no problems with these.

Edit: I just thought, that you might want to try this not in a cup of water first (like in the link above). You might potentially run into not-string-related problems if your water isn't cold enough, or if your ice gets too submerged when you put the salt on it. It's possible your nylon string is fine.


Wieland's method for triphenylamine bromination I'm looking for a Wieland's paper on triphenylamine bromination. I've found other papers referencing it like this but I don't find the actual paper.

Does someone know where could I find it? Or whats Wieland's method all about?

From the reference you supplied the Wieland paper is Chem. Berichte 40 (1907) 4278

The bromination method (3 eq of Bromine in chloroform slowly added to a chloroform soln of triphenylamine) in the experimental section of the reference you supplied appears to be Wieland's method - it is referenced as such


Is ingesting water during swimming harmful? Sometimes I feel the water in the pool is heavily chlorinated. However it is impossible for me to swim without ingesting some water. I am only a learner but even advanced swimmers do concede that they do ingest some water at times.

Could this be potentially harmful depending on the chemical content in the pool? Are there studies which advise on the extent of permissible chemicals that can be added to a pool?

Could this be potentially harmful depending on the chemical content in the pool?

Probably not, unless you managed to drink quite a bit of pool water. Worldwide, millions (hundreds of millions?) of people have collectively ingested millions (billions?) of liters of pool water - with varying chemistries - over many decades (myself included), and nobody seems to have suffered too much. I believe this falls into the "generally regarded as safe" area.

  Are there studies which advise on the extent of permissible chemicals that can be added to a pool?

Numerous ones, Google away. These guidelines, from the State of Indiana, are representative of the recommended levels of various chemicals present in common swimming pools.


Is there a relationship between pH indicator colors and the visible light spectrum? When I examined the colors showed on a pH paper which is red for acids and violet for bases, greenish yellow for neutral and other ranges in between, I noticed that the sequence is exactly same to that of visible light spectrum that is violet, indigo, blue, green, yellow, orange, and red.
Is there a connection between the two, and if yes what it is?

It has to do with the way pH indicators and particularly large-range pH indicators work.

In simplest terms, a pH indicator is simply a weak acid/base which has one color in its acidic form, and one color in its basic form. When the solution is highly acidic, all of the indicator molecules are protonated, and the color of the indicator is the color of the protonated form. When the solution is highly basic, all of the indicator molecules are deprotonated, and the color of the indicator is the color of the deprotonated form.

Where it gets interesting is where the pH of the solution is close to the pKa of the indicator. In those situations, you have some of the indicator molecules in the protonated form, and some of them in the deprotonated form. You don't ever get "fractionally protonated" forms - each individual molecule is either one or the other. Since light absorption happens on a molecular level, you get

You can see this with a titration of something like Bromothymol Blue [image]. The protonated form is yellow, so at low pH, everything is yellow. The deprotonated form is blue, so at high pH, everything is blue. At around the pKa (~7), there's near equal amounts of the two forms, so it acts like you mixed equal amounts of a yellow pigment/dye and a blue pigment/dye. That is, you get a green solution due to the subtractive color mixing model. As you titrate the pH, you get a range of colors as you add and remove the amount of protonated and deprotonated forms, as would be indicated by the Henderson–Hasselbalch equation.

Okay, so that's a single pH indicator, which is normally limited to a small range. (Because more than ±1 pH unit to either side of the pKa you typically have only a single species present.) To get broad range pH indicators, you have to mix multiple different pH indicators, each with their own different colors and changes. But you have to do this intelligently, as the indicators are always present, and will always contribute to the color.

Here's a chart of some of the major pH indicators, their pH ranges and their colors. You can see what might happen with various indicator combinations. For example, with an ethyl violet/ethyl orange mixture, at low pH you'd have an orange solution (yellow+red), which would transition to be purple (blue+red) at around pH 3 and then to be green at high pH (blue+yellow). But those intermediate colors are going to be pretty muddy. At around 1.5 you'll have an orangy-purple, and at around 4.5 you'll have a greeny-purple.  Imagine what things would look like if you mixed orange paint with purple paint, or purple paint with orange paint.

What we need is a layout of color that can gradually transition one to the other and look good all the way through the approximately 7-ish transitons you'd need to segment up a 14 unit pH scale ... right, the rainbow spectrum has that property.

So there's no real chemical reason for the connection. It's more a practical human consideration, where the people combining the indicator molecules to make a large-range pH indicator need to pick particular ones which make a clear, unambiguous color sequence, and the visual color spectrum has the same useful properties they need.


is bleach residue (after evaporation) safe? I have been preparing bleach disinfectant lately in a spray bottle and has been spraying on my couch, door mat, door knots, etc...
I don't rinse it... So I assume there is bleach residue left after evaporation?
Is it safe to leave the residue?

I felt some very mild burn like feeling on my arms when putting my arms on my couch lately; I don't know if this is some kind of placebo effect or if it is real.

Bleach, especially employed in laundry applications, may have, in addition to the active ingredient sodium hypochlorite (NaOCl), an equal concentration of NaCl (as it is prepared from the action of Cl2 on NaOH) and at times, an alkaline, like lye (NaOH) or Washing Soda (sodium carbonate, Na2CO3). The latter can cause skin rashes and irritation (and will be present on drying).

The rationale for the latter is to keep the pH elevated in aqueous sodium hypochlorite as upon exposure to CO2, there is the possible introduction of hypochlorous acid. While HOCl is more powerful (and reportedly nearly a hundred times more potent disinfectant than the associated hypochlorite) is far less stable (a short shelf-life). Further, the decomposition product is O2 and also HCl (and not NaCl).

The created hydrochloric acid can further react with the sodium hypochlorite or hypochlorous acid as follows:

$\ce{HCl + NaOCl -> NaCl + HOCl}$

$\ce{HCl + HOCl <=> Cl2 + H2O}$

leading to a chain decomposition of the bleach, hence the advantage of adding NaOH or Na2CO3. To quote the cited source:

  The pH has a significant effect on the stability of sodium hypochlorite solutions. Below pH 11 the decomposition of sodium hypochlorite is significant due to the shift in the equilibrium in favor of the more reactive hypochlorous acid. A pH between 12 and 13 gives the most stable solution. This equates to 0.4 to 4.0 grams per liter (gpl) excess NaOH. Greater concentrations will not improve the stability. Excessively high improve the stability. Excessively high alkalinity will damage textiles and retard the bleaching and disinfecting actions of the hypochlorite.

I would suggest that you can even dilute your current NaOCl dosing and add some lemon juice, just prior to use, to create (per the source reference noted above) a much more powerful mix (and a more potent bleaching agent, so be careful where you select to employ it) albeit, with a much shorter shelf-life.

To rapidly remove bleach smell (likely Cl2O from the dehydration of HOCl), one can apply a spray of dilute H2O2, which rapidly converts NaOCl to NaCl and oxygen. Caution: this further promotes bleaching action via intermediate radical formation (see 'Hydroxyl radical from the reaction between hypochlorite and hydrogen peroxide'), so again be careful where you apply it.


Why does soup turn purple when lemon juice and salt are added? Well you have probably heard about salpeter. I don't know if it is a myth or not that it is used to reduce men's libido by adding to food especially in boarding schools and remote military campuses.

I didn't know that before but because I studied at boarding school, this topic and its alleged test intrigued me. According to the test, when you add lemon juice and salt to food and if the salt in the food turns to purple/blue, it means salpeter is added to the food. I don't know if it is scientifically proven.

Now I live with my parents but just for fun I wanted to try to test on my homemade food and I shocked because the salt I added to the soup, which was made by mom, turned to be  purple /blue. I've become really baffled and I checked if there is any salpeter in cupboards I found a bottle of white like flour thing. I presumed it was powdered salpeter and tested it with the same method, which I added lemon juice and salt and it turned blue/purple.

But when I asked what that white thing is, she said it was "powdered sugar".
and I tried the same test on new powdered sugar yeah it turned to be blue too.

I hope you are not confused. So far what I am sure about is that when I add salt and lemon juice to powdered sugar it turns to be blue.

My mom said there is neither salpeter nor powered sugar in the soup she made.

Buy how come does soup turn to be blue /purple when I added some lemon juice and a lot of salt?

Here is a picture.The soup is made of lentil, some spices and salt, some potato and carrot.

A possible answer:
(1) Common table salt is usually "iodized" - i.e. it contains potassium iodide.
(2) Something in the soup could oxidize the iodide to elemental iodine.
(3) Iodine plus starch (from the potato in the soup) gives a deep blue color.


Base Peak Intensity different than higher value in Mass Intensity Map I've been working with mass spectrum data analysis (but my field is computer science).

I'm using a library that allows me to get the entire Mass Intensity of the MS1 scan (such as [time, intensity]). However, by searching for the highest value for intensity, it is always a few integers lower than the base peak intensity.

I'm retrieving the information from a mzXML file, and the library that I'm using is ms-data-core-api. The method used to retrieve the intensity map, in case anyone has ever used this library, is this one: public double[][] getMassIntensityMap(). I retrieve the highest peek in the map by iterating the matrix returned and updating a variable with the highest value found in the y axis (given double [x = time][y = mass].

Examples of the difference of data include the following:

scan 1 (MS1): Highest peak in map: 789991.875 / base peak intensity: 793728.0
scan 2 (MS1): Highest peak in map: 714475.5  / base peak intensity: 719125.0
scan 3 (MS1): Highest peak in map: 673152.6875  / base peak intensity: 680567.0

From what I can tell, the instrument used for these scans is a fourier transform ion cyclotron resonance mass spectrometer, using Xcalibur as the  software.

Does this makes sense, and why?

One possibility for the discrepancy is that different peak centroiding / integration algorithms were used at different stages in the data.

For example, the base peak intensity recorded in the plain-text (uncompressed) scan data is probably calculated by the instrument in real-time during acquisition.  The intensities of all detected peaks are also recorded by the instrument, of course, but those intensities may be affected by processes such as centroiding that are not applied to the plain-text recorded base peak intensity.

An exact answer to your question will depend on many factors, such as whether spectra were recorded in profile mode or centroid mode, whether spectra were compressed during conversion to mzXML format, and also details of the (possible) centroiding algorithms and BPC algorithms used by the instrument firmware during data acquistion.  (Good luck trying to find the exact code for vendor firmware...it's probably proprietary.)


Nucleophilicity of alcoxy groups I found a nucleophilicity order as

$\ce{HO-} > \ce{CH3-O-} > \ce{(CH3)2CH-O-} > \ce{(CH3)3C-O-}$

The reason was given that bulky groups reduce nucleophilicity. But why?

The order of the $+I$ effect is $\text{tertiary} > \text{secondary} > \text{primary}$. Due to the greater $+I$ effect, the electron on oxygen must experience greater repulsion and thus the order of neucleophilicity must be reversed. Also, according to the given order of nucleophilicity a tertiary alcohol is more acidic than a primary alcohol. But I have read that presence of more $-I$ effect increase acidity of alcohols like $\ce{Cl-CH2-OH}$ is more acidic than $\ce{CH3-OH}$. Similarly due to more $+I$ effect in $\ce{(CH3)3C-OH}$, it should be less acidic than $\ce{(CH3)2CH-OH}$ which should be less acidic than $\ce{CH3-CH2-OH}$.

$$\Large\color{red}{\text{nucleophilicity} \ne \text{basicity}}$$

Please memorise firmly that nucleophilicity and basicity, albeit related in many ways, are different physico-chemical properties. Most importantly, basicity is a thermodynamic property while nucleophilicity is a kinetic property. Therefore, it is not uncommon for compound $\ce{A}$ to be more basic but less nucleophilic than compound $\ce{B}$.

This is the case here, too. It is unquestioned and supported by the corresponding $\mathrm{p}K_\mathrm{a}$ that tert-butanolate is the strongest base, followed by isopropanolate followed by ethanolate. This is due to the $+I$ effects you correctly outlined.

Nucleophilicity does not parallel this series since the it does not abide by the same rules. For a compound to be nucleophilic, the nucleophilic atom’s lone pairs must be able to actually attack, i.e. reach, the electrophilic site it wishes to attack. This means two molecules interacting and not just one molecule interacting with a proton which is so small it can even tunnel. In tert-butanolate, you have three α-methyl groups that take up space and make said interaction harder. Due to this steric interaction, tert-butanolate is an absolutely lousy nucleophile (while it is a great base).

It is similar for isopropanolate which still has too much steric hindrance to be an effective nucleophile although the conditions are better than for tert-butanolate.


Is atmospheric pressure acting only on the contents or also on the container? $22~\mathrm{g}$ of dry ice is placed in an an empty $600~\mathrm{ml}$ closed vessel at $298~\mathrm{K}$. Find the final pressure inside the vessel, if all $\ce{CO2}$ gets evaporated?

Now by applying $pV = n\mathcal{R}T$, the value of $p$ comes out to be $20.4~\mathrm{atm}$, but in the solution shown, it adds $1~\mathrm{atm}$ i.e atmospheric pressure pressure, so the answer comes out be $21.4~\mathrm{atm}$. Now why do we need to add atm pressure, the question tells us to find the total pressure inside the vessel, isn't the atmospheric pressure acting only on the container, and not on its contents?

According to the way this question is worded, adding 1 atmosphere would be incorrect. The reason is that they specify an empty container - if you have a rigid container and it is truly empty (high vacuum), then the absolute pressure would be zero, and the gauge pressure would be -1 atm (see this wikipedia article for reference to these terms).

Once the $\ce{CO2}$ sublimates, the final absolute pressure would be given by $PV=nRT$ - there is no need to add atmospheric pressure. To find the gauge pressure, you would actually subtract one atmosphere.

It is possible that the question wants you to assume that "empty" means "filled with air at 1 atm," but that is a guess and I would argue that it's not safe to assume that based on the way the question is written.


All-Gauche Alkane: Is It Possible? I've learned about anti and gauche conformations throughout my scholastic journey, and I have heard often that straight-chain alkanes always adopt an all-anti conformation in their lowest energy state. I also know that, in a pure sample of any given straight-chain alkane, (eg. hexane) the anti and gauche conformations are in a certain equilibrium with each other, And that this ratio of gauche to anti shifts towards gauche as the available energy of the sample (i.e. temperature/heat/ pressure) approaches infinity.  My question is as follows: Has anyone in the scientific community to date successfully forced an entire sample of straight-chain alkanes into the all-gauche conformation without significantly raising the conditions above STP? I have scoured the internet but have not surfaced with a definitive answer, so I now turn to the scientific community.

All contributions are greatly appreciated.

At Last! Success!
I found this research paper, which describes a "Self-assembling nanocapsule", which is held together by plenty of hydrogen bonding, and can assemble around a linear molecule of all-anti tetradecane, $\ce{(C14H20)}$ using the energy afforded by the hydrogen bonding to squeeze/twist the alkane into a helical all-gauche conformation. Have a look:

Cool, right?
Sources:
Scarso A, Trembleau L, Rebek J Jr. Helical folding of alkanes in a
self-assembled, cylindrical capsule. J Am Chem Soc. 2004 Oct 20;126(41):13512-8.
PubMed PMID: 15479107.
Note:
Other answers are still greatly appreciated!


How to find molecular charge of fragments using quantum calculations? Assume in a software like Avogadro, I put molecules A and B in a specific distance and I know they will involve in a charge transfer reaction ( as Marcus Theory outer sphere transfer ). Then I run a calculation package as Gaussian or GAMESS.

Now I want to see if any charge has moved from A to B.  One approach is to measure their charges and see if they are partially charged.

The problem is that I don't know how to find that molecular partial charge in a selected part of the system in such softwares (like I select a part of the system on the screen and see its charge )

One approach I found is in Avogadro: View>properties > atom properties and find the charge of each single atom and then add the selected atoms together , but it is very very time consuming in larger systems. I am pretty sure there must be a more effective method.

Your question boils down to "where are the electrons" in a quantum chemical system?

The problem, is if there's any level of interaction between the two fragments, you don't know. The charge will be delocalized across the entire A+B complex.

Put it another way.. Let's say you have a bunch of atoms and you run a calculation and the overall charge on the molecule is +1. Where's the extra positive charge?

Well, you might make some guesses, but it's probably delocalized across multiple atoms. The same thing happens with intermolecular interactions, but to a lesser degree.

There are two approaches:

Constrained DFT: This is a method where you force the charge on a particular fragment to be the way you want. You can then compare this semi-localized charge with the distributed (unconstrained) calculation and compare.
Fragment MO: You can diagonalize the wavefunction onto multiple fragments that you define.

I don't use GAMESS much, but a quick search implies that FMO is supported in GAMESS.


Why am I getting higher conversion values of isopropanol dehydration over a zeolite catalyst than n-propanol dehydration? The two process are run separetly but under the same condition, same activation of the catalyst, same reaction temperature, same flow.
The initial conversion of n-propanol is ~50% and isopropanol ~70%.
My expections was lower conversion values of isopropanol, because of it's branched structure it has a more contrained diffusion to the acid sites of the zeolite.

In a typical acid-catalyzed dehydration, the reaction goes by an SN1 mechanism, and the rate-limiting step is
$$\ce{R-OH2+ -> R+ + H2O}$$
The rate of this step is typically a function of the stability of the cation product. In your case, the secondary cation formed from isopropanol is more stable than the primary cation formed from n-propanol.
I'm using "typical" and "typically" here, because these are results from homogeneous reactions, and it sounds like you are doing heterogeneous catalysis with a solid zeolite. That can certainly change the reaction behavior, but in your case it appears that the zeolite is not selective for one molecule over the other, and the trend is the same as what is observed for homogeneous systems.


How to check for geometrical isomerism in cyclic compounds? I have no problem with identifying whether a given molecule has geometrical isomerism if it's not cyclic. But cyclic compounds are confusing for me. For example, how to check whether the following molecule posses geometrical isomerism?

For a molecule to show geometrical isomerism different molecules should be attached to the double bonded carbons. So, I'm sure that geometrical isomerism (if it is present) is along the double bonded carbons. But how to check whether different groups are attached to it when the double bonded carbons are the part of a cycle?

I hope the statutes of limitation haven't run out. Structure 1 is a (Z)-alkene. Using Cahn-Ingold-Prelog (CIP) protocol, the rings are deconstructed as shown in diagram 2. The larger red arrow marks the higher priority carbon (C,C,C). The smaller red arrow points to the lower priority[carbon (C,H,H). For the blue arrows, sulfur takes priority over carbon. [Addendum: Strictly speaking, the two blue arrows should point to the methylene groups next to the blue sp2 carbon such that C(S,H,H) > C(C,H,H).] The two larger arrows establish a (Z)-configuration. See: What are the CIP rules for cyclic substituents?


If the heat exchanged is different for reversible and irreversible processes, how can enthalpy be a state function? From the definition of Gibbs free energy, we can write for an irreversible process:

$dG=dH-TdS<0$

With $TdS=\pu{q_{rev}//T}$ this becomes:

$dH-dq_{rev}<0$

Since G is defined for constant (p,T) $dH=dq$, in the case of an irreversible process we can write $dq=dq_{irrev}$, so:

$dq_{irrev}-dq_{rev}<0$

Or, by default:

$dq_{rev}>dq_{irrev}$

BUT: since $dH=dq$, this would mean:

$dH_{rev}>dH_{irrev}$

Which is in contradiction with the fact that H is a state function (the path between 2 states shouldn't matter)! What am I missing here?

Even though the OP does not seem to be active in this tread any longer, I'm going to provide an answer anyway, since it seems like such an interesting question.

First of all, even if a closed system undergoes a process in which it is in contact with a constant pressure surroundings throughout the entire process, and this pressure is the same as that of the system in its initial and final states, the heat absorbed Q is still not necessarily equal to the change in enthalpy unless PV work is the only form of work being done. For example, if there is a stirrer doing work to agitate the system (and cause irreversible viscous heating), and no PV work occurs, and the system is in contact with a constant temperature bath throughout, the change in enthalpy will be equal to zero, but Q will be equal to minus the amount of work that the stirrer does on the system.

Secondly, assume that throughout the irreversible process, the system undergoes a change in which it is in contact with a single constant temperature reservoir at the same temperature as the initial and final temperatures of the system, and is also in contact with a constant pressure surroundings at the same pressure as that of the system in its initial and final states.  But consider that a reversible process between the same initial and final end states does not have to resemble the irreversible path in any way whatsoever, as long as it matches the initial and final temperatures and pressures (and species concentrations if chemical reaction is involved).  Even if the boundary temperature for the reversible path is held constant at the same value as the irreversible path, the pressure of the surroundings during the reversible path definitely does not have to be (and will not be) constant throughout the reversible path.  And, given that this is the case, the heat Q for the reversible path would not be required to be the same as that for the irreversible path, even if the enthalpy changes for both paths is the same.  In virtually all cases, it will actually not be possible to devise a reversible path at a constant surrounding pressure between the same pair of end states.  So the heat Q for the reversible path will not be equal to the change in enthalpy for the reversible path (and thus the change in enthalpy for the irreversible path).


Units in stoichiometry $$C_x H_y O_z$$

1.What are the units of x,y,z.I've read that they're termed as 'stoichiometric coefficients'.

2.What is the unit of mass proportion of carbon(c) in the fuel?

$$c=\frac{M_{carbon}}{M_{fuel}} .x$$

The stoichiometric coefficients is something else than you suggest.
stoichiometric coefficients:
The number of molecules of a given component that participate in the reaction as written.
The indexes $x, y, z$ represent quantities of atoms in a compound. So they do not have any units.
@Q2 Do you know what are $c, M$ and $x$ in the equation? If you do, you can answer your question. If you do not, than you should find it out.


Born–Oppenheimer adiabaticity The Born–Oppenheimer aproximation is said to be adiabatic.

What does "adiabatic" mean?
What would be a non-adiabatic approximation to the Schrödinger equation?

$\newcommand{\conj}[1]{\overline{#1}{}}
\newcommand{\braket}[2]{\langle{#1}\,|\,{#2}\rangle}
\newcommand{\bracket}[3]{\langle{#1}\,|\,{#2}\,|\,{#3}\rangle}
\newcommand{\mat}[1]{\mathbf{#1}}
\newcommand{\rel}{\vec{r}_{\mathrm{e}}}
\newcommand{\rnuc}{\vec{r}_{\mathrm{n}}}
\newcommand{\linop}[1]{\hat{#1}}
\newcommand{\Tnuc}{\linop{T}_{\mathrm{n}}}
\newcommand{\Tel}{\linop{T}_{\mathrm{e}}}
\newcommand{\Velel}{\linop{V}_{\mathrm{ee}}}
\newcommand{\Velnuc}{\linop{V}_{\mathrm{en}}}
\newcommand{\Vnucnuc}{\linop{V}_{\mathrm{nn}}}
\newcommand{\Hel}{\linop{H}_{\mathrm{e}}}
\newcommand{\psiel}{\psi_{\mathrm{e}}}
\newcommand{\psieli}{\psi_{\mathrm{e},i}}
\newcommand{\psielj}{\psi_{\mathrm{e},j}}
\newcommand{\Eel}{E_{\mathrm{e}}}
\newcommand{\Eeli}{E_{\mathrm{e},i}}
\newcommand{\Eelj}{E_{\mathrm{e},j}}
\newcommand{\psinuc}{\psi_{\mathrm{n}}}
\newcommand{\psinuci}{\psi_{\mathrm{n},i}}
\newcommand{\psinucj}{\psi_{\mathrm{n},j}}$

Setup

In short, our goal is to separate the coordinates of all electrons $\rel = \{ \vec{r}_{i} \}_{i=1}^{n}$ from that of the nuclei $\rnuc = \{ \vec{r}_{\alpha} \}_{\alpha=1}^{\nu}$. Thus, we write the time-independent Schrödinger equation in the following form
\begin{equation}
    \linop{H} \psi(\rel, \rnuc) = E \psi(\rel, \rnuc) \, , \tag{1}
\end{equation}
where the Hamiltonian operator is given as follows
\begin{equation}
    \linop{H} = \Tnuc + \Tel + \Velnuc + \Vnucnuc + \Velel \, ,
\end{equation}
or, explicitly in atomic units,
\begin{equation}
    \linop{H}
    = - \sum\limits_{\alpha=1}^{\nu} \frac{1}{2 m_{\alpha}} \nabla_{\alpha}^2
    - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^2
    - \sum\limits_{\alpha=1}^{\nu} \sum\limits_{i=1}^{n} \frac{Z_{\alpha}}{r_{\alpha i}}
    + \sum\limits_{\alpha=1}^{\nu} \sum\limits_{\beta > \alpha} \frac{Z_{\alpha} Z_{\beta}}{r_{\alpha \beta}}
    + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
\end{equation}
where $m_{\alpha}$ is the rest mass of nucleus $\alpha$ and $Z_{\alpha}$ is its atomic number, $r_{\alpha i} = |\vec{r}_{\alpha} - \vec{r}_{i}|$ is the distance between nucleus $\alpha$ and electron $i$, $r_{\alpha \beta} = |\vec{r}_{\alpha} - \vec{r}_{\beta}|$ is the distance between two nuclei $\alpha$ and $\beta$, and $r_{ij} = |\vec{r}_{i} - \vec{r}_{j}|$ is the distance between two electrons $i$ and $j$.
Here for simplicity we ignore magnetic interactions and use the non-relativistic treatment.

Exact separation

We know that nuclei are much heavier as compared to electrons, $m_{\mathrm{n}} \gg m_{\mathrm{e}}$, and assuming that the momentum of electrons and nuclei is of the same order of magnitude, $p_{\mathrm{n}} \sim p_{\mathrm{e}}$, we conclude that kinetic energy of nuclei is far smaller than kinetic energy of electrons $p_{\mathrm{n}}^2/(2 m_{\mathrm{n}}) \ll p_{\mathrm{e}}^2/(2 m_{\mathrm{e}})$, or, talking in terms of classical mechanics, nuclei are moving considerably slower than electrons.
For a coupled motion of two subsystems with significantly different speeds it is often assumed that one could treat the motion of a relatively fast moving subsystem (electrons) as if it depends only on the static configuration of a relatively slow moving one (nuclei).
This assumption results in a procedure with two major steps, described below.

First step

First, we solve an equation describing the motion of electrons for a number of static configurations of the nuclei.
The equation is known as the electronic Schrödinger equation and it can be written as follows
\begin{equation}
    \Hel \psieli(\rel, [\rnuc]) = \Eeli([\rnuc]) \psieli(\rel, [\rnuc]) \quad \text{for} \quad i = 1, 2, \dotsc \, .
\end{equation}
where $\Hel = \Tel + \Velnuc + \Velel$ is the electronic Hamiltonian, $\psieli$ is the electronic wave function describing $i$-th electronic state, and $\Eeli$ is the electronic energy of the $i$-th electronic state --- the total energy of a molecule with nuclei being fixed at particular coordinates less the nuclear repulsion potential energy $\Eel = U - V_{\mathrm{nn}}$.

Note that the electronic wave function is a function of electronic coordinates only, though parametrically it still depends on nuclear coordinates. The parametric dependence of $\psieli$ and $\Eeli$ on $\rnuc$ means that for different sets of fixed nuclear coordinates $\rnuc$ we have different electronic Schrödinger equations with different solutions $\psieli$ and $\Eeli$, or, in other words, that the electronic Schrödinger equation written above is a family of equations parametrized by $\rnuc$, rather then one equation.

Second step

Once solutions of the electronic Schrödinger equation $\{ \psieli(\rel) \}_{i=1}^{\infty}$ and $\{ \Eeli \}_{i=1}^{\infty}$ are found for a number of different fixed nuclear configurations $\rnuc$, we took the second step: the task is to solve an equation of motion for nuclei which is obtained as follows.
Note first that since $\Hel$ is self-adjoint, its eigenfunctions $\{ \psieli(\rel, [\rnuc]) \}_{i=1}^{\infty}$ form a complete orthonormal set.
Now we use the assertion that a function of two independent variables $\psi(\rel, \rnuc)$ can be expanded over a complete set of functions of one variable $\{ \psieli(\rel, [\rnuc]) \}_{i=1}^{\infty}$ in the following way
\begin{equation}
    \psi(\rel, \rnuc) = \sum\limits_{i=1}^{\infty} \psieli(\rel, [\rnuc]) \psinuci(\rnuc) \, . \tag{2}
\end{equation}
Substitution of (2) in the Schrödinger equation (1), multiplication of both sides by $\conj{\psi}_{\mathrm{e},j}$, and integration over electronic coordinates $\rel$ only, leads to
\begin{equation}
    \Big( \Tnuc + \Eelj + \Vnucnuc - E \Big) \psinucj
    -
    \sum\limits_{i=1}^{\infty} \linop{\Lambda}_{ji} \psinuci
    = 0 \quad \text{for } j = 1, 2, \dotsc \, , \tag{3}
\end{equation}
where operator $\linop{\Lambda}_{ji}$ is defined as follows
\begin{equation}
    \linop{\Lambda}_{ji} =
    \sum\limits_{\alpha=1}^{\nu} \frac{1}{2 m_{\alpha}}
    \big(
        2 \bracket{ \psielj }{ \nabla_{\alpha} }{ \psieli } \nabla_{\alpha}
        + \bracket{ \psielj }{ \nabla_{\alpha}^2 }{ \psieli }
    \big) \, .
\end{equation}
and arguments $\rel$ and $\rnuc$ were dropped for brevity.
The last equation describes a system of differential equations (one equation for each value of $j$), and once the system is solved and a set of functions $\psinuci$ is obtained, one could write down the exact solution of the Schrödinger equation (1) in the form we introduced above (2).

Approximations

The task to solve the system of equations (3) is extremely difficult and the reason is that equations in the system are coupled in a sense that the solution of any $i$-th equation, $\psinuci(\rnuc)$, enters all other equations as the $\linop{\Lambda}_{ji} \psinuci$ term. One obvious way to decouple equations in the system (3) is by assuming that $\linop{\Lambda}_{ji} = 0$ either for all $i \neq j$, or for all $i$ without any exception. Operators $\linop{\Lambda}_{ji}$ may be regarded as the elements of a square matrix $\mat{\Lambda}$ and of course the fact that we equate these elements with zero is an approximation: in the first case, when we equate with zero all off-diagonal elements $\linop{\Lambda}_{ji}$, we introduce the adiabatic approximation, while in the second case, when we equate with zero all elements $\linop{\Lambda}_{ji}$, we introduce the Born–Oppenheimer approximation.

Both these approximations lead to a system of the so-called nuclear Schrödinger equations
\begin{equation}
    \linop{H}_{\mathrm{n}} \psinucj(\rnuc) = E \psinucj(\rnuc) \quad \text{for } j = 1, 2, \dotsc \, ,
\end{equation}
where $\psinucj(\rnuc)$ is known as the nuclear wave function and $\linop{H}_{\mathrm{n}}$ as the nuclear Hamiltonian, with the only difference being that in the first case the nuclear Hamiltonian takes the following form
\begin{equation}
    \linop{H}_{\mathrm{n}} = \Tnuc + \Eelj([\rnuc]) + \Vnucnuc - \linop{\Lambda}_{jj} \, ,
\end{equation}
while in the second case the last term $\linop{\Lambda}_{jj}$ is absent
\begin{equation}
    \linop{H}_{\mathrm{n}} = \Tnuc + \Eelj([\rnuc]) + \Vnucnuc \, .
\end{equation}

The first thing to notice is that in both the above mentioned approximations the only terms in the nuclear Schrödinger equation for $j$-th nuclear state $\psinucj$ that depend on electronic states, $\Eelj$ and also $\linop{\Lambda}_{jj}$, are related to the same and the only electronic state $\psielj$.
This means that in both these approximations each nuclear state $\psinucj$ depends only on one electronic state $\psielj$, while in the accurate physical picture (3), electronic states are coupled through off-diagonal $\linop{\Lambda}_{ji}$ terms, known as non-adiabatic coupling terms (NACT).

Note also that the first term in the nuclear Hamiltonian in both adiabatic and Born–Oppenheimer approximations represents the kinetic energy of nuclear motion, and consequently, the remaining terms collectively should stand for the potential energy of nuclear motion.
It is said thus that nuclei move on the so-called potential energy surface (PES) defined by the sum of these remaining terms: $\Eelj([\rnuc]) + \Vnucnuc - \linop{\Lambda}_{jj}$ in the adiabatic approximation and just $\Eelj([\rnuc]) + \Vnucnuc$ in the Born–Oppenheimer one.

Non-adiabatic coupling terms usually (but not always) can indeed be neglected, especially around the equilibrium nuclear configuration.
Besides, often the errors due to neglecting these terms are less than the errors resulting from some further approximation introduced when solving electronic and nuclear equations.
However, when two potential energy surfaces come close to each other or even cross, non-adiabatic coupling terms becomes large and are no longer negligible.

Answers

  What adiabatic means?

The term "adiabatic" from Ancient Greek αδιαβατος (α - "not", δια - "through", βατος - "passable") literally means the situation when something "is not passing through" something else.
In thermodynamics adiabatic process is a process occurring without exchange of heat of a system with its environment, i.e. a process in which heat is not passing through system enclosure.
In quantum mechanics adiabatic refers to a process in which no abrupt transition from one state to another with respect to continuous changes of some parameters happens, and thus, energy of a system changes continuously with respect to continuous changes of these parameters.

And the reason why we call the above mentioned approximation adiabatic lies in the fact that in this approximation when solving the electronic Schrödinger equation for different fixed nuclear configurations by continuously varying the coordinates of the nuclei we indeed assume that electronic state of the system continues to be the same. In other words, we assume that we are "staying" at the same PES. This could be the case, of course, only when potential energy surfaces for different electronic states are well separated, so, in the regions of nuclear coordinates where potential energy surfaces come close to or even cross each other adiabatic approximation is inadequate.

  What would be the case of a non adiabatic appox. to Schrodinger eq.?

Non-adiabatic solution refers to the solution of the system of coupled equations (3).


How to determine if a mixing process is spontaneous at constant temperature but variable pressure This is a conceptual question.
Many times we work with the gibbs energy of mixing under conditions of constant temperature but not pressure.
In a problem like this:

  Suppose that 2.0 mol H2 at 2.0 atm and 25 ∘C and 4.0 mol N2 at 3.0 atm and 25 ∘C are mixed at constant volume. Calculate ΔmixG.

Is easy to calculate that the Gibbs mixing energy is -9,7 kJ.
But we also know that we cannot use gibbs' spontaneity criterion because it only works on constant pressure and temperature.

Is there any way to determine the entropy of the universe in this case to establish a criterion of spontaneity?
*For example if we know that the total volume is 10L and the division is even.

Volume of H2 = $\frac{nRT}{P}=\frac{(2)R(298)}{2}$

Volume of N2 = $\frac{nRT}{P}=\frac{(4)R(298)}{3}$

Total volume = $\frac{7}{3}(298)R$

Final pressure = $\frac{nRT}{V}=\frac{(6)R(298)}{(\frac{7}{3})R(298)}=\frac{18}{7}\ atm$

Final partial pressure H2 = $\frac{18}{(7)(3)}=\frac{6}{7}\ atm$

Final partial pressure N2 = $\frac{18}{7}\frac{2}{3}=\frac{12}{7}\ atm$

$$\Delta S=-n_{H_2}R\ln{\frac{\frac{6}{7}}{2}}-n_{N_2}R\ln{\frac{\frac{12}{7}}{4}}=-(2)R(-0.8473)-(4)R(-0.8473)=5.08R$$

For ideal gas mixing, $$\Delta H = 0$$

Therefore, $$\Delta G=\Delta H-T\Delta S$$

Another way of doing this is to (a) first allow the divider to float so that the two chambers equilibrate in pressure and then (b) remove the divider and allow the gases to mix.  You calculate $\Delta G$ for each step, and then add them.  In the end, you get the same answer I got.


Fluorescence Spectra, emission spectrum What exactly does the excitation spectrum tell us about a sample in Florescence Spectroscopy, what information does it tell us? I know that we will get an output of concentration vs intensity but that's it.

I know that we will get an output of concentration vs intensity but that's it.

Actually, that's not what you are getting. Typically, you run an emission spectrum first to determine the maximum of the fluorescence emission (or any other point of interest in the emission spectrum). In the excitation spectrum, you monitor this particular emission line and its intensity while varying the excitation wavelength.


Which reactions are more temperature sensitive: the ones with higher Ea or the ones with lower Ea? And why? Which reactions are more temperature sensitive: the ones with higher $E_\mathrm{a}$ or the ones with lower $E_\mathrm{a}$? And why?

I wasn't able to find much useful content on googling it, however on doing the math I came to a conclusion that the ones with higher $E_\mathrm{a}$ are more sensitive to increase in temperature. But why is it so? I mean math proves it, but why exactly does it happen?

Here are my calculations:

The reaction with the lower activation energy can proceed smoothly without the increase in temperature whereas the the reaction with the higher activation energy cannot. When you increase the temperature, it won't make much difference to the reaction with a lower activation energy since it was already fine with a lower temperature. On the contrary, the reaction with higher activation energy will now gain sufficient energy making the molecules more likely to react and thus, the temperature change affects this reaction much more.


Which one, Mulliken charge distribution and NBO, is more reliable? Sometimes the Mulliken and NBO turn out to be so different that I can't decide which one I can trust. I've heard that Mulliken is inaccurate, but is NBO always accurate?
And should I use Gaussian or other software to calculate charge distribution? Are there methods more accurate than Mulliken or NBO?

And charge distribution doesn't represent charge density, right?

As I do not want to drag this out in the comments, I decided to give a brief answer. Let me first cite the very helpful comments to keep the for (relative) eternity and add my thoughts at the end.

There is no proper, strict and agreed definition of atom border, that's the real problem. Gaussian includes option to fit charges on atoms (assumed for the sake of the procedure as points) so they produced electrostatic field as accurately fitting once calculated from the electronic density distribution as possible, and maybe (I haven't used Gaussian for a long time, I'm not sure) dipoles too. –  permeakra

NBO is more accurate than Mulliken in very many ways. However, NBO is not always accurate. Generally speaking, you would never want to use Mulliken charges for any sort of production level work unless its been calibrated. –  LordStryker

Mulliken is of course the cheapest and fastest way to compute charges. However, this method tends to give qualitative results at best. The reason for this is obvious. It will divide the canonical orbitals equally amongst the participating atoms. There is no polarisation whatsoever. Another problem is, that they are very basis set dependent. And that being said, the worst part is, that the description becomes worse by increasing the basis set.
NBO charges are much more reliable, since the operate on the electron density instead. Localised natural atomic orbitals can be used to describe the computed electron density. Polarisation of bonds is therefore considered. This method obviously only works very well for systems, that can actually be (quite well) described by a Lewis structure. However, they usually give a nice qualitative picture and are quite robust when it comes to increasing the basis set.
There are a couple of more population analysis tools, that can be used and should be considered when it is necessary to derive some conclusions from charges.
Quite common is also the Hirshfeld charges approach, which divides the electron density into spherical basins around the nucleus. Maybe interesting to read is "Are the Hirshfeld and Mulliken population analysis schemes consistent with chemical intuition?" (I can't tell, I do not have access.)
There are also several schemes to obtain partial charges from the electrostatic potential.
Probably the most rigorous approach to atomic partial charges is the Quantum Theory of Atoms in Molecules  (QTAIM). Here the Laplacian of the electron density is used to describe the partitioning in electronic basins. The best feature of this analysis is, that it can be also used for experimentally obtained electron densities. However, as everything, this has also major drawbacks. One of the most important is the computational effort. Unfortunately the integration scheme is not always completely robust when it comes to some particular molecules. However, it is almost completely basis set independent and results could be cross referenced with experimental data.
Since permeakra's initial statement holds true, especially for the agreed part, the best way to deal with partial charges is to be careful and maybe check with different methods. In gaussian there are several schemes implemented, most commonly, Mulliken, Hirshfeld and NBO (3.1) and some others, see the population keyword for more detail.


Flexibility of long-chain aliphatic compounds PubChem is unable to generate the conformers of Z-6-Heneicosen-11-one? The website says that the molecule is too flexible

What does flexibility of a molecule mean in this context? I understand that it is a long chain compound, but I assumed that the bonds are rigid and do not flex? In layman’s terms, how does one quantify flexibility? Is it flexible like a skipping rope or like something else?

PubChem 3D has limits on the number of 'rotatable bonds'

I recently answered a related question about PubChem 3D:

  Beyond the limits of MMFF94 and MMFF94s themselves, PubChem3D had several limits indicated in the accompanying manuscript: Bolton et. al. "PubChem3D: a new resource for scientists" J Cheminf. (2011) v. 3, art. 32)

  Not too large (with ≤ 50 non-hydrogen atoms).
  Not too flexible (with ≤ 15 rotatable bonds).
  Consists of only supported elements (H, C, N, O, F, Si, P, S, Cl, Br, and I).
  Has only a single covalent unit (i.e., not a salt or a mixture).
  Contains only atom types recognized by the MMFF94s force field.
  Has fewer than six undefined atom or bond stereo centers.

Your molecule falls into the "more than 15 rotatable bonds" category. I count about 18 $\ce{sp^3}-\ce{sp^3}$ single bonds.

Basically, the problem is that the search space for conformers with that many rotatable bonds is too big. In principal, it's $\approx3^n$ where $n$ is the number of rotatable bonds.


How does entanglement explain myoglobin's preference for O₂? I was reading this article (full PDF here) about the myoglobin's unusual preference for O2  than CO, in which it was written:

  It turns out that some electrons in the myoglobin involved in binding CO and O2  exhibit a strong ‘entanglement’ effect, which means that their motion cannot be described independently. The all-important strength of this effect is primarily controlled by a property of quantum mechanics (Hund’s exchange) that has been traditionally neglected in such simulations; the team now believe that classical electric repulsion effects are far less important in determining which of CO and O2  is more energetically favourable for binding.

But I cannot understand how entanglement explains this phenomenon. Has it been concluded from this (quantum mechanical) perspective? If yes, can anyone explain how?

So, there are two effects at play here which they account for. One they denote by a $\textbf{U}$ and the other by a $\textbf{J}$. $\textbf{U}$ represents many-body coulomb repulsions between the electrons. That is, they are accounting not just for two electrons repelling each other, but up to three or more. $\textbf{J}$ represents valence fluctuations in spin. Specifically, iron in hemes are in a high-spin triplet state. What they are saying then is that it is necessary to include spin fluctuations which result in changes in the magnetic moment of the system in order to properly capture the binding of ligands. If you look at figure 4 of this paper, you'll find that they have included a comparison of spin-fluctuations in $\ce{MbO2}$ and $\ce{MbCO}$ which are used as model systems and known to be analogous in ligand binding to hemes. Basically what this means is that the increased spin-fluctuations cause $\ce{O2}$ to bind more strongly, but this has basically no effect on $\ce{CO}$. They describe these effects in the following quote:

  In this case, the effect of J on the
  binding energy of O2 may be regarded as a balance between two
  competing effects. The charge analysis (Fig. 3) reveals that
  metal-to-ligand charge transfer is higher for J =0 eV, which is
  expected to enhance ligand–protein interactions for small values
  of J. However, NBO analysis reveals a larger ligand-to-metal
  back charge transfer for J = 0.7 eV, which is consistent with the
  increased occupancy of the Fe $\ce{d3z^2−r^2}$ orbital (Table 1), and is
  expected to cause variational energetic lowering at higher values
  of J due to electronic delocalization.

I won't pretend to fully understand their argument (frankly it seems to me like they are waving their hands at a solution and are basically saying this is some complicated effect having to do with spntaneously changing magnetic moments).

So, in that paper they only ascribe the effect to strong electron correlation, many-body coulomb effects, and spin-fluctuations.

So, they definitely use the word entanglement but don't define exactly what they mean or even use it as an explanatory device. I found in another one of their papers though, Weber, C., O’Regan, D. D., Hine, N. D., Littlewood, P. B., Kotliar, G., & Payne, M. C. (2013). Importance of many-body effects in the kernel of hemoglobin for ligand binding. Physical review letters, 110(10), 106402., where they do actually define what they mean by entanglement and it depends on these parameters $\textbf{U}$ and $\textbf{J}$. You can find their discussion of this on page 11 in an appendix.

Their use of the word entanglement is derived from the Von Neumann Entropy which can be used as a measure of the amount of entanglement in a given system. This is because if two particles are entangled in some way, their entropy will be lower than if the two particles were not entangled. Note that this use of the word of entanglement is related to the usual descriptions of entanglement, but it is more of a measure of the average of spontaneous entanglement and disentanglement. That is, particles can become entangled due to some way in which they interact, but almost immediately interact with something else that breaks their entanglement so that it's very rare to find systems where this has any meaningful effect. This spontaneous entanglement in these heme systems is a result of the many-body effects which I've been describing. This manifests itself in the entropy they calculate. This highlights the importance of using dynamical mean-field theory (which they use) as follows:

  With this definition of the entropy, suitable for DFT
  calculations, we found that L = 0.757 in FeP-p and
  L = 0.765 in Fep-d. The entropy at the DFT level
  is hence much smaller than the DFT+DMFT entropy,
  which is in the range Λ = 2.5 − 4.2. This is explained by
  the absence of the many-body excitations induced by the
  correlations (U and J) which contribute significantly to
  the entropy.

There's no doubt they aren't being very forthcoming in their terminology or definitions, but they are also doing some very complicated and interesting stuff.


How do non-ionic compounds dissolve into water? Into molecules or atoms? I can't seem to find the answer in google, which is surprising.

I'm aware that ionic compounds dissolve into electrolytes. However, when non-ionic sacarose for example dissolves in water, does the sacarose crystals break into single atoms as well or does it stay as a whole molecule in water? What does exactly happen?

Thanks everyone =)

The molecules remain as they are, as discrete particles, they do not break up into single atoms. What can happen to certain compounds when they are dissolved in water is a hydrolysis reaction. But even then, only certain bonds of the molecule are cleaved in a reaction with water, and the molecule does not spontaneously fall apart into the elements it consisted of, because there is no thermodynamic reason to do so.

The same is also true for ionic compounds. Ammonium chloride ($\ce{NH4Cl}$), for example, consists of discrete $\ce{NH4+}$ and $\ce{Cl-}$ ions. $\ce{NH4+}$ is a molecular ion, and it remains as such in aqueous solution and does not break up into single nitrogen and hydrogen atoms/ions. However, the salt can hydrolyze in aqueous solution according to the following equilibrium:

$$\ce{NH4Cl + H2O \rightleftharpoons NH3 + H3O^+ + Cl^-}$$

This is the reason why aqueous solutions of ammonium chloride are weakly acidic.


Melting and boiling point comparison Graphite, silica and diamond are covalent compounds and still they have high melting and boiling points. Why?

The very term "covalent compounds" is confusing, so you may just as well stop using it. The elementary substances you specified are covalent crystals (AKA "covalent solids"). A single crystal of any of these is in fact one huge molecule linked by covalent bonds. To melt that crystal, you have to break the covalent bonds, and they are pretty strong.

On the other hand, we have $\ce{N2}$, $\ce{O2}$, and other covalent molecules. They may be quite strong (in that each molecule by itself is pretty hard to tear apart), but the molecules are linked together by the so-called van der Waals forces, and these are quite weak, so these compounds are gases, and it takes some effort to see them turning into a liquid or a solid.


What commonly availible salts are strong dessicants? I'm looking into building an absorption chiller as a decidedly non-professional DIY project. For safety reasons and because temps. > 0°C are sufficient, I won't be using ammonia but some saline solution as absorber and water as cooling agent.
The obvious choice for the salt would be LiBr, as this is used in comercial installations. However, I'm not sure I can procure it (chemicals suppliers don't sell to private persons in germany) and it's not cheap (~ 30€/100g).
What other salts are strongly hydroscopic and commonly availible?
Alternativly, where can I find a concise list with the physical properties of many common (=household) salts?

Have you looked into silica gel or zeolites? Both are non-toxic, and thus likely obtainable by private persons. Zeolites are used in self-refrigerating beer kegs.