text:
Is change of pH of water with respect to temperature linear?
Only recently I learned that the pH of water varies with temperature, from about 7.47 at 0 °C to 6.14 at 100 °C.
My question is, first, is there a geometric explanation for this, in the sense that water can perhaps arrange itself in a way that partial bonding of oxygen to a third hydrogen can happen without simultaneously producing hydroxide ions to balance things out?
And--related--is the spread of pH above enough to create a detailed profile for the variation of pH with temperature? If so is it linear?

Note that the change in pH does not necessarily mean that the solution is not neutral, only that the dissociation increases at higher temperature.

The variation of pH with temperature is for pure water so $\ce{[H^+] = [OH^-]}$, or in other words the water remains "neutral".

basic means $\ce{[OH^-] > [H^+]}$, acidic means that $\ce{[H^+] > [OH^-]}$, and neutral means $\ce{[OH^-] = [H^+]}$

Water isn't really unique in that sense. The pKa  of acids and the pKb of bases will vary with temperature. We just normally assume everything is at "room" temperature. For water curve see https://en.wikipedia.org/wiki/Water_(data_page)#Self-ionization

Re: basicity means [OH−]>[H]+], i.e., a preponderance of [OH−] in a "neutral" solution. No if $\ce{[OH^-] > [H^+]}$ then the solution is basic. A neutral solution has $\ce{[OH^-] = [H^+]}$. But because of the temperature variation the autoionization constant  of water isn't a "constant" but changes with temperature. But again for all temperatures pure water will be neutral and it will have $\ce{[OH^-] = [H^+]}$.

@MaxW: Yes, I see now you are saying the water is neutral. I will delete some of my comments as a matter of housekeeping. Appreciate the comments.

Don't delete comments or the whole series of comments don't make sense...

Your comments are perfectly understandable and incorporate my (erroneous) assumptions. @Zhe's initial response also indicates the nature of my confusion. At this point I can at least articulate the confusion--that water may be "neutral" but still have a pH above/below 7. Is this correct?

Is the downvote to punish me for deleting comments?! Tough crowd!

"water may be "neutral" but still have a pH above/below 7" I think you've got it now.

@Zhe: Thanks--it is a new idea to me and I still have to process it, but appreciate the help.

$\text{pH} = -\log{ \ce{[H+]}}$ so the $\text{pH}$ doesn't depend on $\ce{[OH^-]}$

To summarize... in aqueous solutions:

Basic means $\ce{[OH−] > [H+]}$
Acidic means that $\ce{[H+] > [OH-]}$
Neutral means $\ce{[OH−] = [H+]}$

There is a for the Wikipedia article for the autodissociation of water. The  chemical equation for the autodissociation is:
$$\ce{H2O <=> H^+ + OH^-}$$
The mathematical relationship between $K_d^t$, $\ce{[H^+]}$, $\ce{[OH^-]}$ and $t$ is: $$K_d^t =\ce{[H+][OH-]}\quad\text{for }t > 32\text{ C}$$
Note that $K_d^t$ changes with temperature $t$, but for pure water at any temperature, $t$,  the water will be neutral, and $\ce{[OH−] = [H+]}$. The temperature dependency of $K_d^t$ also means that pH + pOH is not always 14.
The Wikipedia article shows that the relationship between $K_d^t$ and temperature is a curve. (The curve shown is a bit odd in that water boils at 100 C, so higher temperatures must be taken under pressure.)
Also regardless of temperature the $\text{pH}$ of an aqueous solution is defined as $\text{pH} = -\log{\ce{[H^+]}}$ so the pH does not depend on $\ce{[OH^-]}$.

This is a good summary of the comment discussion. Can you also address the main question of the post? Do we know whether the change in pH is linear over that interval? +1.

Oh--I do have a question--the pH does not depend on $[OH^-]$? But doesn't pH+pOH=14? That is an explicit dependence.

@Danial - No, that is the point. pH + pOH is not always 14 for a neutral solution since $K_d^t$ changes with temperature. If $\ce{[H^+] = [OH-]}$ then $K_d^t = [H^+]^2$ and $pK_d^t = 2 \times \text{pH}$

Thanks, somehow I missed the picture. This is I think a good (and complete) answer. Appreciate your patience with my questions! Will accept this very shortly.


text:
Why H2O is a liquid (water) and H2S is a gas?
I always think about this question.
Is it due to attraction or repulsion or is there something else that makes one of them a gas and the other a liquid (water)?

related https://chemistry.stackexchange.com/questions/63380/how-to-explain-the-periodic-trends-in-boiling-points-in-groups

The reason is hydrogen bonding.
The relatively large electronegativity difference between oxygen and hydrogen (about 2.3) is the reason why the partially positively charged (try to think why they are partially positively charged) hydrogens are attracted to the lone pairs on the oxygen atoms of other water molecules.
The relatively smaller electronegativity difference between sulphur and hydrogen (about 0.5) is too small to polarise the covalent bond to a required extent. Secondly, the lone pairs on the sulphur atom are delocalised in the sp3 orbitals and the d orbitals - an option unavailable to oxygen. So, the “negative charge” of the lone pairs is also distributed and the interaction is lowered.
So, the hydrogen bonding in water causes the water molecules to be associated with each other due to intermolecular forces of attraction, whereas in the case of hydrogen sulphide, these forces of attraction area much weaker.
So, lesser energy is required to overcome the forces of interaction between the hydrogen sulphide molecules than those between water molecules. This energy is available at room temperature and hence, hydrogen sulphide is a gas, while water is still a liquid.

In water molecule the more electronegative oxygen atom has small size and is present as a central atom, so that it forms the intermolecular hydrogen bonding strongly.
However, in $\ce{H2S}$ molecule the central atom sulfur is less electronegative and is bigger in size, so that it is unable to form the intermolecular hydrogen bonding.
Therefore $\ce{H2S}$ molecule can not exist in liquid form.

This is pretty much the same answer as the earlier one with the essential details stripped off.

Well, first of all depends on what temperature you are considering. But let’s take room temperature (20 degrees). In Water, the oxygen atom is highly electronegative and can polarize (partially) the hydrogen atoms, thus hydrogen-hydrogen bonds between the H2O molecules can be formed creating a very high boiling point. In H2S those bonds don’t exist, because sulfur is much less electronegative. Secondly, the lone pairs on the sulphur atom are delocalised in the sp3 orbitals and the d orbitals - an option unavailable to oxygen. So, the “negative charge” of the lone pairs is also distributed and the interaction is lowered.
So, the hydrogen bonding in water causes the water molecules to be associated with each other due to intermolecular forces of attraction, whereas in the case of hydrogen sulphide, these forces of attraction area much weaker.
So, lesser energy is required to overcome the forces of interaction between the hydrogen sulphide molecules than those between water molecules. This energy is available at room temperature  and hence, hydrogen sulphide is a gas, while water is still a liquid.

"In H2S does bonds don’t exist, because sulfur is much electro negative. " That doesn't make any sense: $\ce{O}$ is more electronegative than $\ce{S}$.

@Gert true, I might have a moment there where I skipped the thought

Not "hydrogen-hydrogen bonds".  They,re just hydrogen bonds", said once, and they are between hydrogen and a much more electronegative element with compact atoms (typically N, O, or F).

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text:
What gas has the highest refractive index?
Coming from this answer on the Worldbuilding stack, I'm trying to find a gas with a high refractive index in the visible light wavelengths. The highest index on this site, is chlorine at 1.000773. Is this the highest a gas can be? Or is there a fundamental limit on the refractive index of a gas?
As a note, I am not looking for the densest gas or the densest inert gas. Density does not seem to be correlated with refractive index in this case, as chlorine is not especially dense and the very dense perfluorobutane is not especially refractive.

I don't think there is a fundamental limit. Then again, benzene is not a gas at room temperature.

Well, for a high refractive index (in a simple Lorentz oscillator type of model) both (1) lots of stuff, and (2) lots of electron-photon interactions per bit of stuff. Hard to do with a gas at STP.

(1) For whatever gas it will surely depend on the temperature and pressure. So no real theory to predict it is a matter of looking RI for various gases at that temperature and pressure. (2) Would a supercritical fluid be within scope? For a gas there is not going to be a great density hence fairly small  refractive index. With a supercritical fluid you'd have a lot more matter at the solid surface of a planet.

Venus Gets Weirder: CO2 Oceans May Have Covered Surface

If pressure and temperature are specified (and perhaps a more specific wavelength than simply visible), this question might qualify for reopening.

For those who consider reopening, please see the relevant meta discussion.

The question has to be more specific to get a specific answer.
Even in the table linked in the OP, bromine gas (1.001132) has a higher index of refraction than 1.000773.
At zero degrees C, 5462 Angstroms, and atmospheric pressure, The Refractive Index Dispersion and Polarization of Gases (1936) reports:

$\ce{Si2H6}$ 1.0016574

$\ce{Si2H6}$ , disilane, is a gas at room temperature and atmospheric pressure.
The index of refraction is extremely dependent upon pressure.
As an introduction, see INDEX OF REFRACTION AND DISPERSION OF SEVERAL GASES IN CERENKOV COUNTER USE where indexes of refraction of gases as high as 1.02963 (for 182.79 psig ethane) are measured.

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text:
Local resolution of a crystal structure in PDB File (protein data bank)
I am wondering if it is possible to find the local resolution of a crystal structure. Usually, the resolution of the PDB is assigned the resolution of the highest resolved part of the protein/nucleic acid. However, it is actually a range of resolution (ie. 2-8 A). I am wondering if it possible to extract the resolution of each atom from a PDB file?

If you look at the PDB file, there is a range of resolution. This makes sense because different regions are resolved at different resolutions. Please see below:
REMARK   3   RESOLUTION RANGE HIGH (ANGSTROMS) : 3.00
REMARK   3   RESOLUTION RANGE LOW  (ANGSTROMS) : 20.00

Ah, do you actually want the temperature? http://proteopedia.org/wiki/index.php/Temperature

@gilleain I am aware of the B-factor, but it seems that it is not directly related to the resolution of the macromolecule. Would a larger B-factor necessarily mean a lower resolution? I always thought B-factor was the local flexibility of an atom.

@wanlei I think it's the best you can get. Also B-factor is the uncertainty, so a high B-factor certainly means lower resolution.

It's not directly related since higher uncertainty could come from sidechain flexibility.

The question has false statements:

Usually, the resolution of the PDB is assigned the resolution of the highest resolved part of the protein/nucleic acid.

Usually (in ~90% of the PDB entries) the experimental method used is X-ray crystallography. In such case the resolution refers to a distance between crystal lattice planes: it is assigned the value of the smallest distance between lattice planes that was resolved in the diffraction pattern.
Crystallographic experiment measures reflection. Each reflection corresponds to a set of lattice planes in the crystal. The spacing between these planes, d, is the measure of resolution. "Resolution range 3 - 20 Angstroms" means that the reflections used in the refinement had d between 3 and 20 Angstroms.
The resolution is simply dmin - the smallest distance between crystal lattice planes that resulted in a reflection usable in the refinement. Each reflection contributes to each point in the electron density map, so while we can talk about local quality of the structure, the resolution cannot be local.
Unless the data comes from a Cryo-EM experiment, where the resolution is defined differently and people can talk about local resolution.


text:
Elimination mechanism in alkenes
Do E2 elimination reactions occur in alkenes? If not, what mechanism is used?
If they do, are they limited by any stereochemical rules as alkanes do (antiperiplanar)?
My question is pernitent to the alkyne reaction mechanism, where both the E and Z alkenes undergo elimination to form the alkyne.
My guess is E2 mechanism is possible for the Z alkene, while the E alkene undergoes E1

No E1elimination. E2 in both cases. The E-isomer has a good overlap; syn elimination. The Z-isomer goes through anti elimination.

Thank you. I am guessing no E1 due to the instability of a vinylic carbocation
But I have never heard of syn elimination before
Are there any resources you'd recommend?

You are right about the carbocation. For syn elimination in vinyl halides, See https://chemistry.stackexchange.com/questions/23966/why-can-syn-periplanar-e2-elimination-happen-in-molecules-like-vinyl-bromides-bu


text:
Stability comparison of 2-methylpropene and trans-but-2-ene
Actually, in a book I found that  2-methylpropene is more stable than trans-but-2-ene. But, my teacher and also I am convinced that trans-but-2-ene is more stable than 2-methylpropene. My reasoning is that 2-methylpropene is less stable due to the steric repulsion of the -CH3 on one sp2 hybridised carbon. Also, 2-methylpropene is more reactive for the electrophilic addition reactions involving carbocation intermediate as a 3° carbocation will form and in case of trans-but-2-ene, a 2° carbocation is formed. Please say if I am correct and if not, please explain the correct comparative analysis.

Since these two compounds are constitutional isomers, their heats of formation may be compared directly. 2-Methylpropene (isobutene) has a heat of formation = -4.0 kcal/mol. Trans 2-butene's heat of formation is -2.7 kcal/mol. Isobutene liberates 1.3 kcal/mol fewer kcal/mol than trans-butene (E-2-butene) upon combustion. The more negative the heat of formation, the more stable. The book gets an A.
Resource: http://webbook.nist.gov/chemistry/


text:
Why do certain mixtures foam and others don't? How to predict if a mixture will yield a foam?
I recently acquainted myself with the skill of whipping a mixture of sugar, fresh brewed/ hydrated instant coffee into a pale, fluffy, stiff and stable foam. This, as some might know, is known as Café Cubano or Whipped Coffee in some of parts of the world (gives you a foamy and light textured espresso).
The other day, I tried whipping the mixture, just with the addition of some peanut butter in the start (I had tried it with peanut butter before, but had added it at a later stage) and it failed to stiffen and hold up any air.
I am familiar with how whipping cream works: the mechanism is easily available to be read on several sources over the internet, but am unable to figure out the mechanism for the above mentioned mixture.
Also, I want to know why an impurity (peanut butter, as I've yet explored) inhibits the foaming process. What decides if a mixture can be whipped and how can it be predicted if a given mixture will foam?
All insights are welcome. .)

Peanut butter has peanut oil. Oils kill any kind of milk or cream foam. Fancy chefs use a copper bowl to whip cream since copper can be cleaned of all oil. Plastic bowls retain an oil film and are not good for whipping cream.

@MaxW, as I've gathered, oils are responsible for inhibiting protein-based foams like whipped Egg Whites that depend on their protein networks to retain air while whipped Cream is a fat-based foam, which means that oils can not be expected to have any appreciable affects against foaming (and they don't). Further, copper bowls are used for Egg Whites as copper interacts with the sulphur in the egg whites with the effect of stabilizing the foam to a further degree, cold, heavy metal bowls are preferred for whipping cream as they help in maintaining an optimally cold temperature.

You are right. I confused egg whites with milk/cream foams. Thank you for the correction.


text:
How can copper(II) nitrate hexahydrate be synthesized in the laboratory as well as industrially?
I am researching the hexahydrate and found the same synthesis method being presented for both the anhydrous form and the trihydrate form.
Chemistry of the Elements Second Edition, N. N. Greenwood suggests that dissolving $\ce{Cu}$ into $\ce{N2O4}$ and ethyl acetate will give out the trihydrate by heating at $\pu{90 ^\circ C}$. $\ce{Cu(NO3)2*N2O4}$ is an intermediary product.
Syntheses and Physical Studies of Inorganic Compounds, C.F. Bell claims the same process to form the trihydrate.
And finally, the reaction I got from Ullmann is below:
$$\ce{Cu + N2O4  -> Cu(NO3)2* N2O4  -> Cu(NO3)2}$$
All of this just leaves me with more questions. How do I know if the final product has 3, 6 or no waters?
I would like to obtain the hexahydrate form so I need the proper conditions for this.
There is no mention anywhere of the industrial process or have I been looking in the wrong places? Ullmann just mentions dissolving $\ce{Cu}$, $\ce{CuO}$ and so on in $\ce{HNO3}$, but there is no elaboration on industrial tools and techniques to produce the hexahydrate specifically.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.Have a look at Acta Cryst. 2006, E62, i19-i22.
DOI: 10.1107/S1600536805041851 They report the crystal structure, they probably do also report how they have obtained it. (I can't check now, I have no access currently.)

It appears the authors of the above have been wrong. They now believe the central metal is nickel: Corrigendum. So far I could not find any mire conclusive evidence of the hexahydrate...

Thank you Martin for the corrections, I don't have access to the Corrigendum right now and I didn't understand your last statement. English is not my native language so I understood that you didn't find evidence of the existence of nickel or copper (II) nitrate hexahydrate (which I know should exist).

I'm happy to help out. I meant I couldn't find any crystal structure of copper(II) nitrate hexahydrate, except the one that is wrong. I also could not find a vendor, and I am starting to think this does not exist.

@Martin-マーチン These guys claim they used copper(II) nitrate hexahydrate from Merck, however I only found hexahydrates of Ni and Co nitrates on Merck's website.

Well, here it says that crystallization of a solution of cupric oxide/carbonate in dilute nitric acid yields the nitrate in the form of deliquescent hydrates. At below 24.5° C. the hexahydrate forms as blue, tabular crystals;


text:
How many watt-hours are theoretically necessary to produce 2kg of methane from sea water and air?
Given the composition of sea water in mol:
\begin{array}{lr}
  \ce{H2O}    &:& \pu{53.6 mol}\\
  \ce{Cl-}    &:& \pu{0.54 mol}\\
  \ce{Na+}    &:& \pu{0.469 mol}\\
  \ce{Mg^2+}  &:& \pu{0.053 mol}\\
  \ce{SO4^2-} &:& \pu{0.028 mol}\\
  \ce{Ca^2+}  &:& \pu{0.010 mol}\\
  \ce{K+}     &:& \pu{0.010 mol}\\
  \ce{Ct}~(\ce{CO2 + HCO3- + CO3^2-})
              &:& \pu{0.002 mol}\\
\end{array}
And the composition of air:
\begin{array}{lr}
  \ce{N2}    &:& \pu{78.08 mol}\\
  \ce{O2}    &:& \pu{20.95 mol}\\
  \ce{Ar}    &:& \pu{0.934 mol}\\
  \ce{CO2}   &:& \pu{0.04 mol}\\
\end{array}
My general problem: considering that these concentration do not change during the reaction (excess of reactive) and that the product ($\ce{CH4}$) is taken away immediately, how much energy is theoretically needed to produce $\pu{2 kg}$ of $\ce{CH4}$ from the optimal mix of these two phases?
My specific questions:

How can I be sure of which ions/molecule to definitely "get out" of the equation ?
How can I be sure which reactions are the best ones for my case ?
How do I do a dynamic equilibrium with more than 1 phase ?
How do I do a dynamic equilibrium with added energy ?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This appears to be a homework question, please share your thoughts and attempts towards the solution. If you receive useful answers, consider accepting one.

This is not homework ;-) It's personal development and curiosity.
And thanks for the Latex link, I'll try to learn it ASAP !

Please read the link that I have posted. This kind of question is considered homework here. It will probably be closed, before anyone hesitates to write up an answer.

Is it better now, or should I post the 3 question separately ?

Honestly I don't have any clue what you are trying to do. 2 kg methane equals to approximately 125 mol. The composition you have given is probably in mole fraction, not just mol (or the total amount of substance is missing). The carbon content is almost negligible, i.e. 0.002%, 0.04%. You would need a lot of it just to get some carbon. And how would you force (read: reduce) carbon dioxide to become methane? I really don't understand this.

Let's say we add some algae to the mix. Now we can ask how many watt-hours of solar energy are needed to produce 3 kg of hydrocarbons. Surely that question has been answered under the area of biofuels research.

For the sake of it, say I have unlimited amount of energy at hand (for example, I have a nuclear reactor onboard), and I am moving fast, so there is always enough reactive. What I am trying to estimate is how much energy I need to extract methane/carbon/any sort of fuel from the sea.
Algae do that very well, but they cannot be boosted by electric power.

Mostly because I find the idea of a floating solar platform producing LNG using air and salt water amusing I went ahead and looked up some of the calculations needed for this answer.
First, I would like to thank you for doing your research ahead of time for the composition of air and salt water as well as solving for the bond energies using bond-dissociation.
For your separation of materials you need to go to thermodynamics for the minimum energy needed to get your reactants:
\begin{equation}
   E_\mathrm{sep} \geq \Delta G_\mathrm{mix} \equiv \Delta H - T \Delta S_\mathrm{mix}
\end{equation}
Using most separation techniques we can state that there are no reactions between components and so no enthalpy change, adding an additional simplifications of simple mixing with no interaction between compounds and ideal solutions leaves us with:
\begin{equation}
   E_\mathrm{min} =
     \Delta G_\mathrm{mix} =
     - T \Delta S_\mathrm{mix} =
     -RT \sum_{i}(x_{i, \mathrm{initial}} \ln x_{i,\mathrm{initial}}
       -x_{i, \mathrm{final}} \ln x_{i,\mathrm{final}})
\end{equation}
Using carbon dioxide for an example and assuming magic technology that works at $\pu{298 K}$ to $100\%$ separation (I agree with G M that any modern process will likely be much more energetically expensive than the minimum thermodynamics predicts):
 \begin{equation}
   E_\mathrm{min} =
     -RT[x_{\ce{CO2}} \ln x_{\ce{CO2}} + (1 - x_{\ce{CO2}})
       \ln (1 - x_{\ce{CO2}})]
     = \pu{8.75 J/mol}_\mathrm{mixture} \text{ or } \pu{21.8 kJ/mol}_{\ce{CO2}}
\end{equation}
Using a similar treatment and your values for salt water, water takes $\pu{0.251 kJ/mol}$ water.

I think that's exactly what I wanted, I'll think on it tomorrow and accept it when I understand it fully.
Thanks !

Too much for making it of any practical interest
This question cannot be answered easily, there are too many factors. Bytheway it would not be practically convenient making such a process. With $\ce{CO2}$ as your only source of carbon, you would need an extremely energy dispendious air separation apparatus to get some $\ce{CO2}$. You can get then hydrogen from water electrolysis and use  Sabatier reaction (you need a Nickel catalyst) to get finally methane.

Well, if it is of practical interest for the algae, why can't it be of practical interest for us ? (+1 by the way)

@Nygael you did not specify the usage of micro-organisms. If you looking for a biochemical path you should specify it otherwise people could think you need an industrial process

I did not specify any path. I want the theoretical minimum. Biochemical is not magic, it's chemistry just like electrolysis and Sabatier reaction, so the energy used by algae should, in theory, be the same if the process is efficient.

Sorry, by rereading I think the last comment sounds a bit aggressive. It was not meant to be.

A solution based on bonding energy and on the accepted answer
Considering the potential energy of each of the chemical bonds involved :
\begin{array}{lcr}
  \ce{CH4} &:& 4\times \pu{413 kJ/mol}\\
  \ce{CO2} &:& 2\times \pu{799 kJ/mol}\\
  \ce{H2O} &:& 2\times \pu{366 kJ/mol}\\
  \ce{O2}  &:&         \pu{495 kJ/mol}\\
\end{array}
And the equation:
$$\ce{CO2 + 2H2O = CH4 + 2O2}$$
We have for $\pu{1 mol}$ of $\ce{CH4}$ a potential energy difference of:
$$(\pu{1598 kJ} + \pu{1868 kJ}) - (\pu{1652 kJ} + \pu{990 kJ})= \pu{824 kJ}$$
With the help of the accepted answer:
$\pu{21,8 kJ}$ is $\pu{6 Wh/mol}$ of $\ce{CO2}$ for dissociation.
Then, we need another $\pu{824 kJ}$, or $\pu{228 Wh/mol}$, for "retro-combustion".
To produce $\pu{1 t/year}$, we need $\pu{2.73 kg/day}$, which is $\pu{170 mol}$.
$$\pu{170 mol} \times (\pu{6 Wh/mol} + \pu{228 Wh/mol}) = \pu{39780 Wh}$$
With an average solar panel, we can produce $\pu{1000 Wh}$ per day per square meter. So in a theoretical world we would need $40$ square meter of solar panels to produce $\pu{1 t}$ of methane a year. Impractical, but not formally impossible.

That is completely neglecting that you need to concentrate the carbon dioxide first. That alone will consume much more energy than the final conversion to methane.

Isn't that what Rampalian answer is about ? But I know this is only theory, and that practically we are very far from achieving anything like this.


text:
Inquiry of the Changing of Canned Liquid’s State
Could you heat a can of water to 212 degrees Fahrenheit but stay below the aluminum or the can’s metal’s melting point. Either through putting the can in boiling water or heating it up. Would it be possible to have an unopened can of water vapor/gas state water? Or would the evaporated water condense and precipitate back into the water before I could observe it? In addition, would the pressurized can have an affect? Lastly, If I add more heat (energy) and speed up the movement of the water particles, will I be able to preserve the gaseous state longer so I could experience it?

You seem to imply that water will evaporate in a closed can. It won't.

There are good reasons any chemist advises and is advised against heating closed systems. Do not do that. Nevertheless, let's go through this item by item.

Could you heat a can of water to 212 degrees Fahrenheit but stay below the aluminum or the can’s metal’s melting point Either through putting the can in boiling water or heating it up. Would it be possible to have an unopened can of water vapor/gas state water?

Yes. At ambient pressure, water will boil at temperatures lower than the melting point of most metal cans. If there is a gas phase in the can (e.g. it is not entirely filled with liquid water), some would leave the liquid phase and enter the gas phase upon heating. The result would be a can that would contain an appreciable amount of water vapor. Note that heating a closed can be dangerous because it may turn itself into an open can, spewing hot vapor.
In addition, even in a completely filled can you may be able to obtain supercritical water - but pressurized $\ce{CO2}$ is a better choice for that experiment.

Or would the evaporated water condense and precipitate back into the water before I could observe it?

That depends. At some point, the can will cool down and most of the water vapor will condense to a liquid again. Now the observation: see safety caution above.

In addition, would the pressurized can have an affect?

If you mean: will this have an effect on the can? Yes, it might deform. See safety caution above.

Lastly, If I add more heat (energy) and speed up the movement of the water particles, will I be able to preserve the gaseous state longer so I could experience it?

Not sure how you would want to experience it, but of course, the hotter the can, the longer it will take for it to cool down.


text:
Is propane-1,2,3-tricarbaldehyde the correct IUPAC name of CHO-CH2-CH(CHO)-CH2-CHO?
We were asked this question in class:

What is the IUPAC name of the compound $\ce{CHO-CH2-CH(CHO)-CH2-CHO}?$

I don't know for sure, but I think the answer should either be 3-formylpentan-1,5-dial or 2-(formylmethyl)-butane-1,4-dial. However, I am told that the correct answer is propane-1,2,3-tricarbaldehyde.
How can that be? I searched on the internet and found that we use 'carbaldehyde' as the suffix only when the $\ce{-CHO}$ group is attached to a ring. But here the given compound is an open chain compound, so I don't see why we're using 'carbaldehyde' as the suffix.
My course textbook mentions that this compound is named so "to give identical treatment to all aldehydic groups". But I haven't learnt that such a rule exists in IUPAC nomenclature.
What is the correct name of this compound and why?

Related: When is the carbaldehyde suffix correct?

Related: 1,1-diformyl propanal

From the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, let's cite one of the rules for usage of the -carbaldehyde suffix, the relevant one:

P-66.6.1.1.2 The suffix ‘carbaldehyde’ is used when more than two $\ce{-CHO}$ groups are attached to an alkane.
Example:
$\ce{OHC-\overset{1}{C}H2-\overset{2}{C}H(CHO)-\overset{3}{C}H2-\overset{4}{C}H2-CHO}$
  butane-1,2,4-tricarbaldehyde (PIN)

Thus the correct name is propane-1,2,3-tricarbaldehyde.

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text:
How can this ketone show Cannizzaro reaction?

According to answer key, Cannizzaro reaction of 3-hydroxy-2,2-dimethyl-1-phenylpropan-1-one (a ketone) takes place but how is that possible when we know only aldehydes show Cannizzaro reaction?

There are a number of discussions on this site regarding the Cannizzaro reaction. In strong base there are competing reactions (aldol self-condensation) that can occur with ketones that have at least two hydrogens vicinal (next to) the ketone. Even aldehydes suffer the same fate. That is why the aldehydes are usually aromatic. But I believe in this case reduction may well be facilitated by an intramolecular hydride transfer. Intramolecular reactions are, as a rule, faster than intermolecular reactions owing to the high concentration of the reacting species. The keto alcohol 1 forms alkoxide 2 which reversibly adds to formaldehyde. Intramolecular hydride transfer occurs in intermediate 4. The formate group is saponified (5 ---> 7) forming the diol 9 and sodium formate 8. The deprotonation 7 ---> 8 should not be taken literally. Base and water intercede. Just saving space. See transition state for reduction below reaction mechanism.

So just to be more clear you're saying it's not following Cannizzaro but an intramolecular hydride shift, which happens to behave exactly like Cannizzaro and give the same product.

But the Cannizzaro reduction does involve a hydride addition to an aldehyde as you are used to studying it. To argue why this aromatic ketone reduces, a special mechanism is required. As a control experiment, I expect that tert-butylphenyl ketone (replace hydroxyl group with hydrogen) would not reduce or certainly not as fast. The hydroxyl group helps to deliver hydride by attaching formaldehyde. Ketone ---> secondary alcohol is a 2-electron reduction. Formaldehyde ---> formate is a 2-electron oxidation. The Cannizzaro reaction is a redox reaction.

To sum it up exception of a ketone showing Cannizzaro because of the conditions, right?

Yes! This is a special case.

In the 5th step, isn't the partial positive charge on the carbon of the ester-like system significantly reduced?(where the attack of OH- has been depicted) So would an attack by OH- be feasible there, or is it just due to the concentration of OH-?

@YUSUF HASAN: Do you mean the saponification of the formate ester in 5? Formate esters are readily cleaved with base, faster than most esters. I am not sure what you mean by reduced positive charge at the carbonyl. To me, it goes against experience.

@user55119 Ah yes, I got it when you used the word 'saponification'. I was confused a little there. Thanks!

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


text:
IUPAC naming of isocyanides
What is the IUPAC name of isocyanide : isonitrile or carbylamine?
Are "methyl isocyanide" and "carbylamine" the same thing?

Note that carbylamines is an obsolete and deprecated name for isocyanides (isonitriles).

The most recent IUPAC recommendations[1] states that:

P-61.9: Preferred IUPAC names are formed substitutively using the prefix ‘isocyano’ attached directly to a parent hydride.

This is a change from previous recommendations, and differs from some of your examples above.
To give a concrete example, $\ce{C6H5NC}$ should be named isocyanobenzene - this is known as a PIN by IUPAC (preferred IUPAC name). Phenylisocyanide is also acknowledged as an acceptable trivial name (a name that does not conform to the recommendations but is still commonly used).
[1]: IUPAC Recommendations for Preferred Names 2013, Chapter P-6. DOI:10.1039/9781849733069-00648

Can I get an example of an open chain compound with isocyanide group IUPAC name attaching suffix

Isocyanoethane. The nomenclature system is universal, you just need to know the correct prefix.

Okay that means in case of isocyanide we will not use suffix but in case of cyanides we will still use the suffix "nitriles"


text:
Calcium dynamics and its relationship with nitric oxide
I know that in synaptic transmission nitric oxide release leads to a reduction of calcium inflow. But how does this phenomenon happen quantitatively (for example: I suppose that when the nitric oxide concentration increases, the time derivative of calcium concentration becomes negative and viceversa)? Does a numerical relationship between nitric oxide and calcium concentrations exist, in this sense?


text:
How many carbons in this skeletal structure?

How many carbons are in this skeletal structure? I was told to draw a circle at the start of the structure and at each point the line turns off to get the total number of carbons but doing so would result in 6 carbons which I know is wrong but why is it wrong? And where on the structure are the carbons located?
Thanks for clearing this up for me!

OK, you've described a procedure and immediately told us that it is wrong. How do you know it is wrong?

Because 6 carbons just looks too many for the size of the structure in my opinion.

What a weird way of counting the carbons. Just convert all the points where the lines join into a "C", then you should see how many you have.

So as such this structure should have 3 carbons?

Sorry, I wasn't clear - this is in addition to the ones that are already shown as "C". What is going on here is that 'terminal' carbons (at the end of a chain) can optionally be shown as element symbols (CH3) - you could equally convert these ones to points, and get just a line drawing.

Oh, so 6 would be correct after all?

The IUPAC name is 3-methylpentane. Now, think of how many carbons are there. Actually, I am answering in this way as your question is not clear.
Have you studied about bondline representation of organic compounds? You can find this in books of elementary classes.

@user55394 yes, there are six carbons in the structure: the three explicitly written ones and three more at the points where lines intersect.

Your intuition is not wrong.
There are indeed 6 carbon atoms in the molecule (3-methyl-pentane).
With bondline/skeletal notation, there's a carbon atom at all the vertices; at the ends, and every time the line changes direction.
Counting the carbon atoms then becomes as simple as counting the vertices.
This becomes more apparent if you explicitly write all the atoms like illustrated below:


text:
Why is paper sometimes yellow?
Where I'm from, most paper you find/get at stores has this yellowish tinge. Why is it yellow and not white?
In my chemistry book it's mentioned that paper manufacturers "refine" paper from wood pulp, and that this involves treating the pulp with $\ce{CaCO3, TiO3}$, etc and this is what makes paper white.
But the paper that's commonly available here is yellow(ish). Is it because the manufacturers don't carry out the required treatment with $\ce{CaCO3}$ and $\ce{TiO3}$ (why would they do that anyway?), or could it be due to some other reason?
Also, if treatment with $\ce{CaCO3}$ and $\ce{TiO3}$ is what's used to color paper white, can someone tell me how the color change to white (from what would've otherwise been yellow paper), actually works?

My question is different from the proposed duplicate. In my question, I'd like to know why (relatively fresh) paper is yellow, straight out of the press. I'm not asking for why paper yellows over time.

Possible duplicate of Why does paper turn yellow over time?

@Mithoron  read the question carefully i did not ask why it becomes yellow over time i asked that produced paper itself is yellow from the very beginning of making

Paper is brownish-grey. You only make it white by bleaching or mixing it with various white inorganic powders.

@Karl now i understand , from the origin its yellowish colour then we use different .... to make it white . Thanks

Do you know the name of those inorganic powder ?

You already named them. ?!?

@Karl He probably wants confirmation ;-)

@Mith Think I'll have to agree with the OP here. That doesn't look like a dupe O:)

The one who edited my post really he expressed what is in my heart ... Thanks

@Karl If those powder are used then still why yellow paper come out ...

I see. Well, you're welcome for the edit (I was hoping I didn't deviate too much from what you originally said) O:)

You use titania IF you want white paper. The Umkehrschluss is 100% valid.


text:
If two atoms "fight over" one electron, who will win?
If two atoms of different elements are both missing exactly one electron in their valence shell so that they have a charge of +1
and we place an electron between them, exactly in the middle.
Which atom would exert a stronger force on that electron?

Wouldn't an atom missing an electron have a charge of +1? And I would think this would basically just be a comparison of electronegativity.

@Tyberius Thanks, corrected to +1.   would a comparison of electronegativity tell me the force exerted on that electron if it was hypothetically put at exact distance from both of these Atoms's valence shells ?

This is how Covalent bonds work. In reality, the electron is not really one single object, but a sort of "density of charge". There will be, loosely speaking, bits of charge on both atoms, but probably not in equal amounts.
At a lower level of theory, the electron will go to whichever atom has higher electronegativity.

@iammax I actually arrived with this question after trying to understand why wouldn't two atoms missing exactly one electron have the exact same charge or exerting the exact "pull", regardless of their element

that one which has the largest attractor

@soundslikefiziks while the net charge can be the same, the charge felt by an electron in two different elements will usually be quite different. This is because electrons already on an atom can shield the new electron from some of the positive charge of the nucleus. Conversely, with an electron removed from the element, a new electron would experience more of the nuclear charge.  Related: https://en.wikipedia.org/wiki/Effective_nuclear_charge

If two atoms of different elements are both missing exactly one electron in their valence shell so that they have a charge of +1 and we place an electron between them, exactly in the middle.

There isn't a nice simple answer to this.
Courtesy of quantum theory we know first and foremost that there's no single outcome possible - any possible result of this could happen in an individual case.
So what could happen is :

They could all remain apart
The electron could become bound to just one atom.
The electron could be shared by both atoms in a bond.

What would be most likely to happen depends on the details.  The first option is not very likely in normal conditions, but is possible in more extreme conditions.  The middle option is possible as is the third.
There is no convenient formula for working out energy levels in different atoms, molecules or ions so even working out the most likely configuration is not trivial.

Which atom would exert a stronger force on that electron?

This alone doesn't decide what happens because there is more to determining the most stable state that just the electromagnetic forces.  But there is an important reason why no one could answer this for you : you haven't even given the names of the ions, let alone important details like the temperature or density of the parts of the system.  Even with this info it would be a significant calculation (on a computer) to develop an answer.
There are two related areas for further study :

Molecular Orbital Theory
Valence Bond Theory

So, you are saying that even if one atom has the stronger emf, we would still have no way of being sure it would bound to it ?  what about statistically ?

Yes we do have the way. Give us the atoms, and we'll tell you what will happen.

@IvanNeretin  Any two elements that one is bigger than the other, Let's say Calcium and Magnesium. each missing one electron from it's Valence shell.  i thought there would be a simple calculation for the attraction force of each one.

There is a concept known as effective nuclear charge but it's not a great way to address the problem as it leaves out a lot of effects, but you can read this answer for some more info on that.  I believe the molecular hydrogen ion is the only one we have an exact analytic solution for and it's not a simple to work with solution.


text:
Can a covalent bond exist with one electron shared?
I was doing an exercise with formal charges on anions and it occurred to me that the issues with incomplete octets could be solved with sharing 1 electron between constituent atoms of the ion. To illustrate:

This would be the Lewis structure of a resonance form of the nitrite ion, without bringing in an electron from an external source to complete Nitrogen's octet. Nitrogen basically needs to fill it's octet, so it becomes charged -1 by bringing an electron in from another source.
My question - could the octet be completed if one of the Oxygen atoms shared one of its electrons with the Nitrogen? Like in a dative bond where oxygen contributes its pair of electrons, instead in this case it just donates a single electron? This would satisfy all octets without needing to have an 'extra' electron. What is the need for covalent bonds to share pairs of electrons? Can't covalent bonds exist where only 1 electron is shared between molecules? After all, NaCl is a case where a single electron is used in a ionic bond rather than a covalent bond, so what's the issue with using a single electron in covalent bond?
Hope I've been clear.

The dihydrogen cation, found in interstellar space, has one electron in a bonding orbital that encompasses both atoms.
See https://en.m.wikipedia.org/wiki/Dihydrogen_cation.


text:
Why is the electronic configuration of gold written in this order?
This is an example from my chem text.

Eletronic configuration for $\ce{^79Au}$:
  $$\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 \color{red}{4f^14} 5d^10 \color{red}{6s^1}}$$
  (Emphasis mine)

If I would follow the Linus Pauling diagram, I might write it as "$\ce{5p^6 6s^1 4f^14}$" rather than the example above. Why does the $\ce{6s^1}$ jump to the last position in the line?

I found a solution for myself. It happens because I had to withdraw the remaining electrons from the most energized layers, thus, I have to put it in evidence. I'm not sure if I delete the question or not. It's my first time here.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have prettified your post a little. If you have found the solution, you can add it as the answer below (it should be a little more elaborate than that one line though) and accept it, there is no need to delete your question.


text:
Why is negative charge on phenyl carbanion localized?
Phenyl carbanion has negative charge in conjugation π-bond. So, why resonance is not possible? Why is the charge localized here?

It would help to see it in context. I can only assume the ion was formed from deprotonation of benzene, hence the lone pair will be orthogonal to the pi-system.

https://chemistry.stackexchange.com/questions/50266/why-is-the-lone-pair-of-pyridines-nitrogen-atom-not-part-of-the-aromatic-ring

If resonance was possible, once carbon would have 2 pi bonds and 2 sigma bonds, which means that particular carbon would be sp hybrid. It's orbitals would require 180 angle which is not possible for benzene ring to provide(only 120 is possible because it's a hexagon and all) And this negative charge with have a significant repulsion from the conjugated electrons. So this negative charge, it's existence itself is really unstable.

That is wrong. Very wrong. The molecule is not nearly as rigid as you propose it to be. And it is equally not as unstable as you make it to be.

Provide me some data, saying that it's very wrong is just too vague. It is rigid because the system is planar and it's planarity is the reason for its stability. And you might be talking about some very very specific conditions under which it's stable. But in general case this won't even form. If you think of benzyne it's a partial overlap of p orbitals. That is different because that bond itself is incomplete overlap without harming the consistency of the structure. So before yelling 'wrong very wrong' enlighten me.

The molecule is isoelectronic to pyridine. And I am most certainly not talking about benzyne. There are plenty of sources, including this site, at your disposal. I'm sure I do not need to educate you in the comment section. With wrong, however, I refer to your generally much too simplistic approach to this kind of problem.

The word isoelectronic doesn't mean that it behaves the same way, or in no way have similar stability.
The difference in electronegativity of carbon and nitrogen is a major factor why carbanions are not stable. And when you talk about pyridine, those electrons are not delocalised. In some examples carbanions are stable, but only in specific cases and special conditions. Changing one atom in a structure makes it's properties to change drastically. So isoelectronic molecules and similar molecules are useful to group certain molecules but when it's about stability, I don't see any correlation.


text:
Is there any method to synthesize salicylaldehyde from salicylic acid?
I know salicyclic acid can be produced from salicaldehyde by treatment with $\ce{KMnO4}$.
Is there any way to reverse this, i.e- produce salicaldehyde from salicyclic acid?

There are reducing agents available that would reduce carboxylic acids. Trouble is that aldehydes are usually more reactive than acids, thus you often get the alcohol. But that can be selectively oxidized to the aldehyde and given the cheap starting material, the two-step procedure should be fine.

Salicylic acid can be converted to salicylaldehyde using a triangular or square wave alternating voltage. The process of this conversion is explained in this paper$\ce{^{1}}$but it is behind paywall so here's the abstract:

The effect of superimposing sinusoidal, triangular, and square‐wave
  alternating voltage (aV) on the potentiostatic reduction of salicylic
  acid to salicylaldehyde has been studied. Polarization curves with aV
  modulation were measured with an amalgamated copper rotating disk
  electrode. Enhanced dc current densities were observed with increasing
  magnitude of superimposed aV. However, aV frequency and type did not
  have significant effect on the resulting dc current densities. Batch
  cell electrolysis experiments were conducted with a H‐type cell. The
  conversion of salicylic acid and yield of salicylaldehyde were
  measured as functions of electrolysis time, aV magnitude, type, and
  frequency. The superimposed aV significantly enhanced the reaction
  rate, conversion of salicylic acid, and the yield of salicylaldehyde.
  The energy consumption of ac electrolysis was higher than that of dc
  electrolysis.

doi: 10.1149/1.2100342 J. Electrochem. Soc. 1987 volume 134,issue 12, 3074-3077

It's simple. Reduce with LAH then oxidise it with PCC or PDC.

That would be the first step of the reaction. It won't stop there.

A better oxidation would be to reflux the benzyl alcohol over activated MnO2 in a neutral solvent

I suspect that reduction of salicyclic acid with LAH can afford o-cresol. Salicyl alcohol, as its oxygen bound aluminate salts, can eliminate to form 2-methylene cyclohexadienone, which suffers reduction to o-cresol. Compare mixed anhydride reduction of salicylic acid with NaBH4. Synthesis, 2479 (2005). A fair comparison is the LAH reduction of indole-3-carboxaldehyde to 3-methylindole (skatole).


text:
Can hydrobromic acid perform substitution with propane?
I am a student in the 11th grade, we recently had a chemistry exam, I lost marks because of a question that asks about the reaction between $\ce{HBr}$ and $\ce{C3H8}$. According to our teacher  $\ce{HBr + C3H8}$ is going to react and form $\ce{C3H7Br + H2}$. In my opinion I don’t think that a reaction is going to happen, because $\ce{HBr}$ is an acid and acids can't react with alkanes. Can you please state your opinion and justification?

You are right, they are not going to react, because, er, well, why would they?

I’m not sure , he said that my statement is weak. I am trying to back it up with facts.

Well, technically, your statement is weak, because certain acids do react with alkanes. But HBr doesn't, that's for sure.

So C3H7Br won’t be formed , am I right? No chemical reaction is going to happen?

Yes, I am pretty sure, and if you wait for a while, I believe there are going to be more of us.

I hope so , I’ll be thankful for any contribution.

HBr certainly is not strong enough to oxidise an alkane.

I am trying to prove that , unfortunately I didn’t find any chemical equation that includes both C3H8 and HBr , usually substitution happens when an alkane reacts with Br2 , F2 , I2 etc.

What's unfortunate about that? Of course you didn't find any; that's because no such equation exists!

How can I prove that to my teacher?

This experiment has a +9.9 kcal/mol heat of reaction. No reaction! Read about Hess's Law and Tables of Heat of Formation here: http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#Thermochemistry

I wouldn't think this were an acid/base reaction, but more of a (photochemically initiated) radical process. The issue is that a propyl radical would still prefer to abstract hydrogen instead of bromine from hydrogen bromide... I'd have to think about this more, but that might just be a different way to state the stability arguments other posters have made...

So what am I supposed to do now , how can I support my argument?

I am afraid that you cannot support your argument. Here are two reasons why as I expand on my comment above. If this process were to be a radical chain reaction, the following process should be considered. Hydrogen bromide has a weaker bond then any of the C-H bonds in propane. An initiation step (heat, light?) would be expected to generate atoms of hydrogen and bromine. If you use a bromine atom as a reactant in propagation step 1 (P1), you will produce a hydrogen atom at the end of P2. So we start with a hydrogen and abstract a secondary hydrogen (primary hydrogen is higher in bond strength but it will only make the analysis less favorable) to produce a secondary propyl radical. The bond dissociation energies (BDE's in kcal/mol) are listed over the bonds of interest. The first step is good news. Exothermic by -5 kcal/mol. But P2 is endothermic by +14 kcal/mol. If you add P1 + P2 you get the overall reaction that has a heat (enthalpy) of reaction of +9 kcal/mol. The reaction is endothermic and uphill. The initiation step, the hydrogen atom and the propyl radical do not participate in the computation. The second method is to use heats of formation (in red). Using Hess's Law (ΔHorxn= ΔHfoproducts - ΔHforeactants) gives the same result, give or take a little. However, the reaction will be exothermic in the reverse direction given the right conditions. Sorry not to have better news.

This is a good answer if an initator is specified as part of tbe reactions conditions, however the conditions as reported by the OP do not mention one


text:
How do I balance the reaction equation of zinc and nitric acid?
How would you balance the following reaction using the oxidation number method?
$$\ce{Zn + HNO3 -> Zn(NO3)2 + NH4NO3 + H2O}$$
My approach:
Half reactions:
\begin{align}
\ce{Zn &-> Zn(NO3)2} & \text{Change in oxidation state of }\ce{Zn} &= 2\\
\ce{2HNO3 &-> NH4NO3}& \text{Change in oxidation state of }\ce{N} &= 8\\
\end{align}
Multiplying by the L.C.M of the change of the oxidation number.
\begin{align}
\ce{4Zn &-> 4Zn(NO3)2}\\
\ce{2HNO3 &-> NH4NO3}
\end{align}
Adding them,
$$\ce{4Zn + 2HNO3 -> 4Zn(NO3)2 + NH4NO3}$$
I don't know how to proceed further.
Am I going wrong in any of the above steps?
(Please don't convert the reactants and products to ions, and solve it using the method that I am using, so that I can identify my mistake)

Your half reactions are not balanced, that's the problem.

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

You should convert the reactants and products to ions to identify what actually changes. And you must not not forget the electrons! $\ce{Zn -> Zn^2+ +2e-}$, etc.

But my book says when writing half reactions in oxidation number method, only the atoms whose oxidation number changes should be balanced. Well, that's what I did.

Still, try to balance the half-reactions before adding them together. I believe that's the easier way.


text:
Where can I access H. Landolt's article/book on his Iodine Clock experiments?
I simply want to take a look at Landolt's original writings and data so that I can make notes and reference it for a report I'm writing. I've looked everywhere on Google and I can't find anything other than articles that cite his work. It would be a great help if I can find a translated copy in a digital format.

Landolt, H. Ber. Dtsch. Chem. Ges. 1886, 19, 1317–1365. This is the original in German. A good college library in US should have access. You could also probably get a copy through your local library using inter-library loan.


text:
hygroscopicity vs solubility
Phenols are hygroscopic and form a red liquid, upon contact with air. Yet, Phenol’s solubility in water is said to be moderate. ( Tilted to the low solubility side) This is explained with the benzene ring being non polar.
Then how can it be hygroscopic? Are solubility and hygroscopicity even related at all, then?

Phenol's solubility in water is not all that low, and grows with temperature until at some point phenol becomes infinitely soluble. Then again, there certainly are some compounds which are hygroscopic and insoluble. How can that be? Well, just like that.

related https://chemistry.stackexchange.com/questions/34636/what-is-the-cause-of-hygroscopy

Hygroscopicity is measure of how much the molecule likes to bond  molecules of water, by hydrating/covering charged part of the molecule with molecule/s of water, which relates to it having some polar or charged functional groups, like hydroxyl  group for phenols.
Solubility is about how molecule likes to be surrounded by water, so in this case the structure of whole molecule makes a difference.
Since water is a polar solvent it likes polar solutes. If a molecule despite having one or more polar groups has, for example, a huge part which is not polar, like a carbon chain, then it will not be soluble even though it may be hygroscopic.


text:
How to calculate the pH of the neutralisation of HCN with excess of KOH?

Write the reaction between $\pu{0.05 mol}$ $\ce{HCN (aq)}$ and  $\pu{0.08 mol}$ $\ce{KOH (aq)}$ with $V = \pu{500 mL}$ and then calculate the pH.

So I wrote the reaction:
$$\ce{HCN (aq) + KOH (aq) <=> KCN (aq) + H2O}$$
Begin:
\begin{align}
  n_\mathrm{begin}(\ce{HCN}) &= \pu{0.05 mol}\\
  n_\mathrm{begin}(\ce{KOH}) &= \pu{0.08 mol}\\
  n_\mathrm{begin}(\ce{KCN}) &= -\\
  n_\mathrm{begin}(\ce{H2O}) &= -\\
\end{align}
End:
\begin{align}
  n_\mathrm{end}(\ce{HCN}) &= -\\
  n_\mathrm{end}(\ce{KOH}) &= \pu{0.08 mol} - \pu{0.05 mol} = \pu{0.03 mol}\\
  n_\mathrm{end}(\ce{KCN}) &= \pu{0.05 mol}\\
  n_\mathrm{end}(\ce{H2O}) &= \pu{0.05 mol}\\
\end{align}
The product of the reaction is a salt of a strong base ($\ce{CN-}$) completely dissociated in water:
$$\ce{KCN (aq) -> K+ (aq) + CN- (aq)}$$
And
$$\ce{CN- (aq) + H2O -> HCN (aq) + OH- (aq)}$$
Knowing this I calculated the
$$[\ce{KCN}] = \frac{\pu{0.05 mol}}{\pu{0.5 L}} = \pu{0.1 M};$$
Then I used the equation:
$$[\ce{OH-}]_1 = \sqrt{\frac{K_\mathrm{w}}{K_\mathrm{a}} \cdot [\ce{KCN}]} = \pu{1.27x10^-3 M}$$
Since I still have some $\ce{KOH}$, it dissociates completely creating $\ce{OH-}$ ions, so:
\begin{align}
  [\ce{OH-}]_2 &= [\ce{KOH}]\\
  [\ce{OH-}]_2 &= \frac{\pu{0.03 mol}}{\pu{0.5 L}} = \pu{0.06 M}
\end{align}
Finally
\begin{align}
[\ce{OH-}] &= [\ce{OH-}]_1 + [\ce{OH-}]_2 \\
&= \pu{1.27x10^-3 M} + \pu{0.06 M} = \pu{0.06127 M}\\
\ce{pOH} &= -\log[\ce{OH-}] = -\log(0.0617) = 1.213\\
\ce{pH} &= 14 - \ce{pOH} = 12.787\\
\end{align}
Is it right to calculate the $[\ce{OH-}]$ in this way?
I based my calcualtions on this equation:
Knowing that the amounts of substance in moles are defined by $Z = Z_1 + Z_2$ in $K$ liters of solution, we can consider this concentration to be
$$[X] = \frac{Z}{K} = \frac{Z_1 + Z_2}{K} = \frac{Z_1}{K} + \frac{Z_2}{K} = [X_1] + [X_2]$$

While you've generally done a good job, you've made an important error. The hydrolysis of the cyanide ion should be written as an equilibrium reaction:
$$\ce{CN-(aq) + H2O(l) <=>  HCN(aq) + OH-(aq)}$$
That's because $\ce{HCN}$ is a very weak acid and its conjugated base a weak base (homework: look up the $K_b$ of $\ce{CN-}$).
Now, as you duly noted, there's excess $\ce{KOH}$ in the solution, which dissociates completely. The $\ce{OH-}$ resulting from that dissociation now push the equilibrium:
$$K_b=\frac{[\ce{HCN}]\times[\ce{OH-}]}{[\ce{CN-}]}$$
to the left. As a result, the contribution of the $[\ce{OH-}]$ coming from the hydrolysis of the cyanide is negligible and can be ignored. Only the $[\ce{OH-}]$ from the excess $\ce{KOH}$ is to be counted. So $[\ce{OH-}]\approx 0.06\ \mathrm{mol/L}$.
We can confirm that numerically by assuming $[\ce{OH-}]\approx 0.06\ \mathrm{mol/L}$ and with:
$$\frac{[\ce{HCN}]}{[\ce{CN-}]}=\frac{K_b}{[\ce{OH-}]}\approx 0.0027$$
So neglibibly little cyanide is present as the acid $\ce{HCN}$.

As regards the matter of adding molarities, in some instances that is allowed.
Say we add $n_1$ moles of $X$ to $V\ \mathrm{L}$, that would give a molarity $M_1=\frac{n_1}{V}$. At a later stage we add $n_2$ moles of $X$, that molarity would be $M_2=\frac{n_2}{V}$.
The total molarity would be:
$$M=M_1+M_2=\frac{n_1}{V}+\frac{n_2}{V}=\frac{n_1+n_2}{V}$$
But if we were to mix volumes of solutions it would be:
$$M=\frac{M_1V_1+M_2V_2}{V_1+V_2}$$
Now  they are no longer simply additive.

Since CN- is the conjugate base of a weak acid shouldn't be a strong base?

The $pK_b=4.79$, so quite weak. Thing e.g. acetic acid/acetate.

So considering it a weak base the concentration of OH- is defined only by the KOH concentration, due to the fact that CN- has not enough force to "steal" an H from the water. Right?

That's a poetic way of putting it but basically correct. Better to just look at the equilibrium constant expression, that tells you $[\ce{HCN}]\approx 0$, if excess hydroxide is present. Without excess $\ce{KOH}$ things would be different. The excess hydroxide suppresses the cyanide hydrolysation.

That's the point I was looking for. On the internet and even on my university book there are only examples where no reagent remain at the end of the reaction. If I can ask you, is the cyanide hydrolysation suppressed because of the stronger Kb of the OH- base?

Basically, yes. $\ce{KOH}$ dissociates completely, it is a very strong base. This pushes the hydrolysis of the cyanide completely to the left, so it doesn't hydrolyse at all. It really is just a matter of math i.e. numbers.

You say to look up the $K_\mathrm{b}$ as homework, but later you use it to justify your assumption, why not spare us all work and include the value that you used?

First of all your calculation is based on the assumption that HCN is a strong acid, which it's not. Check for ionisation constant of HCN.
And yes you can add moles directly because moles are discrete particles, even if they are in solution. But you can't add concentration or molarity because it's volume dependent and when you consider volumes, you'll eventually end up adding moles again.

I know that HCN is a weak base (its Ka is 6.2x10^(-3)) but I was thinking that, since it reacts with a strong base, it would be completly neutralized. Am I wrong?

In infinite time, maybe yes. But when you calculate pH with a pH meter, it gives you the instantaneous value. And if I'm not wrong, there's a buffer formation as well

What do you mean in 'infinite time'? Do you realise how fast ionic reactions in water really are? And no, there is no buffer effect here: that would only be the case with partial neutralisation, so there's significant $CN^-$ and $HCN$ present.


text:
Drawing and naming structures
I am trying to learn how to draw and name skeletal chemical structures and would like some opinions and help if you may.
In the second (bottom) structure I can see that there are 6 carbons so hex would appear in the name of the structure somewhere, and I can also see that it is part of the methyl group but I have also seen that it looks very similar to 3-methyl-2-pentene which has thrown me a bit, could somebody walk me through this one step by step? Also this structure is a resonance structure if I am correct?
The other one (top) I am completely lost with, I know there is a carbon ring so that is 6 carbons plus the three which are stated, 9 in total would mean 'mon' would have to appear in the name and is also part of the methyl group, how would I go about naming this structure?
Much appreciated!

6 carbons do not necessarily mean that "hexa-" would appear in the name of the structure.

https://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry

Anybody else help with these? That wiki page looks hella confusing, I'm just beginning with these structure so would like to told how it's done if possible :)

At first you need to identify, if there are any well know functional groups.
https://en.wikipedia.org/wiki/Functional_group
If there are any, then they will be used to create the final name, and this is done according with the rules of IUPAC nomenclature.
https://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry
In your first example.
There is a benzene ring, which is a very common structure in organic chemistry. There are other more complex derivatives of it too, for this example or similar looking molecules the name would refer to either benzene or phenol functional group, this is a little nuanced.
So, in this case we know it is a benzene we also know, there are 3 methanes attached to it or more specifically methane derived groups called methyl groups.(Methane is alkane and a functional group of alkanes is called an alk(-yl) group).
3 methane groups = trimethyl
The last thing is to say where they are attached. This can be found by counting around C atoms on the ring 1, 2 and 3. Ring has no beginning and no end so the starting point is arbitrary, by convention the first group has number 1 and the rest follows.
1,2,3-Trimethylbenzenes
The second molecule.
Here, the idea is to pick the back bone of the molecule and then attach to it smaller and smaller parts like a tree with its branches.
So the trunk of the tree would be 5 C atoms so pentane.
Here the functional groups attached is also methyl group on 3th atom of the backbone pentane, so that would give 3-methylpentane
but..
there is still a double bond that has to be included in the name which  changes pentane to pentene, and lastly its nice to say where is it exactly therefore pen-2-ene
Putting it all together gives us,
3-Methylpent-2-ene

Thank you so much for you're detailed answer, this has helped me loads as I've struggled to get my head around the naming processes!


text:
Why is the periodate anion not found?

Chlorine can be found as the following oxoanions $\ce{ClO–} , \ce{ClO2-}, \ce{ClO3-}, \ce{ClO4-}$ in many chemical stockrooms. However, iodine is never found as $\ce{IO4-}$. Explain this observation.

I didn't really find any reason as to why periodate cannot exist, unlike anions such as flourate where it is limited by flourine's small size. After researching this further, I found that the periodate anion actually exists in compounds such as sodium periodate. Even though it has another form $\ce{IO6^{5-}},$ the periodate anion actually dominates its equilibrium. Thus, I was wondering what the intended/likely answer was?

https://en.wikipedia.org/wiki/Sodium_periodate

Bigger atoms tend to have higher coordination numbers.

Yes, $\ce{IO4-}$ does exist and it is called metaperiodate anion. It can combine with a number of counter ions to form periodates, which may also be regarded as the salts of periodic acid.

Metaperiodates are typically prepared by the dehydration of sodium
  hydrogen periodate with nitric acid.
$$\ce{Na3H2IO6 + 2 HNO3 → NaIO4 + 2 NaNO3 + 2 H2O}$$
They can also be generated directly from iodates by treatment with
  other strong oxidizing agents such as hypochlorites:
$$\ce{NaIO3 + NaOCl → NaIO4 + NaCl}$$

Metaperiodates adopt a distorted tetrahedral geometry with an average
  $\ce{I–O}$ distance of 1.78 Å
The ortho- and metaperiodate forms exist in equilibrium.
$\ce{H4IO6− ⇌ IO4- + 2H2O,      K = 29}$
For this reason orthoperiodate is sometimes referred to as the
  dihydrate of metaperiodate, written $\ce{IO4^-·2H2O}$

One example of salt having metaperiodate ion is sodium periodate(@Gert). Its preparation is mentioned above. In fact, all other alkali metals and alkaline earth metals forms metaperiodate salts. Some examples include potassium periodate, rubidium periodate, calcium periodate etc.


text:
Reaction of bicyclic diol with periodic acid
I'm in an organic 1 class, and we learned about using periodic acid (HIO4) to cleave a glycol into two carbonyl groups, oxidizing the glycol. However, in the given reaction (see picture below), I'm skeptical that an oxidation as presented in our textbook (Organic Chemistry 6th Ed., Brown, Foote, Iverson, Anslyn) actually occurs. The mechanism presented in the textbook is analogous to the first step followed by the step labeled (A). This produces a ring with nine carbons and two carbonyl groups, which to me seems rather unlikely due to angle strain. The step labeled (B) seems like a more likely follow up to the first step, where a ring shift creates a spiro shaped bicyclopentane with a single carbonyl group.
Then again, with option (C), which ignores that first step, a Pinacol Rearrangement could occur, giving the same product as (B).
Which of these is most likely? I would think that B or C would be the mechanism since the final product seems much more stable. Normally we assume reactions are done in water unless told otherwise, so I don't see any reason why C couldn't also work.

I wouldn't expect any significant angle strain in a 9-membered ring with 2 carbonyls.  I see no reason why the reaction shouldn't proceed normally along path A. On page 3 of this link they note that the trans-1,2-diol with one more carbon atom than your molecule does not react with periodic acid - the implication being that the cis analogue does react.


text:
Effective rate constants
I have the following reversible bimolecular reactions
$$\begin{align}
\ce{A_0 +B &<=>[$3k_+$][$k_-$] A_1} \\[5pt]
\ce{A_1 +B &<=>[$2k_+$][$2k_-$] A_2} \\
\ce{A_2 +B &<=>[$k_+$][$3k_-$] A_3} \\
\end{align}$$
that can be written as follows
$$\begin{align}
\ce{A_0 +3B &<=>[$k_1$][$k_2$] A_3} \\[5pt]
\end{align}$$
The initial conditions are $[\ce{A_1}]_0 = [\ce{A_2}]_0 = [\ce{A_3}]_0 = 0$. I know rate constants $k_+$ and $k_-$ and would like to write rate constants $k_1$ and $k_2$ in terms of $k_+$ and $k_-$.
It would be great if you have any ideas. I am quite new to the topic. So far, I can only think of using some fitting algorithms to find $k_1$ and $k_2$.
Thanks!

I'm not getting your point. What data you have?

The slowest reaction in the 'chain' is the so-called rate limiting reaction, a bottleneck, if you like. In this case the third one. So $k_1=k_+$.

Your terminology is sloppy here. $3k_+$ is the reaction rate coefficient. If we assume first order kinetics then the reaction rate, $r$, for the first reaction is $$ r = 3k_+\ce{[A_0][B]}$$

Is it appropriate to assume that all the reactions orders follow the stoichiometry?  For example for the first reaction the forward reaction is second order and the backward reaction is first order? If so then $K_{eq} = 3k_+ / k_-$ for the first reaction. Overall $K_{eq, total} = k_1 / k_2$.

and $K_eq =6k_{+}^{3} / 6k^{3}{-} $ for the chain? so $k_1=6k{+}^{3} $ ?

Finding the equilibrium constant of your overall reaction is straightforward, as you've done in the comments, and we have $$K_\text{tot} = K_1K_2K_3 = \left(\frac{3k_+}{k_-}\right)\left(\frac{2k_+}{2k_-}\right)\left(\frac{k_+}{3k_-}\right) = \frac{k_+^3}{k_-^3}.$$ But this is insufficient to conclude that $k_1 = k_+^3$ or that $k_2 = k_-^3$. Indeed, let us consider what is stated when we have, as in your last reaction, rate  constants for the overall reaction. Such a scenario presumes an overall rate law of the form $$\frac{\text{d}\ce{[A_3]}}{\text{d}t} = k_1\ce{[A_0][B]^3} - k_2\ce{[A_3]},$$ and there is no a priori reason to suggest that this is likely. (Often this is possible in certain limiting regimes, but not in general.) To investigate $k_1$ and $k_2$, we will have to solve our system fully. We can write out the rate equations for our system as
\begin{align*}
\frac{\text{d}\ce{[A_1]}}{\text{d}t} &= 3k_+\ce{[A_0][B]}-k_-\ce{[A_1]}-2k_+\ce{[A_1][B]}+2k_-\ce{[A_2]}\\
\frac{\text{d}\ce{[A_2]}}{\text{d}t} &= 2k_+\ce{[A_1][B]}-2k_-\ce{[A_2]}-k_+\ce{[A_2][B]}+3k_-\ce{[A_3]}\\
\frac{\text{d}\ce{[A_3]}}{\text{d}t} &= k_+\ce{[A_2][B]}-3k_-\ce{[A_3]},
\end{align*}
a coupled system of differential equations, which can be solved using matrix methods---diagonalize the coupling matrix, find the eigenvectors and eigenvalues, and solve---but these calculations are often quite messy, so I will leave them to you if you're interested.
Once $\ce{[A_2]}$ and $\ce{[A_3]}$ are known, we can plug them into the last differential equation and compare it with the presumed form of the overall rate law to determine $k_1$ and $k_2$.

@RodrigodeAzevedo, sorry, I'm not seeing it. Can you write out what you think should be the corrected form of the differential equations?

@RodrigodeAzevedo, oh, yes, that's correct. Thanks!

@ a-cyclohexane-molecule thank you for your reply! Do mean to solve the system in the steady state?

@Sp_J, no, a steady-state assumption is not required, though it would make the calculations simpler.


text:
Run a command at every step of a relaxed scan
In Gaussian, I'm trying to do a relaxed scan and for each step obtain the volume. However, when I use the volume keyword, it seems to only output the volume for the initial and final steps. Is there a way to ensure a keyword runs for every step of a scan? A similar question (How to obtain the Raman spectrum along every coordinate of a scan in Gaussian?) was asked earlier this year, but I want to confirm whether Gaussian doesn't allow any other keywords to run during a scan.

Just as I commented in the question linked, it will be infinitely easier to run do this using an external, self-written script in bash or python.

Guess what: It is possible (in your specific case).
I was wrong. At least a little bit. Most of what is described below is true for the general purpose, and I will not change that part of the answer.
However, in the specific case where you only want to compute the volume, you are able to do that because I think it is part of the population analysis module. You can actually request a population analysis at every step of the optimisation, but be careful what you wish for (running NBO will take forever). That way the optimisation will be a little slower, but that is the most convenient way. Therefore the solution to your very specific problem is adding the following line to the route section:

pop=always volume

Be aware, that the output will be an even bigger nightmare than before.
Here is a complete input of the $\ce{H3B-NH3}$ I have used previously.

%chk=df-bp86-d3.def2svp.chk
%nproc=4
%mem=16000MB
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
pop=always                         ! Perform a population analysis at every step
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Scan

0 1
  N       0.00000        0.00000       -0.80852
  H       0.92634        0.00000       -1.36025
  H      -0.46317       -0.80223       -1.36025
  H      -0.46317        0.80223       -1.36025
  B       0.00000        0.00000        0.19148
  H       0.52348        0.90670        0.85928
  H      -1.04696        0.00000        0.85928
  H       0.52348       -0.90670        0.85928

B 1 5 S 5 0.2

For comparison (I'm not 100% sure I used the right values):

Energies
Step1 SCF Done:  E(RB-P86) =  -82.7531561505     A.U. after    8 cycles
Step2 SCF Done:  E(RB-P86) =  -83.0511089272     A.U. after    7 cycles
Step3 SCF Done:  E(RB-P86) =  -83.1459883448     A.U. after    7 cycles
Step4 SCF Done:  E(RB-P86) =  -83.1661412429     A.U. after    8 cycles
Step5 SCF Done:  E(RB-P86) =  -83.1606918387     A.U. after    8 cycles
Step6 SCF Done:  E(RB-P86) =  -83.1485268431     A.U. after    8 cycles

Volumes
Step1 Molar volume =  491.032 bohr**3/mol ( 43.819 cm**3/mol)
Step2 Molar volume =  415.148 bohr**3/mol ( 37.047 cm**3/mol)
Step3 Molar volume =  401.060 bohr**3/mol ( 35.790 cm**3/mol)
Step4 Molar volume =  406.785 bohr**3/mol ( 36.301 cm**3/mol)
Step5 Molar volume =  514.428 bohr**3/mol ( 45.907 cm**3/mol)
Step6 Molar volume =  453.278 bohr**3/mol ( 40.450 cm**3/mol)

The energies differ very slightly only (from the below approach), but the volumes do differ at some points. Now that could be, that I just used the wrong population analysis, or something else. I don't have the time to investigate. (Leave a comment or edit if you find out.)

I've again looked into that, but without some serious hacking via the external keyword interface, I don't think it is possible. I also couldn't find a suitable iOP to control what gets punched to checkpoint file and when, and then use the result from that for a property/analysis run.
The punch keyword is equally disappointing, only outputting the last state.
While the manual suggests that you can access a given step of a scan in the checkpoint file, I found this to be not true. The documentation is fuzzy on that part and it also warns, that not everything is stored. Since it is a binary file, there is also no real way of checking what is contained.
I guess the real reason for the existence of that option is to restart a calculation that went rogue with different parameters from an intermediate calculation.
I was equally unable to find a way to set a given redundant coordinate to a specific value and then freezing it in order to perform a partial optimisation at that point, and then manipulate that coordinate while reading in the guess and perform another calculation.
Even with the new and generally awesome way to define general internal coordinates in Gaussian 16 (GIC) I was unsuccessful.
From all that trying, I came up with the probably easiest (=lazy) way (not computationally most efficient) to do what you want is to perform the scan, extract the (p'optimised) coordinates and run a series of single point calculations.
A slightly less (probably) computationally heavy approach would be to perform an initial scan on a very low level (like pm6, if your molecule is well-behaved), and the do p'optimisations on the resulting steps. Usually pm6 is able to offer reasonable starting geometries, which in many other cases may also help you reduce computational effort. So this is probably easy and quite efficient.
There are various options to extract geometries from a scan, the following question deals exactly with that: Extract all structures of Gaussian 09 molecular dynamics calculation using babel?
There certainly are more efficient ways to do this kind of task, but in the end, most of them require manual labour.

N.B.: I cannot believe it, but the pm6/popt approach is actually a halfway decent application of compound jobs. Although I have to admit I have not tried it yet.
It works as the following example demonstrates. (I used Gaussian 16 Rev. A.03, but it should work with Gaussian 09 Rev D.01, too. Maybe even earlier.)
I set up an initial scan on the pm6 level of theory for $\ce{H3B-NH3}$ and scanned the $\ce{B-N}$ bond:

%chk=pm6.scan.chk
%nproc=2
%mem=8000MB
#P PM6
OPT(MaxCycle=100)
SYMMETRY(loose)
GEOM(ModRedundant)
Volume

title

0 1
 N     0.000000     0.000000    -0.764994
 H     1.019690     0.000000    -1.164250
 H    -0.509845    -0.883078    -1.164250
 H    -0.509845     0.883078    -1.164250
 B    -0.000000     0.000000     0.235006
 H     0.509845     0.883078     0.634262
 H    -1.019691     0.000000     0.634262
 H     0.509845    -0.883078     0.634262

B 1 5 S 5 0.2

I have extracted the pre-optimised geometries with Chemcraft and added a header section and made use of the --Link1-- separator. After the first optimisation I read the MO for a slight speedup. I'm sure some of this can be further automatised, but for the sake of demonstrating, I did it by hand. The commented input:

%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 1

0 1
  N       0.00000        0.00000       -0.80852
  H       0.92634        0.00000       -1.36025
  H      -0.46317       -0.80223       -1.36025
  H      -0.46317        0.80223       -1.36025
  B       0.00000        0.00000        0.19148
  H       0.52348        0.90670        0.85928
  H      -1.04696        0.00000        0.85928
  H       0.52348       -0.90670        0.85928

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 2

0 1
  N       0.00000        0.00000       -0.90852
  H       0.94043        0.00000       -1.36414
  H      -0.47021       -0.81444       -1.36414
  H      -0.47021        0.81444       -1.36414
  B       0.00000        0.00000        0.29148
  H       0.53853        0.93276        0.86317
  H      -1.07706        0.00000        0.86317
  H       0.53853       -0.93276        0.86317

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 3

0 1
  N       0.00000        0.00000       -0.99853
  H       0.95023        0.00000       -1.38525
  H      -0.47511       -0.82292       -1.38525
  H      -0.47512        0.82292       -1.38525
  B       0.00000        0.00000        0.40147
  H       0.55372        0.95907        0.87762
  H      -1.10743        0.00000        0.87762
  H       0.55372       -0.95907        0.87762

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 4

0 1
  N       0.00000        0.00000       -1.07752
  H       0.95590        0.00000       -1.42049
  H      -0.47795       -0.82783       -1.42049
  H      -0.47795        0.82783       -1.42049
  B       0.00000        0.00000        0.52248
  H       0.56746        0.98288        0.89885
  H      -1.13493        0.00000        0.89885
  H       0.56746       -0.98288        0.89885

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 5

0 1
  N       0.00000        0.00000       -1.14730
  H       0.95649        0.00000       -1.46941
  H      -0.47824       -0.82834       -1.46941
  H      -0.47824        0.82834       -1.46941
  B       0.00000        0.00000        0.65270
  H       0.57849        1.00197        0.92763
  H      -1.15698        0.00000        0.92763
  H       0.57849       -1.00197        0.92763

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 6

0 1
  N       0.00000        0.00000       -1.21413
  H       0.95402        0.00000       -1.53258
  H      -0.47701       -0.82621       -1.53258
  H      -0.47701        0.82621       -1.53258
  B       0.00000        0.00000        0.78587
  H       0.58528        1.01374        0.96868
  H      -1.17057        0.00000        0.96868
  H       0.58528       -1.01374        0.96868

B 1 5 F

For comparison:
Energies

Step1 SCF Done:  E(RB-P86) =  -82.7531561507     A.U. after    8 cycles
Step2 SCF Done:  E(RB-P86) =  -83.0511087360     A.U. after    8 cycles
Step3 SCF Done:  E(RB-P86) =  -83.1459860378     A.U. after    7 cycles
Step4 SCF Done:  E(RB-P86) =  -83.1661471840     A.U. after    8 cycles
Step5 SCF Done:  E(RB-P86) =  -83.1606709617     A.U. after    8 cycles
Step5 SCF Done:  E(RB-P86) =  -83.1484921220     A.U. after    8 cycles

Volumes

Step1 Molar volume =  459.690 bohr**3/mol ( 41.022 cm**3/mol)
Step2 Molar volume =  479.017 bohr**3/mol ( 42.747 cm**3/mol)
Step3 Molar volume =  401.060 bohr**3/mol ( 35.790 cm**3/mol)
Step4 Molar volume =  406.785 bohr**3/mol ( 36.301 cm**3/mol)
Step5 Molar volume =  570.344 bohr**3/mol ( 50.897 cm**3/mol)
Step6 Molar volume =  498.606 bohr**3/mol ( 44.495 cm**3/mol)

Animation


text:
Mass of oxygen in He+O2 mixture

A sample of $\pu{1.42 g}$ of helium and an unweighted quantity of oxygen gas are mixed in a flask at room temperature. The partial pressure of helium in the flask is $\pu{42.5 torr}$, and partial pressure of oxygen gas is $\pu{158 torr}$. What is the mass of the oxygen in the container?

As volume is not stated, I assume it's constant, and temperature is a constant too. Therefore, from the ideal gas law
$$\frac{p}{n} = \frac{RT}{V}$$
For helium and oxygen gas, the above formula is applicable, and the mass of oxygen can be determined:
\begin{align}
\frac{\pu{0.5592 atm}}{1.42/4.0026} &= \frac{\pu{0.20789 atm}}{m(\ce{O2})/31.99} \\
\to m(\ce{O2}) &= \pu{4.22 g}
\end{align}
This is my attempt on finding the mass. However, I believe it's not right. Why is that the case?

Why did you think 4.2 grams was incorrect? You have a case of equationitis using equations instead of thought patterns. Start over taking one step at a time expressing the chemical-math principle involved, ensure correctness, then proceed to the next step, etc.  Dump the calculator and use a slide rule or do longhand arithmetic become intimate with the calculations.

I don't see any flaws in your argumentation. I agree that $R, T, V = \mathrm{const}$, hence
$$\frac{p(\ce{He})}{n(\ce{He})} = \frac{p(\ce{O2})}{n(\ce{O2})}\tag{1}$$
$$m(\ce{O2}) = \frac{p(\ce{O2}) \cdot M(\ce{O2})}{p(\ce{He}) \cdot M(\ce{He}) } \cdot m(\ce{He})\tag{2}$$
Just one minor thing: you incorrectly converted partial pressure of helium: $\pu{42.5 Torr}$ is $\pu{0.056 atm}$, not $\pu{0.559 atm}$, therefore your resulting mass should be 10 times higher.
In fact, there was no need to convert $\pu{Torr}$ to $\pu{atm}$ in the first place. Don't do unnecessary steps if you can avoid them. At the same time don't drop any units and always remember to include them in your final calculation.
$$m(\ce{O2}) = \frac{(\pu{158 Torr})(\pu{31.99 g mol-1})}{(\pu{42.5 Torr})(\pu{4.00 g mol-1})}(\pu{1.42 g}) = \pu{42.22 g}\tag{3}$$


text:
Is this a spontaneous or a non-spontaneous process?
An insulated tank initially contains a liquid-vapor mixture of water. Then, an electric heater is turned on and kept inside the tank until all the liquid becomes saturated vapor.
According to the 2nd Law of Thermodynamics, a spontaneous process occurs when the change in entropy of the universe is greater than zero.
In this case, since the tank is insulated, $Q=0$ and the change in entropy of the surroundings is also equal to $0$; hence, $\Delta S_\mathrm{universe}$ = $\Delta S_\mathrm{system}$. Since adding work to the system (via the electric heater) results in an increase of the thermal energy of water and its evaporation, $\Delta S_\mathrm{system}>0$. Therefore, $\Delta S_\mathrm{universe} > 0$, and the process is spontaneous.
But wouldn't the necessity of adding work to the system (in order for the water to evaporate) mean the process is non-spontaneous?

Two questions. Does it take energy to run the heater? If so, where did the energy to run the heater come from, inside the tank or outside the tank?


text:
Carbohydrates: What does "typical nature" mean here?
I am a 12th grader in India. I have understood basic theory on biomolecules. I know about carbohydates (and structures of important saccharides) and $\alpha$-amino acids, proteins, and peptides.
This is a section given in my book:

I understand what is the meaning of "D"-glucose, and also how to identify the functional group. But, notice the last column in this table. I don't understand what they mean by the "typical nature" in each case.
I searched the net but couldn't find anything. Is my book using obsolete/unknown terminology for an existing concept? What is it? Please tell. Thank you!
References: Page 1, Page 2, Page 3

Please add the citation to the book, so that it can be viewed in context.

I never said you should upload any content. I said citation, big difference. Those pages don't really help at all.

Looks like your book's come up with a "localized" version of the D-L notation.
$\ce{D-L}$ notation in brief: A $\ce{D}$-sugar has the $\ce{-OH}$ group (on the stereogenic carbon atom on the parent chain, furthest away from the aldehyde group) facing right in the Fischer representation.
Likewise an $\ce{L}$-sugar has the $\ce{-OH}$ group facing left.

What your book's attempting to do is extend the $\ce{D-L}$ classification to individual carbon atoms in the parent chain.

The "3rd($\ce{L}$)" bit is supposed to mean, "3rd carbon atom has an $\ce{L}$-type configuration".
The "2,3($\ce{L}$)" bit is supposed to mean, "2nd and 3rd carbon atoms have an $\ce{L}$-type configuration".
You can figure out the rest ;-)

This appears to be some sort of memory aid to help you remember the configuration of the various kinds of 6-carbon sugars. The "typical nature" term was, in all likelihood, concocted by the authors (perhaps they thought it sounds better than "mnemonic"?)


text:
What word is to chalcogen and pnictogen what halide is to halogen?
Would "chalcide" be understood as a group 16 dianion (and something less electronegative)? Would "pnictide" be understood as a group 15 trianion (and something less electronegative)?

https://link.springer.com/chapter/10.1007%2FBFb0119186?LI=true
http://pubs.rsc.org/en/content/articlelanding/1996/cs/cs9962500199/unauth#!divAbstract
I would've expected chalcide, but that might be overloaded. Looks like the preferred term is "chalcigenide." "Pnictide" seems to be correct.

Strongly related: https://chemistry.stackexchange.com/questions/27586/is-there-a-common-name-to-refer-to-the-groups-13-and-14

@Zhe Fairly certain it's because the h is silent in chalcogen (that is, the "ch" in chalcogen makes a /k/ sound, not a /tʃ/ sound), so "chalcide" would sound like "calcide", which would suggest calcium. Pnictide, and halide for that matter, don't have this problem

From what I could find in a brief search, "pnictide"(this link includes some of the origin for the name) is used, but the corresponding term used for group 16 seems to be "chalcogenide", which, while it can be used to describe all group 16 elements, typically is used in a way that excludes oxides.

Pnictogens and chalcogens are the trivial names for group 15 and 16 elements respectively. From here:

Pnictogens – The elements of group 15: N, P, As, Sb, Bi. (Mc had not yet been named when the 2005 IUPAC Red Book was published, and its
  chemical properties are not yet experimentally known.)
Chalcogens – The elements of group 16: O, S, Se, Te, Po. (Lv had not yet been named when the 2005 IUPAC Red Book was published, and its
  chemical properties are not yet experimentally known.)

So, the corresponding ion names should be pnictide(e.g. nitride, bismuthide) and chalocogenide(e.g. sulfide, selenide, telluride).  Note: salts containing oxygen as anion are called oxide salts but salts containing sulfur, selenium as anion are called chalcogenide salts. Since, salts containing oxygen as anion are more common and thus separately given name 'oxide salts' and salts containing sulfur, selenium as anion are less common, so they are collectively given name 'chalcogenide salts'.
Similarly, the trivial name for group 14 elements is crystallogens (for its ability to form crystals?) and for group 13 elements is icosagens or triels. However, these names are very trivial and rarely used.
I have never heard name of group 13 and 14 element ion as crystallogenide or icosagenide. Perhaps, for their inability/difficulty to form corresponding ions(since it is very hard to accept four or five electrons to form ions).