text:
Determine if two molecule structures are the same molecule (perhaps in different conformations)?
I have two .pdb files describing two (small ligand) molecular structures. I need to determine if these two pdbs describe the same molecule, perhaps in different spatial conformations. Is there a program, or a combination of tools that will allow me to do this?
Moreover, if the two pdbs are determined to be from the same molecule, it could be that the atoms are in different orders. So I also need a way to identify the atoms in one PDB with the atoms in the other PDB.
I am dealing with thousands of pdbs, so I need an automatic way to do this comparison.

This is actually a very difficult problem, because a given compound may exist in multiple conformations (have multiple structures), and it might not be obvious what conformation is most relevant.
For example, a seemingly simple compound like glucose has multiple stereoisomers: if you say "glucose", you probably mean alpha-D-glucose, as opposed to beta-D-glucose, or even L-glucose. But a PDB structure file may not specify the stereochemistry to this level, or ignore it completely, and then you have no way of knowing what compound was intended. Conversely, sometimes a specific structure is given, like this open form of D-glucose; but in a water solution this structure spontaneously interconverts with the closed ring form (they are tautomers). And there are other problems with protonation (which depends on solvent and pH), charge distribution, and what not. Chemistry is a mess :)
To sum it up, the fundamental problem is that more than one completely distinct structure can actually represent the same physical compound. Therefore, comparing physical compounds is difficult; it depends on the level of detail of specification you want. I don't think there is a generic solution for  this.
Comparing structures, on the other hand, is perfectly feasible: I would recommend you take a look at the InChi format developed by IUPAC. It is an identifier string that uniquely identifies a molecular structure, so testing for structural identity is equivalent to simply comparing the corresponding InChi strings. The PDB has InChi strings associated with its entries, so this should be straightforward. (But again, this simple procedure will conclude that, for example, the open form and the closed-ring form of glucose are different things.)
I'm not sure about the atom numbering. The InChi format determines atom numbering in some canonical way according to predefined rules, so in this representation the numbering is by definition always the same (and therefore meaningless). But you would need to look into the details of the PDB--InChi conversion procedure to determine which atom gets mapped to which.

If I have a PDB file, what tool can I use to generate a InChi string? And then, how do I compare two such strings for equality?

Openbabel works. Also, you can write a simple Python script to convert the PDB file into a mol block and then the mol block to an InChi string.


text:
What will the results be if I electrolyzed brine with these materials?
I have tried electrolyzing saltwater (brine) with copper electrodes, but I got a rusty sludge and hydrogen being produced at the anode. I read about it and it is just dissolved copper, possibly oxidized. What will the results be, if I used the following materials?

Aluminium foil
Iron
Pencil graphite

First check your assumptions. It is anode, not cathode, where copper gets oxidized and dissolved. Now, other metals are going to be oxidized as well (and even more readily). Graphite may be a good choice, in that it's more inert.

@IvanNeretin I will then do Graphite. I heard that platinum is the best electrode. This is however too expensive. 10 grams for about 300$. I tried aluminum foil recently, but that seemed to produce Hydrogen and Oxygen...

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Propagation of the uncertainty in measurements

The inside dimensions of a box that is cubic is 12.7 cm on each edge with an uncertainty of 0.3 cm.

What is the volume of the box?
What do you estimate to be the uncertainty in the calculated volume?

The first part is simple enough. To obtain volume of the box: $(\pu{12.7 cm})^3 = \pu{2.05E3 cm^3}$
Now for the second part, "estimate the uncertainty in the calculated volume?" – I would imagine that because the uncertainty is $\rm\pm0.3\ cm$ on each side the uncertainty in the volume of the box would be $\rm\pm(0.3)^3 = \pm0.027\ cm^3$? The textbook gives $\rm\pm0.60\times10^3\ cm^3$. I am (excuse the pun) uncertain of how they have reached this answer.

I don't think that the textbook answer is correct. The largest possible volume would be 131313= 2197cm, while the smallest possible volume would be: 12.412.412.4=1907cm. The difference between the largest and the smallest volume is 290cm. The uncertainty can't be bigger than the difference between the two extremes (it should be 145cm).

The volume $V$ can be calculated from the length $x$, width $y$, and height $z$.
$$\begin{align}
V&=x\cdot y\cdot z\tag1\\[6pt]
&=12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times 12.7\ \mathrm{cm}\tag2\\[6pt]
&=2048.383\ \mathrm{cm^3}\tag3
\end{align}$$
If we assume that the uncertainties of length $x$, width $y$, and height $z$ are not correlated (which is not necessarily true when the same ruler was used to measure length, width, and height), i.e. their correlation coefficient is $\rho=0$, the relative uncertainty of the volume may be estimated as
$$\begin{align}
\left(\frac{u(V)}{V}\right)^2&=\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2\tag4\\[6pt]
\frac{u(V)}{V}&=\sqrt{\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2}\tag5
\end{align}$$
Thus, the absolute uncertainty of the volume is
$$\begin{align}
u(V)&=\sqrt{\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2}\cdot V\tag6\\[6pt]
&=\sqrt{\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2}\cdot x\cdot y\cdot z\tag7\\[6pt]
&=\sqrt{\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)^2+\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)^2+\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)^2}\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\tag8\\[6pt]
&=83.809\ \mathrm{cm^3}\tag9
\end{align}$$
Note that the same result may be obtained using the general formula
$$\begin{align}
u(V)&=\sqrt{\left(\frac{\partial V}{\partial x}\right)^2\cdot u^2(x)+\left(\frac{\partial V}{\partial y}\right)^2\cdot u^2(y)+\left(\frac{\partial V}{\partial z}\right)^2\cdot u^2(z)}\tag{10}\\[6pt]
&=\sqrt{\left(y\cdot z\right)^2\cdot u^2(x)+\left(x\cdot z\right)^2\cdot u^2(y)+\left(x\cdot y\right)^2\cdot u^2(z)}\tag{11}\\[6pt]
&=\sqrt{\left(12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\right)^2\times\left(0.3\ \mathrm{cm}\right)^2+\left(12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\right)^2\times\left(0.3\ \mathrm{cm}\right)^2\\+\left(12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\right)^2\times\left(0.3\ \mathrm{cm}\right)^2}\tag{12}\\[6pt]
&= 83.809\ \mathrm{cm^3}\tag{13}
\end{align}$$
However, if we have to assume that the uncertainties of length $x$, width $y$, and height $z$ are correlated, i.e. their correlation coefficient is $\rho=1$, then the equation for the estimate of the relative uncertainty of the volume simplifies to
$$\frac{u(V)}{V}=\frac{u(x)}{x}+\frac{u(y)}{y}+\frac{u(z)}{z}\tag{14}$$
and the absolute uncertainty of the volume is
$$\begin{align}
u(V)&=\left(\frac{u(x)}{x}+\frac{u(y)}{y}+\frac{u(z)}{z}\right)\cdot V\tag{15}\\[6pt]
&=\left(\frac{u(x)}{x}+\frac{u(y)}{y}+\frac{u(z)}{z}\right)\cdot x\cdot y\cdot z\tag{16}\\[6pt]
&=\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}+\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}+\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\tag{17}\\[6pt]
&=145.161\ \mathrm{cm^3}\tag{18}
\end{align}$$
Note that the same result may be obtained using the general formula
$$\begin{align}
u(V)&=\frac{\partial V}{\partial x}\cdot u(x)+\frac{\partial V}{\partial y}\cdot u(y)+\frac{\partial V}{\partial z}\cdot u(z)\tag{19}\\[6pt]
&=y\cdot z\cdot u(x)+x\cdot z\cdot u(y)+ x\cdot y\cdot u(z)\tag{20}\\[6pt]
&=12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times0.3\ \mathrm{cm}+12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times0.3\ \mathrm{cm}\\
&\hphantom{={}}+12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times0.3\ \mathrm{cm}\tag{21}\\[6pt]
&=145.161\ \mathrm{cm^3}\tag{22}
\end{align}$$
The numerical values of an estimate $y$ and its standard uncertainty $u(y)$ should not be given with an excessive number of digits. It usually suffices to quote $u(y)$ to at most two significant digits, although in some cases it may be necessary to retain additional digits to avoid round-off errors in subsequent calculations.
Thus, the result for the volume $V$ according to $\text{(3)}$ and its standard uncertainty $u(V)$ according to $\text{(9)}$ or $\text{(13)}$ may be expressed as
$$V=\left(2048\pm84\right)\ \mathrm{cm^3}\tag{23}$$
or
$$V=2048(84)\ \mathrm{cm^3}\tag{24}$$
Using the uncertainty according to $\text{(18)}$ or $\text{(22)}$, the result may be expressed as
$$V=\left(2.05\pm0.15\right)\times10^3\ \mathrm{cm^3}\tag{25}$$
or
$$V=2.05(15)\times10^3\ \mathrm{cm^3}\tag{26}$$
(According to various international standards, the ± format should be avoided.)

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Is there a non-violent, practical way of making hydrochloride with only Hydrogen and Chlorine?
A way of making HCl gas is to put Chlorine and Hydrogen into a test tube and expose it to UV light.
This reaction is however fast and strong and I would like to store the Hydrochloride from it.
Another way would be to dissolve Chlorine in water, which would make Hydrochloride, but most of it would go in water and make dilute hydrochloric acid. I could however store the HCl fumes from the dilute acid. I do not know about the practicality of this.
What is the best, and exceptionally practical way of making Hydrochloride when the only resources I have are Hydrogen and Chlorine?

The best and practical way is to sell hydrogen and chlorine, and buy yourself some $\ce{HCl}$ (unless it is prohibited in your jurisdiction). However, I would recommend that only if you have some experience in chemistry and understand quite well the dangers thereof.
The technical answer to your question is: build a galvanic cell with platinum anode and graphite cathode, and feed them with hydrogen and chlorine, respectively. It will perform your reaction in a pretty controllable and non-violent way, and also will give you some electricity for free. Not that I expect you or anyone else to be actually doing this, though.
Why would you need gaseous $\ce{HCl}$, really?

In the first paragraph, is it like in the shape of "We will let you use our galvanic cell"? I would like HCl gas for creation of concentrated hydrochloric acid.

Wait, you just asked something about electrolysis - is that your source of hydrogen and chlorine? Are you going to run it at home? Not on a large scale, I hope? Remember that hydrogen in explosive, and both chlorine and HCl are quite poisonous if inhaled. I don't want to be responsible for anyone's death. Again, HCl is easier to buy or make from $\ce{H2SO4}+\ce{NaCl}$.

Electrolysis is the source. Performing in small amounts in a fume hood. I will try the method with the galvanic cell

I don't think it will work without platinum. Really, there are easier chemical ways to HCl (I mentioned one in the previous comment).

How do you plan to store the $\ce {HCl}$ gas you produce? In a balloon? (Don't!) Use $\ce{H2SO4}+\ce{NaCl}$ as Ivan mentioned. It is a method that is used in practice in a professional lab if an HCl lecture bottle is not handy.


text:
Stability of ortho- and para-hydrogen

can the stability of ortho-hydrogen greater than para-hydrogen can be stated as this?

The intrinsic energy of ortho is greater than para-hydrogen and also with rise in temperature %of ortho form increase while para decreases and ultimately at room temperature it's 75% ortho and 25% para-hydrogen. Plus as we know that the everything in nature tries to attain lowest possible energy level, and ortho form with opposite spins counteracts each others magnetic moments, making it energetically stable.

Why is it that pure para can be obtained at lower temperature such as 20 K but pure ortho form is difficult to obtain?
I mean when ortho is more stable it should be ortho which should be available more easily than para form?

Ortho is not more stable.

I have read every where that ortho is the more stable form.

OK, let's abandon the word "stable" altogether, for it is confusing. Para is the one with lower energy.

From "Statistical Mechanics: A Concise Introduction for Chemists" by Benjamin Widom (pp. 35-36):

For the diatom of ordinary hydrogen ($\ce{{}^{1}_{1} H}$), where each
  nucleus is just a proton of spin quantum number 1/2, there are three
  possible symmetric states of the pair of spins (nuclear spin triplet)
  and only one possible antisymmetric nuclear spin state (spin
  singlet). [...]
At high temperatures, which generally means $T \gg
> \theta_{\mathrm{rot}}$ (where $\theta_{\mathrm{rot}}$ is the
  characteristic rotational temperature) and so for $\ce{H2}$ means $T$
  well above 88 K, the three ortho nuclear-spin states and one para
  state are equally populated, so equilibrium hydrogen at high
  temperatures is a mixture of ortho- and para-hydrogen in the ratio
  3:1. In pure hydrogen the magnetic interactions between molecules that
  could cause transitions between nuclear spin states are so weak that
  such hydrogen may be cooled to arbitrarily low temperatures with the
  ortho:para ratio remaining 3:1. That is normal hydrogen. Paramagnetic
  impurities such as $\ce{O2}$, or a catalyst such as $\ce{Pt}$ on the
  surface of which $\ce{H2}$ molecules may dissociate, allowing the
  atoms to recombine with their nuclear spins in a different relative
  orientation, catalyze the ortho-para conversion. Cooled in the
  presence of such a paramagnetic impurity or such a catalyst, then, the
  hydrogen may have its ortho:para ratio continually adjust to that
  which would correspond to thermal equilibrium at the given low
  temperature. [...]
Cooled in such a way, then, hydrogen becomes pure para-$\ce{H2}$ at
  low temperatures, in contrast to "normal" hydrogen, which remains a
  mixture of ortho and para in the ratio 3:1. If the hydrogen is cooled
  in the presence of a catalyst that catalyzes the ortho-para conversion
  and thus allows it to become pure para-$\ce{H2}$ at low T, and if the
  catalyst is then removed and the hydrogen warmed back up to 298 K,
  say, the result is a sample of pure para-$\ce{H2}$ at normal
  laboratory temperatures. Pure ortho-$\ce{H2}$ cannot be prepared in
  this way.


text:
Is there a relationship between an electron's energy and its distance from the nucleus?
I have read that in Bohr's model of the atom electrons in the same atomic orbital have the same energy, and that this energy (as a result of the electron existing in a fixed atomic orbital) is related to the distance of the electron from the nucleus. However, using the quantum mechanical model of the atom, in which an orbital is the area of space around an atom where the probability of interacting with an electron is greatest, it is implied that there is no relationship between an electron's energy and its distance from the nucleus. Is there no relationship, or have I misunderstood something else?
Also, it is often mentioned that an atom is 99.99% empty space etc., but since the probability of interacting with an electron is described by its wave function (which suggests it is possible, though unlikely, to interact with an electron very close to the nucleus), this suggests to me that this isn't necessarily true. What am I misunderstanding?

Erase Bohr's model from your head - it leads down many seedy back alleyways that end in nothingness... The quantum wave function is the electron for all practical purposes.

Also, since the introduction of the wave function, there is no such thing as free space.

I will start off by addressing Jon's comment above. Yes, the Bohr model is flawed. I think it is still worth learning about it just from a historical standpoint, to see how we discovered the quantum mechanical description of the electron, but when you are studying it you absolutely have to remember that all of what he said was effectively rubbish.
Now on to your questions.

Is there a relationship between an electron's energy and its distance from the nucleus?

The first thing that should be said is that in QM, unlike in the Bohr model, there is no well-defined "distance from the nucleus". Electrons exist everywhere (except at nodes) and the probability density is given by $\lvert\psi\rvert^2$. However, it is possible to calculate the "most probable distance from the nucleus" using the radial distribution function, $r^2 R^2$, where $R(r)$ is the radial part of the wavefunction. Essentially, you take the derivative of this with respect to $r$ and set it to 0 - you can read more here. For a 1s electron in hydrogen, you will find that the most probable radius is approximately $52.9 \text{ pm}$, a quantity called the Bohr radius and denoted $a_0$.
(As an aside: oddly enough, this correlates with the flawed Bohr model, which said that the radius of the 1s electron in hydrogen was $52.9 \text{ pm}$. However, what Bohr said was that the electron was always at this distance, which is wrong. QM says that the electron is most likely to be found at this distance.)
Anyway, you will find that there is some degree of correlation between the most probable radius and the energy of the electron. If we consider the 2s orbital of hydrogen, you can perform exactly the same calculation as you did for the 1s orbital, and you will find that the most probable radius is $4a_0$.

Also, it is often mentioned that an atom is 99.99% empty space etc...

That's just a simplified view presented in popular science (or elementary chemistry classes) that ignores the quantum mechanical model. No serious chemistry book would write something like that. I don't know exactly how it's calculated, but I can make a good guess. You take any atom you like, let's say hydrogen. You look up the atomic radius - Google tells you it's $52.9 \text{ pm}$. (No surprises there.) So you can think of the atom as a sphere with that radius, and calculate the volume of the sphere: $V = \frac{4}{3}\pi r^3$. Then you go and look up the volume of an electron and a proton (a physicist would get a headache from just reading the phrase "volume of an electron"), and add it up, and you find that the "atom" is "99.99% empty space".
As Ivan succinctly said - with the QM description of the atom, there is no longer such a thing as free space since the electron "exists everywhere". Apart from that, just like how the distance from the nucleus is no longer well-defined in QM, the "volume" of an atom is also no longer well-defined. We pretty much threw that view out of the window.


text:
Determine polarity of covalent bond with formal charge
In a covalent molecule, can we conclude that if the formal charge between two atoms are +1 and -1 respectively, then the chemical bond acting between them is a polar covalent bond (such as the bond between N and single-bonded O in Nitrate Ion)? On the other hand, if both of the atoms has a formal charge of 0, then the chemical bond acting between them is a nonpolar covalent bond (bonds in sulfur dioxide and sulfur trioxide)? How about bond between atoms with formal charge of +1 and 0 (P with single-bonded O in phosphate ion)?

All bonds between two atoms of different elements are polar to some extent. To determine how polar, you need to check the elements’ electronegativities and only them.
Formal charges are exactly what their name implies: a formality. When drawing Lewis structures you should attempt to generate all-octet structures but often this isn’t possible without moving electrons entirely between atoms. In that case, formal charges are assigned to rationalise e.g. a nitrogen with only four electrons.

Check out these structures of nitric acid, sulphur trioxide and carbon monoxide. (Resonance structures applicable to $\ce{HNO3}$ and $\ce{SO3}$ not displayed.)  All bonds are polarised towards the oxygen, since oxygen’s electronegativity is $3.5$; the second highest amoung the non-noble-gas elements. Note especially that the bonds in $\ce{CO}$ are polarised towards oxygen despite oxygen bearing a formal positive charge. (There is a lot more to $\ce{CO}$ which is shown well in Martin’s answer including the molecular orbitals)


text:
Can I identify an unknown substance measuring density and calomimetric data?
How would I be able to get the identity of a substance (for example, aluminium)?
What do I have to measure? Mass, volume and what else? I HAVE to use the q = m * c * Delta T method.
Any suggestions?

Can you please elaborate on the question a little more? Do you want to identify the substance using laboratory methods, or by thermodynamic properties, or something else?

laboratory methods only

@ToddMinehardt ^^^

Identifying an unknown substance usually involves measuring an intensive property - some property that depends on the identity and composition of the sample and not on its size - and then comparing it with known values.
Density is one of these properties. The density of a substance depends only on the identity of the substance (and the temperature).
Another such property is the specific heat capacity, the amount of heat required to raise 1 gram of the substance 1 degree Celsius. For example, the specific heat capacity for water is $4.184\ \mathrm{\frac{J}{g\cdot ^\circ C}}$. The heat capacity world is a little more complicated than this (for example, there are different heat capacities for constant pressure conditions than there are for constant volume conditions), but that is not something to we need to worry about it.
If we know the heat capacity $c$ and the mass $m$ of an object, we can determine the amount of heat transferred $q$ as a result of a temperature change $\Delta T$.
$$q=mc\Delta T$$
However, if we do not know the heat capacity, we can rearrange this equation to solve for it.
$$c = \dfrac{q}{m\Delta T}$$
Thus, if we can measure $q$, $m$, and $\Delta T$, we can determine $c$. Measuring $q$ is the trickiest part. In lab we usually use a calorimeter. The type of calorimeter used in introductory courses is often a coffee cup calorimeter made from two nested polystyrene foam cups, which insulate the system from the outside. For a quick approximation, you can fill the cups with a know mass of water at a certain cold temperature $T_C$ and plunge a known mass of your unknown sample at a certain hot temperature $T_H$ into the water. The water and the sample reach thermal equilibrium at some final temperature $T_F$. You can determine the heat absorbed by the water using its temperature change, mass and specific heat capacity. Since the heat gained by the water was lost by the sample, you can then calculate the heat capacity of the sample.


text:
Is there a way to concentrate the vitamin content of a particular food?
I'm looking for a way to extract vitamins out of common food sources of, well, vitamins.
Let's take, as an example, Vitamin A which is commonly found in carrots. A carrot (weighing around 60g) should contain around 500µg of retinoic acid (RA). RA is not soluble in water so in principle if I just dehydrate the carrot (which has around 50g of water) I would be left with around 10g of matter that still contains those 500µg of RA. I think that, with this simple procedure, I would be concentrating the amount of RA in my sample.
But maybe (surely) there are better ways to achieve this, to extract/purify a particular vitamin from food. Any ideas?
Also, some Vitamins are, instead, soluble in water, so we would need to apply another method for those.
And, finally, I would like to accomplish this using simple (household, if possible) techniques. It doesn't matter if I don't reach 100% purity, not even 1%, but if I can extract a 1g sample from my original carrot that has most of the RA still in it that would be huge.

You are correct that retinoids are poorly soluble in water. Vitamins A (retinol), D (calciferol), E (tocopherol) and K (phylloquinone) have high hydrophobicity and therefore should be extracted using organic non-polar solvents.
All these vitamins can be identified using UV/Visible spectroscopy.
Since your focus is on retinoids I shall focus on that (but the principles remain the same for other hydrophobic biomolecules).
Note: carrot has beta-carotene and it is a precursor for retinol and not a retinoid itself.

Crush and homogenise the tissue
With some plant tissues you may need to grind the frozen tissue using liquid nitrogen repeatedly. Tissues containing polyphenols and lignins would need additional treatment but not necessary for carrots.
Precipitate proteins (optional; not required in your case).
Extract using solvents such as hexane or chloroform. Isopropanol has also been used for beta-carotene extraction from carrots (Fikselová et al. 2008)[1]. Gundersen (2006)[2] summarizes the methods of extraction and analysis of retinoids from animal tissue.
To purify, you need to perform chromatography. You can differentiate the retinoids using UV/visible spectroscopy. Beta-carotene is easy to detect; it is bright orange in colour and has an absorption maximum at 450nm.

References:
[1] . Fikselová M., Šilhár S., Mareček J., Frančáková H., Czech J. Food Sci., 2008, 26 (4), 268-274.
[2].  Gundersen T. E., J. Neurobiol., 2006, 66 (7), 631–644.

Thank you, do you think the same procedure would work with water-soluble proteins?

@almosnow Extraction of water soluble vitamins may be trickier because in that case you should definitely precipitate and separate out proteins as both of them are water soluble. Same method (solvent extraction) will not work but depending on the type of protein you can use different procedures.


text:
Why do silver nitrate and sodium hydroxide react to produce silver(I) oxide?
I did an experiment in class combining the aqueous solutions of $\ce{AgNO3}$ and $\ce{NaOH}$. I predicted that $\ce{AgOH}$ would form but it didn't, instead a brown precipitate. I found out it was silver oxide. But why does this happen?
I researched and found out that silver is a noble metal and is thus resistant to oxidation, so why is it oxidised in this case?
Also why is $\ce{AgOH}$ not formed? I looked at the reactivity series and apparently silver can't displace $\ce{H+}$ ions out of solution, is that why?

It's not oxidised. This is just a precipitation reaction.

oh. but then why is $\ce{Ag2O}$ formed instead of AgOH?

Silver in silver oxide is no more oxidized than in $\ce{AgNO3}$. So you should ask yourself the same question earlier, even before the reaction. Yes, noble metals are somewhat resistant to oxidation. But still they can be oxidized, and thus can form compounds, of which $\ce{AgNO3}$ and $\ce{Ag2O}$ are the examples.
Even gold can be oxidized, though that requires some extreme measures.
As to the formation of oxide instead of hydroxide, there is no clear-cut distinction between Ag and the rest of metals. Many insoluble hydroxides are not quite stable. $\ce{Cu(OH)2}$, for example, would slowly lose water and turn into oxide; likewise, $\ce{Fe(OH)3}$ turns into $\ce{FeO(OH)}$. Silver does the same, just a little faster.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Is there any law for pH conservation in chemical reactions?
I would like to know (just curiosity) if there is any law for pH conservation in chemical reactions. (like conservation of momentum in dynamics).
EDIT
The reaction I would like to take as example is fuel combustion. Too many sources (mostly media) stating that acid products (NOx and COx dissolved in water) are obtained.
My idea is that acids cannot be created from nothing. Either the inputs of the reaction were acid or the output is not only acid but also basic outputs.
Thus the question whether this reasoning is true.

Welcome to chem.SE! BTW, do you know what pH is, how it's measured and defined? It's very weird to make an analogy of pH and momentum.

sincerely I don't remember from high school

but i remember that base(pH > 7) + acid (pH < 7) neutralises pH (= 7)

There is no such law, much like there is no law of conservation of temperature, or indeed concentration. Surely, when you mix a hot substance with a cold one, you typically get something with temperature between the two. Thus one might be tempted to propose a hypothesis of conservation... But no, there are plenty of reactions when temperature just rises or drops, out of the blue. Same thing with pH.

Conservation of temperature in the example you provided (one hot mass with one cold one) derives from the law of conservation of energy. Meanwhile the number (not concentration) of H atoms is unchanged in chemical reactions.

The very example you gave as part of your question - formation of acidic products from combustion of neutral fuels - reminds me that the name "oxygen" comes from the Greek, meaning "acid-forming", because that is what happens in combining oxygen with many elements, especially the non-metals.

Well, let's take 1l of some strongly acidic solution with pH=0. Dilute it with water till it becomes 10l. Now it has pH=1. What is conserved here, really? Certainly not pH. Number of $\ce{H+}$? Maybe... Then take some water (just neutral, pure water) and add $\ce{SO3}$. All of a sudden it becomes acidic, though we certainly did not add even one atom of H.

@IvanNeretin Second example left hand $\ce{H2O}$ + $\ce{SO3}$ right hand $\ce{H2SO4}$. Number of atoms of H unchanged. I understood that pH is a concentration property (it is not intensive), so it is correct to say that hypothetical law doesn't exist, but turning at the definition of acid: "The Brønsted-Lowry definition is the most widely used definition; unless otherwise specified, acid-base reactions are assumed to involve the transfer of a proton (H+) from an acid to a base." See Arrhenius acid...

No, there is no law requiring pH conservation. And such a law would be chemical nonsense. Why? Read on.
You need to recall what pH actually is. It is defined as the negative base-10 logarithm of the activity (similar enough to concentration for almost all practical purposes) of positively charged hydrogen ions in solution.† Or in equations:
$$\mathrm{pH} = -\lg [\ce{H+}]$$
Note that most conservation laws in physics somehow relate back to the conservation of energy or something related. However, the pH of a solution cannot be easily traced back to any state function or anything else that would warrant its conservation.
In fact, the only thing conserved is the atom itself. Hydrogen, if $\ce{H+}$ is somehow present in a reaction or solution, does not disappear, it needs to be put in on one side of the equation and turn up somewhere else. In reactions that liberate protons of any kind, usually bases are added to capture them. In those that require protons, acids are added. What (if anything) is conserved is the number of hydrogen atoms but if they are not active as $\ce{H+}$ in solution they do not count towards pH. Therefore, there is no sense in postulating pH conservation.

†: $\ce{H+}$ as is does not actually exist in solution. One can treat it as $\ce{H3O+}$ for all intents and purposes, but actually it is better to think of it as a proton being shuffled back and forth by multiple water molecules. I think I recall four being stated as the lowest count in undergrad education.

Yes there is a law , but not exactly for pH conservation as pH is just a notation for concentration ( -log of it , and If you've known that log is basically introduced to make calculations and representations simpler). That is the law of chemical equivalence in a chemical reaction. Any reaction in the whole world holds this law. It states that number of equivalents in a chemical reaction are conserved. For more detailed analysis you can search it over the web.

But it's not a law of pH conservation - which is what the question is about.

I guess I stated it in the starting that pH is basically a notation. It is not an intrinsic property like momentum that it could have been conserved by such law.


text:
Reaction of Diazonium salt with Grignard reagent?
Why doesn't Grignard Reagent undergo Nucleophilic Substitution with a Diazonium Salt? for example if we want to convert phenol to p-cresol. Then first nitrating it, then convert it to a diazonium salt and treating with a methyl grignard reagent. But the reaction dos not happen. Instead we have to use Gilmann Reagent for this process. Why?

Phenolic group reacts with Grignard's reagent.

@Mithoron I'm asking about a Diazonium Salt. Also a Phenol with Grignard Reagent gives us phenoxide which is not my objective.

Well it's like you answered yourself - you have acidic -OH group all the time.

ok I got it @Mithoron . You are saying that the Grignard Reagent will react with the phenolic group

This explains the reactivity of Grignards in more detail: http://chemistry.stackexchange.com/questions/3959/why-do-organolithium-or-grignard-reagents-act-as-nucleophiles-and-not-as-bases-w


text:
Calculating the concentration of a diluted solution
I have this problem:

From a stock solution of initial concentration of $10~\mathrm{mg/l}$, I take $2\,\mathrm{ml}$, and then I add $10\,\mathrm{ml}$ of distilled water.

I am supposed to calculate the final concentration. I have solved the problem in this way.
$$\begin{align}
C_1 \cdot V_1 &= C_2 \cdot V_2 \\
C_1 &= 10~\mathrm{mg/l} \\
V_1 &= 2~\mathrm{ml} \\
V_2 &= (2~\mathrm{ml} + 10~\mathrm{ml})= 12~\mathrm{ml} \\
C_2 &= \frac{C_1 \cdot V_1}{V_2}  \\
&= \frac{(10~\mathrm{mg/l}) \cdot (2~\mathrm{ml})}{12~\mathrm{ml}} \\
&= 1.67~\mathrm{mg/l}
\end{align}$$
Is this correct?

Dear anonymous user, please do not replace mg/l with ppm, as they are entirely different concepts and often not even remotely interchangeable!

The answer is basically correct (see note at the end of my answer about significant figures) but there is a simpler method, using a dilution factor.
The original volume of your solution was 2 mL, and the final volume was 12 mL, so the dilution factor is simply $\frac{2}{12}$.
The original concentration times the dilution factor gives the final concentration:
$$\pu{10\frac{mg}{L}}\cdot\frac{2}{12} = \pu{1.7\frac{mg}{L}}$$
Note that only 2 significant figures were given in the problem and your answer reported 3.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why aren't resorcinol and catechol classified as dihydric alcohols?
Why aren't resorcinol and catechol classified as dihydric alcohols? They contain two alcohol groups.

Note that, according to IUPAC recommendations, resorcinol and catechol are phenols (not alcohols).

"Hello, my name is OH group and I am alcoholic". No, it does not work this way. There are entire classes of compounds (carboxylic acids, for example) that are not considered alcohols, despite containing a hydroxy group.

Related: Differences between phenols and alcohols

As @Loong pointed out in his comment, resorcinol and catechol are phenols, not alcohols. Although both classes of compounds contain an O-H group, they are different in that the oxygen in phenols is bonded to an arene ring such as benzene or napthalene.
Importantly, a lone pair on the phenol oxygen is able to interact with the aromatic $\pi$ system via resonance, which gives phenols characteristically different properties to normal alcohols.

The lone pair delocalisation increases the electron density in the ring dramatically at the ortho and para positions, as indicated by the negative charge at these positions in the resonance structures. This makes these positions much more reactive towards electrophiles than they are in unsubstituted benzene.
The lone pair delocalisation also occurs in the phenoxide ion, the conjugate base of phenol, which makes it much more stable than a regular alkoxide ion, and therefore makes phenols much more acidic than alcohols.


text:
Why is helium a liquid at absolute zero?
Why is helium a liquid at absolute zero? And why is it the only non-solid element at 0K? At 1 atm.


text:
When can a molecule be considered freely rotating at room temperature?
This question sparked from a long discussion in chat about the nature of $\ce{H2O2}$ and whether that molecule can be considered to rotate around the $\ce{O-O}$ axis (and hence display axial chirality) or not.
Considering two rather clear cases:
Ethane is considered to rotate freely around the $\ce{C-C}$ bond. The activation energy for rotation (equivalent to the energy difference between the staggered and eclipsed conformations) is given as $12.5\,\mathrm{\frac{kJ}{mol}}$.
Ethene on the other hand is considered to not freely rotate around the $\ce{C=C}$ bond. The energy difference between the planar and perpendicular conformations (the rotation barrier) is given at $250\,\mathrm{\frac{kJ}{mol}}$. (I am unsure of this value which I found on Yahoo answers. If I understand correctly, this should correspond with the excitation from $\unicode[Times]{x3C0}$ to $\unicode[Times]{x3C0}^*$ which I found elsewhere as $743\,\mathrm{\frac{kJ}{mol}}$.)
Somewhere between these two extreme cases must lie some kind of barrier around which rotation is severely hindered to entirely inhibited.

Attempting to calculate this myself, I remembered the Boltzmann distribution: If two states differ in energy by $\Delta E$, their relative populations $F_1$ and $F_2$ can be calculated as follows:
$$\frac{F_2}{F_1} = \mathrm{e}^{-\frac{\Delta E}{k_{\mathrm{B}} T}}$$
($F_2$ being the higher energy state and $k_\mathrm{B}$ being the Boltzmann constant.)
If I calculate relative populations from the energy differences $12.5, 19, 25$ and $250\,\mathrm{\frac{kJ}{mol}}$ for ethane, butane, $\ce{H2O2}$ and ethene, respectively (which correspond to $0.13, 0.20, 0.26$ and $2.6\,\mathrm{\frac{eV}{particle}}$) I get the following results at $300\,\mathrm{K}$:

Ethane: $6.55 \times 10^{-3}$
Butane: $4.79 \times 10^{-4}$
$\ce{H2O2}$: $4.29 \times 10^{-5}$
Ethene: $2.09 \times 10^{-44}$

Or logarithmic (ln) values of:

Ethane: $5.03$
Butane: $7.64$
$\ce{H2O2}$: $10.1$
Ethene: $101$

Again, somewhere between butane and ethene must be a threshold-ish value, above which free rotation at $300\,\mathrm{K}$ cannot be considered, but what range are we looking at; similar to what energy difference does this correspond to?
Asked the other way round, if I have an energy barrier of $25\,\mathrm{\frac{kJ}{mol}}$, what would I need to put on into the equation to calculate the temperature above which free rotation can be assumed?

The undergraduate organic chemistry text I teach form (Klein, Wiley) lists the barrier in ethene as 63 kcal/mol =264 kJ/mol, but gives no references. Propagations of values like this one in undergrad texts are probably the source of the 250 kJ/mol value you found.

@BenNorris Do you have a different value with a reputable source?

Alas, I do not. I wish I did.

M. H. Wood, Chem. Phys. Lett. 1974, 24 (2), 239-242 calculates 63.2 kcal/mol and references the exp. value as 65 kcal/mol. They also reference a more accurate calculation, i.e. CI, as 63.7 kcal/mol. This is just the first one I opened, there might be more accurate measurements and calculations by now, however, I doubt revolutionary change. (cc. @Ben)

Please be aware that with your Boltzmann approach, you calculated the population of the transition state based on your assumed ground state. This is probably fine when you deal with ethene/ethane, but it might be problematic for butene/butane.

I'm not a transition-state physical chemist, but I think a good approach to this problem is transition-state theory, specifically the Eyring equation:
$$k = \kappa \frac{k_b T}{h} e^{\frac{-\Delta G^{\ddagger}}{RT}}$$
This equation tries to predict the rate constant $k$ from an assumed pseudo-equilibrium between the transition-state and the starting material, which are assumed to differ in energy by $\Delta G^{\ddagger}$.  $k_b$ is Boltzmann's constant, $h$ is Planck's constant, $T$ is temperature, and $\kappa$ is the "fudge factor" or transmission coefficient, the fraction of transition states that react productively instead of converting back to starting material.
In the conformational reactions you are talking about, I think a good starting place would be to assume the eclipsed conformations are the transition state, although I don't know if this has been shown to be 100% true experimentally or computationally.   As a first approximation, let's assume there isn't a huge entropy term in $\Delta G^{\ddagger}$, so that it is well-approximated by the energy level differences you found.  If we also assume that $\kappa$ is 1.0, then we can just multiply the $e^{\frac{-\Delta E}{RT}}$ you've already calculated by $\frac{k_b T}{h} = 6.2\times10^{12}\mathrm{~s^{-1}}$ at 298 K to get (very) approximate rate constants.

Ethane: $6.55 \times 10^{-3} \rightarrow 4.1 \times 10^{10}\mathrm{~s^{-1}}$
Butane: $4.79 \times 10^{-4} \rightarrow 3.1 \times 10^{9}\mathrm{~s^{-1}}$
$\ce{H2O2}$: $4.29 \times 10^{-5} \rightarrow 2.0 \times 10^{8}\mathrm{~s^{-1}}$
Ethene: $2.09 \times 10^{-44} \rightarrow 1.3 \times 10^{-31}\mathrm{~s^{-1}}$

That shows that rotational equilibrium of everything except ethene should be very rapid, but that the half-life for ethene equilibration much greater than the age of the universe ($4 \times 10^{17} \mathrm{~s}$).
Suppose we want to find an energy difference that should lead to a rate constant that is neither very fast nor very slow, say we'd like $k\approx\mathrm{1~hr^{-1}}$.  Inverting our very very crude approximation gives:
$$\Delta E = - RT \ln{\left(\frac{k h}{\kappa k_b T}\right)}$$
Plugging in $k=\mathrm{1~hr^{-1}=2.7\times10^{-4} s^{-1}}$ gives (if I did my math right) 93 kJ per mol.  Some well-known atropoisomers do have estimated energy differences in that range.

If $\kappa = 1$, wouldn't that mean that all transition states react productively? If so, wouldn't a fudge factor of $\kappa = 0.5$ be a much better model? Actually, no it probably wouldn't, because the molecule is rotating towards the transition state, so it's more likely it'll rotate on. Maybe $\kappa = 0.75$? ;) Also, I'll take a large slice of fudge cake, please.

@Jan: In any case, if we're just comparing similar molecules that presumably have similar $\kappa$ values, the actual value doesn't really matter much (since it just scales all the rates proportionally).

Wow, thanks for the reverse calculation! Assuming it to be correct that actually is right what that article said (and what I skimmed across before but chose not to mention)! I would +1 a second time ;)

TST suffers from many more problems, than the "correct" value (choice) for $\kappa$. I think for small energy barriers as they are expected for a free rotation it is quite fair to assume, that all molecules entering in the TS ensemble also convert to the other rotamer. This has probably by far a lesser effect, than ignoring one degree of freedom in the TS.

Martin, not being a TST expert, I'd love to hear a short list of the biggest problems with TST.  Or should that be a new question?

@Curt I think wikipedia gives a brief summary of the general problems. I thought we have a post here, but the closest I could find was this. You can certainly ask about that, if the above does not help you.


text:
Is my reasoning for resonance structures correct?

given this picture, which resonance structure is correct? I'm thinking 1 and 3. You can't have 2 because the carbon which originally had a negative charge would be making 5 bonds, and you can't have 4 because the carbon on the right side of the new double bond would also have 5 bonds.
So is this reasoning correct? I'm just having a bunch of trouble drawing resonance structures.

Well, your answer seems to be correct but not your reasoning. In 2 and 4 there are 2 extra pi electrons in the system which seem to have come out of nowhere. The best way to solve is to draw all the resonance structures and see which ones match with the options.


text:
Concise description of primary organic reaction types mechanisms
Long story short, I'm mostly doing calculations in the past 5 years of my scientific carrier and every year I get less and less confident in my knowledge of organic chemistry which besides was never perfect. Yeah, I still remember pretty well some trivial things like nomenclature, classes of compounds, etc., and I'm pretty comfortable with MO theory, hybridization, etc., but all this are, so to speak, just the foundations of the organic chemistry. What I'm really bad at are the applications, primarily, the reaction mechanisms.
Many times reading a paper on computational studies of a reaction mechanism I caught myself not being able to (relatively quickly) grasp the main idea. I spent a fair amount of time just to recall the very basics of, say, the $S_N2$ reaction. Back many years ago I knew this things. Maybe I knew them not so perfectly and systematically, but even this knowledge of organic chemistry had evaporated almost completely from my head.
So I'm looking for a good resource to refresh and improve my knowledge of organic chemistry primarily (read: almost exclusively) regarding the most common reaction types and their mechanisms. A resource is counted as a good one if:

It is goes straight to the point (reaction mechanism) without discussing the basics of organic chemistry I mentioned, rather assuming that the reader knows them.
It is (relatively) concise and concentrated on the basics. In other words, I have neither time nor interest to read 2000 pages long Advanced Organic Chemistry book with all the details, exceptions, dirty laundries, etc. Just the basics.
On the other hand the basics has to be discussed in sufficient details, i.e. I would not be satisfied with the relevant Wikipedia (or similar) pages.
It is trustworthy. I mean, well established authoritative reference.

I admit I might ask for to much, but who knows, maybe there is something I'm looking for out there.
P.S. The type of a resource does not matter: textbook, lecture notes, MOOC video course, whatever.

It sounds like you're looking for a sophomore level organic book. It would be a few hundred pages, but you don't have to read the whole thing.

Love the question, and wish someone comes up a useful answer. My experience with organic chemistry and calculations is very disappointing: many reaction mechanisms seemed not more than educated guesses at best.

I would be happy to go through my somewhat outdated library and look for the right book for you, but it would help if I had a better understanding of what you wish to do with your calculations. Are you trying to computationally predict rates or outcomes of reactions given the reactants and conditions? If that is the case, you will need experimental data on rates and outcomes to test your methods with. Or someone to collaborate with that can generate such data.

@iad22agp, basically, what I wish is the following: given molecules A and B being able to say what types of reactions are expected and know their comventional mechanisms. I know reality is usually different from these simple models, but often not strikingly different. I need a starting direction anyway.

Maybe Grossman's  book would be helpful: https://www.amazon.com/Writing-Reasonable-Organic-Reaction-Mechanisms/dp/0387954686. You can skip the first chapter about basic stuff, the rest is 100% mechanistic oriented.

I like Clayden, Warren, Wothers, and Greeves for concise, well explained organic chemistry, but it's a big text and you'd have to prioritize areas for reading.

The book you are looking for is The Art of Writing Reasonable Organic Reaction Mechanisms by Robert Grossman.
One of the book's publishers offers this description of the book:

Intended for students of intermediate organic chemistry, this text shows how to write a reasonable mechanism for an organic chemical transformation. The discussion is organized by types of mechanisms and the conditions under which the reaction is executed, rather than by the overall reaction as is the case in most textbooks. The treatment emphasizes unifying principles, showing how common mechanisms link seemingly disparate reactions.
  Each chapter discusses common mechanistic pathways and suggests practical tips for drawing them. Worked problems are included in the discussion of each mechanism, and “common error alerts” are scattered throughout the text to warn readers about pitfalls and misconceptions that bedevil students. Each chapter is capped by a large problem set.
  The author has drawn on his own research and the current literature to ensure that appropriate attention is given to topics across the range of modern organic chemistry. The text is unique in its inclusion of a chapter on reactions mediated or catalyzed by transition metals, an area in which mechanistic understanding is now essential. Relatively new topics such as olefin metathesis and cycloaromatization are covered without giving short shrift to more traditional areas such as carbonyl chemistry. The text assumes a basic knowledge of organic chemistry. It can be used either in a formal course or by students working on their own, and will be particularly useful for graduate students studying for qualifying examinations. It will also be useful to students and researchers in biochemistry, pharmacology, and inorganic chemistry.

Each of the book's chapters are $50 \pm 30$ pages and are packed full of example problems.  Each chapter is organized according to the class of mechanism, not around particular reactants or products.  The chapters are:

The basics.
Polar reactions under basic conditions
Polar reactions under acidic conditions
Pericyclic reactions
Free-radical reactions
Transition-metal-catalyzed and -mediated reactions
Mixed mechanism problems

The author maintains a good web site with additional problems, additional worked examples, and answers to every problem in the book.  You can get a sense of the content of the book by browsing Amazon or by looking some of the answer .pdf files, available for free from Springer.
I found this book very helpful in improving my mechanistic understanding of various types of reactions.

If I were designing a course from scratch, I'd do something like this:

carbonyl addition and conjugate addition
nucleophilic substitution
elimination
reactions of enols and enolates
electrophilic aromatic substitution
pericyclic reactions
radical reactions (?)
carbenes (?)
organometallic chemistry (?)

The topics with (?) are important, but you can probably get by without them.

Here I'm in France so the link is in French sorry, but I turned my phone in english and this application is in english so if you have a phone which can have application from google there is this one which is really good :
Organic Reactions
Why this application is good ?

First because ten of the most used conpounds in chemistry how to synthesis them, what reactions you can do from them to obtain an alcohol for example and then how to protect them. It gives you the mechanism for most of them.
Second it has a name reactions module which have not far from a hundred of reactions for examples Dieckman condensation, Appel's reaction, the mechanism is written full and each molecules involved are in colors so it is really easy to see what is done.
Third, because it has a glossary with (not all it is impossible) but a lot of abbreviations such as for example (DHQ)2PHAL and it gives you what it looks like  which can always be useful.
This application is not all free (in fact I don't remember exaclty) but it is not expensive and you can test it with the free one if it is not.

I hope it can help you, it is easy to carry you can be offline obviously so nothing wrong with this app. It may have but very rarely a typo I may have seen one, one time, but nothing hard to understand.
I hope it can help you.


text:
What causes methane to have 25 times higher global warming potential than CO2
What causes methane to have 25 times higher global warming potential than $\ce{CO2}$?
I think it has to do with $\ce{CH4}$ being a less massive molecule, but I'm not sure.

First we have to look how the greenhouse effect works for all gases:
Light with a wavelength not absorbed by the atmosphere gets absorbed by the soil and heats up the earth. Because of black body radiation for 300K the earth starts to emit light itself with a maximum in the IR spectrum. Light with this frequency gets readily absorbed by vibrational excitation of molecules. But one has to keep in mind that intensity of absorption between different molecules and even between different vibration in one molecule differs a lot. This can not be explained with mass, or just vibration itself you actually have to look at https://en.wikipedia.org/wiki/Transition_dipole_moment or https://en.wikipedia.org/wiki/Selection_rule
But just believe me for a moment, methane is actually very bad at absorbing infrared light. Especially its by far worse than Water or $CO_2$.
But then there is another effect. You can not absorb more light than there actually is. So if you practically absorb 100% of the light for a given frequency then there is no effect in increasing the concentration of the molecule absorbing this frequency.
Both water and $CO_2$ absorb with nearly 100% in their frequency range (compare fig. 1 which shows the IR absorption of the atmosphere).
So in order to absorb more light you have to absorb in other frequency ranges.

This is where methane gets its role. If you look at figure 2 and keep in mind to change the wavenumber into wavelengths you will see that it absorbs in a frequency range that is covered neither by water nor by $CO_2$.

Yeah I found answer
It is because methane is able to absorb the infrared radiation that pass through carbon dioxide, which causes the molecule to vibrate and generate heat

Just keep in mind that all molecules vibrate and hence all molecules absorb some infrared radiation, only in different frequency ranges. Methane just happens to cover the range which is important to us.


text:
Tensile properties of cellulose fibres
In cellulose fibres, such as cotton or hemp, I've noticed different percentages of lignin and cellulose. How does this affect their tensile strength and toughness? Are there any other important factors? Thanks!

Lignin and cellulose in cotton, work together like the epoxy resin and  the glass fibers in  fiberglass composite materials (a fiberglass boat, for example). So, the cellulose is the fibrous component and is the primary load-bearing element.  While lignin is the polymeric matrix that provides stiffness and rigidity.
The more cellulose is in cotton, the higher is the tensile strength.

Does a wooden-stick work in a similar cellulose-lignin combination? Wooden stick has both cellulose and lignin. Mighr work in asimilar way.

I think so. Because it has both cellulose and lignin.


text:
p-value for ions in a solution?
The question in my class notes read as:

Calculate the p-value for each ion in a solution that is $2.00 \times 10^{-7}\ \mathrm{M}$ in $\ce{NaCl}$ and $5.4 \times 10^{-4}\ \mathrm{M}$ in $\ce{HCl}$.

I was confused how to approach it, but my teacher did the solution on the board with the work done out and obtained the values of $\mathrm p\ce{H}$, $\mathrm p\ce{Na}$ and then used the totals of those values to get the concentration of [$\ce{Cl-}$]. From that, she then took the negative log of the concentration of [$\ce{Cl-}$].
For reference, she obtained the following values:
 $$\begin{align}
\mathrm p\ce{H} &= 3.27, \\
 \mathrm p\ce{Na} &= 2.699, \\
 [\ce{Cl-}] &= 2.54 \times 10^{-3}, \\
 \mathrm p\ce{Cl} &= 2.595
\end{align}$$
I understand how one has to take the negative log of a given concentration to obtain the p-value of it in a "rote" type of sense.
But what I misunderstood was the log rules and the calculation of the first $\mathrm p\ce{H}$ value.
Her work for the calculation of $\mathrm p\ce{H}$ was as followed:
(1st step)
$\mathrm p\ce{H} = -\log[\ce{H3O+}] = -\log[5.4 \times 10^{-4}]$
(use the log rule, $\log(a \cdot b) = \log(a) + \log(b)$)
(2nd step)
$\mathrm p\ce{H} = -\log 5.4 - \log 10^{4}$
(3rd step)
$\mathrm p\ce{H} = 3.27$
But why in the 2nd step above is the $\log 10^{4}$ not have a negative value, such as $\log 10^{-4}$?
In her solution of p$\ce{Na}$ it was the case that the second additive had a  negative exponent, which can be seen below in the second step.
(1st step)
$\mathrm p\ce{Na} = -\log[2.00 \times 10^{-3}]$
(2nd step)
$\mathrm p\ce{Na} = -\log2.00 + \log10^{-3}$
(3rd step)
p$\ce{Na}$ $= -0.301 -(-3.00) = 2.699$
Also if anyone could explain to be why we sum the values of $\mathrm p\ce{H}$ and $\mathrm p\ce{Na}$ to obtain $[\ce{Cl-}]$? Does this have to do with the "stoichiometry" of something? My chemistry is not great, nor is my math. Thank you.

You should format both units and functions (such as M and log) in upright type in maths mode. This always works by enclosing them with \mathrm{ and }. I think for logarithms you can also use \log .

Also to answer the first question: Yes, that is a typing error. It should be $-\log \left (2.0 \cdot 10^{-3} \right ) = - \left (\log 2.0 + \log 10^{-3} \right ) = - \log 2.0 - \log 10^{-3} = - \log 2.0 + \log 10^3$

@Jan               So would: $pH = - \log{5.4 * 10^{4}}$, and then
$pH =−\log{5.4} − \log{10^{4}}  $ be correct or no?

Technically the calculation is correct but not realistic. However: $p\ce{H} = - \log 5.4 \cdot 10^{-4} = - \log 5.4 - \log 10^{-4} = - \log 5.4 + \log 10^4$

It seems like the calculation done or copied from the board by you has gone a little wrong.
In your first step when taking $-\log$ of the given concentration you have ignored the signs on the lower of the argument of the log function.
$$-\log 10^{-3} = \log 10^3 = 3 \log 10 = 3$$
since the base of log here is 10. And same correction for next steps.
And to your other question ... We add up the concentrations (adding up p values is irrevalent) because here NaCl is a strong electrolyte and HCl is a strong acid. They both dissociate completely and individually contribute to the concentration of chlorine anion.

Corrected , was a mistake . I'm sorry.


text:
What does all chromatography have in common?
We're learning about chromatography in class, and I'm confused about all the different types (e.g. those that separate liquid-liquid solutions vs. those that separate liquid-solid solutions vs. those that separate solid-solid solutions, etc.).
Could you provide a summary of the similarities between all forms of chromatography? (just an overview, do not require specifics)

The common factor in all chromatographic techniques is slowness, both absolutely and relatively! ;-) (Ducking now!)

Here are some similarities:

There is always a mobile (e. g. Gas - GC, Liquid - HPLC; GPC) and a stationary phase (liquid; gel - GC, solid - LCs, GPC).
Compounds in the sample interact different with both phases and are therefore held back stronger or lesser. This results in the accumulation of compounds that interact similar at some point in the system $-$ with some chromatographys, especially affinity chromatography, one might even get pure substances.
The mobile phase carrys the sample and is commonly less polar than the stationary phase, except for RPCs.
The structure of a chromatograph is basically like this:

transport system: reproductive transport of the mobile phase, adjustable (e. g. piston pump - LCs, compressed gas cylinder - GC)
sample application: import of the sample into the flowing mobile phase (e. g. injector, valve)
column: stationary phase and its mounting (e. g. capillary column in the column oven)
detector: transformation of the chemical or physical properties, which change with substance concentration, into electrical signals (e. g. UV/VIS-detector, thermal conductivity detector)
plotting unit: visualisation of the electrical signals (e. g. integrator, computer)

KhanAcademy video
You need to understand that the basis for this chromatography business is separating a mixture using a "stationary phase" and a "mobile phase."
I wouldn't take those terms as intuition, but column chromatography for instance. A fine powder, usually silica(Stationary phase) is put in solution with a solvent(mobile phase), then you add mixture A. The different compounds in mixture A will have different levels of "attraction" to the floating silica powder, and the part of mixture A that is "less attracted" to the stationary phase will flow out of the funnel.
One of the main issues is that sometimes this can take a very long time because the surface area of the silica may be too great decreasing flow out of the separatory funnel substantially - which is why we use high performance liquid chromatography.
It utilizes high pressure machines that offset the large surface area of stationary phase (pressure = force/surface area) thus reducing the amount of time it takes for separation.

Your answer seems to be limited to liquid chromatography. I agree, however, that the "stationary phase" vs. "mobile phase" thing is really the core of the matter.

The attraction part does not hold for size exclusion chromatography.


text:
How to calculate the boiling point of a solid substance?
Say that I have a solid substance; I know its density, mass, volume, melting point, specific heat, the pressure which it is under, and nothing else. (for now.)
How do I find the heat at which it boils? (So it goes solid -> liquid -> gas)

you can't...it must be found experimentally (it has) P.S Even if you know the BP at 1 atm, you can never calculate the real boiling point at any given BP, you may come close

As in, it cannot be calculated using the information at hand, or it must be found experimentally in any case? If it's the former, what other information do I need in order to find at least an approximation of the boiling point?

It cannot be calculated using the information at hand, not even approximately. As to what other information may help... well, you may ask about the enthalpy and entropy of vaporization, but then why not ask about the boiling point itself.


text:
Which metal hydroxides are soluble?
Alkali metals (Li, Na, K, Rb, Cs) and the alkaline earth metals (Ca, Sr, Ba - sometimes Mg is included also) form hydroxides that are well-known to be soluble or at least slightly soluble and create strongly basic solutions. Apparently these solutions of hydroxides are also known as alkali.
Since the more well-known d metals (Cr, Fe, Mn, Cu, Zn, Ni, Co, Ti etc.) forms no soluble hydroxide, I used to assume that apart from the 9 metals stated at the top, there is no more metals that corresponds to soluble hydroxides.
However, while searching my (non-English) version of Constants of Inorganic Substances: A Handbook by Lidin, Andreyeva and Molochko, I discovered that Europium (II) hydroxide was written as if it is soluble in a chemical equation in the Europium section. Later, I did search for information on the compound with the keywords ""Europium II hydroxide" soluble" , and I found some results:

europium (II) sulfate is scarcely soluble in water, while europium (II) hydroxide dissolves readily and gives an alkaline reaction

(Concise Encyclopedia Chemistry, page 392)
Is there any other research on the compound Europium (II) hydroxide? Is it really soluble?
Which metal hydroxides, apart from those of those group 1 and 2 metals, are soluble? Is there any specific reason why europium (II) is such an anomaly? If there are other such cations, is there a specific reason for that particular cation to form such a soluble hydroxide?

Thallium hydroxide is one example because the size of $\ce{Tl+}$ is appox. same to size of $\ce{K+}$ and thus thallium hydroxide dissolve in water easily and form a strong basic solution.

Thallium (I) hydroxide, of course.

Europium (II) is an anomaly because it is, well, "II".  Most metals in Period 5 or higher, apart from the alkali and alkaline earth metals, favor higher oxidation states which lead to highly insoluble hydroxides or oxides.  Europium is different because the $+2$ ion is stabilized by having a spherically symmetric electron configuration $[\ce{Xe}]\mathrm{4f^7}$.
So we have a $+2$ ion similar in size and spherically symmetric electron configuration to the heavier alkaline earth metals, and $\ce{Eu(II)}$ compound tend to resemble their alkaline earth counterparts -- including the solubility of the hydroxide.


text:
How does solubility of corrosion product affects rate of corrosion?
Why is it that if the corrosion product are soluble, then the rate of corrosion increases while insoluble corrosion products cause rate of corrosion to decrease?(How exactly would insoluble corrosion product protect the material?)

If the corrosion product remains on the surface, how will the corrosive reach new material to corrode? Instead, if the corrosion product floats off into the infinity of the liquid, a fresh surface is exposed for attack.

The process in general is called passivation. It means, that the surface of a metal is corroded and the corrosion product forms a protective layer. If the corrosion products are hardly soluble, it therefore blocks the metal beneath it from further corrosion. If the corrosion products are soluble, this protective layer can be washed away and the material is exposed to further oxidation.
In iron, for example, the initially formed iron hydroxides are somewhat soluble in water. The formed rust is flaky and friable, which basically means it is removable by little or no stress, i.e. rain can break the surface and expose more metal to the oxidising agent. It therefore usually always decomposes completely.
In another example, copper does form such a protective layer of copper oxides. This is also called patina. These layers are hard(er) to remove and protect the underlying metal from further oxidation. This is also the reason, why for example many church roofs and the Statue of Liberty are green.
This method is often used to protect metals from corrosion, i.e. the formation of the protective layer is induced chemically.

If you don't mind me asking, is there a way to determine theoretically if a given metal will undergo passivation?

@Shubham From the top of my head I cannot think of one criterion to be able to tell that. The solubility product certainly has something to do with it, but also how good the oxide "bonds" or sticks to the surface. I think this would make a good next question for the network, you can provide a link to that question for context.

I just posted the question, I referred to the order of passivity (from Ti to Cu) and asked how do we predict that series, and is it possible to predict if some metal would undergo passivation or not, can you suggest something else to add/edit?
this is the link:
http://chemistry.stackexchange.com/questions/37877/how-do-we-determine-if-some-metal-will-undergo-passivation-under-suitable-enviro?noredirect=1#comment58593_37877

Corrosion is nothing but actually the hydration of the metallic compound.
For example if you take rusting of pure iron you have to stick water molecules around it to make it rust or corrode. And we know that solubility is greatly affected by hydration hence the substance with more solubility corresponds to more corrosion as compared to one less solubility.

I don't quite understand your answer here. Perhaps you could clarify how it helps to answer the question.

bon is right, I don't think I understand the explanation either.

I guess I have interpreted the question as to the rate of corrosion of the substrate while questioner meant the corrosion agent's solubility. Can you confirm ?

Consider this example: Formation of PbSO4 in case of Pb in H2SO4 medium supresses further corrosion.

Sorry I don't have an answer to this. Maybe it's just a fact or a experimental analysis.

Well experimental analysis is what I am trying to avoid.


text:
Why are alkali hydrides strong reducing agents?
My book says that alkali hydrides are strong reducing agents. It also says that this is especially true at high temperatures? Is there any particular reason behind it?

Just like I said in the answer to you broader question, it's all about elements and their individual preferences. Hydrides (of any metals, not necessarily alkaline) contain hydride ion $\ce{H-}$, which is basically a hydrogen having one extra electron (or, in other words, a hydrogen in oxidation state -1). Now, hydrogen's electronegativity is relatively low for a non-metal, and so is its electron affinity. It can be said that hydrogen takes that extra electron almost reluctantly, and is ready to give it away to any oxidizer, even a weak one. This makes the hydride ion (and any compound which it is a part of) a powerful reducing agent.
In due time you will learn about the redox potentials which are a quantitative measure of the compound's oxidizing or reducing power. (Yes, they may depend on temperature and other things.) I'd rather not touch on that right now, for it's a long story.


text:
What is the internal energy at constant pressure?

What is the internal energy when a reaction is run at constant volume and $9.19\ \mathrm{kJ}$ of heat is absorbed, when run at constant pressure $8.62\ \mathrm{kJ}$ is absorbed? (for an ideal gas)

I know that internal energy for constant volume is equal to the heat then $\Delta U=9.19\ \mathrm{kJ}$, but what is the internal energy for a reaction at constant pressure? I know that $\Delta H=\Delta U+p\Delta V$ and from this I know that $\Delta H=8.62\ \mathrm{kJ}$ but how does it relate to the internal energy?

it is an ideal gas

If you are working with an ideal gas mixture, for which the internal energy of the reactants and products is not a function of pressure, the change in internal energy in going from reactants to products at constant temperature and volume is the same as the change in internal energy in going from reactants to products at constant temperature and pressure.  With regard to the changes in enthalpy, the same is true.  If the number of moles of reactants and products is the same, then the change in enthalpy for each of these processes will be the same as the change in internal energy.  However, if the number of moles changes, the change in enthalpy will differ from the change in internal energy in proportion to the change in the number of moles.

$$\mathrm dU = \delta Q + \delta W$$For an ideal gas expanding against an external pressure $p$:
$$\mathrm dU = \delta Q -p\,\mathrm dV$$
So at constant volume, $\mathrm dU = \delta Q$. Therefore $\mathrm dU = \pu{9.19kJ}$
If the reaction is identical, then $\mathrm dU = \pu{9.19kJ}$ for the ideal gas at constant pressure. From this we can then work out $\delta w$ at constant pressure as $\pu{9.19kJ} - \pu{8.62kJ} = \pu{0.57kJ}$.
Asking for the 'internal energy' and not the change in internal energy is meaningless here, because internal energy, $U$, is a state-function without a scale (unlike $S$) to define absolute values against.
So really you answered your own question in your question, $\Delta U=\pu{9.19 kJ}$


text:
Is it possible predict the oxidizing/reducing nature of a compound?
I have recently begun with Inorganic Chemistry and am finding it totally difficult to understand things (I am trying to minimize memorization and approach logically).
One of the things I am having problems with is understanding oxidizing and reducing agents.
Given a compound(and a few other necessary conditions in some cases), can I predict whether the compound will behave as a reducing or an oxidizing agent?
Well obviously if there is any such rule, there will be exceptions. But is there anything as such to reason out a few statements and take me partly towards the answer?
For example I have begun with Group 1 and the book says at some point that peroxides and superoxides of alkali metals act as strong oxidizing agents, group 1 hydrides are powerful reducing agents etc. Well I find it difficult to understand these statements and my book offers no explanation.

You may or may not be familiar with the concepts of electronegativity and oxidation states, but ultimately here is the rule of thumb you seek. If your compound contains highly electronegative elements not in their most negative oxidation state, expect it to be an oxidizer. If it contains elements with low electronegativity which are not in their most positive oxidation state, expect it to manifest at least some reducing properties.
On top of that, just like you anticipated, there is a whole lot of exceptions and subtleties which together constitute the magnificent field of elementary inorganic chemistry.

fortunately I do know about basics such as Electronegativity and Ox. states and redox potential and similar things.It would be great if you had a source or a link somewhere on the internet so that I could read and know more about it.

Well, if you are studying inorganic chemistry now, I guess you'll stumble upon a lot of links, sources, and facts to consider.


text:
Why does Avogadro's Law hold true for ideal gases?
Avogadro's Law states that the volumes of ideal gases are proportional to the number of molecules that are present in the gas (or the number of moles in the gas). I've been trying to find out more about it, but I can't seem to find out why this holds true, at least for ideal gases.

Avogadro's Law actually states that the volumes of ideal gases at a fixed temperature and pressure are proportional to the number of molecules that are present. This is actually a sub-case of the more general ideal gas law, which states $PV = nRT$ (or $PV = nkT$, if you're working on a per-particle basis) .
The ideal gas law actually falls out of a basic statistical mechanics analysis of the behavior of a collection of molecules which interact solely by elastic collision. (Here's a 25 minute video of one such derivation.)
Basically, the more particles you have in a given region with a certain fixed speed (i.e. a fixed temperature), the more frequently they collide - not only with each other, but with the walls of the container. This means that putting more particles in the (same size) container (at the same temperature) increases the outward pressure on the container walls. In fact, the rate of collisions (and thus the pressure) is directly proportional to the number of particles (doubling the number of particles at a fixed temperature doubles the rate of collisions).
But we want to hold the pressure constant! If we can't change the temperature, one way to reduce the pressure is to increase the volume (e.g. let the container expand like a balloon.) As the volume grows, each particle has to travel farther before it can collide with the walls of the container, meaning that the rate of collisions and thus the pressure on the walls gets less and less.
It just so happens that this relationship between pressure and volume is inversely proportional (that is, Boyle's Law). So to halve the pressure, we need to double the volume. So if we double the number of particles, to keep the same pressure we need to also double the volume. Or more generally, the volume is increased in direct proportion to the number of particles - that is, each mole of gas takes up a certain volume (at a fixed temperature and pressure). If we try and put it into a different volume, the pressure (or the temperature) will change.
Okay, but why doesn't a fixed number of different particles take up different volumes? For example, why doesn't doubling the mass of the particles make it take up half the space (or double, or a quarter ...)? - Heavier particles might hit the walls of the container harder, but at any given temperature heavier particles are moving slower then lighter ones, so the collision rate is less. When you run the numbers and calculate the velocity of the particles, the rates of collision, and then how hard they hit the container walls, the mass cancels out. At a fixed temperature, the contribution to pressure from a single particle colliding with the walls of the container is independent of its mass.
You can do the same sort of analysis with other properties. They are either 1) are explicitly discounted by the assumption of "ideal gas" (e.g. no particle interactions) 2) don't factor into the calculation at all (e.g the absorption spectra of the atoms) or 3) precisely balance out - leaving you with $PV = nRT$ being the relation for an ideal gas.

I think you could have made a bit more forceful statement -- By definition, ideal gas molecules interact solely by elastic collision which allows basic statistical mechanics analysis to model their behavior.


text:
Is this representation correct although it seems to show a pentavalent carbon?
While I was trying to draw a possible representation of $\ce{(CH3O)2CH=CHCH(OCH3)2}$ I was perplexed  by the fact that from left to right the second carbon atom has a double bound with the third one, while at the same time it is connected to two methyl groups, plus the H element.  That makes five bounds.
I found this as an answer:
$\hspace{25ex}$
But in my opinion, in the image, the double bound is misplaced as it should be on the from the first carbon atom after O.
I am stuck here.

The formula you have there is impossible because as you recognized, there is a carbon with five bonds.

@pigeon Your condensed formula corresponds to $\ce{C7H15O4}$, whereas your structural formula corresponds to $\ce{C8H16O4}$.

That's right, the molecule is symmetric around the double bond, and so should be the formula. One CH is missing.

Yes that is right . i did not notice this one! It is a mistake made in the book i am reading from.

Also the structure you have drawn has 8 carbons but the formula you give only has 7.

So, did you send a bug report to J. Clayden and S. Warren?


text:
do strong bonds have lower energies?
I was looking at a graph of the internuclear distance vs. the potential energy, and the strongest H2 bond had the lowest potential energy.
Is this always the case?

Well, yes, and this is not limited to bonds. The stablest anything has the lowest potential energy, because things just tend to find positions with lower potential energy and stay there. This is the very nature of our world.
On the other hand, in this case the energy required to get things out of this position (that is, to break the bonds, if we are talking about bonds) is the greatest.


text:
Meaning of the 1s orbital diagram
I'm a little confused about the interpretation of the 1s orbital diagram.
I know that the outer boundary is where the wave function of the 1s orbital has a constant value, which means means there's a constant probability of finding an electron on that outer boundary.
The reason I'm confused is because, isn't there an infinite number of distances from the nucleus where there's a constant probability of finding an electron around a cross-sectional slice of the sphere?
What I mean is, if we choose r, meaning distance from the nucleus as 50 pm, you could draw a circle representing that boundary as r = 50pm. You could do the same with an infinite amount of distances from the nucleus and have the same result.
Basically I'm confused as to how/why the boundary is chosen -- does the probability density need to go beneath a certain value for the boundary to be defined? Or is it arbitrary? If r < infinity, there's a finite probability of finding an electron at that distance. So doesn't this mean that the boundary is pretty meaningless?
Graphically what i mean is:

Why not:

?
Your help is greatly appreciated :D

You are right: the outer boundary is quite arbitrary. There is no any intrinsic threshold; the probability just gradually decreases lower and lower, but never reaches 0. You may draw a sphere so as to have the electron inside with 90% probability, or 95%, or 99%, or any other value as you see fit.
This, in particular, is the reason why atomic radii are somewhat arbitrary too.


text:
Entropy Change for mixing water at two different temperatures
Assume that the heat capacity of water is independent of the temperature.  Calculate the net entropy change when 1 mole of water at 0 degrees Celsius is mixed with 1 mole of water at 100 degrees Celsius. Assume that the heat capacity of water is (4.184 J/(K-g))(18 g/mol) = 75.3 J/(K-mol) and that the heat capacity of the calorimeter is negligible.
I was wondering if the following general methodology of my attempt was correct.
In attempting to solve this, I first found that the equilibrium temperature was (273+373)/2 = 323 K.
Then, I calculated the entropy by using the equation S = (heat capacity)*ln(T_eq/T_initial) for both 0 degrees and 100 degrees.
Afterwards, I added the two values for entropy and obtained 1.827 J/(mol-K) as the answer.
.

This looks great to me. I actually did a problem extremely similar to this the other night because it was part of my p-chem homework for the week. Conceptually, your answer makes sense because the change in entropy for this process is positive.

Well, your procedure is entirely correct. There's only one problem: the units. Since the number of moles of water is specified, your final answer should not include $\text{mol}^{-1}$. And since there's exactly $1 \text{ mol}$ of both the cold and hot water, the numerical answer should be the same.
I don't know where your error arises from but I suspect it may be because you used the units of $C_{p,m}$ (molar heat capacity at constant pressure) instead of $C_p$ (heat capacity at constant pressure). The former is an intrinsic property, whereas the latter is an extrinsic one, equal to the former multiplied by the amount of substance: $C_p = n\cdot C_{p,m}$. And their relation to the $\Delta S$ of heating at constant pressure:
$$\begin{align}
\Delta S &= C_p\ln{\left(\frac{T_2}{T_1} \right)} & &\text{Units: J K}^{-1} \\
\Delta S_m &= C_{p,m}\ln{\left(\frac{T_2}{T_1} \right)} & &\text{Units: J K}^{-1}\text{ mol}^{-1}
\end{align}$$
I'm entering nitpicking mode again so I shall also pedantically point out that as far as this equation is concerned, entropy comes with a triangle in front of it. Meaning you write $\Delta S$ instead of $S$.

The triangle is a “delta,” and it appears because you’re calculating the change in a state property (in this case, entropy).


text:
Minimum amount of water in pressure cooker
When cooking I would like to know the minimum amount of water I can add to my 6.2 L pressure cooker, worst case, so without any food. At second ring cooking the pressure is 0.8 bar.
I assume I need to solve $pV = nRT$ and know the density of water vapour at 0.8 bar, but that introduces $m = \rho V$, where I suppose $m$ and $V$ are unknowns?
Question
My high school chemistry is really rusty. Can someone help me in the right direction, and perhaps also tell if it at all something that can be solved without knowing the temperature?

Can't figure out exactly what you are trying to calculate. The minimum "about" (=amount?) of water to do what? What will hapen beneath or above the minimum?

Thanks. Yes, it should have been "amount". When pressure cooking you want as little water as possible, as it only dilutes the food, but some is required in order to get it under pressure.

You might want to post this question on the cooking.SE site. It sounds like this is more of a practical matter that someone who cooks with a pressure cooker can explain without resorting to gas laws and physical chemistry, subjects which will only ruin your appetite anyway.

Find the manual for your pressure cooker; it should tell you.  Hopefully it's online somewhere if you don't have a paper copy.

I think this question is on-topic and fits well here at chem.se.  The initial question as worded is slightly unclear.  I believe the question is, how much water needs to be added to a 6.2 L pressure cooker to ensure that the gauge pressure can reach 0.8 bar (of wet steam)?

In a pressure vessel, the volume $V$ is approximately constant. It is given as $V = 6.2\ \mathrm l$. However, this volume includes

liquid water
steam
air
food

in unknown amounts.
The gauge pressure is given as $p_\mathrm e = 0.8\ \mathrm{bar}$; i.e. the absolute pressure is approximately $p = 1.8\ \mathrm{bar}$.
We may use so-called steam tables to look up the properties of water at the given pressure (in the following, I use parameter values taken from the REFPROP – NIST Standard Reference Database 23, Version 9.0). We may find that the saturation point (equilibrium of liquid water and steam) at a pressure of $p = 1.8\ \mathrm{bar}$ corresponds to a temperature of $T = 117\ \mathrm{^\circ C}$.
At this point, the density of liquid water is $\rho_\mathrm l = 946\ \mathrm{g/l}$ and the density of steam is $\rho_\text{steam} = 1.02\ \mathrm{g/l}$.
If we ignore the volume of the remaining liquid water, air, and food, we may consider the limiting case in which the entire volume $V$ is filled with steam. Thus, the mass $m_\text{steam}$ is given by
$$\begin{align}
m_\text{steam} &= \rho_\text{steam} \cdot V \\[3pt]
&= 1.02\ \mathrm{\frac gl} \times 6.2\ \mathrm l \\[3pt]
&= 6.3\ \mathrm g
\end{align}$$
Therefore, only $6.3\ \mathrm g$ of water (which corresponds to about $6.3\ \mathrm{ml}$ of cold water at normal pressure) are required to fill the entire volume with steam at the given pressure.
However, a significant amount of excess liquid water should remain in the cooker in order for it to function properly. The liquid water is heated in the lower part of the cooker and boils at about $117\ \mathrm{^\circ C}$. The steam is distributed in the entire volume and condenses on colder surfaces (food and in particular the walls of the cooker). The condensed water runs down and it is collected in the lower part of the cooker. This circulation is essential for the uniform distribution of heat within the cooker; hence, the cooker should not dry out. The additional amount of liquid water should be sufficient to wet all surfaces (food and cooker) thoroughly.
Up to now, we have ignored the air that remains in the cooker. At initial (cold) conditions (pressure $p_0 = 1\ \mathrm{bar}$ and temperature $T_0 =20\ \mathrm{^\circ C}$), the density of dry air is approximately $\rho_{\text{air},0} = 1.189\ \mathrm{g/l}$. If we now ignore the volume of the liquid water, steam, and food, we may consider the limiting case in which the entire volume $V$ is filled with air. The mass $m_\text{air}$ is given by
$$\begin{align}
m_\text{air} &= \rho_{\text{air},0} \cdot V \\[3pt]
&= 1.189\ \mathrm{\frac gl} \times 6.2\ \mathrm l \\[3pt]
&= 7.37\ \mathrm g
\end{align}$$
The air cannot escape. At the final pressure of $p = 1.8\ \mathrm{bar}$ and temperature of $T = 117\ \mathrm{^\circ C}$, the trapped air is compressed to a density of approximately $\rho_\text{air} = 1.607\ \mathrm{g/l}$. The corresponding new volume is
$$\begin{align}
V_\text{air} &= \frac{m_\text{air}}{\rho_\text{air}} \\[3pt]
&= \frac{7.37\ \mathrm g}{1.607\ \mathrm{g/l}} \\[3pt]
&= 4.6\ \mathrm l
\end{align}$$
We see that a large part of the volume is taken up by the trapped air. The amount of water that is actually required to create steam at the given pressure is reduced accordingly.

It depends on the temperature at which you want to cook it, because, theoretically, you can also have 1.8 bar absolute pressure with only a few water molecules in there, if you just heat it up enough:
$$p = \frac{nRT}{V}$$
As you increase T at fixed values for n and V also p gets bigger.
So lets take 121 °C (394.15 K) for an example.
And expand our formula by $n = \frac{m}{M}$ and $m = \rho V(aq)$:
$$V(aq) = \frac{pVM}{RT\rho} $$
And insert all into it:
$$V(aq) = \frac{1.8E5~\mathrm{Pa} \times 0.0062~\mathrm{m^3} \times 0.018~\mathrm{kg/mol}}{8.314 ~\mathrm{J/(K~mol)} \times 394.15~\mathrm{K} \times 1000~\mathrm{kg/m^3}} $$
And out comes:
$6.1 \times 10^{-6}$ m$^3$ = $6.1$ mL of Water.
So basically thats the minimum amount of water that will produce this pressure when heated up to 121 °C in your pot. Every additional water will condense, resulting in wet steam and furthermore in liquid water on the inner surface of your pot.
Note: This really is only water vapor, without any liquid water, it would look absolutly transparent. And there is also no other gas pressure included, like the chamber would be free from any other gas (air etc.). Under normal conditions you would therefore need even less water vapor.

1.8 in the equation should be 0.8. Apart from that, it can it really be correct that I can get it to boil with less than 1mL of water? I would expect the right answer to be around 1dL.

@JasmineLognnes Normal ambient or atmospheric air pressure is 1 bar. The value 0.8 bar given for your cooker is the excess pressure (so-called gauge pressure) above ambient pressure. Thus, the total or absolute pressure is 1 bar + 0.8 bar = 1.8 bar.

Found the problem: accidently used the specific gas constant :) 6.1 mL is still not much water but sounds legitimate to me.


text:
Why elemental carbon is solid?
Carbon with 6 protons, the first of its group, is a solid, while the very next elements, nitrogen and oxygen, are gaseous in their elemental form.
Why carbon is a solid? If it's not the molar mass that defines carbon phase (carbon's molar mass is lesser)
In pnictogen and chalcogen group only the second element, phosphorus and sulfur, is solid.

Yet lithium, beryllium, and boron are also solids at room temperature - why not worry about them?

Li, Be, and B are also solids at STP, so the better question is why the transition from C to N is a major matter state change (solid to gas).

related http://chemistry.stackexchange.com/questions/18798/why-does-nitrogen-form-n%e2%82%82-and-not-n%e2%82%84-while-but-phosphorus-forms-p%e2%82%84-and-not-p%e2%82%82 - why nitrogen is gas is better question

Carbon forms large molecules such as the ring structure of graphite, or tetrahedrons of diamond. Boron, like carbon, forms relatively large covalently-bonded molecules.
Nitrogen, oxygen and fluorine are diatomic, so molecules are less massive. The inert gases He and Ne are monatomic and therefore have lower b.p. yet.
In a similar fashion, metallic bonds hold together Li, Be and other metals.

The obvious follow-on question is why carbon forms large covalently-bonded complexes and not diatomic molecules. (That is, why are $\ce{C2}$ molecules not stable?) Or alternatively, why do N and O form diatomic molecules and not large covalently-bonded complexes?

This is an extended comment. First notice that in any period, as we head towards noble gas configuration, we eventually become a gas. As we go down the periodic table, that transition is delayed, so by period 5 we transition from solid (I) to gas (Xe) in one step. In period 1, both elements (H, He) are gases at STP. In period 3, we get to gas at STP once we hit Cl, and the next element is a noble gas (Ar). Period 2 is the "most interesting" in that we get a solid to gas transition from C to O. That this transition eventually occurs in any period is not that interesting. But it is somewhat interesting why the C to O transition occurs some distance from the noble gas configuration (Ne). Given all of the properties I have discussed so far, it may just be a basic periodic property, but I am hopeful that a better answer is out there!


text:
Nomenclature of cycloalkane with substituent having the same number of carbons
What is the IUPAC name of the following molecule? More importantly, what is the rule for cycloalkanes having a substituent with the same number of carbons.

The undergraduate organic chemistry texts that I use when teaching tend to hide from complex nomenclature issues that cannot be addressed by the more common rules. Most texts clearly state that the parent chain in cases like these is the one with the larger number of carbon atoms.
For example, the following is butylcyclopentane.

The following is 1-cyclopentylhexane.

The addition of another substituent also makes it clearer, since we choose parent chains to maximize the number of substituents. The following is 1-cyclopentyl-3-methylpentane.

Related: IUPAC Naming: 2-phenyloctane or 2-octanylbenzene

Related: How do I decide the parent chain of an organic compound that has a ring as well as a side chain while naming the compound?

Many compounds can have two or more systematic names in accordance with several methods recommended by IUPAC. Concerning systems composed of rings and chains, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.
(…) For selection of a preferred IUPAC name, see P-52.2.8.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

The corresponding names for the examples given in the question are:

Method (1): “pentylcyclopentane” (preferred IUPAC name; ring preferred to chain)
Method (2): “1-cyclopentylpentane”

Method (1): “butylcyclopentane” (preferred IUPAC name; ring preferred to chain)
Method (2): “1-cyclopentylbutane”

Method (1): “hexylcyclopentane” (preferred IUPAC name; ring preferred to chain)
Method (2): “1-cyclopentylhexane”

Method (1): “(3-methylpentyl)cyclopentane” (preferred IUPAC name; ring preferred to chain)
Method (2): “1-cyclopentyl-3-methylpentane”


text:
Resonance Forms in Electrophilic Aromatic Substitution
In most diagrams for electrophilic aromatic substitution, the following is represented:

What happens to the other two resonance forms of the sigma complex? It seems like most of the time the first form is simply chosen to illustrate the return of aromaticity.
Suppose I switched the resonance structure in the second step to the following:

I think you are confused about resonance. The intermediate action only has one structure, which can be represented as a resonance hybrid. The individual resonance structures do not exist on their own so talking about switching between them is meaningless.

I am aware that resonance structures co-exist. I am also aware that in some resonance intermediates, one structure may dominate over another, depending on relative stability. Would the consideration of individual resonance forms be meaningless in, say, the bromination of asymmetric alkenes?

Yes? I'm not quite sure what you mean by that last question. My point is that you can draw a mechanism involving any of the resonance structures but the product will be the same because it is the same intermediate reacting in all cases.

Well I'm too lazy to open up chemdraw to make a nice drawing so my handwriting will have to do.

The way I see this kind of question is that, to interconvert resonance structures, electrons need to move. Of course, you'd represent that with curly arrows (top half of the picture).
Now for the deprotonation step, imagine you were using the first resonance structure. You'd draw the first two of the three arrows that I drew in the bottom half of the picture. (Or you might omit drawing the base, although for simple mechanisms like this I prefer having everything accounted for.)
What's the third arrow for (the one going from a double bond to the adjacent single bond)? It's the reverse of the arrow you used to interconvert the resonance structure. So you could think of the first two arrows being the "actual" reaction, and the third arrow being a "bookkeeping" arrow to make sure that everything ends up nicely aromatic.
Of course, that's kind of a flawed view because the reaction as shown has to have all three arrows or else you end up with a product that doesn't make sense. But using this idea of "the original reaction" + "reverse resonance arrow" works for every single instance of the question that you are asking, i.e. how to draw a mechanism involving an alternative resonance form. As long as you are able to draw arrows to show the interconversion between resonance forms, you should never run into a problem drawing a mechanism, regardless of what resonance form you use.
Now just a note on what you said - "It seems like most of the time the first form is simply chosen to illustrate the return of aromaticity." You're absolutely right about that. Using the first resonance form is just the simplest option since it involves the fewest arrows. Otherwise the writer of the book is just giving him/herself extra work... and nobody likes that.

It should be no different. If you label the carbon atoms clockwise as 1,2,3....6 starting at the ipso (or the top/12'o clock position) then as soon as the H in step 2 moves out, the bond pair would come between C1 and C2. Simultaneously, the vertical double bond (between C2 and C3) moves and comes between C3 and C4 (via resonance).