text:
Avogadro Molecule Editor: How to render image from command line?
Given a molecular geometry in the .xyz file format, how can I use Avogadro on the command line to render an image of it?
The website states that there is some command line API available, but I can't seem to find any documentation/information about it. Alternatively, are there any other molecular visualization tools that provide a convenient command line interface for image rendering of .xyz files?

Pymol does a pretty good job and you can do everything using the command line. There's also a lot of documentation on it.

@DSVA Thanks for pointing that out ; ) I think I will go into that direction for now. Do you know if pymol can also directly render to png without starting the UI?

Yes, you can start it in command line mode (https://pymolwiki.org/index.php/Command_Line_Options) or use API and the command you are looking for is "png" https://pymolwiki.org/index.php/Png

That's awesome. I really needed that functionality for a presentation on thursday. I somehow have a lot of geometries to analyze, so I needed some kind of access to rendering geometries from python. I think that really is what I was searching for : ) You made my day, thanks!

There is no way to currently do this using Avogadro, although Python bindings have been "in the works" for a while. The problem is, with volunteer developers, there's not always enough free time to tackle all features.
You can create pseudo-3D images using Open Babel. I developed this for the Pitt Quantum Repository to give SVG and PNG images in the search interface. It's not perfect, but I like the effect.
obabel beta-carotene.xyz -O beta-carotene-2d.svg -xS -xd -xb none

The -xS indicates this 3D "ball and stick" style, -xd prevents the molecule name from being included, and -xb none gives  transparent background to the SVG (useful for presentations).

This current rendering is available in the development version of Open Babel - a slightly different version is available in 2.4.x. The new "tweaks" add some scaling to atom sizes and opacity to give a depth effect, which works well in most molecules. Suggestions are welcome.

I really think that convenient solutions for batch job visualizations of molecules (e.g. md trajectories) are lacking. The only solution I found so far is pymol, which however I quite dislike due to their licensing choice (e.g. real 3d rendering lacking in free version and so on). So this is quite interesting...

Agreed on the need for convenient solutions for batch visualization. I was hopeful that we'd have Python bindings for Avogadro2 by now, but the student tackling that last year in Google Summer of Code disappeared (likely doing some other research).

@deusexmachina You may be able to script VMD to do this -- Geoff, thoughts? It's Tcl/Tk driven, right?

@hBy2Py - that's likely true, although I can't find any documentation on running VMD as a command-line tool.

@GeoffHutchison What is meant by python bindings. Sorry if this is an amateur question. I am very new to python, but we use OpenBabel frequently.

If what you want is a simple 2D structure, OpenBabel can do it for you.
Say you have a structure named beta-carotene.xyz:

The following command issues a 2D depiction of the above:
$ obabel beta-carotene.xyz -O beta-carotene-2d.png -xp 600 -x0 molfile

Its about 3D structures.


text:
Can drinking a lot of water be fatal?
I vaguely recall having heard that drinking too much water can, over time, prove fatal to the human body.
Nothing special about the water; not distilled or de-ionized or anything … just plain ol' water.
Now the reason that accompanied this "fact", was that drinking too much water serves to dilute, and thereby, disrupt the electrolyte balance in all (living) body tissue. This in turn, messes with the sodium-potassium pump present in cells leading to loss of functionality of cells.
Now I can't seem to find the source of that information, but I do recall it wasn't particularly trustworthy.
So what I want to know is:

Is it true, that drinking a lot of water can be fatal?
If so, is it because the electrolyte concentration inside and outside cells are "diluted"?
Does this affect all (living) tissue equally?**

Edit:
I fail to see why this was put on hold as a personal medical question. I vote to reopen this on the grounds that it does not:
a) Ask for advice on treating water intoxication
b) Suggest, promote or request, in any way, alternative forms of medicine.

See also: Death because of distilled water consumption on Biology.SE

Related: Is there anything that is completely non-toxic to humans at any dose? on Biology.SE

Not so much "over time", as "over a short amount of time". That is, drinking 10 glasses of water a day for a month won't kill you... but drinking the same ~18 gallons in a day will lead to the water intoxication as referred to by @paracresol.

Too much of any chemical can kill you.

One woman died after drinking 7 litres of water (http://www.dailymail.co.uk/news/article-1224051/Wee-For-Wii-water-drinking-contest-death-Jennifer-Stranges-family-awarded-10m.html), and another  died after drinking 4 litres of water in under two hours (http://news.bbc.co.uk/1/hi/england/bradford/7779079.stm). Those two cases are just the ones I've seen or heard about on telly, and besides it's just as @Bob said, too much of any chemical can kill you.

Although too much of a biologically inert chemical such as silicon dioxide (pure sand) can kill you only by its mechanical effects (crushing, suffocation etc.)

Based on what I gathered from this Wikipedia article, Yes.
Drinking copious amounts of water can prove fatal. The proper term is "Water intoxication".
When you start taking in a lot of water (by "a lot" I mean more water than your body can excrete via sweat or urine), the interstitial fluid that bathe the cells that form your (living) tissue end up getting "diluted", i.e- the concentration of ions like $\mathrm{Na^{+}}$ is greatly reduced. The result? The concentration of ions in the interstitial fluid is far lower than the concentration of ions inside the cells it surrounds. Thinking of this in another way; the concentration of water in the interstitial fluid is far higher than the concentration of water inside the cells.
Ever soaked dry raisins in water? Over time they begin to absorb water and swell up as a consequence of the concentration gradient set up between the tissue in the raisin (Low concentration of water) and the water in the bowl. This phenomenon, where water flows from a region of higher concentration (of water) to a region of lower concentration (of water) across a semi-permeable membrane is called osmosis.
It's pretty much the same thing in the case of water intoxication. A concentration gradient has been set up between cells and the interstitial fluid that surrounds it. So water begins to flow into the cell, and the cell swells up quite a bit, resulting in a build-up of turgor pressure
This increase in turgor pressure will be seen to varying degrees in all sorts of (living) tissue, especially tissue that is highly vascular. An important example would be neural tissue, particularly that in the brain. This swelling up of tissue results in an increase in intracranial pressure which can lead to loss of functionality over time. This is what makes water intoxication so lethal.
There isn't really much about the chemistry of water that's responsible for this, apart from osmosis that is.

Addendum:
Drinking a ton of water isn't the only way for water to build up in potentially lethal amounts in the body. Hyponatremia, which refers to a condition of low (below-normal) sodium levels in body fluids, can also lead to water accumulating in the body in excess.

Yes.  See Jury Rules Against Radio Station After Water-Drinking Contest Kills Calif. Mom

The husband of a California woman who died after participating in a radio station's water drinking contest said he hopes a jury's $16.5 million compensation award following a wrongful death lawsuit will send a message to other corporations dealing with the public.

...

[She] drank nearly two gallons [7.5 l] of water in over three hours

...

A coroner ruled that her death was due to water intoxication.

For more information on the subject see Fatal water intoxication Journal of Clinical Pathology 2003 Oct; 56(10): 803–804., which discusses and earlier fatal case:

Water intoxication provokes disturbances in electrolyte balance, resulting in a rapid decrease in serum sodium concentration and eventual death. The development of acute dilutional hyponatraemia causes neurological symptoms because of the movement of water into the brain cells, in response to the fall in extracellular osmolality. Symptoms can become apparent when the serum sodium falls below 120 mmol/litre, but are usually associated with concentrations below 110 mmol/litre. Severe symptoms occur with very low sodium concentrations of 90–105 mmol/litre. As the sodium concentration falls, the symptoms progress from confusion to drowsiness and eventually coma.

There have also been several US military deaths due to excessive water consumption as reported in Death by water intoxication Military Medicine 2002 May;167(5):432-4.

It's also been reported to be a factor in drug deaths (MDMA)

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Decrease in Surface area leads to Increase in load capacity for mesoporous material?

So at the bottom of page 86 it says that when SBA-15 is functionalized with Glutaraldehyde it results in a substantial decrease in the surface area and pore diameter of the mesoporous material.
This paper rationalizes this by saying that the bridge chain insertion causes an increase in hydrophobicity that improves its ability in loading ibuprofen.  Can someone please why this is the case!!?? thank so much it just seems very conterintuitive.
http://ac.els-cdn.com/S0927776514001751/1-s2.0-S0927776514001751-main.pdf?_tid=cea489ec-ec87-11e6-a9e2-00000aab0f01&acdnat=1486398051_2d9403232650bb1a93cf30c41e9a1d72

Although I only have access to the abstract of the link you provided (unless you want to spot me $40 ;) they have characterized their glutaraldehyde-functionalized SBA-15 surface using at least 7 different analytical techniques, so I'm pretty convinced that the surface area, pore diameter and hydrophobicity are as they say.
The ~50% increase in surface adsorption in spite of the 13-fold decrease in surface area just illustrates the fact that surface adsorption is a function of two parameters: the affinity of the compound for the surface in question and the area of the surface available for adsorption.  It seems that for the system you've described, these two parameters simply switched from high-surface area / low adsorption affinity to low-surface area / high adsorption affinity, with the adsorption affinity "winning out" over the decreased surface area.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What is the difference in (alc) and (aq) states in reaction mechanism?

In these reactions between alkyl halide and KOH in different states the product is different only because of the change in state of KOH from aqueous and alcohol .
Can someone explain the mechanism which causes this drastic change?

Nucleophilic substitution always competes with Elimination reaction in such cases.
What is the difference between them?
As put there by philipp:

In alcoholic solution the $\ce{KOH}$ is basic enough ($\mathrm{pK}_{\mathrm{a}}$ 15.74) to deprotonate a small amount of alcohol molecules ($\mathrm{pK}_{\mathrm{a}}$ 16–17)), thus forming alcohlate salts $\ce{ROK}$. The alcoholate anions $\ce{RO-}$ are not only more basic than pure $\ce{OH-}$ but they are also bulkier (how much bulkier depends on the group $R$). The higher bulkiness makes $\ce{RO-}$ a worse nucleophile than $\ce{OH-}$ and the higher basicity makes it better at $\mathrm{E}2$ eliminations.

Mechanisms below:


text:
Pattern to determine the maximum ionic charge for transition elements?
According to my textbook, main group elements follow a simple pattern when determining their maximum ionic charge. The maximum cationic charge is always equivalent to their main group number (group number minus 10 for the groups 13 to 18 according to the new IUPAC terminology) while the maximum anionic charge is eight minus that number.
For example, the maximum charge of an aluminium cation (third main group) would be $+3$, the maximum charge of a sulfur anion (sixth main group) would be $-2$.
Some elements form multiple ions, but no charge will be larger than determined by the above method
Is there a similar method or pattern to determine maximum charges of transition metals? Why or why not?

Related question on oxidation states: http://chemistry.stackexchange.com/q/59219

The pattern you observe for main group elements is due to the number of valence electrons present in the s and p subshells of the outermost shell. The most you can do is either lose all those s and p electrons (giving a ‘cation’[1] charged exactly according to the old main group number — new group number minus 10 for groups 13 through 18) or you can fill these up to a total of eight.
For transition metals, a d subshell is in the process of being filled. Since the d subshell of a lower shell is between the s and p subshells of a higher shell according to the aufbau principle, we can imagine that the highest number of ‘usable electrons’ would be 18. This is not typically the case, however.
For the first few transition metals, you can remove as many electrons as are present in the s and d valence subshells (e.g. 4s and 3d). In the 3d series, manganese is the last one that can do this to give manganese(VII), in the 4d series ruthenium can reach ruthenium(VIII) and in the 5d series iridium can reach iridium(IX).[2] For the remaining elements of a certain transition metal period, the highest accessable oxidation state again decreases. You can find an overview of the maximum oxidation states known for certain elements on Wikipedia.
The only elements that again follow a somewhat more simple scheme are the zinc group (zinc, cadmium and mercury). Since these have a completely filled d subshell in ground state, they are reluctant to give away any d electrons; hence their highest observed charge is $2+$, as if they were in group 2. (Indeed, they behave like group 2 metals in a lot of other respects, too; except mercury which is slightly different again.)
When it comes to the highest possible negative charge or oxidation state, things are more complicated yet again. Since all transition metals are metals, they are not commonly found as anions or anionic particles although some are more likely to than others. For some of these — typically the iron group and the two following groups — this corresponds to $12 - \text{group number}$. This small trend is more of an exception than of a rule, though. Note, for example, how zinc, cadmium and mercury are all known as in the $\mathrm{-II}$ state which does not make sense immediately just looking at the respective shells and subshells.
Many of the actually observable oxidation states can be rationalised by doing a complex molecular orbital analysis of the corresponding compounds. However, for most of these there is no simple and immediately obvious pattern that one can use — with the notable exception of the small bit of pattern outlined in the third paragraph.

Notes:
[1]: A better term here would be oxidation state. For many elements the highest possible free cation is not known but a compound with a corresponding oxidation state is.
[2]: Again, none of these are observable as free cations in standard chemistry; rather they are in compounds. Hence I have stopped using the term charge and started using oxidation state.


text:
What is the difference between SBA-15 and MCM-41?
I'm trying to understand the difference between these two mesoporous materials.  They both have the same polys-siloxane backbone and surface silanol groups, the same hexagonal pores, and yet I don't understand why they are different materials.

The difference between your two materials is primarily in pore geometry: for MCM-41, the pores are hexagonal; for SBA-15, they are round. Further, there are apparently differences in wall thicknesses.

Using high-resolution transmission electronic micrograph (HR-TEM) observation, one can clearly see the pore geometry of the MCM-41 and SBA-15 mesoporous silicas to determine that their pore shapes are hexagonal and round, respectively. With the perpendicular orientations of the nanochannels to the electron beam, parallel line images of the (100) and (110) repeating spacings were observed. In the SBA-15 mesoporous silicas, there are byproducts of the granular silica and disordered mesostructures, attributed to the weak hydrogen interactions between Pluronic 123 blockcopolymer and the silica species. There are also many different and significant +π disclination defects in SBA-15 and MCM-41 surfactant-silica composites. The SBA-15 with a thicker silica wall is more stable under irradiation by high-energy electron beams compared to MCM-41, which has thinner wall thickness. Some carbon nanostructure impurities were found in some carbon films on the metal grids.

Reference: Lin, Tang, and Lin. Detailed Structural Characterizations of SBA-15 and MCM-41 Mesoporous Silicas on a High-Resolution Transmission Electron Microscope. Journal of the Chinese Chemical Society 49(6) 981-988 (2002). Abstract here.

Oops, my bad...


text:
Creating Floc in the classroom
I'm the youth ed manager at Denver Water in Colorado. We do a water treatment demo in the classroom, and currently we use the first part of the Winkler DO test to create floc. I'm looking for a better demo that creates white floc and doesn't turn the water brown. I've tried using alum and mixing it with alkaline iodide azide reagent. It creates wonderful, white floc. What I don't know is if it's safe in a classroom setting. Can you help?

alkaline iodide azide reagent is strongly basic. No worse than Drano which is a commercial product in grocery stores. But it certainly isn't something that I'd want to spill on a kid. Could you just ppt CaCO3 by using CaCl2 and NaHCO3?

How about curdling milk with lemon juice or vinegar? Milk goes through coffee filter and you get zilch. Curdle milk then you get the curds on filter. Bonus is that this is something that the kids can do at home. // Learned that the hard way. Tried to be nice to my wife once and cut up a kiwi in her cereal. The acid in the kiwi curdled the milk -- yuck!

Have you assessed the SDS for these materials? Alkaline iodide azide is approx 50% hydroxide, so extremely caustic. Not familiar with the Winkler test, but it appears that it involves lots of KOH also. What risk assessment and safety data have you got for this test in the classroom?

I'm an old dog. Not sure if MSDS (Material Safety Data Sheet) has been renamed to SDS or not. But if the reagents were bought as stock solutions then they would have to have a MSDS. That is an excellent place to start looking at the safety considerations.

@MaxW, another old dog here.  I've just learned in the past few months that an MSDS is now called an SDS, though they are the same thing.  They did it to simplify things LOL.  Take a look here.

Geeze, I thought SDS was Students for a Democratic Society ;-)


text:
Prove that a Change of Volume will not Change the Value of the Equilibrium Constant
According to my book, a decrease in volume will result in an increase in pressure for an equilibrium mixture, which will in turn cause its equilibrium to shift to the side with the least number of gas molecules, given the reaction involves gases (and according to Le Chatelier's Principle).
It also says that Volume and Pressure changes will cause the position of equilibrium to shift, but the equilibrium constant remains unchanged...
If the equilibrium constant (K) depends on the concentrations of reactants and products...
and according to the molarity formula, Volume is inversely proportional with respect to concentration...
... then why does the equilbrium constant remain unchanged?
So far I have this, but I don't know what it means...

Surely you have the details confused. Is the book talking about a specific example rather than generalities?

It's referring to it in general terms. This is exactly what the book says: "A decrease in pressure will cause a shift in the equilibrium position to the side with the larger number of molecules. A different equilibrium position will be achieved, but the value of K will be unchanged, as long as the temperature remains the same"

In your workyou have changed the "equilibrium constant" which is a fixed while keeping the number of moles constant of all species in both instances. Rather its the opposite
Consider the reaction;$$\ce{AB <=> A + B }$$ with equilibrium constant $K_c$ at temperature $T$ and volume $V_0$
Initially let $n(AB) = x_0$ , $n(A) = y_0$ , $n(B) = z_0$ where $n()$ refers to the number of moles.
then $$K_c =  {\frac{{\frac {x_0}{V_0}}{\frac {y_0}{V_0}}}{{\frac {z_0}{V_0}}}}$$
Now the volume is reduuced from $\ce{V_0\bond{->}V}$
Hence all the concentrations increase.
Then the reaction quotient at the moment the volume is changed;$$Q_c =  {\frac{{\frac {x_0}{V}}{\frac {y_0}{V}}}{{\frac {z_0}{V}}}}$$
Now $Q_c > K_c$ hence equilibrium shifts to the right reducing the number of moles of $A$ and $B$ increasing the number of moles of $AB$.Due to this effect $Q_c$ will gradually reduce and at acertain point will becaome equal tp $K_c$ with different number of moles.
$$K_c =  {\frac{{\frac {x_0}{V_0}}{\frac {y_0}{V_0}}}{{\frac {z_0}{V_0}}}} = {\frac{{\frac {x}{V}}{\frac {y}{V}}}{{\frac {z}{V}}}} $$
The only parameter remaining constant(under constant temperature) is the Equilibrium constant is a certain parameter(i.e. Volume here) is changed, other parameters will vary keeping Equilibrium constant a constant.

"A decrease in pressure will cause a shift in the equilibrium position to the side with the larger number of molecules. A different equilibrium position will be achieved, but the value of K will be unchanged, as long as the temperature remains the same"
The book has overgeneralized. Look at this reaction:
$\ce{2BrCl(g) <=> Br2(g) + Cl2(g)}$
There are the same number of molecules on both sides. So decreasing the pressure doesn't always cause a shift in equilibrium.

I found the problem. I was confusing the reaction quotient Q with the equilibrium constant K.     Q changes with all sorts of factors like pressure, volume, catalysts, and temperature, but K only changes with temperature. According to other comments, Q eventually becomes equal to K again after some time when the equilibrium is re-established.

Q isn't the problem...

I didn't include the whole lecture above, but it does mention what you said. If the change in gas molecules between reactants and products is 0, then a change in pressure or volume will not change the equilibrium position or K. However, when the change in gas molecules is not 0, then the equilibrium position will change, but not K. It says K only changes with temperature.

Make Cl diatomic. I couldn't do it for you, evidently there need be 6 or more characters in an edit.

Both the question and your bold quotes refer to examples where one side has more moles of gas. Your example doesn't appear to apply.

@Sam202 you are correct; the issue is mixing up Q and K (a common one). Glad you sorted it out.

@MaxW  "...to the side of with the larger number of molecules..." Where is a problem ? If there is no such a side, shifting rule does not apply.

Le Chatelier talks only about how the and on which side will the equilibrium position shift. Not about the change in equilibrium constant. Let me make two points. :-

Change in volume affecting the equilibrium constant:- $\Delta$n = (sum of stoichiometric coefficient of products)-(sum of stoichiometric coefficient of products). E.G.

$$\ce{aA + bB <=> cC + dD}$$
here $\Delta$n= (c+d)-(a+b)
now,
$$K_c=\frac{[C]^c [D]^d}{[A]^a[B]^b}$$
since $$[A]=\frac{moles_a}{volume(V)}$$
similarly for all .
 $$K_c=K_x*V^{\Delta n}$$
So if
$\Delta n= 0$ then it doesn't affect
$\Delta n >0$ the it is directly proportional
$\Delta n <0$ the it is inversely proportional
E.G.
$$\ce{H_2 + I_2 <=> 2HI}   \Delta n= 0$$
$$\ce{N_2 + 3H_2 <=> 2NH_3}  \Delta n= (-2)$$
$$\ce{ PCl_3 <=> PCl_5 + Cl_2}  \Delta n= (1)$$

Le Chatelier  says that if volume is increased then number of moles per volume decreases then the equilibrium will shift towards the side with more moles which is evident from the previous explanation.

If you still find a doubt then let me know I will like to give another try with more explanation

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


text:
IUPAC nomenclature for many multiple bonds in an organic compound molecule
My previous Chemistry teacher while teaching IUPAC nomenclature said that the double bond is always given greater priority over the triple bond and the numbering of the carbon atom chain is done in such a way that the double bond gets the lowest possible number while my present Chemistry teacher says that the numbering of the carbon atoms is done in such a way that the multiple bond gets the lowest number whether it is the double bond or the triple bond (whichever multiple bond is nearest to the end when both the ends are considered, numbering is done from that end).
On surfing through the various sites on the Internet, i came across the lowest sum of locants rule.
I am confused and cannot understand how to number the carbon atoms for IUPAC nomenclature purposes when many multiple bonds are present in the molecule. What is the actual rule?

Immediately forget the ‘lowest sum of locants ‘rule’’ (note the double inverted commas) which is not a rule but can lead to wrong assignments of locants. The correct IUPAC rule is lowest set of locants. Some answers here have shown how the set rule can lead to different results than the sum.

^ See Loong's answer to IUPAC nomenclature: “Smallest sum of locants”?

Related: How can I find the parent chains in these two compounds?

The individual rules that have been mentioned by your teachers are correct; however, you have to pay attention to the priorities. For any given compound, there may be other relevant rules (e.g. concerning the principal characteristic group that is expressed as suffix, or the longest chain) that determine the parent chain and the numbering of locants first. Only if there is still a choice, the rules concerning multiple bonds are taken into account.
In this case, the principal chain is the chain with the greater number of multiple bonds. If there are more than one chain with the greatest number of multiple bonds, the principal chain is the one with the greater number of double bonds.
The corresponding rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows:

P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. (…)
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)

Likewise, with regard to numbering of locants, low locants are given first to multiple bonds, all considered together as a set. If there is still a choice, low locants are given first to double bonds:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(…)

The “lowest sum of locants rule” that you have mentioned does not exist in the IUPAC Recommendations and can lead to wrong results. The corresponding section in the current  IUPAC Recommendations actually reads as follows:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
(…)

The simple way which I use for solving such question is here.
(1). The numbering of the parent chain should always be done from that end which gives lowest sum for the multiple bonds. For example,

In this case if you will start numbering from LHS then it is $1+3=4$. And If you start from RHS then $2+4=6$. As lowest sum is obtained from LHS, hence name will be 3-penten-1-yne and NOT 2-penten-4-yne.
(2). If, however, there is a choice for numbering, i.e. numbering from both LHS & RHS gives same sum then, double bond is always given preference over the triple bond. For example:

In this case, numbering from both LHS & RHS gives same sum i.e. $1+4=5$. So here you should prefer numbering from that side which will give priority to double bond. Thus name will be 1-penten-4-yne and NOT 4-penten-1-yne.
This is a easy way for answering such questions , just you need to keep those above two points in your mind. I hope this will help you.


text:
How to test for Chromium(III) Oxide
I got some chromium(III) oxide ($\ce{Cr2O3}$) sold as pigment, called "Chrome Green". I've been having a hard time trying to react it with $\ce{HCl}$ to make $\ce{CrCl3}$. I thought it was just hard to break the $\ce{Cr-O}$ bond.
Now I'm just afraid that the pigment that I bought was actually cut with something else, like a silicate-type of pigment. How do I test for chrome(III) oxide, if I can't get it to become a salt in the first place?

I was thinking of roasting it in excess amounts with either $NaCO_{3}$ or $NaOH$ to get sodium chromate or chromite (the chromite I can just further oxidize). Dissolve it all in water and then dry it. If I see any white specs I can assume them to be sodium silicate. But this only tests for silicates mixed in. I want to know if the store I bought it from mixed it with enough actual $Cr_{2}O_{3}$. Perhaps, a test specific to Chromium ions would be better

The test for $\ce{Cr(III)}$ can be performed in the solid state using an oxidative melt. You will need

chromium(III) oxide (obviously)
sodium carbonate
potassium nitrate
a Bunsen or Teclu burner
a magnesia trough or a crucible
suitable pliers to hold the trough

Mix the green pigment with up to the threefold amount of sodium carbonate and potassium nitrate, place the mixture on the magnesia trough and melt it.
A yellow melt indicates the formation of sodium chromate according to
$\ce{Cr2O3 + 3KNO3 + 2Na2CO3 -> 2Na2CrO4 + 3KNO2 + 2CO2 ^}$

Chromium(III) oxide contains $\ce{Cr^3+}$ which can be detected through salt analysis. Only specific test for $\ce{Cr^3+}$ is mentioned.

Source 1*

$$\ce{Cr^3+(aq) + 3OH^{–}(aq) -> Cr(OH)3(s)}$$
Chromium(III) hydroxide is formed when $\ce{Cr^3+}$ is treated with
  ammonia or sodium hydroxide, which is soluble in excess alkaline
  solution (amphoteric, dissolves in acids too) but not soluble in
  excess ammonia . With sodium hydroxide, a dark green soluble
  hexahydroxo–complex ion is formed.
$$\ce{Cr(OH)3(s) + 3NaOH(aq) -> [Cr(OH)6]^{3–}(aq)}$$

Source 2**

Chromium can be taken through a series of colored tests which leaves
  no doubt as to its identity. Chromium(III) forms a steel green
  hydroxide which dissolves in excess strong base to give a deeply green
  colored solution of the hydroxy complex. Treating this complex with 3%
  hydrogen peroxide gives the yellow solution of the chromate ion, which
  upon acidification with dilute nitric acid gives the orange color of
  dichromate. Treatment of the cold solution of dichromate with 3%
  hydrogen peroxide gives the intense blue color of a peroxide of
  chromium. (The actual composition of this peroxide is not known, but
  it is believed to have the empirical formula $\ce{CrO5}$). This
  peroxide readily decomposes to the pale violet color of the original
  hydrated chromium(III) ion. In low concentrations of dichromate the
  blue color is fleeting, and attention must be focused on the test tube
  during the addition of the hydrogen peroxide to avoid missing the
  color change.
$$\ce{Cr(OH)^{4-} (green) + H2O2 + OH^{-}→ CrO4^{2-}(yellow)}$$
  $$\ce{CrO4^{2-} + H+→ Cr2O7^{2-} (orange)}$$ $$\ce{Cr2O7^{2-} + H2O2 +
HNO3→ CrO5 (blue)→ [Cr(H2O)6]^3+ (violet)}$$
The following color changes are all indicative of $\ce{Cr^3+}$. Add an
  excess of 6 M $\ce{NaOH}$ to about one mL of test solution. To this
  green solution add 10 drops of 3% $\ce{H2O2}$. Heat the test tube in
  the water bath until the excess $\ce{H2O2}$ is destroyed as indicated
  by the cessation of bubbles. Acidify the yellow solution with 3 M
  $\ce{HNO3}$. Cool the resulting orange solution in an ice bath. To the
  cooled solution add a drop or two of 3% $\ce{H2O2}$ and observe the
  immediate fleeting blue color.

Reference
*http://www.docbrown.info/page13/ChemicalTests/ChemicalTestsc.htm
**http://www.wiredchemist.com/chemistry/instructional/laboratory-tutorials/qualitative-analysis

Is the acidification with $HNO_{3}$ specific to the test? Is it because you need a strong oxidizing acid? Or would common $HCl$ suffice?


text:
How to test for magnesium and calcium oxide?
Is there a way to test for $\ce{MgO}$ and $\ce{CaO}$?
I want to prove that  $\ce{MgO}$ and $\ce{CaO}$ are created from their respective calcinations.
$$\ce{MgCO3 (s)->[heat] MgO(s) + CO2(g)}$$
$$\ce{CaCO3 (s)->[heat] CaO(s) + CO2(g)}$$

For iron clad proof!?! Use $200,000 x-ray diffractometer.

I assume these are 2 separate solutions, one with MgCO3 and one with the CaCO3, and not a single mixture of the two?  Or do I have that wrong?

@airhuff - ?!? the (s) in the chemical equations means solids not solutions.

@MaxW, brain-skip on my part, slipped right by me, thx.  My first thought for a quick and dirty test is to heat the salts under an inert gas stream, measure the CO2 coming off by whatever means, then weigh the remaining solid and see if it all adds up.  If you really want to be sure, digest the solid, dilute like crazy, and measure by Mg or Ca ICP-AES or AA!

"Proof" is the kicker. You can't analyze $\ce{CO3^{2-}}$ by either ICP-AES ($200K instrument too?) or AA ($50K instrument?). A simple flame test would show difference between Ca and Mg but is that "proof?"

@MaxW, yes, proof is by degrees.  And I said to measure the Mg and Ca by ICP or AA. The liberated $\ce{CO2}$ (g means gas here ;) would simply be measured by any number of classical techniques (titration in cheap $\ce{CO2}$ coulometer, IR detector, adsorb and weigh, etc.).  You could also measure the oxygen in the (undigested) solid by classical pyrolytic Mertz oxygen analysis (~35$ analysis), though the latter could have a 10% + relative uncertainty in this matrix.

Hi - please don't use MathJax in titles on chem.SE, thanks.

Magnesium oxide and calcium oxide can be easily differentiable by its reaction with water. The former is practically insoluble in water(0.0086 g/100 mL (30 °C)), hence magnesium hydroxide will be formed in negligible amount. However, the latter is soluble in water and will form forms calcium hydroxide to a large extent.
You can also differentiate them by forming their sulfates and testing their solubility.
There are selective test like ammonium oxalate test (more details here) and $\ce{Na2HPO4}$ test (more details here) which is described in the other answers to this question. The best test however is Flame test (also described in the other question).

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


text:
Synthesis of tin(II) nitrate
I tried to prepare $\ce{Sn(NO3)2}$ using tin(II) chloride and nitric acid according to the reaction
$$\ce{SnCl2 + 2 HNO3 -> Sn(NO3)2 + 2 HCl},$$
but stannous chloride appears to be poorly soluble (turbid solution). What is the correct method to prepare $\ce{Sn(NO3)2}$?

SnCl2 is pretty well soluble. This reaction is no good for other reasons.

I stay away from all nitrates and cyanides, at all costs. I wanna stay legit in the eyes of everybody.

I don't think that $\ce{Sn(NO3)2}$ exists as a salt. You can evidently make it in solution by dissolving $\ce{Sn(OH)2}$ in dilute nitric acid.

Related: http://chemistry.stackexchange.com/questions/8748/make-tin-nitrate

Stannous Nitrate is obtained in white leaflets having the
composition $\ce{Sn(NO3)2.20H2O}$ by cooling to $\pu{-20 °C}$. a solution of
stannous oxide in nitric acid of density 1.2. It is also formed by the
action of highly diluted nitric acid on the metal, some of the acid
thus suffering maximum reduction to ammonia:
$$\ce{4Sn + 10HNO3 -> 4Sn(NO3)2 + NH4NO3 + 3H2O}$$
According to C. H. H. Walker, tin dissolves in nitric acid forming
stannous and stannic nitrates, the relative proportion of these two
salts produced depending on the temperature and the strength of the
acid; moreover, the yellowish white precipitate which separates when
the somewhat concentrated acid is employed is said to be a hydrated,
ill-defined, stannic nitrate.

Ref.: http://tin.atomistry.com/stannous_nitrate.html
You can also check the post linked by Ivan.

All of this suggests that you get mixture of compounds every time, IMO there should be a better way.


text:
Is this what an orbital means?

In this picture, there is a $1s$ and a $2s$ orbital. Are the orbitals only the dark gray circles ? Or are the light gray circles (which are surrounding the dark gray ones on both sides) also included in the orbitals? Other than these two types of circles , there are the nodes which never include any electrons. And since people define an orbital as a region where electrons are most likely to be found, this must mean that the orbitals are only the gray circles because if they included the light gray circles too, then the probability of finding electrons in an orbital would be 100%.

Also, i have the knowledge of an 11th grader in physics. Not anymore than that so dont come at me with all the complex equations and stuff.

This is a really poor illustration. Look for 2S orbitals on wikipedia, there are many better examples.

This picture is quite bad, since there are patterns inside the gray areas which shouldn't be there but the gray areas are the orbitals. You could think of it like writing down the position of an electron over and over again and every dot is a position where you found the electron. More dots means it was in that area more often (=higher probability of finding it there).
An electron, theoretically, can be everywhere except the nodes. So an electron from an atom right in front of you could be lightyears away. The chance is very, very, very low for that to happen but it is possible. So for having a 100% chance of finding the electron you have to look in the whole universe, of course it has to be somewhere. That's not very useful at all, since every electrons can be found somewhere in the universe so we don't work with 100% chance but rather with 95% ( or 99% or 99.9%,..) since that's already a probability of finding the electron very, very close to the core of the atom and it is there most of the time.

So the orbitals are the grey circles only and the regions where the electrons could be (other than the orbitals) are the light grey regions, right ?

@Sam19KY No. The electron is always in an orbital if it is bound to an atom. The orbitals extend to infinity but the probability of finding an electron at a point decreases as that point gets further away from the nucleus.

That is not a $1s$ orbital and a $2s$ orbital, the two together make up the $2s$ orbital. If you take a look at the radial distribution function versus atomic distance for a $2s$ orbital, you'll see that there are two peaks and a node between them.

The graphic you presented is (rather poorly) illustrating the above graph of the RDF, with darkness corresponding to the magnitude of the RDF at a given radius. It's useless to try and cut it up so cleanly, though. Electrons don't orbit the nucleus in a well-defined path like the planets orbit the sun—they can be literally anywhere except the node. The only 100% chance of finding an electron in an orbital is by integrating over all space (from center of the nucleus to infinity, in all directions).

The orbital is not where the electron is most likely to be found, it is the region of space that it occupies as described by a mathematical function (its wave equation). In order to understand why we treat quantum mechanics the way we do, I would strongly recommend reading up on the the uncertainty principle as well as wave-particle duality (though work through the math you'll need integral calculus, linear algebra, differential equations, and some probability theory). In essence, we can't really know everything about electrons for sure, but what we can know is the probability that they are a certain way at any given time.

Actually nodes are as far as I understand a weakness of describing of electrons behaviour with orbitals, next to infinite volume which doesn't account for presence of electron only in its light cone. Orbital is only an approximate model. also @DSVA


text:
quantum capacitance of supercapacitor electrodes, high or low needed?
quantum capacitance of supercapacitor electrodes should be low or high? to achieve high specific capacitance. The quantum capacitance forms among the layers of low DOS materials when used as an electrode for supercapacitor.

It got the answer myself. High quantum capacitance is favorable.


text:
Degeneracy of second excited state of H-?
This is a question presented in a IIT-JEE 2015 paper I exam. It says,

. Not considering the electronic spin, the degeneracy of the second excited state (n = 3) of H atom is 9, while the degeneracy of the second excited state of H- is-

I saw a solution which says it's 3.
I think by 9 degenerate energy levels the question meant 9 possible orbitals for M shell. But I am not sure how's that 3 for for H-.
Please don't delete this question as I haven't found it anywhere.

I think it is important to understand that for hydrogen atom (or any other one-electron system) all orbitals from the same shell have same energy. For instance, $E_\mathrm{2s} = E_\mathrm{2p}$, $E_\mathrm{3s} = E_\mathrm{3p} = E_\mathrm{3d}$, etc. Thus,

The first excited state of hydrogen atom would be one in which either $\mathrm{2s}$ or one of the three $\mathrm{2p}$ orbitals is occupied and it will be 4-fold degenerate: $\mathrm{1s^0 2s^1}$, $\mathrm{1s^0 2p_x^1}$, $\mathrm{1s^0 2p_y^1}$, $\mathrm{1s^0 2p_z^1}$.
Analogously, the second excited state of hydrogen atom would be one in which either $\mathrm{3s}$ or one of the three $\mathrm{3p}$ or one of the five $\mathrm{3d}$ orbitals is occupied and it will be 9-fold degenerate: $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^1}$, $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^0 3p_x^1}$, $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^0 3p_y^1}$, $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^0 3p_z^1}$, plus 5 configurations with only $\mathrm{3d}$-orbitals occupied.

For hydride the situation is different: it is not a one-electron system, so different orbitals from the same shell do not have same energy anymore. For instance, $E_\mathrm{2s} < E_\mathrm{2p}$, $E_\mathrm{3s} < E_\mathrm{3p} < E_\mathrm{3d}$, etc. Thus,

The first excited state of the hydride would be one in which one electron populates $\mathrm{1s}$ orbital and another $\mathrm{2s}$ one, i.e. a non-degenerate $\mathrm{1s^1 2s^1}$ state.
The second excited state of the hydride would be one in which one electron populates $\mathrm{1s}$ orbital and another one of the three $\mathrm{2p}$ ones, i.e. a 3-fold degenerate state: $\mathrm{1s^1 2s^0 2p_x^1}$, $\mathrm{1s^1 2s^0 2p_y^1}$, $\mathrm{1s^1 2s^0 2p_z^1}$.

So one of the electrons remains in $\ce1s^1$ at all times? Is this a rule? This question was intended for students who only know how to work around with hydrogen-like species (i.e single electron species). PS I am aware about this particular examination, its syllabi as well as the difficult level of its questions. @Wildcat

I am confused about the same that why one of electrons remains in 1s1 all the time?

For hydrogen atom (or any other one-electron system) all orbitals from the same shell have same energy, why is it so? @wildcat

@KhushiLadha The energy difference between 1s and 2s is more than that of 2s and 2p. Thus, in the 2nd excited state, the electron jumps from 2s to 2p.

In a single electron system like H atom or He+ all orbitals with same principal quantum number (n) have same energy
Therefore the degeneracy refers to the 3 subshells 3s 3p 3d which have same energy


text:
With a partition coefficient how do you decide which solute concentration is the numerator?
Kpc = [X(in solvent A)]/[X(in solvent B)]
I realise that Kpc is actuallly a ratio but when you need the numerical value it has to be treated as a fraction.
When it has to be treated as a fraction how do you decide which solvent is A and which is B?

In your last sentence, do you mean solvent, and not solute? If you mean solute, then I don't understand the question.

@airhuff Thanks for pointing that out. Stupid mistake of mine!

The proper way to express the partition coefficient of compound X distributed between an aqueous and organic phase, for example, is:
$$\mathrm{(K_P)_X = \frac{[X](org)}{[X](aq)}}$$
This unambiguously expresses the mathematical fraction to be used in calculating the partition constant for a solute, X between two solvents, aq and org.


text:
Positive electrode of an electrochemical cell?
I had the following problem:

Relevant information from Data Book:

$\ce{Ag+ + e- <=> Ag}$
$E^0 = +0.80$
and
$\ce{Fe^3+ + e^- <=> Fe^2+}$
$E^0 = +0.77$

This means $E_\mathrm{cell}$ is +0.03. (right)
And (I thought) that the left ($\ce{Fe}$) half cell has the positive electrode/anode as the ions lose electrons to the anode. I thought that the half equation with the more positive $E^0$ value went forwards.
However this is wrong. Where did I go wrong in my reasoning?

Your reasoning was correct, except for the part that Fe is positive electrode. As $Fe^{2+}$ lose electron, it produce electron, so that the electrode becomes negative. Ag electrode becomes negative for the same reason.

As the standard reduction potential of Ag electrode is more than that of Fe electrode, Ag electrode will act as cathode (where reduction or gain of electrons takes place). The Ag electrode will be a positive electrode because it is electron deficient and thats why reduction is taking place at this electrode.
The Fe electrode will act as anode (where oxidation or loss of electrons takes place) and it will be the negative electrode because there is excess of electrons on this electrode.
Ecell = (Ereduction)cathode - (Ereduction)anode

      = +0.80 - (+0.77)

      = 0.03V

So the Ag electrode is positive because the electrons flow through the circuit to it and the Fe electrode is negative because electrons flow away from it?

Yes thats the reason. At Fe electrode Fe+3 loses a electron so there is excess of electrons here and at the Ag electrode Ag+ from the AgSo4 solution gets the electron and gets reduced to Ag and deposits itself of the Ag electrode.


text:
Efficient, low overhead means to scrub acid gases without water loss?
This question is pretty much what it says on the tin: what would be an efficient (low pressure drop, reasonable energy consumption), low overhead (simple, lightweight) means to scrub acid gases from a feedstock stream without water loss (and ideally, with water capture)?
Say, the stream is primarily $\ce{CO2}$, a few percent nitrogen, 80kPa / 25°C; the $\ce{H2O}$ partial pressure is ~5 Pa (~20ppm); and there's a variety of acid gases in the dozens to hundreds of ppm range, and you'd like to scrub at least 90% of the $\ce{H2O}$ and acid gases, preferably more.  Obviously if you were to just, say, run it through a packed bed with water you're just going to humidify the gas stream in the process of scrubbing, because the vapour pressure of water is three orders of magnitude higher than the $\ce{H2O}$ partial pressure already in the stream.  Obviously if the gas stream were compressed to several hundred atm or more then you could do that, but such levels of compression would neither be efficient nor low overhead.  You could likewise freeze out $\ce{CO2}$ to increase the partial pressures of everything else, and even distill the remaining gases - but these are anything but efficient, low-overhead solutions.  I thought about the use of dilute sulfuric acid as the scrubbing medium to drop the vapour pressure, but the vapour pressure drop isn't nearly enough until high molarity, which would be relatively useless for scrubbing.
I'm sure there's a "right" solution for such a case, but I'm not sure what it is.  Adsorption into a solid chemical bed, with subsequent regeneration (and if so, what sort of chemical)?  Multistage, with say acid gas scrubbing first, then water scrubbing (and if so, what method for the latter to get down to under a few pascals partial pressure)?  Selective membranes - say, to exclude $\ce{CO2}$, with a partial vacuum drawn on the other side?  Some sort of ion pump?

Are you trying to scrub the carbon dioxide along with the other acid gases? $\ce{CO2}$ is itself an acid gas, so just about anything that'll pull down $\ce{SO_x}$/$\ce{NO_x}$ will also take up $\ce{CO2}$. You'll consume absorption media like crazy.

Not trying to scrub the CO2, just everything else (HCl, SO2, SO3, some H2SO4 mist, small amounts of other acids, etc).
I don't know about other media, but simple Henry's Law calculations show that water can easily scrub everything else without taking a significant amount of CO2, given adequate contact time/area.  The problem is that it would also humidify the CO2 stream.

Taking away aqueous scrubber solutions  is limiting.  Do you need to roughly maintain that 5Pa value, or what is the limit for water vapor?

Ideally even lower than the 5Pa value.  Minimizing water remaining is critical. I'm thinking, perhaps a 3-stage process is best - an aqueous scrubber, than an easily reversible desiccation bed (say, silica gel), then a very aggressive desiccator like phosphorus pentoxide to scrub what remains.  Hmm... come to think of it, I don't know how to calculate how rapidly water would move into or out of the absorber to determine if a given arrangement is plausible...

How about water absorption of acid gases followed by a CO2 membrane separator? IIUC there are some really good CO2-selective membranes starting to reach commercial viability, which I think should be able to pull CO2 and leave behind water and N2.

High throughput and good longevity?  I read a research paper on zeolite-based CO2 membranes a while back but they had very poor longevity, and only recovered partially by thermal regeneration.  Do you have a paper (or manufacturer site) where I could read more?

The specific vendor I'm aware of is MTR; the landing page for their CO2 membranes is here. I've seen talks from Tim Merkel at recent MS&T meetings.


text:
There is a higher energy 2Σ+ state with a bond length of 1.15 Å for CN molecule, explain this

How can there be an excited state for CN where the bond length is lower (stronger bond) than the actual ground state? The molecular orbital diagram suggests that the single electron in the sigma orbital gets excited, but to which orbital? This was a question asked by one of my professors and I don't know where to start, I think that the ground state X 2Σ+ goes to the B 2Σ+ excited state and maybe the electron could go to the 3S orbital? but then how would this affect the bond long is beyond me. In addition, would the vibrational fine structure go from V'=0 to V'=2? Can anyone help?

The inter-atomic potential has a minimum at some distance - this is the ground state bond length. Higher energy states will sample other parts of the inter-atomic potential, both at larger distances as well as small distances.

Do you understand how to derive the term symbols for a diatomic molecule? You should start by 1) drawing a MO diagram for CN; 2) figuring out which electronic excited configuration can correspond to a $^2\Sigma^+$ term symbol. There's no "3s" orbital in a diatomic molecule. As for the bond length, it is quite a straightforward concept: if you fill more bonding orbitals, then the equilibrium bond length decreases, and vice versa if you fill more antibonding orbitals.

However, note that a shorter equilibrium bond length does not necessarily mean more stable. Excited states can easily have shorter equilibrium bond lengths than ground states. It also does not necessarily imply a stronger bond. If you can draw a simple Morse potential curve, you will see that the bond strength (which is a vertical parameter) and the equilibrium bond length (which is a horizontal parameter) are quite different things.

@orthocresol I was thinking exciting one of the the 2s sigma star electrons to the 2p sigma orbital, I assume this would give the desired term symbol; however I'm unsure if this is a viable option

Okay, one thing about the MO labels. The sigma-type orbitals contain quite a good mixture of 2s and 2p orbitals. (For more details consult any text on the MO diagram of CO.) So, it's usually more common (and more accurate!) to label them $1\sigma$, $2\sigma$, $3\sigma$, $4\sigma$ without mentioning which atomic orbitals they arise from. I guess, you are referring to a $2\sigma \to 3\sigma$ promotion; indeed, that seems quite plausible to me. It would not be the first excited state (you already know that is a $^2\Pi$ term); it would be a higher excited state.

Hmm, and was that what you meant by "3S" in your question? I interpreted it as a 3s atomic orbital, sorry... You can typeset mathematical and chemical expressions on chem.SE using MathJax; this post contains further details.

Bond lengths shorter than in the ground state are known in a few other molecules $\ce{C_2}$ and NO for example. The MO construction can only indicate  the number of bonding /anti-bonding/non-bonding orbitals are occupied and  one can then surmise about relative bond length changes when electrons are promoted to higher levels. Note that each excited state has its own set of vibrational & rotation levels. The particular geometry a molecule has determines which of these are excited when an new excited state is formed say by absorbing a photon. Selection rules determine which states are reached.


text:
Effect of sodium hydroxide solution on copper plate
I have a copper etching plate that has over 100 years of dried ink in the lines of the plate. Someone suggested flake sodium hydroxide as a possible solvent for the ink. My concern is preventing damage to the copper itself.
Suggestions? For example, what should the concentration of the solution (in water) be?

Did you consider the use of an ultrasonic bath, possibly with hot water, but without the sodium hydroxide?

How big is the plate? Do you have a scrap plate that can be used for testing? Is the ink itself soluble in water?

The plate is 19" x 29" and due to the age of the plate, I am unsure of the makeup of the ink, but probably oil-based, but certainly not water based. I have tried other solvents, i.e. denatured alcohol, mineral spirits, water, benzene, kerosene. I get some small toned liquid release, but I think this is more removal of dirt and certainly not the ink. Ultrasonic bath is out of the question due to size. Talked to Graphic Ink & Chemical Co. in Chicago, and they suggested sodium hydroxide, but I hesitate to use it until I know the plate will be undamaged.  Will try on the back of the plate to test.

Regarding the copper metal itself, the only down side I see of using a $\ce{NaOH}$ is that it is unlikely to remove some of the oxides that would be covering any exposed copper.  The initial etching process would have mainly produced water-soluble $\ce{CuCl2}$, likely with a bit of insoluble $\ce{CuO}$.  $\ce{CuO}$, $\ce{CuCO3}$ and some $\ce{Cu(OH)2}$ would have likely formed on any unprotected copper surfaces over time.  $\ce{NaOH}$ could remove $\ce{Cu(OH)2}$ and $\ce{CuO3}$, but not $\ce{CuO}$.  It also should be unreactive toward the copper plate itself.
A common method of putting a good shine on tarnished copper is with weak organic acids like citric acid.  Lemon juice and table salt make a favorite home copper cleaning solution.  Again, this is only with regards to the cleaning oxides from the copper metal itself, not the ink residues coating it.
My bottom line advise is to do what I’ve seen in your last comment that you plan to do; test a portion of the bottom side of the plate with the $\ce{NaOH}$ solution to verify that this solution cleans the ink residues without harming the copper.  I think you can do this with confidence that the copper metal will be unharmed.  If the ink residues are removed and you want to remove any remaining oxidation to create a fresh, shiney copper plate, you could dissolve just a bit of salt into lemmon juice and wipe the plate with this solution until shiny!

Exposed copper will develop a patina. It is quite possible that the surface crud is a combination of ink and corrosion compounds.

For what it's worth, I  just came across this related question and answer.


text:
Why are absolute entropies given the units J mol−1 K−1?
I've just started looking at the concept of entropy and can't understand why absolute (or standard molar entropies) are given the units $\pu{J mol^−1 K^−1}$. Could someone explain the rationale for these units?

So G = H-TS still works as expected - remember those 'absolute' entropies are specified at a standard temperature, and come in to play in expressions for Gibbs free energies as a function of temperature.

the OP asked such a basic question that I'm 99.9% sure that the OP has no idea what Gibbs free energy is either.

There are a number of formulas for entropy. Perhaps, from a classical thermodynamics standpoint, you have $\mathrm dS = \mathrm dq_\text{rev}/T$ (the Second Law). Heat has units of J and temperature units of K, so entropy itself has units of $\pu{J K^-1}$. Molar entropies then have units of $\pu{J K^-1 mol^-1}$. From a statistical mechanics standpoint, entropy assumes a more fundamental role, but the units are still the same: $S = k_\mathrm{B} \ln W$. As Jon mentioned: absolute entropies are no different from entropy changes, as far as units are concerned.

It is because $T\Delta S$ is a quantity of heat now usually measured in Joules. The 'per mole' part is not fundamental but because it makes sense to standardise measurements. When an isothermal reaction runs reversibly, $T\Delta S$ is the heat absorbed from, or released to, the surroundings, thus the work done may be either greater or less than the heat of reaction. This was first clearly seen by Gibbs and thus the idea of free energy as $G=H-TS$.

Related: Why can entropy be measured in joules per kelvin?

The best explanation I can give is that in order to measure entropy for a process we can exploit the fact that it's a state function. Entropy doesn't depend on the pathway that we take. So if you take for example ice melting at 273 K, this process is thermodynamically reversible. At 273 K ice and liquid water are in a state of equilibrium, but if we apply heat we can cause ice to melt. So this allows us to measure $ \Delta S$ directly by looking at how much heat we apply to cause this process to proceed. So we look at the amount of heat in joules and compare that to the temperature where we applied the heat.
If you want to think conceptually, think what adding heat will do to the system. We associate adding heat with an increase in entropy. But the magnitude of the change is related to the amount of energy the system currently has (which is directly related to its temperature in kelvin).
Standard entropies of formation are given in molar quantities because they assume the process is taking place to create 1 mole of the substance. Entropy itself is traditionally described with the units of J/K.


text:
Reasoning behind IUPAC nomenclature?

What is the reason for writing prefixes like di and tri in front of prefixes of alkyl or functional groups ?

For example in 2,3-Dimethylpentane we write 2,3 which is the location of the methyl group in the parent chain but then we also write Di to indicate the presence of two methyl groups.
I do not understand why we write Di, isn't it obvious that if there are two locants given then there will be two methyl group ? one methyl group is not going to be present at two places simultaneously.

Ok now lets assume for a second that writing Di is somehow useful, then
in 2,2-Dimethylpentane why we write 2,2 ? for me it is obvious that if there is Di before the methyl and only one locant is given then two methyl groups will be present at that location.

To summarise :

In 2,3-Dimethylpentane why we write Di if we have already written two locants to indicate the location of two methyl groups ?
In 2,2-Dimethylpentane if we write Di then why we write 2,2 ? I think only 2 is suffice here.

Related: Why do we insert a numeric multiplier after group positions when naming an isomer?

@Loong I got the answer for my first question but not for the second question.

This is of course conjecture, but a point would be to make it easier to "say" the compounds. The system would be weakened if, by intuition, people called substances by different nomenclature than what they wrote. If iupac nomenclature forced us to call something methyl-methyl-pentane we would simply say dimethylpentane anyway. Might as well accomodate easy spoken names and include the prefix as well as the numerical. This way there is 1 system for all, no exceptions. Fewer exceptions means stronger system.

@StianYttervik That might explain the first part but what do you think of second part ?

I mean that it is easier for everyone if we just keep both the "oral" tradition (di-, tri-) and the numeric ones as the standard. If there were to be exceptions when it didn't mean anything it would be confusing and had to be decided on a case for case basis whether or not to use the numerics. Better to just always have them, they usually represent an insignificant waste of bytes...


text:
Number of moles of Grignard reagent
How many moles of Grignard reagent required by one mole of the below compound?
That is,

I think it should be 4, two for each carbonyl.
But in mybook it is given as 7 .

So, what's the question?

@Zhe but in my book it is given as 7

Oh, well, looks like we both missed the naphthol moiety on the left, plus the thiol... I think that puts us at 6. Still look for 7...

@Zhe what is naphthol moiety

The OH on the left is acidic. The thiol as well. These will react to destroy 2 equivalents of Grignard reagent.

@Zhe how OH can react . What would be the product when OH is reacted

$\ce{RO-}$? It's exactly like what would happen if you put your Grignard reagent in water.

@Zhe If I remember correctly, I think tBuLi would react with the alkyl chloride (eliminating HCl). I don't know if Grignards are the same.

@orthocresol Had the exact same thought. Unfortunately, I don't think this is a great pathway since that would mean that Grignard generation destroys a lot of the alkyl halide used to make it...

@Zhe true, true...

I don't know about the alkyl chloride either.  It's true that Grignard reagents, like lithium reagents, can disproportionate and exchange, but I expect everything else to happen more rapidly.

In the given compound
1) grignard reagent will act as a nucleophile and attack twice on both carbonyl groups. So 4 moles here
2) it will act as a base and react with the phenol, thiol, and chloride group. So 3 moles here
Therefore a total of 7 moles will be required

Can you please give me an example of acting as a base

This might help https://en.m.wikipedia.org/wiki/Grignard_reaction


text:
Which reaction is this?
In my book there is a reaction that is

$\ce{HCCl2-CH2F}$ in presence of $\ce{RO^- ROH}$ gives $\ce{CCl2=CH2}$

I could not understand my type of reaction it is?
And why does $\ce{F-}$ leave whe $\ce{Cl-}$ should leave as it is a better leaving group?

Which is the more acidic proton? ;-)

@KlausWarzecha the hydrogen attached to the carbon which ha Cl

@KlausWarzecha but the energy required for F- is more

The two chlorine atoms attached to the other carbon have a very strong electron-withdrawing effect that weakens the C-F bond, making it even weaker than the C-Cl bond weakened by said fluorine. Thus, in this context the fluoride ion is a better leaving group than the chloride ion.

Look at that $\ce{RO-}$. It is a base, right? A strong base. A strong base is likely to attack an acid. In this case, the most acidic hydrogen is the one which is the one in $\ce{CHCl2}$ (See explanation below). For convenience, let's call it $\ce{H-\alpha}$ and the other two $\ce{H-\beta}$.

(The base shoud be $\ce{RO-}$, but the drawing tool I'm using does not support placeholder signs.)
Now we have this:

This is called E1cb mechanism. (cb stands for intermediate carbanion)
Why is the hydrogen attacked
$\ce{H-\alpha}$ has totally three -I effect group: two chlorine and a fluoromethyl group. Meanwhile, $\ce{H-\beta}$ only has two -I effect group: a fluorine group and a dichloromethyl group. Therefore, the intermediate is more stable than it would be if $\ce{H-\beta}$ is eliminated.

It appears to be an e1cb mechanism, where formation of the carbanion on the dichloro end is the rate limiting step. This appears to be a far more acidic proton than the ones on the fluorine-bearing carbon. Once the carbanion is formed, it rapidly eliminates fluoride ion to give the stable dichloroethylene product, rather than eliminating chloride (admittedly a better leaving group) to give the carbene, a higher energy product.


text:
How can I separate these two alcohols from each other out of the aqueous solution?
I have the following question for homework and am stumped. I am tasked with separating 3-(2-methoxyphenoxy)propane-1,2-diol and (1R,2S)-2-(methylamino)-1-phenylpropan-1-ol. They are both solvated in water.

I was thinking to work up with acid to cleave the ethers on the left compound, but that won't do much to reduce solubility for crystallization. Esterification with a large hydrophobic carboxylic acid wouldn't do much either because both compounds have hydroxyls. Anyone have suggestions?

If you don't want to protect the diol, maybe reversed-phase chromatography could do the trick.

I am tasked with separating these two compounds.

If you want to keep both compounds, degrading one of them is obviously not an option.

They are both solvated in water.

This can be changed - reversibly ;-)
Your left compound is a diol. React the mixture with a ketone in the presence of an acidic catalyst.
Typically, these reactions are performed to protect ketones by heating them in with ethanediol in toluene in the presence of toluene sulfonic acid. There is no reason not to use it the other way around to protect the diol.
The reaction product is a cyclic acetal (1,3-dioxolane) which is much less  soluble in water that the parent diol.

I think you could also protect the amine using a similar strategy (with different reagents of course).

Yeap but more difficult I think, considering that you also have an alcool function that may react and the stereochemistry to keep.

@Autumn Epimerization of the OH in the amino alcohol during deprotection was my concern too.

Actually, I remember the odd total synthesis (read: Trauner Denksport) that used an acetonide to protect a 1,2-diol. Or a benzaldehyde-acetal to protect a 1,3-diol. So it’s not entirely uncommon.

Neutralize the amine with HCl to pH 4-5 to make it even more hydrophilic. Then add lots of NaCl to salt out the diol. Separate the diol and then basify the amine solution to salt it out.
After neutralizing (and again after adding salt and decanting the diol), cooling the aqueous solution might precipitate the amine hydrochloride as a crystalline material. So if the diol doesn't salt out, maybe the amine hydrochloride will.


text:
Decomposition of Gold Hydroxide
On heating will gold hydroxide $\ce{Au(OH)3}$ decompose to form pure gold or will a gold oxide (e.g. $\ce{Au2O3}$) be stable? Does anyone know the temperature of these reactions?

https://en.wikipedia.org/wiki/Gold(III)_hydroxide

I wasn't sure how much to trust wikipedia... so simply $\ce{2Au(OH)3}$ > $\ce{Au2O3}$ + $\ce{3H2O}$ at 160 degrees? Will $\ce{Au2O3}$ just form Gold and oxygen gas at higher temperatures?

https://books.google.pl/books?id=GuxPDAAAQBAJ&pg=PA148&lpg=PA148&dq=gold+iii+oxide+decomposition&source=bl&ots=DUxMPP-X4V&sig=lJ63GKnZeDJQ-8pF8QzzG00Og5c&hl=pl&sa=X&ved=0ahUKEwjVx_f8_f7RAhUrG5oKHZWmAxgQ6AEIZDAJ#v=onepage&q=gold%20iii%20oxide%20decomposition&f=false It looks like, but hydration seems to complicate this.

There is yet no consensus on what occurs when gold(III) hydroxide is heated. Part of the problem is that the exact structure of $\ce{Au(OH)3}$ is unknown, or dubious at best. It has been inconsistently formulated as "truly" $\ce{Au^{III}-(OH)3}$$^{[0]\ [1]\ [2]\ [\text{a}]}$, oxide-hydroxide hydrate $\ce{AuOOH*H2O}$$^{[3]}$, oxide hydrate $\ce{Au2O3*xH2O}$$^{[0]\ [4]}$, even an acid-like hydrate $\ce{H[AuO2]*H2O}$$^{[5]\ [6]}$. I am going to use $\ce{[AuOOH*H2O]^{*}}$ with an asterisk in equations to keep this controversy in mind.
At temperatures below $130\ ^\circ \mathrm{C}$, the decomposition is probably best described by$^{[3]\ [7]}$
$$\ce{[AuOOH*H2O]^{*} ->[below $130\ ^\circ \mathrm{C}$] AuOOH + H2O}.$$
Around $140\ ^\circ \mathrm{C}$ the oxide-hydroxide decomposes further.$^{[3]\ [7]}$
$$\ce{2AuOOH ->[$140-150\ ^\circ \mathrm{C}$] Au2O3 + H2O}$$
Gold(III) oxide is already considerably unstable from $155\ ^\circ \mathrm{C}$ .$^{[1]\ [3]}$ It is not definitively shown what reaction follows. Some posit $\ce{AuO}$$^{[3]}$, which might not exist or actually  be better described as $\ce{Au[AuO2]}$$^{[3]}$, others propose a combined product of elemental gold $\ce{Au}$ and $\ce{Au2O}$$^{[1]}$. There are studies that rule out these products, at least as isolable intermediates, and instead propose $\ce{-> Au + O2}$$^{[0]\ [2]\ [7]\ [8]}$.
$$\ce{2Au2O3 ->[$150-165\ ^\circ \mathrm{C}$] 4AuO + O2}$$
$$\ce{4Au2O3 ->[$150-165\ ^\circ \mathrm{C}$] 4Au + 2Au2O + 5O2}$$
$$\ce{2Au2O3 ->[$150-165;\ 300\ ^\circ \mathrm{C}$] 4Au + 3O2}$$
The proponents of the bottom reaction, a more direct $\ce{-> Au + O2}$ claim that the oxides were mistaken for a complex mixture of $\ce{Au}$, $\ce{Au2O3}$, and interstitial or dissolved $\ce{O2}$.$^{[9]}$
The authors$^{[9]}$, nevertheless, still note

It  is  not  yet  clear  how  this  “dissolved”  oxygen  should  be  described.  Is it really dissolved in interstitial metal lattice positions,  are there trapped $\ce{Au2O3}$ oxide  clusters in a densely sintered metal  matrix, or encapsulated oxygen gas in lattice imperfections of the gold?$^{[9]}$

If the intermediates $\ce{AuO}$ and $\ce{Au2O}$ exist, then not for too long.  The decomposition of these oxides would most likely consist of multiple interconversions between $\ce{AuO}$ and $\ce{Au2O}$. For example$^{[1]\ [3]\ [\text{b}]}$,
$$\ce{2Au2O + O2 <=> 4AuO},$$
$$\ce{Au + AuO <=> Au2O},$$
$$\ce{Au2O -> Au + O2}.$$
Over $300\ ^\circ \mathrm{C}$ only gold metal $\ce{Au}$ and dioxygen $\ce{O2}$ remain.$^{[9]}$ Another experiment$^{[10]}$ supports this conclusion but differentiates between surface and subsurface oxides. The latter are found to be stable well over $300\ ^\circ \mathrm{C}$$^{[10]}$. More research is required to shed light into what is actually the main decomposition pathway. An additional way of expressing the process at lower temperatures is$^{[9]}$
$$\ce{6AuOOH -> 4Au + Au2O3 + 3O2 + 3H2O.}$$

$[\text{a}]$ $\ce{Au(OH)3}$ is presumed to assume a square planar coordination with four $\ce{-OH}$-groups.$^{[2]}$ A most recent study provides evidence for a four-coordinated linear polymer.$^{[0]}$

$[\text{b}]$ Equations are written on the analogy with copper.$^{[1]\ [3]}$

$[0]$ Daisuke Kawamoto, Hiroaki Ando, Hironori Ohashi, Yasuhiro Kobayashi, Tetsuo Honma, Tamao Ishida, Makoto Tokunaga, Yoshihiro Okaue, Satoshi Utsunomiya, Takushi Yokoyama. 'Structure of a Gold(III) Hydroxide and Determination of Its Solubility'. Bulletin of the Chemical Society of Japan, ($2016$), 89(11), 1385$-$1390. DOI: 10.1246/bcsj.20160228.
$[1]$ H. Karik, K. Truus. ($2003$). Elementide keemia. (p 313)
$[2]$ C. Klanner, N. Weiher, E. A. Willneff, S. L. M. Schroeder. 'In Situ Quick XAFS Studies of the Thermal Decomposition of Gold(III) Oxide'. Hasyweb Annual Report $2000$, ($2000$), part 1, 46, 2986.
$[3]$ R. Ripan, I. Ceteanu. ($1969$). Chimia metalelor. II volume. (p 768$-$769)
$[4]$ I. V. Mironov. 'Properties of Gold(III) Hydroxide and Aquahydroxogold(III) Complexes in Aqueous Solution'. Russian Journal of Inorganic Chemistry, ($2005$), 50(7), 1115
$[5]$ Κ. V. Astakhov, A. G. Elitsur, K. M. Nickoluev. Zhurnal obshcheĭ khimii, ($1951$), 21, 1753.
$[6]$ A. G. Massey, N. R. Thompson, B. F. G. Johnson, R. Davis. ($1973$). The Chemistry of Copper, Silver, Gold. (1975 reprint)
$[7]$ O. Diaz-Morales, F. Calle-Vallejo, C. de Munck, M. T. M. Koper. 'Electrochemical water splitting by gold: evidence for an oxide decomposition mechanism'. Chemical Science,  ($2013$), 4, 2334$-$2343. DOI: 10.1039/C3SC50301A.
$[8]$ C. Klanner, N. Weiher, E. A. Willneff, S. L. M. Schroeder, C. Figulla-Kroschel, M. Jansen. 'Thermal Stability and Decomposition Kinetics of Crystalline  Gold(III) Oxide and Disordered Gold(III) Oxides/Hydroxides'. Hasyweb Annual Report $2001$, ($2001$), part 1, 46, 5440.
$[9]$ M. Peuckert,  F. P. Coenen, H. P. Bonzel. 'On the Surface  Oxidation of a Gold Electrode in $1\ \mathrm{N} \ce{H2SO4}$ Electrolyte'. Surface  Science, ($1984$), 141, 2$-$3, 515$-$532.
$[10]$ L. K. Ono, B. R. Cuenya. 'Formation and Thermal Stability of $\ce{Au2O3}$ on Gold Nanoparticles: Size and Support Effects'. Journal of Physical Chemistry C, ($2008$), 112(12), 4676$–$4686. DOI: 10.1021/jp711277u.

Wikipedia quotes on the decomposition of gold(III) hydroxide:

It is easily dehydrated above 140 °C to gold(III) oxide.

Moreover, the equation of the decomposition is given here:
$$\ce{2Au(OH)3 -> Au2O3 + H2O}$$
If we heat up further, the auric  oxide would decompose:
$$\ce{2Au2O3 ->[160-290 C] 4Au + 3O2}$$
Mind you, you will not become a millionaire by extracting gold in this way. The gold obtained is very impure. :)

Thanks for the chemiday.com link, that's really useful. The same site also lists the decomposition of $\ce{Au2O3}$ at 160 - 290 °C by $\ce{2Au2O3 -> 4Au + 3O2}$ here

@BJH Thank you, i could include that in the answer.


text:
CF and MO theories: seeing connections between the two approaches
I wonder if and how when considering MO theory, we are accounting for the electrostatic interaction / repulsion between electrons on the metal and the ligand. It seems like MO assumes pure covalency, so why is this necessarily more accurate than crystal field?
I am primarily looking for a quantum mechanical explanation of what is accounted for when MOs are being constructed, but any suggestion or indication of the right direction would be great. Thank you!

I guess you're thinking about Ligand field theory not MO in general?

@Mithoron Yes, that's right!


text:
Rate of Reaction for Elementary Reactions
The rate law for an elementary step of a reaction of the form $2A \Rightarrow$ Products is given by $rate=k[A]^2$, while in the form $A \Rightarrow$ products is given by $k=[A]$. Is there some sort of intuition as to why the the reaction with $2A$ is quadratic, instead of double the one with $A$? Essentially, why is the equation $k[A]^2$ instead of $2k[A]$?

First of all rate law for any reaction is experimentally determined and also In general concentrations of reactant & products are raised to their stochiometrical coefficients.

Yes, but can you give a reason as to why there is an exponential relationship rather than a linear one?

Intuitively, a reaction between A and B to give C will go twice as fast when you double the concentration of A (because collisions between A and B will occur twice as frequently). Similarly, doubling the concentration of B will also double the reaction rate. But what about the specific case where A and B are the same substance (A + A -> C)? When you double the concentration of A, the rate is quadrupled; when you triple the concentration of A, the rate goes up by a factor of 9, etc.


text:
Can any solute be used to lower the freezing point of water?
Let's say I want to lower the freezing point of water by 5 degrees Celsius.
I know that for a fixed volume of water, I will need $x$ moles of solute to do so.
Because freezing point depression is a colligative property, does that mean I can use any solute to lower the freezing point of water as long as the molality of the solution is maintained, or are there other properties of the solute that need to be considered?

http://chemistry.stackexchange.com/questions/44979/is-the-freezing-point-of-solution-always-lower

http://chemistry.stackexchange.com/questions/25123/how-to-raise-the-melting-point-of-water

In chemistry it seems there is always an exception...

Not any, an involatile solute is needed and that it does not dissolve in the solid phase. The chemical potential of the liquid solvent  is then lowered and hence the freezing point. see figure in this answer http://chemistry.stackexchange.com/questions/35340/justification-for-freezing-point-depression-boiling-point-elevation-in-solutio/64853#64853

The key is in the terminology- the freeze point suppressant must be a solute. Oil won't suppress the freeze point of water because it doesn't dissolve. A small amount of sodium chloride or ethanol will suppress the freeze point of water to -5°C.
The first exception is when the solute stops dissolving. This is called the eutectic point. For example, you can only dissolve 23.3% NaCl in water (by mass). That mixture will freeze at -21.1°C. After that, adding more salt won't reduce the freeze point further.
The second exception is with large amounts. If you mix water and ethanol in a 50/50 mol ratio, which is the solvent and which the solute anymore? You can calculate the freezing point suppression of each with respect to the other and they won't match. Liquid/liquid mixtures also have eutectic points at the minimum freezing point, which is less than the freezing point of either component.
To get the actual freezing point for non-dilute systems, try searching for "solid liquid equilibria", "eutectic point", or "binary phase diagram".


text:
Entropy as Sum of Kinetic Energy?
Is it possible to view entropy as the sum of the kinetic energy of all molecules in the system? Since a system with 0 entropy would have 0 motion, and as you increase the motion of the molecules, entropy increases as well. If not, could you provide a simple counterexample in which entropy does not make sense as the sum of kinetic energy?

You might want to review the concept of entropy, including but not limited to the third law of thermodynamics and the statistical mechanical view of entropy.

I see the classical explanations of entropy as the 'randomness' of a system, but this seems to be a direct contradiction with the units of entropy (Joules/K).

@AndiGu You have randomness in Boltzmann's description of entropy. S=Boltzmann constant* lnW where W is macrostate multiplicity. But  Clausius first described entropy with dS= dq/T which makes it unit J/Kwhich doesn't incorporate randomness. And Boltzmann introduced randomness to the concept of entropy(and it's a classical concept). But I think he could just do with lnW but to get it consistent with a pre-established theory Boltzmann introduced kB to the equation.

As an example, there is entropy increase upon mixing two substances together or even of mixing up sets of different objects even if both sets are already 'random'. No kinetic energy need be involved.  Randomness is an unfortunate term often used to describe entropy. Its the degree of randomness that is described and what this is  is not clear in everyday usage so its best totally avoided . Entropy is $S=k\ln(\Omega)$ where $\Omega$ is the number of ways of arranging objects, in molecules this means calculating the number of ways of placing then in all their possible energy levels.

@AndiGu Your view is also a direct contradiction with the units of entropy (Joules/K). These do not match the units of energy (Joules).

In a crystal lattice, there are two contributions to the entropy: the vibrational entropy and the configurational entropy. The first is related to the kinetic energy; atoms with higher kinetic energies have tend to have larger displacements from equilibrium, and have a larger "footprint" in phase space, resulting in a higher entropy. The second contribution, configurational entropy, arises from disorder in the crystal lattice, namely the number of ways of arranging the atoms. In a perfect crystal, there is only one way, so the configurational entropy ($k\ln W$) will be 1.
An counterexample to the suggestion you make is any disordered solid, whether it's amorphous, a random alloy or some kind of crystal with geometric frustration. The most obvious example of this is water ice. Each water molecule occupies a lattice position and is tetrahedrally coordinated with its neighbours. In a defect-free ice crystal, the ice rules state that each molecule must accept exactly two hydrogen bonds and donate exactly two hydrogen bonds. This is quite a strong constraint, but allows orientational disorder --- the molecules can be oriented in one of six possible ways at each lattice position, resulting in an orientationally disordered structure. This will have a finite entropy even at absolute zero.

Thanks for the clear explanation - just one question: if 'configuration' is an aspect of entropy, how does that contribute to the J/k notion of entropy? I understand the kinetic energy part - but where does this configurational part come in in terms of energy?


text:
The teacher insists that iodine oxidation number in iodate ion is 7
What started as this question http://chemistry.stackexchange.com/questions/67756/is-it-possible-to-use-the-school-supplied-algorithm-to-build-the-lewis-diagram-o continued with an email from the teacher, where they confirmed that the octet rule has to be followed and also mentioned that the oxidation number of iodine in iodate is 7, not 5 as I believe.
My thinking is based on the fact that oxygen is always 2, therefore iodine has 3*2-1=5. Two double bonds and one single bond with oxygen are present in iodate ion.

Iodine has oxidation number 7 in periodate ion as 4*2-1=7.

Where is my error?
EDIT:
More back and forth with the teacher, and they pointed me at an example of how they think $\ce{[ClO3]-}$ should look in a Lewis diagram:

According to the teacher, both chlorate they have in the course and iodate in the assignment would have to follow the exact same steps when building Lewis diagram.
I have an issue with this methodology, as two $O$ are somehow missing their 2nd bonds. This is absurd and does not jive with any of the textbooks I studied. How should I explain this to the teacher, who does not seem to understand basic arithmetic?

Another example of "alternate facts"... // PS - This isn't worth getting in a argument  with your teacher - especially a public one.

Your extension was a completely separate question that does not touch the original; I have therefore rolled it back. If you wish to discuss how the full Lewis structure of chlorate (or iodate) you should ask it separately — but note that there are almost certainly questions (and answers) on how to draw it on this site already, so please don’t forget to search.

Lewis diagrams usually use different symbols for the electrons from different atoms so that you can keep track of which atom the electrons came from. Like solid dots and circles.

Do you want to be right or do you want to be happy?  Getting into a brewhaha with the teacher isn't necessarily the best way to proceed.

Smart comment, but what I want is a guarantee that the teacher cannot arbitrarily deduct points. If I commit to the wrong process used in the course, there is no such guarantee. Makes sense?

Related: http://chemistry.stackexchange.com/questions/25097/iodate-ion-and-octet-rule

And http://chemistry.stackexchange.com/questions/7560/iodate-ion-io3-is-it-polar-or-non-polar

Some parts of the previous edits were objectively worth keeping: the links were prefixed with https my a bot edit in line with https being used across SE and images should always include an image-describing alt-text stating what they are about.

"oxygen is always 2" While oxygen is -II most of the times, this could be horribly misleading in Hydrogenperoxide or Oxygen Fluorides.

Plus 5 according to: https://en.wikipedia.org/wiki/Iodate
Teachers do sometimes make mistakes.

I would be very hesitant to quote Wikipedia, as it is a resource, which can be edited by anyone with any agenda at any moment. But I do agree that there is no way in hell I could have OS=7 in iodate. Only in periodate.

@ajeh. Quote me then! I added the reference because some users seem to demand them, especially on the Biology S.E. Here's another reference but you'll have to parse it out of a reaction: https://books.google.com/books?id=STxHXRR4VKIC&pg=PA57&lpg=PA57&dq=iodate+oxidation+number&source=bl&ots=EG3wLLc2TW&sig=OzRc1-9iwIcS2NGwMjDybNzYgu0&hl=en&sa=X&ved=0ahUKEwi4spu1jIHSAhWGUrwKHcOmDp8Q6AEIjQEwFA#v=onepage&q=iodate%20oxidation%20number&f=false

@bpedit The link says that the page is unavailable for viewing or I've reached the limit for this book. But I did find a confirmation of OS=5 for iodates in several textbooks, including f.ex. Modern Inorganic Chemistry by Chambers and Holliday, p. 260

@bpedit Can you please check the link you posted as I noted above, it does not open? Thx!

@aiag. Hmmm... it opened fine for me just now. ???

I don't think there's a particular need for a source for this. It's quite straightforward and if anything, an explanation of how the oxidation state is arrived at will be much more useful. Although - to be fair - OP's original question already contains the correct reasoning.

@orthocresol. I know the rules, I taught them, the OP seems to know the rules. Evidently the teacher, who should know the rules, was recalcitrant for some reason. I thought an appeal to authority was the best solution for this particular situation.

Fair enough. (I have no doubt you know the rules!) It was slightly more directed towards the commenter who said that Wikipedia was unreliable.

Be aware that it is dependent on the country you open a google books link whether you see the content or not. The books are in most cases copyrighted and google may not hold all copyrights in all countries. Therefore it is best to quote the relevant passage and cite the source.

There’s a simple way to calculate how many bonds you need to form a Lewis structure, and how many lone pairs remain. It is four simple steps:

How many valence electrons do we have? (add up)
For both chlorate and iodate the calculation is identical:
$$\underset{\text{halogen}}{7} + \underset{\text{oxygens}}{3 \times 6} + \underset{\text{charge}}{1} = 26\tag{1}$$
How many valence electrons are required for full octets/hydrogen dublets?
This equates to $2m + 8n$, where $m$ is the number of hydrogens and $n$ the number of other elements.
$$4\times 8 = 32\tag{2}$$
How many electrons are missing? These will be shared, i.e. form covalent bonds. $(2) - (1)$
$$32 - 26 = 6 = 3~\text{pairs}\tag{3}$$
How many electrons are remaining? These will form lone pairs. $(1) - (3)$
$$26 - 6 = 20 = 10~\text{pairs}\tag{4}$$

Thankfully, knowing the general structure is ‘halogen in the centre, oxygen atoms around it’ and knowing that there are three oxygen atoms, this clearly shows us that all $\ce{X-O}$ bonds ($\ce{X} = \ce{Cl, I}$) are single bonds. The halogen would then need one more lone pair to be satisfied, the remaining nine lone pairs go to the three oxygen atoms (3 each). This leads us to the following, final Lewis structure:

(Unfortunately, the $2+$ on iodine and its remaining lone pair intersect in the image.)
Note that this diagram is that your teacher used, too, except for explicitly noting the formal charges of both iodine and oxygen. If you want to simplify, you can draw square brackets around the molecule, write the total charge next to the closing square bracket and ignore the distribution of formal charges — that is valid (and what your teacher did). I am sorry to say that they are correct in that regard — and since they admitted their error concerning the oxidation state of iodine in iodate, they are fully back on track.

@Greg This question has moved so much since its existence …


text:
Ideal Gas Vs Real gas
If ideal gas equation is formulated using ideal gas concept then why is the same equation used for real gases os they are not the ideal gases

The ideal gas law is an excellent approximation to the behavior of real gases at low pressures and high specific volumes.

For real gases there are some contradictions in the assumptions made while defining real gases.
Therefore some corrections were needed in the equation which were given by van der waals in his real gas equation
The van der Waals equation corrects for the volume of, and attractive forces between, gas molecules:
(P+a(n/v)^2)(V-nb) = nRT
There are two corrective factors in van der Waals equation. The first,   , alters the pressure in the ideal gas equation. It accounts for the intermolecular attractive forces between gas molecules. The magnitude of a is indicative of the strength of the intermolecular attractive force.
The factor - nb accounts for the volume occupied by the gas molecules. b has units of L/mol. Since b corresponds to the total volume per mole occupied by gas molecules, it closely corresponds to the volume per mole of the liquid state, whose molecules are closely layered. b is generally much smaller in magnitude than a . The values of a and b generally increase with the size and complexity of the molecule.

The van der Waals equation does somewhat correct for real gas behavior, but it isn't a particularly good correction. There are other corrections that do a better job.

For starters... https://en.wikipedia.org/wiki/Real_gas


text:
Is it possible to figure out how many carbon-carbon double bonds are present using just the molecular formula?
One of my test review questions reads as follows:

"The prostaglandin precursor arachidonic acid has the molecular
  formula $\ce{C20H32O2}$. Given that arachidonic acid is an acyclic carboxylic
  acid that contains no carbon-carbon triple bonds, how many
  carbon-carbon double bonds are present?"

My original instinct was to draw out the structure, but having just the molecular formula to go off of and nothing else, it seems like it would take a ton of trial and error to figure out the correct structure. Is there a simpler way to approach a problem like this to figure out the amount of double bonds?
The only other thing that came to mind was to calculate the degree of unsaturation from the formula, which came out to 5. So, I know there can't be more than that. But that could represent both double bonds and/or rings, so. Anything I'm missing?
(The answer is 4, by the way. I just need help figuring out how the process of getting there is laid out.)
Thanks.

Not really. Consider $\ce{C6H12}$ can be either a hexene or cyclohexane.

Both you and Rapid99 have missed the word "acyclic" in the problem statement.  So you don't have to worry about rings (because they said acyclic).  [Also no triple bonds.]
You also know that it's a carboxylic acid ("COOH group", formula but not structure of carboxylic acid group), so don't worry about something funky with the oxygens.  The formula is then $\ce{C19H31-COOH}$.
If the part of the compound on left was full saturated and sans $\ce{COOH}$ it would be $\ce{C19H_{(2*19+2)}}=\ce{C19H40}$.  Remember any acyclic alkane has formula $\ce{C_nH_{2n+2}}$.
Accounting for the $\ce{COOH}$ group location means the saturated alkane would then be $\ce{C19H39-COOH}$ (take away one $\ce{H}$ "spot" to attach the COOH group).
So we are really 8 hydrogens short: 39-31 (from above) = 8.  Each double bond takes away two hydrogens.  So if there are 8 missing hydrogens, there must be 4 double bonds.  X=8/2.

Re: ... missed the word "acyclic"... Guilty as charged. I only read the title not the actual question.

Oh! Well this made plenty of sense, thank you.


text:
Is orbital hybridisation real?
Is the concept of orbital hybridisation an accurate description of a physical process that actually occurs in bonding atoms, or is it, like VSEPR, a heuristic tool to reason about the characteristics of classically infeasible chemical bonds? Forgive the naivete of my question: I ask this because I am not actually acquainted with the bulk of the underlying theory of chemical bonds, but only with a somewhat flimsy and superficial description that seems inadequate and overly contorted to me.
Thanks in advance for any answers!

Orbitals themselves are not quite as real as sticks and stones. So yes, better think of hybridization as a heuristic tool.

@IvanNeretin Thank you. I did think it might be something like that.

Orbital hybridization is something people teach in school or universities because it is so straight-forward and easy to understand. I would argue that it is useful for nothing but teaching - if you like, it's a first glimpse at quantum mechanics for the young chemist. It is a great tragedy that it is not made clear by the teachers that in fact it has no applications or connection to the real world and many advanced chemists still try to argue with it. This forum is full of people being confused by it.

Orbital hybridisation is, like most other topics bordering quantum mechanics, just a model of reality (in fact, I would go as far as to say like all topics ever thought of, but lets not get too philosophical). As such, there are limits to its usefulness, so models should never be thought of as a "true picture" of anything.
Models are always simplified and in the end "wrong", but sometimes they are useful. Through models we can gain insight and make new predictions that can be falsified or not and as such we advance our understanding (and the most learned even advance science as a whole). Without models I would not understand anything, but I considert it counter productive for myself to think of them as representing some "truth".
$$\text{Observation} \rightarrow \text{interpolation} / \text{modelling} \rightarrow \text{prediction} \rightarrow \text{experiment} \rightarrow \text{observation}$$

This is a truly outstanding answer that can be applied to almost any field of study. Well put.

Thank you @ringo. The only perfect model of the universe, or indeed any individual component of it - is the universe itself ;)


text:
Why does a carbonyl group form instead of alkene in alcohol oxidation?
I am looking over mechanism of the oxidation of alcohol by chromic acid and I am confused on why $\ce{H2O}$ abstracts hydrogen from alpha carbon and form double bond between $\ce{C}$ and $\ce{O}$ atom rather than beta hydrogen and form alkene like in usual elimination reaction we learned. Like to start from beginning i am confused why in chromate ester, which is an intermediate, entire group of $\ce{O-CrO3H}$ does not leave as learning group. Why does only $\ce{CrO3H}$ leave, leaving $\ce{O}$ atom behind which thus makes $\ce{H2O}$ abstact hydrogen from alpha carbon instead of abstracting beta hydrogen and creating alkene.

Technically, it would be possible for an entire chromate group $\ce{CrO4^2-}$ to be eliminated along with a β-hydrogen to give a double bond. However, that elimination would have to proceed via either an $\mathrm{E2}$ or an $\mathrm{E1_{cb}}$ mechanism as the chromium(VI) centre by itself is much more happy being reduced to chromate(IV) (the intermediate chromium species) in the electrocyclic breakdown of the complex you have been taught.
The entire reaction proceeds in acidic conditions, so any conjugate base mechanism is basically impossible. And likewise, there is no sufficiently strong base present that would allow an $\mathrm{E2}$ mechanism to occur, either.


text:
Is there a compound or element with properties similar to H2O?
Is there an existing compound or element with properties similar to H2O?
( Liquid, non toxic, stable, and with a clear / blue coloration )  I would expect that any such element or compound would not exist in a stable form, however, I curious to see whether something similar may exist.
I need this for a fictional piece I am working on as a substance from an alien planet.

Saturn's moon Titan has lakes of liquid hydrocarbons.

Drinkable by what/whom? Anyway ammonia  could be a possibility (but personally I don't really believe it fits your criteria), see http://worldbuilding.stackexchange.com/questions/3090/what-would-an-ammonia-based-world-look-like for more thoughts.

Why do you need real chemistry? You can make up any liquid you want, just call it a different name.

I've just seen this question on Quora: https://www.quora.com/Im-trying-to-create-a-made-up-planet-How-can-I-create-new-elements