Hybridisation and the Schrödinger equation I am slightly confused about hybridisation and how it relates to molecular and atomic orbitals, despite having pored through many sources online. I was hoping someone could verify whether my current understanding is correct, in particular regarding what hybridisation actually is/does because I have not read this explicitly, but am assuming it is the case from what I have read so far:

The Schrödinger equation can be solved to give the atomic orbitals (at least, some simplification involving the effective nuclear charge can be used to find the outer (and inner?) atomic orbitals.
In hybridisation, we first consider the shape of a molecule and then we consider each atom seperately and how we could combine the atomic orbitals so that the geometry about each atom is correct. Now what I am particularly unsure about is: Are the hybridised orbitals entirely a conceptual construct, or are they mathematical solutions to the Schrödinger equation? I was thinking that maybe the hybrid atomic orbitals that are created are a linear combination of the atomic orbitals, and thus this linear combination also solved the Schrödinger equation and can exist but would have a different energy to the individual atomic orbitals? I am finding it difficult to think how a linear combination of the atomic orbitals could produce something with a completely different shape though (although I suppose the orbitals are orientated and essentially vectorial, so I think it could work). Perhaps  the hybrid atomic orbital form could also be, say, the product of the atomic orbital forms? Although then I do not think this would solve the Schrödinger equation anymore. But I am not sure. And anyway, all of these ideas assume that hybridisation has a mathematical basis in the Schrödinger equation, which I am not sure about at all!
Then molecular orbital theory takes the orbitals of each atom (I think theoretically it takes every orbital in every atom?) in the molecule and combines them to form a molecular orbital for the whole molecule - no longer simply an overlap between two atoms. This certainly also solves the Schrödinger equation. From what I understand, theoretically the calculation to form MOs should involve only the AOs of each atom, but to simplify the situation sometimes hybrid atomic orbitals are used in MO construction? If so, surely the hybrid atomic orbitals must be a linear combination of the atomic orbitals that solve the Schrödinger equation, so that this can be used in molecular obrital theory which is based in the solution to the Schrödinger equation? Also, where in the calulcation are the coefficients for each atomic orbital arising? Is it from the boundary conditions including things like molecule geometry, or perhaps that we know the energy of a molecule and when we put this particular energy in to the Schrödinger equation it gives us the coefficients?

I apologise for the long post and I realise there seem to be many questions within here, however they are all linked and about the mathematical underpinnings of hybridisation/valence bond theory and molecular orbitals, so I thought they belonged in one post.

Mathematically, atomic/molecular orbitals are 1-electron wavefunctions (hydrogen-like wavefunctions) that are used as a basis with-which the total N-election wavefunction is expanded.  The N-election wavefunction is a determinant (or a linear combination of determinants) built from these 1-electron wavefunctions: an anti-symmetric (Fermi statistics) linear combination of Hartree products (usually a product of MOs expanded in a basis of AOs).

For simplicity's sake, lets assume we are interested only in 1-determinant wavefunctions.  The Hartree-Fock method will (when used where HF is appropriate) result in a variationally-optimal 1-determinant wavefunction that is an energy eigenfunction of the applicable Hamiltonian operator.  Not only is the final, self-consistently optimized HF total N-electron wavefunction an energy eigenfunction, but the individual HF canoncial MOs are also eigenfunctions of the Hamiltonian/Fock operator, with eigenvalues that are the so-called 'orbital energies' (canonical orbitals are defined as the optimzed orbitals which diagonalized the Fock matrix).

However, there is nothing special about these canonical orbitals in the context of the total system energy.  Any linear, norm-preserving transformation of the occupied orbital space will result in an N-electron wavefunction that is still an eigenfunction of the Fock operator, while the individual MOs are no longer eigenfunctions of the Fock operator (for example, localized MOs, LMOs).

So, once we have a HF wavefunction and optimized MOs, we can linearly transform the MOs however we want.  Such a linear transformation might have the effect of producing orbitals that resemble hybridized orbitals.  Or maybe we are interested in generating localized MOs.  The point is, a linear transformation within the occupied orbital space results in a wavefunction that has the exact same electon density and energy eigenvalue as the wavefunction HF originally provides us in terms of Canonical orbitals.

Orbitals transformed to look like hybrid-orbitals are on the exact same footing as any other choice of orbital.


High content of potassium reported by spectrometer in silverware items We tried to analyze a few silver items (800/1000 fineness) using a hand-held  ED-XRF spectrometer, which is normally used for analyzing industrial alloys etc. The operating element range of the device was Mg-U.

The most surprising result of the experiment was a very high reported content of potassium, often higher than silver. For example, one of results (listing just elements with content >= 1%):

| element                           | content |
|-----------------------------------|---------|
| K                                 |   36.0% |
| Ag                                |   27.5% |
| (sum of elements lighter than Mg) |   24.7% |
| Cu                                |    5.0% |
| Sn                                |    4.5% |

Such high potassium content was pretty consistent across all our silver samples, but not present in others (like stainless steel)

Clearly it's not possible that the alloy could contain that much potassium. Where could this result come from? Can potassium be somehow bound to the molecules at the thin surface layer? Or is there some explanation for this phenomenon?

To give some numbers to MaxW's comment, you can see on this webpage the x-ray absorption energies of basically all the elements.

For example, the K-edge of potassium is $3.61~\mathrm{keV}$. The L-II edge of silver is $3.52~\mathrm{keV}$ and the L-1 edge of silver is $3.81~\mathrm{keV}$. So, it seems like you are somehow automatically assigning the peaks and scanning over a very large range of energies. The problem is that this will run into problems if you assume that all transitions are out the $1s$ orbitals (K-edge) because there are L-edge transitions for heavier elements which are comparable in energy to those from lighter elements.

So, without more information about how the transitions are being assigned, it seems very likely to me that MaxW was right to say these are L-edge transitions from Ag.


For molecular orbitals, how is an energy of zero set or defined? When labeling the energies of molecular orbitals, often some anti-bonding MOs are shown with a positive energy. For example, MolCalc has the following energies for methane (the anti-bonding MO #8 at +19.47 eV depicted on the right):

I'm trying to understand what these numbers mean. I get that electrons in MOs #2 through #5 help to keep atoms bound (having them occupied contributes to bonding), whereas electrons in MOs #6 through #9 would lead to an increase in bond lengths or the molecule falling apart. On the other hand, MO #1 does not participate in bonding, right (basically, it is the inner 1s electrons of carbon)?

So what defines the zero reference point of these orbitals? It can't be the free electron because I thought at least for the hydrogen atom, even the excited states are states where the electron is still bound (i.e. energies are negative). It can't be the gradient of increasing all bond lengths (i.e. with the extreme case being separate atoms) because MO #1 would be hardly affected by that, given that it hardly participates in bonding.

This question is not limited to one particular software or author. Here is another example (source: https://www.researchgate.net/publication/234126374_Electron_rectification_through_donor-acceptor-heterocyclics_connected_to_cumulenic_bridge_A_computational_study):

I realize that the language in this question is sloppy, reflecting my poor understanding. I encourage anyone who answers to use the correct terminology.

The answer technically depends on the method. In most cases, meaning HF or DFT with standard basis sets, it is infinite separation of all bare nuclei and all electrons. One exception known to me is VASP (a periodic, solid-state code), which uses pseudopotentials to avoid having to model the core singularity and core electrons. There, the zero-point is in reference to some atomic state.

Note that the MO energies are basis-set dependent, which is most pronounced and most easily understood for core electrons.


Order of citation of alkyl and halide prefixes in IUPAC names I've been playing around with ChemDraw and found that it assigns the priorities of substituents in a way I wouldn't have expected, depending on the length of the alkyl substituent. I'm not a nomenclature buff, so I'd like to know if it's correct and if so, why.

$n = 1$: 1-ethyl-2-iodobenzene
$n = 2$: 1-iodo-2-propylbenzene
$n = 3$: 1-butyl-2-iodobenzene
$n = 4$: 1-iodo-2-pentylbenzene

What determines the priorities of the alkyl group and the iodo group?

You correctly figured out that the prefixes that represent the substituent groups are arranged alphabetically. However, note that the order of citation and the numbering are laid down in two different rules.
With regard to the alphabetical order of citation of substituent prefixes, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the alphanumerical order in your examples is correct:
x-ethyl-y-iodobenzene
x-iodo-y-propylbenzene
x-butyl-y-iodobenzene
x-iodo-y-pentylbenzene
The relevant rules concerning the numbering of locants for substituent prefixes are:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
other locant sets, each cited in order of increasing value, has the lowest term at the first
point of difference; (…)

and

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule (f) takes precedence over Rule (g).
In accordance with Rule (f), your first example could be named as 1-ethyl-2-iodobenzene as well as 2-ethyl-1-iodobenzene since both names correspond to the locant set ‘1,2’. However, according to Rule (g), this example is named as 1-ethyl-2-iodobenzene rather than 2-ethyl-1-iodobenzene since ethyl is cited first as a prefix in the name.
Hence, the numbering in your examples is correct:
1-ethyl-2-iodobenzene
1-iodo-2-propylbenzene
1-butyl-2-iodobenzene
1-iodo-2-pentylbenzene


Why does rotating a double bond break the bond? It is well known that double bonds are "stiff" and make particles less likely to rotate around the bond. But why? What makes it so that the bond has to be orientated a specific way?

Consider a carbon-carbon ($\ce{C=C}$) bond.

A double bond consists of a $\sigma$ bond and a $\pi$ bond. A $\pi$ bond is formed by sideways overlapping of unhybridized p-orbitals of two carbon atoms, above and below the plane of carbon atoms.

If now one of the carbon atoms of the double bond is rotated with respect to the other, the p-orbitals will no longer overlap, and the $\pi$ bond should break. But the breaking of the $\pi$ bond requires 251 kJ/mol of energy, which is not provided by the collision of the molecules at room temperature. Consequently the rotation about a carbon-carbon double bond is not free, but it is strongly hindered or restricted.

In carbon-carbon single bond ($\ce{C-C}$) only 12.55 kJ/mol of energy is required (this data is for ethane molecule). At room temperature, the collision of molecules supply sufficient kinetic energy to overcome this energy barrier.


Function of water and the need of two barrels in a biogas Generator For my chemistry assignment, I have to build a biogas generator. I found the following diagram,

This is the website that the image is from: http://www.motherearthnews.com/renewable-energy/other-renewables/biogas-generator-zm0z14aszrob
(Manure goes through the feeding tube, then anaerobic bacteria produces methane. The methane then goes to the second tub and is collected)

On the right side of the diagram,

why does the barrel need to be immersed in water? Does it do something to the methane?
why does the methane not immediately go from the first tank through the tube out to the burner but instead goes into the second tank?
Once the methane is inside the second tank, what's stopping it from rising and going out?

The reasons are as follows:

The inverted barrel must be full of water. Then, when methane is formed, it will displace the water and the barrel will start floating. Methane does not do anything with water. It is almost insoluble (solubility is around 40 mg/L) and therefore does this setup work.
The second tank is storage. Without storage you have to burn the methane when it is formed. With the second tank you can use it when you need it and you can a higher flow rate.
If to much methane is produced and not consumed at the burner, it will come out of the second tank. But you can control that by stopping the feed in the first tank.


Why are diagonal arrows disallowed in Jablonski diagrams? My professor mentioned in one of our lectures that diagonal arrows in a Jablonski diagram are disallowed. As far as I know, horizontal arrows signify conservation of energy, while vertical arrows signify conservation of momentum. Is this true/connected to the issue?

Since nobody has responded to this yet, I'll offer a short answer.

In a Jablonski diagram, the electronic states are grouped together horizontally by spin multiplicity $2S+1$. So, in the simplest example, you see a column of singlet states $S=0$ and a column of triplet states $S=1$. (Quite often, the vibrational levels are also illustrated schematically). Diagonal lines, then, represent transitions between states of different multiplicity.

These are more usually termed "forbidden transitions" rather than "disallowed". An example is phosphorescence, the decay from an excited triplet to a lower-lying singlet, most commonly the ground state. There is a general selection rule, $\Delta S=0$, which in principle makes the transition moment vanish. However, the rule can be violated: one possible reason is that $S$ is not a perfectly good quantum number due to spin-orbit coupling. Typically, the consequence is that the transition occurs very infrequently (which is why phosphorescence occurs slowly, and weakly).

So, it is not "disallowed" to put diagonal lines on a Jablonski diagram,
but in most cases they correspond to weak, "forbidden", transitions.


Explanation for solubility of group 2 halides I understand that the solubility (in terms of moles/volume) of group 2 halides increase with increase in anion size, i.e.

$$\ce{MF2 < MCl2 < MBr2 < MI2},$$

where $\ce{M = Mg, Ca, Sr, Ba},\dots$ due to large decreases in lattice enthalpy.

But what is the explanation for the following discrepancies?

$\ce{BeF2 > MgF2 = CaF2 < SrF2 < BaF2}$
$\ce{BeCl2 < MgCl2 < CaCl2 > SrCl2 > BaCl2}$

In other words, how can I determine the solubility order of $\ce{BX2, MgX2, CaX2, SrX2, BaX2}$ (where $\ce{X = F, Cl, Br, I}$)? The given orders of solubility have been verified with data.

To me it does not make too much sense to try to compare and explain the solubility (mass concentration) of anhydrous group 2 chlorides.

First of all, mass concentration does not seem to be the adequate physical quantity for comparison. Some of the chlorides have such high solubility that their saturated solution contains significant less water per volume than others. Molar fractions of solute and water would be a better choice.

Second, all of the chlorides precipitate as the hydrated salt from water. At normal temperature these are

$\ce{BeCl2\cdot 4 H2O}$
$\ce{MgCl2\cdot 6 H2O}$
$\ce{CaCl2\cdot 6 H2O}$
$\ce{SrCl2\cdot 6 H2O}$
$\ce{BaCl2\cdot 2 H2O}$
$\ce{RaCl2\cdot 2 H2O}$

Besides differences in structure their different stoichiometry makes a comparison difficult. And, most important, the formation of hydrates shows that we probably should not discuss the solubility of anhydrous chlorides at all.

Third, some anhydrous halides are more covalent than ionic compounds ($\ce{BeF2, BeCl2}$). This makes it difficult to argue with lattice energies and ionic radii.


Do polar compounds travel further or less far than unpolar ones in hexane/ethyl acetate TLCs? Would ethyl acetate/hexane be considered nonpolar or polar overall? I know that the ethyl acetate is polar and the hexane is nonpolar. I am trying to understand what if polar or nonpolar analgesics would travel further in TLC. I know that if  ethyl acetate/hexane is polar than a more polar one will travel further, and if it is nonpolar then nonpolar will travel further.

I was looking here and I see that caffeine is the most polar (out of aspirin and acetaminophen) and traveled the least distance (lowest $R_\mathrm{f}$) in my experiment. I thought that ethyl acetate/hexane was polar so I would have thought caffeine would be the least polar since it traveled the shortest.

So am I wrong that  ethyl acetate/hexane is polar or are they wrong that caffeine is the most polar?

Never have I seen a normal phase (i.e. silica gel or alox) TLC where a more polar compounds travels further than a less polar compound — and I have used some extremely (considering normal phase TLC) polar eluents in the past.

The top-running compounds are always the least polar and the slowest are always the most polar in normal phase TLC.


What is it that is "symmetry-adapted" about symmetry-adapted perturbation theory? I think this is a fairly straightforward question which I should know the answer to, but surprisingly cannot find a discussion of online.

Symmetry-adapted perturbation theory (SAPT) is a fairly common method to describe intermolecular interactions using physically meaningful terms (electrostatics, exchange, induction, dispersion), where the interaction is built up. That is, SAPT calculates the interactions directly as opposed to finding the interaction strength by subtraction.

So, quite simply, in SAPT, what does the "symmetry-adapted" specifically mean, and why is it necessary to specify this?

In other words, is there a non-symmetry-adapted perturbation theory which can be used for intermolecular interactions?

Finally, how, if at all, does this symmetry-adaptation relate to the kind we use when constructing molecular orbitals using group theory?

The "symmetry" in symmetry-adapted perturbation theory refers to the anti-symmetry of the wave function with respect to electron exchange:

$$\Psi(\mathbf{r}_1,\mathbf{r}_2) = -\Psi(\mathbf{r}_2,\mathbf{r}_1) $$

The most usual way to enforce antisymmetry in electronic structure theory is to enforce orthogonality between spin-orbitals.  So SAPT is intermolecular perturbation theory that includes terms that depend on the overlap between orbitals in different molecules.

Intermolecular perturbation theory gives rise to concepts like eletrostatic interactions, polarization, and dispersion.  With SAPT you then get additional concepts like exchange repulsion (aka steric or Pauli repulsion).


Internal vs external steam distillation To extract essential oils from plant matter (amongst other purposes), steam distillation is often used. I've seen a lot of different glassware setups for this. I've sometimes seen them broken into two categories:

Internal steam distillation
External steam distillation

Here's my (admittedly shaky) understanding of each:

Internal

With internal steam distillation, you have one boiling flask containing plant matter and liquid water. You simply boil the mixture and send the gas to to the condenser.

External

With external steam distillation, you have two flasks. One contains water only. You heat this one and route the steam into the second flask. The second flask contains a mixture of water and plant matter. The steam from the first flask combines with the liquid mixture in the second flask. Optionally, the second flask also has a heating mantle. The output of the second flask then goes to the condenser.

Differences?

What is the effect of using one or the other setup? External is substantially more complex, so I can't imagine it would be used if it didn't offer some advantages in some circumstances. When would you use external, and why?

The main purpose is to distill the component (essential oil) with quite high boiling point.
You have to think of distillation of immiscible liquids, http://www.chemguide.co.uk/physical/phaseeqia/immiscible.html
where you can take advantage of the equillibria in liquid and gas phase separately.

In short - two very beneficial things happen:

boiling point of the mixture is lower than boiling point of water (and substantially lower than BP of the oil)
the composition of vapor is independent of composition of liquid, i.e. you can distill out even the smallest fraction of oil (c.f. Raoult's law for miscible liquids)

On the other hand, the partial pressure of the oil is small, let's say 5% that of the water, therefore you need to evaporate and condense 20x more of water than of oil. To achieve this, you can either put all the needed water in the flask and do the "internal" distillation, or add just a small amount of hot water and use steam from external source in "external" setup.

In general, it is simpler to just boil pure water (or even have a direct source of steam) and use it for distillation, than to take care of the whole process in one flask.

As such, I would see the "external" setup simpler and more robust, even though it needs little bit more glassware.


Can molten aluminium be used to reduce iron oxide (rust) into iron? Will surface rust on an iron crucible be converted if used to melt aluminium?

Can one convert bog/lake iron ore into iron using molten aluminium similar to the Thermite reaction?

Or perhaps the real question is what temperature would this occur at?

EDIT: I'm hoping one could use the molten aluminium instead of aluminium dust to achieve the same reaction as occurs in the Thermite process.  I know about Thermite steel rail and copper earth connection welding. My goal is to reduce rust to iron by sprinkling it (with attendant sparks) onto a molten aluminium surface. Will the temperature be too low in practice or will the iron oxidise or blow off before consolidating?  Trials today were inconclusive due to low temperature and small sample size.

Short answer: Yes.

Take a look at the Ellingham Diagram.

The line for aluminium oxide is lower than the line for the all the iron oxides. This indicates greater stability of aluminium oxide, or in other words indicates aluminium's greater affinity for oxygen. Thus, based on thermodynamics alone one would expect aluminium to be able to reduce iron oxides.

Based on this figure, one would also expect this reaction to be spontaneous at all temperatures.

Important caveat, the ignition temperature remains critical. Just as a piece of paper will not combust unprovoked, even though the reaction is spontaneous, the thermite reaction needs to reach its ignition temperature.

Additionally, of interest to you would be this paper. Given below is a DSC curve from it

The endothermic peak ca. 660 corresponds to the melting of aluminium, while the exothermic peaks correspond to the thermite reaction.

The activation of the process is reported as 145 kJ/mol, and according to he authors in the study the product is controlled by the diffusion of $\ce{Al}$ in $\ce{FeAl2O4}$


Height of CCP Lattice Are the height of the Cubic Close packing Lattice and the cubic Hexagonal Close Packing the same?
My understanding says that these unit cells are formed by different arrangement of atoms, ABC for CCP and ABAB for HCP and because they are made from two different layers their heights should be different, is it so? And if it is, how do I calculate the Height of a CCP lattice?
(I know how to calculate height for cubic HCP.)

Face-centered cubic (FCC, or closest-packed cubic, CCP) and hexagonal close-packed spheres have the greatest (and equal) density of packed spheres, occupying 74.048% of the actual volume. Ref 1
However, because of the different interlaying of the hexagonal, tight packing in two dimensions, the repeating unit, or unit cell, is described differently for each crystal type. While each layer in a FCC crystal is identical to every other layer, and to each layer in a HCP crystal, the different layering makes it possible to distinguish between them, and to identify repeating units. The smallest repeating unit in a FCC crystal contains 4 atoms and can be described with one size parameter. The smallest repeating unit in a HCP crystal contains 6 atoms. Although its density is the same as that of a FCC crystal, it needs 6 atoms to describe its layering. Interestingly, body-centered cubic crystals can be described with only 2 atoms per unit cell, but have a density of only 68% (they are not close-packed). Ref 2
The construction of FCC and HCP crystals is identical for two layers, but the third is differently placed. HCP has only two positions, a and b, but FCC places a third layer c so that the atoms do not lie directly over any in the first or second layer. HCP places layers as a,b,a,b,a,b...; FCC does it this way: a,b,c,a,b,c,a,b,c...

FCC and HCP unit cells are drawn to show aspects of their packing, so interatomic distances (lattice parameters) specified are between differently placed atoms. Comparing heights isn't really comparing repeating units. If you measure the distance between two adjacent atoms in FCC, you will find the same interatomic distance in HCP. However, comparing the distance between more remote atoms can give you different results. It's a bit like comparing your vertical height at age 6 months with your horizontal dimension now. They're similar, but different because you are looking with a different viewpoint.
It's fascinating to consider that crystal structures are energetically distinguishable not just by next-door atoms, but by atoms down the street by one house. Or maybe more!
Ref 1. https://en.wikipedia.org/wiki/Close-packing_of_equal_spheres
Ref 2. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Solids/Crystal_Lattice/Closest_Pack_Structures


Reaction between phosphorus pentachloride and sulfuric acid? I am a novice in chemistry. While reading a magazine, I stumbled on some information about phosphorus pentachloride and sulfuric acid. Phosphorous pentachloride and sulfuric acid are both good dehydrating agents. What is the reaction between them (at standard temperature and pressure).

First phosphorus pentachloride reacts with sulfuric acid to form the intermediate(thanks @bon) chlorosulfonic acid. See here. Also Wikipedia

$$\ce{H2SO4 + PCl5 → POCl3 + HSO3Cl + HCl~~~~~~~~~~~~~~~~~~(1)}$$

Chlorosulfonic acid is actually an intermediate between  sulfuryl chloride  and sulfuric acid. After the formation of the intermediate, the reaction will proceed forward to form stable sulfuryl chloride. (Here)

$$\ce{HSO3Cl + PCl5 → POCl3 + SO2Cl2 + HCl~~~~~~~~~~~~~~~~~~~(2)}$$

Adding (1) and (2), we get the final reaction:

$$\ce{H2SO4 + 2PCl5 ->2POCl3 + SO2Cl2 + 2HCl}$$

My reaction is same as that of Vishnu JK's reaction and bon.


Why can we still breathe in valleys? Air is 1% argon.  Argon is heavier than air.

Why doesn't the argon concentrate in low-lying areas, choking out life there?

It does.  You would find the average percentage of the atmosphere that is argon is very slightly higher at the floor of valleys.  However, bear in mind first of all it wouldn't be anywhere near a complete stratification -- a layer of pure argon, then another of pure N2, and so on.  A mixture of nearly ideal gases doesn't do that, at least at equilibrium, because it would eliminate the considerable entropy of mixing.  (It can happen in liquids because liquids have strong intermolecular forces that normally favor separation and oppose the entropy of mixing.)  Another way to think about it is that since the atoms and molecules in gases don't (much) interact, there's nothing stopping an individual argon atom going slightly faster than nearby nitrogen and oxygen molecules from bouncing up higher than they do.

What you would get in a theoretical ideal (uniform gravitational field, complete stillness -- no wind -- and uniform temperature) would be an exponential fall of pressure with altitude, and the exponential for heavier gases would be steeper than for lighter gases.  That would result in enrichment of the heavier gases at lower altitudes.  A little work starting from the Boltzmann distribution of gravitational potential energies of each type of atom and molecule would get you an ideal estimate of the argon excess as a function of altitude.

In practice the lower atmosphere has so much mixing due to wind and big thermal gradients that I doubt you could even measure the mild excess of argon and other heavy gases.

There is one fascinating short-term exception, which bears directly on your question.  Sometimes volcanoes will belch out a considerable quantity of CO2, which is significantly denser than air, and this CO2 can accumulate briefly in a thick layer at the bottom of a valley or over a lake, if there isn't much wind.  It can persist for some hours, perhaps days, before it diffuses away and is mixed with the rest of the atmosphere.

Then indeed the valley bottom becomes an invisible death trap for humans and animals: walk into the valley, or be unable to exit fast enough when it happens, and you will suffocate for no reason you can see.  The most famous example of this is the Lake Nyos disaster in 1986 which killed thousands of humans and animals.  I think the government now has mixing devices installed in that lake to prevent any future sudden release of CO2.


Difference between 'homogeneous mixture of several chemical compounds' & 'non-homogeneous system' I just started reading Thermodynamics by Enrico Fermi; he started defining state of a system for homogeneous mixture of several chemical compounds & then non-homogeneous system as:

I thought non-homogeneous system to be a mixture of several homogeneous chemical compounds separated by boundaries. But Fermi seems to have done it otherwise; he has defined the state separately for homogeneous mixture of several chemical compounds & non-homogeneous system that is, mixture of several homogeneous chemical compounds. I am not getting how they are different at all. Can anyone please explain how they are different? What's the difference between  homogeneous mixture of several chemical compounds & non-homogeneous system?

You are right that normally a non-homogeneous system would be one where there is boundary between two internally homogeneous phases (like milk which is basically oil and water mixed at a very fine scale).

But Fermi as also allowing for the possibility that there may be only one phase where the concentration of its components varies continuously in space (if there is more than one phase the variation is discontinuous by definition).

It certainly isn't a common situation if the phase in in equilibrium where we would expect things like diffusion to create a uniform concentration of things over time, at least in liquids and gases). But it is very common in the real world where equilibrium is rarely reached (imagine the state of a poorly stirred cup of tea where the concentrations of tea-stuff will be larger near the tea bag than far away from the tea bag).

It is also very common in places where it is hard to get to equilibrium which is often true in solids (I don't know whether this is what Fermi was thinking of). Pyrex glass, for example, is apparently a single phase but the structure is deliberately made different near the surface than far away from the surface to create tension in the surface layers that protect the glass from thermal shock. But the properties vary continuously.

The examples I can think of may not be relevant to what Fermi was saying as they mostly are metastable and won't be at equilibrium which is often what thermodynamics restricts itself to worrying about.


Are there non-transparent gases? Are there gases that are not transparent at room temperature (i.e. at temperature below the point where the substance starts to radiate visible light due to heating)?

First, a little bit of background. Transparency is not an absolute property of a material. Every substance is opaque, so long as light has to pass through enough of it, and opacity also changes according to ambient conditions.  Some substances, such as most metals, are opaque even in $100\ \mathrm{nm}$ thin films, while many gasses will let a small amount of light through mostly unperturbed even after several kilometres. For example, here is a measure of how far photons of different energies can travel on average in the pure elements before interacting once with their atoms (the graph is actually for high energy x-rays rather than visible wavelengths, but they're all photons nonetheless). It makes intuitive sense that a gas should let more light through, as the atoms in it are much more spaced than in a solid or liquid.

The two general processes that work to make materials non-transparent are photon absorption and photon scattering (reflection being included as an example of scattering). When we say a material is "transparent" colloquially, what we generally mean is that the material, at the thickness observed, does not scatter much light and thus allows a coherent image to pass through the material (basically, you can see through it to the other side without a lot of distortion, that is, the object is "see-through" or clear). Interestingly, this means that a clear material need not be colourless, as correctly pointed out by Geoff, because photons of a given colour may be absorbed while letting most other wavelengths pass through without scattering.

Gasses in general are highly transparent because they are both highly colourless (absorb little light) and very clear (scatter little light) unless you're looking through several kilometres of gas, as I mentioned previously. However, there are examples of highly clear but coloured gasses, as it is possible for them to absorb a significant amount of photons of a certain energy due to low energy electronic transitions, while allowing the rest of the visible spectrum to pass through unimpeded. While some gasses are coloured and others are not, I believe all pure gasses must be very clear (small amounts of gas will scatter almost no light), and they can only scatter a fair amount of light across a short distance in conditions where they are probably not best described as gasses anymore (plasmas, supercritical fluids, etc).

Perhaps the most classic example of a coloured gas is nitrogen dioxide, $\ce{NO2}$, a strongly brown-red noxious gas which is easily formed by decomposition of nitric acid, among several other ways. The unpaired electron in its structure is somewhat uncommon, and it can be easily excited to a higher electronic state upon absorption of photons in the blue region of the visible spectrum. Since blue light is strongly absorbed even in relatively thin samples (a few centimetres of $\ce{NO2}$ gas), the rest of the white light passes through with essentially no scattering, forming a perfect image of the other side but with a strong red tinge.

Even more interestingly, $\ce{NO2}$ reacts with itself at low temperatures or high pressures and dimerizes to form dinitrogen tetroxide, $\ce{N2O4}$, a colourless solid/liquid/gas (depending on conditions). In other words, the two substances are in a reversible equilibrium:

$$\ce{2 NO2 (g) <=> N2O4 (g) + energy}$$

Coincidentally, this equilibrium is characterized by an equilibrium constant which is close to 1 at ambient conditions, and this constant can easily be changed by varying temperature or pressure to higher or lower values. This means that under relatively easily achieved conditions, it is possible to study the reversible change of a coloured gas into a colourless one!

(Source)

Though I focused on $\ce{NO2}$ because it is a remarkable case, there are several other examples of clear but coloured gasses. As Geoff mentioned, the halogens all form coloured gasses, though bromine is a volatile liquid and iodine is a solid at room temperature, so a small amount of heating is necessary. Ozone, $\ce{O3}$, is somewhat blue (source), though its colour is best observed in the liquid phase. The same goes for oxygen gas, $\ce{O2}$, which is slightly blue, though it is not the cause for the blue sky. There are probably a few more examples out there.


How do I convert benzoic acid to m-nitrobenzyl alcohol using 2 steps? The question asks to carry out the following conversion in not more than 2 steps:

Benzoic acid to m-nitrobenzyl alcohol

I first used nitration (conc. $\ce{HNO3}$ /conc.$\ce{H2SO4}$) and then I'm got stuck, as reducing the $\ce{-COOH}$ group would also reduce $\ce{-NO2}$ group (using $\ce{LiAlH4}$).

I tried to reduce $\ce{-COOH}$ group using $\ce{LiAlH4}$ to $\ce{-CH2OH}$ and then proceed to nitration but got confused whether $\ce{-CH2OH}$ is o/p or m-directing. I think it should be o/p directing.

Use the meta-directing effect of $\ce{-COOH}$: nitrate benzoic acid first.
Reduce m-nitrobenzoic acid with $\ce{LiAlH4}$. The nitro group is not attacked, but when you calculate the equivalents of  $\ce{LiAlH4}$ needed, don't forget that you are reacting an acid with a hydride.

Edit 1:
Nitroalkanes or nitroalkyl-substituted arenes can be converted to the corresponding amines by $\ce{LiAlH4}$. This is usually not the case for nitroarenes. Here, catalytic hydrogenation, e.g. over $\ce{Pd/C}$ is the method of choice.

Edit 2:
Let's play, change the order of events and first reduce benzoic acid to benzyl alcohol. The hydroxymethyl group isn't m-directing, but what will happen to it under nitration conditions ($\ce{H2SO4}$ and $\ce{HNO3}$)?

Under these pretty acidic conditions, elimination of water from the starting material furnishes benzyl cations, which might react with benzyl alcohol in different ways, such as (multiple) Friedel-Crafts alkylations or etherification. I'd dare to say that the result is a mess.


How much water evaporates in a vacuum before freezing? I placed a container with 1000 grams (1 kg) of water at room temperature (20 degrees Celsius) inside a vacuum at 0.03 atm pressure. The water boiled for a few seconds then froze solid.
I need to calculate how much water evaporated before freezing occurred. Anyone know a formula I could use?

You will need to know:

the heat capacity of water $C_p$
the latent heat of fusion of ice $L_f$
the latent heat of vaporization of water $L_v$

First determine how much heat must be lost to change the temperature from 20 to 0 using the heat capacity.

Next determine how much heat must be lost to change from 0 degree water to 0 degree ice using the latent heat of fusion.

Finally, add the two heats from above together to get a total heat loss, and use the heat of vaporization of water to calculate the amount of water evaporated that corresponds to this total heat loss.

Note: The above involves an approximation that the amount of water evaporated is relatively small.  In reality, once an infinitesimal portion of water evaporates, the amount remaining that needs to cool will be infinitesimally less than one kg.  If you are expected to know and use intergral calculus for your class, you may need to take into consideration how the mass is varying thoughout the process.

To calculate more accurately, break the problem is into two parts.

First, a mass m1 evaporates, to cool water from 20 degres to 0 degrees.

Let $m$ be the varible liquild mass during the process and temperature be $t$

$mC_pdt=L_vdm$

$C_p\int_{20}^{0}dt = L_v \int_{1}^{1-m1}(dm/m)$

The remaining mass is 1kg-m1

Second, a mass m2 evaporates, to freeze the remaining water.

$L_v m2 = L_f (1kg - m1 - m2)$


What is the black slag formed when sterling silver is melted with borax? What is the black graphite looking slag that is formed when sterling silver is melted with borax?  Is there a way to remove it from the surface of the sterling silver?  It appears to be indestructible!

The melting is related to jewelry casting.  All melting was done in an electric Ney muffle furnace.  Here is a melting dish with some half-melted buttons of sterling silver.

I can't remember why I didn't completely melt this.  I assume the green is salts of copper from the sterling silver.  Notice how the black slag erodes the walls of the melting dish.

Here is a closeup of the erosion.

This erosion property ate through the bottom of my furnace! UGGG! It'll be $600 to replace the muffle.  I heated to 1,900 Fahrenheit for about one hour and this is what I saw when I opened the furnace to get the crucible with the molten sterling silver, but with glowing white-orange light.

Here I have the crucible laying on its side next to the hole.

Here is a close up of the hole and in one of the images I'm lighting it up with a flashlight.  I'd say it's about 1.5 inches in diameter and 1.5 inches deep.  There are 2.25 Ozt of sterling silver down in the hole somewhere because the eroding black slag ate holes through the wall of my crucible.

A closeup of the erosion in front of the main hole.  These two erosion spots were from the previous melt cycle with the same crucible.  I saw that the black slag was penetrating the crucible but I figure it was just minute capillary action and it'd be ok.  Notice that the black slag seems to be working like a catalytic reaction because the amount of borax in the crucible is only about the volume of one grain of white rice.  These two depressions were not depressions at the end of the previous melt cycle they were just flat black areas of the same size and shape.

Here is a view on the outside of the crucible.  I assume one of those holes is where the silver leaked out.

A view inside the crucible.  In person, the black slag has a metallic sheen to it.  It is very similar to the dark grey metallic sheen of graphite.

What is this black slag?

"Indestructible" by what means exactly? How did you try to "destroy" it? Provided that sterling alloy is an alloy of silver and copper, the first guess is that your slag is primarily a copper (II) oxide. Depending on the actual composition of the alloy, it may contain other 3d-metal oxides. The formation of borates (not of silver, but of less noble metals) cannot be excluded, for borax is frequently used as a constituent of the high-temperature fluxes for spectroscopic and calorimetric applications. To what extent the borates are formed depends on the temperature and duration of your melting experiment.

You should be able to dissolve your slug in a boiling mineral acid of your choice. Personally, I'd try the $\ce{HCl}$ solution, or $\ce{HNO3}$ if you do not care that silver dissolves as well.

The following may be irrelevant to the question asked, but, based on my experience and some textbooks, people first try to purify silver and then melt the pure metal, not the other way around. The methods of purification are aplenty, but most of them are based on dissolving the silver-containing alloy in $\ce{HNO3}$. Then $\ce{AgNO3}$ can be transformed into $\ce{AgCl}$, which, in turn, can be reduced by metallic zinc in acidic environment or formaldehyde - in basic.


Phosphate buffer system Can anyone show me how to find the masses of $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$ needed to prepare $1.0~\mathrm{L}$ of a $0.1~\mathrm{M}$ phosphate buffer solution with $\mathrm{pH} = 7.4$ using the simultaneous equations method?

This is what I have so far:

$$\ce{NaH2PO4 + H2O <=> H3PO4 + NaOH} \\
\ce{Na2HPO4 + 2H2O <=> H3PO4 + NaOH + H3O+} \\
\ce{NaOH -> Na+ + OH-} \\
\ce{H3O+ + OH- <=> 2H2O} \\
\ce{H3PO4 + H2O <=> H3O+ + H2PO4-} \\
\ce{H2PO4^- + H2O <=> H3O+ + HPO4^2-} \\
\ce{HPO4^2- + H2O <=> H3O+ + PO4^3-}$$

The species present are: $\ce{HPO4^2-}$, $\ce{PO4^3-}$, $\ce{H2PO4-}$, $\ce{H3PO4}$, $\ce{Na+}$, $\ce{OH-}$, and $\ce{H3O+}$.

I have two equations:

$$\begin{align}
K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}] &= 1.0 \times 10^{-14} \\
[\ce{OH-}] + 2[\ce{HPO4^2-}] + 3[\ce{PO4^3-}] + [\ce{H2PO4-}] &=  [\ce{H3O+}] + [\ce{Na+}]
\end{align}$$

I need 5 more equations for this system. How do I get them?

Things to consider:

Sodium is merely a spectator ion. It will not take part in any reactions, so the initial amount of sodium you weigh in will remain constant. We can thus ignore it.
The concentrations of $\ce{H3PO4}$ and $\ce{PO4^3-}$ will be sufficiently small to entirely ignore them.

This reduces your system to the following relevant chemical equation:

$$\ce{H2PO4- + H2O <=> HPO4^2- + H3O+}$$

With the unknown concentrations of $\ce{H2PO4-}$, $\ce{HPO4^2-}$ and $\ce{H3O+}$. For these three we only need three equations and luckily we have them. Remember that we know:

the final phosphate concentration we want;
the final $\mathrm{pH}$ we want; and
the $\mathrm{p}K_\mathrm{a}$ value of our acidic species.

This gives us three equations that I am going to hide behind a spoiler tag:

   $[\ce{H2PO4-}] + [\ce{HPO4^2-}] = 0.1~\mathrm{M}$
 $10^{-7.198} = K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]}$
 $10^{-7.4} = [\ce{H3O+}]$

We can now put the third equation into the second resulting in exactly the Henderson-Hasselbalch equation, because this exactly is how that equation was derived. The maths of one possible way to approach this problem is shown in the following spoiler tag.

   $10^{-7.198} = \frac{10^{-7.4} \times [\ce{HPO4^2-}]}{[\ce{H2PO4-}]}$
 $10^{-7.198 + 7.4} = \frac{[{HPO4^2-}]}{[\ce{H2PO4-}]}$
 $[\ce{H2PO4-}] \times 10^{0.202} = [\ce{HPO4^2-}]$ ← Here! You have the Henderson-Hasselbalch right here! (In a different form but equivalent.)

 $[\ce{HPO4^2-}] + [\ce{H2PO4-}] = 0.1~\mathrm{M}$
 $[\ce{HPO4^2-}] = 0.1~\mathrm{M} - [\ce{H2PO4-}]$

 $[\ce{H2PO4-}] \times 10^{0.202} = 0.1~\mathrm{M} - [\ce{H2PO4-}]$
 $[\ce{H2PO4-}] \left (10^{0.202} + 1 \right ) = 0.1~\mathrm{M}$
 $[\ce{H2PO4-}] = \frac{0.1~\mathrm{M}}{10^{0.202} + 1} = 0.0385~\mathrm{M}$

The last steps to calculate the required concentration of $\ce{Na2HPO4}$ including calculating the respective masses of the water-free salts should be trivial and can be found in another castle answer.

I can only speculate why you are not allowed to directly use the Henderson-Hasselbalch equation since mathematically, every optimal approach will simplify itself to exactly that equation. Maybe the point of the excersize is to derive that equation; only your prof will know.


What is the exact definition of state functions? How to determine state functions? How do you know if something is a state function? I know Internal Energy (U) is a state function since it is the sum of energies, which is constant at a given time.

However I don't know for sure if $\Delta$U is a state function. I know that $\Delta$U$= U_{final}-U_{initial}$, and so any path from final to initial will yield the same $\Delta$U. So by this definition any $\Delta$ function should be a state function.

As a result I would expect Heat ($q$) to also be a state function since $q=\Delta$TE, but this is not the case.

Where am i going wrong in my understanding? I understand the "altitude" analogy, however i need a more technical definition with chemistry terms.
]

Where am i going wrong in my understanding?

In short: everywhere. I mean all possible ways. Trying to cheat by randomly throwing symbols and words out there without deep thinking is a way to nowhere. Thermodynamics is a tough subject because it requires a strong mental discipline, so to start from, you should know and understand the definitions. Funky pictures won't help unless you understand the actual meaning of the analogies used there.

State function

A state function is a property of a system that depends only on the current equilibrium state of the system, not on the way in which the system acquired that state.

Using this definition we could answer the question

  How do you know if something is a state function?

trivially: if a quantity is uniquely determined by the equilibrium state then it is a state function, otherwise it is not. $\Delta U$ is clearly not a state function, since by its very definition ($\Delta U = U_\mathrm{final} - U_\mathrm{initial}$) it is not determined uniquely by the state of the system: neither by the final, nor by the initial one. $\Delta U$ is not even a characteristic of the state (since it clearly refers to two of them, not just one), rather it is the characteristic of a thermodynamic process of transition from one state into another.

Internal energy

  I know Internal Energy ($U$) is a state function since it is the sum of
  energies, which is constant at a given time.

Sorry, but this sounds like a nonsense to me. What do you mean by constancy of internal energy at a given time due to it being sum of some energies? From a microscopic point of view, internal energy of a system can indeed be subdivided into kinetic energy of motion of all the system's particles and potential energy due to all kinds of interaction between particles in the system,
\begin{equation}
 U = E_{\mathrm{micro,k}} + E_{\mathrm{micro,p}} \, .
\end{equation}
But what kind of constancy follows from this subdivision? Do you want to say that $\Delta U$ is always zero or what?

Note also that such definition of the total energy belongs already to the realm of statistical thermodynamics, not the classical one. In classical thermodynamics the total energy is postulated to be a state function. Essentially, the first law of thermodynamics tells us that there exist a state function called internal energy, $U$, the increase in which for a closed system during a thermodynamic process is equal to the heat supplied to the system plus the work done on the system,
\begin{equation}
 \Delta U = q + w \, ,
\end{equation}
or in infinitesimals for a quasistatic process,
\begin{equation}
    \newcommand{\dif}{\mathrm{d}}
    \newcommand{\difbar}{\text{đ}}
 \dif U = \difbar q + \difbar w \, .
\end{equation}
Note that infinitesimal heat and work transfers in a quasistatic process are denoted by $\difbar$ rather than $\dif$, used to denote infinitesimal change in internal energy, to emphasize the fact that heat and work in contrast to internal energy are not state functions. Mathematically speaking, $\dif U$ is an exact differential, while $\difbar q$ and $\difbar w$ are inexact differentials. For more on that story, check this answer of mine. What is really important for our current purposes is the very final sentence there: essentially it was experimentally found that for a closed system $\oint \difbar q \neq 0$ and $\oint \difbar w \neq 0$, but $\oint (\difbar q + \difbar w) = 0$. This finding implies that there exist a state function $U$ such that $\dif U = \difbar q + \difbar w$. So, the law that postulates that $U$ defined by $\Delta U = q + w$ is a state function is based on experimental evidences.


Dubowski’s Formula vs Henry's law applied to alcohol sensor 'calibration' I'm working with some hobby-level gas sensors and I've run across a "calibration" system I don't understand. A general description of this kind of gas sensor is described a bit more in this question and also here. They have some selectivity between gases, but anything that will react with atomic oxygen at elevated temperature in the presence of a catalyst ($\ce{CO}$, $\ce{H2}$, $\ce{CH4}$, etc.) will register.
Since I don't want to mess with explosive gases, the alcohol sensor is a convenient place to begin. Many hobbyists become voluntary test subjects, and heroically subject themselves to alcohol, then breath on the sensors. Since I'm not so brave, I thought I'd try the recommended calibration test setup below. I'm not interested in doing a careful calibration, but just get some kind of ballpark quantitative agreement.
Henry's law suggests a proportionality between the concentration of a dissolved gas and the partial pressure of the gas above the liquid, and the van 't Hoff equation provides for a temperature dependence within some limited range of variation. However it's of the form $exp(\frac{1}{T}-\frac{1}{T_0})$ and the equation shown below is only a straight exponential of temperature.
This is described more thoroughly in this excellent answer:

Henry's law works for small concentrations of ideal mixtures at equilibrium. Henry's Law constant varies with temperature according to the Van't Hoff equation:

$$
k(T)=k(T_0)exp\left[-C\left(\frac{1}{T}-\frac{1}{T_0}\right)\right]
$$

Here $C$ is a constant related to the enthalpy of solvation for each gas, and $T_0$ represents the standard state, T = 298 K and 1 atm.

Question: Is Dubowski’s formula just a further approximation, or does this method  contain something new and/or different that I should be paying attention to?

below: Screenshot from Adafruit's AN4 Using MiCS Sensors for Alcohol Detection.pdf

Dubowski's formula is a parameterization of the temperature dependence of the water-vapor partition constant (Henry's law constant). It is based on data accumulated over 63 years (see Ref 1). It is assumed that Henry's law holds, in other words, that the vapour pressure over the solution is proportional to the concentration in the solution as $\frac{p}{c} = K_{a/w}=$ constant.
It can be used as a proxy for the concentration of alcohol in the blood although it is not expected to be entirely accurate, as Dubowski comments (Ref 2):

We believe that the conversion of a breath quantity to a blood concentration of ethanol, for forensic purposes, should be abandoned and that the offense of driving while under the influence of alcohol should be statutorily defined in terms of the concentration of ethanol found in the breath in jurisdictions employing breath analysis.

Irrespectively of the details involving breath-alcohol analysis, Henry's law states that the vapor pressure is linearly proportional to the concentration in solution as $p=K_Hc$, and is applicable in practice when the alcohol concentration in the blood is very low (it is in general applicable as a limiting law).The following figure shows the partial pressure of ethanol in dilute aqueous solution at $25 ^\circ C$ as a function of the mole fraction of alcohol in solution, where the dashed line represents the pressure predicted by Henry's law:

For reference, a blood alcohol concentration (BAC) of 0.1%(w/w) translates into a mole fraction of $\pu{3.9E-4}$.
The Henry's law constant is obtained by extrapolating to zero concentration of the solute (here alcohol). In general, water-ethanol solutions are highly non-ideal due to hydrogen bonding and would not be expected to follow Henry's law except for very dilute concentrations of one component in the other. As soon as interactions between solute molecules become significant the law can be expected to break down.
The difference in the functional form of the temperature-dependence in Dubowski's formula compared to Van't Hoff's can be ascribed to differences in units: one uses T in Kelvin, the other in degrees Celsius. With proper application of Taylor series formulas they can be shown to be equivalent within narrow T ranges.
References

K.M.  Dubowski.  Breath-alcohol simulators: Scientific basis and actual performance. J. Anal ToxicoL 3" 177-82 (1979).

M F Mason, K M Dubowski. Breath-alcohol analysis: uses, methods, and some forensic problems--review and opinion. J. Forensic. Sci. 1976 Jan;21(1):9-41.


Numbering starts from the side that has more locants or from the side that has a more preferable group? I came across this question while practicing iupac nomenclature.

Should we number the parent chain from left to right (shown in blue) because there are three cholrine atoms attached to C1 while there are only two alcohol groups attached to C2?

Or should we start the numbering from the right side instead (shown in orange) as alcohol group is preferred over the chlorine group?

Concerning the numbering of locants for substituents, the most important simplified criteria are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The $\ce{-OH}$ group is the principal characteristic group in this case. It is expressed as suffix ‘ol’.
Therefore, low locants are assigned first to the $\ce{-OH}$ groups according to Rule (c), which yields ethane-1,1-diol.
After that, a low locant is assigned to the chloro groups according to Rule (f); however, the only possible numbering left is 2,2,2-trichloroethane-1,1-diol.


Chemistry of toilet bowl mineral stains on porcelain I live in Massachusetts, my house has 3 bathrooms and I live alone.  If I don't scrub my toilet bowls once a week they build up brownish-yellow mineral stains.  These seem independent of whether I've actually used that toilet bowl since the previous cleaning (since I live alone some toilets are seldom used).

Looking on the web there's lots of advice from "folksy" sources like "frugal blonde" or "professional mom" about how to clean toilets -  usually involving vinegar, Clorox, pumice-sponges, etc, and lots of elbow grease, rubber gloves, bathroom fans and time.   One guy uses WD-40!     But being "folksy" none of these involve any discussion of the chemistry.

I thought if I understood the science I could save work and maybe even prevent the stains but I can't find ANYTHING on the web about this!  What are the minerals in my water and what form are they in?   Are they ionized or just some sort of suspension?  What are the principal forms -  I assume calcium, magnesium (and silicate?) compounds -  and what form do they precipitate out in on the porcelain?   What IS porcelain (chemically-speaking) anyway, and what makes such tight bonds at the interface between the minerals from the water and the porcelain surface?

If I knew the answers to these maybe I could even prevent these porcelain-mineral bonds from forming

Many places have water with dissolved iron and manganese, which precipitates out as a brown or black stain when exposed to air.

Massachusetts has areas of podzol, where the fine sandy glacial till had been covered by pine forests. This makes for acid soil, which in anaerobic conditions readily dissolves iron in the ferrous form, $\ce{Fe++}$ from the ground, which is a pale green color, almost unnoticeable. If you dig into a podzol, exposing it to air, the soil turns brownish or orange in a matter of minutes as the iron is oxidized to its ferric, $\ce{Fe+++}$, form. That's what is happening on your porcelain fixtures.

It's hard to dissolve the brown ferric iron stain, but there are two ways to do so:

Reduce the iron (and manganese) to the lower-valence state. Sodium metabisulfite $\ce{Na2S2O5}$ is sold for this purpose under names such as Red-B-Gone. It works well on clothing and light stains on porcelain. Use this with caution, since it can generate noxious sulfur dioxide fumes if it comes in contact with an acid.
Dissolve the oxides in a strong acid, such as $\ce{HCl}$, hydrochloric acid. Though more dangerous than reducing agents, 10%-30% $\ce{HCl}$ can remove thicker layers of stain. Use with great caution, including eye protection and rubber gloves. Rinse immediately with plenty of water if any gets on you or your clothing! $\ce{HCl}$ is available in hardware stores and in plumbing supply shops.

You can also treat the water to remove iron and manganese or to sequester it with a polyphosphate additive.

Do not use alkaline drain cleaners (e.g. $\ce{NaOH}$ and Drano®) on vitreous china, as it etches the glaze and allows stains to soak into the porcelain, making the stains difficult or impossible to remove.


How is enthalpy change equal to heat supplied at constant volume? We know that:

$$\begin{align}
H &= U + pV \\
\Delta H &= \Delta U + \Delta (pV)
\end{align}$$

Now,

If $p$ is constant,

$$\begin{align}
\Delta H &= \Delta U + p\Delta V \\
&= q_p + w + p\Delta V \\
&= q_p \qquad\qquad  \text{(since } w = -p\Delta V\text{)}
\end{align}$$

If $V$ is constant,
$$\begin{align}
\Delta H &= \Delta U + V\Delta p \\
&= q_V + w + V\Delta p \\
&= q_V + V\Delta p \qquad \qquad \text{(since } w = -p\Delta V = 0\text{)}
\end{align}$$
We also know that at constant $V$,

$$\Delta U = q_V$$

Now my question is that: my book says that at constant $V$, $q = \Delta H = \Delta U$, which I think is false, since I obtained $\Delta U = q_V$, but not $\Delta H = q_v$.

Please tell me where am I going wrong?

Here is a link to the chapter.

Please refer to Pg 161, right column, first paragraph.

Link to chapter: Thermodynamics

No the book is not completely wrong.If you read the next 6 lines then you will find that it wants to point that this approximately true only for solids and liquids, not for gases.

From the First Law,

$$\Delta U = q + w$$
and since work done is $0$ in an isochoric process (constant $V$),

$$\Delta U = q_V$$

Furthermore,

$$\begin{align}
\Delta H &= \Delta U + \Delta (pV) \\
&= q_V + \Delta (pV)
\end{align}$$

Since we are considering solids and liquids, the changes in the volume of these matter upon changing pressure are negligible (as they are incompressible matter) when compared to gases, so the $\Delta (pV)$ term can be considered negligible.

Therefore

$$\boxed{\Delta H \approx \Delta U = q_V}$$

for these incompressible solids and liquids. However this fails when we consider gases as the term $\Delta (pV)$ will no longer be negligible.

Comparison between the $\Delta (pV)$ values for different states of matter, being heated from $20\ ^\circ\mathrm{C}$ to $30\ ^\circ\mathrm{C}$:

$$\begin{array}{c|c}
\text{Substance} & \Delta (pV)\text{ / J} \\
\hline
\text{Air} & 2850 \\
\text{Water} & 0.1 \\
\text{Iron} & 0.0004
\end{array}$$


Bond Dissociation Temperatures for Hydrocarbons I'd like to know the temperatures that various hydrocarbons such as methane, butane, and propane, etc. will dissociate into hydrogen and carbon.  Is there a reference where I can look up these temperatures?

Does pressure or partial vacuum effect these temperatures?

When hydrocarbons undergo thermal decomposition carbon-hydrogen and carbon-carbon bonds are broken.
$$\ce{R-CH3 -> R-CH2. + ~H.}$$
$$\ce{R-R' -> R. + ~R'.}$$
Here is a link to a table that shows the energies required to break these bonds (the bond dissociation energy). These reactions are unimolecular decompositions and are governed by first order kinetics.  The rate of the reaction is given as
$\text{reaction rate} = k[\ce{A}]$
where k is the rate constant and $[\ce{A}]$ represents the concentration of the hydrocarbon.  The rate constant can be described using the Arrhenius equation
$k= A e^{-E_a/(R T)}$
where $A$ is the pre-exponential factor, T the absolute temperature in kelvin, ${E_a}$ is the activation energy, and R is the universal gas constant.  In the thermal unimolecular decomposition of a hydrocarbon, $A$ will have a value around $\pu{10^{12}sec^-1}$ and the bond dissociation energy can be used for the activation energy.
The Arrhenius equation tells us that this decomposition will be occurring at all temperatures above absolute zero.  As we increase or decrease the reaction temperature the rate constant will increase or decrease respectively, but the reaction is always occurring as long as we are above absolute zero.
As to pressure, yes it will affect the reaction rate.  Pressure is directly related to concentration, the $\ce{[A]}$ term in our rate equation.  As we increase or decrease the pressure, we are increasing or decreasing the concentration of $\ce{A}$, and thereby increasing or decreasing the rate of the reaction.


Comparing nucleophilicity We have to compare the nucleophilicity of $\ce{OH-,NH2-,CH3CH2-}$ and $\ce{CH#C^{-}}$. I managed to find out that $\ce{CH3CH2-}>\ce{NH2-}>\ce{OH-}$ and that $\ce{CH3CH2-}>\ce{CH#C-}$. But I don't know how to compare $\ce{CH#C-}$ with either $\ce{OH-}$ or $\ce{NH2-}$.

My attempt:

I know that the acidity order of the 2nd period is $\ce{CH4}<\ce{NH3}<\ce{H2O,HF}$. Now the stronger an acid is, the weaker its conjugate base will be. Therefore, the conjugate base order will be:
$$\ce{CH3-}>\ce{NH2-}>\ce{HO-}>\ce{F-}$$

We know that nucleophilicity parallels basicity along the period, therefore the nucelophilicty order will be as the basicity order:
$$\ce{CH3-}>\ce{NH2-}>\ce{HO-}$$

Now we can compare $\ce{CH3CH2-}$ with $\ce{CH3-}$.

Since methyl is an electron donating group, it will increase the negative charge on the carbanion and therefore $\ce{CH3CH2-}$ will be more nuceleophilic than $\ce{CH3-}$.

Since $\ce{CH3-}$ is more nucleophilic than $\ce{NH2-}$ and $\ce{OH-}$. Also $\ce{CH3CH2-}$ is more nuceleophilic than $\ce{CH3-}$. As a result $\ce{CH3CH2-}$ will be more nuceleophilic than $\ce{NH2-}$ and $\ce{OH-}$ and $\ce{CH3CH2-}$.

We can also compare $\ce{CH3CH2-}$ and $\ce{CH#C-}$. Since $\ce{CH#C-}$ has sp hybridization, it will be less nuceleophilic than $\ce{CH3CH2-}$.

But how can I compare $\ce{CH#C-}$ to either $\ce{OH-}$ or $\ce{NH2-}$?

You also may want to consider Hard-Soft Acid-Base Theory. The carbanion is generally said to be a better nucleophile than hydroxide, presumably as measured in a highly controlled setting using a standard electrophile like methyl iodide. I'm under the impression nucleophilicity is very nearly defined as how fast an electron donor reacts with $\ce{CH3I}$. Note that there are no hydrogens available for elimination on this electrophile, which precludes the possibility of the nucleophile acting as a base instead.

In practical reaction conditions (protic solvents, beta-hydrogens on electrophile, etc), we may need to worry about our nucleophile acting as a base. This is where HSAB Theorycomes in. Hydroxide is considered a "hard base", which means it may be more inclined to react with "hard acids" like $\ce{H+}$ rather than "soft acids" like methyl iodide. As a "soft base", a carbanion will probably prefer to bond with methyl iodide (as a classic nucleophile) than with any available protons (as a Brønsted-Lowry base). So in practice, the difference between a carbanion's nucleophilicity and hydroxide's might be larger than expected due to competing "hard base" reactions (Brønsted-Lowry rather than Lewis). Just food for thought.

Qualitative nucleophilicty ratiomale


How to explain the excited states in the dinitrogen cation? The $\ce{N2^+}$ molecule has the molecular orbitals
$$\ce{\sigma(1s)^2\; \sigma^*(1s)^2\; \sigma(2s)^2\; \sigma^*(2s)^2\; \pi(2p_x)^2 \; \pi(2p_y)^2 \; \sigma(2p_z)^1,}$$
which can be seen on the following picture:

So, the ground state has $\mathrm{A_g}$ symmetry in $D_\mathrm{2h}$ point group, i.e. $\Sigma_\mathrm{g}^+$ in $D_\mathrm{\infty h}$.

But what will the excited states look like?

I'd expect, that the first excited electrons will be the ones, which "need" the smallest amount of energy to get into a higher state.

So, the first electron will be excited from $\ce{\pi(2p_x)}$ or $\ce{\pi(2p_y)}$ orbitals to $\ce{\sigma(2p_z)}$ like this:

The first excited state will have $\mathrm{B_{3u}}$ or $\mathrm{B_{2u}}$ symmetry in $D_\mathrm{2h}$.

The second excited state would be

I.e. it would have the $\mathrm{B_{3g}}$ or $\mathrm{B_{2g}}$ symmetry.

Is my expectation correct or will the electrons be excited in a different order? And what will the 3rd and 4th excited states look like?

Excited states make things a bit more complicated. The electron configuration only represent the HF ground state, and changing the occupied orbitals will have an effect on all the orbitals (they are correlated). So this picture becomes more and more inaccurate the higher lying the electronic state. What you do to resolve this is expand the electronic state in many configurations (which all most belong to the same IRREP).

For diatomic molecules there is a lot of spectroscopic data available. So let's check it: $\ce{N2+}$ in the NIST Chemistry WebBook

The states in that table are ordered by energy, with the ground state at the bottom. So scrolling down shows us the 3 lowest lying states: $\rm X ^2\Sigma_g^+$, $\rm A^2\Pi_u$, $\rm B^2\Sigma_u^+$.

Basically, this means you are right about the first state. To explain why we need a bit of symmetry. The first excited state has $\Pi_u$ symmetry, in the $D_{\infty h}$. Since we are working with $D_{2h}$ here, we will need to reduce it. Sparing the details, the result is $\Pi_u$ becomes $B_{2u}\oplus B_{3u}$, which means it will show up once in each IRREP, which is fine because $\Pi$ states are doubly degenerate anyway (just as $\pi$ orbitals are doubly degenerate).

The other state ($\Sigma_u^+$) becomes $B_{1u}$. This one you can for example get by excitation from the anti-bonding $2\sigma_u$ into $3\sigma_g$, with respect to the ground state configuration (see this reference). So there is an electron "jumping over" the $1\pi_u$ orbitals.

Note, that the ordering of the states can change if you start to move away from the equilibrium distance towards dissociation.


pH of aqueous solution of HCl of low concentration What is the pH of $10^{-8}~\mathrm{M}$ $\ce{HCl}$ solution in water?

My attempt:

  pH = $-\log(10)^{-8}$ = 8

But this is wrong because it should be acidic. Where have I gone wrong?

It is known that the equilibrium constant for the following reaction is $10^{-14}$.

$$\ce{2H2O(l) <=> H3O+(aq) + OH-(aq)}$$

That means,

$$\frac {[\ce{H3O+(aq)}][\ce{OH-(aq)}]} {[\ce{H2O(l)}][\ce{H2O(l)}]} = 10^{-14}$$

where the concentration of water $[\ce{H2O(l)}]$ is assumed to be $1$.

Initially, we have $[\ce{H3O+(aq)}] = 10^{-8}$.

Then, assuming that $x$ moles of $\ce{H3O+(aq)}$ is generated. An equal amount of $\ce{OH-(aq)}$ must also be generated:

$$(10^{-8}+x)(x) = 10^{-14}$$

Solving for $x$ gives $x = 9.51 \times 10^{-8}$.

So, the total concentration of $\ce{H3O+(aq)}$ is $1.051\times10^{-7}$.

Therefore, the pH is $6.978$.

The problem with your approach

Let the initial molarity of the hydronium ion be $a$.

Solving $(a+x)(x) = 10^{-14}$, if $a$ is well above $10^{-5}$, gives $x \approx 0$.

The approximation that the pH is equal to $-\log{{\left([\ce{HCl(aq)}]\right)}}$ is only accurate for normal concentrations.


Regarding the infinitesimal form of the First Law of Thermodynamics First Law of Thermodynamics is expressed as[1]

  The internal energy of an isolated system is constant.

If $w$ denotes the work done on a system, $q$ for the energy transferred as heat to the system, and $\Delta U$ for the resulting change in internal energy, then it follows that $$\Delta U = q + w \label{1}\tag{1}$$

$$\mathrm{d}U = \mathrm{d}q + \mathrm{d}w \label{2}\tag{2}$$

What I don't understand is how does $\eqref{2}$ follow mathematically from $\eqref{1}$. Taking the limit $\Delta U \rightarrow 0$ gives the left-hand side but how does $q$ change into $\mathrm{d}q$ and $w$ into $\mathrm{d}w$? I understand the interpretation of both the equations physically. Is this related to the fact that $U$ is a state function whereas $q$ and $w$ are not?

Reference:

[1] Atkins, P.; de Paula, J. Physical Chemistry, 10th ed; Oxford UP: Oxford, U.K., 2014.

The first law of thermodynamics is a statement of energy conservation thus and defines the internal energy E as an extensive state function. In an infinitesimal transformation, the first then law reduces to
$$\mathrm{d}E = \mathrm{d}Q + \mathrm{d}W$$

where $\mathrm{d}E$ is a total (exact) differential for infinitesimal transformation.

However, $\mathrm{d}Q$ and $\mathrm{d}W$ are not exact ($Q$ and $W$ are not state functions); $Q$ and $W$ in a thermodynamics transformation are process-dependent. All of these are properties of functions of more than one variables


Why do the halides run in reverse orders in the spectrochemical series and the trans-effect series? Spectrochemical series: $\ce{F^-}>\ce{Cl^-}>\ce{Br^-}>\ce{I^-}$

Trans-effect series: $\ce{I^-}>\ce{Br^-}>\ce{Cl^-}>\ce{F^-}$

This seems slightly odd to me because their relative ordering surely should be the same because both series depend on sigma donor ability. Does this have something to do with π donor ability?

In what order does σ donor and π donor ability go for halides in transition metal systems?

The two features you cite — trans-effect and spectrochemical series, i.e. ligand-induced crystal field split — are based on two different phenomena, two different structures and two entirely different observations that warrant two different discussions. Thus, the probability for the two trends to be inverse is $0.5$.

Spectrochemical series

The spectrochemical series in itself is already a convolute of multiple different effects and it is defined for octahedral complexes. However, if we stick to small regions of it, we can make out trends based on general observations. Here, we are dealing with four very similar ligands which are all weak-field ligands, σ bases and π bases. To understand the effect of their π basicity and to relate them to each other, we need to take a look at the molecular orbital diagram of a typical $\ce{[ML6]}$ complex accounting for π effects. Figure 1 shows such an MO scheme.

Figure 1: Octahedral $\ce{[ML6]}$ complex including σ and π ligand-metal interactions.

We see that the $\mathrm p$ orbitals (or more precisely: that $\mathrm p$ group orbital with $\mathrm{t_{2g}}$ symmetry) interact with the metal’s $\mathrm{t_{2g}}$ orbitals in a bonding/antibonding manner of π symmetry — thus the interaction is weaker than the $\mathrm{e_g}$ interaction which features σ symmetry. In the case of all halido ligands, the ligand orbitals are fully populated meaning that the metal electrons end up in the $\mathrm{t_{2g}^*}$ orbitals which have been raised with respect to the neutral (or π-effect absent) state. This reduces the energy difference between $\mathrm{t_{2g}^*}$ and $\mathrm{e_g^*}$ and thus the value of $\pu{10Dq}$ or the field split.

The different halides differ only in their electronegativity and thus in the relative positioning of their orbitals. The higher the electronegativity the lower the corresponding orbitals — i.e. fluoride will have the lowest-lying orbitals. Since these are all non-metals with relatively high ionisation energies (even for the anions that we are looking at here), they will all be below the metal orbitals. However, a larger interaction — and thereby a lower value of $\pu{10Dq}$ — is typically obtained if the interacting orbitals are close in energy.

Iodide is the one that is the least far away, meaning that the &#3c0; interaction is largest in an $\ce{[MI6]^n-}$ complex, meaning that $\pu{10Dq}$ is lowest meaning that it is on the very weak end of the spectrochemical series. The higher up we go through the group, the lower the halide’s $\mathrm p$ orbital energies will be, thus the weaker the interaction and the higher $\pu{10Dq}$. Naturally, the orbital energy decreases monotonously and therefore the ligand series of the halides will be as observed.

Trans-effect

The trans-effect is typically observed for complexes of platinum or related elements and is typically referred to for square planar geometries. It concerns ligand exchange which has been shown by a number of experiments to proceed via an associative mechanism:

$$\ce{[MX4]^n- + L -> [MX4L]^n- -> [MX3L]^${(n-1)-}$ + X-}\tag{1}$$

The initial and final complexes are square planar, the intermediate is trigonal bipyramidal. Going by the principle of least movement, the additional ligand $\ce L$ should approach via the distinguished axis (I’m calling it $x$) towards the inital complex and displace two ligands which will now simultaneously occupy the $xy$ plane (meaning that $z$ is the axis of the two non-moving ligands). Then, one of the three $xy$ ligands will get displaced; it has to be the one that is trans to the ligand which exercises a trans-effect and it has to be one of the two in the $xy$ plane — i.e. those that were originally on the $y$ axis.

Since the pentacoordinated transition state is obviously most important here, we should draw up its molecular orbital scheme. This is a scheme much less known than the octahedral scheme shown above so I will point out a few things. The trigonal bipyramidal complex’ point group is $D_\mathrm{3h}$. The metal’s $\mathrm d$ orbitals transform as $\mathrm{a_1' + e' + e''}$ with the orbitals within each irreducible representation being $\mathrm{d}_{z^2}, (\mathrm{d}_{xy}, \mathrm{d}_{x^2-y^2})$ and $(\mathrm{d}_{xz}, \mathrm{d}_{yz})$, respectively. The ligands’ σ-symmetric orbitals transform as $\mathrm{2a_1' + e' + a_2''}$. Broken down, the $z$ ligands are $\mathrm{a_1' + a_2''}$ and the $xy$ ones are $\mathrm{a_1' + e'}$. The π-symmetric $\mathrm p$ orbitals of the $z$-ligands should transform as $\mathrm{e' + e''}$ and the π-symmetric $\mathrm p$ orbitals of the ligands in the $xy$ plane are $\mathrm{a_2' + a_2'' + e' + e''}$. We can see a lot of $\mathrm e$ irreducibles which is where all the fun takes place. For completeness, the central metal’s $\mathrm s$ orbital transforms as $\mathrm{a_1'}$ and its $\mathrm p$ orbitals as $\mathrm{a_2'' + e'}$.

Figure 2: Qualitative molecular orbital scheme of a trigonal-bipyramidal $\ce{[ML5]}$ complex including both σ and π interactions.

The MO scheme is given in figure 2 above. The key to the analysis is to find out why certain ligands preferentially occupy the $xy$ plane but are not displaced themselves. The key lies within the orbitals that are labelled $\mathrm{e'}$: they bond in a σ fashion will all equatorial ligands in the intermediate being degenerate while in the reactant and product state they have different energies — one corresponding to a σ bond, one to a π bond. Thus, in moving from the reactant to the intermediate, we are generating a new interaction of σ symmetry which — again according to symmetry — affects all three equatorial ligands.

As I pointed out before, the lower a halide’s electronegativity the higher its orbitals energy levels and thus the greater the interaction. Therefore, by having the less electronegative halide participate in the new σ bond (i.e. by having it as part of the equatorial ligand sphere and thus by displacing the ligand trans with respect to it) will generate a more stable intermediate and will thus be a more favourable path.

Thus it is the same property of the halogens that create the two line-ups but the final effect is opposite in the two cases.


What is meant by "basis" in "... its basis in particles leads to a material comprising crystallites or grains tightly bound together, ..."? I am currently studying Ultrasonic transducers – Materials and design for sensors, actuators and medical applications by Kentaro Nakamura. In chapter 1.2.1 Ferroelectricity, the author says the following:

$\ce{BaTiO_3}$ is realised as a ceramic, originating as a mouldable paste, which is dried and sintered to form the final material. The properties of the paste result from the mixing of particles with a liquid and its basis in particles leads to a material comprising crystallites or grains tightly bound together, with properties differing between the individual grains and the collection of grains in the bulk ceramic because of the effects of the interfaces between the grains. Following the principle of minimisation of energy, and taking into account the resulting internal dynamics of the relative positions of the ions, $\ce{BaTiO_3}$ divides itself into multiple domains within each grain, each domain having a common polarisation, but with the polarisation differing in direction from that of neighbouring domains by either $90^\circ$ or $180^\circ$.

My concern is with this part:

The properties of the paste result from the mixing of particles with a liquid and its basis in particles leads to a material comprising crystallites or grains tightly bound together, with properties differing between the individual grains and the collection of grains in the bulk ceramic because of the effects of the interfaces between the grains.

What is meant by "basis" in "... its basis in particles leads to a material comprising crystallites or grains tightly bound together, ..."? I don't understand what this means. Is this possibly a typo or incorrectly used term by a non-native English speaker? If so, then does anyone know what they meant (or what it should say for this described process)?

The author is attempting to indicate that the "paste" is a suspension of particles that aggregate as it dries (ie the liquid portion evaporates), in contrast to a solution of a dissolved solid that recrystallizes as the solvent evaporates. Because the particles never dissolve, there is no possibility for a single crystal to form that is any larger than an already existing particle. Therefore, the resulting solid is necessarily composed of many small crystals/particles with boundary zones between them, rather than having a large scale uniform structure.


Intermolecular forces - Covalent bonding Im confused. How to relate permanent dipole dipole moment, temporary dipole induced dipole and Van der waals forces of attractrion?

Permanent dipoles are the result of charge separation within a bond, and normally results from bonds where the two elements have a large difference in electronegativity. Dipole moment is a measurement of the polarity of a compound. For example, the carbon-oxygen bond in acetone gives the compound a permanent dipole, explaining its large dipole moment. A temporary dipole occurs when electrons randomly group together in one part of a bond. These dipoles result in London Dispersion Forces, the weakest king of Van der Waals force. Van der Waals forces are a collective term for all residual attractive and repulsive forces between compounds that are not the result of ionic or covalent interactions. Induced dipoles are dipoles that result from a non polar compound becoming polarized as a result of interactions with an ion or polar compound.


Calculating the molarity of DNA in a cell In the following questions use a value of 3 for $\pi$, $6 \times 10^{23}$ for Avogadro’s number and $660$ for the molecular weight of $\pu{1 bp}$ of DNA. The volume of a sphere of radius $r$ is $4/3\,πr^3$. A bacterium has a single copy of a $\pu{4 \times 10^6 bp}$ circular genomic DNA.
If the diameter of this spherical cell is 1 micrometer, what would be the molar concentration of DNA in this cell?

The volume comes to be $\pu{2 \times 10^{-5} L}$.
Amount of substance is $\pu{6.7 \times 10^-18 mol}$.
Dividing this by the volume, my answer comes $\pu{3.3 \times 10^{-13} M}$, but the answer given in my book is $\pu{3.3 \times 10^{-9} M}$.

There is one particle of dsDNA in the cell. Divide by $N_\mathrm{A}$ to get the amount of substance, and divide by the volume to get the concentration matching the concentration given in the answer.
\begin{align}
  V_{\mathrm{cell}}
    &= 4 \cdot {(\pu{0.5 μm})}^{3}\\[0.5ex]
    &= \pu{0.5 μm3}\\[0.5ex]
    &= \pu{5E-16 L}\\[3.5ex]
  N_{\mathrm{A}}
    &= \pu{6E23 1//mol}\\[3.5ex]
  n_{\mathrm{DNA}}
    &= \frac{1}{N_{\mathrm{A}}}\\[0.5ex]
    &= \pu{1.7E-24 mol}\\[3.5ex]
  c_{\mathrm{DNA}}
    &= \frac{n_{\mathrm{DNA}}}{V_{\mathrm{cell}}}\\[1.5ex]
    &= \frac{\pu{1.7E-24 mol}}{\pu{5E-16 L}}\\[1.5ex]
    &= \pu{3E-9 mol//L}
\end{align}


Skeletal formula for carbon with two double bonds How do I depict an sp-hybridised carbon with two double bonds?

$$\ce{R=C=R}$$

If I just leave it out, like I would for a normal $\mathrm{sp^3}$ or $\mathrm{sp}$ carbon, it's difficult to see that it's there.

In chemical structure diagrams, carbon atoms that are bonded to at least two other atoms are usually left unlabeled. The carbon atom is indicated by the angle in the bonds.

However, according to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), a carbon atom with two double bonds should be explicitly labeled:

  (…), any carbon atom with two identical collinear bonds should always be explicitly labeled, to remove the possibility of the two bonds being misinterpreted as one long bond.

  The use of a dot in place of an explicit carbon atom label is acceptable in allenes and related molecules with three consecutive carbon atoms. It is not acceptable to use a dot to represent a carbon atom when either of its adjacent atoms is other than a carbon atom.

  When a dot is used to represent a carbon atom, it must be depicted clearly and unambiguously. The dot should be large enough to be clearly seen by the reader, and must be visibly separated from the adjacent bonds.


How to distribute the Boltzmann energy to internal and external motion? If we have an isolated pentane molecule at room temperature (RT) the Boltzmann energy $(E=k_\mathrm{B}T)$ is approxately $0.59\ \mathrm{kcal/mol} \overset{\wedge}{=} 207\ \mathrm{cm^{−1}}$. There is not enough energy for an electronic or vibrational excitation, so the molecule is in its electronic and vibratory ground state. However usually 10–20 rotational states are populated at RT. Can we attribute mathematically (to a first approximation) this $207\ \mathrm{cm^{−1}}$ amount of energy to $x\:\%$ translational energy ($E_\mathrm{kin} = \frac{m v^2}{2}$) and to $(100-x)\:\%$ rotational energy?

This question is related to the question about the internal degrees of freedom for a molecule: Rotational degrees of freedom (3N-5 and 3N-6)

The translational energy of a mole of molecules is $\tfrac{3}{2}RT$ which corresponds to an average kinetic energy of $\tfrac{1}{2}M\langle v^2\rangle$.  The average reflects a Boltzmann distribution of speeds
$$f(v)=4\pi\left(\frac{M}{2\pi RT} \right)^{3/2}v^2\operatorname{e}^{-\tfrac{1}{2}Mv^2/RT} $$
i.e. some molecules have very low velocity and some have very high velocities.

Kinetic energy is constantly being transferred, so if you could monitor its velocity over time its average kinetic energy would be $\tfrac{1}{2}M\langle v^2\rangle=\tfrac{3}{2}RT$

In addition one mole of gas has $\tfrac{3}{2}RT$ of rotational energy ($RT$ for linear molecules), which also reflects an average rotational kinetic energy.  For individual molecules at any given time, the amount of translational and rotational energy is independent.  For example, a molecule can move very slowly while rotating very rapidly – until it collides with another molecule and things change.  However, on average there is an equal amount of translational and rotational energy (for non-linear molecules).


Volume change inside a balloon upon decreasing the the outer pressure A balloon is filled with hydrogen at room temperature. It will burst
  if pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies
  2.27 L volume, upto what volume can the balloon be expanded?

Now i can easily figure out that the pressure is 0.2bar at a volume of 11.35L. However, when i check the answer in my book, it says that the volume of the balloon should be less than 11.35L.

However, if you decrease the volume from 11.35L, wouldn't the pressure become greater than 0.2bar, hence bursting the balloon?

The relationship between pressure and volume is given by Boyle's Law.
$$\mathrm{P}_1\mathrm{V}_1 =  \mathrm{P}_2\mathrm{V}_2$$

      P       V
     0.199    11.41
     0.20     11.35
     0.201    11.29

     1.0      2.27

So yes, you're right. Using Boyle's Law, the volume should be more than 11.35L so the pressure stays below 0.20 bar.

However this doesn't make real world sense. When you blow up a balloon, as the pressure increases the volume increases.

Edit - Thanks @Mithoron here is a version of the problem that works...

    A balloon is filled with hydrogen at room temperature (25 C) and pressure (1.00 bar) to a volume of 2.27 L in a chamber. The pressure in the chamber is then reduced isothermally. The balloon will burst when the pressure in the chamber is reduced to 0.200 bar. What will the volume of the balloon be just as it explodes?


Michael reaction or nucleophilic attack at the alpha carbon of an alpha-beta-unsaturated ketone? What conditions promote a nucleophile to undergo the Michael reaction over the nucleophilic attack at the alpha carbon of the carbonyl group in an alpha-beta-unsaturated ketone? I'm looking for an answer that considers organic and inorganic nucleophiles (like organocuprates/organolithium compounds).

An $\alpha,\beta$-unsaturated ketone is electrodeficient at the $\beta$ position. This can be seen if you draw the resonance structures of such a molecule.

The $\beta$ carbon is thus a good site for nucleophilic attack. But, as you know, carbonyls are also prone to nucleophilic attack. To discriminate between the two, you need to look at how the reaction is controlled, either thermodynamically or kinetically.

In a kinetically controlled reaction, the product that is formed fastest predominates. In a thermodynamically controlled reaction, the predominant product is the energetically favored one.

A Michael addition is a 1-4 addition, where a nucleophile attacks the $\beta$ carbon, and produces the thermodynamically favored product. On the other hand, a 1-2 reaction (on the carbonyl) gives the kinetic product, and is obtained at low temperatures.

Why is the 1-4 product thermodynamically more stable? Because the resulting product benefits from keto-enol tautomerism, which results in lowering the energy of the system. Usually, the more resonance forms a compound has, the more its electrons are delocalized, the more stable it is.

Draw the resonance forms of the 1-4 and 1-2 products, and see.

You asked for specific affinities of different organometallics in 1-4/1-2 additions. My knowledge is that organocuprates ($\mathrm{R-CuLi}$) will perform Michael additions, and that organolithians seem to prefer 1-2 addition. Also, according to this source, Grignard reagents do not seem to have a preference.

My take on this is that the cuprate is less reactive, and therefore can form the thermodynamic product, whereas the lithium reagent is so destabilized that it reacts right away.


What is the equation for the reduction of water? I'm a grade 12 student trying to understand redox reactions. What's the equation for the reduction of water?
The back of my textbook says: $\ce{2H2O + 2e- -> H2 + 2OH-}$
I get that this makes sense, but I have no idea how to derive it myself. Any help would be appreciated!

I get that this makes sense, but I have no idea how to derive it myself. Any help would be appreciated!

Good question because such questions were pursued by prominent physical chemists and Nobel laureates (Ostwald/ Nernst and Hoffmann).
The reduction of water can be tested visually and the apparatus to demonstrate it indeed named after Hoffman. Basically, it is an electrolysis of water experiment. What will see at the side connected to the negative terminal of the battery that it produces a gas, which is twice the volume as compared to the side connected to the positive terminal of the battery.
This gas which is very light, burns with a blue flame and it forms water again- so this must be hydrogen. Secondly, another interesting effect is that electrode region which is producing hydrogen becomes highly alkaline. If you put phenolphthalein indicator, the negative terminal becomes pink indicating an excess of OH(-) ions.

So three things can convince us with the following:
$\ce{2H2O + 2e- -> H2 + 2OH-}$
a) The electron rich electrode is the cathode in electrolysis cell. Water can gain electrons here.
b) The cathode becomes highly alkaline during electrolysis
c) The cathode produces hydrogen gas, which is twice in volume as compared to oxygen.
d) You can even calculate the number of electrons used to produce a given volume of hydrogen from the measured current and the time.


large wavelength range and list of x/y numerical data points of elements I'm trying to get a set of FTIR or (FTIR like range) x/y numerical data points for a couple of elements,  namely Silver, Platinum, Palladium, Gold and Sodium.  I've looked on line and found images only of the FTIR but I need the numerical x/y data points with the largest wavelength range.  I tried NIST but they just show the NMR data.  Is there a site where I can get just these or can someone send these/post these for me?  And is it possible to get an FTIR of Sodium just by itself or does it need to be combined with another element.

Thanks

Please note I know metals (due to covalent bonds) don't work well with FTIR but I'm trying to get / see if it's possible to get a large wavelength range analysis of the items above like in FTIR.  Sorry for using the FTIR term I don't know the proper term to use at the moment.

I don't know about the IR spectra of metals. Certainly metals will have vibrational spectra, but I'm sure where to find it (if NIST does not have).

I can tell you how to get x/y data from NIST, if they have the spectrum.

For example, for benzoic acid. At the bottom of the page, there is a link to download the sepctrum in JCAMP-DX format. The .JDX file format is really a space-delimited text file containing x/y data with some markup at the front for the JCAMP plugin.

See all that lovely x/y data? Force the .JDX file open in your favorite text editor, like notepad. Do not use a word processor - they will add additional markup. Do not use a spreadsheet program yet. Spreadsheets can open .JDX files, but they won't always choose the most helpful delineation. Once open in notebook, resave the file as a text (ascii) file. Make sure you type in the .txt to change the extension (alternately, you can just change the extension to .txt using any other means you may feel comfortable with).

Now... open your favorite spreadsheet program, and use it to open your shiny new text file. You may have to tell the program to look for "all files" (MS Excel is bad about this). You should get some dialog/options on how to open the file. Choose "Delimited" if its an option, and when it comes time to choose your delimiters, make sure your choice includes "space":

Now you should have a file with the x/y data. The first column is wavenumber (at least for IR) and the second is transmission ($\frac{I_{\text{out}}}{I_{\text{in}}}$). Some of these files have more than one column of y-data. When I graphed each column from the benzoic acid file, I got the same spectrum (almost).


Explain how to filter Ca and Ba ions separately out of solution using insoluble compounds I had a question on a test that said to use insoluble compounds to make precipitates to filter out Ca and Ba from solution separately, but I couldn't find anything that will react with one and not the other??

You could carry out the standard qualitative analysis procedure. Dissolve the mixture in  hot acetic acid solution.

Then add pottasium chromate ,barium is precipitated as its chromate salt(pale yellow ppt).

Filter it and you could then precipitate Calcium from the resultant filtrate by adding ammonium oxalate.(precipitated as calcium oxalate, white ppt)


How to extract caffeine from coffee To extract caffeine from coffee, I tried reacting it with baking soda to make tannins more soluble. I then put in salt to push caffeine out of solution. I saturated it with salt and then mixed the solids with acetone to get the caffeine out. There is also some salt in the acetone. Washes with water should remove the salt and tannin salts leaving behind the caffeine(not pure). Would this procedure work? Is there another procedure that would be more efficient and not require DCM.

The popular procedure to experimentally isolate caffeine is sublimation.

Slowly heat a teaspoon of coffee or tea from 150°C to 250°C. Let the sublimated caffeine precipitate at a watch glass cooled with ice. The resublimated caffeine forms white needles that can be identified under a microscope.

The sublimation temperature of caffeine is 178°C.


Graham's law of diffusion The rates of diffusion of two gases $\ce{A}$ and $\ce{B}$ are in the ratio $1:4$. If the ratio of their masses present in the mixture is $2:3$, the ratio of their mole fraction is?

My tries:

$\frac{r_1}{r_2}=\frac{1}{4}=\sqrt{\frac{M_2}{M_1}}\to\frac{1}{16}=\frac{M_2}{M_1}$, $M_i$ is molar mass of $i$.

Also given $\frac{w_1}{w_2}=\frac{2}{3}$, multiplying this by above one gives $\frac{M_2\cdot w_1}{M_1\cdot w_2}=\frac{2}{3\cdot 16}\to\frac{n_1}{n_2}=\frac{1}{24}$, where $n_i$ represents number of moles of $i$.

But this is not a correct answer please help.

$$\frac{r_A}{r_B} =\frac{1}{4}\implies\sqrt{\frac{M_B}{M_A}}=\frac{1}{4}$$
$$\therefore \frac{M_B}{M_A} = \frac{1}{16}$$
On dividing the given mass by molar mass, we get the respective number of moles.
Let the number of moles be denoted by $n$
$$\therefore {n_A} = \frac{m_{A}}{M_{A}}$$
$$\therefore {n_B} = \frac{m_{B}}{M_{B}}$$
$$\therefore \frac{n_{A}}{n_B} = \frac{m_{A}\cdot M_{B}}{M_{A}\cdot m_{B}} = \frac{m_{A}}{m_{B}} \cdot \frac{M_{B}}{M_{A}} = \frac{1}{24}$$

I do not see what's wrong.
Do you want the mole fractions of A and B?