text:
Why can't a potassium sulfate anion KSO4(-) exist in solution?
Please explain to me why $\ce{KSO4^-}$ doesn't exist as a molecular ion in (aqueous) solution. I couldn't find a reason why.
I assumed if you took one $\ce{SO4^{2-}}$ and one $\ce{K^+}$ ion, then only that one would form the molecular ion $\ce{KSO4^-}$.

You can make an aqueous solution of $\ce{KSO_4^{-}}$ (and no other cations), in the sense that you can make a regular $\ce{K2SO4}$ solution which lacks some potassium ions. By doing so, you are producing a solution with a net electric charge. The higher the net electric charge, the less stable the solution is, as electrons will start redistributing into the surroundings in order to correct the charge imbalance to a high precision. While a net charge is possible, it seems unlikely to be able to produce a solution which contains even close to a 1 µM excess of an ion, and I do not know if electrically imbalanced solutions can be prepared quantitatively. Take at look at the answers and comments to this previous question for more information.

In solution, $\ce{K2SO4}$ is going to fully dissociate into $\ce{2K^+}$ and $\ce{SO_4^2-}$, see the Wikipedia article on dissociation.
In solution, the ions will orient towards each other so as to functionally neutralize the charges, but you are still going to have two $\ce{K+}$ ions orienting towards each $\ce{SO4^2-}$ ion, unless there is something else in solution that is "distracting" one of the $\ce{K+}$ ions.

Ok but what if I remove the other K^+ ion? There will be only one ion left, oriented towards the SO4^2-.

@guest How are you planning on removing it? Cuz if you were to add something like NaCl (a simple ionic compound) and one of the K+ ions went with the Cl, there would be Na+ which would orient towards the anion. The key principle is that the charges have to balance, even in solution, unless you are applying a charge to the solution.

You can mask $\ce{K+}$ with crown ether removing it from the equilibrium. However, you are right. But consider $$\ce{K2SO4 <=>[H_2O] K+ + KSO4-},\\ce{KSO4- <=>[H_2O] K+ + SO^{2-}4},$$ if it is concentrated enough, you will always have all three sulphate molecules present. (I am completely unsure about the degree of dissociation.)

You could have something like $\ce{KSO4^-}$ in the form of ion pairs, in a suitably concentrated potassium sulfate solution.  But, of course, such ion pairs are easily dissociated.  Add $\ce{HCl}$, for instance, and the ion pair releases its potassium ion as the weakly basic sulfate ion is protonated to bisulfate.


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Why is it a good idea to put some salt in the bread dough?
The salt that you can typically find in a kitchen is the $\ce{NaCl}$, isn't this salt supposed to act as an antiseptic?
I find a lot of recipes that suggest the addition of salt when preparing the mix for the leaven of the bread.
Isn't the salt supposed to act against the formation and the culture of bacteria that are carrying forward the leaven?

Maybe this should go in the cooking SE? I think a cook would know better than a chemist why bakers add salt to bread dough.

I found a question that should answer yours nicely: http://cooking.stackexchange.com/questions/41501/bread-with-no-salt/41507#41507

@qwersjc I asked this here because my focus was on salt and bacteria considering the antiseptic behaviour of the salt, I don't care that much about how to make good bread with this question.

Common table salt is not necessarily an antiseptic in the conventional "sterilizing" sense. What gives it weakly antiseptic properties in biological context is its ability to absorb water from cells through osmosis, should the concentration of the salt outside the cell be too high, the cell would lose too much water through its membrane and die.
However this is strongly dependent on concentration. For example, according to this source, a concentration upwards of 20% is required to kill bacteria; at lower concentrations the salt will only hinder growth, which is most likely the reason it is added (in moderation) to bread.

Nice answer, 1 thing though: Have you ever tried bread without salt? It tastes horrible. So it is likely that the main reason to add the salt is for the taste/texture, rather than any bacteria killing/reducing.

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Voltaic and electrolytic cells
Why more attractive metals will be oxidized? Aren't they being reduced, because they attract and receive electrons from negative polyatomic ions? So they will be anodes, where oxidization occurs. Why anodes are negative in voltaic cells and positive in electrolytic cells? Same for cathodes.

Metals want to lose electrons and become positively charged. When the metal loses the electron, it is called oxidation. The substance that gets the electron is reduced.
Oxidation could occur at the anode if it takes electrons from a metal (i.e. $Fe^{2+}$ to $Fe^{3+}$).
Anodes are negative in voltaic cells since there are excessive electrons available to power the circuit. Cathodes give electrons away to the solution.

I saw cation related cathode but cation is positive ion.

but then why still anodes are positive in electrolytic cells? in other words, why excessive electrons stay in the cathode of a electrolytic cells?

You need to remember that electron flow is the reverse of current flow (someone guessed, he was wrong). Take a moment to think about which way the electrons are moving and what you want to call positive. http://blog.sciencegeekgirl.com/2009/02/27/electron-flow-vs-current-flow/

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Are there any safety guidelines for mixing sulfate with chloride?
Some ceramicists decorate pottery using soluble salts (usually chlorides and sulfates e.g. copper chloride, cobalt sulfate) which are used individually or mixed to produce different colors.
I have been exploring the use of these chemicals on porcelain.  From the MSDS of each chemical I've become aware of the safety concerns and the proper precautions to take when mixing, applying, and firing them in a kiln.
But I don't know much about chemistry and do not understand the reactions when mixing these chemicals.  Are there any chemicals which should absolutely not be mixed together?  For instance, the Wikipedia article on Potassium permanganate says "Concentrated sulfuric acid reacts with $\ce{KMnO4}$ to give $\ce{Mn2O7}$, which can be explosive."  Does this mean I should not mix Potassium permanganate with a sulfate like Copper Sulfate?  In general, what happens when a sulfate is mixed with a chloride?  Are there any general safety guidelines you could offer?  Is there a fairly simple way to understand these interactions without the requirement of years of Chemistry study?
Apologies for my ignorance!
Some chemicals that ceramicists have used before:
Color

gray

copper chloride (heavy application and heavy reduction can give pinks and reds)
palladium chloride
ruthenium chloride
selenium (selenous acid, selenium toner)
silver nitrate
tellurium chloride
vanadium (vanadyl sulfate, vanadium pentoxide)

blue

cobalt chloride
molybdenum (molybdic acid)

green

ammonium chromate
nickel chloride
potassium dichromate
sodium chromate

brown

iron chloride (iron chloride emits heat when mixed with water so the water    should be added gradually in small amounts)

pink/purple/maroon

gold chloride (1-5% solution, adding either cobalt, manganese or
tellurium will give different shades)

yellow

praseodymium chloride (very pale color)

black

cobalt chloride (50% solution) and iron chloride (100% solution)
cobalt chloride (50% solution) and nickel chloride (50% solution)

Are you more interested in the chemistry or the colourants in aqueous solution or in the ceramic glaze?

Since ceramic glazes are usually handled and sprayed, I'm very hesitant to add any of these salts directly into the glaze or even clay body.  I was thinking more of painting the salts and then firing, limiting exposure as much as possible.

Overall, always wear disposable gloves, goggles, and a dust mask.  Add a lab coat, smock, or apron so as not to contaminate your clothing.  Avoid skin contact, don't ingest and don't inhale (this includes the dust).
Manganese heptoxide is a special case with concentrated sulfuric acid.  Avoid mixing strong oxidizers (permanganate, chromate/dichromate, perchlorate) with organics.  NEVER screw around with chlorate salts.  They are powerful oxidizers with very low kinetic barriers.  Finely divided metals (especially titanium) can go to white-hot burning in open air.  Titanium burns under air, carbon dioxide, nitrogen, water...
Palladium is a heavy metal, poison and tissue corrodant.  A microgram of tellurium by any contact gives "tellurium breath"
Local contact with chromates can cause ulceration.  Neodymium gives you purples, cerium yellow to orange.  I seem to remember that cobalt blue (cobalt aluminate after firing) in tin oxide applied to white ware gives bubblegum purple.
Tin Glazing
Gold via the Purple of Cassius is very nice.
Here's a pretty!  Suppose you mixed clay with hollow glass microbubbles, then fired.  Would the clay foam to an interesting surface or whole body?  This is the stuff, sold generally as lightweight resin filler, including hobby stores.  Most prices are outrageous when sold by volume.  Density of 3M Scotchlite K20 is 0.2 g/cm^3.  Goggles and dust mask.  Be careful or the stuff goes everywhere.
http://www.aircraftspruce.com/catalog/cmpages/bubbles.php
Good price.
http://solutions.3m.com/wps/portal/3M/en_US/Energy-Advanced/Materials/Product/?N=5002440+7570097&rt=r3

Hi Al,  This is a wonderful answer, thank you.  I was not aware of the safety issues you mentioned- sometimes just looking at the MSDS of chemicals doesn't make it very clear which chemicals are dangerous vs. super dangerous.

And microbubbles are a great idea.  Lately ceramicists have been experimenting more with additives in clay bodies, from paper pulp to styrofoam beads.  Maybe glass microbubbles would melt and fuse with the clay body at high temperatures, perhaps hardening the resulting ceramic and decreasing shrinkage.  Might even influence ability to absorb thermal shock?

Porcelain is glass-ceramic. Whole foamed body, added thin clay surface or shape decoration, or in the glaze are all interesting. K20 is E-glass borosilicate, 400 psi crush. There are some dozen compositions from fly ash to phenolics, with different size cuts and crush strengths/wall thicknesses. I use a stiff paste of K20 in architectural epoxy as two seal layers for messages in bottles. Sunlight eats organics, seawater eats metals, shingle beaches shatter bottles. A 100-year seal is challenging.

Sorry for resurrecting the topic, but original question wasn't fully answered.
What happens when a sulfate is mixed with a chloride? The answer for glazing is, nothing.
The exception would be mixing reducing and oxidizing salts, but there are no such sulfates or chlorides in your list.
From you list, potassium dichromate, sodium chromate and exotic substances may be dangerous, as stated in selected answer.
Given basic protection from dust and swallowing,                                                                       don't be afraid to experiment with commonly available salts.
No one would sell you tellurium, while chlorates and permanganate are probably pointless in glazing.


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metal salt solutions g/l means hydrate or anhydrous
I want to redo an experiment. They've used
Nickel chloride: 40 g/L

Nickel chloride can be bought in two forms: anhydrous ($\ce{NiCl2}$) and hydrate ($\ce{NiCl2.6H2O}$). The paper doesn't say which one they've used. Which one did they mean in the paper? In one liter of water should I dissolve 40 grams of $\ce{NiCl2}$ or 40 grams of $\ce{NiCl2.6H2O}$? Is there a convention?

Unless the experimental part of the paper provides additional details, such as equivalents or moles or the colour of the nickel chloride used ($\ce{NiCl2*6H2O}$ is green, while the anhydrous chloride is yellowish) you're up to guessing.
However, if the experiment you're trying to repeat is conducted in aqueous solution, it is safe to assume that you can use the hexahydrate.
Btw, gloves and proper waste management are a good idea here.

I would rather interpret it as a concentration, which would imply that there are $40~g~\ce{NiCl2}$ in $1~L$ solvens. It then does not make a difference what kind of salt you use, but its effect on the concentration might be huge...

Thanks a lot. This is the link to the paper http://www2.bren.ucsb.edu/~dturney/port/papers/Modern%20Electroplating/11.pdf
I want to make the second alloy on page 313, 'Iron-nickel Magnetic[37]'. I'd really appreciate it if you can take a look and confirm whether I should assume hexahydrate or anhydrous. Thanks.

@ramda I can only guess, but from comparison with the other entries in the table (such as the one above), I'd say that the hexahydrate is safe. The prices for anhydrous nickel chloride are astronomical anyway, so I'd rather dry it myself by heating the hexahydrate in thionyl chloride, if necessary. But in this case, I'd use the hydrates of all three components in the mixture.


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What's the relationship between Van der Waals forces and hydrophobic interactions?
A question on a past exam for a course I'm studying for asks:

What's the relation between Van der Waals forces and hydrophobic
  interactions?

From what I understand, Van der Waals forces are just a name to summarize a bunch of little weak forces that only come into play when molecules are placed close to one other, based off this previous answer to my question.
I understand that hydrophobic interactions arise because non-polar components of a molecule are unable to form bonds with hydrogen in water, but disrupt the hydrogen, forcing them to make a cage-like structure (although I'm not clear on why), based off this entry in the Chemistry wiki.
Is the relation between these two forces that Van der Waals forces can be encouraged by hydrophic forces? Is there some other relation I'm missing here? Does it have something to do with the interactions denaturing at high temperatures?

Van der Waals forces are the attractions/repulsions - the forces - between molecules or atoms, other than attractions like ionic attractions, and covalent attractions. These forces are:

Keesom Effect - This is an effect caused by two polar atoms interacting with each other. Two permanent dipoles are involved, meaning the molecules/atoms involved are polar. This may be attractive or repulsive, depending on the dipoles involved.
Deybe force - This is an effect caused between a molecule with is polar, and one that is not. Because the one that is polar affects the electrons on the second (non-polar) molecule, it creates an attractive force between the two molecules.
London dispersal effect - This is a force that acts between two non-polar molecules/atoms. Because the electrons around each molecule/atom repel each other, it creates a redistribution of charge, inducing an instantaneous dipole moment. This is the force that acts in liquid noble gases, to make them liquid, or between methane molecules in liquid form.

Hydrophobic forces are caused because molecules like ethane and other hydrocarbons - petrol etc. and oils like sunflower oil etc. are non-polar, and therefore don't "like" being dissolved in a polar solvent like water. In water, there are hydrogen bonds between the individual molecules, which is why water is a liquid - the van der Waals forces are far too small in water to bind the water as a liquid at room temperature. If we introduce non-polar substances into the water - polar solvent - it upsets these hydrogen bonds, and creates an increase in enthalpy because of this. Therefore the lowest energy state is for the hydrophobic hydrocarbons to separate themselves from the water.
The relationship between van der waals forces and hydrophobic interactions is that the van der waals act to bind the hydrophobe - non-polar substance - together, to separate from the polar solvent/water, and these contribute to the energy needed to separate the two substances. The seperation causes a decrease in the entropy of the system. To counter this decrease in entropy, there must be some decrease in enthalpy. Because the hydrophobe disrupts the hydrogen bonding in water, when they separate, the hydrogen bonding then causes an decrease in enthalpy, because of the favorable interactions. This is also where the van der Waals forces come in. They are favourable interactions, so cause a decrease in enthalpy, and this helps the separation of the hydrophobe and the water into two separate phases, because it makes the separation more energetically favourable. They are not necessary, but make separation more favourable, because the enthalpy change is greater for the separation process. This makes it more favourable.

This is a good answer, but the last part about van-der waals forces being required to pull the hydrophobic solvent together is incorrect. If you had particles with no attractive forces between them in a solvent that has attractive forces, as long as the enthalpy drop for phase separation is larger than the entropic penalty, they will separate. Van der waals forces help to make the enthalpy drop larger, but they aren't necessary.

There are also some more vdW forces, like pi-pi-stacking, etc, pp.

@Martin. That's interesting. I didn't know about them!

The definition is a little bit fluffy and you are right, that the named three are the prominent ones. In a more common and/ or general sense it refers to all kinds of weak interactions. This answer has listed some more.

Atoms cannot be described as "polar". Polar molecules, polar bonds, not polar atoms...

I would add also the Kasimir effect https://en.wikipedia.org/wiki/Casimir_effect#Relativistic_van_der_Waals_force

@Tan Yong Boon Atoms can have induced dipole moments as result of influence of external permanent or other induced dipoles. Even in lesser extent than for nonpolar molecules.

Kind of confused, Aren't london dispersion effects produced due to vacuum fluctuations of the electron density?

Could this be answered by microscopic scale,without introducing the themodynamics?

All sorts of dispersion force and polarization interactions are around the 5 kcal/mole of a hydrogen bond. Recent hype is the "halogen bond"
http://cen.acs.org/articles/92/i1/Defining-New-Carbon-Bond.html
DOI:10.1021/cg500094b
DOI:10.1039/C3CP00054K
Long before that, the curious co-crystallization of $\ce{CBr4}$ and $\ce{CPh4}$ was noted.  Tetra-(4-bromophenyl)methane gives the same structure.
DOI:10.1021/ja953373m
http://galoppinigroup.com/wp-content/uploads/2012/06/Weak-Intermolecular.pdf
http://ccaasmag.org/chem_2011/vol1/Darbost_halogenAryl.pdf
Clathrates with water (methane, $\ce{SF6}$, $\ce{CO2}$, xenon).  Benzil is interesting.  Pi-stacking and lattice forces make the solid state conformation very different from gas phase, solution, or molten,

If you are in water, hydrophobic exclusion from the hydrogen-bonded polar continuum affords a lot of energy to play with, then diddle with the medium,
http://www1.lsbu.ac.uk/water/kosmos.html
kosmotropes and chaotropes

Let me see if I'm understanding correctly here, because my experience in chemistry is quite limited. You're saying, that like the Van der Waals force, hydrophobic interactions aren't really as a result of a single force, so it's wrong to talk about them that way. Consequently, saying that hydrophobic interactions and Van der Waals reactions are related is correct, since there's probably something in the two collections that overlaps?

Perhaps it is worth noting that I am not coming from a chemistry background, but an electrical engineering background and that the context of my question is studying for a Biophotonics course?

van der Waals is one class of interactions.  People must publish. Low energy interactions become more minutely differentiated.  http://imm.buct.edu.cn/mlei/wp-content/uploads/2012/08/cen20120827-dl.pdf intenal page 36; http://www.epj.eg.net/articles/2013/12/2/images/EgyptPharmaceutJ_2013_12_2_95_124000_t11.jpg  For biology, view http://www.ncbi.nlm.nih.gov/books/NBK21726/


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what is the name of the molecule that has the greatest number of different atoms?
What is the name of the molecule that has the greatest number of different kinds of atoms?
For example, H2O is water has 2 different kinds of atoms which are H and O.
For example, C6H12O6 is fructose and has 3 different kinds of atoms which are C, H and O.
After warned to be a too broad question. I will rephrase it. The molecule must be stable and can be find in nature.
what is the name of the molecule has the record?

I'm sure that the first question people will ask is "in a theoretical sense or in reality?".

This question could have an interesting answer, but it's quite hard to define the limits that would make it interesting. For a sufficiently large molecule, you could fit just about anything in the periodic table somewhere in it. Throw a ton of different compounds in a flask containing every element, find a crazy way to cross-polymerize and coordinate everything, and you could obtain a mess composed of gigantic molecules, at least one of which would be statistically expected to contain at least one atom of each element (with the possible exception of some noble gasses).

Actually, if you attached endohedral fullerene compounds, even helium and neon would make it into a molecule. In fact you could likely encapsulate any atom or small molecules containing any atom, then attach the fullerenes onto a polymer backbone.

I'm not sure this question is answerable since I'm not sure anyone keeps track of this "record". It would have to be a biological molecule. For example, if we consider chromatin(the DNA-histone complex that makes up chromosomes) to be a single molecule, we would have (so long as we did not forget the various possible counterions) a molecule with $\ce{C,H,N,O,P,S,Cl,Na,K,Ca,Mg}$.

What about metallic bonding. I'm sure we can just dump a bunch of metals to form some sort of super alloy


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The bond in coordination complexes
This is a very basic question and I'm surprised it only just struck me. The nature of the bond in coordination complexes is a coordinate covalent bond. Only the ligand donates electrons for bond formation between the metal and ligand.
Why is it that we never place a positive charge on the ligands? For example, in $\ce{[Pt(NH3)5Cl]Br3}$, $\ce{NH3}$ is attached to three Hydrogens and the lone pair is donated for forming the coordinate bond. So there should be a positive charge on $\ce{NH3}$, right?

In a typical covalent bond, such as is found between carbon and hydrogen in
methane, each atom is considered to contribute one electron to the two-electron,
two-centre bond which is formed. The metal-ligand bond is often envisaged as a second type of covalent bond in which we still have a two-centre, two-electron bond, but where
both of the electrons come from one of the atoms or from a molecule. This type of
bond is known variously as a coordination, a dative covalent or a donor-acceptor
bond.
It was recognized early on that the formality of dative covalency in coordination
compounds presents some difficulties. Many inorganic compounds are conveniently
thought of as ionic salts in which there is an essentially complete charge separation
between cationic and anionic species. Compounds of the group 1 and group 2 metals
tend to be readily categorized as ionic. Dissolution of a salt of a group 1 or a
group 2 metal results in the formation of solutions containing solvated cations and
anions - only in concentrated solutions are there significant cation-anion
interactions beyond simple ion-pairing. However, the ionic model does not appear
to be suitable for the description of the properties of many transition-metal
compounds. For example, the compound $\ce{K4[Fe(CN)6]}$ dissolves in water to give
solutions containing solvated potassium ions and the $\ce{[Fe(CN)6]^{4-}}$ ion, rather than
solvated potassium, iron(II) and cyanide ions. The interactions between the cyanide
and the iron(II) centre appear to result in longer lived species than result from
simple electrostatic interactions of the type observed in sodium chloride. This is, of
course, the sort of argument which led to the development of the description of
coordination compounds in terms of donor-acceptor interactions between the
ligands and the metal centre.
Let's have a look at the consequences of the formation of a donor-acceptor
bond in a little more detail. If the donor-acceptor bond is completely covalent,
then there is a net transfer of one unit of charge from the donor to the acceptor as
a direct consequence of the equal sharing of the electron pair between the two
centres. This result leaves a positive charge on the donor atom and a negative
charge on the acceptor atom. The limiting ionic and covalent descriptions of a
complex cation such as $\ce{[Fe(H2O)6]^{3+}}$ are shown below.

As commented above 'ionic' structures are not in accord with the
chemical properties of coordination compounds. However, there are also a number
of objections to the covalent description. The charge distribution is such that the
iron(III) centre bears a three minus charge, whereas the oxygen atoms of the water
ligands each bear a single positive charge. This would be unrealistic in view of the
electronegativities of these elements ($\ce{Fe}$, 1.8; $\ce{O}$, 3.5), which predict that the $\ce{Fe-O}$ bond should be polarized in the sense $\ce{Fe^{\delta +}-O^{\delta -}}$. The problem was addressed by Pauling, who recognized that, in reality, it was not appropriate to describe most bonds as being purely 'ionic' or purely 'covalent'.
He developed a description of the bonding between a metal and its ligands which included considerable ionic character in the metal-ligand bonds within a basically covalent regime. In order to determine the amount of ionic character within a given metal-ligand bond, Pauling framed his electroneutrality principle. In this, he opined that the actual distribution of charges within a molecule is such that charges on any single atom are within the range -1 to +1.
It can be seen how this works by reconsidering the ion $\ce{[Fe(H2O)6]^{3+}}$. An ionic
description results in a +3 charge on the metal centre, whilst a covalent description
gives the metal a -3 charge. Now the electroneutrality principle suggests that the
ideal charge on the metal centre is zero. That would be achieved if the iron centre
gains a total of three electrons from the six oxygen donor atoms; in other words, if
each oxygen loses one half of an electron rather than the whole electron which the
fully covalent model demands. Pauling describes this situation as 50%
covalent (or 50% ionic).

Pauling's model gives a good indication that the bonding situation in transition metal complexes is rather complex since the bonds are highly polarized (An 'accurate' view of the bonding in transition metal complexes can only be gained by more sophisticated methods such as the molecular orbital method).
Thus, Lewis structures are not particularly well suited to depict the bonding situation in coordination compounds because Lewis structures are intended for covalent compounds only.

So are you suggesting that the coordinate bond is a type of polar covalent bond? Are there any differences in terms of the character of the bond from a covalent bond? ("Covalent bond" here refers to any covalent bond, not necessarily purely covalent)

Also, you have misunderstood the idea of a "formal charge". To a chemist, "formal charges" are theoretical constructs indicating regions of electron-richness and electron-deficiency in a molecule and are not actual charges. Please take note.

Also, I believe in Para 2 Line 10, the charge of the complex ion is "4-" not "4+".

@TanYongBoon Are there any differences in terms of the character of the bond from a covalent bond? That depends on what exactly you mean by "character of the bond", so you might get different answers from different people. I don't see much of a conceptual difference between covalent and coordinate bonds but my view is heavily informed by MO theory and there both types of bonds are described on the same footing. It's just that in coordinate bonds d orbitals play a major role.

"Character" refers to the electron distribution and also, the likelihood of it participating in particular reactions.

@TanYongBoon Also, you have misunderstood the idea of a "formal charge". To a chemist, "formal charges" are theoretical constructs indicating regions of electron-richness and electron-deficiency in a molecule and are not actual charges. What makes you think that? I never claimed otherwise. But one thing that relates formal and actual charges is that all the formal charges within a molecule have to add up to the actual charge. Apart from that I'm well aware that the two should not be mixed up.

I agree with the "d" orbitals playing a great role but that is only for the case of transition metal ion complexes, not the case of simple covalent substances such as carbon monoxide or the ammonium ion.

@TanYongBoon Yes, that's why I wrote that this is the only difference between coordinate bonds and covalent bonds that I see.

@TanYongBoon Also, I believe in Para 2 Line 10, the charge of the complex ion is "4-" not "4+". Thanks, I corrected it in my answer.

The case would perhaps not be so simple in TM complexes where the ligands are not all the same? How would the formal charges distribute then? You mention Pauling's guideline presented here is supported by MO calculations; could you perhaps include some results? It would be interesting to look at some of the MOs, and perhaps see what the partial charges look like.


text:
Determine products of NH4NO3 hydrolysis
What are the products of the following hydrolysis reaction? How do you know?
$$\ce{NH4NO3 + H2O <=> ?}$$

Please add what you have attempted towards solving the problem into the body of your question.  For more information, see the site's homework policy for how to ask homework questions. Thanks!

We've spoken about this issue before.  Including what you already know in the body of the question helps both you and the person answering the question.

@jonsca This is not a homework question.

It will never happen. You can not hydrolyze a salt to any strong acid (nitric acid in your case) or base because the reverse reaction is predominant.

The products will be $\ce{HNO3 + NH4OH}$.
This is because a salt hydrolysis reaction always produces an acid and a base. The $\ce{H+}$ in $\ce{H2O}$ will bond with $\ce{NO3-}$ to form $\ce{HNO3}$, a strong acid. The remaining $\ce{OH-}$ in $\ce{H2O}$ will bond with $\ce{NH4+}$ to form $\ce{NH4OH}$, a weak base.

Saying that is spontaneous for both ions violates thermodynamics.  Bad thing.


text:
Maximum Pressure of Covalently Bonded Material
At high enough pressure, atomic material no longer exists, for example under the conditions of a neutron star.  Is there a different maximal pressure above which covalent bonds no longer exist?

What do you mean by atomic material? Most other stars consist of plasma (which depends on temp and pressure), which could be considered to be an absence of chemical bonds since all particles are ionized.

This seems like it belongs in SE.Physics

Vaguely related: Electron and neutron degeneracy.
http://hyperphysics.phy-astr.gsu.edu/hbase/astro/whdwar.html#c3 and http://hyperphysics.phy-astr.gsu.edu/hbase/astro/pulsar.html#c1 are fun links

@qwersjc, I think I'll have to make my question more specific.  I would consider plasma atomic material until the last ionization is reached.  I'll limit this question to covalent bonds, and maybe ask a different question for metallic or ionic bonding later.

I think you're looking for electron degenerate matter. As far as I understand, individual atomic nuclei still exist in EDM, but all the electrons are stripped, so it's sort of like a metal where all electrons are valence electrons. Maybe something like metallic hydrogen, a gas of electrons held together by a nuclear lattice.

With the twist of thumbscrew, a hand-held diamond anvil cell will exert 300 GPa (43,500,000 psi).  No chemical bonding problem there. If you stack stuff - planetary and stellar cores - it gets more interesting, depending on composition and temperature.  Temperature matters.
http://en.wikipedia.org/wiki/Electron_degeneracy_pressure
Magnetic fields exert pressure.  How much magnetic field diddles with chemical bonding to give you a pressure estimate?
http://en.wikipedia.org/wiki/Magnetar
http://en.wikipedia.org/wiki/Magnetic_pressure
  10^10 teslas is 4.0×10^25 J/m^3 E/c^2 mass density more than 10^4 times that of lead.
That'll put some hurt on chemical bonding.  Scale down to 347 kJ/mole-bond multiplied in a molar volume of diamond.  Atomizing diamond requires about 800 kJ/mol per atom against cohesive energy.  The strongest chemical bonding interaction is roughly 3 eV.  Then you can pontificate with a Lennard-Jones potential.
http://en.wikipedia.org/wiki/Lennard-Jones_potential

Yes, covalently bonded substances become metals in sufficiently high pressures, which vary for different compounds. Unfortunately we still don't know much about really high pressures because they are difficult to generate - you can find details about metallic hydrogen and oxygen on Wikipedia.


text:
Which one really determines how people feel. Heat or temperature?
We use water to cool down some machines because water has high heat capacity, which means it takes more energy to increase 1 degree. Heat transfer occurs only when temperature is different. We cool down the machine because we don't want it to be too hot, but what does the hot mean? The temp or the heat? When we touch water and another liquid at the same temperature but contain different heat. We feel the heat or the temp?

Please add what you have attempted towards solving the problem into the body of your question.  For more information, see the site's homework policy for how to ask homework questions. Thanks!

@jonsca I don't know if this is a homework problem; it sounds too philosophical - "what does the hot mean?"

I think this boils down to "What's the difference between Heat and Temperature" which is googleable (and a frequent high school problem)

@qwersjc I'm with WOPR on this one.  It seems like an assignment to me.  If it is a curiosity question, it still would be helpful to those answering for the OP to put his own ideas in there.

let's put human bodies away. Just talk about the machine

I'm voting to close this question as off-topic because this is more about physics than actual chemistry.

Regarding the last part "Do we feel heat or the temp?" - this is more physiological then chemical, but I want to share this anyway.
We feel heat flux, i.e. how fast heat is removed or supplied to our skin. Since heat flux can be both advection in the surrounding medium (e.g. air) or conduction in the skin it depends on the situation which is dominant.
In general you can write this as
$$ Q=k A \Delta T$$
where $Q$ is the heat flux, $k$ is some transfer coefficient, $A$ is the exchange area and $\Delta T$ is the temperature difference. The value of $k$ depends on the mechanism of transfer e.g. for advection it scales with $\rho C_p v$ i.e. it depends on the heat capacity and flow velocity of the surrounding medium.
The consequence of feeling heat flux is that we can only feel temperature differences not absolute temperature and that advection gives rise to a phenomenon called 'wind chill' that causes us to feel colder when there is more wind, because the higher wind velocity increases the heat transfer coefficient.

Further proof that thermoception relies on the rate of heat exchange, rather than solely the temperature difference, is the fact that metals feel much cooler to the touch than plastics, even if both are at the same temperature (if below body temperature). Metals tend to have a much higher heat conductivity, so they steal body heat faster than plastics. For the same reason, metals above body temperature feel hotter than plastics at the same temperature.

@NicolauSakerNeto indeed, good point about metal vs. plastic

Sensation of heat via the nervous system is complicated, as there are different kinds of proteins which respond to temperature and take part in thermoception.  The proteins themselves respond directly to temperature, but we don't sense the temperature-based conformation directly.  Rather the impact of this change on ion flux leads to a change in electrical potential across the cellular membrane, ultimately resulting in the 'firing' of neural impulses.  As such, higher heat transfer results in the sensation of more heat due to more rapid change in membrane potential even though the proteins are directly responding to absolute temperature.
This is further complicated by inhibitory neurons in the sensory processing system that 'accommodate' for ongoing signals by down-regulating a sensation when it occurs continuously.  This is why there is a qualitative difference between the feeling of initially stepping out of air conditioning into a hot day and how it feels after being out in the heat for a while.
Note that the feedback from the various sensors are further integrated to provide more complex sensory signals.  For instance, the sensation of intense (painful) heat can actually be created by suppressing the sensation of intense cold using cool stimuli.  This is the basis for the thermal grid illusion where contact with alternating warm and cool bars can create the illusion of painful heat.
In summary, our sensation of hot and cold is not a simple direct relationship to heat or heat transfer, though it certainly correlates  in a useful way that helps us to avoid damage to our bodies under 'typical' conditions.


text:
Titration color change duration
Why is it that in titration a color change occurs initially but then the solution quickly reverts to its previous color?
Thanks so much!

You need to be specific in your question. What are you titrating and what colors are seen?

As an example, let's say you're titrating a strong acid with a strong base using phenolphthalein as an indicator. As you initially add the base to the acid, you may not see the color change occurring because the acid so readily neutralizes the added base. This is because of the large concentration of H+. As you approach a basic pH, however, concentration of H+ decreases and the acid is not able to neutralize the base as quickly. As the base disperses in the solution more slowly, the phenolphthalein will have time to react and show a slight color change.


text:
Why NH4OH? Why not NH3(aq)?
The title says it all.
I had this question when helping someone with their high-school chemistry homework. What are the reasons that someone would use an (from my point of view) obfuscated notation?

So that you can send the not-so-bright students out for a quest to search for solid $\ce{NH4OH}$, if they need a base but both $\ce{KOH}$ and $\ce{NaOH}$ would contaminate their sample for further cation testing.

Related: Ammonium Hydroxide Name

The lone pair at the nitrogen is very well suited to form hydrogen bonds. This bond is especially strong, since ammonia has a quite high proton affinity, it is a medium strong base.
Now consider the autoprotolysis reactions in pure water and (liquid) ammonia:
\begin{align}
\ce{2H2O &<=> H3O+ + OH-}\tag1\\
\ce{2NH3 &<=> NH4+ + NH2-}\tag2
\end{align}
Mixing these systems, the reaction of ammonia with water will yield
\begin{align}
      && \ce{2H2O + 2NH3 &<=> H3O+ + OH- + NH4+ + NH2-}\tag3 \\
& (~-~)& \ce{H3O+ + NH2- &-> H2O + NH3}\\\hline
      && \ce{H2O + NH3 &<=> NH4+ + OH-}\tag4\label{eq:ammonia+water}
\end{align}
\eqref{eq:ammonia+water} is incorporated into a huge and strong network of hydrogen bonds. In this case the equilibrium will be very much pushed to the right side, meaning, that almost all ammonia molecules will be protonated.
Therefore it is more appropriate to write:
\begin{align}
\ce{NH4OH (aq) &<=> NH4+ (aq) + OH- (aq)}\tag{4a}
\end{align}
However, it is always advisable to look at the conditions of the whole reaction system.
Note that $\ce{NH4OH}$ is a pretty outdated notation, if you would want to consider one water entity explicitly, $\ce{NH3.H2O}$ would be more appropriate. See the answer by Maurice for some more information: Can acid-base reactions occur in a non-aqueous medium?

$\ce{NH4OH}$ is a terrible notation which makes no chemical sense, as there is no such a chemical species and therefore not a dissociation of a mythical species, but deprotonation of water what is behind the basicity.
But I guess textbooks like to use it for "didactical" purposes (means it is easier to tell the kids that $\ce{NaOH}$, $\ce{KOH}$, $\ce{NH4OH}$ are bases then tell the chemistry behind it).

Well that statement might contain some truth, the part of no chemical sense is brutally overestimated. While $\ce{NH4OH}$ might not completely appropriate, there are crystal structures of $\ce{NH3.H2O}$.

Ammonia hydrates.  A fascinating area of research!

@Martin First of all, the notation $\ce{NH3.H2O}$ codes something completely different than $\ce{NH4OH}$, second, there is no a molecular species like $\ce{NH4OH}$ in water, therefore it is totally misleading notation. Why should one use a notation that has no merit, only to hide the chemistry and confuse kids?

Simple: like every chemical situation we derive, it's a model. I mean, why do we teach kids how to use lewis structures when MO theory is far more precise? It's an instructive oversimplification.

@BreakingBioinformatics Because explaining MO theory and tell people to calculate them in head is far more difficult and then count till eight; while telling people that ammonia is not NH4OH takes about 2 minutes, and an excellent didactic choice to introduce acid-base theory at high school level. Also, if NH4OH is a proper choice of notation, I am really curious how do you handle organic amines, amino acids etc in a consistent way.

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text:
Cathode + Anode + Rechargeable battery
I'm confused by the following from this webpage:

The cathode is a metal oxide and the anode consists of porous carbon. During discharge, the ions flow from the anode to the cathode through the electrolyte and separator; charge reverses the direction and the ions flow from the cathode to the anode.
On discharge, the anode undergoes oxidation, or loss of electrons, and the cathode sees a reduction, or a gain of electrons. Charge reverses the movement.

This says that the metal oxide electrode is always the cathode and the porous carbon electrode is always the anode. As far as I know, that designation should be correct on discharge but backwards on charge. The anode is always whichever electrode is doing the oxidation, and the cathode is the electrode doing the reduction.
My naive understanding would be this:

The cathode is the electrode with the reduction half reaction. The anode is the electrode with the oxidation half reaction. This is true for both charge/discharge.
The anode is the one that produces electrons and cathode receives electrons. This is true for both charge/discharge.
When switching between charge/discharge, the redox reactions reverse, and the cathode/anode designations also switch to preserve cathode==reduction and anode==oxidation.
During discharge, the battery functions as a galvanic cell, where the redox reaction produces electric energy, electrons go down their electrical gradient from the negative electrode to the positive electrode. The anode is the negative electrode, the cathode is the positive electrode.
During charge, the battery functions as an electrolytic cell, where electric energy drives a nonspontaneous redox reaction, electrons go up their electrical gradient from the positive electrode to the negative electrode. The anode is the positive electrode, the cathode is the negative electrode.
In a lithium ion battery, the positive electrode is the metal oxide and the negative electrode is the porous carbon. The anode/cathode designations switch depending on whether the battery is charging or discharging.

Please help me understand this.

I have asked experts -> I am right. The common convention is to label the physical electrodes as cathode/anode based on discharge even though that will be backwards when the battery is charging. That is a poor, confusing, and illogical naming convention, but it is what it is.
The more clear and logical naming convention would be to name the physical electrodes either positive or negative since those labels apply in both charge/discharge directions. Some people use this logical naming convention, but the illogical convention seems more prevalent.

Have to agree with clay. Naming battery terminals based on the chemical reaction is just plain silly. Negative and positive don't change, and if we did it this way there would be so much less confusion.

Okay, so your understanding is correct about rechargeable batteries after taking electrochemistry in class.However, electrons migrate from anode to cathode . The reaction at the anode produces electrons.

Exactly correct. Electrons migrate from anode to cathode. In a rechargeable battery, the electron flow will reverse when switching between charge/discharge, so the anode/cathode designations of the two electrodes will also switch.


text:
Total number of octahedral void(s) per atom in a cubic closed pack structure

The total number of octahedral void(s) per atom present in a cubic closed pack structure is
(a) 1
(b) 2
(c) 3
(d) 4

I answered with (d), but the key on internet says it is (a). Please clarify the dispute.

I assume you've seen diagrams?
http://www.chem.shef.ac.uk/chm131-1999/cha99ngs/intro4.gif & http://www.wou.edu/las/physci/ch412/ccp.jpg ... Seem to make the case for 1 gap pretty clear
(http://www.chem.shef.ac.uk/chm131-1999/cha99ngs/aintro.html)

Finally got it, it was a misunderstanding. I ignored to read the word ‘per atom’ in a hurry. So clearly the answer is 1.

Congratulations, you've just earned a self-learner badge (answer your own question and score +3).

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text:
Why is Cyclodextrin cavity hydrophobic?
Actually this is going to be a list of interconnected questions. Please bear with me.

Wikipedia article on cyclodextrin (CD) says that it is hydrophobic from inside.
(1) How is it hydrophobic when there are so many hydroxyl groups on the interior surface ?
(2) This image
    labels the interir cavity as apolar rather than hydrophobic. I guess
    the apolarity is because of mutual cancellation of the dipole
    moments.
Am I right on this?
In any case I don't understand how the cavity excludes water.
There are reports on trapping organic substances such as naphthalene, toluene and crude oil in the CD cavity. Another report says that nitrosamime is trapped in CD by interacting with the hydroxyls.
(3) How does this trapping exactly occur ?
(4) Do the hydroxyls play a role in restricting the diffusion of the trapped organic molecules out of the cavity ?
All of this, I started investigating because of a recent news about an alcoholic beverage in powder form- palcohol. Initially I didn't believe it but later I found that alcohol is absorbed in CD to make a dry powder.
(5) What prevents alcohol from adsorbing on the exterior surface and not make the powder sticky ?
(6) Other studies say that there is a 1:1 ratio of trapping but palcohol claims that it is 58% alcohol by weight. Even this old patent says that CD absorbs a lot of alcohol. How does that happen ? In all it seems that CD behaves differently with different molecules.

1: http://www1.lsbu.ac.uk/water/images/cyclo.gif looking at this rendition you can get a different picture of the inner loop. While there are O-H in there, be sure to check out the carbon edges also extruding in.
2: I thought apolar == non-polar, which generally leads to hydrophobicity. Linking this to 3 and 4, it might be that water isn't excluded altogether, but these other groups are just preferred over water.
6: I'm not sure if we can assume palcohol is JUST Cyclodextrin + alcohol. Also 1:1 ratio of trapping is by moles of the molecule, right? Typically these aren't shown by weight...

Guessing for 5, preference? At first glance it seems way more favorable for the alcohol to simultaneously interact with many sites than to just hang off of one.

Cyclodextrin hydroxyls are the external surface (hydrophilic and hydrogen-bonding).  They circle the larger open end.  The carbon and hindered ether skeleton is internal (hydrophobic).  A cyclodextrin looks like a bonded hollow detergent micelle in water.

Before you believe advertising hype, remember that underivatized cyclodextrins are nephrotoxic, crystallizing upon concentration, probably as the cholesterol endohedral complex.  Look it up.
Ethanols insert their hydrocarbon tails into the center, then hydrogen bond their polar ends with each other and the rim.


text:
Boilng vodka for A/C cleaning
I hope this doesn't sound too strange.
I have plenty of cheap vodka and want to clean the A/C on my car.
So I thought about boiling 1/2 litre of vodka in my car while I ran the A/C with recirculation on.
The alcohol should get liquid within the air conditioner and desinfect and clean the vaporizer.
I am just not sure if I create an explosive mixture inside the car this way.
David

Visit a private mechanic, not a chain operation. What do you expect your porous high surface area upholstery will be doing as you fill your car with booze vapor?

I have my own garage, I rebuild my whole engine... I don't need a mechanic :)
I think there should be no problem for the interiour (mostly leather) if there is some steam, that will me immediately sucked by the AC...

Alcohol vapors can certainly form explosive mixtures with air, so filling the car with ethanol vapors is potentially dangerous. A source of ignition, for example a spark, could under the right circumstances ignite the vapors. If the fan motor of the AC system is a brushed motor it can create small sparks. So I would say there is a fire hazard.
But there are other problems with your suggested setup. The ethanol and water vapors would condense on every surface of your car's interior, the windows, the roof, the doors, etc. This would mean you loose a lot of alcohol, so you would need more. Ethanol might attack and start to dissolve some of the interior, glued parts may become loose. And of course the car would smell very strongly like vodka for a long time.
So I would suggest disassembling the dash cluster and whatever else you need to get to the AC system, and using the vodka in liquid form to clean it properly.

I would advise against using vodka at all. Alcohol 100% (which you'll get from the vapor, depending how you do it) isn't a good anti-septic. Neither is Alcohol 40% (vodka). Alcohol 70% v/v is much more effective. Want to make good use of that cheap vodka? Drink it.


text:
What is the compression work?
A cylinder contains 2 kg of steam at 200 kPa and 200 $^\circ$C. Now steam is compressed at constant pressure until its temperature rises to 250 $^\circ$C. What is the work done during this compression process?

Welcome to Chemistry.SE. This appears to be a homework question. Unless you show some work, thoughts, or effort, it is likely to get put on hold.

To calculate work at constant pressure, you need the following:
$$w=-P \Delta V$$
where $w$ is work, $P$ is pressure and $\Delta V$ is the change in volume.
You will need to solve some gas laws relations to get both $P$ and $\Delta V$ in terms of the single variable $T$.
Algebraically, it will simplify to $P\Delta V = nR\Delta T$, assuming ideality.
If you need to assume a non-ideal gas (which is a good assumption for water at high pressure). If we use the van der Waals equation, things get dicey and calculus is needed. I'm hoping you don't need calculus:
$$d[(P-\frac{n^2a}{V^2})(V-nb)]=d(nRT)$$
$$d(PV-\frac{n^2a}{V}-nbP+\frac{n^3ab}{V^2})=d(nRT)$$
$$PdV+VdP+\frac{n^2a}{V^2}dV-nbdP-\frac{2n^3ab}{V^3}dV=nRdT+RTdn$$
Since $n$ and $P$ are constant $dP=0$ and $dn=0$
$$PdV+\frac{n^2a}{V^2}dV-\frac{2n^3ab}{V^3}dV=nRdT$$
$$\int_{V_i}^{V_f}(P+\frac{n^2a}{V^2}-\frac{2n^3ab}{V^3})dV=\int_{T_i}^{T_f}nRdT$$


text:
Find the molecular formulas of these alkanas and alkenes

Can Somebody please solve this chemistry question
It came as a part of a previous test and i wasn't able to find a way to use ny/nx since all my answers would be negative. It wasn't a homework problem.

While this was not a homework, it is still a homework-like question. From our standpoint, there is no difference between homework problems, self-study problems, end-of-chapter problems, exam problems, example problems, etc. Whether it was assigned for a grade or not is not the key feature of such questions - they exist so that you can learn how to do these problems.

Non-cyclic (or acyclic) alkanes have the general formula $\ce{C_{n}H_{2n + 2}}$, and corresponding alkenes have the general formula $\ce{C_{n}H_{2n}}$.  So in this problem let's say our alkane (x) is $\ce{C_{n}H_{2n + 2}}$, then our alkene (y) with one more carbon atom will be $\ce{C_{n + 1}H_{2(n + 1)}}$.  The molecular weight of the alkane will be $$\ce{MW(x) = (12*n) + (1*(2n +2 )) = 14n + 2}$$ and $$\ce{MW(y) = ((12*(n + 1)) + (1*(2n + 1 ))) = 14n + 14}$$  We also know that grams(x) = grams(y), and that moles(x)/moles (y) = 7/5.  The number of moles of a substance is equal to its weight in grams divided by its molecular weight.  Therefore $$\frac{N_x}{N_y}=7/5=\frac{G(x)/14n+2}{G(y)/14n+14}$$  But in this problem grams(x) = grams(y) so our equation simplifies to $$7/5=\frac{14n+14}{14n+2}$$ Solving for "n" gives you the number of carbon atoms in the alkane, and "n+1" will be the number of carbon atoms in the alkene.  I'll leave the mathematical solution and the naming of the two moleculesto you.

It's not an assignment btw its is just a practice of a previous test the teacher gave us but didn't give its explanation. I wasn't able to solve it since i didn't understand how to use the mol ratio, and all the answers got to be negative which is impossibly so. But thanks for the answer.


text:
Why is weight of 1 mole of substance equal to atomic/molecular mass in grams?

Why is weight of 1 mole of substance equal to atomic/molecular mass in grams?

According to me, it happens because mole has been defined in such a way. It is defined as the numbers of particles in $\pu{12 g}$ of $\ce{^{12}C}$. If it were $\pu{24 g}$ instead of $\pu{12 g}$, then the weight of 1 mole of substance would equal 2 times the atomic/molecular mass in grams.
Please correct me if I am wrong somewhere (or maybe everywhere).

Yes, you are basically correct.

Why is weight of 1 mole of substance equal to atomic/molecular mass in grams?
According to me, it happens because mole has been defined in such a way.

Yes! That is correct.

It is defined as the numbers of particles in 12 g of C12. If it were 24 g, instead of 12 g, then the weight of 1 mole of substance would equal 2 times the atomic/molecular mass in grams.

Also correct, assuming that the definition of unified atomic mass units (amu) remained the same. @Martin's answer is correct, but we can also arrive at the same conclusion using a simple dimensional analysis approach.
First we need the definition of an amu:
$$\pu{1 atom}~ \ce{^12C} = \pu{12 amu}$$
Now take the real definition of a mole:
$$\pu{1 mol}~ \ce{^12C} = \pu{12 g}$$
Now, divide the first equation by the second:
$$\frac{\pu{1 atom}~ \ce{^12C}}{\pu{1 mol}~ \ce{^12C}} = \frac{\pu{12 amu}}{\pu{12 g}}$$
Cross-multiply and reduce:
$$1~\frac{\pu{g}}{\pu{mol}~ \ce{^12C}} = 1~\frac{\pu{amu}}{\pu{atom}~ \ce{^12C}}$$
What this tells us is that the ratio of g/mol to amu/atom is exactly one - and we made sure it would work out that way by carefully choosing how we defined moles and amus.
Since the masses of all elements are defined in terms of amu, which are ultimately based on the mass relative to carbon-12, this ratio holds for all any atom or molecule.
Let's take it further and put in your hypothetical doubling of the number of grams of carbon-12 per mole:
$$\frac{\pu{1 atom}~ \ce{^12C}}{\pu{1 mol}~ \ce{^12C}} = \frac{\pu{12 amu}}{\pu{24 g}}$$
Now when we cross-multiply and reduce, we get:
$$2~\frac{\pu{g}}{\pu{mol}~ \ce{^12C}} = 1~\frac{\pu{amu}}{\pu{atom}~ \ce{^12C}}$$
Since we kept the definition of an amu the same, but changed that of the mole, what we see is that 1 mole of carbon-12 using these units would have to weigh 24 g (the molar mass), even though its atomic mass would still be 12.
That's a lot harder to keep track of mentally, and so it is a good thing for us that the definitions of amus and moles were chosen so carefully!

Avogadro's constant is the inverse of 1 amu...How is that related to mole concept?

The definitions of moles and a.m.u. in combination, yes.

The use of the atomic mass unit is deprecated, it should be used the unified atomic mass unit. What is the difference? Amu was based on $\ce{^{16}O}$.

@Martin you are right about that - typically today, a.m.u. refers to the unified atomic mass unit. I'll edit my answer to reflect that.

Actually $\mathrm{u}$ refers to the unified atomic mass, just to avoid confusion.

"It is defined as the numbers of particles in 12 g of C12. If it were 24 g, instead of 12 g, then"the $\mathrm{(a.m.u)}~\text{in grams}=\frac{\pu{1 g}}{3.01\times{10^{23}}}$

Shouldn't it be written $\text{mass(1 atom}{}^{12}C\text{)}=12\text{amu}$ and likewise for the next line, considering that one carbon-12 atom is not really equivalent to 12 a.m.u?

In short terms:
Your assumption is correct. If the definition of mole would have been based on 24 grams of carbon-12, all molecular weights would double. This is not the case and I highly doubt, that this definition will ever change.

In long terms:
This question is very definition specific. One should not be confused with the several types of definitions. All these definitions are based on the carbon-12 isotope. Hence the atomic mass (constant) has a specific value:

One twelfth of the mass of a carbon-12 atom in its nuclear and
  electronic ground state,  $m_u = 1.660 5402 10~\times~10^{−27}~\text{kg}$. It is equal
  to the unified atomic mass unit.

The unified atomic mass unit is therefore a derived non-SI unit:

Non-SI unit of mass (equal to the atomic mass constant), defined as
  one twelfth of the mass of a carbon-12 atom in its ground state and
  used to express masses of atomic particles,  $u\approx 1.660 5402 10~\times~10^{−27}~\text{kg}$.

The atomic weight (i.e. relative atomic mass) is a ratio (=number) and therefore has no unit:

The ratio of the average mass of the atom to the unified atomic mass
  unit.

Also based on carbon-12 is the definition of mole, which you stated correctly:

SI base unit for the amount of substance (symbol:  mol). The mole is
  the amount of substance of a system which contains as many elementary
  entities as there are atoms in 0.012 kilogram of carbon-12. When the
  mole is used, the elementary entities must be specified and may be
  atoms, molecules, ions, electrons, other particles, or specified
  groups of such particles.

Derived from this is the Avogadro constant:

Fundamental physical constant (symbols: $L$,  $N_A$) representing the
  molar number of entities:  $L = 6.022 141 79 30 \times 10^{23}~\text{mol}^{−1}$.

Therfore one can derive several other commonly used properties, such as molecular mass and molar mass. In a biannual publication "Atomic weights of the elements" are published by IUPAC. In Pure Appl. Chem., 2013, Vol. 85, No. 5, pp. 1047-1078 (or here) it says:

The atomic mass, $m_\text{a}$, of an unbound neutral atom of
  carbon-12, $m_\text{a}(\ce{{}^{12}C})$, in its nuclear and electronic
  ground states is $12~u$ exactly, where $u$ is the unified atomic mass
  unit. The atomic weight (also called the relative atomic mass) of
  isotope $^i\text{E}$ of element $\text{E}$, symbol
  $A_\text{r}(^i\text{E})$, in material $\text{P}$ is
$A_\text{r}(^i\text{E})
> =\frac{m_\text{a}(^i\text{E})_\text{P}}{\frac{1}{12}m_\text{a}(\ce{{}^{12}C})}=\frac{m_\text{a}(^i\text{E})_\text{P}}{u}$
Thus, the atomic mass of $\ce{{}^{12}C}$ is $12~u$, and the atomic
  weight of $\ce{{}^{12}C}$ is $12~$ exactly. All other atomic weight
  values are ratios to the $\ce{{}^{12}C}$ standard value and thus are
  dimensionless numbers. The atomic weight of element $\text{E}$,
  $A_\text{r}(\text{E})$, in a material $\text{P}$ is determined from
  the relation
$A_\text{r}(\text{E})_\text{P}=\sum\left[x(^i\text{E})_\text{P}\times A_\text{r}(^i\text{E})\right]$
where $x(^i\text{E})_\text{P}$ is the mole fraction of isotope
  $^i\text{E}$ in material $\text{P}$ (also called the isotopic
  abundance).

Therefore the standard atomic mass of carbon is (given in an interval)
$m_{\text{a}}(\ce{C})=A_\text{r}(\text{C})_\text{graphite}\times u = [12.0096, 12.0116]u$, with
$u=\frac{1}{12}m_\text{a}(\ce{{}^{12}C})$.
However, the atomic mass of a single molecule is always an integer multiple of $u$.
The molar mass of carbon can then be defined as
\begin{aligned}
M(\ce{C})&=m_{\text{a}}(\ce{C})\times L\\
&=[12.0096, 12.0116]\times10^{-3}~\text{kg/ mol}\\
&\approx 12.01~\text{g/ mol}~\text{(4 s.f.)}
\end{aligned}
The whole relation becomes a little bit clearer when looking at bromine. There are two important Isotopes of bromine: $\ce{^{79}Br}$ and $\ce{^{81}Br}$ (ref). So in elemental bromine ($\ce{Br2}$) the molecules may have three different masses $m_{\text{a}}(\ce{^{79}Br2})=158u$, $m_{\text{a}}(\ce{^{81}Br2})=162u$ and $m_{\text{a}}(\ce{^{79}Br^{81}Br})=160u$.
The standard atomic mass of bromine is $m_{\text{a}}(\ce{Br})= [79.901, 79.907]u$.
Therefore $M(\ce{Br})=[79.901, 79.907]\times 10^{-3}~\text{kg/ mol} \approx 79.90~\text{g/ mol}~\text{(4 s.f.)}$ (based on the reference value).

When calculating with molar masses $M$ one always has to keep in mind, that the used standard values are based on a (global) statistic.

Related definitions:
amount of substance, $n$, Also contains definition of: number of moles

Base quantity in the system of quantities upon which SI is based. It
  is the number of elementary entities divided by the Avogadro constant.
  Since it is proportional to the number of entities, the
  proportionality constant being the reciprocal Avogadro constant and
  the same for all substances, it has to be treated almost identically
  with the number of entities. Thus the counted elementary entities must
  always be specified. The words 'of substance' may be replaced by the
  specification of the entity, for example: amount of chlorine atoms,
  $n_\ce{Cl}$, amount of chlorine molecules,  $n_{\ce{Cl2}}$. No
  specification of the entity might lead to ambiguities [amount of
  sulfur could stand for  $n_\ce{S}$,  $n_{\ce{S8}}$, etc.], but in many
  cases the implied entity is assumed to be known: for molecular
  compounds it is usually the molecule [e.g. amount of benzene usually
  means  $n_{\ce{C6H6}}$], for ionic compounds the simplest formula unit
  [e.g. amount of sodium chloride usually means $n_{\ce{NaCl}}$] and for
  metals the atom [e.g. amount of silver usually stands for
  $n_{\ce{Ag}}$]. In some derived quantities the words 'of substance'
  are also omitted, e.g. amount concentration, amount fraction. Thus in
  many cases the name of the base quantity is shortened to amount and to
  avoid possible confusion with the general meaning of the word the
  attribute chemical is added. The chemical amount is hence the
  alternative name for amount of substance. In the field of clinical
  chemistry the words 'of substance' should not be omitted and
  abbreviations such as substance concentration (for amount of substance
  concentration) and substance fraction are in use. The quantity had no
  name prior to 1969 and was simply referred to as the number of moles.

relative molecular mass,  $M_r$

Ratio of the mass of a molecule to the unified atomic mass unit.
  Sometimes called the molecular weight or relative molar mass.

relative molar mass
Molar mass divided by  $1~\text{g/ mol} (the latter is sometimes called the standard molar mass).

Like "1 dozen",  "1 mole" refers to an exact number. Molar mass refers to the mass of a very specific number of molecules/atoms: 1 mole. The molar mass is the mass divided by the amount (number of individual entities such as atoms or molecules) of substance measured in moles.
Now 1 mole is equal of course to Avogadro's constant $\pu{N_A}$. To make matters clear
$\pu{N_A = 6.022 \times 10^{23}/mol}$
but it is usually ok to speak of Avogadro's number as equal to the number of particles in 1 mole. It is therefore fair in practice to make the following substitution when performing computations:
$$\pu{1 mole = N_A = 6.022 \times 10^{23}}$$
withthe understanding that $\pu{N_A}$ always refers to the number of entities in 1 mole.
Now lets say we take 1 mole of carbon-12 atoms and determine its mass. It's $\pu{m = 12.0 g}$, which means the molar mass is
$$\pu{M = 12.0 g/1 mole = 12.0 g/mol}$$
But what if we want to know the mass of 1 atom of $\ce{^{12}C}$? Ok, substitute "1 mole" with Avogadro's number. That's just
$$\pu{M_{molecule} = 12.0 g/N_A = 12.0 g/6.022 \times 10^{23} = 1.66\times 10^{-27} kg = 1.66 yg}$$
But $\pu{1.66\times 10^{-27} kg}$ is a really tiny and cumbersome number, and the "yoctogram" is very rarely used. However we can define an atomic mass unit:
$$\pu{1 amu = 1 g/N_A (exactly)} $$
Then
$$\pu{M_{molecule} = 12.0 amu}$$
This is much tidier! But that's the same as before, except that we changed "g/mol" with "amu", so the units are practically exchangeable.
However they are not always interchangeable (sometimes you should be careful): the mass of an atom or molecule refers rather obviously to a single atom or molecule, whereas the mass of a mole is a sum over all species in that mole, which may contain for instance a mix of isotopes. Of course, if you refer to an "average mass" then amu and g/mol are interchangeable.
[3]: Actually, chemical engineers distinguish between lb-mol and g-mol and so forth, but I'm a plain chemist, so I stick to chemist's (or SI) convention.

Take the magnitude of formula mass and append "grams" to obtain the mass of one gram-mole.  If  you have a formula weight magnitude and append "pounds," it is one pound-mole.  One gram-mole contains one Avogadro number of formula units.
Like a "dozen eggs," a mole is simply a bigger basket.  It allows you to take microscopic descriptions to real world scales.  Moles are scaling factors for thermodynamics, as in molar volumes of gases.

I do understand what mole is, but why does 1 mole of an atom/molecule weigh the same as the atomic/molecular mass in grams?

That is how it is defined through Avogadro's number as the scale factor for a gram-mole.  If the kilogram is redefined as a single crystal silicon ball, Avogadro's constant will be a defined value rather than a measured one.

It is about the definitions, but I suspect you are overcomplicating in thinking it isn't obvious.
A mole is an Avogadro number (6.022 x 1023) of things (the things could be anything but being a big number it is only a useful metric for atoms and molecules). Given this definition, if a molecule weighs the equivalent of a carbon atom then a mole of it will be 12g. A mole of hydrogen gas weighs about (about because it might have some deuterium) 2g as hydrogen gas consists of H2 molecules.

But why is the molecular weight of hydrogen 2 and it's molar weight also 2g?

One number is the other divided by avogadro's number. Changing the number of molecules or atoms doesn't change the ratio of their atomic or molecular weights.

The Avogadro number is based and therefore the result of the exact definition given by the OP. http://goldbook.iupac.org/M03980.html That is also the reason, why the molecular weights deviate from natural numbers.


text:
How do I lookup a polymer by its structure?
This is more of a meta-chemistry question, but, here goes...
Given the structure of a polymer, how can I search online for which polymer it is?
Specifically, I have an image like this:

I'd also like to know which polymer specifically this is, but also a method for figuring this out in the future by myself :)

A polymer is generally an organic compound with a base component of repeating units- generally 1000's of units such as CH2 for polyethylene. Nothing repeats in this compound so its not a polymer. An oligomer is like a polymer but has far smaller numbers of repeating units. A polymer as a material is a plastic- it have engineering value whereas an oligiomer behaves like a chemical.

That looks like bilirubin, from hemoglobin catabolized in the liver  Monomer, dimer..tetramer...oligomer...polymer...high polymer.  It is a tetrapyrrole, not even an oligomer.


text:
How to determine the pH of a mixture of two weak acids?
We have two solutions:

Solution 1 is $\ce{HCOOH}$, its concentration is $c_1 = \pu{10^-2 mol/l}$, its volume is $V_1 = \pu{50 ml}$, and its $\mathrm{pH}_1 = 2.9$.
Solution 2 is $\ce{CH3COOH}$, its concentration is $c_2 = \pu{10^-2 mol/l}$, its volume is $V_2 = \pu{50 ml}$, and its $\mathrm{pH}_2 = 3.4$.

How would be the equation and the ICE table, and what is the $\mathrm{pH}$ of the mixture of these two solutions?
I used numbers to differentiate the solutions, and to understand how it works, but for no other reason.
I tried to do $\mathrm{pH}=\frac{1}{2}\left(\mathrm{pH}_1+\mathrm{pH}_2\right)$, but it’s trivial and doesn't work.

Below a more general approach.
Suppose that we have two weak acids $\ce{HA}$ and $\ce{HB}$.
The initial concentrations are $C^0_\ce{HA}$ and $C^0_\ce{HB}$, and their constants are $K_{\mathrm{a},\ce{(HA)}}$ and $K_{\mathrm{a},\ce{(HB)}}$.
Suppose yet that volumes, $V_\ce{HA}$ and $V_\ce{HB}$, are additives.
So we have:

Reactions

\begin{align}
  \ce{HA + H2O &<=> H3O+ + A-} &
  K_{\mathrm{a},(\ce{HA})}
    &= \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}
  \tag{1}\label{eq:KAcidHA}\\
  \ce{HB + H2O &<=> H3O+ + B-} &
  K_{\mathrm{a},(\ce{HB})}
    &=\frac{\ce{[H3O+][B-]}}{\ce{[HB]}}
  \tag{2}\label{eq:KAcidHB}\\
  \ce{2 H2O &<=> H3O+ + OH-} &
  K_\mathrm{w} &= \ce{[H3O+][OH-]}\tag{3}\label{eq:KWater}
\end{align}

Mass balance

\begin{align}
  C_\ce{HA} &= \frac{C^0_\ce{HA} V_\ce{HA}}{V_\ce{HA} + V_\ce{HB}}
  &&=\ce{[HA] + [A-]}\tag{4}\label{eq:MassBalanceHA}\\
  C_\ce{HB} &= \frac{C^0_\ce{HB} V_\ce{HB}}{V_\ce{HA} + V_\ce{HB}}
  &&=\ce{[HB] + [B-]}\tag{5}\label{eq:MassBalanceHB}
\end{align}

Charge balance

$$\ce{[H3O+] = [OH-] + [A-] + [B-]}\tag{6}\label{eq:ChargeBalance}$$
Replacing \eqref{eq:KAcidHA}–\eqref{eq:MassBalanceHB} equations on \eqref{eq:ChargeBalance}, we have:
$$
\ce{[H3O+]} =
  \frac{K_\mathrm{w}}{\ce{[H3O+]}}
  +\frac{C_\ce{HA} K_{\mathrm{a},(\ce{HA})}}{\ce{[H3O+]}
  + K_{\mathrm{a},(\ce{HA})}}
  + \frac{C_\ce{HB} K_{\mathrm{a},(\ce{HB})}}{\ce{[H3O+]}
  + K_{\mathrm{a},(\ce{HB})}}
\tag{7}\label{eq:GeneralEquation}$$
or as polynomial
\begin{align}
\begin{split}
 &\ce{[H3O+]}^4\\
+&\ce{[H3O+]}^3 (K_{\mathrm{a},(\ce{HA})}
  + K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}^2 \left[
  K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  - ( C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
    + C_\ce{HB}K_{\mathrm{a},(\ce{HB})} )
  - K_\mathrm{w} \right]\\
-&\ce{[H3O+]} \left[
  (C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  + K_\ce{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})}) \right]\\
-& K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w}\\
=&\ 0
\end{split}\tag{8}\label{eq:GeneralPol}
\end{align}
This single equation will exactly solve any equilibrium problem involving the mixture of any two monoprotic acids, in any concentration (as long as they're not much higher than about $\pu{1 mol L-1}$) and any volume. Depending of $K_\mathrm{a}$, $K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}$, and $C^0_\ce{HB}$ values, we can yet obtain a simpler version[1].
The \eqref{eq:GeneralPol} equation can simplified considering[2] that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w} \ll \ce{[H3O+]}^4$:
\begin{align}
\begin{split}
 &\ce{[H3O+]}^3\\
+&\ce{[H3O+]}^2 (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}\left[
  K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  - (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
    + C_\ce{HB} K_{\mathrm{a},(\ce{HB})})
  - K_\mathrm{w} \right]\\
-&\left[
  (C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  + K_\mathrm{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
  \right]\\
=&\ 0
\end{split}\tag{9}\label{eq:GeneralPolSimp1}
\end{align}
The \eqref{eq:GeneralPolSimp1} equation can simplified considering[2] that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} \ll \ce{[H3O+]}^3$:
\begin{align}
\ce{[H3O+]}^2 + \ce{[H3O+]} (K_{\mathrm{a},(\ce{HA})}
  + K_{\mathrm{a},(\ce{HB})})
  - (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
    + C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})
= 0\tag{10}\label{eq:GeneralPolSimp2}
\end{align}
The \eqref{eq:GeneralPolSimp2} equation can be solved as usual.
$$\ce{[H3O+]} =
\frac{
  - (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
  + \sqrt{
    (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
    + 4 (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
      + C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})}
}{2}$$
Or using the initial concentrations
$$\ce{[H3O+]} =
\frac{
  - (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
  + \sqrt{
    (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
    + 4\left(%\displaystyle
      \frac{C^0_\ce{HA} V_\ce{HA} K_{\ce{a},(\ce{HA})}
        }{V_\ce{HA} + V_\ce{HB}}
      + \frac{C^0_\ce{HB} V_\ce{HB} K_{\ce{a},(\ce{HB})}
        }{V_\ce{HA} + V_\ce{HB}}
    +K_\mathrm{w}\right)}
}{2}$$
Replacing $C^0_\ce{HA}=\pu{0.01 mol L-1}$, $C^0_\ce{HB}=\pu{0.01 mol L-1}$, $V_\ce{HA}=\pu{0.050 L}$ and $V_\ce{HB}=\pu{0.050 L}$, and using $\text{p}K_\ce{a}=3.75$ for formic acid and $\text{p}K_\ce{a}=4.756$ for acetic acid[3], we have
$$\ce{pH}=3.05.$$
Addendum
@MaxW made some warnings about the assumptions which simplificate ($\ref{eq:GeneralPol}$)(see comment) and ($\ref{eq:GeneralPolSimp1}$)(see comment) equations.
To completeness and with the aim of show the limitations of present approach, consider the following graphics.
The ($\ref{eq:GeneralPol}$), ($\ref{eq:GeneralPolSimp1}$), and ($\ref{eq:GeneralPolSimp2}$) equations were solved using $\text{p}K_{\mathrm{a},(\ce{HA})}=\text{p}K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}=C^0_\ce{HB}$, and $V_\ce{HA}=V_\ce{HB}=\pu{50.0 mL}$.

As we can see, the ($\ref{eq:GeneralPolSimp1}$) equation can be used on any $\text{p}K_\ce{a}$ values, inasmuch as the concentration be major of $\pu{1E-6 mol L-1}$.
The ($\ref{eq:GeneralPolSimp2}$) equation becomes problematic when dilute solutions or strong acids solutions are used.
Thus, when applied to book-examples, which $\text{p}K_\ce{a}$ values from 3 to 6, and the concentrations from $1.0$ to $\pu{1.0E-4 mol L-1}$, the assumptions are valid and the equations can be used, providing good approximations of $\ce{pH}$ values.
Acknowledgement
Thanks to @MaxW for draw attention to limitations of assumptions.
Reference

Edited based on first @MaxW's comment.
Edited based on second @MaxW's comment.
Haynes, W. M.;  Lide, D. R.; Bruno, T. J., CRC Handbook of Chemistry and Physics. 97 ed.; CRC Press: 2017; pp. 5–88.

@GRSousaJrHow assumption "${K_\ce{a}}\left(\ce{HA}\right){K\ce{a}}\left(\ce{HB}\right)K\ce{w}\ll 1$"
simplifies equation ($8$) to equation ($9$)?
How assumption "${K_\ce{a}}\left(\ce{HA}\right){K\ce{a}}_\left(\ce{HB}\right)\ll 1$"
simplifies equation ($9$) to equation ($10$)?

Look the acetic and propionic acids example.
The $\text{p}K_{\ce{a}}$ values are $4.756$ and $4.88$ respectively.
$K_{\ce{a}(\ce{Ac})}K_{\ce{a}(\ce{Pro})}K_{\ce{w}} = \pu{2.31E-24}$ and $K_{\ce{a}(\ce{Ac})}K_{\ce{a}(\ce{Pr})} = \pu{2.31E-10}$.
The contribution of these values to sum can be disregarding and equations simplified.

Thanks, I understand, but how the assumption simplifies the quartic equation to the Cubic equation?

Divide the equation by $\ce{[H3O+]}$.

@KarstenTheis My mistake. 3.77 come from an old (1955) reference. See my answer updated.

If ($\ref{eq:GeneralPol}$) equation is solved exactly using the matlab or octave the pH value is 3.04.

I don't like this solution. You define HA and HB as any weak acids and any concentrations. You then make assumptions that are not true for those assumptions. In order for your assumptions to work, then $\mathrm{K_{a,(HA)}} \gg \mathrm{K_w}$, $\mathrm{K_{a,(HB)}} \gg \mathrm{K_w}$, $C^0_\mathrm{HA} \gg 1^{-7}$, and $C^0_\mathrm{HB} \gg 1^{-7}$. Given those restrictions the autoionization of water can be ignored from the outset.

I would agree that you have a general solution up to equation 8. I don't like the simplifications beyond that...

@MaxW, I made some improvements on answer. See my updates.

In order to get from (8) to (9) you still have made the wrong assumption. You used the general assumption that $$K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w} \ll 1$$ which isn't true. The correct assumption is that $$K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w} \ll \ce{[H+]^4}$$ since you know that the $\mathrm{pH} \approx 3$ and you only will get 2 significant figures in the solution.

@MaxW You are completely right! Answer updated. The assumption from ($\ref{eq:GeneralPolSimp1}$) to ($\ref{eq:GeneralPolSimp2}$) were corrected too.

How you got from (9) to (10) is invalid, or at least incompletely expressed. Sorting that out is too much for a comment.

I understand that correct assumption from ($\ref{eq:GeneralPolSimp1}$) to ($\ref{eq:GeneralPolSimp2}$) must be
$$\left[(C_\ce{HA} + C_\ce{HB}) K_{\ce{a},(\ce{HA})} K_{\ce{a},(\ce{HB})} + K_\ce{w} (K_{\ce{a},(\ce{HA})} + K_{\ce{a},(\ce{HB})})\right]\ll \ce{[H3O+]}^3$$
I'm working on a way to show $C_\ce{HA}$, $C_\ce{HB}$, $K_{\ce{a},(\ce{HA})}$, $K_{\ce{a},(\ce{HB})}$, and $\ce{[H3O+]}$ values for which such expression is true.

Given the following:
\begin{align}
[\ce{CH3COOH}]     &= \pu{1 * 10^-2 mol L-1}\\
[\ce{HCOOH}]       &= \pu{1 * 10^-2 mol L-1}\\
K_{z-\ce{CH3COOH}} &= \pu{1.8 * 10^-5 mol L-1}\\
K_{z-\ce{HCOOH}}   &= \pu{1.8 * 10^-4 mol L-1}\\
[\ce{H3O+}]_1      &= \pu{1.3 * 10^-3 mol L-1}\\
[\ce{H3O+}]_2      &= \pu{4.0 * 10^-4 mol L-1}
\end{align}
Seeing that we could write
$$\ce{HCOOH + CH3COOH + 2H2O -> HCOO- + CH3COO- + 2H3O+},$$
we can derive a equilibrium
\begin{align}
  \frac{[\ce{HCOO-}][\ce{CH3COO-}][\ce{H3O+}]^{2}}
    {[\ce{HCOOH}][\ce{CH3COOH}]}
  &= \frac{[\ce{HCOO-}][\ce{H3O+}]}{[\ce{HCOOH}]}
    \frac{[\ce{CH3COO-}][\ce{H3O+}]}{[\ce{CH3COOH}]}\\
  &= K_{z-\ce{HCOOH}}K_{z-\ce{CH3COOH}}.
\end{align}
I think we still need to know the original concentrations of $\ce{CH3COOH}$ and $\ce{HCOOH}$, we can calculate
$$[\ce{CH3COO-}] = \frac{[\ce{CH3COOH}]}{[\ce{H3O+}]_2} K_{z-\ce{CH3COOH}} = \pu{4.5000 * 10^-4 mol L-1}.$$
For $[\ce{HCOO-}]$ we get $\pu{1.3846 * 10^-3 mol L-1}$ (keeping some significance).
\begin{align}
  n_{\ce{HCOOH}} &= \pu{1.1385 * 10^-2 mol L-1}
    &&\text{(total)}\\
  n_{\ce{CH3COOH}} &= \pu{5.225 * 10^-3 mol L-1}
    && \text{(total)}
\end{align}
Solving for $y$ (neglecting the already present $\ce{H3O+}$ of water)
\begin{align}
  K_{z-\ce{HCOOH}} &= \frac{[\ce{HCOO-}][\ce{H3O+}]}{[\ce{HCOOH}]}\\
  &= \frac{x(x+y)}{(0.5n_{\ce{HCOOH}}-x)}
\end{align}
gives
$$y = -0.00018 - x + \frac{1.02465 \cdot 10^{-6}}{x},$$
using $y$ in
\begin{align}
  K_{z-\ce{CH3COOH}} &= \frac{[\ce{CH3COO-}][\ce{H3O+}]}{[\ce{CH3COOH}]}\\
  &= \frac{y(x+y)}{(0.5n_{\ce{CH3COOH}}-y)}
\end{align}
gives
$$x = 0.000884322 \vee x = 0.00690291$$
(there are more solutions, but these are the only real positive ones). From this we can calculate
$$y = 0.0000943623 \vee y = -0.00693447.$$
$y$ should be positive, thus $x = 0.000884322$ and $y=0.0000943623$. This gives us a $pH$ of
$$\mathrm{pH} = -\log(0.000884322+0.0000943623) \approx 3.01$$
Which seems to me as correct (it will be slightly more acidic than when adding the amounts of $\ce{H+}$).

Checking an online pH calculator, your answer seems correct. You have the right idea of combining both equilibria in a single calculation, though your solution seems a little more complex to read (it could be just me though!). There's a nice,organized and easily extendable way to solve any type of acid-base pH problem, based simply on charge and matter balance equations. If you're curious, check out some of my previous answers to other problems, e.g. here and here.

Another source with the mathematical description of these solutions is given here.

@NicolauSakerNeto That's good to hear. This is actually more simple than it looks like. You can extend this by solving more equations with more unknown. In this example we used x, y, but if I would be including $H_2O$ we would get x, y, x and an extra equation $K_w=[OH^{-}][H_3O^{+}]$. Thanks for the links, I'll look after it.

Also, it seems the links provided do essentially the same thing as me. Setting up equations and then substituting.

@NicolauSakerNeto Can you give the link of your online pH calculator?

I basically clicked on one of the first search results I got for "online pH calculator". Specifically, I used this, though a better calculator may well be out there.

@JoriI think : $[\ce{CH3COO^{-}}]={[\ce{H3O^{+}}]_2}= \pu{4.0 \cdot 10^{-4},mol,L^{-1}}$
and $\ce{[\ce{CH3COOH}]}=\pu{(0.01 - 4.0 \cdot 10^{-4})mol,L^{-1}}$ in  the following equation:

$$[\ce{CH3COO^{-}}]=\frac{[\ce{CH3COOH}]}{[\ce{H3O^{+}}]2}K{z-\ce{CH3COOH}}= \pu{4.5000 \cdot 10^{-4},mol,L^{-1}}$$

Correct me.thanks

I think you neglected to account for the dilution that happens when the two solutions are mixed. I think that explains the discrepancy between your result and other results.

Using concentrations and not activities, user GRSousaJr presents the exact solution to the problem in equations 1-8. However the simplifications don't really seem appropriate. This is a chemistry problem, not a problem in numerical analysis to solve. The gist is that we can make some simplifications based on the number of significant figures and a knowledge of the chemistry.
RANT - I wish book problems would use significant figures consistently. The concentrations are only given to 1 significant figure as are the pH values from which the pKa's can be calculated (rather than looked up). 2 or 3 significant figures would be appropriate, but 4 is definitely too many.
Chemistry - Since both solutions are around pH=3, the final pH must be around 3 too. So  $\ce{[H+]} \gg \ce{[OH-]}$ and the charge balance can be expressed as:
$$\ce{[H+] \approx [A-] + [B-]}\tag{6a}$$
that leads to:
$$\ce{[H+] =}\dfrac{C_{\mathrm{HA}}\cdot \mathrm{K_{a,HA}}}{\ce{[H+]} + \mathrm{K_{a,HA}}} + \dfrac{C_{\mathrm{HB}}\cdot \mathrm{K_{a,HB}}}{\ce{[H+]} + \mathrm{K_{a,HB}}}\tag{7a}$$
solving this leads to:
\begin{align}
\begin{split}
 &\ce{[H3O+]}^3\\
+&\ce{[H3O+]}^2 (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}\left(
  K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  - (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
    + C_\ce{HB} K_{\mathrm{a},(\ce{HB})})
  \right)\\
-&\left(
  (C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
   \right)\\
=&\ 0
\end{split}\tag{9a}
\end{align}
Since $\ce{[H+]} \approx 10^{-3}$, $\ce{[H+]^3}\approx 10^{-9}$
and $\left( (C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}   \right) \approx (0.01)* 2\cdot10^{-4} * 2\cdot 10^{-5} \approx 4\cdot10^{-11} $
Thus the term $\left( (C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}   \right)$ is on the edge of what can be neglected.
We know $C_\ce{HA} = C_\ce{HB} = 0.005$, and let's use $K_{\mathrm{a},(\ce{HA})}= 1.778\cdot10^{-4}$ and $K_{\mathrm{a},(\ce{HB})} = 1.754\cdot10^{-5}$ then:
$\ce{[H3O+]}^3 + 1.953\cdot10^{-4} * \ce{[H3O+]}^2 -9.734\cdot 10^{-7} * \ce{[H3O+]} - 3.119\cdot10^{-11} = 0 \tag{9b}$
Plugging the equation into Wolfram Alpha yields $\ce{[H3O+]} = 0.000910902 \ce{->[round]} 0.00091 $, $\mathrm{pH} = 3.0405 \ce{->[round]} 3.04 $
For yucks solving:
$\ce{[H3O+]}^2 + 1.953\cdot10^{-4} * \ce{[H3O+]} -9.734\cdot 10^{-7} =0 \tag{9c}$
yields $\ce{[H3O+]} = 0.000893781 \ce{->[round]} 0.00089 $, $\mathrm{pH} = 3.0487 \ce{->[round]} 3.05 $

Note - I took chemistry in high school and college from 1968-1973. At that time I only ran into a very very few problems that had a cubic and those could be solved by a couple of iterations. In my day I used 4 place logarithms or my slide rule. I didn't get a calculator until grad school about 1975. Not sure if using equation solvers in chemistry classes is allowed now or not.

How would be the equation and the ICE table, and what is the pH of the mixture of these two solutions?

There are three equations:
$$\ce{HCOOH <=> H+ + HCOO-}$$
$$\ce{CH3COOH <=> H+ + CH3COO-}$$
$$\ce{H2O <=> H+ + OH-}$$
Hydroxide is a (very) minor species, so you can neglect it. As I explain below, at the pH of the mixture, the acetate ion is also a minor species, so you can first neglect the dissociation of acetic acid, and then correct for it later.
I would not use an ICE table in this situation, at least not one that describes all three reactions.
My strategy to determine the pH of the mixture
First, you can determine the equilibrium constants for each acid from the pH values of their aqueous solutions (before mixing). The corresponding pKa values are 3.74 and 4.78 for formic and acetic acid, respectively.
After mixing, the concentration of total formic acid and total acetic acid drop by a factor of 2 (mutual dilution). Formic acid is the stronger acid, and would have a pH of 3.06 at this concentration in the absence of acetic acid (less acidic than pH = 2.9 because of dilution).
At pH 3.06, acetic acid dissociates very little (the difference between pH and pKa is more than 1.5, vs. less than 0.7 for formic acid). As a first approximation, we can say that the pH is determined by the stronger acid.
To figure out how much the presence of acetic acid does influence the pH, we can calculate how much it would dissociate at pH 3.06, and correct the pH accordingly. Our new pH will be more acidic, so we assumed a too acidic pH in our correction, overshooting a bit. After a couple of iterations of this, the pH comes out as 3.04.
So the biggest effect of adding the acetic acid comes from diluting the formic acid (pH would go from 2.9 to 3.06 just by adding water). A secondary effect comes from acetic acid dissociating a tiny bit, giving a pH of 3.04.
Here are results from the spreadsheet used for the calculations:

And here are the formulas used (to dampen the overshooting effect, the pH value used in each step is the average of the updated pH and the previous pH, not shown)

I tried to do $\mathrm{pH}=\frac{1}{2}\left(\mathrm{pH}_1+\mathrm{pH}_2\right)$, but it’s trivial and doesn't work.

To convince yourself once and for all that pH averaging does not work, consider these two extreme examples:

Mixing 5 mL of 1 M HCl with 6 mL of 1 M NaOH (starting pH values are 0 and 14, resulting pH is roughly 13).
Mixing 10 ml of 1 M HCl with 90 mL of water (starting pH values are 0 and 7, resulting pH is 1)

Considering significant figures I'd say that the calculation converged on the second line, certainly by the third. To get a pH as precise as 3.038443 you couldn't neglect the Kw of water.

I updated the significant figures (for an iterative solution, you need to show some guard figures to make sure there is not a slow creep to different values after 1000 steps; or you plug in your solution and make sure it matches the equilibrium constant to sufficient significant figures). @MaxW

Karsten - LOL - We're having a violent agreement... // Some problems would just require solving a cubic equation, but I never saw a book problem that did. At worst I've seen problems like this where a couple of iterations would yield a decent solution.

$-\log[\ce{H}]=\text{p(H)}$ you can calculate concentration of $[\ce{H}]$ for each acid, then add the $[\ce{H}]$ for two acids and calculate $\text{p(H)}$ by $-\log[\ce{H}]$.

I can't understand what are you trying to do, by the way $pH=-\log [H^+]$ doesn't works for weak acids.

pH=−log[H+] work for all acids.you have pH for tow acids then you calculate concentration [H+] by pH=-log[H+] and addition [H+] for tow acids then calculate pH by pH=-log[H+]

This is so obviously wrong that I comment just to prevent a newbie reader to be driven out.


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How do i detect the compounds a plant is made of, automatically?
Let's  say I want to take a random plant , and get the chemical formulas of  the compounds he's made of. Is there an automated method for doing so  and how costly is it(roughly) ?

HPLC-MS (high performance liquid chromatography- mass spectrometry), the latter with a suitable database for known small molecule plant components (alkaloids, terpenes, THC, etc.) would do it.  Cost would be three to low four figures/shot.  Saccharides and proteins are different and more expensive problems.
Consider what you are asking.  Mere pokeberry contains astragalin, canthomicrol, caryophyllene, esculentic acid, jaligonic acid, lectins, oleanolic acid, 3-oxo-30-carbomethoxy-23-norolean-12-en-28-oic acid, 6-palmityl-Δ7-stigmasterol-Δ-D-glucoside, 6-palmytityl-α-spinasteryl-6-D-glucoside, phytolaccagenic acid, phytolaccagenin, phytolaccatoxin, phytolaccosides A,B,C,D,E,F,G (esculentoside E), pokeweed mitogen, tannin, triterpene alcohols, α-spinasterol, and α-spinasteryl-β-D-glucoside and a whole bunch more.  One injection is hardly a beginning.  If you do not know how to cook it, it kills you.

Thanks.i didn't know it was important but i want a list of everything, not just known substances, and including proteins etc. Is it possible ? how at what cost ?. and cooking don't matter. Just thinking if it's possible to "catalog" everything.

You don't get "everything" as a practical matter. How would you characterize "everything" re celluloses and hemi-celluloses, lignins, tannins? You do care about cooking in the cited case, re "poke salad," http://southernspaces.org/2011/mess-poke and "seizures, coma and death."

In general, no it is not possible to catalog all of the compounds a plant is made of automatically. Uncle Al's answer is as close as you could get, but even that is far, far away from getting every single molecule (there would be tens of thousands, at least). HPLC-MS is good first step and would get you most of the small molecular weight substances.
For larger biomolecules, you would have to use gel electrophoresis, which is not very "automated."
We are a long way from cheap, automated separations of biological mixtures, but it is one of the major research goals of the pharmacology, molecular biology, and medical fields.
This link describes many of the currently available methods for protein purification, which is one of the most difficult problems in the field of separations.


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Does Nitrogen dioxide change colour if we put it in a constant temperature of e.g. 100 kelvin?
The changing of the colour of $\ce{NO2}$ is because of a change in temperature, resp. heating and cooling?

The changing color of $\ce{NO_2}$ is because it exits in equilibrium with $\ce{N_2O_4}$ as shown in the following equation $$\ce{2 NO_2 <=> N_2O_4} + \mathrm{58~ kJ/m}$$ $\ce{NO_2}$ is a reddish-brown colored gas, while the $\ce{N_2O_4}$ is colorless.  Considering Le Chatelier's principle,

Taking heat out of the equilibrium will push it to the right
Putting heat in, will push it to the left
Increasing pressure will push it to the right

As we shift the equilibrium we change the concentration of the components, and consequently the color of the gaseous mixture.
Back to your question, in the solid state (which is where it would be at 100°K), the equilibrium is shifted completely to the right.  So if we had a tube of gaseous $\ce{NO_2}$ and $\ce{N_2O_4}$ at room temperature it would have a distinct red-brown color.  If we then plunged it into a bath at 100°K, it would solidify to $\ce{N_2O_4}$ and appear colorless.

Does the reaction shift again to restore the concentrations of NO2 and N2O4 to the left while constant temperature?

Not sure I follow.  If the "constant temperature" is high enough that the substances are in the gas phase, then changing pressure will shift the equilibrium at constant temperature.  Changing the temperature from 100K to room temp will restore the equilibrium that exists in the gas phase at room temp.

no Chatelier's principle says if concentration changes the equilibrium shifts ,,
if the change in temperature changes concentration? does that mean the equilibrium is going to shift again because the concentration changed ..

The concentrations will change with temperature.  If a temperature is held constant long enough, then an equilibrium will be reached.  At each temperature there will be a distinct equilibrium constant - which means distinct concentrations of reactants and products.  At a given temperature, once equilibrium is achieved, the concentrations will not change again unless the equilibrium is disturbed (change in temperature, pressure, adding more reagent, etc.).


text:
Modelling planar molecules and their evolutions
I am trying to simulate a chain of molecules which are planar.
Can anyone please give me an example of the molecular chain which can stay as planar with it's minimum energy state is planar. A published study on the evolution of the geometry of the molecule will be highly helpful.

One-dimensional Fabre and Bechgaard salts, TTF-TCNQ,

http://www.chem.tamu.edu/rgroup/dunbar/Publications/2002-147.pdf
platinum blues
DOI:10.1002/asia.200700370
hexaphenylbenzene discotic liquid crystals
http://www.crcpress.com/product/isbn/9781439811436
"Chemistry of Discotic Liquid Crystals: From Monomers to Polymers," Sandeep Kumar (2010)
http://link.springer.com/chapter/10.1007/978-94-017-2319-0_16
middle
http://physics.aps.org/synopsis-for/10.1103/PhysRevB.82.180401
  Organic magnets
http://pubs.acs.org/cen/topstory/8201/8201notw8.html
Ladderanes

stacks
http://www.google.com/patents/WO2011017711A2?cl=en
http://www.halcyon.com/nanojbl/Images/ConPoly3.gif
and a central hexacoordinate silicon with oxygen bridges

I think you missed graphite.

"simulate chain of molecules" Graphite is 2-D planar not 1-D chain.  I think you missed h-BN, MgB2, zirconium phosphonates, smectite clays, micas, FeO(OH), FeCl2, heavy metal dichalcogenides...Langmuir-Blodgett depositions...DOI:10.1039/C001929A, http://chemistry.aalto.fi/en/research/groups/inorganic_chemistry/ etc.


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Stability of transition metals in aqueous solution

Which is more stable in aqueous solution: $\ce{Cr^3+}$ or $\ce{Mn^3+}$? Why?

My approach: $\ce{Cr^{3+}}$ should be more stable as the $\ce{3d}$ electrons will enter the $\mathrm{t_{2g}}$ orbitals. Due to CFSE (Crystal Field Stabilization Energy) of half filled $\mathrm{t_{2g}}$ orbitals $\ce{Cr^{3+}}$ should be most stable.

Both elements are stable under certain conditions in various oxidation states. In chromium they range from -2 to +6 and in manganese from 0 to +7.
However, some of them are more common than others, especially in aqueous solution. In chromium the most common are +3 and +6. As a "free" ion $\ce{Cr^{3+}}$ is most stable due to a filled s orbital and only one electron in the d shell. If you dissolve chromium(III)chloride in water you will get a complex formation.
$$\ce{CrCl3 + 6H2O <=> [Cr(H2O)6]^{3+} + 3Cl-}$$
Assuming an octahedral ligand alignment in crystal field theory, you are correct in assuming, that the $t_{2g}$ orbitals will be half filled, resulting in a more stable configuration.
For manganese the most common oxidation states are +2, +4 and +7. If you dissolve manganese(III)chloride, then you will also form a aqueous complex, which will slowly transform to manganese(III)hydroxide and precipitate.
$$\ce{MnCl3 + 6H2O <=> [Mn(H2O)6]^{3+} + 3Cl- <=> Mn(OH)3 v + 3H3+O + 3Cl-}$$
The resulting hydroxide may be oxidised by air to manganese(IV)oxide hydroxide.
$$\ce{2\overset{\color{\green}{+III}}{Mn}(OH)3 + \overset{\color{\green}{\pm0}}{O2} -> 2\overset{\color{\green}{+IV}}{Mn}O(OH)2 + 2{}^{-}OH}$$
Over time this will convert into manganese dioxide.
$$\ce{MnO(OH)2 -> MnO2 + H2O}$$

Is it possible to generalize this stability of transition metals in aqueous solution for the ions $\ce{Co^+, Fe^3+, Cr^3+, Sc^2+}$ as well? Or should I post a new question? (on the precondition that you answer it as well ;) )

@GaurangTandon Well, I think you'll be able to find stable "free" ions in aqueous solution for all of them, however, I'm not sure what you'd want to generalise here. I can give no guarantee that I would be able to write an answer to a follow-up question, that depends very much on my schedule and whether or not I can answer it quickly. However, this should not discourage you from seeking out the knowledge of the community.

Okay! I was more hoping to compare their relative stability and generalize that trend to all 3d metal ions. Though I ended up posting a narrower question first instead: Compare stability of transition metal cations in aqueous solution

Cr(III) is common as dirt and stable in solution as any number of different salts.  Mn(III) is exotic, labile, and chemically reactive,
http://www.hach.com/asset-get.download.jsa?id=7639984472
DOI: 10.1039/AN9911600285
http://en.wikipedia.org/wiki/Manganese(III)_acetate
DOI:10.1039/C3RA42480D
interesting free radical one-electron oxidation reagent


text:
Heat of combustion per mole and per kilogram
I need to calculate the heat of combustion per mole and per kilogram of complete combustion of, say, X.

I write reaction : $\ce{X + O2 -> ... + Q}$
Find Q using Hess law

Now, what do I do after that?

Q is the heat of combustion per mole. Next, you would use the formula weight to determine the heat of combustion per g (or kg).