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Heisenberg's Uncertainty Principle's inequality
In the uncertainty principle , whats the significance of the inequality sign ?
Why is it not equal to over there ?
Why do we say greater than equal to instead of just equal to ?

The HUP just gives a lower bound in the uncertainty of $\Delta x$ and $\Delta p_x$, it is of course possible to measure both quantities with a higher uncertainty than given by the HUP, but not less.

Also, this is better moved to physics.SE

@andynitrox So the minimum uncertainty that will be there is given by it and uncertainty can be more than it too ?

http://physics.stackexchange.com/questions/170684/why-uncertainty-principle-on-a-large-scale-doesnt-impose-limitations-on-precisi

@Mithoron Got it ! Thanks . Should I delete this post since it has an answer over there ?

@andynitrox I think that questions about the uncertainty principle are also on topic for this site, even if it is rooted deep inside physics. Because of this I also think the question should not be removed, but closed and linked to the post at physics. (I know that cross network posts are not a reality at the moment.)

I'm voting to close this question as off-topic because it has already has been asked on Physics.SE

In addition to my comment further above I might add that the Heisenberg Uncertainty Principle has nothing to do with an error in a measurement. The HUP does not state that we cannot measure something more accurately than this lower bound, but instead that it is inherently not determined to a better degree.
To illustrate, the two properties of a one-dimensional wave packet , frequency and location, are similarily interrelated.
Imagine a sharp wave packet, a peak, something like this:

You'll have no trouble locating 'the wave packet', simply because it is fairly sharp.
Conversely, for a monochromatic (and therefore technically infinitely long wave 'packet'), it is very easy to figure out what frequency or wavelength it has:

However, trying to find out where this wave is, cannot really be answered. Keeping in mind that a wave packet is essentially made up of different monochromatic waves with different frequencies and amplitudes, you'll see that the sharper a packet is, the more frequencies it contains, so it will become nearly impossible to say what frequency it has.

We can give a lower bound for the accuracy of both measurements: the certainty in the measurement of the frequency and of the location of such a wave packet. We cannot obviously measure both at the same time with arbitrary accuracy. Note that we may very well see exactly what such a wave packet looks like, but we can still not pin-point one single frequency and one single location that desribes such a packet. That is basically the essence of the Heisenberg Uncertainty Principle.
Using the wave number $k$ and as location $x$, we can describe this relation in the case of a wave packet as $$\Delta k \cdot \Delta x \geq 2\pi$$
The case of the wave-particle duality is analog to this purely mathematical example (the factor $2\pi$ comes from the coefficient in the Fourier transform that is used to switch between the $k$-space and the $x$-space).
Images: http://www.newtonphysics.on.ca/heisenberg/image615.gif, https://skullsinthestars.files.wordpress.com/2007/12/sinesnapshot.gif and http://hyperphysics.phy-astr.gsu.edu/hbase/waves/imgwav/wpac5.gif

Please leave a reference or a link to where the diagrams can be found.  Fair use affords these kind of cases, but it's always best to give credit.

Thanks for the extra information. I am accepting your answer and not mine since my answer has been quoted from your comments and you won't get the credit if I do so.

As andynitrox said in the comments :

The HUP just gives a lower bound in the uncertainty of Δx and Δpx, it is of course possible to measure both quantities with a higher uncertainty than given by the HUP, but not less.

The inequality exists to show that, the uncertainty is atleast this much and in real-life cases, its much larger than this . (source)(provided by Mithoron)

Refer this for more information.


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Can weathering change the composition of something so that it appears different?
There's a murder trial ongoing in Ireland at the moment and one of the pieces of evidence is a spade found at the scene along with the skeletal remains. It was contended by the prosecution that this spade was belonging to the defendant but while they were similar

"forensic evidence found the paint splatters on the spade were not the
  same as on the [defendant's] garden fence"

My question is: can something left out in the open be changed either chemically or structurally by the environment such that it appears to be similar but not the same? If so why would this not have been called out as a caveat by the expert forensics witness?
The specifics of the chemical tests weren't gone into in the media (sample article here) so I don't know how accurate it was.

Well, you leave iron bars in the open and receive them rusted from mother nature a while after. ;) Paint is also something that could change, but the test on it apparently was real accurate. BTW, welcome to chemistry.SE!

Sunlight is a serious issue with paints due to UV incidence, and careful formulation is required to ensure a long usable lifetime. But who am I kidding, sunny days in the British Isles? Hah!

Yes, weather can and probably will change the composition of materials and make them different.
However, forensic analysis usually takes that into account. There are some properties that might change because of weathering (you shouldn't compare water content, for example), but others won't. According to the sample article, the difference was found in the chemical composition of the paints.  Since details were not disclosed, one can only guess what these differences are, but some possibilites are:

Presence or absence of metals in the paints
Substances that wouldn't appear or disappear because of wheatering
Different isotopic ratios of elements
Traces of the solvent indicating they were not the same
Probably others

Assuming the analysis were performed correctly, it is usually not hard to say two samples are different. On the other side, when the results are similar, you can hardly say they are equal, but only that they match and that no differences were found.

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Chemical writing: the term "synthesis route"
Can someone tell me, which of the following terms should be used or avoided? (Google shows the presence of all the candidate terms.)

synthesis route
synthetic route
route of synthesis
synthesis pathway
synthetic pathway

Imagine for example a sentence such as "A new ... was developed."

all of these seem pretty acceptable. there is no definitive rule on this

In your example, "synthetic route" or "synthetic pathway" would fit best.  I suppose it is a matter of personal style, but that's what sounds best to me.

I like concision, so I would vote for "A new synthesis was developed."  You don't need any extra words in there; the meaning is clear from "synthesis" alone.

In papers I often see "synthesis route" or "pathway"

Synthetic is an adjective, which seems to more naturally modify the following 'pathway' in English.  While I am not a native German speaker, my understanding is that combining nouns is natural in that language and as such 'Synthesis Pathway, makes more sense.

I also saw "synthesis route/pathway" often in publications, but also "synthetic synthesis/pathway". Thank you all for your answers! You helped me.

If I may comment on "concision": If the synthesis includes nearly the same procedure of what the scientist has to do, but the alteration of a single compound leads to a different pathway (reaction mechanism, scientific context), I would still choose "synthetic pathway/route". However, in most cases I would agree with you, that the addition of "pathway/route" creates redundancy.

Ask this question on English SE.

English SE sent me back for this.

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What is has a higher priority in naming organic compounds, alkyl groups or halides?
2-Chloro-3,3-dimethylbutane or 3-Chloro-2,2-dimethylbutane?
According to these sources, both are correct:

NIST/TRC Web Thermo Tables (WTT)
ChemSpider

The current IUPAC recommendations (2013) read as follows:

Numbering
When several structural features appear in cyclic and acylic compounds, low locants are assigned to them in the following decreasing order of seniority.
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g.
Furthermore,

The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference (…)

Therefore, the example is named as 3-chloro-2,2-dimethylbutane rather than 2-chloro-3,3-dimethylbutane since the locant set ‘2,2,3’ is lower than ‘2,3,3’.

but the iupac naming rules also say that if there are 2 substituent groups at the same position in a compound alphabetical priority comes into play!!!

@user166748 Alphabetical order is used to establish the order of citation of detachable prefixes (that is why ‘chloro’ is cited before ‘methyl’). For numbering, however, all prefixes are considered together (see item f above). Only when there is a choice between equivalent numberings, the lowest locant is assigned to the substituent cited first as a prefix in the name.

yaa so that means both the names are correct one by locant rule and one by alphabetical priority? sorry for the punctuations

No. First, the alphabetical order is used to establish the order of citation of the prefixes; thus, the name is x-chloro-y,y-dimethylbutane (not x,x-dimethyl-y-chloro-butane). Then, low locants are assigned using Rule f considering all prefixes together; thus, the numbers are 2,2,3 (not 2,3,3). Therefore, the name is 3-chloro-2,2-dimethylbutane and we do not need any further rules. – For example, 2-chloro-3-methylbutane would be a different case, which would also require Rule g (and thus implicitly the alphabetical order of prefixes) in order to assign the numbers.

@Loong You should add this last comment into the front of your answer. Reading simple english is much easier than the IUPAC rule book. Thanks!

What I understand in simple terms is:
See that longest chain is butane.
Now we can number the carbon atoms of main chain either:
(A)  1, 2, 3, 4: In this case, the numbering of substituents is 2, 3, 3.
(B) 4, 3, 2, 1: In this case, numbering on substituents is 2, 2, 3.
Now, we apply "First point of difference rule". Using this rule, we see that the first 2's are same, but in the second number, there is a difference. So we select that numbering which has the lower number.
So, we follow numbering (B).
But to write the name, we write the chloro part first and dimethyl later because alphabetically c of chloro comes before m of methyl.
So the correct name is:
3-Chloro-2,2-dimethylbutane
NOTE: The first point of difference rule is used before the alphabetical rule. Like if there was symmetry, say the molecule was 2-Chloro-3-bromobutane, then its correct name would be 2-Bromo-3-chlorobutane.

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Polarity of the nitrate ion
We observe that nitrate ion has a trigonal planar geometry according to its lewis structure and the VSEPR theory. Its resonance hybrid consists of three contributing structure -three probable double bond organisations. According to my knowledge, we would expect that pi bond to be distributed equally over the molecule and thus, the dipole moments would cancel each other out. My question is, why such an ion would dissolve in water even though it is  non-polar?

You realize that nitrate is an ion, right?!

So it dissolves in water just because it is an ion?

A common assumption is "all ions must be polar". Is it true?

I hate courses that provide a short view - no offence. What is polarity compared to a negative charge on species?

It is a result of electronegativity difference.

See the comments in this relevant question. When we mention polarity, we are usually referring to electric dipoles. In that sense, some ions may be electric monopoles without electric dipoles and hence could be called non-polar ions (though they could still be expected to have strong ion-dipole interactions with water and hence be soluble). Benzene, on the other hand, has no electric monopole or dipole, but has a significant electric quadrupole, even if it is also called non-polar.

Just to clearify, nitrate ion is a non-polar specie, right?

I can't recall ever reading anyone refer to ions as polar or non-polar, so I don't know for sure. The term may not be usually applied to ions.

@Pınar If, as I think, the problem here is reasoning its solubility, notice that polarity results in electron concentrations' difference, while an ion totally has an extra charge. I never had to worry about the fact that an ion is polar or not.

Don't overthink it.
The oxygens have a partial negative charge in nitrate from resonance.  The protons in water have a partial positive charge from dipoles.  Positives attract negatives.
Molecule polarity becomes an issue when discussing neutral molecules interacting with neutral molecules.  Molecular polarity is one thing, and bond polarity is another.  Transformations / arrangements / alignments of molecules such that the dipoles (the things that make a molecule or bond 'polar') align to maximize interaction are certainly important for dissolution, but when a species is a bona fide free ion, water will tend to dissolve it.


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Order of priority in hybridization of d-orbitals
During hybridization of d-orbitals why the $d_{z^2}$ and $d_{x^2-y^2}$ orbitals are used up before the $d_{xy}$, $d_{yz}$ and $d_{zx}$ ?

The rotational frame of reference which defined which axis is the z-axis is changed such that $d_{z^2}$ and $d_{x^2-y^2}$ are hybridized first is a convention / practice.
In theory (meaning mathematically), you should be able to get equivalent geometry if you used the other d orbitals for hybridzation.
Basically the donut thing is weird enough we try to fold it in so we can deal with the more normal cloverleafs as leftovers.

Of course hybridisation with d orbitals doesn't really happen, not even in a theoretical way.


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Why is diborane cleaved symmetrically by trimethylamine but unsymmetrically by ammonia?
I know that soft bases cleave diborane symmetrically but hard ones don't. However, what makes ammonia harder than trimethylamine? And why does hardness/softness affect this?

Ooh, this question is a little old but I once read somewhere that this particular case was because of sterics. I can't remember the source - either Massey's Main Group Chemistry or Greenwood & Earnshaw's Chemistry of the Elements.

Greenwood & Earnshaw's Chemistry of the Elements does indeed mention sterics as the reason but doesn't go into much detail. It's on page 165

Excerpts from my answer here.

The mechanism proposed for the cleavage reaction involves an initial attack by the donor on one boron atom in diborane, leading to cleavage of one $\ce{B-H-B}$ bridge.

This is followed by the attack of a second donor molecule, cleaving the remaining $\ce{B-H-B}$ bridge. If the boron atom in the pendant $\ce{BH3}$ group is attacked, two moles of adduct result, whereas attack on the atom already carrying a donor produces cation and borohydride:

Support for this view comes from the fact that the intermediate ($\text{I}$) is detectable by tensimetric titration of amine-borane with diborane and by $\ce{^{11}B}$ nmr.

It can then be interpreted that hard bases like ammonia don't experience the steric effects which softer bases like trimethylamine do. Sterics inhibit the attachment of the second donor at the same boron atom.
Source: Boron Hydride Chemistry, Earl Muetterties


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Do core electrons have molecular orbitals?
Valence electrons are associated with molecular orbitals and hybridizations. Do core electrons have molecular/hybridized orbitals, or the original atomic orbitals?

tl;dr: If you are talking about VSEPR for an undergraduate level class, core orbitals do not mix.  If you are talking about Hartree-Fock theory for a graduate level thesis, core orbitals most certainly do mix.
The answer to this question depends greatly on the level of theory you are looking at.
High-level Quantum Mechanics calculations will typically involve mixing of core orbitals as well as valence orbitals, and give the most realistic picture.
That being said, by making the simplifying assumption that core orbitals do not hybridize (a situation which is almost assuredly untrue), one might make many useful predictions about geometry and reactivity. For example, the entire discipline of Organic Chemistry is based (in part) upon the central assumption that the carbon 1s orbital does not participate meaningfully in hybridization, and yet the discipline still provides many useful tools for discussing structure and reactivity.

What is your definition of core orbitals? Can you give examples where the mixing of core orbitals plays any role in chemical bonding? How would you justify the use of effective core potentials for heavier elements? What is your definition of hybridisation?

Core orbitals are non-valence orbitals.  I don't justify anything QM does with effective core potentials, semiempirical approximations or the rest; Ron's answer handles this part well enough, I see no need to extend this one.  Hybridization is the weighted linear combination of an s and some number of p (and occasionally d orbitals). which are useful for predicting the geometry of compounds.  Given that no exact solution for QM wavefunctions exist, no-one can say with certainty the core orbitals do or do not participate in bonding. Answer to comment depends on context (also noted in my answer)

I am sorry I did not express myself clearly. You state "[...] core orbitals most certainly do mix." and "High-level Quantum Mechanics calculations will typically involve mixing of core orbitals [...]". Could  you please give (an) example(s), where this is true, even if it is just a marginally effect. I am dealing with high level quantum chemical calculations everyday and I have not encountered that to any notable effect. But I am curious to know where I should pay more attention.

I've been retired from physical organic chemistry, focussed on kinetics, for about 6 years now, so my knowledge is getting rusty, as well as this topic being on the edge of my expertise.  I'd have to unpack GAMESS and get some of those verbose coefficient charts to show it, but my dim memories point to some contribution from the 1s; I will happily take some correction if I've misinterpreted the effects, and I will happily concede that such contributions to MO's were smaller for inner orbitals than valence orbitals.

The discussion in Carey and Sundberg, Part A, makes it clear that not including the inner orbitals in the final molecular orbitals is a simplifying assumption to make the calculations more palatable from a time when such things were more difficult.

Not including the inner orbitals is a crude simplification and a mistake. And they do of course form molecular orbitals (linear combinations), after all, they have to represent the geometry and symmetry of the molecule. But they are mainly just symmetry adapted and somewhat polarised. Notable sp mixing (at the same centre = hybridisation) I highly doubt, since energy levels of the core orbitals should be too far apart to contribute significantly. We nowadays only use small core ECPs  mainly for describing some relativistic effects for the heavy elements.

And again we return to the concept of scope of the question. If we consider even modestly complex molecules like say, ethane, the idea of a molecular orbital versus the 'hybridized' carbons and their resulting orbitals appear to be somewhat at odds.  What we tell the sophomores is different than what we tell the graduate students.

When valence atomic orbitals start to interact with orbitals on another atom or molecule (e.g. bond formation), the valence electrons respond by forming molecular orbitals with the electrons on the other atom or molecule.  The core electrons also respond to this bonding interaction.  Their energy levels and spatial distributions are perturbed.  The more deeply buried the core electrons, the smaller the perturbation - but the core electrons will also be affected by the bonding interaction.  If one were able to write out the exact wave function to describe the new molecule that was formed, the wave function would include the core electrons.  Their importance would be much less than the valence electrons, but they would be involved in describing the total bonding situation.

When the electronic structure of a molecule is represented by molecular orbitals, all electrons occupy molecular orbitals.
Molecular orbitals can be approximated as linear combinations of atomic orbitals.  A complete basis set for the molecular orbitals is an infinite set of atomic orbitals.
A finite set of atomic orbitals will be a less accurate approximation of the molecular orbitals.
So if you represent a low-lying molecular orbital as only a linear combination of core atomic orbitals, for example the lowest energy sigma MO of NN as only linear combination of the two 1s-1s orbitals, there will be some error associated with this approximation.  And if you do not include the 1s orbitals in the representation of the the higher molecular orbitals there will be some error associated with this approximation.  However, in the cases that I know of the error is small.
For example, in the following early molecular orbital SCF LCAO papers, the 1s orbital are linearly combined (mixed, hybridized) with the other atomic orbitals:
SCF LCAO MO Study of Li2 J. Chem. Phys. 27, 369 (1957)
An SCF LCAO MO Study of N2 J. Chem. Phys. 23, 569 (1955)
Electronic Population Analysis on LCAO–MO Molecular Wave Functions. I J. Chem. Phys. 23, 1833 (1955)
As the nitrogen paper summarizes "As was expected, the inner-outer-shell mixing
is extremely small".


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Why does sulfur form S8 instead of S3?
P forms $\ce{P4}$ due to its inability to form double/triple bonds. So it makes a sort of tetrahedral structure.
S however makes $\ce{S8}$ cycles w/ $sp^2$ hybridization. Why go all that way to connect the chain? Can't $\ce{S3}$ do the trick? It will still have $sp^2$. What makes $\ce{S8}$ more stable?

please avoid latex in titles due to searchability issues.  http://meta.chemistry.stackexchange.com/questions/253/should-we-revisit-the-url-slug-issue

The ring you propose would have severe issues with ring strain, and would quickly 'pop-open' into a $\ce{S3}$ chain.
All other chains have similar benefits and penalties.  The larger the ring, the greater the entropic cost for formation in a growing chain model.  this is traded off with ring strain: too tight causes one kind of strian and too loose causes another.
$\ce{S8}$ is the 'sweet spot' for handling these competing factors.
As far as phosphorous goes, my understanding is that $\ce{P4}$ is the most reactive form of molecular phosphorous, and other allotropes are thermodynamically preferred.

White phosphorus $\ce{P4}$ is the only molecular phosphorous at standard temperature and pressure. All other modifications are polymer forms $\ce{P_{$n$}}$ with well knows structures, only red phosphorous is amorphous.

Trisulfur does exist, but is not circular, it is ozone-like and unstable at common temperatures.
My guess is that while angle strain does exist in ring form, it is not that harsh. Unlike carbon, having natural bond angle ~109 degree, for bivalent sulfur it is closer to 90 degree (sulfur exhibits little to no hybridization).  However, the S-S bond strength is only 40% of C-C bond and in trisulfur ring  all S-S bonds would have dihedral angle equal to zero, which is unfavorable because of nonbonding pairs interaction. This issue is a lot more severe than in tetraphosphorus, and leads to ring opening.
For further information on sulfur allotropes see the Wikipedia article.

$$\scriptsize
\underbrace{\left[\ce{S_{\alpha} <=>[95.6~^\circ\mathrm{C}] S_{\beta}}\right]}{\normalsize\text{solid}}
\ce{<=>[119.6~^\circ\mathrm{C}]}
\underbrace{\left[\ce{S{\lambda}<=>S_{\pi}<=>S_{\mu}}\right]}{\normalsize\text{liquid}}
\ce{<=>[44.46~^\circ\mathrm{C}]}
\underbrace{\left[\ce{S8<=>S7<=>S6<=>S5<=>S4<=>S3<=>S2<=>S}\right]}{\normalsize\text{gaseous}}$$
The liquid and gaseous phases are temperature dependent equilibria. Holleman Wiberg

@Martin For the sake of completeness we should note that $\ce{S_{6+}}$ ring may be prepared and isolated at relatively low temperatures. Why exactly $\ce{S8}$ is the only one obtainable from common sources is indeed an interesting question without simple and clear answer.

Absolutely right! I was typing up the formula for an answer myself, but then I noticed, you already stated what I had to say, so I just left the hard coding as a comment ;)

Trisulfur or thiozone is significantly present in the gas phase of sulfur at high temperature and has been studied by rotational spectroscopy as reported in The Rotational Spectrum and Geometrical Structure of Thiozone J. Am. Chem. Soc., 2004, 126 (13), pp 4096–4097
The angle of $\ce{S3}$ is very close to isoelectronic $\ce{O3}$ (116.8 degrees), $\ce{SO2}$ (119.5 degrees), and $\ce{S2O}$ (118 degrees).
The above article states that the central sulfur is $sp^2$ hybridized.
Even though the ground state is bent, the cyclic form is only slightly higher in energy, about 5-10 kcal/mol higher.
There is evidence that the cyclic form can exist as a ligand of a transition metal complex, as reported in Transition Metal Complexes of Cyclic and Open Ozone and Thiozone J. Am. Chem. Soc., 2005, 127 (4), pp 1278–1285.  According to this reference, early calculation incorrectly showed the cyclic form to be more stable for both ozone and thiozone "due to the fact that electron correlation plays an important role in these molecules, especially in the open form".


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Can hydride shift and methyl shift happen one after another in a carbocation?
Can a hydride shift be followed by a methyl shift for stabilizing a carbocation?
In case both are possible, which will occur first?

Are you looking at a specific compound? In principle, the answer is yes.

Yes it is possible. But it must fulfil some basic condition for that shift.

The empty p-orbital of the carbocation must be in plane with the $\ce{C-H}$ or the $\ce{C-C}$ bond. When the alignment of a carbon–hydrogen bond with a vacant p orbital takes place allowing for hyperconjugation, a "pseudo-double-bond" develops. As illustrated in Figure 5.13, this can be envisioned as a double bond with a closely associated hydrogen ion.
If, as shown in Figure 5.13, hyperconjugation results in the formation of species possessing both double-bond character and associated hydrogen ions, equilibrium-controlled migration of the associated hydrogen ion can be expected. This transformation, shown in Scheme 5.8, is known as a 1,2-hydride shift and results in the migration of a proton
  from carbon 1 to carbon 2.
While the example illustrated in Scheme 5.8 shows equilibrium between two chemically identical carbocations, there are factors influencing the direction of these transformations when applied to more complex systems. If we consider Scheme 5.9, we notice that the positive charge migrates exclusively to the tertiary center, reflecting the increased stability of tertiary carbocations over primary carbocations. In general, where 1,2-hydride shifts are possible, rearrangement of less stable carbocations to more stable carbocations is expected.
It is important to remember that hydride shifts occur much more readily than the corresponding alkyl shifts. In fact,as a general rule, alkyl shifts will not occur unless a hydride shift cannot take place.
Alkyl migrations occur when the resulting carbocation is more stable than the starting carbocation. But for hydride shift its not such a hard and fast rule.

Source: "Arrow Pushing in Organic Chemistry, An Easy Approach to Understanding Reaction Mechanisms" by Daniel E. Levy; Copyright 2008 by John Wiley & Sons, Inc.; page 110 in chapter 6

I downvoted this answer because the quoted part does not even once mention the possibility/occurrence of multiple shifts.

Multiple rearrangements can happen if the resulting species is stabilized by it. In general hydride shift is more favourable over methyl shift because of its smaller size.
The example proposed by James was not known to me but its truly an amazing example.

James:
  It’s possible for multiple hydride/alkyl shifts to occur. One amazing example is in the biosynthesis of lanosterol.
Multiple shifts are certainly possible, and they could happen, but generally will only happen if each shift generates a successively more stable carbocation.
  For example, you probably wouldn’t see a shift if it involved turning a tertiary carbocation into a secondary carbocation.

Source
From Wikipedia:


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Why doesn't the Bessemer process oxidize the iron?
The Bessemer process for making steel involves blowing air or oxygen thru hot iron. Impurities and carbon oxidize away. Why does the iron itself not oxidize?

I thought you asked a great question, one that I have wondered about myself.
The answer, as best I could determine, is that the iron does oxidize.  A good description of the Basic Oxygen Steelmaking process, today's descendent of the Bessamer process, says so anyway.  The BOS process uses pure oxygen instead of the original Bessamer processes's air.  It reports that some iron does oxidize.
The oxidation of carbon to form carbon monoxide is more thermodynamically favorable than the oxidation of iron, so it occurs preferentially.  The page I found (once again the link is here) mentions using on the order of 1800 scf of oxygen per ton of metal.  A ton of steel is about 18 kmol of iron.  1800 scf of oxygen is about 2 kmol of oxygen.  So there is not enough oxygen to oxidize even close to all the iron in the average BOS run.  That fact, combined with the fact that carbon, silicon, and other impurities in the pig iron oxidize preferentially, mean that not much iron is oxidized during BOS or Bessamer.
But "not much" does not mean zero.  My guess is that iron oxides are more soluble in the slag than they are in molten iron and so are also removed with the slag after BOS.


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What is the mean curvature of nitrates and nitrites?
I am reading the book Soft Matter Physics - Synthetic and Biological Self-Assembling Materials from Hamley and particularly page 201.
I started to think how this packing occurs in case of molecules or molecular patches, particularly in case of nitrates and nitrites because they have highest priority set by WHO in their drinking water guideline.
The curvature can be defined as the amount by which a geometric object deviates from being flat.
You can assume that these molecules clamp together such that there is no other interactions.
How can you approximate the mean curvature of nitrates?


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Why does Si2H4 readily polymerize at RTP but ethene does not (requires catalyst)?
In the case of both Silicon and Oxygen, the double bond is less stable than two single bonds. Consequently, surely both should polymerize, why don't they? What accounts for the difference between Carbon and Silicon in this context?

Please avoid latex in titles due to searchability issues http://meta.chemistry.stackexchange.com/questions/253/should-we-revisit-the-url-slug-issue

Why does $\ce{Si2H4}$ readily polymerize at RTP but ethene does not?

$\ce{Si2H4}$, the silicon analogue of ethylene, is referred to as disilene, a member of the silene family.

The silicon atoms in disilene are $\ce{sp^2}$ hybridized and the orbitals  overlap to form both a sigma and pi bond, analogous to the hybridization and bonding in ethylene. However, the $\ce{Si=Si}$ bond length is significantly over 2 angstroms, whereas the double bond length in an ethylene is much shorter at 1.34 angstroms. This much larger double bond length in silenes results in extremely poor p-orbital overlap which yields an extremely weak silicon-silicon pi bond.  While it takes energies on the order of 65-70 Kcal/mol to cause rotational cis-trans isomerization about a carbon-carbon double bond, it only takes 10-20 Kcal/mol to effect similar isomerization about the silene double bond.
Since the pi bond in a silene is so weak it doesn't require much energy to disrupt it making the molecule very reactive, much more so than an olefin.  The molecule can polymerize or react with whatever else may be around such as oxygen, water, etc.
Replacing the 4 hydrogen atoms in disilene with bulky groups stabilizes the molecule and allows the molecule to be isolated and characterized.  The bulky groups make it more difficult for another molecule to get close enough to react with the disilene pi bond.

Still, even these sterically protected molecules react with oxygen and water at temperatures not far above room temperature.  Here is a link to a full paper that discusses the preparation and reactivity of these sterically protected silenes.


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Aromaticity-Bromonium Ion

Is the bridged ion intermediate aromatic or not?
What is the way to quickly identify aromaticity?

A chemist would say that the reaction proceeds through an "aromatic transition state". The bridged, chloronium ion transition state drawn in your question involves 2 electrons.  No electrons from the electrophile were used, just 2 electrons from the pi-bond that it added to (draw the resonance structures for the bridged intermediate, each has 2 electrons in the $\ce{C-Cl}$ bond and no electrons in the carbocation p-orbital).  If a transition state contains 4n+2 electrons, then it is said to be aromatic. This case has n=0.
Here is another example. The top line in the figure below shows a suprafacial 1,5 hydrogen shift.  6 electrons are involved here, so the transition state is also referred to as aromatic (4n+2, n=1).

image source (look far down on the page)
Like the bridged halonium ion, this reaction is also very common because it too has an energetically favored (aromatic) transition state.
The second line in the figure depicts a suprafacial 1,3 hydrogen shift.  4 electrons are involved here, so the transition state does not fit the 4n+2 rule.  The transition state is said to be anti-aromatic.  Because it is disfavored energetically (anti-aromatic), the suprafacial 1,3 hydrogen shift does not occur (at least not via a low-energy concerted pathway).
Extra Credit: I don't want to jump to far ahead, but let me briefly say, that what I've written above applies to Hückel transition states, opposite rules apply to Möbius transition states.  See this earlier answer, and references therein, for an explanation of the Hückel and Möbius concepts.

Hmmm... Then what is the mechnanism by which the bromination of acetylene takes place?

@TanYongBoon  In the OP's question it is stated that the mechanisms for bromination and chlorination are similar.

I really hope I am counting that correctly; the bridged-ion transition state appears to have 4 pi electrons. It is driving me a bit mad right now. And that would contradict this whole answer, which is puzzling me even more.

Taken from Advanced Organic Chemistry: Part A: Structure and Mechanisms
By Francis A. Carey
The bridged ion may also be considered to be antiaromatic. Thus, while bromonium ions are relatively accessible intermediates for alkenes, the activation energy to reach the corresponding species is much higher for acetylenes. The bromonium ion remains the most accessible intermediate in many cases, but it is energetically relatively less favorable than for alkenes.

Your answer seems to contradict Ron's answer. Could you show how the bridged ion is in fact antiaromatic?

Read it: https://www.masterorganicchemistry.com/2017/03/27/antiaromaticity/

the concept is too simple !!

I fully understand antiaromaticity. So I suppose this is a 4 pi electron system with 2 4p electrons from Br and 2 pi electrons from the pi bond?


text:
Do core electrons in bulk material have discrete energy levels or continuous energy band?
The energy band of valence electrons gets smaller and smaller when the lattice (bulk) atoms are more and more separated from each other, until discrete energy levels are formed, due to the isolation of the atoms from each other (see image below).
What happens with energy levels of core electrons in the bulk, as the lattice constant is changed? Do they form a continuous energy band in the first place? Why?

(source)
If core electrons form a continuous energy band, how come K,L,M,... shells are differentiable in xray spectroscopy? In other words, if all electrons form one continuous band, how can xray have element specificity?

(source)

What happens with energy levels of core electrons in the bulk, as the
  lattice constant is changed? Do they form a continuous energy band in
  the first place? Why?

Using an organic example, as you move from ethylene to butadiene to hexatriene, you have 1, 2 and 3 bonding and antibonding pi molecular orbitals respectively. If you continue this chain extension and wind up with a very large polyene, the energy difference between all of the bonding (or antibonding) orbitals becomes smaller and smaller.  In addition to this compression between energy levels, each molecule is in a slightly different environment.  Some are near the surface, some deep in the bulk, some have a structural deviation or packing non-uniformity close by.  This serves to broaden the energy width of each orbital.  Between the energy broadening of an orbital and the compression of the energy separation between orbitals, one can imagine a limiting case where all of the bonding orbitals are so close to one another that one continuous band, the valence band, exists. The same would be true of the antibonding orbitals and they would merge into a continuous conduction band. The only remaining separation is the separation between the valence band and the conduction band.  This separation is referred to as the band gap.
In the real world, at least at normal pressures, this limiting case of a single band comprised of all bonding orbitals seems unlikely. Instead, some orbitals where the energy separation between them is large likely form separate bands.  In this case, the HOMO would still broaden significantly (as would the other molecular orbitals) and transform into the valence band, remaining distinct from the other bands at lower energies.
Still, it is difficult to probe the bulk and to understand exactly just what the situation is there. X-ray spectroscopy probes to a depth of only a micron or so, so it is really examining the surface states.  Certainly at least some of the molecules at the surface retain their individual orbital identity and are not part of the band structure.  Therefore it is not surprising that XPS (since it probes only the surface) can detect individual orbitals and serve to identify materials.

what does it have to do with core electrons (not valence)?

Like I said, all orbitals, including core electrons like the 1s in an atom or the lowest occupied MO in a molecule, will fuzz out and widen into a band that will eventually merge with other nearby bands (like the 2s).  As the density of states further increases, all of these bands merge into one continuous band, the valence or conduction band.

in light of your answer, could you also explain how come K,L,M,... shells are differentiable in xray spectroscopy? in other words, if all electrons form one continuous band, how can xray have element specificity?

It is difficult to know in advance at what level to write an answer.  Generally, I try to keep it simple.  I've gone back and edited my answer to try and make it more complete.

I'm confused. On one hand XPS can detect individual orbitals and serve to identify materials i.e. core electrons reside in distinct orbitals; on the other hand all orbitals will fuzz out and widen into a band that will eventually merge with other nearby bands i.e. no distinct orbitals.

That's the difference between surface and bulk states.


text:
What substance has the lowest solubility product?
What substance has the lowest $K_\mathrm{sp}$ and what is its value? The lowest I could find is $2.6\cdot 10^{-124}$ for cobalt(III) sulfide $\ce{Co2S3}$.

Could you add a reference for this extreme value of the solubility product of cobalt(III) sulfide?

This is somewhatartificial.  The $K_ {sp}$ value for mercury (II) sulfide involves two ion terms and the one for cobalt (III) sulfide has five.  On a per-ion basis, assuming (not really accurate) a simple dissociation reaction when the solid dissolves, mercury (II) sulfide beats cobalt (III) sulfide.

The $K_\mathrm{sp}(\ce{Co2S3})$ value of magnitude of $10^{-124}$ appears in paper by Goates et al. [1].
The refined "thermodynamic" value of $\pu{2.6E-124}$ that you've listed and is used by numerous textbook up to these days has been proposed by Waggoner [2].
However, thirty years later (late 1980s) there's been another study by Licht [3], which showed significant deviation from the previous studies due to two factors:

A new value of the free energy of sulfide ion formation for aqueous solutions  has been used:
$$Δ_\mathrm{f}G^\circ(\ce{S^2-(aq)}) = \pu{(111 ± 2) kJ mol-1}$$
Previous value [1] was
$$Δ_\mathrm{f}G^\circ(\ce{S^2-(aq)}) = \pu{20.64 kcal mol-1} \approx \pu{83.68 kJ mol-1}$$

Free sulfide activity $a(\ce{S^2-})$ used for $K_\mathrm{sp}$ determination

$$
\begin{align}
\ce{M_xS_y &<=> x M^{$2y/x$+} + y S^2-} &\qquad K_\mathrm{sp} &= a(\ce{M^{$2y/x$+}})^x \cdot a(\ce{S^2-})^y \\
\ce{HS- &<=> H+ + S^2-} &\qquad K_2 &= \frac{a(\ce{H+})\cdot a(\ce{S^2-})}{a(\ce{HS-})}
\end{align}
$$

has also been misinterpreted in early studies by acidification of hydroxyl $\ce{OH-}$ instead, erroneously substituting $K_2$ by $K_\mathrm{w}.$

Table I from [3] lists more recent $\mathrm{p}K_\mathrm{sp}$ values, among which the one for $\ce{Co2S3}$ that has increased significantly $(\mathrm{p}K_\mathrm{sp}(\ce{Co2S3}) = 49.9,$ $K_\mathrm{sp}(\ce{Co2S3}) \approx \pu{1.26E-50}).$
Five least soluble sulfides from that table are:
$$
\begin{array}{lrc}
\hline
\ce{M_xS_y} & \mathrm{p}K_\mathrm{sp} & K_\mathrm{sp} \\
\hline
\ce{Ir2S3} & 196.3 & \pu{5.0E-197} \\
\ce{Bi2S3} & 115.1 & \pu{7.9E-116} \\
\ce{Mo2S3} & 107.8 & \pu{1.6E-108} \\
\ce{Ni3S4} & 104.5 & \pu{3.2E-105} \\
\ce{In2S3} & 96.3 & \pu{5.0E-97} \\
\hline
\end{array}
$$
So that it seems like the new champion (at least among sulfides) is iridium(III) sulfide $\ce{Ir2S3}$ with $K_\mathrm{sp} = \pu{5.0E-197}$.
References

Goates, J. R.; Gordon, M. B.; Faux, N. D. Calculated Values for the Solubility Product Constants of the Metallic Sulfides. Journal of the American Chemical Society 1952, 74 (3), 835–836. DOI: 10.1021/ja01123a510.
Waggoner, W. H. Textbook Errors: Guest Column. The Solubility Product Constants of the Metallic Sulfides. Journal of Chemical Education 1958, 35 (7), 339. DOI: 10.1021/ed035p339.
Licht, S. Aqueous Solubilities, Solubility Products and Standard Oxidation-Reduction Potentials of the Metal Sulfides. Journal of The Electrochemical Society 1988, 135 (12), 2971. DOI: 10.1149/1.2095471.

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


text:
Exercising to use up energy consumed
Suppose I eat a whole $450~ \rm{g}$ chocolate Easter rabbit. The enthalpy of combustion of chocolate is $−22.4~\rm{ kJ/g}$. One kilogram of body fat contains $39000 ~\rm{kJ}$ of energy. Walking for an hour uses up $690 ~\rm{kJ}$ of energy.
For how many hours do I have to walk to use up the energy consumed?
So far I have calculated the energy found in $450 ~\rm{g}$ of the Easter bunny by multiplying $450~ \rm{g}$ by $-22.4 ~\mathrm{kJ/g}$. This gave me $-10080~ \mathrm{kJ}$.
I’m thinking my next step would be dividing $-10080 ~\mathrm{kJ}$ by $690 ~\mathrm{kJ}$ to find the answer. But I feel this is incorrect because I don’t understand the relevance of $1~\mathrm{kg}$ of body fat containing $39000 ~ \mathrm{kJ}$.

I don't think the energy contained in body fat is of any relevance (just like you said). It may just be given to confuse the person attempting the question.

As binary geek said, that information is not required in the calculations. Its probably given to make people realize that by gulping chocolate and walking for just 5-10 mins wont make you reduce your weight. Standard Satire has been employed!

You must understand the extra information has nothing to do with any of your calculations. It is just there to throw you off although it is satirical. You must understand that if you feel that you have arrived at the right answer, and it makes sense, then there is no reason for you to second guess yourself, especially in Chemistry.
Thanks. :)


text:
How to arrange carboxylic acids (phenylacetic acid and acetic acid) by increasing acidity?

Why is $\ce{Ph-CH2-COOH}$ more acidic than $\ce{CH3COOH}$ although the equilibrium of ionization lies mainly backward for both of them?

Attempt:
In the ionized form of $\ce{Ph-CH2-COOH}$ we have a $\ce{Ph-CH2}$-group donating its electron to carbonyl carbon while in case of acetic acid we have a methyl group donating its electron to the carbonyl carbon.
But which one of them is stabilized more?

The $\mathrm pK_\mathrm a$ difference between phenylacetic acid and acetic acid is around 0.5 and the trends become more obvious when other arylalkanoic acids are included in the comparison. Numbers in red are the $\mathrm pK_\mathrm a$ values.

Apparently, phenylacetic acid is a stronger acid than acetic acid. Adding a further phenyl subsituent even increases the acid strength.
The effect levels off with the distance of the phenyl substituent and the carboxylate. Note that 4-phenylbutyric acid is just as strong as acetic acid. It's as if there isn't any phenyl group at all.

Would you think that there is a $-I$ effect of the phenyl substituent on the $\ce{-COOH}$ group?

Phenyl ring is supposed to exert +I effect on carbonyl group


text:
How do I solve for Ksp
The problem gives that the solubility of Silver dichromate is $8.3\times10^{-3} g/100mL$. I need to find $K_{sp}$ which is supposed to be $2.8\times10^{-11}$.
$K_{sp}$ is $\ce{[Ag+]^2[Cr2O7^{2-}]}$
I changed the $8.3\times 10^{-3}$ from g/mL to mol/L and got $1.9\times 10^{-8}$.
Silver dichromate breaks into 2 Ag and one dichromate.
So that would be $(2 \times 1.9\times10^{-8})^2 \times ( 1.9\times10^{-8})$
That's what I did but it's not giving me the right answer. Where am I going wrong here?

The problem gives that the solubility of Silver dichromate is $\mathrm{8.3\cdot10^{-3} g/100\,mL}$.
[…]
I changed the $8.3\cdot10^{-3}$ from g/mL to mol/L and got $1.9\cdot10^{-8}$.
[…] Were am I going wrong here?

Was the solubility given in $\mathrm{g/100\,mL}$?
$$\mathrm{8.3\cdot10^{-3} g/100\,mL = 8.3\cdot10^{-2}\,g/L = \frac{8.3}{431.72}\,mol/L = 1.92\cdot10^{-4}\,mol/L}$$

Thanks. I was doing something stupid going from mL to L.

My pleasure! Blame it on the late night or the early morning ;)

Haha, yea. Think I pull way too many of those (late nights). I'm lucky the internet makes it easy t find a fresh set of eyes to give a second look.

Yes, I'm not on CST. It's 08:00 over here.

It's just about 3:30am here. Kind of the only time it's quiet in here so...


text:
What is more stable: a carbonate ion or a sulphate ion?
The sulphate ion has a higher number of resonance structures than the carbonate ion. Hence, it should be more stable than the carbonate ion.
But my book says that the carbonate ion is stronger than the sulphate ion. Where am I going wrong?

Stronger in terms of what character? Basic?

Do you mean by "stable" that the ion stays in its form and does not re-admit  protons to re-form the acid?

Stability is a type of unicorn. You do not know what it means a priori and, most importantly, you cannot deduce it from a structural formula a priori. The only valid measure of stability is thermodynamic stability according to standard formation enthalpies. The following values are taken from the NIST Chemical Webbook:
$$\begin{array}{ccc}\hline
\text{compound} & \Delta_\mathrm{f}H^0_\text{(liquid)} & \text{source}\\
\hline\ce{Na2CO3} & -1108.51~\mathrm{kJ/mol} & \text{[1]}\\
\ce{Na2SO4} & -1356.38~\mathrm{kJ/mol} & \text{[2]}\\
\hline
\end{array}$$
Gas phase heats of formation would be even better for comparison because then you could consider ‘single’ species, but at least the liquid phase does not contain a significantly skewing lattice enthalpy term.
According to this data, sulphate should be more stable than carbonate. This is not simply because of simple explanations such as:

‘more resonance structures’ (sulphate actually has only one significant resonance structure, carbonate three)
geometrical features such as being tetrahedral
electronegativity (carbon is less electronegative, one would assume its anion to be more stable)
resistance to reduction
an equilibrium with a gaseous compound
whatever else your teacher may tell you.

The thermodynamic stabilities of carbonate and sulphate do not compare well with each other since there are far too many differences between the two molecular ions. You might as well ask whether iron or carbon are more stable.

Sources:
[1]: M. W. Chase Jr. J. Phys. Chem. Ref. Data, Monograph 9, 1998, 1.
Link to the relevant Webbook page.
[2]: M. W. Chase Jr. J. Phys. Chem. Ref. Data, Monograph 9, 1998, 1.
Link to the relevant Webbook page.

Are those data for the solid state? If so they shouldn't be comparable because the enthalpies of formation would include lattice energy terms

@orthocresol Daaa~mn, didn’t think about that. Oh well, even the more reason to say that stability is a unicorn.

Indeed, I agree there's really no basis for comparison, the constituent atoms aren't even the same so it's not even like comparing cis and trans isomers of an alkene..

@orthocresol I’d love to hear your input about how the liquid formation enthalpies don’t compare =D


text:
Can a strong electrolyte be sparingly soluble?
A strong electrolyte is an electrolyte that completely, or almost completely, ionizes or dissociates in a solution.
My book has a question:

In a saturated solution of sparingly soluble strong electrolyte $\ce{AgIO3}$ the equilibrium with sets in is $\ce{AgIO3→Ag+ + IO3-}$

Is the above statement wrong or is there anything as sparingly soluble strong electrolyte?

It means that by just adding it to water wont make it dissociate into ions(sparingly soluble) but if current is passed through its solution, it will easily dissociate and give ions. These cases look rare.
But Curt.F's reasoning looks better :

In the low concentration regime where the salt is "soluble", it completely dissociates, but that the solubility limit is just low.

Do you have a reference for that?  I would naively expect the answer to be different from what you said: that in the low concentration regime where the salt is "soluble", it completely dissociates, but that the solubility limit is just low.  Am I wrong?

@ChemExchange You should write such things as comments not answers


text:
Why is chloroethene said to be a conjugated system?
Why is chloroethene said to be a conjugated system?Is it because of p orbitals present in chlorine or something?

The conjugation is weak though because the Chlorine lone pair is in a 3p orbital but the pi bond is consists of two 2p orbitals

Conjugation can be between pi-bond and another pi-bond, lone pair of electrons and a pi-bond(as in your case) and pi-bond and a vacant orbital(p or d).

Chloroethene is said to be a conjugated system because the lone pair on the chlorine atom in in conjugation with the double bond between the 2 carbon atoms.
See this image for the resonating structures.

Here also: is called resonance and resonance structure. It has nothing to do with resonating.

Chloro ethane is a conjugated system because of the +M effect of chlorine with the double bond of ethene. Chlorine has a tendency to resonate with a double bond by donating its lone pair. What this means is that the lone pair of chlorine is in an unhybridised atomic p orbital in the same plane as the p orbital of the adjacent double bonded carbon atom, due to which, the lone pair "jumps" from the p orbital of chlorine to the adjacent carbon (forming a double bond) and displaces the original double bond to the next carbon.

resonates to

This kind of explanation simplifies the concept beyond the useful. The concept you are trying to explain is resonance, which has nothing to do with resonating. Equally bad is using jump for electrons, also when in quotation, because this simply isn't happening. There is no displacing. Resonance is a superposition of contributors.

I am sorry for the use of non technical terms, i am a little new to chemistry


text:
Stirring the reaction between hydrochloric acid and a powdered substance
For an experiment, I'm adding several substances (as powders) to hydrochloric acid, and measuring the temperature changes to calculate enthalpy. However when adding the powders, the mixture requires some stirring to get all of the substance reacting.
As I am measuring temperature via a temperature probe hooked up to a data logger (to obtain a temp. against time graph), I need a way to briefly stir the mixture in a controlled way that won't affect the change in temperature (reactions are both endo and exothermic). I have tried stirring with the probe itself, but this is not ideal as the probe is pretty sensitive and friction/bumping with the side of the vessel may affect the recording.
I initially looked at a magnetic stirring to obtain uniform and controlled stirring, but then I realized the magnet would absorb heat and potentially increase the temperature of the mixture through kinetic energy. So I'm after a stirring method that will not affect temperature, so that the temperature change I record is solely due to the reaction.
Any suggestions are welcome, because I can't think of any tool or method that would suit.

Since you are measuring variation, stirring will not affect the result, I mean, just take the initial temperature under stirring, then then the difference would be the same with or without stirring.

I'm not 100% sure what you mean. The graph of temp. against time looks like this, and stirring tends to make the curve a bit bumpy and difficult to extrapolate.

I mean, to measure temperature change, you have to make a difference. So if the two parts of the difference are impacted by the same perturbation, they'll cancel.

Do you mean hydrogen chloride or hydrochloric acid (aqueous) ?

@Lighthart Sorry, I meant hydrochloric acid. Fixed it up.

To minimize thermal coupling between the stirring apparatus and the solution, the best choice is a small, slowly stirred, PTFE or equivalent coated magnetic stir bar. Heat generated by friction should be negligible.
A potential source of heat transfer into the system from this configuration is heat from the stirrer motor. To mitigate this, the surface of the stir plate should be covered with a strip of Styrofoam as thermal insulation. Furthermore, the reaction vessel could be physically separated from the stir plate by using a ring stand and clamp or equivalent. It may be necessary to adjust the thickness of the Styrofoam and distance of physical separation in order to keep the stir bar magnetically coupled to the stir plate magnet.
The setup should first be tested by doing several full runs without adding the acid or the powders. This is essentially an analytical blank. If the temperature changes can be considered negligibly small throughout the duration of an experiment, then you are ready to go. If necessary, any temperature change measured from the blank runs can be subtracted from the actual experimental measurements.


text:
Decomposition products of hexaurea-chromic-chloride-trihydrate
I am sitting over a TG diagram of the decomposition steps of hexaurea-chromic-chloride-trihydrate ($\ce{[Cr(CO(NH2)2)6]Cl3 \cdot 3 H2O}$) and cannot really make sense of the products it might decompose into.

It is clear that water leaves first (-3.5%), and the next step might be (with a wide error margin) fumes of $\ce{CO2}$, $\ce{H2O}$ and $\ce{2 NH3}$. What is left is then $\ce{[Cr(CO(NH2)2)5]Cl3}$.
I have not found any resources on what might happen further - the four urea ligands could decompose into $$\ce{4 CO} + \ce{5 NH3} + \ce{3/2 N2} + \ce{1/2 H}$$ and I am left with $\ce{CrCl3}$ and urea, but those steps show up last (17.9% + 17.3% + 5.4% = 40.6%), leaving the 15.5% in the middle unexplained.
What other decomposition could take place after water and two urea ligands are gone? Is there a way to figure that out other than by guessing or experiment?

Did you consider this article on the corresponding nitrate?

Using the article Klaus Warzecha linked to and some trial-and-error, a possible decomposition could look like this and is in accordance with the TG diagram in the question:

Decomposition of $\ce{3 H2O}$ and $\ce{2 CO(NH2)2}$:
$\ce{[Cr(CO(NH2)2)6]Cl3 $\cdot$ 3 H2O -> [Cr(CO(NH2)2)4]Cl3 + 3 H2O \uparrow + 2 CO(NH2)2 \uparrow}$
Decomposition of $\ce{2 CO(NH2)2}$ and $\ce{2 HCl}$:
$\ce{[Cr(CO(NH2)2)4]Cl3 -> CrH6C2N4ClO2 + 2 CO(NH2)2 \uparrow + 2 HCl \uparrow}$
Decomposition of $\ce{2 CO2}$, $\ce{N2}$, $\ce{3 H2}$ and $\ce{1/2 Cl2}$ and oxidation of $\ce{Cr}$ to form $\ce{CrN2}$:
$\ce{CrH6C2N4ClO2 -> CrN2 + 2 CO2 \uparrow + N2 \uparrow + 3 H2 \uparrow + 1/2 Cl2 \uparrow}$

Given the temperature range and pressure (probably standard pressure), the formation of $\ce{CrO2}$ is unlikely.


text:
The acid in soft drinks dissolves your teeth. Which type of chemical could you use to stop the acid?
The acid in soft drinks dissolves your teeth. Which type of chemical could you use to stop the acid?

What you meant is known as erosion. Erosion is the chemical loss of enamel due to acid.When acid continuously attacks teeth, they cannot repair themselves and will gradually begin to turn fuzzy and dissolve. Most soft drinks contain acids such as citric, phosphoric and malic or tartaric acids. A study was done where extracted human teeth were placed in softdrinks. Within two days, the teeth became very soft and the enamel surface lost much of its calcium.
Brush your teeth at least twice a day with a fluoride-containing toothpaste.A mouth rinse containing fluoride can help prevent tooth decay, according to the American Dental Association. For children, sealants can be used.

Just this week I was talking to my dentist about this, and she said that it is recommended to wait at least 40 minutes after drinking acid beverages (like orange juice and wine) before brushing your teeth. If you brush them right afterwards, the teeth are still very fragile and erosion will be more signifcant, whereas waiting some time will provide enough time for some cleaning and recuperation. Also, there are erosion-specific toothpastes for those with these kinds of problems.

Yeah. There are lots of flourinated toothpaste(recommended for minimizing erosion) available. Likewise, as you have mentioned that we shouldnot brush before 40 min of intake of softdrinks and beverages, same goes for our daily meal as well.


text:
How do we predict the order of melting points of group 1 and group 2 halides?
I get really confused by this:
The order of melting points of group 2 chlorides is: $$\ce{BaCl2 > SrCl2 > CaCl2 > MgCl2 > BeCl2}$$ This is completely understandable by Fajan's Rules.
But, the order for group 1 halides is: $$\ce{NaCl > KCl > RbCl > CsCl > FrCl > LiCl}$$
This is what I don't get! Considering Fajan's Rules, $\ce{NaCl}$ is the most covalent, least cationic size, so shouldn"t it have the lowest melting point?

Melting point depends on Lattice energy. NaCl has similar sized ions which make the lattice stronger. LiCl comes last since it has a lot of covalent character.

Thanks for the answer. So we cant just predict their properties simply using fajans rules, isnt it? We need to know the trends experimentally, and then "explain" them somehow...

Fajan's rules doesnt say anything about melting point. Just about polarization.

For group 1 halides polarisation is not a dominating factor(except for lithium) since the size of cations is relatively larger than any other cation therefore we must consider lattice energy for determining melting point. As interionic distance increases lattice energy decreases hence melting point decreases down the group.
Lithium is an exception because of its very small size that brings high degree of polarisation. you may want to refer to lother meyer's curve on atomic volumes of elements to get relative idea about the size of the elements.

Although lattice energy of LiCl higher than NaCl, but LiCl is covalent in nature and NaCl ionic. There after, the melting point decreases as we move NaCl because the lattice energy decreases as size of alkali metal atom increases. Higher lattice energy denotes higher melting and boiling points.


text:
Difference between radial, planar, angular and spherical nodes
What is the difference between them?
I think radial nodes and spherical nodes are the same, and angular and planar nodes are the same.
Reference
Finally, how many spherical nodes are there in a $2p$ orbital?

Does this answer your question? What are angular and radial nodes?

Yes, radial nodes and spherical nodes are the same, and angular and planar nodes are the same.

The wave function $\Psi(r,\theta,\phi)$ of a one-electron (hydrogen-like) atom is seperable as the product of a radial function $R(r)$ and an angular function $Y(\theta,\phi)$
$\Psi(r,\theta,\phi) =  R(r)Y(\theta,\phi)$
If $R(r_1) = 0$, there exists a radial node.  The radial node is a sphere with radius $r_1$. Therefore the terms "radial node" and "spherical node" are the same.
$Y(\theta,\phi)$ is further seperable as
$Y(\theta,\phi) = P(\theta)F(\phi)$
If either $P(\theta)$ or $F(\phi)$ is zero for a given respective angle value, there is an angular node.  However, a angular node is not necessarily a planar node.  An angular node could be a planar node or a conical node.
$F(\phi)$ being zero corresponds to a planar node.
$P(\theta)$ being zero corresponds to either a conical node or a planar node (some think of the planar case as a specical case of conical, with apex angle being 180 degrees)
Overall, there will be $n-1$ nodes.
$l$ nodes will be angular
$n-l-1$ nodes will be radial (spherical)

Does this mean that all p orbitals have planar nodes and in d, only $d_{z^2}$ has conical node while others have planar node. I am clear about radial/spherical nodes. Thanks.

yes, px, py, pz each have one planar node.  Dz2 has 2 conical nodes and the other d orbitals have 2 planar nodes http://winter.group.shef.ac.uk/orbitron/AOs/3d/


text:
Is RS- more basic than RO-?
The basic strength is determined by the ability of an ion or molecule to accept a proton. How do I know whether RSH is more stable than ROH?
(R is an alkyl group)

My recent answer to this question is quite relevant, with the small difference that one hydrogen gets swapped for an alkyl group. Just remember that stronger acids create weaker conjugate bases.

I think it's the other way around. Thiols are more acidic than alcohols so the conjugate base of a thiol is a weaker base than the alcohol conjugate base.

It isn't. ${}$

It may also be helpful to point out that the strength of a base is only as you define it for a Brønsted-Lowry base.

https://chemistry.stackexchange.com/questions/147841/why-is-negative-charge-more-stable-on-sulphur-than-on-oxygen/147846#147846

RobChem has already pointed out in his comment that your assumption is not quite correct.
Take a look at $\mathrm{p}K_\mathrm{a}$ values for $\ce{ROH}$ and $\ce{RSH}$ in water from the CRC Handbook of Chemistry and Physics and/or other online sources, such as this or that.
\begin{array}{lrr}
\mathbf{R} & {\ce{\mathbf{OH}}} &\ce{\mathbf{SH}}\\
\hline
\ce{H} & 15.7 &7.0\\
\ce{Et} & 15.9 & 10.6\\
\ce{(H3C)3C} & 18.0 & 11.7\\
\ce{C6H5} & 9.9 & 6.6\\
\end{array}
In all the cases above, the $\mathrm{p}K_\mathrm{a}$ value for $\ce{RSH}$ is smaller than that of $\ce{ROH}$.

$\ce{RSH}$ is a better acid than $\ce{ROH}$ (as the $\mathrm{p}K_\mathrm{a}$ of $\ce{RSH}$ is lower than the $\mathrm{p}K_\mathrm{a}$ of $\ce{ROH}$, shown in a previous answer). This means that $\ce{RSH}$ dissociates into $\ce{RS-}$ and $\ce{H+}$ more, i.e. the equilibrium below lies more to the right.
$\ce{RSH + H2O <=> RS- + H3O+}$
To compare between $\ce{ROH}$ and $\ce{RSH}$, we must look at the difference, which is the atom: $\ce{S}$ or $\ce{O}$. As they are in the same group, we look at their polarisability. As $\ce{S}$ is larger and hence more polarisable, it means that the negative electron is more stable in $\ce{RS-}$ than in $\ce{RO-}$. Hence the conjugate base, $\ce{RS-}$ is more stable than $\ce{RO-}$, and so $\ce{RSH}$ is a stronger acid than $\ce{ROH}$.
Conversely, $\ce{RS-}$ is a worse base than $\ce{RO-}$.


text:
What happens in an acid and metal reaction where the acid is twice as concentrated and the metal twice as much in quantity?
Experiment A
3cm metal
0.5M acid
5 degrees temperature change
Experiment B
6cm metal
1M acid
How much would the temperature change be, and how much faster or slower would the reaction take place in?
I thought the answer would be either "the same" or "4x as much". I don't exactly understand the reaction yet; would the doubling of each material make the reaction 4x stronger (2x2), or would they cancel each other out?
Of course, it might've been something else I haven't thought of, too.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website. This is a homework question. We have a policy which states that you should show your thoughts and\or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

I wouldn't exactly call it 'homework'. It was a question in a recent test I've taken; one which the test papers are not given back. I've done a bit of googling and found nothing, then remembered about the stackexchange websites so I tried posting it here.
As for the question itself, I thought the answer would be either "the same" or "4x as much". I don't exactly understand the reaction yet; would the doubling of each material make the reaction 4x stronger (2x2), or would they cancel each other out?
Of course, it might've been something else I haven't thought of, too.

Please read the meta page I linked - Homework here isn't only questions that are homework, exam preparations, self-studies, puzzles etc. count as homework too (so does your question) I think if you simply add the info in your comment to your question this Q will not get closed.

Loooks like your data is oversimplyfied - 6cm not cm3 ? I guess that volume of the acid is the same and much bigger than metal ?

Without providing more detailed specifications of the metal such as: state of the metal - powder or solid, shape of the solid form etc., the rate of the reaction will be the same, but the second reaction will be a bit faster because of the bigger temperature change (as it affects reaction rate coefficient).
I assume that you meant 3 and 6 $cm^3$ and not $cm$.
The same metal reacts in both reactions. So we can assume that the density of both metals is also the same ($\rho_1=\rho_2=\rho$). We can further assume that metals were dissolved completely and that the temperature change is actually temperature increase.
$\rho=\frac{m_1}{V_1}$, $\rho=\frac{m_2}{V_2}$
$\frac{m_1}{V_1}=\frac{m_2}{V_2}$
$\frac{m_1}{3}=\frac{m_2}{6}$
$\frac{m_1}{1}=\frac{m_2}{2}$
The bigger piece of metal weights 2 times more than the smaller metal. We can convert weights to moles (molar mass of both metals is the same):
$n=\frac{m}{M}$
$\frac{n_1}{1}=\frac{n_2}{2}$
so we can say $2*n_1=n_2$
The 1st reaction:
$n_1$ reacts with 0,5M of acid.
The 2nd reaction:
2*$n_1$ react with 1M solution and it is the same as $n_1$ reacts with 0,5M
The volume of the solution is the same and the specific heat capacity will be nearly the same. The reaction enthalpy corresponds to the amount of reacted material. Because two times more material reacts in the second reaction,  the temperature change will be nearly two times bigger (nearly - because of the change in specific heat capacity).


text:
What decides the order of priority of functional groups in organic compounds?
Is the order of defined by IUPAC or we follow some special conventions to know the order of priority?

It’s an arbitrary agreement by IUPAC, although note
that there is some correlation between the
oxidation state of the carbon and the priority
(more oxidized groups tend to be higher
priority). Generally, the functional group which shows its characteristics more dominantly even when present with other group gets a higher priority.


text:
Why macromolecule ligands such as Crown ethers form more stable complexes than open chain ligands?
This was in J. Huheey's book and no explanation was given

The reason for crown ethers being more stable than their equally long open chain derivatives is the difference in Gibbs Free energy. (If you are not used to these terms, the Wikipedia article will help I think, though it is quite, long and thorough.)
The main point to take from the use of the Gibbs free energy in this case is this: it is made up of two parts, enthalpy ($\Delta{}H$) and entropy ($\Delta{}S$).
$$
\Delta{}G = \Delta{}H - T \cdot{} \Delta{}S
$$
If only the binding affinity of the (crown) ether chain is considered, only the enthalpy is relevant. The strength of the individual ether moieties is as strong in a crown as it is in a chain. The difference then, lies with the entropy. To have a crown ether form a structure around a ligand 'costs some entropy'. The same entropy difference should be considered for the chain ether, with added to that the cost of pulling the ends of the chain together! That difference in entropy results in the difference in $\Delta{}G$ and therefore yields a more stable complex. (The ends of the crown ether are also pulled together, this cost of entropy however, has been paid in an earlier reaction when making the crown ether, and therefore has no influence here.)
You can see this also from the relation between the Gibbs free energy and the equilibrium constant of the complex (In the imaginary reaction $\ce{A + B -> AB})$:
$$
\Delta{}G = - RT \cdot{} Ln (K)
$$
where,
$$
K = \frac{[\text{AB}]}{[\text{A}] \cdot{} [\text{B}]}
$$
A higher $\Delta{}G$ leads to a higher equilibrium constant.
Hope this helped, and I'd be happy to clarify anything further.

authors: Tʌŋɱʌƴ Jʌŋʛɩʀ


text:
Analysis project to conduct
We have been asked to conduct an analytic project for chemistry. It is important that the project conducted hold social implications and the conclusion achieved help inform people better. Several ideas have come to my mind for such projects, such as determining amount of phosphoric acid in cold drinks, acetic acid in fruits such as guava, amount of insecticides/pesticides in vegetables/fruits in the market. However, I am still yet to find an idea which is striking. I would like to do something unique. Any suggestions?

Amount of pesticides on fruits? That's really hard to find out!

@MARamezani I realized that after I came up with the idea. But at least estimating it would be viable?

I've replace the new tags with one already used here: [tag:everyday-chemistry]. I hope you're fine with that.

As you might have guessed: It depends!

What is the educational level of the study (High School, College, etc.)?
What's the time frame - how many man-hours?
What's the equipment of the lab?
If equipment or material has to be bought: What is the budget?
Is it a plain academic study or is it intended to go public with the results in order to change/improve consumer awareness?

"Something unique" might mean that

there are no reference data
it is too "academic" to have social implications (or at least that it will be difficult to explain to the public that it actually has)

What about something as "boring" as sugar content in local soft drink brands as compared to similar products on the international market? I'm sure you'll find lots of references, such as this.
Go for something that is relevant for the average Joe Bloggs. Go for something that can be bought thoughout the while time of the study, just in can you have to repeat measurements or need to cover seasonal variations. If it is a fresh, seasonal product, you might have to plan ahead and stockpile and deep-freeze material.
In the end, it is up to you to decide - planning and outlining the study is part of the project.
Speaking of relevance for the average consumer: Everybody needs excrete - but you do not want to examine that. Fortunately, everybody needs to eat too and seemingly, a lot of people tend to eat too much of the wrong stuff. The recent report of the US Dietary Guidelines Advisory Committee might give some directions what to look for.

I'm currently studying in 12th grade. We have approximately 6 months to fully conduct the project, but we have to submit the idea in 1 week. Equipment should not be an issue. It is not required to actually go to the public, but during writing the report, I need to mention how the conclusion of my experiment helps to increase consumer awareness. I will have to do something "boring" if nothing else comes to mind.

@Gummybears Don't worry about the "boring", as long it is carefully conducted, related to known data and well presented. Most of the work is 5% inspiration and 95% transpiration ;)

That's the difficult part. I wanted to turn those figures around xD

@Gummybears Rumour has that everybody wants to achieve more with less. Experience tells that days and nights in the lab can be very long and frustrating ;)


text:
Representing woody biomass concentration for kinetic reactions
Woody biomass is composed of cellulose, hemicellulose, and lignin components. The following diagram provides representative formulas for each component:

Biomass can be represented by it's density, ultimate analysis, and composition. For example, beech wood can be represented by its density, $\rho = 700~\mathrm{kg/m^3}$, its ultimate analysis by weight percent such as $\ce{C}=49.05\%$, $\ce{H}=5.83\%$, $\ce{O}=45\%$, $\ce{N}=0.12\%$, and its composition by weight percent such as $\text{cellulose}=48\%$, $\text{hemicellulose}=28\%$, and $\text{lignin}=24\%$.
The following diagram represents a kinetic scheme for cellulose, CELL, which can be represented by the formula $\ce{C6H10O5}$:

The reaction rates for each pathway are first-order Arrhenius equations where $A$ is the pre-factor in $\mathrm{s^{-1}}$, $E$ is the activation energy in $\mathrm{kcal/kmol}$, $T$ is the temperature, and $\mathcal{R}$ is the universal gas constant:
$$
K = A\cdot \exp{\left(\frac{-E}{\mathcal{R}T}\right)}
$$
To apply the cellulose kinetic reaction to beech wood, I used the density of the wood, $700~\mathrm{kg/m^3}$ to calculate the initial concentration of cellulose as 48% of the wood density, $336~\mathrm{kg/m^3}$.  Using the initial mass concentration of cellulose, the reaction rates at a certain temperature can be applied to that concentration. This approach will give the mass concentration for each group (G1, G4, LVG). Notice that each group represents various chemical species on a mole basis.
Since the individual chemical species are represented on a mole basis, it seems like there are two approaches to solve this problem:

Mass Basis Approach

start with an initial mass concentration of the biomass
divide that mass conc. into cellulose, hemicellulose, and lignin components based on the weight % composition
apply the reaction rates on a mass basis for each component to determine the mass concentration of each group
finally use the mass concentration of the group to calculate the individual chemical species (as mass or moles?) in that group

Mole Basis Approach

represent the biomass as moles of C, H, O, N using the ultimate analysis
divide the biomass into cellulose, hemicellulose, and lignin components
apply the reaction rates on a mole basis to determine the chemical species
atomic balances for C, H, O, N should be checked at each step

How do I attempt to solve this problem on a mole basis?
Or is it best to treat it initially on a mass basis then convert to moles to solve for the individual chemical species?

Since now you know (at least roughly) what kind of molecules you have in your sample, you could find the molar mass of each compound (for hemicellulose you might have to use an average or a representative molecule). Divide the 0.48 * the mass of one $m^3$ of beech wood by the molar mass of cellulose and you end up with the molar density for that; repeat and it should add up to 100%. I will work through this and formulate a real answer.

Possible duplicate of Determine rate equations for entire kinetic scheme

First step is finding the molar masses of each compound. Since cellulose, hemicellulose and lignin are collections of similar molecules, I picked the following as a crude estimate:

cellulose: $\ce{C6H10O5}$ with $M_c = 162~\mathrm{g/mol}$
hemicellulose (L-Arabinose): $\ce{C5H10O5}$ with $M_h = 150~\mathrm{g/mol}$
lignin: $\ce{C31H34O11}$ with $M_l = 582~\mathrm{g/mol}$

Using a density of $700~\mathrm{g/dm^3}$ you will find the relative masses of the three compounds to be:

$0.48 \cdot 700~\mathrm{g/dm^3} = 336~\mathrm{g/dm^3}$ (cellulose)
$0.28 \cdot 700~\mathrm{g/dm^3} = 196~\mathrm{g/dm^3}$ (hemicellulose)
$0.24 \cdot 700~\mathrm{g/dm^3} = 168~\mathrm{g/dm^3}$ (lignin)

With $n = m/M$ we now find the number of moles of each compound in the sample ($1~\mathrm{dm^3}$):

$2.074~\mathrm{mol/dm^3}$ cellulose
$1.307~\mathrm{mol/dm^3}$ hemicellulose
$0.289~\mathrm{mol/dm^3}$ lignin

Keep in mind that this is a very rough estimate. You'll probably get a better result by taking the average of all species of hemicellulose, etc. in your sample and cross-checking it to make sure it fits the ratios of the elements (weight percent).

Let us continue this discussion in chat.


text:
What reaction is happening when a lithium polymer battery "is on fire"
When something goes wrong with a lithium polymer battery (as are commonly used to power electric models and robots), and it "catches fire", what is actually happening?
What gas or gasses are being released?
Here's a video of such an event.

Interesting, it's connected with overheating, but what exactly happens... may I'll find out...

I'll take a stab at this one, but I'm not a battery expert.
What is a lithium "polymer" battery?
Wikipedia says that "lithium polymer battery" is now commonly used to refer to nearly any lithium ion battery that is encased in a flexible polymer shell instead of a hard metal casing.  There is a chance for confusion here because it apparently used to mean a battery with a polymeric electrolyte instead of a purely small-molecule liquid-phase electrolyte.  I'm assuming you mean the former "in a polymer shell" definition for the purposes of my answer.
What type of anode, cathode, and electrolyte materials?
There are many types of lithium ion batteries.  Electrolyte composition, cathode materials, and anode materials can all vary.  Perhaps the most common is a lithium cobalt oxide cathode with a intercalated graphite anode.  Organic solvents must be used as electrolyte because water is not stable to the high voltages that lithium ion cells provide.  Solvents stable over a wider potential range must be used instead to dissolve salts that can carry electric current.
What happens when it's crushed:
So for lithium cobalt oxide batteries with liquid electrolyte, I was able to find a good source describing the problem of "thermal runaway":

The temperature of the cell rises due to the
  chemical reactions between the organic solvent and electrode materials leading to cell failure/explosion. The chemical reaction is prompted because of the coexistence of 1)
  combustible organic solvent with inorganic salt electrolyte 2) lithiated graphite anode and 3)
  partially delithiated LixCoO2 cathode in the charged state. The graphite anode intercalate
  fairly well during the initial cycles. As the cycle proceeds, the available intercalating sites in
  the graphite slowly decreases and the lithium deposition over the surface of the anode
  increases during charging process. Such a situation prompts non-even lithium deposition or
  even dendrite growth over the graphite surface. When the battery is in the charged state the
  cathode remains in a delithiated state. Flow of current over and above the
  tolerable/standard charged state causes instability to the cathode which starts to release
  oxygen into the electrolyte. Thus released oxygen reacts exothermally with the lithium
  plated over the graphite anode and increases the temperature of the cell making the cathode
  to release oxygen further. The cell which was in the dormant state in the absence of oxygen
  and heat now becomes an explosive device in the event the cell is met with any abuse.

That describes what happens when the battery is overcharged and leaks oxygen into the electrolyte.  I suspect that when the batteries are crushed, even if they aren't overcharged, that at least two things can happen leading to similar thermal runaway: First, the lithiated graphite is air- and water- reactive, heating the now-crushed battery and eventually lighting the electrolyte solvent on fire.  Second, the cell may be shorted by the mechanical disruption, so that even without air or water contact, the voltage in the battery can discharge very quickly.  This would create large heating from ohmic resistance, also eventually igniting the elctrolyte.


text:
Does Gallium (liquid or solid) corrode all forms of brass?
I know that it corrodes copper, but does it do the same to all brass or bronze compositions?

From here

Gallium is corrosive to all metals except tungsten and tantalum, which have a high resistance to corrosion.

Beta-Brass is embrittled by Gallium
Due to its highly reducing character it is highly corrosive material.

As I understand it, 316 stainless steel is fairly resistant as long as we are not talking structural use

I found it actually does corrode 316 steel here , here. But here, the penetration seems to be only $12 \mu m$. I probably think under no stress conditions, 316 is resistant to Gallium due to passivation.

I had looked at more direct sources from the Wikipedia link that was provided in the previous answer to this question. On page 99 of the 1954 version of "Liquid-Metals Handbook", there is a chart that might answer your question better:

The next two pages from the the same book gives more information about the other chemicals that leads me to believe that the Wikipedia article might be misquoting the information.

Please embed all the mentioned URLs (you can use standard [text](link) Markdown) as well as an exact and complete reference for the illustration, including author(s)/editor(s), publisher, page number(s) and IDs (ISBN/ISSN/DOI/…), preferably using ACS citation style. Currently it's quite hard to follow the timeline and to understand what quotes what in here. Also note that "Wikipedia" is capitalized.


text:
What chemicals help define "taste"?
As we all know that there are 5 tastes : sweet, sour, bitter, umami, salty.
So what chemicals are there that help form these tastes, in other words how do our tastebuds experience these tastes?

Also can there be more tastes than these basic five. I am not saying that there should be a blend of these to make a taste but rather are there any other undiscovered tastes?
Maybe we haven't tasted them yet?

There really are only 5 "tastes" on our tastebuds. The "flavors" from those tastebuds are derived from a combination of tastebuds and mostly olfaction from our noses.

Well the question in your title is "too broad". Chemicals don't define taste. Taste crudely defines some chemicals. Tasting is a weird process, and is indefinite. An animal may taste sweet what you taste sour.

It's not exactly true that there are 5 "tastes". There are 5 generally accepted classes of gustatory transducers, each comprising many different types. The perception of taste is a product of not only the output of these transducers, but also information from the olfactory system, visual cues, textures, sound, etc.
The gustatory system itself is based (like the olfactory system) on overlapping sensors with partial selectivity. So a given compound, sucrose for instance, may stimulate several sweetness sensors (and probably even some from other classes) to varying degrees, and the particular combination of the responses is what gives sucrose its unique taste.
The types of molecules that stimulate the different sensors is not known clearly, but we can make some broad generalizations:
Sour - These sensors are thought to be primarily ion channels sensitive to pH—acidic solutions produce a sour taste.
Salty - Also thought to be mostly ion channels sensitive to small metal cations. Different ions have somewhat difference response. ($\ce{Na+}$ doesn't have the same response as $\ce{K+}$) There is some debate about the actual mechanism though.
The rest are g-protein coupled receptors. They change shape when they bind a molecule which causes a chemical signalling cascade.
Sweet - Not known precisely, but many compounds containing hydroxyl or carbonyl groups trigger these receptors.
Umami - Mostly triggered by amino acids, particularly glutamate. Inosinate and guanylate are known to enhance the effect of glutamate.
Bitter - This one is rather strange. A very wide variety of compounds stimulate these receptors, even with very different structural features. This taste is thought to have evolved to protect us against toxic compounds as many poisonous alkaloids and such taste bitter. Denatonium is an example of a bitter compound (one of the most bitter known).
As for whether any other classes of transducers exist, it's entirely possible. Some researchers have identified candidates for receptors for things like fats. Also, there some receptors like TRPV1 (which responds to capsaicin) are known that respond to compounds found in food, but aren't usually considered taste receptors.


text:
NaClO Concentration
In my previous occupation the usage of bleach--$NaClO$--I found was quite staggering, and based off of the health issues that I and other coworkers were having, I wanted to see how close we were cutting it to being over exposure limits. Taking the definition, $$ 1 \, \, ppm = \frac{1 \,\,mg \,\,solute}{1 \,\,L \,\,solvent} = \frac{1 \,\,g \,\,solute}{1 \,\,m^3 \,\,solvent} $$
I have a $ 946 \,\,mL $ bottle of solution labeled to be $.65\%$ Sodium Hypochlorite. This leads me to calculate the presence of $6.149 \,\,mL$ of the entire mixture being $NaClO$. Subtracting that from the total volume to get the volume of the solvent, I obtain $.940 \,\,L \,\,solvent$. Using the fact that the density of $NaClO$ is $ \rho =1.11 \frac{g}{mL}$, I obtained $6.83\,\,g$ or $6830\,\,mg\,\,solute$. Plugging these values in for the concentration in $ppm$ I obtain $\frac{6830\,\,mg}{.940\,\,L}= 7265.957\,\,ppm$ ($\frac{g}{m^{3}}$).
The American Industrial Hygiene Association only recommends $2\,\,ppm$ for short term exposure, which leads me to believe that my calculations might be in error in some way. Is this calculation correct or do I have some fundamental misunderstanding about the concept of molarity? I just don't feel like something that's $3633$ times more concentrated than the AIHA recommends wouldn't even make it off the drawing board, let alone into workplaces.

The "0.65%" you refer to is ambiguous.  You took to mean volume percent, but it could mean mass percent.  The would mean the solution was 6500 ppm sodium hypochlorite.  It could also mean the notorious and misleading "% weight-volume", i.e., it could mean 6.5 g/L.  In any case the exposure limits are for airborne concentrations, not for liquid phase concentrations, as Loong's answer correctly notes.

The recommended Workplace Environmental Exposure Level (WEEL) refers to the concentration of airborne substances in the workplace air. This value is not applicable to the concentration in cleaning agents or other products.

0.65% NaClO is lower than normal household bleach, which about 5%.
Guidelines for the Use of Chlorine Bleach as
a Sanitizer in Food Processing Operations
maybe helpful to you.
Apparently, in the USA, it is permissible to wash fruits and vegetables in upto 2000 ppm NaClO.


text:
How concentrated can an acid be?
How concentrated can an acid be without it being supersaturated? Is there a certain limit to how concentrated an acid or base can be?
When I mean concentration, I mean molarity; so how concentrated can an acid or base be?

NOTE: Can be a strong or weak acid/base. Because we are just looking for the amount of solvent needed to dissolve to start the dissociation process.

EDIT: No autoprotolysis please. Acids that need a solvent to dissociate.

are you asking for the strongest acid in water? this is $\ce{H3O+}$. see the solvent levelling effect http://en.wikipedia.org/wiki/Leveling_effect

In terms of theory, "how acidic can an acid be?". It depends on how much of hydrogen ions are dissociate in the water. The more the hydrogen ions dissociated in the water, the stronger the acid is which is very acidic. Examples of strong acids are Phosphorus acid, Sulphric acid, Hydrochloric acid and Hydrofluoric acid. Examples of weak acids are ethanoic acid, methanoic acid etc.

I'm afraid you didn't really answer the question but repeated @asker12 misconception - undissociated acid is still acid and in many cases stronger than hydronium.

The problem is that leveling effect is here a source of misunderstanding - is important in diluted not highly concentrated solutions.

http://chemistry.stackexchange.com/questions/15421/why-is-magic-acid-a-superacid

Are you suggesting that if an acid molecule is surrounded by a sufficiently small number of water molecules, then it might display lower than 100% ionization even if it is a "strong acid" ($pK_a<-1.76$)? That is true; some acids we commonly consider to be strong, such as nitric, hydrochloric or sulphuric acid, will not completely ionize in sufficiently concentrated aqueous solutions, even if there is more than one molecule of water per molecule of acid.
So is there an acid capable of protonating water quantitatively even when there is only one molecule of water per molecule of acid? Yes, in fact there are several. Interesting proof of this is that sufficiently strong acids will form solid crystalline salts containing the hydronium cation when mixed with water in a 1:1 molar ratio. The most striking examples are probably the hydronium carborane salts, such as $\ce{(H3O)^+(HCB11Cl11)^{-}}$. I recommend taking a look at this fantastic article from the Reed group for details on the wonderful properties of carborane superacids, with a section describing its several salts when reacted with water in different proportions and environments.

I am talking about dissolving in water since an acid needs some sort of solvent to dissociate.

No, it doesn't. You neglect autoprotolysis where the water-free acid itself is the solvent.
\begin{align}
\ce{2H2SO4 &<=> H3SO4+ + HSO4-}\\
\ce{3HF &<=> H2F+ + HF2-}
\end{align}

But where does the extra H+ come from?

which extra H+ ?

Even without autoprotolysis lack of solvent is not a problem.


text:
Can you find [SO₃²⁻] with just the two acid dissociation constants Kₐ₁ and Kₐ₂?
Sulfurous acid, $\ce{H2SO3}$, is a diprotic acid.
\begin{align}
K_{a1} &= 1.6 \cdot 10^{-2}\\
K_{a2} &= 6.4 \cdot 10^{-8}\\
\end{align}
Is it possible to find the concentration of $\ce{SO3^2-}$ from this?
I would think you would need to know the concentration of at least one thing. It is like trying to solve for 3 unknowns with 2 equations!

No, it is not possible.
For example, as total concentration of acid approaches zero, concentration of sulfate would also approach zero.

You can use approximations like $K_\mathrm{a1}$ is much greater than $K_\mathrm{a2}$ and that first dissociation goes to almost completion. Accurate values cannot be calculated mathematically though.

Here is something I found, on weak polyprotic acids, that I think answers the question:

So then [$\ce{SO3^{2−}}$] $= 6.4 \cdot 10^{-8}$?

There should be a fair warning involved in this. It is approximating, that the dissociation $\ce{HA- + H2O <=> H3O+ + A^2-}$ does not contribute to the hydronium ions. In general it can be assumed that $\ce{[H3O+] \geq  [HA- ]}$ and therefore $K_{a2}\geq [\ce{A^2-}]$.


text:
What is the process or material that can enhance the conductivity of the graphene oxide?
I am doing a research and experiment on using Graphene oxide for deicing. My purpose is to coat graphene oxide on the surface of the ship in the north and south pole where large amount of large size ice cubes and icebergs are present. So when voltage is applied to it, it can generate a large amount of heat to melt the ice but the amount of heat must not be significant enough to vaporize the water to steam.
So, the problem is: As you know graphene oxide has a very high resistance. So there must be some process or material that can enhance the conductivity of the graphene oxide. What is it? In other words, is there a process that can enhance the conductivity of graphene?

Addressing the problem of high resistance, we can identify that the resistivity derives from the oxidation of graphene, changing $sp^2$ carbons to $sp^3$ in the lattice, which increases the band gap, and creates an insulator. There are a number of answers to this question that could be correct, including simply mixing another, more conductive material in the synthesis, but since the question seems to imply that the material has already been synthesized and simply needs to be reduced, removing the oxygen, creating reduced graphene oxide (rGO). One simple approach is through thermal annealing, where GO is reduced by a heat treatment.
You can read more about the reduction of graphene oxide here.


text:
Weak base titrated with a strong acid
If you have a weak base XOH titrated with a strong monotonic acid, where do you start with that? Would you first have the base react with water and break up and determine how much of it does? Or just it breaking up alone, not reacting with water? Reacting just with the acid right away? I'm not sure where to start for this problem type...
I need to find the pH at the equivalence point. I have the concentrations of the acid and base, which are both equal, no volume given.

You asked where to start, but you didn't ask what you're looking for. Do you want to calculate pHs for every addition? The base concentration? Or simply describe the experiment?

Sorry, need to find pH. I have the concentrations of the acid and base, which are both equal, no volume given.

If you have no volumes given, did they give you a ratio or do you have to draw a graph? If you do not have the ratio, it is impossible to determine the pH.

I just have that they're both 0.10M solutions of each. No graph need, just need the pH at the equivalence point.

Equivalence point means equal numbers of moles of acid and base have been added. If $x$ moles of the weak base $\ce{B}$ and the strong acid $\ce{HA}$ have been added, then the $\ce{H^+}$ ions will react away $\ce{OH^-}$ until all the weak base has associated into $\ce{BH^+}$ and $\ce{OH^-}$  (which of course has been reacted with $\ce{H^+}$ forming water). This leaves us with water and $x$ moles of the salt $\ce{BHA}$. This salt is ionic and so soluble in water, so it completely ionises to form $\ce{BH^+}$ and $\ce{A^-}$ ions in solution.
As $\ce{BH^+}$ is the conjugate acid of the weak base we had originally, it can react with water in an equilibrium reaction as follows:
$\ce{BH^+} + \ce{H2O} \rightleftharpoons \ce{B} + \ce{H3O^+}$
As $\ce{B}$ is a weak base, the above equilibrium will lie to the right, and hence there will be a considerable concentration of $\ce{H3O^+}$ in the solution at equivalence, which leads to an acidic pH.
Calculating the pH depends on the fact that $K_b K_a = K_w$ for a base and its conjugate acid. This is derived here.
For the equilibrium
$\ce{BH^+} + \ce{H2O} \rightleftharpoons \ce{B} + \ce{H3O^+}$
the acid dissociation constant has the form
$K_a = \dfrac{[\ce{B}][\ce{H3O^+}]}{[\ce{BH^+}]}$
As we know $K_b K_a = K_w$, we can rewrite this equilibrium expression as
$[\ce{H3O^+}]=\sqrt{[\ce{BH^+}]\dfrac{K_w}{K_b}}$
To find pH you can $-\log$ this. (Note $[\ce{B}]=[\ce{H3O^+}]$ as one molecule of each is produced for each molecule of $\ce{BH^+}$ that reacts with a water molecule).
So in order to find the pH at the equivalence point, you need to know the $K_b$ of the base you're using. And as everything is in the same volume, all the volumes will cancel from the concentration terms.
I hope this helps, and please correct me if I've made any errors.

So for the "then the H+ ions will react away OH− until all the weak base has associated into BH+ and " part, if I have the base NH2OH, does that mean it first does this: NH2OH + H -> NH3 + OH ?

And then NH3 + H2O -> NH2 + H3O?  And then that's it?

So we don't worry about the NH2OH reacting with water first because it's so weak and it won't do much, but in the second equation, NH3 would react with water significantly because it's strong?

So if in a strong base weak acid problem you need to use Kb instead of Ka when at equilibrium, does that mean in this case you'd need to use Ka at the end instead of Kb? No wait, yea, we would need Ka, cause there H3O in the final equation.

I'm not familiar with hydroxylamine so I couldn't say how it reacts specifically. The answer I wrote basically deals with a salt solution of a weak base, which would be produced at the equivalence point. Re $K_b$ or $K_a$ you have to remember that above I was considering $\ce{BH^+}$, which is the conjugate acid of $\ce{B}$, and so an acid dissociation constant was appropriate. In an example with a strong base and a weak acid, you would do everything exactly the same, except you would consider the association of the conjugate base.

Alright, well i should at least be able to get close based on what you've given me. Thanks for the good detailed answer!