text:
Decarboxylation / Elimination type reaction
I'm quite familiar with E1/E2 reactions and usually use those ideas to explain elimination reactions, however I came across a reaction which was a bit different (in the sense it isn't E1/E2) but still an elimination.

Looking at this reaction I ruled out an E2 type elimination because there's no axial hydrogen which is trans to the bromine group, so something else must have happened here. I drew out the lowest energy chair conformation ('locked' in that conformer because of the t-butyl group) that would react to help me understand what happened and this is what I came up with (excuse the poor chair drawings):

I know hydride is a very strong base and the first step would be deprotonation of the carboxylic acid group (Due to it being the most acidic proton). However after this, mechanistically I wasn't so sure what would happen and how it would happen. My thinking was the oxygen which is now quite charged would push the electron density in towards the carbonyl carbon, and hence the bond would break (between the carbonyl carbon and the cyclohexane carbon). That would form a double bond, and $\ce{CO2}$ and bromide would leave. Is this realistically a correct explanation for what is observed? Also what kind of reaction is this? I haven't come across it before during the usual E1/E2 reactions. Are there any other examples of similar elimination reactions?

Yes, absolutely correct.

Yeah, KH seems like an extreme choice here. Anything that deprotonates that acid should work (probably needs heat, though). Your mechanism is what I thought as soon as I saw the starting material.

This type of reaction falls into the category of Grob fragmentation  [Wikipedia]. Conceptually, it is very similar to an E2 mechanism, where an anti-periplanar pair of electrons flows into the sigma anti-bonding orbital of a carbon-leaving group bond resulting in a new pi bond and cleavage of the carbon-leaving group bond. The primary difference between these two is that in an E2, that anti-periplanar pair of electrons comes from deprotonation, and in a Grob fragmentation, the electrons are from breaking some other sigma bond other than a C-H bond.

A syn-oriented version of this reaction can be seen in the generation of benzyne.

Also, decarboxyation is very common with a beta-carbonyl group. This doesn't exactly follow the Grob pattern, because there is no leaving group. The carbonyl plays the role of electron sink.


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Do Grignard reagents deprotonate amines?
Can I add Mg/dry ether to 2,4,6-tribromoaniline followed by $\ce{D2O}$ in order to replace the bromines with deuterium?
The reaction I'm proposing is part of a synthetic route as shown below.

Amine or amide?

Amine. The picture shows the conversion to be done, and I'm with TriBromo anilene trying to replace it's Br s with D

Yes, it will deprotonate tribromoaniline

you can never use Grignards in presence of acidic protons.

@gannex Usually true, but if you have an excess of Grignard and the deprotonated species still reacts with the surplus Grignard, you might be able to get away with it. It is a wasteful way to use Grignard reagents, though. And in this case it risks throwing away a lot of the substrate.

Yes this is true. This seems to be a common reaction with carboxylic acids, but will this also work with amines? I figure it's a question of pKa and whatnot, but I've never read about people doing reactions like that.

If you really wanted to start with the substituted aniline, you could probably oxidize, deuterate, reduce to get the product if you indeed want the amine and not the amide, but the yield would be poor. Various peroxides and peroxybenzoic acids will do this for you. Lui, Li, et al. indicated that oxidation of substituted anilines would work best in dichloroethane or THF with 4-5 equivalents of m-CPBA, with little difference produced by electronic effects. Then you could use your metal reagent, deuterate, and reduce using a method of your choice.

Of course it's probably easier to just nitrate your 1,3,5-tribromobenzene directly with nitric acid.
If this is a question from a book, they're probably trying to make sure you notice that you can't use organometal chemistry when you have acidic protons, so yes, grignard reagents do deprotonate amines.

What about alpha hydrogens? Why don't Grignard reagents deprotonate these acidic protons?

α-hydrogens are not that acidic (pka = 20-30) and Grignard reagents are strong nucleophiles, so if a carbonyl is present in the molecule, nucleophilic attack will be preferred over proton-abstraction.

Li would react with nitro-compound even easier than with amine.

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Software for visualizing molecular force vectors
I would like to visualize molecular normal modes as vectors sitting on each atom, and not as actual vibrations. I am able to do this, in part, with Molden by copying in the vector into a file readable by Molden (extension is .geo.molden, and is generated from a Molcas calculation). The vector is represented as red lines, but they do not look very nice (see image). Also the background is black, and Molden has a limited number of ways to visualize the molecule.
Is there some other software (free) that can do this? Perhaps Avogadro, VMD, or something along these lines? Specifically I would like to be able to change to different molecular visualizations, and show the vectors as arrows with a easily distinguished color.

Figure 1. This is what Molden can do... not very pretty!
Or possibly, after changing the background color to white with GIMP:

Figure 2. Not really better, since the Hydrogen atoms are represented with white by Molden, as is the label text.

I am not sure what are your needs, but here is a program that I came across. I haven't tried it, but it looks very versatile http://www.paraview.org/gallery/

This might help: http://molecularmodelingbasics.blogspot.dk/2009/06/force-is-strong-in-this-one.html

@JanJensen The link to the jmol application is dead, unfortunately.

BTW, you can change the background color (or even choose a gradient) in molden by clicking the palette icon in the main toolbar widget and selecting "Background Color" or "Background Mode".

As long as the output from your harmonic frequency calculation can be read directly using Avogadro, it will do exactly what you want.

Notice that:

The normal mode of interest is selected on the right-hand side.
"Force" is selected on the left-hand side; this box is present when "Display Settings..." is highlighted in the top toolbar.

Avogadro cannot read geo.molden files. "Reading molecular file failed", it says. The .geo.molden file (obtained from a geometry optimization) contains the geometry and the forces, so perhaps I could construct a readable file manually?

I'd be curious to see the Molden file. It should be readable by Open Babel. While no one has asked about reading forces from such a file, it wouldn't be hard to implement - no one has asked before. OB and Avogadro can visualize these for vibrations or for force field calculations already.

@AndersMB Send an e-mail to the Avogadro e-mail list or to me privately.

@GeoffHutchison e-mail sent, not sure whether you received it or not?

@AndersMB I did. I'm knee deep in other work right now - probably take me about a week to get to it.

@GeoffHutchison Is it possible to change the color of the force vectors?

@AndersMB - No, but it's a good suggestion for 1.2.1...

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Naming compound with multiple carbonyl and/or hydroxyl groups
I think I understood the main idea behind naming compounds with carbonyl groups, however I am not sure about a few things.

For the first compound, there is more than one carbonyl group and one of them is in the end of the chain. I am trying to decide between 2-oxopropanal and propana-1,2-dione.
For the second one, I am not sure what happens if there are two oxygens attached to a carbon atom with a double bond. I am trying to decide between methanedione and methanedial.
I am not sure about this because one definition of aldehydes I have heard is when the carbon atom is attached to a hydrogen as well as the oxygen. And the other is if it is in the end of the chain.
And the final compound that I am not sure about is when there are two hydroxyl groups attached to the carbon as well as an oxygen with a double bond. I think it is hydroxymethanoic acid, but I am not completely sure.

The second compound is carbon dioxide anyway, the first one is 2-oxopropanal because it doesn't have 2 ketones and number 3 is carbonic acid

but hydroxymethanoic acid is also correct

@MrLuke370 Yeah, many names are correct. However, I am looking for the IUPAC preferred names.

Why don't you try chemspider to see what options it gives?

@MrLuke370I had never heard of it before, but there doesn't seem to be an option to name it. It just gives me a list of compounds that match mine and they have names but they are not the systematic names.

In the name sectionin light grey it says IUPAC beside the correct names, and I also put those compounds through chemdraw and it gave the names in my first comment, and as far as i know chemdraw gives IUPAC names .However I could stand to be corrected on that last point

Carbon dioxide and carbonic acid are inorganic compounds, and therefore would not be named using the IUPAC nomenclature for organic compounds.
The other compound is named 2-oxopropanal, because aldehydes have higher priority than ketones by IUPAC rules.

And why are they considered inorganic when they contain carbon in its excited state? What are the criteria for deciding if a compound is organic or not?

It's complicated, but generally speaking, $\ce{C-H}$ bonds.

What is "carbon in its excited state"... there is no such thing.


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Can we make glass by heating Martian Soil
I was puzzled by a very interesting question. Can we heat up Martian soil and produce glass just the way we do here on Earth? (the basic way)

For info, Martian sand consists of the following -

$\ce{Na2O}$ — $4~\%$
$\ce{MgO}$   — $14~\%$
$\ce{Al2O3}$  — $18~\%$
$\ce{SiO2}$   — $8~\%$
$\ce{P2O5}$  — $2~\%$
$\ce{SO3}$   — $10~\%$
$\ce{Cl}$ — $2~\%$
$\ce{K2O}$  — $2~\%$
$\ce{CaO}$  — $14~\%$
$\ce{TiO2}$   — $2~\%$
$\ce{Cr2O3}$  — $2~\%$
$\ce{MnO}$    — $2~\%$
$\ce{FeO}$ — $4~\%$
$\ce{Ni}$   — $8~\%$
$\ce{Zn}$    — $6~\%$
$\ce{Br}$    — $2~\%$

I can't see anything out of place here, if we heat it up, the silica
should do it at around $1700~\mathrm{^\circ C}$.
If we can figure out ways to make glass up there then it would be easy for Martian colonization if that is ever going to happen.

I think this would be closer to a brick than a glass block.

Note that the sand does not actually contain chemical compounds like $\ce{Na2O}$, $\ce{P2O5}$, $\ce{SO3}$, $\ce{K2O}$ etc. These are just the reference compounds for the elemental analysis.

Mars was found to have quite large reserves of sulfur, which can be used as a hot binder (at lowly ~200°C) for sand to make a kind of cement. http://arxiv.org/abs/1512.05461

@Loong - The OP called it "Martian sand" but I think "Martian soil" would probably be more accurate. We now know that Mars did have water, but I'd wonder about Mars having any ancient sandy beaches. My point about a brick is that you'd use less heat to sinter the soil to a brick than you would if you really melted the whole block to form a glass (assuming the composition is amenable which i doubt).

The problem is sorting. Ergo, unless you bring your separator with you, the best areas for finding naturally sorted materials are in outflows such as found in Candor Chasma. Semi-translucent larger grains of quartz-rich phenocrysts might be found in useful quantities. Fusing these with a solar kiln would produce a brick of about 0.125 meters thick that would both shield from the rain of radiation and pass some of the weak Martian sunlight, easing the circadian rhythm of colonists.
Time to temperature (1700 degrees C) still makes production slow. And the loaders and other equipment need to be transported and successfully landed.
See
A Fused Regolith Structure
Ethan W. Cliffton
Proceedings of Space 90
Engineering, Construction, and Operations in Space
ASCE 1990
Albuquerque
and
Candor Chasma Camp Ethan W. Cliffton
Proceedings of Space 88
Engineering, Construction, and Operations in Space
ASCE 1988
Albuquerque


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The solubility of different compounds relative to each other in water

Which of the following has the least solubility?
$\ce{NH4HCO3, NaCl, NH4Cl, NaHCO3}$

The answer is given as $\ce{NaHCO3}$, but I don’t understand why.
Solubility is a chemical property referring to the ability for a given substance, the solute, to dissolve in a solvent. It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium.
Here's what I was told:

Solubility decreases with increase in polarisibility
It increases with increase in covalent nature.
It increases with increase in hydration enthalpy and lattice enthalpy.

Which leads me to two other questions:

How do these factors really effect solubility, especially, covalent nature?
I guess polarisibility can be linked with hydration or solvolysis, so I understand it.
How can I find the order of solubility of different compounds relative to each other in water?

"Solubility increases with an increase in covalency" makes 0 sense - look at silver halides.. btw it is "affect" not "effect" and spaces go after commas not before.

Sorry I didn't notice it while typing, it must be typing error. How can be find the order of solubility?

First of all, let's temporarily make a distinction between ionic and covalent compounds. Generally speaking, covalent compounds are more soluble in nonpolar solvents while ionic compounds are more soluble in polar solvents. Of course ionic and covalent bonding exist on a spectrum and some compounds you would expect to be ionic are indeed more covalent in nature. This concept is often visualized using the Van-Arkel Ketelaar triangle of bonding

As described by the Fajans' Rules, highly polarizing cations will distort polarizable anions, pulling their electrons into the inter-atomic space to the extent that they essentially form ionic compounds. This is one of the reasons solubility decreases with increased polarity. Solubility does not increase with increased covalent nature. In water, it would decrease. The best example of this is beryllium. Almost all $\ce{Be^{2+}}$ compounds actually exhibit covalent chemistry, as we can see with  beryllium peroxide in the Van-Arkel Ketelaar triangle above. This is because beryllium has extremely high charge density.
In order to properly understand solubility trends of ionic compounds, we should examine the enthalpy of dissolution using a Born-Haber cycle.

The first step is vaporization of the crystalline lattice, which has a large positive enthalpy. This is affected by a few factors, but it basically represents the stability of the crystal. Generally, dipositive anions/cations will have much greater lattice energy because of stronger electrostatic interactions, and lattice match should also be considered. A significant mismatch in size will lead to lower lattice energy. In the Born-Haber cycle, the ion-ion interactions are broken when the crystal is vaporized, then ion-dipole interactions are formed when water molecules surround the gaseous ions, so the other  factor to consider in the Born-Haber cycle of dissolution is hydration enthalpy, which is negative. Generally, this is increased with increasing charge density. Sometimes net enthalpy is positive, and in these cases, entropy may also come into play. For the most part, you can think of ionic solubility as a balance between lattice energy and hydration enthalpy.

Solubility increases with decreased lattice enthalpy increased hydration enthalpy

and you're right that

Solubility decreases with increased polarizability

Sodium's ionic radius is 227 pm, while ammonium's is 175. Often ammonium is a good substitute for sodium, as a large, low charge density cation (for stabilizing highly polarizable anions like bicarbonate). In fact, ammonium bicarbonate is one of the only soluble non-group 1 bicarbonates. Still, ammonium is still a little smaller than sodium. As a solid, ammonium will stabilize chloride better, but bicarbonate worse.
Since bicarbonate is much much more polarizable than chlorine, the two bicarbonate salts will probably be less soluble than the two chlorides. Ammonium will stabilize chloride better by size match, so ammonium chloride will be a little less soluble than sodium chloride. Due to it's large size and low CD, though, sodium will stabilize bicarbonate better, so sodium bicarbonate will be a little less soluble than ammonium carbonate.
The way I see it, the order should be:

$\ce{NaCl}$ > $\ce{NH4Cl}$ > $\ce{NH4HCO3}$ > $\ce{NaHCO3}$ (decreasing solubility)

Of course solubility is a pretty complex concept and these are only generalizations. To get a better handle on it, I would try looking at the enthalpy and entropy cycles of different types of salts in each of the main groups. Trends exist, but the way they work is a little different depending on the group. I think, for example, that lattice match is mostly important in group 1 salts because of their lattice types. Anion charge becomes an important factor in group 2. Trends can be established but a lot of the time they are only consistent within a given group.

From where did you get the data that the ionic radius of Na is 227 pm? According to wikipedia Na+ has a radius of 116 pm and Na has a radius of 154 pm.

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Determine the concentration of hydrogen cations to prevent precipitation of monocalcium phosphate?

Determine the $\ce{[H+]}$ needed to just prevent precipitation by $\ce{H2PO4-}$ in a $\pu{0.25 M}$ $\ce{H3PO4}$ solution that has $\ce{[Ca^2+]} = \pu{0.15 M}$.
  The $K_\mathrm{a1}$ of $\ce{H3PO4}$ is $7.1 \times 10^{–3}$. The $K_\mathrm{sp}$ of $\ce{Ca(H2PO4)2}$ is $1.0 \times 10^{-3}$.

For this problem, I first found the $\ce{[H2PO4-]}$ present in the solution when it becomes saturated, which is $\pu{8.2 \times 10^-2 M}$. So I interpreted this as at equilibrium, there is at most $\pu{0.082 M}$ $\ce{H2PO4-}$ present in the solution. When I went to set up my ICE chart for $$\ce{H3PO4 <=> H2PO4- + H+},$$ I put $\pu{0.082 M}$ as the equilibrium value of $\ce{[H2PO4-]}$. However, the answer key wrote

$
\displaystyle
K_\mathrm{a1}
= \frac{\ce{[H+][H2PO4-]}}{\ce{[H3PO4]}}
= \frac{x (0.082 + x )}{(0.25 - x )}$

meaning that $\pu{0.0082 M}$ is the initial concentration and the equilibrium concentration of $\ce{H2PO4-}$ is $0.082 + x$. But if this was the case, then wouldn't $\ce{Ca(H2PO4)2}$ precipitate since there's more than $\pu{0.082 M}$ at equilibrium?
I think I'm probably misinterpreting the question, so any clarification on what's happening here would be appreciated.

edited - MaxW
Added image (cut and paste job for needed parts...) of problem from 2002 U. S. NATIONAL CHEMISTRY OLYMPIAD NATIONAL EXAM—PART II

You're correct. $$\ce{[H2PO4^{-}] =} \text{ 0.082}$$ which means that $$\ce{[H3PO4] =} \text{0.25 - 0.082 = 0.168}$$.

@MaxW So is what the answer key wrote wrong then? In this case, the question is kind of pointless since having 0.25 M $\ce{H3PO4}$ and 0.15 M $\ce{Ca^{2+}}$ won't precipitate anyway, so what does "the $\ce{H^+}$ needed to just prevent precipitation" mean?

Think of starting with a phosphoric acid solution but increasing the pH by adding NaOH. At what pH will $\ce{Ca(H2PO4)2}$ start to ppt? // Since $\ce{[H2PO4^{-}]}$ and $\ce{[H3PO4]}$ have been calculated, you can us Ka1 to calculate $\ce{H+}$.

@MaxW But how does this relate to what the answer wrote?

It shows that the answer key is wrong, or there is some interpretation of the problem where $\ce{[H3PO4] + [H2PO4^{-}]} \ne 0.25$ // The gist is that I can't explain an answer that doesn't make any sense...

Even though this is an old questions, if you have a chance it would be great if you could add a source and citation for the exercise.

The question is 3c in 2002 U. S. NATIONAL CHEMISTRY OLYMPIAD
NATIONAL EXAM—PART II // Since exam is sponsored by ACS I suspect that I missed something...

I was looking for something else and stumbled across this old problem. I noticed Martin's comment about the source, and I looked for one. Noticing that this was sponsored by ACS made me wonder if my initial analysis had been correct.
The answer key, the OP, and I agree that the $\ce{H2PO4^-}$ concentration must be $8.2\times10^{-2}$ molar to cause $\ce{Ca(H2PO4)_2}$ to start to precipitate.
Problem 1 - The problem statement lists $K_{\alpha1} = 7.1\times 10^{-3}$ but the answer key uses $7.3\times10^{-3}$
My faith in the infallibility of the ACS is shaken...
Problem 2 - There are really two ways to solve the problem.
A. Neutralizing the acid
The solution starts off as 0.25 molar $\ce{H3PO4}$ and 0.15 molar $\ce{Ca^{2+}}$. You add $\ce{NaOH}$ until $\ce{Ca(H2PO4)_2}$ starts to precipitate. What is the pH?
Mass balance for all phosphate species is:
$0.25 = \ce{[H3PO4] + [H2PO4^-] + [HPO4^{2-}] + [PO4^{3-}]}$
The solution is very acidic so we can assume $0 = \ce{[HPO4^{2-}] = [PO4^{3-}]}$ so
$0.25 = \ce{[H3PO4] + [H2PO4^-]}$
$\ce{[H3PO4] = 0.25 - [H2PO4^-] = 0.25 - 0.082 = 0.168 }$
Now using the equilibrium expression for the first ionization of phosphoric acid
$K_{\alpha1} = \dfrac{\ce{[H^+][H2PO3^-]}}{\ce{[H3PO4]}}$
Rearranging and using $7.3\times10^{-3}$ from the answer key
$\ce{[H^+]} = \dfrac{K_{\alpha1}\ce{[H3PO4]}}{\ce{[H2PO4^-]}} = \dfrac{(7.3\times10^{-3})(0.168)}{0.082} = 0.014956 \ce{->[Rounding]} 1.5\times10^{-2}$
Fortuitously using 7.1 from the problem statement yields the same rounded value.
$\dfrac{(7.1\times10^{-3})(0.168)}{0.082} = 0.014546 \ce{->[Rounding]} 1.5\times10^{-2}$
B. Making a buffer
To the 0.25 molar solution of $\ce{H3PO4}$ you add $\ce{NaH2PO4}$ until $\ce{Ca(H2PO4)_2}$ starts to precipitate. What is the pH?
Let $x$ be the nominal molarity of the $\ce{NaH2PO4}$.
Again the solution is very acidic so we can assume $0 = \ce{[HPO4^{2-}] = [PO4^{3-}]}$ so
$0.25 +x = \ce{[H3PO4] + [H2PO4^-]}$
But $\ce{[H2PO4^-] = 0.082}$, so
$\ce{[H3PO4]} = 0.25 + x - \ce{[H2PO4^-]} = 0.25 + x - 0.082 = 0.168 + x$
The charge balance is given by
$\ce{[Na^+] + [H^+] = [H2PO4^-] + 2[HPO4^{2-}] + 3[HPO4^{3-}] + [OH^-]}$
Since the solution is very acidic we can neglect $\ce{[HPO4^{2-}]}$, $\ce{[HPO4^{3-}]}$, and $\ce{[OH^-]}$. So
$\ce{[Na^+] + [H^+] = [H2PO4^-]}$
$\ce{[H^+] = [H2PO4^-] - [Na^+]} = 0.082 - x$
Now substituting into the equilibrium expression for the first ionization
$K_{\alpha1} = \dfrac{\ce{[H^+][H2PO4^-]}}{\ce{[H3PO4]}} = \dfrac{(0.082-x)(0.082)}{0.168+x}$
$0.0073(0.168+x) = 0.082^2 - 0.082x$
$(0.0073 + 0.082)x = 0.0893x = 0.082^2 - 0.0071(0.168)$
$x= 0.06194$
So the buffer solution would be 0.25 molar $\ce{H3PO4}$ and 0.062 molar $\ce{NaH2PO4}$.
$\ce{[H^+] = 0.082 - 0.6194 = 0.02006 ->[Rounding] 2.0\times 10^{-2}}$

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Tin in silver nitrate, any soln color change?
I have lab question in which a $\ce{AgNO3}$ solution with a tin strip placed in it is mentioned. This isn't a reaction we actually did in the lab, but due to a color change in the solution of one we did do, I was wondering if there would be any for this reaction. Googling I found that the $\ce{AgNO3}$ solution is initially colorless, but that's all I could find. So would there be for this one, or would the solution remain colorless?

It would remain colorless.

The solution will remain colourless until silver particles will participated from the solution.
Silver nitrate solution is unstable, its decomposition is accelerated by light:
$\ce{2 AgNO3(l) -> 2 Ag(s) + O2(g) + 2 NO2(g)}$
That will change the color of the solution to brown. After silver participation, the solution will be colorless again.
A tin strip added to the nitrate solution will dissolve into tin(II) nitrate and will consequently react with silver nitrate accordingly:
$\ce{Sn(NO3)2 + 2AgNO3 + H2O -> Sn(OH)4 + 4HNO3 + 2Ag}$
Analogous silver reactions are described by Tollens' reagent, mirror preparation and gelatin silver process.


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Does this produce the anti-Markovnikov product because of the ROOR?

I know that the anti-Markovnikov product is produced when there is a free radical present. Does the ROOR separate into two free radicals and that changes the reaction? But ROOR doesn't end up in the product at all right? Why? Is my thinking correct?

Yes, having a peroxide will change the regiochemistry to an anti-Markovnikov addition. The reason, as suggested, is because of a change in mechanism to a radical process (below). The homolytic decomposition of the peroxide initiates the radical chain, generating two hydroxyl radicals. The hydroxyl radical abstracts a hydrogen from HBr to give a bromine radical. This is where the cycle really begins, with bromine radical adding to the terminal end of the alkene, leaving a radical at the internal position. The stability of a tertiary radical relative to a primary radical drives this selectivity. The tertiary carbon radical abstracts a hydrogen from the next molecule of HBr, giving the product and a new bromine radical which reenters the radical chain.


text:
Can I remove flesh from seeds with hydrochloric acid?
I have some crenellated seeds with which I make jewellery and I need to remove the remaining fruit flesh from them.  I have tried lots of things including ammonia, vinegar, bleach and a ultrasonic jewellery cleaning machine.  the only thing that works is a wire brush which takes forever and results in grazed fingers.  I wondered if some solution of the hydrochloric acid I use in the pool would work.  If so, what concentration should I use and how long should the seeds be immersed in it?  If not, what do you think would work?

I'd be inclined to try something like a AC waterpik. The AC version has more power than the battery operated one.

It depends highly on the seed and the fruit flesh, but using hydrochloric acid to clean off the "jelly" from freshly collected tomato seeds is a recognized laboratory protocol. That protocol uses a 1:1 mix of concentrated HCl and water for 20 min., followed by a 20 min wash in 10% w/v trisodium phosphate to neutralize the acid.
Depending on what sort of flesh is covering the seeds, it may take different amounts of time and different strengths of acid to accomplish the job. You will likely have to experiment a bit. Start "gentle" (as in, less concentrated acid) and work your way up. - You probably don't want conditions that results in complete flesh removal, but rather ones where gentle agitation (e.g with an old toothbrush) after neutralization will dislodge the flesh. - Viability of the seeds will also vary depending on the exact protocol and how tough the seeds are, though it doesn't sound like you need the seeds to be able to germinate afterwards.
As always, when working with concentrated acids and bases, use of proper safety equipment is highly recommended - even for acids and bases they sell in the hardware store.
For the chemical perspective, concentrated acid solutions catalyze the hydrolysis of peptide, ester and glycosidic bonds of the polymers that make up the flesh of the fruit surrounding the seeds. With the polymers degraded, they more easily dissolve in solution. (This is also the reason that our stomachs have hydrochloric acid in them.) The cellulose that makes up the seed coat is recalcitrant to degradation, and so remains untouched to the "mild" acidic conditions.

Great reference! // One mystery here is how large the seeds are. I was thinking of something like a peach pit. The more flesh is removed prior to the acid bath, the better the acid bath will work. Also 50% HCl isn't really something for casual home use. There are some real safety concerns.

Thanks for your help.  I tried a 1:1 acid/water bath for 20 mins followed by 20 mins in a caustic soda solution to neutralise the acid.  This worked fine, the  dirt and flesh came off easily with a wire brush.  I then tried just putting them in caustic soda for 20 mins followed by a thorough flushing with water.  This also worked perfectly!  When I find the ultrasonic jewellery cleaner I will try that instead of the wire brush.  I appreciate your help. The seeds, by the way, are quandong seeds, they have very deep crenellations.


text:
How can I get rid of 700 million tons of chlorine gas?
I am conducting a thought experiment in which 40 million tons of hydrogen are supplied through saltwater electrolysis each year, creating 700 million tons of chlorine gas (Cl2) and 800 million tons of sodium hydroxide. (Lye)
I think I can get rid of a good chunk of the sodium hydroxide by mixing it with nitric acid and creating sodium nitrate for use as substitute for ammonium nitrate fertilizer, but I am not sure on how to get rid of the chlorine gas.

Well, if you're allowed to use anything you like, just react it with sodium metal and enjoy a lifetime's supply of table salt...

Just sell it.  It is used in industry.

This is a follow-on to a previous question http://chemistry.stackexchange.com/questions/49346/will-hydrofluorosilicic-acid-react-with-chlorine-gas

Simply, sell it (Javel water!)

Current world production of chorine is about 100m tonnes per year. You would seem to be creating a gross surplus. Perhaps giving everyone in the world clean water and lots of sodium hypochlorite would be a start. Or you could make a lot of PVC and distribute it free.

Why not go one step further, you could react the chlorine with the sodium hydroxide to get NaCl + NaOCl and if there's a way of somehow separating the two you can get a lifetime's supply of table salt AND water disinfectant :D

Let the electrolysis produce oxygen instead of chlorine.

mattblack you are correct, this would produce a tremendous surplus of chlorine and sodium hydroxide.  Poly Vinyl Carbonate is a good use for it, but even so, the demand is limited.  Water Treatment is also a good option.  Orthocresol, this is a good idea, but then what would you do with all that bleach?  aventurin, even with a catalyst, the rate and efficiency of hydrogen production using distilled water is very slow and inefficient.

@orthocresol That would be a lot of energy!!!


text:
How does a group 15 oxide react with water?
Given the following reactants:

$\ce{P2O5}$ and $\ce{H2O}$

How does one determine what product the two will form in a reaction?
I instantly tried to treat the problem as a double displacement reaction. However, the compounds that I got by doing so were absurd. Do double replacement reactions only occur with ionic compounds (e.g., the cations of two ions switch)?
The two elements listed above form $\ce{H3PO4}$. How do I even go about determining that? What indicates that such a product forms from those two reactants?

Hint: try an addition reaction instead of double replacement.

The reaction between the two substance you have mentioned above in synthesis
it is an acid-base reaction .
and about the other question you asked above i can say that we have reactions where two ionic substances take part in oxidation reduction reactions instead of double replacement.

Let me outline the answer previously given by eyesun:
In order to figure out how something might react with water, it is helpful to describe and categorize the starting material and to find analogies.

Apparently, $\ce{P2O5}$ is a group 15 oxide, the oxide of a non-metal.
Other non-metal oxides that come to our mind are from the same element ($\ce{P2O3}$), from the same group ($\ce{N2O5}$), or from the neighboring group 14 ($\ce{CO2}$) and group 16 ($\ce{SO2}$, $\ce{SO3}$, $\ce{SeO2}$).

What other compounds do we know that are similar to these oxides?

We do know that a number of oxo acids exist. Mostly, they do contain one atom of a group 14-16 element with a couple of oxygen atoms and some protons attached.

Can we match these?
$$\ce{H2SO3 <=> SO2 + H2O}$$
$$\ce{H2SO4 <=> SO3 + H2O}$$
$$\ce{H2CO3 <=> CO2 + H2O}$$
$$\ce{2HNO3 <=> N2O5 + 2H2O}$$

We conclude that these non-metal oxides are like oxo acid minus water and we therefore decide to call them anhydrides.

Having worked our way through all of these, we decide to apply our knowledge to $\ce{P2O5}$:
$$\ce{2 H3PO4 <=> P2O5 + 2 H2O}$$

to realize that phosphorous pentoxide is the anhydride of orthophosphoric acid.

I would add that the complete name of this acid is the orthophosphoric acid. \ce{H3PO4}.

@Another.Chemist Good point. Fixed.


text:
How to test if white powder is lime fertilizer?
I found two buckets (white and black) containing a very light grey powder in the basement. The label is partially removed, but it probably originally contained lime fertilizer.
However, I am not sure if the current content is still lime or if it got refilled with something else?
How can I proof that the powder is calcium oxide or something similar that I can throw on my lawn to make the grass greener?

I already tried to dissolve the powder of both buckets in water, there are a few grains like sand that stay, but the rest seems to dissolve or rather suspend, the water becomes milky.
Adding normal household vinegar, the powder from the black bucket did not react any differently than with water, but the powder from the white bucket dissolved and created bubbles, so I assume the reaction with acid produces carbon dioxide.
What further tests which I can perform using household stuff would you recommend to make sure I have a bucket of lime fertilizer or anything else?

Yes, it pretty much reacts like a carbonate. In order to test for calcium, a bunsen burner or gas stove to perform a flame test would be helpful.

@KlausWarzecha So I have to expect a bright orange flame when holding a bit of the powder over a "dark" blue flame if it contains mainly calcium? Also, what about the powder from the black bucket that did not react with the acid?

Yes, that's what I would expect too. As for the other bucket: that could be anything from another fertilizer, such as a mixture of ammonium phosphate and potassium nitrate, to de-icing salt. There's just not enough information without the label. If availabel,  I'd probably give it to a communal waste collecting service for proper disposal.


text:
Determining the temperature at which 90% of the reactants in a equilibrium have reacted

I have the following problem I am stuck on, specifically the last bit (iii). The first parts are fairly trivial and I have arrived at the following equation to determine the temperature:
$$T=\frac{\Delta H}{\Delta S - R \ln K}$$
However I am unsure on how to proceed. I do not have a value for $K_p$ and I can't think of how to deduce it. I have tried the following, by assuming that
$n_{\ce{Fe3O4}}=n_{\ce{CO}}$, going to the following expression
$$K = \frac{(0.9 x)^3 (0.9 x)}{(0.1x)(0.1x)}$$
but obviously that doesn't cancel.
Any help? I'm probably barking up the wrong tree with the above expression as I'm aware I'm looking for $K_p$ in terms of partial pressures but it would surely result in the same overall problem.
Thanks.

Just out of curiosity, where is this question from?

Assuming $\ce{FeO}$ and $\ce{Fe3O4}$ are solid throughout the range of temperatures we are interested in, they won't appear in the expression for the equilibrium constant $K$ (which in this case is equivalent to $K_p$). If $n$ is the total number of moles of $\ce{CO}$ that you start with,
$$K_p = K = \frac{p_{\ce{CO2}}}{p_{\ce{CO}}} = \frac{n_{\ce{CO2}}}{n_{\ce{CO}}} = \frac{0.9 n}{0.1 n} = 9$$
Plugging this into the formula previously derived, we find $T = 1388~\mathrm{K}$.

Indeed, although I'm not sure how you reached a value of ~1800 K, I think this is the correct answer. On a side note - please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. I'll help you edit it this time.

The question is from the 2014 exam paper for the IA (First year undergrad) Chemistry course at Cambridge Uni.

Indeed, I thought so.


text:
Which reactions are more temperature sensitive: the ones with higher Ea or the ones with lower Ea? And why?

Which reactions are more temperature sensitive: the ones with higher $E_\mathrm{a}$ or the ones with lower $E_\mathrm{a}$? And why?

I wasn't able to find much useful content on googling it, however on doing the math I came to a conclusion that the ones with higher $E_\mathrm{a}$ are more sensitive to increase in temperature. But why is it so? I mean math proves it, but why exactly does it happen?
Here are my calculations:

hint: see Maxwell velocity or kinetic energy distribution

The reaction with the lower activation energy can proceed smoothly without the increase in temperature whereas the the reaction with the higher activation energy cannot. When you increase the temperature, it won't make much difference to the reaction with a lower activation energy since it was already fine with a lower temperature. On the contrary, the reaction with higher activation energy will now gain sufficient energy making the molecules more likely to react and thus, the temperature change affects this reaction much more.

Plotting a graph between lnK vs 1/T shows that reaction having high activation energy are more temperature sensitive. Since the slope of high Ea is more steeper than the one having low Ea. This shows that rate constant hence rate of reaction will be high for large Ea

Hi Amit Ujjwal, welcome to Chem.SE! Take the tour to get familiar with our site. Regular text can be formatted with Markdown, and mathematical expressions and chemical equations can be formatted using Latex syntax. That said, can you please properly photoscan your image, currently it's a little difficult to interpret. Thanks!


text:
Comparing radii in lithium, beryllium, magnesium, aluminium and sodium ions
Apparently the of last four, $\ce{Mg^2+}$ is closest in radius to $\ce{Li+}$. Is this true, and if so, why would a whole larger shell ($\ce{Mg^2+}$) be closer in radius to $\ce{Li+}$ than its isoelectronic neighbor ($\ce{Be^2+}$) with just one extra proton?

It's a common misconception that a 2s orbital in lithium is the same size as a 2s orbital in, say, fluorine. It's not true at all. That "just one extra proton" has quite a large impact. To directly answer the question, https://en.wikipedia.org/wiki/Diagonal_relationship Second paragraph, second sentence: "On moving across a period of the periodic table, the size of the atoms decreases, and on moving down a group the size of the atoms increases."

Consider the size of the first s-orbital of $\ce{Mg}$. The nuclear charge is 12+. Therefore the first electrons, close to the nucleus, are attracted by a relatively large force. These first electrons the 'shield' the nucleus from some of the charge when adding more electrons to the orbitals, these latter electrons experience a smaller charge than the first. The result of this is that nuclear radius does not change according to number of electrons (only).

Many people may think that orbitals, or ‘shells’ essentially work like planet orbits around the sun; i.e. no matter how many electrons populate ‘Mercury’, ‘Venus’ and ‘Earth’ or how many protons are in the ‘Sun’ Venus’ orbit will always have the same dimension.
This is very much not so when studing atomic orbitals. The principle of hydrogen-like orbitals (which are the only orbitals that can be exactly calculated with the Born-Oppenheimer–approximation) is that we are dealing with an electron free to rotate and translate on a potential energy surface which is formed by the electrostatic potential of a proton or a nucleus with a charge of $z+$. Commonly, taking the 1s orbital as an example, there is a distance from said nucleus where the probability of locating the electron in the sphere defined by that radius is larger than a certian cutoff value, e.g. $90~\%$. If we go from a hydrogen-like system to a $\ce{He+}$ cation, we are doubling the charge thus strongly increasing the attractive force between nucleus and electron. As such, the $90~\%$-radius is much closer to the nucleus than before.
This holds true with limitations when no longer observing hydrogen-like atoms but actual multi-electron atoms: the more protons in the nucleus the more contracted the orbitals become. While that effect in itself is rather small, the second one is substantial: the higher an ion’s charge is, the more contracted its orbitals become. Therefore, $\ce{Be^2+}$, which is even considered to be the smallest atom or atomic ion known (yes, smaller than hydrogen) is substantially smaller than lithium.
When directly comparing two atom sizes of elements that are above and below each other in the periodic table, naturally under otherwise identical conditions the lower one will be larger; even though it has at least 2 (lithium) more protons it has also started populating an additional shell which is further removed from the nucleus. Similarly, if you compare $\ce{Li+}$ and $\ce{Mg+}$, magnesium will still be substantially larger. (In this case, the additional remaining electron in magnesium’s 3s orbital is the deal breaker but note that the same argument applies to $\ce{C+}$ and $\ce{P+}$ where no shell is completely empty.
As noted before, however, if we substract another electron, i.e. if we compare the isolobal $\ce{Li+}$ and $\ce{Mg^2+}$, the latter’s orbitals shrink substantially. By empirical observations scientists have discovered what is known as the diagonal relationship: two ions that are top-left bottom-right diagonally placed in the periodic table, whereby the bottom-right one’s charge is greater by one, have approximately the same radius.


text:
Does increasing electronegativity imply more covalent character?
If there is a larger electronegativity difference between two atoms, shouldn't the compound be more ionic? In an explanation of the diagonal relationship, it states that "on moving across the period, the elements become progressively more covalent" yet their electronegativity increases. What is even meant by an element being covalent? I thought only compounds/molecules could be covalent.

The main force responsible for holding together covalent bonds  and ionic bonds is different. Ionic bonds are held together through electrostatics. A transfer of electrons causes one atom to be positively charged and another one to be negatively charged. In other words, electrons are given away or taken away by atoms. Covalent bonds are held together due to the delocalization of electrons, i.e. electrons are shared by the atoms. For more detail look at What makes the difference between ionic and covalent bonds? (physics.se)
There is no meaning to the statement that an element is covalent. The reason is that the nature of a bond (metallic, covalent, ionic) is determined by the elements participating in the bond. Sodium may form a metallic bond with itself, but it will form an ionic bond with halogens. Halogens will form covalent bonds with themselves but will form ionic bonds with alkali metals. Furthermore, the conditions can determine the nature of a bond. For example, hydrogen is metallic under high temperature and pressure.
Just a side note, compounds can be ionic. I recommend that you read this thread explaining the difference between compound, element, mixture, etc: What is the definition of of 'compound', 'mixture', 'element' and 'molecule'?
Tldr: Wikipedia is not being exact. Saying that an element is covalent is inaccurate because the terms covalent and ionic refer to bonds not elements.

The words covalent and ionic chosen in the Wikipedia article are bad. Better word choices would be non-metallic and metallic or something along these lines.
What the article wants to say is that the elements on the left half and bottom half of the periodic table are mainly metals and thus often (not always, not preferably, just often) bond metallicly to neighbouring elements and typically have a metallic element structure, while elements on the top half and right half are typically non-metals and often (see above) bond covalently to neighbouring elements/themselves. Since the elemental structure of non-metals is usually dictated by covalent bonds, one may excuse the poor word choice there, but metallic bonding is not ionic.


text:
How many electrons does boron have on average in B2H6?
The boron interaction with hydrogen gives it one electron I suppose, but then the octet is not complete. So how many electrons does the boron atom have exactly and why?

Don't use mathjax in titles, also you used it improperly.

Six for each as you can count. See http://chemistry.stackexchange.com/questions/228/what-makes-banana-bonds-possible-in-diborane


text:
Adding Standard Reduction Potentials
I am working the following USNCO problem (#41 from 2002). Based on this question and answer Deriving a reduction potential from two other reduction potentials, it seems that $\Delta G$ must be calculated and then added. However, when I do this, I get an answer that is not one of the given choices.
The problem is:

Use the given standard reduction potentials to determine the reduction potential for this half-reaction:
  $$\ce{MnO4- + 3e- +4H+ -> MnO2 + 2H2O}$$
The given reactions are:
$$\begin{align}
\ce{MnO4- + e-} &\rightarrow \ce{MnO4^2-} & E &= +0.564~\mathrm{V} \\
\ce{MnO4^2- + 2e- + 4H+} &\rightarrow \ce{MnO2 + 2H2O} & E &= +2.261~\mathrm{V}
\end{align}$$
The possible answers are: $1.695~\mathrm{V}$, $2.825~\mathrm{V}$, $3.389~\mathrm{V}$, and $5.086~\mathrm{V}$.

The correct answer is A. Using $\Delta G$, I got $E = 1.928$, but this is not a choice. Am I doing the math wrong, or is my method incorrect? Thanks!

It's somewhat unclear what you mean by "using $\Delta G$", but here's the full working. The information you are given is:

$$\begin{align}
\ce{MnO4- + e-} &\rightarrow \ce{MnO4^2-} & E &= +0.564~\mathrm{V} \tag{1}  \\
\ce{MnO4^2- + 2e- + 4H+} &\rightarrow \ce{MnO2 + 2H2O} & E &= +2.261~\mathrm{V} \tag{2}
\end{align}$$

Converting these to Gibbs free energy changes is fundamentally the correct step. We have $\Delta G = -nFE$. (Standard state symbols omitted for clarity, they are implied throughout.) For half-equation (1), $n = 1$, and for half-equation (2), $n = 2$. So:
$$\begin{align}
\Delta G_1 &= -(1)(96485~\mathrm{C~mol^{-1}})(0.564~\mathrm{V}) = -54417.7~\mathrm{J~mol^{-1}} \\
\Delta G_2 &= -(2)(96485~\mathrm{C~mol^{-1}})(2.261~\mathrm{V}) = -436306.7~\mathrm{J~mol^{-1}} \\
\end{align}$$
You need to now sum up these two equations, which also means summing up the changes in Gibbs free energy. Let's define a new reaction $(3) = (1) + (2)$:

$$\begin{align}
\ce{MnO4- + 3e- + 4H+} &\rightarrow \ce{MnO2 + 2H2O} & E &= E_3 \tag{3}
\end{align}$$

As you can see, all I have done is to add the two half-equations together and cancel out any terms that appear on both sides. We have $n = 3$ for this half-reaction.
$$\begin{align}
\Delta G_3 &= \Delta G_1 + \Delta G_2 \\
&= -490724.4~\mathrm{J~mol^{-1}} \\
E_3 &= -\frac{\Delta G_3}{nF} \\
&= 1.695~\mathrm{V}
\end{align}$$
Last thing, please always quote your units in your answer! You cannot have an electrode potential that is unitless.


text:
What is inhibiting polystyrene's ability to dissolve in traditional solvents like ethanol or acetone?
My friend and I have been struggling with this problem for a couple days. Because polystyrene has a similar structure to many plastics, we assumed that it would dissolve in acetone. However, after running several tests, we found that polystyrene would not dissolve in a measurable quantity even if we let it sit in acetone for a significant amount of time.
Here is the plastic we are attempting to dissolve:
https://i.stack.imgur.com/HF16p.jpg
And here is a plastic more easily dissolved:
https://i.stack.imgur.com/qo2BT.jpg
We understand that most molecular-bonded plastics dissolve via van der Waals dispersion forces, so why isn't polystyrene dissolving? My theory is that the benzene ring is interfering with the dissolving process -- that only half the compound of polystyrene is interacting with the acetone and thus it does not dissolve like a polyethylene (which obviously has twice as much surface area to dissolve).

Are you using bulk polystyrene or expanded polystyrene? expanded polystyrene clearly dissolves very quickly in acetone but bulk polystyrene might take much longer simply because it has a much smaller surface area for the acetone to infiltrate.

We are using extruded polystyrene -- so styrofoam.

Polystyrene is soluble in acetone, MEK and other ketones, but solubility and the speed with which it dissolves depends, in part, on the molecular mass of the polymer.
If the solvent diffuses very slowly into the polystyrene, only the surface will appear to be affected. Applying a bit of energy with heat or mechanical crushing of the plastic greatly speeds the action of the solvent.
For more technical information, see Solubility Parameters: Theory and Application.

Have you considered monoterpenes as polystyrene solvents? Have a look at:
Shikata et al. (2011) Dissolution of polystyrene into cyclic monoterpenes present in tree essential oils. Journal of Material Cycles and Waste Management 13(2):127-130
Gutiérrez et al. (2012) Recycling of extruded polystyrene wastes by dissolution and supercritical CO2 technology. Journal of Material Cycles and Waste Management 14(4):308-316.

Welcome to Chemistry! Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference.


text:
Excitation of electrons
IF an electron is excited, what can it do? Dissipate the energy to the object, emit another photon?
IF a photon hits an object and NO electron is excited, what happened there?
What are the possible causes of electron excitation?

Also, can a non-exact photon (to get the electron to the required energy level) cause electron excitation?

The question of photons with non-exact energies is addressed here: http://chemistry.stackexchange.com/questions/49583/exactly-equal-and-at-least-in-electron-excitation


text:
Why does a buffer work best at the pH closest to its pKa?
I think I'm having trouble understanding exactly how the $\mathrm{p}K_\mathrm{a}$ relates here on a conceptual level. For example I know that carbonic acid works somewhat well as a buffer at $\mathrm{pH}$ near 7, since its $\mathrm{p}K_\mathrm{a}$ is relatively close, but why is that exactly?

A good way to think about pKa is that it is the pH at which the molecule is 50% ionized.

A buffer works best when there is the same amount of weak acid/base and its conjugate. If you look at the Henderson Hasselbalch equation, and set the concentration of the weak acid/base equal to each other, $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$.

I propose this simple graphic answer as a complement to the more thorough replies above (comments welcome to confirm validity of my answer).
Trace pH Vs % [HA] and [A-]. 0% means you only have the acid [HA] and 100% you only have [A-].
Addition of acid or base, to the buffer solution, affects the ratio [A-]/[AH] and consequently, pH. This has been explained quantitatively in previous answers.
On the plot, you see two drops in HA % due to addition of base. Compare the corresponding changes in pH and notice that operating around pH =pKa lowers the impact on the pH..

Why does a buffer work best at the $\mathrm{pH}$ closest to its $\mathrm{p}K_\mathrm{a}$?

For the novices like you who just begin to learn chemistry, this question can be easily explain by using Henderson Hasselbalch equation. Suppose you have an aqueous weak acid solution ($\ce{HA}$). It will be in ionic equilibrium as follows:
$$\ce{HA (aq) <=> H+ (aq) + A- (aq)} \tag1$$
Thus by definition:
$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\tag2$$
Take $-\log$ on both side of the equation $(2)$:
$$-\log K_\mathrm{a} = -\log \left(\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \right) \ \Rightarrow \ -\log K_\mathrm{a} = -\log [\ce{H+}] - \log\left(\frac{[\ce{A-}]}{[\ce{HA}]} \right)$$
$$\therefore \ \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\left(\frac{[\ce{A-}]}{[\ce{HA}]} \right)\tag3$$
The equation $(3)$ is called Henderson Hasselbalch equation, which is valid for calculation of buffer solutions (e.g., a solution of a weak acid and its congugate base or a weak base and its congugate acid). By this equation, if you know the $\mathrm{p}K_\mathrm{a}$ of any weak acid $\ce{HA}$ and molar ratio $\frac{[\ce{A-}]}{[\ce{HA}]}$ of the solution where $[\ce{A-}]$ is the concentration of the weak acid's conjugate base, you can calculate the $\mathrm{pH}$ of the solution.
For example, suppose you have a buffer solution of acetic acid/sodium acetate mixture in $1:1$ mole ratio (say $\pu{0.1 M}$). The $\mathrm{p}K_\mathrm{a}$ is given as $4.75$. What is the $\mathrm{pH}$ of the solution?
You can apply the values to the equation $(3)$: $\mathrm{pH} = 4.75 + \log \left(\frac{\pu{0.1 M}}{\pu{0.1 M}}\right) = 4.75$.
If the molar ratio is $1:2$ (say $\pu{0.1 M}$ in acetate and say $\pu{0.2 M}$ in acetic acid), then $\mathrm{pH} = 4.75 + \log \left(\frac{\pu{0.1 M}}{\pu{0.2 M}}\right) = 4.75 - 0.301 = 4.45$. Thus, as long as if you keep $[\ce{A-}]: [\ce{HA}] \approx 1:1$,  $\mathrm{pH} \approx  \mathrm{p}K_\mathrm{a}$, because $\log 1 \rightarrow 0$ in the equation $(3)$.
In your example, $\mathrm{p}K_\mathrm{a1}$ of carbonic acid (which is a polyprotic acid) is $6.35$, thus $\mathrm{pH}$ of $1:1$ carbonic acid : bicarbonate solution is $6.35$. If you want it at exactly at $7.00$, the mole ratio should be:
$$7.00 = 6.35 + \log \left(\frac{[\ce{HCO3-}]}{[\ce{H2CO3}]}\right)  \ \Rightarrow  \ \left(\frac{[\ce{HCO3-}]}{[\ce{H2CO3}]}\right) = 10^{(7.00 - 6.35)} = 4.47$$

When the pH is equal to the pKa, you have high concentrations of both conjugate acid and conjugate base. You need the weak base to react with added acid, and the weak acid to react with added base to stabilize the pH.
When the pH is very different from the pKa, one of those two species will be at very very low concentration (there is an exponential relationship).
For a more quantitative explanation, see the other answers, which introduce the Henderson Hasselbalch relationship, and look up buffer capacity. The pKa of carbonic acid is 6.35, so it buffers better at pH 6 than at pH 7.

Buffers are made with weak acids or weak bases, that dissociate slower in water. In your example, carbonic acid is a weak acid written as $\ce{HA}$ and will dissociate into $\ce{A-}$ and $\ce{H+}$ when a strong base is poured into its solution. During this titration, the $\mathrm{pH}$ of the solution will increase steadily as $\ce{HA}$ dissociates, because:

Dissociation is forming $\ce{A-}$, which is a conjugate base and $\ce{HA}$ concentration is reducing; and

You are adding a base ($\ce{OH-}$) to the solution, and there is also $\ce{H+}$ being formed through the weak acid's dissociation. But it will react with the $\ce{OH-}$ to neutralize each other and the solution will be resisting a huge increase to $\mathrm{pH}$ but increasing steadily with the increase of $\ce{A-}$.

This will happen throughout the buffer region and in the middle of that region is the point where the amount of $\ce{HA}$ (original weak acid) becomes equal to the amount of $\ce{A-}$ (conjugate base). If you can get a weak acid or weak base in an environment with $\mathrm{pH}$ close to its $\mathrm{p}K_\mathrm{a}$, it is guaranteed you'll be within this buffer region since $\mathrm{p}K_\mathrm{a}$ is in the midpoint of that region.

that dissociate slower in water

It is not about speed, but equilibrium. They dissociate just partially, regardless of speed.


text:
Volume of dibasic acid required to neutralise 0.04 mole of OH-

The volume of 0.1 N dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04  mole of OH− in aqueous solution is:

200 mL
400 mL
600 mL
800 mL

Since it is a dibasic acid, it has two ionisable hydrogen atoms and hence its n-factor is 2. I simply used $$V\cdot 0.1\cdot 2=0.04$$ and got volume as 200 mL. But however according to the answer key it is given as 400 mL. Now the answer key is not of the highest quality and often has errors, so it can be challenged (but you need to pay a decent sum of money for the chance that they may look into it). Am I right to challenge this question or am I missing something?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

Unfortunately you are missing something. The concentration is already given as an equivalence concentration, thus taking the dibasic nature of the acid into account. Therefore you have been given $c(\ce{H+})=0.1~\mathrm{mol\,L^{-1}}$, not $c(\ce{H2A})=0.1~\mathrm{mol\,L^{-1}}$. In the latter case your calculation would be correct.
The use of normality is ambiguous and it is not recommended to be used any more.


text:
Difference between Epimers and Diastereomers
My lecture notes describe epimers as compounds which differ by
configuration at only one carbon and gives the example of D-erythrose and D-threose as being epimers.
However isn't that the same thing as diastereomers? My understanding is that epimers are supposed to be some kind of subset of diastereomers, however I am not sure how or why. I tried searching the web but I got more confused with one site saying that carbohydrates in open form can have two different chirality centres and still be epimers.

epimers as compounds which differ by configuration at only one carbon

Yes, at only one centre.

However isn't that the same thing as diastereomers?

No, not completely. Diastereomers differ at least at one, but at less than all stereocentres. If two compounds would differ at all stereocentres, they would be enantiomers.

Update
As far as epimerism in open form carbohydrates is concerned, examining the Fischer projections using a German mnemonic derived from the signal horn sounds of police cars (ta-tü-ta-ta) might help. Yes, sounds weird ;-)
Let ta and tü be $\ce{OH}$ on either side of the projection.
Then, glucose is ta-tü-ta-ta, whereas galactose is ta-tü-tü-ta. Apparently, they differ at one stereocentre. Consequently, they are  epimers.

Thanks for clearing that up. What about 'carbohydrates in open form can have two different chirality centres and still be epimers' ? Is this statement correct?

This scheme I just drew up specifically for you should answer your question.
Diastereomers are stereoisomers that are not enantiomers of each other. That includes conformers (geometric isomers that derive from single bond rotation; usually interconverting rapidly) and atropisomers (under which I would subsume E/Z isomers; they derive from hindered rotation around a typically single bond and are separable) and anything with differeces in asymmetric carbons.
If you have exactly one asymmetric carbon with inverted stereochemistry, then the diastereomer is a special one: an epimer. Anomers (e.g. α- and β-D-glucose) should be considered special cases of epimers.
Let me use the menthol isomers (image taken from Wikipedia, where a full list of authors is available) to clarify the concept:

Two vertically aligned structures are always enantiomers. Two structures that are not vertically aligned are diastereomers.
(+)-Isomenthol is (+)-menthol’s C4-epimer, while ($-$)-neoisomenthol is (+)-menthol’s C1-epimer. And ($-$)-neomenthol is (+)-menthol’s C2-epimer. All of these are also diastereomers.
(+)-Neomenthol is not an epimer of (+)-menthol. It is still a diastereomer, however.

Thank you so much for that great answer! I would love to accept this answer however sadly I can only accept one answer.

@Nanoputian Well you can change the accepted answer ;þ Or apply a bounty. But I’m fine anyway ^^

You can extend flow chart, by including nob-anomeric epimers in epimers.

@AnubhavGoel Not all possibilities are listed. For example, there are diastereomers that are neither geometric isomers nor epimers.

Yes you are correct that epimers are a subset of diastereomers.
Diastereomers are compounds that have similar configuration at some carbon and dissimilar configuration at some carbons.
Whereas epimers are compounds that differ in configuration at only one chiral carbon.
The point of difference arises in the fact that epimers have only one chiral carbon which is different in configuration whereas diastereomers can have any number of different configuration around its chiral carbons (except all being different).

They are chiral carbons, not achiral carbons (although IUPAC seems to prefer "chirality centre" now)


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How to calculate n factor for salts in this reaction?
My source says : For  salts which react such that no atom undergoes change in oxidation state ,
n-factor for such salt is the total moles of cationic /anionic species replaced in 1 mole of the salt .
For example, in $$\ce{Na3PO4 + BaCl2  -> NaCl + Ba3(PO4)3}$$
The n-factor of $\ce { Na3PO4} $ is 3.
Can anybody please give more examples and explain that to me ?

$$\ce{2 Na3PO4 + 3 BaCl2 -> 6 NaCl + Ba3(PO4)2}$$
Now this is clearly evident that dividing whole equation by 2 we get 1 mole of $\ce{Na3PO4}$ and at the right hand side 3 moles of $\ce{NaCl}.$ Therefore, 1 mole of phosphate ion is replaced by 3 moles of $\ce{Cl-}$ ion.
Hence $n\text{-factor} = 3.$
In reactions like this where there's no change in oxidation number, the $n\text{-factor}$ will always be the net positive charge in 1 molecule of that compound, in this case since there are 3 $\ce{Na}$ ions in sodium phosphate, the $n\text{-factor} = 3 × 1 = 3.$


text:
Naming compounds with multiple -COX functional groups
I know that this compound is named propanoyl chloride.

However I am not sure what to do when here are multiple -COX groups and they are of the highest seniority. There are two cases, when they are the same:

And when they are not:

For the first case I think it is called propanedioyl chloride, but I do not know what to do for the second. Is one of them cited as a prefix and if so which one?

Systematic names for acyl halides from suffix acids are formed according to Subsection P-65.5.1.1 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

P-65.5.1.1 Acid halides in which hydroxy groups of all acid groups expressed as the suffix denoting the principal characteristic group (carboxylic, sulfonic, sulfinic, selenonic, etc. acids) have been replaced by halogen atoms (F, Cl, Br, and I) are named by citing the name of the acyl group (…) followed by the name(s) of the specific class(es) as a separate word(s), in alphabetical order, each preceded by a multiplicative prefix, as needed.

Therefore, the preferred IUPAC name (PIN) for the examples given the question is ‘propanoyl chloride’, ‘propanedioyl dichloride’, and ‘propanedioyl chloride fluoride’, respectively.

However, note that various carboxylic acids were often given trivial names (e.g. reminiscent of their animal or vegetable origin), which are retained for general nomenclature (including functionalization).


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Solubility of succinic acid in water vs ethyl acetate?
Today I completed an organic lab practical test about discontinuous extraction. I had to extract succinic acid from a water solution using ethyl acetate as the solvent extractor.
After the extraction, I calculated the partition coefficientto be 0.3.
So, succinic acid is more soluble in water than ethyl acetate, but why?
Does it have to do with the ionization of the succinic in water that may not be possible in ethyl acetate or its related to the intermolecular interations?

Hydrogen bonding interactions are also much stronger between succinic acid and water as opposed to EA since EA doesn't have any H-atoms that can participate in the hydrogen bonding.
Ionization is also a significant point to consider, but without pH data we can't say for sure.

This reads more like a comment than an answer. Consider adding more detail or it may be deleted.

Unless you solved something basic to your water only a small fraction of your succinic acid will ionize, so my guess is water can have more interactions with succinic acid than ethyl acetate which makes sense because water is small.
The partition coefficient is very pH dependent in this case. If you would ionize succinic acid it would have a near 100% selectivity for water.


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Solubility of compound in acid
I have a substance that is insoluble in water but soluble in hydrochloric. If dissolve this substance in the acid, would neutralizing the solution with sodium bicarbonate then cause the substance to separate from solution?

Yes, you've described a property of amines. When protonated by an acid (e.g. $\ce{HCl}$), an amine is converted into into a water soluble salt. This is actually a thing to do in pharmaceuticals to increase the amount of a drug that can be dissolved in water and be taken up by the body. The next time you have the chance take a close look at the chemical names on a drug label and theres a good chance you might see "hydrochloride."
Just as something can be protonated, it can be deprotonated. For the hydrochloride salt of an amine, $\ce{NaHCO3}$ should be basic enough to deprotonate it, and cause it to precipitate out of solution, as it would no longer be a water-soluble salt.


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Why do we not include water in equilibrium expression calculations?
Why do we not include water in equilibrium expression calculations? And can you go into detail on why pure liquids and solids have a constant concentration as well, please and Thanks.

Related: http://chemistry.stackexchange.com/questions/42008/why-is-water-not-part-of-the-equilibrium-constant

Usually when you have water in a reaction it is taken in excess, i.e in a much larger quantity compared to the other reactants. Ultimately the change in concentration of water is negligible. As an example you can consider the hydrolysis of ethyl acetate to ethanol and ethanoic acid. For better understanding consider the initial amount of ethyl acetate to be 0.01mol and that of water as 10 mol.
This however is valid only when water is taken in excess. If the amount of water and the other reactants is comparable, then water has to be included in the equilibrium expression.


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What solutes will give the largest reduction in volume when dissolved in water
This question is inspired by another question.
If you take some common salt (NaCl) and some water and mix them, then the volume of the solution is less than the volume of the constituents.
I did some calculations with;

5.75 L salt
44.25 L water
total initial volume 50 L

Density of salt is 2.170 kg/L so resulting solution is 22% concentration, which has a density of 1.164 kg/L. Therefore resulting volume is 48.73 L. This means the solution has a volume 1.27 L lower than the constituents.
My question is in two parts;

Are my calculations correct?
Is there another solute that would give a solution with volume 4 L lower than the constituents, given the volume of the constituents must be 50 L.

If you consider all kinds of solutes, “the largest reduction in volume when dissolved in water” applies to gases such as $\ce{CO2}$, $\ce{SO2}$, $\ce{HCl}$, or $\ce{NH3}$ (see also fountain experiment).

(1) Your calculations are correct.
(2) $\ce{CaCl2}$ has a density of 2.16 kg/L
           kg        vol(L)
  CaCl2    18.00     8.38
  H2O      41.62    41.62
           -----    -----
           59.62    50.00 L

But when dissolved the solution has a density of 1.300 kg/L, so the total volume is 45.86 L. (Data about $\ce{CaCl2}$ solutions.)


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What is the absolute configuration of carbon 4 in glucopyranose?
My colleague and I have been wrestling with the assignment of chirality to the 4th carbon atom in glucopyranose. In the linear form of glucose, the 4th carbon is definitely (R); therefore it will retain that chirality as it forms the hemiacetal. However, looking at the ring, we are having a hard time figuring out why the 3rd carbon gets priority over the 5th by CIP priority rules. Both the 3rd and 5th carbons are attached to $\ce{C,H,O}$. The question is where do you go after that to assign priority?

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therefore it will retain that chirality as it forms the hemiacetal.

This statement is incorrect. Many chiral centres retain their CIP-descriptor upon reactions that don’t influence them directly, many others do not.
How do you assign CIP priorities? You basically had the same problem in the linear form of D-glucose. You move along the chain in order of priorities, finding the first priority difference and using that to assign the higher-priority side. On C4, you have (read this from left to right, i.e. everything with the same indention at the same time):

an oxygen (a)
the carbon C3

an oxygen (a)

a hydrogen.

the carbon C2 (b)

an oxygen (a)
a carbon C1 (b) <-- this one is higher priority than on the other side!
a hydrogen (c)

a hydrogen (c)

the carbon C5

an oxygen (a)

a hydrogen

the carbon C6 (b)

an oxygen (a)
a hydrogen <-- less priority than carbon
another hydrogen.

a hydrogen (c)

a hydrogen (d)

Following the same method in the pyranose form we get:

an oxygen (a)
the carbon C5

an oxygen (a)

a carbon <-- this one is higher priority than the other side!

the carbon C6 (b)
a hydrogen (c)

the carbon C3

an oxygen (a)

a hydrogen <-- this is lesser than carbon!

the carbon C2 (b)
a hydrogen (c)

a hydrogen (d)

Therefore, the absolute configuration of said atom changes from (R) to (S).
This is actually a pretty common thing that can happen in highly functionalised molecules with very similarly modified carbons. Because we cannot distinguish at any proximate position we need to go far down the chain and a single distal modification can change a lot.
Other examples that would have changed the absolute configuration of carbon C4 without changing its geometry would be:

methylation at C5
silyl protection at C5
oxidation at C5

All of these would make C5’s priority higher than C3’s.


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Does freezing point depression depend on the charge of the solvated ions?

Assuming an ideal solution, which of these 0.10 M solutions would have the least freezing point depression?
$\ce{HCl}$
$\ce{NaBr}$
$\ce{KNO3}$
$\ce{MgSO4}$

They all produce solutions with two ions per formula unit, but the correct answer is somehow supposed to be $\ce{MgSO_4}$. How is this? I don't think it can be due to the weak basicity of sulfate; that should in fact increase the ions present and have an even greater freezing point depression.

Frankly I think that the problem is posed poorly. First the problem states to  "assume an ideal solution," then wants an answer that requires some sort of non-ideal behavior. // I'd expect that the answer has to do with the hydrated $\ce{Mg^{2+}}$ and $\ce{SO4^{2-}}$ species attracting one another in solution. So the attracted clusters behave more like one molecule of $\ce{MgSO4}$ rather than two ions. This isn't "ideal" behavior.

@MaxW Isn't every ionic compound subject to ion-pair formation?

Yes, but $\ce{MgSO4}$ is the only compound that has a charge of 2 on the anion and cation. All the other anions and cations have a charge of 1.

First let's gather the data from the CRC Handbook of Chemistry and Physics, 92nd edition (mol/L is moles per liter). The data below is reported as molarity rather than molality since the problem is posed in molarity. However in a solution as dilute as 0.1 M the difference shouldn't be large.
         wt%      mol/L      Delta-F
  HCl     0       0        0.00
          0.5     0.137   -0.49
          1       0.275   -0.99
          2       0.553   -2.08
          3       0.833   -3.28
          4       1.117   -4.58
          5       1.403   -5.98
  least squares fit Delta-f = = -3.4683*[HCL] - 0.5624*[HCL]^2
  [HCL] =0.1 M, Delta-f = -0.35

         wt%      mol/L      Delta-F
  NaBr    0       0.00     0.00
          0.5     0.049   -0.17
          1       0.098   -0.34
          2       0.198   -0.69
          3       0.301   -1.04
          4       0.405   -1.39
          5       0.512   -1.76
  least squares fit Delta-f = = -3.4823*[NaBr] + 0.0930*[NaBr]^2
  [NaBr] =0.1 M, Delta-f = -0.35

         wt%      mol/L      Delta-F
  KNO3   0        0        0.00
         0.5      0.05    -0.17
         1        0.099   -0.33
         2        0.2     -0.64
         3        0.302   -0.94
         4        0.405   -1.22
         5        0.509   -1.50
  least squares fit Delta-f = = -3.4264*[KNO3] + 0.9923*[KNO3]^2
  [KNO3] =0.1 M, Delta-f = -0.33

         wt%      mol/L      Delta-F
  MgSO4  0        0        0.00
         0.5      0.042    -0.10
         1        0.084    -0.19
         2        0.169    -0.36
         3        0.257    -0.52
         4        0.346    -0.69
         5        0.437    -0.87
  least squares fit Delta-f = = -2.3466*[MgSO4] + 0.9726*[MgSO4]^2
  [MgSO4] =0.1 M, Delta-f = -0.22

So there is a difference. 0.1 molar $\ce{MgSO4}$ has a freezing point depression of 0.22 degrees Celsius whereas the other 0.1 molar salts are about 0.35 degrees.
As far as the "answer" goes the problem is very poorly worded. First the problem states to "assume an ideal solution," then poses the problem in such a manor that some sort of non-ideal behavior must be assumed to obtain an answer. "Ideal behavior" for electrolytes only occurs in very dilute solutions. A 0.1 molar solution is much too concentrated for ideal electrolyte behavior.
Without any "lookup" of data the only solution seems to be some hand-waving about $\ce{MgSO4}$ being the only salt with divalent ions so the effective dissociation is smaller because of forming ion-pairs in solution.


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Does mechanotransduction of cells play a role in the biocompatibility of titanium?
What exactly is mechanotransduction? The definition seems to vary from source to source. Some indicate that mechanotransduction is the underlying principle where cells pull on the surface they grow on to test their stiffness, so they can stretch and facilitate cell growth and differentiation. Others indicate that mechanotransduction is how cells detect mechanical stimuli and turn them into chemical signals like our four senses (touch, hearing, smell, etc).
Also, is this the main reason titanium is biocompatible to replace bones because it has a low surface resistance and cells can stretch on it?

Mechanotransduction (mechano-relating to mechanics + transduction-to convert into another form) is indeed the second definition you listed. Mechanoreceptors in somatosensory nerve cells the ability to detect mechanical pressure or distortion. These facilitate our touch, hearing, and proprioception, or the sense of how our body is positioned. Olfaction, however, is actually facilitated through chemoreceptors, and depends on how molecules bind to them.
Cellular response to a foreign object is the result of largely chemoreception. Interaction at the implant-cell surfaces determines how an implant will be perceived by the body. Titanium's biocompatability is a result of the protective oxide film that forms on the surface of the metal in the presence of oxygen. This oxide film is strongly adhered, insoluble in water, and chemically impermeable, preventing reactions between the metal and the surrounding environment. Titanium's high dielectric constant allows it to be easily integrated into cells and bone tissue$^{[1]}$.
From Wikipedia:

Titanium naturally passivates, forming an oxide film that becomes heterogeneous and polarized as a function of exposure time to bodily environments. This leads to the increased adsorption of hydroxyl groups, lipoproteins, and glycolipids over time. The adsorption of these compounds changes how the material interacts with the body and can improve biocompatibility.

[1] Mishra, B. Review Of Extraction, Processing, Properties, and Applications of Reactive Metals 1999 TMS Annual Meeting, San Diego, CA, February 28 - March 15, 1999; Wiley: Hoboken, 2013.


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3-step process to make propyl ethanoate from propene
I was asked to make propyl ethanoate from propene in three steps. I have access to the reactions below.

Nucleophilic substitution (Both SN1 and SN2)
Electrophilic addition
Hydrogenation
Dehydration
Hydration
Condensation
Free radical substitution
Oxidation of alcohols

I'm assuming that you can add whatever reactants you want, since propyl ethanoate has five carbons and you can't just chop a carbon off of a propene with the above reactions. I know that the final step has to be condensation, since that's how you make esters. So essentially, I need a two-step process to make a primary alcohol that becomes the propyl in propyl ethanoate.
If you use hydrogenation on propene first, you get propane. Once you have that, it takes two steps to make a primary alcohol: one in free radical substitution (to get 1-bromopropane) and one in nucleophilic substitution (to get propan-1-ol).
If you use hydration first, you get a secondary alcohol in one step. However, you can't make propyl ethanoate with propan-2-ol.
If you use electrophilic addition first, you get a secondary haloalkane (2-bromopropane). That can also only get you a secondary alcohol.
Does anyone know how to do this? I'm convinced that it's impossible to do.

Indeed, with the reactions you have, I'm not sure how you get to the ester in 3 steps. Of course it becomes extremely simple if you use something outside of your syllabus, but well...

Can you suggest something like that? That would be helpful as well.

is it ethyl ethanoate or propyl ethanoate?the heading says ethyl ethanoate and the description has matter dealing with propyl ethanoate...

Sorry about that! It's propyl ethanoate.

My proposal is hydroboration followed by Fischer esterification. It's 2 or 3 steps depending on how many you count hydroboration to be.

Alternatively if you are intent on going through a haloalkane pathway, you could try anti-Markovnikov addition of $\ce{HBr}$ by reacting it in the presence of $\ce{H2O2}$ to prepare 1-bromopropane. Then you could do nucleophilic substitution of $\ce{OH-}$ followed by Fischer esterification. This method would have a lower yield though due to the competing E2 reaction.

Could you explain how one would consider the above procedure to contain either two or three steps? By the way, how do you draw such nice chemical equations?

@DefinitelyNotAPlesiosaur with ChemDraw. And it's two steps if you consider the entirety of hydroboration as one step, or three if you consider borane addition and $\ce{NaOH}$ and $\ce{H2O2}$ addition as separate steps. If you don't want to use my proposed one, the alternative one is definitely three steps.

Essentially, I need to get  a primary alcohol from an alkene. Do you think that I could do this with electrophilic addition in some special solution (like a protic solution) or in some special conditions (like extreme temperature)?

Yes...that is exactly what the alternative method is. Special conditions by which an acid ($\ce{HBr}$) adds to the alkene to form a primary haloalkane, which can then be used to form the primary alcohol.

Here's another way.

Oxidise propene to prop-2-en-1-ol with selenium dioxide. Here's a reference: J. Org. Chem., 1979, 44, 4683

Esterify. The typical Fischer esterification (with carboxylic acid and concentrated sulfuric acid) should probably work, although there are many other possibilities. One is to use the acyl chloride with a weak base. The weak base serves as a nucleophilic catalyst, and also helps to mop up the $\ce{HCl}$ produced in the reaction. A common choice is 4-dimethylaminopyridine, often called DMAP. Another option is to use the carboxylic acid with dicyclohexylcarbodiimide (DCC) and DMAP. The DCC activates the carboxylic acid in what is known as the Steglich esterification.

Reduce the double bond by catalytic hydrogenation. C=C double bonds are generally more susceptible to catalytic hydrogenation than C=O double bonds (disclaimer: this may vary depending on the catalyst and conditions chosen), so this reaction can proceed chemoselectively to give the desired product, n-propyl acetate:


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Why/how is blood red? (colours of hemoglobin)
Oxyhaemoglobin is red, deoxyhaemoglobin is bluish-purple, and carboxyhaemoglobin is a cherry red colour.
Evidently in the porphyrin around the iron molecule in heme there is a conjugated $\pi$-system which could account for the colour, but how exactly does this work? Also, what is the role of the metal ion (specifically $\ce{Fe}$) in the middle, its oxidation state, or the ligand, in changing/determining the colour in conjunction to that? Do the globin chains play any role in determining the colour (if so, how)?
Chlorophyll, for example, also has a porphyrin motif, but it is green due to having magnesium in the middle. Vitamin B12 is also intensely red in colour (according to Wikipedia).

The $\ce{Fe}$-porphyrin complex is a vastly conjugated $\pi$-system, in which the oxidation state of the $\ce{Fe}$ plays a central role. The wavelength of light absorbed by hemoglobin exactly corresponds to the the difference in energy between the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) of the heme $\pi$-system, and is what gives hemoglobin its color. These values are exact and quantifiable. Changes in this color, therefore, are the direct result of changes in the energies of the molecular orbitals of the hemoglobin $\pi$-system.

To understand the color change, it is important to think about the geometry of the molecule. When oxygen is not bound, the iron atom is in the +2 oxidation state. It's slightly too big to fit into the hole in the center of the plane of the immediately-surrounding "heme," so it actually rests just on top of the heme plane. Here is the molecular orbital of the iron species with no $\ce{O2}$ bound:

When $\ce{O2}$ does bind, the $\ce{Fe}$ and $\ce{O2}$ orbitals bond and their energies change. In particular, the $\ce{O2}$ $\pi$* orbitals interact with the $\ce{Fe}$ $\mathrm{xz}$ and $\mathrm{z^2}$ orbitals. A side effect of the $\pi$* accepting properties of all three of these ligands is that iron formally changes from being in the +2 oxidation state to the +3 oxidation state. You can think of it as having "given up" one of its electrons to the ligand via the ligand's $\pi$* orbitals. As a side effect of losing one of its outer electrons, the iron atom is now small enough to align perfectly with heme plane.
Almost counterintuitively, however, formation of oxyhemoglobin is actually energetically disfavorable, which can be explained by its color. Deoxyhemoglobin appears purplish-blue, meaning that it must be absorbing yellow light (lower energy),and is therefore more stable. Oxyhemoglobin appears red, meaning that it must be absorbing at blue-green light (higher energy) and is therefore less stable. This is actually vital to oxyhemoglobin's functionality. Because it is a higher energy, and thereby less stable structure, it easily releases the bound oxygen in the presence of hypoxic cells.
[Source]

Does the color necessarily show the relative stability? The color is correlated to the HOMO-LUMO gap, but you could hypothetically get a larger while making the compound more stable. @ringo


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Conical intersection using GAMESS
I would like to know how to locate conical intersection using a GAMESS calculation. I really appreciate an example alone with this since I am a novice I do not have a clear idea among bunch of examples on internet given. Also I want to know the parameters as well especially for the $DATA part in GAMESS.

Setting up the CAS for a specific system is already a complex task; there is no general recipe for that. What you are asking goes beyond that. You are not even specifying a system that you want to calculate. I'm afraid this questions is too broad to be answerable.

Since I am a novice I have this problem. I am working on a photochromic compound. I need to locate the conical intersection to find the transition state. I have GAMESS software to do it.  I could not write the proper input so far. I will send more information


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Solubility of n-butanol and n-butanamine
I ran across something saying that n-butanamine is miscible in water while butanol is only partially soluble. Is this correct? Can anyone offer an explanation? I would have guessed that the decreased polarity of the amine relative to the alcohol would have resulted in lower solubility for n-butanamine.

An amine has two hydrogens attached to an electronegative atom and so can form more hydrogen bonds with water than an alcohol. I think this is correct but I have no source.

It is the basicity of the amine that makes the difference.

For nButanol --
Theorical solubility for nButylamine --
PubChem value for nButylamine

Out of butanamine and butanol, butanol is more soluble in water due to higher electronegativity of oxygen . And as butanol  have more polar oxygen so it will be more attracted towards water and form H bond.so it is soluble but amine are miscible due to enthalpies that will release out because of electron donating nature of nitrogen.


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How to convert isopropanol to acetone and hydrogen
From Wikipedia:

Isopropyl alcohol can be oxidized to acetone, which is the corresponding ketone. This can be achieved using oxidizing agents such as chromic acid, or  by dehydrogenation of isopropyl alcohol over a heated copper catalyst: $$\ce{C3H8O -> C3H6O + H2}$$

However, how exactly do I do this? Is it possible to do this by simply mixing isopropanol with an oxidizer (e.g. sodium chlorate)? Also, if the copper catalyst is actually required, what temperature is needed to drive the reaction? Neither the Wikipedia article nor the cited source provide much information on this. I want acetone as opposed to isopropanol, as a fuel, because its flash point is lower and it is much easier to ignite.
This is just an experiment, and if this method of producing acetone is impractical, I will just make/obtain it in other ways.

The method is impractical except on an industrial scale. Not sure how hot the copper would be but I'd guess several hundreds degrees C at least. "Pure" isopropyl alcohol would be "blown" over the hot copper so there is a huge fire and explosion hazard with this. Such a process would require careful engineering to maintain control. Definitely not something to try in your basement.

@MaxW What about the method with the oxidiser?

Both acetone and isopropanol are readily available and cheap chemicals. Making your own is a terrible way to obtain them. If you want acetone, just buy it.

There are several ways to convert isopropanol to acetone.

Use of $\ce{Cr2O3}$ as catalyst[1]

The conversion of isopropyl alcohol to acetone has been studied over $\ce{Cr2O3}$ powder at 473 K in a flowing microcatalytic reactor. Appreciable steady-state catalytic activity was achieved only when oxygen was present in the reactant gas flow. The effect of mild pretreatment of $\ce{Cr2O3}$ in $\ce{O2–N2}$ at 473 K on the initial activity, and the observed dependence of the process on $\ce{O2}$ pressure has been related to the oxidation–reduction cycle of the surface chromium species. $\ce{Cr^{(3+n)+}}$ on the surface of $\ce{Cr2O3}$ have been considered to be the active species, where n= 1–3.

2 . Catalytic dehydrogenation of isopropanol
a) Source 1.

Catalytic dehydrogenation of isopropanol can be chosen as an
  alternative    synthetic route when high-purity acetone is required,
  such as in biomedical applications. Turton et.al. mentions that a
  single pass conversion of 85-92% with respect to isopropanol, with
  reactor conditions of 2 bar and 350° C, is generally achieved. A
  molten salt stream will be used a heat source for the endothermic
  reaction:
$$\ce{(CH3)2CHOH -> (CH3)3CO + H2}$$
The acetone produced in the reactor passes into a phase separator and
  then into a separation system that includes one stripping and two
  distillation columns. A recycle stream takes a mixture of unreacted
  isopropyl alcohol and water, with a trace amount of acetone, back into
  a mixer that feeds the reaction system. Using the catalyst which will
  be employed throughout this analysis, the reaction is first order with
  respect to the concentration of isopropanol and has an Arrhenius
  dependence on temperature with E=72.38 MJ/kmol and k=351,000 cubic m
  gas/cubic m reactor sec.

b) Source 2

At the beginning of the process, feed including i-propyl alcohol and
  water, and recycle stream are mixed in feed drum. From here, this
  mixture is send to vaporizer to changestream’s phase as vapour. After
  vaporizer, mixture is heated to reaction temperature in theheater.
  Reactor used is a tubular flow reactor. Acetone, hydrogen gas (H2) are
  produced andwater and i-propyl-alcohol are discharged. The mixture
  with acetone, hydrogen, water, i- propyl-alcohol are sent to cooler
  and then to condenser. After condenser the mixture is sent toflash
  unit. Hydrogen, acetone, i-propyl-alcohol and water are obtained as
  top product. This top product is sent to scrubber to remove hydrogen.
  The bottom product of flash unit which isformed by acetone, water,
  i-propyl-alcohol are mixed with the bottom product of scrubber  before
  acetone column. In acetone column, acetone is obtained from top
  product with 99 wt%. i-propyl alcohol and water and also 0.1% of
  acetone is sent to i-propyl-alcohol column from bottom product. The
  top product of this column is sent to feed drum and bottom product is
  thrown away as waste water.

Using chromic acid [2]

Oxidation at 25' to 40' is accordingly specified, requiring the
  mixture to be chilled in ice after each addition of oxidant. In 1926
  the writer worked out a student experiment which involves the same
  reaction, the oxidation being effected by gradual addition of chromic
  acid solution to the boiling solution of isopropyl alcohol in water.
  The reaction is carried out in a flask surmounted by a short
  fractionating column, through which the acetone is distilled during
  and after the oxidation. The crude acetone (collected below 65'), and
  several subsequent fractions, are redistilled through the column,
  permitting the easy isolation of a high yield (70% to 80% or more) of
  acetone boiling 57' to 60' and over 99% pure, and also the isolation
  of any unchanged isopropyl alcohol (76' to 82').

References:

Ilyas, M.; Shah, S.; Nigar, R.; Khan, H. Conversion of isopropyl alcohol to acetone catalysed by Cr2O3 at 473 K: role of molecular oxygen. Faraday Trans. 1994, 90 (16), 2413. DOI: 10.1039/FT9949002413.
Wagner, E. C. Oxidation of isopropyl alcohol to acetone. A student experiment. J. Chem. Educ. 1934, 11 (5), 309. DOI: 10.1021/ed011p309.

You can consider transfer hydrogenation using isopropanol as a hydrogen source. Many years ago I wanted to make a coordination polymer from trans-1,2-diphenylphospinoethylene and nickel chloride. I put the reagents together in a autoclave and heated it. I allowed the thing to cool slowly and I got some nice orange crystals.
These turned out to be the dppe complex of nickel with two chlorides on the nickel. What happened was that the alkene in the diphosphine was hydrogenated by hydrogen formed from the isopropanol being converted partly into acetone.


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Reaction mechanism of aqueous ammonia with 3,4-dibromopyridine

I am struggling on how this mechanism would occur. I am aware that the bromine will be removed by for some reason I cannot figure out the $\ce{NH3}$ attaches onto the ring, can anyone prove me with any help?
I have also had a feeling that deprotonation occurs and a benzyne is formed, in which a bromine gets kicked off and one of the $\ce{:NH3}$ attaches onto the ring. Is that correct?

https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/heterocy.htm Read the pyridine section.

What mechanisms have you tried?

The reaction between ammonia and 3,4-dibromopyridine is a nucleophilic aromatic substition reaction (SNAr). This is possible because the ring is rather deactivated due to the presence of the electron withdrawing nitrogen atom of the pyridine drawing electron density towards it. The ammonia attacks carbon at the 4-position preferably, rather than the carbon at the 3-position as this position is more deactivated than the latter. This attack results in the breakage of the double bond, resulting in a loss of aromaticity. This results in the formation of a carbanionic Meisenheimer intermediate. The bromine at the 4-position then leaves and the double bond reforms, resulting in the restoration of aromaticity in the ring. This results in a net substitution of $\ce {-Br}$ with an $\ce {-NH2}$ group.

Why is position 4 more deactivated then position 3? I don't get how that comparison was made. Could you please elaborate?

@GaurangTandon Refer to the resonance structures of pyridine. It can be observed from that there is partial positive charge on the 4-position, which attracts nucleophilic attack.

Okay, this was way tougher than I thought. I finally found this question, which sort of seems to explain that very well. I'd suggest adding this into your answer, because it wasn't immediately obvious. Thanks though!

This is a Nucleophilic substitution reaction.In this type of reaction,a weaker base is a better leaving group hence the Nucleophile (NH3) will attack the C containing the Br for the Br to be displaced resulting in the formation of the above product.


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What is a hydrogen bond?
I was doing a little research on the Chemistry of water, and it said that water molecules have a hydrogen bond between them. What is it, and what does it do to water?

This answer should give you a nice short intro.

Firstly you should consider this simple model of atoms: There is a compact nucleus and smeared out electron clouds. In H this cloud is simply spherical. In many other atoms (e.g. O, C, N, S, etc.) the clouds are usually arranged like this (I did not find english site for some reason) (imagine tying 4 ballons together, forming a tetrahedron). This also explains the general geometry of water (and ice).
As you know in water the hydrogen is covalently bonded to oxygen. But due to the different properties of O and H the electrons of these bonds are not distributed symmetrically but much nearer the oxygen, leaving the hydrogen slightly positively charged. (Note that different elements cause different levels of asymmetry, a higher difference in electronegativity corresponds to a stronger asymmetry. This is why hydrocarbons ( C-H-bonds are rather apolar) are not very soluble in water.)
Now these positive H nuclei can bond to "full" electron clouds (e.g. the two of oxygen in water that are not bonded to hydrogen)
For something to be suluble in water you need to have many of those polar bonds and "free electron clouds", so e.g. sugars are easily soluble but e.g. alkanes are not.
These illustrations hopefully make now more sense. Sadly I could not find an english text about the tetrahedron model, which I consider didactically very useful.
Edit: Note that although hydrogen bonds are due to the electrostatic attraction, they are quite distinct from ionic bonds in the aspect that they have a definitive orientation, i.e. in ice the bonds still form tetrahedrons and not a cubic lattice or something like that.
I do not know a very good explanation but I think this is simply because the scale is much smaller; if the angle changes just a bit, the negatively charged part of the hydrogen atom interferes already.

Note that one of the defining features of a hydrogen bond is that it has a significant covalent component. It is not just an electrostatic interaction!

IIRC, the covalent component of a hydrogen bond makes up around 10% of the bond energy.


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Why isn't the ion-product constant the sum of the molar concentrations of H+ and OH-?
The Chemistry texts I have present the ion-product constant as:
$$K_\mathrm w= 1.0 \times 10^{-14} = [\ce{H+}][\ce{OH-}]$$
Additionally I'm told that this constant at (25 °C) is an equilibrium. As one concentration rises, the other declines.
It seems easy to imagine a constant as a sum which the $[\ce{H+}]$ and the $[\ce{OH-}]$ concentrations add up to in a mixture but it's very difficult for me to grasp the idea that they are a product of multiplication.
What am I missing here?