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How do I perform thin-layer chromatography to identify alkaloid content in fruits?
Fruits and as well as other plant parts usually contain alkaloids, which are nitrogenous compounds. I'd like to know how the  Thin-Layer Chromatography method can be used to determine the alkaloid content in a fruit.

Are you trying to isolate? If so, you basically want to do preparatory TLC.
If not, you're in trouble because you either need a standard to compare $R_{f}$ factors or you need to scrap some off for GCMS. Also, TLC is not really that quantitative for analytical purposes.

TLC in alkaloid research
Both analytical and preparative TLC have been used to separate and identify alkaloids from plant material.
Pretreatment of TLC plates and adsorbents
If you are using silica plates, it is often advantageous to wave the plate above an open bottle with aqueous ammonia or pretreat the plates with an organic solvent containing some triethylamine. Otherwise, the $\mathrm{R_f}$ values of the alkaloids might be too small.
Alkaloid detection on a TLC
As pointed out by Zhe in a comment, this is crucial for preparative TLC. Usually, you run an analytical TLC (same material and mobile phase) in parallel and stain the analytical TLC using Dragendorff's reagent.
General hints on the extraction
Process different parts of a plant separately
Different parts of a plants have different functions. The distribution of particular compounds or compounds classes, such as

chlorophylls and carotenoids
quinones
polyphenols
carbohydrates
terpenoids
alkaloids

will vary in flowers, fruits, seeds, leaves, stems (and bark), roots, and tubers.
It is therefore reasonable to process the different parts of a plant separately.
Use dry plant material
The water that you remove in the beginning doesn't disturb the extraction in a later stage. Dry the plant material. If you assume that the compounds you are looking for are air-sensitive, dry the material in vacuum.
Use powdered plant material
Extraction is more efficient on solid materials with a larger surface.
Use the right solvent
Two different strategies are conceivable:

Extract all soluble compounds with a polar protic solvent (methanol, ethanol) at once and partition between solvents of different polarity later
Leach the dry plant material stepwise with solvents of increasing polarity:

cyclohexane
ethyl acetate
chloroform
methanol

Use the right technique
You can

stir/shake the dried plant material with a solvent at room temperature
boil plant the material in a solvent
use a Soxhlet extractor

If necessary, perform the extraction under a nitrogen or argon atmosphere
Specific advice on extraction
Rambutan (Nephelium lappaceum)
In Antioxidants, 2014, 3, 371-386 (Open Access article), the authors write

The fruit samples were washed and air-dried, followed by drying in an oven at 40 °C for 24 h. All the dried samples were ground to fine powder with a grinder, after which they were vacuum packed into a nylon-linear low density polyethylene film before analysis. An extraction with 80%
  ethanol was carried out, and the mixture was shaken at 50 °C for 120 min using an orbital shaking incubator as optimized in a previous study. The residue was subjected to another round of extraction under similar conditions to obtain maximal yield of antioxidant activities. The extracts were combined and centrifuged, followed by filtration and concentration of the supernatant using a rotary evaporator at 45 °C. The concentrated extract was freeze-dried, wrapped with aluminum foil and kept at −20 °C. The crude extract as then partially fractionated using silica packed open column chromatography and using increasing polarity of solvent, ethyl acetate, chloroform and methanol (100:0:0 to 0:0:100 v/v).

In the article cited above, alkaloids were not mentioned.
Ambarella (Spondias dulcis)
In Asian Pac J Trop Biomed., 2013, 3, 682-691, the authors give the following procedure

The pulverized plant materials were soaked in 3.5 L of methanol in separate containers and kept for 1 week with occasional shaking. The whole mixtures were then filtered through sterilized cotton followed by Whatman No. 1 filter paper and concentrated using a rotary evaporator (IKA , Germany). The weight of crude extracts of fruits and leaves were 135 g and 101 g, respectively. A total of 40 g crude extract of each plant materials was then partitioned with chloroform and dichloromethane. The percentage
  yields of fruit and leaves were methanol ( 21.3 % , 15.8 %), chloroform ( 19.5 % , 23.1 %) and dichloromethane (20.9 % , 25.4 %).

The authors mention that extracts both from the fruits and the leaves of S. sulcis gave a positive test with Dragendorff's reagent, but do not provide further information on the amount and/or nature of the alkaloids.

Hi, I'll make sure that. I'm actually not clear on the part where I extract alkaloid from a fruit. What are the procedures required to extract the alkaloids from a fruit before proceeding to TLC ?

@user38146 That's a rather broad topic. Do you have particular plants and their alkaloids in mind?

@KlausWarzecha.Yes, I have multiple of them, particularly the fruit part of the planet, such as Apple,Mango,Ambarella and Rambutan fruits.


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How do I build a home-made apparatus for identifying what minerals are present in water?
While there are many readily available kits in the commercial market for testing the type of minerals present in water,

is there a way we can build one at home?

I am ready to spend some bucks on raw materials, if I can get the knowledge of how to build such apparatus

This probably not what you are looking for, but the cheapest and least time-consuming analysis is the one done by others for free.
Over here (Germany), local water suppliers are required by law (Trinkwasserverordnung - TrinkwV) to monitor the quality of potable water. This includes a large number of microbiological and chemical parameters. The details are given in the annexes 1-5 of the TrinkwV.
The results are published and are are available online at no cost.
Examples (mostly in German) are:

RheinEnergie (Cologne area)
RWW - Rheinisch-Westfälische Wasserwerksgesellschaft (one of the largest suppliers in Germany)

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Calcium carbonate and hydrogen peroxide reaction
In application of hydrogen peroxide to soil and groundwater to remediate petroleum contamination, how will the naturally occurring limestone react with the peroxide, thereby reducing the effectiveness of the application.  Would calcium peroxide be more suitable.

Calcium peroxide is produced when hydrogen peroxide reacts with calcium hydroxide and not calcium carbonate.

The effect of aqueous hydrogen peroxide on natural limestone has been examined in a different context (removal of moss and algae from the walls of frequently visited and therefore illuminated show-caves).
The results were published in Environ Pollut., 2003, 122, 417-422 (PubMed).
The authors exposed etatons of limestone to various aqueous solutions for about four months and examined possible damages using a scanning electron microscope. They found:

Based on SEM study, the polished surfaces of the carbonate etalons did not show any changes after 135-day exposure to rainwater, 15% peroxide, and 30% peroxide, respectively. No etched pits or any other visible
  effects of dissolution [...] were detected.

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What is the difference between a polar molecule and polar bond?
I don't understand the difference between a polar molecule and a polar bond. Please explain what the difference between those two is.

A polar bond is one where the charge distribution between the two atoms in the bond is unequal. A polar molecule is one where the charge distribution around the molecule is not symmetric. It results from having polar bonds and also a molecular structure where the bond polarities do not cancel. For example, water has polar bonds and its bent geometry means that there is a non-zero dipole moment. On the other hand, carbon dioxide has two polar carbon–oxygen bonds, but they are such that the polarities cancel out, so there is zero dipole moment.

But even those non-polar molecules like CO2 do have a transitional dipole moment when distorted by vibrations such that the individual dipole contributions from each bond do no more cancel out. This is the reason why we can even observe a greenhouse effect with carbon dioxide: in order for the molecule to be able to absorb infrared radiation at a certain frequency it requires to have an oscillating net dipole vector upon distortion (e.g. bending  away from linear to angled geometry and back).

@user1573870 No argument there. On time average though, there is no net dipole. Based on OP's question, I don't we need to worry about those factors at the moment.

I agree this is an advanced topic, but it's relevant nevertheless since the question was also about polar /bonds/. E.g. we wouldn't have such an oscillating dipole e.g. in diatomic molecules consisting of only one type of element (N2, O2…), whose bonds are non-polar.

I would also amend that chemists use 'polar' synonymous with dipolar; other disciplines might consider any molecule with any kind of x-polar moment as polar. For example carbon dioxide is a quadrupolar molecule. In general, that way is more accurate in terms of explaining reactivity, but in general, the used terminology has served us well for a long time already.

Just because a bond is polar doesn't mean the molecule will be
All bonds where the atoms in the bond are not the same will be polar (in the sense that the electrons will be unevenly distributed between the two atoms resulting in a dipole). Some bonds (like C-H bonds won't be very polar, but they will have some polarity.
Molecules, though, are not polar just because they contain polar bonds. carbon tetrachloride ($\ce {CCl4}$) is a non-polar molecule that contains 4 polar bonds. But the polar bonds have dipoles that are symmetrically arranged around the carbon as a perfect tetrahedron and those dipoles therefore cancel each other out leaving the molecule with no net dipole moment. Chloroform ($\ce{CHCl3}$) where one of the chlorines has been replaced by a hydrogen, breaking the symmetry, is a fairly polar molecule. Likewise carbon dioxide is non-polar as the structure is linear and therefore the dipoles in the two polar bonds cancel. Carbon monoxide also has a polar C-O bond but there is no second oxygen to cancel out the dipole so it is polar.
So for molecules to be polar they need to contains bonds that are polar but they also need to be asymmetric enough for the bond dipoles not to cancel each other out.

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Why might a planar, conjugated molecule be higher in energy than a nonplanar system?
I have a small molecule (23 atoms) for which I want to verify parameters in a molecular mechanics force field.

My first step is to look at the potential energy as a function of the dihedral angle, defined by these four atoms:

The results for the dihedral scan, rotating the guanidino tail from 0 to 360 degrees in increments of 5, looks like this:

Alright, something is funky around the 0 and 180 degree structures. Honing in on 180, and performing a QM scan with increments of every 0.1 degrees, I see this:
There is a gap between 177.1-178.2 and between 181.5-182.6, where the QM optimization failed. Checking the plateau and either side of the plateau, what I find is that the plateau structures are all planar, whereas on either side of the plateau, some hydrogens come out of the plane (consistent on both sides of the plateau). For example, this is an overlay of two structures: blue = 178.2° = planar = higher energy. Red = 177.1° = nonplanar Hs = lower in energy. (front view and bottom side view)

Let's assume that the QM calculations are to be trusted (this was run using a few different levels of theory). A mere 1° causes this shift between planarity and nonplanarity. Intuitively, if we think about the ring system being the same in both cases, this makes no sense, as a planar molecule should be better conjugated and thus lower in energy, correct? What other factors might account for this phenomenon?

I'm not sure I understand what all these calculations mean, but I can offer something that may be of assistance: by Hückel's rule, cyclodecapentaene should be aromatic, but due to the various types of ring strain, including the steric clash of the hydrogens of carbon-1 and carbon-6, the ring is bent into being nonplanar, and has no aromatic character. If these two carbons are bridged, howerver,  aromaticity is achieved. Perhaps something like this occurring here.

There is no need to go that far into the details. The answer to your titular question is: because of steric problems. Think of biphenyl, or hexahelicene for that matter.

There is a gap between 177.1-178.2 and between 181.5-182.6, where the QM optimization failed. What do you mean by this? Fail as in the run was not completed successfully? From experience, Gaussian 09 gets a bit funky with flat molecules and 180 degree angles, which could cause the calculation to fail - have you tried the optimization with the "nosymm" keyword?

@tetrahydrofuran - yes, failed as in the run was not completed successfully. This was run using the Psi4 software package.

I reproduced your calculations with PBE/def2-SVP (using Orca). The energy "jump" is even smaller than with MP2, however the causes are probably the same. So here is my answer:

The planar structure is a Transition State, with one negative frequency.
The dihedral you are scanning is probably not the best to describe this twisting.

Now in more detail. Here you can see the Minimum Energy Path (calculated with pDynamo using the NEB method) between the two symmetrically related minima.
 You can see that, as expected, the path is smooth.
The main change between the structures is the pyramidalisation of the terminal NH$_2$. This can be described by the N-C-N-H dihedral, which also changes smoothly as you can see here:

I don't think the reason why the terminal NH$_2$ is not planar is steric problems as @Ivan Neretin suggests. I think its more a question of nitrogen hybridisation, which is between sp$^2$ and sp$^3$. It is the same reason why aniline is not planar. This is discussed in these two questions:

Is nitrogen of aniline sp² or sp³ hybridized?
How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine?

Hope this helps!

Ramon, thank you for the in-depth analysis! Questions I had regarding your answer:
(1) The planar structure is a Transition State, with one negative frequency.
Does that mean you believe that you think the planar structure is actually higher in energy and not just an artifact of the calculation?
(2) The dihedral you are scanning is probably not the best to describe this twisting. Can you explain further? (3) I don't think the reason why the terminal NH2 is not planar is steric problems Could it be that steric problems lead to mixed N hybridization?

(1) Yes. I don't think it is an artefact, it is higher in energy as a planer ammonia is higher than the pyramidal structure (in fact, a planer NH$_3$ is also a TS) (2) I mean that this dihedral is not the one that best describes the reaction coordinate from the TS to the minima. And that is why you get a jump in the energy. The other dihedrals I suggest may be a better scan (though I haven't tried) (3) Could be, but I don't thing so. I think it is mainly electronic. It's a competition between sp$^3$ hybridization typical of N and sp$^2$ hybridization from some resonant forms.

Depending on the geometry of the hydrogen on the "imine" moiety of the guanadine, you could be creating a very strong syn-pentane interaction. In either case, either the lone pair or the hydrogen is being quite sterically crowded with with N-H's on the 5-membered ring.
The planar form of this species actually looks pretty bad. I would expect a bit of out plane distortion to relieve some of the strain.

The syn-pentane looks like it could be relieved by the hydrogen pointing to the right, though.

Well, you'll still have a lone pair there... Unlike a methyl group, there's no free rotation to point that out of plane.

But the hydrogen can form a cyclic hydrogen bond between the two nitrogens.

Good point... :/

An artificial jump like this usually points to convergence issues. You could try reading the wavefunction from the previous step to converge to nearly the same SCF solution. If it is not that then it could have to do with how the geometries are obtained. You might want to read the previous wavefunction and tweak the geometry converger to take smaller steps to stay in the minima and not kick out to a higher energy local minimum.


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Blue flashes in rum during fermentation
While fermenting my rum wort (what rum is called before it is distilled) in a plastic fermentation barrel, I noticed little blue flashes appearing over the surface of the liquid. Any idea what could have caused this? I have looked around and the closest thing I could find is triboluminescence, but that doesn't sound right to me...

When you say "flash," are you saying that you can observe this in the dark?

A photo would be much appreciated.

Some microorganisms produce light, which is called bioluminescence.

Another thought - maybe you have created artificial will-o'-the-wisps? https://en.wikipedia.org/wiki/Will-o%27-the-wisp#Attempted_scientific_explanations

A most interesting observation :). You generally need quite a lot of energy to cause something to glow, far more than normal thermal energy. Is it possible that static from the plastic bottle has something to do with it plus alcohol vapour?


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Numbering in phenolphthalein versus ascorbic acid
According to Wikipedia, the recommended IUPAC name for phenolphthalein is 3,3-bis(4-hydroxyphenyl)-2-benzofuran-1(3H)-one, and for ascorbic acid is (5R)-5-[(1S)-1,2-dihydroxyethyl]-3,4-dihydroxyfuran-2(5H)-one.
Why does the numbering in phenolphthalein start from the $\ce{O}$ atom next carbon atom, while in ascorbic acid it starts on the $\ce{O}$ atom? If we want to prioritize the keto group, why doesn't the numbering start on the carbonyl carbon? And other groups attached to the furan on both molecules don't seem prioritize any of the 2 counting.

Related: IUPAC name, phthalic anhydride

The subsituents do not take the highest priority here, rather, it is the parent molecule that determines the numbering. According IUPAC's organic chemistry book for nomenclature, the Blue Book, for heteromonocyclic components fused to a benzene ring:

P-25.2.2.4 (...) The presence of a single heteroatom determines the numbering in a monocyclic compound; the heteroatom has the locant ‘1’.

This gives the following numbering for the isomers of benzofuran:

The parent molecule of ascorbic acid is a furan ring, for which the numbering is dictated by skeletal replacement ('a') nomenclature:

P-15.4.1.2 For naming and numbering purposes, the following decreasing order of seniority is followed: $\ce{F}$ > $\ce{Cl}$ > $\ce{Br}$ > $\ce{I}$ > $\ce{At}$ > $\ce{O}$ > $\ce{S}$ > $\ce{Se}$ > $\ce{Te}$ > $\ce{Po}$ > $\ce{N}$ > $\ce{P}$ > $\ce{As}$ > $\ce{Sb}$ > $\ce{Bi}$ > $\ce{C}$ > $\ce{Si}$ > $\ce{Ge}$ > $\ce{Sn}$ > $\ce{Pb}$ > $\ce{B}$ > $\ce{Al}$ > $\ce{Ga}$ > $\ce{In}$ > $\ce{Tl}$.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), heteromonocyclic parent structures with three through ten ring members and containing one or more heteroatoms are generally named by using the extended Hantzsch-Widman system; however, retained names (such as furan) are preferred for several heteromonocyclic compounds. When a single heteroatom is present in the ring, it is assigned the locant
‘1’.

P-22.2.2.1 Constructing and numbering Hantzsch-Widman names
P-22.2.2.1.1 (…) The presence of a single heteroatom determines the numbering in a monocyclic compound; the heteroatom has the locant ‘1’ and the numbering usually proceeds clockwise, when unsubstituted.
  (…)

Furthermore, the numbering for heteromonocyclic with retained names is explicitly shown in Table 2.2 of Subsection P-22.2.1.
Therefore, the oxygen atom in furan has the locant ‘1’.

Accordingly, the oxygen atom in the furan ring of ascorbic acid has the locant ‘1’, too, which leads to the systematic name (5R)-5-[(1S)-1,2-dihydroxyethyl]-3,4-dihydroxyfuran-2(5H)-one.

In phenolphthalein, the furan ring is fused to a benzene ring. The rules for such compounds read as follows:

P-25.2.2.4 Heteromonocyclic components fused to a benzene ring
Unless listed as a retained name (…), a benzene ring fused to a heteromonocycle of five or more members (a benzoheterocycle) is named by placing the locant(s) indicating the position(s) of the heteroatom(s) at the front of the name consisting of the fusion prefix ‘benzo’ followed by a retained name, a Hantzsch-Widman systematic name, or a name formed by skeletal replacement (‘a’) nomenclature (…). The locants cited correspond to the full bicyclic structure. As in Hantzsch-Widman names, locants are placed in the order corresponding to the order of citation of the heteroatoms in the heterocyclic component. The locant ‘1’ is always assigned to the atom of the heterocyclic component next to a fusion atom. Heteroatoms are allocated lowest locants as a set, without regard to kind; if there is a choice, lowest locants are assigned in accordance with the seniority of the ‘a’ prefixes (…). In general nomenclature locants may be omitted when the name is unambiguous; for preferred IUPAC names locants must be cited. The letter ‘o’ of the ‘benzo’ prefix is elided when followed by a vowel. Indicated hydrogen is placed at the front of the name, when required.
  (…)

Accordingly, the systematic name of phenolphthalein contains the name of the parent structure ‘benzofuran’, which consists of the prefix ‘benzo’ followed by the retained name ‘furan’. The locant ‘1’ is assigned to the atom of the heterocyclic component next to a fusion atom. Therefore, the oxygen atom in the furan ring of phenolphthalein has the locant ‘2’. The locant indicating the position of the heteroatom is placed at the front of the name, which leads to the name ‘2-benzofuran’ for the parent structure in phenolphthalein.

Accordingly, the oxygen atom of the furan ring of phenolphthalein has the locant ‘2’, too, which leads to the systematic name 3,3-bis(4-hydroxyphenyl)-2-benzofuran-1(3H)-one.


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Using table salt as a way to limit humidity in an unoccupied house
We own a static caravan in the UK (Snowdonia region, so rainy and wet all year round, moderately cold winters) and we leave it unoccupied during Winter (Oct-Feb).
We were advised of an apparently common practice among caravan owners which is to leave open boxes of table salt in each room, under the premise that they will capture moisture in the air and lock it away chemically in order to reduce the risk of moisture related decay in the building's structure.
I was skeptical that this would work but in the interests of science I did it anyway last winter.  I took 7 plastic ice-cream tubs and put roughly 1 kilogram of domestic table salt into each and put one in each room of the home (two in the largest room - the kitchen/living room space).
I (stupidly) never took photos of this phenomenon but I have attempted to illustrate my findings...

Fig. B is complete speculation on my part as I did not return to the home during Winter to observe it.  However in Spring when I returned, all of the tubs had experienced a change in their appearance.
All the tubs were now dry again, presumably down to evaporation due to increasing weather temperatures.  And therefore releasing all that moisture back into the building again!
Three of the tubs were largely unchanged with some noticable "caking" together of the salt into crumbly, grainy lumps which returned to normal looking salt grains when crushed.
The most profound change from the remaining tubs was as you see in Fig. C of the diagram.
The salt had actually accumulated on the walls of the tub as a fine sediment.  This suggests that water had accumulated in large amounts in the tub and had in fact risen higher than the original depth of the dry salt grains!  I'd estimate that the tub would have had to accumulate about 0.5kg of water in order for the water/salt solution to reach the depth indicated by the dry sediment.
The salt had solidified into a single, large mass.  The volume seemed to have increased noticeably but the density had also decreased accordingly, so the salt had basically expanded in it's container and solidified.  It was crumbly and brittle and some of it had been reduced to a very fine sediment.
The home is a single storey, about 12m x 4.5m x 2.5m in volume.
My questions then:

Is this a valid technique for capturing excess moisture over Winter?
Are my observations and presumptions reasonable...  Is Fig. B what really happened?
What is the chemistry / physics process that caused the salt to be transformed from Fig. A to Fig. C?
How many times did the tubs cycle between states B and C?  Was it a single cycle that lasted all of winter, or a daily cycle following ambient weather temperatures?  I could not tell just by looking at C on the last day of the experiment.

My guesses: 1. It is up to you to assess the moisture-related damage to the building; the idea seems valid, but is 1kg of salt enough? BTW, certain other salts might do better, but they are also harder to come by. 2. Sounds reasonable to me. 3. Water vapor pressure over any solution is lower than that over pure water, so water would primarily condense wherever there is something well-soluble. 4. It was one huge cycle with small daily oscillations overlaid on it.

@IvanNeretin, Regarding (1): it's a brand new building, I'll get back to you in 30 years :) hehe.

Why, we can do faster than that: leave around some pieces of clean iron, not the stainless kind, and see what happens. This way we'll get the results in two years (one winter with salt, the other without).

@IvanNeretin, nice idea.  I've got some old rusty iron nails in the garage, if I grind away a bit of the surface to reveal fresh, pure iron I can leave them in the caravan next year.

I prefer thin iron plates which you may obtain by disassembling an old electric transformer; they seem to rust easier, so the effect will be more prominent. Meanwhile, we may look up the average temperatures and calculate the total moisture content in the air.

Whatever you use you must seal the caravan completely as there will be a moisture (water vapour pressure) gradient between inside your caravan and outside. To try to re-establish equilibrium, water vapour will always enter and use up your drying agent so you end up trying to 'dry' the whole atmosphere.

@porphyrin, in general the caravan is well sealed but there are two windows that have a small permanently open vent built in to them.  Drafts do enter through these vents in high winds but I think there is some kind of dense permeable filter material to prevent mild gusts from entering.  But it does seem unlikely that such a large structure could ever be fully sealed.  Would such a resistive air vent allow a long-term (months) humidity reduction inside the caravan do you think?  I guess that's more of a physics question.  There is at least some token attempt to stopthe outside air coming in.

The classical chemical to dehumidify rooms is calcium chloride. Sometimes also magnesium chloride is used.

I thought ‘rainy and wet all year round, moderately cold winters’ describes the entire UK and not only Snowdonia …

@Jan, can't argue with you there.  Snowdonia is the UK of the UK - going by rainfall. :)

Because it absorbs water vapour from air and eventually it attracts enough water vapour so that it starts to melt partially and makes clump.

According to Transportation Information Service: Salt:

At up to 74% relative humidity and 20°C, salt does not absorb any appreciable quantities of water vapor. The critical water content of sodium chloride (NaCl) is 0.5% at 74% relative humidity, which is the flow moisture point at which salt begins readily to absorb water vapor, and increases such that, at 75% relative humidity, the salt dissolves.

See this graph for better understanding.
So at high humidity, salt can absorb lots of moisture, enough to become liquid.
When it is liquid, it will creep up the walls of the container by adhesion.
Then, when humidity decreases, the salt will lose almost all its moist, and contract, but leave residue up to the height of the former meniscus.

Is it fair to say that temperature mainly has an effect on the rate of water absorption by the salt, rather than being a prerequisite for it to happen at all?

@Wossname yes, see the second table here http://www.omega.com/temperature/Z/pdf/z103.pdf The 75% value only varies slightly with temperature.

Interesting answer! I've just asked: When it rains, it puddles. Spilled salt cycles between wet and dry with humidity. Is this akin to a phase change?

Bad idea, fresh air with moisture is continuously entering ( and leaving) as  the temperature changes and as atmospheric pressure changes . A continuously running electric dehumidifier would be needed.

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What is the difference between low-density and high-density Polymethyl methacrylate (PMMA) on a molecular level?
I know why Polymethyl methacrylate can be made to have a low or high density but i am unable to illustrate how they look like on the molecular scale. Can someone help me with a sketch please?

This might help : http://www.123rf.com/photo_20143793_methyl-methacrylate-molecule-poly-methyl-methacrylate-or-acrylic-glass-building-block-atoms-are-repr.html

@NilayGhosh no it doesn't that's just a monomer...

@DSVA if you know the monomer unit, you can easily figure out the polymer.

Also, I said "this might help". I did not actually gave the answer.

@NilayGhosh knowing how the (linear) polymer looks still won't explain the difference between LD and HD PMMA. It's not helping at all. This might be more confusing than anything else.

Low density and high density PMMA differ by density, which is given as the ratio of mass and volume.
Given that they are prepared from the same monomer, two polymers with the same number of monomers will have the same mass.
If the polymers differ in density, this must be due to the fact that they occupy different volumes.
Note that we are talking about polymers here and not the ordered packing of molecules in a crystal. Nevertheless, inter- and intramolecular interactions (and thus "packing") is important here as well.
As a rule of thumb, polymers with higher branching will "pack" less tightly and thus occupy a higher volume.
This results in a lower density.


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Oxidation of guanine with singlet oxygen
I'm studying the oxidation of guanine using singlet oxygen and one step of the reaction is not very clear to me.

My problem is with the first step: it is a 4+2 Diels–Alder cycloaddition, and Fukui's rules suggest that this kind of reaction doesn't happen photochemically, only thermally.
However, since the reaction is between guanine and singlet oxygen, the oxygen has been photochemically excited, and I don't think it would happen thermally with oxygen at the ground state because otherwise we would have constant oxidation.
So what exactly is happening there?

You're right that ground state $\ce{^3O2}$ doesn't undergo this reaction.
There is no direct excitation of oxygen either.
The key in these processes is the electronic excitation of tetraphenyl porphyrine (TPP). The excited singlet state of the dye undergoes intersystem crossing to a triplet state. The latter interacts with ground state oxygen and generates singlet oxygen.
$$\ce{TPP ->[h\nu] ^1TPP^*}$$
$$\ce{^1TPP^* ->[ISC] ^3TPP^*}$$
$$\ce{^3TPP^* + ^3O2->[sens] TPP + ^1O2}$$
Singlet oxygen then acts like the typical dienophile in a Diels-Alder reaction.


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Product of the following reaction
Thia is the question and my approach.

I just want to confirm if it is the correct product. I have not gone for elimination as it will give anti-aromatic product.

I have a vague feeling that we're in for something more spectacular.

It's actually not anti-aromatic because the two double bonds aren't conjugated...

@Zhe It would obviously have been an anti-aromatic structure.

@ReeshabhRanjan No, you're missing the point. Aromaticity and anti-aromaticity require that the $\pi$ system is conjugated. Cyclobutadiene can be formed, and it circumvents the problem of anti-aromaticity by not conjugating the two pairs of $\pi$ bonds. It's rectangular and not square as you might expect with resonance. So, it's neither aromatic or anti-aromatic because the system isn't conjugated.

@Zhe This is a new thing for me. In my school, it is taught for it to be plainly anti-aromatic, not going into further discussion. Anyways, do you think of any other product than cyclobutadiene?

Well, it turns out you don't isolate the cyclobutadiene because it immediately reacts with itself. Your job is to figure out what that reaction is.

@Zhe That’s wrong terminology. The double bonds are conjugated, there is no way around that. The distortion is there to remove the degeneracy since that would imply a diradical structure. Nonconjugated double bonds can be found in cyclooctatetra-1,3,5,7-ene.

@Jan Fair, but they're not fully conjugated. If you start increasing the separation, the orbital overlap decreases rapidly.

@Zhe What’s ‘fully conjugated’? They either form a pi system or they don’t. In cyclobutadiene, they do.

@Jan Let's just take this to chat.

Anyone got to a conclusion?

@ReeshabhRanjan the product you suggested is not aromatic since all atoms don't have the same hybridization .Moreover the molecule is non planar.one more thing,why did you go in for substitution .Alcoholic KOH is used only for elimination reaction and aqueous KOH for substitution.

@aniline Yes but according to me, it will give an anti-aromatic product. I don't know what Zhe and Jan are talking about, because I have not learnt of any such exceptions from anti-aromaticity.


text:
Why do homologous series differ by CH2, and why does this not affect the compound’s properties?
OK, so this might be a dumb question, might be not. Im a 10th grader so bare with me please.
So I just learned what homologous series is. According to the definition that the teacher told me:

A homologous series is a series of compounds in which each member differs from the next/previous by $\ce{CH2}$ or 14 mass units. Physical properties change in a homologous series but chemical properties remain almost the same because the functional group does not change.

Now, I got a couple of question on this:

Why $\ce{CH2}$? Why not $\ce{CH4}$? What makes $\ce{CH2}$ so special that we have an entire term dedicated to a series in which every compound differs by $\ce{CH2}$?
It says that there is no change in chemical properties as the functional group does not change. Well … the hydrocarbon chain does change. So does that mean that a hydrocarbon has no unique chemical properties and all its chemical properties comes from the functional groups in it?
This is related the the 2nd question. It says that the functional group remains the same. I drew a chemical diagram thing:

Now I don't know if this is 100% correct but what I imagined is that the $\ce{CH2}$ replaced the $\ce{OH}$ and the carbon bonded with the carbon chain, then it took the $\ce{H}$ of the $\ce{OH}$ and we are left with $\ce{O}$ alone. If I am correct and if this can really happen then doesn't this mean that we are changing the chemical properties of a homologous series (which can't practically happen)?
If this example of mine is incorrect then there may be some other example like this where the $\ce{CH2}$ replaces the functional group itself thus changing the chemical properties?

This 3rd question may be illogical and stuff but I am curious.

Related, but not duplicate: http://chemistry.stackexchange.com/questions/63220/how-are-alkanes-alkenes-and-alkynes-homologous-series-since-they-have-no-funct

I’m going to answer these questions backwards, because there is a serious misconception in the third that must be addressed as soon as possible.
Concerning the third question, $\ce{CH2}$ is not added to an existing molecule. Rather, the idea of homologous compounds is a purely empirical and comparative one. You cannot go from butan-1-ol to pentan-1-ol in one step by adding a mysterious $\ce{[CH2]^\dagger}$ compound. There is no such reaction and therefore there is no chance to introduce ‘side reactions’ like the one you imagined. There are homologisation reactions, e.g. the Arndt-Eistert homologisation for carboxylic acids but they are significantly more complicated than inserting a $\ce{CH2}$ group (and not taught until university).
On the second question, saying that chemical properties will not change is a mild simplification — depending on where you draw the line between physical and chemical properties. You may know already that the heavier a molecule is, the more London interactions hold it together and thus the higher melting and boiling points become. Additionally, assuming only one single functional group, the longer a molecule is the less effect a functional group has on the entire thing. So while methanol (chain length 1) is infinitely miscible with water, pentan-1-ol (chain length 5) is not.
The hydrocarbon chain in itself actually does not display a lot of interesting chemistry except for combustion. However, research efforts are being strongly directed towards what is known as $\ce{C-H}$ activation, meaning forcing a (seemingly arbitrary) carbon–hydrogen bond to react under defined conditions. If that is pushed far enough, the hydrocarbon chain can be understood to have interesting chemistry but at present we are not there yet.
Finally, your first question. I’ll just quickly reiterate that the idea of a homologous series is purely empirical, comparative and descriptive. Taking alcohols as an example, we know methanol $\ce{CH3OH}$, ethanol $\ce{C2H5OH}$, propan-1-ol $\ce{C3H7OH}$, butan-1-ol $\ce{C4H9OH}$, etc. Upon carefully analysing the reactions of each of these compounds, it was soon established that they share a similarity in spite of their different physical properties and their notably different effects on the human body. When elemental analysis was established, it could be shown that they indeed all have different carbon contents and the increase is always $\ce{CH2}$.
And why $\ce{CH2}$ and not something else? Well, remember that carbon is typically tetravalent, i.e. is most stable if it has formed four bonds. In a formal (and paper chemical) exercise, take a random alkane for example ethane. Its structure is given below.
  H H
  | |
H-C-C-H
  | |
  H H

Now assume you want to draw a new structure with one more carbon atom. For example, assume this carbon atom to be added at the very right. I’m first going to formally pull apart the $\ce{C-H}$ bond, then formally put in the carbon, connecting it to the left and right. Notice what will be missing:
  H H          H H            H H            H H H
  | |          | |            | | .          | | |
H-C-C-H  =>  H-C-C· ·H  =>  H-C-C-C-H  =>  H-C-C-C-H
  | |          | |            | | ˙          | | |
  H H          H H            H H            H H H

Note how it requires two hydrogens to satisfy carbon’s valence. This is why the group formally added is $\ce{CH2}$: Assuming a carbon inserted into something it needs two additional hydrogens to keep satisfied.

oh, so reaction like the third question example don't simply happen. Whenever we add a Carbon to a hydrocarbon chain, it needs 2 more hydrogens to satisfy the valency and thus the element "CH2" is chosen and hydrocarbons don't really have any prominent chemical properties of its own so addition of a CH2 does not change its chemical properties much. It does change the physical properties because the molecule grows bigger..am I right?

Historical Note: The concept of a homologous series was introduced by the French chemist, Charles Gerhardt. https://www.britannica.com/biography/Charles-Gerhardt

$\ce{CH2}$ because the chain is 1 $\frac{longer}{shorter}$. Remember that carbon tends to form 4 bonds. The other 2 bonds that would be attached to Hs in methane, for example, are in this case attached to other carbons in the chain. So $\ce{CH2}$ (plus a bond, or 2 half bonds) is really all that you are adding.
Its chemical properties do not change significantly, but its physical properties might. Boiling point, density, etc are physical properties. Formic acid, acetic acid, and propanoic acid are quite similar, but have differing chain lengths.
The $\ce{CH2}$ is just added to the chain. ie $\ce{HO-CH2-CH2-CH3}$ vs $\ce{HO-CH2-CH3}$. $\ce{O2}$ should not be released.

Your initial answer as linked does some additional work beyond what is necessary. No need to change the $\ce{OH}$ at all, just add the $\ce{CH2}$.

Formulas of alkane alkene and alkyne  itself results in the difference of CH2 unit. As if I talk about methane. It has formula of CH4 as it is alkane and following the formula of CnH2n+2 .Similarly Ethane it is having formula C2H6 .And if you see the difference between the two you will automatically get CH2 .The same thing implies to all other members of series alkane . Even if you see alkenes and alkynes you will find the same difference i.e. CH2 between all the members .


text:
Why is (1E,3Z,5E,7E,9Z,11E)-cyclotetradeca-1,3,5,7,9,11-hexaen-13-yne not antiaromatic?

The compound shown below was synthesized by Professor I M Dimm who was astonished to find that it was exceedingly stable. Prof. Dimm had expected world renown for making what he thought would be an unstable compound. Dimm reasoned that with six double bonds and a triple bond in the molecule there would be 16 π electrons and thus, no aromatic character ($4n, n=4$, not $4n+2$). Where was his reasoning wrong?

My reasoning so far:
Two of the electrons in the triple bond are not part of the conjugated π system. They are in a π orbital that is perpendicular to the rest of the p orbitals that make the conjugated π system. The conjugated π system thus only contains 14 electrons. $14 = 4n + 2$ for $n= 3$. So this molecule is aromatic.

Your answer seems correct to me.

Actually, it depends on the molecular geometry; until we know it, we know just about nothing. There are two possibilities. (1) Both sides of the cycle are conjugated to the same $\pi$ component of the triple bond; then your reasoning is correct. (2) Sides of the cycle are conjugated to different $\pi$ components of the triple bond; then there is no cycle, the whole thing is effectively an equivalent of a long linear conjugated polyene, which is neither aromatic nor antiaromatic. Now, the description of the compound as "exceedingly stable" favors the first option.

@IvanNeretin Wow … has that kind of twisted pi system been synthesised so far? I am very much intrigued! I love the idea!

Come to think of it, more twisted systems are known.

The textbook answer to this is that the triple bond consists of two orthogonal π systems whose interaction with each other is symmetry-forbidden. So as the attached figure shows, the two p-orbitals marked in red cannot participate in the aromatic system if the molecule’s structure is as drawn; thus the two electrons do not count towards the Hückel rule, thus $4n+2$ is satisfied with $n=3$ and $4n+2 = 14$.

However, that is not the end of the story; merely its beginning. Professor Dimm should take the time to determine the actual crystal structure of the compound he synthesised. You might want to compare this with cyclodecapenta-1,3,5,7,9-ene. While it may look like a perfect Hückel-aromatic molecule, it does not adopt a planar structure due to ‘inverse Baeyer strain’ (the flat all-cis conformation would have $144^\circ$ angles, far away from the theoretical $120^\circ$ $\mathrm{sp^2}$ angles) or steric hindrance (the (1E,3Z,5E,7Z,9Z)-confomation is not flat due to steric strain of the internal hydrogens). A similar case can be made for (1E,3Z,5E,7E,9Z,11E)-cyclotetradeca-1,3,5,7,9,11-hexaen-13-yne as can be seen in the figure below.

The hydrogens coloured in red would overlap far too much to create a stable molecule. It will adopt a non-planar configuration to relieve this steric stress which is a greater source of destabilisation than giving up an aromatic system. We can only speculate on the molecules structure until we actually saw it. Unfortunately, it is also unknown as far as SciFinder is concerned. However, two geometric isomers of dehydro[14]annulene (as the molecule is also called) have been synthesised and isolated:

Interstingly, these compounds that were discovered in the 1960’s[1] show aromaticity by their chemical shifts. The internal hydrogens are shielded and display a chemical shift of approximately $-1~\mathrm{ppm}$ (symmetric isomer) and $0$ and $-1~\mathrm{ppm}$ (unsymmetric isomer). Similarly, the outer hydrogens appear as a group in the aromatic region.[2] Cresp and Sondheimer used low-temperature NMR studies to assign the correct structures.[2] And in 2013, Santander et al. performed calculations to determine the mechanism of interconversion between the two (‘by allowing a dilute ether solution to stand without protection from daylight, or, more rapidly by irradiation of a refluxing benzene solution with a $100~\mathrm{W}$ lamp.’[2]), which turns out to be via an interesting (in my opinion) Hückel-bond shift.[3]
Let’s hope that Professor Dimm actually synthesises the compound and properly reports its structure.

References:
[1]: F. Sondheimer, Proc. Roy. Soc. A, 1967, 297, 173. Taken directly from reference [2].
[2]: T. M. Cresp, F. Sondheimer, Tetrahedron Lett. 1982, 23, 1731. DOI: 10.1016/S0040-4039(00)87202-0.
[3]: M. V. Santander, M. B. Pastor, J. N. Nelson, C. Castro, W. L. Karney, J. Org. Chem. 2013, 78, 2033. DOI: 10.1021/jo302072p.


text:
What are modern approaches to describe and/or quantify aromaticity?
I know that Schleyer described an approach for evaluating aromaticity using nucleus-independent chemical shifts (NICS, see e.g. Chem. Rev. 2005, 105 (10), 3842–3888) quite some time back. I also somehow have the feeling that there should be a more direct way to describe aromaticity than via chemical shifts.
What are modern approaches to describe/quantify aromaticity? Could anyone point out some links/references on recent developments in the field?

I recall reading somewhere that planarity and electron delocalization could be used to explain whether a compound was aromatic or not. However, aromaticity is not well understood.

There is a review on the topic: Structural Aspects of Aromaticity. Chem. Rev. 2001, 101 (5), 1385–1420.

Also take a look at: What is aromaticity? Pure Appl. Chem. 1996, 68 (2), 209–218.

Here's a really interesting discussion of aromaticity in nBuLi: http://www.ch.imperial.ac.uk/rzepa/blog/?p=9841

Actually, that entire issue of Chem. Rev. is devoted to aromaticity! Vol. 101, Issue 5

I'm more familiar with the Electron Localization Function approach to aromaticity commented by Quantum AMERICCINO, it's been proposed that the average value of the $\text{ELF}_{\sigma}$ and $\text{ELF}_{\pi}$ bifurcations would be used for a general scale of aromaticity of a molecule or molecular cluster. The points of bifurcation can be interpreted as a measure of the interaction among the different basins or, simply, electron delocalization. Although the ELF depends on the density (i.e. no separate information of $\pi$ or $\sigma$ bond), a topological analysis on a separated ELF formed only by the $\pi$ orbitals and other formed by the $\sigma$ orbitals can be done (also you can do a separation for the spin $\alpha$ and $\beta$ contributions to the ELF). [J. Chem. Theory Comput. 2005, 1 (1), 83–86.]
There is another approach based on how a bond responds when its electron density is perturbed by a finite amount. Specifically, they used the interaction coordinate (IC) to see this. Quoting

If the delocalization in the aromatic system is more extensive, for constraining a bond by unit displacement from the equilibrium geometry, it requires more energy. So the response of the other internal coordinates will be greater for constrained optimization and is proportional to the effective delocalization and hence related to the aromatic stabilization. The IC is a measure of the response of a bond (its electron density) for constrained optimization when another bond or angle is stretched by one unit (its electron density is perturbed by a measured amount).

Instead of taking the full molecule, they take only the aromatic skeleton to account the electron density perturbation on the system. Also, it says that this is very general yet, so the IC approach can take new directions.
[J. Phys. Chem. A 2016, 120 (18), 2894-2901.]
Another approach is making a relationship between the amount of the electron density (due to ring size, number of $\pi$ electrons, etc) and the non-nuclear magnetic shielding (cf. NICS).
[Phys. Chem. Chem. Phys. 2010, 12 (39), 12630–12637.]


text:
internal alkyne stability to strong alkali
Internal alkynes are relatively electron rich and are not usually prone to attack by strong (nucleophilic) bases such as hydroxide or alkoxide.
However, this isn't always the case.  A very nice overview of the most common reactions of alkynes says:

The sp-hybrid carbon atoms of the triple-bond render alkynes more electrophilic than similarly substituted alkenes. As a result, alkynes sometimes undergo addition reactions initiated by bonding to a nucleophile. This mode of reaction, illustrated below, is generally not displayed by alkenes, unless the double-bond is activated by electronegative substituents, e.g. $\ce{F2C=CF2}$, or by conjugation with an electron withdrawing group.
$\ce{HC≡CH   +   KOC2H5    —>[in C2H5OH at 150 ºC]   H2C=CH-OC2H5}$

I'm interested in more information about this and similar reactions, particular for internal (not acetylenic) alkynes.
Unfortunately, the linked page doesn't provide a reference for this reaction, and searching in Google Scholar for "internal alkyne" "potassium hydroxide" doesn't provide useful results.
Thus my questions are:

Does hydroxide ever react similarly with internal alkynes, at any temperature?  If so, under what conditions?
How long does the ethoxide reaction indicated above (at 150 °C) take?  Days?  Hours?  Minutes?  Would it be significantly different for an internal alkyne rather than acetylene?
In the case of a hydroxide reaction, would the ultimate product be a ketone (a.k.a. base catalyzed hydration)?

I can't read German, but this paper is probably relevant. From what I can tell, it is describing the reaction between acetylene, an alcohol/phenol (including ethanol), and KOH at elevated pressures to give a vinyl aryl or vinyl alkyl ether.

There are also multiple patents describing how the reaction can be carried out in industry, for example this one. Page 9 states that the reaction with MeOH (to form vinyl methyl ether) was run with KOMe catalyst at 45 psi, 145 deg C, for 40 hours giving a 96% yield.

And I did another search on Reaxys for vinyl ethers of the form $\ce{R^1HC=C(OR^2)R^3}$ but sadly no syntheses from internal alkynes turned up. I'll leave somebody else more qualified to try to rationalise it.

According to A Textbook of Organic Chemistry by V. K. Ahluwalia and Madhuri Goyal

1-Butyne may be isomerised to a more stable 2-butyne in presence of alcoholic potassium hydroxide and the reverse of this reaction may occur with sodamide

(see 17.2.1 of Best Synthetic Methods: Acetylenes, Allenes and Cumulenes for reaction conditions in the 2-alkyne to 1-alkyne direction, actually many relevant reactions are in this book)
So in the present of the strong base $\ce{NaNH2}$ the internal alkyne $\ce{H3CC#CCH3}$ rearranges to $\ce{HC#CCH2CH3}$
My thinking is that with just hydroxide the more stable 2-butyne is favored, while if an even-stronger base is used, that fact that the acetylenic-H can deprotonate pulls the equilibrium toward 1-butyne.
Additionally, see 17.2.11 Isomerisation of 1,4 bis(alkoxy)-2-butynes to the corresponding allenes, which is catalyzed by t-BuOK.
Add see table 17.1 "Base catalyzed isomerisations of acetylenic compounds" about 10 different internal acetylene reactions are listed in the table.

Thanks for this!  I'll probably accept & award in the next day or two (unless a better answer materializes).  I'm especially impressed w/ table 17.1.  It seems like alkynes can readily isomerize to allenes or other slightly more stable isomers under basic conditions.


text:
Can 25mL of 10% phosphoric acid substitute for 10mL of 25% solution?
Could I substitute $25~\mathrm{mL}$ of $10\%$ for $10~\mathrm{mL}$ of $25\%$ phosphoric acid solution?
This is for a measurement of free $\ce{SO2}$ in a sample of wine. The phosphoric acid is added to the wine sample and then more steps take it through a color change as it goes from acid to base. Finally a reagent is added dropwise to measure the amount of free $\ce{SO2}$ in the wine.
The full test is set out at More Wine.com as "Test for free SO2 in wine".

See my edits; if you disagree with any of them, feel free to change them back.  I was unable to find the webpage with the test; morewine.com is not even a live webpage. Can you post a link to the test webpage into your question?  Thanks!

Not reliably. Acid strengths measured in percentages are usually given as percent by weight, and if that's the case just swapping the volume numbers around will not provide you an accurate quantity of acid.
You first need to be sure of how the percentages are expressed.  Assuming they are percents by weight, you would then need to look up the density of phosphoric acid at the two percentages and work out the calculation so that you were adding the same number of moles of $\ce{H3PO4}$.


text:
Does oxygen gas form hydrogen bonds?
I learned that hydrogen bonding is formed from a 'sandwich' of hydrogen and either oxygen, nitrogen, or fluorine where nitrogen, fluorine, or oxygen is the bread.
Will oxygen gas and water form hydrogen bonds? The oxygen-hydrogen sandwich could be made, but oxygen gas has no hydrogen attached. Is it necessary for both molecules to contain hydrogen for hydrogen bonding to occur?
To make it clear, the question is: do both molecules have to contain hydrogen in order for hydrogen bonding to occur?

Does my answer actually answer your question? The specific case of whether $\ce{O2}$ engages in hydrogen bonding remains unanswered. Feel free to wait a few days before marking an answer as accepted -- many questions go several days before being marked as answered.

Yes, the O2 was just an example, I was just trying to find if any Hydrogenless molecule can undergo hydrogen bonding

Semantics - Oxygen gas doesn't form hydrogen bonds with water. Oxygen gas can dissolve in water and once dissolved it will form hydrogen bonds. But dissolved oxygen in water isn't oxygen gas.

To make it clear, the question is: do both molecules have to contain hydrogen in order for hydrogen bonding to occur?

No, as noted by Prakhar in his answer, tertiary amines are one class of compounds that can participate in hydrogen bonding (as an acceptor) when there is no hydrogen atom directly bound to the participating atom.
As an example, trimethylamine can hydrogen bond just fine with water:

More information can be found, e.g., here at ChemGuide (scroll down to "Solubility in water").

Yes both the molecules should have Hydrogen atoms for Hydrogen bonding to take place.
For instance there would be no hydrogen bonding between Water(H2O) and Oxygen(O2).
Since in oxygen both the atoms are equally electronegative therefore there is no partial charge developed.
Whereas in water a partial charge is developed due to higher electronegativity of Oxygen compared to Hydrogen. Thus Oxygen becomes partially negatively charged whereas the hydrogen becomes partially positively charged.
But water and oxygen wont form hydrogen bonds due to lack of partial charge separation in oxygen.

Here since both the oxygens are equally electronegative there is no charge separation.
NOTE:This is mostly true except for very rare cases such as hydrogen bonding between water and tertiary amines.

This is not strictly true -- the hydrogen bond acceptor atom does not need to have a hydrogen bonded to it. Consider tertiary amines, for example.

A hydrogen bond is the electrostaticattraction between two polar groups that occurs when a hydrogen (H) atom covalently bound to a highly electronegative atom such as nitrogen (N), oxygen (O), or fluorine (F) experiences the electrostatic field of another highly electronegative atom nearby.

This is what wikipedia has to say.

Ok... and what I said is fully consistent with Wikipedia.  It says nothing about a hydrogen needing to be bonded to the acceptor atom.

Sorry, edited the answer.

Secondary amines do have a hydrogen bonded to the nitrogen. It's only the tertiary amines that are a counterexample to your stated principle.

Sorry again!!

It's no problem - sorry to pester you about it, just want to help you write as precise an answer as possible. :-)

Well, problem is that you are wrong about need of partial negative charge on acceptor. Also H-bonding isn't limited those three elements.

Can anyone address the question of whether diatomic oxygen (either singlet or triplet) has been shown to be a hydrogen bond acceptor? I would predict that it may be one.

@Mithoron What acceptor atoms are there that bear a positive charge?

Hydrogen bonding is limited to only Nitrogen, Oxygen and Fluorine. And yeah @Mithoron give an example where the acceptor atom bears a positive partial charge.

http://chemistry.stackexchange.com/a/21800/9961


text:
CN* band spectrum in spark emission spectroscopy
An arc between carbon electrodes in air has enough energy to activate the nitrogen in the air.  The nitrogen then behaves as nitrogen in the combining form and by reaction with the carbon from the electrodes, the CN bands are produced. To prevent this, special arrangements, such as using an inert gas  to remove adsorbed nitrogen from the surface of the electrodes, have to be made.
In our experiment, no particular precautions were made, and nevertheless CN bands were not found.
Why?

Does "The nitrogen then behaves as nitrogen" mean "The $\ce{N2}$ then behaves as atomic nitrogen" or something similar?


text:
Material that acts as a "sponge" for sulfuric acid?
I would like help with a materials problem I am looking into that pertains to a product I would like to develop.
What I seek is some kind of substrate that acts like blotter paper does. Ideally, the substrate will be a thin material which can be soaked in sulfuric acid and hold a small amount "suspended" in it without a reaction taking place.
I have looked into which plastics are best compatible with sulfuric acid (HDPE and PTFE) but was unable to find any information about "sponge-like" forms of these plastics.

I think glass wool would work.


text:
Is glycine achiral?
In this video (10:20>) the woman says that the Glycine is not chiral because it has 2 different group only rather than four.
Then I can understand from her things that amino acid can be chiral just in case that it has 4 different groups. But I don't understand why when even it has only two different groups it can not be chiral, here is the group of the left side can be on the right and vice versa and this is an a mirror image as well.

There is a mirror plane that contains the central carbon and bisects the angle between the two hydrogen substituents.

@Jan I don't think this is a duplicate. The answer may be similar, but the question is quite different.

Well, my reasoning is that I can see all of OP’s question being asked in the potential dupe-target (while that asks more).

Due two down-votes on my questions and the attitude that I experienced here as a new member here I have the feeling that this site is for phD students rather than for beginners.  I thought it is a site for studding and asking questions chemistry in any level.

It is not chiral because If the amino group and the carboxylic acid group changed places, the "new" arrangement would be superimposable on the old arrangement.


text:
Magnesium powder + sodium carbonate solution: what happens?
I put magnesium powder into a sodium carbonate solution and it fizzed very slowly. The gas produced put out a lit splint immediately, which implies that carbon dioxide was released, but I cannot work out the equation for this reaction.
I'm fine with an acid plus a carbonate or a metal salt displacing another but this was just metal powder plus a carbonate.


text:
Are the total number of electrons present in the central atom of triiodide equal to 10?
My book states that

the total number of electrons present in the central atom of $\ce{I_3^-}$ is equal to 10.

What if I put a negative charge on the central iodine atom such that it gains a noble configuration and then I say that it is coordinatively bonded with other two iodine atoms? Then the number of electrons which the central atom will have would be 8. Wouldn't it?
Am I wrong somewhere in my analogy?

Maybe you could specify the title of "the book", that could be helpful ; )

Looking at the wikipedia article here, your explanation seems to be exactly right:

The ion is linear and symmetrical. According to VSEPR theory, the central iodine atom has three equatorial lone pairs, and the terminal iodines are bonded axially in a linear fashion, due to the three lone pairs bonding to the central iodine-atom. In the molecular orbital model, a common explanation for the hypervalent bonding on the central iodine involves a three-center four-electron bond.

Thus, the central iodine atom has $3\cdot 2$ valence electrons from the lone pairs and 3 three-center four-electron bonds, so that I speculate you end up with $3\cdot 2$ VE + $4/3 \cdot 3$ VE = 6 VE + 4 VE = 10 VE. Which also fits to the reference given by you.


text:
Why are these ions classified as isoelectronic?
According to Chemistry the Central Science 13 Edition, these ions are isoelectronic:
$\ce{Ca^2+}$ (18 electrons); $\ce{Cs+}$ (54 electrons);  $\ce{Y^3+}$ (36 electrons)
Furthermore, the textbook states that the caesium ion is the largest.
Now, every definition I've looked up, including from the book, states that in an isoelectronic series, the atoms have the same number of electrons. Clearly, these don't. I have noticed, however, that they are in multiples of 18, where the Calcium ion is 18, and Caesium is 3 times 18, and Yttrium is 18 times 2.
A friend, who took chemistry 1 a few years ago, recalls that there was some formula (or something) to determine isoelectronic series.
I suspect that they have something to do with the electrons in their valence.
The book does not give an explanation. It moves on to ionization energy.
But it also states that of these three, caesium is the largest, which compounds my confusion since just a few lines up, the book states that the nuclear charge is inversely proportional to the ionic radius, so an increasing nuclear charge (atomic number) means a decreasing atomic radius. It would seem appropriate that the calcium ion has the largest ionic radius since it has the smallest nuclear charge.
Can anyone help?

Core Electrons don't count. It's the valence electrons that are counted (outermost shell). The valence shell of all 3 is empty. So they all have 0 electrons in their valence shell.

So isoelectronic is defined solely as atoms having the same number of electrons in its valence shell with the exclusion of core electrons?

@DannyRodriguez yeah

1) They all have the configuration of a noble gas with a full P sublevel as the highest sublevel.
2) Nuclear charge is inversely proportional to anion radius within a given period.


text:
Toxicity of metallic lead (Pb)
I have some pieces of lead ($\ce{Pb}$) at home. I keep them in a box in the closet and never touch them.

Can it be harmful?
Should I throw them away?

I would like to keep them, but I am not certain about their toxicity.

Related: http://chemistry.stackexchange.com/questions/16054/toxicity-of-a-sheet-of-lead

Yes they can be harmful if you lick them, or if you handle them and then handle your food. Otherwise, feel free to keep them. There are no toxic gases, that's for sure.

Are children around?

@mcocdawc No, we don't have any children at home. I've packed them in a box and are out of reach, I've found them at a mine, and they are really beautiful and metallic, there isn't any dust and they are like small clods.

@Aref You packed your children in a box? ;-)

Metallic lead is very low risk.
Lead compounds are fairly poisonous: they slowly build up in the body and cause many harmful effects. But lead metal is very inert and you would need to do something fairly risky with it to create much likleihood of generating dangerous lead compounds.
Don't eat it, for example. Don't put it in contact with food (especially acidic foods which can solubilise lead). But normal handling should not be a problem.
Lead is so inert that it has been used to pipe clean water supplies to people for most of history (or at least the parts of history where somebody had access to clean water). This is safe unless the water supply is particularly acidic and leaches lead from the pipes.
In short: don't worry about lumps of lead.

Good answer.  Any metallic looking lead is lower risk (not zero but way down there).  What you need to be careful of is any powdery surface coatings that appear.  These may be white, yellow, red or other colours.  These are various lead salts and some are more soluble and more toxic than others.  Moisture and other reactive substances in contact are the biggest cause, damp wood, vinegar or other chemicals should be kept away.  Kept dry in plastic containers should pose little risk.

As a letterpress operator, I'd add "Avoid breathing lead dust, of any color." As an electronics tech I'd also add "or breathing more fumes of molten lead than you must."

And remember the problems with lead in Flint's water pipes was due to some acidic water, they had been fine before they switched water supplies.

I used to handle lead every day without gloves -- solder spatters, lead shot for casting, handling those cast lead pieces, cutting and installing lead foil for shielding scientific instruments, etc. A few years ago I took a bit of time to evaluate my work area and habits. I was concerned that my habits were getting careless. I decided to clean things up and make an effort to not be so careless. I also had my blood tested, so I something to measure. I went to my doctor and told him my concerns. He sent me to have a blood test for heavy metals. The results came back showing my levels to be quite low (If I recall correctly, they tested specifically for lead and mercury and cadmium, and some others, I think, but don't recall). At any rate, blood levels don't necessarily reflect the total amount of heavy metals carried by the body, but given that the tests were taken when my exposure was at a peak, and given that the results came back so low it seems that I didn't have anything to worry about.
That's just my story. I don't handle lead quite as often now, but I still handle it pretty regularly. I am not sure why you have any concern about the lead you handle. It sounds like you have nothing to worry about.


text:
Why (abundance of M+1)/(abundance of M)=(number of carbon atoms)*(1.1/100)?
In mass spectrometry, why is this equation true?
$\frac{A_{\mathrm{M1}}}{A_{\mathrm{M}0}} = n_\ce{C}\;\;\times\;\;
\frac{1.1}{100}$
Here, $A_{\mathrm{M1}}$ is the abundance of the M+1 peak and $A_{\mathrm{M}0}$ is the abundance of M.
I do not understand the logic behind the equation. Let's say that in mass spectrometer I got $A_{\mathrm{M1}}=1\%$ and $A_{\mathrm{M}0}=99\%$ for methane.
Now, if I use the equation I get, $(1/99)=1 \times(1.1)/100$.  The right hand side does not exactly match the left hand side, but maybe the answer is correct enough if I round it up?
Can anyone explain to me how the equation derived and the logic behind it?

Why the down votes? This is actually a decent question.

@Zhe in its initial form it was very difficult to understand. I retracted my -1.

Let's look at where these numbers come from. For carbon, the main isotopes are carbon-12 and carbon-13. Carbon-13 has an abundance of 1.109%.
Consider a molecule with $n$ carbons in it. The ratio of the $M_{1}$ peak to the $M_{0}$ peak is related to this above fraction, but you have to account for the fact that there are $n$ ways to create this molecule with one carbon as carbon-13 and the rest as carbon-12. This is basically one of the binomial cofficients:
$$\binom{n}{1} = n$$
You can take the total distribution of $p + (1-p)$, where $p$ is abundance of carbon-13. If you have $n$ carbons, we have:
$$\left(p + (1-p)\right)^{n}=(1-p)^{n} + \binom{n}{1}(1-p)^{n-1}p+\ldots$$
Each of $n+1$ terms corresponds to the relative height of one of your MS peaks. The ratio between peak 0 and peak 1 is:
$$\frac{\binom{n}{1}(1-p)^{n-1}p}{(1-p)^{n}}=\frac{np}{1-p}$$
Since $p=0.01109$, this expression comes to approximately:
$$n\cdot\frac{1.1}{100}$$
This means if you have $n$, you can estimate the height of the first peak relatively to the zero-th order peak. Technically, given the method I provided, you can estimate an higher order peak.
Order, if you have the base peak and the first peak, you can estimate $n$. This isn't supposed tobe perfect if you're using 2 significant figures, but your measurement of the MS peak intensity is probably similar in terms of error, but you should find that even in your example, $1/99$ is a lot closer to $1.1/100$ than $n\cdot (1.1/100)$ for values of $n$ other than 1.

(p+(1−p))^n, why it is there?

The is the probability distribution for $n$ carbons. Notice that this value is 1 because $p + 1 - p = 1$, and $1^{n}=1$. But this expansion tells us the exact distribution for all combinations of $i$ out of the $n$ carbons being carbon-13 where $i$ is the exponent on the $(1-p)$ term.

Why can't the formula be 1.1*n/100=abundance of M1 in percent?

Try $n=1000$ and that clearly doesn't work.

Actually, shouldn't it be 98.9, instead of 100?

I have asked a related question recently. Perhaps, you may know how to obtain the answer for it: https://chemistry.stackexchange.com/questions/87654/what-is-the-relative-size-of-the-m2-peak

@TanYongBoon Yes, but rounding there is perfectly fine. It leads to a roughly 1% error, which you won't notice in the peak intensity.


text:
In Simonini reaction how does the $2:1$ ratio of reagents enable formation of ester as major product?

The Wikipedia article on Simonini reaction says:

The ratio of the reagents play an
  important role in the determination of products, namely, if $1:1$
  ratio of salt and iodine is used alkyl iodide is formed. On the other
  hand, a $2:1$ ratio gives $\ce{RCOOR}$ and $3:2$ ratio gives both the
  products.

For $1:1$ ratio I understood why alkyl iodide is formed (found the mechanism here). But for $2:1$ ratio why is ester formed as major product ?

This is the Wikipedia article from which I have taken the quoted lines and the image  in the question: https://en.wikipedia.org/wiki/Hunsdiecker_reaction#Simonini_reaction

If you have the extra equivalent of the carboxylate, then this carboxylate will never react with iodine to form the alkyl iodide. Instead, it will do the next best thing, which is $S_{\mathrm{N}}2$ reaction with the alkyl iodide that you did make.
Note that this probably only works if R is primary.


text:
How does reactions like these proceed?
I encountered this question in the chapter Amines and trust me I've gone through many textbooks but couldn't find the functioning of those reagents, well some of them.

Firstly could you explain what does the 3 reagents under the same arrow signify? Also, can you explain the mechanism step by step?
I'd be greatful.

You could not find out the functioning of the reagents? Come on $\ce{P_4O_{10}}$ removes water, ammonia acts as a base/nucleophile, KOH is a base, Ph-Cl is used as arylating agent and aq. KOH for hydrolysis. That is very standard and really not that difficult. What textbooks where you looking at?

Step 1

Your starting material is o-phthalic acid.
Phosphorous pentoxide, $\ce{P4O10}$, is a strong dehydration agent.
The $\Delta$ symbol in your original image means that heat is applied
What happens here?

 The intramolecular reaction of the two $\ce{-COOH}$ groups yields phthalic anhydride.

Step 2

Carboxylic acids can react with amines to form amides, but what happens here?

 Anhydrides react with amines to form imides.
 Here, the reaction of phthalic anhydride with ammonia yields phthalimide.

Step 3

The product of the previous reaction is first reacted with potassium hydroxide, which is a strong base. What could possible happen here?

The $\ce{N-H}$ proton of an imide can be removed easily. Draw some resonance structures of the anion to see why this is the case!

Chlorobenzene ($\ce{Ph-Cl}$) is added.

 This reaction will furnish N-phenyl phthalimide. Read about nucleophilic aromatic substitutions to understand what is going on here.

The product of the previous step is treated with aqueous potassium hydroxide.

 Read about the hydrolysis of esters, amides, and imides to understand what has happened.

So it's a Gabriel synthesis.

@ketbra Yes, as far as the reaction of the imide is concerned. But I tried to keep it simple and give some more general hints instead ;-)

Yeah it is Gabriel synthesis... Correct

I would've thought the nucleophilic addition wouldn't have been that great...

@Zhe I did prepare a lot of phthalimides, but I never used aryl halides. Typically, I reacted phthalic anhydrides with alkyl or arylamines.

Great answer sir, can you please simplify it and generalise what every reagent is doing and what it's supposed to do otherwise in other reactions?


text:
Water condensation above dew point
Anyone who stores rust-prone instruments knows that even in temperature-controlled air spaces rust can form on uncoated steel.  Since rust requires water, this suggests that water will condense out of air onto surfaces even above its dewpoint.
How is surface condensation of water above the dew point characterized and quantified?
Or is "condensation" not the right term for this?  There does seem to be a non-linear relationship between relative humidity and the observed effects of that airborne moisture on exposed solids.

Water vapour is always in the air and so water molecules will come into contact with any and all surfaces. These will form your rust. You will have to keep the humidity in your room very , very low to prevent this; or rather to make the reaction rate so low that it can be ignored. Temperature controlled does not necessarily mean humidity controlled.

Right, and what I want to see is the relationship between relative humidity and this "reactive contact rate."  Note that temperature controlled in this case means that relative humidity never exceeds 65%, and that no tools or surfaces are ever colder than their surrounding air.

At 65% relative humidity approx 1% by weight of the air is water vapour; this is a lot. Even if surface is not colder than air the water vapour is still coming into contact with it just as the air does. One way to keep the surface clean is obviously to to remove all water vapour, another way   is to heat it so that there is thermal gradient between it and the surroundings. I have kept hydroscopic crystals perfect in this way for years.

@porphyrin , isn't it necessary to have at least a small liquid layer present for some of the steps in the oxidation mechanism, like $\ce{4 Fe^{2+} + O2 → 4 Fe^{3+} + 2 O^{2−}}$? (This Wikipedia article)  I do suspect that you can get enough of an adsorbed layer of water for this to happen at relative humidities below 100% though.

@airhuff which is why heating is probably best if all the water cannot be removed.

@porphyrin, agreed.

@porphyrin: This isn't about countermeasures to rusting.  This question is about vapor/liquid dynamics in non-condensing but humid environments.  For example: If a "liquid layer" is necessary for iron oxidation, then if rust forms in non-condensing conditions that means that there is a mechanism for a "liquid layer" to form in non-condensing conditions.  What is that "liquid layer?"  By what mechanism does it form and evolve?  Of what is its formation and evolution a function?

I think that 'liquid layer' is too strong a statement, water molecules will form a monolayer to fully or partly cover any surface in proportion to the strength of any physisorbtion interaction, which is primarily van der Waals in nature.  In doing so they will be in equilibrium with water molecules in the air. (At high humidity multiple layers may in principle form). Rusting seems to rely on having water molecules present so will be greatest where their concentration is also greatest.


text:
Is phosphoric acid to be considered a hazardous waste?
I've found some English sites proclaiming phosphoric acid to be used for porcelain cleaning and e.g. clogged drains, even though I mainly bought it to use to clean coffee makers and water kettles.
In my country (Finland) phosphoric acid can be found from the list of hazardous waste. However, there are contradictory views regarding whether phosphoric acid can be poured in to the sewer or not. If it's a hazardous waste, then it shouldn't. But I have this English resources saying that it can.
Should I abide by the local rules saying that it's a hazardous waste or is there a reason to believe it can be used safely e.g. in clogged drains or for cleaning the toilet?

Or you could pour it onto your garden as it is an essential element ( phosphorus) for plants. I pour TSP ( tri sodium phosphate ) in my garden ; my soil is too acidic to use phosphoric acid.

Phosphoric acid is a "food" additive because its acidity prevents the growth of microorganisms.  By itself it can cause burns, and high levels of phosphate in the diet have been linked to weaker bones (greater risk of fractures).
If you pour it down the drain, its acidity should not be the issue.  You could neutralize it with baking soda, and it will tend to become neutralized and buffered in nature on its own, but I think the primary issue is that it will yield phosphate ions which act as a fertilizer/nutrient source for algae growth and can cause algal blooms in ponds and streams that can dominate the ecosystem and choke out plant and animal diversity.
For that reason, don't pour it down the drain even though it will likely become neutralized as it acts on the clog.
If you just need to manage it, neutralize it with baking soda, and boil off the water, or let it evaporate to yield sodium phospate salt.  We don't want the sodium phosphate to enter directly into the local watershed, but you could put it on houseplants.  I don't know if phosphate fertilizer is allowed where you live, but if it is, you could put it on your lawn.  It PROBABLY would also be OK to dispose of the solid Sodium Phosphate in the trash.

I think, it depends to a strong extent what the wastewater treatment plant does to the water; i.e. if it constantly monitors phosphate to precipitate it or not.

I know that In some cases, wastewater goes straight to a septic lagoon or sea without any treatment.


text:
How do acids help 'solubilise' compounds?
I've heard that root hair cells in plants secrete acids to help 'solubilise minerals' so that they dissociate and are in ion form... this supposedly allows them to be absorbed.
How so can acids solubilise certain compounds? It must be noted: I am a student and my solubility knowledge is fairly basic. I understand that ionic compounds dissolve in water because of the water molecule and the compound's polarity, but it goes no further than that.
Thank you in advance.

There are multiple ways in which plant root and root hair cells increase the availability of nutrients and positively affect the plant’s ability to absorb nutrients from the soil.
One of these ways is by excretion of $\ce{H+}$ ions from malaic acid. This promotes a cation exchange process that results in cationic nutrients being brought into the cell in order to maintain charge balance.  This is the process to which you are referring in your question.
Additionally, enzymes, for example phosphotases, can convert chemically unavailable nutrients, like organic phosphorus, into bioavailable inorganic phosphates.  Phosphate uptake is a case in which acidic soils actually inhibit the availability of this nutrient to plants. (Annals of Botany, 112: 317 – 330, 2013)
There are also proteins excreted by root hairs that can complex metal ion nutrients like iron, facilitating the uptake of such nutrients.


text:
Why non polar liquids wet non polar surfaces better than polar surfaces?
In semiconductor industry adhesion promoters (such as HMDS) are applied on various surfaces to improve adhesion between the surface and the photoresist:

On SiO2 and many metals surfaces, the air humidity forms polaric
  OH-bonds. Such substrates are hydrophilic, but reveal a bad wetting
  and adhesion of the unpolaric resist. Adhesion promoters make
  hydrophilic surfaces hydrophobic thus improving the resist adhesion.
  [...] The  methyl  groups  of  the  HMDS  fragment  hereby  form  a
  hydrophobic  surface thus improving resist wetting and adhesion.
  (source)

But why should non polar (hydrophobic) substances wet non polar substrates better than polar substrates? (in the opposite case of polar substances it is clear why)

You mean if the polar substance can't move to get away from the nonpolar liquid, what should keep the latter from forcing itself on the former? A static dipole should be able to induce a dipole moment in the hydrophobic, and attraction should follow?


text:
Does electron shielding increase or stay constant moving LEFT to RIGHT across a period?
Does electron shielding increase or stay constant moving left to right across a period?
I have read about both, and I just want to know which one is right.

I believe that electron shielding remains constant because when you move across a period, you are essentially adding more valence electrons, not shielding electrons, in your valence shell.    Therefore, your
valence-electron-count increases from left to right in a period, but your shielding-electron-count stays the same.
So the overall shielding remains constant, just the amount of valence electrons increases

However, my explanation has a flaw:
It does not support the reason why an atom's radius increases when it becomes an anion.
When an atom becomes an anion, it adds more valence electrons to its outermost shell. Now this is controversial to what I said previously, because a gain in valence electrons, but not shielding electrons, wouldn't increase the electron shielding of the atom. So the atomic radius wouldn't increase either!
The only way for an atom's radius to increase when it becomes an anion is if the electrons that are added to the valence shell have some sort of effect on the shielding, right?
This supports the first thought in my question that if we move left to right across a period, shielding increases.

But the idea that electron shielding increases as we move left to right has a flaw as well:
It is controversial to another periodic trend: ionization energy
If electron shielding did increase across a period from left to right, then this would mean that ionization energy is lower as we move from left to right, which is not true.
The only way for ionization energy to increase across a period is if the  only the number of protons and valence electrons have effect on the amount of energy required, not the shielding electrons. So the shielding-effect must stay the same all throughout a period.
This goes with the what I had initially said about electron-shielding remaining constant.

So now we have a flaw for each possible answer to my question!
Please correct me if I am wrong in any of my context.

Valence electrons repel each other and this is independent of shielding and nuclear charge.

In terms of ionization account for the fact that generally more ionizations occur to form the anion moving from left to right.

And 3) as for ionization energy rising with the same degree of electron shielding, account for the fact that the nuclear charge is greater so the core charge is also greater moving from left to right.

@JosephHirsch so does electron shielding increase or stay constant across a period?

@JosephHirsch And also, how does valence electrons repelling each other affect anions and make them larger than their neutral atom?

Electron shielding (as can be seen in the rise in TRUE net effective charge of less than 1 per additional proton) rises from left to right.
This MIGHT be due to repulsive forces between valance electrons.

Regarding the previous comment, I suppose it also is possible that true the net effective charge is closer to the hypothetical net effective charge (protons-core electrons) as you move from left to right not because shielding is greater, but because there are more neutrons in the nucleus.

To answer this question, it's important to define what you mean by shielding. Generally, shielding refers to a reduction in the effective nuclear charge experienced by an electron in a given orbital due to the other electrons on the same atom.
The quantitative degree of shielding for a given electron can be approximated by Slater's rules. According to those rules, electrons within the same group of orbitals (for example 4d) have a coefficient of 0.35 (except for 1s electrons, which have 0.30). So valence electrons do shield each other, just not as much as the lower level electrons shield the valence electrons.
For example, let's consider the elements with increasing numbers of 2p electrons (B, C, N, O, F, Ne). Going from left to right, each addition of a 2p electron reduces the effective nuclear charge experienced by another 2p electron by 0.35. So the amount of shielding is increasing as we move left to right.
The apparent contradiction with the ionization energy comes about because you have not considered the increase in the actual nuclear charge. Each time we add a 2p electron, we also add a proton to the nucleus. So even though 0.35 of that charge is shielded from a valence 2p electron, that electron still experiences an increase of 0.65 effective charge and is therefore bound more tightly than a 2p electron on the element to its left.
As a specific case, let's compare the shielding experienced by a 2p electron in carbon versus neon. The carbon electron configuration is $1s^22s^22p^2$. Using Slater's rules, we can calculate that the shielding coefficient for a 2p electron is 2(0.85) + 3(0.35) = 2.75. So the effective nuclear charge experienced by the 2p electron is 6-2.75=3.25.
For neon, the electron config is $1s^22s^22p^6$, so the shielding coefficient of a 2p electron is 2(0.85)+7(0.35)=4.15. So there's more shielding. But the effective charge is 10-4.15=5.85, so the effective charge is still greater and the energy of the coulombic interacation of the electron with the nuclear charge is greater in magnitude. Hence, the ionization energy is greater.
EDIT: Original version of this answer did not address the question about anion formation (presumably mostly interested in anion formation by addition to an already partially filled shell). When an anion is formed, the additional electron increases the shielding of the other valence electrons, but there is no change in the actual nuclear charge. Thus, each valence electron experiences a lower effective nuclear charge. Since the size of an orbital is a function of the energy of interaction with the nuclear charge, the atomic size increases.

Hopefully by explaining the 'flaws' may shed some light on your question.

It does not support the reason why an atom's radius increases when it becomes an anion.

The valence shell is not only comprised of 1 single orbital position of a fixed radius regardless of the number of electrons, the different orbitals with different geometries fill up as we move right across a period, and the more orbitals, there is more repulsion and range of radii for the orbitals, increasing the atomic radius. These orbitals are governed by the wavefunctions of these orbitals, which can be expressed in terms of the radius(as you usually use polar coordinates anyways) which increases as the number of electrons increase.
For anions it is the same, as the electrons occupy orbitals and pair repulsion and orbital repulsion increases.
But the idea that electron shielding increases as we move left to right has a flaw as well:It is controversial to another periodic trend: ionization energy
Lets take an example:
Ionization energy of neon is higher than that of fluorine, which is one to the left across the same period, this is much because the valence shell with a full S and p orbitals has a higher stability due to something caused by symmetric repulsion (which is partly why palladium has a full d subshell despite the s orbitals of lower energy being available)and thus there will require more energy to ionize the atom as it moves right across a period


text:
Nonflammable epoxy remover that won't chip paint
I was wondering if anyone knew of a nonflammable epoxy remover that also won't chip paint. My professor wants to use it in our lab to remove dried epoxy from the floors and the fume hood.

I find it a bit bizarre that a chemistry professor needs to outsource this query to someone. VTC.

I recommend elbow grease. Very low flammability.


text:
Are there any acid-base indicators whose protonated and deprotonated forms have the same number of resonance structures?
This question asks about the color change of 7-hydroxyphenoxazone, the active compound in litmus paper, and the relation of that color change to the conjugated system in the molecule.

In Klaus Warzecha's answer, he mentions 4-nitrophenol as another, simpler indicating system, analogous to 7-hydroxyphenoxazone.

In both of these species, as noted by Klaus, deprotonation of the hydroxyl group to an -olate moiety enables a resonance between the substituents, through the molecular cores, which is not possible in the protonated species:

I'm interested in knowing whether this property of an "opened-up resonance on deprotonation" and the concomitant influence on the excitation spectrum are necessary characteristics for indicating compounds.
So: Are there any known examples of acid-base indicators (e.g., compounds exhibiting a change in visible-spectrum absorbance upon deprotonation) where the removal of the proton does not "open up" additional resonance structures in this fashion?
Clicking through some of the links on Wikipedia's $\mathrm{pH}$ indicators page, I would guess that cresol red, bromocresol green, and bromophenol blue potentially fall into this category, but I'm not confident enough in my evaluation of resonance structures to be certain.

In contrast to 4-nitrophenol, cresol red, bromocresol green, and bromophenol undergo structural changes upon deportonation because the sultone opens up.
But again, you can draw quinoid resonance structures for the resulting anions.

In the examples you drew, these resonance structures are not ‘opened up’; they are there in both the protonated and the deprotonated form. However, since the protonated form requires charge separation to access these resonance structures, they are less important.
It rarely makes sense to do resonance structure bookkeeping like you are trying to implement in your question.
Indicators need to display a colour change upon protonation/deprotonation. that can either happen by modifying the structure (e.g. bromocresol green) or by the electronic changes induced by the charge occurring. The latter case will mean some shifting in the orbital energies — most importantly in HOMO and LUMO — and can practically always be explained in a ‘resonance way’.

In the latter case, electronic changes induced by the charge occuring, you say the change in HOMO-LUMO transition can be explained in "a resonance way". What do you mean? I have problems explaining the colour change when there is not a blatant change in extent of conjugation upon deprotonation.

@Adroit Best ask a question.

http://chemistry.stackexchange.com/questions/61740/can-the-colour-change-in-litmus-paper-be-explained-by-conjugated-systems


text:
What nuclei need to be NMR active in an HMBC experiment?
I recently had a discussion where I was told that in an $\ce{^1H}/\ce{^13C}$-HMBC experiment coupling through an $\ce{C-O-C}$ ether bond would not be observed with the reasoning that O is not NMR active. That somehow surprised me as I thought only the "end nodes" in the coupling needed to be NMR active. So I have the following two questions:

Do the nuclei in-between need to be NMR-active?
Does the end node X need to be NMR-active?

It is incorrect to say that oxygen block the HMBC coupling because it be magnetically inactive.
The theoretical explanation is that the coupling of spins $^nJ_{\ce{XY}}$ is mediated by the electrons i.e. transferred completely independently of the nuclei on the way. And I can provide you with multiple examples that show how coupling crosses over magnetically inactive nuclei:

I myself in my bachelor’s thesis have distinguished the two different ortho-nitrobenzyl ethers of 1-(4-bromophenyl)-1,2-ethandiol by HMBC coupling. It was always possible to locate a coupling from one side’s hydrogen to the other side’s carbon in HMBC in spite of the magnetically inactive oxygen in-between.
During my Ph.D. thesis, I performed a number of $\ce{^29Si}$-HMBC experiments to determine which hydroxy groups were protected by the silyl group used. Again, this transferred across a magnetically inactive oxygen.
Remember that $\ce{^12C}$ itself is magnetically inactive. In your standard $^3J_{\ce{CH}}$ 3-bond-HMBC-experiment, the coupling and thus the cross-peak is transferred along $\ce{H-C-C-^13C}$. I didn’t spell it out explicitly but considering each carbon has only a $1~\%$ chance of being $\ce{^13C}$, it’s much more likely for the chain to be $\ce{H-^12C-^12C-^13C}$, i.e. involving only two magnetically active nuclei.
Finally, your standard hydrogen spectrum exhibits a lot of $^3J_{\ce{HH}}$ coupling. Again, this crosses over two carbon nuclei; the probability of both carbons being $\ce{^13C}$ is $0.01~\%$. If only magnetically active nuclei were able to transfer coupling, then the uncoupled hydrogen signals should amount to $99.99~\%$ of a signal with the coupling signal having only $0.01~\%$ of the intensity. Clearly not what is being observed. In fact, $98~\%$ of the coupled hydrogen signal is due to molecules without magnetically active carbons between the vicinal hydrogens.

The final node has to be NMR-active, otherwise you wouldn’t be able to observe magnetisation.

I see, nice examples. I also observed HMBC coupling over inactive nuclei. So this basically means that J coupling does not need NMR-active nuclei. Right?

@ketbra Beginning and end must be active, or else you wouldn’t observe anything. Nuclei in between do not need to be active. The explanation is that the magnetic spin information is transferred by electrons.

As Jan as demonstrated, observation of long range HMBC couplings across oxygen is commonplace.
The key factor that dictates the appearance of coupling across multiple bonds the correct alignment of intervening bonds between the two coupled nuclei. The two end nuclei obviously need to be NMR active to be coupled to each other. The intervening nuclei do not.
Coupling arises from the favourable alignment of the σ and π orbitals of the bond pathway between the two nuclei. If this overlap of bonds is favourable, a couplng will be seen. For saturated systems, this long range coupling is usually confined to a well constrained W-coupling. It is also why long range coupling is most commonly observed in highly constrained cyclic systems.
And, although just 0.04% naturally abundant, oxygen-17 is actually NMR active.


text:
How to make alkaline iodide azide
I'm an IB Chemistry student and I'm doing the Winkler's titration for my Internally Assesed Investigation. However, my school doesn't have alkaline iodide azide, so I need to make that myself first. How do you do that?

See: "A Laboratory Manual for Environmental Chemistry" and" Introduction to Environmental Sciences"
They are both in google books. As well as their procedure.
Make sure that you are doing this under supervision. A quick look to the MSDS of sodium azide shows that it can explode.

You might need a blast shield to do this. It's not recommended at the IB level...

I'm quite sure you meant alkaline iodide sodium azide solution. The preaparation of this reagent is given here:

To prepare this reagent, take 700 g of potassium hydroxide/500 g of
  sodium hydroxide and add 150 g of potassium iodide/ 135 g of sodium
  iodide and dissolve in freshly boiled and cooled water and make up tp
  1000 ml.
Dissolve 10 g of sodium azide in 40 ml distilled water and add this
  solution with continuous stirring to the cool alkaline iodide prepared
  previously.

The link you sent in your answer sometimes uses alkaline iodide azide and sometimes alkaline iodide sodium azide. I can't tell if they're using them interchangeably or if they are referring to two different things.

@ILoveJesus There is nothing called alkaline iodide azide. It is  alkaline iodide sodium azide.

the cool alkaline(NaOH/KOH) iodide(NaI/KI) solution they are referring is the 1 litre solution prepared previously.

@NilayGosh  I calculated that the molar ratio between the Iodide and the Hydroxide is 1:1. And then I calculated that the ratio of the masses between NaI and NaOH should be 3.75:1. However, in the thing you sent me and in other labs, they always take more NaOH than NaI, and it's actually with the same ratio, just inverted. Could you explain this to me please?