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How do galvanic cells work?
I was wondering how a galvanic cell works
If you look at the picture below, if I have magnesium metal in sulfuric acid, will the metal react with the acid to be then turned into magnesium ions and in the process give away 2 electrons and that is how electrons are given in any cell: the electrode interacts with the electrolyte in the anode and then gives up electrons? And if this is the case, then why wouldn't the zinc replace copper in copper sulfate at the cathode and also give 2 electrons away?
What am I doing that is wrong?

This galvanic cell is bad in that each half-cell would readily react on its own (zinc would replace copper, magnesium would replace hydrogen). That being said, it still might be able to produce some current.

ok then how should one choose an electrolyte

Choose it so that it would not react with the corresponding electrode. In this case, $\ce{MgSO4}$ and $\ce{ZnSO4}$ would do.

so if the electrodes don't react with the electrolyte, then how do they give electrons?

Like always: ions to the solution, electrons to the wire.

so if would change the electrolyte, what would your different electrolytes be? And again, how are the electrons given if the electrode doesn't interact with the electrolyte

Basically, I am trying to ask is if you take a zinc copper battery, zinc replaces copper sulfate in solution and thus gives away 2 electrons. But this is when zinc is physically in copper sulfate solution. However, when they make a zinc-copper cell, they put zinc in zinc sulfate solution (and copper in copper sulfate solution), then how does the zinc give up its 2 electrons?

When zinc is physically in copper sulfate solution, this is not a galvanic cell, or probably a bad galvanic cell like your initial example. Like I said, the electrolyte should not be able to react with the electrode on its own (without the wire).

Ya so then how does the electrode give up its electron?

Like always: ions to the solution, electrons to the wire.

could you please go into detail and explain what that means?

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

my question is what is responsible for the electrons? The metal electrode just can't magically donate its electrons?

Yes it can, though not magically.

could you explain that? because from what i have read, when you place an electrode in a solution, the metal will interact with the solution

okay so let's say I put sodium bicarbonate with magnesium, would the cell work then?

Yes, it would. But then again, your initial design would work too, just not very efficiently, because some material would be lost to the side reactions. You got rid of the side reaction with Mg, but you still have the one with Zn.

alright so then if I put copper and copper sulfate instead of zinc, would that work?

Yes, it would surely be better this way.

and electrons will be transferred from one electrode to other?

Yes, they will (if you connect them with a wire, that is).

I agree with Ivan it's a really bad galvanic cell and both metals would displace the components of the respective containers .But still one of them is more electropositive than other hence there would be some net flow of electrons .You can also compare cells by comparison of electrode potential of respective half cells which in your case have same sign and hence EMF net would be low .

my question is what is responsible for the electrons? The metal electrode just can't magically donate its electrons?

First of all I think that it would not work. In a galvanic cell they should create an electron potential, but in your cell, half cell would react within itself, so no electron potential will be created. The one with $\ce{Mg}$ and $\ce{H2SO4}$, $\ce{Mg}$ will displace $\ce{H2}$ creating $\ce{MgSO4}$. In a cell it is necessary to have a similar salt in container to the plate we have taken, here it is of $\ce{Mg}$ so that they would not react itself rather react with the one in other half cell. I hope that this will be enough. I am ready to help you further.


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Intermolecular Forces
Do intramolecular forces affect intermolecular forces (other than in polarity)E.g. in a molecule of water, when oxygen and hydrogen are covalently bonded, electrons spend more time near the Oxygen. Firstly, do electrons actually revolve around the atom?
How does this differ if double and triple bonds were in substance that is covalently bonded? Would they also spend more time near the more electronegative section of the molecule?

Possible duplicate of Does the shared electron in Covalent bonds revolve around nucleus?

To answer your first question, Electrons don't revolve around the nuclei like planets in the solar system (as the bohr model may imply). The location of the electron is based upon probability.
In an example The Hydrogen atom has one electron. And according to the bohr model, it follows a fixed orbit like a planet. This however is not accurate! The electron can actually be found almost everywhere, but it is usually found in a very particular radius about the nucleus. This is illustrated in a probability density. (https://universe-review.ca/I12-06-wave.jpg)
It is is actually possible for the electron to be found inside the nucleus! (very very low percentages).
So, Because Hydrogen is much less electronegetive than Fluorine, The Electron density map will change such that it is much more probable for the electrons to be found around the Fluorine atom.
The same idea applies to double and triple bonds. Although the probability densities of those will be different from a single bond, the same general idea holds.
There is a lot of Quantum mechanics to this. If you would like to know the math  you may want to head to the Physics forums.

While electron density etc is useful, lots of people forget that electron has not only wave but also corpuscular character - they indeed revolve around nuclei, it was even directly observed!

@Jess L. - got it, but do these intramolecular forces affect intermolecular forces in a way other than polarity?

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Arrhenius equation derivation
I couldn't find the derivation of Arrhenius equation. So, How can we derive the Arrhenius equation:
$k = A \mathrm e^{-E_\mathrm a/(R T)}$?

Well, the Wikipedia article gives the original Arrhenius references if you would like to follow his thinking. But, what do you mean by 'derivation'? This isn't something that comes out of a pure math proof starting from F=ma. And, sometimes it works and sometimes it doesn't, depending on the system.

As already pointed it out, the Arrhenius equation isn't born out of pure math. Thinking of it more like an empirical equation that fits experimental data in most situations (and sometimes it doesn't). "Arrhenius-like equations" do come out of theoretical considerations when you look at collision theory or transition state theory.

The equation is rooted in experimental observations. It is not derivable from any physical formalism, and it is not the best way to describe many reactions. Similar equations can be derived from Transition State Theory. It is important to note that the similarities are limited to the functional form.

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How to calculate whether a redox reaction is spontanuous?
I have some difficulties to apply the basic concepts of calculating whether a reaction is spontanuous or not.
given:
$$\ce{H_2SO_3(aq) + 2Mn(s) +4H^+(aq) -> S(s) + 2Mn^{2+}(aq) +3H_2 O(l)}$$
and those tables containing the stadnard reduction potentials:

The formula to apply is the following: $E_{cell} = E(reduction)_{cathode} - E(oxidation)_{anode} $
$E_{cell} = 0.20 - (+1.23)$
$E_{cell} = -1.03$
The answer is that the reaction occurs sopontaniously. However this is not what I calculated, according to my calculations it doesn't. What am I doing wrong?

This 0.20 is for completely different reaction...

@Mithoron in that case could you tell me which value I should pick? I though 0.20 was correct.

The correct reduction potential isn't on the table you provided.  For the reduction of sulfurous acid to Sulfur, the reduction potential is +0.45 V.  You also used the wrong reduction potential for the oxidation half reaction as well.  The reduction potential for the Manganese(II) ion to solid Manganese is -1.18 V.

You need to approch the question in a systematic manner.  The first step is to write the reduction and oxidation half reactions:
Reduction Half Reaction: (cathode)
$\ce{H2SO3 + 4 (H+) + 4 e- -> S(s) + 3 H2O}$
Oxidation Half Reaction: (anode)
$\ce{Mn -> (Mn^2+) + 2 e-}$
Now find the reduction potentials for both half reactions (only one was on the tables you provided, but a different set of tables might have both).  Remember they are reduction half potentials so to find the reduction potential of the oxidation half reaction you must reverse the reaction direction:
$\ce{(Mn^2+) + 2 e- -> Mn}$
The reduction happens at the cathode, and oxidation happens at the anode.
$$\ce{E_{cell} = E_{cat} - E_{an}}$$
$\ce{E_{cat} = +0.45 V}$
$\ce{E_{an} = -1.18 V}$
$\ce{E_{cell} = (+0.45) - (-1.18)}$
$\ce{E_{cell} = +1.63 V (spontaneous)}$
Just to clarify again, the cathode's reduction potential (+0.45) was NOT on the tables you provided.  I had to search the internet for a different set of tables which did have the cathode's half reaction reduction potential.


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Is the decomposition reaction of nitrogen monoxide to dinitrogen and dioxygen possible?
Is the reaction $\ce{2NO -> N2 + O2}$ possible? Of course with some kind of catalyst plus high temperature. How can I find the answer to such questions and where can I find more information about this reaction?

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

@Mithoron In catalytic converters, don't they need a reduction agent, ie. hydrocarbons or carbon monoxide? I was clear that I did NOT include these in my equation in my question.

You should look up enthalpy of formation and enthalpy of reaction in your textbook. Also have a look at (Gibbs) Free energy change on reaction. This will tell you  if the reaction is going to be spontaneous, as it is in this case.
The thermodynamic energies never tell you how fast the reaction will be as these relate only to the starting (reactant) and ending (product) energies. The activation energy, the barrier between reactants and product is unknown to thermodynamics.
Secondly, a catalyst will not change the thermodynamics of the reaction, the starting and ending states are the same, but it speeds up a reaction by allowing the reaction to proceed by an alternative pathway with a smaller activation barrier.

Yes, that reaction is possible. It is used in catalytic converters to help convert exhaust into less toxic gases. The catalytic converters are generally platinum, or palladium and rhodium. Platinum can reduce and oxidize, while Rhodium just reduces and Palladium just oxidizes. Other material can be used but they have more limitation.

Erm, since these are catalysts we’re talking about, they’ll leave the reaction unchanged so they will reduce and oxidise ;)

In catalytic converters, don't they need a reduction agent, ie. hydrocarbons or carbon monoxide?

Yes. They do. Sorry.


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What are common procedures for lab absorption spectroscopy of gases?
I'm doing research for applications of absorption spectroscopy. I'm trying to find common practices such as how the gasses are mixed, measured, contained, and discarded.
The gases involved will be those commonly found in the atmosphere or the human body. Top among these will be $\ce{NO2}$, $\ce{CO2}$, and $\ce{CO}$.

I certainly hope that $\ce{CO}$ is not commonly found in the human body...

@Zhe, that caught my eye too, but it does say "or the human body" ;)

Non-dispersive infrared absorption spectroscopy is commonly used to measure $\ce{CO}$ and $\ce{CO2}$ (and other gases, like methane and $\ce{NO2}$). The absorption overlap from water interferes with $\ce{CO2}$ measurements in particular but various algorithms can be used to mitigate this.
$\ce{NO2}$ is more commonly measured by chemiluminescence because of its high sensitivity, but FTIR and other techniques are used for higher levels (i.e. exhaust pipe and stack emissions).


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How many stereoisomers are possible for the compound in picture?

How many stereoisomers are possible for the compound in picture? I'm getting 8. Is that correct?
My approach was as follows: The left hand methyl groups may be both up or one up and one down. For each of these cases the groups on the right may be in $(E{,}Z'),(E'{,}Z),(E{,}E'),(Z{,}Z')$ configurations. So $4\cdot2=8$.
I think it is quite evident that carbon having 2 π bonds must be linear. I apologize for the non-linearity in the picture. I hope you will be able to understand the question otherwise.

Carbon with two double bonds is linear, not bent.

@IvanNeretin I don't have a better picture...

Good to know you understand that. (Also, drawing it the wrong way may earn you downvotes and close votes.) Now to the point. Your E,Z calculation is correct. As for the "up or down" thing, the right-hand groups may also be either way, so it gets more complicated.

I get your point @IvanNeretin...its really getting complicated..

There are 4 chiral centers in the ring. The diene moiety (upper right) has 2 isomers and the lower right moiety also has 2 stereoisomers. So I counted 36 stereosiomers total.

@SteffX Can you explain your method in detail?

@SteffX 36 sounds kinda superfluous. Are you sure you didn't mean 32?

I'm getting something different.There are 6 stereogenic areas.So I just use $2^6$...is that wrong? @IvanNeretin

@SteffX
I'm getting something different.There are 6 stereogenic areas.So I just use 26...is that wrong?

Half of those are enantiomers. Or rather, all those 64 are pairs of enantiomers.

@IvanNeretin But according to Wikipedia "Enantiomers, also known as optical isomers, are two stereoisomers that are related to each other by a reflection: They are mirror images of each other that are non-superimposable. Human hands are a macroscopic analog of stereoisomerism. Every stereogenic center in one has the opposite configuration in the other." SO ENANTIOMERS ARE ALSO STEREOISOMERS.

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

Sorry, it is 64 stereoisomers, aka $2^6$ (I am tired on Friday)

So do you agree with $2^6$ @IvanNeretin i.e $2^5$ pairs of enantiomers?No mistakes in the formulation..right?

OK, then 64 it is.

@IvanNeretin Hi.Can you answer this if possible?Its a similar question. http://chemistry.stackexchange.com/questions/57761/how-many-stereoisomers-are-possible-for-the-compound

The cyclohexane ring contains 4 chiral centers, one for each carbon in the ring bearing a substituent. Also, the ring shows no plane of symmetry.
The diene moiety (upper right) contains 2 double bonds which could each be E or Z, but the terminal double bond has only one substituent, in other words it contains a $\ce{CH2}$ group, so there is no stereoisomer.
Now the fun part. The lower right moiety (which is quite unusual, btw) also has 2 stereoisomers. The $\ce{CH=C=C=C}$ is linear but its bonds cannot rotate freely. In fact, we could consider it as a simple double bond and the terminal methyl group can be E or Z compared to the cyclohexane ring.
There are 6 stereogenic centers so there are $2^6=64$ stereoisomers.

Follow these steps:

Identify all possible sources of stereochemistry. You got pretty far.
Determine how many isomers can derive from each. I think you went wrong here.
Do you have symmetry? Reduce accordingly.
Multiply all the numbers together.

In your molecule:

Double bonds:

the internal double bond in the residue on the top right
the cumulene, which also shows isomerism (E/Z for uneven double bond counts, aR/aS for even ones).

Asymmetric carbons:

the top methyl group
the bottom methyl group
the cumulene-containing side-chain
the other unsaturated side-chain.

Every one of these can be either R/S or E/Z.
The molecule does not have any elements of symmetry, so we cannot reduce.
All we have are factors of 2; two from the double bonds, another four from the asymmetric carbons. This leads us to:
$$n(\text{isomers}) = 2^{(2 + 4)} = 2^6 = 64$$


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Which has the largest bond angle between water, oxygen difluoride and dichlorine oxide?
Which one out of $\ce{H2O}, \ce{Cl2O}, \&\ \ce{F2O}$ will have largest bond angle?
I think it should be $\ce{H2O}$ because oxygen is most electronegative in this case so electrons will be more towards oxygen hence increasing bond angle, but the answer is $\ce{Cl2O}$.  Why?

Related: Bond angle order of SF2, OF2, HOF; Why does bond angle decrease in the order H₂O, H₂S, H₂Se?; Bond angles for the hydrides

There is practically only one effect that determines the angle, but there is an underlying mechanism that you also should understand.
Ideally, a central atom such as oxygen would want to use its p-orbitals exclusively for bonding and have the s-orbital electrons sitting comfortably as a lone pair close to its nucleus. However, that requires a $90^\circ$ angle which can create rather strained molecules — remember that each atom is surrounded by an electron cloud with negative charge that repulses other atoms’ electron clouds. There are now two possibilities of how to relieve this steric strain:

enlarge the bond lengths, so that different substituents are further away from each other while keeping a bond angle as perfect as possible;
enlarge the bond angles so the substituents move further apart from each other while retaining a close distance to their central atom with a more perfect bond length.

In practice, almost all systems stabilise themselves by modifying the angles. Increasing bond lengths decreases bond strengths and thus energy gained much more strongly than tinkering around with angles, because the overlap will inevitably get significantly smaller in absolute value number no matter which way you look at it.
An average bond length can be thought of as the distance which is just large enough for those orbitals that interact to do so favourably while minimising nonfavourable interactions. For certain atom pairs, bond length can essentially be considered almost a constant.
All of this taken together means that the very small hydrogen atoms have a much easier time of crowding together closer to the $90^\circ$ angle than the larger fluorine or huge (in this context) chlorine atoms. Even though the latters’ bond lengths are longer in absolute values their larger radii makes the effective bond lengths shorter and thus makes them require more space. Since chlorine is the largest of the lot, the $\ce{OCl2}$ bond angle is largest.

Even though this is the the accepted answer, this is partially wrong because it does not consider the effect of electronegativity difference between atoms. The last part about "hydrogen atoms having easier time of crowding together closer to the 90 degree angle" is completely wrong: it isn't supported by experimental data. H2O has bond angle 104.45 degree and F2O has a bond angle of 103 degree. So clearly hydrogen is not happy to stay closer to 90 degrees.

There are a few factors which affect bond angle which you must consider.  Loan pairs typically occupy more space than bonding pairs, so you should expect the two loan pairs of oxygen to repel the bonding pairs inward such that the angle between them is less than that of a perfect tetrahedral (109.5°).  This holds true for $\ce{H2O}$ and $\ce{F2O}$, but as for $\ce{Cl2O}$, the opposite is seen ($\ce{Cl2O}$ bond angle is 110.9°).  This can be attributed to a concept known as Ligand Close-Packing.  The ligands (atoms attached to the central atom, in this case oxygen) have a fixed radius, and the lone pairs of the central atom (oxygen) will repel the bonding pairs (between the ligand and the central atom) until the ligands "just touch".  The bond angle is dependent on how far the lone pairs can repel the bonding pairs before the ligands "just touch" (and therefore is also dependent on the atomic radius of the ligand and bond length between the ligand and the central atom).  This explains why $\ce{H2O}$ and $\ce{F2O}$ have a smaller bond angle than tetrahedral, while $\ce{Cl2O}$ has a larger bond angle.

Source

Is this possible that oxygen has 2 lone pairs right, so it makes another bond with one of the chlorine to minimise lone pair lone pair repulsion and hence its bond angle is most

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How can a Zn-Cu galvanic cell have a voltage of 1v in stead of 1.1v
Let s say a galvanic cell based on the reaction of zinc and copper half-cells. How is it possible to get the voltage at 1V in stead of 1,1V?
My guess: maybe by modifying the salt bridge in some way as this influences the charges in some way, but I am really not sure.

Impurity. (6 more characters to go...)

@KennyLau that was what I first thought, but doping was not the expected answer. Other possibilities: electrode size, $Zn(NO_3)_2$ while $CuSO_4$ , different concentrations or volumes

You can get a smaller potential if you alter the concentration of one or both ion solutions away from ideal 1M concentration. This behavior is described by the Nernst equation:
$$E_{cell}=E_{cell}^{o} - \frac{RT}{zF}lnQ$$
where $E_{cell}$ is the actual potential in the cell, $E_{cell}^o$ is the potential of the cell under ideal conditions, R is the ideal gas constant, T is the temperature, F is Faraday's constant, z is the number of electrons in the reduction oxidation reaction and Q is the reaction quotient. By altering Q to be greater than 1, the actual voltage of the cell dips below the ideal value.


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Reaction of a non-metal with alkali to produce hydrogen gas
I want to ask that is there any non-metal which reacts with an alkali (NOT alkali metal) to produce hydrogen gas?

Does silicon count as a nonmetal?  https://www.webelements.com/silicon/chemistry.html.

By "non-metal", are you asking about elements that are not metals (that includes metalloids and non-metals) or non-metals by definition? Anyways, I'll stick to the latter. The non-metals include carbon, nitrogen, phosphorus, oxygen, sulfur, selenium, the halogens (except astatine) and the noble gases.
In short, the answer is: "non-metals that give reaction do not give hydrogen as product". We can find out by listing them:

Carbon: No reaction

Nitrogen: No reaction

Phosphorus: Give phosphine

Oxygen: No reaction

Sulfur: Depending on reaction conditions, it can various polysulfide species along with sodium thiosulfate and water

Selenium: Might give similar reactions to sulfur (couldn't find any reference)

All halogens reacts with sodium hydroxide

Fluorine react to give oxygen(II) fluoride and fluoride ion (webelemets)
$$\ce{2F2(g) + 2OH-(aq) → OF2(g) + 2F-(aq) + H2O(l)}$$
Chlorine reacts with cold hydroxide to give chloride and oxychloride but with hot hydroxide it gives chlorate

$$\ce{2OH-(cold) + Cl2 -> Cl- + OCl- + H2O}$$
$$\ce{6OH-(hot) + Cl2 -> Cl- + ClO3- + H2O}$$

Bromine and iodine esentially give the same reactions forming bromate(V) and iodate(V) respectively (chemguide)

No noble gas reacts with sodium hydroxide

thanks for these but I asked for reaction of a NON-metal with an alkali  though these reactions will be helpful

@Shinjinee the above elements are all non-metal

?!? Al is a metal, as is tin unless you keep it in the cold for too long.

@Karl,  they are non metal in normal conditions.

Who on earth told you that? Both are good conductors (esp. Al), ductile, have a silvery surface. Yes, neither is as perfectly a "metallic" element as sodium or iron, but to call them non-metals is utterly ridiculous.

Since you didn't say "aqueous", boron fused with sodium hydroxide produces hydrogen and sodium borate.

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review

Um... boron is a non-metal, sodium hydroxide is an alkali and hydrogen... is hydrogen.  How can this NOT be an answer to the question?

However, if I'm really picky, I'd have to admit that boron is really a metalloid, and the real non-metals (halogens + O + N + C + P + S + Se) don't produce hydrogen when fused with sodium hydroxide.  So if the OP means 'not a metal including metalloids' I'm ok...

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Do swimming pools deplete the ozone?
I believe that most people know that CFCs deplete the ozone... as does free chlorine. I have read that pool chlorine is not a factor because, it is "rapidly converted to forms that dissolve in water and therefore are removed from the lower atmosphere." and/or "ends up as hydrochloric acid or sodium chloride."

So, the premise is that bleach simply won't react with carbon in the soil to produce ozone depleting chemicals (ODCs)? Or is the mind boggling complexity of chlorine in the environment sufficient to say that it must be gone?
I know that bleach (and/or chlorine gas) reacts with VOCs (like acetone, to make chloroform). Bleach will also release chlorine gas (Cl2) in conjunction with mild acids or even just carbon dioxide. And most importantly, chloramines (which evaporate from the pool) will react with ozone or UVs to produce more bleach.
So I'm wondering, is bleach (especially from swimming pools) really not that bad?

CFCs contain Flourine, no chance they are produced from dirt and simple bleach. And yes, all the lowly chlorinated compounds that can appear from reation of usual environmental substances with chlorine bleach are much too reactive and unstable to have a good chance of diffusing up to the ozone layer.

@Karl wow, duh... I meant ODCs. Boy, I can be stupid sometimes

Bleach in low concentrations will not make chloroform etc., and the lowly chlorinated compounds that do appear are environmentally problematic, but do not diffuse up to the ozone layer, because they are destroyed by reaction with UV/atmospheric dust/O2 etc.  before.

I've heard that question before (which is why I asked). So, to head off any potential urban legends...
There is evidence that the regulations against CFC's are making a difference in the loss of ozone. There's a nice little synopsis of the situation here:  The Ozone Hole

That graph on the right looks like stockbroker chart analysis. Significance: zero. ;-) I'd like to see a good graph, convincingly showing that the ozone hole is indeed recovering (like this years newspapers said).


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Is a compound aromatic if it also has anti-aromatic rings?
Is 5,6,11,12-tetradehydrodibenzo[a,e][8]annulene aromatic?

If I use Hückel's rule for the whole molecule, I get the answer as anti-aromatic. But the compound has two benzene rings which are individually aromatic.

I'd go as far as to say there is no such thing as aromatic compounds. Sure, people say it all the time, and we even have a tag to that effect; so what? This is just a shorthand for "compounds with aromatic rings". Aromaticity is a property of one ring, not of the whole compound. A molecule, as you know, may be arbitrarily complicated. It may involve some anti-aromatic rings together with some aromatic ones. Then you'll be compelled to consider such compound both aromatic and anti-aromatic at the same time, which does not feel well.

The molecule actually optimizes to a planar, delocalized $\pi$ system
Using B3LYP/6-31G(d) optimizations, we see that the molecule optimizes to become planar. (The initial configuration was intentionally bent.)

Moreover, if I plot the HOMO, it's highly delocalized:

Similarly, the LUMO is also delocalized across the entire molecule:

Consequently, I don't see any indication that the "side" benzene rings are localized from the cycloocta-1,5-diyne ring. The bond lengths are in fairly good agreement with the crystal structure:

Here's the crystal structure from J. Am. Chem. Soc., 1975, 97 (3), 658–659. Note there is a profound triple bond character, but the bridging bonds (1.44 Å) are somewhere between a typical aromatic bond length (1.40 Å) and a full single bond (~1.54 Å).

I will admit my initial intuition was that the central ring was non-planar, but figured "why not run a calculation and see?"

That is bloody weird.  Glad you ran the calc.  Would you add HOMO/LUMO +/- 3 energies to your post?

Guys, guys, that's how stabilisation via anellation with aromatic rings works. Benzene rings are destabilised, but central more stable. Cyclooctadienediyn most probably doesn't exist.

@Mithoron - yes, exactly. That's why I wanted to add a better answer for this one.

@Mithoron Cyclooctadienediyn could exist (at least in silico) and I would like to see the contributions to the stabilization. Dissected into destabilization of benzenes by geometric deformation and also some Huckel-like diagram. Could be a nice paper in J. Chem. Education ;)

Could the triple bonds have something to do with it?  They favor a linear bonding geometry and thus tend to make the planar structure more "rigid".  Does anyone know what happens if we put hydrogens on C-5, 6, 11, 12 so there aren't the triple bonds?

@OscarLanzi - that seems like a separate question.

@Geoff it turns out that breaking the nonconjugated pi bond by adding those hydrogens does destroy the planarity of the central ring.  See here.


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Effect of evaporation on the concentration of a saturated solution
When evaporating a saturated solution, does the concentration remain the same?
My thinking is that because the solution is saturated, and the volume is decreased, the excess solute would precipitate out of the solution. Therefore maintaining the original concentration.
On the other hand, would the precipitate be included when measuring the concentration? And therefore evaporation of a saturated solution would lead to an increase in concentration?
For example: if you have a 100mL of 2.5M saturated NaCl solution and evaporate it until the volume is 50mL, what would the final concentration be?

When a solution is saturated, it contains the maximum amount of solute which can be dissolved at that particular temperature. So effectively it is at its "maximum concentration" (for a given temperature).
This means that if your solvent is evaporating and the temperature remains constant, then solute will precipitate out of solution. The concentration of the solution will still be at its maximum as you are concerned with the amount of solute dissolved.
If you are heating up the solution in order to evaporate the solvent then the increase in temperature might mean that your solute solubility increases. This will depend on what solvent/solute you have.
However, a saturated solution which cools down without being disturbed may form a super-saturated solution whereby its concentration has gone beyond the solubility of the solute at that temperature.

As a heated saturated solution cools, excess solute may not necessarily fall out of solution, instead forming a supersaturated solution.  However, the supersaturated solution will be unstable, and insertion of a "seed" crystal or rough object will cause immediate crystallization.  Glassware that has a scratch or rough area on the inside will prevent this from happening though, as the rough/scratched portion provides a surface for crystals to form on, preventing supersaturation.  It's quite interesting to watch: https://www.youtube.com/watch?v=2Kud9eVbmRY&app=desktop

@Ki11akd0g Thank you, I did not consider that, edited my answer to reflect that.

This one is a little more clear https://www.youtube.com/watch?v=HnSg2cl09PI&app=desktop


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Are melting points for mixtures of solids ALWAYS below the melting points of the individual components?
In every chemistry lab we are taught that mixtures of compounds are going to have melting points lower than either of the components pure melting points.
Is this always true? Or can there be ratios of compounds that would generate a melting point above one pure compound and below the other?
For example, lets take acetanilide (MP: 114 C) and mix it with sodium acetate salt (MP: 324 C) in varying ratios. What would the melting points be for these?
1.) 98% acetanilide, 2% sodium acetate (slightly below 114?)
2.) 75% acetanilide, 25% sodium acetate (slightly above 114?)
3.) 50% of each (significantly above 114?)
4.) 25% acetanilide , 75% sodium actate (below 324?)
5.) 2% acetanilide, 98% sodium acetate (slightly below 324?)
NOTE: I chose these two compounds completely arbitrarily. Perhaps there is literature that has examined melting points for varying ratios of two particular compounds?
Follow up question:
Does there exist a mixture and/or ratio of solids that intermolecularly bond so strongly to each other that the melting point of that mixture is higher than the melting point of the individual components? What about a specific azeotrope with a higher boiling point than the liquids that make it up?

related http://chemistry.stackexchange.com/questions/30/why-do-impurities-lower-the-melting-point-of-an-isolated-substance

Depends on the compounds in question. If there are no strong specific interactions involved, then yes, the melting point of the mixture will be below. Otherwise... Well, a mixture of water with perchloric acid (1:1) is solid.

Is this always true?

No. As a general rule of thumb, if two compounds are miscible in each other, that is they exhibit solid solution, then the melting point of a (chemical, not mechanical) mixture will be intermediate. If the two compounds are not miscible and it's only possible to make a mechanical mixture and not a chemical mixture, then you will have a lower melting point.
Let's try to see this with several examples of phase diagrams, taken from FACT.

...mixtures of compounds are going to have melting points lower than either of the components pure melting points.

This is called eutectic relationship, for example sodium chloride and sodium sulfate. There is no intermediate compound, and they have a eutectic relationship, where the melting point of any mixture is lower than the melting point of both compounds. Note that if you are not in the pure liquid field, you will have a pure compound together with a liquid that is a mixture.

Silver and gold are a good example for a case when there is complete miscibility. All intermediate compounds can exist, therefore the melting point of the mixture changes continuously with the composition:

The double line is there because there is a very thin temperature range in which a liquid coexists with the solid of a different composition. See the solid solution page for more about that.
These were two simple cases. More complex cases arise when you have discrete compounds rather than a complete solution solution. A rather good example is cerium and nickel:

This system consists of several eutectic and peritectic points.

The melting of a substance occurs when Gibbs free energy favors liquid.
That implies that entropy times T of the liquid at melting point equals
the binding energy of the solid, on a per-molecule basis.
Entropy of a random mixture is greater than entropy of a pure material, so one expects
the mixture will have a lower melting point.   The only exception, would be
a mixture which reacts or otherwise causes a new crystal structure
to be favored, which has different binding energy than the pure
materials (this happens when calcium and oxygen combine to form lime, for
instance).   Chemical reaction also destroys the 'random' nature of
the mixture, of course.
Some mixtures are not miscible in liquid form, so the liquid mixture doesn't occur.
Presumably oil and water has no increase in entropy due to mixing, and
no melting point depression is expected.


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Platinum Electrodes
In industry, numerous companies state that they use platinum electrodes for the purpose of electrolysis. My question is: are these electrodes pure platinum or could they also be, for example, another element/ compound coated in platinum such as platinum coated titanium mesh?
If they are about 100% platinum, why not use platinum coated titanium mesh in industry?

In most cases, as you mentioned, platinum coated titanium meshes are used (or sheets). I am not aware of any application where pure platinum electrodes are demanded (for electrolysis).


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Why does formic acid smell like getting hit in the head?
I was cleaning up a concentrated formic acid spill and when I was close enough to the hood to get a whiff it smelled exactly like the smell sensation from hitting my head.
We know that olfactory memory is very keen.  While many people identify a vaguely metallic odor when hit in the head, I doubt many non-chemists have had occasion to smell pure formic acid.  Do chemists familiar with the chemical generally agree that formic acid matches the olfactory sensation of head concussion?
If so: Is there a physiological effect of head concussion that would mimic the effects of formic acid vapor on the olfactory system?  (Or maybe even produce formic acid that could be picked up by olfactory nerves?)

I'm voting to close this question as off-topic because it is more about the biological process, hence should be on biology.

I edited to make it clear that this is a general question, not a personal question.  And note that most of the [tag:smell] questions implicate the "biology" of olfaction.  But, when they're answered, such Q&As can be both very popular and highly rated.

I've both (a) smelled pure formic acid and (b) experienced head trauma.  I've never noted a link between (a) and (b).  The only inkling I have ever heard that the two might be related is your question.  My (off-topic) guess is that your association of the two is purely idiosyncratic.

@CurtF. - That's a useful data point. I don't suppose, in the interest of science, we could get a bunch of chemists to hit their heads against a hard object? Or, perhaps more practically, get a bunch of chemists to visit soccer practice fields with a vial of formic acid and ask players if the smell reminds them of anything?  (Note that this depends not only on the severity of the blow – e.g., it can't be so severe as to render the observer incapable of noting the immediate smell sensation – but may also depend on the direction of the blow – see, e.g., nasal impingement of the cribiform plate.)

I've noticed that smell with isopropyl alcohol, but not every the time.  It could be some additive or impurity used when it's sold as a cleaning product for electronics, or a reaction with what it's cleaning, or it could be the strength of exposure.  Maybe it's just the synaesthesic smell of brain damage?

@feetwet I know this is really late, but if you're still interested, I'd suggest you post this on Biology. I think you're more likely to get a good answer there - based on my experience with biochem I'm inclined to think that this is a bit too biological for us.

@orthocresol - It's never too late for discovery!  I'll try posting there shortly.  Thanks!

I'm closing this question as off-topic because it's more inclined towards physiology and therefore belongs to [biology.SE]. It wasn't a personal medical question, I feel that that deserved clarification.

Reposted on Biology.SE.


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What is the signal at around 2090 wavenumbers in the nitroguanidine FTIR spectrum?
There is a signal at around 2090 apparent in the FTIR spectrum of nitroguanidine  from Sigma-Aldrich (first image). It's also apparent in FTIR spectra we've taken in the lab (second image), but when assigning peaks to bonds, we are unclear to what it is.
We are unsure of how Sigma-Aldrich prepared their pellets, but we used KBr. We are thinking perhaps KCN, but then again - how could that possible form?

Its common Nujol, a hydrocarbon oil used to prepare paste-samples for IR from powdered materials. Hence the C-H valence around 2090 cm-1

I doubt if a alkane like nujol or aliphatic kerosene would have a peak at 2090, there is a possibility that a peak in the range of 2090 could be due to the fact that the refractive index of sodium chloride disks and the nujol are different. You sometimes get some weak additional peaks due to this effect (it relates to the thickness of the KBr disk or oil layer)


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Why is pH used as a measure of acidity if not only hydronium ions can act as a acid when in solution?
The Brønsted-Lowry theory tells us that the when an acid such as $\ce{HCl}$ dissolves, the following reaction takes place:
$$\ce{HCl + H2O <--> Cl- + H3O+}$$
It is said that the hydroxonium ion is what functions as an acid when a base is added to such a solution. However, there will indeed remain some $\ce{HCl}$ molecules in solution. So wouldn't they also act as an acid in some instances?
This problem becomes more apparent when we look at weak acids such as ethanoic acid. There are many many ethanoic acid molecules in comparison with the hydroxonium ions. Why is the acid 'weak' if the ethanoic acid molecules can just react with the base anyways?
To further explain where my problem is, consider the scale which we use to measure the 'acidity' of a solution, the pH scale. Consider a scenario where I have 2 different solutions of ethanoic acid and $\ce{HCl}$, both of the same concentration. The $\ce{HCl}$ solution will definitely have a lower pH and will be considered 'more acidic'. But would it really be correct to call the $\ce{HCl}$ solution 'more acidic' just because the $\ce{HCl}$ has dissociated to produce a much higher hydronium ion concentration? The ethanoic acid solution will have a lower hydrogen ion concentration, but the ethanoic acid molecules can still act as an acid … The main problem here is with using pH as a measure of acidity if molecules which haven't dissociated may also act as acids …

If you add enough alkali, your acid will react completely, be it strong or weak. This is not what tells them apart.

I understand that it is how much of an acid dissociates which tells the strength of an acid. I'm questioning why this measurement is used? In the case of ethanoic acid, there will remain many many ethanoic acid molecules which can act as an acid. When a base is added, I'm not saying that the H3O+ will keep decreasing in concentration such that the equilibrium position keeps shifting to the right, but rather that base molecules can directly collide with ethanoic acid molecules to accept a hydrogen ion.

Question has been edited.

Acidity tells you how strong is the acid, and not how much acid do you have. As to why this measurement is used... well, that's a bit of a dead-end question. It is used because we found it useful. Many reactions depend on pH and would go differently with weak and with strong acids.

Please use paragraphs.

Hint: salt hydrolysis

Jan, I will do. Thanks for the comment and sorry about not using paragraphs.

HCl being a strong acid, it is quantitatively dissociated in water, meaning that it fully dissociates into hydronium and chlorides. The hydronium are therefore the main contributors to the acidity of the solution. In a solution of ethanoic acid, the majority of acid molecules are undissociated. This is what "weak acid" means.
The amount of base necessary to neutralise a given acidic substance can be used as a scale of acid "strength" in chemical risk assessment.

If ethanoic acid molecules remain in solution, then do the ethanoic molecules also directly act as an acid?

Question has been edited.

It all boils down to the question which species can protonate which other species in a Brønsted-Lowry acid-base reaction. A lot of compounds have a generally somewhat acidic proton but it would be very wrong calling them an acid. For example, dihydrogenphosphate $\ce{H2PO3-}$ has a $\mathrm{p}K_\mathrm{a}$ value of $7.198$ — it can be deprotonated easily in aquaeous solution by sufficiently basic compounds but I really wouldn’t want to call it acidic. Or, to give another extreme example: furane can be deprotonated by n-butyllithium, but you really wouldn’t want to call it acidic, would you?
The good thing about the oxonium (or hydronium) ion concentration is that it is a pretty good approximation of the acidic effect. If I have a solution of acetic acid with $\mathrm{pH}\ 5$ and a (much more strongly diluted) solution of $\ce{HCl}$ with the same $\mathrm{pH}$-value, the protonating effect they have on different compounds will generally be the same.
And finally, $\mathrm{pH}$ is typically only used to measure the resulting acidity of a solution. When comparing the strengths of acids, $\mathrm{p}K_\mathrm{a}$ values are typically used. The picture being that with a $\mathrm{p}K_\mathrm{a}$-value difference of $\approx 3$, the weaker acid can be considered fully protonated at equimolar concentrations.

But would it really be correct to call the HCl solution 'more acidic' just because the HCl has dissociated to produce a much higher hydronium ion concentration?

Yes, it would be correct to call HCl more acidic because as an acid it is donating more protons compared to ethanoic acid now you argue that undissoaciated ethanoic acid can also react with a base to form salt.
But salt formed will be of a weak acid and strong base(assuming), so this salt will hydrolyse and finally give back the acid again but this not the case if we use HCl , so no matter we have same concentration of different acids, more acidic will be one which gives more protons easily. So undissoaciated etanoic acid will remain again undisassociated.

So are you saying that if a base such as ammonia reacts with ethanoic acid molecules, the equilibrium position will lie very close to the left, such that the reaction can be approximated as non-happening? And then this allows us to very well approximate that H3O+ ions are the only acidic substances in solution? May you please tell me where you learnt this from/cite your source?


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Is activation energy related to reactivity?
I understand that activation energy is a property of a reaction.

If a substance is highly reactive, will it have low activation energy and vice versa ?
How is the activation energy calculated and what are the factors that affect it?

Activation energy is a property of reaction, not of substance.

Yes, I am saying however, is a usual trend in reactions, that substances with high reactivity have low activation energy?

You say you know that activation energy is a property of a reaction, and immediately go on to discuss it as if it were a property of a substance.

If by reactivity you mean, "does the reaction occur quickly", then yes, a highly reactive molecules do have a low activation barrier. The fastest molecules can react bimolecularly is if they do so on first encounter with one another. This effectively means that  there is a very small potential barrier between reactants and products so that they have enough energy, at the prevailing temperature, to always overcome this. In solution the rate constant is determined by how fast the molecules can diffuse together and in mobile solvents such as water or acetone the bimolecular rate constant is around $\ce{10^{10} dm^3mol^{-1}s^{-1}}$ depending on exact solution viscosity and far less in viscous solvents such as glycerol.
If by reactivity you mean that the reaction is very exothermic, i.e. free energy is large and negative, $\Delta G < 0$ then there is generally no way to precisely predict how big the activation barrier will be. It is usually measure using the Arrhenius equation and plotting log(rate constant) vs. reciprocal temperature. I'm not sure if ab initio quantum calculations can predict this in general. In fact many exothermic, thermodynamically spontaneous reactions occur slowly. Most organic substances react with oxygen exothermically, but are quite stable, because the activation energy to do so is so large. You and me for example :)


text:
Why are most hydrocarbons less dense than water?
Why are most hydrocarbons, like alkanes and alkenes, less dense than water?

Mostly because oxygen atom is heavier and also smaller than carbon.

But hydrocarbons have so many carbon atoms. Like pentane quite a large mass number.

So what? True, all these atoms together weigh more than one, but they also occupy more space.

Well, that's like strongest simplification possible. It gets complicated quickly if you delve into it. Starting from hydrogen bonds and VdW interactions.

Density is not only related with mass, but with volume too. Density of a substance is its mass per volume. It means increasing volume will decrease the density.
The molecules of hydrocarbons are large in size, increasing the length of their chain will increase both their mass and volume. On the other hand, water molecules are smaller and simpler in structure, but it is heavier than its size.

There is no truly simple explanation for bulk properties of liquids. But the balance of two factors probably goes a long way to explaining the difference between water and simple hydrocarbons.
The density of a liquid is determined by the mass of the molecules and the efficiency of how they pack in the liquid.
Molecules made of heavier atoms will tend to be denser (e.g. chloroform where the heavy chlorines give the liquid a density about 50% higher than water). Water is made of moderately heavier atoms than a hydrocarbon (oxygen is about 30% heavier than carbon).
The other factor is how well the molecules pack in the liquid. This depends on their shape and the molecular interactions among them. Water molecules have very strong hydrogen-bond interactions causing them to, on average, stay closer than molecules of equivalent size that don't hydrogen bond. Hydrocarbons have only weak van der waals interactions so don't, on average, stay so close.
The interaction of the molecular mass and the forces determine the overall density. Water wins on both compared to hydrocarbons.

To extend the answer provided by @Junaid Jamil, you may borrow a concept used in small molecule crystallography.  For an estimate, the volume of the individual molecule is approximated by the assumption that each atom with an atomic volume.  These values are tabulated (see e.g., earlier answers provided here and here).  Because density is the ratio of mass over volume, the second property you need to know for this estimate is the mass of the individual molecule.
If you plug-in the values, you recognize a trend, matching the pattern of densities of these liquids experimentally recorded in comparison to water. Equally note the jump passing for halogenated hydrocarbons:
|--------------------------------+-------+----------+---------+---------+------------+----------------------|
| compound                       | water | n-octane | benzene | ethanol | chloroform | carbon tetrachloride |
|--------------------------------+-------+----------+---------+---------+------------+----------------------|
| formula                        |   H2O |    C8H18 |    C6H6 |  C2H5OH |      CHCl3 |                 CCl4 |
| mass [amu]                     | 18.02 |   114.23 |   78.11 |   46.07 |     119.37 |               153.81 |
| volume [A**3]                  | 21.55 |   203.28 |  114.36 |   69.83 |      96.46 |               117.18 |
| calculated density [amu/A**3]  |  0.84 |     0.56 |    0.68 |    0.66 |       1.24 |                 1.31 |
|--------------------------------+-------+----------+---------+---------+------------+----------------------|
| experimental density [g/cm**3] |  1.00 |     0.70 |    0.88 |    0.79 |       1.49 |                 1.59 |
|--------------------------------+-------+----------+---------+---------+------------+----------------------|
#+TBLFM: @5$2..@5$7=@3/@4;%.2f

Note the limitations: These estimates err since packing in the (ordered crystalline) solid state from which the atomic volumes are derived leaves gaps between the molecules packed.  Even the dense hcp packing of spheres occupies only about 74% of the volume available.  In many cases, the intermolecular gaps increase further for molecules freely moving in a melt/liquid.  This increment system neither accounts for the eventual molecular shape; in the liquid e.g., n-octane might be in a straight extended conformation, or may curl into a more spherical shape.  Nor does it discern if $\ce{C8H18}$ refers to n-octane, or any other isomer of octane (18 constitutional isomers); both variations are known to affect density (e.g., n-octane $\pu{0.703 g cm-3}$ ref; 2,2,4-trimethylpentane $\pu{0.692 g cm-3}$ ref).

The experimental densities the table states are densities the English edition of Wikipedia reports for the compound in the liquid state, which were rounded to two decimals.  Preference was given to reports for densities at  $\pu{25 ^\circ{}C}$: water, n-octane, benzene, ethanol, chloroform, carbon tetrachloride.

Density depends on mass and volume. Higher mass and lower volume occupied per molecules promote density. Hydrocarbons have high molecular mass (except methane) and molecules take more space due to presence of vander wall force only. In case of water molecules due to presence of hydrogen bond molecules take less space ya they are more compact than hydrocarbons resulting instead of more molecular mass of hydrocarbon they are less denser.


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Melting of Covalent Networks
When you melt a covalent network, you break intramolecular covalent bonds, is that not a chemical change?
Also, is ionisation a chemical or physical change?

This is more or less pointless categorisation and only useful for passing exams IMO. Still if you heat elements and the process is reversible - you can solidify melt to get the same thing or even ionise vapor and get the same thing after cooling, that is considered "physical".

A chemical change is defined as when your substance reacts with another substance to form some products, or decomposition into other products. In a chemical change, there will be formation of new chemicals that are different in terms of chemical formula.
Thus, breaking of intramolecular bonds will not be a chemical change in the case of C-C bonds in diamond/graphite, because there is no formation of new chemical substances with an alternative formula. Similarly, ionization (without formation of complex ions) would be physical change

Uh, there are no molecules in graphite. Or diamond.

@DEAD Intramolecular forces refers to covalent/ionic/metallic bonds between the atoms that make up a molecule. In this case I raised the example of diamond/graphite because it is a covalent network and the bonds are considered intramolecular

Your statement that breaking intramolecular forces is not a chemical change is blatantly wrong.

A chemical change is defined by the formation of new chemicals from the original substance via a chemical reaction. The context in this question is strictly restricted to breaking of bonds of a networked structure (i.e. macromolecular covalent structure), as the breaking of intramolecular bonds in a simple covalent structure (e.g. H-Cl bond) would be trivially a chemical change, due to formation of other chemicals with a different formula.
I argue my case from the fact that there is no formation of new chemical products with a chemical formula different from the original substance.

I respectfully disagree. The products necessarily do not have a different chemical formula than that of the reactants. Could you provide a source on your claims?

If you define ionization as the addition or removal of one or more electrons from a molecule, that is a chemical change.


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Which isomers of C6H4(OH)2 has a higher melting point?why?
At first I thought that the first isomer must have the highest melting point since it's the most polarized isomer. So I looked them up on Pubchem to be sure and the answer was exactly the opposite;the third isomer has the highest melting point and it's not polar.
Any idea why?
Thanks!

Polarity rules the day if there is no other interactions involved. There is quite a lot of potential interactions in organics, that may overwhelm polarity by orders of magnitude.
To name a few.
There is a well known effect of halogen-halogen interactions, that may raise bp/mp dramatically. This is due to so-known dispersion forces, caused by correlation of electron movement in two atoms. Sometimes it is described as interaction of 'momentary dipoles'. To produce strong dispersion forces, atoms involved must be large and have a lot of electrons. Naturally, heavier atoms in a column are typically larger and have more electrons, thus we have strong increase of mp/bp for most haloalcones whith shift from chlorine to iodine.
Another well known effect is formation of intermolecular hydrogen bonds. Hydrogen bonds are common in compounds containg $\ce{O-H}$ or $\ce{N-H}$ bonds, but sometimes may pop-up in other, usually quite exotic settings. The most common form of hydrogen bonds is a bond of form $\ce{R-O-H\cdot\cdot  \cdot O(R)-R}$ . Such a bond may be stronger of weaker. It is stronger if the second oxygen atom has strong negative charge and the hydrogen involved has strong positive charge. Hydrogen bond is stronger if it is linear, i.e. the entier $\ce{O-H\cdot\cdot  \cdot O}$ fragment is linear. Intermolecular hydrogen bonds have competition : intramoleculear hydrogen bonds. Intramolecular hydrogen bonds prevents two potential sweet spots from interacting with other molecules, and thus a compound capable of formation of intramolecular bonds typically has lower mp/bp than compound that cannont, but have same amount of potential hydrogen bonding sites.
And, finally, come electrostatic interactions. They are not limited to just polarity, some compounds are ionic and this typically results in high bp/mp. Ionic bonding is common in inorganics, but may occur in organics as well, say, $\ce{N(CH3)4 PF6}$
Now, about example at hand. All three molecules are capable of formation of hydrogen bonds. So, the one that is better in it wins the contest of bp/mp.
The 1,2-disubstituted compound is capable of forming intramolecular bonds, and the oxygen atoms are a bit crouded. This makes bonding a bit weaker.
The 1,3-compound is better in this regard, but has $\ce{OH}$ groups in meta-position. It is known, that oxygen in such group is capable of donating some of its electron density into aromatic ring. However, if we put two oxygens in 1,3-pattern,  the two will suppress donation of each other thanks to resonance effects. The last compounds would have the most polar hydrogen bond. It also has a high symmetry, which helps. It is well known, that slight asymmetry mar lower mp considerably. Consider, for example, toluene. It has higher bp than benzene but significantly lower mp.
Consequently, compound 3 (hydroquinone) wins the prize.

Thanks! Would you suggest a source for more information about the last part (donation of electron density to aromatic ring) (even a wikipedia article!) somehow i can't get it...thanks again!

The donation as a thing is usually considered in 1) discussion of electrophilic aromatic substitution  and 2) acidity/basisity of aniline and phenol. The orientation thing is pretty obvious once you draw relevant resonance structure.

I don't see why symmetry should be important, all the molecules have 'symmetry' they just belong to different point groups. The melting points are all a lot higher than that for phenol so this suggests that hydrogen bonding is more extensive in these compounds and that this will raise the melting point. The hydroquinone has the highest melting point and probably the ability to make a very extensive hydrogen-bond network. The 1-2 can only make small zig-zag chains before a dimer terminates this, similarly the the 1-3 can make zig-zag chains but will eventually be terminated by a ring, but the 1-4 can make extensive chains.

Symmetry is important because solids that form more regular packing patterns will have higher melting points. Better packing increases the likelihood of all kinds of intermolecular interactions.

@Ben Norris Of course, yes, symmetry is important, but not in some undefined way, which was my point. I then try to explain why the different symmetry the molecules have could lead to the effects observed.

The higher melting point of hydroquinone (3) is most probably due to the high symmetry that together with the ability to form hydrogen bonds leads to a more stable crystal structure.

Thanks;  but how is symmetry defined?

See https://en.wikipedia.org/wiki/Molecular_symmetry


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How does the Mentos-Coke explosion work?
We all might have seen this in some videos or in real life. But how exactly does this work? What is the chemical reaction involved?
And why is the reaction of Mentos with Diet Coke more vigorous?

From left to right: action of five Mentos candies (per bottle) with Perrier, classic Coke, Sprite and Diet Coke
Image source

It's actually not a chemical reaction, but a physical reaction.  The surface of mentos acts as a nucleation site for the dissolved carbon dioxide in soda, allowing the gas to come out of solution quite rapidly.  This rapid release of gas is what causes the coke geyser.  Despite mentos appearing to have a smooth surface to the naked eye, there are actually a multitude of microscopic crevices, providing a large amount of nucleation sites.  As far as diet coke working better than regular coke, I am unsure (perhaps diet coke has more carbonation?  Or perhaps there are chemicals in diet coke which make it form foam much better than the other sodas).  I always thought that it was the mint vs fruity mentos that made the largest difference (mint seems to have a lot more nucleation sites which explains the larger geyser in comparison to fruity mentos).

I will also note that Diet Coke seems to work better than other diet colas.

Diet drinks probably work better for two reasons. One is that they are less viscous than sugary drinks (which can be 10% sugar by mass making the liquid much more viscous than water; diet drinks have fractions of a % sweetners). The other is that some additives (especially aspartame) act as surfactants stabilising bubbles when they form.

Complete speculation: perhaps one of the ingredients in Diet Coke acts as a catalyst for the conversion of $\ce{H2CO3 -> CO2 + H2O}$, and thus more of the carbonation is able to effervesce to form foam?


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sp2 hybridized carbon versus sp3 hybridized silicon
Carbon's most commonly found allotrope is graphite/soot/fullerene/carbon nanotubes which are all $\mathrm{sp^2}$ hybridized forms of it, while silicon being its first periodic twin is mostly found in the $\mathrm{sp^3}$ hybridized form in bulk, while formation of silicene is extremely difficult. While in the case of carbon, diamond is the one which is harder to find, what is the reason for this discrepancy?

Carbon — and all the other elements of the second period, for that matter — is exceptionally good at forming double bonds. A double bond is formed if there is a σ bond and a π bond. Forming a σ bond is not hard: You have two orbitals coming from two different atoms pointing at each other. Forming a π bond is much harder, since the orbitals are pointing orthogonally away from the bond-axis. Thus, the overlap of π bonds is often a lot weaker than for σ bonds.
This gets more important if you consider third-period elements rather than second-period ones. They have a total of eight more core electrons making them much larger in size and the orbitals more diffuse. A σ bond is still formed easily, there will be the point when two diffuse orbitals are close enough to interact in a σ fashion. To arrive at favourable interactions for a π bond would require the atoms to come much closer than what would be comfortable; unfavourable interactions between the core electrons would already be a thing.
In fact, I was taught in my Concepts of Inorganic Chemistry class, that disilene $\ce{Si2H4}$, the heavier homologue of ethene, does not form a classical π bond. Instead, it can be understood as a par of two singlett silene fragments $\ce{SiH2}$ which approach each other so that the empty p-orbital of one silene can interact with the lone pair in an $\mathrm{sp^2}$ orbital of the other silene fragment. Obviously, the symmetry is lower and the bond is much weaker.
So the reason behind it is that $\ce{Si=Si}$ double bonds are hard to create and this gets even harder for heavier homologues while $\ce{C=C}$ double bonds are reasonably stable and easy to generate.

The primary reason why diamonds are not more common is the kinetic control of the graphite to diamond reaction.
The activation energy for a conversion from diamond to graphite is very high as a previous poster on this forum has estimated and sourced:
Are diamonds REALLY forever?
Consequently, a high activation energy indicates that a particular transition state is very hard to reach, and thus a reaction is not feasible. The high activation energy between graphite and carbon is the reason why graphite is not found as common.  Please see the image attached:

In regards to why silicene is not more common than graphite, we must consider pi-stacking.  Pi-stacking refers to the stacking of p-orbitals which allows for the free flow of electrons in pi-bonds allowing for further stability.  Silicon's bond's do not allow for pi-stacking and thus do not form the layered and stable structure that we see in graphite (Jose, Datta, 2014).
Sources:
https://www.researchgate.net/figure/278647847_fig2_Figure-4-4-Energy-diagram-for-conversion-of-diamond-to-graphite
http://pubs.acs.org/doi/full/10.1021/ar400180e


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Heat resistant odor neutralizers
I am looking for a heat-resistant odor neutralizer that can be released into the air at 200 degrees Fahrenheit and still be effective.
Does anyone have any ideas?

If it is just for household and pet odors, why does it need to be effective at 200 degrees Fahrenheit?

Metazene 99% or other Alpha Aromatics high heat compounds.
What is the mal-odor caused by, and the complete application?

Pet odor / household odors. Application is for candle wax.

This isn't really an answer to the question.  Can you explain what metazene is (it's not a standard chemical term), what "alpha aromatics high heat compounds" are, and add more details?  If you have comments or clarifications to ask the OP, those can be done in comments, not in your submitted answer.  Thanks for participating in chem.SE, glad to have you here.

Ozone is widely used commercially and residentially for the destruction of odor causing gases and particulates (i.e. smoke), and a variety of different types of ozone generators are readily available.  According to this Wikipedia article:

Ozone generators are used to produce ozone for cleaning air or
  removing smoke odours[69] in unoccupied rooms. These ozone generators
  can produce over 3 g of ozone per hour.

This is a reference to an extremely high-capacity device. The rest of the above article further describes the array of ozone generators that are commercially available.
Of course, ozone truly destroys malodorous compounds as opposed to covering them up. The rate of destruction of these compounds also generally increases with increasing temperature.


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Selectivity of catalysts
What is the reason behind the fact that methane on controlled oxidation gives methanol in presence of Cu catalysts whereas formaldehyde is obtained if we use molybdenum oxide ($\ce{Mo2O3}$)?
IUPAC says:

The term selectivity (S) is used to describe the relative rates of two or more competing reactions on a catalyst.

What is the reason behind favouring one products?
Would I be correct that in controlled oxidation of methane in presence of Cu catalyst, we get both methanol and formaldehyde?
Should I need to memorise what catalyst yield what products? Or there is any other way?

I can't offer you an answer in as much detail as you'd probably like, but as you probably know, catalysts lower the activation energy by allowing a reaction to take place through an alternative mechanism.  The Cu catalyst likely allows a mechanism of action that results in the formation of methanol preferred over the formation of methanal.  I would assume that methanal could still be formed, but the methanol will form much faster in the presence of Cu and methanal formation would be minimal if present at all.  However, I am only speculating based on what knowledge I do have of catalysts.

In short: yes, you have to memorize. Catalytic mechanisms can be pretty complicated, so in general you won't be able to predict what goes where.


text:
What happens to Aluminium in Autoclaved Aerated Concrete
Aluminium is used in Autoclaved Aerated Concrete in order to produce the holes.
On Wikipedia I'm reading about the ingredients: sand, calcined gypsum, lime (mineral), cement, water and Aluminum powder.
Also it says "Aluminum powder reacts with calcium hydroxide and water to form hydrogen".
I'm wondering what Aluminium ends up as if anyone can help with detailed reactions.

$$\ce{2Al + 7Ca(OH)2 -> Ca[Al(OH)4]2 + 3H2 ^ + 6CaO}$$
or maybe
$$\ce{2Al + Ca(OH)2 + 6H2O -> Ca[Al(OH)4]2 + 3H2 ^}.$$
I think you'd get basically (ha) aluminium hydroxide.  This is effective because bases can dissolve the protective aluminium oxide layer that usually protects aluminium metal from corrosion.  The concrete is foamed by the hydrogen.


text:
How do I determine the amount of Calcium in a substance using EDTA titration?
I need to determine the amount of calcium in a substance (that also contains Magnesium) using EDTA titration. From what I understand, this can be done by adding Eriochrome T indicator to the solution, then titrating it with EDTA. The solution will start out Red then turn blue as the Ca ions bond with the EDTA. Many of the labs I search up online seem to follow these procedures. However, doesn't Magnesium also bond with EDTA? Wouldn't that affect my calculations since I'd be measuring Ca AND Mg as they both bind with both Eri T and EDTA?
Thanks

Yes, EDTA will chelate to both calcium and magnesium. However, since the complex formation constant for $\ce{Ca^2+}$ is significantly greater than that for $\ce{Mg^2+}$ ($\log \beta_{f,\,\ce{Ca^2+}} = 10.65$; $\log \beta_{f,\,\ce{Mg^2+}} = 8.79$), we are justified in assuming that EDTA preferentially chelates to $\ce{Ca^2+}$ and only chelates to $\ce{Mg^2+}$ when no $\ce{Ca^2+}$ is available.

Remark. How significant is "significant"? In principle we could avoid the approximation and solve for the system analytically, but then we would still have error from (i) imprecision in visual determination of endpoint, (ii) imprecision in determination of formation constants, and so on.


text:
What is distorted square pyramidal structure of a compound?
I have recently being studying structures of different types of compounds, when I came across the structure of $\ce{[XeF5]+}$ which have a structure distorted square pyramidal. I could not understand what "distorted" square pyramidal is and how it is different from square pyramidal.

related http://chemistry.stackexchange.com/questions/50176/bonding-in-square-antiprismatic-compounds

And http://chemistry.stackexchange.com/questions/50187/what-is-hybridisation-of-xef6-in-solid-state is useful (what may be surprising it seems).

@Mith It was only surprising to somebody not getting the way in wich $\ce{[XeF5]+}$ is distorted ;)

Going by a standard argumentation, the $\ce{[XeF5]+}$ cation would be expected to have two different bond types: a classical 2-electron-2-centre bond to the axial fluorine and a pair of two 4-electron-3-centre bonds making up the base of the pyramid with the equatorial fluorines. This structure is shown in figure 1.

Figure 1: Hypothetical perfect structure of $\ce{[XeF5]+}$. Image taken from Wikipedia, where a full list of authors is available.
The actual structure of $\ce{[XeF5]+}$, reported by Leary et al.,[1] is not perfect, though. The axial fluorine atom does not sit at the centre of the pyramid, destroying the perfect $C_\mathrm{4v}$ point group. This is probably an effect of core orbitals on the central xenon atom (I cannot access the paper where I am right now). Because the pyramid no longer has four equal faces, it is distorted.
Reference:
K. Leary, D. H. Templeton, A. Zalkin, N. Bartlett, Inorg. Chem. 1973, 12, 1726. DOI: 10.1021/ic50126a004.

Jan, I think it is the other way around.  When all of the angles are 90° (e.g. your Figure 2) you have a regular square pyramid.  When the angles deviate from 90° you have a distorted square pyramid.

Chemistry borrowed the term from mathematics were it was originally defined, so how common a chemical structure is doesn't matter.  2) At least in the solid phase $\ce{XeF5^{+}}$ is distorted, the $\ce{F_{ax}-Xe-F_{eq}}$ angle is 79°, not 90°

@ron Sorry, I misinterpreted what you (and Mith, and probably OP’s textbook, too) meant by distorted, not knowing about the crystal structure (and not reading the other answer in good enough detail). I know get your point and see that your reasoning is correct. Will edit the answer in a sec.

Jan, just read your revised answer.  I think the article says the axial fluorine does sit at the top of the pyramid, in the center, maintaining approximate $\ce{C_{4v}}$ symmetry.  It is just that the 4 equatorial fluorines are pushed up towards the axial fluorine ($\ce{F_{ax}-Xe-F_{eq}}$ ~ 79°) and away from the axial lone pair on the other side of the pyramid. There are still 4 equivalent faces.

@ron Atm I just feel like deleting for simplicity :D well, whatever, I’ll check out the article once I get a chance …

The structure of $\ce{XeF5}$ is approximately described by the $\ce{C_{4v}}$ point group. This means that the four equatorial F atoms may lie in the same plane as the Xe atom and that the fifth, axial, F atom lies above the Xe atom at an angle of 90 degrees to the plane of the other F atoms (and so in the principal axis). If the point group is $\ce{C_{4v}}$ the adjacent angle $\ce{F_{eq} - Xe - F _{eq}}$ has to be close to 90 degrees. The lone pair of electrons occupies the sixth position, opposite to the axial F atom. In the xray - diffraction data of $\ce{XeF5+ PdF6-}$ crystals the  $\ce{F_{eq} - Xe - F _{ax}}$ angle is ~80. In the crystal this may be due to packing or interaction of adjacent ions, as there seems to be no reason, based on symmetry,  why the $\ce{F_{eq} - Xe - F _{ax}}$ angle is not very close to 90 degrees. (The structure has been reported by Leary et. al. Inorg. Chem. 1973, vol 12, p 1726)
Edit. Considering the comments below if the Xe atom is not in the plane of the four equatorial F atoms then the axial F atom can still be on the principal axis, but the angle $\ce{F_{eq} - Xe - F_{ax}}$ will differ from 90. The structure will appear distorted from square pyramidal in the sense that the Xe is not in the plane.

Nope, this is wrong. Even regular square pyramids where the central atom is not in the plane of the pyramid’s bottom for atoms have $C_\mathrm{4v}$ (try it!).

"there seems to be no reason, based on symmetry..."  Since the $\ce{F_{eq}-Xe-F_{ax}}$ interaction is different from the $\ce{F_{eq}-Xe-LP_{ax}}$  interaction (LP=lone pair), symmetry tells us that the $\ce{F_{eq}-Xe-F_{ax}}$ angle cannot be 90°.  It may be 89.999° or ~80°, but it can't be 90°.  Consequently the structure must be a distorted square pyramid.

@jan Yes, of course,  in $\ce{ C_{4V}}$ they can be in or out of the plane but only as long as they are on the principal axis , but in the crystal structure they are in the plane as far as in can see, and we are discussing this molecule.  So my statement is not "wrong", just incomplete. The down vote is misleading to the OP, if not others, as this answer does describe the actual experimental structure not any theorising as to what it might be. Also the equatorial and axial bond lengths are 0.181 and 0.184 nm so almost identical, which suggests very little difference in bonding.

@ron, I cannot see why without any external forces on the molecule why the angle should not be arbitrarily close to 90 $^o$.  I can see how the molecule might distort to make the square containing the four F atoms into a diamond shape but that would not tip the axial F off axis as it appears to be. Any displacement of the axis F to one side would cause more interaction only to be countered by a lessening to another and so restore  the atom to be on axis. With an external force  in one direction a non-right angle is possible, and in fact expected.

I am not saying that the axial fluorine tips off center.  I am just saying that since the $\ce{F_{ax}-Xe-F_{eq}}$ interaction must be different from the $\ce{LP_{ax}-Xe-F_{eq}}$ interaction, that the $\ce{F_{ax}-Xe-F_{eq}}$ angle cannot be exactly 90° and, in fact, it turns out to be 79°.

@ron, Thanks, now I understand, I had assumed that the Xe atom is in the plane of the other four F atoms. If it is not then the eq-Xe-ax angle need not be 90 and the F can still be on axis, i.e. still $\ce{C_{4v}}$


text:
Is there a relationship between density and viscosity of a fluid?
I'd like to know if there is a connection between the two physical quantities of density and viscosity.
e.g. How can you explain low density liquids such as mineral oils have a lower density than water but far greater viscosity?
e.g.2 Instead, for honey (18% of water and 82% of sugar...) a density of about 1.4 kg/litre but a greater viscosity?

As you see there's no simple relationship, as you can see.

I imagined it! Anyhow, I don't need a quantitative answer, I don't need a formula or an equation, it's enough a qualitative answer...  for example, I think it's a molecular issue: density and viscosity depend on how much condensed are molecules, hence on the strength of chemical bonds... this is my thought, at least tell me if it is not the wrong way

I don't understand the meaning of " how much condensed are molecules"- it doesn't sound right; but the strength of the bonds is important.  Viscosity is kind of a measure of resistance to acceleration. The viscosity of solid steel is crazy high. The viscosity of air is really low, water, alcohol, silica and clay powder, and polymers like oil and gum have have viscosities in between. Temperature (kind of a measure of molecular velocity), density (heavy resists acceleration), and concentrations of mixtures of the fluid(s) and solutes affect the resistance to acceleration (viscosity).

with condensed molecules I mean the intramolecular forces which make molecules more or less attractive or repulsive. In light of this consideration (my thought), I expect substances with high intramolecular forces have to show high density AND high viscosity. Hence, are oils an exception?

They are not directly related as such. Density is defined as mass/volume units $\ce{kg/m^3}$. Water has a value 1000, (liquid) iron 7000, (liquid) lithium 516 , ethanol 789 and carbon tetrachloride 1632 all in units of $\ce{kg/m^3}$.
Viscosity is the rate of transfer of momentum exerted on a plane of fluid when a second plane of fluid a parallel distance $x$ away is moved with viscosity $v$.  The force exerted is $F=\eta \frac{v}{x}$, where $\eta$ is the viscosity.
The fundamental process that takes place when a liquid changes shape is that one layer of it has to slide past another. If we suppose that the liquid could be divided up into cells, then one cell has to 'hop' past another so that the liquid can move. The more difficult that this is the greater the viscosity. Thus to 'hop' a cell has to have an activation energy, that is it has to accumulate in itself enough energy to overcome resistance from other cells that prevent it from moving. Now defining just what a cell is is somewhat vague, but it is larger than a single molecule. But leaving that aside, the activation energy does reflect the inter-molecular interaction and shapes of individual molecules. Spheres with little inter-molecular interaction, e.g. small van-der-Waals forces, no hydrogen bonds, will have a lower viscosity than an awkward shaped molecule such as glycerol with the ability to form lots of hydrogen bonds.
Viscosity has units $\ce{Pa~s = N~m^{-2}S^{-1} = kg~m^{-1}~s^{-1}}$ but more commonly centipoise where $\ce{1~cP = 10^{-3} Pa~S}$. Some values  (in CP at 25 $\ce{^oC}$) are water 1, $\ce{CCl4}$ 0.97, iodoethane 0.56,  ethanol 1.07, cyclohexane 0.89, cyclohexanol 57.5, 1-propanol 1.945, ethylene glycol 16, glycerol 934.

OK, so you agree with me on the importance of intramolecular forces, and this is a first result! Then let's make a further step deep into my issue: are intramolecular forces responsible (maybe, likely not the only) of substances density?

I agree that intermolecular forces do hold liquids together and that they cause viscosity. Colloquially put, density measures how much stuff can you get into a given volume, viscosity about how easily this stuff can move about. Suppose that once molecules are packed together in a liquid intermolecular forces are quadrupled (say). This will have almost no effect on density (because electronic repulsion cuts in rapidly with separation), i.e. liquids are 'incompressible' but  the  increase in intermolecular force will increase viscosity.


text:
Why is NH3 a nucleophile in NH3+CH3 -> H3NCH3
In the image below:

$\ce{NH3}$  has no charge as it has a full octet.
$\ce{CH3}$ needs 2 more electrons to form a full octet, so has a positive charge.
Why does $\ce{N}$ of the $\ce{NH3}$ want to bond with the $\ce{C}$, even though $\ce{N}$ already has a full octet and neutral charge, and bonding would make it have a negative charge?
Is it because $\ce{N}$ is more electronegative than $\ce{C}$ ?

In that picture it’s more like C (which is $\ce{CH3+}$ in there) wants electrons so badly that it’ll get them for crying loudly enough.

When working with nucleophiles and electrophiles, it is easier to think of them as "positive-ophiles" and "negative-ophiles" respectively.
With that said, NH3 (ammonia) has a lone pair of electrons that is attracted to positive charges.  The carbon atom with three bonds is electron deficient and positively charged (which your image does not show).  Thus, as a "positive-ophile"/nucleophile, the ammonia through its lone pair is attracted to and forms a bond with the positively charged CH3+ ion.
Yes, nitrogen does have a full octet, but its lone pair of electrons is nonetheless still attracted to the electron deficient and positively charged CH3+ ion.  This is similar to how water molecules (also fully satisfied octets) can bind with positively charged sodium ions in solvation process of salt.

It's all about thermodynamics. The decrease in energy brought about by filling the carbon octet is larger in magnitude than the increase in energy caused by having a positive charge on the nitrogen atom (it is positive since the total charge of the system must stay the same). This means the sum of the energies of the products is lower than the sum of the energies of reactants, so the reaction is spontaneous.


text:
Why are transition metals said to have different valencies?
I understand that the term valency refers to the number of bonds an element would be able to form with either chlorine or hydrogen. If this is directly related to the number of valence electrons an atom has, why do transition metals have variable valencies?

Only maximum valency is, no one says there can't be unpaired electrons or lone pairs. Also valency is rather primitive concept, coordination number is better, but still not perfect.

The concept of valence electrons doesn't really work for transition metals.

I think it's kind of a poor use of terms here. When talking about transition metals, the valency generally refers to the oxidation state of the metal atom. E.g: hexavalent chromium is $\ce{Cr^6+}$, not necessarily a chromium bonded to six other things.

You can (and should!) use mhchem to format chemical expressions: typing $\ce{Cr^6+}$ renders as $\ce{Cr^6+}$. Nitpicking, I would say that hexavalent chromium is Cr(VI), not $\ce{Cr^6+}$ as the latter implies a naked ion with a 6+ charge.

I don't think it's just a poor use of terms. It's the overuse of a concept that was ill-defined in the first place. According to the IUPAC gold book there is only one valence per element. Hence the statement of "variable valencies" is self-contradictory already. As well as "maximum valence" being redundant. Any making sense of that is diving into muddy waters. One should switch to different concepts, like oxidation numbers, and coordination numbers, as they do not only contain more information, but also less ambiguity.

Transition metals have variable valencies because the energies of the 3d orbital and 4s orbitals (or similar orbital comparisons in lanthanides and actinides, etc.) are similar, so electrons are able to bond from the d-shell as well. Kind of similar to how some high atomic number elements are able to violate the octet rule, such as Xenon in xenon tetrafluoride or Sulfur in sulfur hexafluoride, because their 3d orbitals are able to participate in the valence bonding.
This effect also increases because the elements in the transition series also increases in the number of unpaired electrons, which increases the effect of the d-orbital.

Someone good can check over my work... it might be primitive cause I'm not a rocket scientist


text:
When DNA degrades, does any chemical environment preserve sequence information?
Small bits of DNA information can be found in ancient remains.   In analogy to
old bones becoming mineralized, is there a chemical environment in which the
nucleic acid sequence of the original strand might remain readable though in other, more
stable, chemical form?

You wanted to say nucleic acid sequence, right?

Yes, nucleic acid...

A mineralized form might be possible (maybe DNA can form a pattern of missing atoms in stone) but reading it will be next to impossible. We have many techniques to amplify and read tiny bits of DNA, doing this for imprints in stone would be a huge challenge, especially because many (all?) techniques for reading DNA depend on the amplification of DNA.
As for conservation, as far as I know most of the damage making the DNA unreadable is hydrolysis of the backbone. If the DNA is in an environment without water this will not happen. Of course you need an extremely dry environment for DNA to last millions of years.

I was hoping that a 'stain' might attach to some active sites, and perhaps loss of chemical integrity (so, no longer DNA) would not erase the sequence info.   The prospects for SEM microprobe and suchlike readout technology in the future are ... open.


text:
Type of reaction between metal and salt
What type of reaction is the following?
$\ce{ 10Na(l) + 2KNO3 (s) -> K2O(s) + 5Na2O(s) + N2 (g)}$
So far my best guess has been that it's a double replacement reactions with evolution of gas. Is this correct or am I totally lost?
Edit: It's not a double replacement reactions with evolution of gas (don't know why I thought it was), but what is it then?

K2O won't be gaseous. I rewrote reaction to be readable but check your details.

My bad. K2O should be solid...

Now that's funny, if I added proper redox tag I would basically answer via tagging! So, yes it's redox, you could maybe call it also decomposition...

So what you are saying is that, it's a decomposition reaction? But couldn't you also say that a addition reaction is happening? Some of the oxygen has been separated from the Kaliumnitrate Binds with two of the Sodium to form Sodium Oxide?

Seems you're not getting it, there are redox reactions and the rest and no I think it shouldn't be called decomposition as two thing react and definitely not addition. Also I think that trying to put it into other categories than redox is waste of time.

Reactions don't really have types, even though some textbooks might try to convince you otherwise.


text:
Why do the field of study "Chemical biology" often return papers that are in general anything disease related?
For example, if one tries to look up research opportunities or browse papers in the field chemical biology, most often return research fields related to biological sensing, drug activity etc. and which the ultimate goal of all these researches are related to treating diseases of some kind.
http://pubs.acs.org/journal/acbcct
and only occasionally, do the other non disease treatment related sector of chemical biology (those related to proteins, environmental biochemistry, microbial reactors etc.) can be found
http://www.ch.cam.ac.uk/rig/biological/about
http://chemistry.mit.edu/research/biological
http://www.kemi.uu.se/research/molecular-biomimetic/microbial-chemistry/

Why are most chemical biology research dominated by disease treatment goals despite the chemistry of biological systems are not limited to the medical sector, or is it because the non disease related sector of chemical biology are often lumped into the environmental science, biotechnology and engineering sectors?

Because when we try to sell some new molecule to the chemical biology journals' editors and reviewers we think health and drug like actions sell better than a mere signalling molecule or so ?


text:
re/si face of alkenes
How to determine whether the face of the attached alkene is re or si?
If assigning priorities in regard to the left C atom it is re, but for the right C atom it is si. How is it actually determined?

I’m pretty sure that this depends on the atom attacked. Most attacks on double bond are stepwise to a certain extent. And even when they aren’t, the absolute configuration should be discussed on a per-atom basis.
But if you have a concerted reaction and want to label both: (2si,3re).

Is there any other way to label them? If the alkyl chains are too complex, it isn't easy to determine the locants of these C atoms (like 2 and 3 in the given example).

Not that I know of =C @Marko.

So it is obligatory to know the name of the compound with all the locants to be able to name the face properly.


text:
What is the name of this ester?

Can you tell me the name of this ester? It is two $\ce{C8}$ fatty acids ester-linked with a molecule of 1,3-butanediol in the middle. I would be tempted to say (R)-butanediol-1,3-diyl dihexanoate, but I am really not sure.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the names of esters are usually formed by placing the alcoholic component in front of the name as a separate word. The particular subsections concerning esters from a single alcoholic component and multiple acid components read as follows:

P-65.6.3.3.3 Polyesters formed from a single ‘alcoholic’ component
Esters derived from a single ‘polyhydroxylic’ component are named by placing the name of the ‘polyhydroxylic’ component denoted by a multivalent organyl group (alkyl, aryl, etc.) in front of the name(s) of the acid component denoted by the anion names derived from the appropriate acid(s) (…).
P-65.6.3.3.3.1 When anions are identical, functional class multiplicative nomenclature is used. Names are formed by citing the multivalent group, the multiplicative prefix, and the multiplied anionic component name. Multiplicative prefixes ‘di’, ‘tri’, etc. are used when anions are unsubstituted; when substituted, prefixes ‘bis’, ‘tris’, etc. are used.

In this case, the polyhydroxylic component is denoted by the multivalent organyl group ‘butane-1,3-diyl’. The acid component is denoted by the anion name ‘hexanoate’, which is derived from the corresponding acid ‘hexanoic acid’. Since both acid components are identical and unsubstituted, the multiplicative prefix ‘di’ is used in functional class multiplicative nomenclature. Therefore, the preferred IUPAC name for the compound given in the question is (3R)-butane-1,3-diyl dihexanoate.


text:
Is this problem about acid-base titration wrong?
I wouldn't post mere 'problems' here normally, but I will hopefully be starting to give chemistry education to some 10-11-12th graders and I want to make sure that it is not I that has a problem but rather the problem in the book itself. Here is the problem:

$100~\mathrm{mL}$ of $\ce{H2SO4}$ solution is titrated with $5\cdot 10^{-2}\ \mathrm{M}$ solution of $\ce{NaOH}$. Here are the information that are given in the titration graph:
Total volume of $\ce{NaOH}$ solution added: $0$ | pH: $x$
Total volume of $\ce{NaOH}$ solution added: $400$ | pH: $7$
Total volume of $\ce{NaOH}$ solution added: $600$ | pH: $12$
According to what has been provided above, what is the $\mathrm{pH}$ of the initial $\ce{H2SO4}$ solution ($x$) ? (Assume that $\ce{H2SO4}$ fully dissociates in water to give $\ce{2H+}$ and $\ce{SO4^2-}$ ions)
A) 1
B) 2
C) 3
D) 4
E) 5

My answer: $-\log (2\cdot 10^{-1})$
My steps:

The $\ce{H+}$ ions in the acid solution have been neutralized by $400\ \mathrm{mL}$ of $5\cdot 10^{-2}\ \mathrm{M}\ \ce{NaOH}$ solution. That means that the amount of $\ce{OH-}$ ions in $400~\mathrm{mL}$ of that $\ce{NaOH}$ solution must equal the amount of $\ce{H+}$ ions in the acid solution.

The amount of $\ce{OH-}$ ions in $400~\mathrm{mL}$ of $5\cdot 10^{-2}~\mathrm{M}\ \ce{NaOH}$ solution can be found by taking the product of the liter of solution and the molarity of the solution. And that gives us $0.4\times 5\cdot 10^{-2} = 2\cdot 10^{-2}\ \mathrm{mol}\ \ce{OH-}$ ions. (Yes, I didn't use the units, I don't have to, I explained with words what I was going to do)

Since we used $2\cdot 10^{-2}\ \mathrm{mol}\ \ce{OH-}$ ions to neutralize the acid solution, that should mean that there were $2\cdot 10^{-2}\ \mathrm{mol}\ \ce{H+}$ in the acid solution.

To find the pH of the acid solution, we need to find the concentration of $\ce{H+}$ ions in that solution. The volume of the solution is known, $100~\mathrm{mL}$. And the amount of $ce{H+}$ ions too is known, $2\cdot 10^{-2}\ \mathrm{mol}$. The concentration of $\ce{H+}$ ions in the acid solution should therefore be $2\cdot 10^{-2} / 0.1 = 2\cdot 10^{-1}\ \mathrm{M}$.

$-\log[\ce{H+}] = -\log (2\cdot 10^{-1}) =$ none of the choices above.

Your calculations are correct. I also did a second sanity check: Adding $200~\mathrm{ml}\ \ce{NaOH}$ to a neutral solution does indeed give a $\mathrm{pH}\ 12$ solution, so they also wanted you to assume that the neutral point is the equivalence point.
Which answer to select? Well, considering that:
$$-\log(2 \cdot 10^{-2}) \approx 1.699 \approx 2$$
I would probably go with $2$. Along with telling the pupils that this is a badly worded question and in their exam there will be clearly solveable ones.

This could also be viewed as an exercise in significant figures. There is just one significant figure in $5\cdot 10^{-2}\ \mathrm{M}$ solution of $\ce{NaOH}$ and one significant figure in a pH of 7.

@MaxW True, but there is only one sig fig in $2\log 2$, too.

that was my point. For one significant figure 1.699 rounds to 2. So it doesn't seem that the question is really "badly worded."

@MaxW Hmm … When in doubt, I would prefer the single sig fig $2\log 2$ over the other single sig fig … Fell free to add your view of the topic as an additional answer, though ;)


text:
Composition of carbon rods in alkaline batteries
The carbon rods in batteries look somewhat similar to graphite, but they do not leave a significant mark on paper. What form of carbon are they? Are they simply just graphite with a large amount of binder (clay) mixed in?

They could be coated with something or another.  Did you try breaking one in half or smashing one with a hammer and then marking with some part of the interior?

No, thanks for the suggestion. I'll try when I have another used battery, and post the results as a comment


text:
Definition of "Diabatic" vs "Diathermic"
In Atkins' Elements of Physical Chemistry (Sixth Edition) it is written that:

walls that permit heating as a mode of transfer of energy are called diathermic

and that the converse situation is given the label adiabatic.
Do the words adiathermic or diabatic have any meaning, and if so, how do they differ from the two above?

In thermodynamics we use  adiabatic to mean impassible to heat, i.e. a process in which no heat enters or leaves. Diathermy means 'heating through' and does not seem to be that commonly used. We could use non-adiabatic and diabatic instead but its unusual in thermodynamics. The terms diabatic, adiabatic & non-adiabatic  are usually used in the context of reaction dynamics / quantum mechanics and unfortunately their usage is not entirely consistent in the literature. I don't recall seeing 'adiathemic'

Pippard in his amazingly clear exposition of thermodynamics uses the adjective "adiathermic" for an arbitrary process without heat exchange and "adiabatic"  as a process that is "adiathermic" + "reversible".


text:
Cyanide complexes in Molecular Orbital Theory
In molecular orbital theory, please explain why cyanide ion combines with metal from the carbon end and not from the nitrogen end?

It can be other way round, depending on cation. Why this theory not another? You need sth more advanced then HSAB?

agree with Mith, HSAB is the best explanation imo since cyanide is ambidentate, but I guess if you just want an explanation of why it can coordinate via C then I suppose the shape of the HOMO would explain it. That doesn't mean it always coordinates via C.

In this answer of Martin’s, you can find a molecular orbital diagram of $\ce{CO}$. The corresponding diagram for cyanide, $\ce{CN-}$ is essentially identical, there will only be different orbital energies and very slightly different extends of the lobes.
When forming a coordinate bond to a metal centre, cyanide will primarily attack with its highest occupied molecular orbital, the HOMO, since it is a nucleophile (attacking a positively polarised metal centre). You can also see the HOMO depicted as the bottom image of the following set as taken from Wikipedia:

Note the large lobe extending outwards on carbon’s side. An attack with this lobe is greatly preferred, hence why isocyanides are comparatively rare.