text:
Nomenclature problem, when would ring be parent?
Nomenclature problem, when would ring be parent?

[(but-3-yn-2-yl)oxy]cyclopentane or 3-cyclopentoxylbutyne
I used chemsketch and it generated the former one.
The chain has a alkyne group. I thought that would make it has a seniority over the ring.
Also, the another rule said we should choose the one with more carbon as the parent chain so I am confused.
IUPAC bluebook,
P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.
P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

You have already found some relevant rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.
(…) For selection of a preferred IUPAC name, see P-52.2.8.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Therefore, for the preferred IUPAC name (PIN), the ring is generally selected as the parent structure according to P-52.2.8, which also corresponds to the method described in P-44.1.2.2 (1). However, this rule applies to structures with “the same number of characteristic groups cited as the principal characteristic group”. If the ring and the chain have different numbers of substituents corresponding to the principal characteristic group, another rule comes first, which stipulates that the senior parent structure is the one that has the maximum number of characteristic groups cited as the principal characteristic group.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

For example, for compounds like 3-propylcyclohexan-1-ol or 3-cyclohexylpropan-1-ol, the senior parent structure is given according to P-44.1.1 by the principal characteristic group, which is expressed as suffix (here: ‘-ol’). The further rules P-44.1.2.2 and P-52.2.8 are not relevant in these cases.

The compound given in the question does not have any substituents that are cited as the principal characteristic group (by means of a suffix or a class name, or implied by a trivial name).* Thus, no decision can be made using P-44.1.1. According to P-52.2.8, the ring is selecetd as the parent structure to construct the preferred IUPAC name [(but-3-yn-2-yl)oxy]cyclopentane. In general nomenclature, however, the context may favour the the chain according to P-44.1.2.2 (2), which would lead to the name 3-(cyclopentyloxy)but-1-yne.

* Note that some sources try to simplify the nomenclature rules by combining different concepts into a single one. For example, the so-called “Table of Functional Group Priorities For Nomenclature” shown on masterorganicchemistry.com tries to combine the seniority order of suffixes with prefixes and endings. This can be misleading and result in wrong names. In particular, there is no suffix ‘-ane’, ‘-ene’ or ‘-yne’ (these are endings), and multiple bonds (and all groups that follow in that table) are actually not considered in the seniority order. Please do not use such sources.


text:
Are CuCl and Cu2Cl2 equivalent?
I have seen $\ce{CuCl}$ and $\ce{Cu2Cl2}$ used interchangeably; they seem to be equivalent. If that is the case, then why does the latter have the '$2$' subscripts?

Without more context, this is hard to judge. Consider $\ce{Hg2Cl2}$, which has a polycation.

I was looking at Sandmeyer reaction catalysts (high school course).

For this very special case, it might be that the typesetting is off (This is a shot in the dark). In the Sandmeyer reaction, Cu changes its oxidation state between +1 and +2. Could the subscripts erroneously refer to this? Please reevaluate the material with this in mind.

$\ce{Cu2Cl2}$ is actually the dimer of cuprous(I) chloride $\ce{CuCl}$. Cuprous chloride dimerises and also trimerises and tetramerises in gas phase. The molecular structures of the species are measured with Gas-Phase Electron Diffraction and Quantum Chemical Calculations. Some information about the various species of cuprous chloride is given here:[1]

The molecular geometry of gaseous cuprous chloride oligomers was determined by gas-phase electron diffraction at two different temperatures. Quantum chemical calculations were also performed for $\ce{Cu_nCl_n (n=1–4)}$ molecules. A complex vapor composition was found in both experiments. Molecules of $\ce{Cu3Cl3}$ and $\ce{Cu4Cl4}$ were present at the lower temperature (689 K), while dimeric molecules ($\ce{Cu2Cl2}$) were found in addition to the trimers and tetramers at the higher temperature (1333 K). All $\ce{Cu_nCl_n}$ species were found to have planar rings by both experiment and computation. The bond lengths from electron diffraction ($r_\mathrm g$) at 689 K are 2.166±0.008 Å and 2.141±0.008 Å and the $\ce{Cu-Cl-Cu}$ bond angles are 73.9±0.6° and 88.0±0.6° for the trimer and the tetramer, respectively. At 1333 K the bond lengths are 2.254±0.011 Å, 2.180±0.011 Å, and 2.155±0.011 Å, and the $\ce{Cu-Cl-Cu}$ bond angles 67.3±1.1°, 74.4±1.1°, and 83.6±1.1° for the dimer, trimer, and tetramer, respectively.

Reference
(1) Hargittai, M.; Schwerdtfeger, P.; Réffy, B.; Brown, R. The Molecular Structure of Different Species of Cuprous Chloride from Gas-Phase Electron Diffraction and Quantum Chemical Calculations. Chem. Eur. J. 2003, 9 (1), 327–333. DOI: 10.1002/chem.200390027

So, CuCl has no real existence? And is it a covalent molecule?

@user226375 it is a solid at room temperature. It just polymerizes in gas phase. It is ionic in nature and has various crystalline forms.

thanks for replying. Is it ionic in even gas phase?( where it exists in dimer form)

@user226375 no it is not ionic in  gas phase and it polymerizes

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Why is ice cubical in structure at very low temperature?
My book says that at atmospheric pressure ice crystallises in the hexagonal form but at very low temperatures it condenses to cubic form.
Why and how does this happen?

Any number of elements, much less compounds, have multiple solid allotropes - under different temperature/pressure regimes they assume different crystal structures. This means that the Gibbs free energy is lower in that configuration under those conditions. For example, Mn has 4 allotropes (two of which are unique to it), and Pu has no fewer than 6 different crystal structure.

First, a little clarification. The kind of ice you see at ambient pressure, ice I, exists as stacks of bilayers of water molecules. If they are stacked in an ABAB sequence, they form hexagonal ice, and cubic ice in an ABCABC sequence. The difference between the two allotropes is just the stacking of the molecules, so in fact they are structurally quite similar. The structure of cubic ice is analogous to the diamond structure of carbon, with hexagonal ice being analogous to lonsdalite. This picture is complicated by the fact that ice is generally proton-disordered, i.e. the molecules can have different orientations that break the crystal symmetry, subject to certain rules (the "ice rules").
There is evidence that ice crystallises as cubic ice in the upper atmosphere, namely the angle subtended by solar haloes and the angles formed in nascent snowflakes, for example. However, more recent evidence suggests that cubic ice doesn't not exist as a pure phase, but only in stacking-disordered hexagonal ice (http://pubs.rsc.org/en/content/articlelanding/2015/cp/c4cp02893g#!divAbstract). Curiously, there is also evidence that the lonsdalite form of carbon doesn't exist in reality, and is just stacking-faulted diamond (http://www.nature.com/articles/ncomms6447).
This is all very well, but does not tell us why hexagonal ice forms in preference under certain conditions, or cubic ice (?) under others. If you were to perform density functional theory calculation (essentially approximately solve the Schrodinger equation for a many atom system), you would find that hexagonal and cubic ices have almost identical energies, and are essentially degenerate at zero temperature, which makes it even more confusing why hexagonal ice is almost exclusively seen on Earth. I don't think there is a convincing answer to your question, since the best calculations do not show why the hexagonal structure of ice is lower in energy than cubic, but it could be due to a number of reasons: orientational disordering, which complicates modelling ices, zero point energy or free energy considerations, or a breakdown in the approximations used to perform such calculations. In any case, it looks like cubic ice doesn't exist under terrestrial conditions.

This is a difficult question. Apparently, researchers don't know yet (or I didn't manage to find an article which completely solves the problem).
The preference for cubic structure at low temperatures was long thought to be caused by a lower liquid-ice surface tension in cubic ice nucleus (see
G. P. Johari, J. Chem. Phys., 2005, 122, 194504). However, some authors point out that this assertion is not supported by strong evidence, and note that thermodynamically speaking, there is no reason why cubic should be preferred over hexagonal (E. B. Moore and V. Molinero, Phys. Chem. Chem. Phys., 2011, 13, 20008), that is to say that cubic ice is not more stable. These authors performed molecular dynamics simulations, and concluded that a the cubic ordering is kinetically controlled, though they left the elucidation of the kinetic factors an open question.
Of course, this dates back to 2011, any update would be welcome (it is not exactly my field of expertise...).

As explained in the other answers, water ice may not actually become cubic at low temperature.  In fact, if formed at higher temperature as the usual hexagonal phase, water ice will remain so when cooled.  There is no thermodynamic driving force to rearrange the hexagonal structure to a cubic one.
We can expand on the earlier answers in two, make that three, ways:
1)  Ice at very low temperatures can undergo a proton ordering process, forming a lower symmetry phase.
2)  Cubic ice is doubtful as Ice I, but is known in a higher pressure phase, Ice VII.
3)  Update:  Cubic Ice I, as $\ce{D2O}$, has been reported from careful warming of a metastable low-density phase, Ice XVII.
Proton ordering:  Ice XI
The ice structure presented in this answer is somewhat of an idealization.  The hydrogen atoms there are perfectly ordered.  However, in real Ice I (and other molecular phases, which share this characteristic), the hydrogen atoms can migrate around the hydrogen-bonded rings so that the molecules are still $\ce{H2O}$ but the protons are randomized.  Only the oxygen atoms are really fixed in Ice I, and the familiar hexagonal symmetry comes from the arrangement of oxygen atoms in this randomized environment.
However, the proton-ordered form is slightly lower in energy and, at low temperature, this overcomes the favorable entropy of the random arrangement and the idealized structure now does apply.  With this change comes a lowered symmetry of the oxygen atoms so that now the ice is orthorhombic instead of hexagonal (or cubic).  This is Ice XI.  Ice XI is believed to be stable up to 72 Kelvins, so it may exist on colder bodies in our Solar System and in interstellar space.
When the Pressure's On:  Ice VII
While the cubic form of Ice I appears to be of doubtful existence, it does have the property that if we expand it slightly another identical cubic structure can be intertwined with it.  This is what happens in Ice VII and its low-temperature, proton-ordered form Ice VIII (again proton ordering lowers the symmetry, so Ice VIII is tetragonal rather than cubic).  Both are formed  under several GPa pressure.  Ice VII has been identified in diamonds, indicating that Earth's water is not just on the surface; it penetrates deep into the mantle and thus broadly affects the structure and physics of Earth's rock.  The finding of Ice VII on Earth is discussed in more detail in this answer from Space Exploration Stack Exchange.
This Just In:  Cubic $\ce{D2O}$ Ice I
Leo del Rossi at al. report in Nature Materials that purely cubic Ice I may be obtained from careful warming of $\ce{D2O}$ Ice XVII between $110$ and $180$ Kelvins.  Ice XVII itself is metastable and was made in this case as a clathrate with hydrogen (or deuterium).  Carefully removing the hydrogen at low temperature resulted in purified Ice XVII, which when warmed transformed to the cubic Ice I phase before going on to hexagonal Ice I.  Multiple analytical techniques corroborate the existence of the cubic ice.

[PS. I do not know much but I'll still try to answer based on my logic and my prior knowledge]
The molecular structure of ice is Hexagonal always. At higher temperatures however, the molecule condenses to such as extent that it appears cubical. Refer to the picture

Temperature is the average kinetic energy of the molecules/particles in a compound/substance. At very low temperatures, the kinetic energy of the molecules tend towards 0. Also when a material freezes, the intermolecular bond length between 2 water molecules would decrease hence condensing the entire molecule.
Hope thats helpful.
More info at This site
If your question was about the shape of a bigger ice, I think that the shape of the ice also depends upon the container than you put the water in to freeze.

There are many different crystal structures of ice.

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Why do the lone pairs of electrons first occupy axial positions before occupying equatorial positions for species with octahedral geometry?
Why do lone pairs not reside in equatorial positions in species with octahedral geometry?

In octahedral geometry, all positions should be equivalent (before putting a bond or a lone pair there). It is by convention that the lone pairs are put on "axial" positions. However, they will definitely be across each other, consider XeF4.

@TAR86 I don't understand what you mean by "equivalent". Do you mean all positions have the same bond angle to each other?

Can you give an example?

@b3nj4m1n Consider a cube (such as dice). Octahedral geometry means that the central atom is in the center of the cube and the surrounding atoms are on the center of the faces of the cube. Thus, due to the high symmetry of the octahedral geometry, they are all equivalent. If this explanation does not help, you need to build a model.

https://chemistry.stackexchange.com/questions/117585/ab5e-type-molecule Can refer this if you want.

The main reason is repulsion with the electron pairs. Compare $\ce{SF6}$ &  $\ce{SF4^2-}$ as an example. To get from $\ce{SF6}$ to $\ce{SF4^2-}$, we need to replace two fluorines with lone pairs. The first replacement is arbitrary; in an octahedral geometry, all the positions are equivalent. Once we have replaced that first fluorine however, the bond positions are no longer equivalent. We will have four equatorial bonds (90 degree separation from the lone pair) and one axial bond (180 degree separation from the lone pair). Lone pairs are very repulsive, especially with respect to other lone pairs, so we want the next lone pair to be as separated as possible from the first. The axial bond is further from the lone pair then any of the equatorial bonds, so it makes sense for the lone pair to go there.

simply because in octahedral geometry both the equatorial position or base and the axial-equatorial position maintain same bond angle which is 90deg, so at first you can chose any position to place the first lone pair, but for the next electron pair the remaining positions are no more identical in the view of the bond angle.Cause if u chose eq. position for the second one, if it's a lonepair too then according to VSEPR theory you haven't made the right choice, cause you are still having an option to reduce the greatest replusion (LP-LP) between electron pairs, that is the axial position.So that angular distance between the LPs become 180.
If the explanation above for octahedral is clear to you than you yourself can find why it is vice versa for a tetrahedron.:)

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. This doesn't make much sense, first you state that all positions are equal, then they are not. You could put two lone pairs in equatorial positions and maintain 180° (If measuring angles between lone pairs would be possible.) It would also be better, if you avoid abbreviations and do not use emoticons.

In case of two lone pairs in a tetrahedron it creates a dilemma for me too. But for 1 or more than 2 electron pairs it makes sense. And for the first electron pair in an octahedral shape all the positions are equal, but not for the next one. I don't see any problem here.

The VSEPR theory tells us that the order of repulsion is :
lone pair-lone pair >lone pair-bond pair >bond pair-bond pair.
It is clear that you need to place the lone pairs such that they have minimum neighboring bond pairs or lone pairs. An equatorial position in an octahedral geometry has 5 neighboring bond pairs/lone pairs ( 3 in the equatorial plane and two in the axial plane ) while an axial position has 4 neighboring bond pairs/lone pairs (4 in the equatorial plane, the other axial position is not a neighboring position).
That is the reason why lone pairs first occupy axial positions in an octahedral geometry. Hope this helps.

Are we talking about the same thing? https://en.wikipedia.org/wiki/VSEPR_theory see SF6. All F's are equivalent.

wait @TAR86 so are all the positions equal or not? I'm confused now

@b3nj4m1n I am utterly convinced that all positions of the surrounding atoms in an octahedral geometry are equal. This is due to the symmetry of the octahedron.

See, one lone pair isn't any issue, because all six sites are equivalent, as @TAR86 points out. OP is asking about why two lone pairs will go trans to each other. The use of "axial" or "equatorial" is highly misleading in my opinion.


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Is the VSEPR theory correct in determining the bond angle of sulfur dioxide?
Recently when we were doing the VSEPR Theory in class and discussing the bond angles in various compounds. That's when I thought of sulfur dioxide whose structure can be like this:

According to the Internet, the bond angle between the two oxygen atoms = 120°. But mathematics shows a little off. I was taught that the VSEPR Theory states that

Lone Pair-Lone Pair Repulsion > Bond Pair-LonePair Repulsion > Bond Pair-Bond Pair Repulsion

According to this, the angle between the two oxygens should be less than 120º and I can prove it.
Imagine that instead of the lone pair there was a double bond of oxygen with sulfur. This would imply that the bond angle between the oxygens would be exactly 120° on the same plane because each one of them would repel each other with the same repulsion force. But now instead of a bond pair, we have a lone pair which means (according to the VSEPR Theory) that it would repel the two oxygens much more than the imaginary bond pair was repelling. Therefore the angle between the lone pair and the bond pairs with oxygen would be more than 120°. This would result in a smaller than 120º angle between the two oxygen atoms. Then how does the net say that the bond angle is close to 120°? If what the web says is right, it implies that the repulsion of a lone pair and 2 bonds is exactly equal to the repulsion between 2 double bonds. Is it correct?

VSEPR is not much of a theory; it is just an empirical model which happened to work surprisingly well in many cases. Then again, according to Wikipedia, the angle is indeed slightly less than $120^\circ$, so I see no contradiction.

Please clarify your concepts on why lone pair-lone pair repulsions are stronger than bond pair-bond pair repulsions. I think you will get your answer

First off, your structure violates the octet rule on sulphur and is thus not a good representation of the molecule. A better one is this one:

Figure 1: electronic structure of $\ce{SO2}$.
All methods of approximation of bond angles you have learnt or will learn about are no more than educated guesses. VSEPR gives very good approximations for e.g. $\ce{NH3}$ and $\ce{H2O}$ but fails horribly at $\ce{PH3}$ and $\ce{H2S}$. For the latter two, an estimation based on the most likely orbitals used for bonds (all p) does a much better job. To accurately determine a bond angle, one needs to perform a structure elucidation experiment such as X-ray diffraction.
This has indeed been performed with $\ce{SO2}$ (likely multiple times using different methods; I am too lazy to check the references) and the results give a bond angle $\angle(\ce{O-S-O})\approx119^\circ$. This is indeed very close to $120^\circ$ so you can say that the VSEPR method gives a good approximation of the $\ce{SO2}$ bond angle.

Figure 2: Structural data of $\ce{SO2}$. Taken from Wikimedia, where a full list of authors is available.

From nowhere you can cited that the repulsion of a lone pair and a double bond are exactly the same. Instead, the lone pair is little closer to sulfur than the other two, as oxygen is far more electronegative compared to sulfur. As a result, the repulsion of the lone pair is little stronger than the other two; and the molecular angle is a little smaller than 120 degrees.


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Why do heavier transition metals show higher oxidation states?
In p-block elements, higher oxidation states are less stable down the group due to the inert pair effect. This is not the case for transition metals.
Why do heavier transition metals show higher oxidation states than lighter ones? Is the inert pair effect not valid for transition metals also?

http://chemistry.stackexchange.com/questions/43650/why-is-movi-more-stable-than-cr-vi/50223#50223

http://chemistry.stackexchange.com/questions/48971/why-is-wf6-stable-whereas-crf6-is-unknown

http://chemistry.stackexchange.com/questions/7704/plutonium-having-more-oxidation-states-than-samarium

And there are loads of similar question out there in chem.SE. Would you mind searching them?

The inert pair effect is based on the fact that main group elements’ oxidation states depend on s and p orbitals (and only them). When going down the periodic table, the energy difference between s and p orbitals changes leading to some elements losing their valence s electrons more easily than others.
Transition metals’ chemistry happens in the d orbitals primarily — unless you count the copper and zinc groups wherein a significant part of the chemistry is in fact only s orbital chemistry. The d orbitals — at first approximation of the free ion — are degenerate, i.e. they all have the same energy. Thus, these electrons are typically much more accessable. Furthermore, going down the periodic table increases the number of electrons counted as core electrons meaning that the outermost valene electrons experience a weaker effective nuclear attraction. It is therefore easier (i.e. requires less energy) to remove valence electrons and higher oxidation states are much more accessible.
If it weren’t for the inert pair effect, this would also be visible in main group chemistry — except it is, if you compare the relative stabilities of iodate and chlorate.

Highest oxidation state of heavy metal is more stable than lighter metal . That is heavy metals have more electrons as compared to lighter metals so they can oxidize easily by losing electrons and hence become stable but on other hand lighter metals like chromium)+6 is small in size so its valence electrons are properly held by its nucleus so it act as an oxidizing agent whereas heavy metals like tungsten and molybdenum are stable !


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Why caffeine is easily oxidized to uric acid?
I want to understand how caffeine (1,3,7-trimethylxanthine) can be easily oxidized to uric acid.
Is it safe to conclude that caffeine may increase the uric acid level in blood?

Well, I knew the opposite. Caffeine can be made from uric acid. This is called Fischer's synthesis.

I doubt caffeine can be so easily oxidized to uric acid. I'd propose a way by demethylation via cytochrome P450 to the corresponding xanthines and oxidation via xanthine oxidase (but these are enzymatic reactions).
From a synthetic perspective, one will start with guanine (-> guanine deaminase followed by oxidation; or oxidative deamination of Guanine followed by another oxidation)
It's not safe to conclude that caffeine increase uric acid blood level.
Long story short:
Coffee consumption seems to lower uric acid blood levels.
Decaf coffee consumption also seems to lower uric acid blood levels.
Tea consumption seems to have no effect on uric acid blood levels.

Serum uric acid level tended to decrease with increasing coffee intake
  (Figure 1). After adjusting for age and sex, serum uric acid level in
  individuals with coffee intake 6 cups daily was lower than in those
  with no use by
  0.43 mg/dl (95% CI 0.22, 0.65; P for trend  0.001). After adjusting for other covariates, the differences were attenuated but remained
  significant (P for trend  0.001) (Table 2).

"Coffee, Tea, and Caffeine Consumption and Serum Uric Acid Level: The Third National Health and Nutrition Examination Survey", Choi HK, Curhan G., 2007
DOI 10.1002/art.22762
It has to be noted this was the first study I found on the fly and there may be other results out there

now I understand that Caffeine is not easily oxidized to uric acid. On the quote provided, I researched  that "Coffee may also contain substances which inhibit xanthine oxidase, an enzyme converting xanthine to uric acid..."source: https://www.hindawi.com/journals/jnme/2010/930757/. Coffee has a component that inhibits Caffeine to be converted to uric acid.  So I may still conclude that Caffeine  may increase UA level on the blood after being metabolized in the body.

I don't think so. The conclusion that Caffeine may increase UA level is contrary to both publications provided here. As statet in your source: "Age-adjusted and multivariate-adjusted means of serum
UA were progressively lower with increasing consumption of
coffee..." and later "However, caffeine was found to be unrelated
to serum concentrations of UA" and also in the study I provided there is no evidence for an increase in UA level. The opposite is shown.

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Unit conversion of Molar Conductivity
I know this might be a silly question but I am confused with different answers of internet. My question is:
Molar conductivity is given as:
Molar conductivity = $\frac{\kappa}{M}$.
S.I. unit of :
$\kappa$ is $\pu{Sm}^{-1}$  &
$M$ is in $\mathrm{mol \, m^{-3}}$.
Thus, Molar conductivity in S.I. is
$\mathrm{S \, m^2 \, mol^{-1}} \tag{1}$
When I substitute
1 m = 100 cm in (1).
I got new unit as
$[\mathrm{S \, cm^2 \, mol^{-1}}] \times10^4$.
But in a book It was given as
$[\mathrm{S \, cm^2 \, mol^{-1}}]  \times 10^3 \tag{2}$
where they have taken S in S/cm and molarity in mol/litre. But where I was wrong.
If you feel the question too confusing just explain me How I can convert that SI unit given in (1) into unit given in (2).

If M is molarity then yes it is taken in mol/litre and you have taken it mol/${m^3}$ which you can't do because it is defined that way.

But SI unit of M is actually moles/m3. And hence for SI unit of molar conductivity it must include M in moles/m3 & k must be in S/m. Then only we get SI unit of molar conductivity as Sm^2/mol.

Refer to page 77 of Chemistry NCERT text book, class 12, part 1.
You were actually right about the $\mathrm{10^4\ S\ cm^2/mol =S\ m^2 /mol}$. Check if the $\mathrm{10^3}$ you are referring to has molarity included in its equation because molarity is given as moles per litre.
You can refer to the online NCERT book at its official website. Thank you, I hope this was useful and sorry if I got it wrong.

Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. Furthermore, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.


text:
absorption spectrum of water at different wavelengths
I know I can get the visible absorption spectrum of water https://en.wikipedia.org/wiki/Water_absorption But how does one go about getting and testing what the absorption spectrum is for a glass of water in the audio range of (0hz to 20000hz) I know nasa uses water to absorb sound https://youtu.be/08aHEmuzMLI.
How can I measure what the sound absorption spectrum is in a glass of water?

I'm voting to close this question as off-topic because this (1) belongs on physics section (2) OP has to explain why he wants to measure in a glass. How big is the glass? thickness of glass walls diameter of glass Et cetera, et cetera, et cetera!

Music.SE has a tag called "Sound", whose description reads: "Sound is a mechanical wave that is an oscillation of pressure transmitted through some medium (like air or water), composed of frequencies which are within the range of hearing." It seems like this might be more likely to get you a good answer.  There are a variety of sensors for measuring sound intensity at different frequencies; and even a properly tuned piano could be a reliable source of frequencies up to 6000hz anyway, and there are certainly better solutions than that depending on the accuracy of frequency that you require.


text:
Why does the rate of reaction increase less than proportionally to the increase in mass of catalyst?
I conducted an experiment where I wanted to speed up the decomposition of hydrogen peroxide with the use of a catalyst (potassium iodide).
I used 5 different test tubes with 5ml of hydrogen peroxide in each, then added different amounts of the potassium iodide: 0.2g, 0.4g, 0.6g, 0.8g, 1.0g. Then I measured the time it took for the reaction to reach a temperature of 30°C. The results I got was that the more of the catalyst I added, the faster the rate of the reaction became. However, the increase in the reaction rate decreased when more of the catalyst I added, and I don't understand why.
I would think that since temperature and amount of hydrogen peroxide are constant variables and the only variable affecting the rate is the amount of potassium iodide, if I doubled the amount the speed would also be doubled. But why is that not the case? Or does it have something to do with errors in the method?

Surface area of the catalyst is more important than it's mass.

Per @bon 's point, was the reaction stirred or was the KI allowed to settle and possibly not completely mix/dissolve?

Yes, the test tubes were shaken right after the KI was added

@airhuff Doesn't matter as regards the scaling - even with perfect mixing, mass increases proportionately to volume, but geometric surface area increases to only the 2/3 power with volume and thus also with mass.

Don't think this is just surface area. // What concentration was hydrogen peroxide?

3,3%  (diluted 33%)

There is a lot going on here. Part of the problem is poor experimental design. I'll just throw this out without evaluating all the energies. (Too lazy...)
(1) The potassium iodide has to be dissolved into the solution since you added it as a solid. This means there is an energy change and this could be endothermic or exothermic. Adding more salt increases the effect obviously.
(2) It takes some time to dissolve the potassium iodide. So before all the potassium iodide is dissolve some is already reacting with the hydrogen peroxide.
Greater surface area (smaller particles of potassium iodide) would allow the crystals to dissolve faster.
(3) The reactions are:
$\ce{H2O2(aq) + I-(aq) = H2O(l) + OI-(aq)}$
$\ce{H2O2(aq) + OI-(aq)= H2O(l) + O2(g) + I-(aq)}$
Overall the reaction is:
$\ce{2H2O2(aq) =  2H2O(l) + O2(g) + \text{heat}}$
But the heat is the combination of the two reactions.
If $\ce{[H2O2](aq) >> [I-](aq)}$ then both reactions will happen.
If $\ce{[H2O2](aq) << [I-](aq)}$ then only the first reaction will happen. The iodide isn't acting as a catalyst but as a reactant.
3.3 % Hydrogen peroxide is
$0.033 \cdot 15 \approx 0.5$ g $\ce{H2O2}$
Moles $\ce{H2O2}$ = $\dfrac{0.5}{34} = 0.015 \text{moles} = 15 \text{millimoles (mM)}$
0.2 g KI = $\dfrac{0.2}{166}$ = 1.2 mM
1.0 g KI = 6.0 mM
So you're adding so much KI that some might get left as $\ce{OI-}$.
(4) so to do the experiment better:

cut the KI concentration
dissolve the KI in 5 ml of water & equilibrate  temp to that of 3.3% H2O2
mix solutions with stirring


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Technique to measure the volume of a crystal
I am doing a project on crystal growth. I am making potash alum in two different conditions, one in the fridge and the other one in room temperature. I wanted to how can I measure their volume? Techniques?

Submerge in liquid, note change in total volume. 3D laser mapping. Mass and lattice parameter measurements.

weigh 'em. Volume = mass/density

Potash alum is soluble in water (~14g/100mL at 20C) but not in acetone. I would submerge the sample in a known volume of acetone and measure displacement. If you want to be super thorough, cool the acetone first.
I wouldn't use reported density because you don't necessarily know if you made a pure product etc.

Good solution.  I'd also be sure to give the crystal a good rinse on filter paper with acetone to prevent adding water and dissolved potash alum to your volumetric measurement setup; any residual dissolved potash alum will precipitate out to give you an increased mas of precipitate.  For that matter, just the excess water would give you a high-bias in your measurement. Probably negligibly  small, but it doesn't hurt.  Also, since you need to dry the crystals before measuring their volume anyway, an acetone rinsed crystal will be much easier to dry than a crystal soaked in water.

I'd wonder if the OP is going to get one nice crystal or some massive crystal cluster. So measuring the volume is trickier than one might suppose. The dang air bubbles...

@MaxW, I don't know what equipment OP has available, but do you think an ultrasonic bath might help with the air bubbles?  I guess it would also destroy a nice big crystal or otherwise degrade the quality of the crystals in a way OP doesn't want.


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How to calculate the heat released by the combustion of ethanol in excess oxygen?
I have a question in my textbook and I do not really understand their explanation of the answer, it is as follows:

$$\ce{2CH3OH + 3O2  -> 2CO2 + 4H2O}$$
  When 2 moles of $\ce{CH3OH}$ react with 3 moles of oxygen, exactly $\pu{1354 kJ}$ of energy is released. How much heat is released when $\pu{5.00 g}$ of $\ce{CH3OH}$ is combusted in excess oxygen?

Here is the procedure to solve it:

Convert the mass of $\ce{CH3OH}$ to amount of substance, which is $\pu{0.156 mol}$.
$\displaystyle \pu{0.156 mol} \times \left(\frac{\pu{1 mol}}{\pu{2 mol}} ~\ce{CH3OH}\right) \times \pu{-1354 kJ} = \pu{-106 kJ}$

My first question is, why is $\pu{0.156 mol}$ multiplied by $\pu{1 mol}/\pu{2 mol}~\ce{CH3OH}$? Where is the $\pu{1 mol}$ in the numerator coming from, if for every two moles of $\ce{CH3OH}$, 3 moles of oxygen react?
My second question is, why is this expression multiplied by $\pu{-1354 kJ}$? It says in the problem that $\pu{1354 kJ}$ of energy are released. Does this mean I need to infer based on the equation whether the reaction is exothermic or endothermic? I just started learning thermochemistry so I am not quite familiar yet with these types of problems and what I need to do.

Regarding your second question - yes, you have to infer from the question's phrasing the direction of energy "movement". Since the definitions of endothermic and exothermic reaction are pretty straightforward as far as energy movement is concerned, it should be rather simple to deduce that when "energy is released" we are dealing with an exothermic reaction. Remember that the definitions are based on a convention, and the convention is system-centric, i.e., we are interested in what is happening in the system, and not the surroundings.

Regarding your first question - this seems to be a poor choice of conversion factor. In this type of questions you are expected to use the "enthalpy of reaction" (here specifically it is the enthalpy of combustion) as a conversion factor. Enthalpy of reaction has the units of kJ/mol of substance. This way, when you are multiplying a certain amount of moles by the enthalpy of reaction, you should be left with a value in kJ, i.e. an amount of energy only. If you calculate 0.156 mol x -1354 kJ/2 mol CH3OH (given enthalpy), this may simplify things and make more sense. What do you think?

It says that 2 moles of of $\ce{CH3OH}$ release $\text{1354 kJ}$. This is the same as saying that 1 mole of of $\ce{CH3OH}$ releases $\text{677 kJ}$. Here I just divided the 1354 by 2 to obtain the number of the energy released when one mole is burned. This way it is easier to do dimensional analysis. If we were to express this in a more mathematical way,
$$\frac{677\,\text{kJ}}{\text{mol}\,\,\ce{CH3OH}}$$
Note that this is equal to $$\frac{1\ \text{mol}\,\,\ce{CH3OH}}{2\  \text{mol}\,\,\ce{CH3OH}}\cdot 1354\,\,\text{kJ}$$
Basically, this is done to find the energy that ONE mole of $\ce{CH3OH}$ releases. Your confusion stems from a poor track on the units that are involved. You do not need to worry about the oxygen since it is in excess.
As for the sign of the energy (the number in front of kJ) is due to convention. You are right that it is to denote that the process is exothermic or endothermic. But I suggest that you read more of it. This is important, and will come back again in physical chemistry.

This is a great answer. Just note that when you are writing (1mol CH3OH/2mol CH3OH) x 1354 kJ you are left with a value in kJ only (the mol CH3OH may be canceled out and we are left with a value of 1/2), so this cannot be used as a conversion factor and is not equal to 677 kJ/mol CH3OH (which is indeed a conversion factor) as you wrote. Remove the nominator and you're good to go.

I've never done thermochemsitry. But this is how I arrived to the same answer.
In two moles of methanol there is:
$$\pu{2 mol} \times \pu{32.04 g//mol} = \pu{64 g}~\ce{MeOH}$$
Which releases $\pu{1354 kJ}$ of energy. Therefore:
$$\frac{\pu{1354 kJ}}{\pu{64.08 g}}~\ce{MeOH} = \pu{21.1298 kJ}~\text{released per gram of}~\ce{MeOH}$$
This implies:
$$\pu{5 g}~\ce{MeOH} \times \pu{21.1298 kJ//g} = \pu{105.649 kJ}~\text{released }$$
And yes, you are correct about the negative denotation of energy.
If a reaction has a positive value, energy must be provided for the reaction to proceed (endothermic). If the reaction has a negative value, the system is losing energy as heat (exothermic).

In my opinion, when doing mathematics in chemistry, it is best to write things out in a way that is similar to this.


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Lewis Structure of Fe4(SiO4)3
I'm a first year university student and am working on a paper regarding factors that affect how far an asteroid will penetrate into the Earth's atmosphere before exploding.
One of the factors I came across is the composition of the asteroid, and hence, the boiling points of the substances within the asteroid. I am currently comparing 2 types of asteroids in this manner: Carbonaceous asteroids (mostly composed of organic compounds), and silicaceous asteroids (composed mainly of magnesium- and iron-silicates such as $\ce{Fe4(SiO4)3}$). I have been trying to draw the Lewis structure for $\ce{Fe4(SiO4)3}$ for what seems like ages but I don't seem to be getting anywhere.
I feel like I'm missing something but I'm not sure what it is.
So now I have two questions in particular:

Is a structure for $\ce{Fe4(SiO4)3}$ even possible and if so how would you go about drawing it?
Is there an easier way for me to determine (and prove) which compound will have the lower boiling point?

Simply check crystal structure. There are no discrete molecules in silicates.

Hint: compound consists of ferric cationi $\ce{Fe^3+}$ and orthosilicate anion $\ce{SiO4^4-}$.

@Mithoron ,  from what I understand, by saying there are no discrete molecules in silicates, you must be implying that the compound with the silicate will have a higher B.P in comparison to compounds with discrete molecules. So to be able to compare the compositions of the 2 types of asteroids, how can I know whether the compounds in the carbonaceous asteroids will have discrete molecules or not?

That means there's no point in trying to draw Lewis structure, but yes it's affecting b.p.

There are several mistakes in your question:

Carbonaceous asteroids (mostly composed of organic compounds)

No. They are composed mostly of silicates and oxides. They usually have a few percent of carbonaceous material (not necessarily organic, by the way).

$\ce{Fe4(SiO4)3}$

This thing doesn't exist. The formula implies that all iron is oxidised: $\ce{Fe_4^3+(SiO4)_3^4-}$. Asteroids are not oxidised enough for this to happen. Even if they were for some reason, the stable oxidised minerals would be magnetite or hematite ($\ce{Fe3O4}$ or $\ce{Fe2O3}$). The compound you're looking for is fayalite: $\ce{Fe^2+_2SiO4}$, which never occurs in this pure form in asteroids but rather as in solid solution with forsterite ($\ce{Mg2SiO4}$) to form olivine.

I have been trying to draw the Lewis structure for Fe4(SiO4)3 for what seems like ages but I don't seem to be getting anywhere.

That's because you're trying to draw something that doesn't exist, and if it did, it wouldn't be a molecule drawable as a Lewis structure. It would be a crystalline solid with a defined repeating structure.

Is there an easier way for me to determine (and prove) which compound will have the lower boiling point?

Any organic material will boil first. It's pretty intuitive: boiling or melting organic material is easier than boiling rocks. Note that in the case of pure carbon as graphite or diamond this will not happen, as these are the some of the materials with the highest boiling points known. For the Mg and Fe silicates, a rule of thumb is that the Mg-bearing compounds are more refractory. $\ce{Mg2SiO4}$ melts at 1890 °C whereas $\ce{Fe2SiO4}$ melts at slightly above 1200 °C. Intermediate solid solutions (olivine) will (partially) melt at intermediate temperatures.


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Intermolecular forces between carbon dioxide and water
I'm wondering which intermolecular forces will exist between carbon dioxide and water. Instinctively, I would expect there to be only dispersion forces between the 2 molecules, because carbon dioxide is non-polar due to its linear shape. However, even though carbon dioxide is non-polar, it still contains dipoles. I'm curious if its possible that these dipoles in carbon dioxide will allow it to form dipole-dipole forces (or even H-bonds) with water. Could the partially negative oxygen atomspotentially accept a H-bond from water, even if carbon dioxide is non-polar?

According to the this PubChem data table for the physical and chemical properties of carbon dioxide, it contains two hydrogen bond sites (each of the oxygens). This largely accounts for the high solubility of $\ce{CO2}$ compared to that of non-hydrogen bonding compounds like methane, which has about $\mathrm{1/25^{th}}$ of the Henry's law solubility in water as does $\ce{CO2}$.


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Reducing character of group 15 hydrides
It was given in my text book that reducing character of group 15 hydrides increases down the group. I am a bit confused about this. For $\ce{NH3}$ this is fine as I assume that:
$$\ce{NH3 -> N + H^+ + 3e^-}$$
My doubt arises in the higher elements of the group. In $\ce{SbH3}$, $\ce{H}$ is present as $\ce{H^{-}}$ and $\ce{Sb}$ as $\ce{Sb^3+}$. So would $\ce{Sb^3+}$ be oxidised to $\ce{Sb^5+}$ or would $\ce{H^{-}}$ be oxidised to $\ce{H2}$?  I know that for $\ce{Sb}$, $\ce{Sb^3+}$ is more stable than $\ce{Sb^5+}$ . So would  $\ce{H^{-}}$ be oxidised to $\ce{H2}$ ?
Also electronegativity of $\ce{As}$ and $\ce{H}$ is given tobe equal in my textbook. So what would be oxidation product of $\ce{AsH3}$ ?
For $\ce{BiH3}$ it seems like:
$$\ce{Bi^3+ -> Bi^5+ + 2e^-}$$
is more feasible as $\ce{Bi^5+}$ is more stable. Can someone please tell me the oxidation products of group 15 hydrides? Also can someone tell $\ce{E°_(H^{-}/H2)}$ value as I couldn't find it on net?

When your book says that reducing character increases down the group, it PRIMARILY meant in gaseous state. What you have described above are all ionic reactions, i.e., in some solution.
NOW in gaseous state, as we go down the group, the bond energy of hydrides reduces (due to increasing atomic size as we go down the group). So they can be more easily broken to release nascent hydrogen (see my note at the bottom), which is the cause of reducing nature. So reducing ability increases down the group in gaseous state.
NOTE : Nascent hydrogen is referred to the hydrogen which has just evolved post some reaction, and has not yet formed its molecule. This is not the same as hydrogen atom so keep that in mind.
Reducing property in a solution
This is the catch. In aqueous solution, reduction takes place due to the group 15 ion, and not hydrogen ion. So it depends on how easily the group 15 ion can lose an electron (reduction is gain of electrons)
Down the group, electronegativity decreases and hence it becomes easier to remove electrons for reduction. So reduction ability decreases down the group in solutions too.
Now to answer your question, Sb(3+) will be oxidised to Sb(5+) as it is what is causing the reduction. You should be able to answer your following question now.
Btw, the E0 value for (H-/H2)is 0.00. Note that all oxidation reduction reactions of hydrogen gas have this value as 0.00

$E^\circ$ for $\ce{H-}/\ce{H2}$ is most certainly not 0.00.

Forgive me if my electrochemistry is weak. Should you come to know about the value for the reaction, please do let me know. Thanks

But what about the $H^-$ produced in aqueous solution? Would it combine with $H^+$ and form $H_2$ ?

I do not know why your book says that Hydrogen anion is produced in the case of SbH3.  I am pretty sure that H+ is produced even in this case.

Check this link for yourself   - https://books.google.co.in/books?id=fuJV1H18KtEC&pg=PA172&lpg=PA172&dq=SbH3+in+aqueous+solution&source=bl&ots=hsuEeh74F9&sig=sH24lW0FpFGzjkMi2F8OnlWfNmw&hl=en&sa=X&ved=0ahUKEwjfk6DW7fXRAhUhSo8KHZnBBmIQ6AEIIDAA#v=onepage&q=SbH3%20in%20aqueous%20solution&f=false

See how they say that H+ and Sb take three electrons to form SbH3, and not H-


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Ring chain tautomerization of hydroxyacetone
My chemistry textbook says that hydroxypropane-2-one and 1,2-epoxy-2-hydroxypropane are ring chain tautomers, but it did not mention the steps for conversion of one into the other. I tried to figure them out but got stuck.
My attempt:

I disagree that you start with the ketone as the aldehyde is more unstable than the ketone is, since it is far more electrophilic than the ketone. The ketone has a methyl group attached to it which makes it far more stable. I suggest you approach this question the other way around.

Hello. Which textbook is that? I hadn't read about such an interesting example of ring chain tautomers before, so just curious. Thanks in advance.

At first, even though this is not related to the question, I want to say that you're confused between double-headed arrow between the two resonance forms and double harpoon in reversible reactions. (The harpoons should be written as below.)
Your incomplete mechanism has two errors:

The hydrogen atom bonded to carbon is far less reactive than the one bonded to oxygen.
The positive carbon, which is bonded to oxygen, a very electronegative element, is very, very unstable, which is not likely to happen.

This is the right mechanism for the reaction:

Step 1: Oxygen is a good nucleophile, so it is most likely to take the most acidic hydrogen.
Step 2: Then, the negative charged oxygen will attack the electrophile carbon atom, forming epoxy ring.

While I think your answer is slightly better than the author's, it's still unlikely. I think that tautomerization should happen only in the aldehyde form since it is more reactive. Besides, the formation of the geminal hemi-epoxide is quite unstable as well.

Yeah, I agree that hemi-epoxide is quite unstable, but isn't that the tautomers are the two substances, that include the hemi-epoxide?

Starting from the aldehyde,

The double bond of the aldehyde breaks, allowing the oxygen to bond with the H in the alcohol. The alcohol subsequently breaks the O-H and forms a ketone.
Note that this molecule is stabilised due to hydrogen bonding; therefore it may be prudent to add a base or acid to encourage tautomerization.

OP is asking specifically about ring creation, so this doesn't answer the question.

If you see my diagram there is a six-membered ring formed due to the hydrogen bond formed between the aldehyde oxygen and the alcohol hydrogen.  Thus this reaction is more likely to happen then the above. Besides, ring-chain tautomerization happens more for hemi-epoxides which can form five or six membered rings so I doubt that this is likely to happen in the first place.

"hydroxypropane-2-one and 1,2-epoxy-2-hydroxypropane" - does acetylacetone and its enol look like them? No? OP wrote normal tautomerisation instead of this cyclisation and you changed compound to have a ring...


text:
Predict the order of ionic radius
We have arrange the followig sequence  for decreasing order of ionic radius .
$Se^{-2},I^-,Br^{-},O^{-2},F^{-}$
In my book the answer given as
$I^{-},Se^{-2},Br^{-},O^{-2},F^{-}$
But according to me $O^{-2}$ should be greater than F$^-$
As oxygen has two more electrons , so its zeff should decrease more in comparism to fluorine

If you're showing it in decreasing order, as you've said, then oxygen and fluorine are in the correct order. Oxygen, being larger, would appear on the list in the order you've shown.

Could you clarify your question? You expect the oxygen ion to be greater in size than the fluorine ion, which is also what your book is saying. I don't see your question other than confusion in what is meant by "decreasing order"?

You're right about the expected order. From largest to smallest the expected order would be:
$\ce{_{53}I^-}\quad$ = [Kr] $4d^{10}\text{ }5s^2\text{ }5p^6$ = [Xe]
$\ce{_{34}Se^{2-}}$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr]
$\ce{_{35}Br^-}\text{  }$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr]
$\ce{_8O^{2-}}\text{   }\text{ }$ = [He] $2s^2\text{ }2p^6$ = [Ne]
$\ce{_9F^-}\quad$ = [He] $2s^2\text{ }2p^6$ = [Ne]
You expressed the idea poorly though. $\ce{_8O^{2-}}$ and $\ce{_9F^-}$ have the same number of electrons. However $\ce{_9F^-}$ has an extra proton which should draw the electrons in closer.
There are a lot of ways to measure atomic/ionic size. No idea why the book would have $\ce{_8O^{2-}}$ and $\ce{_9F^-}$ flipped.

I don't see how his book has it switched. It appears your ordering is identical to what he's saying his book shows. However, he says he expects the oxygen ion to be greater in size than the fluorine ion, which is also what his book is saying. I don't see a question other than confusion in what is meant by "decreasing order"?


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Combining Chemicals
I wanted to ask purely from a theoretical point of view what would happen if you were to combine Lanthanum Uranium and Radium together? I chose it as my project since it spells my name purely for comical reasons but since i can't find it anywhere i'm genuinely curious.

From a purely theoretical point of view, they would form a trimer consisting out of La, U and Ra. I'm pretty sure there is a minimum somewhere on the potential energy surface. Maybe it exists in space somewhere.

They will perhaps form an alloy. Depends on the atomic radii of the three elements (they need to be similar, you can check on wikipedia).
Or, if the radii are (suitably) dissimilar, you might get some sort of intermetallic phase (like e.g. brass).
However I seriously doubt that anyone has ever investigated the phase diagram of this specific ternary mixture. Radium is just too radioactive for most people to care.
(You might have suspected this, with three metals: There will be no interesting reaction. :-/)

Two radioactive elements made to react????
Doubt the reaction's feasibility based on various factors such as -

Thermodynamic point of view (huge amount of energy may be required).
Kinetic point of view
Most importantly, cost of carrying out such a futile reaction.

1+2 are completely unfounded statements.

How exactly may i know @Karl

Radioactivity and chemical reactivity have nothing to do with each other. An alloy or intermetallic phase just forms from the melt, or doesn't. Very few reactions require a "huge amounts of energy" to take place, because that usually means the product would not be stable. What gives you the idea that this "reaction" would be kinetically hindered?


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How to calculate concentration of unknown saturated stock solution from diluted solution whose concentration is known?
I want to measure the solubility of a compound. I prepared a saturated stock solution of my compound in 100ml of water by dissolving it in excess and filtering out the undissolved sample.
Then I took 1ml of saturated solution, diluted it to 10ml and measured the absorbance on a UV spectrophotometer. After that I calculated the concentration of diluted solution by plotting a calibration curve of various solutions of known concentrations.
Now how can I calculate the concentration of my saturated stock solution from diluted one?
The concentration of diluted solution is 17ppm/10ml.

First off, welcome to Chemistry.SE!  You need to give more detail regarding your question and experimental details, and you also need to show what you've tried yourself so far.  For a homework question, this is a requirement.  Also, people will be more motivated to help you when they see more effort on your part.  For a good outline of the homework policy, see here.

Unless I misunderstood (please give more info), you've already done the most difficult part of the job by determining the concentration of your diluted solution!
And you just diluted your stock solution ten times, so you'd only have to multiply by ten the value that you found for your diluted concentration...

17 ppm is a concentration. It doesn't matter if you have 10 ml of it or 1 L. It is still 17 ppm.


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Why is zinc blende not directly reduced in smelting furnaces?
My question can also be generalized as, why are some sulfide ores (like $\ce{ZnS}$) roasted first to generate oxides, and only then reduced in a blast furnace or similar furnace?

What process would you suggest for the direct reduction? It is not that we have a cozy interface with a nice red button "Reduce".

The standard free energy of formation of most of the sulphides are larger than those of $\ce{CS2}$ and $\ce{H2S}$. Therefore, these sulphides are most stable than those with carbon or hydrogen. So, neither carbon not hydrogen can reduce metal sulphides to metals.
But standard free energies of formation of oxides are much lower than those of $\ce{SO2}$.  Therefore, oxidation of metal sulphides to metal oxides is thermodynamically favorable. Also, the oxides are easily reducible.


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How to determine the permissible values for which solution of the Schrödinger's wave equation exists?
My textbook( A Textbook of Physical Chemistry, Dr. O.P. Tandon ) says that  solution of the  Schrödinger's wave equation exists only for certain permissible values which are called eigen values.( physically possible ones).
How to determine the permissible solutions of the wave equation? Please explain.

Either the textbook or your quote is somewhat misleading. It is akin to saying that there are many solutions to the equation $x-5=0$, but only certain solution (that is, $x=5$) is "permissible" (that is, right).

@IvanNeretin thanks for pointing out, I have corrected it.

My comment still applies. Also, eigenvalues are surely not the solutions to the wave equation, either permissible or otherwise.

@IvanNeretin is this OK now, if not please correct me.

Now it's OK. You have that E (energy) in the equation; it is just some number, but it can't be any number. There are certain permissible values for it, AKA eigenvalues.

Mathematically there is nothing special about the Schrodinger equation as a differential equation. It is the boundary conditions we apply based on the physical situation, say length of box for particle in a box, that leads to quantised solutions, i.e. eigenvalues.


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Reaction of nitrobenzene with potassium ferricyanide
I came across this question in an exercise book. Can anyone give detailed answer (mechanism) of this reaction?

I am unable to proceed. I thought there maybe some redox reaction involved where $$\ce{Fe^3+ -> Fe^2+}$$ maybe taking place but then nitrobenzene would have to be oxidised. Can someone please help ?

I'd go more for an oxidation, but on the ring (hydroxylation).
My guess comes from this paper here : "The Hydroxylation of Aromatic Nitro Compounds by Alkalies" by O. C. Dermer.
This is very harsh (and somehow old) chemistry.
First, for m-dinitrobenzene and sym-trinitrobenzene, hydroxylation is reported in your conditions, that is to say alcaline conditions with the help of potassium ferricyanide as oxidizing agent (reactions A and B, see reference in the paper).
Second, the authors report the reaction C (with less than 4% yield though, according to my calculations), with a small amount of p-hydroxyazobenzene as a side-product. They did this reaction by mixing only nitrobenzene with potassium hydroxide, so I would expect the same product (o-nitrophenol) with maybe a better yield in your conditions.

First, since there are no electrophiles that are formed, the only reaction that comes to mind is nucleophilic aromatic substitution.
$\ce{NaOH}$ does a displacement reaction with $\ce{K3Fe(CN)6}$, yielding the $\ce{CN-}$ ion.
The cyanide ion attacks the benzene ring, giving three possible resonance structures:

The hydrogen attached to the carbon with the cyanide ion is accepted by the strong base $\ce{NaOH}$.
Next, the nitrile undergoes acid-catalysed hydrolysis to form a carboxylic acid:

I thought it was tough to remove ligands inside a complex. Can you explain how/why CN- comes out?

Btw, you left your mechanism at amide, no need to edit it to acid though. I know the mechanism

While $\ce{CN-}$ is possible to be displaced by $\ce{OH-}$, it is a much stronger ligand. Therefore I suspect that this reaction would not happen to a large extent. What's more, nucleophilic aromatic substitution generally occurs with  aryl halides since $\ce{X-}$ is a much better leaving group.

I've edited the mechanism above already, thanks for pointing that out.

I'm afraid there are some real problems with this loss of a hydride. If CN- attacks nitrobenzene, the subsequent step must expel H- not H+. Do you have any reference for this reaction? Nucleophilic substitution of hydrides is not unknown (Chichibabin reaction), but it doesn't make much sense here. cc @AlphaRomeo

Aside from that there are some errors in the mechanisms. In the first mechanism, the first arrow should be a forward arrow $\to$ not a resonance arrow $\leftrightarrow$. The third structure is missing a double bond between C and N. In the second mechanism, the amide has to be protonated first before H2O can attack, and NH2 has to be protonated before it can leave.

@orthocresol Yes, I don't think this reaction is likely to happen in great amount as I've noted. But there's no possible electrophile to undergo electrophilic substitution either. However, it seems unfair to simply put "No reaction". And I've skimped on the amide part. :P Thanks for pointing that out though.

@orthocresol My exercise book is the only place where I saw this reaction. Unfortunately I dont know the answer to the question. There is a good possibility that the author just made this reaction up, as it wont be the first time this has happened. ( not mentioning the book and author as its a local book for exam preparation, most probably you wont be able to find it on net)

I think H attached to the carbon which bonds to cyanide group is a $H^-$, which is a base (and even stronger than NaOH), so how can it be taken away?


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Why is the the rate of a reaction measured in concentration?
Why is the rate of a reaction measured using concentration against time, rather than mol against time?

Because maybe the differential rate laws are in form of concentration. For 1st order kinetics it doesnt matter.

Simply, because if you used an absolute amount (moles, for instance), by changing the quantity of reagents used, you couldn't use the existing data (or you would still need the volumes, to calculate the concentrations, in order to scale up/down the existing data, and fit them with your reaction).
Since the kinetics doesn't change with volumes, but with concentrations (for instance, if you use the same concentrations of reagents in 100 ml, or in 1 liter, the reaction, at least in theory, doesn't change) it's much easier and useful to express it in terms of concentrations.

This is because when you use concentration, you are taking in account of the reactant's abundance in the solution or mixture. This is compounded by the fact that the rate equation is also written in terms of a constant and powers of the concentration of reactants.


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Naphthalene v/s Anthracene reactivity
Both have similar resonance energy per ring, even Naphthalene has a little higher Resonance energy (127.5 KJ per mol per ring v/s 127 KJ per mol per ring )
Can we prove , why there is a small difference between them and Anthracene has lower resonance energy ?
Is there significant difference in their reactivity ?

In naphthalene, both rings are similar. In anthracene, the middle one is significantly different ("less aromatic") from the other two, hence the reactivity.


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What is the difference between the Bohr model and the quantum model?
Specifically, Bohrs atom model gives the radius of an atom as 0.53Å, while a quantum description offers a similar value. What is the difference between these?

Quantum description does indeed offer the same value, but in a very different context. It is not the literal "size" of the atom, to begin with. Atoms don't have definite sizes at all.

Your statement is mostly true for hydrogen - but for other atoms, the Bohr model does not work quantitatively.

May be useful: https://chemistry.stackexchange.com/questions/59096/atomic-radius-definition-disproved-bohrs-model-to-schroedingers-wave-equatio/66676#66676

related http://chemistry.stackexchange.com/questions/28659/usefulness-of-the-outdated-bohr-model

this is because quantum model offers a possibility of finding an electron somewhere, so when the radius of an atom is measured the average distance is taken

The Bohr model is a comparatively simple model, whereas the quantum description is part of the huge theory of quantum mechanics, so to list all the differences including implications would need to explain quantum mechanics itself.
But I think quite a good starting point would be to say, the Bohr model has electrons travel on certain, specific, classical trajectories, picked out by certain rules. The quantum model does not have electron trajectories. Instead electrons are quite different, they are spread out a bit like clouds of gas, in areas called atomic orbitals. I would recommend reading about atomic orbitals, the Wikipedia page is quite interesting. You will see very quickly they are quite different from the Bohr model.

Orbitals are not "areas of occupied space"! Such statements repeatedly lead to wrong impressions about quantum mechanics. They are wave functions describing the electron itself, not just where it is.

@Feodoran yes, orbitals carry more than position information, with the phases you also get momentum information. Could you please give me an idea of the wrong impression this kind of answer leads to? I can then reformulate it to avoid this. Because of the scope of the question and the asker not apparently having any knowledge of quantum mechanics, it doesn't seem to me like the extra complexity of addressing phase properties would be worth the added information. If you could give a phrasing that explains these details clearly to a lay person I may include it in my answer.

There is no easy phrasing, that's why people keep using the "occupied space" thing. What about "In the quantum model electrons are not considered to be particles anymore, but are typically described by an abstract wave functions (orbitals)."? One would need to explain now what a wave function actually is, but this would lead to explaining all the basics behind QM, as you already noted.

About the wrong impressions: A classical particle moving around in some area can be described by a trajectory. Averaging the positions of this trajectory over time will lead to probability distribution of that classical particle. Just stating that in QM orbitals are "areas of occupied space" does not contradict that classical description. And I think many people will imagine it like this, since they are trying to hang on to the particle picture. It should be made more clear that QM does not work like this. The explanation should include arguments that directly contradict the particle picture.

@Feodoran ok, I have changed my wording to hopefully make the situation clearer.


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How to convert ethanol to benzoic acid?
I have tried this method:

Can anyone suggest me a shorter route?

Criteria: Benzene has to be produced, it is not available.

Edit : I don't understand why the question has been put on hold. They provide the reason that there may be possibly to many good answers or the question is very broad. As far as my knowledge goes, I haven't received any correct answer that addresses what I have asked.
I would request the authority to reconsider on their decision.

Not really a better method but you can covert it into n-heptane and make it into toluene by Cr2O3-Al2O3 .

@AlphaRomeo ethanol to n-heptane? How?

your question is good, although there are many ways and synthesis route, there is one in which ethyne can be directly converted to benzene but it requires very high temperatures (4000-degree celsius)

@phenolicdeath I also converted ethyne to benzene in one single step.

@Resorcinol not particularly sure. Cant think of anything short and elegant

@Resorcinol oh I see, didn't notice

LOL - Two steps (1) Sell ethanol to a wino (2) Use the money to buy benzoic acid.

Are we allowed to use benzene as a reagent or do most/all the carbons have to come from ethene?

If it were allowed he wouldn't have gone through all the trouble to use ethyne to synthesise benzene in a hot iron tube. :P

(1) If you want a short way, I would suggest this: oxidise ethanol to acetic acid, use SOCl2 to make acetyl chloride, then do a Friedel-Crafts acylation with benzene. That gets you to acetophenone. A subsequent haloform reaction will knock that down to the benzoate and acidic workup gives you benzoic acid. But if you can't even use benzene as a reagent, then I'm afraid you might not be able to find the short route you desire. (Note that ~20m tons of benzene are produced annually.) Yours might already be one of the better routes.

(2) If you want the question to be reopened, could you at least please specifically tell us, what are the criteria for such a synthesis. Do you have to synthesise benzene from scratch? At least, this will stop us arguing about whether we can use benzene or aromatic reagents in general.

@orthocresol I have done the requisite edit. Please check.

@Resorcinol Given that I dropped the first close vote here is my rationale: You are asking to synthesise one compound from a completely different compound. There are about a gazillion possible routes none standing out in any way as preferable. There is nothing that would instantly suggest itself. This is basically the same as asking us to come up with a plan for the total synthesis of . Too many possible answers → too broad.

@Jan but which one is the shortest?

@Resorcinol What are the options? The point is that because the starting material and the target are so different there are a huge number of ways that you could design the synthesis and so it is impossible to categorically say which is the shortest/most efficient.

http://m.imgur.com/5d8KKWN

@NilayGhosh why are you replying me with my answer?

Watch carefully. I proposed a different method.

Here is how I would do it.
Part I: Blue
First, ethanol 1 is oxidised to acetaldehyde 2, which undergoes nucleophilic addition with $\ce{KCN}$ and $\ce{HCN}$ to form the hydroxynitrile 3, and finally dehydrated and hydrolysed at the same time with sulfuric acid to give acrylic acid 4.
Part II: Red
First, ethanol 1 is reacted to form the Grignard reagent 5, then combines with the aldehyde 2 to form the alcohol 6. It is dehydrated to form butene 7, dihydroxylated to form the diol 8, and dehydrated again to get the diene 9.
Part III: Black
In the final part, 4 and 9 participate in a Diels-Alder reaction with the Lewis acid $\ce{TiCl4}$ to give 10. Selenium dioxide gives the diol 11 and subsequent dehydration gives us benzoic acid, 12.
Although it may be longer, it does not involve things such as high reaction temperatures and therefore may be more feasible in the context of a laboratory.
EDIT: Here is another attempt:

Ethanol 1 is oxidised to acetaldehyde 2, and subsequently with excess formaldehyde and acidic conditions forms pentaerythritol 3, and is oxidised again to form the quadruple aldehyde 4. Strong heating causes 3 to turn into the diol 5. The Grignard reagent 6 is formed and oxidation and reduction of the alkene forms the reactant 7.
Now, both 4 and 7 react to form the ring product 8, and with strong heating thus produces benzoic acid 9.

I guess your method is not shorter than mine. Still you provided me with new methods. Thanks!

I agree, however I daresay it is easier to do since a lot of these reagents and conditions are easily found in the lab and many of these reactions involved dehydration. Your reaction in a hot iron tube is more like an industrial reaction which requires extremely high temperatures as pointed out. Besides, you are dealing with gases like ethene and ethyne which are hard to control whereas all my reactants are liquid at tolerable temperatures.

Yes, I agree with you. But I won't be able to accept your answer because I am expecting a shorter route.

Wonder which method has a better yield.

At 10 can you use Pd directly to get aromatic compound?

In what kind of reaction exactly? Palladium is more for oxidative insertions and reductive eliminations.

You Provided a few very great routes I really like the convergent approach of Linus Choy. I'd prefer a shorter route.

In the first step we are preparing the ethylhalogen. There are different procedures. I'd use $\ce{HCl}$ and $\ce{ZnCl2}$. by heating around 60 mins there should be conversion. My Main Point here is so we can go for a Friedel-Crafts reaction.
Second step is the Friedel-Crafts reaction with $\ce{AlCl3}$ or $\ce{FeCl3}$ onto the Benzene Ring. So we end up with ethylbenzene as product.
For the third reaction I also considered $\ce{KMnO4}$ and $\ce{H2SO4}$ for oxidation, there should be procedures leading to the same product.
But I really liked the idea of the oxidation by $\ce{ZnO}$ in DMF here because the reaction only takes a few minutes in the Microwave and has good yields.

I had this idea. Unfortunately you are unlikely to get a good yield of ethylbenzene because you will get polyalkylation since ethylbenzene is more reactive than benzene. There are ways around it but they will need more steps.

I would suggest using Friedel-Crafts acylation instead, then simply oxidising the mixture with a chromium derivative. Kudos for trying though!

I don't think you can use benzene.i know it's easily available but then why would resorcinol ask such a question?

Considering what @bon said, you can instead do a $\ce{KMnO4}$ oxidation to acetic acid, eliminate water to form anhydride, perform a Friedel Crafts acylation, then do Clemenssen reduction. But of course, that's alot of steps, so it's not much preferred.


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Need an explanation for how they got this compound name out of this structure

My professor posted this question as practice in naming compounds and provided the answer. I have stared at this for 30 minutes and cannot seem to figure out how he got this name out of it... Can someone please explain.

Could you show us what you've tried?

Try watching some Khan Academy videos about naming organic molecules (naming alkenes is in another section, but you could find it easily on your own). Then come back and answer your own question.

Related: How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

@Don_S I do not recommend to use Khan Academy for nomenclature of organic chemistry. Many names that are shown in these videos are based on obsolete rules from 1993 or even 1979, or simply wrong. They do not cite any actual IUPAC recommendations. Even in case an example name is correct, the corresponding explanations are often misleading or completely wrong.

Thanks, @Loong, I wasn't aware of that because the videos were going along the same lines of how I studies first year organic chemistry 7 years ago. I saw your comment above and went through the related question, which made me realize it is time I update my knowledge in this field...

The names that are proposed in the question are in accordance with obsolete IUPAC recommendations:

2-ethyl-5,5-dimethyl-1-hexene (1979 recommendations)
2-ethyl-5,5-dimethylhex-1-ene (1993 recommendations)

The corresponding rule in the 1979 recommendations is Rule A-3.4:

3.4 – Unsaturated branched acyclic hydrocarbons are named as derivatives of the unbranched hydrocarbons which contain the maximum number of double and triple bonds. If there are two or more chains competing for selection as the chain with the maximum number of unsaturated bonds, then the choice goes to (1) that one with the greatest number of carbon atoms; (2) the number of carbon atoms being equal, that one containing the maximum number of double bonds. In other respects, the same principles apply as for naming saturated branched acyclic hydrocarbons. The chain is so numbered as to give the lowest possible numbers to double and triple bonds in accordance with Rule A-3.3.

Therefore, the parent structure is the unbranched chain that contains the maximum number of double and triple bonds (here: one double bond). Since there is still a choice, the parent structure is the longest chain that contains the double bond, i.e. hexene. Numbering starts at the end of the chain that gives the lowest numbers to the double bond, i.e. 1-hexene. The 1-hexene is substituted using the usual principles of substitutive nomenclature, which yields 2-ethyl-5,5-dimethyl-1-hexene.

According to the 1993 recommendations, locants are placed immediately before the part of the name to which they relate. Therefore, the name 2-ethyl-5,5-dimethyl-1-hexene is changed to 2-ethyl-5,5-dimethylhex-1-ene.
The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) incorporates a major change from earlier recommendations given in 1979 and 1993.

In acyclic parent structures the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed. Thus, the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second.

Therefore, the parent structure of the compound that is given in the question is now the longest chain, i.e. heptane. The heptane is substituted using the usual principles of substitutive nomenclature; i.e. the prefixes (here: ‘methyl’ and ‘methylidene’) are considered together in a series of increasing numerical order. This yields the preferred IUPAC name 2,2-dimethyl-5-methylideneheptane rather than 6,6-dimethyl-3-methyleneheptane since the locant set ‘2,2,5’ is lower than ‘3,6,6’.


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Why do Grignard reagents form?
Why does magnesium insert itself between a carbon atom and halogen atom? Can anyone give any vague idea about why this occurs?
Also does a Grignard reagent only form when there is carbon atom and halogen atom with single bond between them? For example, can magnesium insert between the carbon and oxygen atoms of a carbonyl?

Reactions happen because of decrease in chemical potential associated with it. More precisely they happen if activation energy is provided, products are stable thermodynamically or kinetically and there's a viable pathway for it. Are youu asking about pathway or what?

Yes, this is a very interesting question that till today, not everyone agrees the precise mechanism as to how it works. Here is one with homolytic fission:

For your second query, I don't think it is feasible as the $\ce{C=O}$ carbonyl bond is rather stable.

The answer presents a mechanistic hypothesis, but the question is asking why the reaction happens. The answer needs to include some rationale why the mechanism makes sense.


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Symmetry and SALC: choosing coordinate basis
When we construct symmetry adapted linear combinations of orbitals (aka SOs) to build an MO diagram, we choose the z axis as the axis of highest symmetry and all other axis to make symmetry operations of a group easier to perform.
If we choose z-axis to point in some other direction, we will get a representation that is clearly not irreducible, but yet we would be unable to reduce it with the chosen basis because all orbitals mix (e.g. considering the case of 3d orbitals of a TM octahedral complex and having $d_{z^2}$ at a slant to the z-axis). I wonder if someone could explicitly explain why in this case the approach fails and highlight its limitations in general.
Thank you!

If you choose another axis, say the 3-fold instead of the z or  4-fold axis in an octahedral complex you should still be able to form a linear combination but of mixtures of $e_g$ symmetry orbitals and mixtures of $t_{2g}$  symmetry orbitals (but not of both as these orbitals are not degenerate) and these mixtures will be different (but equivalent) to those obtained from choosing the 4-fold axis. (The maths may be far harder in one choice than the other.)

@porphyrin Ok, thank you! I see: in this case to form an IR in this basis, we need non-integer coefficients, unlike all the cases I've dealt with before. Is this correct?

yes there is nothing special about integer coefficients

In general, when dealing with characters tables and with group-operations, it is crucial to choose the right orientation of axes in order to get sensible results. I say in general because it can also occur that, for highly symmetric molecules the "wrong" choice (or better say unconventional) can give the right result. Almost the 90% of characters tables follow the "Mulliken" conventions stated in Mulliken's paper "Report on Notation for the Spectra of Polyatomic Molecules" (The Journal of Chemical Physics 23, 1997 (1955)). Whenever you orient your axes not in agreement with the characters table you are referring for performing operations, there is a high risk to fail and face strange results as you did (all orbitals mixing etc..). I hope my answer is general enough and helpful.


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What does the newest IUPAC nomenclature have against unsaturation?
IUPAC nomenclature 2013 changed a couple of rules very commonly used in naming simple acyclic compounds. As a result, depending on which nomencalture we choose, the answer to this question varies greatly, as demonstrated in the comments. Educational textbooks, hence, would propose an outdated name that is not PIN (Preferred IUPAC Name) according to the newest standards.
This change has been illustrated in the draft, page 442 (page 58 of P-4, section P-44.3), which reads:

P-44.3 The principal chain
In an acyclic compound, or in a compound composed of chains and rings, the chain on
  which the nomenclature and numbering is based is called the ‘principal chain’. When there is a
  choice for the principal chain, the following criteria are applied, in the order listed, until a
  decision is reached.
$$\mbox{A change to the traditional order of seniority criteria is recommended;}\\ \mbox{the length
of the chain is senior to unsaturation.}$$

What was the reason for this change?
Normally, there doesn't need to be a reason for changes. However, a change as drastic and influential as this must have had a reason behind it. I'm thinking a group of [newly discovered] compounds needed a new nomenclature, and I remember I read some news that said vaguely the same thing. Was there a need for revision of the criteria? Why would a nomenclature in which unsaturation is senior be inferior to the new system of IUPAC nomenclature?

Interesting question.  I must admit that I have a very dim view of changes of this sort.  Isn't the "real" reason likely to be that the IUPAC nomenclature committee needed something to debate during their meetings?  But perhaps I'm too cynical and there is a good reason for this one.  I'll be watching this space to learn more...

Yes indeed. This doesn't make much sense. Troublesome for authors all over the world to keep up to date with these

I think it is important to note that these are recommendation and not rules (in the sense of which you must follow) and hence you can choose for yourself whether you like to implement them or not.
A generated unambiguous name still conforms to the general recommendations of the IUPAC, hence what you call outdated is still a valid name, it just isn't the preferred IUPAC name, which is a concept introduced in 2013. Hence names generated according to the 1993 recommendations will still be valid names according to the new recommendations; the new recommendations are backwards compatible.
The reasoning can be found in the Introduction of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) (It is also in the cited draft):

P-10 INTRODUCTION
  [...]
  A major new principle is elaborated in these Recommendations; the concept of preferred IUPAC names' (PINs) is developed and systematically applied. Up to now, the nomenclature developed and recommended by IUPAC has emphasized the generation of unambiguous names in accord with the historical development of the subject. In 1993, due to the explosion in the circulation of information and the globalization of human activities, it was deemed necessary to have a common language that would prove important in legal situations, with manifestations in patents, export-import regulations, environmental health, and safety information, etc. However, rather than recommend only a single unique name for each structure, we have developed rules for assigning preferred IUPAC names, while continuing to allow alternative names in order to preserve the diversity and adaptability of the nomenclature to daily activities in chemistry and in science in general.
Thus, the existence of preferred IUPAC names does not prevent the use of other names to take into account a specific context or to emphasize structural features common to a series of compounds. Preferred IUPAC names (PINs) belong to a 'preferred IUPAC nomenclature'. Any name other than a preferred IUPAC name, as long as it is unambiguous and follows the principles of the IUPAC recommendations herein, is acceptable as a general IUPAC name, in the context of a 'general IUPAC nomenclature'.
  [...]

If nothing more, the new guidelines provide a more robust nomenclature. In my very personal opinion that is a very good change. Let's look at a more practical approach.
Upon a chemical reaction which modifies a compound (in many if not most case) the principle chain will be retained according to the new change. Since unsaturation often provides the possibility of relatively easy modification, similar to a functional group (and some would even consider it one), a reaction at these bond will likely involve that the principle chain changes according to the old recommendations. Let's stick to the example of Loong's answer in the linked post; and let's just add $\ce{HBr}$ to the double bond:

According to the old recommendations the reader is likely to imply that there was a modification hexene to heptane, and without further inspection one could assume that there was an extension of the carbon skeleton, i.e. a carbon-carbon bond formation.
In the new guidelines it is much more obvious what the modification actually is, a simple addition. This obviously is of little concern for these simple compounds, but when it comes to natural product synthesis, it removes some of the ambiguity of what actually happens, highlighting the change of the functional group rather than implying a modification to the carbon skeleton.
There are probably instance where you deliberately want to deviate from PINs. A simple example would be the protection of an alcohol:

It might be a necessary step in your synthesis, it might even be an important one, for the chemistry involved and expressing what you are actually doing, PINs will be a nuisance. I personally would prefer the names in parentheses. However, when you are eventually writing your patent, you might want to stick to a PIN in order to avoid any ambiguity, after all that's why they were introduces in the first place.

The first three-quarters of your answer are spot on. (a.) Newer IUPAC rules do not invalidate those that preceded (though I would steer clear of the ones whose use is strongly discouraged). (b.) The principal chain of a compound changes when there is a change in one of the $\ce{C-C}$ bonds. PS. I also like this alteration in the guidelines, and thus feel triply at easy handing out a (+1).


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Using specific heat at different conditions
Why can the change in energy of an ideal gas system (through a process in which V is not constant), still be calculated using Cv? (as in DE=Cv*DT)

The subscripts v and p refer to how these heat capacities are measured experimentally, not how they are applied in practice.  In each case, the heat capacity can be measured by determining how much heat has to be added to the system in either a constant volume experiment or in a constant pressure experiment.  On the other hand, for an ideal gas, irrespective of whether the experiment is carried out at constant pressure or at constant volume, the changes in internal energy and enthalpy are given by dU=CvdT and dH=CpdT.  This is because U and H are functions only of T for an ideal gas.  The general mathematical definitions of these heat capacities are $$C_v=\left(\frac{\partial U}{\partial T}\right)_v$$$$C_p=\left(\frac{\partial H}{\partial T}\right)_p$$If you are not dealing with an ideal gas, then the volume or the pressure also affect U and H, and this must be accounted for.  But, for an ideal gas, only the temperature change is required to determine the changes in U and H.

Great job understanding the question

@Chester Miller thanks. This might be more math related but when you integrate Cv with respect to T you treat V as a constant (which is the opposite of the partial derivative that defines Cv), but what if throughout the experiment V was not constant?  Can you still treat it as a constant during the integration and why (since it during the experiment it changed)?

(since it did change during the experiment)*

If it is an ideal gas, then it doesn't matter if the volume changed, since the internal energy is a function only of temperature.  U is a function of state, so you could evaluate its change by following an alternate path in which you first held the volume constant and increased the temperature, and then, in a second step, held the temperature constant and changed the volume.  For the second step, the change in internal energy of the ideal gas would be zero.  So irrespective of whether you followed the actual path or the alternate path, the change in internal energy would be the same.

Ah that's what I was looking for. Thanks a lot for this clarification.


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Does purified water corrode copper pipes?
In other words would a reverse osmosis system corrode copper pipes due to it's purification?  Does purified water speed up copper pipe corrosion leading to pin holes, etc.?

You might find this of interest.  It doesn't completely answer you question but it discusses some of the factors contributing to the corrosion of copper pipes.

Although it seems pretty clear that removing potentially corrosive compounds from drinking water will result in a net of no effect or possibly decreased corrosion, there is at least one consideration worthy of mention.
While metallic copper has a solubility that is well below most regulatory standards for drinking water (i.e. 2 mg/L in the UK), enough copper can still potentially dissolve to result in a significant loss of copper mass, such that pinholes and destructive oxidative processes can take place.
However, according to the UK's National Physics Laboratory, Good Practice Guide No. 120, "Avoidance of corrosion in plumbing systems", September 2011:

"...the rate of corrosion and the release of copper ions slow rapidly in the first few weeks of operation as protective films form on the bore of the copper pipes."

Further:

The surface films (often referred to as a patina) are composed of thin brown or black coloured copper oxide ($\ce{CuO or Cu2O}$) overlaid with a thicker green or turquoise basic copper carbonate layer. These layers, although protective, are not true passive layers (like those which form on stainless steels or aluminium and which are much thinner and invisible to the naked eye). It is when these films on copper do not form properly in the first place or become disrupted later that problems of pitting or blue water can occur.

According to this related document from the UK's "Foundation for Water Research", the low alkalinity and buffering capacity of purified water can result in conditions of slight acidity that can prevent the formation of the desired patina, and recommends the addition of 50 mg/L bicarbonate alkalinity in order to promote the development and maintenance of the protective copper oxide and carbonate layers.
To reiterate, my answer to the question is a net "No" with an interesting caveat.


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How to use a spectrometer to determine NPK
I'm trying to determine if it's possible to use a low-grade spectrometer to read NPK values of a solution.
I believe I should be able to take readings of the individual compounds (e.g., calcium nitrate) and then use that to "subtract" from a reading of the mixed/final solution and basically deduce how much of each compound is present - is that correct?

First off, welcome to SE.Chemistry!  Although including links to descriptions of your equipment and terminology is a good start, you still need to provide additional details for your experiment. Some brief descriptions in additions to the links could be helpful.  And most importantly, you need to show some efforts that you have made to solve this problem.  For example, provide an example with calculations for how you said you believe it should be done.  This may lead to you answering your own question, and it will also motivate others to put fort their efforts to help you.

Your spectrometer detection range is 340-780 nm. That is to say mostly visible, and a bit of UV. If your compounds are colorless, it's a first indication that you won't see anything on your spectra. These compounds seem to be inorganic, no transition metal or anything exotic, so it's not a good method.
Apart from this, yes, you would need to calibrate with solutions of known concentration (unless you know the absorption coefficient, which would give you the concentration with Beer-Lambert law).

Autumn's answer has already indicated that your spectrometer might not be suitable for the direct determination of nitrate, phosphate, and potassium.
In Journal of Research of NBS, Section A: Physics and Chemistry, 1972, 76A, 469-482 (free PDF) the uv spectrum of potassium nitrate in aqueous solution is given. The spectrum exhibits a maximum at $\lambda = \pu{302 nm}$ with a molar absorption coefficient $\epsilon$ as low as $\pu{7 l\,mol\,cm^{-1}}$.
Even if this wavelength wasn't below your detection range, the low $\epsilon$ wouldn't get you very far anyway.
Admittedly, I didn't bother to check for the other spectroscopical data, but without derivatization or complexation, that would turn the ions in question into species with a strong and specific absorption in the visible range, you're aout of luck.


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Look for superabsorbent balls whose expansion is reversible
I am glad to find very cheap superabsorbent balls (see image below). However, I found their expansion is not reversible. The wiki entry 'Superabsorbent polymer' also says that the material will 'retain' extremely large amounts of a liquid relative to their own mass.

Is there any other superabsorbent material whose expansion is reversible? If yes, is it available as tiny balls?

Yes. The ones I have are known commercially as Jelly Marbles or Orbeez. They're sold at gardening stores, home decor, and online. They're used as toys but also as gardening products that absorb excess moisture in wet conditions and supply water in dry conditions. Mine grow up to about 1.5cm from a starting diameter of ~0.1cm. I've been told that some grow up to ~7.5cm; and they're advertised as increasing by 100-300 times in volume at full saturation. The process is reversible (e.g. by putting them in a cup with a wad of paper towels). I use them as a drying agent where powdered $\ce{CaCl2}$ etc. aren't ideal.
Note the actual identity of the polymer in the beads is a trade-secret, so I don't think you'll be able to find that information.


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What is pure oxygen (gas) density as a function of T or P?
I'm looking for a set of data (ideally graphed) of $\ce{O2}$ density as a function of $T$ or $p$ (with constant $p$ or $T$, respectively).
The best I've found is this 1948 paper which provides an empirical model but (1) it's not as recent as I'd like and (2) I want solid experimental data, not empirical fits. One can calculate ideal-gas densities by $$\frac{pM}{RT}$$ but I'm looking for real values.
A recent empirical model would also be very useful!
Something like this, which is $\rho(T)$ for air:

I feel a tad foolish for pointing out that the graph is an empirical fit. So what do you really want? Experimental data so you can do your own fit or some empirical equation so that you can calculate any P and T curve that you desire?

And it isn't like the folks in 1948 were incompetent - why does being 'recent' matter?

@MaxW it includes data points, which I would be happy with. @ Jon better equipment, low and high T fit, etc.

What T range, what P range?

0 -> supercritical, ideally

A colleague showed me a website that can render almost any thermodynamic quantity as a function of another for various substances: http://webbook.nist.gov/chemistry/fluid/.
Using NIST's software allowed me to get an isobar graph of $\rho (T)$ from 55K to 1000K (at 1.0atm).

And $\rho (P)$ from 0 - 800atm (at 273.15K):

It even has an HTML data-point option (and downloadable tab-delimited form). Perfect! In the case of $\ce{O2}$ it uses data from papers in the '80s and '90s (experimental and (from the software) empirical)

LOL - Can NIST be trusted or do they too now use "alternate facts"?

Professionals use it for molecular model development so I'd say it's a reliable resource