text:
IUPAC nomenclature: "Smallest sum of locants"?

While naming any compound, the numbering should be done such that position of substituted groups gives smallest sum.

I'm struggling with the IUPAC naming for cyclic compounds. How exactly does this rule work?

The guideline that you have mentioned (‘While naming any compound, the numbering should be done such that position of substituted groups gives the smallest sum.’) does not exist in the IUPAC recommendations.
The corresponding section in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) 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’.
(…)

Therefore, the smallest sum of locants can lead to wrong results.
For example, for 2,7,8-trimethyldecane, the smallest sum of locants would yield the wrong name ‘3,4,9-trimethyldecane’, since $2+7+8=17 > 3+4+9=16$.

Likewise, for 2,2,6,6,7-pentamethyloctane, the smallest sum of locants would yield the wrong name ‘2,3,3,7,7-pentamethyloctane’, since $2 + 2 + 6 + 6 + 7 = 23 > 2 + 3 + 3 + 7 + 7 = 22$.

For 1,6,7-trimethylnaphthalene, the smallest sum of locants would yield the wrong name ‘2,3,5-trimethylnaphthalene’, since $1 + 6 + 7 = 14 > 2 + 3 + 5 = 10$.

For 1,6-dimethylcyclohex-1-ene, the smallest sum of locants would yield the wrong name ‘2,3-dimethylcyclohex-1-ene’, since $1 + 6 = 7 > 2 + 3 = 5$.

Anyway, the IUPAC recommendations have more than 1500 pages. Hence, you should be careful when applying any simplified or abridged version of the nomenclature rules.

I am not able to understand the lines by IUPAC which you quoted. Do you mean that if we have 2 ways in which carbons attached to functional groups can be numbered, then we should arrange both these sets with their elements in increasing order e.g. {7,11,19} and {3,11,15}then first we will 3 & 7 and 3<7 so {3,11} is correct and we don't need to compare the remaining numbers AT ALL?

@NipunKulshreshtha Yes, that is correct.

The way you have stated the rule can be misleading or even incorrect at times.
I think what you are talking about is the lowest locant rule or the first-point difference rule.

The principal functional group is given the lowest possible locant.
The substituents are then assigned locants based on this numbering scheme in such a way as to give the lowest locant number at the first time there is a difference.
In the event that there is no first point of difference in the locants, substituents are numbered based on alphabetization giving the first substituent the lowest possible locant number.

You might find this link relevant.

There are lots of sites,blogs and other online sources where you can learn about the topic in depth. http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkanes/Properties_of_Alkanes/Cycloalkanes/Nomenclature_of_Cycloalkanes. and here is the site where you can learn by watching online http://leah4sci.com/naming-organic-compounds-iupac-nomenclature/   . You can get many videos in YouTube as well. So go for these sources.

Please note that the link http://www.acdlabs.com/iupac/nomenclature/79/r79_61.htm leads to obsolete IUPAC recommendations of 1979.


text:
When do lone pairs on nitrogen contribute to aromaticity?

I've been told that 1,2,4-trimethyl-1H-imidazole is aromatic. As it has four obvious pi electrons (in the double bonds), I assumed that it was anti-aromatic as opposed to aromatic.
I've been operating under the assumption that a nitrogen does not contribute its pi electrons to the aromaticity count for Hückel's rule unless it is attached to a hydrogen. Does the nitrogen with a methyl group here donate its pi electrons to the count?

Both nitrogens in your drawing are roughly $\mathrm{sp^2}$ hybridized, therefore they each have a p-orbital and 3 $\mathrm{sp^2}$ hybdridized orbitals. All of the p-orbitals, including those on the 3 carbon atoms in the double bonds, are perpendicular to the plane of the screen (e.g. they project above and below the plane of the screen). So there are 5 p-orbitals and each contains one electron except for the one p-orbital located on your "bottom" nitrogen (in green). This p-orbital contains the nitrogen lone pair, so there are two electrons in this p-orbital.
If we count the p-electrons we have $1+1+1+1+2 = 6$ electrons in $5$ p-orbitals that form a continuous loop. Therefore, the molecule is aromatic ($4n+2, n=1$).
Let's take a closer look at the "top" nitrogen in your drawing (in purple). Where is its lone pair? It resides in an $\mathrm{sp^2}$ orbital on this nitrogen. The $\mathrm{sp^2}$ orbital lies in the plane of the screen. This orbital is orthogonal (perpendicular) to the 5 p-orbitals; therefore these 2 lone pair electrons are separated from the electrons that make the molecule aromatic. They are not involved in the molecule's aromaticity.
Here's how to approach this situation in the future:

Identify the nitrogen(s)
Determine if the lone pair is in a p-orbital or an $\ce{sp^2}$ orbital.

If the nitrogen is in a double bond, then the lone pair is in one $\ce{sp^2}$ orbital, 1 electron is in the p-orbital and 1 electron is in each of the 2 remaining $\ce{sp^2}$ orbitals.
If the nitrogen is not in a double bond, then the lone pair is in a p-orbital and 1 electron is in each of the 3 $\ce{sp^2}$ orbitals.

Count the electrons in the cyclic p-orbitals to determine aromaticity

The nitrogen with a methyl group does indeed contribute to aromaticity in this molecule. The criteria for aromaticity under Hückel's rule are that $4n+2$ electrons reside in coplanar p-orbitals in a cyclic system. This allows the p-orbitals to overlap and form a set of π-orbitals, which are delocalised over the whole ring system.
In this example all the atoms in the ring are sp2 hybridised. The three carbons in the ring form three σ bonds with their sp2 orbitals and then have one half-filled p-orbital perpendicular to the plane of the ring which participates in π-bonding.
The nitrogen with a double bond forms two σ bonds with two of its sp2 orbitals and uses the third to accommodate its lone pair. The lone pair is therefore in the plane of the ring, facing outwards, and is thus unable to participate in π-bonding. It is left with a half-filled p-orbital in the plane which contributes another electron to the π-system.
The nitrogen with the methyl group attached forms three σ bonds with its sp2 orbitals, and therefore has a full p-orbital remaining perpendicular to the plane of the ring, which can overlap with the other p-orbitals to form a π-system.
In total there are six π-electrons residing in five coplanar p-orbitals. So, Hückel's rule is fulfilled and aromaticity is present.

Here's a crude, qualitative method:
Note that on top, delocalizing the electrons from that particular nitrogen results in a resonance contributor that has an $\ce{sp}$ hybridized nitrogen. $\ce{sp}$ hybridization implies a linear bond angle (180 degrees). Do you think we can get anywhere close to a linear bond angle in a 5-membered ring of all things? Remember, in resonance contributors the nuclei don't move. It's just the electrons that are "moving" (delocalized).
In the bottom I try to delocalize the electrons from the other nitrogen. This is fine, as the resulting resonance contributor now has an $\ce{sp^2}$ nitrogen. Note that this does not mean that the nitrogen in the ring is exactly $\ce{sp^2}$. It's likely somewhere in between $\ce{sp^3}$ and $\ce{sp^2}$. A more or less $\ce{sp^2}$ atom in a 5-membered ring is more plausible than a $\ce{sp}$ atom. So we can conclude that it's this nitrogen's lone pairs that are delocalized throughout the ring. Incidentally, this makes for 6 pi electrons. Additionally, the other Huckel rules are fulfilled (even the planar requirement; both nitrogens have $\ce{sp^2}$ character so they're more or less planar). So we can conclude this molecule is aromatic.

Leads to the correct result, but the reasoning is faulty (I know you said it is crude). The lone pair on the "pyridine-type" nitrogen (at the top) is orthogonal to the pi system, and that is why resonance does not take place, not because the resulting resonance structure is too unstable.

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Identifying rearrangement reaction?
I was asked to solve this problem, and the answer key shows this:

I'm not sure what's going on here, but it looks like some sort of a rearrangement and reduction. Is it an internal reaction between the ether and the alkene?

This is an example of the Claisen rearrangement.  It is basically the oxa-analogue of the Cope rearrangement.  The Claisen rearrangement can occur with aromatic or non-aromatic allyl vinyl ethers.  The reaction mechanism involves a concerted, intramolecular, [3,3]-sigmatropic rearrangement.  In the aromatic case, the rearrangement is followed by a proton shift (keto-enol tautomerization) to yield the final product with the aromaticity restored.
Here's a diagram of the aromatic Claisen rearrangement

Here's a diagram for the non-aromatic counterpart.

From the historical perspective, the Cope rearrangement is a carbon analog of the Claisen rearrangement. ;)

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Applying Avagadro's law: the relationship between moles and volume of a gas

If $\pu{2.4 mol}$ of gas fill a volume of $\pu{175 mL}$, what volume, in mL, will $5.4$ additional moles of gas fill?

This was a quiz question which I got wrong the first attempt, and I think I got it right this time however I would like some confirmation that it is/is not. The unnamed gas is assumed to be ideal.
$$\frac{\pu{2.4 mol}}{\pu{175 mL}} = \frac{\pu{7.8 mol}}{X}, \quad \text{or} \quad X = \frac{\pu{7.8 mol} \times \pu{175 mL}}{\pu{2.4 mol}} = \pu{568.75 mL}$$
Is this accurate?

You are correct in your approach and answer. Assuming the gas to be ideal, one can apply Avogadro's Law which states that for an ideal gas $V \propto n$.
This implies $V_1/n_1 = V_2/n_2$.
Using the given values one does obtain the volume as $568.75\ \mathrm{mL}$

I think that it is right, because I resolve it at the same method

Although correct, an answer with an explanation would help more users. In its current state it is more a commentary statement.


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How to identify the repeating unit in a polymer?

Polyvinyl acetate, PVA, is a useful adhesive for gluing together articles made from wood, paper or cardboard. The monomer of PVA is ethenyl ethanoate.
   PVA is formed from B by the process of addition polymerisation.
  Draw a section of the PVA molecule containing at least 2 monomer molecules and identify clearly the repeating unit.
  (Source: University of Cambridge international Examinations, General Certificate of Education Advanced Level, Chemistry, Paper 4 Structured Questions, May/June 2011, Q4)

Shouldn't the $\ce{C=C}$ double bond break and each of the carbon atoms involved in the double bond are attached to $\ce{CH3OCO}$. so the repeating unit is $\ce{CH3OCO}$.

Elements of backbone also repeat.

It's not that the "double bond breaks," it's that the vinyl C=C bond reacts during the radical addition polymerization to create polyvinyl acetate.
I'd disagree that the repeat unit is $\ce{CH3OCO}$ since that ignores the backbone entirely. Instead, if you look at the Wikipedia article (or any other picture of polyvinyl acetate), I'd call the repeat unit something like $\ce{-(CH2CH(OC=OCH3)-)}$ or something like that.


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Convert kinetic rate constant into a rate?
Can anyone tell me how I can convert a rate constant derived from first order kinetics (units are $\mathrm{h^{-1}}$) into an actual rate (turnovers/h)?
The scenario is that I know my rate constant for a molecule degrading in solution and it is $0.012\ \mathrm{h^{-1}}$. I know that I have $10^{14}$ catalyst molecules in the solution. How can I translate that into a number of molecules converted  per catalyst per hour?

What approaches have you considered so far?

Okay, I think I have it. The key is to recognize that the rate constant in units of hr^-1 was for the entire solution, and I can just multiply the rate by the number of molecules that I start with and divide by the molecules of catalyst. Factor label method to the rescue!


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Which of sodium or lithium will be more reactive with oxygen?
I do not understand how you can work out which element will react more readily with another element.
For example, I have lithium (Li) and sodium (Na) and I am trying to work out which element will be more reactive with oxygen. How do I manage to work that out?

By 'react more', do you mean which reaction proceeds faster and more vigorously?

Reactivity of Group 1 and 2 elements increases as you go down the periodic table.
So sodium is more reactive than lithium. Sodium will react with oxygen forming $\ce{Na2O}$ (sodium oxide). Lithium forms lithium oxide $\ce{(Li2O)}$.
Basically, there are general trends (reactivity with acids, air, etc.) down a group or along a period.

The OP asked "How do I manage to work that out?" whereas you've just stated a fact "Reactivity of Group 1 and 2 elements increases as you go down...". Please consider expanding your answer to include proper reasoning. The below answer instead is much better.

Here is a harder version of the answer above.

Sodium has 3 electron shells while lithium has only 2 electron shells
Sodium can donate its electrons more easily because the valence electron is further from the nucleus and the force of attraction between both is weaker.
Whereas lithium donates its electron less easily because the valence electron is closer to the nucleus and the force of attraction between both is stronger.
This is why sodium will "react more" with oxygen compare to lithium

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Estimating the protein concentration of an unpure protein solution
I'm confused on this problem for estimating protein concentration. Upon first glance, it looks like a simple Beer's law problem. but then there's the issue that the protein is not pure. How do I account for that?

E. coli Maltose binding protein, like alkaline Phosphatase is a periplasmic protein. Audrey, has a solution containing solublized E. coli periplasmic proteins and maltose binding protein and she just read the absorbance of the solution at $280~\mathrm{nm}$ which was $0.84$, path length $1~\mathrm{cm}$.
  Help Audrey estimate the protein concentration of her solution. Also note Audrey knows the extinction coefficient for maltose binding protein ($\mathrm{MW} \approx43,000$) is $32,290~\mathrm{M^{-1}cm^{-1}}$.
  (Hint: not pure protein solution.)

As mentioned above I think it would just be Beer's law
\begin{align}
A &= \epsilon cl\\
A &= 0.84\\
l &= 1~\mathrm{cm}\\
\epsilon &= 32,290~\mathrm{M^{-1}cm^{-1}}\\
\end{align}
Solving for $c$ gives $0.26~\mathrm{mM}$. My guess would be to say that it is less than that. How do I approach this?

If I get you right, your argument is that whatever impurity absorbs at 280 nm will add to the total absorbance of the solution. Consequently, one an upper limit for the concentration of the maltose binding protein can be given. This sounds reasonable to me!

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details.

What are you trying to find? The concentration of MBP alone or the total concentration of all proteins in your solution? If you didn't dilute your solution for measurement, I had c=0.026mM


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Including H⁺ from the dissociation of the acid in a titration pH calculation?
Let's say you are doing a standard titration problem, for example:

Find the pH after adding $10~\mathrm{mL}$ of $0.3~\mathrm{M}$ $\ce{NaOH}$ in the
  titration of $25~\mathrm{mL}$ of $0.3~\mathrm{M}$ $\ce{HF}$ with $0.3~\mathrm{M}$ $\ce{NaOH}$.
  The $K_a$ of $\ce{HF}$ value is $6.6 \cdot 10^{-4}$

Most sources start by considering the reaction between $\ce{HF}$ and $\ce{OH^-}$, finding the pre-equilibrium concentrations of $\ce{HF}$ and $\ce{F^-}$ and then using Henderson-Hasselbalch or the equilibrium expression directly to find $\ce{[H^+]}$ (as seen here).
However, don't you have to consider the dissociation of $\ce{HF}$ first,
$$\ce{HF <=> H^+ + F^-}$$
and then consider the reaction between those produced $\ce{H^+}$ and the added $\ce{OH^-}$? Because whenever an acid reacts with a base isn't it truly the hydrogen ion formed from the deprotonated acid reacting with the hydroxide ion from the base? How can you consider reactions of unprotonated acids/bases directly?

I'm not entirely sure on this but according to me the hydroxide contribution from the strong base (NaOH) is enough that we can consider HF to be completely dissociated.

There are numerous approximations made when pH calculations like these are made.
In a problem like this the approximations that would usually be made are

pH = -log[H+] (as opposed to pH being -log (H+ activity) which is the true definition).
NaOH is an infinitely strong base, with NaOH completely dissociated. (really, [Na+][OH-]/[NaOH] is about 4)
The dissciation of water  can be ignored (really (Kw =[H+][OH-]= 10^-14).
There exists a constant Ka for HF that is [H+][F-]/[HF], (whereas a true equilibrium constant would be a function of activities, rather than concentrations, of these species, and of water.  The Ka expressed as concentrations is dependent upon ionic strength for example in accordance with the Davies equation approximation).
Particular to this example, it is being assumed that the only fluorine species are HF and F- (really there is also HF2- and another equilbrium constant for it).

So you can take away one or more of these approximating assumptions to try to more accurately calculate pH, but it makes the problem very difficult.

However, don't you have to consider the dissociation of HF first, and then consider the reaction between those produced  H+  and the added OH-? Because whenever an acid reacts with a base isn't it truly the hydrogen ion formed from the deprotonated acid reacting with the hydroxide ion from the base? How can you consider reactions of unprotonated acids/bases directly?

At the end of the titration (or at any other point during the titration along as the solution is homogeneously mixed) the system will be in equilibrium.  The correct approach is to consider all the species that are present and all the equilbrium constants amoung the species.
Here, the species would be H2O, H+, OH-, HF, F-, Na+, and NaOH (and if I want to be picky HF2-).
The equilbrium constants are Kw, Ka of HF, Kb(NaOH), (and Ka of HF2-).
You need the same number of independent equations as unknowns to solve the problem.
[F-] + [HF] = (0.1/0.35)0.3M
[Na+] + [NaOH] = (0.1/0.35)0.3M
[H+] + [Na+] = [F-] + [OH-]
[H20] = 55M (or, if that is too much of assumption [H2O] + [OH-] + [NaOH] = constant by conservation of oxygen atoms)
plus the 3 equilibrium constant equations gives you 7 equations and 7 unknowns.   That would be a comprehensive consideration of the problem dropping assumptions number 2 and 3.


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Given the Pair Correlation Functions for a Multi-component Fluid, is There a Simple way to Obtain the Helmholtz Free Energy?
I think it may not be possible.
Given all the pair correlation functions, it seems possible to obtain the internal energy (assuming different species interact through pair-potentials (i.e. potentials that do not involve more than two bodies)).
To obtain the free energy, the "naive" way is to obtain the partition function and take logarithm. Yet it seems not easy (except possibly through cluster expansions? which is not exact.)
Any help will be much appreciated.


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Reverse electro-plating byproducts
Following this plan, I am wondering what the byproducts of this method are? At the site with the q-tip, the solution turned a bright yellow, with the q-tip turning a dark yellow, almost red. Holding the q-tip in place for about 10 seconds results in a small amount of smoke, and bubbles around the contact point. What are these colors, smoke, and bubbles? I don't know if the kettle is steel or aluminum, it's an old Budweiser keg, if that helps.
In case the link above dies, the method is etching metal using a vinegar and salt solution, combined with a 9 volt battery. Connecting the negative lead to a piece of metal, and using the positive lead with the solution results in a type of reverse electroplating, thus removing metal ions and etching the metal.
Not sure on the tags for this question, so feel free to adjust them, and the description of the etching method.

The yellow colour in solution is likely from iron ions from the metal being oxidized and dissolved:
$$\ce{Fe_{(s)} + 2e- -> Fe^{2+}_{(aq)}}$$ (Probably $\ce{Fe^3+}$, too)
The gas bubbles/smoke is the product of the other half of the reaction, the reduction of something in your solution. Because you have chloride in solution, chlorine gas is a possibility:
$$\ce{2Cl^{-}_{(aq)} -> Cl_{2(g)} + 2e-}$$
Hydrogen may also be produced depending on what you're using as an electrode on the q-tip.


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How do I decide the parent chain of an organic compound that has a ring as well as a side chain while naming the compound?
I think that this is decided by the number of carbon atoms. The part which has greater the number of carbon atoms, is considered the parent chain. But this fact is not leading to the correct result always. Are there some rules that govern this fact besides the number of carbon atoms.

Can you give an example in the form of picture

There are some cases where the "number of atoms" rule is not valid. For instance if you have a benzene ring with an ethyl group which in turn has a carboxylic acid group, the main chain would be the ethyl chain and the name of the compound would be 2-Phenylethanoic acid

Related: IUPAC Naming: 2-phenyloctane or 2-octanylbenzene

If you have a simple chain with carbon atoms and an aromatic ring, the priority has the ring, not matter how the chain is long. This rule is NOT valid when you have a functional group: if it is in the linear chain, it has the priority, and the numeration starts from the carbon more near at the functional group, for example
$\ce{CH3-CH2-CO-CH2-CH2-CH3}$. The numeration starts from left, so you write hexan-3-one.


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Testing isopropyl alcohol purity
I'm curious how to test purity of isopropyl alcohol? I simply got very curious today. I bought rubbing alcohol and it doesn't say what's the purity anywhere. It could be 70%, but it could also be 90%, or 99% or even 99.9%. I'm a curious man so I started googling but couldn't find anything.
If I were a chemist, how would I test the purity?

If you have chemistry equipment ready, then you would use a gas chromatograph (wiki-link). With household items is far more difficult, I cannot think of an accurate way, hence the comment ;)

I just had an idea myself - since impure isopropyl would contain water, if I took 10 grams of isopropyl and somehow measured how much water there is, then I could tell its purity. For example if there were 3 grams of water, then it would be 70% pure.

True, though that only works under the assumption that the only impurity is water. Might it also be ethanol, or methanol? Other hydrocarbons?

You're right. I didnt think of that.

If I were a chemist I would use the USP monograph for IPA, which involves a GC fitted with a TCD. No need for a NMR/MS

If the isopropanol mixture was bought commercially, the chances that it contains significant amounts of methanol or ethanol are fairly small, I think.  (I've never seen commercial isopropanol in the US that contains > 3% of these impurities, at least, but the situation may vary in other countries or markets.)
If you can assume that the only other main component is water, the easiest way to estimate purity is simply by measuring the density.  100% pure isopropanol has a density at 20 °C of 0.786 g per mL.  Pure water has a density of 1.00 g/mL.  At intermediate concentrations of isopropanol, the density is in between those values.  If you can accurately measure both (a) the volume and (b) the weight of a portion of isopropanol, you should be able to measure the density.  I know my kitchen scale and kitchen measuring cups would be precise enough for this.
If you can't assume that water is the only other impurity, other methods such as NMR, chromatography, or spectroscopy would probably be required.  I think NMR would probably be the most informative (but perhaps the least accessible to the home chemist.)

Very good answer. Thanks. I think I can use archimede's law to figure out density at room temperature?

Yes, if you use a completely sealed container and are willing to neglect the weight and volume of the container.

If you really want to figure what the “purity” Or rather the volume of isopropyl to water by weight aka $70\%$, $91\%$. And you don’t have a gas chromatograph, this is about the closest your going to get to an accurate Volumetric representation.
This is a fractional distillation column same basic principle as distilling drinking alcohol but much different equipment and processes. However using this column you are able to precisely control the temperature of the liquid or in this case the isopropyl alcohol in the heating mantle and allow it to evaporate and be distilled and collect after passing through a water jacket to condense the vapor and as long as you monitor the temperature at the neck located at the very top where the vertical column branches off to the right and have a thermometer right at that joint and maintain a temperature of $\pu{80 ^{\circ}C}$ ($\pu{176 ^{\circ}F}$) and let the isopropyl evaporate. Once the temperature begins to rise above $\pu{80 ^{\circ}C}$ ($\pu{176 ^{\circ}F}$) to $\pu{82 ^{\circ}C}$ ($\pu{180 ^{\circ}F}$) or higher, you stop the heat and remove the distillate container. Once everything has cooled down, you can measure what is remaking in the first Vessel and the distillate vessel. And with a bit of a margin for error using the original volume used, in my case for this rig I used I was using 70% and a starting volume of $\pu{500 ml}$. So my resulting measurements were, distillate - $\pu{347 ml}$, undistilled liquid - $\pu{148 ml}$. Resulting in a total volume of $\pu{495 ml}$ with a distillate percentage of $70.10\%$ (isopropyl) and remaining liquid (water) percentage $29.89\%$.
I hope this helps!!!

Other things beside, just from a mechanical perspective, this setup relies too much on the single supporting bar on the left (one clamp only?), behind the column, and the Erlenmeyer flask on the right (high centre of gravity).  It does not allow a quick exchange of the receiver flask without demanding a lot to the joints and Keck rings supporting the Liebig condensers, nor quick removal of the heat if needed without compromising the stand.  The garden-hose like click connector between the hoses still to clamp is a nice thing, though.  And keep the bench below the medical cabinet clean.

Dissolve non iodine table salt into a known amount and let it sit, measure the amount of water that sits in the bottom compared to the amount of alcohol that is in the top after about 10-15 minutes of sitting. You can then extrapolate what percentage of alcohol you bought. Example would be about half water and half alcohol if you got 50% isopropyl alcohol.

I guess you mean that the salt won't disolve in alcohol, right? But how do you distinguish the salty water from alcohol? The salt won't change water color, right? (If so they will be both transparent like water).

I did the test following youtube video, alcohol will be clear layer above murky salt water. But I (almost) didn't see the alcohol layer. I'm using table salt (no iodine smell as I know, but not sure), and sit the mixture for an hour without change. I even drop stamp ink hoping the ink will be dissolved by alcohol nonpolar and not by salt water (the ink just drop directly to salt water layer and keep the ball shape, no ink dissolved on alcohol layer). So that means the alcohol content is very low?

For very quick test, you can go for flame test. Pure IPA when poured on a cotton burns upto a certain flame length . when impurities are added , the length of flame decreases.
Do not forget to keep the size of cotton same while comparing

This doesn't help you to assess the purity unless you have a sample of known purity to compare to. And it doesn't seem very precise anyway.

Simple method is measure 100  ml Isopropyl alcohol water mixture heat at 80 degrees centigrade and after distilling left of liquid,cool and measure the volume. You get water content and the balance quantity is IPA.///

Even if you were able to separate two compounds with a difference in boiling temperature of 20 degrees Celsius with a column (bp water 100, bp isopropanol 82.6 Celsius) this answer does not account for the azeotropic mixture with 91 vol% of IPA distilling over at about 80 Celsius and atmospheric pressure.  So no, there is no sudden stop of the distillation with pure water left in the roundbottom flask.

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text:
Consistent atom names PDB
Lets say you have two times the same molecule in PDB format.
PDB1:
HETATM 1916  C1  EQU A1128      22.568 -34.213 -37.632  1.00 19.82           C
HETATM 1917  O1  EQU A1128      21.807 -35.267 -37.474  1.00 17.07           O
HETATM 1918  C2  EQU A1128      22.218 -33.078 -36.885  1.00 19.81           C
HETATM 1919  C3  EQU A1128      23.024 -31.912 -37.017  1.00 21.42           C
HETATM 1920  C4  EQU A1128      24.206 -31.909 -37.929  1.00 24.79           C
HETATM 1921  C5  EQU A1128      24.504 -33.085 -38.661  1.00 20.61           C
HETATM 1922  C6  EQU A1128      23.703 -34.217 -38.516  1.00 17.92           C
HETATM 1923  C10 EQU A1128      22.658 -30.748 -36.242  1.00 22.26           C
HETATM 1924  C11 EQU A1128      23.461 -29.573 -36.367  1.00 21.01           C
HETATM 1925  C12 EQU A1128      24.609 -29.558 -37.249  1.00 24.75           C
HETATM 1926  C13 EQU A1128      24.992 -30.715 -38.034  1.00 24.09           C
HETATM 1927  C16 EQU A1128      25.402 -28.301 -37.351  1.00 23.63           C
HETATM 1928  C17 EQU A1128      26.110 -28.153 -38.728  1.00 25.12           C
HETATM 1929  C18 EQU A1128      26.990 -29.384 -39.036  1.00 23.20           C
HETATM 1930  C19 EQU A1128      26.187 -30.715 -38.965  1.00 23.69           C
HETATM 1931  C24 EQU A1128      24.645 -26.944 -37.282  1.00 25.63           C
HETATM 1932  C25 EQU A1128      25.847 -25.958 -37.473  1.00 24.96           C
HETATM 1933  C26 EQU A1128      26.770 -26.799 -38.426  1.00 26.13           C
HETATM 1934  O26 EQU A1128      27.843 -26.441 -38.854  1.00 29.46           O
HETATM 1935  C27 EQU A1128      25.111 -27.931 -39.875  1.00 24.05           C

PDB2:
HETATM    1  C10 LIG     1      22.735 -30.818 -36.280  1.00  0.00           C
HETATM    2  C5  LIG     1      24.497 -33.076 -38.614  1.00  0.00           C
HETATM    3  C11 LIG     1      23.469 -29.649 -36.412  1.00  0.00           C
HETATM    4  C6  LIG     1      23.704 -34.219 -38.496  1.00  0.00           C
HETATM    5  C2  LIG     1      22.306 -33.076 -36.899  1.00  0.00           C
HETATM    6  C3  LIG     1      23.085 -31.931 -37.022  1.00  0.00           C
HETATM    7  C4  LIG     1      24.186 -31.919 -37.892  1.00  0.00           C
HETATM    8  C12 LIG     1      24.550 -29.604 -37.290  1.00  0.00           C
HETATM    9  C13 LIG     1      24.951 -30.740 -38.009  1.00  0.00           C
HETATM   10  C1  LIG     1      22.602 -34.190 -37.652  1.00  0.00           C
HETATM   11  C26 LIG     1      26.758 -26.817 -38.473  1.00  0.00           C
HETATM   12  C19 LIG     1      26.180 -30.692 -38.934  1.00  0.00           C
HETATM   13  C25 LIG     1      25.881 -26.016 -37.526  1.00  0.00           C
HETATM   14  C18 LIG     1      27.001 -29.371 -38.996  1.00  0.00           C
HETATM   15  C24 LIG     1      24.709 -26.963 -37.185  1.00  0.00           C
HETATM   16  C16 LIG     1      25.367 -28.337 -37.396  1.00  0.00           C
HETATM   17  C17 LIG     1      26.086 -28.150 -38.747  1.00  0.00           C
HETATM   18  C27 LIG     1      25.093 -27.987 -39.938  1.00  0.00           C
HETATM   19  O26 LIG     1      27.821 -26.440 -38.923  1.00  0.00           O
HETATM   20  O1  LIG     1      21.738 -35.208 -37.538  1.00  0.00           O

In the third column you can see the atom names (C1, O20, etc.). They are different for both PDB files even though it is the same molecule. I would like them to have the same names for the same atoms. So instead of C13 and C9 I would like to have them both named C13 or C9.
What is the convention to name these atoms? What method should I use to convert them to a common naming?
ps. I left out most of the PDB to save space

You'd need to have some sort of ligand dictionary. For many PDB ligands, this exists, and you can look up the atom naming.
For example, the PDB ligand expo allow you to search for ligands by 3-letter code (e.g., HEM).
The PDBeChem also lets you find atom names based on 3-letter codes (the link is HEM).
However, as far as I know, there isn't (yet) any sort of pattern matching that would allow you to go through PDB files and automatically perceive your ligands, sort the atoms, and name them systematically.
If you find one, please let me know - it would be a great addition to programs like Open Babel

The convention for the canonical atoms names can be found in the Chemical Component Dictionary (CCD). As for the method, I would use the Vim method of deleting and pasting lines. Another fancy way would be to use a graph approach based on element and do a matching on a per node basis. This would be an excellent Master thesis project, to develop such a fine piece of software.


text:
Reactivity of thioesters with respect to nucleophilic attack
Why are thioesters relatively reactive with regard to nucleophilic attack? Prof says to wait until pchem 3 when we learn about orbital symmetry. He also said that sulfur’s d-orbitals (?!) don't have the correct symmetry to participate in resonance with the carbonyl carbon.
Wait. I thought that sulfur didn’t utilize its d-orbitals. Was it previously taught that sulfur’s lone pairs were in d-orbitals? Why would sulfur even need d-orbitals when in a thioester? Can't we just explain this based on the poor overlap between carbon's 2p and sulfur's hypothetical 3p orbitals?

Apply the concepts discussed here to thioesters

@ron So would you chalk a thioester's reactivity to the 2p-3p bond between the C and S?

Yes, similar to an acid chloride.

@Ron so d orbitals are irrelevant?

In this case I think so.

@Ron also 3d orbitals have the correct symmetry to interact with 2p orbitals correct?

Some do, some don't ($\ce{d_{z^2}}$), it also depends on the relative orientation of the orbitals.  For example an s orbital has the proper symmetry to interact with a p orbital if the interaction is along the long axis of the p-orbital and at the "end" of the p orbital.  On the other hand, due to symmetry, interaction along the long axis but at the center of the p-orbital cannot result in constructive overlap.

@ron what about the specific case of a trimester?

What is a trimester?

@Ron sorry, that was autocorrect. I mean thioester

Yes, see here for an example

P.S.  Note that the relative sizes of the orbitals in that figure are not drawn to a realistic scale for the 2p-3d case.

According to these lecture notes from Gonzaga University:

The question arises over why thioesters occupy a prominent place in metabolism while oxygen esters play a relatively minor role. One answer is thermodynamic. The sulfur in the thioester linkage is less able to participate in electron delocalization through the acyl group (the usual explanation given for this is poor overlap between the 2p orbital of carbon and the 3p orbital of sulfur), and this makes the thioester bond less stable than the ester bond. Hence, the thioester bond has a more negative standard free energy of hydrolysis (−7.5 kcal/mole vs. about −5 kcal/mole for most oxygen esters). In many cases, it appears that thioesters are more reactive than oxygen esters, undergoing more facile nucleophilic displacement reactions at the acyl group. The reactivity of a thioester at the alpha carbon compares favorably with that of ketones.

...

Because resonance structure 2c [see web page - CF] makes a more significant contribution for a thioester than an analogous structure for an ester, the acyl carbon is more positive, hence more susceptible to nucleophilic attack. An attacking nucleophile would be more readily acylated by a thioester than it would be by an ester. Furthermore, the C-S bond is weaker than the C-O bond, and the thiolate (or thiol, if protonated) is a better leaving group than alkoxide (or alcohol). These factors all make a thioester in general a better acylating agent than an ester.

So the answer doesn't seem to involve d orbitals explicitly, but instead the difference between 2p and 3p orbitals (as well as other factors).

Just to put it in some context, a thioester is about the same reactivity as a ketone.


text:
Molarity vs Normality in labeling sulfuric acid
Molarity (M) is defined as moles of solute per liter of solution while Normality (N) is defined as moles of reacting units per liter of solution.  Most texts use sulfuric acid as an example to demonstrate the difference: a 1 M solution of sulfuric acid is 2 N because each mole of sulfuric acid gives two moles of $\ce{H+}$ (the "reacting unit"). Why then is concentrated sulfuric acid labeled as 18 N when it should be 18 M and 36 N (according to my text's example above)?

Just for reference (and archiving), can you give an example of a label of "18 N" for sulfuric acid?

Who labels concentrated sulfuric acid as 18N?  36N sounds right to me and Sigma-Aldrich:

Concentration: The commercial reagent contains 93-98% H2SO4, the balance being water. Its effective concentration is 18 M (36N).

https://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma-Aldrich/Product_Information_Sheet/258105pis.pdf
IUPAC discourages use of normality because it is ambiguous.  The number of equivalents can be ambiguous depending upon the purpose (equivalents of protons, equivalents of electrons, etc.).
See the following IUPAC document: http://old.iupac.org/publications/analytical_compendium/Cha06sec3.pdf

Indeed, I'd guess "18 N" is probably a typo for "18 M"

Alternately, 18 N sulfuric acid is about 50 mol% sulfuric acid in water, which, depending on your view, is still pretty "concentrated".


text:
Why does magnesium have the lowest melting point of all earth alkalis?
Why does magnesium have the lowest melting point of all earth alkali metals?

Perhaps you could give us the data for the rest of the alkali metals and explain why this is surprising to you.

This website seems to make a good argument that the cause of magnesium's low melting point is not understood (see the "Trends in Melting Point, Boiling Point, and Atomisation Energy" section).  The page also discusses why the reasons typically advanced probably aren't valid.


text:
Is there something which will block liquids but is dissolved by blood?
Is there a material which can be applied to a surface that will block liquids but is dissolved by blood?
The idea is to use the liquid to close a circuit, but only if blood is present.
I'm wondering if there is something I can coat on the circuit to act as a barrier preventing other liquids from reaching the circuit, but that would react to the chemical components of blood by dissolving, allowing the blood to reach and close the circuit.
One application might be in a bandage, to detect if a wound has reopened. The number of liquids in that context is pretty limited. A limited number of false positive edge cases is also acceptable

What do you mean by "block" ? And what class of liquids ? I assume if blood dissolves it, it won't "block" blood.

I've elaborated on the question in the body of the post.

There lots of substances, that are soluble in water (and also in blood) but insoluble in some of other liquids.

Excuse my ignorance. I'm wondering if there is something that reacts specifically to the makeup of blood (as opposed to water), like hydrogen peroxide, but with the effect of dissolving, which can be used as a coating.  I've further elaborated the question.

There is probably nothing that will react only to blood that might not be fooled by some other material. How specific must it be (e.g. human blood, mammal blood, arthropod blood)? What is the environment in which it must operate?

I imagine you are right. One application might be in a bandage, to detect if a wound has reopened. The number of liquids in that context is pretty limited. A limited number of false positive edge cases is also acceptable.

You state this material is to be used to form a protective layer over a sensor, and should react to blood by dissolving. Please tell what should be the reaction to: lymphatic fluid, puss, sweat, water, alcohol, iodine solution.

Ideally the barrier would prevent the listed fluids from reaching the circuit, but if prevented some but not all it would still be of interest.


text:
What happens if I have an acid/base equillibrium and remove hydronium ions until there is 50/50 of acid and conjugate base?
Suppose I had an acid HA, and it reaches equilibrium in water. Then I remove the hydronium ions. I don't know how I would do this, maybe adding some hydroxide ions. My question is can I do this and will I be able to eventually create a 50/50 mixture of acid and conjugate base? Is this a very good buffer?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

The quick answer is that it depends what 'A' is.

You can't just remove hydronium ions because the solution needs to remain electrically neutral.
You can't just add hydroxide ions, but you could add sodium hydroxide until the HA/A- ratio was 50/50.

Is this a very good buffer?

It is the optimum ratio of HA and A- for buffering.


text:
¹H-MRS Detection of Dopamine
Why is it difficult to detect dopamine using $\ce{^1H}$ Magnetic Resonance Spectroscopy (MRS) in the brain (in vivo)?

The concentration of dopamine is low
The concentration of other protons is very high
The magnetic field is not homogeneous.
The brain cannot be rapidly rotated the way a laboratory solution state NMR sample is to average magnetic field inhomogenity
Compare to laboratory solution state NMR, the viscosity in the brain would be high causing further line broadening.

If you look at the spectra in 1H NMR Probe for in Situ Monitoring of Dopamine Metabolism and Its Application to Inhibitor Screening J. Am. Chem. Soc., 2012, 134 (30), pp 12398–12401, you'll see that even in tissue samples, there are too many other stronger signals for conventional 1H NMR to be successful, so C-13 labeling is used in combination with 1H NMR.

Thanks for this answer and the reference. Concerning 2, is there a way to suppress other signal sources, as is done with water, or are there too many sources on the same "wavelength"? Concerning 3, is "shimming" not enough because of the low concentration? Concerning 4, the gradient coils are rotated around the head - does the problem concern the main field B0? Concerning 5, could some kind of contrast be used, similar to BOLD, to address this, noninvasively?

For 2, there are signals that are very overlapping, if you can see in the paper. For 3, having spent countless hours shimming, it is never fully successful on a 5mm sample; but to the degree a field is uniform in a 5mm sample (say to 1 ppb inhomogenity), it would be much worse in a meter scale sample (maybe even 1ppm) http://en.wikipedia.org/wiki/Shim_(magnetism).  4. There should be no gradient in the field for the an NMR peak to approach its natural line width.  5. Contrast agents change relaxation rates, which is the basis for MRI, but (continues below)

(continued from about) if you want to detect a particular compound, you need to look at chemical shift not relaxation rate.

Hypothetically speaking, if the dopamine peak would not be overlapping with anything else, would it help to use multiple averages to overcome the problem of low concentration?

@noumenal In the article "Dopamine Reduction in the Substantia Nigra of Parkinson's Disease Patients Confirmed by In Vivo Magnetic Resonance Spectroscopic Imaging", it is stated that be adding spectra of whole groups of patients together, information of dopamine concentrations can be obtained, but not for an individual. http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0084081


text:
How much gas is inside a pressurised gas cylinder?
Given a 19kg Liquid Petroleum Gas (LPG) cylinder, with this specification:

LPG inside the cylinder weighs 19kg.
The internal volume of the cylinder is 44.5 litres.
The LPG is a mixture of propane gas (60%) and butane (40%).

I would like to calculate the volume of LPG contained in the cylinder. This is the volume the pressurised gas will uncompress to, if it escapes the cylinder.
I have looked at the Ideal Gas Law:
Pressure x Volume = Number of Moles x Gas constant x Temperature
But I cannot use it as I do not know about the pressure inside the cylinder and volume is the unknown I am interested in.
How do I proceed with this problem?

Ideal Gas Law only works when a gas behaves like an ideal gas. It cannot be used in this case for finding pressure inside the cylinder as LPG behaves non-ideally at high pressure (it liquefies). Thanks to @steveverrill for pointing this out.
So, we will use Ideal Gas Law to find out what the volume of LPG will be at normal atmospheric pressure (1 atm) and a temperature of 20 degrees Celsius, when it is still a gas:
A 19kg cylinder contains 19kg of LPG.
Mass of propane                    = 60% of 19kg = 11.4kg = 11400g
Mass of butane                     = 40% of 19kg = 7.6kg  = 7600g
Molar mass of propane              = 44.1g/mol
Number of moles of propane         = 11400g / 44.1g/mol = 258.5mol
Molar mass of butane               = 58.12g/mol
Number of moles of butane          = 7600g / 58.12g/mol = 130.8mol
Total number of moles in 19kg LPG  = 258.5 + 130.8 = 389.3mol

Pressure x Volume = Number of moles x Gas constant x Temperature
1 atm x V = 389.3mol x 0.00008205736 m3 atm K-1 mol-1 x 293.15 K
V = 9.365 m3

So, there is 9.365 cubic metres of LPG at normal atmospheric pressure and temperature of 20 degrees Celsius.

The cylinder contains $\pu{19kg}$ of LPG, $\pu{60\%}$ of which is propane (presumably by mass) so we have $\pu{11.4kg}$ propane and the rest is butane.
The molar mass of propane $\ce{C3H8}$ is $\pu{3*12+8*1 = 44 g/mol}$, so we can work out how many moles of propane there are. Then do the same for butane.
Now we use the ideal gas law:

But I cannot use it as I do not know about the pressure inside the cylinder and volume is the unknown I am interested in.

You are not interested in the pressure inside the cylinder. You are interested in what the volume is when the LPG uncompresses. Therefore the pressure you need to use is atmospheric pressure.
If you apply the formula for the ideal gas law $PV=nRT$ you will find that $1$ mole of gas occupies about $\pu{24 L}$ at $\pu{25^oC}$ ($\pu{298K}$) and atmospheric pressure. That is a good shortcut to remember for cases like this.

Note that the ideal gas law only works when the conditions are in the range where the gas behaves like an ideal gas. You know the volume of the cylinder, but you cannot use the ideal gas law to find the pressure inside the cylinder, because the gas behaves non-ideally at high pressure (It liquifies!)

LPG inside the cylinder weighs $\mathrm{19\,kg}$.
The LPG is a mixture of propane gas (60%) and butane (40%).

With other words, you know:

the total mass of the gas mixture
the mass fractions of propane and butane
the molar masses of both gases
the molar volume of a gas under standard conditions

so, how do I calculate the volume then? Do I use the Ideal Gas Law?

@ChirayuShishodiya We have a policy here that says: "Homework questions should be hinted at, and are not to be answered directly." You can learn more here: http://meta.chemistry.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange

Sorry that wasn't enough of a hint for me, hence I was trying to clarify. Besides, this is not a homework question.

Let me give you a simple method of calculation:
Total weight of LPG in the cylinder: 19 kg
Density of LPG in vapour state at atmospheric conditions: 2 kg/m3 approx.
Hence, the volume of LPG in gaseous state is: 19/2 = 9.5 m3

If it was compressed gas cylinder of Argon or Nitrogen or Oxygen.
Then we know that the pressure inside the cylinder is 150bar. (Standard working pressure for industrial gas cylinders)
If internal volume of such cylinder is 45liters water capacity, the total volume of gas inside the cylinder is 45 x 150 = 6750 liter
But LPG is stored in the cylinder as liquid. Here you cannot use ideal gas equation.
LPG is a specific blend of butane, propane and Isobutane, we can use expansion ratio here:
LPG expansion is 270 times the volume of gas to the volume of liquid. So, 1L of liquid LPG (propane) expands to equal 270L of gaseous LPG
44.5 x 270 = 12015 liter
But what if we had liquid CO2 in 45l cylinder instead of Argon?
Fill rate of CO2 is 0.75kg per liter water capacity of cylinder.
45 x 0.75 = 33.75kg of liquid CO2.
To find volume of CO2 inside, when vaporized at 60°F, the expansion ratio is 535:1
535* 45 = 24075l

The only reasonable way is via the LPG mass,  total LPG molar amount and ideal gas equation. V=m/M_eff . RT/p. You wants, can make nonideality correction.


text:
Would it be possible to create a (non-toxic) alcoholic drink that, when zapped with laser, changes color?
Would it be possible to create a (non-toxic) alcoholic drink that, when zapped with a 0.5mW laser pointer, changes color?
This would be mainly for the visual effects (bartender show) so it should not require aiming the laser at the drink for more than a few seconds. The reason for choosing a 0.5mW laser is to minimize the need for eye protection.
It would not matter what color laser is used, but would this be possible?
If it is possible, what kind of chemical(s) would be needed in the drink?

It would probably depend on the color of the pointer

thanks @Mithoron, I have added a note about that to the question.

Is it correct that you are looking for a permanent colour change?

Let me know when you find it - looks like one hell of a party trick!

I see a couple of problems here:
Beam diameter vs volume of the beverage
The volume of the sample that is actually hit by a narrow laser beam is very small, as compared to the total volume of the sample (= the drink). As a consequence, only a very small amount of the dye in drink will undergo a transformation. A wide beam of a switchable light source definitely is the better option.
Is is possible in general?
Yes, a number of compound classes undergo transformations upon iradiation and change their colour (spiropyranes, fulgides, etc.). Often, the reverse reaction can be induce thermally.
So what?
Food dyes are strictly regulated, you can't add a photochromic dye to a drink just because the colour change looks nice in the lab. If the dye isn't approved as a food dye, there's no legal way! To my knowledge, food dyes typically do not show photochromism but are photostable. Consumers usually don't like that their food changes the colour in the light.
In summary, the idea sounds funny but I rather doubt that it can be realized.
The only option to add a dramatic effect would be the use of fluorescent and food-compatible ingedients. Think in quinine (tonic water).

Do you think it might be possible to make use of a photoacid and an indicator? Only small amounts of both would be needed to change the colour of the entire mixture, if the indicator flips around neutral pH. The problem I see would be finding a photoacid safe for consumption, having to use high frequency photons, low photon intensity and the low photon absorption cross-section in a dilute mixture of the photoacid.

@NicolauSakerNeto Photoacids are a nice idea too. But as you mentioned yourself, "safe for consumption" (= approval by legal autorities) is the key.

Klaus has provided an excellent answer.  Basically your desired approach is possible, but fraught with difficulty.
Perhaps there is another way to attack the problem.  Still, I suspect considerable testing will be required.
For reasons mentioned by Klaus, it is necessary to use a food dye that is already approved.  Many food dyes are photobleachable.  For example, azo dyes exist as cis-trans isomer pairs.  Typically, the trans isomer is more stable and provides the dye color.

Often (not always, this is why you'll need to test), when the trans-dye is irradiated with laser light it will convert to the colorless (or differently colored) cis isomer. This is termed "photobleaching" - light has "bleached" the color out of the dye. Generally the cis isomer remains stable as long as you keep it cold. The same concept can be applied to non-azo dyes that can exist as isomeric pairs.

image source
Buy a number of food dyes that provide the color you desire and can exist as isomers (cis-trans or otherwise).  Dissolve each one separately (until you find one that works) in the minimum amount of one of the ingredients in your "Blue Light Special".  Irradiate the (presumably colored) solution with a general lab laser (maybe you have a chemist friend who can do this for you) -  a laser pointer probably won't have enough power.  If the experiment is successful, the color should disappear from the solution.  Keep the solution cold in your bar fridge.
When someone orders the "Blue Light Special" mix all of the other ingredients in a glass at room temperature.  Add some of the cold, colorless (photobleached) dye solution to the room temperature mix.  As the cold dye warms up it should convert back to its colored form and the drink should become colored right before the customer's eyes.

Whatever name OP'd picked out for the drink, yours is better. :-)


text:
Is 2-bromobutane chiral?
My textbook says 2-bromobutane is chiral, but it look like it will form superimposition mirror image

For a molecule to be chiral it must have non-superimposable mirror images.  Here is a drawing of the two mirror images for 2-bromobutane.

The chiral carbon atom is denoted by an asterisk.  In the case of 2-bromobutane there are 4 different substituents attached to it. The molecule is chiral, you can't pick up one of the mirror images and superimpose it on the other.
Note to think about:

4 different substituents may make an atom chiral, but the molecule as a whole may still be achiral

For example meso compounds

image source

some molecules without 4 different substituents attached to an atom may still be chiral .

For example, compounds with axial chirality such as chiral allenes (1,3-dimethylallene), chiral biphenyls, chiral helicenes, etc.

image source

image source

image source

Yes, it is chiral because there are 4 different groups on the second carbon, H, Br, methyl and ethyl.
Maybe you are looking at a planar depiction.


text:
How to determine the sign of the enthalpy of formation?

During the formation of $\ce{Al2O3}$ from 5.4 grams of $\ce{Al}$ and enough oxygen, the temperature of 2 kg of water climbs by 20 degrees. What’s the enthalpy of formation of $\ce{Al2O3}$ (per mole) ?

What I tried:
The balanced reaction:
$$\ce{4Al + 3 O2 -> 2 Al2O3}$$
5.4 grams of $\ce{Al}$ are 0.2 moles.
If 2 kg of water climbed by 20 degrees then 40 000 calories were invested during the reaction.
To get one mole of $\ce{Al2O3}$ we need 2 moles of $\ce{Al}$ , therefore the enthalpy of formation (per mole) of $\ce{Al2O3}$ is 400 000 calories. However, the correct answer is −400 000.
Can someone please explain the negative sign?

$\ce{Al2O3}$, upon the formation of its lattice, releases heat. This heats up water. The increase in the temperature is in water, not $\ce{Al2O3}$. We usually refer to the water as the surroundings and $\ce{Al2O3}$ as the thermodynamic system.

A thermodynamic system is the content of a macroscopic volume in space, along with its walls and surroundings; it undergoes thermodynamic processes according to the principles of thermodynamics. Wikipedia

So, upon the formation of lattice (among with other, mostly, endothermic reactions) a lot of heat is released and that's what you have estimated. Since $\ce{Al2O3}$ is the system, it loses heat. Thus, $\Delta H_f$ is negative.

I guess you are confused by sign convention of physics and chemistry. In chemistry, aluminium oxide is the system, and the reaction releases energy, and that energy is obviously coming out of the system. Since it is a loss of energy for the system it is referred as negative sign. But in physics the sign convention is just the opposite.

Are you sure the sign convention is opposite in physics for heat released? Because if you're referring to the first law of thermodynamics, then the sign conventions for the work done (W) and not the heat exchanged (Q). Can you please confirm? Thanks!


text:
Why is the work done by a gas during free expansion zero and not infinity?
Suppose there's a spaceship that has a hole in it. Gas is expanding through the hole from the spaceship to outer space (vacuum). According to pressure volume work ($P\Delta V$) the work done by the gas is infinity. According to my homework assignment, it's $0$. Can someone please explain why?
Edit:
According to Wikipedia:

$P$ denotes the pressure inside the system and outside the system, against which the system expands; the two pressures are practically equal for
  a reversible process;

How can $P$ denote two different quantities? Can't this rule be applied for situations where the pressures inside and outside the system are different?

In this case external pressure is 0 (since it expands against vacuum). Therefore work done Is 0.What makes you think work done is infinity?

The example of a spaceship is not very good, since gas expanding through a hole in a spaceship will do work by accelerating the spaceship, unless the spaceship is fixed.  This is the principle by which rockets fly and by which inflated baloons move after you let them go.

@CurtF. Is work done? Or is the acceleration of the craft just a consequence of conservation of momentum, without a thermodynamic work interaction?

I don't think its controversial to say that all energy changes involve transfers of either heat or work.  Take the system as the spacecraft (or balloon) only, i.e. it doesn't include the gas.  The transfer of energy that makes it accelerate sure isn't heat.  So it must be work.

@CurtF. Mm, of course -- to tell whether work is done you have to choose your control volume such that its boundary includes the surface at which the putative work interaction occurs.  Gah, this is why thermo always drove me batty!

The external pressure in space is 0 (well, depends where and not quite, but for the discussion let's say that it is). So $P \cdot \Delta V$ is also 0. Gas expanding from a hole in a spaceship is not an irreversible process. Such processes are very slow and gradual, such that the external pressure has time to equilibrate with the internal pressure.
On the contrary - you have the most irreversible process in here. Gas leaking from a hole in a spaceship into a medium (space) that will never equilibrate with the internal pressure.
Think about it from an intuitive point of view. Work done by an expanding system is there because it moves other things around. In space there is nothing to move, so no work done.


text:
How to calculate the enthalpy change for the formation of ethylene epoxide from ethylene?

I get everything up until the very end. We need to identify the bond enthalpies first:
\begin{align}
\ce{C=C} &= 612\\
\ce{O=O} &= 248\\
\text{SUM of these}&= 860\\
\end{align}
\begin{align}
\ce{C-C} &= 348\\
2\cdot(\ce{C-O}) &= 720\\
\text{SUM of these} &= 1068\\
\end{align}
We know that enthalpy change is = products - reactants, which is $1068 - 860 = 208$, but the problem comes here: I know a minus sign is supposed to go there, but
how do we know that? Exothermic reaction? Is it because of the oxygen, suggesting it's a combustion reaction or not?

When you use bond enthalpies to calculate $\Delta H$, you need to subtract the sum of the bond enthalpies for the products from the sum of the bond enthalpies for the reactants.  In other words "reactants - products".  This is because all bond enthalpies are positive, since energy must be put into the bond to break it.  Bond formation releases energy; the minus sign compensates for this.  In short, bonds of reactant molecule must break, and new are formed to form product molecules.  So, your answer should be negative.

Thank you but does this mean that there is supposed to be a negative sign whenever I see that the enthalpy for products is > enthalpy for reactants? I am sorry, but I really need to understand this, because I always mess up questions as such, due to dumb errors.

In other words: why is it "reactants - products". I have always thought it was "products - reactants". Also, how can I see that the bond is broken or not.. is it because the enthalpy is greater for products.. meaning since greater energy is needed to break them, but less to form bonds, which is what we do here, we then have an exothermic reaction?

There is one mistake in your calculation: oxirane is not just two standard C-O bonds plus a standard C-C bond. Due to ring strain there is some extra energy in the three-member ring.

But I'm not sure if the data sheet had the additional info for the extra energy due to the ring strain. If it did not have the data, how can we use non-existent data to do any calculation?

authors: Сергей Святодух


text:
How much water evaporates in a vacuum before freezing?
I placed a container with 1000 grams (1 kg) of water at room temperature (20 degrees Celsius) inside a vacuum at 0.03 atm pressure. The water boiled for a few seconds then froze solid.
I need to calculate how much water evaporated before freezing occurred. Anyone know a formula I could use?

At .03 atm and 20 °C the water will not boil at all. So, your "observations" are doubtful.

Extract the ice only and weight it.
1000 grams - Ice mass = Vaporized Water mass

For a quick and dirty estimate of the evaporated fraction $x$, you might want to ignore the temperature changes as well as the temperature dependencies of the involved quantities (in particular if the initial temperature is already close to the freezing point) and just balance the enthalpy of vaporization $\Delta_\text{vap}H$ of the evaporating water and the enthalpy of fusion $\Delta_\text{fus}H$ of the freezing water:
$$\Delta_\text{vap}H\approx\Delta_\text{fus}H$$
Then, the evaporated fraction may be estimated from tabulated values for the specific enthalpies $\Delta h$ or molar enthalpies $\Delta H_\mathrm m$ as follows:
$$\begin{align}
x\cdot\Delta_\text{vap}h&\approx(1-x)\cdot\Delta_\text{fus}h\\[6pt]
x&\approx\frac{\Delta_\text{fus}h}{\Delta_\text{fus}h+\Delta_\text{vap}h}
\end{align}$$
or
$$\begin{align}
x\cdot\Delta_\text{vap}H_\mathrm m&\approx(1-x)\cdot\Delta_\text{fus}H_\mathrm m\\[6pt]
x&\approx\frac{\Delta_\text{fus}H_\mathrm m}{\Delta_\text{fus}H_\mathrm m+\Delta_\text{vap}H_\mathrm m}
\end{align}$$
For example, simply using the first values that I could find, which are the molar enthalpy of vaporization at $T=25\ \mathrm{^\circ C}$ of $\Delta_\text{vap}H_\mathrm m=43.98\ \mathrm{kJ\ mol^{-1}}$[1] and the molar enthalpy of fusion at $T=0\ \mathrm{^\circ C}$ of $\Delta_\text{fus}H_\mathrm m=6.01\ \mathrm{kJ\ mol^{-1}}$[2]:
$$\begin{align}
x&\approx\frac{\Delta_\text{fus}H_\mathrm m}{\Delta_\text{fus}H_\mathrm m+\Delta_\text{vap}H_\mathrm m}\\[6pt]
&\approx\frac{6.01\ \mathrm{kJ\ mol^{-1}}}{6.01\ \mathrm{kJ\ mol^{-1}+43.98\ \mathrm{kJ\ mol^{-1}}}}\\[6pt]
&\approx0.12
\end{align}$$
Thus, about $12\ \%$ of the water have to evaporate in order to completely freeze the remaining water.

A better result may be obtained from the enthalpy balance for the initial and final state. If the process is fast or the system is thermally isolated, the system cannot exchange heat with the environment and the enthalpy balance is given as:
$$\begin{align}
h_0&=x\cdot h_\text{vapour} + (1-x)\cdot h_\text{ice}\\[6pt]
x&=\frac{h_0-h_\text{ice}}{h_\text{vapour}-h_\text{ice}}
\end{align}$$
where
$h_0$ is the specific enthalpy of liquid water at the initial state,
$h_\text{vapour}$ is the specific enthalpy of the evaporated water at the final state, and
$h_\text{ice}$ is the specific enthalpy of the frozen water at the final state.
(Note that the specific enthalpy $h$ is a state function that depends only on the equilibrium state of the system and does not depend on the path by which the system arrived at that state. Therefore, the exact changes of the variables, i.e. temperature, pressure, enthalpy of vaporization, enthalpy of fusion, specific heat capacities and masses of liquid water, ice, and vapour, during the course of the experiment are not relevant.)
For example, for liquid water at the initial temperature of $T_0=20\ \mathrm{^\circ C}$ and an assumed pressure of $p_0=1\ \mathrm{bar}=100\ \mathrm{kPa}$, the specific enthalpy of liquid water is $h_0=84.01\ \mathrm{kJ\ kg^{-1}}$[3]
At a final temperature of $T=0\ \mathrm{^\circ C}$, the saturation pressure of vapour in equilibrium with ice is $p=0.61115\ \mathrm{kPa}$. The corresponding specific enthalpies of ice and vapour are $h_\text{ice}=-333.47\ \mathrm{kJ\ kg^{-1}}$[4] and $h_\text{vapour}=2500.90\ \mathrm{kJ\ kg^{-1}}$[4], respetively.
Therefore, the evaporated fraction can be calculated as
$$\begin{align}
x&=\frac{h_0-h_\text{ice}}{h_\text{vapour}-h_\text{ice}}\\[6pt]
&=\frac{84.01\ \mathrm{kJ\ kg^{-1}}-(-333.47\ \mathrm{kJ\ kg^{-1}})}{2500.90\ \mathrm{kJ\ kg^{-1}}-(-333.47\ \mathrm{kJ\ kg^{-1}})}\\[6pt]
&=0.147
\end{align}$$
Thus, about $14.7\ \%$ of the water have to evaporate in order to completely freeze the remaining water. Note that this value is larger than the above-mentioned quick estimate since it includes the amount of water that has to be evaporated in order to cool the water from $T_0=20\ \mathrm{^\circ C}$ to $T=0\ \mathrm{^\circ C}$, which has been ignored for the first estimate.

Note that reducing the pressure to $p=0.03\ \mathrm{atm}$ for liquid water with an initial temperature of $T_0=20\ \mathrm{^\circ C}$ (both values are mentioned in the question and indicated as a red dot in the following phase diagram of water) is not sufficient to make the water boil because the boiling point of water at this pressure is $T_\mathrm b=24.3\ \mathrm{^\circ C}$. The pressure would have to be reduced below $p=2.339\ \mathrm{kPa}\approx0.023\ \mathrm{atm}$ in order to make water boil at a temperature of $T=20\ \mathrm{^\circ C}$.
 source: www.wolframalpha.com
(Note that I don’t know how reliable the steam tables of WolframAlpha are. I only use it here by way of illustration but not for my calculations. Also note that the indicated value for STP is not in accordance with IUPAC recommendations.)
Furthermore, the phase diagram of water shows that it is not possible to freeze water at such temperature and pressure. In order to freeze the water using the process that is described in the question, the pressure would have to be reduced below the triple point pressure of water, which is $p_\mathrm{tp}=611.657\ \mathrm{Pa}\approx0.006\ \mathrm{atm}$.

[1] “Enthalpy of Vaporization”, in CRC Handbook of Chemistry and Physics, 97th Edition (2016), William M. Haynes, ed., CRC Press/Taylor and Francis, Boca Raton, FL.
[2] “Enthalpy of Fusion”, in CRC Handbook of Chemistry and Physics, 97th Edition (2016), William M. Haynes, ed., CRC Press/Taylor and Francis, Boca Raton, FL.
[3] REFPROP – NIST Standard Reference Database 23, Version 9.0
[4] 2009 ASHRAE Handbook of Fundamentals, SI Edition

You will need to know:

the heat capacity of water $C_p$
the latent heat of fusion of ice $L_f$
the latent heat of vaporization of water $L_v$

First determine how much heat must be lost to change the temperature from 20 to 0 using the heat capacity.
Next determine how much heat must be lost to change from 0 degree water to 0 degree ice using the latent heat of fusion.
Finally, add the two heats from above together to get a total heat loss, and use the heat of vaporization of water to calculate the amount of water evaporated that corresponds to this total heat loss.
Note: The above involves an approximation that the amount of water evaporated is relatively small.  In reality, once an infinitesimal portion of water evaporates, the amount remaining that needs to cool will be infinitesimally less than one kg.  If you are expected to know and use intergral calculus for your class, you may need to take into consideration how the mass is varying thoughout the process.
To calculate more accurately, break the problem is into two parts.
First, a mass m1 evaporates, to cool water from 20 degres to 0 degrees.
Let $m$ be the varible liquild mass during the process and temperature be $t$
$mC_pdt=L_vdm$
$C_p\int_{20}^{0}dt = L_v \int_{1}^{1-m1}(dm/m)$
The remaining mass is 1kg-m1
Second, a mass m2 evaporates, to freeze the remaining water.
$L_v m2 = L_f (1kg - m1 - m2)$

The question was, how much water did evaporate until freezing. So, heat of fusion is not necessary.

@Georg Because the question says "froze solid" I was picturing all the water freezing into a solid block.  If the person asking is only interested in how much water needs to evaporate to get to zero degrees C liquid, they can ignore the second step of the calculation.

I understand the question to be implying that water evaporates until all the ice is frozen.


text:
What chemical elements does a nuclear explosion release into its mushroom cloud?
I'm interested in the mushroom cloud produced by a nuclear explosion: What is it made of?

I know that the uranium fission reaction is the following (from this web page)
$$\ce{{}^{235}U + {}^1_0n -> {}^{95}Sr + {}^{139}Xe +2~{}^1_0n + 180~MeV}$$
I know that these elements are not stable and decay, so what is left over?

This topic seems much more complex than this - there are lots of possible products even in first fission (ones you added are only example), also most of the mushroom doesn't come from the bomb itself but from it's surroundings.

Maybe a rough estimation of the element that is on average most present in the cloud

OK, I'll bite on the rough estimate - nitrogen from the air.  For elements not in the air to begin with, mainly silicates from the ground.

https://en.wikipedia.org/wiki/Mushroom_cloud

Fission or fusion ?

According to The Preparatory Manual of Explosives, the mushroom cloud is produced by the partial vacuum left after initial expansive shock wave. That implies that it is primarily the condensation of water that makes it visible.  These clouds form in high explosive as well as in nuclear detonations near the ground. See The Preparatory Manual of Explosives. See also this video.

During the cold war the americans did some work on the chemical form of fallout, the great problem is that shortly after a surface burst or a burst close to the gound the bomb will stir up whatever was close to the detonation or on the ground below it.
The heat of the bomb will cause the dust and other matter to rise up into the firey hot cloud of smoke mixed with fission and activation products together with unused fuel. The nature of the "dust and other matter" depends greatly on the detonation site.
When the detonation site was a coral island then the fallout tended to contain calcium oxide particles from coral plus some iron from the towers which held the bombs. Some of the particles collected were dicalcium ferrite (Ca2Fe2O4). A moderate yeild tower shot in the Nevada test site would give instead glassy spheres. Small yeilding surface shots in the Nevada site gave particles which were partly melted mineral grains.
The chemistry of fallout from underwater explosions was different again, if you want to read more then see C.E. Admans, N.H. Farlow and W.R. Schell, Geochimica et Cosmochimica Acta, 1960, volume 18, pages 42 to 56.


text:
How to run crystal simulation on GAMESS?
I know GAMESS is not best suited for crystal simulation but I am not ready to start learning another package as VASP. As I heard GAMESS is able to do some calculations in a "terminated" crystal (I mean using an ion in the periphery instead of repeating it multiple times).  Mostly I need to calculate HOMO-LUMO gap / hessian matrix and transitional state of a surface reaction .
I looked in the web but couldn't find a good worked example of using GAMESS for crystals.  Is there anywhere I can find such an example ?

Did you consider to have a look at DFTB+?

As you said, there are many packages capable of handling calculations of crystals, slabs, etc. with periodic boundary conditions. VASP is only one of dozens. (That's probably a separate question in its own right.)
GAMESS does not include treatment of periodic boundary conditions, so you would need to run a calculation on a "cluster."
Since you mention wanting to treat a surface reaction, you'd use a builder (e.g., Avogadro) to create a large cluster (e.g., Build->Supercell). So rather than one primitive unit cell, you create maybe a 2x2x2 or 3x3x3 slab. If needed, you'll terminate dangling bonds (e.g., adding hydrogens to cap oxygens in $\ce{TiO2}$ for example.. notice the terminal "uncapped" atoms on the back and right of the unit cell).

Finally, after running calculations in GAMESS, you'd need to check for convergence. So you'd want to run calculations with increasing system size and see if the properties converge.
Now, this isn't as strange as it sounds. In some cases ($\ce{MgO}$ and $\ce{TiO2}$ are good examples) the properties require more advanced calculation methods like multi-configuration QM, and this may be easier to do in a quantum chemical package with a cluster, rather than a solid-state crystal code.

Thank you Geoff.  I am yet struggling with that Hydroxyapatite crystal!  As you suggested in another question, I found a decent .cif file for it in COD( thank you for the suggestion !). I opened it with Avogadro and next step is normally finding a good geometric start point, and I ran UFF optimization but the whole crystal system burst !   I don't know how should I avoid such a heavy energy that destroys the whole system. Here is the file: http://www.crystallography.net/cod/9011091.html?applet=jmol

I forgot to mention that I made a 3x3x1 supercell before running the UFF.


text:
Why aryl halides are very less reactive towards nucleophilic substitution reactions?
I've often herd that aryl halides are very less reactive towards nucleophilic substitution reactions. One of the reasons for this I found was that there is resonance effect in the compound, but how is  that responsible for it to less reactive towards S.N. reactions and in addition to resonance effect are there other factors also responsible ?

http://chemistry.stackexchange.com/questions/5690/aromatic-nucleophillic-substitution?rq=1

No, that is a very incorrect explanation for why (unsubstituted) aryl halides do not participate in nucleophilic substitution reaction. Conjugating effect between the halide substituent and the benzene ring is not responsible for this. The problem is that there is simply no available mechanism to perform the substitution:

$\ce{S_{N}2}$ is impossible since  inversion would give an incredibly weird strained ring (even more weird than a bridgehead carbon that is already forbidden by Bredt's rule).
$\ce{S_{N}1}$ would leave a very unstable empty $\ce{sp^2}$ orbital. This can happen, but only for the very best leaving groups, like $\ce{N2}$.
There is no electron stabilizing group that is conjugated with the halide group. This renders an addition-elimination type mechanism impossible.

What this weird strained ring would like? Thanks in advance.


text:
What is the reasoning behind the resonating structure of halo-benzene?
Can anyone explain the resonating structure of halo-benzene? How does the transfer of an electron from the halogen atom to the α-carbon and from there to ß-carbon and so on take place?

What do you mean with electron transfer? This isn't about $\ce{Ar-Hal + e- -> [Ar-Hal]^{-} -> Ar +\ Hal-}$, or is it?

I meant the transfer of electron from the halogen atom to the α-carbon and from there to ß-carbon and so on...

You can [edit] the question to include more details, so that it is easier for everybody to be understood.

You're simply asking about mesomeric structures, aren't you?

There are so many questions about resonance structures, where people think there is some kind of process going on. I think it is time to stop teaching this stuff and start talking about molecules as multinuclear atoms.

I think the misunderstanding here (it's a common one) is over the interpretation of resonance structures.
What you have to understand about resonance structures is that they do not actually exist in the molecule. The separate resonance structures are never formed on their own but they can be used to represent a delocalised bonding situation, more accurately described using molecular orbitals. So electrons do not actually move from the halogen to the $\alpha$ and $\beta$ carbons but what the resonance structures represent is that, only considering mesomeric effects, there is a greater electron density on the ortho and para carbons and than on the meta carbons, the carbon attatched to the halogen and the halogen itself.

However, this is not the full picture because halogens; especially fluorine and chlorine, have a very strong inductive effect and so there is actually considerably more electron density on the halogen than on any other atom.
Electrostatic potential map for chlorobenzene:


text:
What is the most accepted explanation of aurophilicity?
Gold has a special property called aurophilicity. Inorganic chemistry by Huheey has some explanation, but that was not enough to clear the conception.

Have a look at The aurophilicity phenomenon: A decade of experimental findings, theoretical concepts and emerging applications by Hubert Schmidbaur for some further directions and references.


text:
when do we consider the moles in bomb calorimeter?
When dealing with different problems in thermodynamics
I noticed that some of the problem consider the number of moles in calculating the heat absorbed or released, whereas some don't and they deal with the heat,capacity,difference in temperature WITHOUT considering the number of moles
example1:       1 gram of a sample burns in .....
they divide the grams over the molar mass to get the moles, then multiply by the enthalpy and solve for the needed.
example2:       1 gram of a sample burns in ....
they get one one of (q,C,delta H) directly without considering the number of moles.
what's the difference?

It depends on what you are measuring. Sometimes for instance you'll use "molar" enthalpy so you'll normalized for a mole. Sometimes you'll just use enthalpy then you don't need to normalize.


text:
Why is hydrogen gas so highly reactive?
Why is Hydrogen so reactive? What makes it combustible?

Please note that the homework policy on this site requires you
to show some personal effort.

Hydrogen is not particularly reactive.
For example, just mixing hydrogen and oxygen gas will not cause a reaction at room temperature, but many metal elements oxidize at least on the surface in air.
The most reactive elements in my opinion are fluorine in the non-metals, and the alkali metals like sodium and potassium.
It is combustible because oxygen wants electrons and takes them from hydrogen to form water.

Hmm, Dave, looking at the reactivity from the general point, shouldn't Cs be more reactive than Na?

yes, francium and cesium are the most electropositive, I was just lumping the alkali metals together, and givng sodium and potassium as examples of alkali metals


text:
when do we consider the samples in a calorimeter as a solution in calculation?
I noticed some problems consider the samples in a calorimeter as a solution thereby calculating the mass of the combining samples, whereas in others I found that they separate the samples into different calculations by using the distinct mass and specific heat for each.
could you explain why?
Thanks|!


text:
What is the IUPAC nomenclature for Intermetallic Compounds?
IUPAC has a system for naming organic compounds, ionic compounds, and inorganic compounds. I have seen something vaguely suggesting there is one for intermetallic compounds, such as: $$\ce{Fe-Co}$$ What is the system, if it even exists? If it doesn't exist what is the generally accepted way of naming intermetallic compounds?

Do organometallic compounds count? I sometimes run to folks that have a vague definition of metallic compound. Would be nice if you add an example.

@MARamezani No, I meant something like alloys, which may contain metals and non-metals. So I said metallic compounds because I don't know what else to call them. I wasn't aware this would be something misunderstood.

https://en.wikipedia.org/wiki/Intermetallic

In your example you have given two metals and a bond-indicating symbol.
Grammar
Hyphens, Plus and Minus Signs, 'em'Dashes and Bond Indicators
'em' Dashes (ALT+0151, 'en' dash would be ALT+0150)
a) Used to indicate metal–metal bonds in polynuclear compounds (e.g. "[Mn2(CO)10 bis(pentacarbonylmanganese)(Mn—Mn)").
b) Used to separate the indvidual constituents in names of (formal) addition compounds (e.g. "3CdSO4·8H2O cadmiumsulfate—water")
Alphabetical Order
a) Intermetallic compounds are sorted alphabetically within the group of cations and anions. Deviations are allowed in order to express structural information.
b) In coordination complexes, ligands are sorted alphabetically and the donating atom is put first (e.g. OH2 instead of H2O).
Further Reading
Therefore, I would write your example as "Co—Fe".
There are a lot more rules that you can read here: https://web.archive.org/web/20181005064417/https://www.iupac.org/fileadmin/user_upload/databases/Red_Book_2005.pdf


text:
How do you figure out the dissociation energy for ions?
Specifically $\ce{KF}$ into the $\ce{K+}$ and $\ce{F-}$ ions. I can find tables on the DE into atoms, but where would I go to find the DE between ions?
I am also given that the bond length is 2.17e-10m. Can I use that at all?

Wait a second... I think I just figured it out! I just use Coulomb's Law, correct?

And then multiply by distance. That would return energy.

you would have to integrate Fdx because the force changes as you move the ions apart

So it wouldn't be as simple as saying the attractive force * bond length = energy of the bond, and the opposite of that would be the dissociation energy?

no because the attractive force is not constant as the ions are separated. however, integrating the force over the distance with appropriate limits would give you a rough estimate of the bond energy

also what exactly are you trying to find. the dissociation energy for a gaseous $\ce{KF}$ molecule or the negative of the lattice energy

I assume gaseous, but I don't really know the difference. I'm taking MIT's OCW 5.111 (Freshman Chem.). The answer key says that the energy required to break the bond is just the opposite of energy of the bond. I understand the force integral, but it also makes sense that the energy stored in a bond would be equivalent to the energy required to break the bond. I thought that was how enthalpy was determined!

Maybe it is lattice energy... Neither the question nor the answer says anything about it one way or another!

If you are trying to find the heterolytic dissociation energy for a gaseous $\ce{KF}$ molecule then as a rough estimate you can model the ions as point charges and calculate the energy required to separate them from their equilibrium separation (the bond length) to infinity, using Coulomb's Law: $$E = -\int_{2.17\times10^{-10}}^\infty\frac{Qq}{4\pi\epsilon_0r^2}dr$$
Note that modelling the ions as point charges is a major simplification and so this is probably not a very accurate estimate. Unfortunately I can't find any data for the actual energy so I can't say how accurate this is.

Okay, awesome. You're right; it is a simplification, but since it's for freshman chemistry, it's probably good enough. So that would be kQq/2r for each atom, so the whole bond energy would be twice that. Thus, kQq/r. So I guess we were saying the same thing!

when you integrate the function shown you get $\frac{Qq}{4\pi\epsilon_0r}$ which equals the bond energy when evaluated between the limits shown. not sure where your thing about energies for each atom is coming from?

That is the integral of the force on one ion. There are two. So you have to double it. Or at least that's my understanding.

no this is the integral of the force between two charged objects with respect to distance which equals the work done in separating the two charges

Wait, you're totally right. I just did the integral incorrectly, and then used some mental gymnastics to fit it into what I knew the correct answer to be. The negative indefinite integral equals +kQq/r. Edit: I can't see LaTeX in comments on mobile, so I couldn't see what you wrote before.


text:
How many critical points? Why?

Critical points are defined (by wikipedia) as the "end-point of the phase equilibrium curve".
I would say this diagram therefore has three critical points. But this is unnatural. I doubt its true.
Other curves are different from the ones with critical points because "they can run to infinity". But I think thats ridiculous because its physically impossible.
I don't know if I'm interpreting PDs correctly but I think that what defines that you are on a phase in these diagrams is that if you travel along a horizontal line you meet with a phase equilibrium curve.
If you don't then you were either on a single phase the whole time or on a sort of "critical phase".
I don't like it though. For example if you travel along a horizontal line that wont intersect the L-V curve, how would you know when the LV critical phase is present? Regions are not clearly defined here, except for monoclinic.
Whats wrong with my ideas?
How many critical points does this diagram really have and why?
Thanks.

reg your example , i think the problem is a practical one, creating a large P and/or T is not easy and might be useless. So the curves shown usually have the portion thats useful for the study. 3 is the right amount of critical points in given curve. Also this becomes very obvious ,say in a binary PD .

This phase diagram has one critical point - it is where the line separating the liquid and gas comes to an end - it is now one fluid. So, you can go from liquid to gas (or vice versa) by swinging T,P around the critical point without undergoing a first order phase transition.
The points with arrows pointing to them are triple points, which are also special points, but different from the critical point. Movement away from a triple point means that one (or more) of the phases have to undergo a first order phase transition to the stable phase(s).
Thermodynamically, these are very different things.

Hey, but why are the other curves cut off too? Is it not supposed to be this way?

Sloppy drawing of the figure.  They extend to infinity and beyond.  Probably should have been drawn to the surrounding box, or have arrows on them.  Often the critical point is a filled circle big enough that it is obvious that the liquid/gas line is really supposed to end there.


text:
What are the best chemical liquids to clean metallic surfaces?
I'm just a simple man curious about things. I've no big understanding of chemistry. I had this question today in mind. What chemicals could I use to achieve the cleanest metallic surface?
For example, I could first start by using isopropyl alcohol to remove all dirt and oils. But there are other liquids such as acetone and acids.
If you were a chemist, how would you clean a metallic surface?

Well, what firstly comes to mind is that how "strong" the stain I'm dealing with is.

Let's say I scrape everything off first.

I'm going to assume clean as to kill pathogens

Sorry no - clean to make it thermally more conductive - clean from dirt, oils, rust, oxidation, etc.

Assuming clean means anti-pathogenic, I would use Hydrogen Peroxide "oxyclean" since it can "clean" metals by removing pathogens. $\ce{H2O2}$ is marketed in various strengths. 1% dilution can kill up to 99.99% bacteria in 30 seconds.
Any butyl based cleaning product is a very good degreaser but it may harm rubber products.
Check here and here for more.

Sorry no - clean to make it thermally more conductive - clean from dirt, oils, rust, oxidation, etc.

Thanks for the great return after like 3 months?

:) :) :) :) :) :)

Cleaning metal surfaces in industry is a very common problem.  Common enough, in fact, that an industry has grown up around the solution.  Halocarbons, either singly or in combination, are excellent at removing contaminants from metal surfaces (see the "uses" section in this link).  Typically a perforated wire basket is filled with the parts to be cleaned and then the basket is suspended above a bath of the halocarbon liquid.  The liquid is brought to reflux (typically less than 100 °C) and the vapors dissolve all impurities and the part comes out clean.  The dirty vapors condense, the impurities eventually precipitate out and clean vapor is ready to repeat the cleaning step.  The vapors are contained in the unit so atmospheric contamination is minimized. This approach works well for intricate parts or parts with hard to access cavities.
Simple ultrasonification, using either an organic solvent or water containing a cleaning agent, is another possible cleaning method.

Hmm, if like thinking of a heat-sink mounting surface for a processor IC I think you may be getting a bit over zealous.
The form of heat transfer does matter, if radiated, a dirty black surface is better, if using convection a rough finish might assist and if as you say conducted the intervening gap filler will dictate more than a surface contamination unless the surfaces are flat enough to atomically adhere.
An air gap between two rough clean surfaces will have poor conduction.
Soldering with suitable alloy or placing a conforming powder filled spacer or grease are the usual methods to maximise thermal conduction.  For this you would use one or two practical solvents to strip oils, an ionic cleaner to remove salts and mechanical scrubbing to remove oxides and entrained particles.


text:
Why does basic peroxide react via conjugate addition rather than through direct addition?
The following reaction occurs:

The $\ce{HOO^{-}}$ ion reacts as a nucleophile at carbon number three and consequently the reaction is conjugate addition. However, Why would a peroxide not react through direct addition. I know that if the double bond wasn't there it would attack the carbonyl directly and then a rearrangement would occur (Baeyer-Villiger reaction: http://en.wikipedia.org/wiki/Baeyer%E2%80%93Villiger_oxidation). Why must it react at the alkene?

I have no way to prove that, but i think there is a chance that the ketone forms a hydrate, deprotonated or not, blocking the most electrophilic platoon. Hence there is only the beta carbon left.

The reason that is given by Clayden et al. (chapter 23) is that because the peroxide has two adjecant lone pair on oxygen it exhibits an alpha effect, which is more or less a delocalization interaction between the lone pair orbitals. For reasons not entirely known (the basis of the alpha effect is an open problem in chemistry) this results in an increase in energy of the electrons in the resulting delocalized HOMO, making it more nucleophilic. You can imagine that this delocalized HOMO is also softer than the single localized oxygen lone pair and hence it tends to react at the soft double bond instead of the hard carbonyl site.
Excerpt from Clayden et al.

Your answer is contradictory. When the peroxide ion is more nucleophilic, why wouldn't it primarily react at the carbon with the highest positive charge? I understand that the alpha effect is not yet elucidated, but how and in what way does Clayden explain the stereoselectivity based on increased nucleophilicity, while it reacts at the very opposite centre?

@Martin Nucleophilicity has nothing to do with it, I just mentioned that because it is at the core of the alpha effect. I attached the excerpt from the book. Also, why would a better nucleophile attack more easily at the carbonyl group? I don't see the relation. Iodine is a good nucleophile, yet  rather soft. A bit harsh to immediately downvote without giving me time to comment. I'm just trying to help and answer questions and I think I used an OK source. But ok, it's your vote...

A downvote is just a downvote, it is nothing personal. Also your source is of course ok. The orbital picture, however, I do not agree with. It is a massive stretch to describe any oxygen orbital as sp3, and then there is the problem, that the depicted HOMO has no pi symmetry. I also would not be too sure about that the pi orbital is the HOMO. But it is the weekend now and i don't want to think about chemistry anymore.

@Martin No problem. It was a bit late and I was a little agitated and then seeing those stingy red -2 :) . Efin, there is only so much I can do with an undergraduate text which doesn't cover MO theory thoroughly. This isn't the first time you come and complain about some MO theory effect that I do not understand haha. Could you direct me to some books that cover MO theory more extensively? Anyway, when I have some time next week I'll ask Clayden about it, see what he thinks.

I really think the Clayden is an excellent organic chemistry book, I like it a lot. I know that he is more thorough with MO theory than most other books, and when it comes to the alpha effect, it is absolutely understandable that things get fuzzy. We have a brilliant question on the network reminding us about the very discourse we have with the whole matter: http://chemistry.stackexchange.com/q/7460/4945 I also removed my vote, the context you posted certainly helped a lot with understanding what is actually meant. (I still don't quite agree, but let's save that for another time.)

Why would a peroxide not react through direct addition […]

Nobody says that it doesn't, but apparently that isn't a productive route: there are no products found that would result from such an attack. Consequently, this reaction, which cannot be ruled out, probably is reversible.

I know that if the double bond wasn't there it would attack the carbonyl directly and then a rearrangement would occur  […]

True indeed, but the double bond is there.

As a result, attack on the $\beta$-carbon of the enone, formation of a hydroperoxy enolate and subsequent formation of the epoxide are seemingly favoured.

What chemical drawing program do you use? I really like the drawing style.

@Jori I use BKChem under Linux (Ubuntu 14.04), installed from the repositories. In the settings, I enabled automatic cropping of the SVG drawings. For Chemistry.SE, I export to transparent PNG (Cairo engine).

You might want to switch to a non-transparent background, I have experienced that the image hosting platform of SE likes to misinterpret the alpha channel and making the background black instead.

@Martin For small line drawing like the one in my answer, imgur seems to work flawless and does not alter the file. A diff on the hexdump of the original png and the file on imgur shows no difference. It is however true that larger files are compressed and/or converted to jpg.

First off, it may be useful to go back and take a look at this earlier question of yours and the answer to it. The concept of hard and soft electrophiles and nucleophiles was discussed.  Basically, it was said that

hard nucleophiles are generally small, highly charged atoms. Because
  of this they tend to be more reactive. They prefer to react with
  similar (hard) electrophiles. Their reactions are usually kinetically
  controlled with early transition states and are governed by
  electrostatic interactions. Soft nucleophiles are generally larger
  systems with a more diffuse charge. Because of this they tend to be
  less reactive (more discriminating). They prefer to react with similar
  (soft) electrophiles

The "take away" is that a hard electrophile prefers to react with a hard nucleophile while a soft electrophile prefers to react with a soft nucleophile.
Let's examine the system at hand, the nucleophile is the hydroperoxide anion ($\ce{HOO^{-}}$).  You can draw two resonance structures to describe the anion, delocalizing the negative charge over both oxygens.
$$\ce{H-O-O^{-} <-> H^{+}~ [O-O]^{-2}}$$
Therefore, this anion is a softer nucleophile than $\ce{HO^{-}}$
The carbonyl group is a hard electrophile, while an $\ce{\alpha,\beta}$ unsaturated carbonyl, where the electron density is distributed over 4 atoms is a softer electrophile.
Soft reacts with soft, so would you expect the soft hydroperoxide anion to preferrentially react with a hard (carbonyl; 1,2 addition) or soft ($\ce{\alpha,\beta}$ unsaturated carbonyl; 1,4 addition) electrophile?

@Jori  I've never really understood the basis of the alpha effect, so if you'd like to write up an answer based on it please go ahead.