text:
Why does Carbon-13 NMR not have splitting patterns?
I read about splitting patterns in proton NMR and found that this is due to spin-spin coupling, where reference to Pascal's triangle can explain the splitting patterns found in doublet, triplet and quartet.
My book mentions this later:

There are also no splitting patterns caused by the spin-spin coupling, i.e. interaction of neighbouring protons in the 1-H atoms within molecules.

Advanced Chemistry for You, Third Edition (Ryan)
Why do you not find splitting patterns in Carbon-13 NMR?

There is splitting in 13C. Think about solvent signals or compounds with CF bonds.

In 1H NMR, spin-spin coupling leading to multiplet structure most commonly arises due to coupling to other 1H nuclei. This is most often taught in terms of neighbouring 1H nuclei having "up" and "down" spins, both of which lead to the nucleus of interest experiencing a slightly different magnetic field, and hence having a slightly different chemical shift. While this is only a simplification, it is probably sufficient for the purposes of this question.
In 13C NMR, though, proton decoupling is often employed. As the name suggests, this removes all effects of coupling to protons (i.e. 1H). How is this achieved? Essentially, while the signal from the 13C nuclei is being recorded, the NMR spectrometer simultaneously and continually irradiates the 1H nuclei with energy. This means that each 1H nucleus is rapidly undergoing "spin-flips" between up and down spin, and therefore, what the 13C nucleus sees is actually a magnetic field that is very rapidly oscillating between the two possible values. What is actually observed in the NMR spectrum is an average of the two possible values, which is equivalent to what would be seen if there was no coupling to begin with.
Why do we do this, though?

One concern is sensitivity: when peaks are split due to spin-spin coupling, the intensity of the peak is decreased. Since 13C has a naturally very low abundance, it is of interest to prevent the peak intensity from being spread out due to coupling.
Another concern is simplicity: carbon spectra which have C–H coupling can often be extremely difficult to interpret. For an example I refer you to Glenn Facey's blogpost where an inset of a proton-coupled spectrum of toluene is shown.
A subtle point that is often neglected is the fact that the decoupling process also enhances the intensity of 13C peaks via the nuclear Overhauser effect.

All of these factors make proton-decoupled carbon spectra – or in NMR lingo, 13C{1H} spectra – the preferred experiment.

If you're interested in coupling to nuclei other than 1H:

coupling to 13C is all but invisible, considering the low natural abundance of 13C. Recall that in a 13C spectrum you are only observing 1% of all the carbons present. How many of those carbons are coupled to another 13C nucleus? Or in other words, what are the chances of finding two 13C nuclei next to each other? Pretty low is the answer – so much so that an NMR experiment designed to measure C–C couplings was given the acronym INADEQUATE.
coupling to other abundant spin-1/2 nuclei is in fact very often seen. Two simple examples are 19F and 31P, both of which are close to 100% natural abundance – in compounds containing F or P, multiplet structure will be observed in the carbon spectrum. Here's the spectrum of 1-chloro-2-fluoroethane, $\ce{FCH2CH2Cl}$, courtesy of SDBS. The two peaks on the left are from the carbon next to fluorine; coupling to fluorine leads to a 1:1 doublet being observed. Likewise, the two peaks on the right are from the carbon next to chlorine. It is still coupled to fluorine, but because of the greater distance, the coupling constant is smaller.

There is also probably a wealth of examples in organometallic chemistry (for example 103Rh is also spin-1/2), but I don't currently have the energy to find a good example.
coupling to quadrupolar nuclei (nuclei with spin > 1/2) can be seen, depending on the symmetry of the molecule and on some other physical parameters. For example, since deuterium has spin 1, the carbon peak in $\ce{CDCl3}$ is observed as a 1:1:1 triplet (a familiar sight to organic chemists). But this is not always the case: for a more in-depth explanation see, e.g. Symmetry and NMR linewidths of quadrupolar nuclei

You mentioned that proton decoupling is by averaging the values where the ¹H are up and down spin. However, doesn’t irradiating it only ensures that all of the ¹H is up spin? (where it opposes the magnetic field from what i understand) How does this method ensure that we have a signal where all ¹H is down spin? (where it has an enhancing effect on the external magnetic field)

@ImJustKawaii It's not possible to place all spins in the higher-energy state through continuous irradiation. The explanation for this requires some slightly more complicated theory (see: Einstein coefficients),  but the short answer is that the absorption (i.e. excitation) is balanced by stimulated emission (i.e. decay). That is, for every spin you flip from down to up, there will also be a spin you flip from up to down. Eventually the up and down populations are equalised.

If it’s not possible to place all spins in the higher or lower energy state, what then is the two possible values that we average to remove coupling effect? and what happens at these two values?

@ImJustKawaii The chemical shift of the 13C nucleus depends on whether the 1H is spin up or down. So if the 1H is continually flip-flopping between different states, the 13C chemical shift you see is the average of two different chemical shifts, i.e. a chemical shift in the middle. I thought I described this part already in my answer; what exactly are you confused about? [Note that this description is rather simplified, but it works. QM is needed for a full explanation.]

I think my confusion stems from what two values we are averaging. Initially, I was thinking that to remove the coupling effect we average the chemical shift values of ¹³C when all the protons oppose the external magnetic field and when all the protons enhance the external magnetic field. However, you then mentioned that it’s not possible for all protons to be in the higher energy state (opposing the field I think) by irradiation. So I am now confused as to what two values we are averaging and how irradiation achieves this two values.

@ImJustKawaii I'm afraid I don't understand your question, or at least, I don't understand how it isn't already answered by what I wrote above, in both the answer as well as my comments. If I were to attempt to answer it directly, I would probably end up rewriting exactly the same thing. You might want to consult some other resource. "What two values we are averaging" - chemical shifts, as previous comment says. "How irradiation achieves this two values" - irradiation causes flip-flops between 1H spin-up and spin-down. 13C chemical shift depends on 1H spin. Ergo, two chemical shifts.


text:
Biomolecular energy functions and entropy
From what I've been reading, it seems that entropy (rather than enthalpy) is the biggest driver of protein folding (especially the burying of hydrophobic residues). However, popular energy functions like AMBER, Charmm, etc. don't model entropy. They are however able to identify the native fold within a few angstroms (i.e. the minimum energy structure per the force fields is not more than a few angstroms away from the native structure) (http://onlinelibrary.wiley.com/wol1/doi/10.1002/jcc.20720/full).
If entropic changes are the primary driver of folding, then how is it that energy functions that don't even model entropy are still able to get close in terms of identifying the native structure (in particular, the burying of hydrophobic residues as well)? Is the entropic information incorporated in the solvent models?

Entropy is indirectly related to energy through probabilities. So it's not a separate term in the energy function.  Basic explanation of entropy: https://youtu.be/P9eksvw2e6w. Calculation of entropy from MD simulations: http://www.compchemhighlights.org/2013/03/calculation-of-molecular-entropies.html

@JanJensen I'm not referring to MD but the ability of the energy function + solvent model to identify the native structure as the energy minimum. You can minimize energy even without MD (and softwares like Rosetta and Tasser do so).

Well, if you only want structure, you just need appropriate scoring function. It doesn't even need to explicitly involve energy.

I am sure that a more definitive answer could be provided by someone more familiar with the field, but here are a couple of observations.
First, I think it is a bit of an exaggeration to say that protein folding is almost entirely the result of entropy because the protein will have larger entropy in the unfolded state than in the folded state. The main entropic increase on protein folding will be the entropy of the solvent (and the system solvent + protein is what matters thermodynamically). This is the so-called hydrophobic effect which is what you're referring to by saying folding buries the hydrophobic residues.
Notice, however, that an energy function which contains solvent-solvent, solvent-residue, and residue-residue interactions implicitly contains this entropic effect. This comes by maximizing solvent-solvent and solvent-residue interactions. In order for this to happen, the hydrophobic residues will naturally become buried because this will increase the average number of hydrogen bonds between solvent molecules (I'm assuming the solvent is water) and strengthen the solvent-residue interactions at the exterior of the protein.
Second, the reason an energy-only model of protein folding might get "lucky" by ignoring entropy is that even if the net enthalpy change for protein folding is roughly zero, this does not mean that energy did not play a significant role in the folding process. People often talk about how proteins cannot simply be randomly walking the conformation space because it would take an endless amount of time to find the energy minimum by doing this. This is exactly why intermolecular interactions between residues get discussed so frequently.
So, an energy-only approach will very likely identify the residue-residue interactions which are important structural features of the final folded state. It is entirely possible that there are steps which are energetically destabilizing, but when these steps take place (if they are to take place in the real world), there must be a very constrained conformation space for the step to take place by randomly sampling states. Nonetheless, each step in the protein folding process must hit an energy minimum at some point, so just searching the energetics is a perfectly reasonable first approach. Furthermore, when this approach fails, it tells you that whatever is going on must be quite sensitive to the entropy of the system, so it is not like you are flying blind.
Third, it could easily be the case that energy-only approaches sometimes arrive at the correct structures completely accidentally. That is, there is more than one path from the unfolded state (which unfolded state?) to the folded state. The "real" path, if there is a single path that proteins take, might be quite different from the one that a model like this will take. In fact, if the protein is being sort of randomly perturbed and energy minima are being sought out, the pathway is likely complete nonsense, but this doesn't really matter as long the as protein is folding. It would be interesting to see how models like this do with proteins that have very energetically similar but distinct folded states. If what I'm saying in this part is true, one would expect the model randomly ends up in either state when you run the simulation. This would indicate that in the real world, there is a kinetic effect which forces the folding to follow a certain path and the model does not capture this.

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text:
Does fluorine ever form a double or triple bond?
Does fluorine ever form a double or triple bond?
I wonder if seeming lack of such higher order bonds is related to the electronegativity of fluorine.

Who said it doesn't?

Oxtoby-principle of modern chemistry

It was a rhetorical clickable question. There is $\ce{F=N}$ bond known to exist.

The compound $\ce{NF}$, which is isoelectronic to $\ce{O2}$ is known, has been isolated in matrices and characterised and subjected to calculations. Like for dioxygen, three different states of this molecule are known: one triplet and two singlet states. Without performing any sophisticated analysis of its orbitals, we can expect a bond order of 2, and thus a double bond.
Harbison performed calculations on this compound. He came to the conclusion that the most stable triplet state is best described using only a single $\ce{N-F}$ bond. The two singlet states which require full electron pairing, however, display a much shorter $\ce{N-F}$ bond distance and are thus better described by an $\ce{N=F}$ double bond. Adding formal charges would lead to:
$$\ce{\overset{-}{N}=\overset{+}{F}}$$
Reference:
G. S. Harbison, J. Am. Chem. Soc., 2002, 124, 366–367. DOI: 10.1021/ja0159261.

Going down this line, one wonders what would happen as the atomic number of $\ce{X}$ in $\ce{XF}$ decreases. Apparently boron monofluoride $\ce{BF}$ still has only partial double bond character. Beryllium monofluoride $\ce{BeF}$ would also be an interesting albeit more complicated case...

@NicolauSakerNeto possibly the bond order in BeF is less than in BF. Multiple bonds between nonmetal atoms correlate with completing their atoms, but beryllium has less tendency to take that route.

For most purposes, i.e. writing Lewis Structures and such things, fluorine always forms single bonds. In higher level chemistry, it may be possible (like in the comments). Fluorine, like you said, is very electronegative, and therefore it doesn't like to "share" its electrons, leading to almost always making single bonds.

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text:
Calculation for buffer of pH 5 in sodium fluoride solution
From a journal:

$\pu{500 mL}$ of $\pu{50 ppm}$ fluoride at $\mathrm{pH}~5$ was prepared by mixing $\pu{0.0552 g}$ of sodium fluoride with $\pu{4.1057 g}$ of sodium acetate and $\pu{1.65 mL}$ of acetic acid.

May I know how to get the solution of calculation for amounts of sodium acetate and acetic acid which are $\pu{4.1057 g}$ and $\pu{1.65 mL}$, respectively?
For sodium fluorid calculation is
$$\frac{\pu{50 mg}}{\pu{1 L}} \times \frac{\pu{1 g}}{\pu{1000 mg}} \times \frac{\pu{41.98 g mol-1}}{\pu{18.988 g mol-1}} \times \pu{0.5 L} = \pu{0.0552 g}$$

I want to know the formula to get 4.1057 g of sodium acetate and 1.65 mL of acetic acid

I mean how to know the amount of sodium acetate must be 4.1057g and acetic acid is 1.65ml? I don't know where 4.1057g and 1.65ml came from

The journal entry is sloppy.
 It only specifies the final "concentration" of sodium fluoride. Note that the parts-per notation is highly discouraged! They neither specified the buffer capacity, nor the concentration of the other components.
Without knowing what kind of chemicals were used, you have to guess. So in this case I would guess they used anhydrous sodium acetate, which is a solid, $M=\pu{83 g/mol}$. With $m(\ce{NaOAc}) = \pu{4.1057 g}$ as given in the journal, we calculate that they used about $n(\ce{NaOAc})=\pu{0.05 mol}$.
We can use the Henderson-Hasselbalch equation (ignoring sodium fluoride) to determine the amount of substance of acetic acid we need, $\mathrm{p}K_\mathrm{a}=4.75$, $\mathrm{pH}=5$.
\begin{align}
\mathrm{pH} &= \mathrm{p}K_\mathrm{a}
               + \log_{10} \left(\frac{c(\ce{NaOAc})}{c(\ce{HOAc})}\right)&
\Longleftrightarrow&&
\mathrm{pH} - \mathrm{p}K_\mathrm{a} &=
                 \log_{10} \left(\frac{n(\ce{NaOAc})}{n(\ce{HOAc})}\right)\\
\end{align}
Therefore we need about $n(\ce{HOAc}) = \pu{0.03 mol}$.
Now we have to guess again that they have used glacial acetic acid. This is quite reasonable because it's a small volume and the purest form of acetic acid. According to Sigma Aldrich it has a concentration of about $c(\ce{HOAc})=\pu{17.4 mol/L}$. Therefore we need about $V(\ce{HOAc}\text{, glac.}) = \pu{1.7 mL}$.
I hope you learned from this that you should not leave incomplete notes in your lab journal.

Thanks so much! But can I know that how to know n(NaOAc)=0.05 mol? and amount for sodium acetate =83g/mol x 0.05mol = 4.1015?

@jess Like I said, I guessed. I simply took the values from the entry and reverse-engineered it.

Acetic acid is a liquid. It has a $\mathrm{p}K_\mathrm{a} = 4.75$. A mixture of acetic acid and sodium acetate form a buffer solution which means that the solution resists $\mathrm{pH}$ changes. If the acetic acid and sodium acetate are present in equal molar quantities, then the $\mathrm{pH}$ of the buffer solution will be $4.76$. Since $\mathrm{pH}$ of buffer was $5$, there are more moles of acetate ($\ce{OAc^-}$) than acetic acid ($\ce{HOAc}$).
$$\mathrm{p}K_\mathrm{a} = 4.76$$
$$K_\mathrm{a} = 1.74 \times 10^{-5} = \frac{\ce{[H^+][OAc^-]}}{\ce{[HOAc]}}$$
$$\mathrm{pH} = - \log{\ce{[H^+]}}$$


text:
Stereochemistry of product in Grignard-Gilman addition reaction
I would like to know whether the stereochemistry given in the following reaction is correct or not?

Which attack is more likely? axial or equatorial? I think I should say pseudo axial or pseudo equatorial....since there is a double bond in the ring.

Am I missing anything here...?
Ok. I am adding what I got from other forum, though I do not follow it properly.

You don't seem to be considering the relevant factors. After conjugate addition, the stereochemistry is under diastereoselectively controlled. You should consider the protonation step instead.

I have an answer prepared for you but it won't be posted until you make an effort to address the problem. What does the enolate look like after addition as Zhe has hinted?

Look for a relaxed cyclohexene structure an go from there.

ok user5519, I have added an image, though it is not my work.

I have supplied a diagram that is more legible than the small one you have provided. The cis stereochemistry of the product is a function of axial, but not equatorial attack of the reagent. The stereochemistry is established after the addition occurs. Structures 1 and 2 illustrate the "chair-chair" interconversions of the cyclohexene ring. As the C-C bond is being formed by axial addition in 1, the carbon bearing the red p-orbital will rehybridize from sp2 to sp3 during the transition state (TS) for bond formation. This transition state (TS) will lead to enolate 3 through a low energy, chair-like pathway.
Alternatively, TS 2 would proceed through a higher energy, twist boat-like TS. Both routes occur through axial attack with 1 ---> 3 prevailing over 2 ---> 4. Enolate 3 is in equilibrium with enolate 5, which is unfavored by what is known as an allylic A1,3 interaction.1 The critical step is the kinetic protonation of enolate 3 on workup. Protonation of enolate 3 on carbon from the underside of the double bond (remote from the phenyl group) again occurs through a chair-like TS. Protonation on the top of the double bond would occur through a twist boat-like TS. This pathway is higher in energy and would have provided the trans isomer 9. Structures 6 and 7 are the two chair conformers of the cis-isomer. If one uses to vigorous a set of conditions upon workup, enolization of 6 and 7 can lead reversibly to enol 8 and eventually to the thermodynamic isomer 9.

1) Johnson, F., Malhotra, S. K., J. Am. Chem. Soc., 1965, 87, 5492.

There is no such thing as rehybridisation. 1, 2 are twist conformers, they don't resemble either boat or chair. And a few paragraphs certainly would help readability.

Yes, 2 is a twist. Let's remember who the audience is. Be nice!

It does not matter who the audience is, what is wrong is wrong. And are you really telling me to be nice? What was it in my comment that was offensive to you?

The point of talking about rehybridization was to illustrate that during the TS the carbon goes from sp2 to sp3. Peace.

What if attack of H on either 3 or 5 from the top rather than from the bottom as shown by you.........?

If protonation of 3 or 5 (equatorial?) were to occur on the "top", the trans isomer 9 would form. The kinetic product is cis, not trans.


text:
Kinetics of A + 2 B -> C reaction

A reaction $\ce{A + 2 B -> C}$ is carried out in constant volume at $\pu{227 ^\circ C}$. the volume is $\pu{2 L}$. There is no $\ce{C}$ at the beginning. $[\ce{A}]_0 = \pu{0.035 mol L-1}$. $\ce{B}$ concentration changes over time.
\begin{array}{l|cccccccc}
t~(\pu{min})& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
[\ce{B}]~(\pu{mol L-1}) & 0.09 & 0.066 & 0.05 & 0.039 & 0.033 & 0.028 & 0.024& 0.023
\end{array}
a) At $t = 5$, what is the conversion rate?
b) Calculate the reaction rate of $\ce{C}$ by using graph method according to $\ce{C}$ at time $0$ and $6$.
c) Calculate the total pressure at time $3$.
d) If this reaction should have done at constant pressure what would be the volume change rate?

I answered the a, c and d questions, but I am not sure about it. I can not solve question b, graph method, these were my exam questions and I want to know correct answers. Here are my answers:

Part a.
$$t = \pu{5 min}$$
$$\ce{\underset{0.035}{A (g)} + 2 \underset{0.09}{B (g)} -> C (g)}$$
$\ce{A}$ is a limiting reactant. At constant volume
\begin{align}
[\ce{A}] &= [\ce{A}]_0 (1 - x) \\
[\ce{B}] &= [\ce{B}]_0 - 2x[\ce{A}]_0
\end{align}
$$\pu{0.028 mol L-1} = \pu{0.09 mol L-1} - 2 \cdot \pu{0.035 mol L-1} \cdot x \to x = 0.886$$

Part b.
$$-\ln{(1 - x)} = kt$$
$$t = 6$$
$$0.024 = 0.09 - 2 \cdot 0.035 \cdot x \to x = 0.943$$
$$-\ln{(1 - 0.943)} = 6k \qquad \color{red}{?}$$

Part c.
$$p_T = C_T RT$$
$$0.039 = 0.09 - 2 \cdot 0.035 \cdot x \to x = 0.728$$
\begin{align}
[\ce{A}] &= 0.035 (1 - 0.728) = 9.52 \cdot 10^{-3} \\
[\ce{B}] &= 0.039 \\
[\ce{C}] &= [\ce{A}]_0 \cdot x = 0.035 \cdot 0.728
\end{align}
$$C_T = 0.074$$
$$p_T = \pu{0.076 mol L-1} \cdot \pu{0.082 L atm mol-1 K-1} \cdot \pu{(273 + 227) K} = \pu{3.034 atm}$$

Part d.
$$\epsilon = \frac{1 - 3}{3} = -\frac{2}{3}$$
$$V = V_0(1 + \epsilon x) \qquad \color{red}{?}$$

I edited your question using MathJax and Markdown according to the guidelines (more or less). Please feel free to check the correctness and improve further (see formatting tips here)

@andselisk I'm not sure if you putting a ton of effort into formatting everybody's questions is a flaw... (if you nominate yourself for the next Mod elections, whenever it is, you'll have my vote :D)


text:
Will nitrate be able to oxidise mercury?
We have the following data:
\begin{align}
\ce{NO3- + 3 e- &-> NO} &\quad E^\circ_1 &= \pu{0.96 V}\\
\ce{Hg^2+ + 2 e- &-> Hg} &\quad E^\circ_2 &= \pu{0.86 V}
\end{align}
By the reasoning given in my book, nitrate oxidises mercury:

... of two substances, the one whose reduction $E^\circ$ value is greater, will be able to oxidise the other substance the other.

Then I tried calculating $E^\circ$ for overall reaction also taking into account number of electrons in each reaction. Or that we calculate $\Delta G$ for
$$\ce{NO3^- + Hg -> NO + Hg^2+}$$
$$\Delta G^\circ = -0.96 \cdot 2F + 0.86 \cdot 3  = 0.66F$$
So, $\Delta G^\circ$ is positive and so nitrate does not oxidise mercury. Which is correct?

It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:
\begin{align}
\ce{NO3− (aq) + 2 H+ + e− &<=>  NO2 (g) + H2O} &\quad &E_0 = \pu{+0.80 V} \\
\ce{Hg2^2+ + 2 e− &<=> 2Hg (l)} &\quad &E_0 = \pu{+0.80 V} \\
\ce{Hg^2+ + 2 e− &<=> Hg (l)} &\quad &E_0 = \pu{+0.85 V} \\
\ce{2 Hg^2+ + 2 e− &<=> Hg2^2+} &\quad &E_0 = \pu{+0.91 V} \\
\ce{NO3−(aq) + 4 H+ + 3 e− &<=> NO (g) + 2 H2O (l)} &\quad &E_0 = \pu{+0.958 V} \\
\end{align}
The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is somewhere between you get both $\ce{NO}$ and $\ce{NO2}$.
Now if you also consider the Nernst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.

The question given in the book does not mention anything at all. But I think the last equation of reduction of nitrate applies. The third equation for Hg. And since it is standard reduction, so the concentrations must all be 1 molar right? So why can't we conclude from there? Thanks a lot.

@samjoe - This is sort of odd case. It really depends heavily on $\ce{[H^+]}$ as well as the concentrations of the Hg and nitrate. You can assume anything you want. I'd be sure to explicitly add my assumptions to the answer.

@andselisk - Thanks for the edit! Your edit really formats better and is much more readable.

@MaxW You must be a huge Quake fan, I haven't seen that many \quad's anywhere else:D You are very welcomed, no big deal:)

@andselisk - Just didn't know how else to do it, so I brute forced it. I was last in school 40 years ago. Never used Latex/MathJax until I started on this site. Done a zillion markup languages though.

Thank you for your reply, my question was basically that we cannot predict spontaneity of reaction simply by looking at $E^o$ values of half cells as my book does right? Nitrate was just an example :)

@samjoe - You can't be entirely sure unless you know concentrations of reactants. But if there is a +0.5 Volt difference then you can be almost certain that the reaction will go. The big problem here is the 0.16 V difference that the acid concentration makes on the nitrate half cell.


text:
Confusion in n-factor calculation
How do we calculate the equivalent weight of a compound, in a case where certain fraction of an element of the compound is getting reduced, while the other fraction is unaffected (no change in oxidation state)? For example, consider the following reaction:
$$\ce{Zn + K4[Fe(CN)6] -> K2Zn3[Fe(CN)6]2 + K}$$
I would like to calculate the equivalent weight of potassium ferrocyanide, given its molecular weight $M$. Here some potassium gets reduced to oxidation state (0) from (I).

"Here, some potassium gets reduced to oxidation state (0) from (I)" That reaction equation makes no sense whatsoever. $\ce{K+}$ does not get reduced that way.

I'm sorry, but that reaction actually happens. The question is, how do we calculate the equivalent weight?

https://chemistry.stackexchange.com/questions/86309/calculating-n-factor-of-so2/86315#86315 is an answered question about calculating the equivalent weight in the case of redox reactions.

That particular link doesn't answer my question!

"I'm sorry, but that reaction actually happens." Please provide a credible reference.

The link shows how equivalent weights are arrived at. Apply the same principle to any redox reaction.

Had it answered my question, I'd not have asked it here again. Please try to understand that the general method is not applicable here as only a certain fraction of potassium atoms are reduced, not all!

Your reaction isn't even balanced. Please provide a reference for it. There no reducer strong enough in there to reduce $\ce{K+}$ to elemental K. The only reducing agent present is $\ce{Zn}$.

Also K isn't stable in contact with water. It will react to give $\ce{K ->[H2O]KOH + H2(g)}$

@MaxW: to be fair, he didn't specify aqueous medium.

The OS of $\ce{Fe}$ in that complex doesn't change. It's +III on both sides. Only $\ce{Zn}$ and (cough!!) $\ce{K+}$ does.

I found numerous sources which suggest the following reaction: $$\ce{3Zn^2+ + 2[K4Fe(CN)6] -> K2Zn3[Fe(CN)6]2 + 6K+}$$ which is also performed in aqueous media only. Sorry, no potassium metal among products, not by a chance.

@andselisk: yes, I saw these too. OP's refusal to provide a reference speaks volumes.

This should do your work:
3Zn + 2K4[Fe(CN)6] = K2Zn3(Fe(CN)6)2 + 6 K
Now you can calculate the n factor.
On a second thought Zn may be ionic with +2 Oxidation state releasing K+.

I don't understand why this got upvoted. All you've done is balance a non-existing reaction. We can all do that!

As I see, he wanted to solve some question and the exams many a times commit this mistake. So, it's better to ignore it.


text:
Molecular Orbitals. Relative phase vs Bonding Anti-Bonding
Most intro explanations of molecular orbital formation go something like this:
as two 1s orbitals of H atoms approach each other, their wave functions begin to overlap, and due to constructive and destructive interference, a bonding and an anti-bonding orbital is formed.
Since initially there can be an arbitrary difference in phase between the two 1s orbitals, does this mean that the energy of the molecular bond is a function of the phase difference and that a bonding orbital will form only for a range of phase differences ?

There is no range. The phases can be either similar or opposite.

And either a bonding MO forms (AOs are in phase) or an anti-bonding MO forms (AOs are of opposite phase). And what @IvanNeretin said.

OK.... but why?

Because all other combinations do not produce eigenfunctions.

When electrons get entangled in the bonding orbital, the wave function collapses, so must collapse into an eigenstate/function of the system, and only that combination is an eigenstate?  So the phase difference collapses....

@IvanNeretin what about partial interference ?
http://www.grandinetti.org/molecular-orbital-theory

@M.ghorab.  From that article:  In general, the energy difference between a bonding and anti-bonding orbital pair becomes larger as the overlap of the atomic orbitals increase.

However, I can see the argument that two 1s orbitals of opposite phase form an eigenbasis.  Each 1s orbital is in a linear superposition of that basis.  When they begin to interact a system with four eigenstates is formed, two corresponding to the bonding orbital and two to the antibonding one.  The system becomes a superposition of four states.  But measuring the energy results in a collapse, and the most likely eigenvalue is the energy that corresponds to opposite phases.

Correction:  But measuring the energy results in a collapse, and the most likely eigenvalue is the energy that corresponds to orbitals in-phases.


text:
Niobium has a density of 8.57 g/cm3 and crystallizes with the body-centered cubic unit cell. Calculate the radius of a niobium atom
Here is my question

Niobium has a density of $8.57 \pu{g/cm^3}$ and crystallizes with the body-centered cubic unit cell. Calculate the radius of a niobium atom.

Here is my solution

The thing I cannot get is there are too many different answer for this atomic radius questions. I mean like I've seen three different answer on websites. I've rechecked my question but couldn't find anymistake. Take a look here: http://www.gordonengland.co.uk/elements/nb.htm and here https://www.periodni.com/nb.html Here is anohter different answer too: http://www.elementsdatabase.com/Niobium-Nb-41-element/ Are those all radius correct?

That answer is clearly wrong because you do not give units!  Please do so, or else we may think that your answer is $100$ times as large as you likely intended (assuming length is in S.I. units).

The answer is not certain. As I said take a look that sites. You will see how right I am. Everyone finds different answers. According to Wikipedia, it seems $1.46 \text {pm}$.

According to Wikipedia, answer is $1.46$, on other sites, $1.45$ and $1.44$. on my textbook, $1.43$. Moreover, I rechecked my solutions twice. I'm sure about I didn't do mistake.

using crystallographic density does not account for free space in the material due to vacancies or voids at the grain boundary. Any answer, while close will be off from the true value by a small amount.

Uh ... how is the radius of an atom roughly $1$ picometer?  I seem to recall that the Bohr radius of the hydrogen atom is $51$ picometers and that is just hydrogen.  I would guess that the decimal points in the answers quoted here are not really there!

@A.K. What do you mean?

@Mascular your equation assumes that you are working with a single crystal with all atomic sites occupied. real materials have many crystals and vacancies in the lattice where atoms should be.

@A.K. - Huh?!? the only way to do the problem is to assume a perfect crystal. Also I highly doubt that vacancies are cause of variations in tabulated values for the radius. Frankly anyone who believes that an atom has one "real" radius is kidding themselves. The diameter of an orange depends on how hard you squeeze. So it is with atoms.

@Mascular  - You did make two minor mistakes in your calculations. First you used 93 for the atomic weight instead of 92.906. Second in the next to the last step you truncated to $r^3 = 0.29\times10^{-23}$. You should have carried more significant figures and only rounded to 3 when you got to $r$.

@MaxW I'm sure that the tabulated values are from gaseous species or TEM observations, I was talking in regard to the OP's value. I know it may seem trivial but the results are real. If you do the arithmetic for the density of iridium and osmium, based on atomic radius you find that iridium should be denser, though empirically this is not observed; due to a high valency concentration in iridium, osmium actually ends up being a denser bulk material than iridium.

Do atomic physicist know this?

@A.K. - The assumption that atoms are hard spheres is a bad one which was my point about the orange. So I would expect values calculated from the bulk density, gases species, and TEM to be different. Also if you want to look at trends iridium and osmium crystallize differently.

@MaxW yes they do crystallize differently but FCC and HCP have the same atomic packing factor so that is moot.

@A.K. - That is the point. It is not moot. There is a reason that some metallic atoms prefer FCC or HCP. How the atoms pack effects how they interact (bond) which in turn effects the distance between them. Atoms are not ball bearings which would have absolutely defined packing fractions.

All answers are correct. They are quite smilar.

This is how I would do the calculation...
Given data:
$$\text {Weight of crystal} = \pu{92.90638 g/mole} \ce{->[Round] \pu{92.906} g/mole}$$
$$\text {Density} = \pu{8.57 g/cm^3}$$
Ok, the density is good to only 3 significant figures so the answer shouldn't have any more than that. But doing the whole problem, I'll carry 5 significant figures throughout all the intermediate calculations to try avoid rounding errors within  the multiple calculations. I'll round the final result to 3 significant figures.
$$\text{Volume of mole} =\text{V}_{mole} = \frac {92.906}{8.57} = \pu{10.841 cm^3/mol}$$
In BCC, there are 2 atoms in one unit cell. Also note that the accepted figure for Avagadro's constant to five significant figures is now $6.0221\times10^{23}$
$$\text{Volume of unit cell} = \text{V}_{cell} = 10.841 \times \frac {2}{6.0221 \times 10^{23}} = 3.6004 \times 10^{-23}\text{ cm}^3$$
A unit cell is a cube with each side being $a$
$$\text{V}_{cell} = a^3$$
$$\therefore a = \sqrt[3]{3.6004 \times 10^{-23}} = 3.3021\times 10^{-8}\text{ cm}$$
But the BCC crystal, spheres of radius, $r$, packed inside a cube with side, $a$, the geometric relationship between $a$ and $r$ is: $$a = \dfrac{4}{\sqrt{3}}r$$ so
$$r = \dfrac{\sqrt{3}\times(3.3021\times 10^{-8})}{4} = 1.4299\times 10^{-8}\text{ cm}$$
Now rounding $r$ to 3 significant figures gives
$$r = 1.43\times 10^{-8}\text{ cm} = 143\text{ pm}$$
You shouldn't chase values through websites and literature. You can only calculate given whatever data you're suppose to take as given. For instance no temperature was given for the density which certainly changes with temperature. Also, to make the point again, with a calculator you should carry extra significant figures in the intermediate calculations to try to avoid round off errors in the final result(s).


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Is each hybridized orbital made up of multiple complete orbitals?
So, I was wondering that how can a methane molecule form four $\mathrm{sp^3}$ hybrid orbitals, with each hybrid orbital containing one $\mathrm s$ and three $\mathrm p$ orbitals.
In four of such $\mathrm{sp^3}$ hybrid orbitals, it makes a total of four individual $\mathrm s$ orbitals and twelve individual $\mathrm p$ orbitals; whereas, we only have, numerically, one $\mathrm s$ orbital and three $\mathrm p$ orbitals available that constitute in hybridization.

Because $s^{0.25}p^{0.75}$ is a mouthful.

The first rule of converting atomic orbitals to  hybrid orbitals is that the number of the former must equal the number of the latter. If you start with four (1s + 3p) orbitals, then you wind up with four sp^3 orbitals. Saying sp^3 is not a mouthful.

@user55119 I don't understand what you mean by four (1s + 3p) orbitals. That implies you have 4 s orbitals, which is the exactly the issue that the OP is confused about.

(1s + 3p) is in parentheses. It reads four (----) orbitals. 1 + 3 = 4.

As has been pointed out in the comments, an $\mathrm{sp}^3$ orbital is a linear combination  not of one s orbital and three p orbitals, but rather of one part s orbital and three parts p orbital. Stated differently, the numbers in $sp^3$ denotes the ratio and not the number of orbitals we combine.
Oddities such as the $\mathrm{sp^5}$ orbitals in cyclopropane can therefore be rationalised. It does not mean that there are five p orbitals which form hybrids, which is obviously impossible. It simply means that that particular orbital has ~1/6 s orbital character and ~5/6 p orbital character.

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text:
Atomic radius and nuclear effective charge in a group
Why even though the effective nuclear charge increases in gruop from top to bottom , the atomic radius increases too? Shouldn’t it decrease with more effective charge on valance electrons and eventually more force affecting them?

you have to take into account the electronic repulsion forces as well.

Possible duplicate of Why exactly does atomic radius increase down a group?

Just so that we are on the same page before I attempt to answer this question, the definition of effective nuclear charge we are considering Zeff=Z−σ, where Z is the nuclear charge (number of protons) and σ is the shielding factor of the inner shell electrons. The overall effective nuclear charge experienced by the valence electron in question is Zeff.
You are right in thinking that the effective nuclear charge increases down as a group as the increasing nuclear charge has a greater effect on Zeff than the shielding effect of more inner electron shells. This, as you quite rightly mention, should suggest that because Zeff increases down a group, a greater force of attraction would be experienced between the outer shell valence electrons and the nucleus, thus decreasing the atomic radius. However, what we have failed to consider is the size of the electron orbitals in question. The principal quantum number, n, of electron orbitals affects their size. The quantum mechanical description shows that the 1s orbital of a hydrogen atom is a lot smaller in radius than the 2s orbital of a lithium atom for example. This comparison can be further extended where the 3s orbitals are larger in radius than 2s and these are larger than 1s. Thus we have a quantum mechanical picture whereby the principle quantum number of electron orbitals affects their radius.
Essentially we have two competing factors that affect the trend of atomic radii in question:

The effective nuclear charge, Zeff, increases down a group which draws electrons closer towards the nucleus, decreasing atomic radius.
The principal quantum number, n, of electron orbitals that increases down a group and due to the quantum mechanical nature of electrons, the radius of these electron orbitals increases with increasing n, thus increasing in size as you go down the group.

Simply put, the factor of increasing n down a group is greater than the effect caused by the increasing Zeff, thus causing the atomic radius of atoms to increase down a group in the periodic table.


text:
Is the Haber Process here proceeding at positive Gibbs free energy change?
$\ce{3H2 + N2 -> 2NH3}$ is the forward reaction used in the Haber process, the industrial production of ammonia. After studying Gibbs energy and how it is just another way of saying that the 2nd law of thermodynamics is obeyed if $\Delta G < 0$, I decided to apply it to some common reactions that I know, just to confirm that my understanding of the topic is correct.
When applying it to the Haber process forward reaction, I worked out change in entropy to be $\pu{-199
J K^{-1} mol^{-1}}$. The enthalpy change for the forward reaction is exothermic and has a value of $\pu{-92 kJ mol^{-1}}$. Looking up common reaction conditions for the Haber process, I found that a compromise temperature of $\pu{650K}$ is used.
Applying this to the Gibbs energy equation $-92 - 650(-199/1000)$, I calculated a positive change in Gibbs energy for the reaction of which had a value of $\pu{37.35 kJ mol^{-1}}$.
As a result of this, I am now quite confused as to how the forward reaction can be allowed to proceed by the 2nd law of thermodynamics at a temperature of $\pu{650K}$, and I was wondering where my understanding of the Gibbs energy and thermodynamics falls short. I need help to explain the pitfalls in my calculations, understanding or even whether there is another factor that I am not considering.

This source has it as -32.7 kJ/mole N2: http://surfguppy.com/thermodynamics/caluclate-gibbs-free-energy-nitrogen-and-hydrogen-to-create-ammonia/

Did you adjust the values for $\Delta H$ for the higher temperature?

@Gert That source uses 298K as the temperature which results in change in Gibbs energy being negative. However in industry, a higher compromise temperature of 650K is used for the forward reaction to achieve higher rates of reaction. At this higher compromise temperature, change in Gibbs energy is positive. I assumed that the value for ΔH does not change with temperature. I know that it does but I thought it would only be very negligible. https://www.chemguide.co.uk/physical/equilibria/haber.html This source takes ΔH = -92kJmol-1 and states conditions used are 673K-723K.

Sure but did you adjust you $\Delta H$ values for the higher temperature that you use? Both your reagents and reaction products are now hotter. It makes a difference in the calc, not just $T\Delta S$. Just try it.

Right, I will look into the effect that the increased temperature has on the value of ΔH and try to find one for a temperature of 650K.

You need to do the same thing for the $\Delta S$.

Just because the change of standard free energy for a reaction is positive, that does not mean that the reaction will not go at all.  It just means that the equilibrium conversion will be low.

It's all about the pressure! The Haber process requires a pressure of about 200 atm. This is how the process becomes favorable, since LeChatelie's principle says that increasing pressure shifts toward the side with less moles of gas.
As long as Q is smaller than K, the reaction will go forward. For example, if nitrogen and hydrogen are supplied at stoichiometric ratio and 1% of a given reaction mixture is product, we would have the following partial pressures:
$$ P_\ce{N2} = 0.2475 \times \pu{200 atm}$$
$$ P_\ce{H2} = 0.7425 \times \pu{200 atm}$$
$$ P_\ce{NH3} = 0.01 \times \pu{200 atm}$$
Accordingly, Q would be
$$ Q = \frac{2^2}{50 \times 75^3} = \pu{1.9e-7} $$
In the Haber process, the concentration of product is kept small by continuously removing product (liquifying ammonia at low temperature) and feeding the remaining reactants back into the reaction mixture.

This is nice, could you expand on your explanation?

The tendency of a reaction to proceed in a certain direction depends on the state of the concentration of each reactant. If you calculate the standard Gibbs free energy of the reaction and it's a positive number, that only means that - if the composition of the reaction medium corresponds to all components at standard concentration ($1 \ \textrm{mol/L}$) - then the reaction will shift in the reverse direction (to which you supposed when you calculated $\Delta G$).
I am not sure though if this is the case for the Haber Process as you presented.
It must be noted that often one will increase the temperature of a reaction in a way that is non-intuitive from a thermodynamic perspective: one might conclude that the reaction is being made less spontaneous in the desired reaction. However, in a practical setting, the kinetics are also important, that is, how fast the reaction proceeds towards equilibrium. A temperature increase will always increase the reaction rate, which may get us to an acceptable yield earlier, even if the maximum achievable yield (dictated by thermodynamics) will be compromised.

the ΔH and ΔS don't change too much on temperature, but they change on pressure! when the pressure is 200 atm, the ΔH and ΔS are changed and the calculated ΔG is negative! I couldn't find the equation for ΔH and ΔS to P. But I check the results from Aspen Plus directly.

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Calculating a test statistic?
Please help me with this question. I'm pretty sure I know the null hypothesis is that the two means are the same but I'm not sure how to calculate it.
James and Sally looked at sulfide content in seven coal samples. James's results were on average .40 ppm higher than Sally, with a standard deviation of difference being +/- 0.21 ppm. Is the difference between the two significant?
A. Null hypothesis?
B. Calculate test statistic
C. Compare with critical values at the 90 and 99% confidence levels. What conclusions can you draw?
D. What kind of tests can be carried out with multiple trials of a single sample by each person?

I know the null hypothesis is that the two means are the same but I'm not sure how to calculate it.

Not really. Statistics is a monster. You really need to decide what test you're doing carefully.
If you are testing that the results from James and Sally are the same, then you'd reject if James$>$Sally or James$<$Sally so you'd be doing a two-sided test. For 99% confidence this would mean 0.5% James>Sally and 0.5% James$<$Sally.
You already know that the numbers are such that the results have James$>$Sally, so you want to do a one-sided test.

Okay, that makes sense. So, is the equation I'm suppose to use: t=(x̄-μ)/(s/√n)  If so, is .40 ppm x̄ or μ?

See Wikipedia article for instance $$t=\dfrac{\bar{X}{James} - \bar{X}{Sally}}{s_p/\sqrt{n}}$$


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How to determine the total pressure of a gaseous mixture of xenon tetrafluoride and hydrogen?

Gaseous xenon tetrafluoride at partial pressure of $\pu{2 kPa}$ and hydrogen at partial pressure of $\pu{10 kPa}$, are exploded in an enclosed container producing xenon and hydrogen fluoride:
  $$\ce{XeF4 (g) + 2 H2 (g) -> Xe (g) + 4 HF (g)}$$
  Find out the total pressure of mixture of gases in the enclosed container.

Assuming $T$ and $V$ are constant, we can assume that the pressure is directly related to the amount of substance, hence $\pu{1 kPa}$ of $\ce{XeF4}$ reacts with $\pu{2 kPa}$ of $\ce{H2}$.
This means that only $\pu{4 kPa}$ of $\ce{H2}$ is needed to react with $\pu{2 kPa}$ of $\ce{XeF4}$.
Now, to find total pressure of mixture of gases I did
$$p = \pu{2 kPa}~\text{(from Xe)} + \pu{8 kPa}~\text{(from HF)} = \pu{10 kPa}$$
Why is this wrong? Why must I add $\pu{10 kPa}$ by $\pu{6 kPa}$ which is the unreacted hydrogen?

Pressure is an intensive property. An expression "$\pu{1 kPa}$ of $\ce{XeF4}$ reacts with $\pu{2 kPa}$ of $\ce{H2}$" makes no sense.

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

You are correct up to the last part. The total pressure would be the sum of all the partial pressures in the reaction vessel.
We know that $$P=\left(\frac{RT}{V}\right)n$$ and therefore $n$ varies directly with $P$ as $T$, $R$, and $V$ are constant. For the sake of the problem (it won't matter) we can just say that $x=\left(\frac{RT}{V}\right)$. Therefore, $P=xn$ and the starting amounts of substance are equal to $\frac{2}{x}\pu{ mol}$ $\ce{XeF4}$ and $\frac{10}{x}\pu{ mol}$ $\ce{H2}$.
From the balanced equation and the starting materials and stoichiometry, we know that $\frac{2}{x}\pu{ mol}$ $\ce{XeF4}$ would react with $\frac{4}{x}\pu{ mol}$ $\ce{H2}$ to form $\frac{2}{x}\pu{ mol}$ $\ce{Xe}$ and $\frac{8}{x}\pu{ mol}$  $\ce{HF}$ to make $\frac{10}{x}\pu{ mol}$ product.
However, you only reacted $\frac{4}{x}\pu{ mol}$ $\ce{H2}$ of the $\frac{10}{x}\pu{ mol}$ total, so there is $\frac{6}{x}\pu{ mol}$ unreacted $\ce{H2}$ left in the reaction vessel.
$$\begin{multline}
  \frac{10}{x}\text{ amount of substance of product }
+ \frac{6}{x}\text{ amount of substance of unreacted }\\
= \frac{16}{x}\text{ total amount of substance}
\end{multline}$$
From $P=xn$, the fact that this occurred instantly, and $T$, $R$, and $V$ are constant, we then know that the final pressure is $\ce{16 kPa}$.

I would suggest to have a look at How can I format math/chemistry expressions here? regarding formatting guidelines. And again, pressure is an intensive property, one cannot say that much kPa reacts, it's like counting green and red apples by color.

Thank you for the link, I'm new here and didn't know where to look. In this case, because the reaction is happening in an explosion, and T and V are constant, the number of moles relates directly to the pressure by some constant $$P=(RT/V)n$$ For the sake of the problem, we can carry $RT/V$ through in our stoichiometry, then re-convert it out at the end. In the answer, I merely used this shortcut (which I should have made known) and did not say this.

I get that, but the terminology you and OP are using is just wrong, even though the answer seems to be correct. Comparing quantities using intensive parameters in this context makes little sense. That's all.

Please note that the proper term for "(number of) moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".


text:
The coldest (the most endothermic) reaction
What chemicals can you mix that causes the coldest (the most endothermic) reaction on a $\pu{kJ /kg}$ basis?

Do you mean a chemical reaction, or can it be any reaction? ie Are nuclear reactions acceptable? If not, please clarify by editing the title to include the word chemical.

Hm, I thought by "mix" you were only referring to a spontaneous reaction. It'd be nice if you could clarify if that's what you want. That could be a nice question on its own.

There is an endothermic nuclear fusion for the elements heavier than iron. On Earth it's most likely a nuclear fission of light elements, like the one occurring in early "dry" thermonuclear bombs, or a fission processes preceding D-T fusion reaction:
$$\ce{^7_3Li + ^1_0n -> ^3_1H + ^4_2He + ^1_0n - \pu{2.467 MeV}}$$
It's probably not the answer you'd expect, but the absolute values of thermal effects of any chemical reaction would be much lower in comparison.
Special thanks to David Richerby for correcting me!

+1 Well, I suppose the OP was hoping for something of that magnitude, albeit, he was looking for chemical reaction that does it (non-existent). The OP ought to add some more details IMHO

@paracetamol I suspect that too, but the it's been asked about just a reaction, so I assumed it would be safe to mention nuclear fusion as #1 option.


text:
Arrows used in chemical reactions
A point was raised about arrows used in a chemical reaction. I think we can all agree that $\ce{->}$ arrow in the reaction
$$\ce{A + B -> C} \tag{1}$$
means that the reaction goes in one direction. But what of
$$\ce{A + B <-> C} \tag{2}$$
$$\ce{A + B <=> C} \tag{3}$$
$$\ce{A + B <--> C} \tag{4}$$
$$\ce{A + B <=>> C} \tag{5}$$
$$\ce{A + B <<=> C} \tag{6}$$
Is there any IUPAC or American Chemical Society guideline about how double arrows are supposed to be used?

Examples of usage:

(1) Used for example:
  "Removal of Triphenylphosphine Oxide by Precipitation with Zinc Chloride in Polar Solvents",
  Donald C. Batesky, Matthew J. Goldfogel, and Daniel J. Weix,
  J. Org. Chem., 2017, 82 (19), pp 9931–9936
"Cation Radical Accelerated Nucleophilic Aromatic Substitution via Organic Photoredox Catalysis",
  Nicholas E. S. Tay and David A. Nicewicz*
  Department of Chemistry, University of North Carolina at Chapel Hill, Chapel Hill, North Carolina 27599-3290, United States
  J. Am. Chem. Soc., 2017, 139 (45), pp 16100–16104
(4) Used for example:
  "Cation Radical Accelerated Nucleophilic Aromatic Substitution via Organic Photoredox Catalysis",
  Nicholas E. S. Tay and David A. Nicewicz*
  Department of Chemistry, University of North Carolina at Chapel Hill, Chapel Hill, North Carolina 27599-3290, United States
  J. Am. Chem. Soc., 2017, 139 (45), pp 16100–16104

Previous questions in Chemistry:
In his answer user Chemgoat noted that this question had been asked before:
What are the correct equilibrium arrows?
A comment on that question points to yet another
What is the difference between "reaction in both directions" and "equilibrium"?

The IUPAC Green Book (Quantities,
Units and Symbols in Physical Chemistry, Third Edition, 2007) reads:

2.10.1 Other symbols and conventions in chemistry
  [...]
(iv) Equations for chemical reactions
(a) On a microscopic level the reaction equation represents an
  elementary reaction [...] A single arrow is used to connect reactants
  and products in an elementary reaction. An equal sign is used for the
  "net" reaction, the result of a set of elementary reactions. [...]
$\ce{H + Br2 -> HBr + Br} \quad$ one elementary step in HBr formation
  $\ce{H2 + Br2 = 2 HBr} \quad$ the sum of several such elementary steps
(b) On a macroscopic level, different
  symbols are used connecting the reactants and products in the reaction
  equation, with the following meanings:
$\ce{H2 + Br2 = 2 HBr}\quad$ stoichiometric equation
  $\ce{H2 + Br2 -> 2 HBr}\quad$ net forward reaction
  $\ce{H2 + Br2 <--> 2 HBr}\quad$ reaction, both directions
  $\ce{H2 + Br2 <=> 2 HBr}\quad$ equilibrium
The two-sided arrow $\ce{<->}$ should not be used
  for reactions to avoid confusion with resonance structures [...]
2.12.1 Other symbols, terms, and conventions used in chemical kinetics
  [...]
(ii) Composite mechanisms
  A reaction that involves more than one elementary reaction is said to occur by a composite mechanism. The terms complex mechanism, indirect mechanism, and stepwise mechanism are also commonly used. Special types of mechanisms include chain-reaction mechanisms, catalytic reaction mechanisms, etc.

Examples
    A simple mechanism is composed of forward and reverse reactions
    $\quad\quad\quad\quad\quad\quad\ce{A -> B + C}$
    $\quad\quad\quad\quad\ce{B + C -> A}$
    It is in this particular case conventional to write these in one line
    $\quad\quad\quad\quad\quad\quad\ce{A <--> B + C}$

However, it is useful in kinetics to distinguish this from a net reaction, which is written either with two one-sided arrows or an "equal" sign
  $\quad\quad\ce{A <=> B + C}$
  $\quad\quad\ce{A =  B + C}$
When one combines a composite mechanism to obtain a net reaction, one should not use the simple arrow in the resulting equation.
  $\quad$Example
  $$\begin{align}
\ce{A &\to B + C}\quad\text{ unimolecular elementary reaction} \\
\ce{B + C &\to D +E}\quad\text{ bimolecular elementary reaction} \\
\hline
\ce{A &= D + E}\quad\text{net reaction (no elementary reaction,no molecularity)}
\end{align}$$

The ACS Style Guide reads:

Many kinds and combinations of arrows can be used. For example, two
  full arrows in opposite directions ($\ce{<-->}$) indicate a reaction that is
  proceeding in both  directions. Two arrows with half heads in opposite
  directions ($\ce{<=>}$) indicate a  reaction in equilibrium. A single arrow
  with heads on both sides ($\ce{<->}$) indicates  resonance structures, not a
  reaction.

PS. I realize that the arrow display degraded a little bit. I will improve that with the next release of MathJax/mhchem.

The second one, $(2)$, $\ce{<->}$, indicates resonance structures.
The difference between the second two is explained in another post -
What are the correct equilibrium arrows?
Loong basically says that $\ce{<-->}$, $(4)$, indicates forward and reverse elementary steps happening in equilibrium, that is that those are the only two things actually reacting with each other. On the other hand, $\ce{<=>}$, $(3)$, indicates a net reaction, that is that there may be multiple steps in between, leading to this net equation. He does a more in depth explanation of this in his post.
Case $(5)$ indicates a net equilibrium where products are favored, and $(6)$ indicates a net equilibrium where reactants are favored.


text:
Mass of monomer needed to obtain a polymer of a given molecular weight

In a condensation polymerization between two different monomers, the average molecular weight of the product depends on the ratio of the monomers used. This question illustrates the effect of using different ratios of ethylenediamine ($\ce{A}$) and 1,4-butanoic acid chloride ($\ce{B}$):

The reaction of $\ce{A}$ ($\pu{2 mol}$) with $\ce{B}$ ($\pu{1 mol}$) gives a compound with molecular weight $\pu{202.3 g mol-1}$ as the major product.
The reaction of $\ce{A}$ ($\pu{1 mol}$) with $\ce{B}$ ($\pu{2 mol}$) gives a compound with molecular weight $\pu{297.1 g mol-1}$ as the major product.
What mass of $\ce{B}$ (in grams) should be combined with $\pu{400 g}$ of pure $\ce{A}$ in order to obtain a high molecular weight polymer?

This question is part of a sample test for my course. I calculated the answer as follows.
For the higher molecular weight polymer, we'll need a $1 : 2$ molar ratio between $\ce{A}$ and $\ce{B}$.
That means we'll need $\pu{2067 g}$ of $\ce{B}$. However, this doesn't seem to be the right answer, which is given as $\pu{1032 g}$.
I'm really not sure what I'm doing wrong at this point.

I cannot see anything wrong with the maths of it. Why do you think it is not the correct answer? It would be nice, if you could cite the source of the exercise.

@Martin-マーチン It's part of a sample test for my course. They give the answer as 1032g. I've been trying to figure out how they got that for a while.

I am not an expert on polymer chemistry, so there might be something very obvious I'm missing. I cannot see anything wrong with your reasoning. Your question is still young so hopefully someone else will enlighten us.

@Mithoron Duh, now that you say it... would you be so kind and add a few words and post as an answer.

@Mithoron Is that a property of polymerization? According to the info they give us in the question, it's completely different.

For high weight 1:1 proportion is needed. 1:2 gives primarily low mass products as mentioned in your citation, there are even molecular masses given for main products, so no idea why you thought otherwise. Also this is homework, write reactions yourself and see how it works.

You have a $2:1$ ratio. You need a $1:1$.
The $\bar{X}_m$ resulting (the number average degree of polymerisation) is derived from the Carothers equation:
$$\bar{X}_m = \frac{1+p}{1-p}$$
where $p$ is the mole ratio. For a $2:1$ this is $0.5$, and the effective $\bar{X}_m = 3$. If you mix at $2:1$, you'll have the $\ce{BAB}$ trimer as the major product with a $M_m$ of $\pu{297.1 g/mol}$.
If you have one component in $1\%$ excess ($1.01:1$) the $\bar{X}_m$ will be $199$, at $.1\%$ excess the $\bar{X}_m$ will be $1,999$ etc.

I edited your answer a bit to improve the readability. If you are interested in typesetting on Chemistry.SE, please check out this page, this page and this one on how to format your future posts better with MathJax and Markdown.


text:
Where does magnetic inequivalence come from?
As the question says, I'm having trouble understanding where does magnetic inequivalence in $\ce{^1H}$ NMR come from. All the sources I've found say that two protons are magnetically equivalent if all $J$ values to all other spins in the molecule are identical. So for example, in 1,2-dibromobenzene, the $\ce{H^3}$ and $\ce{H^6}$ are not magnetically equivalent because $J(\ce{H^3, H^4}) \neq J(\ce{H^6, H^4})$ and $J(\ce{H^3, H^5}) \neq J(\ce{H^5, H^6})$. I understand how to apply the rule, but I don't understand why does this affect the spectra. For example, assuming a not unusual $J_\mathrm{ortho} = \pu{7 Hz}$ and $J_\mathrm{meta} = \pu{2 Hz}$, then:
$J(\ce{H^3, H^4}) = \pu{7 Hz} = J(\ce{H^5, H^6})$, and
$J(\ce{H^3, H^4}) = \pu{2 Hz} = J(\ce{H^5, H^6})$. So they are both split the same way, but in reality they also split each other because of their magnetical inequivalence. This is what I find hard to grasp: if they are being in overall affected the same way, by the same type of atoms, in the same positions, how do they "know" that they are, in fact, different.
I guess related to this would be, why do all protons in ethene give the same signal? If they are also related differently between each other? ($\ce{H^1}$ and $\ce{H^{$1'$}}$ have different coupling constants to each $\ce{H^2}$ and $\ce{H^{$2'$}}$, cis and trans, respectively).

Firstly, thank you for asking a very interesting question that certainly pushed my limits of knowledge (and others' as well). I apologise if I have made mistakes, it is a real possibility. All references are to Levitt's Spin Dynamics, 2nd ed. (an excellent book, by the way).
From a quantum mechanical point of view, the complexity in the spectrum of magnetically inequivalent nuclei should be considered the "norm". Fundamentally, in any strongly coupled system where the difference in Larmor frequency between two spins $\Delta \omega$ is much less than the coupling constant $J$, we expect to see complex spectra (see second half of this answer for more discussion).
Magnetic equivalence is a special case which leads to simplifications in the Hamiltonian and hence in the spectrum. As I mentioned in a comment, it can be shown that any coupling between magnetically equivalent nuclei "disappear". The word "disappear" merits further explanation, though. It is not true that the coupling constant $J$ goes to zero, for example. Neither is it true that it is averaged to zero by some sort of physical process, e.g. molecular rotation. It only means that the coupling term in the Hamiltonian has no impact on the expectation value of any quantum mechanical observable, or its evolution. (Formally, this is because the coupling Hamiltonian between two magnetically equivalent nuclei
$$H = 2\pi J_{ij}(I_{ix}I_{jx} + I_{iy}I_{jy} + I_{iz}I_{jz})$$
commutes with every other relevant Hamiltonian—supposedly a 'well-known' fact, according to the famous product operator paper: Prog. Nucl. Magn. Reson. Spectrosc. 1984, 16, 163–192.) Anyway, since an NMR spectrum is simply us humans reading out some expectation values, the NMR spectrum itself is therefore unaffected by the coupling term between magnetically equivalent nuclei.
Having established that it is magnetic equivalence, not inequivalence, that is the special case, I would like to rephrase your question to: "Is there a physical explanation for why magnetic equivalence leads to simplification of the spectrum?"
The mathematical reason is given in Appendix A.9 of Levitt. I won't reproduce it here, so I will just mention it has to do with some operators which commute with others, and these only commute under very specific circumstances, i.e. the rules for magnetic equivalence that you have surely been taught (they must be chemically equivalent, plus either they have the same coupling constant to all other nuclei, or there are no other spins in the molecule).
Given the (relative) complexity of the mathematical explanation, I suspected that a physical explanation would not be particularly instructive. Levitt, though, does offer us something (p 461):

The apparent disappearance of J-couplings between magnetically equivalent spins is quite surprising. A physical explanation runs as follows. For magnetically equivalent spins, the local magnetic fields, whatever their source, are exactly the same on the two spins. It follows that the motion of the two spin polarizations is identical. Whatever happens, the polarizations of the two spins are locked in the same relative orientation. There is, therefore, no need to incorporate a term in the spin Hamiltonian that takes into account the dependence of the energy on this relative spin orientation.

What this means is: it does not matter what the relative orientations of the spins are. For example, in ethene, it does not matter whether two spins are up and two are down, or whether all four are down, or whether all four are somewhere between up and down (i.e. a superposition of $\alpha$ and $\beta$, which is in fact the most likely case by far). The relative orientations can be anything, but the point is that the time evolutions of all four spins are synchronised. Therefore, the coupling terms between them – which are terms, scalar products, that depend on the relative orientation of spin vectors – may be omitted from the Hamiltonian without any impact on the observables of the system.
Returning to a mathematical interpretation, another sign of magnetic equivalence between nuclei $i$ and $j$ is when the labels $i$ and $j$ can be permuted in the spin Hamiltonian without changing it. This implies that $[\hat{\vec{I}}_i,\hat{H}] = [\hat{\vec{I}}_j,\hat{H}]$. From the standard formula
$$\frac{\mathrm d}{\mathrm dt}\langle Q \rangle = -\frac{\mathrm i}{\hbar}\langle[\hat{Q},\hat{H}]\rangle$$
it is immediately obvious that $\mathrm d\langle\vec{I}_i\rangle/\mathrm dt = \mathrm d\langle\vec{I}_j\rangle/\mathrm dt$, thus proving that the spins $i$ and $j$ move in concert.
Regarding ethene: you are correct that the cis and trans (and vicinal) couplings in ethene are different. However, all four of the protons in ethene are chemically equivalent and must be considered together as a group. It is not correct to look at two out of the four protons and call them magnetically inequivalent, without also taking into account the remaining two. In this case, it turns out that if you do the maths, the relevant commutators don't cancel out if you only consider two out of the four nuclei. However, when you consider all four nuclei together, the relevant operators fully commute.
Now, none of the simplifications above apply to magnetically inequivalent nuclei. These experience very tiny instantaneous differences in magnetic fields – hence the name – and therefore all the couplings in the system end up being reflected in the spectrum, as we should have expected a priori. Returning to your analysis ("the coupling constants are the same, so shouldn't they be the same?"), the issue is that the coupling constant $J_\ce{AX}$ is only one part of the coupling term in the Hamiltonian, which is $2\pi J_\ce{AX}(\hat{\vec{I}}_\ce{A}\cdot\hat{\vec{I}}_\ce{X})$. Although $J_\ce{AX}$ is indeed equal to $J_\ce{A'X'}$, the dot product $(\hat{\vec{I}}_\ce{A}\cdot\hat{\vec{I}}_\ce{X})$ is in general not equal to $(\hat{\vec{I}}_\ce{A'}\cdot\hat{\vec{I}}_\ce{X'})$.
Warning: as with the ethene case, one cannot look at individual A and A' nuclei, because they are symmetry-related. You have to look at both of them together as a group. Once you do that, though, the spectra become possible to analyse.

What ethene misses compared to a benzene derivative, is a ring current. There would be no one in the ground state of an isolated molecule, but it will arise in magnetic field. The ring current in the molecule with (artificially numbered) atoms can have one of the two directions, which makes, for instance, H4 and H5 magnetically inequivalent.
In other words, ethene in magnetic field retains it point group symmetry (D2h), while 1,2-dibromobenzene loses their C2v because of asymmetric look of the ring current.

I like this idea. But couldn't ethene might as well have a left or right direction of the electrons on the $\pi$ bond?


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Uncertainty in measuring the mass of the palm oil and the base in saponification
I am a student coming from the domain of mathematics, and I am working on error analysis and uncertainty, but I have a question  in the domain of chemistry.
I am  studying the effect of the measurement error in saponification.  According to different research references (this is one) we need $\pu{142 g}$ of $\ce{NaOH}$  to react with  $\pu{1000 g}$ of palm oil to produce soap.
What if we use  $\pu{140 g}$ of  $\ce{NaOH}$  with  $\pu{1000 g}$ of palm oil? Would this change the mass of the obtained soap? Or would it just change the quality of the soap?

"mass of the obtained..." is called yield and in this case doesn't depend linearly on proportion of reagents.

It would effect primarily the mass obtained (yield) and probably the quality, too, but that depends on how you measure this. Given that you use some procedure to purify your crude product that is equivalent in all cases independent on the mass of sodium hydroxide used, you simply will get less purified product.
Palm oil is a natural product and composed of a mixture of compounds. These compounds are know as fats, which are esters of propane-1,2,3-triol (glycerine) and fatty acids. Saponification is the hydrolysis of these esters. In that case sodium hydroxide is the strong base used, and the obtained soap is a mixture of glycerine, and the sodium salts of the fatty acids.
In principle for complete saponification you need one equivalent of base for each ester group.
$$\ce{RCOOR' + NaOH -> RCOONa + ROH}$$
Extending this scheme to palm oil, for the fat or triglyceride you need three equivalents of base:
$$
\ce{\underset{triglyceride}{H5C3-(OOCR^{1/2/3})3}
   + 3 \underset{base}{NaOH}
   -> \underset{glycerine}{H5C3(OH)3}
   + 3 \underset{salts of fatty acids}{R^{1/2/3}COONa}}
\tag{1}\label{eq:fat}$$
Since it is a natural product you probably have a number of environmental factors influencing the composition of your specific sample of palm oil. Wikipedia cites the following approximate composition:
\begin{array}{lr}
\text{Type of fatty acid}    & \text{Abundance}/\%\\\hline
\text{Myristic saturated C14}       &  1.0  \\
\text{Palmitic saturated C16}       & 43.5  \\
\text{Stearic saturated C18}        &  4.3  \\
\text{Oleic monounsaturated C18}    & 36.6  \\
\text{Linoleic polyunsaturated C18} &  9.1  \\\hline
\text{Other/Unknown}                &  5.5  \\\hline
\end{array}
As you can imagine this composition of the palm oil has an uncertainty. You probably want to look up more accurate values and the errors of the employed techniques. However, the mass of base needed can therefore be only a crude recommendation. In one case be too much, resulting in leftover (unreacted) base, while in another case it might be too little, resulting in only partially or not at all hydrolysed fat.

A crudely (over)nsimplified calculation:
Assuming palm oil is made out of 100% triglyceride of palmitic acid, we know the molecular mass is $M(\text{palm oil})\approx\pu{807 g/mol}$, we also know $m(\text{palm oil})=\pu{1 kg}$, $m(\ce{NaOH})=\pu{142 g}$.
\begin{align}
  M &= \frac{m}{n} &
  \Leftrightarrow&&
  n &= \frac{m}{M} &
  \implies&&
  n(\text{palm oil}) &= \frac{\pu{1kg}}{\pu{807 g/mol}} \approx \pu{1.24 mol}
\end{align}
From $\eqref{eq:fat}$ we know we need three equivalents of the base to produce three equivalents of soap. Ergo $n(\text{soap}) = n(\ce{NaOH}\text{ needed}) = \pu{3.72 mol}$. With $M(\ce{NaOH}) = \pu{40 g/mol}$ we know we need $m(\ce{NaOH},\text{ needed}) = \pu{149 g}$. We have $m(\ce{NaOH})=\pu{142 g}$, or $n(\ce{NaOH})=\pu{3.55 mol}$, ergo we only produce that much of soap. The molecular mass of our soap is $M(\text{soap})=\pu{278 g/mol}$. Therefore we will obtain $m(\text{soap}) = \pu{987 g}$.
I'll leave it to you to do the more precise calculations, since you are the mathematician of the both of us.

Thank you very much for this thorough expalanation.  Can you please tell me how many grams of soap are obtained if  $1000g$ of  pure palm oil are hydrolysed with $142g$ of  NaOH ?

@Nizar Not really, but I included a very oversimplified example calculation.


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Is Inositol chiral or achiral?
I thought Inositol was clearly chiral since it has 6 asymmetric C atoms. But friends of me argue that it is achiral and i've found websites say the same (http://bip.cnrs-mrs.fr/bip10/Inositol-EJB.pdf). So what am I missing? Is it chiral or achiral?

Suppose all atoms have the same configuration. Then each atom has two identical substituents; then how are they asymmetric?

Look here for the chirality of the inositols: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/inositol.html

Having asymmetric carbon atoms does not in itself make a compound chiral.
A compound is chiral if and only if it is not super(im)posable on its mirror image.
If there's a plane of symmetry, then reflecting in that plane gives the same thing, by definition, and that meets the criterion for super(im)posability - the mirror image is superimposable on the original and therefore the molecule is not chiral.
Only D- and L- chiro-inositol are chiral - all the other stereoisomers have planes of symmetry and are therefore not chiral.


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Why is the difference between atomic radii of Na and K greater than that of Rb and Cs?
I am aware that this question has already been asked here: Why does the "gap" between atomic radii get smaller down the periodic table?.
However, I was wondering if there is a way to answer this question using simple periodic trends instead of a formula to calculate atomic radius.
Since there is not only an s and p-block between Rb and Cs, but also a d-block, wouldn't the increase in shielding between Rb and Cs be higher than that of Na and K? Thus, shouldn't Rb and Cs have a larger difference in atomic radii?
I got this question in my chemistry class while learning about periodic trends, and it apparently also appeared on the 1993 AP Exam, so I would think that this problem could be explained using simple periodic trends.

Electronic orbitals do not come in fixed sizes.  A given orbital will shrink due to greater nuclear charge as more protons are addded to the nucleus.  Going from $\ce{Rb}$ to $\ce{Cs}$ you are adding more orbitals than when you go from $\ce{Na}$ to $\ce{K}$.  But you are also adding more orbital-shrinking protons to the nucleus.
The shrinkage effect on each orbital with more protons gets stronger as we go to heavier atoms, when relativistic effects kick in.  Put in very rough terms: when you solve the Schroedinger equation for electronic wavefunctions the size of the orbital would decrease if the mass of the electron were increased.  In nonrelativistic quantum mechanics this does not occur, but in relativistic mechanics the energy needed to squeeze the wavefunctions (which in this case comes from electrostatic attraction) acts like mass, and thus allows for even more shrinkage.  For alkaline earth metals we end up with almost no change at all going from $\ce{Ba}$ to $\ce{Ra}$.  See https://periodictablegroups.wordpress.com/alkaline-earth-metals-2/.


text:
Difference between ginger essential oil and ginger powder
I'm planning to prepare natural toothpaste.
Some recipes include ginger essential oil while others include ginger powder.
What is the difference between them? Will ginger powder be as effective as ginger essential oil for my intended use?
Here in my country, ginger essential oil costs about 11-12x more than ginger powder.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I'm not sure this is the right place to ask that. Maybe health, maybe even cooking. The difference is painfully obvious, too. The one is dried powdered ginger (basically letting the water evaporate), the other uses some form of extraction to get to the oil. Unless you are interested in the extraction of a specific compound from ginger, I believe this is off-topic here.

Gingerol and gingerone are the phenolic compounds that give ginger its kick. https://eic.rsc.org/soundbite/ginger/2021241.article

I'm voting to close this question as off-topic because we cannot state on the effectiveness as preparing toothpaste is outside our scope and the difference between powder and oil is clear enough to not warrant a question.

The main constituant of ginger essential oil seems to be zingiberene, which is what gives ginger its peculiar flavor.
So if the ginger is added for flavoring, I would say that the oil will be much more potent than the powder. I don't know if it will be 10x more potent, though.
I would suggest trying both and seeing which is more economical.


text:
Cation falling off electrode after electrolysis
I'm trying to find the electrochemical equivalent for zinc in an experiment.
I am carrying out electrolysis of zinc sulfate solution.
According to Faraday's law, m=zq.
I made a 0.1 M solution and kept time for electrolysis constant (1 min).
To get a value for z I plotted the mass over the charge (I*t). The current was varied. I'm getting a linear trend (mass deposited increases with increase in current) but a huge discrepancy of 100%.
Also in the experiment, after the electrolysis I noticed that the zinc separated and got suspended in the solution when I disconnected the electrode to measure the new mass. This could have affected results.
Any reasons for this? I might be able to redo the experiment if someone knows how I can solve this or get better results.
The metal ions should undergo reduction at the cathode. Maybe that's not happening?
P.s I also did it for copper sulfate and it was worse. Almost no copper stayed on the electrode after electrolysis
Also I used graphite electrodes for this. Before this, for another experiment I electroplated zinc with copper and all of it deposited. Maybe get better results with zinc/copper electrodes for this experiment instead of graphite?
Regards,
Haseeb

Try reducing the current density, by increasing the electrode's size or by reducing current.

@Gert But I am already varying current from 1 to 5 A in steps of 1A. But do you think it would be possible to the same experiment and get the same relationship if I use copper electrodes instead of graphite. Because I didn't get an issue with the copper electrodes.

$5\ \mathrm{A}$ is very high current for a 0.1 M solution. Graphite really isn't suitable. Better use the respective metals.

Ok, I will try that then. Do you know if the relationship still holds, m=zQ? (keeping time constant of course.)

Yes, of course. Faraday's Law, basically.


text:
Difference required for compound to be a phenol vs benzene compound
When naming organic compounds, how should one decide whether a compound should be named as a benzene compound, or as a phenol?
For example, the compound 3,4-dihydroxybenzoic acid has an -OH group, so why is it not named as a phenol derivative? I feel like I'm missing something here, but none of the numerous sources I've looked at have provided a clear-cut rule to differentiate with.

Because of priority rules? O:)

In organic nomenclature, when you have a compound with multiple functional groups, you follow the priority rules that have been nicely summed up on Wikipedia:

When compounds contain more than one functional group, the order of precedence determines which groups are named with prefix or suffix forms. The highest-precedence group takes the suffix, with all others taking the prefix form. However, double and triple bonds only take suffix form (-en and -yn) and are used with other suffixes.

These rules apply to benzene derivatives as well as aliphatic and other cyclic compounds.
Continuing,

Prefixed substituents are ordered alphabetically (excluding any modifiers such as di-, tri-, etc.), e.g. chlorofluoromethane, not fluorochloromethane. If there are multiple functional groups of the same type, either prefixed or suffixed, the position numbers are ordered numerically (thus ethane-1,2-diol, not ethane-2,1-diol.) The N position indicator for amines and amides comes before "1", e.g. $\ce{CH3CH(CH3)CH2NH(CH3)}$ is N,2-dimethylpropanamine

Here's a shorter (and quite handy) version of the Wikipedia table (shamelessly copied from masterorganicchemistry.com, I've taken a liking to their color-scheme :-))

So in the example you've posted:

You give the carbon atom that has $\ce{-COOH}$ attached to it the first locant (because $\ce{-COOH}$ takes precedence over $\ce{-OH}$, according to the priority rules), and you continue numbering clockwise so the the two $\ce{-OH}$ groups take locants 3 and 4 respectively (Lowest set of locants rule).
That's how you end up with 3,4-dihydroxybenzoic acid :-)

The so-called “Table of Functional Group Priorities For Nomenclature” shown on masterorganicchemistry.com tries to combine the seniority order of suffixes with prefixes and endings. This can be misleading and result in wrong names. In particular, there is no suffix ‘-ane’, ‘-ene’ or ‘-yne’ (these are endings), and multiple bonds (and all groups that follow in that table) are actually not considered in the seniority order.

@Loong Hmm, but I can't find a clear enough table (from Google Images) apart from this :/ Every other table's an eyesore. If you know where I can find a better (neat and brief) table  online, I'd appreciate it if you could give me a link. Thanks! O:)


text:
Generalizations regarding functional groups on benzene ring
While studying several organic mechanisms (high school level), I came to an interesting generalization.
I claim that negative charge on phenoxide ion first ALWAYS delocalises onto the ring and then attacks through carbon atom (like in Reimer Tienman reaction - Wikipedia). This happens because negatively charged carbon atom is less electronegative and hence a better nucleophile than a negatively charged oxygen atom.
But a normal lone pair like that on aniline or phenol attacks by itself without delocalisation (as in Schotten Bauman reaction - Wikipedia). This is because delocalised resonance structures have a formal positive charge on the more electronegative oxygen/nitrogen and are thus less stable.
My questions are:

Does this generalization always work with regards to phenols and anilines?
Is my reasoning correct for claiming these generalizations?
Can this logic be extended to other functional groups on a benzene ring?

PS: Please also justify your answer with explanation. Even a simple counterexample will help.

I have a good counterexample. When try to do reimer tiemann reaction with aniline instead of phenol then another reaction happens instead, called the carbylamine test leading to formation of phenylcyanide instead of ortho formyl aniline

Generalisation while often is a very useful tool at the zeroth approximation level, – that is crude, very crude, barely resembling the truth, – is a temptingly dangerous game to play. It often boils down complicated concepts beyond the point of usefulness and leads to wrong conclusions.
Generalisations are oversimplifications.
An example for this is the ever so annoying inclusion of d orbitals into hybridisation schemes. This was a result from the post rationalisation of VSEPR theory with molecular orbitals. It has been disproved many times, yet it still is used as a common explanation in textbooks. It is a myth that continues to live, and live, and live...
Another example for generalisation taken too far is resonance. The concept most often employed, yet understood so little about it; especially why it is neccessary in the first place. As a result we hear phrase like "most stable resonance structure", which are complete garbage.
As such, while making generalisations you might not yet realise what damage that may cause later. Some concept simply are complicated and should not be broken down further.
It is quite common to draw reaction mechanisms with Lewis structures and electron pushing arrows. The problem with these is the extremely simplified nature of these drawings. Molecules are (usually) not planar, and yet we ignore this for the purpose of representation and depiction. A picture may sometimes be worth a thousand words. However, another, probably worse, problem with these schemes is the depiction of discrete electron pairs; a contributing factor to the resonance problem mentioned above.
When using these mechanisms it is always implied, that these are proposals, simplified to highlight certain aspects of a reaction.
Let's analyse your statements a little and see where you might derail.

I claim that negative charge on phenoxide ion first ALWAYS delocalises onto the ring and then attacks through carbon atom (like in Reimer Tienman reaction).

That statement has already one big problem, it's wrong. The charge in the phenoxide ion is never localised. The electronic structure of a molecule will always adapt much more easily and fast than any given change in the molecular geometry. When phenoles are deprotonated, while the hydron leaves, the electronic structure instantaneously adapts.
You are thinking in terms of discrete resonance structures, which simply do not exist. The negative charge is delocalised. That's it. The corresponding observable is the electron density, which is concentrated at certain positions due to the excess electron. In the approximation of molecular orbital theory, this can be partially explained with the HOMO and lower orbitals. In the context of valence bond theory we need to employ resonance, i.e. the superposition of various electronic structures that feature localised electron pairs. Both representations are approximations, but currently our best way to understand complicated mechanisms.
The way how molecules react with each other depends on many things. The product which is formed depends on a fragile system of well balanced energy levels. You need to understand, that molecules react in all possible ways, but only the most stable products will eventually be retained.
In that regard you statement is also wrong, the phenoxide ion is a nucleophile, it will act through every position with increase electron density, most likely through the oxygen though.

This happens because negatively charged carbon atom is less electronegative and hence a better nucleophile than a negatively charged oxygen atom.

The pure notion of better or worse nucleophiles is highly dependent on the context. Just because formed carbon-carbon bonds tend to be a little more stable than the carbon-oxygen bonds, it is not useful to generalise that. I also highly doubt the mechanism presented on Wikipedia for the Reimer–Tiemann reaction is complete enough. As has been stated before, this is not a clean reaction, hinting at a variety of semi-stable intermediates.

But a normal lone pair like that on aniline or phenol attacks by itself without delocalisation (as in Schotten Bauman reaction). This is because delocalised resonance structures have a formal positive charge on the more electronegative oxygen/nitrogen and are thus less stable.

I don't quite understand where you are going with that, but the lone pair of nitrogen in aniline certainly has overlap with the aromatic system. In general amines and phenols are quite alright nucleophiles.  There are other factors determining why they react the way they do.
Your statement suffers from the most-stable-resonance-structure-syndrome, i.e. you take resonance structures much too seriously. Please see: What is resonance, and are resonance structures real? If there is resonance, then this effect is always stabilising, because otherwise it would adopt a molecular structure where resonance would not be possible (compare cycloocta-1,3,5,7-tetraene).
Additionally formal charges are exactly what they are called: formal. There is little to no truth to them other than keeping track of all the electrons.
One of the most important things is also to realise that stability always is a relative concept. More often than not it highly depends on the chemical environment, if an intermediate is "stable" and if it would be more "stable" than another intermediate.
In conclusion,
as a high school level student, you will learn a lot of useless or even wrong stuff. Some concepts are still taught even though they are known to be wrong. Some concepts are often not correctly understood by teachers themselves. Even at the university level you will encounter hardliners who teach chemistry like it was still the 20th century. Unfortunately that can't be changed (in the immediate future).
However, critical thinking is the way to gain knowledge. Unfortunately that means letting go of generalisations. Many things are much more complicated that they appear at first sight, only a detailed look will yield a better understanding. And once you have obtained a deep and fundamental knowledge, you won't need these generalisations or oversimplifications.


text:
Emission spectra
I am a highschool student and I am really confused about the following:
We studied that electrons cannot be between main energy levels therefore in an emission spectrum of an element lets say hydrogen we see single lines which totally makes sense.
But then we took sublevels and orbitals, taking the second main level for example which consists of the (s,p sublevels) we find that their orbitals cover a large space in the atom and the elctron can exist anywhere in that space which theoretically means that if we find its emission spectrum it will be continous and we wouldnt be able to differ between different elements..
Please correct my misunderstanding.
Thank you.

Jumps between energy levels will indeed yield line spectra not a continuous spectra. Students are first taught the planetary model then shown the more exact models of orbitals using distributions due to  quantum mechanics. To really understand why jumps between the distributions still give you line spectra you'd have to go through the math of quantum mechanics. For now just accept that atoms do behave that way. Truthfully a lot of chemistry is about modeling behavior rather than absolute mathematical rigor.

@MaxW Thank you very much! This is what I was looking for, I try sometimes to try understanding the concepts further than what our teacher or the book offers to us but I always end up into mazes..

In the Bohr theory of the atom, electrons travel in fixed orbits, and do not exist between these orbits.
However, the Bohr theory was disproved, and instead it is known that for each orbital, the electrons may be anywhere in space, except at nodes.
Nonetheless, though the electrons may be essentially anywhere in space, the energy of an electron in a particular orbital is a specific energy.


text:
Do some electrons transfer if I connect with a conductor 2 solid pieces of 2 different elements with different electronegativity?
Making some researches to answer my last post, I found this article.
It explains that when a piece of zinc and a piece of copper are connected with a conductor, since copper is more elctronegative than zinc, electrons start to flow form zinc to copper till too many electrons has reached the piece of copper and they start to  repel the others.
Is this explanation accurate?
Can you generalize this observation for other elements connected through a conductor?
Does it work for solid and liquids?
**EDIT:
I'm not stating that there will be an electricity flow, I'm asking if there is an amount of electrons that will converge from the zinc to the copper. Even if it happens for a millisecond.

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

Well, your elements should be conductive to begin with, otherwise they'll have no idea they are connected with a wire. Then again, you can simply rub them against each other. Triboelectric effect is a pretty broad thing, it is not limited to elements.

https://chemistry.stackexchange.com/questions/20084/why-do-electrons-leave-the-zinc-in-a-galvanic-cell

Also I'm quite disappointed with quality of the blog, on scientificamerican domain no less, which you linked it's self contradictory and you pulled one of worst things from it.

@Mithoron I've no strong background in chemistry, as I stated in the post. I'm here to discuss the accuracy of what was reported in the article and as often I get downvoted without explanation. This site is becoming hell for who's not an expert in the field. Even showing your researches and effort doesn't help anymore.

Well, I told why I I downvoted... Also if you hover cursor over downvote button you can see typical reasons. You made research but it lead you astray. If there's no closed circuit there's no current, that's it. There may be electric potential, but it doesn't mean electricity flows.

@Mithoron I don't see how "You made research but it lead you astray" is in any way a reason to downvote. Researches are made to learn, you don't make researches on what you know, so how I am supposed to know that the "Scientific American" is leading me astray? Moreover you could have just answered the question with the last few lines of your comment, it would have been much more helpful

@Mithoron I have to agree with Gabriele. Without a chemical background, its tough to know whether or not a resource like this is trust worthy and it seems harsh to down vote for being confused by what I imagine many outside chemistry would consider a reasonable source.

Question on hold with no explanation from anyone who decided to put the question on hold. Please explain how the question is unclear in order for me to edit it, because for me it's very clear what's being asked.

@Tyberius Downvote should be treated just like upvote - nothing special. Without downvotes why people would even strive to make better questions? Upvotes are more or less random and all too often reflect rather "popularity" then quality.

@Mithoron  Strange conception of how the site works. Popularity  is the most important index for questions, since it expresses the interest of the community. Quality is the most important index not for questions, but for answers.


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Storing alcohol in a thermos bottle
My thermos bottle's instructions say:

Füllen Sie keine kohlensäurehaltigen Flüssigkeiten oder Trockeneis in das Produkt. Füllen Sie keine alkoholischen oder öligen Flüssigkeiten in das Produkt.

This translates to:

Do not fill carbonated liquids or dry ice into the product. Do not fill alcoholic or oily liquids into the product.

I suppose the reason you're not supposed to fill carbonated liquids or dry ice into the thermos bottle is that the bottle isn't able to handle the pressure which builds up if you close the lid. In the case of dry ice, it might also not be able to cope with the temperature of it. I further suppose that oil can wear down the seal.
But why are you not supposed to store alcohol in it? Sure, some alcoholic beverages are carbonated, but those already are excluded. What about beverages like mulled wine?

What makes you think you shouldn't store alcohol in it? Except for carbonated alcohols as you stated.

For all thermos I had alcohol wasn't a problem , I guess. Might some o-ring  get brittle/ harder upon prolonged alcoholic storage ?  I would say might be with very alcoholic drinks. Zum Voll :)

One issue is what material comprises the inner lining of the thermos. Typically, the inside of a thermos bottles are silvered glass. But that's not a guarantee and even if it is, there's no guarantee that there isn't a coating on the glass. The coating may degrade in the presence of alcohol. Also, oily substances are hard to clean out, and the manufacturer may not want people to think badly of their product because they couldn't clean it effectively.

Storing alcohol in copper is supposed to be a bad idea, because the oxide layer on the copper could oxidise ethanol. Not sure on what timescale that would happen. Stainless steel should be inert i think, and certainly glass is inert. There is no coating in a thermos, because that would certainly degrade when you fill in hot water.

At the temperature of dry ice, most common steel modifications become very brittle, the thermos would instantly break i think.

The only reason I have a thermos is to fill it with boozed tea...

As @Alchimista states, ethanol might attack the O-ring gasket... and also the plastic stopper. Another, less likely, reason is that at alcohol has a higher vapor-pressure than water, and at high temperatures that might burst the inner wall of the vacuum bottle.

Lower surface tension of alcoholic beverages could possibly allow excessive leakage. Leakage would/could also be a problem with carbonated beverages because of pressure buildup. Maybe the seal is not very snug (plastic rather than rubber). Maybe the thermos is just cheap junk.

@JamesGaidis My thermos is just cheap junk, I guess. If I leave anything hot in it, after an hour it always makes a high-pitched noise as air is trying to force its way in. If I take it anywhere in my backpack, I need to make sure it's standing upright or it'll leak.

The disclaimer was probably written by an attorney and may or may not have any physical chemical justification.