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Organic Chemistry..What is the order of importance when naming a hydrocarbon?
What is the order of importance when naming a hydrocarbon?
How do you know what becomes the suffix, and what becomes the prefix between the functional groups?

are you sure you are talking about hydrocarbons, when talking about functional groups? by the way, welcome to Chem.SE

maybe related?: Functional group naming order?

It's rather long story. Here is a relevant part of almost official transcript of IUPAC Organic chemistry recommendations: Guide to Name Construction, R-4.1 General Principles


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Melting point of obsidian
I have tried 4 experiments at various temperatures to melt obsidian.
At varying temperatures it just bubbles up and I don't seem to reach a melting point.
Do you know what gases obsidian might contain and would they be the cause of the bubbles?

Obsidian doesn't have a particular chemical formula. It is a mixture. So there is no definite melting point.

When you say that it "bubbles up" that sounds like melting to me, so I'm hoping you can clarify your meaning. Also, do you know at about what temperature you first see this bubbling, and also what was the highest temperature you used?

I have worked with casting glass(45%lead) in the past which flows into a given shape when melted at 900°c. So I tried melting the obsidian at a simular temp. It did nothing at 1000° but at 1020° it doubled in size due the bubbling and it was grey in colour.I held it for 4 hours hoping it would then flow. It didn't and stayed in its 'frozen' bubble form. Other firings at 1050° and 1100° formed larger bubbles and the colour remained black?

Obsidian
This is a volcanic glass, formed when rhyolitic lavas cool too quickly for crystals to form.  From a chemistry point of view, it's a mixture of silicon dioxide, aluminium oxide, with sodium ,potassium, calcium and iron oxides in various quantities. These lavas will also contain significant volatiles (water and $\ce{CO2}$) held in solution by pressure, although they should be lost at the pressures required for glass formation (higher pressures means slower cooling due to more insulation).
Melting
Note that when molten, rhyolite lavas can have a very high viscosity. So it's possible that you have melted your sample, but just didn't give it enough time to flow. Indeed, with substances like this you will see partial melting - some components will melt at a lower temperature than others. If you are getting bubbles, this would indicate that your glass formed under enough pressure to retain some $\ce{CO2}$ and/or $\ce{H2O}$ in its lattice, and you are now allowing this to be released by heating at surface pressure.  This is all a bit speculative, without a closer inspection.

Thanks Andrew for all your information. I think my next attempt will be to refire a piece that has bubbled at 1020°. I'm hoping that now co2 or h20 has been released it may flow?

Might need to be a bit hotter, the volatiles lower the melting point.

I'll add that the volatiles are probably not CO2, because it is lost way before the lava erupts to become obsidian. Most likely dominated by H2O, and even then it's possible it's being held up by minerals like micas and amphiboles, and not in the glass itself.

I cast using Obsidian (experimenting for 7yrs) an amazing material it has so many different reactions
depends on point of origin of material
Size of pieces
Speed of ramp temps
Hold times.
Temperature range 1000-1130c
You can make light large pieces that float or heavy dense work  my experience is to treat more like a ceramic material than glass when firing as very different to Gaffa glass ](https://i.stack.imgur.com/R1fI4.jpg)

Obsidian is mostly silicon dioxide (about 70%), with a good bit of aluminium oxide and then about 10-20% various other oxides.
Melting point for silicon dioxide is 1,710 °C, and for aluminium oxide 2,072 °C. You're going to need a lot of heat for this.

Welcome to Chemistry.se! While your answer is true, it does not fully address the question. It also does not offer much more than the already given answer. It would be nice if you could expand it a little more.

What you say is correct but likely irrelevant. Mixtures often have radically different melting points to their components so speculating about obsidian based on pure components isn't helpful.

Obsidian melts between 700 and 1050 deg C
http://www.swxrflab.net/blmpaper.pdf

I don't think that is what the linked paper says. Unless you mean something different by melting.

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Why is $\Delta G = -T \Delta S_{\mathrm{total}}$ valid only at constant pressure?
According to my book(Elements of Physical Chemistry by Atkins and de Paula, 5th ed.), $\Delta G = - T \Delta S_{\mathrm{total}}$ is valid only for constant pressure and temperature.
Enthalpy is defined as $H = U + PV$. So, when pressure and volume both are not constant, $\Delta H = \Delta U + \Delta (PV)$. So, enthalpy can be defined even if the pressure is not constant.
Now the criterion for spontaneity is that $\Delta S_{\mathrm{total}} > 0$. But $$\Delta S_{\mathrm{total}} = \Delta S_{\mathrm{sys}} - \frac{\Delta H}{T}$$ where $T$ is the temperature of the surroundings. Now, if any heat enters the surroundings, the temperature of the surroundings does not change provided that the surroundings are large. I think that due to this the temperature of the system has essentially to be the same or infinitesimally greater/lesser than the surrounding temperature as the heat transfer has to take place reversibly.
But now, as $G = H - TS_{\mathrm{sys}}$, $$\Delta G = \Delta H - T\Delta S_{\mathrm{sys}}.$$
Thus, from comparison of equations, $\Delta G = -T \Delta S_{\mathrm{total}}$. So, I did not need to specify that pressure is constant and why does my book do so?

You have actually assumed in your equations that both pressure and temperature are constant. Firstly, we have that
$$\mathrm{d}S_\text{total} = \mathrm{d}S_\text{system} + \mathrm{d}S_\text{surroundings}.$$
By definition, differential of entropy is
$$\mathrm{d}S = \frac{\delta q_\text{rev}}{T}.$$
From the first law of thermodynamics, and the definition of entalphy, for a reversible process
$$\mathrm{d}H = \mathrm{d}U + P\mathrm{d}V + V\mathrm{d}P= \delta q_\text{rev} + \delta w + P\mathrm{d}V + V\mathrm{d}P.$$
When all work done is to expand, then $\delta w = -PdV$, and thus
$$\mathrm{d}H = \delta q_\text{rev} -P\mathrm{d}V+ P\mathrm{d}V + V\mathrm{d}P = \delta q_\text{rev} + V\mathrm{d}P.$$
Only if pressure is constant, do we get the relationship
$$\mathrm{d}H \overset{\mathrm{d}P=0}{=} \delta q_\text{rev} \to \mathrm{d}S = \frac{\mathrm{d} H}{T}.$$
Therefore,
$$\mathrm{d}S_\text{total} = \mathrm{d}S_\text{system} - \frac{\mathrm{d} H}{T}\tag{constant pres., exp. work}.$$
Additionally for constant temperature, or $\mathrm{d}T = 0$,
$$\mathrm{d}G = \mathrm{d}H - S_\text{system}\mathrm{d}T - T\mathrm{d}S_\text{system} \overset{\mathrm{d}T = 0}{=} \mathrm{d}H - T\mathrm{d}S_\text{system}.\tag{constant temp.}$$
Hence the result $$\mathrm{d}G = -T\mathrm{d}S_\text{total}$$ is true when pressure and temperature are constant (in addition to possible other constraints).

It is probably not the best practice on my part to write things like
$$\mathrm{d}H = \delta q_\text{rev} $$
which has the impression of sloppy (but probably not wrong) notation. Interpret this as that, given our assumptions, the number $1$ becomes an integrating factor for $\delta q$, equivalent to the inexact differential becoming exact under special conditions.
A better option might be to write this expression with definite integrals, i.e.,
$$\int_{\text{state 1}}^{\text{state 2}} \delta q_\text{rev} = \int_{H_1}^{H_2}\mathrm{d}H.$$

sir is there a expression for $dS_{system}$

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What simple formal systems are there that have behavior similar to actual chemistry?
I've been doing reading about formal/axiomatic chemistry and am wondering: are there attempts at making formal systems (well defined rules that can be simulated on a computer) that have similar emergent behavior to real world chemistry in some way?
Clearly this would be very difficult for reproducing all of the physics of matter but I haven't been able to find any papers about attempts to make simple systems that have meaningful 3D interactions besides cellular automata which are very different.

Well, there's lots of stuff. I guess you're talking about macroscopic behaviour? CA is only one of ways and not so different from modelling for example with differential equations.

Yea I know at microscopic/quantum levels they've had to get more formal and there are some detailed things like this so yea I'm really asking about introductory chemistry type things you might study, things on a more macroscopic scale

The system is inherently chaotic and probabilistic. There is no way to model it in the way you would like to. But in the case of statistical mechanics, we compute and derive macroscopic quantities from microscopic ones in aggregate. This is probably as close as you can get.

You may be interested in searching for Synergetics by H. Haken roughly defined as emergent phenomena as found in non-equilibrium phase transitions and self organisation in physics , chemistry and biology.

This is a really great question, but it's unfortunately too broad to be a good fit to the format of the site.

@hBy2Py Where do you recommend asking it/how could it be tweaked to fit this format?

I don't know if would be possible to modify it to fit, unfortunately. Per the logic laid out on meta here, questions of the "I'm looking for a thing that is X" is intrinsically open-ended, leading to numerous possible valid answers and/or opinion-based discussion, rather than a targeted, closed-scope, definitive answer. It would probably be best raised on chat, either in the main room or in a new room dedicated to the topic.

While I see what you mean, this question has enough votes that signifies that people are interested in it, and chat doesn't provide a good resource for googlers or people that have similar questions. In my mind, if a question has enough interest and doesn't have somewhere it can be asked that more fits that environment, it should be allowed to stay because it will be useful to others that have similar questions since that is the point of the site.

I'm okay with turning this into, say, a community wiki though

The Lewis model (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3285721/) meets those criteria.

I went to the site of one of the authors of that paper and it provided a ton of great similar things

For the benefit of those who don't have access to the article, could you provide a summary of its contents? As relevant as it may be, on Stack Exchange we try not to encourage answers that only contain a link.


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Overall polarity of polychlorinated benzofuran
I have been struggling over this weekend to find an answer for the overall polarity of this molecule: 2,3,7,8-tetrachlorobenzofuran

In order to find intramolecular forces, I determined the difference in electronegativity:
C-C = 0 and C-H = 0.4 --> non-polar covalent bond
C-O = 0.8 --> polar covalent bond
C-Cl = 0.6 --> polar covalent bond
As the molecular shape is not symmetrical, and there are polar bonds present (unbalanced), I thought it is a polar molecule. However, the research suggests that it is "non-polar molecule, therefore insoluble in water".
Can I ask the reason why it is considered a non-polar molecule?

The molecule is symmetric, along the axis that "splits" the oxygen in the furan and the 2 carbons at the base of the pentagon the defines the furan. There's a permanent dipole moment, that's for sure. Whether or not it has been determined to be "soluble" is something that, experimentally, usually is defined by detection limits.

If you want to calculate the dipole accurately you need to have the atom's coordinates and add the dipole vectors along each bond (except CC as this dipole is zero).  You can see that these vectors will not add up to zero. By the reflection symmetry the x components do add to zero however. Alternatively you can use tracing paper and draw vectors along bonds in proportion to dipole length then you must  keep orientation & cut out & superimpose  base of each vector. Add up to find dipole; messy but possible :)

It has more nonpolar moieties than polar ones, and it has been empirically observed not to be soluble in water. That's a nonpolar molecule for you, relatively speaking.


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How do I make 3d molecular graphics similar to those shown on Wikipedia?
I want to make 3d molecular graphics similar to those found on Wikipedia. In the page for benzene, you can see the structure looks like this:

However, if I go to the interactive image as shown in here:

I get an interactive image that looks like this:

To me, the second image doesn't look nice, and I prefer the first one. However, I cant seem to find a way to get the image of first type for any molecule after converting from SMILES. I want to draw the structure of, say for example, Bismabenzene? How do I produce a graphical image for that if it's not available on Wikipedia?

PyMol comes to mind. There are a ton of freebies out there. In the day (and still) there's RasMol.

@ToddMinehardt Any thats online? I'm not really up for downloading software for now.

Maybe eMolecules? Don't take my word for any of it, I'm old and not as informed as others on this site.

@ToddMinehardt C'mon, to me you seem to be quite up to date! Anyway,  just in case you need one: Here's a brotherly hug from a fellow chemist who has celebrated his 20th birthday more than once too ;-)

The person who made the benzene image (and many other molecules) has instructions for how they do it on their user page, if that helps: https://en.wikipedia.org/wiki/User:Benjah-bmm27/MakingMolecules

@KlausWarzecha - :)

A little odd representation like the second one may happen if the structure originally was drawn and saved as *.mol file with explicit hydrogen atoms.  If not switched off, Jmol will then compute the bonds automatically which however may be disabled (edit ->preferences -> bond).  Yet in favour of Jmol, the same index card has two rulers defining bond tolerance and minimal bond distance, which offers quite some advantage (compared to CCDC Mercury, for example) in the display of crystal structures with positionally disordered chains.

If you are by chance running a Linux based OS, there is a good chance your distro has Jmol in it's repos. It's large (~150M), but I had it downloaded, installed and running in about a minute. I like it. It takes common names, chemical names, SMILES, InChI and CAS, and of course create your own. And it is totally free (product and license) and guaranteed no adds or crapware included. This is the only *mol 3D type of product I've tried (I'm old too), so I can't say if it's better or worse than others.

@PrittBalagopal It looks like it's just a demographic survey on the linked Discovery Studio Visualizer download page, rather than a limitation on who can download. I'd recommend something like Industy:Other JobFunction:Research JobRole:Student CompanyName: -- or whatever you feel is "closest" to your actual situation. Try for the best match, but don't worry too much if it's not exact.

@R.M. Oh, that's nice, I'll try it.

Have a look at Molcalc. It's fully online and uses JSmol as a visualiser.
Alternatively you can browse the Pitt Quantum Repository for structures that have already been calculated.

I want it to look like the one on Wikipedia

@Pritt If you want them to look like the ones on Wikipedia, I suggest you follow the link Carmeister has posted. You'll need to download and install programs though.

You can generate the geometry for your molecule of choice by drawing it in Avogadro. More often than not, you want to refine the geometry through a force field calculation.
Avogadro supports multiple ways of visualization, which all can be configured according to your preferences. A ball-and-stick model, which you seem to prefer, is included.
If you want to generate an animation from the input file, QuteMol might be one option.
The interactive visualization on the wikipedia page of benzene is generated using JSmol, which is a JavaScript implementation of Jmol.
If you are only interested in online tools, MolView could be an option.

Anything that's online? I'm not up for downloading software for now

Wow, MolView is a really cool toy. It's so much faster than Molcalc, unfortunately less accurate, but nevertheless a good starting structure generator.

@Martin-マーチン Honestly, I wasn't aware of the tool until 2 hrs ago :D Online searches are a blessing ;-)

That Benzene molecular model you found on Wikipedia is uploaded by Ben, (a chemistry PhD student at the University of Bristol). He uses  Avogadro and/or CCDC Mercury. Both of them are free, cross-platform molecule editors which can be used to make 3-D Ball-and-stick models.
He previously used to have Accelrys Discovery Studio using which he made the original benzene molecule, but it is no longer supported.

See this tutorial on his user page about making molecules. Hope this helps. Have fun making molecules!

You have pretty much repeated the comment above from Carmeister. Please expand on this answer, otherwise it will be deleted and changed to a comment.

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p-nitrophenol and sodium bicarbonate test
I was doing an experiment in lab and found that about one-third of p-nitrophenol dissolved in $\ce{NaHCO3}$ solution with liberation of $\ce{CO2}$ in small amount. This is contradictory to what most textbooks say. I was surprised as p-nitrophenol is less acidic than carbonic acid yet it gave positive $\ce{NaHCO3}$ test.
Can anyone give any possible explanation? I thought that one reason maybe that since their $\mathrm{pK}_\mathrm a$ differ by about $1$, the reaction maybe in equilibrium to a significant extent, but this kind of thing isn't mentioned anywhere.

Perhaps the textbook is wrong. Seeing is believing. :)

Experimental details matter here, but are lacking from your question.


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translation of a sentence involving antigen/buffer/solution
I'm trying to translate a sentence about biochemical testing into English.
So far, I can figure out the following:
Make a 10,000 pg/ml stock solution from[?] antigen into[?] 7.5 % BSA-TSA buffer.
The prepositions are the literal translations of what the foreign language says. However, despite knowing very little about biochemistry, I'm doubtful that either translation is correct in English.
If you were to read this sentence, would you be able to guess how to replace the highlighted words so that the phrase would sound idiomatic in English?

I'm not sure why "Deutsch" shows up in my profile (I'm not fluent in German), but the original sentence was in Finnish. If you can make sense of it, the sentence is Tee antigeenista 10 000 pg/ml kantaliuos 7.5 % BSA-TSA puskuriin.

I've done some more Googling since posting this, and it looks as though the correct formulation (in this context) might be "from the antigen ... in the buffer". What do others think?

You actually need to prepare a stock solution of 10,000 pg/ml antigen in 7.5 % BSA-TSA buffer.


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What do final zeros to the right of decimal point indicate?
Final zeros to the right of the decimal point are considered significant. What do those zeros indicate and why are they significant? For example, in 2.000 there are four significant figures.

They are treated just the same as any other numbers; the uncertainty is in the next place and assumed to be $\pm 0.0005 $.

@porphyrin But unlike leading zeros to the right of decimal and final zero after a digit (1-9), they are considered as significant. I want to know if these zeros are placeholders then why zeros in .0001 and 10 are not considered significant?

$10, 1000, 12340$ etc. are integers so are absolute numbers. $0.001$ can be written as $10^{-3}$ or $1/1000$ so the zeros are necessary to indicate the magnitude of the number.

@porphyrin Then what does zeros in 2.000 indicate.I am confused

it means that this is a real number not an integer and that its values lies somewhere between $1.9995$ and $2.0005$ but that we don't know exactly where between these two numbers and as a result of our ignorance we write the best guess which is $2.000$ .

Try not to answer a question in the comment session guys.

Related question on Mathematics.SE.

I chose $\pm 0.0005$ but having checked in some textbook it seems that $\pm 0.001$ would be better, i.e. twice the value I gave. The example given below by @DavePhD deals with a different case where the error is already known and some rounding is then done in the standard way.

"$2.000$" does not mean $2.000 \pm 0.0005$.
"$2.000$" does not mean the interval [1.9995,2.0005].
"2.000" means that there is an unspecified amount of uncertainty in at least the last, and possibly the last two digits.
See the NIST Good Laboratory Practice for Rounding Expanded Uncertainties and Calibration Values:

Example 5
The correction for a weight is computed to be 285.41 mg and the uncertainty is 33.4875 mg.
First, round the uncertainty to two significant figures, that is 33 mg. Then, round the correction
to the same number of decimal places as the uncertainty statement, that is, 285 mg

So in other words, for people who follow the NIST standard of stating uncertainty to two significant digits:
2.000 can mean anything from 1.990-2.010 to 1.901-2.099
Others only express uncertainty using one significant digit, in which case:
2.000 can mean anything from 1.999-2.001 to 1.991-2.009

This might be the irrelevant question but I have to ask. In above example 5 you rounded off the number 33.4875 to two significant figures as 33 but my teacher taught me that if you want round off a number to some desired sig fig you must do it in different steps like 33.4875 = 33.488 = 33.49 = 33.5 = 34. So which is correct one?

@AksaK I didn't write the example, I'm just quoting from NIST.  NIST is correct.  Teacher is wrong.  "33.4875" is closer to "33" than "34" so you round to "33".

Though I didn't understand NIST but I understand rest of your answer.

@AksaK In physics publications it's more common to see two uncertain digits reported.  For high school or undergraduate science, it's more common to see just one. I was told in college to use one digit, unless the digit would be "1", in which case to use two digits.

Please note that my answer is dealing with precision of measurement, not uncertainty! The two are different! In terms of precision 2.000 definitely means that the measurement is in the interval [1.9995,2.0005[ as you would otherwise state a different measurement value! The uncertainty of a measurement is only relevant when you do more than one measurement!

@JeppeNielsen I disagree, because, taking the NIST example, the measurement is 285.41 mg.  The measurement interval is 285.405 to 285.415 mg.  The reported value is 285 mg.

Perhaps you should use international standards instead of national ones then: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3387884/

The example the OP has given  is different to yours in that no error is given so there has to be some assessment as to what that might be. I chose $0.0005$ but having checked in some textbook sit seems that $0.001$ would be better, i.e. twice the value I gave.

And the international method of stating the above measurement would then be 285.41(+- 0.005)mg.

@JeppeNielsen NIST follows this international standard: http://www.bipm.org/en/publications/guides/gum.html

Dave, if you measure something with 5 significant digits, then you either round it to 4 significant digits or report 5 digits including the uncertainty. You do not throw away a perfectly good digit of precision eg. the reported value in the NIST example should be 285.4 mg or 285.41 (+-0.005) for maximum precision and clarity.

@JeppeNielsen In that example, the value is measured to 5 digits, but not all 5 are "significant".  The uncertainty in the value is over 10%.  I would throw away at least 2 digits like NIST does, or maybe even 3 digits and say $290 \pm 30$

And that is were our disagreement is. You are concerned about the uncertainty of the mean of the measurements. I am concerned about the precision of each individual measurement. I will naturally agree that each measurement when compared to the mean may be different, but if your measurement device is capable of five significant digits for each individual measurement then you should report that as a measurement. The uncertainty and systematic/random errors should be determined after the measurements have been performed.

@JeppeNielsen No, what I'm saying is not limited to a mean.  "A Type B evaluation of standard uncertainty is usually
based on scientific judgment using all the relevant
information available, which may include
– previous measurement data,
– experience with, or general knowledge of, the
behavior and property of relevant materials and
instruments,
– manufacturer’s specifications,
– data provided in calibration and other reports, and
– uncertainties assigned to reference data taken from
handbooks." https://www.nist.gov/sites/default/files/documents/2017/05/09/tn1297s.pdf

The zeros to the right of the decimal point denotes the expected precision of a measurement.
Thus a value of 2.0 indicates that the the measurement falls in the interval [1.95,2.05[. The value 2.00 correspond to the interval [1.995,2.005[ and 2.000 corrospond to the interval [1.9995,2.0005[.
Therefore the amount of zeros are significant for indicating the precision of a measurement.

Why is there uncertainty of +/- x.5 ? Why not x.1?

@AksaK: Standard rounding rules. <.5 is rounded down, >.5 is rounded up.

I think it's worth noting that the number of 0s is not guaranteed to indicate significance/precision, as it's possible the author did not follow standard conventions.  For instance, they might have reported the number of digits that some measurement device displays, which may or may not correlate with the overall measurement precision in the context of the particular experiment.  When in doubt, look for a discussion of the context of the measurement and an analysis of possible errors; if that's not present, consider asking the author to confirm whether they did, in fact, mean for the number of digits to reflect the precision.


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Can I use gasoline as a non-polar solvent?
I was wondering if I can use gasoline for a non-polar extraction of organic substances instead of prescribed diethyl ether, which is a bit hard to get in my country. If not, are there any products that can be found at hardware stores for example and can be used as a non-polar solvent?

Gasoline is a mixture of several compounds, paraffins, cycloalkanes, and olefins (alkenes), the purification can be a challenge.

Then again, maybe you don't need any purification.

Is there anything you would suggest?

Actually, I will not. The extracted solution will be thrown away. Can I use it then?

Try ethyl acetate or acetone. Or you use gasoline and see what happens. Try with samll amount so you don't have to synthesis again and again.

This sort of mixture used to be called pet. ether and was common in labs for e.g. the non-polar component of chromatography eluent. There will be some things that are ether soluble but insoluble in gasoline but otherwise I don't see why you should not use it.

@Waylander If I'd use acetone, will it dissolve some polar compounds too, as it is a bit polar? Because I need to keep it for the next phase of extraction.

Acetone is more polar than ether and certainly there will be substances dissolved by acetone that are ether insoluble. Guess you won't know until you try whether the materials you're working with are among them.

If you are extracting with diethyl ether because of its non-polarity and not because of specific interactions with the ether group, and if you have access to it, try chloroform. Its polarity is very very similar to that of dimethyl ether.
Alternatively, toluene and turpentine are non-polar solvents that you can buy off the shelves.


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"Cage-like" Formazine particles, how are molecules arranged?
Question: What would the  "cage-like" arrangement of Formazine molecules look like in a Formazine particle while it is still suspended in water? Would it be totally random, or have some kind of imperfect but recognizable regularity?
Formazine is used as an easily defined and prepared calibration or reference standard for turbidity measurements.

above x2: Formazine from PubChem.
From Wikipedia:

Formazine (formazin) is a heterocyclic polymer produced by reacting hexamethylenetetramine with hydrazine sulfate.
The hexamethylenetetramine tetrahedral cage-like structure, similar to adamantane, serve as molecular building block to form a tridimensional polymeric network.

From Formazine:

The readings shown by turbidimeters are not scaled in the measured light intensity, but in concentration of a reference suspension. Because the accuracy of this calibration solution determines the reliability of subsequent turbidity measurements, it is of crucial importance.
The internationally established turbidity standard for calibration is formazine. This standard can be reproduced at any time with the recipe taken from standard ISO 7027 (Water quality - Determination of turbidity).
It is extremely important to observe the prescribed preparation temperature, because it affects the particle size of the formazine particles perceptibly. The following Figure illustrates this. Errors caused by temperature variations are on the order of 1..2% per °C. Consequently, Sigrist keeps the preparation temperature constant to within ± 1°C.

above: "Particle size distributions of formazine at various preparation temperatures" from here.

above: Enlarged from TurbidityStandards.

above: Slide #10 from What is Turbidity? John Daly, ISA NorCal President, South Fork Instruments, Inc. Click for full size.

First, note that

formazine structure attached from PubChem is a different compound;
note that there are also, related, formazan $\ce{HN=N-CH=N-NH2}$ and “formaldazine” $\ce{CH2=N-N=CH2}$ (dimethylidenehydrazine);
formazin (polymer) does not necessarily has to have cage-like structure, as this characterization belongs to hexamethylenetetramine in the quoted citation.

Hexamethylenetetramine (urotropine, $\ce{(CH2)6N4}$) can be prepared from formaldehyde and ammonia. In water it also somewhat decomposes to its precursors:

Formazin (also called formalazine) structure might be still unknown, as a related publication from 1976 states:[1]

… This substance, an insoluble condensation product of uncertain composition, is prepared by mixing solutions of hexamethylenetetramine and hydrazine sulfate. …

Sometimes empirical-like formula $\ce{(C2H4N2)_n}$ is claimed.[2]
Sometimes, claimed structure like following one is presented:[3,4]

It was (with several other structures, less nice or likely), hypothesized (well, rather more complicated, with some ether linkages $\ce{-CH2-O-CH2\bond{-}}$ and $\ce{-NH\bond{-}}$ groups present), based on spectroscopic studies.[5,6]

References:

Rice, E. W. The preparation of formazin standards for nephelometry. Analytica Chimica Acta 1976, 87 (1), 251–253.
Ziegler, A. Issues Related to Use of Turbidity Measurements as a Surrogate for Suspended Sediment. Turbidity and Other Sediment Surrogates Workshop; Reno, NV, USA, 2002.
Kaur, N.; Kishore, D. An Insight into Hexamethylenetetramine: A Versatile Reagent in Organic Synthesis. J. Iran. Chem. Soc. 2013, 10 (6), 1193–1228.
Sadar, M. J. Stabilized Formazin Composition. US Patent 5,777,011, July 1998.
Mashima, M. The Infrared Absorption Spectra of the Condensation Products of Formaldehyde with Hydrazine. Bull. Chem. Soc. Jpn. 1966, 39 (3), 504–506.
Bondybey, V. E.; Nibler, J. W. Infrared and Raman Spectra of Formaldazine. Spectrochimica Acta Part A: Molecular Spectroscopy 1973, 29 (4), 645–658.

Thank you for taking the time to put together such a thoroughly sourced answer and clarification! It's been a while now, so I'm going to take some time and have a look at your references and think about this a bit.

I was able to track most of these down and I have a better understanding of the situation now, thanks!


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Isomers through Nucleophillic substitution
If a single active optical isomer of 2-chlorobutane were to undergo a nucleophillic substitution reaction with iodide ions, how many optical isomers are present in the product? A mixture? Or just a single isomer?
My attempt:
Iodine can only approach the carbon atom from the opposite side of chlorine(the point at which chlorine is bonded)due to repulsion. So a single isomer is obtained as a product.
However, many of my friends are saying my logic is incorrect. Can somebody please clarify this?

How is the logic incorrect?

They stated that two isomers will be formed- laevorotatory and dextrorotatory isomers.

(i) there is a single isomer ... (ii) this can undergo either a SN1 reaction or a SN2 reaction.  An SN1 reaction will give two products; an SN2 reaction will give one product.  You need to know whether the reaction is SN1 or SN2 to be able to answer the question.

I used SN2 reaction(the one in which carbon gets bonded to five groups for intermediate?). Btw is the reasoning I provided correct for SN2?

@Faiq Raees You may have some misconceptions about Intermediates versus Transition States, and Pentavalent Carbons. SN2 is the reaction you are thinking of, but there is no Intermediate (the reaction occurs in a single concerted step) and the carbon in the transition state does not actually have five bonds, two of them are depicted as "partial bonds" in the imaginary process of breaking/forming.


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Why is the addition of methanol during the prepartion of the liquid crystal cholesteryl benzoate necesary
Here's one way to prepare cholesteryl benzoate crystals:
Step 1 Dissolve 1.0 g of cholesterol in 3 cm3
 of pyridine in a conical flask.
Step 2 Add 0.40 cm3
 of benzoyl chloride.
Step 3 Heat the mixture on a steam bath for about 10 minutes.
Step 4 Cool the mixture, and add 15 cm3
 of methanol.
Step 5 Collect the solid cholesteryl benzoate by suction filtration. Rinse the flask and the crude crystals with a little cold methanol.
Step 6 Recrystallize the cholesteryl benzoate using ethyl ethanoate as the solvent.
The methanol is said to be added to react with the benzoyl chloride. Why is this required?
Usually, in a preparation such as that of 1-bromobutane, some substance such as sodium carbonate is added in order to allow for separation of the aqueous substances from the organic layer. That makes sense. However, in the case of the above preparation, by adding methanol the substances within the conical flask go from benzoyl chloride, pyridine and cholesteryl benzoate to HCl, pyridine, methyl benzoate and cholesteryl benzoate. What's the point? Either way, both mixtures are going to eventually be separated from the cholesteryl benzoate by suction filtration.

Pyridinium hydrochloride and methyl benzoate are both soluble in methanol and thus will be washed through the filter leaving behind clean cholesteryl benzoate. If you don't use the methanol the solid you produce will be impure and sticky; the methyl benzoate and pyridinium hydrochloride will not pass through the filter but will bind to the product crystals.. Methanol also removes any excess of benzoyl chloride by reacting with it.

There is a pattern in all organic preparations that there is some kind of intermediate substance that is usually added in order to allow for 'better separation'. In the british education system the reason for adding this 'intermediate substance' (not an official name- just a term that I am using) is always just said to be added to react with some substance within the solution. Should I assume that the necessity of this is ALWYAS to allow you to separate away your desired product from most of the stuff within the mixture?

The usual phrase is "work up" by addition of....whatever. This can be as simple as ammonium chloride soln to neutralise anions, or can be something to react with excess reagent and stop the reaction. However as you move forward into more complex procedures the problem of removing excess reagents and unreacted starting materials becomes far more significant.

It seems like just saying that we want to remove a certain substance within the mixture is a cruel oversimplification. As illustrated by your answer, this is only a proxy to the true reason, which is not to leave behinde a sticky substance that contains a complicated mixture that is not  the cholesteryl benzoate that we want.

It is certainly a simplification. Having a well-worked out procedure that delivers pure product is a rarity in synthetic chemistry. All too often getting hold of the clean desired product is far more challenging than running the reaction.


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Will the dextrorotatory and laevorotatory enantiomers always have a set of stereocenters with opposite absolute configuration?
I was faced with a question where the skeletal structure of of D(+)-glucose was given and I was asked to give the right structure of L(-)glucose. I noticed that the correct answer was the one where all the stereocenters had opposite absolute configuration of those of D(+)glucose.
But my question is: Will the dextrorotatory and laevorotatory enantiomers always have a set of stereocenters with opposite absolute configuration?

Precisely, these are based on the configurations proposed by German chemist Emil Fischer, who established the structures of many sugars.

Emil Fischer elucidated these configurations for the aldohexoses in
  1896. According to the Fischer convention, D sugars have the same absolute configuration at the asymmetric center farthest removed from
  their carbonyl group as does D-glyceraldehyde. The L sugars, in
  accordance with this convention, are mirror images of their D
  counterparts,

Similary,the  Fischer projection for glucose  is shown below.

It is very important not to confuse these enantiomers with epimers   (two sugars that differ only in the configuration around one carbon atom e.g D-glucose and D-mannose, which differ only in
the stereochemistry at C-2)

References

Voet and Voet Biochemistry
Lehninnger Principles of Biochemistry
Biochemistry (Campbell and Farrel)


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What is an easy way to reduce a boronic acid to borane?
I have a selection of boronic acids that I want to reduce to borane forms. Some of the boronic acids in question are phenylboronic acid (C6H5B(OH)2), tolylboronic acid (C7H7B(OH)2), and p-phenylenediborinic acid (C6H4(B(OH)2)2). I want to convert them into phenylborane (C6H5BH2), tolylborane (C7H7BH2), and p-phenylenediborane (C6H4(BH2)2).
I think I can do this by dissolving the boronic acid in ether and adding Lithium Aluminum Hydride (LAH). Giving a reaction that looks something like:

However, I am afraid that, given how aggressive a reducing agent LAH is I might get something like:

So I guess I have two questions:

Will LAH attack the boron-carbon bond shown in the second reaction diagram?
Can the reduction of the boronic acid be accomplished via a less aggressive reducing agent, perhaps Sodium BoroHydride?

Just a note: I am well aware that the boranes produced will be oxygen sensitive, either converting back to boric acid on exposure to oxygen, or catching fire.

There are quite a lot of reducing agents out there. $\ce{DIBAL-H}$ is one of them. $\ce{Sn + HCl}$ is yet another. So is $\ce{NaBH4}$. So many out there.

I know there are a lot out there, but are they strong enough to reduce boronic acids? As I said in one of my questions: Can the reduction be accomplished with something less aggressive than LAH, like NaBH4.

@PrittBalagopal Just because reducing agents exist doesn't mean they will be able to affect a particular transformation

LAH will attack the boron-oxygen bonds without going after the boron-carbon bond, see here (Eq. Z, p. 142).
But it also deprotonates the hydroxyl groups as a strong base like a Grignard reagent. You need to prevent that.  Convert your acid to an ester (OH -> OR), remove the water and then apply the LAH.

Upvote for the reference and the useful note, I hadn't thought about what would happen to the hydroxyl group once it was removed. Out of curiosity, if I do not convert to the ester first, will it just require more LAH and leave me with Aluminum oxide and Lithium oxide? Or will something more insidious happen?

What's insidious is if you deprotonate the hydroxyl groups the boron loses its electrophilicity and you don't get the hydride transfer.  Compare with a Grignard reagent which converts a carboxylate ester to an alcohol but the parent acid just gives a salt.

Ah, I see, thanks. As a physicist with a little more chemistry than most physicists under my belt I have gotten mixed up in liquid scintillator development but sometimes that little bit of extra knowledge isn't enough because of how hazy it is.


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Why is the solid product washed with ice-cold solvent during in the last stages of recrystallisation?
Why is the solid product washed with ice-cold solvent during in the last stages of recrystallisation? Is this to remove soluble impurities? (for example) Well, what if the impurities were also insoluble in the cold solvent?
Recrystallisation is weird in the sense that we have to make some assumptions in order for the process to fulfill the 'ideal' purification of the solid product that we're trying to get to.
For example, we have to assume that all soluble impurities are washed off during suction filtration, and that no impurities manage to creep up within the molecules of the crystals when they form. Would I be correct in saying this?

Recrystallisation may be done in a multitude of ways, depending on your particular situation; the product may be washed with ice-cold solvent, or with another solvent that would supposedly dissolve the impurities but not the product, or not washed at all. You may call it weird, but then again, pretty much all chemistry is weird in the same sense, or worse. It is an experimental science, after all.

So is the main goal of this step to kind of 'clean off' the surface of the crystals?

Yeah, kind of. See, at this stage the crystals are wet with mother solution (admittedly, a tiny amount of it, but it is still there), which contains the very impurities we wanted to get rid of in the first place.

To avoid the collected crystals become melted or liquid again. The cold water will keep the collected crystal stays in crystal form.

Possible sources of impurities in the recrystallization process can be described in terms of phenomena occuring during coprecipitation and which are the basis of gravimetry:

inclusion (impurity occupies positions in the crystal lattice);
occlusion (impurity is physically captured within the crystal as it has been growing);
adsorption (impurity is adsorbed on the surface layer of the crystal);
postprecipitation (impurity forms secondary solid phase once recrystallization of the target stopped/slowed down).

Washing crystals with a fluid only addresses the last two phenomena, adsorbance and postprecipitation. Postprecipitation is rarely a problem as it occurs on a large time scale or when a really poor solvent has been chosen.
Properties of the washing liquid vastly depend on the properties of the compound being recrystallized. In the simplest case it's the same solvent (as in your example), but it can also be an electrolyte, hydrolysis suppressor or another solvent (e.g. distilled water) altogether.
As for the temperature of the washing liquid, it's actually better to use many small portions of the hot solvent as filtration of the hot solutions happens much faster, surface adsorption decreases with the temperature and removal of mother solution from the surface is faster. However, if solubility of the recrystallized product is significantly increasing with the temperature (and this is typically the case), then one has to sacrifice the efficiency of the washing process for the better yield and use cold solvent instead.
Impurities, insoluble in the cold solvent, are practically not an issue. First, recrystallization starts from a homogeneous solution. Second, before crystals are washed, they are separated from the mother solution so that only minor amount of it is left on the crystals' surface, and if there were any impurities, they don't get a chance to nucleate as their concentration in washing solvent is going to be negligible.
As for your perception of the recrystallization as of a "weird" process, I'd say it's probably more complicated than one would expect from looking at the beaker with some crystalline mess. In fact, underlying processes are described with more or less appropriate models, and once a person
 gets experienced enough, it all starts to make much more sense.


text:
Difficulty in cis, trans isomers in Alkene
I have a difficulty in appointing cis trans isomer to an Alkene compound .
I have looked it up and drawn out a few -

Which "point" should I look at to understand in order to allocate cis and trans ? I know it's between the first and last line . Thanks!

The last name is wrong. There is no cis/trans isomerism because the two methyl groups are identical.

You may think of the two lines representing the double bond like the bird's eye view on a door of a saloon:  If the two wings, the one on the left of the double bond, and the one of right flap in the same direction, then the double bond is in cis-configuration.  On the contrary, if the two flap in opposing directions, than the double bond is in trans-configuration.
For future reference -- taking into account the third picture shown by you -- cis and trans may be ambiguous; and triple and quadruple substituted double bonds' configuration are described only with E- or Z-configuration (cf. here).  Eventually, as correctly pointed by @Zhe, in your last example, the double bond is neither E-, nor Z-configurated.


text:
Why does ethanol and water mixture show positive deviation from Raoult's Law?
So, my teacher said, "For a solution to show positive deviation from Raoult's Law, it must have a compound which is lacking hydrogen bond and a compound which has hydrogen bond."
This is because the compound not having a hydrogen bond breaks the hydrogen bond of the other compound.
Now, why does ethanol and water mixture show positive deviation? Both of them have hydrogen bonds.

In more generality, Raoult's law is stated as $$P_i = x_iP_\text{vap}$$ and read "the partial vapor pressure of species $i$ is its (pure) vapor pressure multiplied by its mole fraction in the liquid phase." It assumes an ideal-gas gas phase and an ideal-solution liquid phase; an ideal solution is one in which all intermolecular attractions are of equal strength. More explicitly, the $a{-}b$ interactions are of equal strength to the $a{-}a$ and $b{-}b$ interactions, where $a$ and $b$ represent different chemical species in solution.
A positive deviation from Raoult's law occurs when the partial vapor pressure is greater than would be expected from Raoult's law. From a microscopic perspective, this implies that the $a{-}b$ interactions are weaker than the $a{-}a$ and $b{-}b$ interactions.

You should see now that your teacher's statement is a general rule-of-thumb following from the concepts outlined above, but does not necessarily hold true for all situations. Here, water forms hydrogen bonds of greater strength and magnitude than does methanol, and one expects that a water-methanol mixture will therefore possess weaker intermolecular interactions than those in a solution of pure water.

To OP. The statement of the teacher can be made more general replacing hydrogen bonds with attractive interactions. S/he was referring to cases where the deviation is particularly strong, and thus thumbnail rules particularly useful. Plus one to this answer, too.


text:
What is the volume occupied by the atoms and the edge length in these cubic lattices?

Consider a cubic unit cell of crystal that is composed of a single kind of atom. By placing atoms at every corner of the cube, a simple cubic lattice is formed. Assume that the atoms are perfect hard spheres with a radius $r$ and that the atoms are in close contact to minimize the volume of the cube.
In a simple cubic lattice, the volume occupied by the atoms is $(?)$ % that of the cube, and the length of the edge of the cube is $(?)r$.
To the simple cubic lattice described above, atoms are added to all face-center positions of the cube. In the resulting lattice, the volume occupied by the atoms is $(?)$ % of the cube, and the length of the edge of the cube is $(?)r$.

For the first one, I tried drawing a cube with quarters of spheres at each of its corners and finding the % by calculating the total volume of spheres over the volume of the cube, yet the answer isn't accurate. I get the length of the edge as $2r$.

This is the case of CCP (simple cubic closed packing) and FCC (face centered cubic closed packing)
For CCP
Imagine sitting in a cubical room. Now place some big spheres on each of the vertices of the room such that the centre of sphere and the vertices coincide.
You will see that some part of sphere goes outside the room itself. In chemistry, this is called sharing.
If your room is surrounded in all directions by other rooms, you share one sphere with 7 other rooms. (top 4 and bottom 4)
Thus a single particle gets shared by 8 rooms. The part in your room is 1/8
And the number of such particles are 8 (as cubic room has 8 vertices)
Net particles in your room $= \frac{1}{8} \cdot 8 = 1$
So in CCP the space occupied by all eight particles is basically equal to that of a single particle.
Now as assumed in the question, the particles are tightly packed, the
Radius of sphere equals half of the edge length. In other words edge length is twice the radius of the sphere.
$$\pu{Packing efficency = \displaystyle \frac{\text{Volume of sphere}}{\text{Volume of cube}}}$$
$$= z \cdot \frac{\frac{4}{3} \pi r^3}{(2r)^3} $$
(Here $z$ is net particle contribution; for the current example it is equal to 1)
Calculating this you'll get packing efficiency $\approx 52.4%$

For FCC
Now imagine putting one more sphere in the room, but this time, at the centers of the face. (Do this for each face). Now the particle in center gets shared only by 2 rooms. Yours and the room which is opposite to you.
Therefore, each particle has net contribution of half.
Since a cube has six faces, the net contribution of all spheres at the center the of sphere are
$$\frac{1}{2} \cdot 6 = 3$$
Add the already existent corner particles to it (whose net contribution is 1 as we know)
Thus the total number of particles in FCC are $3 + 1 = 4$

If you're good at imagining things you might have observed that now spheres won't fit in until you reduce their sizes. So we must falsify the assumption that twice the radius of sphere is the edge length. Spheres have shrunk.
I want you to imagine the face of the room now.
There is little space between two spheres on corner but not between the center sphere and corner spheres. Draw a diagonal line in the sphere passing through the diameter of central sphere. Now using Pythagorean theorem:
$$ \pu{(\text{Edge of cube})^2 + (\text{Edge of cube})^2 = (\text{Diagonal of cube})^2}$$
Let edge be 'A':
$$(A)^2 + (A)^2 = D^2$$
$$2(A)^2 = D^2$$
Since the center spheres have no space with corners ones. Therefore
4(radius of sphere) = Diagonal length
$$\sqrt{2}A = D$$
$$\sqrt{2}A = 4r$$
$$\sqrt{2}A / 4 = r$$

There you go. You've found the radius of sphere. Now put it in packing efficiency formula and take $z$ = 4
You will find it somewhere near 72%
On a sidenote: Observe that simple cubic has only 1 sphere which takes up nearly 50% of space of cube. While face centred has 6 + 8 = 14 smaller spheres which take up nearly 70 % of all space in cube. It's like comparing apples and bananas.
You might think that FCC is better than CCP as it uses space judiciously. But since in CCP, the particle size is large therefore CCP is best suited for particles of that size and FCC for smaller sizes.
Although we can discuss hexagonal packing form with CCP.

Thank you, these information are very helpful and I learnt more too!


text:
Why does zinc turn dark?
So I casted a zinc ingot around a week ago and I also casted 2 zinc ingots yesterday. After observing the two side by side I noticed that the older zinc ingot was darker colored, then I went and looked at all of my old ingots containing zinc. They where all dark. Here is a picture of the old one and the new one side by side

The old ingot is on the left and the new one is on the right.
Here is another photo
Is this oxidation possibly? What causes this? Any answers are appreciated.

Give this a read: https://en.wikipedia.org/wiki/Zinc#Reactivity Apparently the dark film is that of zinc carbonate ;)


text:
Why does H2 react with O2?
We know $\ce{H2}$ is formed with covalent bonds (just as $\ce{O2}$ is formed via covalent bonding). Hence, these molecules gain some stability. In spite of this stability, why do they react with each other to form $\ce{H2O}$ or $\ce{H2O2}$?

Burning hydrogen in oxygen is a hugely exothermic reaction. Hence, they gain even more stability by combusting.

That is, they gain even more stability by breaking those bonds and forming other bonds instead.

Indeed, much of the chemistry that takes place in the universe involves the breaking of covalent bonds and the formation of new covalent bonds. The key is that the energy of the products needs to be in a lower state than that of the reactants. This is the thermodynamic requirement anyway. Lets consider one of your examples:
$$\ce{2H2(g) + O2(g) -> 2H2O(g) + \pu{572kJ}}$$
Although we had to break the covalent bonds of hydrogen and oxygen, the new bonds that were formed are even stronger, thus the reaction is thermodynamically favored.
The complicating point of the above example is the fact that there is a large kinetic barrier to the reaction. This basically means that there is an intermediate product in the steps of this reaction that is energetically very high (unfavorable), so some catalyst is required (say, a spark) for this reaction to get over that energetic barrier and proceed on any reasonable time scale.
Summary, TL;DR:
The breaking and creating of covalent bonds plays a major role in chemistry. The criteria for covalently bound compounds to participate in a chemical reaction is that the products have a lower energy state than than the reactants. If this is not the case, the reaction will not happen. Even if it is the case (that the products are lower energy / more stable than the reactants), there may be kinetic limitations requiring a catalyst in order for the reaction to proceed. But, no catalyst can make a reaction proceed if the above criteria is not met, that the products be more stable than the reactants.

You suggest in your summary paragraph that endothermic reactions do not happen, 'if this is not the case, the reaction will not happen'.

@porphyrin , thx for the input. What I'm shooting for is a statement referring to the Gibbs free energy of the reaction needing to decrease for the reaction to have a chance. I hope the way I've re-worded gets that across.

I suppose it depends on what you mean by 'stable' in your last line. The point is that the Gibbs energy must be negative for a spontaneous reaction, rather than just the enthalpy, so entropy change is crucial also. The rate does not matter as far as the thermodynamics goes, and, as you write, the catalyst does not change the reactants or products but only the mechanism of  how one gets from one to the other.


text:
Understanding colour changes of systems in dynamic equilbrium
Consider an aqueous reversible reaction of two chemicals $\ce{A <-> B}$, where $\ce{A}$ is red and $\ce{B}$ is yellow. The reaction is in equilibrium when there is $\pu{1 dm^3}$ of A with concentration $\pu{1 M}$ and  $\pu{0.5 dm^3}$ of $\ce{B}$ with concentration $\pu{1 M}$, such that $K_C = 0.5$ and the solution is orange (due to the equal parts of red and yellow).
If I were to add $\pu{1 mol}$ of $\ce{A}$ to the system without increasing the total volume of the solution, Le Châtelier's principle predicts that the equilibrium position would move to the right, hence there would be more of $\ce{B}$ than $\ce{A}$ and the solution would become more yellow, however this does not seem to agree with my understanding:
$K_C = 0.5$, and is constant when the concentrations of chemicals are changed, so some of $\ce{A}$ must be converted to $\ce{B}$ by the forwards reaction to keep $K_C$ as $0.5$. This will make the concentration of $\ce{A}$ $\pu{\frac{5}{3} M}$ and the concentration of $\ce{B}$ $\pu{ \frac{5}{6} M}$ at equilibrium. The chemicals are still in the same ratio, so shouldn't the colour remain the same instead of becoming more yellow as Le Châtelier suggests?
This is a simplified problem, and I am wondering if the logic applies to more complicated cases (if it's correct in the first place of course!) Essentially, in my understanding, if you increase the concentration of a chemical in a dynamic equilibrium, when equilibrium is re-established, some but not all of the added chemical would be converted into the products on the opposite side of the reaction. This means the amounts of chemical on both sides of the reaction have increased, so it doesn't make any sense to say that the equilibrium position moves to one side of the other - both sides have increased in concentration, and in terms of coloured chemicals, the colour would remain the same.
Please use this more complicated reaction for demonstrative purposes:
$$\ce{Fe^{3+}_{(aq)} + 3\,SCN^-_{(aq)} <--> Fe(SCN)3_{(aq)} }$$
where $\ce{Fe^{3+}}$ is yellow, $\ce{SCN^-}$ is colourless and $\ce{Fe(SCN)3}$ is red. How would the reaction behave if the concentration of $\ce{Fe^{3+}}$ ions is increased (e.g. add iron III nitrate).
Apologies in advance if this question is a little convoluted, I'm not a professional chemist (I'm studying A-level chemistry), and I'm quite confused, so I'm struggling to form a coherent question.
Addition (edit):
In the case of the more complicated reaction, my understanding is as follows:

If I add $\ce{Fe^{3+}}$, the forwards reaction will convert some of the added $\ce{Fe^{3+}}$ and some $\ce{SCN^-}$ (already present) into $\ce{Fe(SCN)3}$.
Not all of the added $\ce{Fe^{3+}}$ will be converted (the equilibrium cannot completely counteract the change, as this would involve converting all the $\ce{Fe^{3+}}$ into $\ce{Fe(SCN)3}$, and this would be an increase in the amount of $\ce{Fe(SCN)3}$ which the reaction would have to counteract).
Therefore the concentration of reactants and products increases, because the added $\ce{Fe^{3+}}$ is split between some which is converted into $\ce{Fe(SCN)3}$, and some that stays as $\ce{Fe^{3+}}$.
Since this gives an overall increase in the concentrations of both chemicals, neither has increased relative to the other, and the equilibrium position hasn't moved, so the colour doesn't change.

I think that this question is interesting. May also have to do something with optics. Here's a video to make you think: https://www.youtube.com/watch?v=LgJGTH5Xp0o
Adding Fe 3+ made the solution darker.

Thanks for pointing me towards that video :) @QuantumAMERICCINO. The video clearly contradicts my reasoning (which I expected to happen; Le Chatelier's is obviously going to be correct), so I'll have to do some more digging to figure out what I'm misunderstanding.

Below is the final bullet point in your question (emphasis mine):

Since this gives an overall increase in the concentrations of both chemicals, neither has increased relative to the other, and the equilibrium position hasn't moved, so the colour doesn't change.

I will show that this is incorrect for the reaction $$\ce{A + B <=> C},$$ even though it is correct for the reaction $$\ce{A <=> B}.$$

Let $K = 1$, $\ce{[A]}=0.2$, $\ce{[B]}=5$, and $\ce{[C]}=1$ (dimensionless concentration units). Then $\ce{[C]/[A]}=5$. Now increase the concentration of A to $\ce{[A]}=1$, leaving the concentrations of B and C unchanged. We can solve the resulting quadratic equation to find the equilibrium concentration of all three species: $$1 = K = \frac{1+x}{(5-x)(1-x)} \quad \Longleftrightarrow \quad x = 0.63,$$ so that $\ce{[C]}=1.63$, $\ce{[A]}=0.37$, and $\ce{[C]/[A]}=4.40\neq5$.
I'm not sure how relevant the above point is, but it's something worth considering.

It's also important to note that the color is dependent not solely on the relative concentration of one species to another, but also on the absolute concentration of a species: consider a solution of water to which red food coloring is added. The more drops of red food coloring we add, the stronger the color. We start out with a solution that looks vaguely pinkish and end up with a solution that looks very red. I think this might be the more important point.

Thanks for your answer, I was stuck thinking about this for hours, so you've been really helpful! I was basically using the colours as a tangible way to understand the movement of the equilibrium position, so the above point was very relevant to me, despite colour not being an exact indicator of equilibrium position. (I would up vote your answer but I haven't earned that privilege yet).

In the general case of $\ce{A + B <=>C}$ if  there are initially $a_0, b_0,c_0$ amounts present , then if an amount x of A and B react to produce C then the equilibrium constant $K_e$ is given by
$$ K_e=\frac{k_1}{k_{2}}=\frac{c_0+x_e}{(a_0-x_e)(b_0-x_e)}$$
where $k_1, k_2$ are the rate constants for the forward and reverse reaction and $x_e$ is the amount reacted at equilibrium.
Solving the equation and  for (algebraic) simplicity let $c_0=0$, gives the amount at equilibrium
$$x_e=\frac{1}{2}\left(a_0+b_0+K_e^{-1}-\sqrt{(a_0-b_0)^2+2(a_0+b_0)/K_e +K_e^{-2}}\right)$$
Using this the amount of each species can be obtained as for example the equilibrium amount of A is $a_e=a_0-x_e$. However, in your question you ask about the colour not the concentrations and this alters things a little. The quantity that is measured optically as in a spectrophotometer is the optical density of for example species A is $D=\epsilon_\lambda[A]l$ where $\epsilon_\lambda$ is the extinction coefficient of the molecule at wavelength $\lambda $ and is effectively how strongly the molecule absorbs at that wavelength, $\ce{[A]}$ is the concentration and l the path-length of the sample through which the light passes.
The total absorption is the sum of that for all species so depending on the values of $\epsilon$ and how the concentrations change the absorption may go up, down or remain constant at any given wavelength.
Furthermore, if you do not use a spectrophotometer and measure optical density but you are looking through the sample you will observe the transmitted light which is by the Beer-Lambert law given by $I_{trans}=I_0e^{-D}$ for each species present at each wavelength, so what you may see becomes pretty complicated.

Thank you very much for this great answer, I've got my head round this problem now :). If I could tick two answers, I would, but since I had to choose, I ticked "a-cyclohexane-molecule"'s, since I found their use of actual numerical values helpful, and it was a little more accessible to me.

I would up vote your answer but I haven't earned that privilege yet.


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What are the applications of quantum field theory to chemistry?
I have a physics background and I learned how quantum field theory is applicable to condensed matter physics, and since some parts of condensed matter physics overlap with chemistry I was wondering if there are any direct applications of QFT to chemistry as well. Also what are the research areas in chemistry that use QFT, because I'm considering doing my graduate work in chemistry and it would be cool if I could use QFT in chemistry as well.

There's a subfield of chemistry called relativistic chemistry that deals with using relativistic corrections for energy levels. It's not quite directly QFT though.

Have a look at this article: "The Physics behind Chemistry and the Periodic Table"
Pekka Pyykkö Chem. Rev. 2012 112 (1), 371-384
(http://pubs.acs.org/doi/abs/10.1021/cr200042e)
I think it answers your question and has most of the references to work done in that particular field. For most questions in chemistry the non-relativistic Schrödinger Equation is good enough. Even for systems with an appreciable amount of relativistic effects, the Dirac Equation is often not solved directly. So, the number of applications where Quantum Electrodynamics is used is really limited.

Thanks this looks like it answers my question, I'll read the full article at my university library to get access later this week. +1

Quantum field theory is sometimes used in chemistry to derive relativistic corrections to non-relativistic quantum mechanical energies. And this is useful when computing energies of molecular matter with heavy atoms, due to electrons moving at high speeds near a massive nucleus.
In reality one does not use quantum field theory. One starts with a Schrodinger-like wavefunction expression
$$H \Psi = E \Psi$$
where the Hamiltonian operator is the sum of a kinetic part $K$ and a potential part $V$, and then one seeks for a potential, sometimes named an "effective potential", so that $V$ gives the same scattering amplitudes than quantum field theory. I.e. quantum field theory is used to verify that the chosen potential $V$ is physically admissible. Note that scattering equivalence does not imply dynamical equivalence. And sometimes incorrect potentials have been proposed in the belief that compatibility with quantum field theory was enough [1].
Basically there is no known expression for $V$ at arbitrary orders in powers of $(v/c)$. The problem is not only computational, it is technical, because quantum field theory cannot really describe bound states; the only rigorous states in quantum field theory are those of free particles. This limitation is fine for particle physics and the study of scattering events, but it is not for molecular sciences as chemistry.
An attempt to unify quantum field theory (specially QED) with the traditional many body methods used in chemistry is that from Lindgren and coworkers. Their research is summarized in book [2]. The point is that QED does not consider all electron-electron correlation effects, which limits its applicability to systems where electron correlation effects are small. Lindgren and coworkers have developed a so named "covariant-evolution-operator method" which they claim

forms a suitable basis for a combined QED–MBPT procedure. This leads to a perturbative procedure that is ultimately equivalent to an extension of the relativistically covariant Bethe–Salpeter equation, valid also in the multireference case and referred to as as the Bethe–
  Salpeter–Bloch equation.

Their approach, however, doesn't solve fundamental questions associated to  Bethe–Salpeter equations; they simply mention the difficulties and move away! Also, from a numerical point of view, renormalization of the radiative effects is only partial in their procedure, with hope to improve in future:

Schemes have been developed for this process but so far not
  been implemented in a QED-MBPT procedure. When the procedure is more developed, critical tests can be performed to find out to what extent the new effects will improve the agreement between theory and accurate experimental data.

The article from Pekka Pyykkö mentioned in another answer is freely available here. There are several objections to this article. For instance, in page 4 he claims that electron-electron interactions can be taken as instantaneous in the Coulomb gauge. This is not true. He seems to believe that electron-electron interactions are only given by the Coulomb term, when the Breit term also describes electron-electron interactions. Pyykkö then claims that for certain uses the CB potential has to be surrounded by projection operators. He omits to say that without those projection operators the standard CB potential gives invalid answers. Pyykkö  only surrounds the interaction part of the DCB Hamiltonian, but in reality the Dirac Hamiltonian $h_i$ (equation 2 in his paper) also has to be surrounded  like $Ph_iP$. That is why in other texts you will find projection operators around the full DCB Hamiltonian (check equation 2.113 in [2]) and not only about the CB term. There are similar objections to the rest of his paper, but the biggest objection, in my opinion, is that he never mentions that quantum field theory is not needed to explain the phenomena he covers.
REFERENCES
[1] https://arxiv.org/pdf/hep-ph/9706219.pdf
[2] http://www.springer.com/gp/book/9783319153858

All the time-dependent quantum chemistry methods I am aware of treat the fermions (electrons) quantum mechanically while treating electromagnetic radiation (bosons/laser-photons)  classically.  A complete QM description of the radiation+matter interaction seems like a natrual extention of quantum chemistry to QFT.  I think the the classical treatment of EM breaksdown at wavelengths that approach to size of the interacting QM-treated matter.  So, the QFT treatment would be needed to describe matter interacting with very high energy photons.


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What is the best way to evaporate toluene?
I decided to use toulene as my non-polar solvent because it's easy to get it. My question is, what is the best way to evaporate the toluene from the soaked organic material?

It would be helpful to have a bit more information. What are you really trying to do? Do you have some organic product (what, how much) that was synthesized/purified in toluene (how much)? A sponge? ;) You get the idea. Also, welcome to Chemistry.SE!

Hello and welcome to StackExchange. Have a look at the [help] to learn how to post good questions, and hopefully get good answers.

Well if your product were a volatile liquid, than toluene's $\pu{110 ^\circ{}C}$ boiling temperature (of the pure solvent at atmospheric pressure) may be dangerously high.
If you aim for a solid of low volatility, evaporate it under the reduced pressure of a membrane pump and gentle warming by a water bath.  If accessible, a rotary evaporator is a good thing for this: lower the pressure gradually to about $\pu{70 mbar}$, then increase the bath temperature to $\pu{60 ^\circ{}C}$.

If you have an acid or amine, you can attempt to extract into water, neutralize, and back extract into ether.


text:
Online, free SMILES interpreter
A while back, I used chemspider.com at some point to get structures from a SMILES string. I found that it wouldn't get some strings, so I switched to chemicalize.org. The site appears to have been taken over and doesn't read SMILES strings anymore. I found this site in this answer, but yet again there are some strings it won't read.
I want to do a SMILES demonstration in a comp lab, so I'd rather have a website instead of having to install a program. Is there any online, free SMILES interpreter?
Edit: I'm looking for an entirely online, no-install-required translator.

Have you tried this site?

Assuming you need the structures for further computations, consider openbabel (http://openbabel.org/wiki/Main_Page).

@airhuff I like that visualizator, but it doesn't like salts or stereochemistry, nor gives the name of the molecule.

@DSVA I am looking for an entirely online solution.

@Buttonwood Not right now but I'll have a look for further use.

@Kurzd For structure <-> SMILES, this Marvin js demo (as in Elsevier's Reaxys) may be something: https://marvinjs-demo.chemaxon.com/latest/demo.html  Does however not offer the structure -> name conversion you aim for.

@Kurzd the first two answers in the linked post got both a solution which is entirely online. And if it's a comp lab I really don't understand why it would be a problem to use actual software. All you need is a server (even a raspberry would do it), throw open babel and jmol or pymol on that and you are done. Structure to name is a whole different beast.

@Buttonwood Marvin will do for now :D. It's from the company that took over chemicalize.org, so I hope it stays available for a long time.

@DSVA Installing software is a hassle, and the Cactus site is cumbersome. The raspberry suggestion is neat but it's impossible for me to do it.

Go to chemapps.stolaf.edu/jmol/jmol.php?model=["ENTER SMILES HERE"]

@PrittBalagopal I tried with linear and cyclic molecules, but all I got was a pink sphere and seemingly no way to rotate it. What am I missing?

@Kurzd I get that installing software might be cumbersome, but openbabel is extremly easy to install on both windows and linux and it can produce pngs of SMILES extremly easy. It's also an very helpful program if you ever deal with any kind of 3D structures. And I'm really wondering what kind of comp lab doesn't have the opportunity to have software installed on the clients or on a linux server (you can use the open babel method from command line, you don't even need X-server)

@DSVA I'd have to make a request for software to be installed. I've requested TeX Live twice, but nothing has happened. As far as I know there's no linux server either. I'll suggest open babel to my classmates in case they want to do some testing on their own computers.

@Kurzd TeXLive has a portable version. You can set it up yourself assuming your PC isn't designed to reset at every logon. (Similar to steady state in the past.)

And here is another website that understands SMILES: http://rmg.mit.edu/molecule_search


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What makes ABS plastic turn yellow?
If you owned old kitchen appliances or old computer you might notice that they the once cream white plastic become increasingly yellow, old and disgusting. I have read that UV lighting can damage plastics as they disrupt the structure of plastics but what really is at play here? What chemical reactions are happening? Are such damages reversible?

What you're specifically describing is typically known as UV degradation.  The color change is due to the oxidation of the polymers that make up the plastic.  These chemical reactions are demonstrated in satisfying detail here.  Under the section titled "Photodegradation" is where you can find the intricacies of the polymer-oxidation process.  And yes, such damages are reversible.  As you will read, the oxidation process results in the formation of free radicals, which are "atoms or groups of atoms with an odd (unpaired) number of electrons," to quote Wikipedia's definition.  The formation of these radicals is the reason for the color change (since certain light gets reflected because of these new molecules, in many cases yellow light, etc...)
To get rid of the radicals, you can add hydrogen peroxide, which allows hydrogen to bond with the radical (hydrogen has an additional electron), thus de-yellowing the plastic. This is demonstrated here.
As a further note, UV degradation is not the only process that makes plastics turn yellow, as you will probably read in that paper.  There are also processes like biodegradation and mechanical degradation that are similar to photodegradation but still have their own cool processes.
Your initial question was "What makes ABS plastic turn yellow."  I should point out that the first few words of the second link read "Manufacturers add bromine to ABS plastic to act as flame retardant. Over time, those chemicals react to the plastic's basic polymers and turn yellow."  Ergo, there is probably more to the yellowing of ABS plastic in particular than just photodegradation.
Lastly, I should point out that the yellowing of plastic needn't happen in the presence of oxygen.  But again, that is addressed beautifully in the paper.

Photodegradation is not reversible. Bonds are broken, oxydised, and chains get crosslinked. To cite  from your first link above Photooxidation of organic materials is a major cause of irreversible deterioration for a large number of substances.

"add bromine" is a little ambiguous. "bromine containing compounds" would be better. But it is correct that photodegradation of them is also a source of colour in light-exposed plastics.


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Find reaction heat of a reaction
1) A(s)+O2(g) →Ao2; ΔH^0=-394kj/mol
2)AO(g)+ $\frac{1}{2}$O2(g) →AO2; ΔH^0=-283kj/mol
3)A(s)+ $\frac{1}{2}$O2(g)  →AO(g); ΔH^0=-Xkj/mol
Find the value of X? I want to know How to find this value.

You need to reverse the 2nd reaction.  This leaves you with Ao2 ---> frac{1}{2}O2(g) + AO(g).  And since you're reversing the reaction, you have to switch the sign of delta H, so it becomes 283 kj/mol (endothermic).  Once you do this, you can cancel out the AO2 and frac{1}{2}O2 with the 1st reaction.  This leaves you with your desired reaction, that is, A(s) + frac{1}{2}O2 ---> AO(g).  That is the reaction you want.  To find delta H for that reaction, simply add the delta H's from the previous reactions.  So, -394+283 = -111 kj/mol.  So your X-value is 111.  This process works because of Hess's Law.  In fact, it is a classic example of how Hess's law would be implemented in chemistry.
To wrap up:  The reason we reversed the 2nd reaction was because doing that allows us to cancel out some stuff with the 1st reaction in order to get the 3rd reaction.  Our goal is to get the third reaction you described.  The reason we switched the sign of delta H (from -283 to 283) is because if a reaction is exothermic (negative delta H) in one direction, it is endothermic (positive delta H) in the reverse direction.  That is just a rule of thumb; you can obtain further understanding of such concepts by learning about bond energies, activation energies and the like.
You can get further understanding of why Hess's Law works, as well as practice, here.


text:
Why is use of calcium carbide as an artificial fruit ripening agent banned?
Fruits like mangoes can be ripened using agents like ethylene, acetylene or calcium carbide. Of these, the latter is dangerous and is banned in most countries. Wikipedia says it is because calcium carbide has traces of arsenic and phosphorus. Is there any other reason?
Will it still be unsafe if pure calcium carbide is used?

I have merged your two questions into one. I normally wouldn't do that, but in this case they seem too related to keep them apart and the current answer partly already addresses your second concern.

Thanks for the edit! I was under the impression double questions were not allowed.

If they are that closely related, I believe it's okay, and even better to ask them together. You are addressing the same people who are answering any way, and the whole process is not too broad to be considered together. When they are conceptually more loosely related, it's preferable to keep them separated. That's why I said I normally wouldn't do this, so your impression is right, this is an exception.

Presumably  $\ce{CaC2}$ is only used to produce acetylene anyway by reacting with water.  I suspect this refers to atmospheric water which isn't well-controlled so could lead to a build-up of acetylene.  If the  $\ce{CaC2}$ is used in an acetylene generator that presumably counts as using acetylene not  $\ce{CaC2}$.

My guess is that pure CaC2 is costlier to synthesize and people would use the industrial grade version which contains harmful chemicals. Thus it is banned, not due to chemical reasons.

Calcium carbide is typically used by traders who can't afford a proper infrastructure for fruit ripening (sealed storage locations and gas generators). Carbide stones are put in permeable packets (typically paper bags) and hanged above the fruit boxes, where they absorb water from the air to produce acetylene gas.
This leads to two issues:

the concentration of acetylene is uncontrolled, and given enough moisture can reach explosive levels
permeable bags release fine particles of reaction byproducts which land on the fruits

Of course, pure calcium carbide would be safe, at least from the nutrition perspective. You'd still have the risk of explosion though, and because carbide cannot be used safely for fruit ripening, nobody is producing it in food grade in the first place.

Calcium carbide is in no way safe for consumption - it reacts with water to produce a gas and an alkali.

Why does acetylene explode? @gsurfer04. I guess the other product of hydrolysis is Calcium Hydroxide. Is that bad for fruits?

@Ashwin (1) acetylene is a combustible gas which, like most such gases is explosive when mixed with air. As for (2), pure byproducts would be mostly harmless in small concentration (though CaO may still damage the fruit skin in some cases). But nobody bothers to produce pure carbide, so you're likely to get some arsenic along with CaO.

Again, chances of poisoning would be slim to none if everyone washed their fruits properly, but food is supposed to be safe in all cases. It's like keeping rat poison next to your salt dispenser: it doesn't make your salt poisonous, but is very unsafe nevertheless.

Nice analogy to make me understand.

According to Anwar et al. the use of $\ce{CaC2}$ is discouraged due to

dangers of explosion and carryover of toxic materials like arsenic and phosphorus to consumers, thus making the healthy fruit poisonous. Since no technical knowledge is considered necessary for its anomalous use, higher quantity of calcium carbide needed to ripen immature fruit, makes them
  tasteless.

For references and more information, see: Anwar, R., A.U. Malik, M. Amin, A. Jabbar and B.A. Saleem, Int. J. Agri. Biol. 2008, 10, 35–41. Available at researchgate.net.

I have no idea what the second sentence of the quote is supposed to mean!

It is directly from the linked paper and seems to be badly written. This is what I understood. Since Calcium Carbide can be used by anyone, there is a chance that people will use a higher quantity as it should hasten ripening. This leads to the fruit becoming tasteless, in addition to it being poisoned.

Yeah, it's a bit poorly formatted, but I interpreted it as Ashwin did.  People with little to no understanding on the effects of the chemical use large quantities of it, and end up spoiling the taste.


text:
Calcium carbide as a ripening agent
Calcium Carbide is banned as a ripening agent in many countries because it contains traces of arsenic and phosphorus. Will it still be unsafe if pure Calcium Carbide is used?

Hello there! You've asked a very similar question a while back https://chemistry.stackexchange.com/questions/74623/artificial-ripening-of-fruits. Since what you're asking here also fits well in your previous question, you'd be well advised to to insert your query "Will it still be unsafe if pure Calcium Carbide is used?" in your previous post. As it stands, this particular post will be closed as a "duplicate" of your previous one. We look forward to your future contributions on Chem.SE. Cheers!


text:
One half of a nail is exposed to vinegar. Un-exposed half develops a brown film. What is it?
I have recently done an experiment for my chemistry EEI (Year 12) where we are testing what factors effect the rate of corrosion. I used vinegar (acetic acid) as one of the acids and Hydrochloric acid as the other.
In the one with the HCl, the results were as expected – the area exposed to the acid will corrode, the unexposed part won't (in the time frame I did).
However, for the vinegar, the results were different. The part exposed to the vinegar didn't show any rust however the part of it that wasn't exposed had a brown coating. Now I understand that the vinegar reacts with the rust, hence the reason we cant see it on the iron nail, but my question is: What is the brown coating on the top of the nail, and what is it caused from? Was there some sort of gas from the vinegar that caused it? If so, what is this gas called and how does it cause corrosion?

It could be simply that the moisture in the air rusted the unexposed part ... Though to confirm it you'll probably need to do more experiments.....

They were in a container with a lid. So i feel like this isn't the case. Especially since it didn't do it to any of the ones. (HCl, water, NaOH). I don't have time to do more tests because the final is due on Friday. hmmmm. Thanks anyway

Do you have an images? A Visual aid could make identification easier.

The nail is in two solutions. The lower solution is 5% acetic acid, saturated with air (20% $O_2$, 80% $N_2$). At 0.2 atmospheres, $O_2$ solubility in water is 40 mg/L x 0.2 = 8 mg/L, or about 8 parts per million.
The upper solution is a thin film on the nail (perhaps 10 microns thick - or less!) of $H_2O$ with some acetic acid (probably about 3-4%) and 8 ppm $O_2$. This film may have deposited from the lower solution by vapor deposition due to temperature differences, or by capillary action, as the meniscus crept up in a thin layer.
In the lower solution, $Fe^0$ goes into solution as $Fe^{++}$ and electrons go to $O_2$.
4$e^-$ + $O_2$ + 2 $H_2O$ --> 4 $OH^-$
The 8 ppm $O_2$ in the lower solution doesn't go very far, but the meniscus, and any film going higher up, will be continually refreshed with $O_2$, and will develop a higher pH and will precipitate iron hydroxide. Any iron dissolving in the lower solution will stay in solution because of the acidity, and the nail may not even show obvious signs of corrosion damage like pitting. The only obvious sign is the brown film of precipitated iron hydroxide.
This is an example of a differential aeration cell, where the upper solution is an oxygenated cathode, and the lower solution contains the anodic portion of the nail. http://www.corrosion-doctors.org/Corrosion-Factors-Cells/corrosion-cells-aeration.htm
We consider this corrosion in acetic acid to be an electrolytic reaction, and also the reaction of the nail with $HCl$, but the scale of cathodes and anodes in the case of $HCl$ is so small that we can't detect them. And in $HCl$ the $H_2$ is generated so fast that it bubbles off and doesn't need to be depolarized by dissolved oxygen.
Adding a picture suggesting the state of the nail:

https://www.ddcoatings.co.uk/2343/what-is-differential-aeration-corrosion.

Help! What is the source of discontent for this answer? If there is some confusion, let me address it.


text:
How does radiocarbon dating work?
I'd like to better understand radiocarbon dating, however, I do not know what resources I must use in this regard. Could someone walk me through the process of radiocarbon dating with a few examples? (I'm already well-versed with the basics of nuclear chemistry).

@para-nitrophenol https://chemistry.meta.stackexchange.com/q/549/4945

The radiocarbon method is in principle fairly simple. $^{14}\ce{C}$ is an isotope of Carbon that has a relative abundance of around $10^{-10} \%$ on earth.
Addition by @MaxW:

$^{14}\ce{C}$ is formed in the upper atmosphere from $^{14}\ce{N}$ so
  that the relative amount of $^{14}\ce{C}$ in the atmospheric
  $\ce{CO_2}$ is reasonably constant. Since plants get most of their
  carbon from atmospheric $\ce{CO_2}$, the relative amount of
  $^{14}\ce{C}$ to $\ce{^{12}C}$ is thus constant.

It is also not stable and decays $^{14}\ce{C ->}^{14}\ce{N} + e + v$.
Now a living organism based on carbon (like us) always has a stable amount of $^{14}\ce{C}$ in his body, because he has to eat and so there is a stable balance between $^{12}\ce{C}$, $^{13}\ce{C}$ and $^{14}\ce{C}$.
Now if that organism dies, it does not eat anymore, so there is no new income of $^{14}\ce{C}$. Since that decays and the other $\ce{C}$-isotopes don't, over the years, the percentage of $^{14}\ce{C}$ is getting lower and lower at a constant rate (you can calculate that rate with the half-time of $^{14}\ce{C}$).
Now when you measure the relative amount of $^{14}\ce{C}$ in a skeleton, you know since when it hasn't eaten anymore, so you know how old it is. The method can be used around 600-50k years to the past.
There are different methods of measuring isotope amounts. You can measure how fast a radioactive decay still is and calculate from that the amount of $^{14}\ce{C}$, which is the method from Libby.
You can also use Mass-Spectroscopy, because $^{14}\ce{C}$ is heavier than  $^{12}\ce{C}$ and $^{13}\ce{C}$ you can "easily" measure the amount of $^{14}\ce{C}$.
There are other methods, you can look this up in Wikipedia.

I think you missed the real hook. $^{14}\ce{C}$ is formed in the upper atmosphere from $^{14}\ce{N}$ so that the relative amount of $^{14}\ce{C}$ in the atmospheric $\ce{CO2}$ is reasonably constant. Since plants get most of their carbon from atmospheric $\ce{CO2}$, the relative amount of $^{14}\ce{C}$ to $^{12}\ce{C}$ is thus constant.

@MaxW hmpmpdfl maybe. I'll add that to the answer.

Tysm :) @MaxW your answer was really helpful


text:
Arrhenius equation in Kinetics and Thermodynamics
It is always stressed that one should never mix Kinetics with Thermodynamic. How is it then possible to calculate the free energy of activation with the Arrhenius equation? As I understood, the Arrhenius equation and thus the activation energy of a reaction "belongs" to kinetics, whereas the free energy represents a term of Thermodynamics.
An example of the calculation I refer to can be found on page 7 of this link:
http://cbc.arizona.edu/classes/bioc460/summer/460web/lecture/EnzIntrod_11.pdf

Free energy of activation ( = free energy of TS minus free energy of reactants) is still a kinetic concept. Compare this with free energy of reaction which is free energy of products minus free energy of reactants - that is a thermodynamic concept.