text:
Geometry of buta-1,3-dien-1-ol

Consider the molecular formula $\ce{CH2=CH-CH=CH-OH}$ $(\ce{C4H5OH})$. Tell the types of bond, hybridizations, bond angles and geometry of this compound.

I didn't have any trouble with the first 3 topics.
Carbon hybridizes in $\mathrm{sp^2}$ and oxygen hybridizes in $\mathrm{sp^3}$.
Each carbon makes 3 $\sigma$ bonds and 1 $\pi$ bond and the oxygen makes 2 $\sigma$ bonds. The angles of the bonds $\ce{H-C-H}$, $\ce{H-C-C}$, $\ce{C-C-C}$ are 120 degrees and the angle of bond $\ce{H-O-C}$ is 109 degrees.
My problem is with the geometry. How can I deduce it? It doesn't seem to be one of the standard geometries.

Well, there are not awfully many possible options. Just draw the molecule like that, planar (because of conjugation) and with the angles like you said. (In fact, they may be slightly different; also, this whole thing does not quite exist, but who cares.)

Let’s start off with the things you thought you got right: Oxygen will not be $\mathrm{sp^3}$ hybridised but somewhat closer to $\mathrm{sp^2}$. This is because that is how the oxygen lone pair in a p-type orbital can take part in π conjugation. I’m not sure what $\angle(\ce{C-O-H})$ I would expect but it’ll be somewhere between $109^\circ$ and $120^\circ$, that’s for sure.
To the entire molecule’s structure: Yes, at first glance it seems like a potato. However, there is still something that you can say: for proper conjugation of the double bonds (usually favourable), you need a planar structure. Thus, the entire molecule will be planar with all atoms in the same plane. That is about all that you can safely derive, since there can still be rotation around the $\ce{C-C}$ single bond.
Also, as Ivan pointed out, the molecule is likely to prefer the keto-tautomer rather than the enol one. The conjugation argument may apply, but it is rather weak.

The molecule is a 1,3-butadiene derivative. The butadiene has a planar structure as conjugation from the pi bonds spreads out along the carbon backbone so the carbon carbon angles would be close to 120 degrees, $\mathrm{sp^2}$ hybridization.

 In your structure the $\ce{-OH}$ group replaces just one of the $\ce{H}$ atoms at the end of the molecule. Thus there are two isomeric structures depending upon which $\ce{H}$ atom is replaced. I have very crudely placed these onto the structure close to the $\ce{H}$ atoms but the positions should be clear. (Sorry, I don't have software at hand just now to make proper pictures). The rotation about the 2-3 (middle) bond will be very, very slow because of the pi conjugation holding the structure planar so the two isomers $(E)$ & $(Z)$ will be distinct. The $\ce{C-C-O}$ bond angle will be close to 120 degrees. Try and look up the crystal structure then you will be able to get accurate angles and bond lengths.


text:
Haloform reaction with limited bromine

We were asked the reaction between 1 mole acetone and 1 mole $\ce{Br2}$ in basic medium. Since only a limited amount of bromine was given I got the final answer as 1-bromopropan-2-one (mono bromo acetone). However, in the given solution the entire haloform reaction was shown with $\ce{CHBr3}$ and acetate ion as the final product.
I am not sure how a single mole of $\ce{Br2}$ is sufficient for this(at least 3 moles are needed, I believe).

Let's take a look at the first product you propose: 1-bromopropan2-one.
The mechanism includes deprotonation by hydroxide to form an enolate. The resulting enolate is better stabilized with more electronegative atoms (Br in this case) and will form more readily than another molecule of acetone turning into an enolate ion. Thus, as the reaction proceeds, the product distribution will be determined by the most stable enolate anion. This will be the tribromoacetyl intermediate enolate ion and will drive the reaction forward. So a whole mol of bromine will be converted completely to the tribromoacetyl intermediate with a 1/3 mol of acetone. Notice how only a 1/3 will react to give the monosubtituted intermediate, than 1/3 mol of bromine will react to give disubstituted product, and lastly the last 1/3 mol of bromine will react to form the trisubstitued product. The last step is displacement by hydroxide via the nucleophilic acyl substitution mechanism (not SN2!) to form acetate and release bromoform. This reaction isn't really great if bromine isn't in excess.

I understand why the carboanion formed is more stable (inductive effect of bromine) however since the question specified 1 mole bromine how is the attack by the enolate ion taking place the second and third time around , these products should be formed in ''excess Br2'' not '' 1 mole Br2''

@Arn_kap One-third of the acetone will react fully to give acetic acid. The other two-thirds will not react.

@orthocresol So am I right in interpreting it in terms of molecules that one molecule of monobromo acetate formed will go ahead and form di and tri substituted as opposed to considering in moles 1:1 ratio used (acetone and bromine) wherein actually it was 1/3:1

$3$ $\ce{H}$ atoms out of total $6$ in a molecule of acetone is brominated.
So $\frac{3}{2}$ moles $\ce{Br}$ molecules is needed to react with 1 mol acetone.
Hence $1$ mol $\ce{Br}$ molecule will react with $2/3$ mol acetone.
Thus acetone is the limiting reagent and the reaction is completed.

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text:
Unusual triplet in 13C-NMR
Why is there a triplet at 46.7 ppm in the following spectrum of 2,2,2-tribromoethan-1-ol?

The spectrum is of really poor quality. We can’t even see if the triplet is a $1:2:1$ or $1:1:1$ one. From the other (poor quality) spectra I can see on SciFinder, it looks as though that should be a singlet — but I can’t see too much on $75~\mathrm{MHz}$ spectra at that resolution. Finally, it seems weird that both signals should be of similar height even though one is quarternary and the other is secondary …

I bet this is just a typo on the website.

So, there actually isn't a triplet in the spectrum shown. This is a 13C{1H}, and so proton coupling to carbons is not visible. Hence, the two peaks at 79.9 and 46.7 ppm are singlets. If you read the question carefully, on the left hand side it indicates what the multiplicity of that peak is i.e. how that peak would appear in the absence of proton decoupling. The peak at 46.7ppm, in this case, would appear as a triplet. This information helps you work out how many protons are attached to a particular carbon centre.
So for this example, one of your carbon centres will have no protons attached, and the other will have 2 protons attached, CH2. A CH2 carbon will split into a triplet, due to the normal 2nI+1 rule.
Multiplicities used to be determined by running off-resonance decoupled experiments, so that multiplicity could be seen clearly for 1J coupling. These days, multiplicity would usually be determined via a DEPT experiment, or preferably a phase edited HSQC.
Of course, that peak at ~77ppm is a 1:1:1 triplet, and that is because it is CDCl3, and a proton-decoupled carbon experiment won't remove coupling to deuterium, and so the chloroform carbon splits into 3 lines, again due to the 2nI+1 rule (2H has spin=1).
Important Edit
I've just had a quick look at the solution on your page link, and it actually appears that there is an error in their assignment of the multiplicities. The -CBr3 peak will come at 46.7ppm, and would be a singlet. The -CH2OH peak wil come at 79.9ppm, and this would be a triplet. They have these two multiplicities switched on the question page.

No, for some reason the website agrees that $46.7~\mathrm{ppm}$ is the $\ce{CBr3}$ group. Which would make sense if that is just a singlet. If you click on solution, it gives you the correct assignment.

@Jan - yes, this is what I mean. The error is on the question page, with the multiplicities. The solution page is quite correct.

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text:
Predict the product of reaction of quinone and hydrochloric acid
This is the question:

And this is my solution:

In the last step, I have tautomerised, and in the second last step, there is no double bond between carbon attached with Cl and OH (just some cutout work)
The answer given is B, but my answer comes out to be C.

You've already been told about improper formatting...

Arrows flow from negative to positive charges in general. In other words, nucleophiles attack electrophiles. The arrow going from the H+ is incorrect, all arrows involving H+ should point toward H+, as it can only be an electrophile.

@Mithoron, I really don't remember. :(

Doesn't it look better now?

@Mithoron, yes it is. Probably while putting it in bold, I though that whole of it was important. Still thanks!

I think your answer is correct, although I haven't seen this exact reaction reported in SciFinder. If this is from an instructor, ask how B can be produced.

@Dissenter Sorry, my bad.

The product of the addition of one chloride ion (C) is easily oxidized by atmospheric oxygen (radical mechanism). The obtained 2-chloro-1,4-benzoquinone is susceptible to conjugate addition of another chloride ion, this happens chemoselectively at the more electron deficient double bond which has already a chlorine attached to one carbon atom.

Which leads to the next question: Why is 2,3-dichlorohydroquinone not readily oxidised by atmospheric oxygen?

@Jan because it is already oxidized. Chlorine is a moderately electron-withdrawing group, and two chlorine atoms make ring electron-deficient enough to not being readily oxidized. Moving further this road, I point out that DDQ (dichloro-dicianot-quinone) is a rather strong oxidizer by organic chemistry standards.

@permeakra Thanks, I suspected something like that but confirmation is always nice.

Well, aren't electron deficient rings more likely to be oxidised?

No, because the first step of the radical mechanism would be a single electron transfer from the aromatic ring to oxygen, producing a radical-cation. Cations are destabilized by electron-withdrawing groups.

@Marko Can you give a link to the radical mechanism you are talking about?


text:
What is the difference between concentration and effective concentration?
What exactly is effective concentration in simple terms? From what I can understand it is the 'real' concentration taking part in the reaction instead of the concentration of the solute. Also, if I am right, should the effective concentration always be less than the concentration?

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I,ve read effective concentration is often higher than the actual concentration, i don´t understand why. It says it represents a ratio of equilibrium for inter and intra molecular reactions in this paper: Urea Dependence of Thiol-Disulfide Equilibria in Thioredoxin: Confirmation of the Linkage Relationship and a Sensitive Assay for Structure? Tiao-Yin Lin and Peter S. Kim* Whitehead Institute for Biomedical Research, Nine Cambridge Center, Cambridge, Massachusetts 021 42, and Department of Biology, Massachusetts Institute of Technology, Cambridge, Massachusetts 021 39 Received January 10, 1989


text:
Can't a redox reaction occur in one half cell alone without flowing through the external circuit?
Say you have two half cells. The electrolytes are connected by a salt bridge and the electrodes by a wire.

Half cell 1’s electrolyte is copper(II) nitrate and its electrode is solide copper.
Half cell 2’s electrolyte is nitric acid and its electrode is inert (graphite).

I’ve deduced that the reduction reaction will be
$$\ce{2 NO3- (aq) + 4 H+(aq) + 2e- -> N2O4(g) + 2 H2O(l)}$$
while the oxidation reaction will be
$$\ce{Cu(s)->Cu^2+(aq) + 2e-}$$
Now, the electrode (anode) of half cell 1 is supposed to release electrodes into the external circuit, which end up at the inert cathode in half cell 2. There, nitrate ions will be reduced. Then you’ve got the reordering of ions through the salt bridge etc. etc.
However, I do not understand how such a voltaic cell would be viable. There are nitrate ions already in half cell 1’s electrolyte. Why must electrons flow through the wire then? Can’t the copper electrode directly react with the nitrate ions in the solution it’s submerged in? Can’t the whole redox reaction take place in half cell 1?

You are overgeneralizing. Yes, some redox reactions can occur in one half-cell; make an electrode out of aluminum or even more active metal and you will see.

@IvanNeretin But what about this specific reaction that I described? It was given to me on a test and it got me so confused.

This one will not (well, almost). The oxidizing properties of nitrate ion are strongly pH-dependent.

Right. Water does not have enough $\ce{H+}$ ions to furnish this particular reaction, but an electrolyte of $\ce{HNO3}$ certainly would.


text:
Is water acidic enough for an redox reaction involving H+?
If you stored $\ce{Pb(NO3)2(aq)}$ in a $\ce{Cu(s)}$ container, would any reaction occur? Lead(II) ions will not react spontaneously with the copper solid as the $E^0_\mathrm{net}$ of the reaction would be negative. However, the $E^0_\mathrm{net}$ of the reaction between nitrate ions and copper solid is positive. The reduction reaction for nitrate ions is
$$\ce{2 NO3- (aq) + 4 H+(aq) + 2 e- -> N2O4(g) + 2 H2O(l)}\qquad{E^0=\mathrm{+0.80\ V}}$$
As is obvious, the reaction requires not only nitrate but hydrogen ions. However, water (the solvent) does have hydrogen/hydronium ions. Would this be enough to furnish the reaction? If not, why?

"Hydronium ion" and "$\ce{H^+}(\mathrm{aq})$" refer to pretty much exactly the same thing.

@hBy2Py I know. That’s the underlying assumption of my question. Perhaps I should rephrase: "would the hydrogen ions in water be enough to furnish the reaction?"

Gotcha. My guess is "Yes," but I'm not sure enough to put an answer down on it.

You are overgeneralizing. Yes, water contains enough $\ce{H+}$ for some redox reactions; throw a little sodium in it and you will see.

Yes the reaction would be spontaneous at standard conditions.  However standard conditions for electrochemistry are at pH 0, i.e. an $[\ce{H+}]$ concentration of 1 molar.
Solutions of lead nitrate by themselves are not going to give you pH 0.  Apparently very very concentrated lead nitrate solutions will reach pH 3 or 4, due to the weakly acidic nature of hydrated lead ions.
The electrochemical potential of the reaction you are interested in at pH 3 can be calculated from the standard (i.e. pH 0) potential via the Nernst equation.
$$E_\mathrm{cathode} = E_0\;– \frac{RT}{nF}\ln{Q}$$
Here, $Q$ is the "half"-reaction quotient, given by
$$ Q = \frac{\mathrm{p}_{\ce{N2O4}}}{[\ce{NO3-}]^2[\ce{H+}]^4}$$ Now assume that nitrate concentration and the partial pressure of dinitrogen tetroxide are in their standard states.
If you go through the math (see examples here or here), then you will see that there is a 118 millivolt-per-decade decrease in the potential, i.e. every increase in the pH by 1 unit decreases the potential by 118 mV.  So already at pH 3 the reaction will no longer be favorable.
Copper is unstable to concentrated nitric acid, but the acid needs to be concentrated.  Dilute acids cannot oxidize copper so easily.

Thanks! I’ve done some calculations and I understand it fully now. However, did you make a mistake with the expression for Q? Doesn’t the Nernst equation apply for the entire cell? Wouldn’t Q need to have [$\ce{Cu^2+}$]?

Yeah apologies my answer was confusing.  You can write Nernst equations for an entire cell, or for each half-cell separately.  The $Q$ I wrote was for the half-reaction you put in the OP, but the Nernst equation is mislabeled as "$E_{cell}$".  It should really say $E_{cathode}$.  I'll edit to correct.

It depends on your redox reaction. For example, if you add permanganate to a neutral solution containing e.g. sulphite, you will observe a redox reaction in which manganese(IV) oxide is created. If you do your redox caluculations, you will observe the following:
$$\begin{align}\tag{red}\ce{MnO4- + 3 e- + 4 H+ &-> MnO2 + 2 H2O}& | \cdot 2\\
\tag{ox}\ce{SO3^2- + H2O & -> SO4^2- + 2 e- + 2 H+} & | \cdot 3\\
\tag{redox}\ce{2 MnO4- + 2 H+ + 3 SO3^2- &-> 2 MnO2 + H2O + 3 SO4^2-}&\end{align}$$
This reaction still consumes protons but proceeds under neutral conditions. (Under acidic conditions, permanganate would reduce to $\ce{Mn^2+}$, und strongly basic conditions it would end up as $\ce{MnO4^2-}$.)


text:
Hydrogen sulfide and sodium carbonate reaction
What is formed in the reaction between hydrogen sulfide and aqueous sodium carbonate? My guesses are sodium sulfite and/or sodium hydrogen sulfite. However, the equations don't seem to balance properly whenever I try to do it. For example:
$$\ce{H2S(g) + Na2CO3(aq) -> Na2SO3(aq) + H2(g)}$$
It seems quite strange that hydrogen gas would be produced in this kind of reaction. Whatever I do, I end up with a bunch of hydrogen atoms that can't combine with any oxygen, and I'm sure I'm missing something.

When an equation wouldn't balance at all, this is a subtle hint that something is really wrong with it. Where would carbon go, what do you think?

To whoever downvoted, can you please explain what's wrong with the question? Otherwise, I can't fix it.

The actual reaction is given here.
$$\ce{H2S + Na2CO3 → NaHS + NaHCO3}$$
Conditions:

Hydrogen sulfide react with sodium carbonate to produce sodium bisulfide and sodium bicarbonate. Hydrogen sulfide should be saturated solution.

While writing a reaction, you must balance the charges and atoms. In your proposed reaction, where did the carbon go? Indeed, the carbon from sodium carbonate gets converted in the form of sodium bicarbonate.

Do not forget to vetify that the reaction is thermodynamically favored.  Among the acids --hydrogen sulfide on the left, bicarbonate ion on the right -- the weaker acid is favored.  We can verify by looking up dissociation constants that bicarbonate ion is in fact the weaker acid here.


text:
What chemical that when cured chemically cross-links to already vulcanized rubber
I'm looking into ways to repair rubber materials and I was wondering if there are any chemicals that when cured will chemically bond to the already vulcanized rubber material.
I know this might involve having to break the strong carbon-carbon or carbon-sulfur bonds to have it re-bond to the chemical, so said chemical itself might have to be volatile.
So, is there a chemical like that?

What type of "rubber"? Many articles are made of blends of various rubbery polymers, such as Buna S, polyurethane or EPDM, some of which can be partially dissolved commercial rubber glues and cements, providing good adhesion. For example, "Shoe GOOP" adheres well to rubber soles.

Various elastomers, such as ethylene propylene diene terpolymer (EPDM), can be cured (crosslinked) using ionizing radiation.
The general radiation effects on polymers are

formation of gas,
cross-linking of polymer chains, and
scission of polymer chains.

However, EPDM rubber contains side chains, ring systems and π electrons, which results in a delocalisation of excitation energy and causes a greater stability of the excited states. Therefore, radiation chemical yields for all dissociative radiation chemical reactions on EPDM are decreased.
The mechanism of cross-linking involves the cleavage of a $\ce{C-H}$ bond on one polymer chain to form a hydrogen atom. The free radical released in the initial radiolysis reaction has enough kinetic energy to abstract a second hydrogen atom from a neighbouring chain to produce molecular hydrogen. Different recombination pathways follow the irreversible loss of hydrogen. Double bond formation results from two events occurring on the same chain; the recombination of two separate macroradicals leads to cross-linking. Cross-linking can also result from the reaction of free radicals with double bonds (comparable to free radical addition polymerization). The radiochemical ageing of EPDM shows the formation of unsaturated bonds (trans-vinylene, vinylidene, and vinyl) as well as the consumption of double bonds by radical addition reactions. Since EPDM already contains double bonds which originate in the diene, the last-mentioned reaction predominates.
The overall effect of cross-linking is that the molecular mass of the polymer steadily increases with radiation dose, leading to branched chains until ultimately a three-dimensional polymer network is formed when each polymer chain is linked to another chain. In contrast, scission is the opposite process to cross-linking in which the rupture of $\ce{C-C}$ bonds occurs. Under irradiation in absence of oxygen, EPDM shows some increased cross-linking with a plateau at roughly $10^5\ \mathrm{Gy}$.
However, further exposure to ionizing radiation makes elastomers hard and brittle until disintegration occurs.


text:
How may the water from rain be tested to determine it is natural, or from cloud seeding?
This article lists several mechanisms that may be used to control the weather and/or induce rainfall. Apparently many states, other than the one listed in the headline, have experimented with such techniques. The article also lists concerns over contamination from such artificial rainfall.
Say I, a layman without access to a laboratory, collect the rain in a glass. What test/tests may I apply to determine whether it is natural rainfall, or the outcome of cloud seeding?

It would be totally impractical to do such testings. The "contamination" levels  would be much too low to be detected. You have to realize that air isn't pure. When a raindrop falls it would wash out all kinds of gunk that was in the air.

True, air and water are not pure anyway, but silver iodide is usually not a natural contaminant; a well-equipped laboratory may be able to detect it. A layman might just as well sit down and pray.

Could it be tested obliquely? Does the presence of silver iodide imbue water with any features/properties that would not normally be there?

While extensive lab measurements have been conducted to measure "seed" concentrations, as @MaxW commented the levels are just too low and methodologies too complex for routine home measurements.  As one example of background concentration problems, in the US only about 1% of the atmospheric emissions of silver come from cloud seeding.

The AgI would act as a cloud condensation nuclei (CCN) on which water vapour would condense allowing cloud droplet formation. Presumably, you could use scanning electron microscopy to determine the composition of the CCN and determine whether cloud seeding has taken place.
Apparently, his has already been done with cloud chamber (so, artificial scenario) measurements (Zimmerman et al., 2007 - "Environmental scanning electron microscopy (ESEM) as a new technique to determine the ice nucleation capability of individual atmospheric aerosol particles").
However, considering that there has been ESEM measurements in field situations before (Pöschl et al., 2010 - "Rainforest Aerosols as Biogenic Nuclei of Clouds and Precipitation in the Amazon") to determine CCN composition, I suppose a similar set up could be performed.


text:
In powder x rd, if the crystals are randomly oriented ,why then do we get Debye rings and not random patterns?
In powder x-ray diffraction ,as I have learnt the sample is multi-crystalline.So if x-ray is diffracted through these randomly oriented crystal grains, aren't we supposed to get a random pattern?

That's because no matter how you would rotate a crystal grain, it would diffract only at certain special angles. Also, welcome to Chem.SE.

Firstly thank for your answer and warm welcome.

Can you please describe why it is so.I am not able to visualise it.

Check out this site for an exhaustive explanation of powder diffraction:- http://pd.chem.ucl.ac.uk/pdnn/powintro/powdiff.htm


text:
Calculating Optical Absorption Spectra using Ab Initio Computations
My main question is, How does one compute the optical absorption spectra of a molecule by using say, Configuration Interaction energy and associated transition dipole moments.
I am trying to use an ab initio package, Columbus to calculate transition moments (for $\ce{H2}$ as of now), however, there is some old code, called MELD, that I am supposed to refer to, that somehow takes in 4 non-zero such moments and their energies and gives a spectrum.
The input looks like this:
Linear absorption from the 1Ag; qci
Linear absorption from the 1Ag state; qci   1
How many components of dipole, and which ones?   3
1,2,3
Total no. of Bu excited states   4
Line width, frequency range, step
0.1,0.0,15.0,0.01
Energy of the 1B1u state
-1.1724565
Energies of the Bu states
-0.677388000000000      -0.262044700000000      -0.245699100000000       -0.245699100000000
Scale the energies  SCALE
 0.036749d0
Next come the dipole moments 4
           1           1  0.000000000000000E+000  0.000000000000000E+000   -1.20320100000000
           1           2  0.000000000000000E+000  0.000000000000000E+000
  0.547801000000000
           1           3  -1.05742200000000      -0.291179000000000
  0.000000000000000E+000
           1           4 -0.291179000000000        1.05742200000000
  0.000000000000000E+000

I am not sure what exactly is happening there? A reference to such calculations would be great too.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. You have to be more specific than that, for example start with what software you are using. Then we need to know which level of theory you are employing. That "old piece of code" could also really be anything. As of now this is much too vage to be answered.

Thanks, I have added details, is that fine? I am new to this.

Please have a look at the links I posted earlier, especially check the formatting help. My edit might have unwanted side effects, please check carefully. If you have any more information on MELD, please supply it, too. I don't think, that the top google result is what you are actually using for this task. Is this the columbus  you are using? It would be good if you could post the input to this calculation, too, so that it is easier to reproduce.

Yes, the MELD code is not available online for some reason. The closest I come to finding it is here and yes, you are right about Columbus. I just got the set of worked out files and there are a lot of them, my only question, as I edited it in bold up there is about 'how to compute the optical absorption spectra of a molecule by using say, Configuration Interaction energy and associated transition dipole moments.'

Do you mean how to plot a continuous spectrum? If so, you will need a Lorentzian and Doppler broadening.


text:
Which is more electronegative: chlorine or nitrogen?
Why nitrogen is considered to be less electronegative than chlorine. Nitrogen should be more electronegative as I was reading some content about this and I read some reasonable arguments:

$\ce{NCl3}$ in $\ce{H2O}$ gives $\ce{NH3}$ and $\ce{HOCl}$, which proves that nitrogen is negatively charged because it attracts the positively charged hydrogen in $\ce{H2O}$ to form $\ce{NH3}$ and chlorine combines with the negatively charged $\ce{HO-}$ radicals.
$\ce{HNO3}$ is stronger than $\ce{HClO3}$, though chlorine and nitrogen have the same oxidation state ($+5$): $\mathrm{p}K_\mathrm{a}(\ce{HNO3}) = -1.4, \mathrm{p}K_\mathrm{a}(\ce{HClO3}) = -1$.

But in spite of that electronegativity value for nitrogen is considered to be $3.04$ while that of chlorine is $3.16$. Why?

Question. It appears as if you've retrieved your values from Wikipedia, however, the measure of dissociation for chloric acid in the article is approximate. The one for nitric acid isn't. I would recommend finding a solid value for chloric acid, first. A cursory search shows little documentation for it.

Depends on the scale used. Allen's definition of electronegativity has nitrogen (3.06) more electronegative than chlorine (2.87). It's a borderline case to compare.

Not only is the strength of chloric acid not precisely known, but comparison with nitric acid is not straightforward because the central atoms have different valence electron counts.  So where nitric acid has a pi bond, chloric acid has a nonbobding pair on chlorine.

So @OscarLanzi should I compare HNO3 with HClO4 as central atom has utilized all the electrons in each case?

No such comparison really works.  You can't cleanly attribute differences in reactivity to differences in electronegativity alone when the elements are in different groups.

Does this answer your question? Scale to be considered for comparing electronegativities of nitrogen and chlorine


text:
Acidity order of nitrophenols
As in other posts I have seen and learnt that, if one is given four compounds, namely para-nitrophenol, ortho-nitrophenol, meta-nitrophenol and phenol and is told to arrange them in order of acidity, then one has to take into account the $\text{-R}$ and $\text{-I}$ effect.
Now, for the inductive effect, the greatest intensity is observed at the ortho-position, followed by meta-, and then para-.
For the resonating effect, the intensity is greatest for para and lowest for ortho, while meta has no such effect. Now keeping in mind, that $\text{-R}$ and $\text{-I}$ both tend to withdraw electrons away from the phenol group (and thus increases the acidity), and the $\text{R}$ effect has more contribution than the $\text{I}$ effect, we can thus deduce that para-nitrophenol is most acidic, followed by ortho-nitrophenol, then meta-nitrophenol and since phenol has no ring activating group, it comes out to be least acidic in the given compounds.
The facts are the proof of my theory, where the $\mathrm{p}K_\mathrm{a}$ for:
$$
\begin{array}{c|c}
\text{Compound} & \mathrm{p}K_\mathrm{a} \\
\hline
para\text{-nitrophenol} & 7.16\\
ortho\text{-nitrophenol} &7.2\\
meta\text{-nitrophenol} & 9.3\\
 \text{phenol}   & 9.95
\end{array}
$$
So far so good.
But one of my colleagues, when I asked the problem, tried to give the theory in different way. He said, in para-nitrophenol, hydrogen bonding is strongest, so it can't be most acidic, which in his view port should be least. I knew this is not the case, but I just can't explain or point out where he's been wrong. Can anyone of you make it little easy for me to understand it?

Hydrogen bond has do direct relation to $pK_a$; also, I see no indication as to why it should be the strongest in case of p-nitrophenol.

@IvanNeretin Doesn't the largest separation between two groups in para position makes H bonding in para strongest?

No, it doesn't; why would it?

@IvanNeretin I'm not sure sir, I'm just asking. So my definition of the case of highest acidity of p-nitrophenol is highest is right, isn't It?

Yes, your reasoning is about right.

I can’t follow your colleague. I would dismiss his argument as downright wrong.

Even I'm confused with that thing. :/ In ortho nitro and para nitro benzoic acid para form is lesser acidic cause of H bonding.why is that not in corresponding services phenol compounds? Also I understand that para nitro phenol is more acidic but why is that H bonding in it doesn't make it lesser acidic? Please explain

Related: https://chemistry.stackexchange.com/q/7683/4945

The para isomer has stronger intermolecular hydrogen bonding while the ortho isomer has stronger intramolecular hydrogen bonding

pKa 7.16 and 7.2? Aren't they both the same? The difference, if there is any, is insignificant.

You have correctly identified that m-nitrophenol should be the weakest acid among the isomers as the resulting anion cannot be stabilised by the $\text{-M}$ effect of nitro group owing to its meta location.
The acidity order of the remaining two nitrophenols can be explained in this way:

Hydrogen bonding is possible in both of them. But note that:

$$\text{H-bonding} \propto \delta^+ \text{charge on $\ce{H}$ atom}$$
Inductive effect (here, $\text{-I}$) is distance dependent. So $\delta^+$ charge on the $\ce{H}$ of ortho-nitrophenol's -$\ce{OH}$ is going to be more than the $\delta^+$ charge the $\ce{H}$ atom of para-nitrophenol's -$\ce{OH}$.

Also,  quoting from this comment:

Usually the $\mathrm{p}K_\mathrm{a}$ of a substance is determined in dilute solution. So just due to dilution, there won't be many/any inter-molecular H bonds, but there can still be plenty of intra-molecular ones.

The answer above the comment relates the strength of these two types of hydrogen bonds to kinetic motion of molecules:

Inter-molecular interaction, such as inter-molecular hydrogen bonding only take place for a short time when two molecules are in each other's vicinity. Shortly after, the molecules separate and the interaction does no longer exist.
Intra-molecular interactions on the other hand are much stronger since the two atoms forming the hydrogen bond remain in each others proximity due to structure of the molecule. In other words, they exhibit strain.

These three factors help us conclude that:

Strength of intra-molecular hydrogen bonding in o-nitrophenol $\gt$ Strength of intermolecular hydrogen bonding in p-nitrophenol

Thus, p-nitrobenzoic acid is more acidic than the o-isomer.

...and since phenol has no ring activating group, it comes out to be least acidic in the given compounds.

This reasoning is incorrect. Phenol has no group that stabilises the negative charge on the resulting phenoxide ion through $\text{-M}$ or $\text{-I}$ effect (a deactivating group). Therefore, it's the weakest acid of all under investigation.

Due to $\text{-R}$ effect of nitro group at ortho- and para-positions, the conjugate base gets stabilised more than that at meta-position. Now at the ortho-position, H-bonding is present, which makes removal of $\ce{H+}$ ion difficult. Therefore correct acidic order of nitrophenols are  para- $\gt$ ortho- $\gt$ meta-.

More the stability of phenoxide after removal of $\ce{H}$ more will be  acidity.

ortho-Nitrophenol phenoxide ion is stabilised by  -R and -I effect where both -I  and -R effect are maximum.
meta-Nitrophenol phenoxide ion is stabilised by  -I effect only.
para-Nitrophenol phenoxide ion is stabilised by  -R and -I effect where both -I  and -R   effect are minimum.

Thus according to stability of anions o-derivative will be more acidic than p-derivative which will be more acidic than m-derivative.
But due to H-bonding in ortho-nitrophenol it is reluctant to remove hydrogen due to which it's acidity decreases and order becomes:
p-derivative $\gt$ o-derivative $\gt$ m-derivative $\gt$ phenol
(acidity in decreasing order)
Source- Jagdamba Singh: A logical approach to Modern Organic Chemistry. Revised edition 2009.  Pg 151

In both ortho- and para-nitrophenol, there is one more resonance structure. So due to resonance effects both ortho- and para-nitrophenol are more acidic than the meta-nitrophenol.
Now we compare the acidity of ortho- and para-nitrophenol. In ortho-nitrophenol, interamolecular H-bonding takes place because of attachment with adjacent carbon atom. But in para-nitrophenol, there is no interamolecular H-bonding taking place. So that's why para-nitrophenol is more acidic than ortho-nitrophenol. So, by increasing order of acidity,
para-nitrophenol $\gt$ ortho-nitrophenol $\gt$ meta-nitrophenol.

By "interamolecular", do you mean intermolecular or intramolecular? This needs to be edited.

Or both :P. Should be intramolecular as it takes place in ortho- but not in para- nitrophenol.

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


text:
Structure of phosphorus pentachloride
Why is the shape of $\ce{PCl5}$ trigonal bipyramidal? All the $\ce{Cl}$ atoms must be at equal distances forming a star like shape. However, in the actual shape, the distance between the $\ce{Cl}$ atoms in the equatorial position is less than the ones in the axial position.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. Have a look at this question, it might already contain an answer.

see the last para of the answer to this question. It basically covers up your question.

I can’t believe that this is not a dupe! =O

The short answer is stated on the Wikipedia page for trigonal bipyramidal:

… there is no geometrical arrangement with five terminal atoms in equivalent positions.

This website from the School of Mathematics and Statistics at UNSW explains why distributing points on a sphere is tough and that perfect distribution is possible only in certain cases $(n=1,2,4,6,8,12,\&\ 20)$. These cases have been known for a long time — they correspond to the platonic solids (plus a point and a line segment). Even trigonal planar does not represent an even distribution on the sphere.
Trigonal bipyamidal then just happens to be the minimum energy conformation. The repulsive interactions between the chlorine atoms cannot be optimized but they can be minimized.

One could say that the positions are indeed optimized - they just aren't all the same positions geometrically.

The geometry of $sp^3d" hybridisation, i.e: trigonal bipyramidal happens because of unequal hybridization of atomic orbitals.
The hybrid orbitals of a trigonal bipyramidal structure consist of two sets namely axial and equatorial orbitals. The two axial orbitals are made by inter-mixing of $p$ and $d$ atomic orbitals whereas the three equatorial orbitals are formed by intermixing of one $s$ and two $p$ orbitals.
Since the equatorial orbitals have a higher percentage of $s$ character compared to the axial orbitals, they are shorter than axial orbitals.
This is why you have two set of distances in $PCl_5$. This is NOT true when there are more than 1 atom except for one specific case which I will add to the answer if requested (did not add it here to avoid complication since the OP appears to be new to the topic)

This is a little bit backwards as orbital hybridization is a mathematical construct developed to rationalize observed geometry with the shapes and orientations of the canonical hydrogenic atomic orbitals. It is more appropriate to say that $\ce{PCl5}$ is $sp^3 d$ hybridized because it has trigonal planar geometry.

$PCl_5$ is trigonal bipyramidal.  Of course, hybridization is a mathematical construct developed to explain observed geometry but I believe that the OP isn't through with the basics yet. So I thought an answer based on VSEPR theory would be sufficient.

Most, if not all, trigonal bipyramidal structures I know, including $\ce{PCl5}$, have no contribution of d-orbitals. $-1$.

huh? You can't form 5 bonds with just s and p orbitals. $PCl_5$ is a hypervalent molecule.

https://en.wikipedia.org/wiki/Orbital_hybridisation#Hybridisation_of_hypervalent_molecules

I get the -1 because I used the hybridization theory? The OP doesn't even use the proper tag, says lewis structure... you can't give a complicated answer.


text:
Using IR to detect / quantify a molecule at low concentrations
I want to detect / quantify a specific terpenoid molecule from a microbiological growth medium. The molecule will however be present at very low levels approx. 10-100 mg/L.
Is it viable to use the IR spectrum to detect this molecule at such low levels? If not quantitatively at least in a qualitative sense? The idea is to have some sort of method that is amenable to reasonably high throughput screening.
A GC / GCMS which is the typical assay method is too time consuming for screening a lot of samples.
The IR spectrum is shown below. Can I just monitor a specific wavelength with max absorbance, (say) 3000 cm-1? That way I won't have to scan the whole wavelength range.
Any other ideas?

There isn't enough information to answer the question with certainty.  We'd need an IR spectrum of the culture medium (possibly dried/lyophilized medium)
I suspect that the peak at 3000 $\mathrm{cm}^{-1}$ is an alkyl CH stretch.  You're going to see similar absorption peaks in lipids and proteins, so I don't think that particular peak is going to be too promising for you.
You might look into headspace mass spectrometry.  Many terpenes are volatile and can be analyzed directly from headspace air (i.e. the air directly above your cultures).  You don't necessarily need to do gas chromatography, so the method can be quite fast.  It won't be as diagnostic as to the identity of the particular terpenes formed, however.

Thanks! Apart from the CH stretch do you see other signatures that might be more unique?


text:
Why assume a neutral amino acid is given for acid-base reaction?
The questions is:

$1.812\ \mathrm{g}$ of a crystallized α-amino acid ($\mathrm{p}K_\mathrm{a1} = 2.4, \mathrm{p}K_\mathrm{a2} = 9.7$) has a $\mathrm{pH}$ of $10.4$ when dissolved in
  $100\ \mathrm{mL}$ of $0.1~\mathrm{M}\ \ce{NaOH}$. Calculate the molecular mass of this amino acid.

As usual we can obtain the ratio of the conjugate base and acid around $\mathrm{p}K_\mathrm{a2}$ by using the Henderson-Hasselbalch equation. However, once we get the ratio, the solutions key says that we know $c(\text{Base}) = 0.1$ because all of the $\ce{OH-}$ reacted with our acid. While I totally agree that all the $\ce{OH-}$ reacted with the acid, since it is a strong base, how do we know there was no $c(\text{Base})$ before the addition of $\ce{NaOH}$? Thus making the concentration $c_0(\text{Base}) + 0.1$.
Is this because it came from a solid crystal where all of the amino acid was in the acid form?

There are a few observations to make.

The amino acid only has two $\mathrm{p}K_\mathrm{a}$ values. Thus, it cannot have a side chain that is in any way acidic or basic. We only have three different possibilities: $\ce{H2AA+, HAA, AA-}$.
It was crystalline when added. Therefore, it must have been in its neutral $\ce{HAA}$ state. Otherwise, it would have been given as ‘the $\ce{HCl}$ salt …’ or ‘the $\ce{NaOH}$ salt …’

If we had any excess base in there, it would have to have been given in the question as the sodium hydroxide crystal adduct or something.


text:
Metaphosphoric acid and water in cold
Consider this redox reaction:
$$\ce{3PCl3 + 2HNO3 (conc.) + 5H2O → 2NO + 3[HPO3] + 9HCl}$$
The metaphosphoric acid is taken in square brackets, because I have doubts about it. As I know, this reaction is endothermic. Is it possible the metaphosphoric acid is formed in this reaction (there is a water in left part of equation, metaphosphoric acid may react with it, but the temperature is decreasing)? Are there some real environmental condition where this reaction is possible? Or it's only the orthophosphoric acid is formed?

Don't add water (or rather, add less water), and this thing will have no choice but to go to $\ce{HPO3}$ and even less hydrated forms.

Can you clarify, is this an actual reaction or one that you are proposing?

That one was written by me on exam. I do not really propose what is the real  yield of that reaction, but i'm interested in conditions that may move the balance to right side and realize the metaphosphoric acid (not orthophosphoric).


text:
Gaussian scan function help for constructing input file
I have the cartesian coordinates for two molecules that are in a stacked position and I want to be able to calculate the potential energy at 0.2 angstrom steps from when the top molecule is 3.0 angstroms to 7.0 angstroms away in the z direction. My professor told me gaussian has a scan function to do this but I'm having trouble figuring some things out. I have the basis set/ correlation methods I need to use and the cartesian coordinates for all the atoms when they are arranged 3.0 angstroms apart. I tried following this site's advice;
 but I'm not sure I constructed the input file correctly. Here is what I have
.
From my understanding, this should just move up all of the atoms of the top molecule by 0.2 angstroms in the Z direction.
Questions:
Do I just throw this stuff anywhere into the input file? What about the original cartesian coordinate array? Surely it has to stay since the "step" array calls back to atoms 101-112. Do I have to change any other commands? The only parameters set are the resource limits and "#P BP86/6-311++G**"
Update:
I think I may have been unclear. I have already optimized my two molecules. One is a nanotube and the other is an arene and my task was to basically generate this plot. I'm trying to find the distance apart where the two molecules are stabilized. Before, I just ran the "#P BP86/6-311++G**"" with my arene @ 3.0 angstrom apart and made new input files changing the arene z coordinates in 0.5 angstrom increments but I can't capture that potential energy well. I decided to try smaller increments to increase the resolution but I have quite a few more situations to test so I wanted to just have Gaussian change the increments and recalculate the potential energy.

Thankfully a full relaxed scan is not necessary in most cases. Oftentimes a nice little rigid scan will give you a decent result. Have a look at the scan keyword. The only downside to this approach is, that you need a z-matrix to make it work. And putting your brain to that task is also a bit tedious.

That's what I initially came across but with ~100 atoms, I just let Chemcraft generate the z matrix from my cartesian coordinates. The thing is, I'm not sure quite how to edit the z-matrix (relevant part below) with the scan commands.
Specifically I want to keep all bond angles the same but move atoms 101-112 in 0.2 A steps in the z direction, and calculate the potential energy.
I guess my question is now if this is a plausible approach. I'll try to look up how to construct the proper z-matrix if that is indeed the best way to do things.
Thanks @Martin

Please note that it is discouraged to significantly alter the question after an upvoted answer has been given. It is by far better to ask a new question, linking back to the original for context.

Firstly, the Gaussian format, while annoyingly unlabeled, has a strict ordering which you can find here.
(Aside from your ordering problem, I highly recommend you check out the Link 0 commands in the Gaussian manual, particularly %OldChk=file which means that it copies the checkpoint file before using it.)
Secondly, I would use (and, actually, have used with an SWNT+arene system) an IRC scan for this: I know it's not quite what you asked about, but if you can start from a reasonably aligned geometry that you think is close to somewhere on the potential valley, it should automatically generate points backward and forward on that valley, without you having to manually specify any coordinates. Using IRC(Report=Read) will spit out the coordinates along with the energy, and from one or two of those sets of coordinates that look close to the minimum, you can perform an optimisation to get the actual minimum. Depending on what you're doing, you may also want to fix many of the nanotube carbons in place before doing this -- it'll dramatically speed up your gradient calculations if I recall correctly. (This is much easier to do if you skip the Z-matrices and just use Cartesian coordinates.)

This is (probably) not going to work the way you intend it to work. Constructing a meaningful modredundant calculation is quite a difficult task, as you sometimes cannot anticipate what the program actually does. Using this interface you will perform a relaxed scan, i.e. you will step one variable, fix the values of the scanned coordinates, and run a full optimisation at this point. This allows you to systematically, brute force a potential energy surface. In you case you specify 12 coordinates to scan with 20 points each, resulting in a whopping $N=\frac{20!}{12!}\approx5\times10^8$ optimisations.
Here are a few pointers when constructing a modredundant input:

don't scan more than two parameters at a time, that roughly equates to a 3D graph of a PES
don't scan xyz coordinates, scan bond lengths and angles instead; scanning xyz coordinates results in fixing the coordinate system in parts disallowing the algorithm to rotate the molecule/ use coordination transformations to speed up the calculation
keep constant what you need not optimised; fixing things will result in a faster calculation

Let's make this a little bit more visual. As an example I'll choose $\ce{(CO2)2}$ and I am using the ridiculous low level of theory HF/3-21G.
First up optimise the equilibrium geometry.

%chk=opt.chk
#P HF/3-21G opt

title required

0 1
C       -1.345383226      0.000000000     -1.107741226
O       -0.793543854     -0.126832008     -0.084830947
O       -1.897222598      0.126832008     -2.130651504
C        1.651522395      0.000000000     -1.243964208
O        2.203361766     -0.126832008     -0.221053930
O        1.099683023      0.126832008     -2.266874487

The result will be the following geometry:

    6
scf done:  -373.128185
 C    -1.546398     0.008833    -1.171593
 O    -0.884661    -0.109039    -0.227710
 O    -2.221473     0.124768    -2.098858
 C     1.852537    -0.008833    -1.180113
 O     2.527612    -0.124768    -0.252847
 O     1.190800     0.109039    -2.123995

Now we want to move the two molecules closer together. If we would use the approach you have come up with, this would be really tedious. First we need to transform the coordinates in a way that we can understand which molecule to move where. Luckily this is not necessary at all. We can simply scan over the "bond length" of the carbon carbon distance. I chose five steps with an increment of -0.2 angstrom.

%chk=modered1.chk
#P HF/3-21G
opt(modredundant)

title

0 1
 C    -1.546398     0.008833    -1.171593
 O    -0.884661    -0.109039    -0.227710
 O    -2.221473     0.124768    -2.098858
 C     1.852537    -0.008833    -1.180113
 O     2.527612    -0.124768    -0.252847
 O     1.190800     0.109039    -2.123995

B 1 4 S 5 -0.2

That will give us the following graph and animation:

You can see that the OCO bond angles changes. The CO bond lengths changes, too, but that is not as visible. This is expected behaviour in a relaxed scan. Only one coordinate is fixed, which is the distance between the carbons. If you are interested in keeping at least the bond angles constant, freeze them:

%chk=modered2.chk
#P HF/3-21G
opt(modredundant)

title

0 1
 C    -1.546398     0.008833    -1.171593
 O    -0.884661    -0.109039    -0.227710
 O    -2.221473     0.124768    -2.098858
 C     1.852537    -0.008833    -1.180113
 O     2.527612    -0.124768    -0.252847
 O     1.190800     0.109039    -2.123995

B 1 4 S 5 -0.2
A 2 1 3 F
A 5 4 6 F

With the result, we'll see again something unexpected happening. There is a sudden rotation at 2.8 angstrom. For the given set of parameters, this is the lowest energy. This is a side effect of a relaxed scan. not everything will run smoothly and sometimes the calculation turns rogue and fails. It really is a delicate matter.

Just for shits and giggles, let's do one that will make us cringe. We will scan also both of the oxygen oxygen distances. Theoretically this should give us 20 data points. Spoiler alert: This will fail at step 8. Try it yourself:

%chk=modered3.chk
#P HF/3-21G
opt(modredundant)

title

0 1
 C    -1.546398     0.008833    -1.171593
 O    -0.884661    -0.109039    -0.227710
 O    -2.221473     0.124768    -2.098858
 C     1.852537    -0.008833    -1.180113
 O     2.527612    -0.124768    -0.252847
 O     1.190800     0.109039    -2.123995

B 1 4 S 5 -0.2
B 2 5 S 5 -0.2
B 3 6 S 5 -0.2

You can see, that with these small molecules it is already hard to predict what happens when you run the calculation. There are traps and pitfalls lining your way. More often than not it is trial and error.
Thankfully a full relaxed scan is not necessary in most cases. Oftentimes a nice little rigid scan will give you a decent result. Have a look at the scan keyword. The only downside to this approach is, that you need a z-matrix to make it work. And putting your brain to that task is also a bit tedious.
I have tried to convert the specifications to a z-matrix putting constants where needed and introducing the scan variable. I'm not sure I succeeded in the way I wanted to, but here we go:

%chk=scan.chk
%nproc=2
%mem=8000MB
#P HF/3-21G scan

title required

0 1
 c
xx   1  1.40000
xx    1  scan    2   90.000
 o    1 oc4         3   90.000      2  88.981
 o    1 oc5         3   90.000      2  -90.00
 c    3  1.93591    1   89.994      4 178.983
xx    6  1.40000    3   90.000      1  90.000
 o    6 oc8         7   90.000      3 180.000
 o    6 oc9         7   90.000      3   0.000

scan        2.794000  5  -0.20
oc4         1.152818
oc5         1.158751
oc8         1.152817
oc9         1.158750

The result is two rods, moving towards each other:

I expect that since you have more than 100 atoms, accomplishing a meaningful scan calculation is a very, very tedious task. It might be much simpler creating the points you want to scan by hand with a molecule editor. But without more information about the actual system, sound advice is difficult.

I'd also add that with large numbers of atoms, it's often easier to write a short script that places the two molecules and writes out a Cartesian input file for each step of the scan.  This is my preferred approach, since you can orient each molecule, and e.g., simply increment Z coordinates (+1.0, +1.2, +1.4, etc.)

@Geoff I agree. At the time I wrote this answer the focus of the question was different and I did not know much about what the OP wanted to achieve. I would most likely have chosen the path you just suggested.


text:
Is there a way to counteract the problems with weighing hot objects?
I am currently performing an experiment in which it would be greatly beneficial to be able to accurately measure the mass of a hot object without waiting for it to cool. The current plan is simply to place the hot object in a crucible with a lid and attempt to measure the mass quickly, before the crucible has time to heat up. The thought is that the convection currents around the object won't matter as much if they are contained within the crucible.
I will be testing this method in the next few days for accuracy and reproducibility, but would appreciate any other ideas to mitigate the problems with weighing hot objects, or insights on how this idea could be better.

Some idea of how hot might be a good thing to add. Are we talking liquid lead, or liquid tungsten? (Or, if posting from Pluto, liquid nitrogen?)

Or are we talking rotavap water bath?

Depending on geometry, mass and temperature (hot is pretty vague), you might be able to use an inertial balance.
If the device used to hold the sample is rigid, you could calibrate its resonant frequency with and without the sample and with known weights.
See this high school lab and a video of the procedure.

The issue here, is, even inside a sealed crucible, the mass fluctuations would still be an issue. The object won't go anywhere, true, but you'll get fluctuations up and down depending on how hot the object is (which, I assume, has an upper limit of about 3400 degrees Celsius). An issue with trying to take force exerted by convection is that you'll have to take into account the local pressure level, humidity, and air composition, as each of these will effect how much pressure the air will be exerting. For example, taking this measurement inside will mean the air will probably have more carbon dioxide in it, because there's people. Taking it on a humid day might increase how much convection affects whatever you're weighing. How much the air is pushing will definitely be different from the top of a mountain versus sea level.
TL;DR? Just wait for it to cool. It'll save you a lot of additional calculations and considerations, and ultimately be more accurate. If you want to do the measurement quickly, consider ways of cooling the object down without contaminating it. I would suggest placing it in a watertight container and dunk the whole thing in a bath of water, if that wouldn't cause stress fractures to form on whatever it is you're measuring.


text:
Paracetamol: does oxygen participate in the delocalization of p-orbitals
I'm analyzing the molecular structure of paracetamol and one of the things I have to do is to make a commentary about the delocalization of p-orbitals in this compound. Well I know that in the cyclic structure of carbons there is indeed delocalization of p-orbitals. I wonder if the oxygen that makes the double covalent bond participates in this delocalization or not... Can you help me figuring this out?
Also if only the 6 carbons have the delocalization of p-orbitals then we will have 6 molecular orbitals $\pi$ delocalized and so we will have 3 $\pi$ bonding orbitals full 3 $\pi^*$ antibonding orbitals empty, right? I saw this on a book (about other compound) and I want to know if this is related to the fact that molecular orbitals number needs to be the same as atomic orbitals number? But for example, if we had 8 $\pi$ molecular orbitals would we have 4 bonding and 4 antibonding? Does that happen?

Which oxygen you're talking about?

Right! I forgot to mention it. I meant the one that makes the double bond of course. Sorry, thank for spotting it

You will have one large pi system including the phenyl ring, the phenolic oxygen, the amide nitrogen, carbon and oxygen. So basically, all non-hydrogen atoms except for the terminal methyl group’s carbon. I’ll leave the formal answer to somebody else who feels like drawing/calculating/whatevering.


text:
Why does clearance increase after phenytoin overdose?
I am finding it hard to envision how an overdose of phenytoin ultimately leads to an increase of the clearance of phenytoin after the enzymes that metabolize phenytoin stop getting saturated.
I know it has to do with zero order and first order kinetics, but I just can't put it in perspective.
Would appreciate if someone could help out.

Isn't it suppose to decrease? Could please provide me with your reference, I have an answer but seems contradictory to your findings hence I wish to see reference before posting the answer.

Okay it's been a while decided to post, perhaps we  get more insight from the community

Hello thanks for the answer. Sorry I only remember during which class I read this, but cannot remember where exactly anymore.

Phenytoin is primarily eliminated by hepatic metabolism (>95%).
  Hepatic metabolism is mainly via the CYP2C9 enzyme system with a
  smaller amount metabolized by CYP2C19. About 5% of a phenytoin dose is
  recovered in the urine as unchanged drug. Phenytoin follows
  Michaelis-Menten or saturable pharmacokinetics.

This is the type of nonlinear pharmacokinetics that occurs when the number of drug molecules overwhelms or saturates the enzyme’s ability to metabolize the drug. When this occurs, steady-state drug serum concentrations increase in a disproportionate manner after a dosage increase:

In this case the rate of drug removal is described by the classic Michaelis-Menten relationship that is used for all enzyme systems: rate of metabolism =$\ce{(Vmax ⋅ C) / (Km + C)}$, where Vmax is the maximum rate of metabolism in mg/d, C is the phenytoin concentration in mg/L, $\ce{Km}$ is the substrate concentration in mg/L, and where the rate of metabolism = $\ce{Vmax /2}$.
The clinical implication of Michaelis-Menten pharmacokinetics is that the clearance of phenytoin is not a constant as it is with linear pharmacokinetics, but is concentration- or dose-dependent. As the dose or concentration of phenytoin increases, the clearance rate (Cl) decreases as the enzyme approaches saturable conditions: $\ce{Cl = Vmax / (Km + C)}$. This is the reason concentrations increase disproportionately after a phenytoin dosage increase.
However, since clearance is dose or concentration-dependent, half-life also changes with phenytoin dosage or concentration changes. As doses or concentrations increase for a drug that follows Michaelis-Menten pharmacokinetics, clearance decreases and half-life becomes longer for the drug.

For drugs such as phenytoin with a low hepatic extraction ratio
  (≤30%), the numeric value of liver blood flow is much greater than the
  product of unbound fraction of drug in the blood and the intrinsic
  clearance of the compound (LBF >> fB)
Plasma protein binding
  displacement drug interactions cause major pharmacokinetic alterations
  but are not clinically significant because the pharmacologic effect of
  the drug does not change.

Because the clearance of the drug is

dependent on the fraction of unbound drug in the blood and intrinsic clearance for a low hepatic extraction ratio agent, a
  decrease in plasma protein binding and increase in unbound fraction
  will increase clearance.

Currently there is no substantive evidence of phenytoin auto-inducing its own metabolism (the theory of phenytoin displacing itself from plasma proteins really isn’t convincing at all but perhaps in a drug interaction with a drug which is predominantly plasma bound). The reason being that the information regarding phenytoin auto-induction is sparse and conflicting. The results of this study aren’t convincing enough when taken in perspective of its ability to auto-induce itself after an overdose.
References

Applied Clinical Pharmacokinetics: A. Bauer, Phenytoin: Basic Clinical Pharmacokinetic Parameters

Does this solve your question?


text:
Evaporate propylene glycol
I tried to make a cannabis infused e-liquid. I made a tincture with marijuana and vegetable glycerin, but it became to thick so a mixed in more propylene glycol than I'd originally planned. Now the e-liquid is to weak, not enough THC per milliliter.
My question is: can I just heat up the e-liquid to 115-123 degrees celsius and let the glycol evaporate over a long time, or is that temperature too low? I don't want to destroy the psychoactive chemicals in the e-liquid.
Best wishes.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. Personal medical questions are off-topic on Chemistry. We can not safely answer questions for your specific situation and you should always consult a doctor for medical advice.

Let me reword the question:
Does $\Delta^9$-tetrahydrocannabinol (THC) significantly degrade when its solution in propylene glycol is heated to 115-123 °C?
This was somewhat answered in a study by Michael A. Repka et al. (DOI) on the stability of THC embedded in a  hydroxypropylcellulose polymer matrix. When heating this system to 120 °C for 75-90 min, about 2.7% of the initial THC degraded. With other words, 97.20% remained unchanged.
Heating to higher temperatures did not lead to significantly higher degradation, but among the degradation products, the amount of cannabinol (CBN), the oxidation product of THC, was much higher here.
Technically, heating to ~120 °C seems to be no problem.
You might want to check the reference given, have a look at the articles cited, compare toxicological data of THC and CBN and decide for yourself.


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Why should animal fats not be used for cooking?
Why we are advised not to use animal fats for cooking oils? I know that vegetable oil contain long unsaturated carbon chains and animals contain saturated carbon chains. What’s the difference if we use this?

I really wouldn’t know why that should be the case. Could you provide a reference to your claim?

For one thing, saturated fats tend to have higher melting points, such that at room or body temperatures, they are often solid (though exceptions obviously exist). Therefore, unsaturated fats and oils will be easier to use since they can be poured.

There is also a longstanding, though possibly unjustified, link between saturated fats and oils, and cardiovascular disease. Avoiding saturated fats is supposedly healthier.
Because livestock are more expensive and slower to grow than plants, vegetable or plant oils are easier and faster and cheaper to obtain.
Finally, animal fats often absorb flavour from the animal meat, giving them an unwelcome taste in many contexts. By contrast, vegetable and plant oils either don't absorb anything or the result is attractive pr characteristic in most cooking contexts. So vegetable or plant oils are preferred for versatility and for flavour.


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Under which criteria HNO3 is a concentrated or a diluted acid?
I would like to buy nitric acid to remove impurities of copper on an aluminum plate. I read that a 10% nitric acid will do the job. However I've also read that concentrated nitric acid is very corrosive to human skin.
So is a 10% nitric acid solution is still very corrosive to human skin. What am I risking if one drop falls on my arm?
Is there any tables that shows the reactivity of acids under various concentrations to human skin?

https://www.youtube.com/watch?v=XeVZQoJ5FdE

https://pubchem.ncbi.nlm.nih.gov/compound/nitric_acid#datasheet=lcss&section=Top Just read it up to find out

According to the European regulation on classification, labelling and packaging of substances and mixtures, the classification of nitric acid depends on concentration.
At a concentration of 20 % ≤ c, nitric acid is classified as skin corrosive category 1A. (For a pure substance on the basis of the results of animal testing, this category corresponds to a substance that produces destruction of skin tissue, namely, visible necrosis through the epidermis and into the dermis, in at least 1 of 3 tested animals where responses are noted
following up to 3 minutes exposure and up to 1 hour observation.)
At a concentration of 5 % ≤ c < 20 %, nitric acid is classified as skin corrosive category 1B. (This category corresponds to a substance that produces destruction of skin tissue in at least 1 of 3 tested animals where responses are described following exposure between 3 minutes and 1 hour and observations up to 14 days.)
Furthermore, at a concentration of 65 % ≤ c, nitric acid is classified as oxidising liquid category 3.
Therefore, nitric acid is labelled in accordance with the Globally Harmonized System of Classification and Labelling of Chemicals (GHS) as follows:
Concentration 65 % ≤ c

H272: May intensify fire; oxidiser
H314: Causes severe skin burns and eye damage
Concentration 20 % ≤ c < 65 %

H314: Causes severe skin burns and eye damage
Concentration 5 % ≤ c < 20 %

H314: Causes severe skin burns and eye damage

I am unaware of the table you are looking for, but I can explain "concentrated". I can also remind you that the better strategy is to prevent exposure.
"Concentrated" means that the reagent is present in its highest concentration possible, which happens in one of three ways:

The solution is saturated
The reagent is essentially pure
The reagent is unstable at higher concentrations

This website from Sigma-Aldrich lists the concentrated reagents, the percent by mass concentration, and the molar concentration. Here are some of the acids:
$$\begin{array}{|c|c|c|c|}\hline
\mathrm{acid} & \mathrm{formula} & \mathrm{\%\ w/w} & \mathrm{molarity}\\ \hline
\mathrm{Acetic} & \ce{CH3CO2H} & 99.8  & 17.4 \\
\mathrm{Formic} & \ce{HCO2H} & 90.0 & 23.6 \\
\mathrm{Hydrochloric} & \ce{HCl} & 37.2 & 12.1 \\
\mathrm{Hydrofluoric} & \ce{HF} & 49.0 & 28.9 \\
\mathrm{nitric} & \ce{HNO3} & 70.4 & 15.9 \\
\mathrm{phosphoric} & \ce{H3PO4} & 85.5 & 14.8 \\
\mathrm{sulfuric} & \ce{H2SO4} & 96.0 & 18.0 \\ \hline
\end{array}$$
So, 10% nitric acid is one seventh as concnetrated as concentrated nitric acid. Is it corrosive to skin? Yes. Is one drop going to be terrible? Not if you wash it off quickly and thoroughly.
What should you do? Use the appropriate personal protective equipment! Obtain a pair of butyl rubber gloves. There are available in sizes that go all the way up to your elbow.


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Kipp's apparatus with powders
A Kipp's apparatus has three distinct bulbs. The topmost one contains a liquid and a tube to the bottommost one. The middle one contains a solid to react with the liquid.

This is very understandable when large chunks of solid are used. However, what do people do when they need to react a powder (e.g. sodium sulfite)? It seems that the powder would just fall through and keep reacting when the valve is closed.

According to the Wikipedia page on the Kipp's apparatus, if the reactant is a powder, and soluble in water, you can soak pieces of pumice in the solution, dry the pumice, and use it in a Kipp's apparatus. There is also a variant of the apparatus that allows two liquids to be used together rather than a liquid and a solid reactant. In that case, you would pre-dissolve the sodium sulfite in water.

There is an inner separator (a conical piece of glass emanating from the top of the base reservoir into which the stem from the top reservoir sits) which is perforated and acts to retain the solid in the middle chamber while allowing the acid to react (or not react) with the solid as the pressure in the system rises or falls.
See this detailed description with images for reference.


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Why does entropy increase when the difference in temperatures is decreased?
From the Wikipedia article about Entropy (Source view via The Internet Archive):

In irreversible heat transfer, heat energy is irreversibly transferred from the higher temperature system to the lower temperature system, and the combined entropy of the systems increases.

If entropy were a measure of disorder, or randomness, or the amount of information needed to describe the microstates of a system, then it seems that when the difference in temperatures is high, entropy should be high. A temperature gradient in the systems is a form of ‘disorder’ hence should be a sign of higher entropy.
So why does entropy increase after the heat transfer, as the combined system approaches a uniform temperature? It seems to me like a state with less randomness or disorder.

Source of the quote?

I changed the quote to another one from Wikipedia which expresses the same thing

I thought that a temperature gradient would be a form of order rather than disorder, which is consistent with entropy.

@hb20007 that is only correct if the system is adiabatically closed. If not, then there is no guarantee that the system entropy after equilibration is larger than before, and the increase of entropy in an adiabatically closed system is the fundamental law of nature that cannot be reduced to simpler statements. It is true of all kinds of equilibration not just of the temperature.

Temperature is a tricky concept, because a "uniform" temperature does not mean the system has a lot of order.  Uniform temperature merely means the statistical distribution of particle velocities is homogenous.
Before the systems are combined, you know a fair amount about the system.  For instance, you know particles on the "cold" side are going to stay on the cold side, and particles on the "warm" side are going to stay on the warm side.  Informally, this means you know something about every particle in the system.  Once you combine the systems, you no longer know where the particles are going to be, so there is more disorder to the system.

Consider two tanks of water. One hot, one cold.
You open a valve between the two. The temperature between the two bleeds, and the hot tank loses entropy, and the cold tank gains entropy. However, entropy is also gained just by this process happening, as it would be impossible to force the 'cold' tank to retransfer the heat back to the 'hot' tank.
Also, you must consider something else. The two conjoined systems now have more molecules than either system alone, increasing the 'randomness' as a whole. Considering dice, it's like having two sets of two, then one set of four. The set of four has more possibilities than just the set of two.

Because the same amount of heat has different effects on the two systems in terms of entropy. The difference in temperature is just an arrow for the transformation, but for entropy, it's the individual temperatures and the total exchanged heat that matters.
The situation is equivalent to having a large room filled with people chatting, taking a small group out of it and moving their conversation to a public library. The room will pretty much stay noisy, but the library will have a huge increment in noise. Same with heat. You moved some heat from one entity to another, but one had a lower temperature. That heat has a huge impact, increasing the entropy of the low temperature body more than the decrease in entropy of the higher temperature body.

During the time that there are temperature gradients present, the combined entropy of the system is increasing with time.  If the initial state is two blocks of material at different temperatures, and the final state is the same two blocks both at the same temperature, then there are a hugely greater number of quantum mechanical energy states available in the final state than in the initial state.  The final state includes all the energy states that were available in the initial state (since these are still available to the final state within the framework of conservation of energy) plus many more states that were not available in the initial state.


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Why are gas laws only applicable to gases?
Why are the gas laws (such as Boyle's Law, Charles's Law, and Avogadro's Law) only applicable to gas and not liquids or solids?

They're gas laws by definition.

By their nature, gases will be different from liquids and solids. The main point is that liquids and solids are controlled by atom-atom (or molecule-molecule) interaction potentials - they are bound together in a potential well. Gases generally are not - the species in the gas spend most of their time zipping around freely, with the only interactions being scattering off of other gas molecules. The details of such scattering just don't matter that much. (Well, ultimately they do, which is why you can't assume a gas is ideal).

@JonCuster Looks like you have the better part of an answer there...

The gas laws are also based on the Kinetic Theory of Gases, so it takes into account the properties specified by @JonCuster

In addition comments by Jon Custer, unlike the condensed phase, gases are highly compressible, this leads to the laws mentioned.

Gases, liquids and solids have different interactions between their components
Ultimately they are applicable to gases because they only apply to gases. The reasons why they work are worth understanding, though, because they are critical to the development of chemistry which might not exist if not fro some of the convenient properties of gases.
The most important law is that a given number of atoms or molecules as a gas always occupies the same volume if temperature and pressure are fixed. This was what allowed the generation of data that formed the basis of modern chemistry: things combine in fixed ratios. This led to the experimental proofs of the existence of atoms which would be a lot harder to do if we had to rely on measuring the volumes of solids or liquids where the volume tells us nothing useful about how much stuff makes up the substance. It is easy to measure the volume of a gas.
But why is this true? The essence of the reason why gases make up a fixed volume whatever they are made of, is that, to a good approximation, in gases the component parts are small, far apart and have only elastic interactions. If you do the statistics of such a system a whole bunch of apparently relevant factors cancel out (like the mass and velocity of the particles) to give a system which obeys the key gas laws independently of the molecule or atom making up the gas.
But this only works in gases. Liquids have strong interactions between their molecules; not strong enough to prevent individual molecules wandering around randomly but strong enough to hold them to a fixed volume without needing a container (gases will expand to infinite volume without a container to prevent them diffusing into the atmosphere). Solids have even stronger interactions that hold their individual molecules or atoms in a well defined location. The specific properties of solids and liquids depend on the details of those forces and, since there are a wide variety of forces and force strengths, the collective properties are a lot more complex.
In gases, the ideal gas-law approximations are good because the molecules of the substance have enough energy to break the attractive forces that would otherwise make them solids or liquids. Once free of those attractive forces they tend to occupy a much larger volume and their behaviour becomes better explained by the (simpler) gas laws as we can start ignoring the complexity of the attractive forces.
But the gas laws are an approximation. If they were not, nothing would ever liquefy or solidify. But the hotter a gas gets the more it behaves ideally as the less the attractive forces are relevant to the behaviour of the gas.

These laws, by definition, apply only to an ideal gas, that is, gasses that behave in a manner interpretable via. a one-size-fits-all mathematical model. Some characteristics specific to an ideal gas include:

No intermolecular interactions between particles/molecules.
They occupy a negligible volume.
The particles/molecules travel randomly.
Completely elastic collisions occur between particles.

In the case of conditions 1 and 2, liquids and solids cannot fulfill these criteria due to the presence of intermolecular attractions between atoms, which provide their characteristic shape and properties (as opposed to gasses).
Even still, in reality, it is usually only low-molecular-weight or high-temperature gasses which predictably follow the stoichiometry of these laws.

"follow the stoichiometry of these laws"? I think there should be sth like behave approximately like those laws predict...


text:
Does the synthesis method used to make graphene oxide composites matter?
I'm doing a research project to test the photothermal conversion performance of graphene oxide composite materials.
We're using a number of different metals and metal oxides ($\ce{Au}$, $\ce{Ag}$, $\ce{Cu}$, $\ce{TiO2}$, $\ce{SnO2}$, etc.) and for each material the literature suggests different methods of synthesis.
We are only using chemical methods; the ones suggested include microwave-assisted reduction, redox reactions, and hydrolysis. Will using different methods of synthesis for each compound have an effect on the outcome of the experiment? Should the same (or similar) method be used for all of the composites?


text:
Chemical Reagent Test sensitive at low concentrations
Would there be a spot test/tests that can detect either/both of the following molecules at low concentrations?
The likely concentrations are in the $10{-}\pu{50 mg/L}$ range in an aqueous solution. Ideally, I'd want something that does not need an exotic/expensive instrument.  e.g. a color change or precipitation or conductivity/pH change.

The background interference: The sample is Terrific Broth which has the following composition:
$\pu{12 g}$ Bacto tryptone
$\pu{24 g}$ Bacto yeast extract
$\pu{4 mL}$ Glycerol
$\pu{100 mL}\ \pu{0.17M}\ \ce{KH2PO4}$ and $\pu{0.72M}\ \ce{K2HPO4}$, sterile, to be prepared separately from the tryptone, yeast extract, and glycerol solution.

Any sort of such a spot test would depend heavily on what else might be in the solution. So what other "stuff" could be in the water? I wouldn't expect either of those molecules to be very water soluble.

Thanks @MaxW I've added info about what else could be in the sample. And yes you are right these are not very water soluble although at the low conc. I am expecting they very well might be.  This is a conc. of less that 0.01%.

In  any case, I want to brainstorm possible test options. At which point I can test them out on control samples of the broth to see what reaction I get.

Bacto yeast extract || yeah, forget about any meaningful spectroscopy for such low concentrations with this unholy mix in the media. Theoretically speaking, a highly selective molecular recognition agent such as specifically tailored antibody may work, but the R&D required would be quite costly, so unless your needs do not exceed thousands of samples, it is not even worth considering. Other than thatn, GS/GSMS is your best bet.

@permeakra Thanks for the insight. That's what I feared. GCMS works well, just that it is slow as number of samples scale.

@permeakra One question motivated by the GCMS idea. Typically we extract into 1:2 Hexane or Ethyl Acetate & then inject into the GCMS. Would that pre-extraction make spectroscopy any more feasible? I suppose most of the unholy mix from the media won't get extracted into the organic phase. The GCMS does come out clean. Thoughts?

@curious_cat Highly doubt it. Living matter contains a lot of minor hydrophobic compounds plus lipids that should extract easily by the mixture, and IC and UV specters usually have wide peaks. Furthermore, some of said compounds are so heavy that they may irreversibly contaminate GC column, some of them (like metal-porphyrins ) leaving something that cannot even be burnt out completely. So, I would look for liquid chromatography solution and filtered the extract using some extremely fine filter before analysis.

Do you have the pure compounds on hand?  If you do, the two target compounds can be crudely isolated by extracting the broth with diethyl ether, concentrating it down, and spotting on SiO2 TLC plate. Compare RF with that of the pure compounds.
Alternatively, the same general extraction methodology could prepare a sample for GC-MS or crude NMR, if such instrumentation is available


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Bond Energy of Ions of Molecular Oxygen

Three of the bond energies of $\ce{O2,\ O2^{-},\ O2^{2-},\ O2^+}$ have the values approximately $200, 490$ and $625\ \mathrm{kJ\ mol^{-1}}$. The value for one of the species is uncertain and, therefore, not given. Write the bond energies of each compound.

My Attempt
I know that higher the bond order of a compound, higher its bond energy. So the bond order of $\ce{O2,\ O2^{-},\ O2^{2-},\ O2^+}$ are $2,\ 1.5,\ 1,\ 2.5$  respectively.
However I am not sure which one's bond energy is not known. I am thinking that it might be $\ce{O2^{+}}$ since it is unstable, however it is just a total guess.
Is there some way to logically deduce which one it is, or is this one of these things that we are just suppose to know?

That would be quite strange; why the one with the strongest bond of all would be unstable? Indeed, there are compounds with $\ce{O2+}$, like $\ce{O2+[PtF6]-}$. Also, there surely are compounds with $\ce{O2-}$ and $\ce{O2^2-}$, and I feel some $\ce{O2}$ around as I write this. They are all stable and pretty well known.

a) small dianions tend to eject an electron in vacuum b) bond energy is well defined (and easily measured) for free molecule/ion in vacuum

There's no reason to speculate on reasons why they didn't put that specific one. It's just homework and you're supposed to find correlation between bond order and energy.


text:
Balancing by oxidation numbers method multiple atoms
In my Chemistry textbook, the rules for balancing a chemical equation using the oxidation-Number method are as follows:

Assign oxidation numbers to all the atoms in the equation.
Identify the atoms that are oxidized and the atoms that are reduced.
Determine the change in oxidation number for the atoms that are oxidized and for the atoms that are reduced.
Make the change in oxidation numbers equal in magnitude by adjusting coefficients in the equation.
If necessary, use the conventional method to balance the remainder of the equation.

I understand how to do this for simple equations, such as $$\ce{SnCl4 + Fe -> SnCl2 + FeCl3}$$
Following the steps:

$\ce{Sn^{4+} + 4Cl- + Fe^0 -> Sn^{2+} + 2Cl^- + Fe^{3+} + 3Cl^-}$
Oxidation: $\ce{Fe^0 -> Fe^{3+} (Fe)}$ reduction: $\ce{Sn^{4+} -> Sn^{+2} (Sn)}$
oxidation: +3; reduction -2
$\ce{3SnCl4 + 2Fe -> 3SnCl2 + 2FeCl3}$
(no change necessary, already balanced)

The Problem: Notice how Sn and Fe (the reduced and oxidized atoms) all have subscripts of 1. However, when their subscripts differ, I'm not sure how to balance it.
For example, I can easily balance the following equations by inspection, but not so easily by this method of oxidation numbers:
$$\ce{KClO3 -> KCl + O2}$$ (O is oxidized, subscripts of 3 and 2)
$$\ce{NH3 + NO2 -> N2 + H2O}$$ (N is oxidized and reduced, oxidation numbers of -3, +2, and 0)
And this gets even more complicated when the subscripts differ and there are more than two elements whose oxidation numbers changed (are reduced or oxidized), e.g.,
$$\ce{NH4ClO4 + Al -> Al2O3 + HCl + N2 + H2O}$$
(Al and N are oxidized, Cl is reduced)
For all of these, I get stuck on the fourth step. Determining oxidation numbers and calculating the change is not the problem, but figuring out how to do it with larger or different coefficients as in the examples above is the problem.

The five general steps in my textbook don't help me understand this method much with more complicated methods such as this. Could you give me a more thorough explanation - a better explanation or a general rule - of how to solve all of these?

related http://chemistry.stackexchange.com/questions/36970/how-to-balance-more-complex-redox-reactions

look at my answer here http://chemistry.stackexchange.com/questions/55477/oxidation-number-of-phosphorus-in-ph3/55486#55486 where calculating oxidation numbers is explained with some examples

In some (many?) cases you get a more parsimonious description by considering an entire reactant as an oxidant or reductant.  Separating out individual atoms makes you miss the forest because you get confused by the trees.
Take the ammonium perchlorate-aluminum reaction. Since aluminum is being oxidized let us render the ammonium perchlorate as the oxidizing agent, ignoring the fact that only some of its atoms are oxidizing.  You then have a reaction:
$\ce{NH_4ClO_4}\rightarrow \ce{N^0}+4\ce{H^I}+\ce{Cl^{-I}}+4\ce{O^{-II}}$
where the Roman numerals indicate oxidation states in the products.  These add up to $-5$ for all the product whereas the ammonium perchlorate, as a neutral compound, started with the states adding up to zero.  So five electrons must be added to the reactants:
$\ce{NH_4ClO_4}+5 e^-\rightarrow \ce{N^0}+4\ce{H^I}+\ce{Cl^{-I}}+4\ce{O^{-II}}$
Thus although some atoms might be oxidized, others reduced, and still others are just along for the ride, we identify a net oxidizing effect of the ammonium perchlorate as a whole, numerically five electrons per formula unit.
Now we know that the redox stoichiometry will be $3\ce{NH_4ClO_4}+5\ce{Al}$ and we can balance the reaction accordingly.  (We will need to double the coefficients to eliminate a fraction, that's just algebra.)


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Effect of pH on volatility (odor) of organic acid
Based on two assumptions:

Ionic solutes can generally be considered nonvolatile
As the pH of a solution of organic acid is increased, an increasing percentage of it is deprotonated, or ionized (Henderson-Hasselbach eqn).

I think I should be able to decrease the volatility of an organic acid solution by increasing pH. However, in several rough experiments with acetic acid, I don't observe a very strong reduction in odor with increasing pH. Even a theoretically neutralized solution retains a significant vinegar odor.
Any ideas why this might be?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] for unanswered questions about it that still remain.

Acetic acid is a weak acid (pKa 4.8). Even if you neutralize it, some acetate ions will react with water to form acetic acid and hydroxide ion (hydrolysis). Since acetic acid has a quite strong smell, we are able to detect its presence in neutral acetate solutions. If you will increase pH to a high value with a strong base, the concentration of acetic acid will drop (Le Châtelier's principle).


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Removal of water from sulfuric acid
Suppose I have a dehydration reaction of methanol to produce methoxymethane:
$$\ce{2 CH3OH(l) ->[H_2SO_4] C2H6O(g) + H2O(l)}$$
Usually, people will use sulfuric acid to dehydrate the methanol. Assuming all the methanol was reacted, how do I get rid of the water in the sulfuric acid? Will simply heating the sulfuric acid be sufficient?

You can use heat to concentrate sulfuric acid.  This is a dangerous process, due in part to evolution of SOx from the boiling acid.

You should be adding stoichiometric amounts of diluted acid. Then distilation to remove the methanol by destilation. Alternatively can also use molecular sieves, or add sodium sulfate to dry the methanol


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How does impurity of distilled water vary with time in air?
If I place distilled water in a glass beaker, how long is it expected to stay distilled?
What are the causes of it's impurity through air or other sources?

Gases like oxygen and nitrogen from air are absorbed quite easily. As you pour the water to the beaker it is most likely already saturated with those gases. Other air components and contaminants (carbon dioxide, sulfur dioxide, nitrogen oxides, ammonia) will also get quickly absorbed. How fast they will contaminate your sample depends on their concentration in the air, temperature of the water, size of the container and so on. Other possible contaminants are dust particles, impurities present inside the beaker, and even glass itself (depending of its type) is slightly soluble in water. The concentration of solid impurities will usually not change very quickly. Of course, as the water evaporates, solid impurities will concentrate.

Air contains anorganic gasses (HCl, HNO3, HNO2, SO2) and particles (aerosols) that usually consists of the following anions: CL-, NO3-, SO42-, NH4+, K+, Ca2+, Mg2 and organic matter (bacteria etc...).  The latter is not taken into consideration now but can influence the long term quality of water due to growth.
On the water surface absorption of these substances will take place.  Through diffusion these substances will be transported further into the water.
Absorption speed and concentration of gasses and aerosols depend on the local temperature/humidity and air quality.


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On what principle do electric alcohol detectors work?
I know that normal alcohol detectors work on the principle that potassium dichromate oxidizes alcohol . On what principle does an electric alcohol detector work?

I believe you are referring to breathalyzers. When you breathe into a breathalyzer, any ethanol present in your breath is oxidized to acetic acid (the main component of vinegar apart from water) at the anode (there is a diagram below).  At the cathode atmospheric oxygen is reduced. The overall reaction is the oxidation of ethanol to acetic acid and water. The electric current produced by this reaction is measured by a microprocessor and displayed as an approximation of overall blood alcohol content (BAC).

This website can give you more information.
Hope this helps!

What's the source of picture?

http://i.stack.imgur.com/IEQ9u.jpg (I found it while searching Google Images)

They were already on imgur... there may be problems with rights, licence etc.

The other important factor that makes these devices acceptable by the courts, is that in the body alcohol is removed by a zero'th order reaction, i.e. linearly in time. (except at v low conc'n below any legal limit). This means that it is possible to calculate exactly what blood alcohol levels are, and to back calculate to earlier times.

Yes, that is very true. Thanks for adding that!


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phase differences between conical intersections and light-induced conical intersections?
I've read recently about light-induced conical intersections, a phenomenon  where conical intersections can be artificially introduced to molecules, and that this can be observed even in diatomic molecules (see for example here and here ). I understand that there are similarities in the way you sketch the crossing of the curves, but I dont understand where and how does the wavefunction changes sign near the light-induced conical intersection version?


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Receiving milk for working in harmful environment
I just got a job in a chemical/biological lab (post-Soviet Country). Chemists here receive milk for "harmfulness of the work" (i .e. their work involves working with all kind of substances/reactives). I'm not a chemist; I do their analysis. The tradition of receiving 0.5 liters of milk a day comes all the way from the Soviet Era. I found this question here: Does milk drinking prevent long-term chemical poisoning?
However, I would like to know if milk is not that great for preventing chemical poisoning (as seen from the accepted answer), why is there still such a tradition of giving milk? And why did this tradition appear in the first place? And do people working in chemistry labs in other countries also get milk? (I'm not sure, but I think in UK they do.)

I guess this question is better suited for History.SE. The chemical part is pretty much covered in the other question you've found.

I'm not sure it belongs to History SE. I would like to know the "reasoning" behind the tradition of giving milk which I guess lies in the structure of milk protein binding with toxic substances. The question in other link shows that milk is pretty weak/does not help in long term exposure however it still was/is used. So there maybe something more to it..ALso I want to know if similar practice exist somewhere else

Reasoning behind the tradition very much belongs to the domain of history. What is a tradition, after all? Something we do the way we do for no good reason, just because we've been doing it for quite a while and got used to it. As usual, there is a grain of truth in the beginning: milk would indeed neutralize some toxins (not all, and not very efficiently).

Possible historical explanation: the calcium in milk would help to protect the body from absorbing strontium into the bone structure--this would be a good idea for someone working in a lab with--or otherwise exposed to--the dangerous isotope strontium-90. This isotope was released from above-ground nuclear tests and became a huge concern leading to the limited test ban treaty in 1963. The milk itself would have to be low in strontium-90 to have a beneficial health effect.

Yeah, that's the problem. If you need some low-background steel, you may just go and salvage one of them WWI battleships, but where can we find low-background milk these days?

Not a scientific answer but a personal comment. I was a chemistry student in Spain when graduating (from 1977 to 1982). Not a time in which food was any kind of problem in my country. Yet, after our work at the labs my friends and myself used to drink a glass of cold milk just to calm our stomach of being in an ambient of smelling acids and bases. It just worked! Notice that I live in northern Spain, where people drinks cow milk until the end of our days. Best regards, Dr. Víctor Luaña

After my research about this topic I can simulate the benefits of milk in factories as an important drink where it can work as buffer solution as it contains a lot of huge chains of proteins which have ability to convert acid base to alkaline and reverse so it prevents from poisonous from heavy metals in our blood and also big chains of proteins good environment to carry metals to exertion organs.

I know that in India there was a very similar law extant at least back till the 1990's when I had encountered it. Not sure if it is still on the books.
The association with the USSR makes sense: during the cold war a whole bunch of Indian factories were constructed with Soviet aid. And the conventions, technology, etc. came along.
I had heard of several alternative explanations for the milk-law; not sure which (if any) were true. Maybe partly all were.

The Vitamin D or Calcium in milk competing against absorption of Lead or some other nasty element, I forget which the workers were exposed to in chemical factories.
The Milk just contributing to the general health of factory workers in an otherwise grossly malnourished nation (remember, this is cold war era India). Presumably due to the attendant health hazards of chemical factory work, keeping a worker there healthy was more crucial.
Just having the stomach full of a competing innocuous substance (e.g. milk) reduced the propensity of competitive absorption of other nasties in accidents, fugitive exposures etc.


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Difference between black and brown eumelanin / red and pink pheomelanin
Wikipedia says the following:

There are two types of eumelanin—brown eumelanin and black eumelanin—which chemically differ from each other in their pattern of polymeric bonds.

What are those differences in polymeric bonds and which bond produces what color? Are the bonding differences also the reason for the different colorations in pheomelanin?

First of all, melanines are mixtures of complex polymers, without highly ordered structures and precisely defined composition. It means that unlike e.g. polypropylene or PET, melanines are not composed of regularly repeating identical units (monomers).
In a great simplification, brown and black eumelanines are composed of units of DHI (5,6‐dihydroxy‐1H‐indole) and DHICA (5,6‐dihydroxy‐1H‐indole‐3‐carboxylic acid). The order of these units is random and the ratio of DHI to DHICA depends on the origin of the pigment. Darker eumelanines are enriched in DHICA, which form branched structures (each DHICA unit can be bonded to three other units). Light-brown eumelanines contain more DHI units. Pheomelanines are build of different units - alanylhydroxybenzothiazines.
Now to the color. The color of melanines results from the absorption of light by electrons of chromophores (i.e. fragments of a molecule that strongly absorb light at certain wavelengths). In eumelanines the chromophores are dihydroxyindole units, and in pheomelanines benzothiazine rings. The color of melanines does not result directly from how are these bonded with each other, but from how many chromophores are in a melanine molecule, how densely are they packed, and from a number of different factors (e.g. presence of metal ions, oxydation). Black eumelanines absorb more light than brown eumelanines, because they have more chromophores (they are more densly packed due to the branching of the polymer) and that is why they appear darker. And pheomelanines have different chromophores that absorb light of a bit different wavelength (blue region). Different bonding patterns affect the optical density of melanines, but not directly their color.

I have found an answer to the color difference from pheomelanin (from Wikipedia):

Pheomelanins impart a pink to red hue, depending upon the concentration.