Strange batch of potassium nitrate with lower than expected melting point I'm part of a rocket building team and I'm working with propellants. We use saltpeter as the oxidizer, bought as fertilizer and purified by recrystallization. Recently we bought it from another source, that identified the chemical as potassium nitrate, as usual, but when we heated it to make the propellant it simply melted at a temperature of $231~^\circ\rm{C}$, $100$ degrees lower than expected.

It works as an oxidizer, as making the propellant with this chemical gives good results, so it must be something similar to $\ce{KNO_3}$. Maybe there is something working as flux? It tastes like saltpeter (I eat a lot of materials we have in the workshop) and has that characteristic high temperature-dependent solubility.

Do you have any idea of what could it be, and what can I do to find out? Have in mind that our workshop doesn't have much in terms of lab equipment, only thermometers, various things to heat stuff and some solvents for a variety of applications. Also, note that I'm a telecommunications engineering student, so I don't have too much of deep knowledge of chemistry.

It might contain $\ce{NaNO3}$ as well as $\ce{KNO3}$. This paper by Foong et. al. cites some thermal measurements, such as melting point and heat capacity, that you could make with a thermometer and heater. Note that there are inflections in the heat capacity curve as the different components ($\ce{NaNO3}$, $\ce{KNO3}$ and the eutectic of the two) melt.

If you have access to a spectroscope, or even a hand-held diffraction grating, you could compare the spectrum with that from samples of known substances, such as $\ce{NaNO3}$, $\ce{KNO3}$, $\ce{NH4NO3}$ and $\ce{Ca(NO3)2}$.


Change in pressure or volume of molten salt I am doing a project in which I fill a cylinder with a volume of $636cm^2$ (made of stainless steel) with molten salt: 60% Sodium Nitrate ($NaNO_3$) and 40% Potassium Nitrate($KNO_3$).

The container might reach (worst case scenario) -35 degrees Celsius in a cold winter night and in the summer (with ideal conditions) up to 700 degrees Celsius.

I'm trying to find out if I should add a relief valve (like in old water boilers) or just seal it tight. This of course depends o how much the pressure increases between those 2 states.

I know that molten salt  melts at 221 degrees Celcius and starts  freezing when it reaches 238 degrees Celcius.

I can't find any relevant data, so I don't know how I could find the result for this. Any ideas or suggestions?

EDIT: My worry is that it would blow up due to pressure or the lid would blow open...

Venting initiates from blockage - anywhere.  All your lines must be hot-traced.

Nitrate melts are powerful oxidizing agents.  If containment fails onto organics (e.g., wood) you've got a catastrophe.  Nitrate melts significantly shrink on cooling and solidifying, and they bond to polar surfaces (glass, metal).  If you have  a half-liter of solidified salt and heat the bottom of containment, the melt will expand against the plug of solid salt above it, possibly deforming or rupturing containment.  The very hot and oxidizing liquid shoots out.  Heat a stripe up the side to start.

Nitrate melts thermally decompose.  Even if the average temp is OK, heating hot spots may give you oxidizing, corrosive, and toxic by inhalation nitrogen oxides.

http://link.springer.com/article/10.1361%2F10599490417542
"Corrosion of stainless steels and carbon steel by molten mixtures of commercial nitrate salts"
http://prod.sandia.gov/techlib/access-control.cgi/2001/018518.pdf
http://www.electrochem.org/dl/ma/197/pdfs/0209.pdf

The articles say your hot end is out of the safety zone.  Welds.  Chloride.  Chloride corrosion, especially pitting, can be aggressive. Halide in general may destabilize nitrate.  Ammonium nitrate is rather tame.  Ammonium nitrate plus chloride contamination is Texas City, Texas 16 April 1947. 2300 tons.

http://www.energystoragenews.com/Molten%20Nitrate%20Salt%20for%20Solar%20Energy%20Storage.html
1600 tonnes of molten nitrate eutectic.

You might put it behind grout-filled cement block or cinder block with #5 rebar weaving vertical and horizontal, and into the foundation, when you build the wall.  Have a large area vent direction.


Is it more precise to measure a substance as a solid or as a liquid? If I have a substance and want to obtain an exact quantity of it, is it better to measure it as a solid or as a liquid?

In other words, if I have, for example, NaCl crystals I can weigh the crystals in an analytical balance, or alternatively I can dissolve them in water and then determine the concentration by, for example, titrating with silver nitrate.

Which method would be more exact?

Weighing is always far more accurate than any volume measurement. The extreme example is that of atomic weight (mass) measurements, which required accuracy to 4-5 decimal places. All the work was always done on the basis of gravimetry or by weighing the substances on high quality balances.


Why does tin form tin (II) compounds? The stability of the +II oxidation state in lead has been widely attributed to the so-called “inert pair effect”: the 6s subshell is stabilized through relativistic contraction due to the near-c speed of its electrons. Now, Pekka Pyykkö states in his 1988 article (p. 585) that

  In addition to the 6s2 inert pair discussed above, analogous compounds (such as GeO and SnO) are known for rows 4 and 5. While the relativistic 5s stabilization is not entirely negligible, the 4s stabilization due to the “d-block contraction” probably overweighs relativity on row 4.

How does the d-block contraction cause a change in valency? I know that due to poor shielding by d electrons, atomic radii across the fourth period don’t increase particularly. But how is it responsible for the increasing stability of the +II oxidation state in group 14 (before getting to lead, where relativistic effects play a significant role)?

Would I be wrong if I said that the cause for the increasing stability of the s electrons in germanium and tin was the existence of a full d subshell (absent in, say, silicon), which, however slightly, shields the nuclear charge for the p electrons better than for the s electrons? Or what is it that happens, exactly?

Finally, according to the orbital potential energies listed on page 16 of Appendix B of Miessler and Tarr's Inorganic Chemistry (5th ed.), the 5s electrons of tin aren’t more stable than those of silicon – their energy (–14.56 eV) is higher than the energy of the 3s electrons in the Si atom (–14.89 eV), even if only by a small amount. Compare this to germanium and lead, whose valence s orbitals are both more stable than the silicon’s 3s (–16.05 eV and –15.12 eV, respectively). How is it, then, that tin forms divalent compounds while silicon doesn’t? Is Pyykkö correct in associating the Sn valency with 5s stabilization?

I feel very ashamed to answer my own question, but having found a possible answer, I don’t see why I shouldn’t share it with the community.

I would like to start by bringing in this table of orbital energies (from the appendix I linked to in the question) as a reference point for further explanations:

$$\begin{array}{ccc}
\hline
\text{Element} & \text{Energy of outer s orbital / eV} & \text{Energy of outer p orbital / eV} \\
\hline
\text{Carbon} & -19.43 & -10.66 \\
\text{Silicon} & -14.89  & -7.78 \\
\text{Germanium} & \mathbf{-16.05} & -7.54 \\
\text{Tin} & -14.56 & -7.01 \\
\text{Lead} & \mathbf{-15.12} & -6.81 \\
\hline
\end{array}$$

As you can see, the energies of the p orbitals get higher and higher as you go down the group, while the energies of the s orbitals don't.

The reason why, in general, orbitals rise in energy when going down a group is that even though the nuclear charge is bigger (which has a stronger effect than the rise in the quantum number $n$), the increased screening effect of added electrons prevails over this rise.

Germanium has such a low 4s electron energy because of the so-called d-block contraction. Since it has 10 added protons due to the existence of the d-block, and the d electrons don’t shield this increased nuclear charge well, the s electrons are stabilised. The p electrons don’t feel this effect as much because they are shielded more than the s electrons (less penetration), thus experiencing the rise in the nuclear charge less, while being shielded additionally by the stabilised s electrons (and presumably by the d electrons as well, to a lesser extent).

The lowered energy of the 6s orbital of lead is the result of the equivalent f-block contraction, amplified by the relativistic contraction of the s shell. The p electrons don’t have a noticeable change for the same reason they didn’t in germanium.

Now, tin has d electrons too, so it will experience the d-block contraction – but to a far more limited extent, since the change in nuclear charge is not as big in percentage. To illustrate this concept, it is useful to bring in this formula:

$$E \approx -13.6 \left(\frac{Z}{n}\right)^2 \mathrm{~eV}$$

where $E$ is the energy of the electron (in the hydrogen-like atom!), $Z$ the nuclear charge and $n$ the principal quantum number. Since it refers to hydrogen-like atoms, this formula obviously doesn’t bring in the screening effect. To see why the d-block contraction doesn’t have such a strong influence on the tin atom, we can use this formula to calculate how the non-screened, non-relativistic energy grows from Si to Sn:

$$\begin{array}{cc}
\hline
\text{Element} & \text{Non-screened energy of outer orbital / eV} \\
\hline
\ce{Si} & -296.178 \\
\ce{Ge} & -870.4 \\
\ce{Sn} & -1360 \\
\hline
\end{array}$$

As you can see, the percentual change from Si to Ge is much higher than the change from Ge to Sn – thus, the increase in nuclear charge from Ge to Sn doesn’t have an enormous effect on the electron energy, while the screening effect between Ge and Sn grows steadily: the orbital energy is lowered in the many-electron atom. Hence, the Sn s shell is stabilised, but to a lesser extent.

Now, why does Sn form divalent compounds so easily, when Ge and Si have lower s energies and don’t? There’s a passage in Greenwood’s Chemistry of the Elements (2nd ed), pp 226-7 that explains it:

  The term “inert-pair effect is somewhat misleading since it implies that the energy required to involve $n\mathrm{s^2}$ electrons in bonding increases in the sequence $\ce{Al} < \ce{Ga} < \ce{In} < \ce{Tl}$. […] This is not so […]. The explanation lies rather in the decrease in bond energy with increase in size from Al to Tl so that the energy required to involve the s electrons in bonding is not compensated by the energy released in forming the 2 additional bonds.

Indeed, that is what happens: the $\ce{Sn-X}$ bonds are longer than the $\ce{Si-X}$ bonds, and thus have smaller energy, as permeakra pointed out in his/her comment. However, I believe this is not due to the interference of the d-subshell, but simply to the larger Sn nuclear charge. This increased nuclear charge amplifies the repulsion to the nuclei of the other species (say, oxygen), thus lengthening the bond. Since the s electrons have a very low energy compared to the p ones, they do not participate in this bond – they cannot overlap.

In short, tin forms divalent compounds because:

Its s subshell is more stable than the p subshell, because of screening differences;
Its bonds are longer, due to the increased nuclear charge: thus, the s electrons don’t bond.

The combined effect is smaller in germanium compounds (Ge is more tetravalent), because even though the s shell is more stabilised, the bonds are shorter; it is stronger in lead compounds (Pb is more divalent) because the 6s shell is more stabilised than the Sn 5s shell and the bonds are even longer.


How to explain (non-/anti-) aromaticity in fulvene with the help of resonance structures? Can someone please explain why the resonance structures of fulvene 1 is non-aromatic and 2 is anti-aromatic?

Why is fulvene non-aromatic, even though it has $4\pi$-electrons and no $\mathrm{sp^3}$ carbons?

TL;DR: You cannot assign aromaticity based on a couple of resonance structures. Penta-fulvene has negligible (anti)aromatic character, which is supported by computational and experimental investigations.

Introduction

Aromaticity is a complex and still not fully understood phenomenon. Active investigations are experimentally and computationally challenging. Unfortunately in schools and university it is often taught as something quite simple to fathom, which can be explained by looking at Lewis structures and counting electrons. This maybe true for many common compounds, but when you dig deeper you'll very soon find the limitations. (See the notes below.) It is certainly not helpful in the case of fulvenes.

Penta-fulvene's resonance and (anti)aromaticity

The resonance structures you have drawn are correct, but the set is missing one member, coincidentally the more important one. (Please see the notes below on resonance.) There are more, but those are with more charge separation and likely have only little contribution.

In general you cannot judge one resonance structure on its own. In this case it is not helpful at all. In all resonance structures, the π-system is fully conjugated, and delocalised over the whole molecule.
Penta-fulvene has C2v symmetry, and we see deviations in the single and double bond lengths. The values are from a quite extensive study on substituted fulvenes: K. Najafian, P. von Rague Schleyer, and T. T. Tidwell, Org. Biomol. Chem. 2003, 1, 3410-3417 (DOI: 10.1039/B304718K). Unfortunately, they use the unsubstituted fulvenes as comparison. From the abstract:

  The fulvenes (1a–4a) have modest aromatic or antiaromatic character, and are used as standards for comparison.

Another study basically comes to the same conclusion, see E. Kleinpeter, and A. Fettke, Tetrahedron Lett. 2008, 49 (17), 2776-2781 (DOI: 10.1016/j.tetlet.2008.02.137). Citing quite liberally from various parts and omitting any literature references:

  Fulvenes 1–4 have previously been synthesized (triafulvene 1, pentafulvene 2, heptafulvene 3 and nonafulvene 4), and were studied with respect to their dipole moments and NMR spectra. The 1H and 13C NMR spectra of triafulvene 1 (both protons and carbon atoms of the 3-membered ring moiety display resonances in the region of aromatic compounds) evidence a significant contribution of the resonance form 1b [aromatic charge separation]; the corresponding NMR spectra of 2–4, however, display typical olefinic compounds with strongly alternating bond lengths and only a small extent of charge separation (corroborated by the relatively small dipole moments).

  [...]

  Dependent on the criterion employed, 1–4 were reported as partially aromatic, non- or even antiaromatic.

  [...]

  [...] However, the expected partial aromaticity of the 3-membered ring moiety of 1 was not observed (vide supra).
  Similar conclusions can be drawn for the presence of partial aromaticity in 2: even if the occupation of πC=C of the exocyclic double bond is lowest in the series (which can be realized with the participation of 2a, corroborated by the correct direction of the dipole moment), both ICSSs [iso-chemical-shielding surfaces] at ±0.1 ppm [2: ICSS = −0.1 ppm (5.0); ICSS = +0.1 ppm (6.2)] are far away from reference benzene 7 [7: ICSS = −0.1 ppm (7.2); ICSS = +0.1 ppm (8.9)] or even from cyclopropenylium cation 6 [6: ICSS = −0.1 ppm (5.9); ICSS = +0.1 ppm (7.2)]—pointing to 2π electron aromaticity. Again, if there is partial 6π electron aromaticity in 2, due to the contribution of 2a, then it is only very small.

  [...]

  Compared with the corresponding fulvalenes, studied previously, which are genuine push–pull olefins and exhibit partial (anti)aromaticity in the corresponding 3-, 5- and 7-membered ring moieties (in the latter if structurally planar), the 3-, 5- and 7-membered ring moieties in fulvenes 1–4 reveal very small, if not negligible (anti)aromaticity only.

From all of the above I hope I was able to make clear how complex the concept of aromaticity is. Only because of thoughtful investigation and interplay between experiment and theory, penta-fulvene can be described as having negligible (anto)aromatic character.

Notes on aromaticity

The original definition of aromatic (gold book) only states is very broad and may include any and no compounds:

  In the traditional sense, 'having a chemistry typified by benzene'.
  A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1HNMR spectrum.
  See also: Hückel (4n + 2) rule, Möbius aromaticity
  The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

Much more rigorous is Hückel's (4n + 2) rule and therefore includes much less compounds. The main problem here is, that its application is often taught carelessly or even wrong. When one considers whether a compound is aromatic or not, it is probably one of the worst rules to follow. For fulvenes it certainly leads to the wrong conclusions.
The main problem is that this rule often gets reduced to counting π-electrons, but that is only a small part of it. Even if we include more recent developments and extensions of the rule, there is a lot more to it. (Originally only valid for a couple of hydrocarbons from which it was derived.) I like to encourage you to read up on the full definition (and links within) in the gold book:

  Monocyclic planar (or almost planar) systems of trigonally (or sometimes digonally) hybridized atoms that contain (4n + 2) π-electrons (where n is a non-negative integer) will exhibit aromatic character. The rule is generally limited to n = 0–5. This rule is derived from the Hückel MO calculation on planar monocyclic conjugated hydrocarbons (CH)m where m is an integer equal to or greater than 3 according to which (4n + 2) π-electrons are contained in a closed-shell system. [...]

There is a more updated version on aromaticity in the gold book, which allows a more rigorous approach to the whole subject. Unfortunately it is not as simple as what was there before. You will need to understand a lot more about quantum chemistry, especially how to construct molecular orbitals. While Hückel MO calculations (which you could probably still do with a pencil and a [few] paper[s]) still provide a good entry point and approximation, it is more convenient using modern electronic structure programs and density functional theory (or similar) to elucidate aromaticity.
For the sake of completeness, here is the newer definition:

  The concept of spatial and electronic structure of cyclic molecular systems displaying the effects of cyclic electron delocalization which provide for their enhanced thermodynamic stability (relative to acyclic structural analogues) and tendency to retain the structural type in the course of chemical transformations. A quantitative assessment of the degree of aromaticity is given by the value of the resonance energy. It may also be evaluated by the energies of relevant isodesmic and homodesmotic reactions. Along with energetic criteria of aromaticity, important and complementary are also a structural criterion (the lesser the alternation of bond lengths in the rings, the greater is the aromaticity of the molecule) and a magnetic criterion (existence of the diamagnetic ring current induced in a conjugated cyclic molecule by an external magnetic field and manifested by an exaltation and anisotropy of magnetic susceptibility). Although originally introduced for characterization of peculiar properties of cyclic conjugated hydrocarbons and their ions, the concept of aromaticity has been extended to their homoderivatives (see homoaromaticity), conjugated heterocyclic compounds (heteroaromaticity), saturated cyclic compounds (σ-aromaticity) as well as to three-dimensional organic and organometallic compounds (three-dimensional aromaticity). A common feature of the electronic structure inherent in all aromatic molecules is the close nature of their valence electron shells, i.e., double electron occupation of all bonding MOs with all antibonding and delocalized nonbonding MOs unfilled. The notion of aromaticity is applied also to transition states.

Notes on resonance

I will not go into a lot of detail here, because bon did an excellent job explaining it in: What is resonance, and are resonance structures real? However, allow me to make one point very clear: You cannot treat resonance structures on their own. You always have to treat them as a set, a superposition. There is no such thing as a most stable resonance structure, as well as there is no such thing as one of these structures dictating the reactivity. From a pencil and paper approach you hardly ever can judge which structure is most important for the description of the total bonding. Also, from a simple Lewis-type drawing you can almost never judge the properties of the compound.


Is real water made up of heavy water and light water? I need to know, is real water made up of heavy water which is formed with deuterium, and light water which is water by protium?
We know that hydrogen has three major isotopes, including protium and deuterium.
We also know that water, $\ce{H2O}$, is made up by hydrogen and oxygen.
All this might mean that real water itself can naturally contain some heavy water - Deuterium oxide. Is this correct?

Water has formula H2O. Oxygen has 3 stable isotopes (99.76% 16O, 0.039% 17O, 0.201% 18O), and hydrogen has two (99.985% 1H, 0.015% 2H). Thus, there are 9 natural isotopic configurations for water: 3 possibilities for oxygen, multiplied by 3 possibilities for 2 hydrogens with 2 possible isotopes.

Out of those 9 possible configurations, only 4 have a natural abundance larger than 1 ppm:

So, pure water is a mixture of 9 different isotopic configurations. You will note that D2O, which is what is usually called heavy water, is not actually among the 4 more common isotopic configurations given by natural abundance.

Also of note: water that is not “heavy water” is sometimes called “light water”, but never “soft water”. “Soft water” is a term used for a chemical process of natural (non pure) water that has been depleted from certain metal cations. (It is used in contrast to “hard water”.)


Why does the copper anode dissolve? In such a setup, the copper anode is known to dissolve. My question is, why? Does it not receive electrons from the $ \ce{OH^-}$ ions. And if you say that those electrons just flow to the battery, let me provide the counterargument: the electrons flowing to the cathode do not prevent the copper cathode from dissolving because copper ions in solution accept the ions, so there is actually nothing stopping the cathode from dissolving as well. So, very simply, my question is this:  Why does the copper anode dissolve?

The anode is where oxidations happen (the cathode is where reduction takes place).

In the vicinity of your anode, you have the following compounds: $\ce{H2O}$, $\ce{Cu}$, $\ce{Cu^2+}$, $\ce{SO4^2-}$. If you check oxidation states, you will realise that oxygen is always $\mathrm{-II}$, hydrogen is $\mathrm{+I}$, sulphur is $\mathrm{+VI}$ and copper is present in both $\mathrm{\pm 0}$ and $\mathrm{+II}$. Thus, the only species available for oxidation are copper($0$) and oxygen($\mathrm{-II}$).

Checking standard potentials:
\begin{align}
\ce{O2 + 4 H+ + 4 e- &<=> 2 H2O} \qquad &E^\circ = \pu{+1.23 V}\tag{1}\\
\ce{Cu^2+ + 2 e- &<=> Cu} \qquad &E^\circ = \pu{+0.35 V}\tag{2}
\end{align}

It should be clear why copper is oxidised.

On the reduction side, hydrogen, copper and sulphate all could accept electrons. Copper is again the happiest of the three to accept electrons.


Endothermic and exothermic reactions Today, in class, I learned about the two reactions.
I know connecting bond releases energy, and breaking it requires and aborb/ consumes energy.
But on my reference table I, 2C+H2 yielding C2H2 realeasing 227.4 kJ of energy. My point is why it releases energy? Also, where the activation energy comes from?

The paradox exists in that lower energy level means more stability, but nature also likes to form bonds thus to increase stability. However,  sometimes forming bonds increases the energy level.

Your idea is in principle correct. The reaction you mentioned works well, you break one H-H bond and create 1 C-C bond (triple bond, in fact) and two C-H bonds. The energy is released. And in most crude approximation it also says where the activation energy barrier comes from - you have to break a H-H bond.

As for the paradox - that is the principle of chemistry, to make bonds we wish for. If it means increasing the energy level, thus being less favored by nature, you have to pay for it. By sacrificing other bonds. Those are the highly reactive chemicals humbly stated above reaction arrow, which quite often do all the magic.

As the most revealing example in this I consider the Nucleophilic Acyl Substitution reaction, where you have a series of derivatives of increased stability. You can always and simply make the more stable out of the less stable, having wide choice of reactions. And there are just a few to climb up, using very nasty chemicals.


When a piston cylinder system that evolves isothermally and quasistatically is irreversible? Suppose we have two states: $\ce{A}$ and $\ce{B}$.
Considering a quasi-static isothermal process of an ideal gas, (and another condition, that is $T = T_\mathrm{surr}$) that join those states.

Then the Poisson's relation $P_\ce{A}V_\ce{A} = P_\ce{B}V_\ce{B}$ is fulfilled.
that means after integrating

$$W = nRT\ln\left(\frac{V_\ce{B}}{V_\ce{A}}\right)$$

$$\mathrm{d}S_\mathrm{s} = -\mathrm{d}S_\mathrm{surr}$$

because $T = T_\mathrm{surr}$. Then

$$\mathrm{d}S_\mathrm{u} = 0.$$

So my guess is that no, that irreversibilities on those conditions come only on finite temperature difference with surroundings.

But then I remember friction. If piston with friction has $x$ joules lost by friction, how does that change everything?

I need someone to "shed light" on this issue. There is always a lot of confusion when we consider a quasi-static but irreversible process.

If it is done quasi statically with friction involved and, if the system is in contact with a constant temperature bath at the initial system temperature T, then, for the same volume change, the change in entropy of the system will be the same.  That means that all the entropy generated as a result of frictional irreversibility will be transferred to the surroundings (the constant temperature bath).  This will be equal to the frictional work divided by the temperature T.  So the entropy of the universe will increase by this amount.


Benefits of a Buchner funnel w/ a fritted disc instead of Quantitative Filter Paper Circles I was wondering if someone could enlighten me on some example us cases where one would be better of using a Buchner funnel with a fritted disc (example), as opposed to a Buchner "style" filter (example) that requires a filter paper (example).

A few times in the past when I was looking to filter a specific substance and asked about fritted discs, most of the replies were "don't use a fritted disc, just use a funnel with the filter papers instead," and this got me wondering when using a fritted disc would be preferable over a filter paper. Originally, in my chemistry naivety, I mentioned that I may end placing a filter paper inside a Buchner funnel with a fritted disc, to "get the best of both worlds" I suppose, but this idea was shot down pretty quickly.

Heres the pros/cons list for both filter types that I have thus far, hoping some can add to it:

Buchner "style" filter with filter papers

Pros

The filter papers are cheaper
Changing the filter paper pore sizes are much easier, as it's just a matter of getting the filter papers with the desired pore size
One can easily cut the filter papers to the proper diameter if needed

Cons

??

Buchner filters with fritted discs

Pros

Filters are reusable if cleaned properly
???

Cons

Fritted discs are much more expensive than filter papers
Fritted discs don't seem to have specific pore sizes
Fritted discs seem like they can be a pain to clean properly

Thanks in advance for any input!

Buchner + filter paper cons:

The filtering surface is not uniform. It filters only where the holes are, so most of the surface is not working.
Some solid can slip between the paper and the filter.  Then the filtrate must be filtered again.
You have to cut the paper.

Fritted glass filter pros:

Uniform filtering surface.
Pore sizes from number 1 (coarse) to number 4 (fine). In organic chemistry, number 3 is the standard.
Can be cleaned easily when the solubility of the product is known.
Can be sterilized at high temperature.

I use them almost for everything except for filtering activated carbon and very dirty unknown samples. If the solid does not dissolve in solvents, we try with concentrated caustic, concentrated sulfuric acid or strong oxidants; in that order.

It is better to buy the ones without the vacuum connection.
They are cheaper and have less points where they can break.


Interaction between a pencil and rubber (eraser) Is there a interaction between a graphite pencil and and a eraser. For example, if I drew a line on a piece of paper and then erased that, what interaction is occurring?

Is it a Van der Waals interaction with intermolecular forces or is something else?

There was much discussion in the comments to the OP regarding whether erasing a pencil mark is a chemical process or merely a mechanical process, akin to removing an oil spill with high pressure water (my analogy). But I think the answer to "Is it a Van der Waals interaction with intermolecular forces" is clearly yes.

Although the classic "pencil head" eraser generally requires a vigorous rubbing action to rapidly remove a pencil mark, even this can be done in a manner that shows pencil marks that had been bound to the paper via intermolecular forces are transferred to a material having a stronger affinity for the pencil mark than does the paper.

This Wikipedia article discusses several types of erasers that clearly illustrate a chemical process rather than a mechanical one is dominating the removal of the pencil mark. For example, regarding the "poster putty" type of eraser the article states:

  "...poster putty works much the same as traditional kneaded erasers,
  but with a greater tack or lifting strength...Repeatedly touching the
  putty to a drawing pulls ever more medium free..."

A "gum" eraser also clearly works by adsorbing graphite from paper. As the gum is dabbed or lightly rubbed on the pencil mark, the intermolecular forces holding the graphite to the paper are weaker than the intermolecular forces between the graphite and the erasure. The following photo from the above Wikipedia article shows that the graphite in the used eraser has accumulated over time. There is no evidence of paper or other material having accumulated along with the material from the pencil mark.

Furthermore, one can imagine using different materials that should make for good erasers if the pencil marks are simply being removed by the mechanical action of an eraser. For example, if that's the case, shouldn't you be able to remove the pencil mark by rubbing it off with a piece of wood, one's finger, another piece of paper, a fingernail, etc.?


Some substances appear white when crushed while others do not. Why? Calcite crystals are semi-transparent but turn white when crushed. I used to think that it has something to do with the scattering of all wavelengths of light from small particles. But that cannot be the whole story. By that explanation crushed carbon should also appear white (which is not the case).

So my question is why do some substances appear white when crushed while others do not.

It depends if the substance in bulk is transparent, highly reflective (metallic) or highly absorptive.

Carbon in its graphite allotrope is highly absorptive, and even finely divided is still black.

Carbon as diamond is transparent. Crushed diamonds are white -- look at a diamond file or abrasive disk. What happens is that incident light is scattered as it encounters particle after particle, refracted or reflected repeatedly, based on particle size, shape and index of refraction (nD). A higher nD increases the amount of light reflected from particles. For that reason, titanium dioxide, $\ce{TiO2}$, which has a comparatively high index of refraction (~2.5), makes an excellent white pigment.

Metals reflect light because the "cloud" of electrons on the surface return a reflected wave out of phase with the incident wave. However, finely-divided metals have fewer electrons to share, and electrons start behaving as discrete, localized charges. Finely divided silver appears black, for example. This is the basis for traditional silver halide photography. Powdered $\ce{AgCl}$ appears white, but where hit by light, it yields metallic silver nanoparticles, which appear black (as a negative image).

If your interested, look up the effect of surface plasmons on metallic reflectivity, which can yield surprising colors, such as gold in ruby glass.


Is it possible to build or embed the SMILES representation of compounds in 3D? I would like to know if there is a proper way to get the 3D information from the SMILES string of a molecule.

Is there a standard way to embed a SMILES string in 3D space?
Are there other representations of compounds which include their spatial information too?

SMILES is insufficient

SMILES strings do not encode 3D structure information.  They only convey atom type, connectivity and bond types.  InChI is like SMILES in this regard.

Thus, you will need either (a) an algorithm to infer or guess a plausible 3D conformation of a molecule or (b) a file type that has already specified the 3D arrangement of the molecule.

File types for storing, reading, and showing 3D conformations

Probably the most standard way to represent the 3D conformation of a molecules is with a *.mol file.  There are many tools to read such files.  You can read more about the format on Wikipedia.

Estimating a conformation from SMILES

You can also use computational tools to estimate a 3D conformation from a SMILES string. Note I say a conformation rather than the conformation; molecules can in general have many valid conformations.  Also, tools for generating conformations rely on molecular force fields, etc.  These have many implicit assumptions; there is no guarantee that a computationally generated conformation will be the real conformation of a real molecule in the real world.

Here is some code for generating a plausible conformation from a SMILES string using rdkit.

from rdkit import Chem
from rdkit.Chem import AllChem
from rdkit.Chem import Draw
from rdkit.Chem.Draw import IPythonConsole

my_mol = Chem.MolFromSmiles('NC(=N)N1CCC[C@H]1Cc2onc(n2)c3ccc(Nc4nc(cs4)c5ccc(Br)cc5)cc3')

my_mol

my_mol_with_H=Chem.AddHs(my_mol)

AllChem.EmbedMolecule(my_mol_with_H)
AllChem.MMFFOptimizeMolecule(my_mol_with_H)

my_embedded_mol = Chem.RemoveHs(my_mol_with_H)

my_embedded_mol

print(Chem.MolToMolBlock(my_embedded_mol))

The printed result is:

    RDKit          3D

 33 37  0  0  0  0  0  0  0  0999 V2000
   -8.0789   -0.7261   -1.9565 N   0  0  0  0  0  0  0  0  0  0  0  0
   -8.3618   -0.9375   -0.6556 C   0  0  0  0  0  0  0  0  0  0  0  0
   -9.4453   -1.5737   -0.3799 N   0  0  0  0  0  0  0  0  0  0  0  0
   -7.4690   -0.4468    0.2422 N   0  0  0  0  0  0  0  0  0  0  0  0
   -7.8136   -0.1283    1.6244 C   0  0  0  0  0  0  0  0  0  0  0  0
   -6.7632    0.8908    2.0392 C   0  0  0  0  0  0  0  0  0  0  0  0
   -5.5246    0.3855    1.3227 C   0  0  0  0  0  0  0  0  0  0  0  0
   -6.0688   -0.0733   -0.0461 C   0  0  1  0  0  0  0  0  0  0  0  0
   -5.2554   -1.2432   -0.6177 C   0  0  0  0  0  0  0  0  0  0  0  0
   -3.8658   -0.8320   -0.9216 C   0  0  0  0  0  0  0  0  0  0  0  0
   -3.6647   -0.1417   -2.0770 O   0  0  0  0  0  0  0  0  0  0  0  0
   -2.3059    0.1587   -2.1237 N   0  0  0  0  0  0  0  0  0  0  0  0
   -1.8139   -0.3885   -1.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -2.7692   -1.0082   -0.2227 N   0  0  0  0  0  0  0  0  0  0  0  0
   -0.4078   -0.3427   -0.6136 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.0488   -1.0902    0.4772 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.3984   -1.0569    0.8486 C   0  0  0  0  0  0  0  0  0  0  0  0
    2.3307   -0.2688    0.1543 C   0  0  0  0  0  0  0  0  0  0  0  0
    3.6731   -0.3282    0.5615 N   0  0  0  0  0  0  0  0  0  0  0  0
    4.8291   -0.0477   -0.0843 C   0  0  0  0  0  0  0  0  0  0  0  0
    5.9334    0.1757    0.5968 N   0  0  0  0  0  0  0  0  0  0  0  0
    7.0123    0.4129   -0.2413 C   0  0  0  0  0  0  0  0  0  0  0  0
    6.7153    0.3213   -1.5854 C   0  0  0  0  0  0  0  0  0  0  0  0
    5.0623   -0.0682   -1.7942 S   0  0  0  0  0  0  0  0  0  0  0  0
    8.3378    0.7031    0.3040 C   0  0  0  0  0  0  0  0  0  0  0  0
    9.3324    1.3464   -0.4485 C   0  0  0  0  0  0  0  0  0  0  0  0
   10.5913    1.6060    0.1057 C   0  0  0  0  0  0  0  0  0  0  0  0
   10.8633    1.2259    1.4171 C   0  0  0  0  0  0  0  0  0  0  0  0
   12.5638    1.5736    2.1593 Br  0  0  0  0  0  0  0  0  0  0  0  0
    9.8883    0.5951    2.1844 C   0  0  0  0  0  0  0  0  0  0  0  0
    8.6313    0.3380    1.6284 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.8659    0.4742   -0.9343 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.5160    0.4406   -1.3141 C   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  1  0
  2  3  2  0
  2  4  1  0
  4  5  1  0
  5  6  1  0
  6  7  1  0
  7  8  1  0
  8  9  1  1
  9 10  1  0
 10 11  1  0
 11 12  1  0
 12 13  2  0
 13 14  1  0
 13 15  1  0
 15 16  2  0
 16 17  1  0
 17 18  2  0
 18 19  1  0
 19 20  1  0
 20 21  2  0
 21 22  1  0
 22 23  2  0
 23 24  1  0
 22 25  1  0
 25 26  2  0
 26 27  1  0
 27 28  2  0
 28 29  1  0
 28 30  1  0
 30 31  2  0
 18 32  1  0
 32 33  2  0
  8  4  1  0
 14 10  2  0
 33 15  1  0
 24 20  1  0
 31 25  1  0
M  END

A semi-interpretable 2D image of this 3D conformation, also generated by rdkit, is shown below.  For comparsion, the "un-embedded" molecule, optimized to look nice on a 2D display, is also shown.

From the admittedly not-great 2D depiction of the embedded molecule, you can at least tell that the various aromatic rings are not coplanar.  For better visualization of 3D conformations, you would want to use a tool like py3dmol.

RDKit is just one example of a software with these kinds of capabilities.  I used it here because it's the one I know.  OpenBabel is another one, as Martin rightly mentions in the comments.


Equilibrium reaction where the number of moles of reactants and products are the same Le Chatelier's principle says that for an equilibrium reaction where all reactants and products are gases a change in pressure will change the number of moles of the product such to counteract the change in pressure.
My textbook also states that this only affects reactions in which the number of reactants and products are different.
I'm trying to think of a gaseous equilibrium reaction where the number of moles of reactants and products are the same.
All suggestions welcome.

A simple example:
If $T > \pu{100 °C},$ and $p \leq \pu{1 atm},$ then water will be in the gas phase and:
$$\ce{CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)}$$
Note, however, that while this reaction's equilibrium will be highly insensitive to pressure changes (within the regime that all molecules remain in the gas phase), it will not be entirely insensitive to pressure.
That's because, even though there are the same number of moles of gas on the LHS and RHS, these are real gases, and thus their molar volumes will differ slightly. This is a small effect compared to the effect of actually having different numbers of moles of gases on the LHS and RHS.


What is the difference between activity and reactivity when discussing elements? I screwed up on a true/false question that stated that lithium is the most active alkali metal. I answered false, because I remembered that the lower you go in this group, the more explosive the element is.

It seems, however, I was thinking of reactivity and not activity. I know about the activity series, and it's something I wouldn't like to memorize, but I would a deeper explanation on the difference between activity and reactivity. Why do we have these two terms?

Activity is an ordered series, showing which element will replace another element. For example, in a solution of copper sulfate in water, a piece of iron quickly becomes coated with copper as the iron displaces copper from the solution because iron is more active. No explosion took place.

$\ce{Fe(solid iron) + CuSO4 (dissolved in water) -> Cu(solid copper) + FeSO4 (dissolved)}$

Reactivity is less clearly defined. In general, it means how easily some chemical change takes place. For example, chlorine, oxygen and sodium are all very reactive, easily attaching to other elements or replacing elements in compounds. Argon and helium are not reactive, and rarely, if ever, combine with other substances. Nitrogen triiodide, $\ce{NI3}$ and acetylene, $\ce{C2H2}$, are very reactive, but in a way opposite that of sodium or oxygen: rather than attaching to other substances, forming molecules, these compounds want to break up their molecules. In fact, even touching $\ce{NI3}$ will make it decompose violently... Now there's your explosion.


Why is the 2nd ionization energy of chloride similar to that of sulfur? The first ionization energy of chloride is significantly greater than that of sulfur. I suggest that this is very expected as the effective nuclear charge of chloride is higher and additionally, sulfur has even more electron-electron repulsion due to advent of paired electrons in 2p orbitals. However I am not too sure on why the 2nd ionization energies are almost the same.

The second ionization of chlorine leaves $[\text{Ne}]3s^23p^3$ where the $3p$ subshell is half-filled with one electron per orbital and gives a favorable exchange energy.  With sulfur this relatively favorable configuration is achieved instead after the first ionization, the second ionization of sulfur breaks the half-subshell and is thus less favorable.

The effect of the half-filled subshell can be seen elsewhere, as well.  The chart below (reference) shows trends in first ionization energies.  Counting backwards from the third peak (argon) we see that phosphorus exceeds sulfur violating the usual rules based on effective nuclear charge.  The first ionization of sulfur gives the half-filled $3p$ subshell after removing the electron, whereas the first ionization of phosphorus would have to break that configuration.  The half-filled shell effect beats effective nuclear charge here.


Would sodium carbonate from carbon dioxide plus hydroxide be formed in solution or as a solid? If normal breathable atmosphere were to be reacted with aqueous sodium hydroxide, what form would the resulting sodium carbonate take, would it be dissolved in the solution or solid?

Note:

I'm defining air here as the atmosphere of earth at between sea level and 200m.

(Not a homework question, just interested.)

This depends on a lot of things: the volume of solution, the time carbon dioxide has to get dissolved and the temperature are probably the most important.

The actual reaction taking place is happening in solution. Thus, carbon dioxide must first dissolve in the solution (1) before dissolved carbon dioxide can react with the hydroxide base (2). You should view these steps as distinct even though the second is much faster than the first.

$$\begin{align}\ce{CO2 (g) &-> CO2 (aq)}\tag{1}\\
\ce{CO2 (aq) + OH- (aq) & -> HCO3- (aq)}\tag{2}\end{align}$$

Since these happen in solution, the counterion — a spectator ion — is irrelevant and the same would happen with $\ce{KOH, Ca(OH)2}$ and others.

At some point you will reach saturation, because either too much carbon dioxide dissolved or (much more likely) too much water evaporated. Thenceforth, sodium hydrogencarbonate or sodium carbonate (I never performed a formal analysis) will precipiate as a white crust. Incidentally in Munich in the inorganic teaching labs, there were 5 litre reservoirs of $6~\mathrm{M}\ \ce{NaOH}$ solution from which smaller bottles could be filled. If there was a spill (and there always was — it was undergrads in that lab after all) it would show up days later as a white crust. When I myself was a TA, part of my job preparing the labs for the next student generation was to clean those areas. Diluted hydrochloric acid worked wonders.


Solvent effect on gold nanoparticle and silver nanoparticle in UV spectroscopy? Can anyone explain to me that how solvent factor affect the wavelength and intensity of gold nanoparticle and silver nanoparticle in UV spectroscopy?

Attenuation of light related to the properties of material is governed by beer-lambert law Many molecules absorb ultraviolet or visible light. The absorbance of a solution increases as attenuation of the beam increases. Absorbance is directly proportional to the path length, b, and the concentration, c, of the absorbing species. Beer's Law states that
A = ebc, where e is a constant of proportionality, called the absorbtivity.

Different molecules absorb radiation of different wavelengths. An absorption spectrum will show a number of absorption bands corresponding to structural groups within the molecule. For example, the absorption that is observed in the UV region for the carbonyl group in acetone is of the same wavelength as the absorption from the carbonyl group in diethyl ketone.

here is beer law description:http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/beers1.htm


Can MCM-41 (mesoporous materials) be used for ion exchange? Wikipedia, and this journal article talk of a microporous and mesoporous material MCM-41. Although MCM-41 has dynamic sized pores which can be changed, but it is hydrothermally unstable.

Can this microporous material be used for ion-exchange similar to zeolites? What is the property that typically allows a substance to act as an ion-exchange resin?

Ion-exchange resins need to have charged groups that can bind your ions of interest. If a certain ion-exchanger will work for your problem depends on what kind of ions you want you to exchange or separate. The pores must be larger than your molecules of interest. E.g. polystyrene resins which are fine for water desalination (to remove e.g. calcium and magnesium ions) won't work well for separating proteins, as their pores are too small for biomolecules to enter the beads.


What causes that foul taste with combination of toothpaste and orange juice? Ugh, I drank orange juice too soon after brushing my teeth and my mouth feels awful!

What causes that foul taste with this combination of toothpaste and orange juice? Is it a reaction between the oil of wintergreen in the toothpaste and the citric acid in the OJ? Is it something else? What's the explanation for this?

As a second part to this question, is there a way to stop this reaction from happening, aside from not drinking orange juice after brushing?

Why do juice taste awful?

Awful taste is due to sodium laureth sulfate, also known as sodium lauryl ether sulfate (SLES), or sodium lauryl sulfate (SLS)- depending on which toothpaste you use.

SLES and SLS are surfactants(wetting agent). Both chemicals are added in toothpaste to create foam and make the paste easier to spread around your mouth.

Both chemicals they suppress the receptors on our taste buds that perceive sweetness, inhibiting our ability to pick up the sweet notes of food and drink. Also break up the phospholipids on our tongue.  These fatty molecules inhibit our receptors for bitterness and keep bitter tastes from overwhelming us, but when they're broken down by the surfactants in toothpaste, bitter tastes get enhanced.

Thus, anything you eat will be less sweeter and more bitter!

Source: mentalfloss

  How to prevent awful taste after brushing?

After a bit research I found few ways to avoid awful taste:

Avoid brushing. (not recommended)
Wait for about 30 minutes.
Look for SLS-free toothpaste.
eat a little bit of cheese first.(suggested by repurposer)
hot water will do the trick.


Which one is the most stable canonical structure?And why? According to my book the answer is given that structure III is the most stable because all the atoms have complete octets.

But I think it should be II because of,

(a) Structure III has positive charge on a highly electronegative atom and

(b) In structure II, the positive charge can be stabilized by the Indicutive effect of Oxygen as well as resonance.

Can anyone provide a more convincing explanation as to why III is more stable?

You're supposed to recognize that oxygen which has more valence electrons than carbon can take on the positive charge without giving up its octet, thus favoring Structure III.

Now, is it really true that the oxygen atom has more positive charge than the formally positive carbons in the other two structures, when oxygen is also more electronegative versus the octet rule?  I suspect that is beyond the scope of your course material.  You have to accept a higher priority for the octet rule in this situation whether that is real or not.

I hate oversimplified textbook problems :-( .


Purification of the Hünig Base My group uses a lot of Hünig base which needs to be dry and pure. The common procedure we use is to first distill from ninhydrin then from calcium chloride.

Apparently the problem is that Hünig base decomposes to give primary amines (which are removed by the ninhydrin). What I don't quite understand is why we need the second distillation (why can't we just distill from ninhydrin in the presence of a mild drying agent) but also why for triethyl amine we can just distill off calcium hydride (in told this drying agent causes Hünig base to decompose but I can't see how this would work given that the calcium hudride isn't a strong enough base to take a proton from the groups on the amine).

Ref: Purification of Laboratory Chemicals, 7 ed.

In the first case, your labs 'common procedure' isn't the 'common procedure' that most people tend to follow. Generally, it should be carried out as follows

Distill the DIPEA from ninhydrin at atmospheric pressure. The ninhydrin reacts with any primary/secondary amines, removing these impurities from the DIPEA. This ninhydrin step liberates water, as it works by condensing the primary/secondary amines.  Since the boiling point of DIPEA isn't that distinct from that of water, people tend not to try fractionally distilling the two away from one another. The DIPEA that comes across in this first distillation is free of primary/secondary amines, but contaminated with up to ~1% water (which is far too much to get away with using in moisture sensitive chemistry).

Ref: Chemistry LibreText

Re-distil the 'wet' DIPEA from KOH, again at atmospheric pressure, which removes water and provides clean and dry base.

The reason why calcium hydride is often not used is because it is thought to promote the degradation of DIPEA by cleavage of the isopropyl groups (they can eliminate to give stable products. Certainly, distillation from calcium hydride is usually fine, but the DIPEA shouldn't be stored over it for prolonged periods.


Thermal and photochemical excitation of electron in photochemistry One of the basic rules of photochemistry is that the electronic configuration of a conjugated polyene does not change when excited thermally, but changes when photochemically excited, i.e. an electron moves from HOMO to LUMO.

Can anyone please explain this to me in terms of quantum mechanics?

I have done some calculations, which I am enclosing here.

(P.S.: Please do not provide explanation like thermal energy is very less to excite than photochemical, since it would imply that at very high temp (~1,00,00,000 C) the reaction is possible, which is not really the case).

But what you say in your last sentence is true, and not specific to polyenes; it is quite general.  The energy of a visible photon at $500$ nm is $20000~ \pu{cm^{-1}}$ but at room temperature the average thermal energy is only $\approx 210 \pu{~cm^{-1}}$. The probability of a molecule with $20000~ \pu{cm^{-1}}$ of thermal energy is $\approx e^{-100} = 3.7\cdot 10^{-44}$ so small! Thus there in effectively zero chance that by thermal excitation an excited state at 500 nm will be produced. (By excited state I mean an electron is transferred from the HOMO orbital into the LUMO or another, higher, unoccupied orbital).

In your calculation you cannot use $C_V\Delta T$ as its not an operator but in a collision transferring energy there has to be an operator so you cannot just remove it from the integral. The properties of the operator determine the selection rules for transitions.

It is possible by collisions between molecules to move up and down in energy, but because thermal energy is small usually this normally involves vibrational and rotational levels. (The effect is immediately obvious by cooling or heating samples in the appearance of hot bands and in relative rotational line intensities.) If there are very low lying electronic states then I can see no reason why these also could not be excited thermally.


What is the pH of a solution made by mixing 10.00 mL of 0.10 M acetic acid with 10.00 mL of 0.10 M KOH? What is the pH of a solution made by mixing 10.00 mL of 0.10 M acetic acid with 10.00 mL of 0.10 M KOH?
  The $K_a =1.8 × 10^{-5}$ for $\ce{CH_3CO_2H}$. Assume that the volumes of the solutions are additive.

I've done the following:
Write the chemical reaction.

$\ce{CH_3COOH + KOH <=> CH_3COOK + H_2O}$

Then I calculated the number of moles of the reactants.

$\ce n({CH_3COOH})= c \cdot V = 0.001 \text{mol}$

$\ce n({KOH})= c \cdot V = 0.001\text{mol}$

Then I assume the mistake lies here:

I thought because 1 mmol of $\ce{KOH}$ reacts with 1mmol of $\ce{CH_3COOH}$ this will form 1 mmol of $\ce{CH_3COOK}$ which leads to no formation of $\ce{OH^-ions}$. Because there is no excess. Another thing I thought is that $\ce{OH^-}= 1 \text{mmol}$. So I could calculate $\ce{pOH= 3}$ but to no avail because of the answer I later saw that is shown below.

The final outcome is $\ce{pH = 8,72}$.

A hint of where my thought process doesn't make sense would be appreciated? Thanks in advance.

Part A:  The conceptually part: Scientific concepts and conception

1-"I thought because 1 mmol of KOH reacts with 1mmol of CH3COOH this will form 1 mmol of CH3COOK"

yes ,the reaction reached at the equivalent point, the weak acetic acid in mixture has been fully neutralized by the stong($\ce{KOH}$) base (no unreacted acid or base present anymore), the solution will only contain $\ce{H2O}$ and the basic salt ($\ce{ CH3COOK}$),so the concentration of the salt :
 $$[\ce{CH3COOK}] =\frac{\text{moles of salt}}{\text{Total volume}} =\frac{1\text{m}mol}{(10+10)\text{ml}}=0.05\text{M}$$
 2-"which leads to no formation of OH−ions. Because there is no excess."

No, the result, at the equivalence point, will be the same as dissolving $(\ce{CH3COOK})$ in water at the same concentration(0.05${M}$) as you have at the equivalence point.
the salt ($\ce{ CH3COOK}$) is  electrically neutral substance formed by cation $\ce{K+}$ (an acid)and  anion$\ce{ CH3COO−}$ (a base), completely dissociated or ionized in an aqueous solution as :
  $$\ce{CH3COOK   -> K^+_{(aq)}   + CH3COO^−{(aq)}}$$
a)  Since $\ce{K+}$  is the conjugate acid of a strong base, it won't be strong enough to react with water; $\ce{K+}$  actually spectator ion.

b)  Meanwhile, since $\ce{ CH3COO−}$ is the conjugate base of a weak acid, and  therefore strong enough to be able to hydrolyze and accept ions$\ce{ H+}$ from water, so water act as an acid leaving a hydroxide ion $\ce{OH−}$ as :

$$\ce{CH3COO^− +H2O <=>CH3COOH + OH^−}$$

Part B:: Calculating the  approximate pH of $\pu{0.05 mol L-1} \ce{CH3COOK}$ with neglecting water autoionization:

The equilibrium equation of the hydrolysis of the conjugate base $\ce{CH3COO-}$:

$$ K_\mathrm{b}=\frac{K_\mathrm{w}}{K_\mathrm{a}} =\frac{[\ce{OH-}][\ce{CH3COOH}]}{[\ce{CH3COO-}]}$$

Assuming :${[\ce{OH-}]=[\ce{CH3COOH}]}$, ${[\ce{CH3COO-}]=[\ce{CH3COOK}]_0}$

Substiute $K_\mathrm{a}\ and\ [\ce{CH3COO^-}]$ in the equilibrium equation and solve for $[\ce{OH-}]$.

$$[\ce{OH-}] = 5.27\times{10^{-6}} , \mathrm{pOH}=5.278 , \mathrm{pH}=8.72$$


DFT Functional Selection Criteria I have a very very general question:

In DFT functional selection , mostly people speak about the most recent ones. For example my professor always asks: " which DFT Functional did you select ? "  and if I say B3LYP, he says : " No ! that's too old ! " but if I answer: M06 , he says : " hmm ... sounds promising, that's a modern functional ".

I think it is too naive to select functionals just based on their chronologic sequence. I want to ask if there is any good and reliable criteria for functional selection. For example a criteria that says for an alkene with some specific characteristics, go for M06-L and for an alkane with other characteristics, use B97xxx family and so on..

Is there such a criteria ?   I hope this topic can become a good guideline for future reference. !

What's up with all that magic? (A chapter formerly known as Introduction)
The hunt for for the holy grail of density functional theory (DFT) has come a long way.[1] Becke states in the introduction of the cited paper:

Density-functional theory (DFT) is a subtle, seductive, provocative business. Its basic premise, that all the intricate motions and pair correlations in a many-electron system are somehow contained in the total electron density alone, is so compelling it can drive one mad.

I really like this description, it points out why we use and need DFT, and as it also points out the flaws, that every computational chemist has to deal with: How can something with such a simple approach be correct?
Something that is often forgotten about DFT is, that in principle it is correct. It's the implementations and approximations, that make it incorrect, but usable. Becke states this in the following quote:

Let us introduce the acronym DFA at this point for “density-functional approximation.” If you attend DFT meetings, you will know that Mel Levy often needs to remind us that DFT is exact. The failures we report at meetings and in papers are not failures of DFT, but failures of DFAs.

I sometimes here that the abbreviation DFT is often used in the wrong context, since we are not talking about the theory itself any more, but about the implementations and approximations of it. One suggestion I heard was, that it should rather be used as density functional technique.
With that in mind I would like to state, that I absolutely agree with the previous answer by user1420303 and it's subsequent comment by Geoff Hutchison. Since you asked for a somewhat more practical approach, I like to offer the advice I usually give to students new in the field.
Old is bad, isn't it?
Some of the functionals are now around for about thirty years. That does not make them bad, maybe even the opposite. It shows, that they are still applicable today, giving reasonable results. One of my personal favourites is the conjunction of Becke 1988 and Perdew 1986, often abbreviated as BP86.[2] It's a pure functional which is available most modern quantum chemical packages.[3] It performs usually well enough for geometries and reasonable well for energies for simple systems, i.e. small organic molecules and reactions.
The magical functional B3LYP was one of the first hybrid functionals, and it was introduced by Gaussian's very own developers.[4] A lot of people were surprised how well it worked and it quickly became one of the most popular functionals of all time. It combines Becke's three parameter functional B3[5] with Lee, Yang and Parr's correlation functional.[6] But why are we surprised it works? The answer is quite simple, it was not fitted to anything. Frish et. al. just reworked the B3PW91 functional to use LYP instead of PW91. As a result, it heavily suffers or benefits from error compensation. Some even go as far as to say: “It is right for the wrong reasons.”[7-9] Is it a bad choice? No. It might not be the best choice, but as long as you know what you are doing and you know it is not failing for your system, it is a reasonable choice.
One functional is enough, is it?
Now that we established, that old functionals are not out of fashion, we should establish something very, very important: One is never enough.
There are a few things, where it is appropriate to do most of the work with one functional, but in these cases the observations have to be validated with other methods. Often it is best to work your way up Jacob's ladder.[10]
How do I start?
It really depends on your system and what you are looking for. You are trying to elucidate a reaction mechanism? Start with something very simple, to gain structures, many structures. Reaction mechanisms are often about the quantity of the different conformers and later about suitable initial structures for transition states. As this can get complex very fast, it's best to keep it simple. Semi-empirical methods and force fields can often shorten a long voyage. Then use something more robust for a first approach to energy barriers. I rely on BP86 for most of the heavy computing. As a modern alternative, another pure density functional, M06-L is quite a good choice, too.[11] Some of the popular quantum chemistry suites let you use density fitting procedures, which allow you to get even more out of the computer. Just to name a few, without any particular order: Gaussian, MolPro, Turbomole.
After you have developed a decent understanding of the various structures you obtained, you would probably want to take it up a notch. Now it really depends on what equipment you have at hand. How much can you afford? Ideally, more is better. At least you should check your results with a pure, a hybrid, and a meta-hybrid functional. But even that can sometimes be a stretch.[12]
If you are doing bonding analysis, elucidation of the electronic structure, conformation analysis, or you want to know more about the spectrum, you should try to use at least five different functionals, which you later also check versus ab initio approaches. Most of the times you do not have the hassle to deal with hundreds of structures, so you should focus of getting the most accurate result. As a starting point I would still use a pure functional, the worst thing that could happen is probably, that is reduces the times of subsequent optimisations. Work your way up Jacob's ladder, do what you can, take it to the max.[13]
But of course, keep in mind, that some functionals were designed for a specific purpose. You can see that in the Minnesota family of functionals. The basic one is M06-L, as previously stated a pure functional, with the sole purpose of giving fast results. M06 is probably the most robust functional in this family. It was designed for a wide range of applications and is best chosen when dealing with transition metals. M06-2X is designed for main group chemistry. It comes with somewhat built in non-covalent interactions and other features. This functional (like most other though) will fail horribly, if you have multi-reference character in your system. The M06-HF functional incorporates 100% Hartree-Fock exchange and was designed to accurately calculate time dependent DFT properties and spectra. It should be a good choice for charge transfer systems. See the original publication for a more detailed description.[14]
Then we have another popular functional: PBE.[15a] In this initial publication an exchange as well as a correlation functional was proposed, both pure density functionals, often used in conjunction.[15b] I don't know much about it's usefulness, since I prefer another quite robust variation of it: PBE0, which is a hybrid functional.[15c,d] Because of its adiabatic connection formula, it is described by the authors as a non-empirical hybrid functional.[15d]
Over the years there have been various developments, some of the are called improvement, but it often boils down to personal taste and applicability. For example, Handy and Cohen reintroduced the concept of left-right correlation into their OPTX functional and subsequently used it in combination with LYP, P86 and P91. Aparently, they work well and are now often used also as a reference for other density functionals. They went on and developed a functional analogous to B3LYP but outperforming it.[16]
But these were obviously not the only attempts. Xu and Goddard III extended the B3LYP scheme to include long range effects. They claim a good description of dipole moments, polarizabilities and accurate excitation energies.[17]
And with the last part in mind, it is also necessary to address long range corrections. Sometimes a system cannot be described accurately without them, sometimes they make the description worse. To name only one, CAM-B3LYP, which uses the coulomb attenuating method.[18] And there are a couple of more, and a couple of more to come, head on over to a similar question: What do short-range and long-range corrections mean in DFT methods?
As you can see, there is no universal choice, it depends on your budget and on the properties you are interested in. There are a couple of theoretical/ computational chemists on this platform. I like BP86 as a quick shot and answer questions relating to MO theory with it, shameless self-promotion: Rationalizing the Planarity of Formamide or Rationalising the order of reactivity of carbonyl compounds towards nucleophiles. And sometimes we have overachievers like LordStryker, that use a whole bunch of methods to make a point: Dipole moment of cis-2-butene.
So I picked a functional, what else?
You still have to pick a basis set. And even here you have to pick one that fits what you need. Since this answer is already way longer than I intended in the first place (Procrastination, yay!), I will keep it short(er).
There are a couple of universally applicable basis sets. The most famous is probably 6-31G*. This is a nice ancient basis set that is often used for its elegance and simplicity. Explaining how it was built is easier, than for other basis sets. I personally prefer the Ahlrichs basis set def2-SVP, as it comes with a pre-defined auxiliary basis set suitable for density fitting (even in Gaussian).[19]
Worth mentioning is the Dunning basis set family cc-pVDZ, cc-pVTZ, ... . They were specifically designed to be used in correlated molecular calculations. They have been reworked and improved after its initial publication, to fit them to modern computational standards.[20]
The range of suitable basis sets is large, most of them are available through the basis set exchange portal for a variety of QC programs.
Sometimes an effective core potential can be used to reduce computational cost and is worth considering.
*Sigh* What else?
When you are done with that, consider dispersion corrections. The easiest way is to pick a functional that has already implemented this, but this is quite dependent on the program of you choice (although the main ones should have this by now, it's not something brand new). However, the standalone DFT-D3 program by Stefan Grimme's group can be obtained from his website.[21]
Still reading? Read more! (A chapter formerly known as Notes and References)

Axel D. Becke, J. Chem. Phys., 2014, 140, 18A301.
(a) A. D. Becke, Phys. Rev. A, 1988, 38, 3098-3100. (b) John P. Perdew, Phys. Rev. B, 1986, 33, 8822-8824.
Unfortunately this functional is not always implemented in the same way, although the differences are pretty small. It basically boils down as to which VWN variation is used in the local spin density approximation term. Also see S. H. Vosko, L. Wilk, and M. Nusair, Can. J. Phys., 1980, 58 (8), 1200-1211.
P. J. Stephens, F. J. Devlin, C. F. Chabalowski, and M. J. Frisch, J. Phys. Chem., 1994, 98 (45), 11623–11627.
Axel D. Becke, J. Chem. Phys., 1993, 93, 5648.
C. Lee, W. Yang, and R. G. Parr, Phys. Rev. B, 1988, 37, 785–789
Unfortunately the B3LYP functional suffers from the same problems that are mentioned in [3].
The failures of B3LYP are known and often well documented. Here are a few recent papers, but there are many, many more. (a) Holger Kruse, Lars Goerigk, and Stefan Grimme, J. Org. Chem., 2012, 77 (23), 10824–10834. (b) Joachim Paier, Martijn Marsman and Georg Kresse, J. Chem. Phys., 2007, 127, 024103. (c) Igor Ying Zhang, Jianming Wu and Xin Xu, Chem. Commun., 2010, 46, 3057-3070. (pdf via researchgate.net)
Just my two cents, that I am hiding in the footnotes: “Pretty please do not make this your first choice.”
John P. Perdew and Karla Schmidt, AIP Conf. Proc., 2001, 577, 1. (pdf via molphys.org)
Yan Zhao and Donald G. Truhlar, J. Chem. Phys., 2006, 125, 194101.
Note, that not always full recomputations of geometries are necessary for all different functionals you apply. Often single point energies can tell you quite much how good your original model performs. Keep the computations to what you can afford.
Don't use an overkill of methods, if you already have five functionals agreeing with each other and possibly with an MP2 calculation, you are pretty much done. What can the use of another five functionals tell you more?
Y. Zhao, N.E. Schultz, and D.G. Truhlar, Theor. Chem. Account, 2008, 120, 215–241.
(a) John P. Perdew, Kieron Burke, and Matthias Ernzerhof, Phys. Rev. Lett., 1996, 77, 3865. (b) The exchange functional was revised in Matthias Ernzerhof and John P. Perdew, J. Chem. Phys., 1998, 109, 3313. (c) Carlo Adamo and Vincenzo Barone, J. Chem. Phys., 1999, 110, 6158. (d) Kieron Burke, Matthias Ernzerhof, and John P. Perdew, Chem. Phys. Lett., 1997, 265, 115-120.
(a) N. C. Handy and A. J. Cohen, Mol. Phys., 2001, 99, 403-12. (b) A. J. Cohen and N. C. Handy, Mol. Phys., 2001, 99 607-15.
X. Xu and W. A. Goddard III, Proc. Natl. Acad. Sci. USA, 2004, 101, 2673-77.
T. Yanai, D. P. Tew, and N. C. Handy, Chem. Phys. Lett., 2004, 393, 51-57.
(a) Florian Weigend and Reinhart Ahlrichs, Phys. Chem. Chem. Phys., 2005, 7, 3297-3305. (b) Florian Weigend, Phys. Chem. Chem. Phys., 2006, 8, 1057-1065.
The point where the use of more basis functions does not effect the calculation. For the correlation consistent basis sets, see a comment by Ernest R. Davidson, Chem. Phys. Rev., 1996, 260, 514-518 and references therein. Also see Thom H. Dunning Jr, J. Chem. Phys., 1989, 90, 1007 as the original source.
DFT-D3 Website; Stefan Grimme, Jens Antony, Stephan Ehrlich, and Helge Krieg,
J. Chem. Phys., 2010, 132, 154104.
Have fun and good luck!


What makes metals malleable and ductile? Two well-known Physical Properties of metals is that they are malleable and dutile

I was just wondering about what causes metals to be malleable and/or ductile and non-metals to be brittle? What factor determines this?

Thanks!

The metals are malleable because they don't make molecules : they exist as individual atoms. In a metallic piece, the atoms are identical, and they are piled up like layers of beads in the bottom of a flat box, each bead touching its neighbors. The first layer of beads covers the entire surface of the surface. You may imagine a second layer identical to the first one covering it, then a third layer, and so on. Hopefully you see that is not very difficult to push a whole layer, or a whole pile of such layers from the side. A whole layer can slide easily on a lower layer. This would not break the whole piece, which gets distorted but not broken. It would be much more difficult to do it if suddenly one or some of these beads are replaced in a given layer by bigger beads or by non-spherical objects. ${}{}{}{}{}$


Synthesis Golf II: Erythronolide B A full FAQ post has been written on meta.chem.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this if you haven't already.

This round of synthesis golf concerns the Corey synthesis of Erythronolide B (J. Am. Chem. Soc. 1978, 100 (14), 4620–4622).

In order to narrow the scope of the challenge, only a small fragment of the natural product will be considered. Fragment 2 was an intermediate on the Corey route, and provided C10-C13 of Erythronolide B.

Crucially, 2 was needed as a single enantiomer and diastereomer. In the original Corey report, 13 steps were needed to accomplish the synthesis of the fragment, a length which prevented significant amounts of material from being made.

The challenge is to propose a synthesis of 2 which sets up the required 1,2-stereochemistry with absolute configuration. You may start from anything commercially available in the Sigma–Aldrich catalogue, including chiral building blocks.

I have established the stereochemistry via Evans aldol addition:

Then, after protection of the hydroxyl group, the chiral auxiliary can be transformed to a Weinreb amide which will be attacked by Grignard reagent:

The methyl ketone is then transformed to an alkyne which is methylated:
Finally, hydroboration (should be very regioselective) of the alkyne will provide the trans alkenylborane which on treatment with NaOH and iodine will provide target 12:

Length (total steps and longest linear sequence): 7 steps


Determine moles of limiting reagent In the reaction involving 0.1 g of calcium nitrate and 1.5 g of sodium phosphate, what is the number of moles of the limiting reagent?

Here is what I have so far:

Balanced equation:
$$\ce{3Ca(NO3)2 + 2Na3(PO4) -> Ca3(PO4)2 + 6NaNO3}$$

$\ce{Ca(NO3)2}$ has MW = 164, mass = 0.1 g and determined moles using n=mass/MW = 0.000609
$\ce{Na3PO4}$ has MW = 164, mass = 1.5 g and determined moles using n=mass/MW = 0.0091496

I was able to identify limiting reagent by:
g $\ce{Ca(NO3)2}$ --> n $\ce{Ca(NO3)2}$ -->n $\ce{Na3(PO4)}$--> g $\ce{Na3(PO4)}$
    0.1 g --s1-> 0.000609 --s2--> 0.000406 -s3--> 0.06658 g

S1 = n = 0.1/164 = 0.000609
S2 = want/have x $\ce{Ca(NO3)2}$ = 2/3 x 0.000609 = 0.000406
S3 = m = nxMW = 0.000406 x 164 = 0.06658g

Therefore as all $\ce{Ca(NO3)2}$ did not use all the $\ce{Na3(PO4)2}$ that means the $\ce{Ca(NO3)2}$ is the limiting reagent.

Now to answer the question. What is the number of moles of the limiting reagent. I am not sure if it is 0.000609 OR 0.001828 (which is moles multiplied by the stoichiometric coefficient, which is 3).

It's just adding up to the existing answer since I don't have enough rep to comment. 0.000609 is the no. of moles of limiting reagent. You don't multiply it by it's stoichiometric co-efficient because it is the no. of moles you were given. Stoichiometric equations just signify the proportions in which compounds react.


Does the field of organic chemistry cover all group 14 elements or only carbon? Does the field of organic chemistry cover all the other group 14 elements or does it strictly deal with Carbon chemistry?  Since the other group 14 elements have similar chemical properties, such as silicon, wouldn't they fall under the same discipline? Why or why not?

The other group 14 elements don't have similar chemistry. In general, the elements in the 2p block are "unusual" compared to the 3p/4p/5p blocks, just as how the elements in the 3d block are quite different compared to the 4d/5d blocks, and just as how hydrogen ("1s block") is very different from the alkali metals.

The obvious difference in chemistry comes from the unparalleled ability of carbon to catenate, or in other words, to form chains of single bonds. Silicon does not do this to any appreciable extent, mainly due to bond energetics (the Si–Si bond, for example, is much weaker than C–C).

Many of the functional groups familiar to organic chemistry have little parallel in silicon chemistry. Carbonyl groups are ubiquitous, but Si=O compounds are barely stable (see Wikipedia), which is a result of the poor π overlap between Si (valence 3p orbitals) and O (valence 2p). The same applies for C=C and C≡C bonds. Silicon analogues are known, but are very difficult to make and generally need to be stabilised using a variety of tricks (see: How come there aren't any silicon analogs of alkenes or alkynes? and links within).

So, traditionally, the chemistry of Si/Ge/Sn/Pb is within the realm of inorganic chemistry. Of course, it is never a clear-cut distinction, since these elements can coexist with carbon in the same molecule. But that is a story for another time.


Generalizations regarding functional groups on benzene ring While studying several organic mechanisms (high school level), I came to an interesting generalization.

I claim that negative charge on phenoxide ion first ALWAYS delocalises onto the ring and then attacks through carbon atom (like in Reimer Tienman reaction - Wikipedia). This happens because negatively charged carbon atom is less electronegative and hence a better nucleophile than a negatively charged oxygen atom.

But a normal lone pair like that on aniline or phenol attacks by itself without delocalisation (as in Schotten Bauman reaction - Wikipedia). This is because delocalised resonance structures have a formal positive charge on the more electronegative oxygen/nitrogen and are thus less stable.

My questions are:

Does this generalization always work with regards to phenols and anilines?
Is my reasoning correct for claiming these generalizations?
Can this logic be extended to other functional groups on a benzene ring?

PS: Please also justify your answer with explanation. Even a simple counterexample will help.

Generalisation while often is a very useful tool at the zeroth approximation level, – that is crude, very crude, barely resembling the truth, – is a temptingly dangerous game to play. It often boils down complicated concepts beyond the point of usefulness and leads to wrong conclusions.
Generalisations are oversimplifications.
An example for this is the ever so annoying inclusion of d orbitals into hybridisation schemes. This was a result from the post rationalisation of VSEPR theory with molecular orbitals. It has been disproved many times, yet it still is used as a common explanation in textbooks. It is a myth that continues to live, and live, and live...
Another example for generalisation taken too far is resonance. The concept most often employed, yet understood so little about it; especially why it is neccessary in the first place. As a result we hear phrase like "most stable resonance structure", which are complete garbage.

As such, while making generalisations you might not yet realise what damage that may cause later. Some concept simply are complicated and should not be broken down further.
It is quite common to draw reaction mechanisms with Lewis structures and electron pushing arrows. The problem with these is the extremely simplified nature of these drawings. Molecules are (usually) not planar, and yet we ignore this for the purpose of representation and depiction. A picture may sometimes be worth a thousand words. However, another, probably worse, problem with these schemes is the depiction of discrete electron pairs; a contributing factor to the resonance problem mentioned above.
When using these mechanisms it is always implied, that these are proposals, simplified to highlight certain aspects of a reaction.

Let's analyse your statements a little and see where you might derail.

  I claim that negative charge on phenoxide ion first ALWAYS delocalises onto the ring and then attacks through carbon atom (like in Reimer Tienman reaction).

That statement has already one big problem, it's wrong. The charge in the phenoxide ion is never localised. The electronic structure of a molecule will always adapt much more easily and fast than any given change in the molecular geometry. When phenoles are deprotonated, while the hydron leaves, the electronic structure instantaneously adapts.
You are thinking in terms of discrete resonance structures, which simply do not exist. The negative charge is delocalised. That's it. The corresponding observable is the electron density, which is concentrated at certain positions due to the excess electron. In the approximation of molecular orbital theory, this can be partially explained with the HOMO and lower orbitals. In the context of valence bond theory we need to employ resonance, i.e. the superposition of various electronic structures that feature localised electron pairs. Both representations are approximations, but currently our best way to understand complicated mechanisms.

The way how molecules react with each other depends on many things. The product which is formed depends on a fragile system of well balanced energy levels. You need to understand, that molecules react in all possible ways, but only the most stable products will eventually be retained.

In that regard you statement is also wrong, the phenoxide ion is a nucleophile, it will act through every position with increase electron density, most likely through the oxygen though.

  This happens because negatively charged carbon atom is less electronegative and hence a better nucleophile than a negatively charged oxygen atom.

The pure notion of better or worse nucleophiles is highly dependent on the context. Just because formed carbon-carbon bonds tend to be a little more stable than the carbon-oxygen bonds, it is not useful to generalise that. I also highly doubt the mechanism presented on Wikipedia for the Reimer–Tiemann reaction is complete enough. As has been stated before, this is not a clean reaction, hinting at a variety of semi-stable intermediates.

  But a normal lone pair like that on aniline or phenol attacks by itself without delocalisation (as in Schotten Bauman reaction). This is because delocalised resonance structures have a formal positive charge on the more electronegative oxygen/nitrogen and are thus less stable.

I don't quite understand where you are going with that, but the lone pair of nitrogen in aniline certainly has overlap with the aromatic system. In general amines and phenols are quite alright nucleophiles.  There are other factors determining why they react the way they do.
Your statement suffers from the most-stable-resonance-structure-syndrome, i.e. you take resonance structures much too seriously. Please see: What is resonance, and are resonance structures real? If there is resonance, then this effect is always stabilising, because otherwise it would adopt a molecular structure where resonance would not be possible (compare cycloocta-1,3,5,7-tetraene).
Additionally formal charges are exactly what they are called: formal. There is little to no truth to them other than keeping track of all the electrons.

One of the most important things is also to realise that stability always is a relative concept. More often than not it highly depends on the chemical environment, if an intermediate is "stable" and if it would be more "stable" than another intermediate.

In conclusion,
as a high school level student, you will learn a lot of useless or even wrong stuff. Some concepts are still taught even though they are known to be wrong. Some concepts are often not correctly understood by teachers themselves. Even at the university level you will encounter hardliners who teach chemistry like it was still the 20th century. Unfortunately that can't be changed (in the immediate future).
However, critical thinking is the way to gain knowledge. Unfortunately that means letting go of generalisations. Many things are much more complicated that they appear at first sight, only a detailed look will yield a better understanding. And once you have obtained a deep and fundamental knowledge, you won't need these generalisations or oversimplifications.


How to determine whether a compound is gas or liquid or solid? My book says, methyl chloride, methyl bromide, ethyl chloride and some cholorofluromethanes are gases at room temperature, whereas higher members are liquids or solids. Why?
With this question, I got another question for which the former would be subset. How to determine whether a compound is gas or liquid or solid (assume room temperature)? Thinking about this question, I thought molecular force between the molecules is going to be the key for answer of this query door. But how to determine, whether a compound has stronger intermolecular force between the molecules or not?
We have another way to check if a compound is gas or solid or liquid, i.e. entropy (randomness), which is given by heat gained or lost divided by temperature. In general, compound which has greater randomness will be gas. We can expect most of the compounds to be gases, which have standard molar entropy greater than $200\ \mathrm{J K^{-1} mol^{-1}}$ at $298\ \mathrm{K}$.
Entropy concept brings many questions under light, for example, $\ce{H2}$ gas has standard molar entropy of $130.7\ \mathrm{JK^{-1}mol^{-1}}$, which is lesser than $\ce{Br2}$ liquid, which has standard molar entropy of $152.3\ \mathrm{JK^{-1}mol^{-1}}$. We would had expected $\ce{H2}$ gas to have greater randomness than $\ce{Br2}$ liquid, because gases should have greater randomness than liquids. Anyway it becomes difficult, to determine a compound as solid or liquid or gas using entropy.

LINKS

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gases1.html#roomt (This link helps to determine a compound as gas, but leads to still more questions like: Why most non metals are gases? Why all the gases are covalent bonded that contain two or more non metals? Why gases are defined to have low molecular weights, in spite of knowing that molecular interactions matter? Can’t ionic bonded compounds be gases?)
I will try to find list of solid and liquid compounds, but now I have just the above link. If any one have links related to this particular topic, they are welcome.

As the molecular weight increases (more halogens or carbon atoms), so does the electron density and consequently the van-der-waals forces(dispersion forces) between the molecules rises. These forces increase intermolecular attraction and therefore increase the boiling point. T^his is valid in case of most other organic compounds' homologous series, not just alkyl halides.


Planarity of given molecules in all conformations Which of these have all its atoms in one plane in all the possible conformations (if any):
A) buta-1,3-diene
B) but-1-en-3-yne
C) $\ce{CH2=C=O}$
D) $\ce{CH2=C=CH2}$

While I was attempting this question, I got stuck in option B!
I was sure about option A that since it has conformational change possible between C2-C3 bond, atoms will be present in more than one single plane.
Next, I am also convinced that option D can't be the answer as allenes are non planar and they have H-C-H planes perpendicular to one another.
Then, for option C also, I know that since all the atoms are $sp^2$ hybridised, it has to be a planar straight molecule.
But, I am having a doubt in option B; how should we determine if the molecule is planar or not and whether all of its atoms are present in a single plane?

To answer this you need to consider the hybridization and geometry around each atom but also think about all of the single bonds present. Single bonds will (generally) allow rotation along their axes as this can happen to a sigma bond without breaking the orbital overlap. If rotation about a single bond moves other atoms in space then you must consider conformational changes that may take atoms out of the plane.

You are correct with A and D. Option C is a ketene and the central C is sp hybridized, it is very similar to the allene in D. However, the terminal oxgyen has no other atoms bonded to it so it doesn't suffer from atoms out of the plane like D. So, C is completely planar (so far as atoms are concerned!), but not for the reason you stated.

For B, which you have shown, it is planar as drawn. Now consider what would happen if you rotated any of the single bonds. Would you break planarity? If not, it is also completely planar. Interestingly, if you rotate the C2-C3 bond, you rotate the alkyne group compared to the alkene but, as the alkyne group has infinite rotational symmetry along its axis, you can't tell the difference, all atoms remain on the plane!


Why can't ligands with "locked structures" undergo chelation? The following quote is from the book Concise Inorganic Chemistry by J.D. Lee (Adapted by Sudarsan Guha), from the chapter "Coordination Compounds", page 177:

  ...it must be noted that $\ce{NH2NH2}$ and $\ce{N(CH2CH2)3N}$ cannot act as chelating ligands due to the formation of a three membered ring and locked structure respectively.

But, as per the accepted answer to the question Are all complexes with a polydentate ligand examples of chelation?, when a ligand forms more than two bonds to the central atom, the result is a chelated complex. This is also inline with the IUPAC Gold Book's definition of Chelation:

  The formation or presence of bonds (or other attractive interactions) between two or more separate binding sites within the same ligand and a single central atom.

As per my understanding, it doesn't seem to exclude any ligands as mentioned in my textbook.

I can understand that since $\ce{NH2NH2}$ forms a three membered ring, there would be ring strain in the structure and hence the process of chelation is not energetically feasible compared to four membered (or higher) rings. I think this would be the same for all such ligands with donors adjacent to each other, as I have not encountered any such species so far.

However, I'm unable to understand why $\ce{N(CH2CH2)3N}$ (1,4-Diazabicyclo[2.2.2]octane) cannot undergo chelation.

From its molecular structure, I can imagine that the three ethylene groups act as springs to accommodate any ring strain while bonding with the central atom and so there must be no hinderance to undergo chelation. The following molecular structure is what I think might be if $\ce{N(CH2CH2)3N}$ undergoes chelation with cobalt:

I felt given the size of the cage it would be difficult to fit the metal atom inside it and hence thought that it might be off-centred as shown in the image above.

So, it would be helpful if you could clarify whether my reasoning for the inability of $\ce{NH2NH2}$ to chelate is correct or not and why is it not possible for $\ce{N(CH2CH2)3N}$ to undergo chelation? Further, it would be helpful is you could tell why is there a contradiction between the book's statement and IUPAC definition? Is the statement from the book incorrect (because IUPAC can't be incorrect)?

My copy of Concise Inorganic Chemistry, 5th edition (J.D.Lee, Wiley publishers)1 has nothing contrary to say about hydrazine or 1,4-diazabicyclo[2.2. 2]octane (henceforth referred to as DABCO). It is most likely that the author of your adaptation has used his liberty to add these facts in your copy.
This then raises the question: is the author correct? Hydrazine derivatives such as Benzhydrazides and Salicylhydrazides 2 have been proven to show chelation. Pure hydrazine, however, is unlikely to show chelation due to excessive ring strain. Also, no literature has been found which shows hydrazine undergoing chelation. DABCO, on the other hand, forms complexes with Mercury halides3 with a denticity of 1. This ligand also does not show chelation, and fitting a large metal atom even off-center is highly unlikely due to large steric hindrance and the geometry of the lone pair, which is perpendicular to the structure and not favourable for chelation. Most complexes formed by DABCO tend to be linear in nature4 and no rings are formed.
In short, the author is correct in stating that hydrazine and DABCO do not form chelates.
EDIT: You may also want to read more on a class of compounds called Cryptands, specifically Cryptand-222. This compound displays the 'cagelike' ligation you were looking for in DABCO.
References:

Lee, J. D. Concise Inorganic Chemistry. 5th ed, Wiley publishers &
Blackwell Science, 1996.
Issa, R. M., et al. Coordination compounds of hydrazine derivatives
with transition metals. I. Metal Chelates with Benzhydrazide and
Salicylhydrazide. Zeitschrift fur anorganische und allgemeine
Chemie, vol. 354, no. 1–2, Sept. 1967, pp. 90–97.
doi:10.1002/zaac.19673540118.
Shan, Zeng-Mei, et al. Synthesis, Crystal Structures, and
Characterization of Three Mercury(II) Halides Inorganic–Organic
Hybrid Compounds with 1,4-Diazabicyclo[2.2.2]Octane Ligand.
Inorganica Chimica Acta, vol. 366, no. 1, Jan. 2011, pp. 141–46.
doi:10.1016/j.ica.2010.10.023.
Thorp-Greenwood, Flora L., et al. Three-Dimensional Silver-Dabco
Coordination Polymers with Zeolitic or Three-Connected Topology.
Crystal Growth & Design, vol. 14, no. 11, Nov. 2014, pp. 5361–65.
doi:10.1021/cg501231v.


Van der Waals equation at high temperature and high molar volume At high temperatures the Van Der Waals equation has the form
  $$P\approx\frac{RT}{V_\mathrm{M}-b}$$

  What would happen with the equation if in addition to high T, molar volume of the gas ($V_\mathrm{m}$) will get very large too?

The only sense I was able to make out of this was that the pressure would be reduced, but if $V_\mathrm{m}$ stands for the volume occupied by a mole of a substance, than wouldn't a larger $V_\mathrm{M}$ mean less available void space, less volume, and thus a greater pressure?

First, let's look at this from a purely mathematical standpoint. This isn't going to be rigorous at all, but I'm happy to clarify anything if asked.

If we assume we have 1 mole, and we say that $V_\mathrm{m}$ and $T$ become very large, then the question becomes what happens to $P$ (our only remaining variable)? Well as $T$ and $V_\mathrm{m}$ become very large with respect to (w.r.t) our excluded volume parameter, $b$, (also written as $T$, $V_\mathrm{m} >> b$), the -b term in the denominator becomes negligible and we approach the conventional ideal gas law:

$$
P \approx \frac{R T}{V_{\mathrm{m}} - b} \implies P \approx \frac{R T}{V_\mathrm{m}} = \frac{n R T}{V}
$$

If this is not clear, try plugging in some numbers to the equation and calculating what $P$ would be for small $P$ and $T$ (same order of magnitude as $b$) vs. large $P$ and $T$ (100 times larger than $b$). I will let you determine if this leads to greater or less pressure.

Now let's switch gears to a chemistry point of view. What Johannes van der Waals did, is add this little $b$ parameter in the denominator of the ideal gas law. This parameter is a correction to account for the actual volume that each mole of particles can occupy. You can think of it equivalently as the actual volume each particle can occupy, since a mole is just a really big number of particles.

This $b$ correction is taking into account the fact atoms can repel each other at a relatively large distance. If we approximate these atoms to be hard sphere bouncing off of each other, the radii of these hard sphere is what we call the van der Waals radii - the distance at which atoms "bounce" off of each other if they come into contact. So the  parameter $b$ in the denominator is derived purely from the van der Waals radii of the atoms in question. So in the case of large $T$ and $V_\mathrm{m}$, we find that this correction is no longer necessary to make a reasonable approximation, because its presence makes such a small difference.

I recommend looking at the Wikipedia page for more information.


What additives could make a masonry unit water-repellent but vapor-permeable? I'm curious about some of the wondrous materials available today are able to repel liquid water but permit water vapor to pass through. I've heard that housewrap-type plastics accomplish this by being perforated with microscopic holes that are slightly too large for liquid water, but large enough for the vaporous form to pass through. I have no idea how silicate mineral paints do it, though.

Housewraps and paints are thin films, though. I'd like to learn about how such properties could be imparted to a masonry unit like a compressed earth block. Could you cast such a block with a mixture of soil and some additive that would grant this kind of water repellency but allow water vapor to pass though?

The answer turned out to be waterglass. That's the carrier for silicate mineral paints.