Microwaving a glass of water, what happens? Most of us here would already know the simplified idea behind microwaving food: Microwave radiation hits the water molecules present in food, which excites them and causes 'em to vibrate rapidly in situ resulting in the heating up of food. Very straightforward.

But what if you decide to microwave a glass of water?

I can't seem to find any reliable literature that deals with this. A lot of highly unreliable sites spout nonsense from "The water becomes poisonous" to "The water becomes radioactive", so I've immediately dismissed those. However, some sources suggest that the "water will explode".

Now I was under the impression that the water would perhaps come to a boil a bit faster than it would by simply lighting a fire underneath it.

So after reading the last few sources it wasn't really difficult to believe that an explosion might occur. At first I simply extrapolated my theory of 'rapid boiling' to the theory of 'Very rapid boiling' and was content with that. It made sense to me since I presumed that, just as in the conventional method of boiling water, bubbles of water would be formed at the bottom of the glass albeit at a much faster rate, build up, rise, thus resulting in an explosion. But after some reflection, I noticed a problem with this 'extrapolation'.

Why do the bubbles even form in the first place when you conventionally heat water? Simply lighting a fire below a beaker of water does not ensure that the entire content o the beaker is uniformly heated at the same time. Obviously the layer of water molecules in direct contact with the bottom of the beaker will get heated rather quickly. They rise up, colder water descends, etc etc, you know the deal: convection. It's the sizable temperate difference between the bottom region and the rest of the water, that enables the water at the bottom to get vaporized at some point forming bubbles while the water immediately above it is still in the liquid state.

The problem I see that prevents me from extending this theory to microwaving water is that the mode of heating is different in the case of the microwave: Radiation

Now since radiation is faster at heating a given volume than convection is; for all practical purposes, I think it'd be safe to accept that the water in the beaker will be heated uniformly (The beaker's pretty normal sized...it's dimensions aren't of an order of magnitude greater than that of the speed of light/electromagnetic radiation, so it'll be alright to take the apparatus as being 'uniformly heated'. Had to say this before someone inevitably points out in the comments that all the water being heated to constant temperatures like this at the exact same time is impossible).

So if the beaker is heated uniformly, then I see no reason for bubbles to form...in other words, I don't see an explosion coming.

So what really happens if you microwave a glass of water?

Will it quickly and steadily vaporize? Will it explode? Or does something else happen? And why does it happen?

Heating water on a hot plate is safe, because the hottest point is at the bottom of the pot. A lot of relatively small bubbles appear there without much overheating of the water, because there is a lot of nucleation at the uneven phase boundary steel-water.

In a microwave, the hottest place is IN the water. The glass does not get heated by microwave (at least not much), and radiates off some heat to the surrounding.

Problem: In clean water, there are few good nucleation points to form bubbles, only some dust particles perhaps. So the water gets overheated rather strongly, and a first bubble that appears can grow a lot before it has cooled its surroundings down to 100 °C. That one huge bubble can throw most of the water out of the glass. It boils over, violently.

Btw. a microwave does not heat its content uniformly. It forms a standing electromagnetic wave (that's not radiation, strictly speaking) in the oven, like a rope swung quickly between two people, or a guitar string. The wave pattern has knots at a distance of $0.5c/f\approx6\ \mathrm{cm}$ (with $f=2.45\ \mathrm{GHz}$ and speed of light $c$ which of course is a bit less in your chicken), where there is very little heating. That's why the microwave oven has the rotating plate, to generate some uniform heating. For pure water, that doesn't matter, because convection sets in anyway and distributes the heat.


Question about liquid-phase in P-T Diagram Currently, I have been learning about P-T diagrams and vapour pressure and I have a question that seems like it should have a simple explanation yet I am not able to understand it.

Taking the P-T diagram of water for example,

Suppose we choose a point that corresponds to the liquid region in the P-T diagram. From what I know of vapour pressure, every pure liquid in a closed container has a vapour pressure and will eventually reach a liquid-vapour equilibrium. If this is the case, what is the meaning of a "liquid region" in a P-T diagram if there must always exist a vapour-liquid equilibrium? And by extension why is there even a coexistence curve between liquid and vapour in the first place?

The gas phase is unique in its ability to expand to fill space, so if there is space above or around a solid or liquid, molecules will eventually enter the gas phase in order to fill that space. That's the vapor-liquid equilibrium you described.

The liquid region of the phase diagram is the region where, in a closed space, some of the material will remain indefinitely in the liquid phase, even though some is also in the vapor phase. For example, in a sealed but expandable container (like a piston) maintained at an external pressure of 1 atm, the equilibrium state at 80 C has some water and some water vapor. But at 110 C, it will be all vapor.

The "coexistence curve" is a misnomer for the curve you are referring to. It does not mean that the liquid and vapor simply coexist (which as you pointed out can happen throughout the liquid region). It means that the equilibrium vapor pressure is equal to the external pressure. At that point, you cannot increase the temp of the liquid any further if the pressure remains constant. All heat added will go towards converting liquid to gas. Only after it is all gas will the temp increase. In order to get liquid at a higher temp, you have to increase the pressure. We define this as the boiling point.


How Fittig reaction gives good yield if phenyl cations are unstable? I am confused,after reading from NCERT text that phenyl cations are unstable due several reasons like the carbon in phenyl group is sp2 hybridized,resonance structures do not stabilizes the phenyl cation.But then,in Fittig reaction we do have phenyl cations as intermediate.For example in a reaction between two molecules of chloro benzene and sodium metal we get a diphenyl molecule.Here we have two phenyl cation generated (if i am right).Then what factor stabilizes the phenyl cation?If there is any error in the theoretical aspect of the question please help me correct it as i am not quite sure about the mechanism.

I would be very much glad to accept an answer with some thermodynamical data like the bond enthalpies of C-Cl bond and newly formed product's bond enthalpy that is C-C bond enthalpy(in the case of diphenyl)and the total realease in energy.

The most probable mechanism for Wurtz reaction or Fittig reaction involves formation of organic free radicals and not cations. Like in Wurtz reaction, phenyl free radicals are formed in Fittig too.


Why does the local density approximation (LDA) overestimate atomization energy? I have read numerous times that the local density approximation (LDA) overestimates atomization energy. For example, here is a quote from Dr. Burke's book:

  LDA typically overbinds molecules by about $30~\mathrm{kcal\, mol^{-1}}$.

What is the cause? Is it the same problem that Hartree-Fock theory has when compared to the exact Schrödinger solution (lack of correlation between electrons)? Also, how can inclusion of gradient can help solve this problem?

What is the cause?

The LDA exchange hole is spherically symmetric and centered on a reference electron, which is a fairly good approximation for the case of molecules, but not for the case of atoms. For instance, in an atom the exact exchange hole is displaced toward the nucleus, while the LDA exchange hole is always centered on its reference electron. This inability of LDA exchange hole to appropriately describe the situation in atoms leads to significant deviations in energies upon bond formation or breaking.

  Also, how can inclusion of gradient can help solve this problem?

In general, the electron density in atoms is more inhomogeneous than in molecules, and GGA is ntroduced exactly in order to account for the non-homogeneity. In particular, the gradients in the separated atoms are usually bigger than the gradients in the corresponding molecule, and therefore gradient corrections lower the energy of the atoms more than the energy of the molecule, which leads to a reduction of the overbinding.

See, e.g. Ernzerhof, M., Perdew, J. P. and Burke, K. (1997), Coupling-constant dependence of atomization energies. Int. J. Quantum Chem., 64: 285–295. DOI: 10.1002/(SICI)1097-461X(1997)64:3<285::AID-QUA2>3.0.CO;2-S


What is a suitable acid to remove limewash from masonry brick Diluted Hydrochloric acid is often touted as the only thing to remove limewash from masonry but the process is dangerous and can adversely affect the bricks, color changes can readily occur.

Is there a viable alternative? I noticed that a kettle descaler product used formic acid and wondered if that could be used instead.

Glycolic acid (hydroxyacetic, HOCH2CO2H) is a mild organic acid, not very expensive, and the calcium salt is quite soluble. It is essentially a non-smelly vinegar! It has been used for brick washing.

One issue is that whereas HCl will dissolve CaCO3 until the HCl is all gone, the pH of glycolic acid solution increases as it dissolves CaCO3, so the dissolution reaction slows down and finally stops at maybe 50% completion.

Citric acid and tartaric acid are very soluble, but their calcium salts are not. Calcium citrate and tartrate may come down as fine crystals in the pores of the brick and be difficult to remove. Calcium formate is moderately soluble and so would be useful, but formic acid is a vesicant, so it must be used with as much or more care than HCl.

Muriatic acid is readily available at 32% and 20% concentrations. Diluting by a factor of 50 would give a pH about 2, which is safer for people and brick, but that's a lot of water to clean a brick wall.


Treatment of 2-vinylbicyclo[2.2.2]oct-5-en-2-ol with potassium hydride What happens when this bicyclic compound is treated with KH in THF?

Does $\ce{KH}$ simply act as a base here? $\ce{H-}$ from $\ce{KH}$ could potentially act as a nucleophile, but I don't find any electrophilic centre in the compound. What happens?

The reaction in question is one of the early examples of an oxy-Cope rearrangement [1] in which a 3-hydroxy-hexadiene undergoes a [3,3]-sigmatropic rearrangement to afford a ketone.

The general mechanism is shown below:

Source: Organic Chemistry, Oxford University Press, 2 ed. Clayden, Warren and Greeves

In a general sense, the reaction is a [3,3]-sigmatropic rearrangement (if you look at the curly arrows, you'll see that we're essentially transposing a sigma bond from one end of the pi system to another).

The presence of the oxygen provides a driving force for the reaction, since the enol(ate) formed during the rearrangement is able to tautomerise, yielding the ketone product.

In the specific case you've given, the presence of a base means that during the sigmatropic rearrangement, the oxygen is deprotonated and therefore anionic. This specific case, using an alkoxide, is a specific kind of reaction known as an anionic oxy-Cope. The anion is thought to help speed up the reaction, to the extent that minimal heat is usually required.

References:

[1] J. Am. Chem. Soc., 1964, 86,5019.  DOI: 10.1021/ja01076a067


Carbon decrease from 1-octyne to 1-hexene I am to turn 1-octyne to 1-hexene.

Method 1

1) Reduction using sodium metal in pure ammonia to an alkene.

2) Halohydrin formation; the -OH installs at the more substituted carbon due to its greater ability to stabilize positive charge (this always confused me; the more substituted carbon is both more positive in character and has more electron-donating methyl groups?)

3) Conversion of OH to a good leaving group (-OTs) using a sulfonate ester and a weak base, pyridine.

4) E2 elimination using NaOH and heat. Because of resonance and sheer bulk and negative induction the OTs(-) should make for a better leaving group than Br-. So the internal alkene is formed.

5) Two-part one step ozonolysis; creation of aldehyde.

6) Reduction of carbonyl with hydrogen and transition metal to terminal alcohol.

7) Conversion of terminal alcohol to a better leaving group using tosyl chloride and pyridine again.

8) Use of LDA to deprotonate and create the terminal alkene.

Method 2

Also we can take another route with the brominated alkene; we can use OsO4 and basic hydrogen peroxide to create a 1,2 (vic) diol, which can be cleaved by periodic acid, and this creates a primary alcohol, which we convert to a good leaving group using tosyl chloride and pyridine, and we can eliminate. Same number of steps, but just something different that I thought of. Well, perhaps slightly simpler since there isn't a need for a two-step process (ozonolysis).

In principle, this can be realized in two steps:

Oxidative cleavage using catalytic amounts of $\ce{RuO2}$ and excess of oxone converts 1-octyne to heptanoic acid. (DOI)
Heating a mixture of acetic anhydride and a carboxylic acid in the presence of $\ce{Pd}$ catalyst turns the acid to a terminal alkene lacking one carbon atom (DOI). Admittedly, the high temperatures needed might be a problem here and may force you to perform the reaction in a sealed tube in an autoclave.


How to convert an organic carbonate to its formate ester? I'm currently working on a chemistry project in which I need to convert ethyl methyl carbonate to ethyl formate.

I only have a simple background in organic chemistry (and I haven't seen anything about reaction mechanisms yet), but I was guessing water could be used in this reaction, to produce $\ce{CH3OH}$ and replace the $\ce{-OCH3}$ group by an $\ce{-OH}$ group. Then the next step would be to remove the oxygen from the  $\ce{-OH}$ group. I've seen that compounds such as $\ce{LiAlH4}$ can do this, but I'm looking for a catalyst to make the process more economically viable, in case the proposed synthetic path is correct.

Does anyone know how to perform the conversion, or even by following another synthetic path? I've got no idea whether this process is correct. I'm also trying to keep the process as low energy-consuming as possible.

I agree it would be better to start from ethanol and formic acid, but I have to start with this molecule.

First use $\ce{EtO-/EtOH}$, then lithium-tri-tert-butoxy-aluminium hydride. This will convert it to diethyl carbonate and as the reducing agent is hindered you can manipulate reaction conditions to get the desired product.

Or another way is to hydrolyse it in basic medium to get $\ce{H2CO3}$ and then evaporate $\ce{H2O}$ and $\ce{CO2}$ then after separating it will open a large number if ways as you have got ethyl alcohol and methyl alcohol.


Is reforming and isomerization the same thing? Okay, just to give an idea of my current understanding. Reforming is a step one can do after one has cracked some hydrocarbons. As such, the process of reforming is dependent on a separate process, cracking. What it involves is cracking the hydrocarbons, then putting them together in new ways, whilst still having the same chemical formula. In a sense, one is reversing the work done by the cracking, since that is a process that changes the chemical formula, however, when one puts the molecules back together, one changes them structurally. Basically, one is making isomers.
However, I do wonder if reformation also includes the saturation of hydrocarbons; as in, making alkenes and alkynes into alkanes. This obviously requires cracking, since double and triple bonds must be broken down into single bonds. If this is a part of reformation, that would mean the term applies to a wider process than just making isomers, since an alkane cannot be an isomer of an alkene/alkyne.
Perhaps isomerization is the creation of isomers, but within any field of chemistry, whereas reforming applies specifically to petrochemistry?

TL;DR: Isomerization is a subpart of reforming process. There are four major reforming reactions. Isomerization is one of them.

Reforming is a processing technique by which the molecular structure of a hydrocarbon is rearranged to alter its properties. The process is frequently applied to low-quality gasoline stocks to improve their octane number and thereby improving combustion.
There are four major types of reforming process:

Dehydrogenation of naphthenes

Dehydrogenation and aromatization of paraffins to aromatics (dehydrocyclization)

Hydrocracking of paraffins

Isomerization of normal paraffins to isoparaffins

You can see that during all the process except isomerization, there is a loss of atoms either carbon or hydrogen. However, there is no change of atoms during isomerization process.


Euler's Theorem and Extensive properties I read in a textbook (Mathematics for Physical Chemistry by D. McQuarrie) an example of Euler's Theorem used in Thermodynamics. The author states "The [variables] are all extensive [quantities], and so [Euler's theorem is applicable]" (I'll provide more details below)

My question: why do variables being extensive indicate that this theorem is relevant?

The theorem starts by stating that a function is homogeneous to degree $N$ in some set of variables if those variables always form terms such that their powers sum to $N$. For example:

$f(x,y) = yx^2 + y^2x + \frac{y^4}{x} + x^3$

that would be homogeneous degree 3 for $x$ and $y$.

The theorem concludes with:

If $f(\lambda x,\lambda y) = \lambda^N f(x,y)$

Then $Nf(x,y) = x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y}$

This
is the wikipedia page for the same theorem.

The example uses the above theorem to derive:

$U = S(\frac{\partial U}{\partial S})_{_{V,n}}+V(\frac{\partial U}{\partial V})_{_{S,n}}+n(\frac{\partial U}{\partial n})_{_{S,V}}$

The author's solution immediately states: "The entropy, volume, and number of moles are all extensive thermodynamic quantities, and so $U(\lambda S,\lambda V,\lambda n) = \lambda U(S,V,n)$." After that statement the solution is simple.

I know that extensive variables means that they scale with the size of the system, but the relation to this theorem is unclear to me. I suppose that if the system is "twice as big" ($\lambda$ is doubled), then energy, entropy, volume, and amount are doubled. That still seems vague to me if its even a good way to think about this.

Thank you.

It really is that simple. Stating that a thermodynamic system observes Euler's Theorem can be considered axiomatic if the geometry of the system is Cartesian: it reflects how extensive variables of the system scale with size. In a homogeneous system with a Cartesian geometry all extensive properties scale identically and so any extensive property can be written as a homogeneous function of degree 1 of the remaining properties.


Why are protons in the same position in 1-chloro-4-nitrobenzene magnetically different? Protons in ortho positions towards the nitro group are said to be magnetically different, even though they're chemically equivalent. Same goes for protons ortho towards the chlorine. Why is that so?

The hunch lies in the definition of magnetic equivalence:

  For two protons to be magnetically equivalent they not only have to have the same chemical shift, but they must also each have the same $J$ coupling to other magnetic nuclei in the molecule.

(Quoted from Hans Reich’s NMR course online syllabus; emphasis present in the original.)

What does that mean for our specific example? Compare the molecule we have in figure 1 below.

Figure 1: 1-chloro-4-nitrobenzene with labelled protons.

Needless to say as you already figured that $\ce{H^1}$ and $\ce{H^3}$ are chemically equivalent (one is transformed into the other by $C_2$‡) meaning that they will display the same chemical shift; likewise for $\ce{H^2}$ and $\ce{H^4}$.

However, for magnetic equivalence we need to consider their coupling constants to each other. If $\ce{H^1}$ and $\ce{H^3}$ are magnetically equivalent, then $J_{\ce{H^1 H^2}} \overset{!}{=} J_{\ce{H^3 H^2}}$. But, $J_{\ce{H^1 H^2}}$ is an ortho coupling with a value of $J \approx 7~\mathrm{Hz}$ while $J_{\ce{H^3 H^2}}$ is a para coupling with $J \approx 0~\mathrm{Hz}$.

Thus we can easily see that $J_{\ce{H^1 H^2}} \ne J_{\ce{H^3 H^2}}$ and therefore the nuclei $\ce{H^1}$ and $\ce{H^3}$ are magnetically non-equivalent.

The same argument can be made for $\ce{H^2}$ and $\ce{H^4}$.

‡: They are also transformed into each other by the $\sigma_{\mathrm{v}1}$ plane. However, planes of symmetry create enantiotopic protons while rotations create isotopic ones. In this case — a non-chiral environment and a non-chiral solvent — enantiotopic protons cannot be distinguished and give a common signal, and thus are sometimes incorrectly called chemically equivalent. The argument is not of relevance here though, since the $C_2$ axis renders the protons isotopic thus equivalent.


What is the structure of 2-methylimidazolate? Looking for the structure of 2-methylimidazolate, I find conflicting results:

this page on sodium 2-methylimidazolate gives:

this other page on cadmium 2-methylimidazolate gives another anion, namely a carboxylate derivative of 2-methylimidazole:

Which is it? Where in IUPAC nomenclature rules can I get the definitive answer?

As far as I understand the IUPAC nomenclature, 2-methylimidazolate would not be the correct name for either structure.

The term imidazolate refers to the anion of imidazole, though it is not really the correct suffix according to the proposed revision of the IUPAC blue book (P72.1):

  Method (1) leads to preferred IUPAC names. Some names and some
  contracted names are retained as preferred IUPAC names and for use in
  general nomenclature. The following suffixes are used:

  ‘ide’ (corresponding to removal of a hydron, H+),
  ‘uide’ (corresponding to the addition of a hydride, H−),
  ‘elide’ (corresponding to the addition of an electron)

  The endings ‘ate’ and ‘ite’ are used to indicate removal of a proton
  from a −OH group of acids and hydroxy compounds.

The correct term for the imidazole anion would be imidazolide, and not imidazolate as imidazol is neither an acid nor a hydroxy compound.

In both database entries the IUPAC name mentioned is not imidazolate, this seems to be a trivial name for those structures. So my suspicion would be that both have been called imidazolate at some point, but this does not seem to be the correct IUPAC nomenclature for either of them.


Filtration methods at removing ions present in water I'm doing a lab where I have to first find out what ions are present in the water (easy part). The next part is to filter out those ions from the water. Here are some methods that are stated on the paper: Coffee filter, Your own design, Britta filter. I've done my research and just can't seem to find what ions coffee filters and Brita filters filter out.

My question is: What are some own designs that i can create to filter out some ions ($\ce{Ca^{2+}}$, $\ce{Pb^{2+}}$, $\ce{Ba^{2+}}$, $\ce{Zn^{2+}}$, $\ce{SO4^{2-}}$, $\ce{PO4^{3-}}$, $\ce{Cl-}$, and $\ce{OH-}$).
Or do those filters actually filter any of those ions?

Cations and anions of water-soluble salts cannot be removed from water using only mechanical filters (which rely on particle size) such as coffee filters.

In order to remove such ions, ion exchange resins can be used. These resins consist of an insoluble matrix, e.g. polystyrene that is crosslinked with divinylbenzene to vary the porosity of the resin. To convert the resin into an ion exchanger, the polymer structure is functionalized with many functional groups. Typical functional groups are

sulfonate ($\ce{R-SO3-}$) for strong acid cation exchange resins,
quarternary ammonium (e.g. $\ce{R-N(CH3)3+}$) for strong base anion exchange resins
tertiary amine (e.g. $\ce{R-N(CH3)2}$) for weak base anion exchange resins

The counterion of new cation exchange resins used for water treatment is usually $\ce{H+}$ or $\ce{Na+}$ (other forms can be used for special applications, e.g. $\ce{Li+}$ in pressurized water reactors), i.e. $\ce{R-SO3- H+}$ or $\ce{R-SO3- Na+}$. These counterions can be replaced by another cation from the water. Thus, a cation exchange resin can remove various cations from the water and replace them with an equivalent amount of $\ce{H+}$ and $\ce{Na+}$, respectively, e.g.
$$\ce{2 (R-SO3- H+) + Ca^2+ <=> (2 R-SO3- Ca^2+) + 2 H+}$$

The counterion of new anion exchange resins is usually $\ce{OH-}$ or $\ce{Cl-}$. Thus, an anion exchange resin can remove various anions from the water and replace them with an equivalent amount of $\ce{OH-}$ and $\ce{Cl-}$, respectively.

In order to remove both kinds of ions, the water has to be passed through a cation exchanger and through an anion exchanger. Alternatively, a mixed-bed ion exchanger (which contains cation exchange resins and anion exchange resins) can be used. They are even more effective but cannot be easily regenerated.

Typical Brita water filter cartridges contain ion exchange resin beads to remove undesired ions and activated carbon to absorb other undesired compounds.


Has it been observed that the dihydrogen molecule dissociates into ions? As we know from the improved methods valence bond theory (VB), as well as molecular orbital theory (MO-LCAO), the wave function of the $\ce{H2}$ molecule consist of two terms, which belong to covalent and ion structure, respectively:
$$\psi =\psi_\mathrm{cov} + \lambda\psi_\mathrm{ion}.$$

According to quantum mechanics, there is a nonzero probability $\propto\lambda^2$ for the ionic state wave-function term. Such reaction should lead to the $\ce{H2}$ molecule dissociation products $\ce{H+}$ (proton) and anion $\ce{H-}$:
$$\ce{H2 -> H+ + H-}.$$

Have such processes been observed in nature?

Surely1. At the end of the day, it is just the matter of supplying enough energy: the homolysis requires approximately 4.5 eV per hydrogen molecule, while for heterolysis more than 17 eV per molecule must be provided.

1) G. Busca, A. Vaccari, Heterolytic dissociation of hydrogen on high-temperature methanol synthesis catalysts, Journal of Catalysis, 108(2), 1987, 491-494, http://dx.doi.org/10.1016/0021-9517(87)90198-9.


What makes a good antioxidant? I am doing a school project on antioxidants and I would like to find out what makes a good antioxidant. I know examples of antioxidants such as Vitamin A, C and E but I was wondering what commonality exists between them in terms of chemical structure and bonding.

I am a novice in chemistry so a simple but informative answer would be appreciated (little to know nomenclature).

Antioxidants are molecules that help pair up electrons in the cell.

The way I see it, antioxidation in the body is all about getting electrons to pair up. dioxygen, $\ce{O2}$ is a molecule with an overall even number of electrons (12), but the two highest energy electrons are in different molecular orbitals, so it is actually a diradical. Species with unpaired electrons, like ground state oxygen, cannot react with the body's even electron count organic molecules. This is why the body must employ metals, which have unpaired electrons in order to reduce oxygen to water to form cellular energy in the mitochondria. On its way to being reduced to water, oxygen can form reactive species which can react with organic biomolecules, like $\ce{O2-}$, $\ce{HO-OH}$, and $\ce{HO.}$ These are known as Reactive Oxygen Species (ROS).

ROS cause oxidative stress, which is a process that damages important biomolecules like lipids and nucleic acids through oxidation. Unstable ROS like peroxides can initiate radical chain reactions, leading to oxidation of important molecules in the cell. Two good examples of ROS damage are lipid peroxidation, which damages the cellular membrane and nucleobase oxidation, which leads to DNA damage. I will show mechanistic examples for both:

Lipid peroxidation:

$\hskip1.8in$ (1)

formation of 8-oxoguanine:

$\hskip0.9in$ (2)

This creates a hydrogen bonding situation where guanine can match with adenine instead of cysteine:

$\hskip1.4in$ (3)

This is an example of base pair mismatch, a common form of DNA damage, potentially leading to genetic mutations.

Because of these possibilities, it is very important for the body to have safety mechanisms to protect itself against these (and other) forms of oxidative damage. There will always be some concentration of ROS in the body's cells, but the body's anti-oxidation systems should help minimize this.

"Antioxidant" is a broad class of chemicals, including both small molecules and peptides. The general idea is that they are molecules that prevent or reverse oxidation of important biomolecules, or that terminate radical chain reactions. In the lipid peroxidation reaction shown above, the final step is propagation of the radical chain reaction. Since radicals always form new radicals when reacting with non-radicals, the chain reaction will only terminate in one of two ways: either the ROS concentration gets soo high that two radicals bind (very bad), or an antioxidant like vitamin E or Vitamin C will "quench" the chain reaction. Vitamin E will "scavenge" ROS in a process coupled with reduction by ascorbate (vitamin C anion):

$\hskip1.6in$ (4)

Since vitamin E is fat soluble, it can help to protect the lipid bilayer. The semidehydroascorbyl radical that's produced is fairly stable, with a half-life of $10^-5$ s. Now, we have a less reactive radical. Processes like this are known as radical "quenching". two molecules of semidehydroascorbyl radical will quickly disproportionate to dehydroascorbate and ascorbate. This is a process where two odd numbers of electrons turn into two even numbers. 1 + 1 -> 0 + 2

Another sort of antioxidant is the sort that dimerizes. Because one electron needs to be donated to each radical to form a non-radical, one way of removing them is to convert two radicals and two non-radicals into three non-radicals. This is a process where two odd and two even numbers of electrons turn into three even numbers. 1 + 1 + 2 + 2 -> 2 + 2 + 2

Glutathione is probably the most important antioxidant in the body's anti-ROS arsenal. Glutathione can convert a molecule of of $\ce{HO-OH}$ to water in a processes catalyzed by the enzyme glutathione peroxidase. A peroxide is not a radical, but it's a high energy ROS which can easily form two $\ce{HO.}$. The overall equation for the reaction is as follows:

$$\ce{2GSH + H2O2 → GS–SG + 2H2O}$$

Glutathione can also turn the dehydroascorbate we mentioned previously back into ascorbate (vitamin C) through catalysis by the enzyme dehydroascorbate reductase. All oxidized (dimerized to GS-SG) glutathione is finally converted back to reduced glutathione by glutathione reductase.

Depending on your level of knowledge of chemistry and biochemistry, some of the specifics here may be difficult to understand, but the key is simple: The commonality between all antioxidants is that they are able to convert odd numbers of electrons into even numbers of electrons in processes like these:
1 + 1 -> 0 + 2 or 1 + 1 + 2 + 2 -> 2 + 2 + 2 and probably many others.

Antioxidants are just molecules that help pair up electrons.

ROS comes into play mostly as exhaust from the mitochondria, the location where oxygen is turned into water. Antioxidants take care of it, and many of them are regulated through the action of glutathione, which is dimerized on oxidation, then split apart again during enzymatic reduction (along with the cofactor NADPH). Through a system of enzymes and coupled reactions, the body tries to maintain a low ROS concentration.

(1) Wikipedia: https://en.wikipedia.org/wiki/Antioxidant#/media/File:Lipid_peroxidation.svg

(2) Jena, N. R.; Mishra, P. C. "Mechanisms of Formation of 8-Oxoguanine Due to Reactions of One and Two OH. Radicals and the H2O2 molecule with Guanine: A Quantum Computational Study" J. Phys. Chem. B, 2005, 109, 14205-14218

(3) Wikipedia: https://en.wikipedia.org/wiki/8-Oxoguanine#/media/File:8-oxoG_forming_Hoogsten_base_pair_with_dA.svg

(4)Valko, M.; Izakovic, M.; Mazure, M.; Rhodes, C. J.; Telser, J. "Role of oxygen radicals in DNA damage and cancer incidence" Faculty of food technology, Slovak Technical University, 2004


Database with organic reaction mechanisms (curly arrows) Is there any database or book that has an index of organic reactions where the mechanisms are written out, complete with curly arrows? It would really help if I could look up a reaction and see a proposed mechanism with curly arrows.

As far as free online resources go, I've found the following useful:

Organic Reactions (organic-chemistry.org)
Name-reaction.com
List of organic reactions (en.wikipedia.org)

I would also strongly recommend acquiring a copy of March's Advanced Organic Chemistry at some point. I think it's fair to say that it's widely regarded as one of the most essential reference works in the field of organic chemistry. It's concise, more comprehensive than one would've thought possible, beautifully organized, and contains thousands of primary literature references. While not every reaction is presented with an accompanying mechanism, many are, and useful exposition is typically provided.

Another book you might find useful is Strategic Applications of Named Reactions in Organic Synthesis. More than a mere reference text, it includes an overview describing important applications for the reactions it covers, examples of uses in practical syntheses, as well as comprehensive stepwise mechanisms.


Does gold always become gold chloride when dissolved with HCl and an oxidizer and can it always be precipitated with sodium metabisulfate? I tried to extract gold from computer parts by STRIPPING the gold using $\ce{HCl}$ and $\ce{H2O2}$ by dissolving the copper & nickel underneath but i added FAR too much $\ce{H2O2}$ by a stupid mistake and I ended up dissolving the gold... I didn't know that could happen? I am correct in understanding the $\ce{HCl}$ dissolves the gold, but it needs both energy and oxygen to do so? Nitric acid (commonly added to $\ce{HCl}$ to form aqua regia which dissolves gold) is $\ce{HNO3}$ which has lots of oxygen and Hydrogen Peroxide is $\ce{H2O2}$ which has oxygen also, albeit less than Nitric Acid.....

My mixture is VERY dark red... It's fuming very lightly, and makes you choke if you get near it. It's not see-through but rather very dark and a bit murky, but could be transparent if you added 200% of water.

I am afraid to try anything because if I screw it up I will lose $200 of gold but am I correct in understanding that adding sodium metabisulfate will drop the gold? What if it is too acidic? Will I have to add Urea or Sodium Bicarbonate to neutralise the pH??? HELP!!

You may find to your dismay that getting dissolved gold separate from dissolved copper and nickel is dicey (to give you some idea, I still have a dissolved solution of gold and copper sitting in my garage ca. 2 years ago because I hesitate to mess with it; also it looks kind of cool).

You should use urea to neutralize first, just going on experience. I've done the same procedure you describe here to try to recover gold from RAM cards etc. Add urea slowly and check pH regularly. I wouldn't try sodium bicarbonate only because I've never done that and it seems you ought to go with the most well-used method if you're risking a big loss.

Sodium metabisulfate is indeed a common reagent for precipitating the gold. It smells pretty terrible before and after adding it (because of sulfur content). I think $\ce{SO2}$ gas is released when you react with gold so you should have a well-ventilated area for this. You should get a brownish precipitate. I would keep adding it to the neutralized solution until no more precipitate is formed, then let it settle.

This point is probably your best hope for separation of the gold. Recover the precipitate by filtering and leave off as much of the fluid as possible (wash with DI $\ce{H2O}$, and melt the brown salt to remove metabisulfate and get the pure metal.

In the future I would not use any oxidizer when stripping the gold from $\ce{Cu}$/$\ce{Ni}$. I found that a much better method is to stir/air-bubble the material in $\ce{CuCl2}$ which can be made from pure copper, $\ce{HCl}$, and careful additions of $\ce{H2O2}$. It will take longer but it is much safer. You'll recover solid bits of gold instead of dissolving it undesirably. i.e.

$\ce{CuCl2}$ > $\ce{HCl} + \ce{H2O2}$ > aqua regia

A lot of this procedure is difficult to improve because (even academic) literature is nebulous on the topic. I once bought a ~$250 book on the topic. "The Properties of Electrodeposited Metals and Alloys: A Handbook" by Safranek. It's hard to get a hold of.

P.S. it does not form gold chloride when dissolved in aqua regia ($\ce{HCl}$ alone won't effectively dissolve it). It forms a complex with $\ce{Cl-}$, chloroauric acid $\ce{HAuCl4}$.


How many stereoisomers of 1-methylidene-2-(4,4,6-trimethylhepta-2,5-dien-1-yl)-1,2-dihydronaphthalene are there? What is the total number of stereoisomer of this compound?

2
4
8
16

As there is only 1 chiral carbon, and 2 out of 4 C=C double bonds that contribute to geometric isomerism (cis/trans), I think there are 2^(1+2)=8 stereisomers. But the answer says 4. Can someone please explain the reasoning? Thanks!

The text is correct. You are partly correct about the alkenes, but mistakenly assume 1) all are stereocenters. 2, it is incorrect to assume that even if stereogenic, alkenes have two stereocenters simply because they have 2 carbons. What makes them stereogenic is that there are geometric differences relative between the two carbons. Thus stereogenicity comes from that relationship, not the number of carbons. 3) While everyone know what you mean by cis and trans, those are not acceptable terms in organic chemistry unless you are referring to sides of rings or in relative terms only.

Looking at just olefins and going from left to right:

Next, the endocyclic alkene must be cis. You cannot place an E olefin in a ring smaller than 8 eight carbons but that would make the name change...this is asking for configurational (stereo) isomers...not constitutional variants.

Next, the exocyclic alkene is terminal. It has two of the same group (H) on the a carbon, it is therefore not a stereogenic center. How do you know? Switch the two hydrogens...does anything change? no, so it can't be.

Next, the E-olefin next to the quaternary carbon is the only stereogenic alkene...it could be Z. So that is a stereocenter (both carbons are)...though that would be incredibly high energy (the largest substituents (and neopentyl is enormous) are sterically congested).

Finally, the last olefin double bond is not a stereocenter. There are two of the same group on one of the carbons. Transposing them changes nothing.

But we're not finished. There is a chiral center. Do you see it? The maximum and actual number of stereoisomers is two. 2^2 is 4, the answer is 4. Here they all are.


How does temperature and other factors affect sigmatropic rearrangements? The image below shows a [3,3] sigmatropic rearrangement. My question is why does the increased temperature favour the right side? The arrangement is technically the same except for the methyl groups on different carbons.

Here is another example:

The increased temperature is necessary to overcome the activation energy of the reaction. Many [3,3] sigmatropic rearrangements have very high activation energies. The products are favored at equilibrium because they are thermodynamically more stable. For example, in the first case, you are exchanging monosubstituted alkenes for disubstituted alkenes.


Theoretical comparison of dipole moments of N,N-dimethyl-4-nitroaniline and N,N,2,3,5,6-hexamethyl-4-nitroaniline These two compounds are given in my workbook, and I am supposed to compare their net dipole:

The solution given in the workbook is that:

Molecule A (N,N-dimethyl-4-nitroaniline) has a "net shifting of electrons" in the downward direction - due to $+M$ effect of $\ce{NMe2}$ and $-M$ effect of $\ce{-NO2}$;
Molecule B (N,N,2,3,5,6-hexamethyl-4-nitroaniline) - due to steric hindrance - does not undergo resonance to an appreciable extent, hence, there is negligible "net shifting of electrons" in any direction. The lone pair of $\ce{NMe2}$ almost remains in its place.

Hence, molecule A has a higher dipole than molecule B. I am content with this explanation.

However, my question is:

  Is this explanation actually correct, or is it just a theoretical mask to explain an experimental observation (i.e. they just made up a random theory on their own to explain an observation)?

Are such types of explanations common, universally accepted, etc?

The workbook's explanation could have been more specific. If you build a 3D model, you may see that the orbital interaction between the lone pair on the $\ce{NMe_2}$ and the $\pi$-system on the ring is somewhat hindered by bad overlap because the steric interaction of the methyl groups (on the ring and the $\ce{N}$) prevents perfect alignment. Same for the nitro substituent. A higher interaction would lead to more charge shift.


How to tell whether lone pairs in heterocycles contribute to aromaticity? How can I decide whether lone pair electrons take part in the continuous overlap of electron or not?

For example, the lone pair in pyridine is not a part of the conjugated system, whereas the lone pair in pyrrole is.

I had read that this was due to the lone pair in pyridine being localised in an sp2 orbital, while the lone pair in pyrrole is delocalised.

How do we identify this?

For the most common situations use this rule: a conjugated ring atom needs either a ligand or a localized electron pair to replace a missing ligand.  Any remaining electron pairs are then part of the conjugated pi system.

Thus pyridine needs a localized lone pair on nitrogen to replace a missing ligand, and that leaves only three bonding pairs for the pi system.  Pyrrole has a ligand on nitrogen so the lone pair is conjugated into the pi system along with two carbon-carbon bonds.  Furan has two lone pairs on oxygen but a missing ligand, therefore oxygen contributes one pair to the pi system.


Potassium permanganate to analyze iron compounds In my lab, I used potassium permanganate in order to analyze the amount of iron in a compound with iron in the +2 oxidation state using titration. My question is how would this analysis be different if we were analyzing iron in a different oxidation state like +3?

The Standard Methods for the Examination of Water and Wastewater 3500-Fe is a method for determining total $\ce{Fe}$, $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ in water using a colorimetric detection technique.  As in your titrimetric method, this method can only directly measure $\ce{Fe^{2+}}$.  In order to measure $\ce{Fe^{3+}}$ (the only other reasonably stable oxidation state of $\ce{Fe}$) you must first reduce the $\ce{Fe^{3+}}$ to $\ce{Fe^{2+}}$.  I believe you could use the same reduction technique to prepare a solution of $\ce{Fe^{3+}}$ for your titration method.

The procedure sited above states that:

  "[the $\ce{Fe}$ in the sample is] reduced to the ferrous state by boiling
  with acid and hydroxylamine."

The hydroxylamine solution is prepared by dissolving 10 g $\ce{NH2OH⋅HCl}$ in 100 mL water.  50 mL of the water sample is acidified with 2.0 mL of concentrated $\ce{HCl}$, and 1 mL of the $\ce{NH2OH}$ solution is added to the acidified water sample.

As $\ce{Fe^{3+}}$ species are frequently insoluble, the procedure described above calls for boiling the solution down to 15-20 mL to get all of the iron both dissolved and reduced before bringing the solution back up to 50 mL for analysis.  If the solution were prepared from a soluble $\ce{Fe^{3+}}$ salt like $\ce{FeCl3}$ for example, simply boiling the solution for 5 minute or so would likely be sufficient.

So, I suspect you could measure the $\ce{Fe^{3+}}$ in your solution using your titration method if you first reduced the $\ce{Fe^{3+}}$ to $\ce{Fe^{2+}}$ using the hydroxylamine reduction described above.


Is HCl (l) + H2O (l) equal to HCl (aq)? The Wikipedia article about aqueous solutions claims that:

  An aqueous solution is a solution in which the solvent is water

So basically, is writing $\ce{HCl(aq)}$ equal to: $\ce{HCl(l) + H2O(l)}$?

$\ce{HCl{(l)}}$ is not the correct notation in this case. Arrhenius acids and bases (chemicals that interchange $\ce{H^+}$ with water to form $\ce{H_3O^+}$ or $\ce{OH^-}$) break apart in water, forming ions. In your case, there is an equilibrium $\ce{HCl{(aq)} + H_2O{(l)}<=> H_3O^+{(aq)} + Cl^-{(aq)}}$ which very strongly lies to the right; virtually all of the $\ce{HCl{(aq)}}$ is broken apart in this way (which is why $\ce{HCl}$ is known as a strong acid).

Now, not everything dissolved in water has to dissociate into ions. For example, glucose $\ce{(C_6H_12O_6{(s)})}$ dissolves in water to form $\ce{C_6H_12O_6{(aq)}}$. Weak acids are acids that do not entirely dissociate into ions (for example, $\ce{CH_3COOH{(aq)} + H_2O{(l)} <=> CH_3COO^-{(aq)} + H_3O^+{(aq)}}$ still has some undissociated $\ce{CH_3COOH{(aq)}}$ in solution.)

A final point is that some dissolved species can, in fact, be solvents on their own. In this case, the definition of whether the species is aqueous or liquid is not well-defined and usually depends on the context. For example, a 1:4 mixture of $\ce{MeOH}$ and $\ce{H_2O}$ is typically written as $\ce{MeOH{(l)} + {H_2O{(l)}}}$, but a solution where a small amount of $\ce{MeOH}$ (maybe used as a reactant) would be denoted $\ce{MeOH{(aq)}}$. This has to do with the definition of the thermodynamic activities of the species and if you have questions about this, there's probably someone else on here who is a better expert than I to explain.


is ph at eq point dependent on acid concentration at all Know for a weak acid strong base titration that the ph will be above 7, but does the concentration of the acid affect the ph at the eq point at all?

The concentration of both acid and base solution matters - or rather the ratio of concentrations matters. Let's look at some simple examples titrating acetic acid, which has a $K_a = 1.74 \times 10^{-5}$ with sodium hydroxide.

$$\ce{HC2H3O2 + NaOH -> H2O + NaC2H3O2}$$

  Example 1 - 20.0 mL of 0.100 M acetic acid titrated with 0.100 M NaOH

In this example, since the concentrations are the same and the stoichiometric ratio is 1:1, we will need 10.0 mL of NaOH solution to reach the end point. The new volume is 40.0 mL and the concentration of acetate anion is 0.0500 M (volume doubled, concentration halved). Some of that acetate ion reacts with water in the following equilibrium:

$$\ce{C2H3O2- +H2O <=> HC2H3O2 + OH-}\\ K_b=\dfrac{[\ce{HC2H3O2}][\ce{OH-}]}{[\ce{C2H3O2-}]}=5.76\times 10^{-10}$$

To figure out how much hydroxide is formed (and thus eventually the pH), we need to do an ICE problem:

$$\begin{array}{|c|c|c|c|}
 & [\ce{C2H3O2-}] & [\ce{HC2H3O2}] & [\ce{OH-}] \\ \hline
I & 0.0500\ \text{M} & 0\ \text{M} & 1\times 10^{-7}\ \text{M} \\
C & -x\ \text{M} & +x\ \text{M} & +x\ \text{M} \\
E & 0.0500-x\ \text{M} & x\ \text{M} & 1\times 10^{-7}+x\ \text{M} \\
\end{array}$$

Then we plug in the concentrations into the law of mass action:

$$5.76\times 10^{-10} = \dfrac{(x)(1\times 10^{-7}+x)}{(0.0500-x)}$$

Solving for $x$ (since I am not making any approximations, I am letting Wolphram Alpha do the grind:

$$x=5.32\times 10^{-6}\ \text{M} \\
[\ce{OH-}] = 5.42 \times 10^{-6}\ \text{M} \\
[\ce{H+}] = 1.85 \times 10^{-9}\ \text{M} \\
\text{pH} = 8.734$$

  Example 2 - 20.0 mL of 0.100 M acetic acid titrated with 0.200 M NaOH

In this example, the NaOH solution is twice as concentrated, so it only takes 10.0 mL to titrate the acid. The concentration of acetate anion is 0.0667 M.

$$\begin{array}{|c|c|c|c|}
 & [\ce{C2H3O2-}] & [\ce{HC2H3O2}] & [\ce{OH-}] \\ \hline
I & 0.0667\ \text{M} & 0\ \text{M} & 1\times 10^{-7}\ \text{M} \\
C & -x\ \text{M} & +x\ \text{M} & +x\ \text{M} \\
E & 0.0667-x\ \text{M} & x\ \text{M} & 1\times 10^{-7}+x\ \text{M} \\
\end{array}$$

Then we plug in the concentrations into the law of mass action:

$$5.76\times 10^{-10} = \dfrac{(x)(1\times 10^{-7}+x)}{(0.0667-x)}$$

$$x=6.15\times 10^{-6}\ \text{M} \\
[\ce{OH-}] = 6.25 \times 10^{-6}\ \text{M} \\
[\ce{H+}] = 1.60 \times 10^{-9}\ \text{M} \\
\text{pH} = 8.796$$

  Example 3 - 20.0 mL of 0.200 M acetic acid titrated with 0.100 M NaOH

In this example, the NaOH solution is half as concentrated as the acetic acid solution, so it takes 40.0 mL to titrate the acid. The concentration of acetate anion is 0.0333 M.

$$\begin{array}{|c|c|c|c|}
 & [\ce{C2H3O2-}] & [\ce{HC2H3O2}] & [\ce{OH-}] \\ \hline
I & 0.0333\ \text{M} & 0\ \text{M} & 1\times 10^{-7}\ \text{M} \\
C & -x\ \text{M} & +x\ \text{M} & +x\ \text{M} \\
E & 0.0333-x\ \text{M} & x\ \text{M} & 1\times 10^{-7}+x\ \text{M} \\
\end{array}$$

Then we plug in the concentrations into the law of mass action:

$$5.76\times 10^{-10} = \dfrac{(x)(1\times 10^{-7}+x)}{(0.0333-x)}$$

$$x=4.33\times 10^{-6})\ \text{M} \\
[\ce{OH-}] = 4.43 \times 10^{-6}\ \text{M} \\
[\ce{H+}] = 2.26 \times 10^{-9}\ \text{M} \\
\text{pH} = 8.646$$


Why is it OH- and not HO-? I am told that in a chemical equation the metal comes first and then the non metal. for example MgO, ZnSO4,etc. But when both the elements are non metals or metals, the one with the lower atomic number would come first. for example:- H2O,CO2,HF,etc. But why is it that hydroxide is OH- and not HO-? Also why is methane CH4 and not H4C?

OH- and HO- are both acceptable.  The charge is on the oxygen, of course.  One can argue cation then anion (salts) versus alphabetical (e.g., AuBe, which is brittle and weak rather than ductile, suggesting ionicity).

Organic formulas are C, H, then alphabetical.  Order may be altered to conform to structure (e.g., propane as $\ce{H3CCH2CH3}$) or convenience $\ce{H3C^{-}Li^{+}}$ (which is nowhere near the real structure of methyllithium, but is how it reacts).


van der Waals coefficients for helium and neon The values of the coefficient $b$ for helium and neon are

$$\begin{array}{cc}
\hline
\text{Element} & b\text{ / }\mathrm{dm^3\ mol^{-1}} \\
\hline
\ce{He} & 2.38 \times 10^{-2} \\
\ce{Ne} & 1.67 \times 10^{-2} \\
\hline
\end{array}$$
(source: Atkins & de Paula, Physical Chemistry, 9th ed., p 916)

This doesn't make sense, however, considering that neon should have a larger atomic radius than helium. Why is $b_\ce{He} > b_\ce{Ne}$?

The question intregues me so I looked up the $a$ and $b$ values in Wikipedia on the web page Van der Waals constants. This doesn't answer the question, but adds more data to focus the question.

$$\begin{array}{cc}
\hline
\text{Element} & a\text{ / }\mathrm{L^2 \ bar\ mol^{-2}} & b\text{ / }\mathrm{L\ mol^{-1}}\\
\hline
\ce{He} & 0.0346 & 0.0238 \\
\ce{Ne} & 0.2135 & 0.01709 \\
\ce{Ar} & 1.355  & 0.03201 \\
\ce{Kr} & 2.349  & 0.03978 \\
\ce{Xe} & 4.250  & 0.05105\\
\ce{Rn} & 6.601  & 0.06239 \\
\hline
\end{array}$$

A plot of the Van der Waals $a$ and $b$ values.

So is the b value for He high, or is the b value for Ne low ?!?

I also looked up for each element what was listed for the covalent radius and the Van der Waals radius on each elements individual Wikipedia webpage.

$$\begin{array}{cc}
\hline
\text{Element} & Covalent\text{ (pm) } & Van~der~Waals \text{ (pm)}\\
\hline
\ce{He} & 28 & 140 \\
\ce{Ne} & 58 & 154 \\
\ce{Ar} & 106±10  & 188 \\
\ce{Kr} & 116±4  & 202 \\
\ce{Xe} & 140±9  & 216 \\
\ce{Rn} & 150  & 220 \\
\hline
\end{array}$$


Is "density per unit volume" something? I was reading articles on the preparation of shellac-based composites when I encountered the following statement (in ref. 1):

  Wood flour and fibres are interesting because of their low cost, good specific  strength, low  density  per  unit  volume,  along  with  their  renewable  and degradable  features. The  drawbacks of  natural  organic  fillers include poor  adhesion between the filler and the polymer matrix, and poor thermal and water resistance (Saba et al. 2014).

Note the same phrase is used in the reference cited in the passage quoted above (Saba et al. 2014) but then the trail goes dead.

Is "density  per  unit  volume" a typo or other error that's been propagated during citation or a real concept of concern for material scientists? If it is, what is the meaning of this concept?

References

Obradovic et al. (2017). “Cellulose-shellac composites,”BioResources12(1), 1943-1959.

Density usually specifies a ratio (or equivalent derivative) of the type "amount/unit dimension" where dimension is usually one of (distance, area, volume, hypervolume). Based on further examples of usage (see links in Todd Minehardt's answer) of the expression "density per unit dimension" where "dimension" might be "distance", "area", "volume" or other spatial dimension, the point with writing "density per unit volume" seems to be an attempt to be explicit about the dimension in the divisor (denominator). For instance, when writing  "density per unit dimension"  it is meant that "density=amount/dimension". Thus "density per unit volume" means "density=amount/volume". One could equally well say "volume density" or "area density" to be specific, but there might be ambiguity there as well, as one might confuse usage with expressions such as "number density" or "energy density" where the property in the numerator is being specified. Then "volume density"  might be interpreted as meaning "density of volume", even if farfetched. One could specify dimensions by placing the units in parentheses, for instance, as density (per unit volume) or density (per unit area) etc.


Producing soap from sodium carbonate and oil Can we get soap by mixing $\ce{Na2CO3}$ from seaweeds with oil?

I am not good at chemistry and saw this in the anime Dr. Stone. I just want to know if it is true or not.

In nature fat is stored as triglycerides (TGs): Three fatty acid (FA) chains bound together by a glycerol backbone. Olive oil is mainly composed of TGs, though it also contains small amounts of free fatty acids (FFAs).

An example of Triglyceride (from Wikipedia)

Soap is commonly made via the reaction of triglycerides  with a strong base: usually Sodium Hydroxide ($\ce{NaOH}$) or Potassium Hydroxide ($\ce{KOH}$). Though $\ce{Na2CO3}$ is a base, it is too weak of a base to be efficiently used in the traditional production of soap: It may saponify triglycerides but only at a very slow rate, if at all. So you are unlikely to produce much soap by mixing $\ce{Na2CO3}$ (of any source) with oil.

Soap can however also be made by reacting a weaker base with free fatty acid (FFA) chains; that is, FAs that have already been broken away from the TGs. $\ce{Na2CO3}$will form salts with FFAs, and it will do so at a much faster rate than it would saponify TGs. It can thus be used to make soap that way instead (and will, for instance, react with the small amount of FFAs in your oil).

You could try to find a way to separate the FFAs from the rest of the oil (though it would be a highly inefficient process as you'd be left with the dominant TG portion), you could in theory also separately break down TGs into FFAs prior to saponifying them (using enzymes for instance), or (highly recommended) you could go the traditional route of using a strong base which will efficiently and (relatively) rapidly saponify both TGs and FFAs in your oil.

Splitting triglycerides is the "hard work" from a chemical point of view.

Sources

https://en.wikipedia.org/wiki/Saponification
https://en.wikipedia.org/wiki/Triglyceride
https://en.wikipedia.org/wiki/Fatty_acid
https://en.wikipedia.org/wiki/Sodium_carbonate
https://www.soapmakingforum.com/threads/how-much-sodium-carbonate-to-use-for-soapmaking.60798/


Does bleach (without rinsing) leave residues and can vinegar, baking soda neutralize them? It's very difficult to get accurate information concerning household bleach residues. Imagine, for instance, that you pour pure (or diluted) bleach onto fabric or synthetic foam and let it dry without rinsing (as a scientific experiment). Some sources say: it will totally "vanish" (with air and/or light), leaving water and salt, and some others say it will leave some toxic compounds behind, such as sodium hydroxide (which is added to the bleach to slow its decomposition).

Question 1: Will the bleach only degrade into water and salt ($\ce{NaCl}$), or will it leave other substances? If so, which ones? If the first case is true (i.e., the degradation products are water and salt only), why does the solution still smell like bleach for a long time?

Question 2: If bleach (a base) does leave toxic residue(s), can we use vinegar (an acid) to neutralize it? $(~\ce{OH- + H+}$ should create $\ce{H2O}$)

White vinegar: I'm confused since there is contradictory information on Wikipedia. Some says vinegar will neutralize harmful residues from bleach, while another source says it "could result in much more caustic and dangerous chemicals, such as hypochlorous acid" being produced.
Will an antioxidant such as vitamin C work as neutralizer? (as suggested by Brian)

It seems that everybody agrees about the use of peroxide, bisulfite, metabisulfite or thiosulfite, and sodium thiosulfate to neutralize bleach residues, but I wonder if more common products such as vinegar or vitamin C can be used instead: What do you think about these two? Any other suggestions?

EDIT: I removed the washing example which was confusing since it implied rinsing + I updated my question since I thought bleach was an acid.

EDIT2:

  Household bleach is, in general, a solution containing 3–8% sodium
  hypochlorite and 0.01–0.05% sodium hydroxide; the sodium hydroxide is
  used to slow the decomposition of sodium hypochlorite into sodium
  chloride and sodium chlorate. wiki

Brian and Drmoishe's answer suggest several ways to way to neutralized bleach but what about the sodium Hydroxide residues? (I check on a french bleach, it shows only on the composition: "3.7% of sodium hypochlorite", nothing else)

After a thorough wash, rinse and dry cycle, little or no bleach should be left in the fabric, but there is possible concern about the effect of chlorine bleach on organic compounds. However, if disinfection is needed, then the infinitesimally small risk of VOC* is outweighed by the elimination of microorganisms.
If any remaining smell of chlorine bothers you, an extra water rinse should be helpful. @Brian suggests a reducing agent, and $\ce{Na2S2O3}$, sodium thiosulfate, is inexpensive and is commonly used to remove residual chlorine for aquaria. However, leaving garments to air-dry or airing them for a day would be as effective.
(*) "Volatile organic compounds"


Calculating the volume of 2 M solution needed to make a 0.1 M solution I need help with a basic concentration problem.
If I have a $\mathrm{2~M}$ solution, and want to make a $\mathrm{0.1~M}$ solution of $\mathrm{100~ml}$, how would I go about that?
My work:
$$M_1V_1 = M_2V_2$$
Where $M_1 = \text{initial concentration}$, $M_2 = \text{final concentration}$, $V_1 = \text{volume needed from initial}$, $V_2 = \text{volume needed from final}$
$$2~M \times V_1 = \pu{0.1 M \times 100 ml}$$
$$V_1 = \frac{\pu{0.01 M\cdot L}}{\pu{2~M}} = \pu{5 ml}?$$
Therefore I need $\pu{5 ml}$ of my $\pu{2 M}$ solution, increased to $\pu{100 ml}$ to make a $\pu{0.1 M}$ solution of$\pu{100 ml}$?

If you have a simple solution of say $\pu{0.1 M}$ sodium chloride, you have $\pu{0.1M}\, \ce{NaCl}$ in $\pu{100 ml}$.
Which is equivalent to "$\pu{100 mL}$ of $\pu{0.1 mol}\,\ce{NaCl}$ per litre", or,
$$n_\ce{NaCl} =\frac{100}{1000} \times 0.1 = \pu{0.01 mol}$$
$\pu{5 mL}$ of $\pu{2 M}\ \ce{NaCl}$ is equivalent to "$\pu{5 mL}$ of $\pu{2 mol}$ $\ce{NaCl}$ per litre", or,
$$n_\ce{NaCl} = \frac{5}{1000} \times 2 = \pu{0.01 mol}$$
So congratulations, you did it right. If ever you aren't sure, just write out all the steps and follow it through.
You could also look at it like this - $\pu{2 M}$ is $20\times$ more concentrated than $0.1\times\frac{2}{0.1}$, so you expect to need $20\times$ less of the $\pu{2 M}$ than the $\pu{0.1 M}$.
$\pu{5 ml}$ is $20\times$ less than $\pu{100 mL}$.


Does ozone has dipole forces Considering ozone in liquid state, certainly there are london forces, but does the position of its molecules induce dipole intermolecular forces?

Ozone has a bent structure with a partial positive charge on the central oxygen atom and partial negative charges on the two terminal oxygens. It has a permanent dipole moment of 0.53 D, according to Wikipedia.


How to minimise solution concentration changes due to evaporation? Lets say I have a precise solution, and would like to store it long term (up to two years) with as little as possible change in concentration. The volume is 10-100ml. What container should I choose? What temperature? Is there any other factors? I would prefer not to use sealed glass container. Is there a way to approximate a solvent loss?  What if I stored some water in exactly the same conditions together with my sample, would I be able to approximate water loss in my sample from the water loss in this reference container, or does the difference in the concentration of the solute affect the rate of evaporation too much?

What if I stored some water in exactly the same conditions together with my sample, would I be able to approximate water loss in my sample from the water loss in this reference container, or does the difference in the concentration of the solute affect the rate of evaporation too much?

That would be a gross approximation, not suitable for a standard solution. Your solution and pure solvent do have different vapor pressures according to Raoult's law.

  What temperature?

It depends on solvent, solute, and their interaction. I'd check two main things: i) solubility at the storage temeprature, which must be as close as possible to the solubility at the working temperature; ii) vapor pressure of the solvent, that gives you a rough indication of the storage temperature. You could take a look at the temperature dependence of the solvent vapor pressure and solute solubility - or, if seriously in doubt, even try simulating the thermodynamics of your solution using a software.

  What container should I choose?

Volumetric flasks are usually the best choice. They minimize evaporation compared to other glassware (bottles, beaker, etc.) and, most important, they have the most accurately placed graduation mark. Use the appropriate rubber cap and close accurately with laboratory film. PTFE band between cap and flask cone might also help preventing losses.

At times, a slight overpressure above the liquid can be desirable. In this case, a gas valve (or a quality glass faucet) will be placed on top of the flask and nitrogen will be flushed inside. The flask can be stored capped by the sealed valve. Also consider the opportunity of degassing your solution by placing sieves on the bottom of the flask or by bubbling nitrogen through the liquid. Finally, consider the need for a dehydrating agent in the case of a moisture-sensitive solution.

But, most of all, don't keep standard solutions for two years but replace them every ~6 months.


Why do these two IUPAC naming examples not begin numbering the chain closest to the alkyne? I am currently reading my chemistry book which states "The chain is numbered from the end closest to the triple bond." However, then it proceeds to give two examples which I do not believe follow this rule. What is the reasoning for this?

The oversimplified statement “The chain is numbered from the end closest to the triple bond.” applies to simple alkynes with one triple bond. It is not necessarily correct for compounds that include several multiple bonds or other characteristic groups.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), low locants are assigned first to the principal characteristic group that is expressed as a suffix and then to multiple bonds.

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (…)

Therefore, the correct name for the second example given in the question is ‘but-3-yn-2-ol’ (not ‘but-1-yn-3-ol’) since a low locant is given first to the suffix ‘-ol’.

If the compound contains several multiple bonds (double or triple bonds), low locants are given to multiple bonds as a set. If there is still a choice, low locants are given to the double bonds.

  P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

  (…)

Therefore, the correct name for the first example given in the question is ‘2-methylpent-1-en-3-yne’ (not ‘4-methylpent-4-en-2-yne’) since the locant set ‘1,3’ for the multiple bonds is lower than ‘2,4’.


Ozonolysis conditions for benzene What are the precise conditions under which benzene undergoes ozonolysis? All anyone can tell me is that it undergoes ozonolysis under "vigorous" conditions. But, what vigorous conditions?

What are the precise conditions under which Benzene undergoes ozonolysis?

In Über das Ozobenzol, an early German publication on the reaction of benzene with ozone, Ber. Dtsch. Chem. Ges., 1904, 337, 3431-3433 (DOI)
, the authors write

Leitet man in solches Benzol 1–2 Stunden lang bei einer Temperatur von 5–10° einen 5-procentigen Ozonstrom, so scheidet sich eine gelatinöse Masse aus, welche stark opalisirt. […]
Lässt man das unangegriffene Benzol verdunsten, so bildet es eine weisse, amorphe Masse von furchtbar explosiven Eigenschaften. Selbst beim Uebergiessen mit warmem Wasser detonirte diese Substanz heftig.
Uebergiesst man aber das gelatinöse Product mit eiskaltem Wasser,
so entsteht eine krystallinische Modification, die wohl möglich noch
heftiger als die amorphe bei der geringsten Berührung, ähnlich wie
Jodstickstoff, explodirt.

To sum this up:

The authors prepared the primary triozonide of benzene by passing a stream of ozone though it at a temperature of 5–10 °C.

Once the remaining benzene evaporated at room temperature, a white, amorphous mass with frightfully explosive properties was obtained.

When warm water was poured over it, this mass detonated.

When cold water was poured over the amorphous, jelly-like mass, a crystalline modification was obtained, which detonated on the slightest touch, similar to to nitrogen triiodide.

In the 1950s, a group of Dutch researchers published a series of articles about their kinetic experiments on the ozonization reactions.
The second article in the series, Recl. Trav. Chim. Pays-Bas, 1951, 70, 1005–1019, (DOI) notes that the ozonisation of benzene was performed in chloroform at a temperature of 242 K (–31 °C).

In the late 1960s, ozonization of benzene again gained some interest as a possible source for glyoxal in industrial synthesis.
In Glyoxal from Ozonolysis of Benzene, published in Ind. Eng. Chem. Prod. Res. Dev., 1969, 8, 89–92
DOI), the authors describe the ozonolysis of benzene in a mixture of acetic acid and nitromethane at temperatures between –20 °C and 15 °C.

Summary

All anyone can tell me is that it undergoes ozonolysis under "vigorous" conditions.

Apparently, your sources never bothered to check the literature! While the ozonide itself may detonate on touch, claims on the "vigorous" conditions necessary for its synthesis seem to be fully unsubstantiated by the literature!


Drying Clothes Outside Below Freezing It is commonly accepted that you can hang your wet clothes outside to dry even when the winter temperature is below freezing. It is mentioned here
http://grist.org/living/ask-umbra-can-i-line-dry-my-clothes-in-the-winter-too/
that water in the clothes will freeze to ice, then sublimate into water vapor, making the clothes dry.

Question: Why does the ice in the clothing sublimate, while the ice on the roads don't sublimate? Why is there a difference between what happens to ice on the clothing versus ice on the roads?

As long as there is less than 100% relative humidity, the ice will sublime to some extent.

Just in the past week it has been in the -12 °C to -5 °C range in my area, and the snow and ice on the ground did sublime, on the order of 1 cm per day, more in the sun less in the shade.

For quantitative information see "Sublimation from a seasonal snowpack at a continental,
mid-latitude alpine site" Hydrol. Process. 13, 1781-1797 (1999).


What is thermal isomerization? Context: a retinal molecule the isomerizes from all-trans to 13-cis form, then "reisomerizes thermally".

I tried googling it but couldn't find a simple, straightforward explanation.

The term "thermally" is used here to contrast with a photochemical isomerization. In the chemistry of vision, 11-cis retinal absorbs light and is isomerized to the all-trans isomer. This is a photochemical process because it involves the absorption of a photon to promote one electron into an antibonding orbital. In this excited state, the geometry can rearrange to the more stable all-trans form before the electron relaxes back to the ground state.

In a "thermal" process, no photons are absorbed, no electrons are promoted, and no excited states are needed. The molecule takes advantage of ambient or supplied kinetic energy to reorganize.


Stability of carbocations: CF3+ vs CH3+ Is $\ce{CF_3+}$ more stable than $\ce{CH_3+}$?

In $\ce{CF_3+}$, there is $\ce{C-F}$ back bonding that increases  stability of the carbocations. But -I effect of $\ce{F}$ dominates +R effect and this decreases the carbocation stability. How can I compare these two competing effects to find the more stable carbocation?

Also, is $\ce{CF_3+}$ more stable than $\ce{CH3-CH+-CH_3}$ and tert-butyl carbocation?

$\ce{CF3+}$ is more stable than $\ce{CH3+}$. The comparison between -I and +R effect is made while talking about activation or deactivation of halogens when substituted on benzene. Fluorine always stabilises a carbocation to a large extent because of a very good overlap of $\ce{2p -~2p}$ orbitals.


Ellingham Diagram for carbon Consider the following reactions:
$$\begin{align}\ce{C + O2&->CO2}\tag{1}\\[1em]
\ce{2C + O2&->2CO}\tag{2}\\[1em]
\ce{2CO + O2&->2CO2}\tag{3}\\[1.5em]
[(2)+(3)]/2 &= (1)\end{align}$$

Since $G$ is a state function, why aren't the straight lines corresponding to these reactions in the Ellingham diagram concurrent?
I have checked the graph from other sites but found none to be concurrent.
Is there some kind of error in data or extrapolation of the graphs or any other specific reason for this?

You are indeed right, I confirmed that it is the fault in the diagram itself. See the diagram by wikipedia:


Finding specific activity of an enzyme from Km values for a reversible enzyme reaction I have an enzymatic equation in the form:

$$E + S\xleftrightarrow[k_{-1} = 2.6\text{ x }10^{-2} \text{s}^{-1}]{k_1 = 3 \text{ x }10^8 \text{M}^{-1}\text{s}^{-1}} EX \xleftrightarrow[k_{-2} = 3.8\text{ x }10^{-8} \text{M}^{-1}\text{s}^{-1}]{k_2 = 1.45 \text{ x }10^3 \text{s}^{-1}} E + P$$

For the forward reaction, I have calcluated the $K_m$ value to be:

$${K_m}_f = \frac{k_{-1}+k_2}{k_1} = 4.83\text{ x }10^{-6}\text{M}$$

And for the reverse, I calculated $K_m$ to be:

$${K_m}_r = \frac{k_{-1}+k_2}{k_{-2}} = 3.82\text{ x }10^{-6}\text{M}$$

From this info, I need to somehow calculate the specific activity of the enzyme in both the forward and reverse reactions.

The weight of one enzyme unit is 50,000 kDa.

Specific activity is defined as the number of 'Units' per mg of protein.

One enzyme unit is defined as the amount of enzyme that catalyzes the formation of 1 μmol of product per min under optimal assay conditions, which are assumed for this question (saturation of substrate in forward reaction / saturation of product in reverse reaction).

Can anyone point me in the right direction? I have a feeling that this is related to the $V_{max}$ term in the Michaelis-Menten equation:

$$v = \frac{V_{max}\text{[S]}}{K_m + \text{[S]}}$$

In that $K_m = \text{[S] at}\frac{1}{2}V_{max}$.

So, $$[S]_{f} \text{ at } V_{max} = 2(4.83\text{ x }10^{-6}\text{M}) = 9.66\text{ x }10^{-6}\text{M}$$

So, $$[S]_{r} \text{ at } V_{max} = 2(3.82\text{ x }10^{-6}\text{M}) = 7.64\text{ x }10^{-6}\text{M}$$

but I dont know how to relate it to specific activity.

If the enzyme is staturated with substrate, the reaction rate will be $k_2$.

1450 moles per second, per mole of enzyme.

87,000 moles per minute, per mole of enzyme.

87,000,000,000 micromoles per minute, per mole of enzyme.

Based upon "50,000 kDa" (which looks like a typo, 50 kDa sounds more reasonable), one mole of enzyme is 50,000,000 grams or 50,000,000,000 milligrams.

So 87/50 = 1.74 units per milligram.

Then for when product is staturated, use $k_{-1}$ instead of $k_2$.


Can I use potassium acetate instead of potassium carbonate? In the synthesis of N-ethoxycarbonyl L-proline methyl ester from L-proline and ethyl chloroformate, I use $\ce{K2CO3}$ to neutralize $\ce{HCl}$. Can I use $\ce{CH3CO2K}$ instead of $\ce{K2CO3}$? I think I cannot use it because it reacts with $\ce{HCl}$ forming acetic acid which can hydrolyse the ester. Is it correct?

Stay with the carbonate.
In the course of the reaction $\ce{HCl}$ is formed. The aim is to remove this from the reaction mixture both as soon as well as complete as reasonably possible.  Compare

The addition of sodium acetate will increase the complexity of your reaction mixture.  $\ce{HCl}$ is soluble in methanol (reference), and upon dissociation ($\ce{HCl <=> H+ + Cl-}$), the $\ce{H+}$ and the acetate $\ce{OAc-}$ join each other to generate the buffer system of $\ce{HOAc <<=> H+ + OAc-}$.  Think about the $\ce{HOAc}$ as a reservoir, it may release $\ce{H+}$ at any moment you probably do not intend; a prominent example may be an aqueous workup.
The quantity of $\ce{HOAc}$ generated may be small -- in comparison of the methanol used as solvent of reaction.  Still it may complicate an aqueous workup and following extraction, because this organic substance solutes so well in water, but fairly enough in organic solvents such as hexanes, or chloroform, too.  So your extraction of the product may be affected.  A distillation (bp 118 C (pure acetic acid) instead of 65 C (pure methanol) at ambient pressure) ahead of the workup may remove this from your reaction mixture.

with the alternative

addition of $\ce{K2CO3}$.  Darn efficient removal of $\ce{H+}$ for once and forever, because the quench along  $\ce{K2CO3 + 2H+ -> 2K+ + H2O + CO2}\uparrow$ is practically irreversible.  In addition, it allows you to monitor the advancement of the reaction since $\ce{K2CO3}$ is not well soluble in methanol.


Is liquid oxygen non flammable? Every one knows that liquid oxygen is flammable, but why this tank labeled like this:

so is it really liquid oxygen is non flammable?

Oxygen is not flammable, as it does not burn, it supports burning. For burning to happen, a strong oxidizer (for example, oxygen) and a strong reducer (for example, carbon) must be present.

No, the problem with oxygen is, that while not being flammable, it is much more efficient in supporting burning. In air oxygen is mixed with nitrogen, meaning that for the same amount of reducer a quadruple amount of oxidizer is required and a lot more heat is lost with inert gases (nitrogen). For example, iron wire can burn in oxygen, but not in air. It is also a lot easier to lit things up in oxygen. A classic experiment is to move a smoldering splinter into a vial with pure oxygen. It immediately starts to burn with a bright flame

Now, liquid oxygen is an entirely new level of crazy. In addition to constantly releasing gaseous oxygen... If an organic material (for example, sawdust) is mixed with liquid oxygen, it becomes an explosive.


High pressure modification to the van der Waals equation I get that a gas behaves non-ideally under high pressure and low temperature conditions. At high pressures, the density of a gas increases and attraction forces operate among molecules due to short intermolecular distance. This allows us to somehow modify the ideal gas equation. If a molecule is approaching the wall of container in order to collide, the neighbouring gas molecules soften their impact.

$$p_\text{ideal}=p_\text{real}+\frac{an^2}{V^2}$$

How was the term $\frac{an^2}{V^2}$ derived?
My thoughts about that term, $\frac{n^2}{V^2}$ may be the extra pressure thats why the $V$ lies on the numerator (inverse relationship)

The most rigorous way to derive corrections to the real gas equation is the virial expansion. At second order this gives Van der Waals law, but this derivation is completely general and can be pushed to higher order.

My answer assumes a basic knowledge of classical mechanics, statistical physics and analysis.

Hamiltonian of the system

We consider a system of identical particles described by the following Hamiltonian:
$$H=\sum_i\frac{\vec{p}_i^2}{2m}+\sum_{i\lt j}U(|\vec{r}_i-\vec{r}_j|)$$
If you don't know what a Hamiltonian is, you can consider it as the energy of the system.

Canonical partition function for the perfect gas

For a perfect gas, you consider that the particles are non-interacting, so that
$$U(|\vec{r}_i-\vec{r}_j|)=0,\quad\forall i,j.$$
In this particular case, life is easy! In fact you can easily compute the Canonical partition function (this is a definition)
$$Z_\mathrm C=\frac1{N!h^{3N}}\int\mathrm d^N\vec{p}\,\mathrm d^N\vec{r}\,\mathrm e^{-\beta H}$$
which, in the case of a non-interacting gas, is simply
$$Z_\mathrm{PG}=\frac1{N!h^{3N}}\int\mathrm d^N\vec{p}\,\mathrm d^N\vec{r}\,\mathrm e^{-\beta\sum_i\frac{\vec{p}_i^2}{2m}}=\frac{1}{N!}\left(\frac{2 \pi mk_\mathrm BT}{h^2}\right)^{\frac{3N}{2}}$$
In order to compute this integral, you can simply use the definition of Gaussian integrals. This is the partition function of a perfect gas and you can now derive all the properties of this system.

Canonical partition function for an interacting gas

If you want to consider an interacting gas, your partition function becomes
$$Z_\mathrm I=\frac1{N!h^{3N}}\int\mathrm d^N\vec p\,\mathrm d^N\vec r\,\mathrm e^{-\beta\left(\sum_i \frac{\vec{p}_i^2}{2m}+\sum_{i\lt j} U(|\vec{r}_i-\vec{r}_j|)\right)}$$
The integration over momenta can be easily computed as before (using Gaussian integrals). In contrast, the integral
$$Q_N=\int\mathrm d^N\vec r\,\mathrm e^{-\beta\sum_{i\lt j}U(|\vec{r}_i-\vec{r}_j|)}$$
is particulary difficult (even for simple forms of the interacting potential $U$). We can write temporarily the partition function of the system as
$$Z_\mathrm I=Z_\mathrm{PG}Q_N=\frac1{N!}\left(\frac{2\pi mk_\mathrm BT}{h^2}\right)^{\frac{3N}{2}}Q_N$$

Grand Canonical partition function for the interacting gas

Now, as you sais, the critical parameter is the density $\rho$: at low density you recover the perfect gas behavior. Therefore, it seems a good idea to develop our formalism in powers of $\rho$. It is easier to work in the Grand Canonical ensemble (you don't have to worry much about this), where the partition function is given by (this is a definition)
$$Z_\mathrm{GC}=\sum_N^\infty\mathrm e^{\alpha N}Z_\mathrm C$$
This is just a generalization of the Canonical partition function $Z_\mathrm C$. With this definition we have, for the interacting gas,
$$Z_\mathrm{II}=\sum_N^\infty\mathrm e^{\alpha N}\frac1{N!}\left(\frac{2\pi mk_\mathrm BT}{h^2}\right)^{\frac{3N}2}Q_N=\sum_N^\infty\frac{z^N}{N!}Q_N,$$
where we defined for simplicity
$$z=\mathrm e^{\alpha}\left(\frac{2\pi mk_\mathrm BT}{h^2}\right)^{\frac32}$$

Relation between the Grand Canonical partition function and the equation of state

From $Z_\mathrm I$ in the Canonical ensemble we can now compute the ratio $p/(k_\mathrm BT)$. The Grand Canonical potential is defined by
$$\phi=-k_\mathrm BT\ln(Z_\mathrm{II})$$
and this Grand Canonical potential is linked to volume and pressure by
$$\phi=-pV,$$
so that you finally have a link between the Gran Canonical partition function and the pressure:
$$\frac p{k_\mathrm BT}=\frac{1}{V}\ln(Z_\mathrm{II}).$$

Since $Z_\mathrm{II}$ is a development in powers of $z$, the same is valid for $\ln(Z_\mathrm{II})/V$. This gives
$$\frac1V\ln(Z_\mathrm{II})=z+\frac{a_2}{2!}z^2+\frac{a_3}{3!}z^3+\cdots$$

Density

Using an easy argument that I will not show here (unless explicitly required), you can then write the density as
$$\rho=\frac{\langle N\rangle}V=z\frac{\partial}{\partial z}\left(\frac1V\ln(Z_\mathrm{II})\right),$$
where $\langle N\rangle$ is the average number of particles in your gas.
Therefore, for the density we can write
$$\rho=z+a_2z^2+\frac{a_3}{2!}z^3+\cdots$$

Developement of the state equation in powers of $\rho$

Since we said that $\rho$ is our critical parameter we can write
$$\frac p{k_\mathrm BT}=\rho+b_2\rho^2+b_3\rho^3+\cdots.$$
In fact the perfect gas law was the first-order expansion with respect to the density. For a diluite gas, $\rho \ll 1$ and therefore we simply get
$$\frac p{k_\mathrm BT}=\rho.$$

Virial coefficients

It is now easy to link the coefficients $a$ to the coefficients $b$, by equating the powers of $z$. In our case you find:
$$b_2=-\frac{a_2}2,\qquad b_3=-a_2^2-\frac{a_3}3.$$
It now remains to link these coefficients to $Q_N$. By definition of $Z_I$ we can write
$$\ln(Z_\mathrm I)=\ln\left(1+Q_1z+Q_2\frac{z^2}{2!}+Q_3\frac{z^3}{3!}+\cdots\right)$$
By Taylor expansion of the logarithm we get
$$\ln(Z_\mathrm I)=Q_1+Q_2\frac{z^2}{2!}+Q_3\frac{z^3}{3!}-\frac{1}{2}\left(Q_1z+Q_2\frac{z^2}{2}\right)^2+Q_1^3\frac{z^3}{3}+\cdots$$
For $a_1$ and $a_2$ we have
$$a_1=\frac{Q_1}V\qquad a_2=\frac1V\left(\frac{Q_2}2-\frac{Q_1^2}2\right)$$
which gives
$$b_2=\frac1{2V}\left(Q_1^2-Q_2\right).$$

Computation of $Q_i$

Now, if we compute $Q_1$ and $Q_2$ explicitly (which is much easier than computing $Q_N$ for a general $N$) we get
$$Q_1=\int\mathrm d^3\vec r=V$$
$$Q_2=\int\mathrm d^3\vec{r}_1\,\mathrm d^3\vec{r}_2\,\mathrm e^{-\beta U(|\vec{r}_1-\vec{r}_2|)}=Va+V^2$$

Van der Waals law

With the expression for the virial coefficients and the analytical expressions of $Q_1$ and $Q_2$ we finally write
$$b_2=-\frac a2$$

Therefore we finally get Van der Waals law
$$\frac p{k_\mathrm BT}=\rho-\frac a2\rho^2.$$

At the second order in $\rho$ we get Van der Waals law, but this procedure is completely general and can be used to compute higher order corrections to the perfect gas law. Computing the coefficients $Q_N$ is quite cumbersome, but can be performed using the cluster expansion.


Is it possible to create weaker version of guncotton by replacing cotton wool with wood chips or linen cloth? If the cotton wool was replaced by wood chips or linen (flax) cloth, would the mixture of nitric and sulfuric acids still react with the cellulose in it to produce nitrocellulose?

If it would, would it be reasonable to assume that the power of such explosive would be (very approximately) proportional to the content of cellulose in your substitute for cotton wool?

Would such explosive produce smoke (as besides nitrocellulose it would contain other flammable substances and no oxygen source)?

Would hemicellulose go through a similar reaction with the acid mixture as cellulose?

This article mentions hemicellulose from wood chips being hydrolyzed, not nitrated by a mixture of ethanol and nitric acid, while cellulose does get nitrated: Isolation of cellulose from poplar wood by nitric acid-ethanol treatment and its effect on the quality of films cast from ionic liquid

I also found a patent for Preparation of wood pulp cellulose for nitration which from the name alone suggests that cellulose from wood can be nitrated (the preparation described is purely physical, no chemicals applied).

Regarding linen and cotton fabric, they will work same as cotton wool. Cotton fabric is how nitrocellulose was first accidentally discovered (wikipedia article) and linen is made of cellulose fiber just like cotton (these fabrics have 94% and 95% cellulose respectively: flax, cotton).

Tldr:

Linen fabric, cotton fabric and cotton wool work equally well. It is possible to nitrate the cellulose in wood chips. Hemicellulose will (most likely) not get nitrated. No idea about smoke production and effectiveness so far though.


Lowering the boiling point of wax Wax comes in all kinds (paraffin, bee wax, palm oil wax, gel wax, ...) but since it's usually a mixture of different carbonated chains, its physical properties vary a lot from one manufacturer to another.

A good candle is usually one made of a wax with a high melting point, such that it takes longer to burn.

I am trying to make a bad candle, more specifically I am trying to make a candles that produces as much wax vapor as possible. To make it happen, I am looking for a wax that has the lowest possible boiling point.

How can I lower the boiling point of a given wax (paraffin wax for instance since it's the cheapest and easiest to buy) by adding other substances inside ? One can easily find on the internet how to increase the BP (by adding stearin for instance), but not the reverse.

Ideally the additional substance should not be added in too great proportion, since that would lower the wax composition and thus the amount of wax vapor emitted.

Whenever you just mix two substances (dissolve in each other), you lower the melting point and raise the boiling point. That's a fundamental principle. If it was the other way round, the substances would not mix, or at least not mix well.

(Azeotropes of compounds with very different polarity (water/ethanol) are the exeption confirming that rule, but the acchievable raise in Bp. won't be large. And I doubt you find one you can make a candle from.)

That sounds bad, but: Paraffin wax already is a mixture, so the desired effect might be reached by fractionating it, instead of adding sth. else.

You might try to slowly crystallise a molten sample of paraffin wax, and pour off the residual liquid when the sample is about half solid. Now the residual solid should have a higher melting point than the original wax.

(Note: This is purely my educated guessing, no idea if the effect is large enough or if some secondary kinetic effect might completely preclude it.)


Adding acid to hydrates What would happen if I added $\ce{Na2CO3 \cdot n(H2O)}$ to a solution of $\ce{HCl}$? Could I consider the reaction the same as just $\ce{Na2CO3 + HCl -> NaCl + CO2 + H2O}$?

There is no difference in properties of salt hydrates and non-hydrates after they are dissolved in water, as they are hydrated to same state. So, yes, you can.

However, there is difference in reaction of salt hydrates and non-hydrates outside water solutions.


Interplanar distance given Miller indice of the planes Say we've a cubic crystal of unity edge length.
A set of planes in such a crystal is specified by their miller indice as $(3 2 0)$. One of these planes then has intercepts on the axes as $(2 ,3 ,\infty)$.
It's well known that the interplanar distance between two adjacent planes in a cubic lattice is given
$d_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}}$,in our case this gives us
$d_{320} = 1/\sqrt13$.
Now if I look at the planes which form the Miller indices of $ 3 2 0$ I find that apart from a plane having intercepts as $(2 ,3,0)$ an adjacent parallel plane to this plane having the same indices will pass via origin $0,0,0$ and the distance between them is
$d_{320} = 6/\sqrt13$
Which contradicts the above result!
Can anyone please point out what's wrong in my reasoning?

Your adjacent plane isn't.

Worth a thousand words, they say.


How to calculate the pressure of an ideal gas that undergoes isothermal expansion? A 2.35 mole sample of an ideal gas, for which $C_{\mathrm{m},v}=3R/2$ initially at $\pu{27 ^\circ C}$ and $\pu{1750 kPa}$, undergoes a two stage transformation. For each of the stages described in the following list, calculate the final pressure as well as $q$, $w$, $\Delta U$ and $\Delta H$. Also calculate $q$, $w$, $\Delta U$ and $\Delta H$ for the entire process.
(a) The gas is expanded isothermally and reversibly until the volume triples.
(b) Beginning at the end of the first stage, the temperature is raised to $\pu{105 ^\circ C}$ at constant volume.

Since the process is isothermal, $\Delta H = 0$. In order to calculate the final pressure. I use the formula,
$P_\mathrm{i}V_\mathrm{i}^{x} = P_\mathrm{f}V_\mathrm{f}^{x}$,
where $x = C_{\mathrm{m},p}/C_{\mathrm{m},v}$.
Hence $x = 5/3$. I then substitute this into the formula to get
$$P_\mathrm{i}V_\mathrm{i}^{5/3} = P_\mathrm{f}V_\mathrm{f}^{5/3}.$$
Since the volume triples, $V_\mathrm{f}/V_\mathrm{i}=3$, hence manipulating the formula I got
$$P_\mathrm{i}/P_\mathrm{f}=3^{5/3}$$
and since $P_\mathrm{i}=\pu{1750 kPa}$, I substitute this into the above equation to get $P_\mathrm{f}$, which answers the first part of the question.
However I don't seem to get the right answer, hence I could not continue doing the question. Could anyone explain?
Also, am I right to say that we use
$P_\mathrm{i}V_\mathrm{i}^{x} = P_\mathrm{f}V_\mathrm{f}^{x}$
for an reversible adiabatic process only.
$w = -nRT\ln\left(T_\mathrm{f}/T_\mathrm{i}\right)$ for an isothermal process and
$\Delta U = nC_{\mathrm{m},v}(T_\mathrm{f} - T_\mathrm{i}) = -P_{\text{external}}(V_\mathrm{f} - V_\mathrm{i})$
for a non-reversible adiabatic process.
Im rather confused as to decide under which conditions should I use each of these three formulas.

I am going to use the physics conventions of work according to which, work done by the gas is positive. So if you want the answers as per chemistry conventions than just flip the sign of the work done, keeping the magnitude same.
Part (a): Reversible Isothermal Expansion
\begin{align}
\Delta U_a &=nC_v\Delta T\\
\Delta U_a &=\pu{0 J}\\
\end{align}
\begin{align}
\Delta H_a &=nC_p\Delta T\\
\Delta H_a &=\pu{0 J}\\
\end{align}
\begin{align}
W_a&= \int_{V_i}^{V_f} P.dV\\
W_a&= \int_{V_i}^{V_f}  \frac{nRT}{V}.dV\\
W_a&= nRT\ln\frac{V_f}{V_i}\\
W_a&= \pu{2.35 \times 8.314 \times 300 \times \ln3 J}\\
W_a&= \pu{6439.37 J}\\
\end{align}
\begin{align}
q_a&= W_a+ \Delta U_a\\
q_a&= \pu{(6439.37 + 0)J}\\
q_a&= \pu{6439.37 J}\\
\end{align}
By Boyle's Law,
\begin{align}
P_iV_i&=P_fV_f\\
P_f&=P_i\times \frac{V_i}{V_f}\\
P_f&= \pu{\frac{1750}{3} kPa}\\
P_f&= \pu{583.33 kPa}\\
\end{align}
Part (b): Isochoric Heating**
\begin{align}
\Delta U_b &=nC_v\Delta T\\
\Delta U_b &=\pu{2.35 \times 1.5 \times 8.314 \times 78 J}\\
\Delta U_b &=\pu{2285.93 J}\\
\end{align}
\begin{align}
\Delta H_b &=nC_p\Delta T\\
\Delta H_b &=\pu{2.35 \times 2.5 \times 8.314 \times 78 J}\\
\Delta H_b &=\pu{3809.89 J}\\
\end{align}
\begin{align}
W_b&= \int_{V_i}^{V_f} P.dV\\
W_b&= \pu{0 J}\\
\end{align}
\begin{align}
q_b&= W_b+ \Delta U_b\\
q_b &= \pu{(0 + 2285.93)J}\\
q_b&= \pu{2285.93 J}\\
\end{align}
By Gay-Lussac's Law,
\begin{align}
\frac{P_i}{T_i}&=\frac{P_f}{T_f}\\
P_f&=P_i\times \frac{T_f}{T_i}\\
P_f&= \pu{\frac{1750}{3} \times \frac{378}{300} kPa}\\
P_f&= \pu{735 kPa}\\
\end{align}
Overall Process:
\begin{align}
\Delta U&= \Delta U_a+ \Delta U_b\\
\Delta U &= \pu{(0 + 2285.93)J}\\
\Delta U&= \pu{2285.93 J}\\
\end{align}
\begin{align}
\Delta H&= \Delta H_a+ \Delta H_b\\
\Delta H&= \pu{(0 + 3809.89)J}\\
\Delta H&= \pu{3809.89 J}\\
\end{align}
\begin{align}
W&= W_a+ W_b\\
W&= \pu{(6439.37 + 0)J}\\
W&= \pu{6439.37 J}\\
\end{align}
\begin{align}
q&= q_a+ q_b\\
q&= \pu{(6439.37 + 2285.93)J}\\
q&= \pu{8725.30 J}\\
\end{align}


why cookware does not melt down? why cookware does not melt down
On stove supllied with butane gas which has a flame temperature of 1970°C?

http://www.engineeringtoolbox.com/flame-temperatures-gases-d_422.html

The answer is that cookware transfers the energy from the flame into whatever is being cooked. For instance a pot of boiling water will stay unmelted until all the water has evaporated, after which the pot will start heating up towards the melting point of the metal it is made from.

This is called heat conductance and is used in both cookware and liquid bi-propellant rocket engines (which burn closer to 3000°C). In fact some rocket engines have been made from aluminium with a melting point as low as ~460°C depending on which alloy is used.


Determining rate constants for parallel reactions from experimental data I have a question regarding the rates of parallel reactions. I have gathered experimental data for the overall reaction (i.e. conversions and selectivities and specific times) but due to the two reactions running parallel to each other for this reaction means I can't use a simple integrated rate plot.

Say for

$$
\ce{A ->[$\large k_1$] B} \\
\ce{A ->[$\large k_2$] C}
$$

By substitution and integration, I have found that:

$$
[A](t) = [A](0) \mathrm{exp}(-(k_1+k_2)t)
$$

and then the following equations can be derived in terms of the products formed:

$$
\frac{d[B]}{dt} = k_1 [A](0) \mathrm{exp}(-(k_1+k_2)t)
$$

which gives

$$
[B](t) = \frac{k_1[A](0)}{k_1+k_2}(1-\mathrm{exp}(-(k_1+k_2)t)
$$

when integrated. Also,

$$
[C](t) = \frac{k_2[A](0)}{k_1+k_2}(1-\mathrm{exp}(-(k_1+k_2)t)
$$

is the equation for the other parallel reaction.

My initial thoughts were to plot the graphs (as I know the values for $[C]$, $[B]$, and $[A]$) and form some simultaneous equations, but this is an exponential plot and so i'm not sure how this would work.. or if I could even plot it, as I don't know $k_1$ and $k_2$ - that is what I need to determine.

One way you could get $k_1$ and $k_2$ is to first find their sum using your equation for $[A](t)$ at a known $t$.

Once you have done that, you can solve for $k_1$ using your equation for $[B](t)$ at a known $t$, since you now have the denominator of your first factor. You could alternatively find $k_2$ using $[C](t)$.

Once you have $k_1$, then $k_2$ is easily found from the sum.


Carbon with two double bonds and oxygen name I have a compound like this:

$$\ce{CH2=C=O}$$

I wonder if it is a functional group or a normal group. As a functional group I couldn't find out.
Is it a normal compound or a functional group and what is it's
IUPAC name

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) the name of the given compound is the systematically constructed name for ketones.

Therefore, the preferred IUPAC name (PIN) is ethenone.

The name ketene is retained, but only for general nomenclature.


Reference Request - bond lengths in PCl2F3 While writing an answer here, I needed references for bond lengths of $\ce{PF3Cl2}$ and $\ce{PF4Cl}$. The latter was easily found in reference (1), whereas I could not find exact references for bond lengths in the former molecule.

References (2) and (3) contain information about the Raman spectra and $\ce{F^{19}}$ NMR spectra, but do not contain information about bond lengths. Reference (2) has information about the bond lengths of $\ce{PCl3F2}$, $\ce{PF5}$ and $\ce{PCl5}$, but not for $\ce{PF3Cl2}$.
A reference for exact bond lengths in $\ce{PF3Cl2}$, similar to the bond lengths shown above, is appreciated.
References:

Leiding, J., Woon, D. E., & Dunning, T. H. (2013). "Bonding in $\ce{PF2Cl}$, $\ce{PF3Cl}$ and $\ce{PF4Cl}$: insight into isomerism and apicophilicity from ab initio calculations and the recoupled pair bonding model". Theoretical Chemistry Accounts, 133(2). doi:10.1007/s00214-013-1428-7
Griffiths, J. E., Carter, R. P., & Holmes, R. R. (1995). "Molecular Structures of $\ce{PCl4F}$, $\ce{PCl3F2}$, $\ce{PCl2F3}$, and $\ce{PF5}$: Infrared and low-temperature Raman Vibrational Spectra". Phosphorus, Sulfur, and Silicon and the Related Elements, 98(1-4), 11–31. doi:10.1080/10426509508036938
Holmes, R. R., Carter, R. P., & Peterson, G. E. (1964). "Molecular Structures of $\ce{PCl4F}$, $\ce{PCl3F2}$, and $\ce{PCl2F3}$: Pure Chlorine Nuclear Quadrupole Resonance and Low Temperature $\ce{F^{19}}$ Nuclear Magnetic Resonance Spectra". Inorganic Chemistry, 3(12), 1748–1754. doi:10.1021/ic50022a021

To complete your literature survey, include this publication cited at least 11 times so far: «Dichlorotrifluorophosphorane (PCl2F3): molecular structure by gas-phase electron diffraction and quadratic force field» by French et al. in Inorg. Chem. 1985, 24, 2774–2777, doi 10.1021/ic00212a014.  Despite the paywall, ACS's «In lieu of an abstract, this is the article's first page.» informs you about some key properties you search, values including ($2\sigma$ estimated) e.s.d.

How did I find this publication?  Because the English Wikipedia has an entry about it (don't overlook the Chinese including complementary information).
But how to know if Wikipedia indexed such a compound?  Way better than trial-and-error is to address this question to the Wikipedia Chemical Structure Explorer allowing to query there by structure or string, which may lead you to the corresponding article (in Wikipedia).  It is backed by a public database (GitHub repository) initiated by Ertl et al.
Source: Ertl, P.; Patiny, L.; Sander, T.; Rufener, C.; Zasso M J. Cheminf., 2015, 7, 10, doi 10.1186/s13321-015-0061-y, (open access).