text:
IUPAC name for 1,2,3-trichlorocyclopropane?
What is the IUPAC name for the two isomers of 1,2,3-trichlorocyclopropane? Similarly what is the name for isomers of 1,2,3,4-tetrachlorocyclobutane, 1,2,3,4,5-pentachlorocyclopentane, etc?
1,2,3-trichlorocyclopropane has two isomers, one of them with all the chlorine atoms facing in one direction and the other with one chlorine facing the opposite direction of the other two. In CIP rule, if two substituents on an atom differ only in their absolute configuration, R takes priority over S. However I have difficulty applying the rule here: two of the substituents are exactly the same. Nevertheless the two molecules are clearly distinct.
In practice the problem could be circumvented by using trans- and cis- prefix. But it does not address the key problem and fails on longer chains --- 1,2,3,4-tetrachlorocyclobutane has 4 possible arrangements, though not necessarily all of them have the problem.

Here are the isomers of cyclobutane and their mirror images. I use + and - to denote a chlorine atom above or below the plane. The number in the parenthesis is the number of distinct molecules.
+ + | - -    (1)
+ + | - -

+ - | - +    (1)
+ + | + +

+ - | - +    (1)
+ - | - +

+ - | - +    (1)
- + | + -

For the second set of stereochemistry, there is only one distinct compound. Rotate one of them 90 deg and it matches the other. Also note that all of these compounds have planes of symmetry, so none are chiral.

@jerepierre thanks for pointing out and particularly for giving insights on the $\sigma$ symmetry.

For a related question, see: Number of diastereomer pairs of 1,3-dichloro-1,2,3-diphenylpropane

Achiral cyclic compounds like 1,2,3-trichlorocyclopropane may contain pseudoasymmetric centers.
Pseudoasymmetric centers have distinguishable ligands (“a”, “b”, “c”, “d”), two of which are nonsuperposable mirror images of each other (enantiomorphic). The lower case stereodescriptors “r” and “s” are used to designate the absolute configuration of pseudoasymmetric centers, for example (2R,3r,4S)‑2,3,4‑trichloropentane:

Note that the stereodescriptors “r” and “s” describing a pseudoasymmetric stereogenic unit are invariant on reflection in a mirror (for example “r” remains “r”, and “s” remains “s”), but are reversed by the exchange of any two ligands (“r” becomes “s”, and “s” becomes “r”).
The configurations (“r” or “s”) at pseudoasymmetric stereogenic units are determined by application of the Cahn-Ingold-Prelog (CIP) priority system and the sequence rules. In particular, note that an atom or group with descriptor “R” has priority over its enantiomorph “S”. In order to establish the order of precedence of ligands in a stereogenic unit, the atoms of the stereogenic unit are rearranged in a hierarchical diagram, called a “digraph”. Descriptors specified in digraphs may correspond to the final descriptors or to temporary auxiliary descriptors used only for ranking ligands and never appearing as final descriptors.
Similar to the above-mentioned case of (2R,3r,4S)‑2,3,4‑trichloropentane, the structure of (1r)‑1,2,3‑trichlorocyclopropane contains one pseudoasymmetric center, which has two ligands that are nonsuperposable mirror images of each other:

The other isomer of 1,2,3‑trichlorocyclopropane is (1s,2s,3s)‑1,2,3‑trichlorocyclopropane, which contains three pseudoasymmetric centers. Each one has two ligands that are nonsuperposable mirror images of each other:

It may not be obvious how to choose the stereodescriptors ‘r’ or ‘s’ for 1,2,3-trisubstituted cyclopropanes. The ranking is a complex process and requires the use of auxiliary descriptors.
Generally, the stereodescriptors ‘r’ and ‘s’ are assigned by applying the Cahn-Ingold-Prelog (CIP) Priority System as for the stereodescriptors ‘R’ and ‘S’. The order of precedence is reached by the construction of a hierarchical digraph and the application of the Sequence Rules.
The constitutional formula of a cyclic molecule is transformed into an acyclic digraph according to the methodology that is described in Subsection P-92.1.4.3 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) and which has been recommended by Prelog and Helmchen.
In order to obtain the acyclic digraph of a stereogenic unit, the didentate ligand (i.e. the ring) is transformed into two monodentate ligands by leaving intact in each case one bond with the core of the stereogenic unit and cleaving the other bond. At the end of each of the two branches thus obtained, a duplicate atom of the core is attached. (Note that this methodology is similar to the case of multiple bonds.) For example, for the pseudoasymmetric stereogenic unit at the atom numbered ‘1’ of (1r)‑1,2,3‑trichlorocyclopropane:

where $\ce{(C)}$ is a duplicate representation of the core of the pseudoasymmetric stereogenic unit.
As the digraph is acyclic, it is now possible to specify the stereodescriptor ‘R’ or ‘S’ for the stereogenic unit at the atoms numbered ‘2’ and ‘3’ in each branch. These are auxiliary descriptors that are different in each branch.

Since the auxiliary descriptors are different, they allow the branches to be ranked using Sequence Rule 5, which states that ‘R’ has priority over ‘S’.

P-92.1.3.5 Sequence Rule 5

An atom or group with descriptor ‘R’, ‘M’, and ‘seqCis’ has priority over its enantiomorph ‘S’, ‘P’ or ‘seq Trans’.

Therefore, the configuration of the pseudoasymmetric stereogenic unit at the atom numbered ‘1’ of (1r)‑1,2,3‑trichlorocyclopropane can be determined to be ‘r’.

In the same way, the configuration of each pseudoasymmetric stereogenic unit of (1s,2s,3s)‑1,2,3‑trichlorocyclopropane can be determined as ‘s’.

In the (1r)-1,2,3-trichlorocyclopropane, why is r assigned to the carbon? I would infer that the left C has S configuration and the right C has R configuration, but why is this the case? Another question is whether (1r,2r,3r) would be an equally valid configuration for the 2nd isomer.

@א_ϵ Please ask a new question if you have a new question. Comments are not intended for that.

Contrary to what I believed earlier, cis- and trans- notation is unambiguous for cyclic compounds and is accepted by IUPAC, though not without some tweak. According to the rule of assigning cis- and trans- in gold book:

The appropriate reference plane of a double bond is perpendicular to that of the relevant σ-bonds and passes through the double bond. For a ring (the ring being in a conformation, real or assumed, without re-entrant angles at the two substituted atoms) it is the mean plane of the ring(s) … If there are more than two entities attached to the ring the use of cis and trans requires the definition of a reference substituent.

In the isomer of trichlorocyclopropane which has all three chlorine atoms on the same side, I can choose any one as the reference and the other two should thus be cis. The name should be r-1,c-2,c-3-trichlorocyclopropane. The other isomer is named r-1,c-2,t-4-trichlorocyclopropane.
However I'm still not sure whether and how CIP rule could be used. This page, which gives additional examples of c- and t- naming, merely states CIP could be used without substantiation.

The preferred absolute stereochemical configuration notation using the R/S, r/s, … stereodescriptors and CIP rules (as in Loong's detailed answer) might be in some cases quite difficult for human chemist to construct or decipher, like in this case.
There's also a non-preferred, but simpler method using cis/trans notation (IUPAC BlueBook 2013 P-93.5.1.2) for cycles with two stereocenters, and r/c/t (reference, cis, trans) notation (for more ring stereocenters), already mentioned by ℵ_ϵ.

P-93.5.1.3 Relative configuration; the stereodescriptors ‘r’, ‘c’, and ‘t’
P-93.5.1.3.1 When one substituent and one hydrogen atom are attached to each of more than two positions of a monocycle, the steric relations of the ligands are expressed by adding ‘r’ (the reference ligand) after the locant of the lowest numbered of these ligands, and ‘c’ for ‘cis’ and ‘t’ for ‘trans’ (as required) followed by a hyphen and the locant of another ligand, thus expressing the relationship to the reference ligand. The relative configuration is expressed by these stereodescriptors. Furthermore, racemates may also be denoted by this method. For preferred IUPAC names, the preferred stereodescriptors are those used in the CIP priority system described in P-91 and P-92, With stereodescriptors such as ‘R’ and ‘S’ to describe absolute configuration and the prefixes ‘rel’ to express relative configuration and ‘rac’ for racemates, as described in P-93.1.3.
  The notation consisting of adding ‘r’ (for reference substituent) followed by a hyphen and the locant of the lowest numbered of these substituents and ‘c’ (for cis) and ‘t’ (for trans) (as appropriate) followed by a hyphen and the locant for another substituent, as used in the 1993 Guide to IUPAC Nomenclature of Organic Compounds (ref. 2), is no longer recommended.
  (…)
P-93.5.1.3.2 When two different ligands are attached at the same position of a monocycle, then the lowest-numbered ligand named as a suffix is chosen as reference ligand. If none of the ligands is named as a suffix, then that the ligand of the pair of ligands having the lowest number and Which is higher ranking in the Sequence Rule is chosen as a reference ligand. The relationship of the Sequence-Rule-higher ranking at geminally substituted positions, relative to the reference group, is cited as ‘c’ or ‘t’, as required.
  (…)
Example 3:

(1R,2R,3s,4S,5S,6s)-1,2,3,4,5,6-hexachlorocyclohexane (PIN)
  1r,2c,3c,4t,5t,6t-hexachlorocyclohexane
Explanation: In this compound, different numberings are necessary in accordance with the precedence of ‘R’ ligands and ‘cis’ arrangements for lowest locants (…)

The notation is for relative stereochemical configuration. But, since each of the six structures in your question is achiral as a whole (superimposable with its mirror image, i.e. a meso form), the relative stereochemical configuration completely describes the absolute configuration as well.
We get the following names

(1s,2s,3s)-1,2,3-trichlorocyclopropane (PIN)
1r,2c,3c-trichlorocyclopropane

(1r)-1,2,3-trichlorocyclopropane (PIN)
1r,2c,3t-trichlorocyclopropane

(1s,2s,3s,4s)-1,2,3,4-tetrachlorocyclobutane (PIN)
1r,2c,3c,4c-tetrachlorocyclobutane

(1R,2r,3S,4s)-1,2,3,4-tetrachlorocyclobutane (PIN)
1r,2c,3c,4t-tetrachlorocyclobutane

(1R,2R,3S,4S)-1,2,3,4-tetrachlorocyclobutane (PIN)
1r,2c,3t,4t-tetrachlorocyclobutane

(1r,2r,3r,4r)-1,2,3,4-tetrachlorocyclobutane (PIN)
1r,2t,3c,4t-tetrachlorocyclobutane
Please note that

PIN states for ‘preferred IUPAC name’;
the numberings for the name pairs do not match (the PIN names are just for reference or comparison), the second one is generally shifted and/or reversed (one can easily deduce that e.g. for the mentioned IUPAC P-93.5.1.3.2 example 3 structure, hexachlorocyclohexane, the two pseudoasymmetric carbons (3s,6s) from the PIN name correspond with carbons 2 and 5 from the non-PIN name illustrated by the formula), as for the r/c/t notation, atom with locant 1 (reference r) and the numbering direction is chosen so that as much as possible subsequent descriptors are c (cis), then t (trans), it's quite easy;
the r stereodescriptor has completely different meaning in both stereochemical notation systems;
the r/c/t descriptors are not in parentheses before the complete name, but directly after the locants in the name;
there are also older obsolete/incorrect variants for this notation with dashes, like

r-1,c-2,c-3-trichlorocyclopropane (IUPAC 1993, R-7.1.1),
1-r,2-c,3-c-trichlorocyclopropane (IUPAC BlueBook 2013, first uncorrected release, some parts).


text:
What is the reason for the anomalous electronic configuration of palladium?
The p block elements generally have the general valence shell configuration as $$n\mathrm{p}^{1-6}\,n\mathrm{s}^{1-2}.$$ Expected electronic configuration for palladium is $$\ce{^46Pd} = \mathrm{1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^8}.$$ But I find that it's valence shell instead is $$\mathrm{4d^{10}\, 5s^0}$$ I first thought, for achieving fully filled d-subshell, 2 electrons go to the d-subshell from s-subshell. This electron transfer should happen for all members of its group, yet only palladium demonstrates this. Why is this so?

It's double electron promotion.

You may check the answer given here: http://chemistry.stackexchange.com/questions/2469/how-can-one-explain-niobium-s-weird-electronic-configuration

I strongly suggest the linked question's title be renamed from being about niobium to a more general one, because the answer to that question is a very general answer, and covers all cases neatly. Most probably, any answer to this question will almost repeat what has already been said. Hence, I've cast a close vote as duplicate.

The Aufbau principle states that, hypothetically, electrons orbiting one or more atoms fill the lowest available energy levels before filling higher levels (e.g., 1s before 2s). In this way, the electrons of an atom, molecule, or ion harmonize into the most stable electron configuration possible.
Electron behavior is elaborated by other principles of atomic physics, such as Hund's rule and the Pauli exclusion principle. Hund's rule asserts that even if multiple orbitals of the same energy are available, electrons fill unoccupied orbitals first, before reusing orbitals occupied by other electrons. But, according to the Pauli exclusion principle, in order for electrons to occupy the same orbital, they must have different spins.
The order in which these orbitals are filled is given by n+l rule known as Madelung Rule or diagonal rule.Orbitals with a lower (n+l) value are filled before those with higher (n+l) values where n represents principal quantum number and l represents azimuthal quantum number.
The fact that most of the ground state configurations of neutral atoms fill orbitals following this n + ℓ, n pattern was obtained experimentally, by reference to the spectroscopic characteristics of the elements.
The Madelung energy ordering rule applies only to neutral atoms in their ground state,even in some cases there are examples which show different electronic configuration than those of experimentally determined.Copper,Chromium and Palladium are common examples of this rule.
For Palladium,according to Madelung's rule, 5s^2 has lower (n+l) value i.e. (5+2=7) and 4d^8 has higher (n+l) value i.e. (4+8=12),so the 5s orbital  fills up first before 4d orbital.
Hence, Palladium shows anomalous electronic configuration.

For 5s isn't the n+l value 5+0=5? As the l value of s subshell is 0?

The answer fails to clear my question, read it carefully. It is the only element besides Niobium to transfer 2 electrons from the s subshell to d subshell..but the link by @Marko cleared my doubt a bit, and it seems I have to study it little more. Thanks @Marko! But, you would have liked to answer it, as it has a bounty!

@Aneek, if you think that Marko's answer helped you than that of mine, then go for it! :)

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text:
Is Hydrogen Bonding a Type of Dipole Dipole Interaction?
I understand that dipole dipole forces is due to the attraction of the different partials charges of atoms in different molecules due to their different electro-negativities.
For hydrogen bonding, my understanding of it is that they are just a special case of dipole dipole forces. In certain molecules in where the difference in electro-negativities between the different atoms is high enough, then even stronger dipole dipole forces will be form which are known as hydrogen bonding. For example, in $\ce{H_2O}$ where the difference in electro-negativity of H and O is 1.24, hydrogen bonding occur. This is why hydrogen bonding can only occur between hydrogen and nitrogen, fluorine or oxygen.
However, my school teacher says that hydrogen bonding in fact is not a type of dipole dipole force as she says that hydrogen bonding occurs differently to dipole dipole forces. She said that hydrogen bonding occurs to the attraction of the lone pairs in oxygen, fluorine and nitrogen atoms to atoms with a positive partial charge. Hence this is why hydrogen bonding is much stronger than conventional dipole dipole forces.
But this doesn't make sense because if hydrogen bonding is due to the attraction  of H to the lone pair of electrons, shouldn't hydrogen bonding also form between H and Cl or I or any other atom that has a lone pair and is more electronegative than H?
Edit 1
From the comments below, it appears that the actual answer is more complicated than the simple explanation of hydrogen bonding given above. After doing some research, I have became aware of the Advanced Theory of the Hydrogen Bond which states that the hydrogen bond actually contains 10% covalent nature. However none actually explain how they are formed. Could someone please explain how hydrogen bonding actually in reality form.
Edit 2
So in response to the excellent answer by jheindel, is the reason why hydrogen bonding is considered partly covalent in nature is because the let say in water, oxygen's MO partly overlaps with the MO of the hydrogen atom in the other molecule. Also this is why they are much stronger than conventional dipole dipole forces?

@Martin -If you don't have time to answer, could you please tell me what area of chemistry the answer to my question involves. Also if there are any, is there any links you could give me that explain my question. Sorry if I sound really annoying and desperate, but I really want to know and I have been scanning the internet for days for an answer and I couldn't find anything that was remotely close to answering my question.

related http://chemistry.stackexchange.com/questions/24157/is-a-hydrogen-bond-considered-to-be-a-van-der-waals-force

I hope this will help you. http://www.adichemistry.com/general/chemicalbond/hbond/hydrogen-bonding.html

Here is some experimental and computational data on very strong hydrogen bonds with covalent character: https://science.sciencemag.org/content/371/6525/160

Well, it turns out that this is a very active area of research. I will only summarize what I understand to be true about the covalent nature of the hydrogen bond, so I'm sure the explanation could be more detailed and potentially more accurate in some places (I hope someone gives a more detailed answer), but here's what I've got.
As you said, it has been found that the hydrogen bond (specifically in water) is about 90% electrostatic and 10% covalent. See this link for a sort-of-review/overview of hydrogen bonding and some relevant research.
Upon following the paper trail, I found the following research article: E.D. Isaacs, A. Shukla, P.M. Platzman, D.R. Hamann, B. Barbiellini, C.A. Tulk, J. Phys. Chem. Solids, 2000, 61, 403-406. (Mirror)
Essentially, they use some method which I don't really understand and find quite conclusive evidence of the (word of the day) anisotropy of hydrogen bonds. That means that the bond is indeed directionally dependent. Bond direction is one defining characteristic of covalent bonds.
That raises the question of what is going on with the orbitals that allow for hydrogen to participate in more than one covalent bond at a time.
A simple explanation of the orbital interactions for a water dimer is to view the oxygen 2p,2s hybridized orbital as having superposition with the hydrogen 1s orbital and then interacting covalently with that same 2p,2s hybridized orbital on the other oxygen in the dimer. This superposition which includes the hydrogen then explains the way that hydrogen is participating in one covalent bond and another partially covalent interaction.
Furthermore, this explanation of what the orbitals are doing fits well with what we would expect from a hydrogen participating in two different covalent interactions, which is that the energy curve has two minima where one is much deeper than the other.
Further evidence for the covalent nature of the bond is the fact that the O-H bond elongation and O---H hydrogen bond length always change together. That is, as the O-H bond lengthens, the O---H interaction length shortens. This implies a shift in electron density towards hydrogen which is most easily explained by a delocalization of electrons participating in a covalent interaction.
So, I hope that's a good start toward understanding the covalent nature of the hydrogen bond--it was for me!
But to answer the simple part of the question which you clearly already know is that yes dipole-dipole is a significant part of hydrogen bonding, and it certainly should be, but as is always true of chemistry, there's a lot more going on than what one might initially think.

EDIT (06/01/2016)
Since I posted this answer, I think I have developed a better understanding of the situation, and feel obliged to update my answer. First, let me split hydrogen-bonding into two obvious parts, electrostatic and covalent characteristics. This is an obvious split, but it is not completely clear yet what we mean by covalent, so let us try to define that by looking at some examples. Furthermore, we must be careful when talking in about hydrogen-bonding because in every hydrogen bond there is a donor and acceptor, and these two roles are often played best by different chemical species.
Hydrogen-Bonding in thiols
Here we go breaking yet another rule. They told us that sulfur couldn't hydrogen bond. Well, that's not exactly true. Okay, but that hydrogen-bonding has got to be basically negligible right? Not exactly.
The first thing we need to do here is get some information. First, sulfur has an electronegativity of 2.58 on the Pauling scale. This is roughly the same as the electronegativity of carbon which we do not consider to hydrogen-bond well. So, if there are hydrogen-bonds involving sulfur, we should expect that they will be primarily covalent as any dipole interactions will be pretty small.
As it turns out, sulfur is a much better hydrogen-bond acceptor than it is donor, and at times is a comparable acceptor to oxygen. Here is a table comparing certain dimers from a great source you can reference at the bottom of the page$^3$.

As you can see, when the sulfur in these dimers (which are all named in the caption of the table) acts a hydrogen bond acceptor, the shift in vibrational frequency of the $\ce{O-H}$ and the binding energy are comparable to when an oxygen is the acceptor atom.
Looking at other research$^4$ involving sulfur hydrogen bonds reveals that approximately 70% of the interaction energy between sulfur and oxygen when sulfur is the acceptor comes from dispersion. That is the key physical insight in all this. Despite the loss of electronegativity, sulfur gains something in its greater dispersion interactions which makes it a good hydrogen bond acceptor. It should be noted, however, that the electronegative component of these hydrogen-bonds maintains their directionality.
Also, to compare, according to that wikipedia article I linked, the interaction energy of a water dimer is only 24% dispersion. In water electrostatics dominate and induction and exchange play meaningful roles.
Conclusions
I have more I could say, but it's perhaps a longer explanation than is worth going into. To conclude, what is meant by covalent interactions in hydrogen-bonding varies based on the species you are considering. In something like water, we can think about this in the usual way by saying that there must be some orbital overlap and hence electronic exchange which will be stabilizing. With other species, we can see that what we mean by covalent interactions might be different. That is, there is still some orbital overlap, but we must also consider non-electrostatic interactions like dispersion.
3 Noncovalent Forces-Chapter 2
4
Wennmohs, F., Staemmler, V., & Schindler, M. (2003). Theoretical investigation of weak hydrogen bonds to sulfur. The Journal of chemical physics, 119(6), 3208-3218.

I am little confused when you say, ' view the oxygen 2p,2s hybridized orbital as having superposition with the hydrogen 1s orbital and then interacting covalently with that same 2p,2s hybridized orbital on the other oxygen in the dimer'. Do you mean at the end of the sentence it is interacting covalently with the other hydrogen atom or am I missing something here?

What I mean is that the acceptor oxygen sp3 (if you still want to use that idea) does, to some extent, overlap with the hydrogen 1s of the donor molecule. This is the covalent nature of the bond. Sorry, I definitely worded that poorly.

I would also point out that this is actually much more complicated than my answer and I don't feel like I understand all the details well enough to lay them out clearly. Hopefully Martin answers.

Hydrogen bonding may be understood in different ways. My own preference is it us a form of delocalized covalent bonding. This is discussed in this answer.
This molecular orbital description does not require the bonding to be polar, although the polarity commonly associated with hydrogen bonds (hydrogen primarily attached to a much more electronegative atom in both the primary bond and the hydrogen bond) favors more bond delocalization and thus a stronger hydrogen bond. Answers to this question explore the possibilities, including one where that cites hydrogen bonding to carbon atoms.

Hydrogen bond is defined as the electrostatic force of attraction which exists between the covalently bonded hydrogen atom of one molecule and the electronegative atom of other molecule. These are special type of dipole-dipole interactions. Here the hydrogen atom is linked with one atom through covalent bond and linked with other atom through hydrogen bond.however hydrogen bond is much weaker than the covalent bond. When hydrogen atom bonded to a strongly electronegative atom the bond pair of electrons displaced towards electronegative atom. Thus the only electron of hydrogen lies away from it, its nucleus get exposed and behaves as a bare proton and exert a strong electrostatic force of attraction on the strong electronegative atom of adjacent molecule this interaction is referred as hydrogen bond.

So what does the lone pair of the oxygen, nitrogen or fluorine atom got to do with hydrogen bonding?

@Nanoputian - its a sink of negative charge which attracts the somewhat 'bare', positively-charged proton (i.e. the hydrogen involved in H-bonds).

Sorry, but telling that h-bond is purely electrostatic is a fairytale for little students, told because truth is hard to understand.

@Mithoron- Could you please explain how the h-bond is really formed then. I am really curious on how hydrogen bonding works.

Don't worry about being too complicated for a high school student to understand. I am familiar with 1st and 2nd year university chemistry and should be able understand how h-bonding occurs in reality.

@Nanoputian It seems H-hond needs at least sth like SCF (quantum mech. method to be modelled and I'm no expert in these. Btw your ping didn't work.

This answer is technically wrong, but it's not delete worthy, VLQ, or NAA. Please don't flag it for no reason.

@Mithoron Its quite an effective fairytale no matter how incomplete it may be.  Even better, take your sentence and change the word 'purely' to 'similar to' if that makes you feel better.

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text:
Percent concentration of 500 μM of H2O2
What is the percent concentration of hydrogen peroxide which has a concentration of $500~\mu\mathrm{M}$?
Is $500~\mu\mathrm{M}$ $\ce{H2O2}$ toxic to the eyes?
A particular honey has a concentration of $\ce{H2O2}$ of $500~\mu\mathrm{M}$ after a 24-hour incubation time, after it has been diluted. I am trying to figure out what the concentration is in percent (%).
I recalculated and the result is actually $0.00153\%$ for $500~\mu\mathrm{M}$. $0.0012\%$ is the equivalent to $400~\mu\mathrm{M}$, but I need someone to confirm this. Someone on researchgate wrote that $30\%$ of $\ce{H2O2}$ is $9.8~\mathrm{M}$.

Note that in the first line you have given the concentration of hydrogen peroxide, not the molar mass.

I will assume you are asking for percent mass/volume, $\mathrm{g/ml}$, since to calculate percent mass/mass you need the density of the liquid (unless you assume it is water which is $1~\mathrm{g/ml}$ so will give the same answer).
We have $500~\mu\mathrm{M}$ of hydrogen peroxide, which means we have $500~\mu\mathrm{mol}$ of hydrogen peroxide per 1 litre of solution. $500~\mu\mathrm{mol} = 0.0005~\mathrm{mol}$. Now you can work out the molar mass of hydrogen peroxide and multiply it by the number of moles to get the mass of hydrogen peroxide per litre of solution. This gives $0.017~\mathrm{g}$. There are $1000~\mathrm{ml}$ in 1 litre. So you have $0.017~\mathrm{g}/1000~\mathrm{ml} = 0.000017 = 0.0017\%$.
For your question about whether hydrogen peroxide is toxic to the eyes, you can search the internet for the Materials Safety Data Sheet (MSDS) of hydrogen peroxide (Sigma-Aldrich is a good site to search). This contains all the hazards associated with this chemical, and when you do laboratory work you always have to look these up for each chemical before you start a reaction. The MSDS states that hydrogen peroxide can cause severe eye damage, among other hazards.

Thank you for your answer it seem like I wasn't far off. I do not know how H2O2 in honey is measured but I assume it's per litre. So it's about a 1000-fold less than standard disinfectants.

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details. There is also a second question included by the OP, I am vey happy to upvote this answer if you could include a little statement about that.

thanks, I'll try to next time. I've included an answer for the other question

Regarding toxicity: Honey solutions are actually sometimes used in ophthalmology. There is a company called Optimel that sells 16% manuka honey eye drops OTC for treating dry eye syndrome but note that very little or almost nothing H2O2 is released in manuka honey. Another example is this case report in which they used a 25% w/v silver fir honey solution as eye drops for the treatment of a corneal ulcer with success.
I found an article in which the use of H2O2 in the eyes in various concentrations is outlined. At the concentration below 60 ppm (the lowest safe point) the eye’s natural enzyme mechanisms can break down the small amount of H2O2 making it safe.
So a 25% w/v (silver fir) honey solution contains around 0.00085% or 8.5 ppm of H2O2 which would make it safe for the use in eyes.


text:
Are all higher cycloalkenes chiral?
Are all higher trans-cycloalkenes chiral as trans-cyclooctene is?
This question supplements my previous question: Why is trans-cyclooctene chiral? See also the follow-on question: Do more double bonds cause a bigger number of stereoisomers in cycloalkenes?

No, and one can tell that from stuff you get from original question.

Well, as I was told to ask that question in a separate thread, I expect here an answer.

@Mithoron The answer 'no' to this question is not immediately obvious to me based upon the answers to the original.  (The fault is of course perhaps mine.) A separate elucidation seems warranted, and will avoid muddying the original question.

@Brian Hmm, unfortunately those additional questions are rather bad and stem from that the original wasn't answered clearly. Such situation shouldn't take place.

@Mithoron I disagree -- my understanding of the SE model is that each Question should be held to a single question, with the scope held as tightly as is sensible. The initial question is specific to trans-cyclooctene; this question is general, about the series of cyclic monoalkenes; the following question is about cyclic dienes.

Background
One can draw conformations of n-butane with the carbon-carbon bonds oriented in certain directions and the methyl hydrogens pointing in certain directions that are chiral.  However, since rotation about single bonds is typically fast at room temperature, these chiral conformers of n-butane would only be resolvable at extremely low temperatures.  Therefore we do not typically consider n-butane to be a chiral molecule.
Take-away lesson: If a chiral conformation can readily interconvert with an achiral conformation, then the molecule cannot be resolved into enantiomers and will be optically inactive.
However, there are molecules where rotation about single bonds is restricted (for example due to steric interference) and the rotational barrier is high enough that the molecules can be resolved into their respective enantiomers at room temperature.
Molecules where the rotational barrier about a single bond is high enough to allow different conformations to be isolated at room temperature are called atropisomers.
The biphenyl pictured below (in a chiral conformation) is an atropisomer due to restricted rotation about the central single bond connecting the two benzene rings.  The steric interference that would develop between the 4 ortho groups if the molecule were to approach a planar (achiral) conformation prevents the chiral conformer from equilibrating with the achiral (planar) conformer.  Hence this biphenyl has been resolved into its two enantiomers at ambient conditions.

Trans-cyclooctene
Trans-cyclooctene is also capable of being resolved into its two enantiomers due to restricted rotation about single bonds - it also belongs to the class of atropisomers.
In the figure below trans-cyclooctene conformation (a) is planar, therefore this conformation is achiral. The two conformers pictured in (b) are chiral - these conformations are enantiomers of each other. Trans-cyclooctene will be achiral if the two enantiomers pictured in (b) can readily interconvert through achiral (planar) conformer (a).
Build a model or draw trans-cyclooctene and notice the two hydrogens on the trans double bond; one of them is "inside" the ring in the planar conformation and would cause severe steric problems as the methylene chain tries to flip through the planar conformation to the other face of the double bond as pictured below. This steric interference makes it difficult for the molecule to achieve the planar conformation.

(Image source: John D. Roberts, Marjorie C. Caserio, Basic Principles of Organic Chemistry. Excerpt available at http://www.organicchemistry.com/cycloalkenes-and-cycloalkynes/.)
Because of this destabilizing transannular interaction, trans-cycloalkenes with up to around 11 carbons do not exist as planar molecules. They exist in a twisted conformation as pictured above for trans-cyclooctene.  In your model of trans-cyclooctene, notice how you have to pull and stretch the methylene chain in order to convert trans-cyclooctene into its mirror image.  This rotational movement is highly restricted due to the internal hydrogen.  It is this barrier to rotation about singles bonds that allows trans-cyclooctene to be resolved into its two mirror image forms at room temperature.
The barrier to racemize trans-cyclooctene (e.g. pass through the planar conformation (a)) is 36 kcal/mole.  As you might imagine, as we make the chain longer it should get easier and easier to loop the methylene chain around the internal (protruding) hydrogen on the double bond.  Indeed, the barrier to interconvert the two enantiomers of trans-cyclononene is only 20 kcal/mole. Optically resolved trans-cyclononene must be kept below 0°C to prevent racemization.  The barrier to racemization in trans-cyclodecene is even lower at 10 kcal/mole.

Are all higher trans-cycloalkenes chiral as trans-cyclooctene is?

First off, these rings can have chiral conformations, just like butane.  I think you are really asking, "Are all higher trans-cycloalkenes optically active as trans-cyclooctene is?".  That is, do these larger rings also have difficulty reaching the planar, achiral geometry?
So the answer to your original question is - it depends.  It depends on the temperature you specify.  At room temperature, trans-cyclooctene is the only trans-cycloalkene where the two enantiomeric conformations are stable enough (e.g. the barrier to reach the planar conformation is high enough) to be isolated.  If we lower the temperature to 0°C, then we can isolate the enatiomers of trans-cyclononene.  If we go low enough in temperature, we could isolate enantiomers of n-butane.

At low temperatures, n-butane only has one conformation,  the linear, achiral one. So while your point is geometrically illustrative, it also is chemical nonsense.

I need to take exception with the term "nonsense".  I wrote about resolving chiral conformations of n-butane at extremely low temperatures.  Chiral conformations of n-butane certainly exist at ambient and at extremely low temperatures, say 100°K.  One could even argue that they could be present at 0°K if an ambient sample was rapidly quenched to absolute zero.

a little bit off-topic, but I just wanted to mention that Kelvin is free from "degree" since the late 60s

I insist, emphasising the adjective "chemical". ;-)  For single molecules, OK, but that's physics.

I don't follow.

You say you'd be able to isolate, resolve, i.e. characterise chiral conformations of n-butane. That's not true. Heat capacity clearly shows they exist, but you won't be able to pin them down. Never in the bulk, and I strongly doubt it for a single molecule. You're over-stretching your argument. It'd be stronger without.

I don't really get the last paragraph. Just because the barrier of recemisation is very low, it doesn't really mean, that the molecule is achiral. For that to be true, the achiral form must be predominant I thought. Which is not the case for the rings with less than 11 (?) carbons, the achiral form is very likely still a transition state only.

On another note, I followed the image source link of your "borrowed" picture and could not find any evidence of reuse rights of their images on other pages. I don't know how strict that should be handled, but I inserted a proper attribution just in case.

@Martin-マーチン "Just because the barrier of recemisation is very low, it doesn't really mean, that the molecule is achiral". I agree, conformations are chiral or achiral.  If a chiral conformation is in rapid equilibrium with an achiral conformation, then the molecule will be optically inactive.  I changed some wording to make this clearer in my answer.  Are we on the same page now?

Chirality is lost (see ron's answer/comment) when the ring get so large that the most stable conformation becomes planar.
Otherwise: Yes they are, unless the ring is so large that it can circumvent the double bond without a noticeable energetic inversion barrier. Strictly speaking, the cycloalkene then is chiral still, only it would racemise instantly. Single crystals would still be optically active.

"Strictly speaking, the cycloalkene then is chiral still...Single crystals would still be optically active". In larger trans-cycloalkenes the planar form is the most stable conformation.  I don't think crystals of such molecules would be optically active.

That's right, they aren't. Same reason why n-butane can't be chiral.


text:
Do more double bonds cause a bigger number of stereoisomers in cycloalkenes?
If trans double bonds cause cycloalkenes to be chiral, does a second trans or maybe cis double bond induce another chiral plane and hence four stereoisomers are possible for the given cyclodiene?
The question is based upon these previous questions:
Are all higher cycloalkenes chiral?
Why is trans-cyclooctene chiral?

If trans double bonds cause cycloalkenes to be chiral

Keep in mind that a trans double bond in a cycloalkene does not necessarily produce a chiral cycloalkene.  Only when the chain connecting the ends of the double bond is short enough so that the trans double bond is twisted will a chiral conformation exist.
So in large trans-cycloalkenes the double bond is not twisted and cannot produce chirality.
Remember too, that even if the trans double bond is twisted and a chiral conformation results, the molecule may still be optically inactive if the chiral conformation rapidly interconverts with its enantiomeric conformation. This latter point was discussed in one of your earlier questions where it was pointed out that

At room temperature, trans-cyclooctene is the only trans-cycloalkene
  where the two enantiomeric conformations are stable enough (e.g. the
  barrier to reach the planar conformation is high enough) to be
  isolated.  If we lower the temperature to 0°C, then we can isolate the
  enatiomers of trans-cyclononene.

Back to your question.

does a second trans or maybe cis double bond induce another chiral
  plane

The key here is that, as mentioned above, if the second double bond is also twisted, then there will be the potential for an increase in the number of chiral stereoisomers (of course inserting a double bond will always increase the number of stereoisomers because we have additional cis and trans double bond possibilities, but I think your question is focused on chiral stereoisomers, not geometric stereoisomers).

Look at the above figure on the left, of a trans double bond in a small ring.  We are sighting down a trans double bond. To alleviate ring strain, the double bond if forced to twist.  It is this twisting out of plane that infers chirality upon the double bond.  Its insertion into the ring will produce an additional chiral plane and the R or S chirality of the new double bond will  need to be specified in the name of the compound.
Note the figure on the right of a cis double bond in a small ring. To alleviate strain here, the $\ce{C_{ring}-C=C}$ angle is pinched down; there is no out of plane twisting motion, the plane of the double bond still exists.  Therefore, in the cis double bond case, there is no chirality associated with the double bond; no additional chiral isomers will be produced.

and hence four stereoisomer are possible for the given cyclodiene?

Not always.  Just as in more common compounds with 2 chiral centers, meso stereoisomers are possible.  As an example, if the known E,E-cycloocta-1,5-diene
adopted a conformation as pictured below, then it has a plane of symmetry and would be an achiral meso compound where one trans double bond has the R configuration and the other double bond the S configuration.

(image source)


text:
How can I measure the amount of CO2 in grams in a known volume of soda based on its weight?
We have a soda machine that can fill either still water or soda. In both cases the amount is 400 ml in the beaker, but the soda weighs a little less.
How can we use the volume X = 400 ml and the weight Y = ?? grams to determine the amount of $\ce{CO2}$ in the solution?

You can't. The soda likely already contains a few bubbles when it is measured into the beaker, so its average density will be less. The density difference due to dissolved $\ce{CO_2}$ is minute, i doubt you will be able to make a sample that is so exactly 400ml that your weight measurements make any sense.

Btw.: I'm sure the density of dilute carbonic acid is actually higher than distilled water. That's true for practically all kinds of solutions: If the density would sink upon dissolving, it wouldn't.

Great :) Why didnt you put that in an answer?

I don't think this is necessarily a homework question.

Actually rhis is work related. We are developing a soda dispenser.


text:
Physical intuition behind negative values for wave function?
So a positive and a positive wave function create a bonding orbital where the probability of finding an electron is summed while a positive and a negative create an anti-bonding orbital with a lower electron probability in the region between them leading to a repulsion. My confusion stems from not having any idea as to what a negative wave function is representing - can anyone give me some physical intuition on how this negative wave function could be correlated to something in reality?
(I tried asking this question on physics.SE but it would seem that in physics you just don't talk about or use the wave function until after you square it)

And imagine, the wavefunction can easily be even imaginary! (Or, better, complex-valued function).

Don't think about "negative" as being different than a "positive" wave function. What's really important is that the signs are either the same or opposite. It's the difference between signs in wavefunctions that give rise to the interesting stuff.

@ssavec, and it is not even that it can, rather that it is a complex-valued function.

The wavefunction of a particle actually has no physical interpretation to it until an operator is applied to it such as the Hamiltonian operator, or if you square it which gives its probability of being at a certain place. So having a negative wavefunction doesn't mean anything physically. However, let's say for a particle in a box, if you solve the momentum operator for the $n^\text{th}$ stationary state, you would get two solutions: $+\hbar k$ and $-\hbar k$, where $k = n\pi/L$. The plus and minus signs mean that there is a $50$% chance the particle could be moving from the left to the right and a $50$% chance of the particle moving from right to left.
In the case of summing of two wavefunctions to get the wavefunction of a molecular orbital, a negative wavefunction doesn't mean anything at all. The reason why when a positive and negative wavefunction is added, an antibonding orbital is formed is simply because when you add a positive and negative number, you get $0$ or a really small number. This is known as destructive interference.  Therefore the new wavefunction will contain a region where the y values equal $0$. So when you square this new wavefunction of the molecular orbital you are going to get that there is very 0 probability of finding the electron in that position (known as a nodal plane), and hence it is called an antibonding orbital.However note that just because there is negative region in the wavefunction, doesn't mean that it will always add to form an anti-bonding MO. For example, if you add 2 negative wavefunction values (known as constructive interference), you get a large negative value. When you square becomes a very large positive value. Hence this means that there is a large probability of the electron being in that region and hence it will be a bonding MO.

but how can it make sense to say "a negative wavefunction doesn't mean anything at all" when it is the distinguishing feature between the combination of two wavefunctions being attractive or repulsive?

@norlesh The phase of the wavefunction (i.e. the sign) doesn't mean anything. Let me ask you this about the $2p_z$ orbital (can be any p orbital, but I choose this one to avoid potential nitpicking). In this orbital, there is a positive lobe and a negative lobe. What is the difference between them? Do you think there is a "correct" way to label which lobe is positive and which lobe is negative?

@norlesh I'm going off soon so I'm just going to write, maybe a somewhat suitable analogy would be the two mirror images of a molecule, (R) and (S) isomers. They're different, but they react exactly the same way (let's assume their environment is achiral; our universe does not distinguish between positive and negative wavefunctions at all, just like an achiral environment does not distinguish between (R) and (S) isomers).

@norlesh Let's say you react these two isomers with a different, enantiomerically pure, molecule to form (R,S) and (S,S) diastereomers. These are "different" molecules, they have different physical and chemical properties. These are analogous to the antibonding and bonding molecular orbitals. Only the relative phase matters; the absolute phase doesn't. As Nanoputian already said, the wavefunction $\psi$ itself actually has no physical interpretation; Max Born actually won a Nobel Prize for finally interpreting $|\psi|^2$ as the probability density (amongst other things, of course).

In addition to @Nanoputian's excellent description of constructive and destructive interference in the formation of MOs, I want to provide a more mathematical explanation for why the phase of the wavefunction does not matter.

Finding the wavefunction
The time-independent Schrödinger equation, in one dimension, reads:
$$\hat{H}\psi(x) = E\psi(x)$$
It can be shown that, if a wavefunction $\psi = \psi(x)$ satisfies the above equation, the wavefunction $k\psi$ (with $k \in \mathbb{C}$) also satisfies the above equation with the same energy eigenvalue $E$. This is because of the linearity of the Hamiltonian:
$$\begin{align}
\hat{H}(k\psi) &= k(\hat{H}\psi) \\
&= k(E\psi) \\
&= E(k\psi)
\end{align}$$

There are several conditions that a wavefunction must satisfy for it to be physically realisable, i.e. for it to represent a "real" physical particle. In this discussion, the relevant condition is that the wavefunction must be square-integrable (or normalisable). In mathematical terms:
$$\langle\psi\lvert\psi\rangle = \int_{-\infty}^{\infty}\!\lvert\psi\rvert^2\,\mathrm{d}x < \infty$$
This means that there has to exist a constant $N \in \mathbb{C}$ such that $N\psi$ is normalised:
$$\int_{-\infty}^{\infty}\!\lvert N\psi\rvert^2\,\mathrm{d}x = \lvert N \rvert^2 \!\!\int_{-\infty}^{\infty}\!\lvert\psi\rvert^2\,\mathrm{d}x = 1$$
From this point onwards, we will assume that we will already have found the suitable normalisation constant such that the wavefunction $\psi$ is already normalised. In other words, let's assume $\langle\psi\lvert\psi\rangle = 1$, because we can. Now let's consider the wavefunction $-\psi$, which is equivalent to $N\psi$ with $N = -1$. Is this new wavefunction normalised?
$$\begin{align}
\int_{-\infty}^{\infty}\!\lvert -\psi\rvert^2\,\mathrm{d}x &= \lvert -1 \rvert^2 \!\!\int_{-\infty}^{\infty}\!\lvert\psi\rvert^2\,\mathrm{d}x \\
&= \int_{-\infty}^{\infty}\!\lvert\psi\rvert^2\,\mathrm{d}x \\
&= 1
\end{align}$$
Of course it is. So, what I've written so far basically says: if $\psi$ is a normalised solution to the Schrödinger equation, so is $-\psi$.
In fact, you could go one step further. Using exactly the same working as above, you could show that if $\psi$ is a normalised solution to the Schrödinger equation, the wavefunction $(a + ib)\psi$ would also be one, as long as $a^2 + b^2 = 1$. (If you like exponentials, that's equivalent to saying $a + ib = e^{i\theta}$.) I've illustrated this idea on this diagram:
$\qquad\qquad\qquad\qquad\qquad\qquad$
If $\psi$ is a real-valued, one-dimensional wavefunction, you could plot it on a graph against $x$. The wavefunction $i\psi$ would then be exactly the same shape, just coming out of the plane of the paper ($\theta = 90^\circ$). You could have the wavefunction $(1+i)\psi/\sqrt{2}$. It would be pointing outwards of the plane of the paper by $\theta = 45^\circ$, exactly halfway in between $\psi$ and $i\psi$, but exactly the same shape. However, physics doesn't know where the plane of your paper is, so all these wavefunctions are equally admissible. From the point of view of the system, they are all the same thing.

Using the wavefunction

"But wait! If the wavefunction is negative, what about the values of momentum, position, and energy that you calculate? Will they become negative?"
"Good question, myself!"

Well, for starters, one thing that you use the wavefunction for is to find the probability density, $P(x)$. According to Max Born's interpretation of the wavefunction, this is given by $P(x) = \lvert \psi \rvert ^2$. Let's say that the probability density described by the negative wavefunction $-\psi$ is a different function of $x$, called $Q(x)$:
$$\begin{align}
Q(x) = \lvert -\psi \rvert ^2 &= \lvert -1 \rvert^2 \lvert \psi \rvert ^2 \\
&= \lvert \psi \rvert ^2 \\
&= P(x)
\end{align}$$
So, the probability density described by the negative wavefunction is exactly the same. In fact, the probability density described by $i\psi$ is exactly the same as well.

Now let's talk about observables, such as position $x$, momentum $p$, and energy $E$. Every observable has a corresponding operator: $\hat{x}$, $\hat{p}$, and $\hat{H}$ respectively (the Hamiltonian has a special letter because it's named after William Hamilton). You use these operators to calculate the mean value of the observable. I'll give an example regarding the momentum. If you want to find the mean momentum, denoted $\langle p \rangle$, you would do the following:
$$\begin{align}
\langle p \rangle &= \langle\psi\lvert\hat{p}\rvert\psi\rangle \\
&= \int_{-\infty}^\infty\!\psi^*\hat{p}\psi\,\mathrm{d}x
\end{align}$$
I'm going to call the value of that integral $p_1$. Now, let's do the same thing. Let's assume that the mean momentum for the negative wavefunction is not necessarily the same value. Let's call the new mean momentum something else, like $p_2$.
Before we go on, I'm going to establish that the momentum operator $\hat{p} = -i\hbar\frac{\mathrm{d}}{\mathrm{d}x}$ is also linear. If you doubt it, you can test it out using the definition of linearity in the very first link I posted. In fact, all quantum mechanical operators corresponding to observables are linear. Therefore $\hat{p}(-\psi) = -\hat{p}\psi$ and so:
$$\begin{align}
p_2 &= \langle -\psi\lvert\hat{p}\lvert-\psi\rangle \\
&= \int_{-\infty}^\infty\! (-\psi)^*\hat{p} (-\psi)\,\mathrm{d}x \\
&= (-1)^2\!\!\int_{-\infty}^\infty\! \psi^*\hat{p}\psi\,\mathrm{d}x \\
&= \int_{-\infty}^\infty\! \psi^*\hat{p}\psi\,\mathrm{d}x \\
&= p_1
\end{align}$$
So, if we talk about the ground state of the particle in a box of length $L$, no matter whether you use the positive wavefunction
$$\psi_1 = \sqrt{\frac{2}{L}}\sin{\left(\frac{\pi x}{L}\right)}$$
or the negative wavefunction
$$-\psi_1 = -\sqrt{\frac{2}{L}}\sin{\left(\frac{\pi x}{L}\right)}$$
or the complex wavefunction
$$i\psi_1 = i\sqrt{\frac{2}{L}}\sin{\left(\frac{\pi x}{L}\right)}$$
you'll get exactly the same values for average position $(= L/2)$, average momentum $(= 0)$, and average energy $(= h^2/2mL^2)$ (the word average is redundant here, since this is a stationary state, but whatever).

Everything that I have said so far can be easily generalised to three dimensions. It can also be generalised to linear combinations of stationary states, i.e. solutions of the time-dependent Schrödinger equation.

A note about molecular orbitals

"Okay, but what happens when you combine atomic orbitals to make molecular orbitals? You have constructive interference from the positive + positive, and destructive interference from the positive + negative, but what about the negative + negative combination?"
"Good question, myself!"

Let's talk about the $\ce{H2}$ molecule. The proper way to find the molecular orbitals is to solve the Schrödinger equation for the entire system, which is really difficult to do. One way to find approximate forms of the MOs is to make linear combinations of atomic orbitals; this method is called the LCAO approximation. Let's call the 1s orbital of the hydrogen on the left $\phi_1$ and the 1s orbital of the hydrogen on the right $\phi_2$. From the previous sections, we have already established that as far as the hydrogen atom is concerned, the individual phases of $\phi_1$ and $\phi_2$ do not matter. So, let's assume for simplicity's sake that their phases are both positive.
Now, from what you already know, you can get two molecular orbitals $\psi_1$ and $\psi_2$:
$$\begin{align}
\psi_1 &= \phi_1 + \phi_2 \\
\psi_2 &= \phi_1 - \phi_2
\end{align}$$
These are the bonding and antibonding orbitals respectively (at least, to within a normalisation constant, which I'm not going to care about here because the details are irrelevant). Now let's talk about those combinations that we missed out.
$$\begin{align}
-\phi_1 - \phi_2 &= -\psi_1 \\
-\phi_1 + \phi_2 &= -\psi_2
\end{align}$$
We already said that $\psi_1$ and $\psi_2$ are (approximations of) solutions to the Schrödinger equation. That means that, from what we've talked about earlier, $-\psi_1$ and $-\psi_2$ must also be (approximations of) solutions to the Schrödinger equation. They must have the same energies as $\psi_1$ and $\psi_2$. In fact, as far as the molecule knows (and cares), they are the same thing as $\psi_1$ and $\psi_2$.

Now, since the individual phases of the atomic orbitals do not matter, if you really wished to, you could declare to the whole world that you define:
$$\phi_3 = \phi_1 \text{ and } \phi_4 = -\phi_2$$
i.e. left hydrogen 1s orbital, $\phi_3$, is positive and right hydrogen 1s orbital, $\phi_4$, is negative. In that case, you can construct the molecular orbitals:
$$\begin{align}
\psi_1 &= \phi_3 - \phi_4 \\
\psi_2 &= \phi_3 + \phi_4
\end{align}$$
The coefficients of the atomic orbitals would have to be different, since you insisted on having them in different phases - however, the outcome is the same! You get one bonding MO and one antibonding MO.

Upvoting simply for "Good question, myself!" (well, and it's a good answer, too!)

My confusion stems from not having any idea as to what a negative wave
  function is representing - can anyone give me some physical intuition
  on how this negative wave function could be correlated to something in
  reality?

Your confusion for its most part stems from two things:

You are using the wrong terms such as "positive" or "negative" wave function which shows that you don't quite understand what is going on here mathematically. This is especially important taking into account the second point.
You are looking for some physical intuition behind a purely mathematical model which has very little to do with physical reality.

Let us for the sake of argument define a positive function of a single real variable $x$ as a real-valued function that takes only positive values, i.e. such that $f(x) > 0$ for all $x$, and a negative function as a function that takes only negative values, i.e. such that $f(x) < 0$ for all $x$. I think this is something OP has in his mind talking about "positive" and "negative" wave functions. But then the way OP uses these adjectives to describe what he has trouble with is surely wrong, since given a wave function $\psi(x)$ how do you know is it a positive or a negative one? Obviously, you don't know that, and consequently, you can't tell either if $-\psi(x)$ is positive or negative.
These whole LCAO-MO business has simply nothing to do with wave functions being positive or negative, rather it is all about forming two orthogonal linear combinations of two atomic orbitals (AO) to form two essentially different molecular orbitals (MO). Nanoputian described this simple LCAO-MO formalism for the particular case of $\ce{H2}$ molecule in  details in his answer, and here I just want to warn OP (and others) about the danger of taking this picture literally.
MOs are formed from AOs in a mathematical and not a physical sense. It is just a game of numbers: take AOs and write MOs as linear combinations of them. One should not think that this primitive LCAO-MO model with bonding and anti-bonding MOs formed as a result of constructive and destructive interference of the corresponding AOs describes real physical processes. As I said, this model has very little to do with reality.

In addition to other answers and very briefly, a bond is formed when there is a lot of electron density between the nuclei, this lowers the overall potential energy. Conversely when there is not much electron density between atoms the potential energy is higher and this is anti-bonding. We identify anti-bonding by looking for nodes between atoms. As a rule of thumb, the more nodes the higher the energy hence more anti-bonding.
When adding or subtracting or multiplying wavefunctions together the normal rules of maths apply because wavefunctions are represented by normal mathematical equations. We have to remember that often they are presented in polar coordinates instead of the usual x,y,z so may not seem so familiar, e.g. shapes of s, p, d orbitals.
The sign of a wavefunction $\psi$ is just part of its mathematical description; some wavefunctions have positive and negative parts, e.g. spatial part of p orbitals, some are represented by complex numbers e.g. some d orbitals. We can let $-\psi$ be the same as $\psi$ as the probability of finding the particle in a small region of space $\tau $ to $\tau+d\tau$ is $\psi^2d\tau$. The measured value of some property,  X, e.g. position, is given by the expectation (or average) value; $<X>=\int \psi^* X \psi d\tau$, so again the sign of $\psi$ does not mater. ($\psi^*$ is the complex conjugate, which is only important if $\psi$ is a complex number.) When adding wavefunctions so as to make a linear combination e.g. $\psi = s_1-s_2+s_3$ is the same as $-\psi=-s_1+s_2-s_3$.
Chemists often use diagrams with the sign of the wavefunction labelled as $\pm $ or coloured in a particular way to help understand bonding. Its a very useful shortcut compared to calculating things out. The figure below shows some examples.

The p orbitals add 'in phase' as $p_z+p_z$ which is bonding and out of phase as $p_z-p_z$ which is anti-bonding.  The shading on the orbitals shows the 'sign' of the wavefunction. In $p_z-p_z$ the difference is not zero everywhere because the orbitals have different origins since they are displaced along the bond axis. The difference is zero at the midpoint, however, and this is called a node. The rules are that the same shading adds, different shading subtracts. The $s+p_z$ is non-bonding as the s orbital has overlap with both parts of the p orbital and the two overlaps cancel.  The three s orbitals can add, all three the same shade make bonding (grey or white its the same, $s_1+s_2+s_3$ is equivalent to $-s_1-s_2-s_3$) or subtract orbitals say as $s_1-s_2+s_3$ which is anti-bonding.

You have this misconception because you forget that a wave function is a wave!!!. Waves can be either in phase or out of phase. All these terminology of negative and positive waves drove your mind to mix unrelated concepts.
Imagine an isolated H atom, you can ask, what is the phase of his wave function?, and the answer will be, who cares or What phase are you talking about?. The phase only has physical meaning if it can be observed, like the superposition of waves. Then, asking the meaning of a "negative" wave functions is senseless.
Think about the superposition of EM waves or string waves. You don't ask for the meaning of a "negative" string wave, you speak about the string wave out of phase with respect another. The above is exactly the same but instead with quantum mechanical waves.
The use of "+" and "-" signs are a handy mathematical tool to describe the phase difference, but you can use another thing like colors.
I expect that the above clarification will help you to really understand how LCAO-MO theory works.


text:
Yellow stains after cleaning silver
I've recently moved to a old house and found some stuff in the attic, along that I found some really beautiful silverware(??) that unfortunately was VERY tarnished, one of the pieces was totally black while some had small places where I could see, in the back of one of them there's a label stating Silver 90.
I googled and found that using a solution of hot water, baking soda, salt, vinegar and strips of aluminium foil would help clean them. I submerged them and after some time I took them out and started to scrub them and the gross black stuff was going out, showing some silver and some yellowish. I them took a sponge and kind got carried away.
The pieces now all have this yellowish stains.
The one that was completely black had no silver-color in any part of it after cleaning, except the bottom, while the rest is all yellow now.
Here's one of the pieces

(the back is a white paper, white-balance is crazy with this piece)
My question is: Is this yellow stain any kind of dirt that wasn't removed in the process, or is this the underlayer of the piece(meaning that the silver is gone in those parts)? And above all, is the method used the best way to remove dirt from silver?

Did find a source and the nature of reaction apparently involves some standard chemistry, surface chemistry, some radical activity and electrochemistry resulting, upon warming, in the demonstration of some yellow elemental sulfur!
Here is a description of the reaction pathways, starting in alkaline conditions, surface chemistry on the Aluminum metal (which can even very slowly be attacked by water liberating hydrogen):
$\ce{Al + OH- -> Al(OH)3 + 3 e-}$
Of consequence, NaHCO3 is amphoteric and a source of H+ ions. Also, in the presence of solvated electrons, the creation of the hydrogen atom radical (per the source) along with hydrogen gas:
$\ce{H+ + e- ⇌ •H }$
$\ce{•H + •H -> H2 (g) }$
Also, per a galvanic cell between the Aluminum and the Silver metal, the half-cell reaction at the anode being given by:
$\ce{Al -> Al(3+) + 3 e- }$
At the cathode, the proposed half-cell reaction explaining the removal of Silver sulfide:
$\ce{Ag2S(s) + 2e- + 2H2O -> 2Ag(s) + H2S(g) + 2 OH- }$
See same source, Equation (6).
And, for the observed yellow color, the cited reaction:
$\ce{H2S + 2 OH- -> 2 H2O + S(s) }$
which is apparently confirmed by the cited source to quote:
"However, after the cleaning of heavily tarnished samples (immersed for 30 and 60 min) some small quantities of sulfur (~0.3wt.%) are detected on the surface by EDS analysis, and the colour is changed to light yellowish."
In total, this is constitutes an albeit complex and advanced explanation for the cleaning of silver from sulfide stains with Baking Soda and Aluminum. Note, also this supporting  source states to quote:
"The nascent hydrogen may also act as a reducing agent as it evolves from the metal surface. "
Unfortunately, the above mix of advanced chemistry is not generally suitable for all audiences acquainted with the removal of Ag2S by the Al/NaHCO3 protocol.

I suggest to do the following:
Line the bottom of a pan with aluminum foil. Put the silver piece on top of the aluminum foil. The silver piece and aluminum foil must be in contact with each other.
To about 2L of hot boiling water add about a half cup of baking soda (Be careful!). Then add the mixture to the pan. Make sure to cover the whole silver piece. The tarnish will disappear within several minutes.
The chemistry underlying this treatment is the following:
The silver tarnish is the chemical product of the reaction of silver with sulfur-containing substances in the air, $\ce{Ag2S}$. Silver sulfide is black. In this process, we convert silver sulfide to silver according to the following reaction:
$$\ce{3Ag2S + 2Al->Al2S3 +6Ag}$$ In fact, aluminum has a greater affinity to sulfur than silver.  The reaction between silver sulfide and aluminum takes place when the two are in contact while they are immersed in a baking soda solution. And it becomes faster when the solution is hot.The silver piece should touch the aluminum because a small electric current flows between them during the reaction. The formed aluminum sulfide may adhere to the aluminum foil, or it may form tiny, pale yellow flakes in the solution.
The advantage of this method of cleansing is that we do not remove the tarnish with some silver. We simply transfer silver sulfide to silver.

That's exactly what I did in the first piece I tried, but it had so much tarnish that it didn't do much. Than I found that adding salt and vinegar would speed up the process.  I started to see some flakes in the solution after sometime, but still after some minutes it was still black, I then took it out of the solution and with a sponge scrubbed a dark goo from the piece. Now I have the piece with silver spots and yellow stains, and some minor black tarnish(where it's hard to scrub).

I wonder if what I scrubbed was the silver sulfide, but it's weird that it didn't budge a single bit when I left it in the solution.

I think that the silver piece was badly tarnished, you would need to reheat the baking soda and water mixture, and give the silver several treatments to remove all of the tarnish. The process is not immediate.


text:
What are analytical expressions for the isothermal change of Gibbs energy?
The change of Gibbs energy at constant temperature and species numbers, $\Delta G$, is given by an integral $\int_{p_1}^{p_2}V\,{\mathrm d}p$. For the ideal gas law, $p\,V=n\,RT$, this comes down to
$$\int_{p_1}^{p_2}\frac{1}{p}\,{\mathrm d}p=\ln\frac{p_2}{p_1}.$$
We find this logarithm in many formulas in chemistry.
I tried and failed to find a general formula for $\Delta G$ for the Van der Waals gas (see this question on the physics board).
It makes me wonder:
Is there even another case of a chemical situation, other than the ideal gas, where $\int_{p_1}^{p_2}V(p)\,{\mathrm d}p$ or $\int_{V_1}^{V_2}p(V)\,{\mathrm d}V$ is known analytically?

My questions are mostly motivated by the desire to understand the functional dependencies of the chemical potential $\mu(T)$, that is essentially given by the Gibbs energy. For the ideal gas and any constant $c$, we see that a state change from e.g. the pressure $c\,p_1$ to another pressure $c\,p_2$ doesn't actually affect the Gibbs energy. The constant factors out in $\frac{1}{p}\,{\mathrm d}p$, resp. $\ln\frac{p_2}{p_1}$. However, this is a mere feature of the gas law with $V\propto \frac{1}{p}$, i.e. it likely comes from the ideal gas law being a model of particles without interaction with each other.

Hi, I think this might be a rather lousy solution as it does not explicitly address the van der Waals state equation, but Atkins' Physical Chemistry talks a little about the fugacity of a real gas, how it can be derived from the compression factor $Z = pV_m/RT$, and how it can be used to find $\Delta G$ for the process you are asking about. Because I'm too lazy to type it out in a proper answer, I've just uploaded the relevant section from the 9th ed here, you can take a look if you want: https://drive.google.com/file/d/0B0GATVY1ly6cOXMtX1JubTlhdm8/view?usp=sharing

I tried the integration and I see what you mean, I couldn't make $V_m$ into a function of $p$ so I have no idea how to evaluate the integral $\int V_m,\mathrm{d}p$. Not sure if what I wrote above is useful because it also involves an integral which is probably equally difficult to carry out. Hopefully someone else will have some idea.

@orthocresol: No, I don't think so. The problem with this, as I understand it, is that the fugacity is exactly introduced to take into account that some gases $\Delta G$#s do diverge from the $pV=nRT$-model. That's ad hoc and begins with what I want to end up with. It's not like there the real volume $V_m$ in your link has a theoretical model, an equation of state that's different from $=\frac{nRT}{p}$ in a specified sense. I'm asking for a material theory, giving $V(p)$  that's different from $\propto \frac{1}{p}$ and still let's me compute $\int V(p),{\mathrm d}p$.

Oh I see, you want something specific... Try the virial equation of state: $pV_m = RT(1 + B'p + C'p^2 + \cdots)$ I think that one is integrable. Or do you find that it's still too close to the ideal gas law? The most common form in which the virial expansion is written is $pV_m = RT(1 + B/V_m + C/V_m^2 + \cdots)$ but I suspect that's not going to work.

@orthocresol: Before you post you think it's integrable, why not try it? This virial equation of state doesn't contain positive powers of the pressure, in any case.

I'm just trying to help you... Of course it's integrable. $V_m = RT(1/p + B' + C'p + \cdots)$ means $\int_{p_1}^{p_2}V_m,\mathrm{d}p = RT[\ln(p_2/p_1) + B'(p_2 - p_1) + (C'/2)(p_2^2 - p_1^2) + \cdots]$. And I already did mention that the virial expansion is usually written the other way, and you probably know that virial coefficients are tabulated for the expansion in $1/V_m$... This is just another form, which is equivalent, but the coefficients $B'$ etc. are of course different. I don't have any other ideas. Sorry.

With regard to the Van der Waals equation, have you considered integrating by parts:  $V\mathrm dp=\mathrm d(pV)-p\mathrm dV$.  That can certainly be integrated analytically.
The answer will be in terms of $V$, but that’s not a big limitation.


text:
Are photons ejected from atomic nuclei?
I have heard that electrons absorb or eject photons when transitioning from one orbital to another. Is this correct? Can atomic nuclei eject photons?

Yes, gamma rays, but it's about physic not chemistry.

What about photons?

Well, you should know that gamma rays are photons

I have heard that electrons absorb or eject photons when transitioning from one orbital to another. Is this correct?

Not exactly.  The atom as a whole emits or absorbs the photon.  There is no reason to single out the electron versus the nucleus in such transitions.

Can atomic nuclei eject photons?

Yes there are two ways a nucleus in particular (as opposed to an atom taken as a whole) can emit photons.
First, nuclei can exist in excited states. When an excited nucleus transitions to the ground state, a photon in the gamma or X-ray range of the electromagnetic spectrum is emitted.  Electrons do not have an analogous mechanism for emitting photons, because electrons are fundamental particles, while nuclei are not.
Second, if a nucleus has a nuclear spin (S), then in a magnetic field there will exist  2S + 1 quantized energy states of the nuclei.  Transitions of nuclei between these states emit or absorb photons, the energy of which depends upon the strength of the field, for example in the radio frequency range.  This is the basis of NMR and MRI.  The electron has an analogous mechanism for emitting photons, known as ESR or EPR.

At least in my area of physics/materials science, we consider gamma rays to come from nuclear transitions, and x-rays to come from electronic transitions (which is why the energy ranges overlap).

@JonCuster maybe I should say "high energy" or "KeV-MeV" instead of  "gamma or X-ray range".


text:
Confused by Alternate Names
I am highly allergic to propylene glycol and polyethylene glycol.  I am finding many hidden sources of both and MANY synonyms for PG and PEG.  Is there a practical way to tell how closely related another chemical might be to my allergens using such resources as the U.S. Household Products data base?  For example:  glycerin's alternate name is propandiol.  Propandiol is also an alternate name for propylene glycol.  I'm very confused about sodium laureth sulfate in particular.  It has PEG listed as an alternate name, but is not an alcohol like the glycols.  Thank you for any clarification you can give me!

When I searched U. S. Household Products database for "sodium laureth sulfate," I got a hit for "diethylene glycol monolauryl ether sodium sulfate," which has "PEG-(1-4) Lauryl ether sulfate" as a synonym: that is not the same as "PEG." Just to clarify, is that what you're referring to above?

Also, glycerin is propane-1,2,3-triol and is not an alternative name for propylene glycol (propane-1,2-diol). You might want to edit your question or reevaluate some of your thoughts in order to clarify it.

You'll probably be better served asking an allergist about this.
While many things contain PG, that is not the same as a compound similar to PG, which could have PG (or part of its IUPAC name, propane-1,2-diol, not to be confused with propane-1,3-diol, which would be its possibly harmless-to-you although structurally similar isomer) as part of its name or could interact with your body in a similar way. Case in point, your confusion over sodium laureth sulfate, which is nothing like PG, although that's an irritant for everybody. However, it looks like PG allergy is sometimes co-sensitized with corticosteroids - which are themselves structurally nothing like PG!
Thus, the confounding variable here is the way your body decides what will or will not set off your immune system, which can have very little to do with the chemical behavior of the allergen and everything to do with what mechanisms your body is using to make that distinction.


text:
What is the difference between reacting and dissolving?
When one chemical is added to another how would we come to know whether the compound dissolves in it or reacts with it? For example $\ce{AgNO3}$ dissolves in $\ce{CCl4}$ whereas it could also have been a double displacement reaction. And what is the major difference between the two  processes at atomic level?

Dissolving is a type of reaction. Double displacement is a different type of reaction. So something which is dissolving is also reacting.

@bon Does this mean that new compounds are formed even when one compound dissolves in another?

@user456 New species may be formed, like solvated ions for instance but these don't count as new compounds.

@bon that sounds a bit misleading. Dissolving is a physical process, while double displacement is a chemical process. So technically speaking, no new chemical compositions are being made when one substance is dissolved in another.

CCl4 is a covalent compound (having no free ions). Thus it cannot enter into an ionic reaction/double displacement reaction.
Basic Requisites for checking reactivity

Reactivity Series (Elements higher up in the series replace the lower elements)
Checking whether bonds are covalent or ionic (You might just need to remember which compounds are which type. Some compounds can have both type of bonds)

Bonds

Ionic = "Ionic bonding is a type of chemical bond that involves the electrostatic attraction between oppositely charged ions. These ions represent atoms that have lost one or more electrons (known as cations) and atoms that have gained one or more electrons (known as anions)." Wikipedia
Thus the elements can take or give electrons. In a solution they exist as ions.
Covalent = "A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs and the stable balance of attractive and repulsive forces between atoms when they share electrons is known as covalent bonding." Wikipedia
Thus it has no free electrons to give or take.

Then how does it dissolve in agno3?

As a matter of fact, when we say dissolve, it means that when you put something into e.g. water, that chemical might break down into tiny pieces or particles -- you will not be able to see those pieces, and we call this "dissolving".
And on the other hand, when we say react, it means that the chemicals  might bond with each other and then produce a new chemical, and this new chemical will then be called as product also.
Consider when you mix sugar in the water present in your cup, then you will first see pieces of sugar in a big size, but after you mix the water, then the water will become clean, but the water has turned those pieces of sugar into very tiny pieces that our eyes cannot see.


text:
What's the reaction type of the Hydrogen shift?
Rearrangement of cycloheptatrienyl carbene yields heptafulvene via a hydrogen shift.

I see that the reaction is favorable because there is a conjugative effect with the $\pi$ system of the ring in the product.
Does this rearrangement happen by a radical transition state (TS) or is the mechanism more of a sigmatropic rearrangement?
My arguments for

radical TS: The electron of the carbene (let's assume a singlet carbene here) lies close to the hydrogen, it could be a planar cyclic transition state.

[1,2]-Hydrogen shift: From my chemical intuition I think the $\ce{C-H}$ bond is more a polar bond and therefore I would describe hydrogen more as a proton (positive partial charge). I would say it's a [1,2]-sigmatropic rearrangement, 2 electrons are involved, and therefore suprafacial and describing the carbene $\mathrm{sp^2}$ as HOMO and the hydrogen as LUMO.

What's the correct mechanism for this rearrangement?

This reaction can be viewed as a 1,2-sigmatropic hydrogen shift or as a 2+1 (if we are counting atoms, 2+0 if we are counting electrons) cycloaddition (adding the carbene's empty $\mathrm{p}$-orbital across the $\ce{C-H}$ bond - analogous to its addition to an olefinic pi bond to produce a cyclopropane). However the reaction is most commonly viewed as and referred to as an "insertion" reaction because the carbenic carbon is inserting itself in between the carbon and hydrogen atoms that comprised the $\ce{C-H}$ bond.

(image source)
Here's what is known.

Thermal decomposition of most diazo compounds produce singlet carbenes.

Singlet carbenes add stereospecifically to olefins. That is, addition of a singlet methylene carbene to trans-2-butene produces only the corresponding trans-dimethylcyclopropane

(image source)
Further, insertion of singlet carbenes into stereogenic centers results in retention of stereochemistry at the chiral center.  These facts suggest that both the cycloaddition and insertion reactions of simple singlet carbenes either proceed in a concerted fashion, or if any intermediate is formed it is extremely shorted-lived in order that stereochemical information be preserved.

Simple carbenes are typically mildly electrophilic. The HOMO (the filled $\mathrm{sp^2}$ orbitals) and LUMO (the empty $\mathrm{p}$-orbital) for a singlet carbene are shown on the left side of the following figure.

(image source)
Therefore, we would expect the interaction of the carbene LUMO (the empty $\mathrm{p}$-orbital) and the $\ce{C-H}$ bond HOMO to be controlling, as the energy separation here is much less than the other possible HOMO-LUMO interaction between the carbene HOMO (the filled $\mathrm{sp^2}$ orbital) and the much higher in energy $\ce{C-H}$ antibonding orbital, the LUMO.
This controlling HOMO-LUMO interaction involving the empty carbenic $\mathrm{p}$-orbital in the rate determining step is consistent with the observed electrophilic nature of the carbene.
All of this information taken together suggests that interaction of the empty carbenic $\mathrm{p}$-orbital with the electrons in the $\ce{C-H}$ bond is the critical step. A geometry like that shown below is likely preferred. As the insertion proceeds, the trigonal carbenic carbon rotates.  The rotation starts to bring the carbenic $\mathrm{sp^2}$ orbital into alignment with the p-orbitals in the 7-membered ring. Also during the rotation the $\mathrm{sp^2}$ orbital starts its rehybridization into a $\mathrm{p}$-orbital. When completed we have created the final product, heptafulvene. The reaction is likely concerted and certainly does not involve any long-lived intermediates. Note too that the heptafulvene is formed with increased electron density on the former carbenic carbon in accord with the dipolar nature (dipole moment ~$\pu{0.5 D}$ due to resonance stabilization) of the molecule.

Thank you for your comprehensive answer! I understand your answer in the way that the carbene makes an insertion into the CH bond and the empty p orbital helps to stabilize this conformation with the hyperconjugative effect. I thought that the empty p-orbital is already perfectly aligned with the π system in the planar ring in your drawing, therefore I'm wondering why we still need a rotation. I would prefer a bisected conformation of the p & carben sp2 with the $\sigma_{CH}$ bond. Is the roation in clockwise or anticlockwise direction if we look from the carbenic carbon onto the CHR2 carbon?

In the early stage of the reaction, the carbene p-orbital is separated from the ring's pi system by an insulating sp3 carbon so there is minimal interaction between the carbenic p-orbital and the ring's pi system.  The conformation I drew up above with the p-orbital eclipsing the C-H bond seems best aligned for the insertion.  The p-orbital becomes a C-H bond and then upon rotation the filled sp2 orbital rehybridizes and becomes the terminal p-orbital in the exo double bond.  The direction of rotation does not matter; because of the p-orbital symmetry,

rotation in either direction produces the same result.

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


text:
Does oxygen catalyze the decomposition of sucrose?
When a bottle of sports drink that contained sucrose is opened, is it expected that the sucrose contained therein will split into its constituents (fructose and glucose) more rapidly than it will via acid hydrolysis?
I understand that in acid hydrolysis there is nucleophilic substitution (provided the acid is protic). Does a similar phenomenon occur when the contents of the bottle are exposed to oxygen?
I am trying to interpret some unexpected results I obtained from an experiment, and I assume what I have detailed here is what happened. However, I cannot find any relevant information to confirm or deny my hypothesis.
RESULTS:
The experiment showed 0.72 g of glucose was present per 100 mL of solution. The bottle label states that it is 0.5 g per 100 mL.

Hmm, there might exist reaction that is catalyzed by O2 but generally it doesn't catalyze anything imo

The manufacturers are allowed to round their numbers.

Acid present in drink could catalyse hydrolysis of sucrose

If any sucrose hydrolysis were occurring due to oxygen as a factor, it's more likely due to acids formed as it oxidizes species in the solution (for example, alcohol groups to carboxyls), which conceivably could include sucrose itself. But, it seems more likely that the $\ce{CO2}$ in air would be a factor because $\ce{CO2}$ dissolution increases the acidity of the solution, in addition to any other acids present (citric, for example).
However, sucrose hydrolysis is really slow unless the acidity and/or heat is significant ($\mathrm{pH} < 0$, or heat near boiling), so I doubt either of these explanations, unless these are really old, and/or had been heated.  More likely the glucose was there in the first place.
The way to properly answer this question would be to do control studies in which known amounts of sucrose and acid are initially present and tested with or without exposure to air having varying amounts of oxygen and carbon dioxide.


text:
Why is acidic hydrogen peroxide a potent oxidizer?
When hydrogen peroxide acts as an oxidizer it itself is reduced. Broadly this means that the peroxide bond is broken and that water is formed.
However, why is there a tendency for hydrogen peroxide to be such a potent oxidizer in acidic media?

In acidic solutions, H2O2 is one of the most powerful oxidizers known—stronger than chlorine, chlorine dioxide, and potassium permanganate.

Why is it not as potent without added acid, i.e. 3% hydrogen peroxide, the stuff you find in medicine cabinets, isn't a very good oxidizer?
The "duh" answer is that hydrogen peroxide is especially unstable in acidic media. But why can't acidic media stabilize hydrogen peroxide, and specifically the peroxide bond?
Simple electrostatics tells me that having a bunch of positively charged hydronium ions around the positively polarized oxygens in the peroxide bond isn't a "good" situation. Of course there are anions from the dissociation of the strong acid but these anions necessarily have diffuse negative charge, which is probably not enough to stabilize the polarized peroxide bond to significant extent. Is there anything else?

Just consider the Nernst equation for the half-reaction
$$\ce{H2O2 + 2e- -> 2OH-}$$
You see the potential is clearly pH-dependent. Increased acidity will decrease the concentration of the reduced form, thus shifting the equilibrium to the right and consequently making peroxide a more potent oxidizer.
(BTW, even pH-neutral hydrogen peroxide is far from harmless. It can leave white marks on your hands and gnaw a hole in your jeans, albeit not instantly.)

The immediate idea that jumps to mind is that acids, especially concentrated acids, have the ability to protonate the hydrogen peroxide. The then protonated hydrogen peroxide can then loose a water molecule either in an $\mathrm{S_N 2}$ type mechanism with rear-side attack on the other oxygen or in an $\mathrm{S_N 1}$ type mechanism to create a hydroxyl cation — although I would deem the latter highly unlikely due to the extreme electrophility of a potential hydroxy cation.
$$\ce{HO-OH + H+ -> HO-OH2+}$$
In the rear-attack version, this would directly void the attacking atom of two electrons formally (unless fluorine attacks which is even less likely). In the other case, the cation would probably jump to anything that is not oxygen and has electrons to share.
Additionally, hydrogen peroxide is stabilised by interactions of the free p-orbital of one oxygen and the $\sigma^*(\ce{OH})$ orbital on the other oxygen. (And vice-versa, of course.) When protonated, the proton will likely be added to the p-type lone pair rather than to the sp-hybrid type lone pair, inhibiting this stabilising interaction and further decreasing the strength of the $\ce{O-O}$ bond.


text:
How can I synthesize an (E)-configured enone using the Wittig reaction?
I have been asked to synthesize the compound (E)-5-methylhex-3-en-2-one:

How can this be achieved using the Wittig reaction?

Does it have to be done with a Wittig reaction? The structure cries out for a aldol condensation in my mind.

The Wittig reaction is used to synthesise double bonds usually from an aldehyde/ketone and an alkyl halide. The following scheme might be one way to achieve your target molecule. Most likely you will have to modify the Wittig reaction, because it is usually (Z)-selective. See also: Why does the unstabilised Wittig reaction selectively form cis alkenes?
If you consider methylglyoxal no good starting material, then you might need to come up with a synthesis for this, too. For the sake of selectivity, you might need protection groups along the way.

This is only one possibility and I am not sure if it actually works, or if the Wittig reaction is the right tool to employ. I am certain, that there are a many other ways of synthesising your target molecule and an extensive literature search would be appropriate.

Thanks you've helped me and provided me with some more reading and things to think about but I think as the ylide you've suggested is an unstabilised ylide I believe it gives anti-addition which would be needed for this wittig reaction. Thank you :)

@Michael Unstabilized ylides tend to give the cis alkene as the major product.

Old question, but if you just reverse the phosphorus ylid + carbonyl group then you get the correct (E)-selectivity and it also doesn't require the use of a protecting group. (Of course, jerepierre's answer already mentions this.)

The Wittig reaction allows for the construction of alkenes from a phosphorus ylide and an aldehyde (or ketone). Shown in the retrosynthetic sense, keeping in mind that there are two possible Wittig disconnections:

Applying this idea to (E)-5-methylhex-3-en-2-one gives the following possibilities:

Which option is chosen often comes down to which gives the most readily accessible starting materials. However, in this case, the stereochemical outcome is critical, and that is controlled by the nature of the ylide used. Looking now in the forward direction, the top option uses an unstabilized ylide, which tend to give the cis-alkene as the major product. The bottom option uses a stabilized ylide, which changes the stereochemical outcome to favor the trans-alkene.

In practice, many chemists would opt for the Horner–Wadsworth–Emmons variant, which gives the trans-alkene as the major product and avoids the production of triphenylphosphine oxide, a byproduct that can be difficult to remove.


text:
What makes water a liquid (when it is)?
I'm curious about what the minimal conditions for some collection of water molecules to be liquid water are. What is the minimum number? What sort of bonds must hold? What other sorts of things I don't know to ask about need to be true?
In my peeking around a subject I have little familiarity with, I've seen mention of a tetrahedron, suggesting a lower bound of 4 molecules for water to be a liquid. But then I also stumbled across the following (from p. 23, here):

Whatever the model concerned, the presence of a fifth
  water molecule is necessary to be considered in explaining
  the molecular mechanism of  defect formation and Hbond
  network rearrangement.

So it seems that although only four molecules are in this bond network undergoing constant change, there must actually be a fifth "neighbor" molecule for the bonds to function appropriately. Is this fifth molecule part of the liquid, or merely a helpful neighbor who makes the four molecules' liquification possible?
In short, what makes water a liquid (when it is)?

Please re-tag and suggest edits as appropriate. Thanks!

Hmm, your problem is rather unclear and it seems you didn't understand  your article well - it's simply describing hydrogen bonding in liquid (normal, big amount) water - how hydrogen bonding works there.

Let me venture an answer as well as I can.
I would point out that the article you are referencing is speaking of the arrangement of molecules in bulk liquid water when it is talking about this "fifth water" being present. The key is to note that they say a "fifth water molecule in the first coordination shell (23)." That means they are looking at how water molecules are coordinated with each other in liquid water. They are studying a massive (compared to a single water) amount of water, so the question of its liquidity is no question at all.
The "fifth water" thing is curious because it is common belief that waters arrange themselves in a tetrahedral arrangement. This is known to be true for ice, but is in question for liquid water. For instance, this article about the liquid water hydrogen-bonding network indicates that water, as it is heated, begins participating in only two hydrogen bonds rather than four.
Now, to answer the question of what makes liquid water be liquid water (when it is indeed liquid water). I think from the outset, it should be clear that this is a very difficult question to answer and any answer will likely be debated. I say that because you must define how many molecules it takes to make a liquid which is a very difficult thing to do.
I will tell you, however, that I have some useful information from research I am doing on water clusters and eventually liquid water (as a computational chemistry research assistant). The largest cluster we have studied is called DD*(20,1) which means (distorted dodecahedron with a water inside the cluster).
An idea of what $\ce{(H2O)21}$ can look is this: (taken from this very interesting article) water 21 http://pubs.acs.org.ezproxy.spu.edu/appl/literatum/publisher/achs/journals/content/jpcbfk/2006/jpcbfk.2006.110.issue-38/jp056416m/production/images/medium/jp056416mf00001.gif
The DD*(20,1) arrangement is image C.
As you can see from these different isomers, they are very distinct and no doubt each monomer behaves dramatically different from the next monomer. 21 monomers is quite a lot for a cluster, but even this does not mimic the properties of liquid water at all particularly because one would still consider this a gas-phase cluster. The fact that we even have isomers is indicative of not being near the point where something is considered liquid water. There are water clusters whose energy minima have been identified theoretically all the way from n=2 to (I've seen an article that says) n=60. Read more about some of that here.
One point I would like to make is that in structure C, for the research I'm doing, we have identified the vibrational properties of many of the monomers as being significantly different from those of molecules in liquid water.
So, all that being said about water clusters, we are certain that many more than five or 21 waters must be present to truthfully call something a liquid. We should then start looking for a system which is just large enough to accurately predict the properties of liquid water which are found experimentally.
As it happens, it is very difficult to do this on any scale, theoretically, at least, and identifying how many water molecules one is dealing with during an experiment to a quantifiable amount of precision is quite difficult. There are, however, good approximations of liquid water which are generated using Molecular Dynamics simulation software (here's a wikipedia page about MD software). The various models used for approximating the behavior of water do quite well at mimicking water's properties from an MD perspective. You can read about the TIP4P model of water if you want.
From the paper I just linked about TIP4P, they say that they approximate the properties of liquid water using 360 water molecules in their MD simulations, so that can act as a reasonable baseline for how many monomers are necessary before a system behaves like liquid water.
I say baseline for an important reason. The results found in that paper use MD simulations which pull a clever trick. Rather than having 360 waters that essentially float around in a box with nothing at all beyond the edges of the box, the software will actually mirror the behavior of a molecule beyond the walls of the box in which you are simulating. That means if you pick a single water, you would find that exact same water in the same place a full "simulation box" away from where you originally found it. So, if this says 360 waters, the number of waters for which interactions are being calculated is actually much larger than the amount of interactions 360 waters alone would have.
All that leads me to conclude that water behaves like water probably somewhere around 500 monomers. It must be noted, however, that the monomers behaving like liquid water as we know it will be the monomers on the inside of those 500 (but likely more) waters. Once you get towards a boundary, things start getting weird.
Hope that helps.

Thanks so much for this answer. It helped me a lot in thinking about these sorts of things. It winds up being relevant to assumptions people make in formal semantics, so thanks for the info.