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Intuition for electrode potential
If I connect a zinc half cell (Left hand side) and a copper half cell (right hand side), I get an E cell value of 1.1V taken from the voltmeter. If I replace the voltmeter with a light bulb, I will get light and heat energy of the bulb.
I learned that the above set up is equivalent to adding Zinc to copper sulphate solution. However, in this case, there is no wire for the electron to flow...
MY QUESTION:
1.Then, where does the energy go when I add Zinc to copper sulphate solution (that in the first scenario I converted to electrical energy)? Is it released as heat?
2.As the electrons flow from the Zinc half cell to the copper half cell through the light bulb, will the voltage between the electrodes decrease as the charges on the electrodes are supposed to decrease. Then what would the formula for total energy released be?
Please correct me if any of my concepts are wrong. Thanks in advance.

I might be able to help you with your first question more rigorously, if you are familiar with overpotential, chemical potential and elementary thermodynamics. Are you?

@SatwikPasani: I just know a bit on the topics

Ad 1) I am not sure what do you mean by "equivalent". But you add Zinc in a copper sulphate solution, the "energy" will be consumed through cementation.
to a recation Cu$^2$$^+$(aq) + Zn(s) → Cu(s) + Zn$^2$$^+$(aq)
Ad 2) Please see Equilibrium Constant of an Electrochemical Cell Reaction Example Problem where you can find ansewer to your second question.


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Werner's theory and complex salts
According to Werner's theory, metal ions have two types of valency-primary and secondary, where the primary valency is said to be satisfied by negative ions only, and  the secondary valency can be satisfied by positive ligand, negative or neutral molecule.
My question here is:
While formation of complex compounds, which of these valencies is satisfied?
What I'm not able to interpret from this is whether there is a 'possibility' for both the valencies being satisfied or is it necessary that both are satisfied?
Also if the later is true how is it possible? Can it be explained taking some example as reference.
I think I'm not able to interpret it and understand its details.

A certain number of ions, atoms or molecules closely associate around a central atom leading to the formation of distinct entity called coordination complex.
The groups atoms or molecules linked to the central atom are said to be coordinated with the latter and their number gives the coordination number.
The non-ionisable units lie in the primary zone or sphere, whereas the  ionisable  units lie in the secondary sphere and appear as ions in the solution.
In hexammine platinic chloride $\ce{[Pt(NH3)6]Cl4}$  and in potassium ferrocyanide $\ce{K4[Fe(CN)6]}$ four chlorine ions of the former and four potassium ions of the latter ionize in the aqueous solution, whereas six ammonia molecules of the former and six cyanogen groups of the latter belong to the primary sphere or non-ionizable sphere and thus fail to respond to their characteristic tests or fail to ionize. Or you could also say that the chlorine ions in the former and the potassium ions in the latter occupy the primary valencies and hence are ionizable and the ammonia and cyanogen group occupy the secondary valencies  and are non-ionizable, while the $\ce{Fe}$ atom in the latter is present in the primary Sphere and thus fails to ionize.
As per your question during the formation of a compound say $\ce{[Pt(NH3)5Cl]Cl3}$, five molecules of ammonia and the chlorine ion (linked by secondary valencies) are present in the primary Sphere and the rest of the three chlorine atoms lie in the secondary sphere. Thus the coordination number of the compound would be  six as five ammonia molecules and one chlorine atom are attached to the central metal atom.
Now, every element tends to satisfy both its primary and secondary valencies. A negative ion tends to satisfy both these valencies i.e. primary as well as the secondary valencies.
The presence of a negative ion in the coordination sphere reduces the amount of charge on the complex ion by the amount of charge present on it while the negative ions in the primary sphere or the coordination sphere are not ionized.

Thanks this was a great help. But I still have a doubt and that is: according to werner's theory primary valency can be satisfied only by negative ions, but in many cases just as in potassium ferrocyanide, it is satisfied by a positive ion, isn't it? Then how is this explained? Also the last part of your answer would be true only if the complex ion would be carrying a negative charge, right? Please correct me if I'm wrong.

Primary Valency: It is the no. of charges present on the complex ion.Usually a complex ion possess positive charge.
And thus negative ions satisfy this charge. But as in the cases of potassium ferrocyanide the the charge on the coordination sphere is negative owing to the -1 charge of each CN ion. Here the charge on iron would be +2 and thus the complex as a whole would posses -4 charge which is balanced by four potassium ions.

Werner's theory proposed by alfred werner  in 1898.He was awarded by noble prize in 1913. It explains about the formation of coordinate complex compounds, bonding of coordinate complex compounds, stability of coordinate complex compounds and isomerism of coordinate complex compounds.
According to Werners theory each coordinate complex compound exhibits two types of valencies.The are a) primary valency and b) secondary valency.
 Primary valencies are satisfied by only negative charge and it is equal to oxidation state of central metal ion.
Secondary valencies are satisfied by either negative charge or neutral species and it is equal to the coordinate number.
Primary valencies are denoted by dotted lines and ionisible.
Secondary valenices are denoted by dark or straight lines and non ionisible.
In coordinate complex compound formed  bonds are coordinate covalent bonds.
In coordinate complex the central metal atoms acts as the Lewis acid and the ligand acts as the Lewis base.
If coordinate complex shows six ligand in it then the structure is octahydral and four ligand in it then the structure is tetrahydral.

Well... while almost all six-coordinate complexes are octahedral, not all four-coordinate complexes are tetrahedral. Many are square planar.

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Belousov-Zhabotinsky reaction
The Belousov-Zhabotinsky reaction is a very fascinating one. The Chemistry behind the reaction concerns thermodynamics and equilibrium. I want to perform the experiment, but I'm not able to find detailed instructions about the reaction. Like, proportions of the chemicals, explanation of the reaction, duration of the reaction, sequence of mixing the reagents, etc.
The Belousov-Zhabotinsky reaction is a video I found on YouTube.
But that does not give any details at all. Can anyone give me a detailed reaction procedure?

That is indeed a fascinating reaction!  The following is a part answer, providing links to some resources that may be of use.
The following resources may be of assistance:
The Belousov-Zhabotinsky page from the Kansas Academy for Mathematics and Science Education provides a bit about the background, chemical processes and most relevant to your request, an experimental procedure.
Mary Newton of the Georgia Regent University has provided a reasonably detailed procedure on the webpage Procedure for B-Z reaction.
Slightly different, is the information provided in the Belousov-Zhabotinsky Reaction (Leigh) page from the Flathead Valley Community College takes a different approach, with a focus on the mathematical derivations relevant to the experiment.

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Mercury assisted deprotection of dithiane
1,3 dithianes are used in umpolung reactions and can be deprotected by mercury containing compounds - but could anyone help explain what the exact mechanism is? Many texts seem to skim over this by explaining it is similar to acetal hydrolysis, but I'd really like a more detailed explanation as to the role of the compounds. A reaction I have seen involves $\ce{HgO}$, $\ce{HgCl2}$, $\ce{CH3CN}$ and $\ce{H2O}$. If it is similar to acetal hydrolysis, the $\ce{H2O}$ is probably to complete the final step of attacking the $\ce{S=H}$ bond, but what about the initial step?
I am assuming one sulfur atom forms a bond to the $\ce{Hg}$, but which Hg compound would it be in this case, the $\ce{HgO}$ or the $\ce{HgCl2}$? What is the role of the $\ce{CH3CN}$ in this case? Thanks!

$\ce{Hg^2+}$ has a high affintity to thiols and other sulfur functional groups, and therefore $\ce{HgCl2}$ forms a Lewis acid-base adduct with the dithiane. The cyclic sulfonium cation undergoes ring opening, and after nucleophilic attack by water, a second Hg-S bond under ring closure is formed. Deprotonation finally yields the deprotected carbonyl compund and the cyclic mercury dithiolate.

Ok, I'm speculating a bit here: The mixture of acetonitrile and water is probably just used to keep all components in solution. HgO serves as an oxidant; alone it might be able to convert the 1,3-dithiane to the corresponding 1-oxide, which doesn't necessarily results in its cleavage. $\ce{ Hg^{2+}}$ on the other hand has a high affinity to (di)thiols.


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Why is buffer capacity at a maximum when the ratio of its components is 1?
A buffer consists of a weak acid and its salt or weak base and its salt.
When the ratio of weak acid and its salt in a buffer (or the ration of weak base and its salt) is equal to 1, we say that the buffer capacity is maximum. How does this tell us that buffer capacity is maximum?

This wikipedia page, and this page are good places to learn this. It addresses all elementary questions about buffer capacity.

@SatwikPasani Feel like you could give an overview of what is in the links and provide an answer?

With a buffer, you're trying to maximize your ability to neutralize acid or base. The log dependence of the pH on the ratio of interest in the Henderson-Hasselbalch equation is why the buffer works well:
$$\mathrm{pH} = \mathrm{pK}_{a} + \log\frac{\ce{[A-]}}{\ce{[HA]}}$$
Now, your ability to neutralize acid is determined by the amount of the conjugate base $\ce{A-}$. Your ability to neutralize base is determined by the amount of the acid $\ce{HA}$. When the ratio between these is 1, you have equal ability to neutralize either. If you're skewed too far one way or the other, then you will not have much of the respective species around to neutralize added acid or base.

Shouldn't the ability to neutralize base is determined by by amount of  $H^+$.Why have you written "ability to neutralize base is determined by the amount of the acid $HA$"?

Everything is in equilibrium, but you can still only provide as much proton as there is acid. @pranjalverma


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What are the limits of analytical chemistry?
As I understand it, the branch of chemistry dealing with identifying substances present in a sample is called "analytical chemistry". This branch looks vast and complicated, with lots and lots of individual techniques applicable in specific circumstances. I'm hardly even a beginner in chemistry, so it's very difficult to get a good grasp of what can be done with current technology.
So, in general, how well can one identify the substances in a completely unknown sample? Does one need to have at least a vague idea of what's in it?
Differently put, let's say I give you a container with a mixture of ten randomly chosen chemicals from the entire CAS Registry, and suppose they didn't react with each other. How difficult would it be to identify all ten of them?

The trouble with your question is the use of the phrase in general. Most analytical chemistry is directed towards specific goals and isn't about trying to find what is in a sample in general.
This means that the range of techniques applied are highly dependent on the problem you want to solve. If you want to detect, for example, pesticide residues in foodstuffs you probably already know the list of things you want to detect and the range of techniques appropriate to identifying them. If you seek the active ingredient of a herbal remedy, a different range of techniques apply. Sometimes you want to verify that a pure substance you have made is what you think it is, this involves a different range of techniques again.
One common thread in much analytical chemistry is separation: if you seek to identify something in a mixture you want to separate the mixture into its different components. So a variety of types of chromatography may be used in the bulk or on a micro scale. Identifying very small amounts of contaminants in food (e.g. pesticides) probably involves some form of extraction into a solvent followed by separation using analytical chromatography (gas or high pressure liquid chromatography (HPLC)). Where the range of compounds is not known in advance this can be supplemented by mass spectroscopy (the chromatography separates the components, the mass spectrograph helps identify the fragments of the molecules in each component).
If you don't know what you are looking for in advance, it isn't easy to elect the right separation technique and a great deal of experimentation is likely to be required.
Another consideration is that some things are much easier to detect than others. Chromatographic and detection tools like the electron capture detector can identify organohalides to below parts per trillion levels. Atomic emission spectroscopy can identify some metals easily and at low levels. Other techniques can help identify structure; NMR is (sometimes) good at identifying structures, but usually needs pure compounds in relatively large quantities. X-ray techniques can be useful for minerals. Some compounds have very distinct features in their IR spectra or their UV absorption. And so on.
The bottom line is there are no general techniques. There are a wide range of tools to apply depending on the circumstances and the problem you are trying to solve. Almost everything starts with separation. Then, for simple problems, you may be able to apply a single known technique to find out what you want. For harder problems, a rage of tools may be required but the tools will be fairly specific to the problem.

I would prefer questions that are more specific, rather than broad. Nevertheless.
It depends. I think you would first need to separate the compounds from each other.You could try    chromatography (I believe that some exotic chromatographs can even separate enantiomers). You could then try to use NMR to determine its connectivity. You could also use atomic emission spectroscopy along with mass spectroscopy.
How much of the sample you have is important as well. Many of the analytical tools destroy the sample. In addition, not knowing how many compounds are present in a mixture makes the process more difficult since you wouldn't be able to tell whether the separation process was effective.
Regardless, it would be awfully expensive to do so.


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What's the reddish powder that oozes out of a black rock?
As an artist, some of the work I do involves the use of seashells, sea rocks/stones, etc.
I picked up a black rock from shore at one of the beaches here in Florida and it looked somewhat clumpy with other rocks and debris encrusted on it. As I do with everything I collect, I thoroughly washed all the rocks with fresh water and then placed them in a bucket filled with water and a good dose of bleach.
After drying the rocks, I noticed that the black one had small, reddish, powdery pimples all over it. It looked odd, so I broke the rock in half with a hammer,but there was nothing red inside. It actually looked like carbon, but the rock isn't brittle or powdery. I stuck the two halves back in a bucket, filled it half way with water, and them put more bleach in it. The water turned dark, almost muddy, and a reddish, almost gelatinous precipitate appeared and it was quite a bit of it. Any idea what the heck is in that rock that can cause that? I didn't notice any odor other than the bleach, albeit the rock did smell a  bit sulfury. Any idea what I have soaking in the bucket?

Is your rock magnetic?  It might not be, but what you are reporting sounds like iron oxide (i.e. rust), and so your rock could be an iron ore like taconite, or even magnetite.  Such an ore could look like carbon.  Even if not magnetic, it could still be an iron ore.

Thank you for your response. The rock is not magnetic nor does it have a metallic look or feel to it. Is bleach reactive with iron to form a reddish precipitate? The gelatinous looking precipitate looks quite intimidating. I may actually filter it out and see if a local chem teacher can check it out. Thanks again.

Iron ores do not appear metallic, they look like rock. I agree with your intuition to further probe the gelatinous participate.  A quick check indicates that there may not be many stable transition metal hypochlorites, but that's no guarantee as to the composition of your participate.  So it is still possible you have an iron ore, as the reddish participate does sound like the oxide.  Do you have access to hydrochloric adic (muriatic acid)? If reacting the acid with the red substance produces a yellow substance, that is a strong indication that your cation is iron.

I will try to run the test with the acid. Think I may have to pay a visit to the local high school and see if the chem teacher can spare a couple of drops of HCL. I will post on here what I have once I figure it out. Thanks for the qualitative analysis suggestion.

I will be simpler to identify the rock, rather than the precipitate. It sounds like iron ore, according to your experiment. Check the ores mentioned at http://en.wikipedia.org/wiki/Iron_ore

Your rock most likely had minerals with Fe(II) in them. These minerals are usually dark coloured, so I wouldn't be surprised that the rock was black. One famous example is basalt - a black rock with lots of Fe(II) in it. Here's a picture of a basalt-like rock:

Some of these minerals are olivine ($\ce{(Mg,Fe)2SiO4}$), pyroxene ($\ce{Ca(Mg,Fe)Si2O6}$) and amphibole ($\ce{Ca2(Mg,Fe)5Si8O22(OH)2}$). You also state that the rock smells a bit sulfury. This is a bit hint that some of the minerals you had in the rock are sulfides such as pentlantide ($\ce{(Fe,Ni)9S8}$), pyrrhotite (ideally $\ce{FeS}$) or chalcopyrite ($\ce{CuFeS2}$). By adding bleach (an oxidiser), you're basically taking this Fe(II) and oxidising it to Fe(III), which binds with abundant water in the bucket to form a compound commonly known as rust. The sulfur is then oxidised to gaseous $\ce{SO2}$ that gives that odour.
You're basically making a small scale version of a very serious environmental problem, acid mine drainage:

source: Jack Pearce at flickr


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How many electrons can each shell in an atom hold?
I know that the first shell can hold up to 2, the second and third can hold up to 8, and the fourth 18. But what about after that?

Remember that search engines are a powerful tool. Searching for electrons per shell in Google yields the following section on Wikipedia among the first results.

@KarstenTheis you're right. I was improving a suggested edit and must have had a mental lapse.

$$\text{Electron Capacity for a shell} = 2n^2$$
where, n represents the Principal Quantum Number, a.k.a. energy level.
Keep in mind that an energy level need not be completely filled before electrons begin to fill the next level because beyond the fourth energy level, you would get into theoretical subshells.
See this.

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Types of radioactive decay
Besides alpha, beta, and gamma, are there any other types of radioactive decay? If so, what are they? Is there any type of radioactive decay that is more powerful than gamma?

Here's a list of types of radioactive decay. The most notable types of decay that are not among the classic three involve the direct emission of a free proton or a neutron, the emission of atomic clusters other than helium nuclei (alpha particles), the absorption of one of the innermost shell electrons into the nucleus, or spontaneous fission of the unstable nucleus.
Regarding the question about gamma ray energies, it is first important to notice that "gamma ray" is a generic term that refers to any electromagnetic energy (photons) coming from a nucleus or subatomic particle decays, and has no direct reference to the energy of the photon (though in the past it did). For example, a nuclear excited state of thorium-229, $^{229m}\ce{Th}$, decays through the emission of gamma rays with an energy of 7.6 eV, which corresponds to photons in the ultraviolet region of the electromagnetic spectrum.
Also, the answer depends on exactly what you mean by "power". Assuming you're comparing kinetic energies, Wikipedia says gamma rays from nuclear decay rarely have energy above 10 MeV, and another source says electrons from beta decay usually have energies up to 4 MeV. The energies are all of the same order of magnitude, which makes sense because all the particles are coming from the same source (an unstable nucleus). Theoretically gamma rays can have a slightly higher kinetic energy than other decay particles because photons have zero rest mass, therefore all the decay energy is converted into kinetic energy, whilst for electrons 511 keV is "wasted" on creating the rest mass (and even higher "wastes" for heavier particles such as protons/neutrons/alpha particles etc). I think one can also consider that some energy is used up in "climbing out" of the electrical potential and residual strong nuclear potential wells of the nucleus, neither of which applies to gamma ray photons (though positively charged species would in fact gain energy, being electrically accelerated away from the nucleus by repulsion, but this contribution is probably smaller than the nuclear force one). However, different types of nucleus have different allowed decay paths, and depending on complicated details the most energetic beta particles or proton/neutron emissions might have a slightly higher energy than the most energetic nuclear decay gamma ray.
As a final note, it is also possible to create even more energetic photons than any that could be produced via nuclear decay by putting electrons in a sufficiently powerful particle accelerator. Any charged particle under acceleration/deceleration loses some energy in the form of electromagnetic radiation. For ultrarelativistic charged particles, the photons emitted can be of extremely high energy. These are a type of x-ray termed "hard x-rays" (more energetic than so called "soft x-rays").
Edit: Correction regarding energy lost in rest masses of particles from decay. Energy is used up in the creation of electrons, but protons/neutrons/clusters already exist inside the nucleus, and so are not created. These particles could never be produced directly as a single one would have upwards of 930 MeV, far above typical nuclear decay energies.


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Why does ammonia have sp hybridization?
Apparently it has $sp^3$ hybridization, but I don't understand why. Ammonia ($\ce{NH3}$) seems to me to not require $sp$ hybridization because all of its bond lengths are already equal. It has 3 hydrogens bonded to the $p$ orbitals. Why can't the lone electron pair in nitrogen's $2s$ shell stay put where it is?

https://socratic.org/questions/why-does-ammonia-have-4-shells-that-are-sp3-hybridized-rather-than-having-the-s-#:~:text=The%20other%20reason%20why%20hybridization,are%20not%20the%20same%20orbital.

Well, I can think of these reasons that can justify the hybridization in ammonia molecule.

Bond angle: If the molecule did not have hybrid orbits, and instead had unhybridized p-orbitals taking part in the bond formation, then the bond angle between the orbitals would be 90 degrees. And as for the real situation the bond angle is nearly 107 which makes the molecule more stable, decreasing the bond pair-bond pair and bond pair-lone pair repulsion.
Energy: Recalling the definition of hybridization, it is the mixing of the atomic orbitals having slightly different energies to form new orbitals that have equal energies. This stabilizes the molecule. As for ammonia molecule due to hybridization the energies of the lone pair of electrons and the bond pair of electrons becomes almost equal, thereby increasing the stability of the molecule.
Geometry: Thinking about the spatial arrangement of the atoms in the molecule of ammonia if there is no hybridization in the molecule then the size of the orbital containing lone pair of electrons would be different from that of the orbitals containing bond pair of electrons. Also if hybridization would not be taking place, explaining the geometry of the molecule, that is it's trigonal pyramidal shape, would not be possible.

I'll mark this as the answer, I found the answer in the meantime after a lot of looking around. It seems that all atoms with 4 or more valence electrons will form sp3 hybrid orbitals (sp2 or sp if there is a double bond or triple bond respectively). I'm still not sure exactly why the s orbital can't stay as an s orbital, I guess because it's needed to shift everything to an sp so that the electrons as a group can be farther away from each other. In any case, thanks for a great answer.

This question has been answered before. Look here: http://answers.yahoo.com/question/index?qid=20130803221208AAYOCBr
If you look at the MO diagram for $\ce{NH_3}$ you'll find that the orbital which the lone pair resides in is higher in energy (in the $2s$ antibonding orbital http://www.d.umn.edu/~pkiprof/ChemWebV2/Bonding/MO-ammonia/index.html). Also, if I recall well, hybridization is not entirely correct. In hybridization, you typically assume that all of the hybridized molecular orbitals are degenerate. This is not always the case due to the different types overlap of the orbitals needed to create the molecular orbitals. Some hybridized molecular orbitals (even the ones of the same "type" such as $sp^3$ or $sp^2d$, etc) are higher in energy even though all of them are bonding molecular orbitals. You'll learn more of this in the second semester of inorganic chemistry.  I think , ammonia is not really $sp^3$ due to the poor overlap (methane has a better overlap between the $s$ orbitals and $p$ oribitals). Take it with a grain of salt. It has been a while since I took anything related to MO's.

I appreciate the help, this question was really giving me difficulty. I found out that Ammonia does in fact have sp3 hybridized orbitals. It seems that atoms with 4 or more valence electrons usually have sp3 hybridization. It allows the electron shape to become tetrahedral so the electrons can be farther apart. It also accounts for the bond angles in NH3 of 107 (slightly different than normal tetrahedral because of the lone pair of electrons). Cheers.

sp3 hybridization does NOT lead to bond angles of 107. Good sp3 hybridization leads to bond angles of 109.5 degrees.

It does in this case because of the lone electron pair. Really it does.

http://en.wikipedia.org/wiki/Ammonia#Structure

The bond angles are an indication of the type of hybridization. In this case, we say that the hybridization is sp3 since it resembles sp3. Ammonia has a poorer orbital overlap compared to methane. Also, MO theory and VSPER theory are not the same. They complement each other, but there are subtleties to it. If you are interested in MO theory, I recommend DekocK and Gray's "Chemical Structure and Bonding"


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Is aromaticity of the 4 aromatic amino acids affected by pH?
During a discussion of NMR, my lecturer told me to think about which amino acids contain protons in an aromatic ring, and whether any are likely to be affected by changes in pH.
The 4 aromatic amino acids are His, Phe, Tyr and Trp. I've marked Histidine as such in my notes, but can't remember why I thought this at the time. Contrary to this, Wikipedia states explicitly that “the imidazole ring of histidine is aromatic at all pH values”.
I also marked Histidine as "+ve" (as in, positive) but I'm stumped as to what I meant by it
I can't find any information on the others, but am guessing perhaps the fact that these have phenyl (rather than imidazole) rings means it's possible they respond differently to changing pH?
Any advice or hints would be appreciated, as I can't find any further information online and nothing in the structures jumps out at me to suggest an answer.

There are two nitrogen atoms in the imidazole ring in histidine, but the $sp^2$ nitrogen is significantly more basic than the $sp^3$ nitrogen, as the electron pair in the latter is delocalized over the ring. Protonating the $sp^3$ nitrogen would destroy the aromaticity, but that would require very strongly acidic conditions, possibly not attainable in aqueous solution. Protonating the $sp^2$ nitrogen does not affect aromaticity because the electron pair is approximately perpendicular to the plane of the ring, meaning it cannot overlap with other orbitals in the ring and therefore does not participate in mesomerism.
Tryptophan also has a $sp^3$ nitrogen atom in an aromatic ring, but like in the imidazole ring it is not very basic.

In proteins, amino acid residues of histidine have pKas near 6 and tyrosine side chains near 10. The protonated form of histidine is positive, and there can be a substantial fraction protonated at physiological pH in a given protein depending upon the exact pK value.
NMR has been used to determine the pKas of particular residues in particular proteins, with interest in deviation from the usual values. In metalloenzymes, the oxidation state of a metal ion can be coupled to the protonation state a side chain.
This is probably what the lecturer wanted you to think about, which pKas would be interesting to try to observe by NMR? not which ones change aromaticity necessarily.


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Does using stonger soaps help me lose weight?
Soaps are amphiphilic molecules that, practically, removes body oils from the surface of the skin. The body, on the other hand, always replenishes sebum through the sebaceous glands after a certain amount of time.
Applying Le Chatelier's principle to this complex chemical equilibrium, where the equilibrium shifts towards the right hand side, where oil is being extracted from the equation, essentially so:
(body · oil)  ⇌  body + oil       # RHS preferred

Because the process of sebum regeneration must be endothermic, this means the body is losing energy in this overall process:
body + (lipids + heat)  ⇌  body + oil  ⇌  (body · oil)

My questions are:

Are my assumptions correct?
Does removing oil from the body more readily, by means of a stronger detergent/soap, reduce body mass?
Does the temperature of the water bath used in the experiment affect the rate of sebum dissociation?

Haha, I have to say this is some original thinking, I like it! I suspect this won't happen because bodily functions tend to have strict kinetic controls (both reactional and diffusional) alongside thermodynamic ones, and so your "equilibrium" wouldn't shift so freely; perhaps sebacious glands are always working at essentially full throttle. If you don't find a satisfactory answer here, perhaps the question can be moved to Biology.SE, as the mechanism by which sebum is produced is likely important.

You may be correct in principle but wrong that the amount could be significant. The issue isn't whether this effect happens but whether it could possibly be large enough to be even measurable in the real world where you weight fluctuates significantly hour by hour according to what you eat and the amount you sweat or excrete.

@matt_black Oh no, I am in no way concerned about its efficacy or practicality (it ruins your skin, for example), and do not encourage anyone bathing in laundry detergent -- just whether or not my thought experiment is on the right track.

I find your assumption that the body will just get some fat out of existing storage to use in sebum production highly amusing. The amount of fat in fatty tissue is regulated by complex hormonal systems and is not just a "dumb storage" for excess calories. For all we know, it could be that the need for fat on the skin kicks the fat-demanding hormones into higher gear, creating more fat than before. So without knowing how fat is made (and I doubt the relationship you mentioned has been studied yet), any relation is possible.

I'm afraid no: you are applying a chemical principle to a complex biological response.
The Le Chatelier principle state:

If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or partial pressure, then the equilibrium shifts to counteract the imposed change and a new equilibrium is established.

We start with the definition of chemical system at equilibrium :

the temperature dependent condition in which chemical species are present at concentrations which have no further tendency to change with time.

So what you need are chemical species: molecules, ions. Although fatty acids could be a product "body" is not a compound or a molecule. So what you have is not a chemical system but a biological one. You could go a little more close if you think something like this:
$$\ce{Fatty acid <=> Chemical  \space Precursor_{Y} + Chemical \space Percursor_{X}}$$
However even this is a wrong oversimplification of many complex biochemistry reactions and equilibria.
Moreover you can't interact with these equilibria, perturbing them and so applying LC principle because sebum, which is a complex mixture of lipids, is stored inside sebaceous glands and is formed by the complete disintegration of the cell it self. So to from new sebum you have to from new sebaceous glands.
Soap simply remove sebum not seabaceous glands, so using stronger soap by the way don't affect sebum production. At this point is clear that is no more chemistry but physiology. In this case an equivalent of Le Chatelier's principle is the homeostasis. However from what I know the diet is one of the main factors that induce sebaceous glands to produce sebum so eat better seems to be the best way to lose weight and sebum!

References
Pub Med Sebaceous gland lipids


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The weedkiller paraquat can be produced from methyl-pyridinium chloride using a cyanide catalyst: what is the mechanism?
Paraquat is 4,4'dipyridinium chloride. There are two common ways to synthesise it starting with pyridine.
In one reaction that can be done on a large scale in water or liquid ammonia, methyl-pyridinium chloride is reacted with a base and catalytic amounts of sodium cyanide in the solvent to give a dimeric intermediate that oxidises easily into paraquat.
What is the chemical mechanism of the dimerisation reaction?

While the mechanism is not 100% certain (or wasn't when I worked on it some time ago), the current best guess is that it is similar to the well known Benzoin condensation:

This is based on the following intermediate:

Applied to the paraquat conditions this might go like this:

The neutral quinone-like intermediate crystallises out of solution and can be separated and then oxidised (it is very oxygen sensitive) to the desired paraquat product.
The details are partially guesswork, though, as it isn't the easiest reaction to investigate mechanistically.

I remember PET (photoinduced electron transfer) works by the Mariano group in which N-dimethylallyl pyridinium perchlorate underwent intramolecular radical cyclizations upon uv irradiation and in fact, I've seen N-methyl chinolinium salts undergo dimerization under these conditions.
Imagine that the N-methylpyridinium cation is irradiated at $\lambda$ = 300 nm. The resulting $S_1$ state oxidizes cyanide to a cyanogen radical.
The N-methylpyridinium cation is thereby reduced to a pyridyl radical.
Radical dimerization furnishes a 4-4'-dihydrobiaryl (see drawing).

Aromatization to paraquat requires the loss of two hydrogen atoms and two electrons; the latter are being used to regenerate cyanide.
The mechanism is a bit speculative but consistent and reflects the catalytic use of cyanide.

Note: I would have liked to add a sketch for the mechanism but BKChem died on me several time while doing so.

I have no idea whether that is possible, but it isn't what happens here as the cyanide is used in catalytic amounts (certainly much less than the pyridinium salt). I've clarified the question to include this fact.

I think the proposed mechanism is (now) compatible with a catalytic use of cyanide.

Doesn't your mechanism require an oxidant?

Yes, it does. The excited singlet state of the N-methylpyridinium cation with its low-lying SOMO is a much stronger oxidant for the cyanide than its ground state.


text:
how to increase Keq(equilibrium) with temperatrue
$$\ce{C_{(s)} + H_2O_{(g)} <=> CO_{(g)} + H_{2(g)}}$$
$$\Delta H = +131\ \mathrm{kJ}$$
If the reaction is in a state of equilibrium under standard conditions, what can be done to increase the value of $K_\text{eq}$?
My question is would you have to increase or decrease the temperature to increase $K_\text{eq}$?
I know the basic principle that you have to decrease the temperature to increase $K_\text{eq}$, but in my notes that was for $\ce{A +B<=>C}$ and the reaction above is not in that form, Instead it has two products. Would this still apply?

For exothermic reactions, you decrease the temperature to increase $K$ and vice versa. This holds good in all stoichiometries.

You can easily see how the equilibrium constant varies with temperature using the Van’t Hoff equation:
$$\frac{\mathrm{d}\ln{K_\text{eq}}}{\mathrm{d}T}=\frac{\Delta H^\circ}{R{T^2}}$$
Rearranging the equation a little bit, we get that:
$$\mathrm{d}\ln K_\text{eq} = \frac{\Delta H^\circ}{RT^2}\mathrm{d}T$$
Now we can integrate each side of this equation:
 $$\int_{K_1}^{K_2} \mathrm{d}\ln K = \int_{T_1}^{T_2} \frac{\Delta H^\circ}{RT^2}\mathrm{d}T$$
Because $\Delta H^\circ$ and $R$ are constants, this will be equivalent to :
$$\int_{K_1}^{K_2} \mathrm{d}\ln K = \frac{\Delta H^\circ}{R} \int_{T_1}^{T_2} \frac{1}{T^2}\mathrm{d}T$$
Solving the integral we will get this :
$$\ln\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$
For $\Delta H^\circ > 0$ we have two cases: when $T_2>T_1$ and $T_2<T_1$.
If $T_2>T_1$ then $\frac{1}{T_2}<\frac{1}{T_1}$ and hence $\frac{1}{T_2} - \frac{1}{T_1} < 0$ . Knowing that $\Delta H^\circ>0$, this means that $\frac{-\Delta H^\circ}{R}$ will also be negative, and the product of these 2 will be positive:
 $$\frac{-\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)>0$$
And from the Van’t Hoff equation we get that : $$\ln\left(\frac{K_2}{K_1}\right)>0 \Leftrightarrow \frac{K_2}{K_1} > 1 \Leftrightarrow K_2 > K_1$$
i.e. The equilibrium constant increases as the temperature is increased.
Using the same principle for $T_2 < T_1$, we get that $K_2 < K_1$

As we know:
$$K=Ae^{-\Delta H/RT}$$
Now as this is a monotonously increasing function in the variable T for the positive value of $\Delta H$. So you need to increase temperature to increase the equilibrium constant.


text:
Equilibrium constant calculation from concentrations
Consider the following equilibrium: $\ce{A2\ _{(g)} + 2B_{(g)} <=> 2AB_{(g)}}$
A chemist puts 8.0 moles of $\ce{A2}$ and 10.0 moles of $\ce{B}$ into an evacuated 1.0 L container. The system reaches equilibrium at a certain temp according to the process outlined above. When the system is at equilibrium the chemist determines $\ce{[AB]}$ is 8.0 M. What is the $K_c$ for this process at this temperature (278 K). I set up a nice chart and everything and I cant figure this out. My answer choices are 10, 3.0, 1.0, 4.0, and .50
Initially I thought that The Kc was [AB]^2/([A2][B^2]) but I put in the values and didn't get any of the answers...
Then i set up an ice chart
It's hard to show on here but I then ended up with [8+2x]^2/([8-x][10-2x]^2) but that didn't really get me anywhere

It would probably be helpful to include the chart in your question so that those answering can point out where you were incorrect.

Well, I haven't done that in a while, but we seem to agree that the equilibrium constant is expressed as
$$ K_c = \frac{[AB]^2}{[A_2][B]^2}$$
What else do we know?

Initially, $[A_2]$ = 8.0M, $[B]$ = 10.0M
In equilibrium, $[AB]$ = 8.0M
For every mole of $B$, one mole of $AB$ is produced.
For every mole of $A_2$, two moles of $AB$ are produced.

With just one cup of coffee so far, I get
$$ K_c = \frac{8^2}{(8 - \frac{8}{2})(10-8)^2} = 4$$

OK! I didn't know you had to subtract 8/2 so there's my problem. Thank you!

You're welcome! If the formation of AB would consume A2, this wouldn't be chemistry but magic ;)


text:
How to get defect-energy level diagram from PL data
I am reading this article Phys. Chem. Chem. Phys. , 2011, 13, 14902-14905 and seen that the authors have drawn a image of defect level using PL data

Can you explain me how this diagram can be drawn from PL data. And if possible can you explain this using some other simplest PL data.


text:
Electron Affinity trends
Imagine an arrangement of atomic orbitals in an alternate universe, in which the s subshell contains 2 orbitals instead of 1, and the p subshell contains 4 orbitals rather than 3. In which groups would you expect the anomalously low electron affinities to occur?
My thoughts are groups 3 and 7, due to the fact that they are "one away" from filling up their subshell. The groups would look like:
s: _ _ _ _  ... p: _ _ _ _ _ _ _ _
However, the answer says 4 and 8. Is there something I'm missing?
Edit: in terms of electron configuration blanks, it would be
s: _ _ ... p: _ _ _ _

I am confused. Are you assuming that the energy ordering is the same? In other words, can you assume that it is 1s2s2p3s3p4s3d...etc?

Yes, but there's only one "shell" in this question. The order would still be s then p so ex. 2s4 2p7 for group 11.

After rediscovering my own question, I'm posting it as an answer for anyone else to reference or to correct.

Electron affinity (EA) refers to "the energy released when an atom in the gas phase accepts an electron." (Burdge, Atoms First, p.118)
Lower EA means that the process of accepting an electron is not energetically favorable. In other words, lower EA = less likely to accept an electron.

Groups 4 and 8, in this case, already have filled subshells. Adding an electron would place it in a new subshell that is higher in energy. The EA would be lower for these groups because of the "reluctance" of the atom to accept an additional electron.
Groups 3 and 7 have the anomalously high EAs, which might have been where the confusion was.

Since the periodic table (only maingroups) would then contain of 12 Groups, you are forgetting 2 (half s) and 12 (full p).


text:
Nature of bond in oxyacids of phosphorus
What is the nature of the bonds between P and O in oxoacids of phosphorous?
Is it a double bond or coordinate bond?
Wikipedia says that it is "double bond" but many times I encountered "coordinate bond".

Could you perhaps link to the specific wikipedia page that says it's a double bond? Just because there are many pages for different oxoacids of phosphorous and to avoid any confusion.

In general, I'm not sure double bond and coordinate covalent bond are mutually exclusive labels and depending on the specific chemical you're looking at, they might both apply.
My Reasoning:

Coordinate bonds are bonds where both electrons were donated by one atom. Usually, the concept of a double bond is used in the context of valence bond theory and coordinate bond is used in the context of crystal field theory. There is evidence for pi bonding in coordinate complexes (i.e. back bonding) leading to what could be called a double bond in valence bond theory.
If you look at bond length data you can usually determine if there is a double bond. (Or if you have some of the acid and access to an IR, you can do this yourself.) The CRC should contain reference bond lengths.

I 'm not an expert in phosphorous chemistry, but these could be helpful:

Structural calculations for many compounds, including a few acids
Study on structures of phosphates, not acids

The phosphoryl group is usually written with a double bond, i.e. P=O, which is easier to use to describe chemical mechanisms (and also clearer on a scheme). It is also drawn like a coordinate bond sometimes. The reasoning is that it is formed from a trivalent phosphorus compound (PR3) which has a lone pair of electrons on the phosphorus, and an atom of oxygen which misses 2 electrons to complete its outer shell; and this is the definition of a coordinate bond.
The most admitted use in the scientific literature is to draw a double bond.

Numerous oxoacids of phosphorus their respective ions or salts are known.Some points which could be enumerated from all the structures are:

All phosphorous atoms in the oxoacids and the oxoanions are 4-coordinated and consists of at least one P=O bond.
All the phosphorous atoms in the oxoacids contain at least one P-OH bond which is ionizable.
Some oxoacids have one direct P-H bond which is non-ionizable
Catenation takes place via P-O-P links or direct P-P bonds. The former either exists in linear chains or cyclic species.

my question is regarding to first point. Sometimes I saw also coordinate bond instead of double bond, can't decide which is correct.


text:
Similarity between Ecell and equilibrium constant Kc
$$\ce{Cu^{2+} + Zn<=>Zn^{2+} + Cu}$$
$K_c=1.9*10^{37}$
$E_{cell}=1.1V$
I know that they both suggest that the forward reaction is going to completion but can someone please tell me a bit more about what both these values are suggesting and how these two values concord with each other.
Also, many books quote such reactions in only one direction and gives a value for $E_{cell}$. However, isn't $E_{cell}$ just another way of illustrating the condition of an equilibrium (i.e. in which direction the reaction would occur if left alone)... so how can a one way reation be shown

See here for a derivation of the relationship between equilibrium constants and potentials.

$E^\circ_{cell}$ and $K_c$ are related. Both indicate which direction of the reaction is spontaneous. Both tell you something about the position of equilibrium.
However, $E^\circ_{cell}$ is a directly measurable value. If you construct an electrochemical cell as using a zinc electrode immersed in a 1.0 M solution of say zinc chloride and a copper electrode immersed in a 1.0 M solution of copper (II) chloride with a salt bride of sodium chloride connecting the solutions, a voltmeter connected between the two electrodes will read the value of $E^\circ_{cell}$, at least initially. The voltage will change as the reaction progresses, but the initial voltageis $E^\circ_{cell}$ which is defined for standard conditions of 1.0 M concentration, 1 bar of pressure, and 25 $^\circ$C.
$$\ce{Zn|ZnCl2}(1.0 \text{ M})\parallel\ce{CuCl2}(1.0 \text{ M})|\ce{Cu} \ \ \ \ E^\circ_{cell}=1.1\text{ V}$$
The relationship between $E^\circ_{cell}$ and $K_c$ is the Nernst Equation. Note the use of both $E^\circ_{cell}$, the standard cell potential and $E_{cell}$, the observed cell potential under other conditions. The remainder of the alphabet soup is defined below.
$$E_{cell}=E^\circ_{cell}-\frac{RT}{zF}\ln{Q}$$

$E^\circ_{cell}$ - Standard cell potential
$E_{cell}$ - nonstandard cell potential at other conditions
$R$ - the ideal gas constant (which appears here because it is equal to Avogadro's number times Boltzmann's constant
$T$ - the temperature in kelvins
$z$ - the number of moles of electrons being transferred in this redox process. If we wanted to use the number of individual electrons, then we would need to replace $R$ with $k_B$
$F$ - Faraday's constant
$Q$ - the reaction quotient

We need to use $Q$ in the Nernst equation, because this equation can relate the cell potential to any range of conditions, not just equilibrium.

At equilibrium $Q = K_c$ and $E_{cell} = 0 \text{ V}$ (there is no net change in the reaction composition, so there is no potential difference across the half-cells.

Thus, the Nernst Equation becomes:
$$0=E^\circ_{cell}-\frac{RT}{zF}\ln{K_c}$$
$$E^\circ_{cell} = \frac{RT}{zF}\ln{K_c}$$

To determine the values of $E^\circ_{cell}$ and $K_c$ for the reverse reaction...

For $E^\circ_{cell}$:
$$E^\circ_{cell} = -E^\circ_{rev}$$
For $K_c$
$$K_{rev} = \frac{1}{K_c}$$

Great post as always. Just a small thought regarding $z$. The quotient $\frac{RT}{zF}$ needs to have a dimension of volts, and by dimensional analysis, the $mol^{-1}$ from $R$ would cancel out with the $mol^{-1}$ from $F$, clearing the quotient of any dimension in quantity of matter. If $z$ had the dimensionality of moles, then it would reintroduce a unit that could not cancel out. I imagine that means the $z$ is only the adimensional stoichiometric coefficient for electrons in the reaction, and as such it can't be converted to number of electrons by substituting $R$ for $k_B$.

Rather, if $k_B$ is used, then I think that instead of $F$ it would be necessary to use $e$, the fundamental electric charge.


text:
Does the number of H+ ions in solution go up on dilution of a weak acid?
In my textbook, a footnote says:

In case of weak acids, on dilution the total number of $\ce{H^{+}}$ ions in solution increases because dissociation of the weak acid increases

This didn't make sense to me however. Consider the general dissociation reaction of a weak acid:
$$\ce{HA (aq)  + H_2O \rightleftharpoons H_3O^{+} (aq) + A^{-}(aq)}$$
Now, the Chatelier's principle tells us that if we add more reactants ($\ce{H_2O}$), the reaction will proceed more to the right, and so the dissociation of the acid will increase, but the part about the total number of $\ce{H^{+}}$ ions in water being higher was still unclear to me.
Delving into the math a bit, we can see that
$$K_a = \frac{[\ce{H_3O^{+}}][\ce{A^{-}}]}{[\ce{HA}]}$$
Since the concentrations of the two products are equal,
$$[\ce{H_3O^{+}}] = \sqrt{[\ce{HA}]\cdot K_a}$$
Now, let us assume that we have a $1 \ M$ solution of the acid in question dissolved in 1 litre of water, plugging this into the equation, we have a $\sqrt{K_a}$ molar solution of $H^{+}$ ions. Since only 1 litre of solvent is there, we have $\sqrt{K_a}$ moles of ions.
Now, say we dump 1 litre of water into this solution. The concentration of our acid will be halved. Plugging this into the equation, we get the concentration of $\ce{H_3O^{+}}$ ions as approx. $0.7 \sqrt{K_a} \ M$. We have two litres of solvent, so we have $1.4 \sqrt{K_a}$ moles of ions, which is considerably higher. So, the numbers all work out, but what's the intuition behind it?

Haven't you pretty much answered your own question?  This looks like a great answer to me.  The intuition is Le Chatelier, and you proved the intuition with math.  Win!

Let's say we have got a solution of an acid with a pK$_a=5$. The resulting pH dependent on the concentration of that acid would be like:
$\hskip2in$
Examples: $0.1~\ce{M}$: pH$=3.0$; $0.001~\ce{M}$: pH$=4.0$; $0.00001~\ce{M}$: pH$=5.2$; $0.0000001~\ce{M}$: pH$=6.8$
Depending on the pH resulting from these example concentrations we can have a look at percentage composition of the whole acid base system. (Degree of association $\alpha'$ in blue, degree of dissociation $\alpha$ in light orange.)
$\hskip2in$
Examples: pH$=3.0$: $\alpha\approx 1.0~\%$; pH$=4.0$: $\alpha\approx 9.5~\%$; pH$=5.2$: $\alpha\approx 61.8~\%$; pH$=6.8$: $\alpha\approx 98.4~\%$
One can see that low acid concentrations lead to higher pH values than higher acid concentrations which automatically result in an increasing degree of dissociation.
So for me this is no big intuition thing but rather a small sequence of actions.

Lets say it in math:
The proton concentration for your system can be described by $$c(\ce{H3O+}) = \sqrt{c_0(\ce{HA})~k_a+k_W}$$ (adding $k_W$ for very low concentrations) and the equation for the degree of dissociation is $$\alpha = \frac{k_a}{k_a+c(\ce{H3O+})}$$ which can be combined to give $$\alpha\left(c_0(\ce{HA})\right)=\frac{k_a}{k_a+\sqrt{c_0(\ce{HA})~k_a+k_W}}$$

One could expand this system a little more (using the exact pH equation including the self-ionization of water) and come up with a thorough scheme like the following.
$\hskip2in$

Would you please elaborate how you got that equation for disassociation constant?

The first equation for $\alpha $?

Yeas, I meant that.

I’m not sure if such a question exists on chem.se. But if not, it would be nice for others, if you’d ask it.

I am just telling you to work out the math of your first equation. And basically disassociation constant equation isn't like that.

Your question didn’t sound like you know the solution but that you want to get an answer. But as you say, that it does not look like this, you seem to know the answer. So, please, how is it? And btw, that "I am telling you to ..."-formulation is quite rude, as you can ask but not tell me to do sth.

Let us continue this discussion in chat.

Intuition behind this concept is thermodynamically based.  If you think of $K_a$ as a thermodynamically favorable constant that describes the analyte concentrations associated with acid dissociation then it is not too difficult to reconcile the increase in $\ce{H+}$ ions in solution as a necessary means to maintain a thermodynamically favorable state when water is added to solution.

This is one of the common problems of ionic equilibrium. Generally questions are asked to calculate the $\mathrm{pH}$ of a diluted weak acidic solution and if the $\mathrm{pH}$ calculation turns out to be $7$ or more than that, we have to add the contribution of water also. I am giving you an example that will resolve your matter completely:
Calculate the $\mathrm{pH}$ of $\pu{1.0E-8 M}$ solution of $\ce{HCl}$.
For the reaction
$$\ce{HCl + H2O <=> H3O+ + Cl-}\tag1\label{hcl-dissociation},$$
if you calculate the pH as
$$\text{pH} = -\log(10^{-8}),\tag2$$
then $\text{pH}=8$.
Now you can see that the $\mathrm{pH}$ of $\ce{HCl}$ can not be $8$, it has to be less than $7$, therefore you have to include the contribution of water.
Consider
\begin{align}
\ce{2H2O &<=> H3O+ + OH-}\tag3\label{selfionisation}\\
K_\mathrm{w} &=\ce{[OH-][H3O+]}\\
K_\mathrm{w} &=10^{-14}
\end{align}
Therefore the $\ce{H3O+}$ or $\ce{H+}$ concentration increases from:

Ionization of $\ce{HCl}$ dissolved \eqref{hcl-dissociation},
Self ionization of water \eqref{selfionisation}.

So in very dilute solutions like this solution we have to consider the contribution of both sources of $\ce{H3O+}$; this also explains why the concentration of hydronium ions increases upon dilution of a weak acidic solution.
now going back to problem
\begin{align}
\text{let } x = [\ce{OH-}] &= [\ce{H3O+}]\\
[\ce{H3O+}] &= 10^{-8} + x\\
K_\mathrm{w} &= (10^{-8} + x)(x) = 10^{-14}\\
0 &= x^2 +10^{-8}x -10^{-14}\\
[\ce{OH-}] &= x = \pu{9.5E-8}\\
\mathrm{pOH}&=7.02\\
\mathrm{pH}&=6.98
\end{align}
Now you can see that we get $\mathrm{pH} = 6.98$ after adding the contribution of water, otherwise we would have gotten the absurd answer of $\mathrm{pH}=8$, which can not be true.


text:
How to increase equilibrium of Cl₂
$\ce{Heat\ +\ PCl5(g)\ \rightleftharpoons\ PCl3(g)\ +\ Cl2(g)}$
According to the process above, which of the following can be done to increase the number of moles of $\ce{Cl2}$ in a sealed container at equilibrium?

Reducing temp
Increasing the pressure at a constant temp
Adding more moles of $\ce{PCl5}$

I'm pretty sure it's not 1. but I'm stuck between 2. and 3. Are they both true or only one of them?

Hint:
For 2. Try to apply Le Chatelier's principle. Increasing pressure will make the system try to decrease it and there is one side of the equilibrium with an unequal number of moles.

I just looked up the pressure part of Le Chatelier's principle and it said that an increase of pressure would shift the equilibrium toward the side with fewer moles and in this case its the side opposite the CL2. So i'll take it that 2 is not true because the equilibrium is shifting in the opposite direction?

@sloth1111 That's right! :)

I'm guessing the the answer is III because it's not I, if you increase the pressure, the equilibium will shift toward the PCL5 because there's less moles on that side, and III is true because adding more PCL will shift equilibrium to the other side, increasing CL2. Am I correct?

@sloth1111 Yup, you are correct and you're welcome :)

yest the answer 3. is correct. It will be even more effective to adding more $\ce{PCl5}$ and increasing temperature too.

Ok thank you!! I assumed it would be 3 because of what you said above. Adding PCL5 would have the greatest impact.


text:
Relationship in equilibrium constants
$\ce{2CH4(g) + O2(g) \rightleftharpoons 2CO(g) + 4H2(g)}$

Equal numbers of moles of $\ce{CH4}$ and $\ce{O2}$ are drawn into a
  sealed container and the system is left until it reaches equilibrium.
  Which of the following relationships pertaining to this particular
  system are true at equilibrium?

$\ce{[CH4]<[H2]}$
$\ce{[CH4]>[O2]}$
$\ce{[H2]>[CO]}$

I know that the equation for $\ce{K_{eq}}$ is
$$\ce{\dfrac{[H2]^4[CO]^2}{[CH4]^2[O2]}}$$
I think that 3. is wrong because if $\ce{CH4}$ and $\ce{O2}$ are equal, then $\ce{H2}$ and $\ce{CO}$ have to be the same. Now I'm confused about 1 and 2. I'm not sure how to distinguish between the two.

Let's say that the amount of $\ce{CH4}$ and $\ce{O2}$ at the start is $x$.
After the reaction is complete, let's say that $y$ moles of $\ce{O2}$ reacted.
The amount of moles of $\ce{CH4}$ left is $x-2y$;
The amount of moles of $\ce{O2}$ left is $x-y$;
The amount of moles of $\ce{CO}$ formed is $2y$;
The amount of moles of $\ce{H2}$ formed is $4y$;
That should make one of the choices clearly false, one clearly true and one might not be so obvious, unless you assume that the forward reaction path is less favourable ($\ce{CO}$ is not very stable)

ok after looking at and using the equations I think I have the answers. Please correct me if i'm wrong. 1 is correct because 4y is bigger than x-2y, also looking at the equation at equilibrium, 4 moles is bigger compared to 2. I believe that 2 is clearly false because since x-y is bigger than x-2y, O2 has to be bigger CH4. I'm confused about 3 but i'm guessing that it's false because 1 and 3 is not an answer choice and 1 is true

@sloth1111 Well, 1. is the choice that is not obvious. Try putting x = 5 and y = 2. Then try using x = 100 and y = 1. But 3. is always true because 4y > 2y.

oh boy sorry! Since you get different answers for 1 under certain conditions would that mean its incorrect?

@sloth1111 It's just that there might not be enough information to clearly say whether it's true or not, but since it's a multiple choice, you usually have to pick the best possible answer.

I think i'm going to go with 3 because it is the most clear since 1 and 3 isn't an option. Thank you for your help! All my teacher does is give us worksheets and i'm trying to understand the concepts

Don't use algerbra to help- simpler would to be simulate what is happening in the container. Equilibrium is almost irrelevant for this question- it only says that a reaction is occurring. The answer is correct for any time after the methane and oxygen enter the container.


text:
Why is the enthalpy change not zero in an adiabatic process?

Which of the following must be true for adiabatic processes?

$C_V = C_p$
$\Delta H = 0$
$\Delta U = 0$
$\Delta S = 0$
$q = 0$

(Source: Chemistry GRE)
The answer is $q = 0$. From what I can find, an adiabatic process is when there is no transfer of heat, but then why is the enthalpy change not zero?

This may be useful for you http://mutuslab.cs.uwindsor.ca/schurko/introphyschem/lectures/240_l07.pdf

The Enthalpy $H$ is defined as $H=U+PV$. Therefore,
$$\Delta H=\Delta U + P\Delta V +V\Delta P$$
For an adiabatic process, $q=0$. therefore from the first law of thermodynamics,
$$\Delta U = q +w =q-P\Delta V$$
$$\Delta U=w=-P\Delta V$$
Substituting this in the first equation you get,
$$\Delta H=V\Delta P$$
If $\Delta P$ is zero during the process (Isobaric) then the enthalpy change will indeed be zero, but it will be non-zero if the process is not isobaric.

Isn't enthalpy defined for systems at constant pressure, so the V∆P part in the second line is itself zero. That gives us ∆H=0 for an adiabatic process.

@jyotiproy It is defined for all systems, and is a state variable. Change in enthalpy does depend on the change in pressure. Anyway, if we deal with infinitesimals, we need not worry about having P and $\Delta P$ in the same equation.

If you have an ideal gas in a constant volume adiabatic chamber, with the gas initially occupying only half the chamber, and vacuum in the other half, with a barrier in between, and you remove the barrier and then let the system re-equilibrate (i.e., free expansion), the work done on the system will be zero (rigid container) and $\Delta U = 0$.  Therefore, the temperature change will be zero, which also means that $\Delta (PV)$ will be zero.  So, $\Delta H=\Delta U+\Delta(PV)=0$ even though the pressure change is not zero.
Even if there is gas present initially in the other half of the chamber, the $\Delta H$ for the system is still zero. Initially the enthalpies of the gas in each of the two chamber halves of volume $V/2$ are:
$$H_1=n_1u(T)+P_1V/2=n_1u(T)+n_1RT$$
$$H_2=n_2u(T)+P_2V/2=n_2u(T)+n_2RT$$
where u(T) is the molar enthalpy at temperature T, and the $n$'s are the number of moles in the two chambers. So the initial enthalpy of the system is:
$$H_{\text{init}}=H_1+H_2=(n_1+n_2)u(T)+(n_1+n_2)RT$$
But, since no work is done and no heat is transferred to the system, the final internal energy of the system is the same, as is the final temperature.  And, of course, the total number of moles doesn't change.  So, that means that
$$H_{\text{final}}=H_{\text{init}}$$ and $\Delta H=0$.  (Of course, the final pressure is $(P_1+P_2)/2$).

(+1) I find that you have very aptly demonstrated the specific cases where $\Delta H=0$. However, in a general situation, we can't say that $\Delta H=0$, as Satwik demonstrated above. Is my interpretation of your answer correct?

For an adiabatic process on a closed system, the enthalpy change from the initial thermodynamic equilibrium state to the final  thermodynamic equilibrium state is $\Delta H=\Delta (PV)-\int{P_{ext}dV}$If the process is adiabatic and also reversible, this equation reduces to $\Delta H=\int{VdP}$.

How can we integrate V here with dp. I am asking this because I am solving a very similar question and I am stuck on using your expression. The volume is not constant in that reversible adiabatic

$nC_pdT=\frac{nRT}{P}dP$

why is the enthalpy change not zero?

When only PV work is possible, and pressure is constant, we can write
$$q_p = \Delta H \tag{1}$$
that is, the enthalpy change at constant pressure is equal to the heat exchanged. If the pressure is not constant all bets are off and you should resort to some other expression such as
$$\Delta H = \int_1^2 d(U+PV)$$
For a reversible process,
$$\Delta H = \int_1^2 VdP$$
This integral cannot be zero unless $V=0$ (which is not possible) or $dP=0$ (which means $P=$constant, bringing us back to Eq. 1, which says that for an isobaric adiabatic process $\Delta H =0$).
Since we have found a case that violates the stated condition, $\Delta H =0$ need not be true.


text:
Determine compound acidity based on structure
In an exercise I have the following structures
Moclobemide:
Paracetamol:
Hexobarbital:
I know for example by heart that paracetamol is an acid. What is the best approach in determination of acidity of a compound based on it's structure?
I'd like to know for example if these compounds are:

Basic
Acidic
Amphoteric

I know for example that many metals (Zn, Pb, etc) can be amphoteric (act as $H^+$ donors or acceptors). But I don't know how to determine what a complex structure is.
Please offer a detailed explanation, with links if possible.
Best Regards,

Welcome! I think you can find some information in this question. In organic chemistry you should always take in account the stability of the compound resulting from a de-protonated or protonated (or accepting or donating electrons).  If the resulting molecule is more stable (e.g. some resonance effect occur) the reaction is favorite. This is the case of acidity of the phenol!

Panagiotis, in order to solve the exercise, one needs to identify the relevant functional groups and explore their environment a bit.
Supposed that we're talking about "standard conditions", i.e., what is likely to happen if the dissolve/suspend the compounds in water, the situation is as follows:
In the structure of Moclobemide, we spot an amide, and - in the morpholin moiety - an ether and a tertiary amine. Amides do have a lone pair on the nitrogen atom, but due to the conjugation with the carbonyl group, the electron density on the N atom is strongly reduced, as compared to an amine. Ethers are rather lousy bases, in fact, the $pK_a$ of protonated tetrahydrofuran is as low as -2.05.
The remaining tertiary amine thus is the functional group that influences the pH: tertiary amines show a $pK_b$ around 3.3.
In Paracetamol we identify an amide (that we've already excluded as irrelevant) and a phenol group. The latter, once deprotonated, will stabilize the negative charge by resonance with the $\pi$-system (G M has already mentioned this). Therefore, phenols are far more acidic than alkanols. Consequently, the compound definitely acts as an acid.
I'm sure that you've already noticed two imide units in the case of Hexobarbital and in fact, it is possible to abstract the proton from the nitrogen atom. However, you'd need a base such as a carbonate under anhydrous conditions to deprotonate here. As a result, Hexobarbital will hardly do anything at all in water.


text:
What route to take in converting ethanol to crotonic acid?
The question asks to write the theoretical transformation steps to convert ethanol to crotonic acid in considerable yield. One constraint is that you cannot use any external organic compounds but inorganic catalysts and reagents are allowed and aldol condensation is prohibited. I do not know the answer, but I came up with the following steps:-

Treatment with $\ce{HBr}$ to form $\ce{CH3CH2Br}$ via $S_N2$.
Treatment with $\ce{Mg}$ in ether, to form corresponding Grignard reagent.
Oxidation of ethanol by Pyridinium chlorochromate to ethanal, and then treating it with the formed Grignard reaction to form $\ce{CH3CHOHCH2CH3}$.
Acid catalysed dehydration to form $\ce{CH3CH=CHCH3}$.
Using NBS (N-Bromo Succinimide) as an allylic halogenating agent, followed by treatment with aq. $\ce{KOH}$ to give $\ce{CH3CH=CHCH2OH}$.
Oxidising this with PCC followed by Tollen's reagent to selectively oxidise alcohol to acid, to finally yield crotonic acid.

I am unsure if any of these steps are incorrect, i.e if any of these steps will not give appropriate yields of the desired product owing to side reactions. Another problem is that this transformation is too long. Is there any shorter way to carry out this conversion in proper yields?

You need to also think about stereochemisrty, making the trans product. Since aldol condensation is not allowed maybe some form of Wittig would be suitable. For example use 1 ethanol to make acetaldehyde, another ethanol to make ethyl brmoacetate etc. (will maybe try to make a drawing later)

Allylic bromination and subsequent hydrolysis might be replaced by an allylic oxidation of the butene using selenium dioxide - if that's inorganic enough for you ;)
But I'm wondering whether there's a shorter way. What about the Grignard reaction you described, followed by oxidation (PCC on silica or manganese dioxide) to 2-butenone. Subsequent bromination with bromine in hydrobromic acid should yield an $\alpha,\alpha'$-dibromoketone which, upon treatment with base (Favorskii rearrangement) and, eventually acidic workup should yield crotonic acid.


text:
Does deprotonation of a hexaaquaion continue until neutral charge
I learned that deprotonation is the process in which a proton is lost from ligand $\ce{H2O}$ molecules surrounding a central metal ion (I am referring to hexaaqua ions). Also, the process continues until the charge of the complex ion is neutralized.
My question:
Why does it only continue until the charge is neutralized, why don't all the six ligand water molecules lose a proton?

Let's look at an example: Hexaaqua aluminium, or $\ce{[Al(OH2)6]^{3+}}$
It is acidic and will undergo deprotonation of a ligand water molecule spontaneously:
$$ \ce{[Al(OH2)6]^{3+} <=> H+ + [Al(OH)(OH2)5]^{2+}} $$
You can even get it to deprotonate twice:
$$ \ce{[Al(OH)(OH2)5]^{2+} <=> H+ + [Al(OH)2(OH2)4]^{+}} $$
But, as indicated, both reactions are equilibrium reactions. So if the concentration of protons in solution is high enough, the hydroxo ligands will be reprotonated and become neutral aqua ligands.
If you add some hydroxide to the solution, you can precipitate aluminium hydroxide, since it is not soluble in water. While this might be a driving force for the deprotonation, it doesn't seem to be the case here.
So we have shown that the process does not necessarily continue until the charge of the complex ion is neutralized.

Now, concerning your question why this doesn't happen until all aqua ligands have lost a proton:
Again, a part of the solution is the equilibrium idea brought forward in the upper part. But there's another way to think about it: If you have a negatively charged ion, you also need a positively charged counterion. The only counterion that exists in the solution of water and the aluminium compounds is the freshly abstracted proton, which happily will bind back to the hydroxoligand.
There are also $\pi$-donor effects that are stronger in the hydroxo than the aqua ligand. So it might be favourable to have no more than three hydroxo ligands in an octahedral complex so no two $\pi$-donors are across each other.


text:
Hard coating for soft surface
Say I have a soft, possibly porous object. I want to coat it with a material,

Resistant to boiling water and some scraping
Not hazardous, at least after the coating
Largely transparent

The idea is to take, say, a fruit with the insides removed and to make a cup or a vase out of it.
Is there something I could buy or make at home for this?

"I need a material that does X" is off-topic.

@Aesin, why is that?

@Aesin - I'm not certain that "I need a material that does X" is off-topic. Certainly, there is nothing in the [help] center to forbid it. Other sites in the network may have explicitly or implicitly forbidden these types of questions, but I am not sure we have yet.

@KarolisJuodelė - This question, however, does not explicitly match our on-topic list either, since you are not necessarily trying to understand the material/phenomenon. Your question could be improved. See http://meta.chemistry.stackexchange.com/questions/235/should-chemical-request-questions-be-on-topic-here

@BenNorris We've been very iffy on such questions, allowing a few and disallowing a few. I think a meta post on formalizing a small policy for these is in order.

@Aesin see above

@BenNorris, I'm not sure I understand. In stackoverflow an equivalent "What tool to use for X" would be closed as opinion based. I don't think that's the case with chemistry. Isn't enumerating compounds and their properties a part of the science?

@KarolisJuodelė - You'll note that I advocated on keeping the question open because we have not settled on a policy. However, our on-topic list centers on understanding of chemical phenomena with questions that may be useful to more than just the OP instead of "I need something to plaster my deck with that is wind, water, and ice resistant and that bird scat bounces right off of and I don't care what it is as long as I can buy it on the internet." This second type of question does not seek understanding, and its answers will only be useful to the OP and not to anyone else.

http://meta.chemistry.stackexchange.com/questions/312/what-is-our-policy-on-material-recommendations /cc @BenNorris

also /cc @Aesin

For me the body question is okay, the title could be better is too generic at the moment why not "How to coat organic materials for food usage?".

@Aesin for me these are "Questions about experimental technology"

I'm voting to close this question as off-topic because ‘I need a material for X is off-topic’

You could vacuform heated thin PET film over it.  That would not stop biology.

+1 for "that would not stop biology" - cutting off the oxygen supply to a fruit changes its metabolism, but many enzymatic processes will continue, as well as bacteria-caused decay. You will need to preserve the fruit first, and the usual methods (formaldehyde, dessication) are not suited to your purposes.

If recipients or children believe it is edible and it is not - by composition or history - that is a bad thing. One could give it a dose of Bitrex denatonium benzoate to keep it out of mouths.  It overall seems like a bad path to pursue.  Edible bouquets are no more sanitary than their handlers.

This probably doesn't count as a real answer since the underlying technology isn't coating, but:
What is possible to turn deceased people and their organs into durable objects should (in principle) work for a pumpkin or a melon too.
The process of plastination, as used by Gunther von Hagens for his Body Worlds exhibitions involves the following steps:

dissection and embalming: cut the fruit open and remove the "flesh", soak with aqeous formaldehyde to stop decay
removal of soluble fats: place object in an acetone bath
impregnation: place object in a vacuum chamber (exsiccator) filled with a reactive polymer and apply vacuum
curing: depending on the polymer, apply visible or uv light to harden the object

Admittedly, this is more of a mind game, but it should work :)

In my opinion this is a (at least) two steps process:

Make the fruit's peel more durable.
Coating the inner part of the object.

Make the fruit's peel more durable.
The conservation of fruit peels (seems strange but is true) is a well know problem in the field of restoration of contemporary art.
Zoe Leonard "Strange Fruit (for David)" is the classical case of study about this topic.

Zoe Leonard after the death of a friends of hers begun to sews the peels and do with different material to recreate the original shape of the fruit. This work of course was very degradable so a German conservator Christian Scheidemann developed  a method to treat the fruit peel to make them more durable. He used the Paraloid B72, a well know ethyl-methacrylate copolymer used in conservation of cultural heritage, under vacuum. I'm quoting from the link above:

Early on, her dealer, Paula Cooper, suggested the possibility of preservative intervention for the sculpture. Leonard was amenable and worked for two years with German conservator Christian Scheidemann to devise a way to arrest the decay of the fruit surfaces. After much testing, Scheidemann developed a solution that consisted of shock-freezing the pieces and then penetrating them with Paraloid B72 under vacuum. This solution was complicated by the need to protect the wires, threads, and other elements from the Paraloid B72; in other words, the piece presented the intricacies typical of any mixed-media work. But Scheidemann succeeded in this as well.

A generic more detailed research about food conservation can be found in this here.
Coating the inner part of the object
As other user stated generally plastic glasses and other for drinking are made of polystyrene, PET or polycarbonate. The third one is more durable and use for stuff used more times. However coat a cup or a vase with plastic on you own is quite dangerous (there are chances of poisoning if done in the wrong way) and expensive and difficult.
If you don't use vacuum you should use blow molding over a mold of the interior of the fruit (so you have to make a first mold and then another "negative" mold) and the attach the treated peel in a second moment with some adhesive, very expensive process. I could go to a Murano glassmaker with your mold! In this case you can add a plastic or glass edge to your cup I don't think is very nice to drink from a plastic peel.
There is an alternative cheapest way, using Food Grade Silicone Mold Rubber this is probably what I would used you can find in internet a lot of types of silicone rubber that can fit your needs.
However I don't think you can later wash it in the dishwasher!
Edit
As stated Jaroslav Kotowski you can use even a more expensive epoxy resin however you should pay attention to use a food grade epoxy resin. Non food grade epoxy resin could be very toxic.

you can use epoxy resin. It is transparent, can be boiled, easy to use and obtainable.


text:
Is the reaction thermodynamically favored or not?
$\ce{CO(g) + 2H2(g) +O2(g) <=> 2CH3OH(g)}$,
$\Delta H=-128\ \mathrm{kJ/mol}$
$\Delta S=-409.2\ \mathrm{J/(mol \cdot K)}$
Determine if the reaction if thermodynamically favored.
I just wanted someone to see if my logic is right. Since there are 4 moles on the reactants side and and 2 on the products side the reaction would not be thermodynamically favored because the reaction decreases the amount of entities, which decreases the entropy.

Your reasoning is correct but there is more to it.
Look at it from the point of view of Gibbs Free Energy,
$\Delta G = \Delta H - T \Delta S$
For a given reaction if:
$\Delta G >0$, the reaction is nonspontaneous in the forward direction, not thermodynamically favourable
$\Delta G < 0 $, the reaction is spontaneous in the forward direction, thermodynamically favourable
If we substitute the values for $\Delta H$ and $\Delta S$:
$\Delta G=(-128\ \mathrm{kJ/mol}) - T\left(-0.409\ \mathrm{kJ/(mol \cdot K)}\right)$
Mathematically, $\Delta G$ will become positive only when $T$ is greater than 313K. Because of this,  the reaction is thermodynamically favorable at any temperature less than 313K (since $\Delta G$ will be negative), but is not thermodynamically favorable at any temperature greater than 313K (since $\Delta G$ will be positive).

thank you! I did not think about the reasoning using delta G or H to figure out that delta G was negative. This helped a lot!

Just one more thing, in your last sentence you mention about "when $T$ becomes very negative", but that doesn't make sense, as absolute temperatures are always positive except in certain systems while using a certain definition of absolute temperature.

I meant from a mathematical perspective, but I should make that more clear

What @pingOfDoom is saying is that only way that delta G can become positive is if T is negative. Because this is physically impossible, delta G can never be positive.


text:
A student is provided with 500 mL of 600 ppm solutions of fructose. What volume of this solution in millilitres contains 0.15 fructose?

A student is provided with 500 mL of 600 ppm solutions of fructose.  What volume of this solution in millilitres contains 0.15 fructose?

Does this mean that I just have to use 600 ppm as the concentration and the 0.15 as the weight? Like this:
\begin{align*}
{600 \space ppm}
&=\frac {0.15 g}{x}\cdot 10^{6} \\
x&=\frac {0.15 }{600}\cdot 10^{6}  \\
x&=250 \space mL\\
&
\end{align*}
And when can 500 mL be used?

0.15 whats of fructose? grammes, mols?

I think if you don't provide the units of 0.15 this question will be close because is unclear what you are asking...

Assuming it's $0.15gm$ of fructose,$$ppm=\frac{Weight\;of\;solute_{(in\;gm)}}{Weight\;of\;solution_{(in\;gm)}}10^6$$
Since density of water is $1gm/ml$, $500ml$ of water would weigh $500gm$
$$\therefore600=\frac{W_b}{500}10^6$$
$$\therefore W_b=0.3gm$$
Thus, $500ml$ of water contains $0.3gm$ of fructose. You can find out what volume will contain $0.15gm$ of it.


text:
What would cause a sudden drop in humidity in my home?
I have a ventless propane heater newly installed in my home, which has been an increasingly source of paranoia for me (constant condensation on the windows, worries about carbon monoxide poisoning).  It's been a interesting few weeks of trying to find the right balance of windows to crack; needless to say I'm looking for a new heater.
Today, I noticed that NONE of my windows had condensation on them, even though the heater had been running for hours and the relative humidity outside was 100% (~30F) with fog and sleet.  I would expect it to feel like a swamp in here.
Now for the chemistry question- I know that the incomplete combustion of propane in a low oxygen environment would produce carbon monoxide, but is it typical for the air to become more dry as well?  Does the lack of oxygen also produce less water, and maybe more H2 or something like that?  What would cause the unusual drop (or perceived drop?) in humidity in my house?

the humidity is caused mainly from sources of water in your home such as plants, pets and of course humans. As you breath out water is coming out of you. And the condensation of water on windows is caused by different temperature outside/inside especially at the interface glass/air.
There is no way that heater can reduce humidity in sealed room (in means of decomposition of water as you suggested). Imho if you have CO alarm you have nothing to be afraid of.


text:
Is there an easy way to find number of valence electrons"
I want an high-school level answer. What I mean with "valence electrons" is the outermost electrons in that atoms' electronic arrangement? For example, 3 in an atom with an electronic arrangement of 2, 8, 3.)
Without actually writing the electron configuration, or orbitals, and knowing electron number, is there any way to find this number easily?
Also, is there an easy way to find if certain atom fits "atomic/orbital symmetry" given the numbers of electrons? (I don't know the english words for this and couldn't find them. So I translated it literally, but what I mean is, if orbital is half-full or full i.e  $\mathrm{s^1}$ or $\mathrm{2p^3}$ or $\mathrm{2p^6}$ and so on... hope you understand what I am trying to say)
I am asking this because sometimes questions give me electron numbers in a range of 40+ and just writing it out doesn't seem like a good way.

You can get the valence electrons in an atom's electronic arrangement by consulting the periodic table:

The Group 1 atoms have 1 valence electron.
The Group 2 atoms have 2 valence electrons.
The Group 3 atoms have 3 valence electrons.
The Group 4 atoms have 4 valence electrons.
The Group 5 atoms have 5 valence electrons.
The Group 6 atoms have 6 valence electrons.
The Group 7 atoms have 7 valence electrons.

The atom with electric configuration 2, 8, 3 is Aluminium and you can find it in Group 3 of the periodic table.
The periodic table is usually given in exams so...

The first two groups of the periodic table are the groups concerned with the filling of the $s$ orbital. So, all Group 1 atoms will have an outermost $s$ orbital filled with a single electron.
Similarly, Group 3 - Group 8 (Group 8 is also referred to as Group 0) are where the $p$ orbital is being filled up. All the Group 5 atoms thus all have an outermost $p$ orbital filled with 3 electrons.
The Periods in the periodic table can be used as an indication of the number of shells the atom has. Sulfur for instance is in Period 3 and thus has 3 shells.

periodic table isnt given in exams here. second part of your answer is really helpful for me, thx for that :) but about first part, I cant really know the group number without writing down electron configs can I ? is there a way to do this without going -2 -8 or 1s1 1s2 was what I was asking

@user4117 Well, that's an unusual case... If your exams aren't too soon, I would suggest using your periodic table as much as possible. This way, it will start getting in your memory ;) Otherwise, there's no really any other way to go about this (besides just remembering it), unfortunately :(

@answerer The table you put in contradicts your whole statement! e.g. the table clearly puts aluminium in group 13 and not 3!

@Nihilist_Frost The number you are looking at is the column number, not the group number.

Also, the picture I had was simpler, with less stuff to confuse. I don't know how the link got broken. I'll try find a simpler one that the one above.

Thanks for replacing that image. Unfortunately groups are defined as columns in the periodic table; the numbering confusions are due to different people using different group naming conventions entirely.

@Nihilist_Frost By that statement, you would need to give numbers to the F-block elements as well... which is not particularly helpful to OP. Furthermore, my education is based on UK's syllabus, which still uses this notation at high school level (Group 1 to 7 and 8/0 skipping the d-block). I don't know how other countries' syllabus differ, but in my opinion, putting Aluminium in group 13 does not help a high schooler at all and instead makes things more confusing. I tried to make things simple with the diagram so that the convention I was using was clear. If it didn't, I welcome suggestions.

I'd admit, Canadian high schools also don't listen to IUPAC. (F-block elements lack group numbers, even by IUPAC).Your statement illustrates the confusion right there.

Post has been inactive for long, but I think I should put down what is correct: The column numbering is an old fashioned way of using roman numerals and alphabets (Like group III A for group 13). Now IUPAC recommends Indo-arabic numerals (like 13) in continuous fashion from left to right. And f-block doesn't need numbering. The Lanthanides and Actinides are all considered to be in the same group as La and Ac, group 3 or IIIB.

Yes there is! For all elements in the following groups;
Group 1A= +1
Group 2A= +2
Group 3A= +3
Group 4A= +4 or -4
Group 5A= -3
From group 5, the valency descends in this order;
-3, -2, -1, 0.
Group 8A elements have a valency of Zero.
I hate complicated stuff too lol.

I think that's oxidation state, not valency.


text:
Chemistry Self Education
I want to learn chemistry. I did not have the benefit of studying it when I was in highschool, and I was always intimidated by it (I feel it's more rote memorization rather than understanding - I'm more of a Physics and Math kinda guy).
Anyway, what are the concepts that I need to learn?
How, and what resources are you going to use to homeschool Chemistry to a teenager?
In that studying it would put the student on par with a normal highschool student.
Some Nice Stuff:
Topics for a Normal Highschool Chemistry Curriculum
and Maybe AP and Honors Program Curriculums?
Comprehensive Textbooks
Exams and Self Testing Resources

This is too broad for the site as it stands.

@jonsca are the answers no good either?

@Amaterasu I am leaving everything intact, but in the future it's best to flag or vote to close questions that are too broad.

The edit doesn't really change the breadth of this topic - there are already several answers to get you going and a better use of Chem.SE would be to ask specific questions about problems that arise in your understanding of the material in the answers' links.

You might want to have a look at free online courses, e.g. at Khan Academy. To my opinion, this can however not replace the real experience of endless, often frustrating hours in the lab ;) - chemistry is an experimental science.

There are several concepts you need to learn, and the direction these will take will depend on your area(s) of interest.
To that end, the following are some resources that could be of help:
Khan Academy tutorials, these start with the fundamental concepts of the atom, periodic table, orbitals and electrons, chemical reactions, stoichiometry, states of matter, gas laws etc.
ChemTutor has some basic concepts to start of with, including measurement and pretty much all that Khan Academy has.
Something, as someone in education, I need to emphasise, and that is to test yourself - there are some online self-testing sites, such as this one from the University of Toronto.
Most of all, don't fear chemistry, it is a lot of fun and requires substantial problem solving and experimentation - you can buy some of these at toy stores and many experiments can be done at home - such as these from Home Science Tools.


text:
Ortho-effect in substituted aromatic acids and bases

When comparing o,m,p-toluidine basicities, the ortho effect is believed to explain why o-toluidine is weaker. But when comparing o,m,p-toluic acid basicities, the ortho effect is stated as a reason why o-toluic acid is stronger acid. I was told that the ortho effect is a phenomenon in which an ortho- group causes steric hindrance, forcing the $\ce{-COOH}$, $\ce{-NH2}$ or some other bulky group to move out of the plane, inhibiting resonance. Then, if the ortho effect inhibits resonance, why is o-toluic acid the strongest and o-toluidine the weakest?
Where am I going wrong in my understanding of the ortho effect?

Because due to being out of plane the resonance hybrids will decrease and in case of carboxylic acid the negative will always be on the oxygen atom, so better resonance because negative charge will be more on more electronegative atom.

Your comment contradicts the quantum mechanical calculations of Martin @YashasvPrajapati (see comments on his answer)

I'd like to throw a tentative explanation for the ortho effect into the ring:

In the molecules in question, an interaction between the protons of the methyl group and the lone pair of the amine nitrogen and the negative charge on the carboxylate, respectively,  can be assumed.
In the first case, the electron density on the N atom is (slightly) reduced and thus the basicity of o-toluidine.
In the latter case, a similar interaction provides additional stabilisation of the carboxylate. As a result, o-toluic acid is more acidic than the isomers.

You mentioned about an interaction. I would be grateful if you could what interaction it is. Is it H-bonding, or field effect? Or any other type of interaction?

Yes, one might call it H-bonding. C-H...X interactions are frequently found, although mostly in crystals. Whether it's really a through-space interaction or through-bond, "mediated" by the backbone is beyond my knowledge.

But can't intramolecular H-bonding also take place in the unionised carboxylic acid?

@InternetGuy They can and do; this probably has an opposing effect, and decreases acidity. doi: 10.1039/b601875k

@Klaus-DieterWarzecha In the end, were you able to finalize your "tentative explanation"? ;-)

I would like to back up Klaus' answer with some Quantum Theory of Atoms in Molecules (QTAIM) results, based on a DF-BP86/def2-SVP calculation. Note that these are results, obtained without the consideration of solvation or condensed phases. I believe they still prove a valid point in the case of electronic structure theory.
I revisited this question in order to answer another, similar question. While putting more effort into this, I realised, that the here treated structures are actually transition states. This does not mean, that the addressed issues are invalidated. Even if the only exist for very short moments, they still exist and have to be considered.

In o-methylaniline you can clearly see the suggested intramolecular $\ce{H}$ bond. The distance $\mathbf{d}(\ce{N-H})=239.3~\mathrm{pm}$ is only little shorter than the sum of the van der Waals radii, $\mathbf{r}(\ce{N})=155~\mathrm{pm}$, $\mathbf{r}(\ce{H})=110~\mathrm{pm}$, but neglecting it is also wrong. Even if this interaction does only exist for very short periods of time, it still means, that it stabilises this state. It will however not be the dominant feature.

(Laplacian distribution, solid blue lines indicate charge depletion $\nabla^2\rho<0$, dashed blue lines indicate charge accumulation  $\nabla^2\rho>0$,
Red spheres are bond critical points, purple spheres are ring critical points, black lines are bond paths, red lines are zero flux surfaces)
Steric effects are usually electronic or dispersive effects in disguise, hence they also refer to an intramolecular hydrogen bond. The average $\ce{N-H}$ bond is only about $\mathbf{d}_\text{av.}(\ce{N-H})\approx99-105~\mathrm{pm}$.
The solvation point made by user4604 should still be considered.
This interaction has to lower proton affinity and or lewis acid affinity and therefore decreases also basicity. Not so much because its electronic effects stabilise one particular conformation, but especially, making the lone pair unavailable for short periods of time.

You can also analyse this for o-methylbenzoic acid and here the effect changes direction.
In benzoic acid there is already some intramolecular hydrogen bonding from the ortho hydrogens, one of these is still present in the substituted case. The distance $\mathbf{d}(\ce{O-H_{o'}})=226.2~\mathrm{pm}$ is only little shorter than the sum of the van der Waals radii, $\mathbf{r}(\ce{O})=151~\mathrm{pm}$, $\mathbf{r}(\ce{H})=110~\mathrm{pm}$, but neglecting it would still be wrong wrong.
The distance $\mathbf{d}(\ce{O-H_{Me}})=210.8~\mathrm{pm}$ is significantly shorter than the sum of the van der Waals radii. You can again see the interaction via a bond path, and ring critical points. In the optimised geometry, the methyl moiety is slightly rotated, giving rise to two equidistant interactions. The rotation of this group can be considered as a free rotation at room temperature.

It is noteworthy, that the bond critical point of the $\ce{O-H_{acid}}$ is almost at the position of the proton, which indicates also that most of the electron density already belongs to the oxygen.
It is also obvious, that the charge concentration in this bond is significantly lower than in any other $\ce{E-X}$ bond, which can indicate a weak bond.

Can we comment, from your calculations, that the nitrogen atom in toluidene ring is planar with the benzene ring? And hence its lone pair would still be in resonance with the ring? Also, have you done similar calculation to consider whether the $\ce{-COOH}$ is planar with the benzene ring in toluic acid? Thank you! (am sorry I am unable to interpret your calculations in any way at all; am only asking for the inference you got from it)

@GaurangTandon Yes, as shown in the graphics, the nitrogen is in the same plane as the ring, as is -COOH. The structure is, however, a transition state, there are rotations in the methyl group. The lone pairs of the nitrogen, as well as the oxygen in the other case, are part of the pi system, at least sometimes.

It can be explained another way: o-toluidene is less basic than aniline cause of a different reason. See, what happens is: when aniline acting as a base becomes $\ce{NH3+}$ (on top of a benzene ring), it is usually stabilised by solvation. But if there is a substituent on the ortho-position, it inhibits solvation. Thus the tendency to act like a base is reduced.

Yours is actually the more correct explanation.Due to steric hindrance a bulky NH3+ group is unstable, hence the acid base reaction between ortho substituted anniline and a proton has a backward tendency.

But, can you explain ortho effect in benzoic acid through solvation effect?

@user4604 this is not entirely true I believe. For example, if the Ortho substituent was -OH or -OCH3 instead of -CH3 then the basicity would actually I due to Ortho effect

In  the presence of bulky group that is methyl in this case causes steric hinderence making the plane the $\ce{NH3}$ out the plane thus preventing resonance which could have  helps it in delocalising the + charge thus making the conjugate base very stable but because of hindernce the + charge gets localized and makes the conjugate base unstable and so the acid weaker.
But in case of o-toluic acid if one draws the resonating structures of the conjugate base we see that benzene ring exerts +m (mesomeric) effect, which is not so in case of o-toluidine, which indeed is inversly proportional to acidic strength and so the non planarity caused due SIR now benefits the substituent by making it free from the +M effect of the ring and so increasing its acidity. Now the substituent can freely make equivalent resonance which can will support the acidity and will not be harmed by the rings +M.

Conjugate base can't be stabilised by resonance. You'd need two o-methyl groups to force losing of planarity with steric effects.

User4604 is correct with the solvation argument.
Put another way, the acidities and basicities of these complexes are usually measured in water, where the free energy of hydrating a charged species can be very high (i.e. it is favorable to dissolve charged molecules). If nonpolar substituents are placed near the charged moiety, it will prevent that charged group from being effectively interacting with and becoming solvated by water. In doing so, the nonpolar substituent will influence the pKa (basicity and acidity) of the charged group. Whether the nonpolar group makes the moiety more basic or acidic depends on specific case (2-methyl benzoic acid will be more basic/less acidic than 4-methyl benzoic acid; however 2-methyl aniline will be less basic/more acidic than 4-methyl aniline).

I analysed the problem in different manner . Basicity implies tendency of protonation. In the first case methyl group exert steric hindrance and avoid protonation which in turn makes it less basic as compared on aniline. On the other hand in second case let us consider the acidic strength of ortho substituted and non substituted nearly equal but due to steric hindrance again tdency to loose electrons is more in unsubstituted making it less acidic as compared to substituted.Thus overall it is supporting acidic nature.

To compare, we have to use something called the second ortho effect. From wiki:

The protonation of Substituted aniline is inhibited due to steric clashes. Upon protonation the hybridization of Nitrogen in amino group changes from sp2 to sp3 making the group non planar. This causes the steric hindrance between the ortho substituted group and H atom of amino group which makes the conjugate acid less stable, hence decreases the basicity of substituted aniline (see the figure on right).

Source

Hindrance (note the spelling; not "hinderance") is about blocking something from approaching, e.g. the nucleophile in an SN2 reaction. Here, nothing is approaching anything, the problem is better described as "steric clashes" or "steric repulsions".