Are the triply-bonded carbons in pyridyne (and benzyne) sp-hybridised?

TL;DR It is sp-hybridised, but maybe not in the way we normally think of alkynes as being.

There are a couple of things to say before going into the actual answer.

Firstly, hybridisation is derived from geometry, not from the bonding pattern (in other words, how many σ- and π-bonds there are). In most cases, the bonding pattern dictates the geometry: for example, most saturated carbons are tetrahedral, and a tetrahedral geometry usually signals sp3 hybridisation, so saturated carbons are usually described as being sp3-hybridised. So, effectively, the bonding pattern does dictate hybridisation. This is all fine, but only up to a certain point, because it can break down in exceptional cases such as cyclopropane. Because of the geometrical constraints of a three-membered ring, there is not quite enough room for the carbon to be perfectly tetrahedral, and so it is not correct to label the carbons as being sp3-hybridised in the usual sense.

Secondly, hybridisation is usually more nuanced than "sp", "sp2" or "sp3". It is individual orbitals that are hybridised, not atoms; so for example, we can talk about an sp orbital, which is a linear combination (a sum) of one part s-orbital and one part p-orbital. When we say that "a carbon atom is sp-hybridised", what we really mean is that the bonding of that carbon atom can be described using four valence orbitals: two sp-hybridised orbitals, and two unhybridised p-orbital. Likewise, an "sp2 carbon atom" uses three sp2-hybridised orbitals and one unhybridised p-orbital; and an "sp3 carbon atom" has four sp3 orbitals.

Again, in the vast majority of cases, these three categories are sufficient to accurately classify all carbon atoms in a molecule. However, as you may guess, this breaks down in exceptional cases: cyclopropane is the most well-known example. Here, each carbon has two sp5(ish) orbitals and two sp2(ish) orbitals. It uses the sp5 orbitals to form σ-bonds to the carbons next to it, and the sp2 orbitals to form σ-bonds to the two hydrogens on it. "How is it possible to have five p orbitals?" sp5 does not literally mean one s orbital and five p orbitals; it just means the ratio of s-orbital and p-orbital contribution to this particular hybrid orbital is 1:5.

With that out of the way, here is how benzyne works (PBE0/cc-pVTZ gas phase optimisation + NBO analysis). We're going to look at the orbitals that the triply bonded carbon (C1) uses for bonding to its neighbours (C2 and C6, see structure below):†

One sp orbital (50.02% s, 49.75% p) for forming a σ-bond to C2;
One p orbital (0.39% s, 99.38% p) for forming the π-bond to C2 in the plane of the molecule;
One p orbital (0.00% s, 99.84% p) for forming the π-bond to C2 perpendicular to the plane of the molecule;
One sp orbital (49.64% s, 50.17% p) for forming a σ-bond to C6.

On their own, the numbers don't tell you which direction the p-orbitals are pointing in, but you can use visualisation software to figure this out. From left to right, these are the (σ- or π-, not the individual hybrid) orbitals, in the same order as listed above.

Based on the above, the orbitals can be crudely illustrated as such:

This is, in fact, pretty much what an organic chemist would describe as an "sp-hybridised carbon atom". However, it is not sp-hybridised in the same sense as an ordinary alkyne. In an ordinary alkyne, the σ-bond between the two triply bonded carbons comes from overlap of sp-orbitals. This is also the case here, but there is a distinction. In an ordinary alkyne, the sp-orbitals point directly towards each other. Here, they are not quite aligned in the same way; they instead form what is sometimes called a "bent bond". The same can be said of the C1–C2 π-bond.

This is fundamentally because of the geometry of the ring. Since we have already allocated two p-orbitals to two other bonds, the two sp-orbitals must be derived from the same p-orbital. As a consequence, the two sp-orbitals must point in diametrically opposite directions, 180° away from each other. If C1 were to use an sp-orbital pointing directly towards C2, then the remaining sp-orbital would point rather far away from C6, significantly weakening the corresponding bond with C6. The orbital geometry that you see is a compromise.

The triply bonded carbons in pyridyne, for all intents and purposes, can be described in the same way as those in benzyne. The effect of nitrogen on the orbitals above is tiny; qualitatively they look the same (the colours are flipped for the first and last ones relative to the benzyne orbitals earlier, but this is of no physical significance).

† The numbers don't add up to exactly 100% because the tiny d- and f-orbital contributions are neglected.


What is the definition of porous materials?

I have used the following definition in my presentations:

"Porosity is where a material has accessible voids which are permeable to liquids or gases".

Another definition:

"a  solid with pores, i.e. cavities, channels or interstices, which are deeper than they are wide."

IUPAC also defines different types of porosity based on the diameters of the pores:

Micropores: d < 2 nm
Mesopores: 2 < d < 50 nm
Macropores: d > 50 nm

M. Thommes, K. Kaneko, A. V. Neimark, J. P. Olivier, F. Rodriguez-Reinoso, J. Rouquerol, K. S. W. Sing, Physisorption of gases, with special reference to the evaluation of surface area and pore size distribution (IUPAC Technical Report), Pure Appl. Chem. 2015, 87, 1051.


Reason Behind Decrease In Potential Energy

The potential energy that changes during reactions is electrostatic energy (+/- attraction, +/+ or -/- repulsion) from the charged particles (electrons and protons) within the atoms. The theory is described by Coulomb's Law (involving the potential energy of charged particles):

http://guweb2.gonzaga.edu/faculty/cronk/CHEM101pub/energy.html

Generally, this comes from the attraction of electrons to nuclei and the repulsion between electrons and between nuclei. When atoms/molecules/the "system" becomes more stable, we will see either an increase in the attraction present (electrons are closer or less shielded from nuclei) or a decrease in repulsion (electrons or nuclei are further apart).

http://ch301.cm.utexas.edu/atomic/#bonding/bonding-all.php

More advanced theories will involve the changing of the allowed potential energy levels of the electrons either within atoms (atomic orbitals) or within molecules (molecular orbitals). Whichever theories are used, the net result is that there is more attraction and/or less repulsion in a more stable/lower potential energy state.

Interestingly, by the first law of thermodynamics, if the potential energy decreases, another type of energy must increase (so you don't end up with less energy than you started). It is normally thermal energy that increases and then is transferred as heat. This process ends up being one the most important ones in understanding why some processes happen and some do not.


standard reference for thermodynamic properties?

A reference that I've used quite a lot (as a student) is SI Chemical Data, which I guess is quite common. It should contain the data you ask for, among other things.

If you are looking for an online source, the NIST-JANAF Thermochemical Tables or NIST Chemistry WebBook should be good places to start. I think both should be respectable enough. The Chemistry WebBook might be a bit more comprehensive, but I suspect that much of the data is from the same source anyway, since NIST is behind both services.

Hope you can use one or the other!


How to interpret intervals on a FT-IR spectrogram of soil samples?

You are almost certainly looking at bands corresponding to functional groups in the humic acid fraction of your sample.

The band at 1750 - 1820 cm$^{-1}$ is likely the $\ce{C=O}$ stretching mode in $\ce{COOH}$ groups in humic acid, but this band is also used to differentiate the fulvic fraction from the humic fraction (see quoted text below). The band at 2820 - 3020 cm$^{-1}$ is likely aliphatic $\ce{C-H}$ stretching.

The composition of humic and fulvic acid fractions in soils varies quite a bit, and how the samples were prepared can significantly alter the composition of the fractions as well. Keep that in mind when interpreting your results. Your bands appear to be shifted from those in the reference I chose to cite, but that is to be expected depending on the composition of your soil and the organic matter contained therein (and can also indicate greater or lesser degrees of $\ce{H}$-bonding, resulting in red/blue shifts). The soil organic fractions referenced in the study below are sandy loams from Ibri, Oman, in an agricultural region. Your samples will differ in composition based on geography, geology, climate, and other factors.

Reference: Helal, Murad, and Helal. Characterization of different humic materials by various analytical techniques. Arabian Journal of Chemistry 4(1), 51–54 (2011) (link to free PDF from ScienceDirect):

  The IR spectra of the three humic fractions are shown in Fig. 2. They have a diversity of bands more or less typical to those distinguishing the humic materials (Stevenson, 1994 and Aiken et al., 1985). Major absorption bands are in the regions of 3400–3300 cm$^{-1}$ ($\ce{H}$-bonded $\ce{OH}$ groups), 2940–2900 cm$^{−1}$ (aliphatic $\ce{C–H}$ stretching), 1750–1720 cm$^{−1}$ ($\ce{C=O}$ stretching of $\ce{COOH}$), 1620 cm$^{−1}$ (aromatic $\ce{C=C}$, $\ce{COO−}$, $\ce{H}$-bonded $\ce{C=O}$), 1280–1230 cm$^{1}$−1 ($\ce{C–O}$ stretching and $\ce{OH}$ deformation of $\ce{COOH}$) and 1040 cm$^{−1}$ ($\ce{C–O}$ stretching of polysaccharide or $\ce{Si–O}$ of silicate impurities). The spectra evidently show predominance of $\ce{OH}$, $\ce{COOH}$ and $\ce{COO−}$ groups which are the most characteristic features of soil humic materials. It is clear from the spectra that fulvic acid is characterized by stronger absorption near 1720 cm$^{−1}$ which implies the high carboxylate capacity, and that of humic acid is stronger than that of humin. This is exactly the case in the results of potentiometric titration (Helal, 2007). The spectrum of fulvic acid is also characterized by the absorption at 1400–1390 ($\ce{OH}$ deformation and $\ce{C–O}$ stretching of phenolic $\ce{OH}$). The results of IR spectra indicate that fulvic acid is more aliphatic, and humic acid and humin are more aromatic. It is obvious that the IR results are in good agreement with the other characterization findings.

Figure 2, from the above-referenced article. FA is fulvic acid and HA is humic acid.


Why does burning wood produce methane and other hydrocarbon chains?

Under the proper conditions, the hydrocarbons present in wood are completely oxidized to $\ce{CO2}$ and water upon combustion.  In fact, this happens every day in analytical chemistry laboratories for example: a sample of wood is dried, ground and placed in a combustion chamber of oxygen with a combustion catalyst at ~$\mathrm{950^oC}$. The evolved $\ce{CO2}$ and $\ce{H2O}$ (along with other gases and ash) from the combustion can then be analyzed to quantitatively determine how much carbon and hydrogen were originally in the wood.

Now lets consider a campfire.  Do you dry and grind your firewood, mix it with a catalyst and apply a heat source of $\mathrm{950^oC}$ under an $\ce{O2}$ atmosphere? Of course not, so why do we go through all these rigors to get a quantitative burn?  So that it all gets burned, no partial combustion allowed. And partial combustion is the key to most of the unburned hydrocarbons that you refer to. Parts of the wood are barely hot enough to burn at all, but still can thermally break down the hydrocarbons into shorter, volatile organics which are are carried away by wind or convection from the fire before the complete combustion to $\ce{CO2}$ and water can be completed.

You also mentioned pyrolysis in your question, and the answer given in your link is certainly relevant here, and it represents one of the processes that I'm talking about. For pyrolysis, you need an oxygen-poor environment.  Going back to the campfire scenario, this largely describes the situation down in the smouldering embers on a calm night.  It's very hot, there is plenty fuel, but no process bringing in sufficient oxygen for combustion. Under these conditions, thermal decomposition rather than just partial oxidation plays a major role in producing unburned hydrocarbons.  Think about methane for example; it is certainly not partially oxidized,


They caught fire because partly because they were milled in toluene and partly because they were such fine powders. As you probably know aluminum forms a passivating oxide on the surface. When you mill the powder, you increase the surface area which will oxidize when exposed to air. You milled it in toluene so oxidation did not occur during and after you were milling as long as it was covered in toluene.

$$\require{cancel}\begin{align}\\
\ce{Al (s) + O2 (g) &->  Al2O3 (s)} \qquad \qquad{\mathrm{\Delta H} = \pu{-1675.7 kJ mol-1}}\\
\ce{Al (s, toluene) + O2 (g) &\cancel{\ce{->}} Al2O3 (s)}
\end{align}$$
Once you filtered off the toluene parts and let it sit the toluene started evaporating and the aluminum started oxidizing. Once you turned the fan on you allowed the reaction to happen much faster to the point that it could not cool as it oxidized and the powder ignited. This is similar to smoldering coals, they are not glowing, but hot and still combusting. When you blow air on them the rush of oxygen will make them hot enough to glow again. This is essentially what turning the fan on did to your powder.


Pseudohalides are "fake halides."   Their chemistry bears some resemblance to true halides ($\ce{F-}$, $\ce{Cl-}$, $\ce{Br-}$, $\ce{I-}$).

Examples of pseudohalides include $\ce{CN-}$ and $\ce{N3-}$. These are generally weak Lewis bases that bear a formal negative 1 charge.

In my opinion, it is rather subjective as to what can be labeled as a pseudohalide. In general, it is in the eye of the beholder, and for a certain set of chemical circumstances.

In short, there is no "dividing line" for being a pseudohalide. If we assume that halides undergo certain (characteristic) types of chemistry, and the species X in question also generally does those same types of chemistry with a similar mechanism, then you may be able to label X a "pseudohalide."

(Note that within the true halides themselves, they can have very different characteristics, and so there is a bit of a moving target for what constitutes a pseudohalide.)


Does supersaturated caffeine polymerize?

Google Images "caffeine crystals"  They are beautiful fibrous masses.  The solvent can influence crystal shape by making various faces grow faster or slower.  Benzil slowly crystallized from hot optically resolved pinene is slender needles.  Crystallized from room temp supersaturated mixed xylenes with a needle seed crystal, it is single crystal blocks.


How do i detect the compounds a plant is made of, automatically?

HPLC-MS (high performance liquid chromatography- mass spectrometry), the latter with a suitable database for known small molecule plant components (alkaloids, terpenes, THC, etc.) would do it.  Cost would be three to low four figures/shot.  Saccharides and proteins are different and more expensive problems.

Consider what you are asking.  Mere pokeberry contains astragalin, canthomicrol, caryophyllene, esculentic acid, jaligonic acid, lectins, oleanolic acid, 3-oxo-30-carbomethoxy-23-norolean-12-en-28-oic acid, 6-palmityl-Δ7-stigmasterol-Δ-D-glucoside, 6-palmytityl-α-spinasteryl-6-D-glucoside, phytolaccagenic acid, phytolaccagenin, phytolaccatoxin, phytolaccosides A,B,C,D,E,F,G (esculentoside E), pokeweed mitogen, tannin, triterpene alcohols, α-spinasterol, and α-spinasteryl-β-D-glucoside and a whole bunch more.  One injection is hardly a beginning.  If you do not know how to cook it, it kills you.


Any molecule can absorb UV light. What the question is probably going for is why those three molecules absorb at longer wavelengths than other amino acids. This has to do with the conjugated pi bonds from aromaticity. Looking at a list of amino acids, we can see that most of them don't have conjugated pi ystems.

amino acids (source)

Conjugated pi bonds decrease the HOMO-LUMO energy gap the more they are conjugated. This allows longer wavelength (lower energy) light to excite the molecule by promoting an electron from its HOMO to LUMO.

UV light is in the range of about 10-400 nm. A single bond (sigma bond) would absorb around 130 nm. Ethene's single pi bond is at 165 nm. A benzene ring's conjugated double bonds peak primarily at 180 and 200 nm. Further conjugation can absorb longer wavelengths and, like anthracene, begins edging into visible light, which as a result has a yellow color and more transitions in the UV-VIS spectrum. NIST has UV-VIS spectra for both benzene and anthracene which you can compare.

Benzene

Anthracene

The more conjugated molecule has more transitions and absorbs at longer and longer wavelengths. This principle is why carrots are orange for example, because they contain a highly conjugated molecule, with 11 pi-bonds, that absorbs well into the visible light range. This molecule is beta-carotene:

All that said, the other amino acids do have double bonds and lone pair electrons (carbonyl groups) that can absorb light in the UV range (remember any bond, including single bonds, will absorb UV light if you go low enough). The lone pairs on a carbonyl group can absorb at higher wavelengths than benzene. These involve electronic transitions between a non-bonding electron to the LUMO; the transition energy is so low because a non-bonding electron does not have its energy lowered like a bonding orbital. The gap between a non-bonding orbital and anti-bonding orbital is therefore much smaller than usual.

There was a good post on pi-conjugation elsewhere on the site, here.
You can read more about carbonyl excitations here. For general excitation values, this page was useful.


Does the prefix of an inverted unit also get inverted?

You should read $\pu{ms-1}$ as $(\pu{ms})^{-1}$ and not as $\mathrm{m}(\pu{s-1})$. Your teacher is correct.

  [...] since $\pu{Hz} = \pu{s-1}$ then $\pu{ms-1} = \pu{mHz}$

This is incorrect.

If something happens once per hour, it happens less often than when it happens once per minute. This is because an hour is longer than a minute. In a similar manner, if something happens once per second, it happens less often than when it happens once per millisecond. This is because a second is longer than a millisecond.

So $\pu{ms-1} = \pu{kHz}$, i.e. the frequency of something happening every millisecond is larger than the frequency of something happening every second.


I do believe something is wrong with this spectrum. I find it highly improbable that a compound would exhibit a single strong absorption peak at low wavenumbers, while showing no other absorbance whatsoever. The fact that there is no overtone of your band also shows me that there was some experimental error when recording this spectrum.


Hygroscopicity is measure of how much the molecule likes to bond  molecules of water, by hydrating/covering charged part of the molecule with molecule/s of water, which relates to it having some polar or charged functional groups, like hydroxyl  group for phenols.

Solubility is about how molecule likes to be surrounded by water, so in this case the structure of whole molecule makes a difference.
Since water is a polar solvent it likes polar solutes. If a molecule despite having one or more polar groups has, for example, a huge part which is not polar, like a carbon chain, then it will not be soluble even though it may be hygroscopic.


What is this glassware between flask and condenser to collect liquid?

This glassware is used to separate /remove the light phase from a biphasic (immiscible) liquid mixture.  This can be used primarily to enhance liquid-liquid extraction from the light phase to the denser phase. As a mixed distillate collects,  the denser phase will eventually siphon back through the side-arm to the flask below and the light phase remains in the collector.


What is the origin of 'tenting' or 'roofing' in NMR spectra?

Not just coupled, but strongly coupled; these roofing effects tend to be seen when the coupling constant $J$ is on the order of $\Delta \nu$ (the difference in resonance frequencies between the two spins).

The origin is quantum mechanical in nature. I will try to give a simplified overview of what is happening without going into the explicit details. For a pair of coupled spin-1/2 nuclei, labelled 1 and 2, the Hamiltonian in a magnetic field is

$$\hat{H} = \omega_1\hat{I}_{\!1z} + \omega_2\hat{I}_{\!2z} + 2\pi J(\vec{I}_{\!1}\cdot\vec{I}_{\!2})$$

If the spins are not coupled at all, then

$$\hat{H} = \omega_1\hat{I}_{\!1z} + \omega_2\hat{I}_{\!2z}$$

and the eigenstates of the Hamiltonian are exactly

$$\begin{array}{ccc}
\text{Eigenstate} & \text{Value of }m_1 & \text{Value of }m_2 \\ \hline
|\alpha\alpha\rangle & +1/2 & +1/2 \\
|\alpha\beta\rangle & +1/2 & -1/2 \\
|\beta\alpha\rangle & -1/2 & +1/2 \\
|\beta\beta\rangle & -1/2 & -1/2
\end{array}$$

NMR theory dictates that only transitions with $\Delta M = \pm 1$ are observable, where $M$ is the sum of the quantum numbers $m_1$ and $m_2$. For example, $|\alpha\alpha\rangle \leftrightarrow |\alpha\beta\rangle$ is an observable transition ($\Delta M = \pm 1$), but $|\alpha\alpha\rangle \leftrightarrow |\beta\beta\rangle$ is unobservable ($\Delta M = \pm 2$). From the table above one can see that there are four observable transitions:

$$|\alpha\alpha\rangle \leftrightarrow |\alpha\beta\rangle; |\alpha\alpha\rangle \leftrightarrow |\beta\alpha\rangle; |\alpha\beta\rangle \leftrightarrow |\beta\beta\rangle; |\beta\alpha\rangle \leftrightarrow |\beta\beta\rangle$$

Each transition corresponds to a fixed energy, and hence each observable transition gives rise to one peak in the NMR spectrum. The area under the peak - or loosely speaking, the height of the peak - is related to how much "$\Delta M = \pm 1$ character" there is in each transition. Of course, there is a way to formalise this, but for the sake of simplicity that will have to be omitted.

For the transition $|\alpha\alpha\rangle \leftrightarrow |\beta\beta\rangle$, this is entirely $\Delta M = \pm 2$. So, the amount of "$\Delta M = \pm 1$ character" is exactly zero, and the transition is entirely unobservable.

On the other hand, for all four of the allowed transitions, the amount of "$\Delta M = \pm 1$ character" is exactly one, and hence all transitions have equal intensity. [Of course, because of the lack of coupling, there will be two degenerate pairs (i.e. there are two pairs of transitions with the same energy), and so you get two singlets in the spectrum.]

Now, as the coupling term $2\pi J(\vec{I}_{\!1}\cdot\vec{I}_{\!2})$
becomes larger and larger, the eigenstates are gradually shifted away from these. These can be expressed in terms of linear combinations of the four original eigenstates. Without proof I will present the results

$$\begin{array}{cc}
\text{New eigenstate} & \text{Expression in terms of old eigenstates} \\ \hline
|1\rangle & |\alpha\alpha\rangle \\
|2\rangle & \cos\theta|\alpha\beta\rangle + \sin\theta|\beta\alpha\rangle \\
|3\rangle & -\sin\theta|\alpha\beta\rangle + \cos\theta|\beta\alpha\rangle \\
|4\rangle & |\beta\beta\rangle
\end{array}$$

where $\theta$ is defined by $\tan(2\theta) \equiv J/\Delta\nu$. As the ratio of $J$ to $\Delta\nu$ goes from zero to infinity, $\theta$ goes from $0$ to $\pi/4$. It is easy to verify that the deviation from the old eigenstates increases as this ratio $J/\Delta\nu$ increases.

Let's look at the transition $|1\rangle \leftrightarrow |2\rangle$. How much "$\Delta M = \pm 1$ character" is there here? Since $|2\rangle$ is a mixture of $|\alpha\beta\rangle$ and $|\beta\alpha\rangle$, the transition $|1\rangle \leftrightarrow |2\rangle$ can be thought of to be a "mixture" of $|\alpha\alpha\rangle \leftrightarrow |\alpha\beta\rangle$ and $|\alpha\alpha\rangle \leftrightarrow |\beta\alpha\rangle$.

So, there is $\cos\theta$ amount of "$\Delta M = \pm 1$ character" arising from the $|\alpha\alpha\rangle \leftrightarrow |\alpha\beta\rangle$ contribution; and there is also $\sin\theta$ amount of "$\Delta M = \pm 1$ character" from the $|\alpha\alpha\rangle \leftrightarrow |\beta\alpha\rangle$ contribution, for a total of $\cos\theta + \sin\theta$.

For quantum mechanical reasons it is necessary to take the square of this quantity, i.e.

$$\begin{align}
(\cos\theta + \sin\theta)^2 &= \cos^2\theta + \sin^2\theta + 2\sin\theta\cos\theta \\
&= 1 + \sin(2\theta)
\end{align}$$

and working through the list of transitions, one finds that the amount of "$\Delta M = \pm 1$ character" for each transition is

$$\begin{array}{cc}
\text{Transition} & \text{Amount of “}\Delta M = \pm 1\text{ character"} \\ \hline
|1\rangle \leftrightarrow |2\rangle & 1 + \sin(2\theta) \\
|1\rangle \leftrightarrow |3\rangle & 1 - \sin(2\theta) \\
|2\rangle \leftrightarrow |4\rangle & 1 + \sin(2\theta) \\
|3\rangle \leftrightarrow |4\rangle & 1 - \sin(2\theta)
\end{array}$$

Aha! So, as long as $2\theta = \tan^{-1}(J/\Delta\nu) \neq 0$, the intensities of the peaks will be different. There will be two strong peaks (those with plus signs), and two weaker peaks (the minus signs).

When the ratio $J/\Delta\nu$ is small, i.e. the weak coupling limit, then $\sin(2\theta)$ is small and the difference in intensity is hardly noticeable - not something that you can tell by eye. For example, in a $\pu{400 MHz}$ spectrometer, two protons that have a chemical shift difference of $\pu{1 ppm}$ would have $\Delta \nu = \pu{400 Hz}$. If we take $J = \pu{7 Hz}$, then $2\theta = \tan^{-1}(7/400) = \pu{0.0175 rad}$ and we have $\sin(2\theta) = 0.0175$, a difference which you'd be hard pressed to spot!

But when this ratio increases, the value of $\sin(2\theta)$ will increase, and the difference in intensity will become more and more pronounced. I won't go into the maths, but it is possible to show that the middle two transitions have the higher intensity, and the outer two transitions have the lower intensity - which, of course, gives rise to the "roofing effect" which you speak of.


Why doesn't the vapor pressure of water depend on the humidity?

The partial pressure of water vapor is equal to the equilibrium vapor pressure at the temperature of the liquid only at the surface of the water.  Above the surface of the water, the partial pressure of the water vapor decreases with distance, and far from the surface, approaches the partial pressure consistent with the relative humidity of the room air.  So there is a concentration gradient of water vapor in the air above the water surface.  Based on Fick's law of diffusion, this provides a driving force for diffusion of water away from the surface.  We call this diffusive flux of water vapor evaporation.  It is driven by the difference between the equilibrium vapor pressure of water at the surface and the partial pressure of water vapor in the bulk of the room air.


Identification of lithium and strontium salts

If your solution contains only $\ce{Li+}$ or $\ce{Sr^2+}$, you can differentiate them by adding a sulfate solution (for example, $\ce{Na2SO4}$ or diluted sulfuric acid). If strontium is present, a white, fine-crystalline precipitate of $\ce{SrSO4}$ will form. In the case of lithium, the solution will remain clear. This is because $\ce{SrSO4}$ is much less soluble in water than $\ce{Li2SO4}$ ($\pu{0.0132 g}$ versus $\pu{34.8 g}$ in $\pu{100 g}$ of $\ce{H2O}$ at $\pu{20 ^\circ C}$ (reference)).

Adding diluted $\ce{HCl_{aq}}$ and $\ce{BaCl2}$ to a solution is commonly used as a test for $\ce{SO4^2-}$ (if it is present, $\ce{BaSO4}$ precipitates). Maybe this could help you to distinguish between $\ce{Li+}$ and $\ce{Sr^2+}$ given the different solubilities of their sulfates. If your unknown salt readily dissolves in water and $\ce{SO4^2-}$ is found to be present in the solution, then it is likely that you have lithium sulfate. Strontium sulfate as unknown salt would not readily dissolve and therefore not give a positive test.

However, if your solution can contain more ions besides $\ce{Li+}$ or $\ce{Sr^2+}$, I would suggest to try the separation with ammonium carbonate which is already mentioned in the comments to your question.


Why does water boil vegetables, but oil does not?

The maximum temperature for water boiling is around 100 Celsius and there is a great deal of water present.  Oil can reach greater temperature and the absence of water allows other reactions, such as the Maillard reaction, to occur.  That's why you never see browning in boiled food.


Equivalence point means equal numbers of moles of acid and base have been added. If $x$ moles of the weak base $\ce{B}$ and the strong acid $\ce{HA}$ have been added, then the $\ce{H^+}$ ions will react away $\ce{OH^-}$ until all the weak base has associated into $\ce{BH^+}$ and $\ce{OH^-}$  (which of course has been reacted with $\ce{H^+}$ forming water). This leaves us with water and $x$ moles of the salt $\ce{BHA}$. This salt is ionic and so soluble in water, so it completely ionises to form $\ce{BH^+}$ and $\ce{A^-}$ ions in solution.

As $\ce{BH^+}$ is the conjugate acid of the weak base we had originally, it can react with water in an equilibrium reaction as follows:

$\ce{BH^+} + \ce{H2O} \rightleftharpoons \ce{B} + \ce{H3O^+}$

As $\ce{B}$ is a weak base, the above equilibrium will lie to the right, and hence there will be a considerable concentration of $\ce{H3O^+}$ in the solution at equivalence, which leads to an acidic pH.

Calculating the pH depends on the fact that $K_b K_a = K_w$ for a base and its conjugate acid. This is derived here.

For the equilibrium

$\ce{BH^+} + \ce{H2O} \rightleftharpoons \ce{B} + \ce{H3O^+}$

the acid dissociation constant has the form

$K_a = \dfrac{[\ce{B}][\ce{H3O^+}]}{[\ce{BH^+}]}$

As we know $K_b K_a = K_w$, we can rewrite this equilibrium expression as

$[\ce{H3O^+}]=\sqrt{[\ce{BH^+}]\dfrac{K_w}{K_b}}$

To find pH you can $-\log$ this. (Note $[\ce{B}]=[\ce{H3O^+}]$ as one molecule of each is produced for each molecule of $\ce{BH^+}$ that reacts with a water molecule).

So in order to find the pH at the equivalence point, you need to know the $K_b$ of the base you're using. And as everything is in the same volume, all the volumes will cancel from the concentration terms.

I hope this helps, and please correct me if I've made any errors.


The oxidation/reduction reactions of

ethanol <-> ethanal <-> acetic acid (ethanoic acid)

are at equilibrium. Oxidation (presence of $\ce{O2}$ or other oxidizing agent) and reduction (addition of $\ce{H}$ to the $\ce{C-O}$ carbon) are in constant flux. The system can be pushed to the right if wine is exposed to air for long periods of time, as you found, but at a certain point it will reach equilibrium, in which case the excess of acetic acid will reduce (marginally) back into ethanal and/or ethanol.

A better explanation may be that since the B.P. of ethanol (78 dec. C) is significantly lower than acetic acid (118 deg. C), the vapor pressure of ethanol is greater and it has a higher affinity to convert to a gaseous state. This also would shift equilibrium to the left to produce a lower post-equilibrium concentration of acetic acid.

Acetic acid does have the possibility of decarboxylating, but this is usually in excess heat. In this case, a $\ce{CO2}$ molecule breaks off of the acetic acid leaving $\ce{CH4}$, which would also be generated as a gas.

Finally, it is possible that you're mistaking "loss of acetic acid" for an actual increase in pH, irrelevant to the presence of whole acetic acid molecules. It could be that other materials in the wine "soak" protons to increase pH and leave ethanoate counter ions ($\ce{CH3COO-}$). You should clarify how you're actually quantifying the concentration.

I wouldn't expect the "decrease" of acetic acid to be significant, compared to the amount that was produced in the first place. I would also expect your system to re-equilibrate after a while longer. It would be interesting to see what happens next.


Are imine or amine bonds similar to ethereal bonds, from the perspective of an enzyme?

This does depend a lot on what the enzyme does and what it expects. Take the following three structures, one from an ether (e.g. diethyl ether), one from an amine (diethyl amine) and one from an imine (N-ethylethanimine).

Scheme 1: Ether linkage, amine linkage and imine linkage (from left to right) including hydrogen atoms on adjacent carbons.

When looking for a ether, an enzyme may well attempt to form a hydrogen bond to the ethereal oxygen. This might work in the case of an amine — there is a spare lone pair on nitrogen that can accept hydrogen bonds, but the additional hydrogen present can present ordering and space problems —, but would in the case of an imide.

However, note that the carbons next to ether and amine bonds have four atoms attached to them (two hydrogens and another carbon each), while the imine has only one hydrogen. The $\ce{C=N}$ bond is planar, meaning that the configuration of the neighbouring atoms is also relatively fixed. Thus, the imine may have a pretty different shape — and may not fit into the enzyme’s active pocket. This might actually also mean that the imine cannot accept the hydrogen bond offered by the enzyme because it cannot orient itself favourably.

As I said above, it depends on the enzyme whether it cares about hydrogen bonding more or whether the overall shape is more important. In general, I would consider an amine to be closer to an ether structurally, and therefore more likely to be accepted. Possibly, the amine hydrogen can point away, allowing for the amine to accept a hydrogen bond nonetheless. In that case, the amine would be the clear winner.

All these heteroatom bonds are clearly different from $\ce{C-C}$ bonds due to the bonds’ inherent polarity (electronegativity of $\ce{O}$ and $\ce{N}$).


How to determine the composition for a sucrose/ zinc nitrate solution from its freezing point?

First find the molality of solute particles.  Assume i = 1; we're just counting how many things are dissolved and it doesn't matter whether they are ions or molecules at this point.  I'm using m for mass and b for molality throughout to avoid confusion.

$\Delta T = ibk_f\implies\frac{\Delta T}{k}=b$

$b=\frac{0.768}{1.86}=\pu{0.413 molal}$

Next find how many moles of particles were dissolved.

$b = \frac{n_{\text{solute}}}{m_{\text{solvent}}}$

$n_{\text{solute}}=b\times m_{\text{solvent}}=0.413\times 0.150 = 0.0619~\text{moles of particles}$

Now determine the mass of each species based on this total number of moles.
$n_{\text{particles}}=3n_{\ce{Zn(NO3)2}}+n_{\ce{C12H22O11}}$

$n_{\text{particles}}=3\frac{m_{\ce{Zn(NO3)2}}}{M_{\ce{Zn(NO3)2}}}+\frac{m_{\ce{C12H22O11}}}{M_{\ce{C12H22O11}}}$

You know the total number of moles of particles, 0.0619, and you know the molar masses of both zinc nitrate and sucrose, but you don't know either the actual mass of zinc nitrate or of sucrose.  However, you do have one other piece of information that allows you to set up a system of equations.

$m_{\text{solute}}=m_{\ce{Zn(NO3)2}}+m_{\ce{C12H22O11}}=\pu{4g}$

Solving this system of equations should give you the masses of both solutes.


1-5 dicarboxylic acid to lactone with SOCl2?

The simplest 1,5-dicarboxylic acid is glutaric acid (pentanedioic acid in a more systematical naming). If you react it with $\ce{SOCl2}$, you will indeed intermediately arrive at a compound that has one carboxylic group and one acid chloride group. Then, as you suggest, the other carboxylic acid can attack the acid chloride and $\ce{HCl}$ is eleminated.

However, the resulting product is not a lactone but rather glutaric anhydride; a cyclic acid anhydride. To convert it to a lactone you would need to remove exactly one carbonyl group. Whether this is done before or after the reaction with $\ce{SOCl2}$ does not matter, but before sounds easier to me.


Why does enantiopure sec-butyl alcohol retain its optical activity over aqueous base but forms a racemic mixture over dilute sulphuric acid?

Although sulphuric acid is a notable dehydrating agent, it is concentrated rather than dilute sulphuric acid that would be typically used for such a purpose.

Here, it is due to the improvement of $\ce{H2O}$ as a leaving group compared to $\ce{OH-}$, due to the lack of any charge as a leaving group.

Under acidic conditions, the sec-butyl alcohol will be protonated in a small amount creating a much better leaving group.  The protonation also makes the alcoholic carbon centre more electrophilic and vulnerable to $S_N2$ attack by other water molecules in solution, leading to inversion of the chiral centre.

This in not possible under neutral conditions as the unprotonated alcohol would have $\ce{OH-}$ as a much poorer leaving group, preventing elimination or substitution reactions.

Under acidic conditions, with neutral water as a nucleophile $S_N2$ substition will be the preferred path.

A secondary alcohol on a small, un-branched carbon chain represents a relatively accessible spot for nucleophilic attack.
$E_1$ eliminiation or $S_N1$ substituion would require the formation of a secondary carbocation with relatively poor stabilisation compared to a tertiary carbocation.
$E_2$ elimination would require the removal of a hydrogen of low acidity from the carbon chain by a very weak base, water.

This allows a equilibrium converting molecules back and forth between the $R$ and $S$ forms of the alcohol to be set up and leads to racemisation in solution.


How does a mixture not retain its chemical identity?

A good example of the difference might be to look at water, $\ce{H2}$, and $\ce{O2}$.

If you have gaseous $\ce{H2}$, and $\ce{O2}$ in a container you have a mixture; you can change it to have $99\%$ oxygen and $1\%$ hydrogen or vice versa or any ratio in between, but it is still a mixture of the two gases. These mixtures will have different properties depending on the ratio of the two gases. Each gas still has its own properties and the mixture exhibits some combination of them depending on the ratio.

If they are allowed to react to form water, now we have a substance. Regardless of how we make water, it will always have the same composition (two hydrogens for each oxygen) and the same properties. Importantly, these properties are completely different from hydrogen, oxygen, or any mixture of the two.


How would I find the molarity/concentration of a gas using pV = nRT?

Supposing the dissociation process is isochoric, the total volume $V$ required for calculation of molarity at equilibrium of $i$th component $c_{\mathrm{eq},i}$

$$c_{\mathrm{eq},i} = \frac{n_i}{V}$$

can indeed be found from the ideal gas law using initial amount of undissociated sulfur trioxide:

$$V = \frac{n_0(\ce{SO3})RT}{p} = \frac{\pu{1.00 mol}\cdot\pu{8.21e-2 L atm K-1 mol-1}\cdot\pu{1000 K}}{\pu{1 atm}} \approx \pu{82.1 L}$$

You have found amounts of substances at equilibrium correctly, the rest is just math:

$$c_\mathrm{eq}(\ce{SO3}) = \frac{\pu{0.25 mol}}{\pu{82.1 L}} \approx \pu{3.0e-3 mol L-1}$$

$$c_\mathrm{eq}(\ce{SO2}) = \frac{\pu{0.75 mol}}{\pu{82.1 L}} \approx \pu{9.1e-3 mol L-1}$$

$$c_\mathrm{eq}(\ce{O2}) = \frac{\pu{0.375 mol}}{\pu{82.1 L}} \approx \pu{4.6e-3 mol L-1}$$


Synthesis of 2-benzylcyclohexan-1-ol from cyclohexanone

Making a carbon-carbon bond at the alpha position of a carbonyl suggests using some kind of enolate chemistry. In contrast to an SN2 type reaction with an alkyl halide (which can give over-alkylation and competitive O-alkylation), an aldol condensation is a reliable way to construct the requisite C-C bond. Here cyclohexanone reacted with benzaldehyde under basic conditions gives an enone. Two subsequent reduction steps, hydrogenation followed by hydride addition to the ketone, give the target compound.


In chemistry we assume a mixture of two solids as two different phases, which do not interfere with each other and thus, the solid solute in the solid mixture will not interfere with the solid solvent. So, the vapour pressure of the solid solution and solid solvent will always be same. In a similar way, when two immiscible liquids are mixed together, the total vapour pressure is counted as the sum of Vapour pressures of the liquids if kept separately.
Hope this helps!!


Kinetic isotope experiment used to distinguish radical and concerted mechanisms for the Diels Alder reaction

What Fleming is likely referring to is work by Edward Thornton (J. Am. Chem. Soc. 1972, 94, 1168) in which he attempted to make use of secondary kinetic isotope effects (KIE) to prove whether the Diels Alder reactions being studied were indeed pericyclic, or occurred via a radical pathway.*

The phrasing in your post (which I assume is from the book itself) isn't particularly helpful. By 'geometric', Thornton was actually talking about the geometric mean and the arithmetic mean as ways of considering the isotope effects.

The reaction they actually studied was the retro Diels-Alder of 9,10-dihydro-9,10-ethanoanthrace, with varying levels of deuteration (none, di, tetra) as shown in the figure below.

In order to discuss whats going on, we need to define some terms:

In the reaction, two bonds are being broken, lets call those bonds A and B. Breaking bond A has a KIE of fA and breaking bond B has a KIE of fB
Each molecule (no deuteration, di-, tetra-) has an associated rate constant, lets call these k0, k2, k4 (with the subscript indicating how many deuteriums we have).

We can now consider the calculation that Thornton went through to determine if the reaction is indeed a concerted one:#

In a completely concerted reaction, intuitively fA = fB, since the TS is symmetrical. If the reaction is non-concerted (aka stepwise), fA cannot equal fB unless there is symmetry to the TS.
Considering the di-deuterated molecule, two pathways are possible (breaking A or B first) to give a non-symmetrical intermediate. Thornton defines rate equations for these two as fAk0/2 and fBk0/2. We can do this since k0 is the same no matter which bond is broken in the undeuterated starting material (with the f modifying it to take into account the KIE).
k2 (the rate constant for the retro Diels-Alder of the di-deuterated starting material) can therefore be thought of as the sum of two pathways (break A then B or B then A). Mathematically: k2 = k0(fA + fB)/2. Crucially, this relationship is valid independent of whether the reaction is concerted or stepwise.
k4 (the rate constant for the retro Diels-Alder of the tetra-deuterated starting material) is able to be obtained similarly. Mathematically: k4 = fA fBk0.
At this point, Thornton provides the result of solving the two previous equations simultaneously, which provides a relationship as below:

$$f_\mathrm{A},f_\mathrm{B} = \frac{k_2}{k_0} \pm \left[\left(\frac{k_2}{k_0}\right)^2 - \frac{k_4}{k_0}\right]^{1/2}$$

It is from this relationship that we are able to consider whether the reactions are concerted or stepwise:

if concerted, fA = fB = k2/k0, which means that (k2/k0)2 == k4/k0. This now allows us to measure the rate constants, plug in the numbers, and determine if the reaction is concerted.

*: There are many examples of Diels Alder reactions that are stepwise, via radical mechanisms. The KIE's measured are necessarily small, meaning its often difficult to know if what you're looking at is actually correct. One thing that does hold up generally is that the reactions aren't ionic, due to the lack of any real solvent dependance (if the reactants are soluble in the solvent and you heat it hot enough, they generally proceed at the same rate, suggesting no stabilisation of 'intermediates' in polar solvents).

#: I'm paraphrasing the original paper, which will hopefully name it more digestible, but I still encourage you to read the actual thing, despite it being fairly dense


Here's an answer without complexity which I read in my book. When you heat a liquid in open vessel, its molecules turn to vapours, but the atmospheric pressure above is pushing these vapours down. Some vapours would have enough energy to tackle this atmospheric pressure and would get away from the liquid leading to evaporation. Only some particles at the surface would have enough energy but when vapour pressure reaches atmospheric pressure, all liquid molecules have enough energy to counteract the pressure on them and thus liquid can form vapours from the whole bulk of the liquid leading to boiling.


Why can a diamond be broken using a hammer if it's the hardest natural substance known?

Hardness and toughness are not the same

Hardness and toughness are very different qualities in materials and are weakly related.

Hardness is strongly related to the more well-defined quantity of stiffness which measures how easily a compound can be deformed under stress. Glass and diamond are very stiff materials, for example. If you try to poke them with something they resist deforming to accommodate your poke. (Hardness is not perfectly aligned to stiffness because of small scale microstructures in many materials, but this is good enough for now).

Toughness is a vaguer term for materials and there isn't a simple way to measure it. This is partially because it varies depending on circumstances in a way that stiffness does not and it is a property of the overall structure and not just the materials that make up the structure. Understanding it requires some insight into why things break (and why other things don't). we need to know a lot about the small scale structure of materials not just the substances involved. For example, stainless steel (used in knives and forks) is a well-known tough material but cast iron is brittle. Both are mostly made of iron. The differences are in the crystalline structures.

One key property of tough structures is that cracks don't propagate. So a stick of glass will break easily as will a glass fibre. But a bundle of glass fibres embedded in an epoxy resin can be very tough (because the cracks in individual glass fibres are not propagated through the epoxy resin). Some tough metals can adjust the micro defects in their crystalline structures to absorb the strain that would otherwise propagate cracks. Some very soft compounds are very tough because the deform so easily that it is hard to start cracks, nylon rope for example.

Diamond is very very stiff. But it has no protection mechanism against cracks. So, like glass, once a crack has started it doesn't take a lot of energy to cause it to spread, so it may be stiff but it isn't tough.

A fuller explanation of this took material scientists and engineers a long time to work out would require a whole book to do it justice. Luckily that book has been written and is called The New Science of Strong Materials. It is well worth a read.


Why is it necessary to heat an esterification reaction?

Basically the use of heat is to liberate the water molecule from the mixture of acid and alcohol that is $\ce{OH-}$ ion from alcohol and $\ce{H+}$ from acid that results in formation of ester.Moreover it helps in evaporation of water since esterification reaction forms an equilibrium.The removal of water from mixture forces the reaction to move forward.It helps us to get a large amount of product.Alternative to heating some dehydrating agents like conc. $\ce{H2SO4}$ can also be added.


Why is BDE low in Iodine, but not behaves like Fluorine?

Where have you found the information that the iodine free radicals are unstable? It does not sound right for me.

BDEs of chlorine, bromine and iodine decrease down the group as the size of the atom increases: 243, 192, and 151 kJ/mol. The BDE of fluorine, is however, lower (158 kJ/mol) than those of chlorine and bromine. Here you were right! But the free radical stability depends also on other factors. It increases with increasing the atom size and with decreasing the electronegativity. Additionally the stability of the free radical decreases as the orbital is held closer to the nucleus. Thus, as we can expect, the iodide radical will be the most stable, and the fluoride - the least./

Btw correlation does not mean causality! (:


How to I identify the diene and dienophile through retrosynthesis of a given compound?

The $\ce{CO3}$ substituents are questionable. Let's assume that this is $\ce{CO2Me}$ instead!

Since this is homework, I won't provide a full solution but:

Note that double bond patterns may isomerize (in the direction of conjugation).
Alkynes can serve as dienophiles too!


If all molecules have basically the same KE at a given temperature, why do lighter isotopes evaporate more readily?

Evaporation depends on molecular velocity not on the energy
The simple mistake in your assumption is that evaporation is driven by molecules having enough energy to escape the liquid phase. This is wrong. Evaporation depends on whether the molecule is travelling fast enough to escape the liquid.
The molecules in the liquid will have, on average, the same kinetic energy. But this means that, in practice, the molecules will have a range of different kinetic energies across a wide distribution. But, since kinetic energy is a function of both molecular mass and molecular velocity, there will also be a distribution of velocities. And, crucially, the more massive molecules (in this case those with 18O) will have slightly lower velocities at the same average kinetic energy. So the lighter isotopes will be slightly more likely to have a high enough velocity to escape the liquid phase.
You can work through the mathematical details (taking into account the statistical distribution of kinetic energy and converting that to the distribution of velocities and then calculating the proportion of molecules with a velocity above the threshold for evaporation) and check that the proportion of light molecules is higher than the proportion of heavy molecules.

Clarification: why is the force-based view wrong
The argument that applying a fixed force leads to a smaller effect on a heavier molecule so heavier molecules don't need to moves as quickly to escape the liquid is flawed for several reasons.
It sounds relevant but isn't because the vast majority of all collisions don't lead to escape from the liquid. There are so many momentum (and KE) redistributing collisions that all we have to consider is the distribution of energy/momentum across the ensemble of molecules in the liquid. And what we need to know is the average proportion of molecules having enough velocity to escape after their last collision.
Yes, the force argument matters for individual collisions (if they are not purely elastic), but, if you simulated a drop of liquid down to the molecule level, you would find that the distribution of energy/momentum would match the statistical one expected from theory and you would have gained no new knowledge. Follow an individual molecule and, yes, the thinking about force will apply. But follow the distribution of velocities across the ensemble of molecules and what matters is the proportion of molecules having a high enough escape velocity and that is governed by the statistical distribution. Think about the properties of the ensemble of molecules not the individual molecule.
And the force argument ("the escape velocity will also be lower") doesn't work at the ensemble level. All that matters is the velocity after the last collision and the only way to know that is to look at the statistics of the whole system not the individual forces on individual molecules.


FCI of N2 with about 10^10 CSFs must be the largest such calculation...

Ref:
A full-configuration benchmark for the N2 molecule,
Elda Rossi, Gian Luigi Bendazzoli, Stefano Evangelisti, Daniel Maynau,
https://doi.org/10.1016/S0009-2614(99)00791-5


Could muscarine be an antidote to atropine poisoning?

In short: no.

It is not that simple. According to Wikipedia, atropine works as a muscarinic antagonist, which blocks the action of acetylcholine at the muscarinic acetylcholine receptor. So, the mode of action as an antidote is to prevent muscarine from binding to the receptor, thus inhibiting its effects on the body.

The poisonous effects of atropine cannot be reversed by muscarine, though. That may be due to the fact, that it is able to displace muscarine and not the other way around.


The best thing would be if you went back to the mass action law, considered all reactions that take place ($\ce{Ca}$ buffer!) and derived a formula for your specific problem. The Henderson-Hasselbalch equation actually does exactly that for a one-component acid/base system with a single protonation stage. Remember that the concentrations that occur in the mass-action law are those at chemical equilibrium, not the initial ones; i.e. for any component $C$, the eq. concentration is $[C] = [C]_0 - \Delta_C$. The problem at hand is then to find how the changes in concentration depend on the added volume of titer.

If you just took the Henderson-Hasselbalch equation (equating the change in concentrations with the added concentration of $\ce{HCl}$), you do not include the $\ce{Ca}$ buffer, which will probably play an role in this. (You get a $pK_a$ of around 0.7, which I guess is a bit on the strong side ...)

One more thing I just noticed: You apparently added 100 mL of titer to 100 mL of sample. That's kind of bad, because you'll also have to consider how the change of overall volume affects your concentrations (effectively the pH) during titration.


Looking for a non-volatile liquid less viscous than water

I guess you could go with Peppermint essential oil. Its density is 0.898 g/mL at 25 °C. It obviously is volatile to some extent if left uncovered however its viscosity is considerably lower than water. Hope this helps you out!


Sulfuric acid fumes from cleaning shower drain

If the workman did use concentrated sulphuric acid (which is not the best choice for drain cleaning) then the best solution if to flush the drain with large quantities of water and wash the surfaces with water.

Sulphuric acid is not toxic but is very corrosive when concentrated. But it is highly water soluble and not harmful when it is dilute enough. So dilution with water will solve whatever problems it seems to cause. If it got splashed around, wash everything thoroughly and there will be no further problem. If you use enough water then it won't even be necessary to neutralise the result as any remaining acid will be swept through the drain and highly diluted by the flow of water.


What kind of fire extinguisher for chemistry at home?

You need to investigate the safety and handling of every chemical you use in your experiments individually. The handling and safety precautions can be very different depending on what substances exactly you use.

In terms of fire, the one thing you must know before each experiment is which methods won't work or even fail catastrophically. Trying to extinguish a sodium fire with water is a really, really bad idea, and that is something you must know in advance.

Metal fires are the most common kind of fire where you have to be careful about choosing the right method. Some metal fires can't even be extinguished by carbon dioxide. There are special fire extinguishers for this (class D), other methods might not work.

A bucket of sand is the most useful method for small fires in a lab. It works for almost everything and doesn't cause a big mess. But it is obviously useless if you cause a bigger fire and can't reliably cover it with sand anymore.

Powder extinguishers cause quite a mess, you'll spend a lot of time cleaning up the room afterwards. Obviously this is a lot better than burning down your home, but I wouldn't use them for small fires you could stop with some sand.

Foam and powder are probably the most useful general-purpose extinguishers. Carbon dioxide is a bit trickier to use from what I heard, I probably wouldn't recommend it in your case.

Thinking about a fire extinguisher is a good first step. Another thing I very strongly recommend is that you read up on general safety procedures in the lab. Simple things like not storing large amounts of solvent directly in your fume hood (or workplace) might prevent a small fire from becoming a dangerously large one. Knowing that you need to inactivate any reactive reagents before disposing them might save you from throwing away some oxidizing agent and accidentally setting your waste on fire. There is a lot more to chemical safety than just choosing the right fire extinguisher.


Why do alkanes have higher boiling point than their ether counterparts?

I suspect London forces dominate in the longer hydrocarbons, whereas they are disrupted by the ether bond. For shorter hydrocarbons, the dipole of the O-C bond is greater than the London forces, but this is reversed as the # of carbons increases.


There is, as far as I know, no shorthand for that, but you might be able to omit it completely in a first order approximation. The closest might be adding $\ce{(aq)}$ when working in aqueous environment.

Adding $\ce{.$x$\,H2O}$ to the formula usually indicates, that it is a solid and that there are $x$ water molecules enclosed in the crystal structure. It depends on what you are looking at, which kind of reaction, what you might be able to omit.

For example, when you are describing a drying process, it is not possible:
$$\ce{\overset{blue}{CuSO4.5H2O} ->[\Delta] \overset{white}{CuSO4} + 5H2O}$$

When you are working in aqueous solution it is usually not important, as there is much more water around it, hence
$$\ce{\overset{blue}{CuSO4.5H2O} + Na2S -> \overset{black}{CuS} v + 2Na+ + SO4^{2-} + 5H2O}$$
is essentially the same as
$$\ce{CuSO4~(aq)  + Na2S~(aq) ->CuS~(s) v + 2Na+ ~(aq) + SO4^{2-}~(s)}$$

If you are describing a redox reaction, it might be the same, hence the following should be fine
$$\ce{CuSO4~(aq) + Zn~(s) -> Cu~(s) v + Zn^{2+}~(aq) + SO4^{2-}~(aq)}$$
Often you can omit the $\ce{(aq)}$ or $\ce{(s)}$ all together.

So depending on what kind of chemistry you are talking about, some information can be implied and therefore be omitted. If you add an example of what kind of reactions/ substances you are working with, I can expand my answer to take that into account.


Is the variation in carbonation of soft drinks the variable that affects the effectiveness of the mentos reaction?

It's my understanding that the two key variables here are carbonation, as you suggested, and viscosity.

People performing this sort of demonstration usually use drinks completely sweetened with artificial sweeteners like aspartame or Sucralose, because they're much less viscous than their full-sugar equivalents, and especially than the high-fructose corn syrup variants.

Common artificial sweeteners are much sweeter than sucrose and glucose [1], so they're added in much smaller quantities and so affect the properties of the liquid less.

The viscosity affects the ability of the liquid to have the rush of bubbles from rapid nucleation required for the fountain effect, and the carbonation affects how much impetus there is behind it.

For an good comparison, you could try three variants of (popular cola product), including a newer one partially sweetened with steviol (look for the ones with green labels: Pepsi Next and Coca-Cola Life are two examples): the wholly artificially sweetened one should jet best, then the part-steviol, part-sugar one, then the full-sugar one.

"Sucralose is about 320 to 1,000 times sweeter than sucrose, three times as sweet as aspartame and twice as sweet as saccharin." -- Wikipedia (I'm not sure how that's measured, but the sucralose figure comes from the FDA approval document cited on the Wikipedia page.)


why do we see characteristic colors of alkali metals in oxidizing flame?

I don't think it is a good idea to connect ionization energy with the color of flame emission, especially in a Bunsen burner.
Ionization energy means that you have separated the electron out of the nucleus attractive field. However, electronic transitions, corresponding to visible wavelengths, take place while the electron is still bound to the nucleus. Thus the flame color of alkali metals might correlated with the IP  but not the causation.
Let us look at the resonance wavelengths of emission of Li, Na, K, Rb, and Cs. Resonance wavelengths means the transition from the $\mathrm{1s}$ to $\mathrm{2p}$ orbital or $n\mathrm s$ to $(n+1)\mathrm p$ orbital in the case of alkali metals.
$$
\begin{array}{r|ccccc}
\hline
\text{Element} & \ce{Li} & \ce{Na} & \ce{K} & \ce{Rb} & \ce{Cs} \\
\hline
\lambda/\pu{nm} & 670 & 589 & 769 & 794 & 894 \\
\hline
\end{array}
$$
You cannot even see these wavelengths with your eyes after 700 nm which are in the deep deep red. Some people can see the red lines of K 769 nm. However, you can see the energy gap is indeed decreasing as we go down the group. The ionization energy also goes down the group. However, you can see, the colors which you see in the Bunsen burner have nothing to do with ionization energy.
The wavelengths have been extracted from a old book titled Graphische Darstellung der Spektren von Atomen und Ionen mit ein, zwei und drei Valenzelektronen by Dr. W. Grotrian, 1928. I think NIST atomic spectra database will be better today but I am not sure if they list resonance lines clearly. This book has graphically labelled transitions for almost all elements known in 1928.


Why doesn't N2O give a positive brown ring test?

The brown ring nitrate test is based on forming ferrous nitrosyl cation, with nitrous oxide formed from nitrates by iron oxidation in acidic solution:

$$\begin{align}
\ce{3 [Fe(H2O)6]^2+ + NO3- + 4 H+ &->3 [Fe(H2O)6]^3+ + NO + 2 H2O} \\
\ce{[Fe(H2O)6]^2+ + NO &->3 [Fe(H2O)(NO)]^2+} + H2O\\
\end{align}$$

Positive results are given also by nitrites and nitrogen oxides able to form nitric or nitrous acid, directly by hydration, or indirectly by disproportionation.

$$\begin{align}
\ce{ [Fe(H2O)6]^2+ + NO2- + 2 H+ &-> [Fe(H2O)6]^3+ + NO +  H2O} \\
\ce{N2O3 + H2O &-> 2 HNO2}\\
\ce{2 NO2 + H2O &-> HNO2 + HNO3}\\
\ce{N2O5 + H2O &-> 2 HNO3}\\
\end{align}$$

Therefore, $\ce{NO}$ and all higher nitrogen oxides oxidizing $\ce{Fe^2+}$, forming $\ce{NO}$, react positively in the brown ring test.

$\ce{N2O}$ cannot do that, because it cannot form $\ce{NO}$ at given conditions.


How to measure amount of oxygen (or carbon dioxide) in air?

This is untested, but I think it might work. You could light some fuel on fire inside the closed container and see how much of it burns (just put the cup of fuel on a kitchen scale) before it dies out and than back calculate how much oxygen was present. Preferably you should take a fuel that burns easily to make sure that the total combustion reaction is the only reaction that occurs. Also if you want to repeat the experiment every hour or so, you want to be sure that the fuel does not evaporate (substantially) because that might cause the entire container to light up when you try to burn the fuel.

I am not really sure which 'household' fuel would be a good option, but maybe one of the other people on this site have suggestions?!

You can probably use one of those candle lighters and stick the long side through a small hole in your container and than seal the rest of the hole to be able to light a fire inside, without having air flowing in or out.

Just be careful not to set the plant (or anything other than the fuel) on fire!


Yes your logic is basically correct.

Gibbs free energy is really derived is from the Clausius in equality which is: $$dS\geq \frac{dq}{T}$$ Basically is states that in a reaction, the change in entropy must always be equal (in the case of an ideal gas) or greater than the transfer of heat energy divided by temperature. Therefore, for a reaction to be spontaneous, the following relation must be satisfied: $$dS-\frac{dq}{T}\geq0$$ Now, lets consider two cases so we can simplify this relation.

Constant Pressure

Enthalpy is given by: $$\Delta H_p = \Delta U + p\Delta V$$
and internal energy by: $$\Delta U = q + w$$
Now by substituting $\Delta U$ in the expression for enthalpy we get: $$\Delta H _p= q + w + p\Delta V$$ Now, since w = $-p\Delta V$, the above equation simplified so that: $$\Delta H_p = q$$ So, now in the Clausius inequality, we can substitute dq with dH and multiply by $T$ to get: $$TdS \geq dH$$ Now by integrating both sides and putting all the variables on one side, we get the question for gibbs free energy: $$\Delta G = \Delta H - T\Delta S$$ and hence this is why $\Delta G$ has to be negative for a reaction to be spontaneous

Constant Volume

At a constant volume, there is no expansion work done, therefore $dU = dq$
Therefore by using the same processes as before, we are able to get the expression for Helmholtz energy, A, which is defined as: $$\Delta A = \Delta U - T\Delta S$$ Here, for a reaction to be spontaneous, $\Delta A$ must be less than 0


Beside that the proportional relationship proposed could be discussed further, as r rearrange, all this statement are related to the sums of forces. They stated that there would be a greater attractive force because they discuss ionization and this is the case. But the attractive force is the result of all attractions and repulsions. Any distribution of electric charges would lead to this situation and is generally an exercise treated after Coulomb law. Any reader can do that providing the dynamics is not involved.


You should not regard this answer as medical advice. If concerned about your health consult a physician.

The glue you link to (TAMIYA  EXTRA-THIN  CEMENT, made by IKEGAMI PAINT INDUSTRY COMPANY), according to a material safety data sheet (MSDS) I was able to retrieve online, appears to be a 1:1 mixture of acetone and butyl acetate. Such a mixture works in a similar way to dihloromethane as a solvent to weld together plastic parts. Quoting the wikipedia,

  Acetone is believed to exhibit only slight toxicity in normal use, and there is no strong evidence of chronic health effects if basic precautions are followed.

OSHA classifies acetone under the following health codes ("principal effect(s) of exposure to each substance [...] used in determining if a violation of an air contaminant standard is serious"):

  HE16     Irritation-Eyes, Nose, Throat, Skin---Mild

OSHA classifies butyl acetate as follows:

  HE8    Nervous System Disturbances---Narcosis

  HE15   Irritation-Eyes, Nose, Throat, Skin---Moderate

For completeness, I include a screenshot from the beginning of the MSDS I found for the adhesive. For more on the GHS classification (divisions or categories) you can refer to the method employed in classifying substances.


Why is the maximum % for a gold alloy 92%?

It's not that you can't, it's that you usually don't.

The majority of Gold goes into jewellery with some in coinage (more in the past, we make gold coins now but they don't circulate much). But pure gold is too soft to be useful in most applications (coins have to last, rings have to not scratch too much and they aren't much use if they bend too easily). It is also expensive. So the most used alloys have to make a compromise between usability and cost. There are also some that try to achieve a particular aesthetic effect like a different colour.

This imposes limits of practicality. At the cheap end alloys that look like gold might be mostly other metals. But at the high end the issue is to get the correct aesthetic effect while retaining the appeal and value of gold without having something  too soft or weak to be practical. Many coloured golds are only about 75% gold (18 karat) because that gives a wide range of colours. But top end "pure" gold objects can be made at 22 karat (92% isn). We could make richer alloys, but they would be too soft.

So, mostly we don't. We can make many alloys with more gold, including many that would be useless as jewellery like the AuBe alloy mentioned in another answer (it is far too brittle), but you will rarely see them as they are not useful for anything.


Best way is sending a sample for Atomic Adsorption Spectroscopy (AAS). You can get the concentration for $\ce{Mn, Pb, Fe, Cr, Ni}$, and other water goodies. Contact a local university or lab and get a price quote for a test.
Or get a flame spectroscope and DIY https://web.pdx.edu/~atkinsdb/teach/427/Expt-AtomicSpec.pdf


Tool that generates electrostatic maps of molecules?

Most computational chemistry software packages geared toward students/education should be able to do this. Examples include Scigress, formerly CAChe, Spartan, which has a 'lite/student' version that can probably do it, etc. Generally, any modeling software, for example, Gaussian or GAMESS should generate a file with the electrostatic potential map provided you give it the correct instructions. The output files of these methods can be visualized using many types of software (like Scigress and Spartan) and open source software like Avogadro.

If you are looking for an inexpensive solution, GAMESS is free for research/education and Avogadro is open source.


Are all corrosives either acids or bases?

Wikipedia lists many corrosive compounds.  While a lot of them function as strong acids or bases, there are also these (not an exhaustive list):

Strong oxidizers including concentrated hydrogen peroxide
Fluorides (they say "fluoride ion", so salts as well as the acid are meant)
Organic compounds that can act as alkylating agents such as methyl sulfate

The moral of the story: Hazards can't be pigeonholed into specific types of chemicals.  Best to find and read the safety data sheet with any chemical.


Whats the difference between ionization energy and orbital energy?

First off, the idea that the orbital energy is the negative of the ionization energy is actually Koopmans' theorem. It works qualitatively because of cancelation of errors.

But I digress...

Orbitals are single-electron constructions. That is, in Hartree-Fock theory, an orbital is a single-electron wavefunction assuming the average electrostatic potential of the remaining electrons.

Ionization energy trends are experimental observables and you see effects of multi-electron interactions.

So the short answer is that orbital energies are not physical quantities (i.e., being single-electron mathematical results) and ionization energies are based on multi-electron interactions.

Probably the simplest way to understand the trends in ionization energies vs. orbital energies is to look at Slater's rules and also remember Hund's rules. That is, when we add electrons, we are partially shielding the nuclear charge, but we also have to consider the stabilization due to exchange of electrons and destabilization when we pair electrons in an orbital. These are multi-electron effects, and only some of that can be recovered in a simple one-electron picture.

For the $2p$ orbitals, as you mention, there's a decrease in the ionization potential when you put two electrons in the same orbital. This happens because this new electron does not benefit from the exchange effect (Hund's rules) and has an electrostatic repulsion with the electron in the same orbital. Thus it's easier to remove it than might be expected from a one-electron orbital picture.


Determination of percentage of Methanol in an Alcohol based hand sanitiser

Russell, agree with your down voting frustration but it is a common problem of any public forum. The bad news is that "simple" distinguishing test do exist, but they generate more toxic products (including a carcinogen) in a so-called iodoform test. Good for school labs only. The problem is not in distinguishing methanol versus ethanol, the problem is the matrix (as analytical chemists call it), i.e., these alcohols are present in a complex mixture.
The only reliable test test to determine methanol in a complex mixture like a hand sanitizer is gas chromatography. You need to consult a professional certified service to ask them to analyze the sanitizer. There is no home-kit or simple test to differentiate the two in a mixture.


Reaction of silver nitrate and hydrochloric acid

One of the most difficult parts of chemistry is learning to recognize a type of reaction based solely on its reactants. This has to be done before you can apply a reaction pattern to the problem, and so it is critical to get this step right in the beginning.

In this case, you are looking at the reaction:

$$
\ce{AgNO3 + HCl ->  AgCl + HNO3}
$$

You already know the products, but are questioning how these products can be formed since the activity of H is greater than Ag, implying that Ag is easier to oxidize than H. This is true - but take a look at the oxidation numbers of each species in this reaction:

$\ce{Ag: +1 -> +1}$

$\ce{NO3: -1 -> -1}$

$\ce{H: +1 -> +1}$

$\ce{Cl: -1 -> -1}$

Since there is no overall transfer of electrons, this can't be a redox reaction, and that means activities won't matter here.

That means there must be another driving force for this reaction - another reaction pattern that fits better.

I don't want to give you the answer directly, but I will give you some advice that should help:

Write the full equation - including the phases.

$$
\ce{AgNO3(aq) + HCl(aq) ->  AgCl(s) + HNO3(aq)}
$$

See if you can find another reaction pattern that fits this equation better. Later, I'll update this answer to show how you can identify the correct pattern for aqueous reactions using only the reactants.

Update

As I mentioned, one of the hardest parts of chemistry is learning to identify reaction patterns based only on the reactants. This is a double-displacement (or metathesis, or precipitation) reaction. It is easy to see now, given that the reactants are aqueous and at least one of the products is solid (the precipitate). Formation of the solid is the driving force for this reaction - the quick explanation is that the forces attracting silver and chloride ions together are stronger than the solvation forces between those ions and water, as well as the forces holding them to sodium and nitrate ions.

This is good, but how do we predict that this is a precipitation reaction ahead of time?

The key is to have a good understanding of the three common types of salt/acid/base reactions that occur in aqueous solution, and to learn to recognize clues in the reactants.

Briefly, the three types that are most commonly seen are:

Precipitation
Acid/base (Arrhenius definition)
Redox (single displacement)

You can recognize each of these by the reactants if you know what to look for.

Precipitation - Two soluble salts (ionic compounds), or a salt with an acid or base.

Acid/base - An Arrhenius acid and base (compound containing hydroxide)

Redox - a salt or acid and an elemental metal

There are many other types of reactions that occur in aqueous solution, and many variations of the acid/base and redox category, but these three cover the cases most commonly seen in a classroom.

Once you have identified the likely pattern that the reaction will follow, the next step is to predict the products using that pattern to see if they make sense. If they do, then you have likely chosen correctly.

You can use this algorithm for more advanced chemistry as well - in organic chemistry, for example, one of the major goals is to learn to predict reactions based on functional groups. Once you can identify functional groups and have memorized reaction patterns for them, it becomes possible to predict a huge range of reactions.


May I treat units (e.g. joules, grams, etc.) in equations as variables?

Yes, you may. It is quite common to convert units into each other. The simplest conversion might be the prefixing of units, e.g.
$$\mathrm{\frac{km}{m}}=1000 \Longleftrightarrow \mathrm{1~km = 1000~m}.$$

Another example is the interconversion of units of energy. In some parts of chemistry, it is still quite common to use calories, in others Joule, as a SI unit, has taken its place.
$$\mathrm{\frac{cal}{J}}=4.184 \Longleftrightarrow \mathrm{1~cal = 4.184~J}.$$

In principle, all units are defined to be constants, special numbers in a wider sense. You can treat them as such.


The crystal structure of the metal holds the atom in place. BTW, since electrons are shared in a "sea" of electrons or conductivity band, no one atom is necessarily ionized. The crystal structure is also important for nuclear physics, e.g. using Mössbauer spectroscopy to measure gravitational effects.
Individual electrons don't actually move around the whole circuit, even in DC applications. Instead, think of it as a large stack of electrons; as one is pushed on the stack at one end, electromagnetic forces pop one off the other end. Though electrons do move a bit, the drift velocity is much less than the speed of light (in fact, it is somewhat slower than the last-placed snail in a race).
Compare the mass of a single electron and the mass of an atom, or the mass of an ion in a plasma. Note that acceleration of a particle is inversely dependent on its mass. This is important in fusion research, where confining electrons and ions is a bit tricky.

See this PPT presentation on metallic conduction.


Bicyclo[1.1.0}butane is an achiral molecule.

Any molecule containing a plane of symmetry must be achiral and bicylco[1.1.0]butane has 2 planes of symmetry.  One plane bisects the molecule and passes through the 2 tertiary carbons, the other plane bisects the molecule and contains the 2 secondary carbons.

Even replacing one of the secondary or tertiary hydrogens with a substituent does not make the molecule chiral.  Such a substitution will remove one plane of symmetry, but not the other one.

In the above bicylco[1.1.0]butane where we have substituted a chlorine for one of the tertiary hydrogens, the plane containing the 2 bridgehead carbons and the chlorine still remains.  However, if you replace one of the tertiary hydrogens and one of the secondary hydrogens, the resultant disubstituted-bicylco[1.1.0]butane will be chiral since both symmetry planes have been destroyed and no other plane or axis of symmetry exists.


What's the energy density of liquid nitrogen?

Just skimmed over the links:

Highview:

  Using ambient heat to warm it, the process recovers around 50 per cent of the electricity that is fed in, says Highview's chief executive Gareth Brett. The efficiency rises to around 70 per cent if you harness waste heat from a nearby industrial or power plant to heat the cryogen to a higher than ambient temperature, which increases the turbine's force, he says.

  Batteries under development .. have efficiencies of around 80 to 90 per cent, but cost around \$4000 per kilowatt of generating capacity. Cryogenic storage would cost just \$1000 per kilowatt because it requires fewer expensive materials, claims Brett.

Here's some evidence that people who know about gas and liquid air are interested in the technology: strategic partnership between Messer and Highview

Actually Highview has a presentation on their web-site that contains a lot of information. Note that they propose to run a 4-stage process against heat (ca. 100 °C), not ambient temp.

Here are some values for energy storage efficiency comparison found after a very quick search:
Batteries (Wiki)
German Wikipedia has lower efficiency values
Energy conversion table (Wikipedia)

Pump-stored hydroelectricity: 70/75 - 80% (Wiki + de.Wiki)
(Note the difference between technology that is actually in use, so efficiencies that are achieved in practical use are known and prototypes/scientific studies about achievable efficiency.)

Dearman claims 70% efficiency for the engine.

Unfortunately they don't say % of what (remember the low-temp heater manufacturers claiming > 100% efficiency because they used achievable efficiency with high temp exhaust gas as their denominator?).
Neither do they say something about the power at which they have this efficiency (ideal thermodynamic efficiency is at P $\rightarrow$ 0 ($t_{cycle} \rightarrow \infty$))
Neither can we assume that they start their calculation taking into account the efficiency of producing liquid nitrogen.
Nor storage loss.

=> so far, Highview's stationary plant concept looks more sound.

Let's have a look what thermodynamics says:

ideal efficiency
boiling T of liquid air: 77 K
ambient T: assume 290 K => $\eta_{ideal} = 1 - \frac{77}{290}$ = 73%
waste heat: assume 370 K => $\eta_{ideal} = 1 - \frac{77}{370}$ = 79%
(in other words, Dearman claims close-to-ideal efficiency)
energy density of liquid $\ce{N2}$: 620 kJ/l = 712 kJ/kg (from presentation at Highview)
The Linde process is said to need approx. 500 kWh / t = 1750 kJ/kg of air
(that would be ≈40% efficiency; no real source, just something someone answered in a German-language chemistry forum)
Here's short report that gives ≈20% efficiency for producing liquid air in Egypt. They also give storage loss (ca. 0.17 \%/d)

Thus, the 50% (achieved by Highview) to 70/80% (short term goal Highview against waste heat, quote for Dearman's engine) mean that

either the liquefaction was not part of the efficiency calculation,
or they have a super efficient air liquefaction that is about 2 - 4 times as efficient as the two numbers I found.
(Anyone around here who knows? Linde, Messer, AL?)

Update: just had a closer look at the Highview presentation. Their 50 - 80% are cycle efficiency based on "Available enthalpy for cold cycle ~300kJ/kg", so less than half of the energy density they give. Calculating with 1750 kJ/kg to produce the kg liquid air, we're at 17% (close to the "Egyptian" 20% efficiency for producing liquid air), of which they recover 0.5 - 0.8.

Conclusion: electric -> liquid air -> electric cycle with an overall efficiency in the order of magnitude of 10%


Which base pair is more stable, U:G or I:C?

One can discuss the strength of H-bonds, but that would be half of the picture. It is more important how well (or poorly) this pair fits in the whole helical structure. Stacking interactions (arene-arene interactions) are playing important role in stabilization of DNA. If the pair disturbs backbone shape the binding will be poor. Google scholar search for relevant papers might be the most productive way to go.


Inversion of Configuration - Opening of Epoxide

It's slightly difficult for me to see clearly, but your drawing appears to be correct, and it corresponds to a $180^{\circ}$ rotation about the $\ce{C_1-C_2}$ bond axis relative to the textbook rendering ($\ce{C_1}$ being the carbon bearing the oxygen atom and $\ce{C_2}$ the sulfur atom, respectively). An unambiguous method of determining whether the two representations are equivalent is to assign the absolute configuration for the chiral carbon(s) in question. It may also help to use other projections (Fischer or Newman) if visualization of bond rotation and stereochemistry in the skeletal formula becomes difficult, as you did with the Sawhorse projection.


Why is the boiling point of heavy water more than that of normal water?

The hydrogen bonds in deuterium oxide are slightly stronger than those in water. This is due to a quantum mechanical effect; the bonding interaction has a lower zero point energy due to the greater mass of the deuterium atom. It, therefore, requires more energy to excite the bonding electrons from the ground level to the dissociation point, and a higher boiling point is observed.

You can find a good explanation of the physics behind these effects on this page.


I have the following comments about nickel steels:

Tensile strength is different than hardness. I assume you are specifically asking about hardness, not strength.
Hardness has two different aspects: resisting wear and machinability. Different steels will exhibit different degrees of the two aspects. Ideally, you want a steel that is hard wearing, yet highly machinable.
There are different alloys involving nickel, notably the nickel alloys and chromium-nickel alloys. These two types of steel are different.
Different alloys have very different hardening characteristics. Even very minor changes to the amount of carbon or nickel in an alloy can have dramatic effects on its hardening characteristics.
The main effect of nickel when present in a ferritic or pearlite matrix is to increase the fineness of the grain structure.
Nickel slows the growth in size of ferritic/pearlite grains. This makes it easier to obtain a consistent carburization on the surface of a workpiece. For this reason, small amounts of nickel are valuable in alloys which will be case hardened.
Nickel preferentially dissolves into the ferritic crystals of certain steel phases, therefore it can potentially increase the hardness of a steel without harming its ductility.


What does a fractional order of reaction mean for the mechanism?

Here's an example.

Let's consider a reaction of the generic form:

$$\ce{3A -> 2C}$$

Now, let's say we have some knowledge that this reaction proceeds by the following two-step mechanism:

  Step one is fast and reversible.

$$\ce{A <=>[k_1][k_{-1}] 2B}$$

  Step 2 is slow and irreversible.

$$\ce{A + B ->[k_2] C}$$

The rate of the reaction (expressed as the change in concentration of the product per unit time) can be expressed in terms of the concentrations of A and B.

$$\mathrm{rate}=\dfrac{\Delta [\ce{C}]}{\Delta t}=k_2[\ce{A}][\ce{B}]$$

However, we would prefer to not use intermediates in our rate law. Since the first step is fast and reversible, we can use the law of mass action to create a relationship between the concentrations of A and B:

$$K_1=\dfrac{k_{1}}{k_{-1}}=\dfrac{[\ce{B}]^2}{[\ce{A}]}$$
$$[\ce{B}]^2=K_1[\ce{A}]$$
$$[\ce{B}]=\sqrt{K_1[\ce{A}]}$$

If we substitute into the  rate law:

$$\mathrm{rate}=\dfrac{\Delta [\ce{C}]}{\Delta t}=k_2[\ce{A}]\sqrt{K_1[\ce{A}]} =k_2 K_1^{1/2}[\ce{A}]^{3/2}$$

The term $k_2 K_1^{1/2} =\dfrac{k_2 k_1^{1/2}}{k_{-1}^{1/2}}$ becomes $k_{\mathrm{obs}}$, the observed rate constant of the overall reaction. The fractional order comes from the second step being rate-determining, but the first step being an equilibrium with stoichiometry that generates the fractional tern.


Dissolving Tylenol Science Experiment - why is it fastest in distilled water?

Water can only dissolve a certain amount of solids at a given temperature before it reaches saturation. Distilled water has been boiled and captured in gas phase, almost completely eliminating the presence of dissolved solids.  This lack increases the number of ' water to solid' molecular interactions that will take place and so speeds up the disolving.  The same would be true of any solute.


Most probable point for finding an electron in the 1s orbital of a Hydrogen atom

It makes a difference if you look in a point or in a volume element.

For the radial distribution we sum up the probabilities within (very thin) shells at different distances from the core.
If the radius get's bigger the volume in this shell gets bigger.
Remember that the surface of a sphere goes by $4 \cdot \pi \cdot r^2$ so the volume of the shell is essentially for all practical purposes $4 \cdot \pi \cdot r^2 \cdot \Delta r$.
At the core the radius is zero which means the volume is zero and thus the probability is zero too.

At the same time the probability of finding the electron in a point at a certain distance goes down, but the volume goes up.
Because of the scaling of the volume and the "decay" of the probability to find it in a point further away it turns out that there's a maximum of finding the electron at a certain distance away from the core (but not a single point at that distance).


What does HClO2 aqueous solution decompose to in an open environment?

Per Wikipedia on Chlorous acid to quote:

  The pure substance is unstable, disproportionating to hypochlorous acid (Cl oxidation state +1) and chloric acid (Cl oxidation state +5):

  $\ce{2 HClO2 → HClO + HClO3}$

Also, here are some interesting comments on the decomposition of chlorous acid in "Kinetics and Mechanism of the Decomposition of Chlorous Acid" from J. Phys. Chem. A 2003, 107, pages 6966-6973, to quote:

  Of the many mechanistic models tested, the one that fit best included the following reactive intermediates: HOCl, Cl2O2, Cl2O3, •ClO, •OH. The stoichiometric ratio of ClO2 produced to Cl(III) consumed varies with pH and [Cl-]. Reaction of Cl2O3 with Cl(III) yields chlorate exclusively. Reaction of Cl2O3 with Cl- favors ClO2 over chlorate, but does not entirely exclude chlorate, because it is produced by hydrolysis of Cl2O2. Invoking Cl2O3 explains the variation in stoichiometric ratio as well as the maximum observed in the initial rate of ClO2 formation as a function of pH. The kinetics of chlorous acid decomposition cannot be quantitatively fit through the last stages of the reaction without postulating a first-order decomposition. Scission of chlorous acid to give short-lived hydroxyl and chlorine-(II) monoxide is a plausible route for this process [...]

  Several groups of investigators[5-7] have found
  that in the absence of chloride ion the stoichiometry of the
  decomposition of chlorous acid is given by reaction A:

  $\ce{4 HClO2 -> 2 ClO2 + ClO3- + Cl- + 2 H+ + H2O}$ (A)

  The stoichiometry of the decomposition of chlorous acid in the
  presence of chloride ion is given by reaction B:

  $\ce{5 HClO2 -> 4 ClO2 + Cl- + H+ + 2 H2O}$  (B)

Also, to quote:

  Earlier studies,[9,18] in agreement with our present results, have also found the formation of more chlorate than predicted from reaction A. Reaction C

  $\ce{3 HClO2 -> 2 ClO3- + Cl- + 3 H+}$  (C)

  also plays a role in determining the stoichiometry at higher $\ce{HClO2}$ concentrations."

So, several intermediate products and depending on chloride presence possibly $\ce{ClO2}$, which is a problematic explosive and toxic gas, along with the strength of the $\ce{HClO2}$, which can introduce chlorate at higher chlorous acid concentrations.

Not a particularly good path to an acid as you are also inquiring about. As a disinfectant, the transient creation of the powerful disinfecting HOCl and associated radicals, may actually make it weaker, albeit more stable (considering end products) than hypochlorous acid, in my opinion.

[EDIT] To answer a comment question, as to whether is ClO2 safe, here is  a statement from the CDC, Public Health Statement for Chlorine Dioxide and Chlorite, to quote:

  If you are exposed to chlorine dioxide or chlorite, many factors will determine whether you will be harmed. These factors include the dose (how much), the duration (how long), and how you come in contact with them. You must also consider any other chemicals you are exposed to and your age, sex, diet, family traits, lifestyle, and state of health.

Interestingly, ClO2, which is a stable free radical, has found commercial application as an odor removal agent, likely due to its interaction with organics to create volatile organic chlorides (I would suspect that the presence of light would be catalytic). These VOCs are much more of a long-term health danger (as in carcinogenic), in my opinion.


Short answer: one should not consider the absolute value of a potential as long as the potential difference drives the ions in the appropriate direction.

Longer answer:
The potential difference is needed to accelerate ions in the appropriate direction of the mass spectrometer.

One should consider that in an atmospheric pressure or moderate vacuum ions need to be driven by the electric potential difference towards the MS high vacuum region by a constant decrease in potential (due to loss of kinetic energy through collisions with the gas) whereas in the high vacuum region, there is no need to keep this evolution true as long as the potential difference remains positive relative to the initial (entrance, near thermal kinetic energy) potential.

Just as an illustration, considering a simplified ESI atmospheric pressure introduction with only an ESI capillary, a cone to rough pumping region and a skimmer to the UHV region, one can have:

ESI Needle: + 4,000 V
Cone: + 100 V
Entrance skimmer: 0 V

Or

ESI Needle: + 4,200 V
Cone: -100 V
Entrance skimmer: -200V

The ions will see the same electric field in both cases, as it is the potential difference which creates the electric field and not the absolute value of the potential. There is just one issue: sometimes it is easier in instrument design to have one region of the instrument operated at ground potential (or close to ground potential), and it would be this part of the design which will decide on the absolute values of the potentials that will be applied.


What is the balanced equation for the reaction of potassium permanganate, iron sulfate, and sulfuric acid?

Your first half reaction, for the reduction, is correct:
$$\ce{10 e- + 10 H+ + 2 KMnO4 + 3 H2SO4 -> K2SO4 + 2 MnSO4 + 8 H2O}\tag1\label{red}$$

For the second half reaction, the oxidation, start by balancing iron:
\begin{align}
\ce{FeSO4 &-> Fe2(SO4)3 }\tag{2a}\\
\ce{2 FeSO4 &-> Fe2(SO4)3 }\tag{2b}
\end{align}

Add $\ce{H2SO4}$ on the left so you can balance sulfur:
\begin{align}
\ce{2 FeSO4 &-> Fe2(SO4)3 }\tag{2c}\\
\ce{H2SO4 + 2 FeSO4 &-> Fe2(SO4)3 }\tag{2d}
\end{align}

Now balance the protons and electrons:
\begin{align}
\ce{H2SO4 + 2 FeSO4 &-> Fe2(SO4)3 }\tag{2d}\\
\ce{H2SO4 + 2 FeSO4 &-> Fe2(SO4)3 + 2 H+ + 2 e-}\tag{2e}\label{ox}\\
\end{align}

Now add \eqref{red} and five times \eqref{ox} so that the electrons are equal on every side and coincidentally the protons also balance:
\begin{align}
\ce{10 e- + 10 H+ + 2 KMnO4 + 3 H2SO4 &-> K2SO4 + 2 MnSO4 + 8 H2O}\tag1\\
\ce{5 H2SO4 + 10 FeSO4 &-> 5 Fe2(SO4)3 + 10 H+ + 10 e-}\tag{$5\times$2e}
\end{align}

And the final result is:
\begin{align}
\ce{2 KMnO4 + 8 H2SO4 + 10 FeSO4 &-> K2SO4 + 2 MnSO4 + 8 H2O + 5 Fe2(SO4)3}\tag3
\end{align}


Recoil velocity of atom when electron is excited

There are three things to conserve during an electronic transition: energy, momentum, and angular momentum. As always, these things are related to some extent. Let's forget about the angular momentum because it isn't relevant to this question.

The energy of a photon is given by,$$E_\gamma=h\nu.$$This energy is related to the momentum of a photon through the mass-energy equivalence,$$E^2=m^2c^4+p^2c^2.$$ Remembering that the mass of a photon is zero, we have,$$P_\gamma=\frac{E_\gamma}{c}.$$

The energy of an electronic transition between two states in a hydrogen atom must equal the energy of the photon being absorbed, which conserves the energy upon either absorption or emission. Thus, we have,$$E_\gamma=\frac{m_ee^4}{8h^2\epsilon_0^2}\left (\frac{1}{n_1^2}-\frac{1}{n_2^2}\right ),$$where we have used the solution to the energy states of a hydrogen atom coming from using a Coulomb potential to describe the potential energy of the system.

Then, we impose the conservation of momentum condition that $P_\gamma=P_{atom,f}-P_{atom,i}$. When I write $P_{atom}$ we have to note that this is given by the sum of momenta of the nucleus and electron.

Let's define the recoil velocity as, $v_{recoil}=v_f-v_i$

I don't wanna deal with the special relativistic correction to the momentum, so I'll leave that to you if you wanna figure it out.

Now, we can write,$$P_{recoil}=\frac{E_\gamma}{c},$$which can be written in terms of the recoil velocity according to,$$v_{recoil}=\frac{E_\gamma}{(m_e+m_p)c}.$$In the last expression, $m_p$ is the mass of a proton (i.e. the mass of the nucleus).

The expression would get uglier if you used $P=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$ instead of $P=mv$.

Using the above expression, the recoil velocity ends up being,$$v_{recoil}=\frac{10.2\cdot10^{-5}MeV}{938.272\frac{MeV}{c}}=3.26\ \frac{m}{s}.$$

So that's a reasonable velocity I think.

It's totally possible I made a mistake somewhere because I've never thought about this question before, but I think that is all correct. It's a little bit weird to think about whether the momentum is being transferred to the electron or the nucleus, but it seems like it must be transferred to both somehow?

One way to check the work (which I don't wanna do) is to try to get the same expression by only balancing energies and using $\frac{1}{2}mv^2$.


Significant figures used to imply stated error?

The use of significant figures always implies some error, that is not explicitly stated. It is therefore irrelevant how big this error is. I merely states that the last digit is rounded.

In the first case the quantity is given as $2.3\pm0.001$, which gives you the interval $(2.301;2.999)$, which is expressed as $2.300$ in significant figures.

The same applies to the second quantity, which is given as $1.989\pm0.0003$, which in turn gives you the interval $(1.9893;1.9887)$, which is expressed as $1.9890$.


Show that there can be 18 electrons in a "g" subshell

Simple mnemonic: each orbital, s, p, d and f, has double the consecutive odd numbers of electrons, i.e. 1, 3, 5 & 7 --> 2, 6, 10 & 14.

BTW, from spectroscopy, s, p, d & f refer to splitting of lines, the sharp, principal, diffuse and fine; see Atomic orbital. As Tom Lehrer stated, "These are the only ones of which the news has come to Hahvahd, and there may be many others, but they haven't been discovahd." Although element 121, unbiunium, begins the theoretical g-block...


Ok, there are two reactions for the 2 moles of A

$$\ce{A ->[rate\space =\space k_1] 2B}$$

$$\ce{A ->[rate\space =\space k_2] 2C}$$

$$\dfrac{k_1}{k_2} = \dfrac{1}{2}$$

The question asks for the total number of moles of A+B+C when the reaction is 75% complete.

At that point 25% of A remains or 0.5 moles of A
It does matter what the ratio of B and C produced is since either reaction of A creates 2 moles of product.  So there is 2*1.5 = 3 moles of product (B+C).

Thus there is 0.5 + 3 = 3.5 moles of A+B+C in the solution when the reaction is 75% complete.

There are 4 moles when the reaction is 100% complete.

So the book answer is wrong.


According to this post at www.reference.com/science

  Boiling chips are generally porous, trapping heating bubbles and
  allowing solvent vapors to form in the air cavities. They should
  always be added to a solvent prior to the beginning of a heating
  cycle; adding to a hot liquid may cause it to violently boil over.
  Boiling chips and stones are only used once, as the absorptive
  material becomes impregnated with the first solvent used in.

And Wikipedia confirms the basic idea of why they shouldn't be used more than once:

  The structure of a boiling chip traps liquid while in use, meaning
  that they cannot be re-used in laboratory setups.

From personal experience however, I have to disagree with these absolute "cannot be done" sort of statements.  I used to rinse the used chips in an appropriate solvent and then heat them at some temperature well above the solvent's boiling point overnight and don't remember ever having problems.

I realize that the boiling point isn't completely relevant when you can have porous, high surface area material with adsorbed solvent.  Also, as I stated in the comments, I don't remember using PTFE coated chips.