text:
UV-Visible Spectroscopy in the analysis of sodium chloride in potato chips
Here is the question and answer out of an exam paper:

Firstly, I thought UV-Visible can also use radiation in the visible spectrum. Also when analyzing sodium chloride (a molecule), then UV-Visible would be more appropriate than AAS, because AAS would be used for just Na (Sodium). Would I be wrong to have said UV-Visible?

NaCl solution as you know is almost transparent in the VIS region, UV-VIS spectroscopy can be used to determinate concentration of soluble salt see J. Phys. Chem. A 2008, 112, 2242-2247 however UV part of the spectra is used instead of VIS part, so this does not fit the restrictions of your question.
Surely AAS has a greater sensibility and so is the best technique to use in this case.

how would i know what "type" of compounds are transparent in the VIS region, for UV-VIS spectroscopy?
Or is this type of question just expecting you to say AAS on the basis that UV-VIS may use the UV region of the emission of the spectrum... ?

@confused yes it's a legitimate question, you can't know it from the structure if you don't have a Gaussian install in your brain. The rationale behind this question is that you are searching for 1 technique that suite for elemental analysis: AAS has a greater sensibility and you know for sure that can detect sodium.  So however if for hypothesis NaCl could have a VIS spectra AAS is a better choice.


text:
Change in pressure or volume of molten salt
I am doing a project in which I fill a cylinder with a volume of $636cm^2$ (made of stainless steel) with molten salt: 60% Sodium Nitrate ($NaNO_3$) and 40% Potassium Nitrate($KNO_3$).
The container might reach (worst case scenario) -35 degrees Celsius in a cold winter night and in the summer (with ideal conditions) up to 700 degrees Celsius.
I'm trying to find out if I should add a relief valve (like in old water boilers) or just seal it tight. This of course depends o how much the pressure increases between those 2 states.
I know that molten salt  melts at 221 degrees Celcius and starts  freezing when it reaches 238 degrees Celcius.

I can't find any relevant data, so I don't know how I could find the result for this. Any ideas or suggestions?
EDIT: My worry is that it would blow up due to pressure or the lid would blow open...

If you are worried about pressure building up, it may be more important to know what the container is made of. Anything that can withstand temperature range you posted can probably withstand the pressure change the occurs with the phase change at fixed volume (unless the phase change is concomitant with a massive density change).

Good point. I updated the question

Venting initiates from blockage - anywhere.  All your lines must be hot-traced.
Nitrate melts are powerful oxidizing agents.  If containment fails onto organics (e.g., wood) you've got a catastrophe.  Nitrate melts significantly shrink on cooling and solidifying, and they bond to polar surfaces (glass, metal).  If you have  a half-liter of solidified salt and heat the bottom of containment, the melt will expand against the plug of solid salt above it, possibly deforming or rupturing containment.  The very hot and oxidizing liquid shoots out.  Heat a stripe up the side to start.
Nitrate melts thermally decompose.  Even if the average temp is OK, heating hot spots may give you oxidizing, corrosive, and toxic by inhalation nitrogen oxides.
http://link.springer.com/article/10.1361%2F10599490417542
"Corrosion of stainless steels and carbon steel by molten mixtures of commercial nitrate salts"
http://prod.sandia.gov/techlib/access-control.cgi/2001/018518.pdf
http://www.electrochem.org/dl/ma/197/pdfs/0209.pdf
The articles say your hot end is out of the safety zone.  Welds.  Chloride.  Chloride corrosion, especially pitting, can be aggressive. Halide in general may destabilize nitrate.  Ammonium nitrate is rather tame.  Ammonium nitrate plus chloride contamination is Texas City, Texas 16 April 1947. 2300 tons.
http://www.energystoragenews.com/Molten%20Nitrate%20Salt%20for%20Solar%20Energy%20Storage.html
1600 tonnes of molten nitrate eutectic.
You might put it behind grout-filled cement block or cinder block with #5 rebar weaving vertical and horizontal, and into the foundation, when you build the wall.  Have a large area vent direction.

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

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Zinc Picolinate - Positive, negative or neutral charge?
I'm going to be using a finings agent with a positive charge to remove negative trub from a solution (beer).
I would like to keep some chelated zinc in the solution - ideally zinc picolinate - so wondering what its charge is? I'm hoping it's neutral...
Depending on the answer, are there any zinc chelates that carry a neutral charge?
It's all so zinc can remain in the beer after fining to reduce the estrogenic effects of hops.

Zinc picolinate with two ligands per metal ion ($\ce{Zn(C6H4NO2)2}$) is a neutral chelate complex. Deprotonated picolinic acid has one negative charge at the carboxylate group ($\ce{NC5H4-COO^-}$). Two picolinate ligands thus compensate the two positive charges of $\ce{Zn^2+}$, forming a neutral complex in which they act as bidentate ligands. The nitrogen of the pyridine ring and the carboxylate coordinate the Zn center ion.

Image source: http://www.lookchem.com/Zinc-picolinate/

Could you expand a little bit on your answer, giving the reasons for why the complex is neutral?

the complex is neutral because of the 2 negatively charged oxygens and Zn^2+. Thus while 2 oxygens are negative and the zinc is positive, the molecule itself is overall, neutral.

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


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Why can temperature changes have units of either degrees Celsius or kelvin?
One equation to calculate a heat change is $q = c_p \cdot m \cdot\Delta T$. I know that $\Delta T$ can be in either degrees Celsius ($^\circ{}\mathrm{C}$) or kelvin ($\mathrm{K}$). However, one thing that confuses me is how the units cancel if $\Delta T$ is in degrees Celsius, since $c_p$ has units of $\mathrm{J\ g^{-1}\ K^{-1}}$. How does this work?

Welcome to Chemistry.SE! While in this case it is easy to grasp what you're asking, please include the definitions of all the quantities you are using in formulas in your future questions since this makes it easier for other people to understand your question and answer it.

The temperature in kelvin is equal to the temperature in degrees Celsius plus 273.
So, $\Delta T$ will be the same whether the temperature was reported in $\mathrm{K}$ or $^\circ\mathrm{C}$.  As an example if $T_\mathrm{initial} = 0\ ^\circ\mathrm{C}\ (273\ \mathrm{K})$ and $T_\mathrm{final} = 100\ ^\circ\mathrm{C}\ (373\ \mathrm{K})$, $\Delta T$ is 100 degrees no matter whether you used $^\circ\mathrm{C}$ or $\mathrm{K}$ in your computation.

Alright now I would want the person who flagged this as VLQ enlighten me of the issue this answer has. This answer is what the OP looked for!

Elaborating Ron’s answer through a concrete example, we start by looking at the Celsius scale
$$ T_\text{C1} = 25\ \mathrm{^\circ C} ,~ T_\text{C2} = 100\ \mathrm{^\circ C}$$
Thus, we obtain
$$ \Delta T_\text{C} = T_\text{C2} - T_\text{C1} = 100\ \mathrm{^\circ C} - 25\ \mathrm{^\circ C} = 75\ \mathrm{^\circ C} $$
In a similar fashion, using the Kelvin scale
$$ T_\text{K1} = 298.15\ \mathrm K = (273.15 + 25)\ \mathrm{K},~ T_\text{K2} = 373.15 = (273.15 + 100)\ \mathrm{K}$$
Thus, we obtain
$$ \Delta T_\text{K} = T_\text{K2} - T_\text{K1} = (273.15 + 100)\ \mathrm{K} - (273.15 + 25)\ \mathrm{K} = 75\ \mathrm{K} $$
As you can see, the constant offset of 273.15 between the Celsius scale and the Kelvin scale cancels, and the difference is exactly the same. The units remain, but the numerical value of the difference is the same.

Or, to put it another way - when using temperature differences $\Delta T$, the degree Celsius and the kelvin are equivalent units since $1\ ^\circ \text{C}$ and $1\ \text{K}$ represent the same temperature interval. $1\ ^\circ \text{C} \equiv 1\ \text{K}$

Good clearification!

If your specific heat has units of degrees Celsius in it then use the temperatures in degrees Celsius to find $\Delta T$. Although, the difference would be exactly the same as if you used kelvin as 1 degree difference is the same on both scales. Nevertheless, I would discourage using degrees Celsius as the SI unit is kelvin and is the one that's normally used.

Note that both the kelvin and the degree Celsius are SI units.


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Calculate accelerating voltage required to separate two particles in a magnetic sector analyser
So, in mass spectrometry we have the following equation to describe a magnetic sector analyser,
$$\frac{m}{z}=\frac{B^2r^2e}{2V}.$$
Lets say that I have two particles with different mass/charge ratios,
$$\frac{m_1}{z_1} = 120.9,\quad \frac{m_2}{z_2} = 121.0,$$
and I have $B=2\text{T}$, $r=25\text{cm}$. For example, what accelerating voltage $V$ do I need to apply to separate these two particles? What criteria do I have to determine whether two particles are separated?
** ** EDIT ** **
I don't know where to start with this. So even though you don't generally answer homework-like questions on this forum, some pointers in the right direction would be greatly appreciated!

We don't answer homework like questions. You are required to attempt to find the solution and ask where you need help.

Okay, fair enough. But would it be acceptable to give some guidance on the problem so I can know where to start?

You are given B, r, and m/z. The unknowns are $e$ and V. Isn't $e$ a constant? Which means you can solve for V.

Separation involves somehow the use of a slit, with a slit width, so the question is probably misformulated. You could consider the infinitely thin slit limit (i.e. find V for both m/z values), but it will not mean that they are separated, as all the dispersion factors would have to be taken into account.

Baseline separating C-12 from C-13 from C-14 requires accelerator mass spectrometry,
http://en.wikipedia.org/wiki/Accelerator_mass_spectrometry
Your example requires 100 times the finesse.  I'm not optimistic for reduction to practice vs. a paper solution.  Your target could be a charge-coupled device or a channel plate amplifier.  Observe the spatial separation of signals above noise.  A Faraday cup works differently.  Time-of-flight is not optimistic.
You might diddle FT ion cyclotron resonance in a high-Q cavity.


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How to calculate m/e value for the base peak in mass spectroscopy?

My working out:
Propan-1-ol : CH3CH2CH2OH
This can be broken up into:

CH3 + CH2CH2OH
CH3CH2 + CH2OH
CH3CH2CH2 + OH

Base peak = most abundant fragment formed (highest peak)
How do i know which fragment is the "most abundant?"
After I know the fragment structure, i can work out the molar mass, and then deduce the m/e value on the assumption of a (+1 charge).
Here is the exam paper solution:

The answer is right in front of you. What is the total mass of propan-1-ol? What is the mass of (propan-1-ol -18)?

molar mass of propan-1-ol = 60.10 ??

Now subtract 18.

42 ? why am i subtracting 18 ??

You are subtracting the mass of water from the alcohol. Water is likely to be lost when the molecule is fragmented.

that is 42, not 31. How does this give me the value of the base peak - most abundant fragment??

42 is the mass of a possible fragment. Since it is not one of the choices, we go to the next possible fragmentation. The next fragmentation is 31 and 29. Of these 2, which is more likely to have the positive charge?

how do you know which could have the  charge? I thought both were equal possibilities.

They should be, but it appears that one is more likely. If we look on the web for the mass spectra of propan-1-ol (or 1-propanol), we will see what is correct.

I did a search for mass spectra of 1-propanol and found that the base ion has indeed a $m/e$ value of 31. My guess is that the molecular ion loses an ethyl radical, which would give the cation $\ce{H2C=OH^+}$ ($m/e = 31$)

I would say that the question is not quite well formulated and that the answer that you quote is even further misleading in giving the proper answer.
Finding the answer
To start, let's consider fragmentation of primary alcohols, which is the question at hands here. First of all, can we rule out any of these answers?

Propan-1-ol undergoes complete fragmentation in a mass spectrometer.

This sentence means that we can rule out the molecular ion $M^{\bullet +}$ as a base peak in the spectrum. Propan-1-ol has a monoisotopic peak at $m/z$ 60 so it rules out answer D.
Now we apply the standard fragmentation rules for alcohols: consider ionization on the oxygen lone pair, we have two possible $\alpha$-cleavage leading to $m/z$ 31 and $m/z$ 59. An additionnal side reaction could also be hydrogen transfer to the oxygen atom, followed by water loss. This would lead to an ion at $m/z$ 42.

In the options above, should we thus completely rule out $m/z$ 29? I think that in the spirit of those who asked the exam question, the answer is yes, and we should rule out answer A. But see below...
We are now left with two options which can be explained by the fragmentation chemistry. Which should we choose? Technically, the problem here is that now we rely on abundances and not on the simple occurence of these fragments anymore. This is much more tricky, as there are no absolute rules. If we have a look at good textbooks, we find that loss of the largest radical is favored. Therefore, we have here to compare loss of $H^{\bullet}$ vs loss of $CH_{3}CH_{2}^{\bullet}$. Thus the answer is B.
Is the above answer complete?
The problem in such a question (and in the answer that is given below) is that it assumes that in mass spectrometry one could correctly predict fragment ion abundances based on very simple rules. And, as I tell my students, currently, even if you were able to map out the complete potential energy surface of the fragmentation pathway and do some RRKM calculations on the transition states, you would probably be off by a factor of 2 from the experimental results in terms of abundances. So asking for predicting the base peak (i.e. the highest abundance peak in a spectrum) can be quite wrong when two fragmentation channels are quite close to each other.
Let's have a look at the experimental electron ionization spectrum. Well OK, the base peak is indeed $m/z$ 31, so we are not too bad.
But let's have a deeper look. Remember we crossed-out $m/z$ 29 as a possible base peak, and it happens to be the second most abundant peak in the spectrum, ahead of $m/z$ 42 and $m/z$ 59 which followed the basic rules!
$m/z$ 29 could be either $HCO^{+}$ or $C_{2}H_{5}^{+}$. The former could be a secondary fragmentation of $m/z$ 31 after $H_{2}$ loss, whereas the latter could be produced as a heterolytic bond cleavage of the C-C bond. (In fact it is the same bond elongation leading to $m/z$ 31, except that they electron remains on the opposite side.) Energetically, these two channels differ by the difference in ionization energy between the $C_{2}H_{5}^{\bullet}$ and $^{\bullet}CH_{2}OH$ radicals, i.e. 8.11 eV vs 7.56 eV. Although it amounts to almost 100 kJ/mol, with the fairly large energy distribution in electron ionization, this is still possible.
And for the second rule that we applied, is it always true that the largest radical if formed? This empirical rule works out quite well, even more so when the comparison is between a hydrogen radical (which is quite difficult to produce) and a carbon atom centered radical.
And how about the exam paper solution
Here I am very highly surprised by what they seem to affirm. OK as we just saw above, $m/z$ 31 is a quite characteristic peak of primary alcohols. And loss of water is also characteristic of alcohols having side chains longer than 3 carbons. But does it mean that these should be the base peaks?
Let's have a look at the series of small primary alcohols:
Alcohol              m/z 31        Water loss           Other base peak
Ethanol              100            10
Propan-1-ol          100            15
Butan-1-ol            95           100
2-methylpropan-1-ol   38            10                  m/z 43
Pentan-1-ol           38            50                  m/z 42

So I do not see the rational behind this answer. It starts failing as soon as the small alcohol sizes reaches four carbon atoms. Of course there is a rational behind each of these base peak (for instance, in the case of 2-methylpropan-1-ol, the ionization energy of the isopropyl radical is of 7.37 eV, i.e. lower than that of $^{\bullet}CH_{2}OH$ and thus it becomes the most abundant ion, with exactly the same bond cleaved), but you have to go through the mechanisms, the thermochemistry and sometimes even the transition states to get to the global picture.


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Calculating molar conductance of CaCl2
Ok, so the following question is given in my text book.

Of the molar conductance value of $\ce{Ca^2+}$ and $\ce{Cl^-}$ at infinite dilution are respectively $118.88\times10^{-4}$ and $77.33\times10^{-4}$ then that of $\ce{CaCl2}$ is  (all have same unit)

So what I did was I added the molar conductivity of $\ce{Ca^2+}$ and added $2\times$ molar conductivity of  $\ce{Cl^-}$. I got the answer. It was one of the options. So I basically wanted to know: Did I do it correctly and is my process correct? And if I did it wrong then what is the correct process?

Yes, the process is correct, but do you understand why?

As per my knowledge there is one mole of $Ca$ and 2 mole of $Cl$ in $CaCl_2$ that's why am i correct

@Akash I guess you were just lacking confidence about your answer when you had it solved all along.

You are calculating limiting molar conductivity

limiting molar conductivity  =  V+ A+ +
V- A-
V+  =  number of cations per formula unit of the
electrolyte
V-  =  number of anions per formula unit of the
electrolyte
A+ = molar conductivity of the cations
A- = molar conductivity of the anions

So , your process is correct.


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What was the misconception about the search of Philosopher's Stone and how does modern chemistry deals with this topic?
I just read about Alchemy. And it seems to me that many of their principles are closely related to chemistry.
So, here is my doubt - Why is not possible to make that alchemist stone called Philosopher's Stone? What was the misconception about the research of Philosopher's Stone and how does modern chemistry can prove, or at least consider, that actually it is a Dead-end research?
Has any modern chemist ever tried to reopen the research on this topic?

I don't think this is a serious question.  There are countless refutations of this on the net.  Turns out you can do it with a particle accelerator, but that's cheating.  If you can ask a constructive question about that in particular, I'll reopen.

This question appears to be off-topic because it is about pseudoscience.

@GM I completely agree with you. This was what I had thought many a times before asking the question. I hadn't expected this would be put on hold. Anyways, I would leave it to the community.

If empirical reality can be falsified by wishes, then the burden is upon the wisher to reduce to practice upon demand.  The OP is ordered to explicitly state an experiment.

@UncleAl ROFLing at ordered to explicitly state an experiment. I have no intentions of turning copper in gold man...

@AwalGarg however I think your title cold be better (it seems that Philosopher stone is supported by chemistry...) jonsca simply put it on hold to force you to improve  your question, in my opinion try to accomplish this task! :-) P.S. see this related question http://chemistry.stackexchange.com/questions/8455/can-we-goldify-metals/8478#8478

@GM,jonsca Edited. Better?

@AwalGarg I know what you mean, I voted to reopen your question, the problem is that there is no "principle" behind Philosopher's Stone and chemistry in fact "does't not support it", so your assumption are wrong, maybe for this reason jonsca close it, I think you could try something like this (check my english): What was the misconception behind the Philosopher's Stone and how does modern chemistry deals with this topic?and rearrange your question in this way, because actually it seems that you believe the philosoper's stone really existed.

@GM I am pleased to know you understand what I mean. Thank you for your vote. Umm, would you mind editing the question accordingly. I do not have much experience at Chemistry.SE and thus, I think you can better convey the means...

@GM Thanks for the edit. I myself now feel more comfortable with it...

@AwalGarg you're welcome, thanks for the patience I hope you understand jonsca simply try to allow you to pose the question in the best way, you know I really like the question from the beginning, now it sounds more serious. I will try to contact jonsca.

@GM Yes, I understand jonsca. I respect each and every community member, no matter what there actions be on anyone's posts... We are all here to help flourish this site.

I had almost forgot about this question. I wonder why is it not yet reopened. It seems completely perfect now...

I think the alchemy tag is inappropriate for this site and that is why I removed it from the question. Alchemy is a part of the [tag:history-of-chemistry] and therefore this tag is more than appropriate. Also this is the only question with this tag and it is also flagged as off-topic.

@AwalGarg I agree with Martin, I don't think that tag is appropriate.  Please don't edit it back in again or I will have to lock the post.

@jonsca Ok... Sorry, Martin I didn't know about that.


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How is Humic substance destroyed in nature?
Humic substance consists of many, to my knowledge poorly understood, species of organic matter, mostly carbohydrates chained in intersting (and also poorly understood) ways. Humic substance is an endproduct of digestion of other organic matter, and is in itself known to be hardly biodegradable. Industrial processes to destroy humic substance or humic acid in wastewater involve UV and strong oxidizers. But how is humic substance destroyed in nature? Given that at least part of the carbon embedded in life ends up as humic substance at one point or other, there must be a natural pathway to destroy it, or several. What is it?

Sunlight is aggressive, as is microbiology.  Ozone, hydroxyl radicals (including Fenton chemistries with peroxides), air pollution in general eventually degrade and destroy near anything organic.  Examine sediments downstream of rivers, including outflows, especially for heavily colored waters.  Sampling vs. linear distance and depth will give you histories of degradative chemistries and biologies.  Note that phenolics are overall good microbiocides re lignin.
Terra preta soils demonstrate the survivability of charcoal.  Humic substances are more functionalized and more reactive.  One can start at the other end, too - weathering of coal.  Coal contains dispersed iron pyrite and other sulfide minerals that speed degradation.  Powdered coal heaps can self-ignite from cumulative oxidation heat.  Carbon dioxide, water, and elemental nitrogen are deep thermodynamic holes.


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How to determine source of air pollution
It should be determined by some metal markers (meaning originating from traffic etc.) but I cannot find any source which would describe it.
Thanks.

Hi, Can you be more specific? What do you want to do? What is your budget? What type of pollution?

The main idea is the following: you will need multiple sampling points and record enough data to obtain a representative dataset. Then when drawing wind roses with high percentiles (let say P95 or above) you will be able to easily point out an unique source of a relevant and exclusive pollutant.
But of course, reality is much complex... And there are some adverse effects that you must address before pointing out the source with a reasonable confidence.
First you better be not so close to the source, in order to measure immission instead of emission. Then environment and local conditions will average out. And wind speed, wind direction and concentrations become meaningful. In this case high percentiles likely originate from your source.
Second, your analyte must be relevant and if possible exclusive from your source. Measuring CO2 is not a good choice when monitoring traffic (it is stable, widespread thus global and not exclusively produced by traffic), NO is a better one because it is local, reactive and strongly related to high compression ratio thermal engine combustion source.
Third, if there are multiple isotropic sources, you will get nothing useful using wind roses. And your problem becomes really tricky.

All catalytic converters faintly emit heavy metal-containing microparticles continuously detached from the catalytic wash bonded upon their ceramic matrix.  Correlate your pollutant(s) with those particles and you establish proximity (though not causality).
Isotopic labeling of emissions (C-13 in fuel, O-17,18 in air) then expensively establishes causality.  You could use conceivably use inert taggants like $\ce{CF4}$, $\ce{SF6}$, $\ce{NF3}$ injected into the exhaust stream, but then Enviro-whiners will attack (Greenhouse Effect, Global Warming, Climate Change).  There is no postulated loss so small that social advocacy will not demand vastly greater real world losses to possibly prevent it.
http://www.ncbi.nlm.nih.gov/books/NBK98071/
Particulates


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endothermic dissolution process
Can anybody give me an example of an endothermic dissolution process, preferably one in which name of the substances involved are easy to remember. I have searched the Web thoroughly but could not find an example, only definitions. Thanks in advance.

An example of an endothermic dissolution is dissolving potassium iodide in water.
Added: Here is a table of Enthalpy of Solution for various compounds. Since this is the enthalpy of solution, some compounds will release energy when dissolve when the table indicates it would require energy to dissolve, and vice versa. If the number is negative, it is likely to be exothermic, and positive number are likely to be endothermic.

If $\Delta_{\text{sol}}H$ is negative, it is exothermic. If it is positive, it is endothermic. Both kinds of reactions can be overall exergonic (spontanious), endergonic (hindered) or in equilibrium.


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Instantly separate oil and water
I have a solution of water and oil mixed to as much as possible, since they won't actually mix. The supply is coming down a pipe at a relatively fast rate and I need to separate it so that the water goes down a pipe and the oil goes through another pipe.
How would this be done, without waiting days for it to occur?

If you mix oil and water really well, you get something that looks like mayonnaise. It could take years to separate.

Maybe if you add a highly polar salt in large quantities it will help break the emulsion? Dump a bunch of sodium chloride in the mixture under gentle stirring and see if it separates faster. You could also possibly selectively filter out one of the liquids by forcing the mixture through a highly polar or non-polar membrane with tiny pore size, but I don't think it would work perfectly in a single stage. Mere heating also promotes the separation of emulsions, I believe.

Dairy farming: cream separator.

Graphene oxide

I'll just point out that for any industrial type application there are bounds to be laws about the amount of contamination in the waste water that would be acceptable. You'd probably be required to do some sort of sampling to insure that the process stays in control too.

Just poking around and I recently heard the freezing an oil & water mixture is a simple way to separate the two. The separation is not immediate so I'm leaving this as a comment instead of an answer. Cheers!

Oil-water separation is a big money industrial process, and there are a number of methods for successfully separating the two. So the answer to your question is best approached in two parts: a theoretical answer, and a practical answer.
Theoretically, yes. A number of systems exist and the best are described below. There are off-the-shelf applications, as well as major capital infrastructure for major water treatment plants. Whether these solutions are for you will depend on your application, as well as budget. Hence, the practical answer may be different if you are looking for a cheap, backyard-employable device to solve the problem created by Granny filling the radiator through the sump oil inlet again.
Wikipedia has a very nice summary of water treatment, and includes a lot of information about the API (American Petroleum Institute) oil-water separator, which has been in effective use for decades. It is a very effective device, albeit costly, and can treat upwards of 50000 litres of water per hour. You can read about it in detail here, but I don't doubt it is what you are after.
Something more in the lines of what you describe is the Vertical Gravity separator. It is a pressurized system which promotes separation through convection mixing as the liquid is pushed though a complex path. The differences in liquid densities encourages separation of the two components. VGS systems can treat up to 3000L per hour. A document describing the operation of this system can be found on a link to a pdf document on this page. There are some problems associated with this type of device, in that the 'filter' mesh can get blocked if the material is very dirty.
Of course, any treated water is not 100% pure, and may need additional treatment depending on your requirements.

There are a number of interesting web resources where you can read more about oil-water separation methods. Pielkenrood have an excellent page of various methods, past and present. This blog also has a number of useful pages, although it is a commercial site promoting their own preferred devices.

Well then, it turns out there was significantly more depth to this question than I had first assumed. Thanks for this answer.

Membrane technology can certainly separate immiscible solvents from water. One commercial (lab) solution is from Zaiput, but other technologies are available.  Another option could be centrifugal separation. If you look up the type BXP by Rousselet Robatel it will tell you what's possible and it mentions oil-water mixtures.  They do a small lab one which scales to their process ones. Again, other suppliers are available.


text:
Given the volumes: determine the pH and the final temperature of a mixture knowing only the initial pH and the temperature of the un-mixed components
I have a pretty basic question but the last time I took Chemistry was in 2007.  I am studying the Navy's Nuclear study guide for their interviews and one of the question I am faced with is below.

Determine the final pH and temperature when the following two solutions are mixed together in a $\pu{3L}$ container.
Sol A: $\pu{2L}$, pH = 3, and $\pu{80 ^\circ F}$
Sol B: $\pu{1L}$, pH = 5, and $\pu{40 ^\circ F}$

Would this be as simple as $(2/3) \cdot 3 + (1/3)\cdot5$ for the pH and $(2/3)\cdot80 + (1/3)\cdot40$ for the temperature?

For the temperature, this method of calculation will give you a rough estimate of the final temperature, but it will not be exactly right. The reason is that mixing two solutions comes at an energy cost/gain due to the interaction of the two solution molecules. For some mixtures this enthalpy of mixing yields a temperature rise (i.e. energy is released from the molecular potential energy), for others it yields a temperature decrease (i.e. energy is taken up to 'make the mixing possible'). For many mixtures this difference is not more than a few degrees, which makes your linear estimate reasonable, but it is good to keep in mind that it will not be exact.
For the pH, the shortcut you are taking is incorrect and will at best give you a very crude estimate. The reason is that the pH scale is logarithmic, it is correlated with the concentration of protons in the solution as $\ce{pH}=-\log_{10}\; [\ce{H+}] $, where $[\ce{H+}]$ is the proton concentration. (I will make the assumption here that you have simple liquids, which are not buffered.) Therefore, you would first need to calculate the proton concentration in both solutions from this equation, yielding $10^{-3}\,\ce{M}$ and $10^{-5}\,\ce{M}$ for sols A and B resp. This means that the amount of $\ce{H+}$ in sol A is $2\cdot10^{-3}\,\ce{mol}$ and that in sol B $10^{-5}\,\ce{mol}$. Mixing the two you get 3 liter of solution with $2.01\cdot10^{-3}\,\ce{mol}$ protons, which is $6.7\cdot10^{-4}\,\ce{M}$ which works out to $\ce{pH=}3.17$. Indeed, very far off from the 3.67 predicted in your linear method.
Short discussion on the pH: because the scale is logarithmic and your 2 $\ce{pH}$ values are actually not that far from each other the linear estimate is still ok-ish. If, for example, you would have used $\ce{pH}$ 0 and 7 you would have found $\ce{pH}=0.17$ vs. $2.33$ from your estimate.

The temperature would be correct. And the pH would be correct for this exercise, but not in reality. Since most solutions contain weak acid and base, it is difficult to determine the pH unless you knew which buffers and quantities are in the system.

Can you explain what you mean by knowing which buffers and quantities are in the system?

The pH portion of the answer is not correct. Dilution of species is linear in concentrations, but the pH scale is logarithmic in concentration. For example, 1 L of non-buffered pH 3 solution + 9 L of pure water (pH 7 solution) makes 10 L of pH 4 solution.

Does this answer give you enough of an idea of how to calculate pH after mixing?

@NicolauSakerNeto that is too complicated.  I need something simpler as an example.

@LDC3 - your answer for the temperature would actually have to be the same as that for pH: correct in this exercise, but not in reality, because in reality you have to account for the enthalpy of mixing which could change the temperature of the final mixture by some degrees

@Michiel If both of the solutions are mixtures of water, the enthalpy of mixing would be minor. But since it is not stated, there could be a significant enthalpy of mixing.

@LDC3 - I agree, with water mixtures it wouldn't be significant, but solutions with a pH don't necessarily have to be mainly water.

@michiel The pH is defined as $\text{pH}=-lg[a(\ce{H+})]$ in aqueous solution (meaning in first approximation it behaves like water). You can extent the concept towards non aqueous solutions, but then you would face some other difficulties as well.

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text:
pH and materials selction
I working on the Navy study guide for their nuclear engineering programs and I am not a Chemist.  Thus, I have come here to try and develop a better understanding of the subject matter.
Why is pH important in materials selection?

The pH scale is merely a way of keeping track of the concentration a species (the hydrogen ion $\ce{H^+}$, $pH = -log\  a_{H^+} \simeq -log\  [H^+]$) in a medium (usually liquid, not necessarily aqueous). You can define arbitrary scales for other species, too. Analytical chemists are more familiar with scales such as pCl, with $pCl = -log\  a_{Cl^-} \simeq -log\  [Cl^-]$, and that's fine, but they're much rarer. The reason chemistry pays so much attention to hydrogen ions specifically is because the pH scale keeps track of what is arguably the most reactive species in all of chemistry (one could make a case for alpha particles, i.e. bare helium-4 nuclei $\ce{^4_2 He^{2+}}$ instead, but hydrogen ions are nevertheless comparable in strength and far more common).
If you stop to think about it, a (non-solvated) hydrogen ion, $\ce{H^+_{(g)}}$ is a bare proton. Let me explain why this is an important realization. In chemistry, we talk a lot about how electrons get shoved around when different molecules meet (also known as chemical reactions), and the exact way in the electrons shuffle about is dependent on complicated electromagnetism at the quantum level. However, to a simple but enlightening approximation, one can imagine how substances behave based on how much electrical charge they concentrate in a region; simply, concentrated charges will display stronger electromagnetic effects.
In most of chemistry, we deal with charges of order unity relative to the fundamental electrical charge ($e=1.6022 \times 10^{-19}\ C$) spread in the volume of a few atoms (of order $10^{-3}\  nm^3$). The resulting charge density can be found by dividing the charge by the volume. Now here's the thing - the nucleus of an atom is tiny compared to the size of the entire atom (with its electrons). A lone hydrogen atom has a radius of approximately $0.1\ nm$, but a lone hydrogen nucleus has a radius of about $1\ fm = 0.000001\ nm$, a difference of five orders of magnitude. The volume difference is thus fifteen orders of magnitude. Therefore, the bare nucleus which comprises a $\ce{H^+}$ ion has a charge density 15 orders of magnitude greater than a similarly-charged atomic ion, for example $\ce{Li^+}$. In effect, this means that a bare $\ce{H^+}$ has such a huge positive charge density that it will tear opposite charges (electrons) from absolutely anything. Helium and neon are notoriously unreactive, but both react quite happily (and rather exothermically) with gaseous $\ce{H^+}$ forming hydrohelium(+) and hydroneon(+) as the respective products.
So why doesn't water (or any ready source of $\ce{H^+}$ ions, i.e. protic substance) immediately dissolve absolutely everything it touches? It so happens that, in condensed phases, whatever molecules close to a hydrogen ion are so attracted that they effectively encase the hydrogen ions into solvated hydrogen ions, forming a shell which protects everything else from the aggressive reactivity of the hydrogen ion. For example, in water, $\ce{H^+}$ is actually not a very correct description of reality. Rather, one finds $\ce{H(H_2O)_{n}^+}$ (often alluded to by formulae such as $\ce{H3O^+}$, $\ce{H5O2^+}$, $\ce{H9O4^+}$ and others), or more concisely (but less distinguishably) $\ce{H^+_{(aq)}}$. After being surrounded by water molecules, the hydrogen ion loses almost all of its reactivity, as the original charge is now effectively dispersed over several molecules, greatly dimishing the charge density of the solvated hydrogen ion compared to the non-solvated hydrogen ion. Even after the massive handicap, solvated hydrogen ions are still quite reactive. Thus, it is important to keep track of their amount, to figure out if certain materials are capable of withstanding the onslaught.
After all I said, it seems like it would be a good idea to minimize the amount of $\ce{H^+}$ in water, which can be done by going to high values in the pH scale. However, by doing so you begin to significantly increase the amount of hydroxide ions $\ce{OH^-}$ present, in order to obey the autodissociation reaction equilibrium of water. In some sense, hydrated $\ce{OH^-}$ ions in solution are about as reactive as solvated $\ce{H^+}$ (even though non-solvated $\ce{H^+}$ is much, much more reactive than non-solvated $\ce{OH^-}$). Thus, going to either end of the pH scale can be a bad idea. One could say that water solutions are least reactive* when their pH is close to 7 (including pure water), as one minimizes the sum of $\ce{H^+}$ and $\ce{OH^-}$ present.
* Least reactive in the specific sense that if a material resists acidic aqueous solutions or basic aqueous solutions, then it must resist pure water and neutral aqueous solutions, assuming that whatever else anything else in the water incapable of altering the pH value is also unreactive; a pH 7 sodium chloride solution for example can be quite corrosive to metals, but not because of the water itself.

A few links that should shed some light:
Effect of pH on the corrosion behaviour of high nitrogen stainless steel in chloride medium
 i.e. ~seawater, the introduction should prove manageable even if some of the rest of the article gets overly technical.
Wikipedia also has general articles on corrosion and pH that may provide some extra background, though I would suggest just using them to find key terms to combine in google searches to try and find an introductory level university course, they are typically more focused in their explanations.

Metals are overall are attacked by acid, especially plus air.  Aluminum and zinc are attacked by both acid and base.  Cooling and heating water, and especially boiler water, must be delicately adjusted for pH and solutes to avoid chemical and galvanic corrosion of metal containment.
Seawater is nasty stuff. Seawater plus air is nastier.  Acid seawater plus air is very bad indeed.  How big is your ship's paint locker?
One could build things from Inconel 686 or Hasteloy C-2000. They would not corrode, not even in hot Green Death, hot Yellow Death, or any variation of seawater. Alas, they would be expensive, very heavy (dense), and allergenic to some people (the nickel).


text:
Difference between an absorbance, emission, and excitation spectrum for uv-visible spectroscopy

From the list below, circle the wavelength that would be the best setting for the UV-visible spectrophotometer used to determine the FITC concentration in the sample.
\begin{array} {}
\mathrm{wavelengths:} & \pu{400 nm}& \pu{450 nm}& \pu{500 nm}&\pu{550 nm} &\pu{600 nm}
\end{array}

The exam paper answer is $\pu{500 nm}$, because we need the wavelength of "maximum absorbance." But this graph shows two curves, and the absorbance curve is omitted.
But what is the difference between an emission, absorbance, and an excitation spectrum? What are they measuring? How do you interpret/read these graphs - those with multiple curves.

When you have an excitation spectrum and an emission spectrum, it is for fluorescence spectroscopy. To obtain these spectra, you hold either the emission wavelength or the excitation wavelength and scan the other to gather the information. The excitation spectrum shows at what wavelengths the solution uses to produce its fluorescence. The emission spectrum shows what wavelengths are given off from the solution.
For absorbency spectrum, the solution has all the wavelengths shining on the solution, and you monitor the wavelengths of interest. The absorbency shows which wavelengths are absorbed by the solution.
The excitation spectrum has the same peaks as the absorbency spectrum; although the absorbency spectrum may have more peaks.

i see... so in this case id be reading the excitation spectrum to find maximum absorbance?

It would be a good indication. There might be another peak in the absorbency spectrum that is greater, but you would have to have the spectrum to tell.

The question is bogus! Fluorescence in quantitative analysis uses the intensity of the emitted light as a measure of amount. The lack of light at 500nm is a filter cut off not a sample absorbance.

This explains why the emission wavelength is always longer than the excitation wavelength. Absorbance wavelength has the same value as excitation wavelength.

The energy gap of absorption is greater than that of emission. The gradient in between is termed stoke's shift. This because when the excited electron looses energy, it first gets to zero point level of excited state, emits and hits the potential well of the ground state after which it looses energy again to retire to the zero point level of the ground state. Emission takes longer time than absorption

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text:
HF and water miscible?
According to Wikipedia, HF is miscible with water. But isn't hydrogen fluoride a gas at RT and a sensible pressure? What would happen if you just kept adding more HF gas to some water?

Aqueous hydrofluoric acid of commerce is 48 wt-% HF, which averages tight to hydronium fluoride, $\ce{H3O^+ F^-}$.  You can snug 100% HF as $\ce{3 HF}$ being in equilibrium with $\ce{H2F^+ + {FHF}^{−}}$.  However, there is no concentration of HF that will separate into two liquid phases or a solid phase with solution at ambient temperature and pressure.
Hydrofluoric acid is exceptionally aggressive and lethal toward tissue by contact.  Effects can be delayed but not reversed.  Topical or infused calcium gluconate can ameliorate further damage.

What do "averages tight" and "snug" mean in this context? Also, does this really answer his question?

@Cohen_the_Librarian The answer gives information on the dissociation of hydrogen fluoride in water, describes further species characteristic for HF (and not found for HCl and HBr) and provides a well-meant health warning for less-experienced participants. If this is not an answer, what would be a better answer you would give?

@KlausWarzecha The health warning is well appreciated. However, the vocabulary questions I raise are because those phrases are incomprehensible to me and Google does not provide any top listing with those words + "chemistry". As a result of that incomprehensibility, I doubted the OP's ability to understand either (perhaps too much of an assumption). The next reason for my comment is that the OP's question of "What would happen if you just kept adding HF?" is not clearly answered. Perhaps it should read: "The continuous addition of HF will..." (I don't know exactly how to answer at this point).

@Cohen_the_Librarian I see the product of both autoionization and homoassociation of HF at high concentrations (when remaining water becomes just a trace) clearly mentioned.

Uncle Al is a make believe James Joyce. HF in water is not a solution it is a reaction and cannot be treated as a molecular solution. Also the equilibria usually involve the gas phase. The health warnings are a nonsequitor.

But isn't hydrogen fluoride a gas at RT and a sensible pressure?

Actually, it depends on the room in question. $HF$ boils at 19.5 C, meaning one can check if it mixes well with chill water or at relatively hight pressure.

What would happen if you just kept adding more $HF$ gas to some water?

Learn about binary diagrams (thermodynamic chemistry) and  find one for $HF/H_2O $ system

HF is miscible with water. But isn't hydrogen fluoride a gas at RT and a sensible pressure? What would happen if you just kept adding more HF gas to some water?

HF vapors miscible in water due to Hydrogen bonding due to Which HF make azeotropic mixture with water.


text:
Empirical and Molecular formula: Homework
I'm working on a high school chemistry book and want to double check one of the homework assignments. The question is for students to write the empirical and molecular formula for 1,2-dichloroethane.
I can't find this online. The molecular formula is given as C2H4Cl2, but could this also be the empirical formula? I don't find any sites as listing the empirical formula as CH2Cl.

The molecular formula and empirical formula are right. As we known, empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.
If the compound molecular formula is C2H4Cl2, so we divide it with 2, 4, and 2 GCD / greatest common divisor. The GCD is 2, so the empirical formula is CH2Cl.


text:
Bonding, multiplicity and quantum chemistry for FeP2
I’m trying to do some quantum chemical calculations for the linear molecule $\ce{FeP2}$ in the gas phase as well as its crystal (orthorhombic symmetry like here). I am lead to believe that in both cases iron has an oxidation state of +6, and phosphorous has an oxidation state of −3. My questions are:

Are such oxidation states a reasonable assumption for this molecule?
Would the molecule normally be a singlet (multiplicity of one)?
What quantum chemistry methods would be appropriate for this system?

While I find your question interesting, I'd also very much like to see your thoughts on the questions you pose and where exactly you have problems.

P.S. Welcome to Chem.SE!

Are my problems not clear from the question? I am mainly concerned with the oxidation state and spin configuration that could be assigned to this system, and aslo what minimal method from quantum chemistry I should use to tackle it.

The reference you provided states that FeP2 has marcasite-like structure. Marcasite is FeS2 polymorf with S2 units and Fe(II), so marcasite-type FeP2 likely has Fe(IV) This is a high value, but reasonably common for Fe-Ni in sulfide neighborhood, so FeP2 may have said formal oxidation state (since P is close neighboor of S). I would be very hesitant to use it for any consideration, though, since P is not very electronegative, so the real charge of Fe is likely very low.

Your reference also specifically states that Fe has $d4$ configuration, it corresponds to +4 oxidation state.

Ah, sorry, I misread. You are conserned about the molecule. Did you tested triangle geometry? It may yet prefer this form. As for QC method, in my experience gaussian pbe/6-31g* is enough for qualitative calculations.

@permeakra I am pretty certain, that PBE is not enough for this system, not even qualitatively, especially when you try to find the correct spin state. A multi reference method is well advised. At least a hybrid functional it advisable in any case.

@Martin-マーチン For in-depth investigation with multiplicity search I'd prefer at least MP2 with triple-split basis and even it maybe not enough. For qualititative investigation in my experience involving d-elements, pbe works well enough and incresing basis abouve 6-31g* is pretty useless for this imprecise functional. I fully admit, I the systems I have experience with maybe not representative for this particular case.

@permeakra Unfortunately MP2 will fail horribly for this system, because small HOMO-LUMO gap can be expected (pretty common for the iron group). In other cases I would agree with you on the qualitative level, but whenever it comes to transition elements, one has to be a lot more careful. *sigh*

Admittedly it is vacuum phase, but are you certain about your oxidation state assignments?  "nearly temperature independent susceptibility of -(8.8±0.7)centerdot10-6SI (at room temperature)"
doi:10.1016/j.jpcs.2013.04.015
https://sciborg.uwaterloo.ca/~lfnazar/publications/Electrochem_Solid-State_Lett_6_8_A162-A165_2003.pdf
doi:10.1021/ja301173q
http://www.staff.uni-bayreuth.de/~bt220235/refs/HPR_29_235.pdf
doi:10.1088/0031-8949/4/3/010 (given)

No, I am not sure at all. That is why I would like to do some calculations, and also why I raised the question... any other suggestions?

You might write an orbital diagram for the elements and push   electron population into reasonable configurations. I'm happier with Fe(II). Elemental phosphorus is not a strong oxidizing agent. However, Vacuum phase has its own rules.

Can phosphorous have an oxidation state of -1 in such phosphides?


text:
Why there exist no void with co-ordination number 5
Here
      oh = octahedral
      td = tetrahedral
      CN = co-ordination number

My problem is that, that why it is not possible to have a void with CN = 5 ?
My reasoning is:
According to the definition of CN i.e. the number of nearest neighbors of a particle (here a void because we are talking about it).
In the following figure if we treat silver sphere as the void then we say that it is a td void.
But if we place a sphere over this td void then we generate a void whose distance from center of void to the center of all the 5 spheres is equal. And so making the CN = 5
So kindly help me in understanding that why a void with CN = 5 not possible.

Trigonal bipyramidal complexes are abundant.  However, if you want a periodic crystal, you must tesselate volume with translation - the unit cell of the crystal lattice.  A five-fold symmetric figure cannot do this, for the angles do not add up without gaps.  2-, 3-, and 6-fold symmetries are allowed, offering 230 unique 3-D space groups overall.
If you allow two kinds of unit cell, you can get approximate periodicity, as in quasicrystals and gas-water clathrates.  Both can have 5- or 10-fold symmetry overall and give sharp x-ray or neutron diffraction spots (the IUC definition of a crystal).

"2-, 3-, and 6-fold symmetries are allowed". Not 4-fold?


text:
Chemical Equation: Sulfuric acid and powdered sodium carbonate
I was asked to write a balanced chemical equation for the reaction between sulfuric acid and sodium carbonate. The balanced equation isn't the problem. The question asks to also show all the state symbols.
All the equations I find online have sodium carbonate as a solution, not a solid. I'm also aware there are two forms of the balanced equation:
$$\ce{H2SO4(aq) + Na2CO3(s) -> H2CO3 + Na2SO4 -> H2O + CO2 + Na2SO4}$$
So do I say the sodium sulfate is a solid or in solution?

I guess Carbon dioxide is not the problem? So ask yourself what would happen if you put Sodium sulphate in water. Also $\ce{H2CO3}$ is never stable in water.

Thanks, that helps :-) It just feels better to double check.

As best practice share always what's your hypothesis. What do you think: is a solid or goes in solution? For us will be easier to help you to gasp the concept...

Try to follow this reasoning. (mouse over for answer)
Which parameter is mandatory to determine if a substance precipitate or not?

 Solubility.

How does this parameter behave with sulphates?

 Many sulphates are soluble in water. Generally the sulphates become less soluble as you go down the group. So sodium sulphate is surely soluble, confirm with Wikipedia.

Additionally think about others parameters that can have some sort of influence in this to improve your understanding of chemistry. So which parameters can affect this behavior?

 In fact as best practice you should always take in account the temperature and the amount of substance, and if the solution is saturated the excess will precipitate.

What's the answer?

 (aq)

The answer depends on the reaction stoichiometry, reactant concentrations, and temperature.  As written and unless concentrated, $\ce{CO2}$ fizzes off and everything stays dissolved.  With excess acid you'll get sodium bisulfate.  Concentrated and cooled, you can precipitate sodium sulfate decahydrate (Glauber's salt), M.P. = $32.38 ~\rm^{\circ} C$.  That is an interesting M.P. standard and constant temperature bath during the phase transition.  There is also a heptahydrate.


text:
Animate molecular vibrations as multiple xyz coordinate files
I have a frequency calculation by Gaussian09. Now I would like to extract vibrational modes as a set of xyz coordinate files. I would like to manually recompute numerically the points along a transitional vector, since IRC calculations all fail. (I suspect it is no real saddle point, flat PES.)
Consider the formation of the Lewis-Acid-Base complex $\ce{H3N\bond{-}BF3}$ with a transition state given in $$\ce{NH3 + BF3 -> [H3N\cdots BF3]^{TS} -> H3N\bond{-}BF3}$$ where this procedure would equal a (rigid) scan of the $\ce{N\cdots B}$ bond. This would be easy to accomplish.
Things get more difficult considering the pseudo vibration of ammonia, pyramidal to planar to pyramidal. Here it is not only one variable that has to be changed. (With dummy atoms it is still manageable.)
If molecules get more complicated, the manual specification of the bond parameters is too complicated and a source of many errors.
In the days when some of the gui programs were not able to calculate the vibrations and showing them in that fashion it was common to produce a set of multiple xyz files, each of them a snapshot of the coordinates along a specified vibrational vector. (This is now done internally and you can look at fancy pictures.) One of those programs capable of doing so is XVibs. For various quantum chemistry packages they calculated the distorted geometries. Unfortunately that is now ancient. (It was possible with a Gaussian09 output.)
For Turbomole there is a tool distributed along called screwer.
Unfortunately I have no access to that program package and I also do not want to change all the computations so far done to this software.
Lordstryker already posted in the comments, that it might be done by hand with chemcraft. I tried that and it is not very satisfying. One produces too many errors.
Are there any other programs/scripts that might be able to produce these files?

mmm I don't know with Gaussian, but in the output there should be a part with xyz frequencies coordinates, however you can see the animations using Avogadro see this video.

@GM That is absolutely correct. It will give you all the numbers you need to visualize the vibration. I want to take snapshots of a mode and recompute the energies at every point.

I'm not sure I've understood right, you mean that you want perform the calculation in an non optimized geometry e.g. during a stretching mode of a molecule?

@GM yes, I want to (approximate) map the PES around a stationary point along calculated frequencies. Basically distorting the molecule along a mode. For a diatomic molecule, this would equal to scanning the bondlength around the equilibrium. For more atomic molecules this is not so easy to accomplish, as more coordinates (bond lengths and angles) have to be changed.

@Martin I know Chemcraft will animate vibrational modes and will display the corresponding XYZ coordinates.  I just checked and there isn't a way that I'm aware of to save a sequence of XYZ coordinates automatically.  One could do this by hand however.  As with any program, you'll want to be careful with how you scale the vibrations before you decide which xyz's to save.

@LordStryker Thanks for the hint. It is a bit complicated though, but I could make it work.

@Martin You can do with avogadro simply using the cartesian editor and plug in the values of the new inter-atoms distances during the vibration for each atoms. You can then export the new 3D non optimized geometry in .xyz format. Keep in mind that Gauss optimize automatically the geometry so you have to disable this feature if you want to run a simulation in a non optimized molecule.

@GM that seems like an awful lot to change for a complex of roundabout 100 atoms. If it is not an (at least semi-)automatic solution then it is not useful to me. Manually changing the coordinates is what I want to avoid. [Gaussian does not optimized by default. A lot of interface (gui) programs, like Avogadro, specify the opt keyword when creating an input. I do it by hand.]

Ah yes I was wrong...I'm sorry. I've read "manually" in your question so I thought you are dealing with a small molecule...

As of today I decided to add an answer to this problem to declare it as unresolved. The answer of Klaus only answers the part of visualisation, which is not what I want or need. There are at least ten different programs that can do that.
While turbomole comes with the built-in functionality of screwer, other program packages might not have that. The mentioned XVibs is more than outdated.
The solution LordStryker added in the comments, doing it with the ChemCraft software is workable but highly ineffective and may introduce many errors.
I will certainly come back, whenever I find a program capable of this, but until further notice: *There is no such thing.

If I remember correctly, frequency data are in the stdout clusterfuck usually named log file, as well as in the (formatted) checkpoint files. There shouldn't be any need to prepare a cube file for this.
If GaussView is not available, GM's comment has the answer: Avogadro can visualize vibrations and I'm sure I've done this. Unfortunately, I don't have an output file at hand here.

I do not want to visualize, I want to create xyz files of the vibration and recompute energies statically at these geometries giving me an approximate PES around the saddle point.

If you are performing these computations for non-commercial purposes, one program package you might consider is ORCA (free registration required).  Binaries are available for Mac, Win and Linux.  Check out, e.g., Sections 5.9.3.5, 6.14.8.8, and 6.14.8.9 in the current manual (available at the above link) to see if the 'orca_pltvib' helper module or the 'mode trajectory run' (MTR) features will do what you're looking for.
'orca_pltvib' generates 'canned' OpenBabel XYZ multi-geometry animations for one or more vibrational modes of a Hessian. It may not provide sufficient configurability for your needs, though, as it only creates animations for single modes, and the displacement sizes are fixed.
The MTR feature is a lot more flexible.  It is meant to compute single-point energies at all of the points in a given mode scan (which can scan multiple modes at once) but also spits out a multi-structure XYZ file for each point, Once the TS geometry and modes are calculated, if you don't really care about the energies you can probably run the MTR at a very low level of theory, and just grab the geometries.  One possible hitch with scans over multiple modes is that it only generates the XYZ files for pairs of mode displacements, and each XYZ file only contains an animation of displacements along a single mode (see p406 in the current manual for a more complete explanation).

Thanks for the suggestion, I might have a look in it. Unfortunately I am quite limited in my choice of quantum chemistry programs. But I might be able to switch for a different project.

If all you're doing is generating these animations, it's something that could readily be performed on a personal computer (if, say, you're limited to programs available on a multi-user cluster).  The big challenge there might be getting the initial Hessian in ORCA's format, if your reference structure and Hessian are from, e.g., a highly correlated level of theory. ORCA has a lot of methods implemented, but some will be prohibitively slow on a personal machine.

@Martin-マーチン -- Just recalled QMForge -- not sure if it does multiple-mode animations, though.

Thanks for the additional input. I have lost track of this entirely, I will have to really take the time to investigate the matter again.


text:
Laws of Chemical Combination
I'm doing revision questions and want to double check something. The opening information reads:

Two chlorides of iron were prepared. One was prepared by reacting iron with dry chlorine
  gas. 4.50 g of iron reacted with chlorine gas to produce 13.01 g of the chloride.
  The other chloride was prepared by reacting iron with dry hydrogen chloride; 2.80 g of iron reacted with hydrogen chloride to produce 6.35 g of the chloride.

The questions are then:

(i) Calculate the mass of iron that reacted with 2.80 g of iron in each chloride.
  (ii) Find the ratio of the different masses of chlorine that combined with 2.80 g of iron.
  (iii) Which of the chemical combination laws is demonstrated using this data?

Is that first question right? I would've thought they're asking to calculate the mass of chlorine, which you'd simply get by subtracting the mass of reactant iron from the mass of the end product.

Two chlorides of iron were prepared.

I do not read that as: $\ce{FeCl_{n}}$ was prepared by two different methods, but as:

Reacting iron with dry chlorine gas yields $\ce{FeCl_{n}}$
Reacting iron with hydrogen chloride yields $\ce{FeCl_{m}}$

Apparently, iron is oxidized in both cases.

What are the oxidants (what is reduced)?
Which of them is the stronger oxidant?
What is the stoichiometry (what are n and m)?

Could it be helpful to divide the amount of chloride produced by the amount of iron used? Would different numbers indicate which of the products has a higher chloride content?

(i) Calculate the mass of iron that reacted with 2.80 g of iron in each chloride.

Is it possible that they meant the same as indicated above, i.e.:
Calculate the amount of iron chloride that would have been produced from 2.80 g (instead of 4.50 g) iron in the first experiment and compare with the outcome of the second experiment.


text:
Spectrochemical Series - Sigma Donor Capability
I fully understand and recognise the manner and which the series works.
I cannot find any references for the change in sigma donor capability moving up the series.
For example, comparing $\ce{Cl-}$ and $\ce{NH3}$. Which of the two would be more a more willing sigma donor ?
Does this have something to do with how diffuse the orbitals are ?

The spectrochemical series is an ordering of relative ligand interaction strength with metal ions. It can be used to infer relative $\sigma$-donor strength, if $\pi$-interactions are not possible for those particular ligands. Examples of this type ligand with no $\pi$-interactions include $\ce{NH3}$, $\ce{CH3-}$, and $\ce{H-}$.
Most ligands will exhibit $\pi$-interactions, operating as $\pi$-donors or $\pi$-acceptors, and these interactions dominate the ordering of the spectrochemical series. The $\sigma$-donor interactions are obscured and heavily affected by the $\pi$-interactions. Experimentally all of these effects are important, and it is difficult to separate these interactions from one another.
For example, the $\sigma$-donor interaction of a $\pi$-acceptor ligand is drastically increased from some initial level. $\pi$-backbonding reduces electron density at the metal center, leading to increased $\sigma$-donation (which increases $\pi$-backbonding, and so on). This will make a given $\pi$-acceptor ligand act as a better $\sigma$-donor than would be expected. The converse effect is observed with $\pi$-donor ligands reducing $\sigma$-donor strength.
The complexity of these interactions tempers the question of gauging $\sigma$-donor interactions only. The combined bonding effects influence one another.
Theoretically separating out $\sigma$-donor interactions from all $\pi$-interactions could be possible computationally, but I am unaware of any research on this specific topic. From a very general standpoint, improved metal-ligand orbital overlap will increase $\sigma$-donor strength, which may help estimate these characteristics qualitatively.
Most inorganic textbooks should be able to provide further detail on this topic under sections on ligand field theory.

can relative Bronsted basicity be used gauge sigma donor capabilities?

I would say that Lewis basicity would be a better approach to evaluating sigma-donation as this is more accurate in how ligands coordinate to metals. Gauging Lewis basicity as only sigma-donation generalizes somewhat the complete electronic structure. Nevertheless, this seems like a good way to determine approximate sigma-donation strength. Below is a link to a book that has more information on this, where it appears they have computational and thermodynamic values for complexation enthalpies of various Lewis bases to Lewis acids such as BF3: https://books.google.com/books?id=KsF44nB9Ex8C


text:
To model early fluid flow which types of pumps should be used?
I have I microchannel test rig , I want to push a 150-300 ml/hr from water in my test , I would like to know which pump type i can use ?


text:
Formula for hypo
Can someone please tell me the chemical formula for just "Hypo"?
This question came up in previous exams and I don't know what is the general /chemical formula for just "Hypo". I can find the formula for hypo chlorite and hypo bromide, but I am unable to find formula for just " Hypo"

Could it be that "hypo" is actually an abbreviation of something that is entirely unrelated to chemical formulae? For example, I think "hypo" can be used to refer to hypotonic solutions used in clinical rehydration.

@NicolauSakerNeto In this case, a hypotonic solution means below tonic solution. In essence, what I said about having one less oxygen in the ion complex.

@LDC3 As far as I understand, hypotonic rehydration solutions have nothing to do with hypo-oxyanions. Rather, they're just specially prepared aqueous solutions with total osmolarity lower than that of blood (so there is a net addition of water to the body), while containing essential ions which may be dangerously low in dehydrated patients, such as potassium. Providing water with an inadequate electrolyte balance can actually help kill a severely dehydrated person faster.

@NicolauSakerNeto You took my comment literally. What I am pointing out is that hypotonic (below tonic, lower salt content) is similar to the ion complexes where the prefix hypo- indicates less oxygen than the ion without the prefix. Sorry about the misunderstanding.

Back in the day when photographs were processed chemically rather than digitally, "hypo" was the chemical used to remove unexposed, undeveloped silver chloride, bromide and/or iodide from the emulsion to fix the image.  The chemical was a solution of sodium thiosulfate, $Na_2S_2O_3$.  The common name at the start of photography was sodium hyposulfate...

Right, sodium hyposulfate, $Na_2S_2O_3$ has one less oxygen than sodium sulfate, $Na_2SO_4$ (although it exchanged the oxygen for a sulfur).

I agree with @User58220 My teacher had told us to write formulas for some of the common names of compounds like - blue vitriol... and hypo was one of them which is sodium hyposulfate.

First, hypobromide does not exist, hypobromite does.
Hypo- is not a formula per see. It is an indicator that the oxide has one less oxygen than the -ite oxide.
Examples:

chlorite ==> $ClO_2^-$
hypochlorite ==> $ClO^-$
bromite ==> $BrO_2^-$
hypobromite ==> $BrO^-$

In chemistry, the prefixes hyper and hyper are often used to denote the number of oxygen atoms in oxoacids as compared to the normal oxoacid. The prefixes are taken from Greek.

hyper ($\upsilon\pi\epsilon\rho$) = above in the sense of more than or super
hypo ($\upsilon\pi o$) = below in the sense of less than

Consequently, there is no general formula for these prefixes.
The prefixes are used in medicine too, e.g. to refer to conditions resulting from vitamin levels above (hypervitaminosis) or below (hypovitaminosis) an average and healthy level.


text:
Examples of electronically conductive foods?
There are many foodstuffs that are ionically conductive (for example, anything moist and salty). What about foods that have a high electronic conductivity?
The only thing I can think of would be something which has been charred – presumably the char, being mostly carbon, would have a fairly high conductivity.

If it is charred until it conducts electricity - is it still food?

I wonder why nobody mentioned the famous vodka with gold flakes. Admittedly, this hardly qualifies as food, and the amount of gold is minuscule, but in the absence of any solid examples with true metallic character I think it may deserve a curious look.

Many foods conduct electricity. The charge carriers are ions, and the conductivity is a function of temperature and ion mobility. Liquids, therefore, are generally better conductors. The electrical properties of foods are important parameters to consider in food processing operations such as ohmic heating, microwave heating, pulsed field irradiation and induction heating. This has implications in a number of important areas, such as, the industrial microwave pasteurisation of egg-whites, and high temperature short exposure sterilisation of certain food products. A quick search of the Thomson Reuters Web Of Science database returns over 350 hits for "electrical conductivity of foods". Indeed there are whole book chapters dedicated to the subject. See for example this and this.
So, what about some numbers then?
Conductivity for some general materials at 20 °C are given on this page in Wikipedia. (All values in S/m)

Copper  $6 \times 10^7$
Sea Water $4.8$
Drinking Water $10^{-4}$ to $10^{-2}$
Deionized water $10^{-6}$
Glass $10^{-15}$
Carbon $10^3$
The following values for foods have been taken from a variety of published literature sources. Again units are S/m. Most temperatures are at or near 20 °C, where I can find the value stated.
Beer 0.1
Coffee 0.18
Coffee with milk 0.36
Orange juice 0.36
Apple Juice 0.24
Cranberry Juice 0.09
Milk 0.5
Carrot Juice 1.3
Tomato juice 1.7
Beef 0.26
Chicken 0.19
Tomato paste 0.35-0.82
Gatorade 0.2

My question is specifically about electronic conductivity, not ionic conductivity. That is, I'm wondering about foods where charge transport is mediated by electrons rather than ions.

I found this question when contemplating this recent news article about 3d printing and Vegemite, a naturally electrically conductive spread for bread that is consumed in Australia.
http://munchies.vice.com/articles/3d-printed-vegemite-could-be-the-future-of-edible-electronics

Char need not be complete.  It need only be sufficient to form a conducting network, and so is a legitimate answer for electronic conductivity of food.  An organic conductor generally pi-stacks and preferably has unpaired spins.  Look up the conductivity of melanins (if you like eating long pig or naked goat).  DNA is a pi-stack!
http://www.drproctor.com/os/amorphous.htm
https://www1.ethz.ch/lbb/Education/Biosensors/DNA_conductivity.pdf
http://mike.zwolak.org/publications/DNA_Electronics.pdf
http://physicsworld.com/cws/article/news/2007/dec/17/conduction-seen-in-dna-backbone
http://www.physics.ucla.edu/research/biophysics/news/pdf/outlook.pdf
http://www.ceesdekker.net/files/PW_DNA.pdf

You may want to add that DNA's conductance is minimal in macroscopic sense, and therefore may not qualify as an answer.


text:
Could someone please explain the difference between m/z and m/e in mass spectroscopy
The definition:

The abbreviation m/z is used to denote the dimensionless quantity formed by dividing the mass number of an ion by its charge number. It has long been called the mass-to-charge ratio although m is not the ionic mass nor is z a multiple or the elementary (electronic) charge, e. The abbreviation m/e is, therefore, not recommended. Thus, for example, for the ion $\ce{C7H7^2+}$, m/z equals 45.5.

Could someone please explain the definition? I found it a bit difficult to comprehend.

What is "ionic mass"?
What do they mean by "z a multiple or elementary charge"?
$\ce{C7H7^2+}$, m/z equals 45.5 -> what did they do here?

There are no differences simply the second notation is not recommended by IUPAC. The explanation is below:
What is ionic mass? The ionic mass is the real mass of the ion that takes in account the weight of the electrons in the overall weight. However electrons are much more lighter (0.548 mDa) then nuclei,  so according to IUPAC definition m is intend to be A:

Total number of heavy particles (protons and neutrons jointly called > nucleons) in the atomic nucleus. Also called nucleon number. Symbol  m
  in mass spectrometry.

And not the ionic mass (that in fact would more accurate).
What do they mean by "z a multiple or elementary charge"? This is another approximation they suppose that we could use an elementary charge +1, +2, +3, -1, -2, -3 etc. etc. that is not a real quantification in Coulumb of the electrical charge of the ion as stated Jannis Andreska.
$\ce{C7H7^2+}$, m/z equals 45.5 -> what did they do here? They simply divide m (see definition above) by z in this case (2+=2).

Addendum: This definition silently says that the m/z ratio is an approximation m is not the real weight of the ion (in fact is the weight only of the heavier particles and z is not the real charge of the ion) so it says: "don't use 'e' because this is an approximation we are not using the real charge nor the real weight, and the m/z ratio is is a dimensionless quantity!". I believe that in the future when even the lighter electrons, due the increasing accuracy of the instrument, will matter we will use $m_{i}/q_e$ where $m_i$ is the ionic mass and $q_e$ is the real charge... Only matter of time (and matter!)!
Reference here!

Thank you, but I have one last question - Normally in my exam papers they use m/z and m/e interchangeable. When the paper uses m/e is the calculation going to change? Because I normally simply divide the mass number by the elementary charge.

@confused yw no, is the same thing however the second notation is discourage because of the facts mentioned above.

I do not completely agree with your definition: in a mass spectrum, m should not refer to the total number of nucleons but really to the actual experimental mass (expressed in a.m.u.) of the molecule. Unfortunaltely, there are multiple definitions of the masses used in mass spec (see http://en.wikipedia.org/wiki/Mass_(mass_spectrometry)), and the notation m/z can apply to most of them. One should in addition to the m/z notation also indicate which mass definition is being used.

@PDL thanks a lot for your comment, I've rearranged the question to remove the misconception, please add your answer so the question could be more complete to the others users. More answers make this site better!

$m$ is the atomic mass of the ion (in units ($u$); $1u=1.665402\times10^{-27} \mathrm{kg}$), $z$ is the number of its positive/negative charges. $z$ should not be used in interchange with $e$ because it is dimensionless, while $e$ has the unit $\mathrm{C}$ (Coulomb). For $\ce{[C7H7]^2+}$, $m/z$ is given by (the unit of $m$ is omitted):
$$m(\ce{C7H7})/2=(7 \cdot 12+7 \cdot 1)/2=91/2=45.5$$

One should note that the IUPAC definition for $m/z$ has changed in the latest recommendation (Definitions of terms relating to mass spectrometry (IUPAC Recommendations 2013)) which helps clarifying the question asked above.
The question of what the labeling of the x-axis in a mass spectrum is supposed to represent is listed as the most prominent Controversial term in the introductory section of this definition document. Thus the question remains somewhat open. The definition of $m/z$ has slightly changed compared to the 1991 one cited in the above question:

Abbreviation representing the dimensionless quantity formed by dividing the ratio of the mass of an ion to the unified atomic mass unit, by its charge number (regardless of sign). The abbreviation is written in italicized lowercase letters with no spaces.

The major change from the 1991 definition (quoted in your question) is the replacement of the mass number by the actual ratio of mass to the unified atomic mass unit. This follows the deveolopment of high resolution instruments and of ESI, which means that the $m/z$ scale has become abundantly used for multiply charged ions and that one cannot approximate anymore the ion mass divided by $1 u$ as the total mass number of the atoms present in the ion.

+1 so the gold book definition seems obsolete, it depends from which definition the chemist choose...

m/z is chemical jargon that no longer complies with the modern rules of metrology. Therefore: just foreget it.
Instead follow this:

Mass spectrometers measure the quantity mass/charge.
The recomended symbol of mass is m
The recomended symbol of charge is Q
therefore, the best symbol of the derived quantity mass/charge is m/Q
the unit of mass that is usually used in mass spectrometry is the dalton Da
the unit of charge that is used in mass spectrometry is the atomic charge unit e
therefore the unit of m/Q is Da/e which is also referred to as a thomson Th

To go back to your questions:

e is not a symbol of charge, it is a unit of charge
ionic mass is again chemistry jargon. It is like saying "my humic mass is 80 kg". Nobody would do this. It should say "m is the mass of the ion ..."
the mass m(C7H7^2+) = 91 Da
the charge Q(C7H7^2+) = 2e
therefore m/Q(C7H7^2+) = (91 Da)/2e = 45.5 Da/e = 45.5 Th


text:
What kind of equilibrium constant we use for Gibbs free energy and Van't Hoff equation?
We know that the Gibbs free energy is related to the equilibrium constant by the following equation:
$$\Delta_\mathrm{r}G^\circ=-RT\ln K$$
We also know the Van't Hoff equation:
$$\ln\left(\frac{K_2}{K_1}\right)=\frac{-\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$
$\left(R = 8.314\ \mathrm{J~mol^{-1}~K^{-1}}\right)$
My question is: what kind of $K$ do we use for these equations? Concentration quotient ($K_c$) or pressure quotient ($K_p$)?

You use the equilibrium constant $K$ that is formulated in terms of activities. This are relations connecting this $K$ to $K_{c}$ and $K_{p}$ and in the limit of ideal behaviour $K = K_{c}$. Have a look at this answer of mine. It explains the derivation of $K$ and gives the relation to $K_{c}$ ($K_{p}$ is not included but I could add it if that would be helpful).

Second equation you posted looks like the Clausius-Clapyeron equation, not the van't Hoff equation.

I would disagree to the last comment, @Dissenter. As far as I can see, the posted equation is indeed the Van't Hoff equation

@Philipp As you use van't Hoff for determining the equilibrium for $\ce{aA + bB <=> cC + dD}$ shouldn't it be all the same, since you would use the same definition for $K^\circ_{T_1} = \frac{a^{c}(\ce{C})\cdot{}a^{d}(\ce{D})}{a^{a}(\ce{A})\cdot{}a^{b}(\ce{B})}$ and possible constants would cancel each other in the quotient?

@Martin I'm not 100 % sure but looking at the relation between $K$ and for example $K_{c}$, \begin{equation}
K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ ,
\end{equation}
I would say that the quotients $\frac{K(T_{1})}{K(T_{2})}$ and $\frac{K_{c}(T_{1})}{K_{c}(T_{2})}$ will be equal only for small temperature differences since $\left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}}$ contains $T$ directly and the fugacity $\varphi_{i}$ is temperature-dependent as well.

@Philipp Ahhh, of course, you are right. I forgot to think about that temperature dependency. I guess the temperature dependency of $\varphi$ is the least problem, ideal gas is far worse an approximation, but $\left(\frac{T_1}{T_2}\right)^{\sum\nu}$ might have a significant effect for all reactions where $\sum\nu\neq0$.

The first equation actually contains the definition of the standard equilibrium constant:
$$K^\circ = \exp\left\{\frac{−\Delta_r G^\circ}{R T}\right\}$$
With this definition the equilibrium constant is dimensionless.
Under standard conditions the van't Hoff equation is
$$\frac{\mathrm{d} \ln K^\circ}{\mathrm{d}T} = \frac{\Delta H^\circ}{R T^2},$$
and therefore uses the same constant. The integrated variant is therefore already an approximation and may be correct using a different definition of the equilibrium constant.
$$\ln \left( {\frac{{K_{T_2} }}{{K_{T_1} }}} \right) =
  \frac{{\ - \Delta H^\circ }}{R}
  \left( {\frac{1}{{T_2 }} - \frac{1}{{T_1 }}} \right)$$
Now the ordinary equilibrium constant may be defined in various forms:
$$K_x = \prod_B x_B^{\nu_B}.$$
Probably one of the best representations for the standard equilibrium constant involves relative activities, for an arbitrary reaction,
$$\ce{\nu_{A}A + \nu_{B}B -> \nu_{C}C + \nu_{D}D},$$
this resolves in
$$K^\circ = \frac{a^{\nu_{\ce{C}}}(\ce{C})\cdot{}a^{\nu_{\ce{D}}}(\ce{D})}{a^{\nu_{\ce{A}}}(\ce{A})\cdot{}a^{\nu_{\ce{B}}}(\ce{B})}.$$
The concentration is connected to the activity via
$$a(\ce{A})= \gamma_{c,\ce{A}}\cdot{}\frac{c(\ce{A})}{c^\circ},$$
where the standard concentration is $c^\circ = \pu{1 mol/L}$.
At reasonable concentrations it is therefore fair to assume that activities can be substituted by concentrations, as
$$\lim_{c(\ce{A})\to\pu{0 mol/L}}\left(\gamma_{c,\ce{A}}\right)=1.$$
See also a very detailed answer of Philipp.
The partial pressure is connected to the activity via
$$a(\ce{A}) =
\frac{f(\ce{A})}{p^{\circ}} = \phi_{\ce{A}} y_{\ce{A}} \frac{p}{p^{\circ}},$$
with the fugacity $f$ and the fugacity coefficient $\phi$ and the fraction occupied by the gas $y$, the total pressure $p$, as well as the standard pressure $p^\circ=\pu{1 bar}$ or traditional use of $p^\circ=\pu{1 atm}$.
For low pressures it is also fair to assume that you can rewrite the activity with the partial pressure $p(\ce{A})$, since
\begin{align}
\lim_{p\to\pu{0 bar}}\left(\phi_{\ce{A}}\right) &=1, &
p(\ce{A}) &= y_{\ce{A}}\cdot{}p.
\end{align}
Of course concentrations and partial pressures are connected via the ideal gas
\begin{aligned}
pV\ &=nRT\\
p\ &\propto \text{const.} \cdot c,
\end{aligned}
and therefore it is valid to write:
$$K_c\propto \text{const.} \cdot K_p.$$
It is important to note, that the two expressions are not necessarily being equal, and can scale by powers of $(\mathcal{R}T)^{\sum{}\nu}$.
While using these equations it is always necessary to keep in mind, that there are a lot of approximations involved, so it depends very much on what you are looking for. Either use might be fine, as all these functions are related - some might lead to simple, some may lead to complicated solutions.

Here is a simple and detailed explanation.
For a general reaction $aA+bB ⇌cC+dD$, $K_{eq}$ is defined as:
$$K_{eq} = \frac{(a_{\ce{C}})^c(a_{\ce{D}})^d}{(a_{\ce{A}})^a(a_{\ce{B}})^b}$$ Without going into detail, the activity "$a_X$" is basically :

magnitude of concentration in molarity(if solution)
magnitude of partial pressure in bar(if gas)
$1$ (if solid or pure liquid)

Observe, the repetitive use of "magnitude". This is to emphasize that activity is a dimensionless quantity. It trivially follows that $K_{eq}$ is also a dimensionless quantity.
The above expression for $K_{eq}$ holds for any general reaction, no matter how many or which phases are involved. For example, for $aA(s)+bB(aq)+cC(g)⇌dD(l)+eE(g)+fF(aq)+gG(aq)$, the value of $K_{eq}$ is:- $$K_{eq} = \frac{(P_{\ce{E}})^e[\ce{F}]^f[\ce{G}]^g}{[\ce{B}]^b(P_{\ce{C}})^c}$$
Note here that I have taken the magnitudes of partial pressures and concentrations.
So, this in itself answers your question as to whether to take $K_p$ or $K_c$. The answer is you take neither-you simply take $K_{eq}$ which can be calculated for any reaction as explained above. This way, you'll never need to worry about taking $K_p$ or $K_c$.
Addendum: So, what's all the fuss about $K_p$ and $K_c$? Well, it so happens that sometimes you may be given the value of $K_p$ and $K_c$ for a reaction which you may need to use. Although, this is never useful over simply giving $K_{eq}$ (in my experience), it happens to be the way things are done sometimes. So, it's worth knowing a few important things about $K_p$ and $K_c$. I'll proceed by breaking down the concept into a number of small points:

Like $K_{eq}$ , $K_p$ and $K_c$ are defined in terms of activities and are dimensionless.
$K_p$ is typically only used when the reactants and products have only solids, pure liquids and gases(i.e. no solutions). Thus, we can say that $K_p$ is another term for $K_{eq}$ for reactions not containing solutions.
$K_c$ can be used for any general reaction, including those with all of solids, pure liquids, gases and solutions. $K_c$ is defined as-

$$K_{c} = \frac{(a_{\ce{C}})^c(a_{\ce{D}})^d}{(a_{\ce{A}})^a(a_{\ce{B}})^b}$$ where "$a_X$"is :

magnitude of concentration in molarity(if solution )
magnitude of concentration in molarity(if gas)
$1$ (if solid or pure liquid)

Observe, unlike $K_{eq}$, here "$a_X$" for gases is not magnitude of partial pressure in bar and is instead the same definition as solutions. This is also why I didn't take $a_X$ to refer to activity here, because technically "magnitude of concentration in molarity" is not the activity of a gas. This "re-definition" has the effect that $K_c \neq K_{eq}$(in general).
Now, what do we mean by "magnitude of concentration in molarity" for a gas? Well, rewriting the ideal gas equation $P_XV=n_XRT$, we get $\frac{P_X}{RT}=\frac{n_X}{V}$=[X]. so "magnitude of concentration in molarity" is $=\frac{P_X}{RT}$. So, now instead of $a_X=P_X$(in $K_{eq}$), we have $a_X=\frac{P_X}{RT}$ (in $K_{c}$). For example, for $aA(s)+bB(aq)+cC(g)⇌dD(l)+eE(g)+fF(aq)+gG(aq)$,the value of $K_{c}$ is:-
$$K_{c} = \frac{(\frac{P_E}{RT})^e[F]^f[G]^g}{[B]^b(\frac{P_C}{RT})^c}=K_{eq}(RT)^{c-e}$$
Note here, I have taken the magnitudes of partial pressures and concentrations and also the magnitude of temperature.

We can extend the last result to $K_c=K_{eq}(RT)^{ - \Delta n_g}$ or $K_{eq}=K_c(RT)^{ \Delta n_g}$. Several sources cite that $K_p=K_c(RT)^{ \Delta n_g}$ which is true, but only when $K_p$ is valid i.e. for reactions not involving solutions. So, it doesn't show the whole picture. $K_{eq}=K_c(RT)^{ \Delta n_g}$ is true for any general reaction.We, can also see from this result that $K_c=K_{eq}$ only when $\Delta n_g=0$ which is usually seen in reactions only involving solutions but may also be seen in some gaseous reactions where the amount of gas is same on both sides.

Strictly speaking these equations should contain some correction of activity (concentration quotients) or non-ideal behaviour of gases (pressure quotients).
If you don't need to deal with those and you can treat your system as ideal, the concentration of a gas and pressure are equivalent. Be careful with the units, though!

Even with ideal gases, $K_p$ and $K_c$ are not always equivalent - they are related, but one might be scaled relative to the other by powers of $(RT)$, since $C=\frac{n}{V}=\frac{P}{RT}$.

Exactly. Kc does not necessarily equal Kp.


text:
Why are high pressures used in cracking of long-chain hydrocarbons?
If we have a long-chain hydrocarbon, such as decane, and we split it through thermal cracking (say in an industrial plant), we use high temperatures, and high pressures. Cracking produces smaller molecules - alkanes, and alkenes. It might look like this:
$$\ce{CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH3}$$
Which is cracked into products like this:
$$\ce{CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH3-> 2CH2=CH2 + C3H6 + C8H18}$$
Why do we use high pressure in this case? Le Chatelier's principle states that the equilibrium will shift to resist the change — in this case it would move the position of equilibrium to the left, not the right. In cracking we want maximum amount of products (ie. the right) — therefore surely a low pressure would shift the equilibrium that way.
The reaction is not dependent on molecules colliding — merely enough thermal energy has to be provided to enough molecules to cause them to split — homolytic fission — and become free radicals which react to form smaller molecules. Therefore a pressure requirement to increase the collisions doesn't seem necessary.
Why then do we use high pressure when cracking hydrocarbons?

First of all, cracking is not an equilibrium process, therefore La Chatelier-Brown principle doesn't apply! (irrelevant)
Also, the mechanism is much more complicated than this:

In gas phase, radical reactions generally have extremely complex reaction networks. E.g. burning of methane with oxygen contains thousands if not more elementary reactions. I am pretty sure that gas phase cracking is not a pure first order reaction.
In industry, cracking is generally done on heterogeneous catalyst, therefore not a gas phase reaction. Heterogeneous catalysis is pretty much controlled by the absorbed amount of molecules on the catalyst surface: high pressure, faster reactions.

If you were a refinery, you'd throughput hundreds of tonnes/day from one reactor.  To keep process volumes manageable, high gas pressures and fast throughputs.  If you thermally cracked long chain alkanes, you would need red heat and get coking.  If you did it over acid catalyst, typically zeolite whose pores define product by fit, you'd get coking.  The way to do it is to reform under high pressure and modestly high temperature over noble metal catalyst-doped zeolite plus added hydrogen.  Coking is then deeply suppressed - "Platformate."
Ethylene is better made otherwise; alpha-olefins overall are assembled catalytically.  You do not want straight-chain products.  You want a maximally branched ~$\ce{C8}$-cut for gasoline, with maximum octane number.  Linear ~$\ce{C10}$-cut and up is cheap diesel and kerosene, with maximum cetane number.
Government intervention tortures your access to liquid fuels.  The most recently built refinery is ancient.  Enviro-whinerism has carefully lowered the energy content/gallon.  "Sustainability" means you pay for it big time but don't have a right to receive it.  How is it bottom of the barrel diesel costs so much?  It's the law! written by idiots and crooks.

I think its worthwhile defining what coking is.

I suspect there may, in fact, be a good answer to this question hidden in your first paragraph above, and perhaps the second one too, if you just unpacked it enough so that even people who are not industrial chemists could understand it. As for the last paragraph, whether valid or not, it doesn't seem to have any relevance to the question (or to the topic of this site) whatsoever.

Chair parade. Given the answer, look up its sources. Pyrolyzing organics absent air produces coke (graphite) as by-product. This wastes carbon in light streams and ruins catalysts. Low value still bottoms are terminally pyrolyzed to recover lighter ends plus coke. Part (most?) of the National Petroleum Reserve is refinery waste purchased as Saudi light crude then disappeared in Louisiana solution-mined salt domes, re third paragraph.


text:
Product of treating 1,3‐dibromocyclobutane with alcoholic KOH at elevated temperatures

Passage-II

$$
\ce{
A ->[\text{1.}~\ce{NH2-}(\pu{2 mol})][\text{2.}~\ce{CH3-I}(\pu{2 mol})]
B ->[\ce{HgSO4}, \ce{H2SO4}][\ce{H2O}]
C ->[\text{dil.}~\ce{OH-}][Δ]
D [Mixture]
}
$$
31. "A" gives
A) Lucas test
  B) Silver mirror test
  C) White precipitate with ammoniacal $\ce{AgNO3}$
  D) Yellow precipitate with ammoniacal cuprous chloride
32. How many isomeric products are existed in the mixture "D"?
A) 2
  B) 4
  C) 6
  D) 8

What would be 'A' in the reaction? Any help would be deeply appreciated.
I know that alc. KOH does elimination, but I was confused whether it would eliminate one bromine each from two rings and create a bond between both the rings or it would simply eliminate both bromines from same ring and create two conjugated double bonds (cyclobut-1,3-diene).
EDIT: There is one more possibility: both the bromines get eliminated and a diagonal bond is formed in the ring (cyclobutane with any two diagonal carbons joined).

The two equivalents $\ce{HBr}$ are eliminated from two different molecules, one from the starting material and one from an intermediate. In the first step, $\ce{HBr}$ is eliminated under cleavage of the cyclobutane ring, which is favored by the release of ring strain. Elimination of $\ce{HBr}$ from 1-bromo-1,3-butadiene finally yields but-1-en-3-yne.

When 1,2-dibromocyclobutane is heated with KOH, 2 moles of acetylene are formed, probably via a cyclobutadiene intermediate. 1,3-dibromocyclobutane reacts differently.
Here, the final product is vinylacetylene (but-1-en-3-yne, $\ce{HC#C-CH=CH2})$.


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What are some good research topics for a high school student?
I absolutely love chemistry and I really want to take part in science fairs. I am also taking an AP chemistry class. Cleaner fuel and energy sources and air pollution (especially due to smog and vehicle exhaust) interest me. My initial idea was to make solar panels work along the lines of Chlorophyll, but that is a bit too advanced for my level. Where could I get inspiration for some science fair ideas or chemistry research topics (high school level)?

I would recommend asking professors at local colleges and universities. A lot of ISEF qualifiers and winners work with university professors on their research. Find topics that are interesting already being pursued, then reach out to them with your interest.

Green chemistry is a highly advanced topic. It ain't for the weak of heart, so it is probably not for below university level.
I personally favor pyrotechnics as relatively safe iff every bit of safety rules is strictly followed and stunningly spectacular topic that is pretty easy and cheap to do by own hands (most materials are inexpensive as well as most tools). However, it is extremely time-consuming and has questionable legal status in some countries (and for a good reason: safety may be easily compromised and potential for disaster is high).
Another interesting topic is fluorescence. Many compounds shine under UV light, and there are families of such compounds that are not exhaustively studied and yet are relatively easy to make. However, a deep study is required, but it is probably advanced school level.
With access to advanced materials/tools/labs more topics become available, but a full list of such tools is required to recommend anything.


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Is there a metal that melts and solidifies quickly?
Is there any kind of metal that melts very quickly with very low heat or temperature, and at the same time, becomes hard very quickly?

Mercury is liquid at room temperature and gallium melts in your hand: https://www.youtube.com/watch?v=HMekQGnnQjI . But this just addresses melting points, I'm not sure what you mean by 'quickly' - the heat dispersion rate? Solder works good for this, if you don't mind the 300 degrees or so it takes to melt it.

The melting points of some eutectic mixtures are tabulated in an article by French.1 Some options for low-melting point metals/mixtures are:

Gallium-indium eutectic, mp 15.7 °C (Sigma-Aldrich catalogue)
Galinstan, mp –19 °C (Wikipedia article)
Various gallium-based Indalloy® alloys, mp between 7.6 and 29.78 °C (corporate page)
A general search for "fusible alloys" brings up several sources with many choices:

Wikipedia page
Rotometals sales page
Belmont sales page
Bolton Metal sales page

Reference

French, S. J. Melting Points of Eutectics. Ind. Eng. Chem. 1936, 28 (1), 111–113. DOI: 10.1021/ie50313a026.

Some bismuth alloys have rather low melting points.
Wood's metal (50% bismuth, 25% lead, 12.5% cadmium and 12.5% tin) melts around 70 °C.
Field's metal (51% indium, 32.5% bismuth and 16.5% tin) melts around 62 °C.
An alloy of 44.7% bismuth, 22.6% lead, 19.1% indium, 8.3% tin and 5.3% cadmium melts around 47 °C.

There is a game, which is a tradition in Germany around New Year, to predict good fortune for the upcoming year: Molybdomancy (german - Bleigießen) While this is unscientific jibber jabber, there is actually some real physics involved.
Tin has a (very) low melting point ($\approx \pu{230 ^\circ C}$) which can easily be achieved with a candle. If you pour the molten tin into water, it will immediately become solid again. (This also works with lead $\approx \pu{330 ^\circ C}$, but that is very toxic.)


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Controlling the shape of metal
Is there a way to control the shape of metal? Suppose I have a fast melting metal, is there a way to make it take a specific shape? Of course I don't mean to like put it in a template to take the shape of a template, but rather to control the shape using a computer or something. Is something like this possible or done? If not possible, then why?

Molten or solid at the end?  It can be shaped in an induction field (liquid)  or dimagnetically levitated (liquid to solid).  Magnetoforming, hydroforming.  Spin molten into into an oblate ellipsoid (solid).
http://www.capturedlightning.com/frames/shrinkergallery.html
Z-pinch and such.
Munroe effect for jets and planes (angle charges), and self-forging munitions.