text:
Energy level of the product in a cell reaction
I'm very new to chemistry, so please try to be as plain as possible with your answers. I'm studying the chemistry of alkaline cells:

The cell I'm interested in has the following chemical reactions going on:
\begin{align}
&\text{Anode (oxidation):} &\ce{Zn (s) + 2 OH- (aq) &-> ZnO (s) + H2O (l) + 2 e-} \\
&\text{Cathode (reduction):} &\ce{2 MnO2 (s) + H2O (l) + 2 e- &-> Mn2O3 (s) + \color{red}{2 OH- (aq)}} \\
\end{align}
The part that I do not understand is marked in red. As far as I know, chemical reactions occur only if it results in a lower energy level than the reactants.
However, the second reaction creates a hydroxide that has a negative charge, wouldn't that violate that principle? How or why is that a lower energy state?

Redox reactions and electrochemistry are but two unnecessary levels of complexity; let's remove them altogether. Throw some NaCl in the water; do you know what will happen next?

The two hydroxides you produce are consumed by the anodic oxidation process. Overall, no hydroxides are required.

@IvanNeretin yeah they will dissociate into Na and Cl ions and float freely in the water because water is polar.

@Jan I see that they are inmediately consumed, but why do they form in the first place?.

also those who decided to downvote, it would be nice to explain why.

Well, those $\ce{OH-}$ will float freely in the water all the same, because water is polar.

My comment was also the explanation of my downvote.

@IvanNeretin i totally get that the OH will float because it's polar, but my question is why does it form in the first place

@Jan i dont get it, how does your comment justify that the question does not show reaserch effort, is not clear or not usefull?.

Your question seems to implicitly assume that free-floating ions can't be a low energy state which comes about spontaneously. The example of NaCl dissolution in water shows that in fact it can.

That does not mean that the question does not show reaserch effort, is not clear or not usefull. It just shows that, as i said, I'm new to chemistery, that's why I ask questions.
I didnt assume that the ions cant be a low energy state, I asked how or why the ion is in a lower energy state.
Even if the question implicitly assumed that flee floating ions cant be in a lower energy state i still fail to see the reasoning behind the downvote.

To quote from the question:

As far as I know, chemical reactions occur only if it results in a lower energy level than the reactants.

This is correct in principle and it is correct if your energy meter is in fact Gibbs free energy $G$. Reactions will occur spontaneously (are exergonic) if $\Delta G < 0$ and will not occur spontaneously if $\Delta G>0$.
However, you are applying that principle wrongly. You are looking at only one single product of one half-reaction to which you arbitrarily slap the label ‘high in energy’ — the hydroxide ion. First, it should be stated that ‘high in energy’ is noninformation unless you compare its energy to something else. For example, the enthalpy ($\Delta_\mathrm f H$, the measure typically associated with the energy content of a compound) of a hydroxide ion will be higher than that of water but lower than that of an oxide ion. Yet oxide ions are well known and form freely; sometimes even in aquaeous solution.[1]
The enthalpy of individual compounds that are formed or consumed in a reaction is, however, mostly irrelevant. What counts is the overall equation. Often, very reactive compounds with a high enthalpy content can be generated easily just be the liberation of something else that happens to be particularly stable.[2] The principal corresponding driving force in your reaction is the formation of zinc oxide from zinc metal.
Furthermore, you claim that hydroxide ions are ‘produced’ at all. That claim may be founded on a microscopic level, but as I mentioned these cancel out macroscopically. While it is common practice to separate oxidation and reduction half-reactions in education, this separation does not happen in real life. Rather, to determine enthalpy or Gibbs free energy, the full reaction $(1)$ must be examined. If you do that, you will notice that the hydroxides cancel out.
$$\begin{align}\ce{Zn + 2 OH- + 2 MnO2 + H2O + 2e- &-> ZnO + H2O + 2 e- + Mn2O3 + 2 OH-}\\
\ce{Zn + 2 MnO2 &-> ZnO + Mn2O3}\tag{1}\end{align}$$
When calculating enthalpy and entropy, the total hydroxide ion concentration remains the same and the formed hydroxides thus do not contribute to the final values at all.

Notes:
[1]: For example, if you dissolve an aluminium salt in hydrochloric acid an then slowly increase the pH for example with sodium hydroxide until you have reached a pH of just above seven, an aluminium cluster complex $\ce{[Al13(μ4{-}O)4(μ{-}OH)24(H2O)12]^7+}$ will be formed. (Reference in German)
[2]: For example, consider the peroxide dimer of benoic acid $\ce{PhCOO-OOCPh}$. The high-energy peroxide bond can be cleaved to give two benzoyl radicals. These will liberate carbon dioxide and form the very highly reactive, very unstable phenyl radical. This reaction $(2)$ is possible because of the enthalpy and entropy gain associated with the formation of $\ce{CO2}$
$$\ce{PhCOO-OOCPh -> 2 PhCOO^. -> 2 Ph^. + 2 CO2}\tag{2}$$


text:
Can I use chromatography to compare the amounts of chlorophyll in different leaves?
I have been unable to find a clear answer online about the effect that concentration of a solute has on the retention factor.
If it is as I suspect, I should see plants with more chlorophyll having higher retention factors for chlorophyll. Is this true, and if not, what is the relationship?
Sorry if this is a terrible question, thanks for the help!
Feel free to correct my question, I'm not very familiar with chromatography.

Retention refers to the system stationary phase / eluent / eluate(s) not to the ability of your leaves to contain chlorophyll. In principle you can. Most of the work will be the quantitative extraction of C. from the leaves and doing so in a meaningful way.

In theory, the concentration of the analyte does not affect its retention factor in chromatography. That is the entire idea of having a constant retention factor: that you can immediately identify a compound based on where on the TLC/chromatogram it turns up.
In practice, very high amounts can lead to overloading which in turn leads to tailing. Each column or TLC plate has a certain capacity up to which the compound will travel normally and after which it will no longer. The top of the analyte will still reach the same position but it will draw a lengthy tail behind it. However, this typically happens way above the detection limit so you can half your sample size without any problems. (For liquid or gas chromatography, this might even be required because otherwise you may have left the linear region of the sensor.)

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text:
Why is there only one bond length for sulfur hexafluoride?
In sulfur hexafluoride, there are two 3-centre 4-electron bonds, according to the book I am reading (Essential Trends in Inorganic Chemistry by D. P. M. Mingos). The other two fluorines are bonded to sulfur, by interacting with its sp-hybrid orbitals. Therefore, we should expect to observe two sets of bond lengths, one set for those involved in the 3-c 4-e bonds and one set for those involved in the bonds involving the hybrid orbitals. However, only one set of bond length (156.4 pm) is reported on Wikipedia. Why is this the case?
In comparison, we observe two different sets of bond lengths in sulfur tetrafluoride, one set of values for the axial bonds and another set of values for the equatorial bonds.

SF6 has symmetric bond angles, so it sort of makes sense that it also has symmetric bond lengths. Maybe the wavefunction is a rotationally symmetric superposition of the different orientations of the structure you describe.

Yes that’s precisely it: one resonance form is not sufficient to describe the structure of SF6. Alternatively, why is there only one C–O bond length in the carbonate anion even though the conventional Lewis structure has one double bond and two single bonds? The answer to that question is exactly the same as the answer to this, it just needs to be adapted to the present context.

@orthocresol Why is it that we can describe PCl5 and SF4 with just one resonance form and yet describe SF6 with multiple resonance forms? Does molecular symmetry determine the resonance forms?

Yes, exactly. In SF6 when you stick to one single resonance form, you have arbitrarily decided which atoms are participating in a single bond and which are involved in the 3c4e bond. However, because all atoms are symmetry equivalent, one resonance form cannot be sufficient. Same goes for carbonate. The follow up question is presumably why do we assume it is symmetrical? and truthfully I don’t know a full answer. There are cases where it is more stable to not be symmetrical, cf Jahn–Teller distortion, but SF6 is not one of them.

@orthocresol Do you know the answer to the follow-up question you proposed? If you do, please feel free to write an answer. I would think that the symmetry is justified by experimental data (i.e. exp. bond lengths and angles)?

I guess you could argue that experimentally we observe an octahedral structure (NMR, vibrational analysis, etc). But from a theoretical point of view I’m not sure if there’s anything deeper than plain old VSEPR theory.

There might also be the argument that $\ce{SF4}$ is fine with having a sulphur lone pair in an s orbital, two 2c2e p orbitals and one 4e3c p orbital. In $\ce{SF6}$ we cannot do that, the s orbital cannot participate in bonding. If you were to assume two sp and two p orbitals, you would have two different bond energies with the stronger bond (2c2e) having more s contribution to an electronegative substituent. Meaning that the multiple resonance structures turn out the more stable solution. (CC @orthocresol)

as far as SF6 is concerned is it not the case that the $O_h$ symmetry means that there is, in effect,  no preferred direction for x, y or z so that all bond lengths must therefore  be equivalent.

This is an old question and is largely answered in the comments, but here's a full answer for the sake of completeness.
There are really two questions here:
1) is the given bonding orbital description correct for SF6?
2) why is SF6 octahedral?
The first question is easier: no, your description is incomplete as you've described it, but perhaps correct as intended by the author (I don't have the textbook to see the exact wording). It is correct if you explain further that there are three possible representations with two $sp$ bonding orbitals and two 3c/4e bonds. The three different representations look identical except that the $sp$ orbitals are aligned on different Cartesian axes (ie one for $x$, one for $y$ and one for $z$). The true structure is a composite of these three, meaning that each S-F bond is 1/3 $sp$ and 2/3 3c/4e (which is essentially 1/2 of a $p$ orbital) which adds up to a total of 2/3 of a bond with a contribution of 1/6 from $s$ and 1/2 from $p$.
I suspect that the textbook uses this representation because it connects readily to earlier discussions of integer hybridization, i.e. "$sp$" orbitals are familiar and easy to visualize. And presumably the concept of a 3c/4e bond has also been previously introduced. Making a composite of familiar concepts is perhaps easier than introducing the idea of delocalized molecular orbitals, but in this case they are simple enough that it's worth going through them to show that the result is the same.
First, the $s$ orbital is spherically symmetric, so it necessarily interacts the same with all six identical ligands. The lowest energy molecular orbital is thus represented by the $s$ orbital on sulfur and the six fluoride ligand orbitals all in the same phase (images below). That's where the 1/6 $s$ comes from. The $p$ orbitals each are able to interact with two of the ligands, one on each end of orbital. The other four ligands do not contribute at all, because they are exactly on the nodal plane of the $p$ orbital. Since each $p$ interacts with two ligand orbitals, each S-F bond gets 1/2 $p$ to go with the 1/6 $s$, exactly the result we got from the composite of hybrids. Below are drawings of the four bonding molecular orbitals:

Now on to the second question: why is SF6 octahedral if it only makes a total of 4 bonds?
Given that there is a lot of negative charge on the ligands and positive charge on S, a simplistic representation is to make the atoms solid charged spheres. A simple electrostatic potential energy minimization would then lead to an octahedral arrangement of the ligand spheres around the sulfur sphere. That arrangement minimizes the repulsive interactions between the ligands and maximizes the attractive interaction between ligand and sulfur.
Without a lot of empirical data about atomic interactions, though, we cannot rule out the possibility that there could be favorable bonding interactions with a strong distance dependence that would make it more favorable to, for example, push four of the ligands even closer to the sulfur and push the other two farther away. However, it turns out that the distance dependence of the bonding interaction is similar enough to the distance dependence of the electrostatic interaction that the geometric shape of the minimum energy conformation is the same - all six ligands at equal distance from the sulfur and spaced evenly around it.

"The true structure is a composite of these three" Do you have anything to support this claim? Because I have never encountered this before when I was reading up the literature.

That statement is supported by the MO diagrams, as I tried to explain. To be clear, I mean by  "composite" that it is one structure that is the average of those three, not that it switches between them in any way. The average of those three is the symmetric octahedral structure that is well known for SF6, and the average of those orbitals is the same as the four well-established MOs of SF6 that I've shown.

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text:
Bond orders in the 3c-4e bond and 3c-2e bond
According to D. P. M. Mingos, in his book Essential Trends in Inorganic Chemistry, the bond order of the three-centre four-electron bond is approximately 0.717, which is one divided by the square root of 2. However, in the case of the three-centre two-electron bond, the bond order is 0.5. This would imply that the former is stronger than the latter. However, both of them have similar-looking molecular orbital diagrams. In the case of the former, the nonbonding MO is filled while in the case of the latter, only the bonding MO is filled. Thus, from the MO perspective, why would these two bonds have different bond orders?

Not sure about this one, but I suspect it depends on how one defines the bond order. See eg IUPAC Gold Book for a complicated definition. Does Mingos not give any kind of justification?

Mingos has indeed justified the bond order for the 3c-4e bond mathematically (by using the definition of bond order to be the product of the coefficients of the orbitals contributing to the bond times the number of electrons occupying the orbital). However, he did not justify the bond order of o.5 for the 3c-2e bond.

Oh, ok. I have a hunch I might be able to answer this one, but again, not right now... Good questions, though.

I would guess that at least in the simplest cases of three-center bonding, the bond order is the same with two or four electrons should be the same; the second electron pair is in an essentially nonbonding orbital.

Revisiting this I don't see how the maths works out. I have seen this definition before, but if I assume the usual form of the MOs for three-centre bonding, the population of the middle MO should not affect the bond order between adjacent atoms because the coefficient of the central atom is zero. Probably need more details from the book.

@OscarLanzi Doesn't bond order reduce when there is the filling of the nonbonding orbitals. Ron writes "Notice how the four electrons are distributed - there are two electrons in the HOMO which is a non-bonding MO, so the bond order in this bond is reduced." in https://chemistry.stackexchange.com/questions/18427/why-does-f-replace-the-axial-bond-in-pcl5/18544?noredirect=1#comment154935_18544

What bonding is taken away by filling a _non_bonding orbital?

@orthocresol Unfortunately, there were not many other details from the book. This book isn't so much based on a mathematical understanding of bonding. So I can't help you with that...

@OscarLanzi I agree with you too. The book says that a 3c-4e bond and a 3c-2e bond have the same MO diagram, just that in the case of the former, the nonbonding MO is also filled. However, I was thinking shouldn't there be more fundamental differences between these two bonds. The 3c-4e bond is used to rationalise hypervalency while the 3c-2e bond is used to rationalise hypovalency. These are two very different uses.


text:
Temperature of chlorine gas at given pressure and volume

Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters. Suppose that the volume of a particular sample of $\ce{Cl2}$ has is $\pu{9.22 L}$ at $\pu{1124 torr}$ and $\pu{24 ^\circ C}$. At what temperature will the volume be $\pu{15.00 L}$ if the pressure is $\pu{8.76e2 torr}$ ?

I found the moles of the sample to be $\pu{0.559 mol}$. However, now I'm stuck at the last part of the question. Can I use $\pu{0.559 mol}$ to solve for the temperature when the volume is $\pu{15 L}$ and the pressure is $\pu{8.76e2 torr}$?

Assuming at these conditions it's safe to use ideal gas law ($pV = nRT$) for chlorine, and given the same gas sample ($n = \text{const}$, hence $nR = \text{const}$), let's express the unknown temperature $T_2$:
$$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \quad \to \quad T_2 = \frac{p_2V_2}{p_1V_1} \cdot T_1$$
Generally, I would recommend not to do intermediate calculations, especially if you are not asked to do so. Construct the mathematical expression first, convert the units if necessary, and plug in the values when you are done. Note that in this case you only need to convert temperature units.

 $$T_2 = \frac{\pu{876 torr} \cdot \pu{15.00 L}}{\pu{1124 torr} \cdot \pu{9.22 L}} \cdot \pu{297 K} = \pu{377 K}~\text{or}~\pu{104 ^\circ C}$$

What is the reason behind why is n is constant , then nR is constant ?

@user175089 The problem says "particular sample", which, I presume, implies to its composition and amount. A product of multiplication of two constants is a constant too.


text:
Extraction of Activated Carbon from Lignin
I'm currently working on a research project which involves the extraction of activated carbon from lignin. I'm already aware of the method of preparation using an alkali along with lignin products but I'm looking for more offbeat ideas that haven't been done before. Some suggestions would be great.

Extraction is the wrong word.


text:
Is ionic charge just a label?
We say that when an atom loses an electron it becomes positively charged ion, and similarly when an atom gains electrons it becomes a negatively charged ion. My question is that are 'positive' and 'negative' just labels indicating more protons than electrons or vice versa, or do the ions actually display an increased interaction towards other charged particles?

Is the sum printed on your bank account statement just a label indicating more or less dollars, or do you actually display an increased ability to buy goods?

Are you familiar with electroplating ?

Ummm, and here I was thinking for a moment that this is question "If ionic bonding is just a label?" which may be worth these upvotes. And here it looks like joke...

These ionic charges actually do behave as you would expect from electric charges: cations will be drawn towards the negative pole in an electrolytic cell while anions will be drawn towards the positive pole. The same feature can be observed with α and β radiation (which are $\ce{^4_2He^2+}$ and electrons, respectively); these are deflected in an electric field as expected.
On a more philosophical level, the entire concept of one charge being ‘positive’ while the other is ‘negative’ is of course meaningless. We might as well have called them ‘East’ and ‘West’ charges akin to the North and South poles of a compass. The only physically observable is that the two act as opposites; two particles of like charge will repel, two of opposite will attract. Having positive and negative charges makes calculations much easier though.


text:
What are the molar and mass concentrations of 18% HCl?

A solution of hydrochloric acid of mass fraction $\pu{18.0\%}$ has a density of $\pu{1.0878 g cm-3}$. What are the molar and mass concentrations of this solution?

I got $\pu{5.37e-3}$ and $\pu{195.8}$, but it looks wrong.

No matter what you got, returning it without units is wrong.

@andselisk You voted to keep this open, probably because the OP says he isn't sure if his answers(s) are right. But I'm under the impression that simply stating the final "answer(s)" beneath the question doesn't necessarily imply the OP put effort in it (perhaps the OP's comparing two options in a question). If the OP includes his working/calculations here, then that counts as effort IMO.  Just thought I'd mention it (why I voted to close), no biggie O:). Crisp answer you've put up ;-)

@paracetamol I'm still under the impression of recent close statistics. I decided that OP actually put some effort to solve the problem after I realized the values are not copy-pasted from the answer in textbook or some internet forum and the approach to solve the problem was not entirely wrong. But thanks for investing your time in explanation and kind words:)

@andselisk < Nods > I read ortho's post too... I lamented over it for a while, and then kinda forgot about it. But it's admirable that you're actually working towards the change that ortho (and the other Mods) envisioned, by being more tolerant (then again, you always were) to posts that most of us would close without a second thought. Kudos to you! < salutes >

The values you've obtained signify that you are aware of the mathematical expressions for molar and mass concentrations, but your attitude towards handling units is somewhat frivolous. I would recommend to think about how these three ways of expressing concentrations are interconnected, and always perform calculations with respected units.
So, in order to find molar concentration $C_i$ and mass concentration $\gamma_i$ defined as
\begin{align}
C_i &= \frac{n_i}{V} \\
\gamma_i &= \frac{m_i}{V}
\end{align}
one needs to express the amount $n_i$ and mass $m_i$ via volume $V$, density $\rho$, mmass fraction $\omega_i$ and molecular weight $M_i$:
$$\omega_i = \frac{m_i}{m} \quad \to \quad m_i = \omega_i m$$
$$m = \rho V$$
$$m_i = \omega_i \rho V$$
$$n_i = \frac{m_i}{M_i} = \frac{\omega_i \rho V}{M_i}$$
Finally, the expressions are ready to plug the known values in:
$$\require{cancel}
C_i = \frac{\omega_i \rho \cancel{V}}{M_i \cancel{V}} = \frac{\omega_i \rho}{M_i} = \frac{0.18 \cdot \pu{1.0878 g cm-3}}{\pu{36.48 g mol-1}} = \pu{5.37e-3 mol cm-3} = \pu{5.37 M}$$
$$\gamma_i = \frac{\omega_i \rho \cancel{V}}{\cancel{V}} = \omega_i \rho = 0.18 \cdot \pu{1.0878 g cm-3} = \pu{0.196 g cm-3} = \pu{196 g L-1}$$


text:
What's the evidence that it is the nitronium cation (NO2+) acting as the attacking electrophile in mixed acid nitration of aromatic compounds (SeAr)
As per the title, I'm looking as to what is the evidence is that it is the $\ce{NO2+}$ cation acting as the attacking electrophile in an electrophillic aromatic substitution reaction involving the 'mixed acid' reagents of $\ce{H2SO4 + HNO3}$

I have considered the following although I feel as if other explanations exist and doesn't really specifically tackle the question:
The generation of the electrophile is as follows
$$\ce{2 H2SO4 + HNO3 -> H3O+ + NO2+ + 2 HSO4-}$$
Of which has been shown to be valid via spectroscopy methods & being consistent with the expected freezing-point depression. The $\ce{NO2+}$ cation can be trapped within unreactive anions (such as $\ce{BF4-},\ce{PF6-}$) to form salts which can then themselves act as nitrating agents to obtain the same nitration reaction product.

Water will be protonated to $\ce{H3O+}$ in this medium.

Can you be more specific? What are some alternatives that you have considered?


text:
How do I calculate the molarity of a 40% w/v HCl solution?

A 40% hydrochloric acid (w/v) is found to have density of 1.20g /mL. The molarity of the solution is nearly?

I think that you're missing something

@creek Well, if you can assume the molar mass of $HCl$ is 36.5g (or any value for that matter... just as long as you're permitted to calculate the molar mass on your own, since it isn't given in the question), then you're all set to solve it (it's anything but "straightforward" though... took me a quite while to figure it out) O:)

As someone said: "less is more"

First we need the molecular weight of $\ce{HCl}$:
$$\pu{1.008 g//mol} + \pu{35.453 g//mol} = \pu{36.461 g//mol}$$
Now the tricky part is that $40\%~\mathrm{w/v}$ means $\pu{40 g}$ in $\pu{100 mL}$ solution, not in $\pu{100 g}$ solution, so you don't need the density. If the question asked for molality instead of molarity, density would have been required.
So now we write down what $40\%~\mathrm{w/v}~\ce{HCl}$ means and use unit factors to walk our given units to the desired units, canceling the bad units and replacing them with the good as we march inexorably along:
$$\require{cancel}\left[\frac{40\text{ }\color{red}{\bcancel{\color{black}{\text{g HCl}}}}}{100\text{ }\color{blue}{\bcancel{\color{black}{\text{mL}}}}\text{sol'n}}\right]\left[\frac{1000\text{ }\color{blue}{\bcancel{\color{black}{\text{mL}}}}}{1\text{ L}}\right]\left[\frac{1\text{ mol HCl}}{36.461\text{ }\color{red}{\bcancel{\color{black}{\text{g HCl}}}}}\right]=\frac{10.97\text{ mol HCl}}{1\text{ L sol'n}}=11.0\text{ M}$$

I'd agree that you have worked out 40 % w/v HCl correctly. After seeing your work I deleted my answer which was wrong.

@MaxW Correctness is not as important as the test of illustrating the unit factors method via MathJax. I think it makes it much more readable to be able to cancel out in matching colors, and I believe that the unit factors method should be taught to posters who are green enough to want help on their general chemistry homework. Do you think the layout of my answer in MathJax looks readable?

It is a fantastic answer. The MathJax is beautiful.

I down-voted this answer, because it is far from being exceptional enough to keep questions like these on the site. Also, the first sentences of any answer should be: "The concentration % (w/v) does not make any sense. Don't use it!"


text:
Synthesis of cycloserine: What is the mechanism of this step?
I want to know the mechanism or some explanation as to how the last step here (with hydroxylamine) works:

But why for example hydroxylamine doesn`t attack chlorine as nucleophillic substitution to give OH group back there? It would make much more sense as chlorine seems to be more reactive then some ester bond. @Waylander

I can't definitively answer that except to observe that the major product is the hydroxamic acid. What I don't know is what the yield of this reaction is, attack on the chloromethyl could be a byproduct.

C-Cl bond is strong, chloride is poor leaving group - these are factors that may contribute to its relative lack of reactivity under these conditions. Obviously, these are only back-rationalisations.

The hydroxylamine attacks the methyl ester through nitrogen giving the intermediate hydroxamic acid. The oxygen to the hydroxamic acid then attacks the chloromethyl to displace chlorine giving cycloserine.


text:
Calculating n factor of SO2
I'm having a hard time trying to figure out n factor of $\ce{SO2}$ in the reaction
$$\ce{FeS2 + O2 → Fe2O3 + SO2}\qquad \text{(unbalanced)}$$
To balance it, I saw that in 1 mole of $\ce{FeS2}$, it looses a net total of 11 moles of electrons, while $\ce{O2}$ gains 4 moles of electrons per mole of it. That means they will have to react in a $4:11$ ratio, from this, the equation balances out to
$$\ce{4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2}$$
The question was (wrt to this equation), what will be the equivalent weight of $\ce{SO2}$, if the molecular weight of $\ce{SO2}$ is $M$?
This means I will need the $n$ factor of it to divide by $M$,
I tried to reverse this reaction
$$\ce{2 Fe2O3 + 8 SO2 → 4 FeS2 + 11 O2}$$
But the problem is that now it's hard to tell from which $\ce{O}$ atoms (of the now-products), the oxidation state is getting changed to $\pm0$ (zero). It's now confusing me.
The answer is given as $M/5$. I don't know why, according to the solution they only concidered the oxidation number change of $\ce S$ while ignoring the others, I don't think it's right.

All oxygen atoms undergo the same change in oxidation state.

I was thinking of the definition of n as the moles of electrons supplied or used up by a mole of substance in a reaction. For SO2, i saw the change in ON of S but for O, I got confused, even that will give electrons when the ON changes from -2 to 0. Why are we not considering that change in ON? Also which ones, as even Fe203 was formed from the same O2?

All oxygen atoms on the $\ce{O2}$ side of the reaction have an oxidation state of $\pm0$. All oxygen atoms on the other side of the reaction have an oxidation state of $\mathrm{-II}$. I am confused where you are confused. (Incidentally, is that iron(IV) sulphide or iron(II) disulphide? That is confusing me.)

It's pyrite Fe is in the +2 state and S_2 has 2-

So iron(II) disulphide (or should I call it bissulphide). Thanks!

This is confusing me: The n is the total number of moles of electrons a substance tranfers in a reaction, right? Here in the case of FeS2, the Fe changes from +2 to +3 while  both the S's go from -1 to +4, so that means for 1 mole of FeS2, 11 moles of electrons were taken from it for this reaction to proceed. Thinking in a similar way for SO2, I thought that in one mole of SO2, the S goes from +4 to -1, but even the O changes from -2 to 0. Like in the last case, we calculated the net total changes, even here we should consider the changes in ON of O to calculate n, right? But why don't we?

As this is clearly a homework question, I will only point you in the right direction.
Balance the oxidation reaction:
$$\ce{S2^2- -> 2 SO2 + $n$e^-}$$
Balancing the charges will give you $n$. (I'll leave the balancing to the OP, as this is homework).
The equivalent mass of $\ce{SO2}$ in this reaction is then:
$$N=\frac{2M}{n}$$
The factor $2$ is needed because the electronically balanced reaction has two molecules of sulphur dioxide in the equation.

Thanks , but even Fe^2+ is getting oxidized to Fe^3+, right? Why aren't we considering that? In your equation, balancing charges should give n = 2, but that's not matching.

You're looking at the oxidation of $S_2^{2-}$ to $SO_2$, not the ferro to ferric. The half reaction requires 6 electrons but you need to half that because there are 2 $SO_2$. So you get $n=3$ and $N=\frac{M}{3}$. You might want to look into what $n$ and the equivalent weight really stand for.

The $5$ comes from the fact that $S$ in $S_2^{2-}$ has $OS=-1$ and in $SO_2$ has $OS=4$, so the difference is $5$. But $S_2^{2-}$ contains 2 $S$, so you also need 2 $SO_2$. That then give  a difference of $2\times 4-2\times(-1)=6$. So $5$ is wrong.

Well, actually you made a slight calculation mistake.. that last difference will be 10 so 5 per molecule. Thanks alot, I understood it now. I was confused about why we aren't involving the changes in OS of the other atom O, I guess it's because.. it's kinda like a spectator.. but it isn't right? , it's the one taking electrons in the reaction to oxidize FeS2. Why are we not considered it? (Sorry if this sounds like a stupid question... I mean, at one time I feel it's obvious since we are writing an oxidation half reaction , but I'm not able to think of an exact reason.)

@Rick: damn, you're right! ;-) Yes, for equivalences, just concentrate on that half reaction and ignore the rest. It's about the sulphur (in this case) and nothing else. You can also calculate $n$ for the iron: in that case just concentrate on that element.

For the half reaction of $\ce{S_2^{2-}}$ you still need to add the oxygen. The atoms must balance in a valid half reaction. So it should be $$\dfrac{1}{2}\ce{S_2^{2-} + O_2 -> SO2 + ne^-}$$


text:
Energy balance for a system

A hydropower plant uses a river to generate electrical energy. It has a drop of $\pu{107 m}$ and the flow rate of water is $\pu{275 m3 s-1}$. The flow rate of the water is the same in and out of the system. Determine whether the system is open or closed.

The answer is: the system is open.
I am a bit uncertain as to how to simplify the general energy balance for the system:
$$Q + W = \Delta H + \Delta E_\mathrm{k} + \Delta E_\mathrm{p}$$
I think that only $\Delta H$ can be removed from the equation since the enthalpy won't change. Is that correct?
$$Q + W = \Delta E_\mathrm{k} + \Delta E_\mathrm{p}$$
If so, how can I use that to calculate the maximum amount of electrical energy that can be generated?

This is a physics question.

This a question we were given in our chemical engineering class @Gert

Maybe so, but there's no chemical content in it. Migrate it to physics.stackexchange.

I did and now it was put on hold as off-topic

@J.Se: I know, I saw it. Maybe I'll answer here, later.

@paracetamol 'consider posting it there instead" is what should never be told under homework, I hope you see it now as OP admitted cross-posting of closed question...

The only three terms that need to be taken into account are $\dot{W}$, $\Delta \dot{E_p}$ and $\Delta \dot{E_k}$, as there's no heat involved. I use the 'dot' notation to indicate these are energies per unit of time (powers).
For potential energy power we have:
$$\Delta \dot{E_p}=-\dot{m}gH$$
Here $\dot{m}$ is the mass flow ($275\ \mathrm{kg/s}$).
For kinetic energy power we have:
$$\Delta \dot{E_k}=\frac12 \dot{m}v^2$$
$v$ is calculated from:
$$\dot{m}gH=\frac12 \dot{m}v^2$$
 I.e.:
$$v=\sqrt{2gH}$$
Now calculate $\Delta \dot{E_k}=\frac12 \dot{m}v^2$.
Then apply the energy (power) balance, as:
$$\dot{W}=\Delta \dot{E_p}+\Delta \dot{E_k}$$
This is the power the plant extracts from the falling water, as work done on the turbines. This then converted to electrical energy.


text:
How do I find the number of grams of the product when given reactants?
I have the following chemical equation
$$\ce{H + O2 -> H2O}$$
which balances out to
$$\ce{4H + O2 -> 2H2O}$$
How would I find out how many grams of $\ce{H2O}$ and $\ce{H}$ I have, if I have $\pu{6 g}$ of $\ce{O2}$?
What if I know I have $\pu{6 g}$ of $\ce{H2O}$, how would I find grams of $\ce{H}$ and $\ce{O2}$?
What if I know I have $\pu{6 g}$ of $\ce{H}$, how would I find out how many grams of $\ce{H2O}$ and $\ce{O2}$?
I found the molar mass of $\ce{H2O}$ to be 18, $\ce{H}$ to be 1, and $\ce{O2}$ to be 32.

I'd double check the problem. Hydrogen won't exist as the atom but rather as the diatomic molecule. It could be that that an "artificial" problem was made-up using just an atom of hydrogen, but that would be weird.

Double-check the reaction and don't omit molar mass units.

@MaxW yea, it is an artificial problem, I just want to know if there is a way to solve it.

@andselisk Is there a way to solve it, and what do you mean molar mass units, isn't that amu or g/mol?

Using your second balanced equation gives the reaction as four hydrogen atoms plus one oxygen molecule yields two water molecules. The ratios of atom counts hold. A mole is just a big counting number like a million or a billion. Thus you have to figure out how many moles of oxygen molecules there are in 6 grams of oxygen molecules and work out the rest of the ratios in moles. You'd then convert to grams as needed.

@Rixwaan I don't know what units there are as you haven't typed any. You can solve for any reaction, the question is how meaningful the result will be. Stray hydrogen looks very suspicious.

We know that $$\ce{H2O molar mass is approximately 18 g/mol}$$
$$\ce{H molar mass is approximately 1g/mol}$$
$$\ce{O molar mass is approximately 16g/mol}$$
So $\ce{2H is 2/18 = 1/9 of the H2O mass}$ and
$\ce{O is 16/18 = 8/9 of the H2O mass}$
Since you have 6g of $\ce{H2O}$ that means you'll have
$$\ce{6/18 = 1/3 mol of H2O}$$
Then
$$\ce{H mass would be approximately 1/3 * 1/9 g}$$
$$\ce{O mass would be approximately 1/3 * 8/9 g}$$
N.b: results will be very approximate


text:
pH value of a buffer solution of HClO and NaClO
So,
I have 200mL of HClO 0,64M.
I mix it with 0,1mol of NaClO.
How should I calculate the pH?
My question is about this: should I keep attention about changes made to the solution volume after adding NaClO?
So far this is what I've done.
Considering:
$\ce{NaClO + H2O -> Na+ + ClO-}$
With n(NaClO) = n(ClO-) = 0.1mol,
I calculated the molarity of the conjugate base:
[ClO-] = 0.1mol/0.2L = 0.5M.
Then I applied the Henderson-Hesselbalch equation:
pH = pKa + log([ClO-]/[HClO]) = 7.53 + log(0.781M) = 7.422
In this case I didn't consider the variation to the solution volume due to the addition of NaClO. So, is this correct?

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

I'm a college student, this is not a homework question. I've already solved it but I'm not sure about the result. I would like to compare my result with someone who know exactly how to solve it.

Please see the homework link in my above comment to learn what qualifies as a homework type of question and how to ask one. You can get help with this here, you just need to follow the guidelines. Best of luck.

In addition to the problem that this would be considered a homework question, it also qualifies as an amirite question (seeks to consider confirmation or clarification on an already derived answer) which we have determined to be a bad fit for the site.

You're close.  Everything is correct, except that when you take the ratio of concentrations in the H-H equation that ratio is not in moles.   It's just a number, because you divide moles by moles .  Rule of thumb: logarithms and exponential should never involve anything with units.  So don't include the molar unit under the logarithm and you're good.


text:
What structure is this?

What is the name of this structure?
It's close to Leucine but I'm not sure.

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

First of all we need to count the number of carbons in the main chain.
In this case there are five atoms so the name prefix will be "penta-".
The main functional group is a carboxyl group on the C1.
So the methyl group will be on the C4 the molecula suffix will be "-oic acid".
Knowing this is pretty simple:
4-methylpentanoic acid

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


text:
Do serotonin drugs contain serotonin or a precursor to it, e.g. tryptophan?
With serotonin drugs, is it actually serotonin in the drug, or is it tryptophan which converts into serotonin through the biochemical conversion process after the drug is consumed? Serotonin can be consumed through foods, whereas tryptophan cannot, but the biochemical conversion process is what has brought this question to my mind about the actual drugs which can increase serotonin in the body.

I think you've got largely (if not exclusively) misconceptions here.  It might be best to do some basic research on something like Wikipedia and then ask a follow-up question to that.

No, the drugs that interact with serotonin receptors do not contain serotonin nor are they serotonin precursors. There are a lot of serotonin receptor subtypes in the body and the drugs are designed to interact selectively at a single subtype (with greater or lesser success). Many of them bear only a passing resemblance to serotonin as extra functionality is added to enhance selectivity.


text:
When comparing boiling points of two ionic compounds, what bonding factors should you assess?
When comparing boiling points of two ionic compounds, should you look at their electronegativity difference, lattice energy, or strength from LDF forces (or all 3 factors)?
For example, between LiCL and NaCl, which one has the higher boiling point?
If we look at EN difference and LDF forces, it would be NaCl, because the EN difference here would be $2.1$ compared to $2.0$ for LiCL; also, since Na has an extra energy level, it has a larger electron cloud and thus stronger LDF's.
However, if we look at it in term of lattice energy or Coloumb's law, LiCl has a larger boiling point because high lattice energy is a combination of large charge and small ions. Therefore, even though Li and Na have the same charge, Li has a smaller atomic radius than Na.

To compare 2 ionic compounds' melting point, there are generally 3 determinants:

Ionic Radii, smaller the ionic radius, closer the ions are to each other, therefore stronger the electrostatic attraction between them. (analogous to stronger attraction when closer to a magnet) A good example would be melting points of cations from the same group with the same anion (NB: LiF does not follow the trend, potentially due to its more covalent nature):

$\ce{NaF}\ -\  \pu{993 ^\circ C ( 1819 ^\circ F)}$
$\ce{KF}\  -  \pu{858 ^\circ C ( 1576 ^\circ F)}$
$\ce{RbF}\  -\  \pu{795 ^\circ C ( 1463 ^\circ F)}$

Charge of ions, larger the charge an ion has, more positive or negative the ion is, therefore there is a stronger attraction between ions. (analogous to stronger attraction with two stronger magnets in opposite poles) An example would be $\ce{NaCl}$ and $\ce{MgO}$:

$\ce{NaCl}\ -\ \pu{801 ^\circ C ( 1,474 ^\circ F)}$
$\ce{MgO}\ -\ \pu{2852 ^\circ C ( 5166 ^\circ F)}$

Ratio of ions, in most cases, closer the ratio is to 1 : 1 between anions and cations, stronger the attraction is. larger or smaller the ratio is, the more the anions or cations will have to 'share' or 'divide' the electrostatic attraction from the oppositely-charge ion, decreasing the melting point. An example would be $\ce{NaCl}$ and $\ce{MgCl2}$:

$\ce{NaCl}\ -\ \pu{801 ^\circ C ( 1,474 ^\circ F)}\ [1 : 1]$
$\ce{MgCl2}\ -\ \pu{714 ^\circ C ( 1317 ^\circ F)}\ [1 : 2]$

The determinants for melting points would generally also apply to the boiling points of ionic compounds. However, it should be noted that other factors would affect the boiling point as well.

Hello there, thank you for your contribution! Your first two points are quite correct, but the third one doesn't seem to work so well. For example, $\ce{KCl}$ and $\ce{CaCl2}$ boil at 1420 °C and 1935 °C, respectively. Also, I would be wary of your final statement; there are factors which affect melting points but do not affect boiling points, and vice-versa (e.g. benzene and toluene have relatively close boiling points, but wildly different melting points). Therefore, the comparisons and arguments really should be made directly regarding the boiling points.

Welcome to Chemistry.SE!  Please note that formulas can be better expressed with $\ce{ }$ for chemical formulas/equations, $ $ for math term/equations, and $\pu{ }$ for units. More information is available in this meta post Also, it is suggested that all new users take a minute to look over the help center and tour page to better understand our guidelines and question policies if you haven't already.

Point 3 tries to address one of the harder parts of this, how to compare compounds with fundamentally different structures. For this one could also look at the Kapustinskii equation, see for instance https://en.wikipedia.org/wiki/Kapustinskii_equation


text:
How does Rosuvastatin 5 S-lactone form in human body?
I've searched lots of free resources online but cannot find the answer.

the top compound is Rosuvastatin.
Some websites say it is metabolized by the enzyme CYP2C9,
but I can't figure out how a CYP450 mediated reaction will produce Rosuvastatin 5 S-lactone. And I tried thinking about an intramolecular esterification, but it does not look like the answer. Perhaps I missed something?

It seems like you have drawn two excess double bonds. Maybe this is the source of your confusion. Besides this intramolecular esterification is ok.

Oh I compared the picture of pubchem and drugbank, I found that they are different. If pubchem is right then it can be form through intramolecular esterification but I'm not sure. DB: https://www.drugbank.ca/reactions/2249 PC: https://pubchem.ncbi.nlm.nih.gov/compound/66631534#section=2D-Structure

Rosuvastatin is metabolized to a limited extent by CYP2C9 to form an N-desmethyl metabolite and the other metabolite is the rosuvastatin lactone.

The acid moeity (highlighted part) is likely to undergo glucuronidation first and subsequently lactonisation to form the rosuvastatin lactone
Unlike other statins, rosuvastatin undergoes relatively little metabolism by hepatic CYPs with no significant involvement of CYP3A4.

The metabolism of statins is complex, involving the inter-conversion
  of acid/lactone by various pathways. The acid forms are converted to
  the corresponding lactones via a glucuronide intermediate.

Here is the proposed, generalised metabolic pathway of statins. The acid forms of statins are converted to the corresponding lactones via a glucuronide intermediate via an elimination reaction.

Sidenote for some statins: (After metabolism by CYP3A4 some of the resulting lactone metabolites are converted to the corresponding acid forms:
References

Focus on Statin research. B. A. Wong: Drug Metabolism and Transporter profiles of Statins.
Foye’s Principles of Medicinal Chemistry 7th ed

Really appreciate your detailed answer!


text:
What is the relation between pressure and chemical potential?
I know that in a two-component gas-liquid system (for simplicity) the chemical potential $\mu_i$ ($i=1,2$) and total pressure $p$ can be defined by
$$\mu_i=\left(\frac{\partial U}{\partial N_i}\right)_{S,V,N_j}$$ and
$$p=-\left(\frac{\partial U}{\partial V}\right)_{S,N_1,N_2}.$$
where $U$ and $V$ are the internal energy and volume of the system, $N_i$ is the mole number of each component, and $p=p_1+p_2$ with $p_i$ being the partial pressure.
But how the partial pressure and chemical potential are related? Intuitively, I think $\mu_i\propto p_i$. Thank you!

Are you looking for an explicit equation relating the two quantities, $p = p(\mu)$ or $\mu = \mu(p)$? What system are you looking at---a gas, a liquid, a solid?

@a-cyclohexane-molecule Thanks for your reminding. I'm looking for an explicit expression relating chemical potential and partial pressure for a gas-liquid system.


text:
How do potentiometric cells do not proceed through a spontaneous reaction like in a galvanic cell?
It seems to me that the cell setup of a galvanic cell is pretty much the same as in a potentiometric cell. However, galvanic cells just proceed the spontaneous direction of the net redox reaction of the anodic and the cathodic species in the cell, while in a potentiometric cell, it seems that the equilibrium activity ratios of the measured solution are not to be affected.
I've read that in a galvanic cell, current flows as to follow the natural tendency of electron flow--from the anode to the cathode. But in a potentiometric cell, it's said that the current of electrons flowing is kept to a minimum. How does a potentiometric cell does that? How does it avoid for reduction at the measured cell or oxidation at the reference electrode to happen?
P.S. I know that the question looks uncohesive, but it's because I'm really confused how these concepts occur. Please try to clear up the misconceptions that might have been apparent from my question.

First, I wouldn't classify cells as potentiometric or galvanic.
If you take any electrochemical cell with non-zero total cell potential and connect a wire between the two electrodes it will discharge. This will happen spontaneously in the galvanic direction.
In potentiometric measurements of electrochemical cells, what you do is a potential measurement which ideally happens at zero current(definition of a voltmeter), and practically at picoampere or lower current. Therefore you are not allowing the electrons to flow from the outside circuit and hence your cell doesn't discharge.

Ahh, but then you said that when you connect a wire between two electrodes mediating for a redox couple (that are practically possible to undergo spontaneous reaction), the reaction will automatically proceed to the more spontaneous direction, hence, electrons will be transferred and current will flow. So how will that current be minimized to picoamperes or lower magnitudes in a potentiometric setup?

That is the job of the instrument. The input impedance of the meter used is very high and doesn't allow appreciable currents to flow(this is also why standard voltmeters don't work in pH measurements). You may want to look into opamps with with FET inputs if you would like to see further details.

Thank you!! That clears up much of my confusion regarding them. So it's really because of the voltmeters that potentiometric set-ups don't end up proceeding like a galvanic cell :) (I'm so mad at my books that they don't tell us those essential concepts,)

No problem. In their defense, at least one example does talk about this. 9th edition of Analytical Chemistry by Skoog, West , Holler and Crouch has a diagram that talks about this, and even shows what the potential does when connected through a wire(a resistor in their case).

Really? In what page was that? I can't advance reading from the potentiometry's opening paragraphs to the second section because they didn't explain why it doesn't proceed like in a galvanic cell.

In the hardcover 9th edition, Figure 18-2 was what I was thinking of, but 18-4 and 18-5  also talk about this(pg 447-455)


text:
Why the valence shell configuration of Lead is 6s2 6p2?
Acoording to energy rule the electronic configuration is in order of 6s 4f 5d 6p but when writing valence shell configuration we write 6s after 4f and 5d.Why is it ?
Take Lead as an example.

This is convention. To be perfectly honest, this order is not always kept up. But it makes sense to think of the orbitals by shell (first 1, then 2, …) and not by energy level.
For lead and all elements of the p block, whether the d and f orbitals are fully populated or not does not matter, they do not take part in chemistry (only the s and p orbitals of the outermost shell do). Considering that, we can write the d and f orbitals right at the beginning where it is ‘clear’ they are not needed.
For elements of the d block, we actually are doing chemistry with the d electrons. So it is maybe more customary to actually put 3d behind 4s, just to state that they are important.
There is no physical meaning behind the ordering in either example.

thanks for the clarification.So I should focus on the block to which a particular element belongs, right ?


text:
Why does water attack the more substituted carbon in oxymercuration?
In hydration of an alkene by oxymercuration-demercuration, for an asymetrical mercunium ion, the more substituted carbon atom is attacked by water.
Why is this the case? Since R groups are electron donating, it should make the more substituted carbon less electrophilic, also water would like to attack the carbon having less steric hindrance but the reverse is true.

Using 1-methyl-1-cyclohexene as an example, the mercurinium ion is not symmetrical. If it were, the positive charge would all be on mercury. The ion is unsymmetrical with part of the positive charge on the more substituted carbon. Why this carbon? Follow Markovnikov's Rule: tertiary carbocations are more stable than secondary carbocations. This bias is manifested in the transition state permitting water to attack at the more substituted carbon. The addition is anti. I chose 1-methyl-1-cyclohexene as an example because adding a tertiary butyl group to the 4-position produces more subtle aspects of the reaction mechanism.

While structures 1 and 2 above are both tertiary alcohols, only alcohol 1 is formed. This selectivity is controlled not only by the anti-addition but also by the more favorable chair-like transition states (red arrows) over the boat-like transition states (blue arrows). These issues are provided in greater detail at
http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#Mechanisms as PowerPoint and pdf files.

What if an electron withdrawing group is present at one side of the double bond, will condition change or it will remain same?

If the methyl group were a methyl ketone, then it would a more electron-deficient double bond and less reactive. If the mercurinium ion formed, my best guess is that the mercury would wind up at the alpha-position and water add at the beta-position. If this is the case, then the hydration should be catalytic in mercury (Cf. oxymercuration of alkynes). The product would be a beta-hydroxy ketone.

Oxymercuration is very regioselective and is a textbook example of reaction obeying Markovnikov's reaction. This phenomenon is explained by examining the three resonance structures of the mercurionium ion formed at the end of the first step during reaction. In one stage you will have a tertiary carbocation, which of course is a very reactive electrophile. The nucleophile will then attack the mercurinium ion. Therefore, the nucleophile attacks the more substituted carbon because it retains a more positive character than the lesser substituted carbon.
And about steric hinderance: oxymercuration is an anti-addition.


text:
Why doesn't water evaporate in a balloon filled with water?
I know for sure that if you fill a balloon with water (without leaving any air inside) and you start heating it, the balloon won't expand until the temperature reaches 100 °C.
I think it is because while the temperature is below 100 °C, the vapor pressure of the liquid is <760 mmHg so any expansion of the balloon is prevented by the atmospheric pressure.
However, let's suppose to put some water in a water bottle and close it. Initially, there was no water vapor in the bottle, there was only air. But, after a certain amount of time, water will evaporate until the pressure of the water inside the bottle reaches the vapor pressure of the water at that given temperature.
So, my question is: why can water evaporate if there is some air over it but it cannot without air, although in both cases it has to overcome the atmospheric pressure? What is the difference between a closed container with no air inside (only liquid) and a closed-one with some air inside?

I'm not sure I'm following you here, I just want to remind that water always contains dissolved air, which solubility decreases as the temperature rises so there is always water+air in the balloon if water hasn't been deaerated beforehand.

You are right... I should have stated that water has actually been deaerated before being put in the baloon.

The difference is the empty space in the latter case and lack of it in the former. By "empty" I mean "filled with air", but that's irrelevant; it could have been truly empty with the same result. Air pressure is also irrelevant; it could have been 10 atm, and that still wouldn't prevent water from evaporating.
See, when several gases occupy the same volume, they do not even know about each other. They do not care if another gas enters the same volume and starts to share it with them. But they care a great deal about the volume itself, and try to increase it with all their might; that's what we call pressure.
So it goes.

So, in a bottle which contains some water and some air, water can evaporate because the pressure of the air INSIDE the bottle is irrelevant, while water in a baloon (completely filled with water) cannot evaporate because the pressure of the air OUTSIDE the baloon prevents water from evaporating?

@MarcoPagonia No, the water only feels the pressure on itself. In a half-filled bottle, twice as much water is in the air than in a three-quarter-filled bottle, and this linearly goes to zero the more water and less airspace is in it. Fill a bassin to the top and put an even lid on it, and no water will ever evaporate.

@MarcoPagonia Yes, that's right.

@IvanNeretin No, the pressure outside the ballon is irrelevant, unless it is so low the resulting pressure inside the ballon is below the vapour pressure of water.

That's precisely the extent to which the outside pressure is relevant.

I think the other answers don't touch the crux of your confusion.
The problem is, you mix up two forms of vaporization:
Evaporation (surface phenomenon): water will fill the entire available volume with its molecules, regardless of pressure due to other vapors.
Boiling (volume phenomenon): bubbles will only form and survive if vapor pressure overcomes pressure.
So…

why can water evaporate if there is some air over it but it cannot without air

Because evaporation needs an available volume, which, in the latter case, it doesn't have.

although in both cases it has to overcome the atmospheric pressure?

Wrong. Evaporation doesn't need to overcome any pressure (water molecules on the surface thoughtlessly jump into air, and that's it*). Evaporation also doesn't increase the vapor phase volume -- that's why your baloon would not expand if you filled it with air and water and left it at room temperature.

What is the difference between a closed container with no air inside (only liquid) and a closed-one with some air inside?

The former does not have available volume for molecules in the liquid state to jump into. The latter does. Therefore, the former doesn't see evaporation. The latter does.
However, any container will accomodate boiling bubbles. That's why your baloon expands. That's why badly designed pressure vessels burst.
Always keep in mind the important distinction between these two modes of vaporization.
* water molecules in the air also jump back to the liquid phase's surface; it's just that, below vapor pressure, more liquid evaporates than vapor condenses back.

The individual water molecule does not "know" about the external pressure. If there already is a free space it can evaporate into, it will do so, to the point where the partial pressure in the free room is equal to the vapour pressure of the water body.
With an infinitesimally small free space above the water (like in your ballon), the amount of water that evaporates is also infinitesimally small.
If your water bottle was a cylinder with a piston that is permeable to air (not necessarily to water), that piston would of course sink to the water surface over time.

The total evaporation is the same regardless of how much empty space there is above the water. Once it evaporates, it fills the space above the water. Once there is no more empty space, the vapor has nowhere to go except back into the water. Thus, in a close container, an equilibrium will be reached in which as much water is evaporating as is condensing, and there will be no net evaporation. The amount of water vapor that's present in that equilibrium depends on how much empty space there is.
The term "vapor pressure" is a term that is used to describe this process, and draws on normal pressure as an analogy. As more water evaporates, the vapor pressure increases. With normal pressure, the larger the volume, the more gas you need to reach a fixed volume. Similarly, the larger the volume, the more vapor you need to reach a fixed vapor pressure. Thus, we can compare the vapor to volume ratio of the balloon and bottle and find that we get the same number in both cases, and we can think of this constant as being a "pressure". When this "pressure" reaches a critical value, equilibrium is reached, and net evaporation ceases. Given this formulation, your statement "water will evaporate until the pressure of the water inside the bottle reaches the vapor pressure" is true for the balloon as well, it's just that the vapor pressure is reached much more quickly.

The total evaporation is the same regardless of how much empty space there is above the water. Could you please clarify this, since evidently 1 m³ of empty space will require more water vapor to reach the same partial pressure than 1 mL of empty space would?

In both cases, the number of water molecules evaporating per second is the same. However, with only 1mL of space, molecules that evaporate will quickly condense back into a liquid. With 1mL of space, however, the rate of condensation will start small, and increase as the partial pressure increases. Thus, the first case has smaller net evaporation, but both have the same total evaporation.

I understand now. But I think such terminology is confusing. I took it to mean total volume evaporated before you explained, not rate of evaporation. Is it mainstream? If not, I suggest a change in the wording to become clearer.


text:
Is lanthanum a lanthanoid?
Is lanthanum classified as lanthanoids? If yes, then can somebody explain in detail why La has 6s2 5d1 as its electron configuration and Ce has 6s2 4f1 5d1?

Yes, Lanthanum is classified as a lanthanide although it does not have any electrons in its f orbital.
La does not obey the general rule of lanthanides ([Xe]4f^n 6s^2), but is included in this group due to its characteristics (particularly trivalency) being in accordance with the rest of the group. Such as its atomic radius & ionisation energies.
Ce is another exception to the rule (as are Ga & Lu). It ought to be [Xe] 4f^2 6s^2, but the effective nuclear charge, which is still relatively small, is not great enough for f-orbital contraction to be sufficient for 2e in the f orbital. Thus the energy required to fill the f orbitals with 2e is still greater than the energy required to fill the d orbitals.
That being said, it is equally as difficult to fill the 5d orbitals with both electrons, as the f orbital contraction has lowered the energy required enough such that 1 electron can occupy the f orbitals and the other can occupy the d orbitals.
I understand if this isn't the exact answer you are looking for as many areas of periodicity have discrepancies, but reasons can essentially be derived back to a balance between the stability of the species & the energetics arising from the occupation of the electron in available orbitals.


text:
Order of second ionization energy for cations
Why is the correct order of second ionisation potential of $\ce{Li , Be , B , C }$ such as: $\ce{Be} <\ce C <\ce B <\ce{Li} $?
I understand that $\ce{Li+}$ has a stable inert gas electronic configuration and $\ce{B+}$ has one fully filled s orbital. Hence these will have a high value of ionisation potential. But on what basis do we then arrange the ionisation potentials of carbon and beryllium? What is the general steps of rules to solve such kinds of questions?

You are correct about $\ce{Li^+}$: the second electron to be extracted would be one of the two extremely tightly bound $\text{1s}$ electrons, which explains the very high second ionisation energy of that ion.
As for the comparison $\ce{Be^+}$ to $\ce{C^+}$, the main effect here is effective nuclear charge, $Z+1$, which in the case of $\ce{Be^+}$ is $5$ and in the case of $\ce{C^+}$ is $7$.
Equivalent electrons (same or similar orbitals) are more tightly bound to the nucleus, hence harder to extract, for higher $Z+1$. This results in a trend of higher second ionisation energies for higher $Z$, across a period.
This is a general trend across each period.

What is the general steps of rules to solve such kinds of questions?

Chemguide is a good place to start.


text:
Find the final temperature and volume of the gas

$\pu{10 L}$ of a monoatomic ideal gas at $\pu{0 ^\circ C}$ and $\pu{10 atm}$ pressure is suddenly released to $\pu{1 atm}$ pressure and the gas expands adiabatically against this constant pressure. What are the final temperature and volume of the gas? (Answer: $\pu{174.9 K}$, $\pu{64.04 L}$)

There are some points I don't get:

First of all, it is not given whether it is reversible of irreversible expansion.
If it is reversible, I am at wits end deciding whether to use one of the following (without success):
$$ pV^\gamma = \mathrm{constant} \quad \mathrm{or} \quad T^\gamma p^{1 - \gamma} = \mathrm{constant}$$
If it is irreversible, I tried using
$$W_\mathrm{irreversible} = \Delta E$$
i.e.
$$-P_{ext} \left(\frac{nrT_2}{p_2} - \frac{nrT_1}{p_1}\right) = nC_\mathrm{v}\Delta T$$
I am having trouble using $r = \pu{0.0821 L atm mol-1 K-1}$ and $R = \pu{8.314 J mol-1 K-1}$.

Just a note: IUPAC recommends to use lowercase $p$ for pressure to avoid confusion with power $P$. ACS allows both. I personally prefer lowercase $p$ as the safest option, hence my edit.

The process should indeed be considered irreversible, as the problem statement gives a rather explicit hint stating that the pressure was released suddenly, so that your 3rd suggestion is correct. With an irreversible expansion against a constant external pressure, the adiabatic equation $pV^\gamma = \text{const}$ is inapplicable, so one must use the first law of thermodynamics. Work is done at the expense of the loss of internal energy:
$$W = - \Delta U = nC_\mathrm{v}(T_1 - T_2) \tag{1}$$
For ideal monoatomic gas the following is applied:
$$n = \frac{p_1V_1}{RT_1} \tag{1.1}$$
$$C_\mathrm{v} = \frac{3R}{2} \tag{1.2}$$
so that (1) can be rewritten as such:
$$W = nC_\mathrm{v}(T_1 - T_2) = \frac{3p_1V_1R}{2RT_1}(T_1 - T_2) = \frac{3p_1V_1}{2} - \frac{3p_1V_1T_2}{2T_1} \tag{1.3}$$
At the same time work by gas expansion is
$$W = p_2(V_2 - V_1) = p_2V_2 - p_2V_1 \tag{2}$$
and from the ideal gas law the reduced product is
$$p_2V_2 = nRT_2 \tag{2.1}$$
so that (2) can also be rewritten:
$$W = p_2(V_2 - V_1) = nRT_2 - p_2V_1 \tag{2.2}$$
I would leave for OP the task to to equate (1.3) and (2.2) and express the final temperature $T_2$ from there:
$$\frac{3p_1V_1}{2} - \frac{3p_1V_1T_2}{2T_1} = nRT_2 - p_2V_1$$
Here I bring up the final result:
$$T_2 = \frac{3p_1V_1 + 2p_2V_1}{5p_1V_1} T_1 = \frac{3 \cdot \pu{10 atm} \cdot \pu{10 L} + 2 \cdot \pu{1 atm} \cdot \pu{10 L}}{5 \cdot \pu{10 atm} \cdot \pu{10 L}} \cdot \pu{273.15 K} = \pu{174.7 K} \tag{3}$$
Volume after expansion is also to be found using ideal gas law and (1.1):
$$V_2 = \frac{nRT_2}{p_2} = \frac{p_1V_1RT_2}{RT_1p_2} = \frac{p_1T_2}{p_2T_1}V_1 = \frac{\pu{10 atm} \cdot \pu{174.7 K}}{\pu{1 atm} \cdot \pu{273.15 K}} \cdot \pu{10 L} = \pu{64.0 L} \tag{4}$$

Your book has a printing mistake (or you might have typed it out wrong), as Gert pointed out, such a process doesn't even make sense. The part "against this constant temperature." should be replaced with "against this constant pressure."
Now, those words "suddenly" and "against this constant pressure" tell us the process has been performed irreversibly, it's like a heavy block was sitting on top of that piston, and now suddenly the block just vanished away, all hell breaks loose for the gas molecules during such an expansion, there's no way we can define stuff like pressure or temperature for the gas during this period. In reversible processes, parameters like P,V,T could have been defined all through out, and that's why we can draw such a nice sweet curve on a PV graph.
So, now that we know the process has been performed irreversibly, we cannot use those formulas involving $\gamma$, you can use them only if the process was reversible. We can go back to the basics like what the first law of thermodynamics says.
You're doing great so far with the equation you wrote. Now try substituting known values in that. I really don't think you should have a problem with R, because on substituting the Cv of an ideal monoatomic gas, you'll see you can just divide R out from both sides of the equation.

I think you're right, so +1 from me. 'In my day' we solved most of these problem assuming reversibility, maybe that's changed now?

Sorry, typing error.

The gas cannot "expand adiabatically against this constant temperature" (of $\pu{0 {^\circ}C}$) and then end up at $\pu{174.9 K}$, at the same time. That's a contradiction in terminis. Adiabatic expansions cannot be isothermal, in any case. Adiabatic processes mean there is no heat transfer. Fast processes are often adiabatic because heat transfer takes time and fast processes don't allow that time.
The fact that the pressure is lowered "suddenly" makes me believe the process is irreversible because mainly slow, equilibrium processes without friction, turbulence, etc are reversible. But I believe the spirit of the question will allow you to simply use:
$$P_0V_0^{\gamma}=PV^{\gamma}$$
to find final volume $V$. If the process was irreversible that would have been specified but it is common for problems like this to use reversibility as a simplifying approximation, just like most gasses are assumed ideal (when few are).
You could then use the Ideal Gas Law to find final temperature $T$.

As said in the comments, the numbers don't stack up.
For example:
$$V=V_0\Big(\frac{P_0}{P}\Big)^\frac{1}{\gamma}=10\times 10^\frac{3}{5}=39.8\ \mathrm{L}$$
So the question is either badly formulated or the process is irreversible or polytropic.
However, the Ideal Gas Law is respected:
$$\frac{P_0V_0}{PV}=\frac{T_0}{T}=\frac{10\times 10}{1 \times 64.04}=\frac{273.16}{174.9}=1.562$$

I am getting $ V_2  $ as $ \pu{39.83 L} $ that way.

Let me try. 5 minutes, plse.

What value for $\gamma$ did you use?

Using $\gamma=\frac{7}{5}$ (diatomic gas) the numbers don't stack up for me either. Nothing works out at all. Either this question is badly formulated or this process is VERY irreversible. Or the expansion may be polytropic: https://en.wikipedia.org/wiki/Polytropic_process. Quitee common in industrial processes.

Ooopsie: it's a mono atomic gas, so $\gamma=\frac{5}{3}$ but still no joy.


text:
Cytotoxicity of Ru-complexes
I have a basic question: why are Ru(II)-complexes cytotoxic?
What I know is that depending on the used ligand some of them interact with the DNA either via DNA-pi-stacking (intercalation) or binding in one of its grooves (minor or major groove). Some might even bind irreversibly. In any case, a high affinity of this interaction is necessary as otherwise those "mistakes" were corrected by the DNAs correction mechanism. But why not use other metals? Whats the advantage of Ru(II)?
Sure, the complex should be inert and the metal non toxic, but what is the advantage about Ru? (I know about cis-Pt complexes and its mimicks...)
Further one could read much about N-containing ligands like bpy for those complexes and about their nice photophysical properties. But all the literature doesn't show up the connection between cytotoxicity tests and the interaction with the DNA and the choice of the ligand.. Please help me in getting a deeper understanding in this!


text:
Basic amino acids
Amino acids having more number of amino groups than carboxyl groups are basic amino acids.
Histidine has a 'NH' attached to 2 carbons and it is called basic amino acid.
However, tryptophan (having >NH) is called neutral amino acid. Why is it so?

related https://chemistry.stackexchange.com/questions/48499/basicity-of-heterocyclic-compounds

Not my area of expertise, but isn't it a function of which groups are protonated/deprotonated at physiological pH? I.e. the imidazole group of histidine is partially protonated whereas the indole side chain of tryptophan is not?

@airhuff Indeed it is. OP doesn't get how and when lone pairs are delocalised and loose most of their basicity.

@airhuff,sorry,  I didn't get you.

Only one nitrogen of histidine’s side chain is basic. Hint: it is not the protonated one. For more, please go back to your books.

According to PubChem, indole (the nitrogen containing ring system of tryptophan) has a pKa value of -2.4. That means that it is a very weak base: at pH 0 (strong acid) it will be less than 1% protonated, and at neutral pH (physiological conditions) it will be neutral for all practical purposes. Notice that the indole nitrogen lone electron pair is conjugated as part of the pi system of the indole ring--hence it is unable to act as a base. The imidazole ring of histidine, which superficially resembles the indole group in tryptophan, has two nitrogens--one is similarly conjugated, but the other has a lone pair in the plane of the ring and is available to be protonated.


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What's a good way to find out how much lead is in parts being used in toys?
We are using parts from China in small-scale manufacturing of toys.  We used an abotec "lead inspector" test kit on the same part from 3 sources, using abotec's leach test, in which we soaked each part in white vinegar for more than 4 hours, and then added a few ml of the vinegar used to a drop of reagent on a swab.  The swab turned black each time, indicating that the part had at least 50 ppm of lead.  The vinegar in each test tube became a little cloudy, but none turned dark.  Any thoughts about how much lead, and whether that might be unsafe?  PBGC's standard is 100 ppm.

If that amateur kitchen test already finds lead, then you are most probably in hot water, but that's definitely not an adequate test. Find an analytical laboratory, get yourself a lawyer.


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Order of Solubility
Why solubility of alkaline earth metals decreases down the group?
In my book, it is given that solubility of earth metals increases down the group
$$\ce{BeSO4 > MgSO4 > CaSO4 > SrSO3 > BaSO4}$$
I couldn't really understand how is this possible.
Can anybody explain?

the size (atomic radius) of the element increases. Mg 2+ is a much smaller in size than Ba 2+ because it contains less protons and neutrons in its nucleus. As a result, larger cations form STRONGER bonds with larger anions (CO3 is considered a decent size anion). The rule here is ions of similar size attract one another strongly, and thus do NOT dissociate or become soluble in solution. (used CO3 as example)


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Why is a set of p-functions added to the minimum basis sets of Li and Be?
The minimum basis set consists of the smallest number of functions necessary to contain the electrons of the neutral system. In the book Introduction to Computational Chemistry, Second Edition, by Frank Jensen, it is stated that "a set of p-functions" usually is added to the minimum basis sets of Li and Be atoms, despite them containing 3 and 4 electrons respectively and thus would require 2 s-functions. I can not, however, understand why.

A minimal basis includes a single basis function for each orbital assumed to be relevant to the atom (or molecules that include that atom), which are usually those in the same shell. Think about Be; you might be able to describe a lone Be atom with just the 1s and 2s orbital, but you would describe bonding very poorly without including 2p orbitals.

By your definition, any minimal basis set would only include as many orbitals (including spin state) as there are electrons. This implies that each atom would be described as a single determinant and therefore the entire system would be described by a single determinant. Without unoccupied orbitals, there is no binding. Hence, a minimum basis set has to fulfill a few more constraints. A common requirement is the inclusion of the full valence shell and invariance under spin-flip. For second period elements, this includes p-orbitals.


text:
Suggest why melting point of sodium iodide is lower than the melting point of sodium bromide
I would assume this has something to do with bonding and perhaps density, but I haven't the faintest where I'd begin.

Related: https://chemistry.stackexchange.com/questions/82246/which-has-the-highest-melting-point-naf-nacl-nabr-nai/82254#82254

The charges are further separated in the iodide, so ...

Reactivity decreases as you go down the halogen group. This is because when you go down the halogen group, you add another shell of electrons and increase the amount of electron shielding. So the valence electrons feel less of a pull from the nucleus, and are not tightly bound. Note that this is the same logic for why reactivity increases as you go down the alkali metal and alkaline earth metals group.
So, $Br$ is more reactive than $I$ with $Na$, because $Br$ is higher up the group (the atomic number of $Br$ is $35$, while the atomic number of $I$ is $53$). Since $Br$ is more readily reactive with $Na$ than $I$, the bond between $Na$ and $Br$ is stronger than the bond between $Na$ and $I$. Thus, the boiling point of $NaBr$ is higher than the boiling point of $NaI$ .
To check, the boiling point of sodium bromide ($NaBr$) is $~1,396^{o}C$, while the boiling point of sodium iodide ($NaI$) is $~1,304^{o}C$.


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Why is dinitrogen tetraoxide less stable at high temperature?
I have calculated the enthalpy of dimerisation for the reaction at low temperature:
$$\ce{2NO2 <=> N2O4}$$
$$\Delta_\mathrm{dimerisation}\ce{H}= \pu{-13.87 kcal mol^{-1}}= \pu{-58.03  kJ mol^{-1}}$$
Also for this reaction, $\ce{\Delta S}$ is negative due to presence of more gaseous moles on LHS.
My book mentions that $\ce{N2O4}$ is stable only at low temperature. Why is it so? I couldn't find any reliable sources on the internet.

By Le Chatelier's principle, exothermic reactions tend to shift their equilibrium position to the reactants when the temperature is increased.

Enthalpy should be given in $\pu{kJ/mol}$ or $\pu{kcal/mol}$ if you must use those outdated nonstandard units. This doesn’t matter here but your answer does not work if you ignore the mol.

I figured out the answer using Van't Hoff Equation:
$$\ln\left(\frac{\ce{K_2}}{\ce{K_1}}\right)=-\frac{\ce{\Delta H}}{\ce{R}}\left({\frac{1}{\ce{T_2}}-\frac{1}{\ce{T_1}}}\right)$$
Let's assume that $\ce{\Delta H}$ is constant for a limited range of temperature.
Since $\ce{\Delta H}$ is negative for the reaction $(=\pu{-13.87 kcal mol^{-1}})$ if we increase the temperature i.e. $\ce{T_2>T_1} \to\ce{K_1> K_2}$. Thus, equilibrium shifts towards the left as the temperature increases implying the decreasing stability of $\ce{N2O4}$ with increasing temperature.


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Isotopic mass to density
I want to figure out/calculate the density ($\pu{g/cm3}$) of the various isotopes, here $\ce{^109Ag}$. I know it's isotopic mass is $\approx109$ and calculated the number of particles/gram. But I couldn't find out how to get the density.
From what I could find out over some mole maps I need to multiply the molar mass with a factor to get to the density but is that even possible or am I wrong?

o.O It seems you don't know what you are talking about. Density is found experimentally, is variable and does not depend on other physical parameters in any simple way; it's other parameters that depend on it.

Uhm, yes. Thats the reason I'm asking it here. My thoughts were that one could calculate it by: number of particles/cm³ multiplied by the atomic mass.
So, the density is like a approximation through testing then?

Number of particles/cm³ can be derived from particle mass and material density, it can't be measured directly.

I see. Well thats no good for me I need a exact solution. Will keep on looking, thanks anyway.

It's not a solution, it's me telling you that you're going backwards. Density can be approximated from atomic mass and parameters of crystalline cell, but it's imprecise and rather dumb.

@Mithoron There is unit cell and crystalline solid, "crystalline cell" doesn't make much sense. I don't understand how the crystallographic density determination is "dumb". It's no dumber as any other method, and typically provides quite good results.

This is possible, just use the density equation for a crystal lattice
$$ d = \frac{z\times M}{V_c \times N_\mathrm A}$$
Where $d$ is density; $z$ is the number of atoms per unit cell; $M$ ist he molar mass of the material (isotope in this case); $V_c$ is the volume of a unit cell; and $N_\mathrm A$f is avogadro's number.  Since silver is FCC, $z= 4$ ($\frac{1}{8}\times 8 + \frac{1}{2}\times 6 = 4$) and $V_c$ is equal to the lattice parameter squared ($a^3; a = \pu{0.409nm}$).
This method will give you a very close approximate value but is limited in accuracy to the experimentally observed density as it does not account for free space in the crystal due to vacancies in the lattice or voids at grain boundaries. For example, if you calculated the densities of iridium and osmium you find that iridium should be denser, but experimentally you will not observe this result because iridium has a sufficiently high concentration of vacancies that osmium is actually denser.

Thanks for your answer. I have tested that equation and came the expected result.


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Notation for specific/volumetric entropy
I'm working on a problem related to entropy, and for this particular use case it's advantageous to write down properties in 'specific' units. In the case of entropy (with units Joules over Kelvin, I think is commonly referred to as: $[S] = \pu{J K-1}$), I want to know how to communicate volumetric entropy, i.e. Joules per cubic meter Kelvin: $[\color{red}{?}] = \pu{J K-1 m-3 }$.
Is it something like $\bar{S}$?

Are you looking for an analogue to specific volume or the specific gas constant?

No I'm wondering if the notation is different for the intrinsic property of entropy. Specific entropy is useful in fields such as chemistry and thermodynamics to compare substances. Here's a link to a thermodynamic table using specific entropy (in that case, kg in the denominator)

I'd definitely go with the IUPAC answer as a default, but just so you know, a lot of times people will use a \hat character over the (capital) symbol to denote "specific". That's pretty common in introductory thermodynamics textbooks. Koretsky's text comes to mind.

Specific entropy
According to IUPAC "Green Book" Quantities, units, and symbols in physical chemistry, specific entropy is denoted as lowercase latin "s": $s$ [1, p. 56], whereas $\bar{S}$ would refer to molar entropy:

\begin{array}{lll}
\text{Name} & \text{Symbol} & \text{Definition} & \text{SI unit} & \text{Notes} \\
\hline
[...]\\
\text{molar quantity}~X & X_\mathrm{m}, (\bar{X}) & X_\mathrm{m} = X/n & [X]/\pu{mol} & 5,6 \\
\text{specific quantity}~X & x & x = X/m & [X]/\pu{kg} & 5,6 \\
[...]\\
\end{array}
  [...]
$(5)$ The definition applies to pure substance. However, the concept of molar and specific quantities (see Section 1.4. p. 6) may also be applied to mixtures, n is the amount of substance (see Section 2.10, notes 1 and 2, p. 47).
$(6)$ $X$ is an extensive quantity, whose SI unit is $[X]$. In the case of molar quantities the entities should be specified.
Example $V_\mathrm{m,\ce{B}} = V_\mathrm{m}(\ce{B}) = V/n$ denotes the molar volume of $\ce{В}$.

Just as specific heat capacity $c$, specific entropy $s$ is measured in $\pu{J K-1 kg-1}$ [1, p. 90].
General note [1, p. 6]:

The adjective specific before the name of an extensive quantity is used to mean divided by mass. When the symbol for the extensive quantity is a capital letter, the symbol used for the specific quantity is often the corresponding lower case letter.

Volumetric entropy
Quite often there is no special notation used, and volumetric entropy is denoted with $S$. Specifically, volumetric entropy generation rate for the convective heat transfer in flowing viscous fluid is often denoted with S-prime notation, e.g. from classical paper "A Study of Entropy Generation in Fundamental Convective Heat Transfer" [2]:

\begin{align}
& S' & [\pu{W K-1 m-1}] \\
& S'' & [\pu{W K-1 m-2}] \\
& S''' & [\pu{W K-1 m-3}]
\end{align}

References

IUPAC “Green Book”. Quantities, units, and symbols in physical chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
Bejan, A. J. Heat Transfer 1979, 101 (4), 718–725. DOI: 10.1115/1.3451063.


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Are all ionic compounds salts?
According to Wikipedia:

A salt is an ionic compound that can be formed by the neutralization
  reaction of an acid and a base.

Are all ionic compounds salts? Are all salts ionic compounds?

Interestingly, IUPAC states that a "salt" is "a chemical compound consisting of an assembly of cations and anions". Under this definition, all ionic compounds are salts, and all salts are ionic compounds.
Therefore, something like sodium hydroxide ($\ce{Na+OH-}$, definitely an ionic compound) could actually be correctly called a salt. This clashes with the commonly taught high-school level definition of a salt ("the product of an acid-base reaction"), unless you consider very general definitions of acids and bases such as the Usanovich definition, whereby sodium metal $\ce{Na^0}$ is an electron donor (and therefore a base) and water is an electron acceptor (and therefore an acid).
That said, the high-school definition is too simplistic. It is common for compounds to be an acid, a base and a salt all at the same time; consider for example sodium bicarbonate ($\ce{Na+HCO3-}$). It is made of cations and anions, and therefore is definitely a salt. Furthermore, it can act as both a Brønsted–Lowry acid ($\ce{NaHCO3 + OH- -> H2O + Na+ + CO3^2-}$) and as a Brønsted–Lowry base ($\ce{NaHCO3 + H+ -> Na+ + H2CO3}$). Another amusing example is hydrazinium sulfate, a salt, acid and base, where both the cation and anion are also both acids and bases!

Sodium hydroxide is a potential acid-base product in ammonia solvent, produced from water (acid) and sodium amide (base).  Thus the concept of a salt as an acid-base reaction product is a case of "it depends".

I think the better way to produce sodium hydroxide in an acid-base reaction would be to generate sodium oxide and then neutralise that with the acid water. That way, you don’t introduce any additional contaminants.

The logic of the first paragraph is flawed. The definition requires anything called a salt to be an ionic compound. The converse doesn't follow - it doesn't require all ionic compounds to be salts.

@IanBush Indeed, strictly speaking you are correct. That said, I think the "spirit of the law" is to establish bidirectional equivalence between the terms. I can't think of a particularly good reason to carve out some definitional space for "ionic non-salts".

What about zwitterionic compounds?

@matt_black The less German name for zwitterions is "inner salts", so I guess they count, and they can be indubitably ionic, even when their net charge is zero. It's just the special edge case where the cation and the anion happen to be located on the same chemical species. That said, I suppose there are still hyperedge cases which could be a gray area - for example, are mesoionic compounds like sydnones and münchnones ionic, and are they salts? Hopefully one realizes the limitations of absolute definitions outside of mathematics.


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Vapour pressure in an enclosed container with liquid at normal conditions
Can vapour pressure happen in an enclosed container with liquid at normal conditions?
I guess the answer is yes, but I don't understand why some liquid on the liquid surface may process sufficient kinetic energy to escape into gas phase.

Does the liquid initially fill the entire container?  Is that what you're saying?

https://chemistry.stackexchange.com/questions/59541/is-a-liquid-in-a-container-always-in-equilibrium-with-its-vapour

Yes, there is vapour pressure in a closed container.
The molecules in a liquid or gas don't all have the same energy: some have a little, other have a lot. There is a distribution formula in thermodynamics to describe that range of energies (in ideal gases its is called the Maxwell-Boltzmann distribution; liquids are a bit more complicated). The average energy of a molecule in a liquid isn't enough to get a molecule into the gas phase, but a proportion of the molecules have higher energies than that and they will have enough energy to escape the liquid phase.
The same is true for the molecules in the vapour, but in reverse. Some of them don't have enough energy to be in the vapour phase and they will end up back in the liquid. In equilibrium the number of molecules leaving the liquid equal those returning to it.
As temperature rises, a higher proportion of the molecules in the liquid have enough energy to escape and the vapour pressure rises.
The key to understanding this is realising that the average energy of the molecules is not representative of all the molecules. In reality energy is distributed unevenly with some molecules having more and other having less. When you realise this, there is no mystery why some have enough to escape the liquid.

The kinetic theory of gases is based on the idea that particles are moving randomly. Even if all particles started motion with exactly the same speed, as they interact with each other and the walls of the container their velocities would become randomized. The kinetic energy of particles at a given temperature is an average; some particles are moving much faster, and can escape from a liquid. These faster-moving particles, removed from the liquid, leave the liquid a bit cooler. Thus, evaporation puts a liquid into a gas and chills the liquid.
For a more complete explanation, see this video.


text:
Interference and absortion filters
I am learning a bit about Uv-vis spectroscopy filters. As far as I know there are two commonly used: interference and absotion filters.
The first one works with the interference of light principle and the second is a glass with some pigments which absorbs light (of some specific frequency).
Question
Have you got any idea if there is any further classification of absortion filters? Any help will be greatly welcome.

There are a great many of each type of these filters.
Interference filters are optimized for different wavelengths, different bandwidths, transmission vs. reflection etc. by controlling the materials and the thicknesses of the layers. A commercial supplier, Edmund Optics, shows how to specify and order filters.
In a similar manner, absorptive filters depend on the chemicals used to make them. For example, a solution of $\ce{CuSO4}$, copper sulfate, in water removes wavelengths between ~600-750 nm, but letting blue light through..

Possibly I haven't been clear enough. I wonder if there is any further classification of absortion filters...Thanks for the answer anyway!

Sure: you could classify them any number of ways: Solid, liquid or gaseous? Absorption due to chemical formula or due to colloidal properties (e.g. CuSO4 blue vs. colloidal-gold-in-glass red)? Solid filters could be subdivided into single vs. multilayer, glass vs. plastic... What is the point of your question? Classification is arbitrary.


text:
How much yeast is necessary to produce the same amount of carbon dioxide that I exhale?
I think I might have bed bugs. I read that bed bugs are attracted to $\ce{CO2}$ and heat, I also know that yeast produces $\ce{CO2}$ as a waste product. Knowing that, I did some research and found this article on intructables.com which creates a "trap".
My gut instinct is that a 1/2 teaspoon of yeast will not produce anything close to the amount that a person exhales, but I don't know My question is, how much yeast would I need to create the same amount of $\ce{CO2}$ that I exhale on average in a day, per day?
What I know so far:

We exhale $\pu{0.02 m^3/h}$ while resting or low activity (Oscar W. Richards and Florence W. Haynes, Plant Physiol. 1932, 7 (1), 139–144.)
This graph (Figure 3 and 4 of above citation) shows the average $\ce{CO2}$ production per minute by yeast, but I don't know enough about chemistry to know what the units are or how to convert them to something useful (in my case, "useful" would be I need to buy x grams of yeast).

The trap definitely works as I've caught 2 in 24 hours. However, catching them isn't my interest (I wouldn't be able to get rid of them all with a trap).

However, please don't be distracted by the bugs. The bugs themselves aren't the problem of this question; I'm trying to determine the quantities. I'm fairly naive when it comes to chemistry, but in my mind this shouldn't be a hard problem to solve. Unfortunately, I don't know how do the conversions or read the paper's to arrive at the answer but I'm still actively trying to figure this out.

Might help. https://www.engineeringtoolbox.com/co2-persons-d_691.html

This was a pretty interesting post. But I still feel this should've been posted at the Biology.SE instead O:)

Some pointers: (1) I haven't  checked the links yet, but if you'd rather have the bugs heading for the "trap" you mention than your body, then you'll probably need to generate more $CO_2$ for your trap than the amount you exhale. (2) One way you can get rid of bedbugs is to spread copious amounts of dry ice over and under you bed (make sure you get the corners too), in your cupboards (bedbugs hide out in clothes too) and dump whatever dry ice you have left over into every nook and cranny you see... then spend a day camping outside before you head back in ;-)

Thanks paracetamol. I think people are distracted by the bed bug thing however and I might edit that out. The obvious answers to that problem is to just call an exterminator. What I'm extremely curious about, to the point that I'm still actually teaching myself this stuff 3 days later, is to just know how much yeast is needed to create the same CO2 that I breath, bed bugs aside.

Please don't use edit statements, this is what the edit history is supposed to be for. Make your case for a person that reads it for the first time, include everything relevant into the question body.

Actually, I recommended posting at the Bio.SE because of this involves the action of yeast (on some substrate) not because of the bed-bugs (I'd probably not recommend editing out the bed-bugs... context helps) ;-P

I think this is a chemistry question, so I will answer it here.
First, I would question whether the amount of $\ce{CO2}$ is relevant or the concentration (or if more convenient, partial pressure). There is no way for the bed bugs to integrate over the entire trap vs. the bed. If you have a concentration gradient from bed to trap, it should work.
Second, yeast is a catalyst. The amount of $\ce{CO2}$ made depends on the amount of available food (the yeast will multiply to keep up). In the simplest terms, if you want the yeast to produce as much $\ce{CO2}$ as you do, you have to feed it the same amount of food that turns into $\ce{CO2}$.
Because yeast and humans have a slightly different diet, you have to interpret "same amount of food" a bit. In human metabolism, all carbon from foods with calories (i.e. not fiber) is used in biosynthesis or ends up as exhaled $\ce{CO2}$. In an adult who doesn't gain or lose a lot of weight, you can estimate the amount of $\ce{CO2}$ produced from the amount of food taken in. In yeast metabolism, depending on the conditions, either most carbon turns into $\ce{CO2}$ (when there is plenty of oxygen) or into ethanol (when there is a lack of oxygen). For bed bug traps or baking, you want the former, while for brewing, you would want the latter.
For practical purposes, though, I think the total amount of $\ce{CO2}$ is moot (see my first point). As long as you have a source of $\ce{CO2}$ in the trap and remove the source of $\ce{CO2}$ in the bed (i.e. sleep elsewhere) and keep the room warm, there should be a nice $\ce{CO2}$ concentration gradient for the bugs to follow.


text:
Enthalpy calculation

What is the value of enthalpy change for the following reaction
$$\ce{X + 2Y -> 2Z}$$
given
\begin{align}
\ce{W + X &-> 2Y} &\qquad &\Delta_\mathrm{r}H_1 = \pu{-200 kcal} \tag{1}\\
\ce{2W + 3X &-> 2Z + 2Y} &\qquad & \Delta_\mathrm{r}H_2 = \pu{-150 kcal} \tag{2}
\end{align}

I have the answers, and the correct answer is $\pu{250 kcal}$, so I'm assuming you do $400-150$, but not sure how you would go about it.

FWIW, a direct implication of Hess's Law is that you can superpose those equations, which allows to treat those equations as linear equations of chemical energy; mathematically, you can said that $w + x  - 2y = - 200$, $2w+3x - 2y - 2z = -150$, which gives you a nice $\begin{matrix}1 & 1 & -2 & 0 & -200\2 & 3 & -2 & -2 & -150\end{matrix} $ matrix, which you'd need to transform to get $\begin{matrix}0 & 1 & 2 & -2 & result\end{matrix}$ form - if it's impossible (the usual caveats from https://en.wikipedia.org/wiki/System_of_linear_equations apply), then the set of equations is unsolvable.

$$\ce{X + 2Y -> 2Z}$$

\begin{align}
\ce{W + X &-> 2Y} &\qquad &\Delta_\mathrm{r}H_1 = \pu{-200 kcal} \tag{1}\\
\ce{2W + 3X &-> 2Z + 2Y} &\qquad & \Delta_\mathrm{r}H_2 = \pu{-150 kcal} \tag{2}
\end{align}
$(-1) \times 2$ + $(2)$ gives:
$\ce{4Y +2W + 3X->2W + 2X + 2Z + 2Y} \equiv \ce{2Y +X -> 3Z}$
Using Hess's Law,
the enthalpy change for the required reaction is given by  $-2\times \Delta_rH_1+\Delta_rH_2= 400-150 = \pu{250 kcal}$

Keep in mind you are allowed to perform basic math operations (+,–,×,÷) with the balanced equations of chemical reactions. I'd recommend to start with undesired component(s) ($\ce{W}$) which should not appear in final equation, and try to cancel them out.
Here, to obtain target reaction and eliminate $\ce{W}$, you need to subtract doubled first reaction from the second one (e.g. "$(2) - 2 \cdot (1)$"), which results in
\begin{align}
\require{cancel}
\ce{\cancel{\ce{2W}} + 3X - \cancel{\ce{2W}} - 2X &-> 2Z + 2Y - 4Y} &\qquad &\Delta_\mathrm{r}H = \Delta_\mathrm{r}H_2 - 2\Delta_\mathrm{r}H_1 \\
\ce{X + 2Y &-> 2Z} &\qquad &\Delta_\mathrm{r}H = \Delta_\mathrm{r}H_2 - 2\Delta_\mathrm{r}H_1
\end{align}
$$\Delta_\mathrm{r}H = \Delta_\mathrm{r}H_2 - 2\Delta_\mathrm{r}H_1 = \pu{-150 kcal} -2 \cdot (\pu{-200 kcal}) = \pu{250 kcal}$$