text:
Separating nickel coin into copper and nickel components
I'm trying to create a high school level lab in which we separate the metals in an alloy. The objective is to understand and apply standard reduction potentials, and also to do some extra redox reactions to reclaim solid metals. Plus I think it might be fun!
My plan was to simply submerge nickels in an acidic solution. Theoretically, as long as I don't use nitric acid, the nickel will dissolve and leave the solid copper. I tried it out in the fume hood (1 nickel, 100mL 5M HCl solution), but didn't see the telltale blue/green of nickel ions. I tried sanding the nickel, in case there was an inert oxide layer. For a few minutes, I saw a tiny bit of hydrogen evolution, then the sanded surface turned black.
I read that magnesium is sometimes added to coins for corrosion resistance, which is probably why I'm having problems. Does anyone have suggestions to make this lab possible? Maybe some electrochemistry?

Compare solubilities of Nickel and Copper salts. In particularm consider acetate. Also consider reduction of copper salts by iodide-ion

Such destruction of coins may by prohibited by law also some coins have added stuff to prevent obtaining pure copper electrolytically as pure metals may be more valuable than coin.

It is not nickel and copper, it is an alloy, rather like a compound. You must dissolve the total and then precipitate or electroplate to separate.

I recommend to dissolve the copper-nickel alloy in nitric acid and determine the dissolved $\ce{Cu(II)}$ by iodometric titration.

Start by weighing the coin, then dissolve it by placing it in nitric acid solution (under a fume hood). Continue to boil the solution for 20 min after complete dissolution of the coin (to remove any traces of remaining $\ce{NO2}$ in soution. The volume of the solution is about $40 \,\mathrm{mL}$.
$$\ce{Cu(s) + 4HNO3(aq)->Cu^{2+}(aq) + 2NO3^- (aq) + 2NO2(g) + 2H2O(l)}$$
$$\ce{Ni(s) + 2H3O^+(aq)->Ni^{2+}(aq) + H2 (g)  + 2H2O(l)}$$
Prepare a fresh solution of sodium thiosulfate of precise concentration $\approx 0.04\,\mathrm{M}$
Add $1.00 \,\mathrm{mL}$ of the diluted copper-containing solution to $15 \,\mathrm{mL}$ of $10\% (W/V)$ potassium iodide solution.
$$\ce{Cu^{2+}(aq) +  5I^- (aq)->CuI(s) + I3^- (aq) }$$
Titrate the yellow-orange suspension with sodium thiosulfate solution until the solution color has become yellow. To detect the end-point add about $1 \,\mathrm{mL}$ of starch solution.
$$\ce{I3^- (aq)  +  2S2O3^{2-} (aq)->S4O6^{2-} (aq) + 3I^- (aq) }$$
Repeat the titration to precisely determine the end-point.

Add coin into nitric acid and sulfuric acid. Nitric should be 60% weight of the coin and sulfuric should be 1.6 times. It will dissolve in minutes. Heat it up to boil unused nitric. Add some urea to neutralize acids. Add iron fillings to remove copper. Filter and melt the copper powder. You will have pure copper. Take rest of the solution, dilute it a bit, Add hydroxide peroxide twice of the weight of copper in the coin. Add calcium hydroxide or soda ash and bring pH to 3.5-4. That will precipitate all the iron. Filter and wash. The liquid has nickle. Add caustic soda and bring pH to 8.5. Take out nickle hydroxide. Wash it. Roast to 300C. You will have nickle oxide. Either go by carbothermic reduction or go for thermite. Thermite is easy. Add aluminium powder  in nickle oxide put it in a crucible and and heat it with torch. It will start to reduce and you will have pure nickle.

Yomen Atassi has given you a anayltical method for the measurement of the copper content. I will give you a chemical method for the separation of the two metals.
If you want to separate the metals then I would use the following method, I would dissolve the coins by immersing them in hydrochloric acid with air present. This will convert the nickel / copper alloy into a solution of copper chloride and nickel chloride.
This might take a while as you need to wait for the coins to undergo oxidation in the acid.
Next dilute the mixture to a suitable concentration and filter it to get a nice and clear liquid. To separate the copper, I would suggest that you shake the copper / nickel mixture in hydrochloric acid with 30 % aliquat 336 diluted with aromatic kerosene (or ethyl benzene). Do this in a fume hood.
The copper will be extracted by this solution of an ionic liquid in the aromatic solvent and the nickel will be left behind in the aqueous solution. To recover the copper from the organic layer, shake it with dilute hydrochloric acid or an EDTA solution.
In a school lab the use of toluene and aliquat 336 might not be viewed as reasonable for health and safety reasons. If you want to do a method which avoids volatile organics and a toxic ionic liquid. Then use DOWEX 1, this is a cross linked polystyrene which has tetralakyl ammonium salt groups on it. This will function in the same way as aliquat 336 in toluene.
When the chloride concentration is high the copper will bind to the resin in the form of [CuCl4]2- anions. So by passing the copper / nickel solution through a pad of resin you can remove the copper onto the resin. To obtain the copper, I would first wash the resin with hydrochloric acid to wash off the nickel solution, then I would elute the column with water to get the copper back.
I my lab I would use ICPOES to measure the copper in the solution, but I think you could use the titration method of Yomen.
To get ready to do the elution separation method, you may have to do some batch experiments. In these you put a known amount of the copper containing solution in a vial with a known amount of resin. After allowing the system to come to equilibrium you measure the copper content of the solution. You also compare the copper content with the original solution.
In this way you determine how much copper was adsorbed per gram of resin, you should divide the concentration of the copper on the resin (moles Cu per gram of resin) by the concentration of the copper in solution (moles Cu per ml of solution) to obtain the Rd (Kd) value.
I would want to make this measurement at differnet resin to liquid ratios with a moderate amount of copper present, this would allow you to determine the amount of copper which the resin is able to bind. This is an important thing to know.
With a small amount of copper, I would want to make the measurements for copper with different amounts of sodium chloride added. This will allow you to measure Rd as a function of the chloride concentration. It will allow you to better design the copper separation process. You could use the titration method to make the copper measurements.
For a dilute solution of hydrochloric acid you can assume that the acid is totally dissociated, but for very strong hydrochloric acid the chloride concentration is lower than the anayltical acid concentration.


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What topics to review before starting organic chemistry?
It's been over a semester since my most recent chemistry class, (general chemistry 2) because I could not take organic at my community college. It is only considered a sophomore level class unless taken in-residence at the university I am transferring to (which is understandable.)
I was curious if I could narrow down what topics I could begin reviewing before the start of the semester that would help me in organic chemistry.

Bonding, hybridization, and resonance seem the most vital to me. Welcome to chem.SE, but this raises my alarms as primarily opinion-based. We can never know how much time you have, how much you're gonna spend on them etc.

Thanks for your answer. I don't really need a study tutorial or anything, just topics in lower level general chemistry that apply to organic chemistry, it seems like that lies in the realm of the factual.  Sorry for the confusion.

Well, FTW pretty much everything in chem101 and chem102 will be needed for organochem studies, depending on how deep you wanna dig in.

Understanding Aromaticity, huckel's rule and conjugation will be helpful.

@LiamH, OP is asking about prerequisites for an organic chemistry course, you're talking about the actual content of it. The concepts you mentioned are undoubtedly important, but they are introduced in the course itself, OP does not need to know them before entering the course.

As @inɒzɘmɒЯ.A.M pointed out, the usual prerequisite for an organic chemistry course is that a general chemistry course is already taken and successfully passed. Topics include:  atomic theory, periodic table, nomenclature, stoichiometry, bonding, molecular geometry, basic types of reactions, acids and bases, solubility, thermodynamics, kinetics, coordination chemistry.

@Wild "Meow" Cat - That's one good answer IMO.

As @inɒzɘmɒЯ.A.M pointed out in his comment, the usual prerequisite for an organic chemistry course is that a general chemistry course is already taken and successfully passed. The general chemistry is, of course, a rather large topic, so I'll just name a few subtopics which I feel are absolutely essential to start a venture into the world of organic chemistry.

Structure of matter.

Modern atomic theory (Schrödinger equation, atomic orbitals, spin,  Pauli exclusion principle, electron configurations).
The periodic table (periodicity, group trends).
Chemical bonding (covalent and ionic bonding, Lewis structures, resonance, VSEPR theory , introductory Valence Bond Theory and Molecular Orbital Theory).

Chemical reactions.

Nomenclature & stoichiometry (empirical and molecular formulas, chemical equations, mole, limiting reagents, yield).
Solutions (solubility, concentration, properties of solutions).
Acid-base reactions (definitions of acids and bases, pH, strong/weak acids and bases, trends).
Oxidation-reduction reactions (oxidation numbers, balancing oxidation-reduction reactions).

Thermodynamics.

Basic concepts (energy, work, heat, enthalpy, entropy, Gibbs free energy ).
The first and the second laws of thermodynamics.
Hess's law.

Kinetics.

Reaction rates (temperature dependence, Arrhenius equation).
Reaction orders (first and second order reactions, molecularity, rate laws).
Catalysis.

Chemical equilibrium.

Basic principles (the notion of equilibrium, the law of mass action, equilibrium constants, the direction of a reaction).
Shifting the equilibrium (changing amounts, pressure, volume, temperature).

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text:
Phlorizin content in grapes
I work in a winery and we have some problems about a new process involving phlorizin. We know that phlorizin is the major dihydrochalcone found in apples.
I have been reading that this compound is also found in a grape named Albariño, a white variety of grape. What else is known about the phlorizin content in grapes?

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

https://www.bio-conferences.org/articles/bioconf/pdf/2016/02/bioconf-oiv2016_02006.pdf

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Ag-Sn-Cu Phase diagram to formulate homogenization temperature / duration
I am working on Ag-Sn-Cu alloy phase diagram to figure out the best temperature / duration to homogenize the alloy (annealing). The %age of elements in the alloy are

Ag - 40 %
Cu - 27.8 %
Sn - 32.2 %

(Composition can be varied if necessary but not more than 5 %)
Now the main goal is:

Find out the temperature / time duration at which the ingot of the above said alloy should be annealed to get the maximum Ag3Sn phase in the final ingot. Ag3Sn is the most desired phase with Cu3Sn being the second desired phase. Anyhow the main goal is to make as much Ag3Sn as possible.
The time duration for which the ingot should be placed at the desired temperature. (The total ingot is usually 6–7 kg)
[Supplementary question] Should the temperature be gradually reduced to the desired level of maximum Ag3Sn yield (or gradually increased)?

The phase diagram of Ag-Sn-Cu is as follows:

I'm away from my relevant databases now, but two comments: phase diagrams will not help with kinetic questions, and that particular phase diagram is not the best for your question. Early next week I'll have s real answer for you.

thanks @JonCuster .. Any tip to lead me in the right direction is much appreciated.

Much of this answer is based on a document available from NIST at Properties of Ternary Copper-Silver Systems [originally J. Phys. Chem. Ref. Data, 6(3) 621-673 (1977)] - this has an equivalent to the diagram in the question (which is just the surface of the liquidus). Now, since you are talking about balances of solid phases, it is also worth looking at the binary diagrams to discuss those.
For Ag-Sn, one gets:

For Cu-Sn there is:

Now I'll throw out a few observations:
Ag$_{3}$Sn is a line compound, which in the binary system is not stable above 480C. In the NIST report this phase is referenced as $\theta$, but called Ag$_{3}$Sn($\epsilon$), causing great confusion since the $\zeta$ Ag-Sn phase is designated $\epsilon_{2}$, and Cu$_{3}$Sn($\epsilon$) is designated $\epsilon_{1}$. So, watch it...
The Cu$_{3}$Sn line compound converts to the fcc $\gamma$ phase at around 676C. The $\gamma$ phase is sometimes also referred to as Cu$_{3}$Sn (yet more confusion possible). The Cu$_{3}$Sn and Ag$_{3}$Sn line compounds are different crystal structures, and do not form a solid solution across the ternary.
For both Ag-Sn and Cu-Sn, another thing to notice is the various other phases that are in close proximity to the Ag$_{3}$Sn and Cu$_{3}$Sn line compounds. Going into the ternary system, you get questions like how adding Cu into $\zeta$Ag-Sn impacts the free energy.
So, what now? In the NIST report, on page 664 is a pseudo-binary diagram where the Sn wt% is held at 20%, and Cu wt% is varied from 0 to 80 (balance Ag). Personally I hate weight percent, but...  The point that one can eventually pick out is that the Ag$_{3}$Sn phase, designated $\theta$ appears almost nowhere - there is just a little sliver way over near 0 wt% Cu and under 250C. It appears that with the addition of Cu, the $\zeta$Ag-Sn phase rapidly replaces the Ag$_{3}$Sn phase.
So, as far as I can tell, I'm not sure that the kinetics even matter here. You basically can't get much Ag$_{3}$Sn to form within the ternary parameters you have chosen - you will get $\zeta$Ag-Sn instead.

Insightful. That explains a lot. I have been looking into other documentations and here is something which I found can be a little importance. This paper  on page 735 explains the behaviour and existence of phases. Ag3Sn & Cu3Sn & liq would exist between 385 - 435.  Though the Ternary parameters are not exactly wo what I'd like but still .... What is your take?

And the main goal of the question is to maximize the Ag3Sn .. We can compromise on the Cu3Sn phase.

Interesting. There would appear to be some contradictions between your reference and mine.  The NIST ternary and pseudo-binaries are quite complicated compared to the Fima paper you cite. On the other hand, all of their data (on, say Figure 3) are in the simplest part of the ternary, away from ugliness of the low Sn content areas of the Ag-Sn and Cu-Sn diagrams.  But, take their Figure 3 as fact - where does that leave you? If you try to solidify at ~25 at.% Sn and Ag rich, you will first form a liquid+$\zeta$ mix, and then hope that you can convert the $\zeta$ to Ag$_{3}$Sn. Not easy.

Well, the original ingredients %wt would be 32.2 % Sn, with Ag 40% n Copper 27.8%  .. (not exactly Fig 3 but somehow close, fingers crossed).  My brainwave is to cool down the melt to around 400 C and keep it there, middle point to avoid the zeta & high enough to sustain acceptable Ag3Sn precipitation. Your thoughts ?

You can't avoid passing through the liquid-zeta region unless you splat cool (perhaps). So you will get some zeta forming. Then the liquid will get richer in Sn. Once you finally hit the top of the Ag$_{3}$Sn line you will start forming it and stuff to the Sn-rich side of the curve.  To convert whatever zeta you made will require Sn to diffuse back into the zeta and have the zeta nucleate and transform into zeta.  A tough problem.

Well the method tried right now quenching the ingot once its 700 - 800 to room temp and then heating it again. Another point, with some maths, the amount of Sn (33.2 %) is used up in creating Ag3Sn & Cu3Sn (if both are formed completely), so while we are creating Ag3Sn & Cu3Sn, Sn is depleting as well.

To simplify the solidification issue - assume pure Ag-Sn, starting with 25 at.% Sn, just perfect to end up with pure Ag${3}$Sn. You melt it all together and start cooling.  At around 950K you nucleate fcc Ag with ~12 at.% Sn in it. As you continue cooling, the liquid moves down the liquidus line to the right, and at 900K you have ~30 at.% Sn in the liquid, and the stable solid phase jumps from fcc Ag(Sn) to zeta Ag-Sn. You continue cooling down to ~760K where Ag${3}$Sn becomes the preferred stable solid phase, and the liquid has 45 at.%Sn in it.

So, you have made a good percentage of solid, but not the right phase, and if that won't transform quickly (and it won't) you have to wait for diffusion.  Your approach is about as good as it gets for an ingot, but it might take a long time to convert.

You are right, but this doesn't yield better results (tested) .. I have another approach in mind. What if we plot the precipitation kinetics? I came across a tool called pandat which can plot the precipitation of an alloy phases over time. But it requires kinetic databases for certain parameters. To build that we need the following parameters for Ag-Sn-Cu alloy.
Molar_Volume,  Interfacial_Energy, Atomic_Spacing, Nucleation_Site_Parameter, Driving_Force_Factor.

As per my discussion with their customer service,

Molar_Volume and Atomic_Spacing of a certain crystal structure. You do not need the exact values, but the right magnitude.
Driving_Force_Factoris usually given as 1
Nucleation_Site_Parameter, which is related to defect, is usually given a value in the order of 10^-4 to 10^-6
Interfacial_Energy is usually not available, you will need some experimental data on the precipitates, such as evolution of particle size and volume fraction to optimize this kinetic parameter.

Do you know of any knowledge base which can help in getting these details for Ag-Cu-Sn ternary system?

From another question it looks like you have tried CompuTherm with no luck. The other big player is ThermoCalc (thermocalc.com). I don't believe it comes cheap, but you could at least check with them if their software could do the job (or if they would consult on it).

Interestingly, CompuTherm Pandat demo allows you to analyze alloys with upto 3 elemental compositions ...

Thanks for recommending Thermo-Calc btw, great tool. I was looking at property diagrams for the alloy composition above & this is what I get for different temperatures. Up till 435 Ag3Sn phase is produced.

One more thing, The actual phase composition for AG3SN_L60_CU3TI is (Ag,Cu)0.75(Ag,Sn)0.25 & CU3SNSB_GAMMA_IM is (Cu)0.75 (Cu,Sn)0.25 .. How do we find out how much Ag3Sn total exist in all of this system?

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Is there any experimental evidence for OClF5?
On page 88 and 89 of this document, the author laments that $\ce{OClF5}$ has still not been discovered. He says it would be "just about the ultimate possible storable oxidizer" (for rockets).
(Note: it's pages 88 and 89 as labeled by the writing on the page, not the pdf metadata.)
The text seems a little dated. Has anyone ever found/produced any $\ce{OClF5}$? A related compound, $\ce{OClF3}$, was made in 1970 (footnote on page 89). So this seems to indicate that one of those weird loopy pairs of electrons can be jammed/bonded with an atom of oxygen.

In ClF5 there's an axial lone pair so your compound may be possible.

The isoelectronic ClF6+ ion is known, but you have to hit ClF5 pretty hard, with PtF6 or [KrF][AsF6] according to Housecroft and Sharp. Greenwood and Earnshaw say "F5ClO has been claimed but the report could not be confirmed", but doesn't give the reference.

Yes there has been experimental evidence published.  Reaction of CIF5,  with OF2: Preparation and Properties of Chlorine Oxide Trifluoride and the Existence of Chlorine Oxide Pentafluoride, Angewandte Chemie, Internation Edition, vol. 11, pages 1094-5, purports to show a series on F-19 NMR spectra of the compound, at temperatures ranging from 34 to -76 degrees C.
The series shows a relatively sharp peak at 34 degrees C, which progressively broadens and shifts upon cooling to until about -40, upon which further cooling sharpens to a peak with much different chemical shift than the room temperature peak.  The authors conclude that at room temperature there is fast exchange among the 5 F atoms, and at low temperature 4 equivalent equatorial F atoms are observed, while the peak for the axial F atom is too broad to be observed.
However, the above claim is considered refuted by "Chlorine Oxyfluorides" Advances in Inorganic Chemistry and Radiochemistry vol. 18, pages 319-98.
See On the XeF+/H2O System: Synthesis and Characterization of the Xenon(II) Oxide Fluoride Cation, FXeOXeFXeF+ J. Am. Chem. Soc., vol. 131, pages 13474–89 for further discussion and references:

Chlorine oxide pentafluoride, ClOF5, would be the highest performing earth-storable, liquid rocket-propellant oxidizer. Although the synthesis of ClOF5 had been claimed in 1972,(1) this claim was subsequently refuted.(2) Continued efforts to prepare this compound by oxidative fluorination of substrates such as ClOF3(3-5) or ClOF4−(6, 7) have failed because the oxygen ligand is more easily oxidized than the central chlorine atom. This failure prompted a search for alternate synthetic approaches, such as the reaction of ClF5 with an oxidative oxygenator. Potential candidates for oxidative oxygenators include hypofluorous acid, HOF, a well-known unstable compound,(8, 9) and its protonated form, the H2OF+ cation,(10, 11) if indeed it should exist.


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SN1 reactivity order
The question is to compare the SN1 reactivity for the following compounds. According to me it should be like R>S>Q>P.
But the correct answer is Q>P>R>S. How can a lone pair donating into a vacant p be more stable than allyl/benzyl carbocation?
These are the four compounds in the order PQRS respectively:

The most crucial pair to compare is the methoxymethyl (P) with benzyl (R) and then it's easier to devise Q>P and R>S (as you did). Why do you think (R) should give a more stable cation than (P) ? I suggest you to spend some more time pondering on that question - it's going to pay off later. btw this is a nice practical question. All these reagents would be routinely used in a med chem project for example to optimise an ether molecule.

@abhirikshma Yes, such cations are more stable, why you think they wouldn't be?

I dont know, but I tend to have a general notion that benzyl cations are very stable because of the extent of delocalisation of the positive charge. On the other hand, lone pair donating into a vacant p does not have any such attributes.

A cation derived from Q can be thought of as having a double bond $\ce{N=C}$ and plus on N. It is basically the same as ammonium cation with some organic substituents. These are stable, like real compounds; you may take some of them off the shelf. As for benzyl cation, it is indeed very stable... for a reaction intermediate. That's a wholly different degree of "very stable". Well, with suitable anion like $\mathrm{SbF_6^-}$ you may be able to isolate one as well, but that requires some effort.

Benzyl and allyl cations are stabilised by the neighbouring $\unicode[Times]{x3C0}$-systems. The electrons in the $\unicode[Times]{x3C0}$-system are actually required for (formally) form double bonds to ensure all-octet structures as well as possible. The cationic structures will contain an electron sextet somewhere (most often drawn on the benzylic carbon for the benzyl cation).
The $\unicode[Times]{x3B1}$-oxygen and $\unicode[Times]{x3B1}$-nitrogen systems on the other hand contain free lone pairs on the heteroatoms. These are not used for double bonds, so they can be shared with the cation to form a new double bond. The resulting mesomeric structure is an all-octet structure which are usually much more stable than non-octet structures (at least for main group elements).
The nitrogen-containing carbocation could even, after the double bond has been formed, loose its proton to create a neutral all-octet structure, the formal $\ce{HCl}$ elimination product.

In carbocation stability, mesomeric effect has highest priority, then hyperconjugation and then resonance.
Mesomeric effect of NH in 'Q' is greater than O in 'P'.
'R' has resonance and in 'S' carbocation has least stability.

What is the difference betwen mesomeric and resonence effect     ?

There's no difference and they are all stabilized mesomerically, only more ore less strongly.


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Inductive effect of phenyl ring
I completely understand how phenyl group 'releases' or 'donates' electron through resonance, but how is that a phenyl group shows a weak -I effect, despite having electrons literally smeared on itself?

Related: Why is phenyl weakly activating?

As the $\mathrm{s}$ characteristic of an atom increases, so too does its electronegativity. Proof of this can be seen when comparing the $\mathrm{p}K_\mathrm{a}$'s of 2-carbon hydrocarbons. The $\mathrm{p}K_\mathrm{a}$'s of $\ce{CH3 -CH3}$, $\ce{CH2 =CH2}$, and $\ce{CH ≡CH}$ are $51$, $44$, and $25$ respectively.
When considering phenyl group for what it is, a ring of six $\mathrm{sp^2}$-hybridized carbons, it is easy to see why it shows a strong electron withdrawing effect, and a weak electron donating effect. The electronegativity of the carbons make benzene a nucleophilic structure that does not easily give up its electron cloud.

in phenyl group the carbons are sp^2 hybridised which are having less s character and are thus electronegative in nature hence shows -I effect

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text:
How can I light a fire in this case?
Is there any gas that contains oxygen so that it doesn't require oxygen from the environment in order to burn?
What I am trying to do is use LPG gas, which is fed through a pipe to a burner that is placed in an environment that has no air, somewhat like a vacuum. Is there any way to light the burner inside that vacuum environment?
A few wild ideas that I had included finding some gas that contains oxygen in itself. I may be wrong.
EDIT - The question doesn't end here. Please read the comments section below for any doubts that you might have. And if it isn't answered in comments section then ONLY comment.

you mean you want to burn fire in vacuum?

The burner would be kept in partial vacuum. Somewhat like a closed vessel. I was thinking that if there is something that contains oxygen in the gas itself? in other words, the oxygen/air would be mixed in the LPG/any other gas, itself.

Why dont you simply mix oxygen with the gas in the required amount?

Is that even possible? Considering the gas to be LPG (commonly used in Indian households).

You really don't need oxygen. Vacuum means there should be nothing there, so this is somewhat unanswerable. Not the downvoter though.

@inɒzɘmɒЯ.A.M Vacuum in practical sense means, lower pressure than atmospheric presssure. What you're saying is in theoretical sense. I don't think meant by the word "vacuum" that nothing should be there. Then the question itself becomes meaningless!

Well, @Curious I'm pretty sure the reason for at least 2 of the downvotes is this thing's not being clear. How could we simply assert they meant the practical sense? We can't. That's the info the OP has to give, preferably in the original revision. Furthermore, it's also unclear what confusion the OP had in mind. Considering this comment, I can deduce they do not know what they're talking about.

What do you get inside a pressure vessel? It's not entirely vacuum but "somewhat like vacuum" and do notice that I've mentioned somewhat. I didn't mean to say complete vacuum. Anyway, now I guess the doubt is clear? If it still isn't then let me know and I'll try to clear the doubts. :) I'm just trying to know that is there any mixture/gas that doesn't require air from environment and can contain the air within itself. That's it. If there's something like this, then let me know. Thanks :)

Would you simply try to make a rocket or sth? All rockets take oxidizers with them. Also explosive often use added oxidizers.

If you burn stuff inside a vacuum it isn't really vacuum anymore, is it?

@Jori Please go through the above comments. Moreover, I'm trying to burn fire inside an environment with no air  (I don't mean that the pipe from which is gas would come can't contain air). The condition is that the environment doesn't have air but you CAN provide air through the pipe from which the gas is coming. So, I'm basically asking that is it possible? Is it possible to mix gas with air/equivalent so that it burns in such environment? ~ Made it bold because people are asking same thing over and over.

@Anoneemus After going through the giant comment chain, I'd like to ask: is the vacuum effectively finite in size? Basically, if you have a box (sitting on your desk) that's a vacuum and you want to know if you can burn something in there, the answer is "Of course. Fill the box with air first." If the box is effectively infinite in size (like outer space), there are some more interesting problems that pop up.

Please add all relevant information to the question itself. Going through that comment chain is a nightmare and I still do not really understand what you are trying to achieve. Comments are expendable and this question has now attracted so many, that they need to be cleaned soon. The post is unclear and the comment thread is too long, this is probably the reason for such a controversial voting (+4/ -5) and yes, I did downvote.

If you're really looking for heat in a vessel with no air, you should instead think about an electric burner. You don't need to worry about fuel mixing ratios, etc, because you do not require fuel or oxidizer. And you can keep it completely sealed if you want (even have a complete vacuum), with no affect on the heat delivered.

The most convenient solution for your question can be the Hydrooxy gas (also sometimes called the Brown’s gas). Simply put its water split into hydrogen and oxygen. Hydrogen and oxygen can be combined back by ignition and can create a maximum temperature up to 2800 °C (around 600–700 °C hotter than burning hydrogen in air), which makes it a good fuel for metal welding and cutting.
Though 2:1 hydrogen and oxygen ratio is enough to produce water via combustion, but on a practical solution you will need around 3:1 to 5:1 ratio to avoid oxidizing flames. The temperature you can achieve by burning hydrogen oxygen mix varies, depending on the ratio of both gases used.
Hydrogen and oxygen can be obtained via simple electrolysis.
$$\ce{2H2O + Energy -> 2H2 + O2}$$
and combined back as
$$\ce{2H2 + O2 -> 2H2O + Energy}$$
It might be worth noting that for all practical purposes, the energy you use to split hydrogen and oxygen will always be greater than what you can get by combining them back (like what happens in every combustion engine, humans have ever created).
If you are planning to develop an actual application, there are many precautions that you would need to consider, the most important of which would be back-fire protection (a common problem with gas based welding), so that the flame doesn’t reach back into the gas tank, which of course will explode.
So goes without saying, but you might want to adopt some precautionary measures if you do use it in any practical application. Temperatures this high can seriously hurt a person even if the contact was for some milliseconds. It can also damage other materials if the flame is directed at them.

Yes. Oxyhydrogen is a mixture of oxygen and hydrogen, and has been or is being used in applications ranging from lighting to welding. The oxygen is mixed with hydrogen in a ratio such that the hydrogen is burned with the oxygen supplied in the gas mixture. This should meet your requirements.

TNT(trinitrotoluene),Potassium permanganate,Potassium dichromate,hydrogen peroxide,fluorine are some of the oxidising agent which you can use light up LPG without requirement of atmospheric oxygen.


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Improving copper contact and preventing oxidation
I have a model railway in which I use pure copper on the tracks and in the engine to power the train with electricity.
Is there any solution to prevent the copper surfaces from corrosion, and maybe improve the conductivity?

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Though I have not tried them DeoxIT products claim to do what you want, available as liquid and spray.
If anyone has used these, please add a comment as to effectiveness.

It seems promising, but 7USD for 2ml just doen't worth it...


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Is my understanding of the Frasch Process correct?
So basically, the whole point of this process is to extract elemental sulfur from underground deposits. Is this process essentially correct or am I missing out certain details?

Firstly, three concentric pipes are placed into the underground sulfur deposit
Then, water that has been superheated is pressurised into the sulfur deposit through the outermost pipes.
The result of the above is melted sulfur stuck in the ground in an emulsion. The pressure of the chamber is also increased (i don't know why?)
To get it out of the ground, compressed air is injected through the innermost pipe: this forces the sulfur-water emulsion to the surface through the middle pipe
When this mixture is cooled, the solid sulfur separates from water and 99.5% pure sulfur is obtained

Is this basically the process? Also, are these the properties of sulfur that this process relies on:

Sulfur has a low density, hence the compressed air is able to lift the sulfur-water emulsion to the surface
Sulfur is insoluble in water and not reactive. Hence, sulfur can be recovered more easily after the sulfur-water emulsion cools as it does not mix with the water, nor react with it.

There are also some problems with this process if I am right:

The hot water used in this process causes thermal pollution, and may also contain some dissolved minerals. This may disrupt the chemical composition of local ecosystems, so one must recycle it.
Depending on the make up of the area, remaining crypts can lead to earth subsidence. These are also harder to refill.
This process releases sulfur, which must be quickly cooled otherwise it oxidises to sulfur dioxide or may be reduced to hydrogen sulfide, both of which are serious air pollutants.

Is this process economically viable? There is a huge glut of elemental sulfur on the market now due to the fact that it is produced as a by-product of petroleum refining. Pyramid-sized piles of sulfur are growing in Canada and the Middle East, among other places.


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How can we tell whether a phenomena was a chemical reaction or not?
I mean in principle, in general.
How can we tell if some phenomena was a chemical reaction(s)? Change in physical properties cannot be since large-scale structural change can change physical properties as well, like heat treatment of steel. And what about changing diamond into graphite by heating. Hence, assuming there are only chemical and physical properties, only a change in chemical properties implies chemical reaction. But how do we know if chemical properties changed? Observing chemical reactions of the result(s). But how do we know what happens is a chemical reaction? A different instance of the same problem, which we started with.
How can anyone sure claim that a chemical reaction happened, or equivalently that some sample is a different substance and not just an allotrope or has larger "grains" for example?
Maybe mass or energy change can rule out those cases?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. || I know there was little actual experiments going on before Boyle, so the scope of this question might be somewhere between Boyle and Dalton, or maybe Faraday. IIRC, many believed that when you see a change in odor, color or luster, there was a chemical reaction going on. Since the elements' idea was not really established, it's hard to reckon whether they actually recognized the products.

It's a difficult, quite possible impossible, question. First you have to say what, exactly, you mean by chemical properties and a chemical reaction. You indicate the problems, but to play devils advocate and take deliberately extreme examples I've seen A+BC(v=0)->A+BC(v=1) where v shows the vibrational quantum number called a reaction. Or does forming a clathrate count as a chemical reaction? Or an endo-fullerene compund? And how about the difference between chemi- and physisorbtion. And formation of charge transfer complexes, is that a reaction? It's all too tricky - I'm off too the pub!

For your level it would suffix to say that you start off with some chemicals and after they react that you end up with different chemicals. // Chemical reactions do not create mass differences (I won't go much much deeper and confuse you...). So you start off with 100 grams of chemicals (reactants) and you end up with 100 grams of chemicals (products).

For some chemical reactions, it would be very difficult to prove without chemical analysis of the materials before and after the reaction. However, with modern analysis techniques, you could show whether a reaction took place. For example, if you tempered steel or converted diamond to graphite, analysis would show that it was a physical change or change in the crystal structure that occurred.
For a low-tech example, consider the combustion of some condensed-phase hydrocarbon in oxygen. If you ignite the hydrocarbon under oxygen in a container sealed except for an oxygen source and an exit to a condenser, you would see the hydrocarbon vanish and water would appear in the condenser. For this we'll just ignore carbon dioxide production. You then easily confirm that the water is not just another form of the hydrocarbon by measuring it's freezing and boiling temperatures. You could then repeat the experiment without ignition of the hydrocarbon and see that you get nothing.
This would confirm beyond any reasonable doubt that you've had a chemical reaction that converted the hydrocarbon to water.


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Titration of reducing sugars on honey
So I'm trying to figure out how to titrate the reducing sugars on honey without generating iodine.
The reaction so far I think I could do is do a Fehling reaction with $\ce{Cu}^{2+}$and then back-titrate the remaining with EDTA, but I'm not sure that the EDTA wouldn't solubilize the $\ce{Cu2O}$, or be able to displace the tartrate.
Some sources I've looked say that this can be done as a direct titration using methylene blue as an indicator, but they haven't given a pH range, even if the reaction is fast enough to be done as a direct titration.
Any thoughts?

The usual titration of glucose with Fehling's Reagent is to place the Fehling's Reagent in a flask and titrate the boiling solution with glucose. At the end point, all the $\ce {Cu^{2+}}$ ions are converted to $\ce {Cu^{+}}$; this makes the solution colorless. That is difficult to see since the red copper(I) oxide is floating in the solution.
With methylene blue, after the copper has been reduced, the glucose can then reduce the methylene blue (in a basic solution) to a colorless solution. This makes the end point clearer since the methylene blue is much darker than the $\ce {Cu^{2+}}$ ions. The pH of the solution does not change much in this reaction.


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Why can't I reproduce this photolytic reaction of ferrocyanide?
I was so intrigued by the reported observation that sunlight precipitates the iron from $\ce{K4Fe(CN)6}$ (originally attributed to Matuschek, 1901) that I wanted to see it myself.  I put a saturated aqueous solution, uncovered, in direct sunlight for four hours at mid-day.  Absolutely nothing precipitated from the solution!
My understanding from the given reference is that in solution the $\ce{K4Fe(CN)6}$ gives $\ce{Fe(CN)6^{4−}}$ and then:
$\ce{Fe(CN)6^{4−} + 2H2O <-> Fe(CN)5 + (H2O)^{-3} + HCN + OH-}$
The photolysis causes $\ce{Fe(CN)6^{4−} ->[h\nu] Fe(CN)6^{3−} + e-}$,  so then we also have
$\ce{Fe(CN)6^{3−} + 2H2O <-> Fe(CN)5 + (H2O)^{-2} + HCN + OH-}$
Is it correct to assume that if a $\ce{Fe(CN)6^{4-}}$ anion absorbs a photon of adequate energy (in this case it appears to require $\lambda < 313nm$) then with 100% probability the anion will photolyze the Fe bond?
And are there models that predict (at least order of magnitude) the absorption probability of a photon by an aqueous anion (being irradiated with a given spectral flux)?
Furthermore, what is the reaction that would precipitate iron in this case?  When we bring the K back into the equations it looks like the HCN would react to produce KCN.  I don't know what should happen to $\ce{Fe(CN)5}$.
In fact, after several days in open sunlight the solution is subjectively getting darker and I do see a small amount of red sediment at the bottom of the dish that appears to redissolve on swirling.  Could be an iron oxide, but would also be consistent with $\ce{K3Fe(CN)6}$.  In either case I don't have any indication from the source what should happen to the $\ce{(CN)5}$ groups if iron is precipitating, or what else might take one of the K atoms from the initial compound.
(Ultimately I had hoped to quantify the rate of photolysis by weighing the precipitate.)

You do realize that the glassware (unless it's quartz) strongly absorbs UV light below 300 nm? If you want to use UV light below 350 nm, I suggest you shine it directly on the solution and not have it pass through glass.

@LDC3: Yes, it's a shallow glass container that I left uncovered.  Presumably the heating of the solution by the sunlight and elevated ambient temperature is offsetting any evaporation that might be occurring that would cause the initial solute to precipitate.

If your solution is basic, then $\ce {Fe(OH)_3}$ will precipitate.

@LDC3: I confirmed its pH is 8.6.  The amount of red sediment that has settled on the bottom is miniscule.  Also, curiously, what little is there does not follow a magnet.  So if some form of Fe is precipitating but staying in suspension there goes my idea of magnetic separation....

8.6 is barely basic. I was thinking 10 or higher.

I don't really understand the question, but I just noticed you are talking about an iron (II) species in the first sentence, i.e. $\ce{K4[Fe(CN)6]}$ and an iron (III) species later, i.e. $\ce{[Fe(CN)6]^{3-}}$. Does the source you reference in any way specify as what iron is precipitated? What photolysis should take place? A reaction equation certainly would be very helpful and it is necessary if you want to have an attempt on the quantification of the rate. (The question could also benefit from a more specific title.)

@Martin-マーチン: The source says that the Fe(+II) is oxidized to Fe(+III) by absorption of photons < 313nm.  I'll clarify the title and try to update the question with some equations.

@feetwet Good edit! Unfortunately, google does not let me check the page any more, but I kind of remember reading $\ce{Fe(CN)5.(H2O)^−3}$ (There was no plus). Style advice: You can write something above the reaction arrow with \ce{...->[h\nu]...} to get $\ce{...->[h\nu]...}$ Using ->[above][below] is also a shortcut.

@Martin-マーチン: Thanks for the notation help!  What is the restriction on the Google book access?  I'm still able to see a bunch of that book.  Maybe I can post a snapshot of the relevant page.

I just saw this was retagged as inorganic chemistry.  Given that we're talking about aqueous cyanides does that make sense?

I would put the solution under a blacklight exposure source or a equivelant such as a mercury arc lamp at close range. The referenced writing makes no specifics about the the timeframe of prolonged. This could have been 3 months of exposure to sunlight.

Good point.  I'll have to check sub-300nm luminance ratings on shortwave UV lamps and compare with peak daylight.  I had a short period in mind because on the next page of the reference in my question it notes that a recent paper (which I've requested but not yet received) showed complete photolysis by sunlight within 30 minutes, but that was of a miniscule concentration of $\cf{Fe(CN)6^{-4}}$....


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What are the non-aqueous solvents for sodium polyacrylate?
I am attempting to use sodium polyacrylate as a sodium transport mechanism in a sulfur-bromine cell. I can upset the sodium balance with sodium salts, but the contamination introduced by the non-metallic ion from that salt is unacceptable. Water is a solvent that can be ultimately coerced into a solution compatible with the acrylic polymer I wish to matrix the polyacrylate within.
I have tried (with no success) alcohols, acetone, and ketones. Are there non-aqueous solvents I should consider that suit my needs that I have not yet tried?

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

This is a link to the theoretical machine, but it has nothing to do with dissolving Sodium PolyAcrylate in other than salt laden water.

It's not useful for you but may be for guys who don't know about these batteries. More to the point, maybe solvents you used weren't polar enough - maybe sth like DMSO would work.

Interesting compound, this DSMO. Thanks for mentioning it, it is readily available and not very expensive. Worth a try!

Dissolve the powder in Methanol. You'll get your results

Sodium polyacrylate can locally organize solvent molecules to become soluble. Since the backbone is huge compared to water, the water aligns with its more positively charged end towards the carboxylate site while the sodium ions end up on the negative end of the water. I can't find the good papers right now, but if you search google scholar for using polyacrylic acid as a draw solute for forward osmosis you can dig way deeper into the mechanisms.
What this means practically is that you need really polar solvents that can line up with the ions. Wikipedia has a good page explaining solvent classification in more detail with a table of solvents including their dipole moment (quantifying the polarity). As suggested by Mithoron, DMSO seems to be a good option. Unfortunately, after that you start getting into much nastier solvents like acetonitrile and DMF.


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Synthesis of oxalyl chloride that doesn't involve phosphorus
I have been synthesizing TCPO (bis(2,4,6-trichlorophenyl) oxalate) for chemiluminescent reactions, using 2,4,6-trichlorophenol and oxalyl chloride.

Since I am tired of paying for oxalyl chloride (and hazard class 3 shipping), I wish to find a synthetic route to produce my own stock.
The most common synthesis of oxalyl chloride uses oxalic acid and phosphorus pentachloride. This would reduce the cost, but is also problematic because phosphorus chlorides are considered to be drug precursor (21 CFR 1310.02), still too expensive, and hazardous to ship.
Can anyone recommend a synthetic route to oxalyl chloride that does not use phosphorus? Or an alternate (safer / easier / less expensive) route to TCPO?

I believe you can try to find a route via transesterification. Excess of TCP heated with dialkyloxalate in presence of catalitic amound of strong base should give the desired product. The produced alcolohol should be distilled off to enforce the reaction. I highly discourage use of PCl5 or similar compounds outside a professional lab.

I would like to point out to you that TCP, or 2,4,6-trichlorophenol, is  carcinogenic and I would not recommend using this.

@reluctantmathematician there are several notions of synthesis of acetyl chloride using $\ce{S2Cl2}$.  This compound is not much better than $\ce{PCl5}$, but can be prepared from reasonably available and unregulated chlorine and sulfur (Chlorine can be obtained by reaction of hydrochloric acid and bleach, but it should be dried by bubbling through sulfuric acid). $\ce{S2Cl2}$ should work for oxalyl chloride as well. That said, I think that buying oxalyl chloride is preferable, and transesterification is even better.

T3P (n-propylphosphonic anhydride) is also another useful coupling agent for esters

Most amide coupling agents are suitable for esters. Feast your eyes on: Amide bond formation - beyond the myth of coupling reagents.
This lists old favourites such as CDI, carbodiimides, etc. A lot of it is overkill. The water soluble carbodiimides might be preferred for ease of urea removal later.
The likes of CDI or $\ce{Ac2O/DMAP}$ can react with your phenol when over-stoichiometric amounts are used.
Alternatives could be: boronic acid derivatives e.g.  phenylboronic acid catalysis; scandium or yttrium triflate catalysts (all requiring water removal, but a Dean-Stark could be used if a water immiscible solvent is used); dimethyl oxalate (already suggested)  but in conjunction with Takasago's ZnTAC24 catalyst. MeOH removal by distillation is required, which is quite trivial for such a volatile component.

i think you should use oxalic acid and convert it to oxalyl chloride by thionyl chloride.take a few mL of thionyl chloride and put oxalic acid into it and reflux for about three to four Hr, evaporate the remaining thionyl chloride. i used this method for the preparation of benzoyl chloride


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What is the relationship between the two definitions of the difference between basicity and nucleophilicity?
(Clarification: this is a question about terminology, not so much about when to use different concepts.  I think I do understand which characteristics of a species to use when trying to determine something about a reaction in some circumstances.  I just don't know what word to call it.)
When someone on the internet (here or elsewhere) is asked the difference between "basicity" and "nucleophilicity", they typically give one of the following two answers:

Basicity is about how well a species donates electrons to a proton, while nucleophilicity is about how well a species donates electrons to a more general atom or ion (especially carbon).
Basicity is about thermodynamics (i.e. equilibrium constant) while nucleophilicity is about kinetics (i.e. activation energy).

While these two answers may be related, they do not seem identical!  So I am wondering which one is actually correct, or whether there is some way that they are both correct.  In particular, I'm wondering what happens when you have a situation that puts those two descriptions at odds with each other.

We could have a reversible reaction involving carbon (though I know that's not the usual situation) that does in fact have some equilibrium.  When we're talking about that reaction's equilibrium, do we still use the word "basicity"?
In a reaction with a proton rather than a carbon, when we're talking about kinetics, do we still use the word "nucleophilicity"?

I hope I've gotten my question across OK.

The second set of definitions is the correct one.
Basicity is a thermodynamic property. In essence, the basicity is measured by determining the equilibrium constant of the following (generic) reaction:
$$\ce{B- + H2O <=> B-H + OH-}\tag{1}$$
The proton transfer that is the gist of this is an extremely fast process — faster than diffusion control due to the Grotthuss mechanism. You may have witnessed this during acid-base titrations: equilibrium is reached extremely rapidly. Another way to observe this is by adding carbonate bases to an acidic solution; $\ce{CO2}$ evolution is basically instantaneous. Therefore, there is no way to observe just one step (forward or reverse); you basically always have them both.
Nucleophilicity is a property that is related to the rapidness of the following, again generic, reaction:
$$\ce{Nu- + R-LG -> Nu-R + LG-}\tag{2}$$
As similar as this may seem to the above, the lack of a reverse arrow may already show that this is tied to only one directiong; that it is not an equilibrium process but rather a kinetic one. Most of the time, there will be one species that is clearly a good nucleophile ($\ce{Nu-}$ in the scheme) and one that is only a good leaving group ($\ce{LG-}$ in the scheme). The reaction will be driven to completion and stay there.
To exemplify the difference, consider what would happen if you added hydroxide to the acid-base equilibrium $(1)$ versus additional leaving group to the nucleophilic substitution $(2)$. In $(1)$, the equilibrium would readjust; if the base is less basic than hydroxide it will be deprotonated, i.e. the reaction will be shifted back to the reactants. In $(2)$, typically nothing will happen, e.g. if you add $\ce{OPPh3}$ to a Mitsuobu reaction; showing the kinetic nature of the process.
Another way to look at the problem — and to separate the parameter basicity from the parameter nucleophilicity — is to look at the influence of steric effects. For example, consider ethanolate, propanolate and deprotonated isobutyl alcohol. In all three cases, the negative charge is unambiguously located on the oxygen atom. Thus, we would expect a thermodynamic property to be relatively equal for the threeof them. At the same time, there is a clear increase in steric hindrance from ethanolate to the isobutyl alcohol. This steric effect should manifest itself in reaction kinetics and lead to the latter being less nucleophilic than the former. Indeed, sterically hindered bases are often described as non-nucleophilic while the same structure with less hindrance often works fine as a nucleophile.
Finally, let’s work at debunking the claims laid out in the first set of definitions. First, basicity. After the Arrhenius and Brønsted-Lowry theories of acids and bases came the Lewis theory. In this, the reaction of a Lewis base with a Lewis acid — not necessarily a proton — is an acid-base reaction. One of many possible examples would be the formation of a transition metal complex such as $\ce{[Cu(NH3)4(H2O)2]^2+}$. Here, the ammine ligand is acting as a base, but it is donating to a non-hydrogen atom (namely copper). The claim that basicity only goes towards hydrogen is obviously false.
Unfortunately, it is not possible to fully debunk the second part since a nucleophilic attack on a proton is not really possible (that would indeed be the thermodynamic acid-base reaction). However, we may remember that most of the non-metals can act as the recipient of a nucleophilic attack; for example sulphur in a Swern oxidation, phosphorus in a Horner-Wadsworth-Emmons reaction or iodine in a Dess-Martin oxidation.

I would rather say that nucleophilicity always somehow describes an attack (followed by an addition) whereas basicity describes the ability of taking away positive charge (like a proton). Nucleophilicity and basicity is quite similar with the antagonism of 'adding negative charge' and 'removing positive charge' which de facto is only a difference in regarding the same phenomenon from two different points of view.


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The reason for bicarbonate acting as a buffer
My question is: what is the primary reason that bicarbonate ($\ce{HCO3^-}$) can act as a buffer in a solution? The two (possibly) relevant equations that I can think of are
$$ \ce {H2CO3 + H2O <=>HCO3- + H3O+ }$$
$$ \ce {HCO3- + H2O <=> CO3^{2-} + H3O+} $$
Using these two equations, I can come up with two 'reasons' as to why $\ce{HCO3-}$ can act as a buffer. The first reason is that $\ce{HCO3-}$ can create $\ce{H+}$ ions in the second equation, but can also use up $\ce{H+}$ ions in the first equation. This means that if I put an excess amount of $\ce{H+}$ ions into the solution, the $\ce{HCO3-}$ can act by Le Chatelier's principle to remove the $\ce{H+}$ ions. If I put an excess of $\ce{OH-}$ ions into solution, they will react with the $\ce{H+}$ ions, and cause a shortage. The $\ce{HCO3-}$ can then react with water to replenish the $\ce{H+}$ ions in the solution. The only requirement for this explanation to occur is that enough $\ce{HCO3-}$ exists within the solution to be able to balance out whatever changes may occur.
This was the picture I had in my head as I began to study my textbook today. However, my textbook differed, listing only the first equation as the reason $\ce{HCO3-}$ can act as a buffer. In the first equation, an excess of $\ce{H+}$ ions would drive the equation to the left, removing the excess ions from the solution. An excess of $\ce{OH-}$ ions would react with the $\ce{H+}$ ions, and drive the equation to the right, replenishing the $\ce{H+}$ ions, and restoring neutrality. This is a much simpler explanation, but it requires that there be adequate amounts of both $\ce{HCO3-}$ AND $\ce{H2CO3}$ in the solution to maintain neutrality.
Both 'reasons'seem to operate on the same fundamental rules, and yet they require different things to operate, and I am pretty sure my textbook is right and not me. One reason I can think of why is perhaps that the second equilibrium lies far on the left, so far left that barely any $\ce{CO3^{2-}}$ is ever present in solution, which would mean the second equation could never help with buffering the solution, or if it did it would be negligible help. Is this line of reasoning correct?

That line of reasoning of yours  seems to be correct.  Those two pKa difference of about 7 units. So, the second equilibrium can be neglected.

Have you come across weak acids/bases and their conjugates?

@bonCodigo Yes, but only in reading, haven't really learnt it in depth


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How can I predict the product of the reaction of 2-methyltetrahydrofuran with HI?

I am supposed to explain why treatment of 2-methyltetrahydrofuran with hydrogen iodide leads to the formation of 5-iodopentan-2-ol. However, I am getting stuck as it seems that the less stable primary carbocation is being preferentially formed, leading to the product. Is my mechanism incorrect?

The reaction proceeds as either an $\ce {S_{N}1}$ or an $\ce {S_{N}2}$ mechanism. You need to review those mechanisms and determine which mechanism $\ce {I^-}$ would use.

Initially, the oxygen in the ring gets protonated. However, there is no formation of a carbocation. The ring opening proceeds by an SN2 pathway, where iodide ion attacks at the less hindered carbon:

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


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Is There an Acceptable Potential Which Can Represent the Interaction between Finite-sized Ions?
The question is as mentioned in the title. Take for example the interaction between $\ce{Ar^+}$ and $\ce{Ar^+}$. In general, is there an acceptable potential which can represent the interaction between $\ce{X^{m+}}$ and $\ce{X^{n+}}$, where $\ce{X}$ is a given element and $m,n=0, \cdots, Z$, where $Z$ is the atomic number of the element $\ce{X}$?
In many cases the coulomb potential (e.g. $V(r) = 1/r$) suffices. However, when density increases (e.g. at liquid/solid density), the size of the positive ion might become relevant.

The Lennard-Jones and Morse potentials are both "acceptable," depending on just what you're simulating - are you focusing on liquid and solid phases?

Thanks for your reply. Lennard-Jones is fine for inter-atomic interactions. Morse seems to be used for particles bound in a molecule. Indeed I'm interested in a plasma model in which charged species, $X^{+}, \cdots, X^{Z+}$, may exist. In the absence of screening (by free electrons), the potential should have a Coulomb tail (i.e. $\sim m/r$, for some $m$). What I'm looking for is something like the Coulomb-Buckingham potential (https://en.wikipedia.org/wiki/Buckingham_potential).

I'd also like to note that for highly charged species, the coulomb interaction should dominate in short ranges. On the other hand, I'm not so sure about whether there is a distance (which is comparable to the "size" of the ion) at which the repulsion due to the "size" of the ion will dominate over the coulomb repulsion.


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Is there open source software for SEM EDX data analysis?
This can't be unique to me and would apply to anybody who has to use equipment in other departments has had departments move etc.
What open source packages exist to read and potentially manipulate SEM EDX (electron dispersive x-ray elemental analysis) data after the data has been collected?
(In my case I have data from an Oxford Instruments EDX but there are many more manufacturers.)

You could use HyperSpy library for Python.
Some examples of what you can do with it are available here.

brr, Python - but its something :) - thanks - I wonder what the others will suggest? (+1)

Hehe :) I guess you have tried with Originlab,  the free version is pretty useful.

@davcs86 did hyperspy work for you? as i work mainly on TEM-EDS platform and it gives out ipj format, and there is nothing that i could find to analyse data other than the bundled software. https://groups.google.com/forum/#!topic/hyperspy-users/YbM5F7cN950
here it is mentioned that hyperspy works on opensource data format which is supported by bruker but not oxford :(


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Free Open Source Gas Chromatography Software
Collecting data is one thing, however sometimes one needs to plot or analyse data at a later date when access to the original software may no longer be feasible or possible.
Open Source software is at times a good alternative to commercial products. In this case, please list open source and/or free software for gas chromatography.
(Having said that, the databases for mass spectrometry are not open as far as I am aware, though suggestions are also welcome.)

One software that I have come across and used is OpenChrom https://www.openchrom.net/
At least for basic tasks it performs well as far as I can tell.
For those seeking pretty chromatograms, OpenChrom can be used in conjunction with R. Open a plot in OpenChrom, export it to a csv file and then plot the data using R.
A script for plotting two overlapped chromatograms in R is below:
filename = "Dodecane-Neat"
filename2 = "Dodecane-Stressed"
PlotTitle = "Dodecane Neat Chromatogram - Pre and Post Test"

infilename = paste0("./",paste(filename,"csv",sep="."))
infilename2 = paste0("./",paste(filename2,"csv",sep="."))
outfilename = paste(filename,"pdf",sep="-rect.")

## Read Data -> File is called 14mielczarek11.csv
undiluted <- read.table(infilename, sep=";", header=TRUE, nrows=10000)
undiluted2 <- read.table(infilename2, sep=";", header=TRUE, nrows=10000)
## Work out Total Peak (want the Chromatogram, not the mass spec)
undiluted$peaks <- rowSums( undiluted[,3:404] )
undiluted2$peaks <- rowSums( undiluted2[,3:404] )

GCxx <- c(undiluted$RT.milliseconds./1000/60, rev(undiluted$RT.milliseconds./1000/60))
GCyy <- c(rep(0, nrow(undiluted)), rev(undiluted$peaks)/max(undiluted$peaks))
GCxx2 <- c(undiluted2$RT.milliseconds./1000/60, rev(undiluted2$RT.milliseconds./1000/60))
GCyy2 <- c(rep(0, nrow(undiluted2)), rev(undiluted2$peaks)/max(undiluted2$peaks))

library(survival)
library(Hmisc)

# A4 is 8.27 × 11.7
pdf(outfilename, height=5, width=8)
par(mar=c(5,5,3,2))

plot("",
tck=-0.02,lab=c(10,10,4),las=1,
xaxs="i",xlab="Time (min)",
xlim=c(0,15),
yaxs="i",ylab="Detector Count Fraction",
ylim=c(0,0.15)
)

polygon(GCxx2+0.017, GCyy2, col= rgb(0,0,1,0.5), border=NA) #Blue - Shift the blue ever so slightly to properly overlap
polygon(GCxx, GCyy, col= rgb(1,1,1,1), border=NA) # White (so we don't get violet)
polygon(GCxx, GCyy, col= rgb(1,0,0,0.6), border=NA) # Red

lines(undiluted2$RT.milliseconds./1000/60+0.017,undiluted2$peaks/max(undiluted2$peaks),
type="l",col=4,lwd="0.05",lty=1) #Blue
lines(undiluted$RT.milliseconds./1000/60,undiluted$peaks/max(undiluted$peaks),
type="l",col=2,lwd="0.05",lty=1) # Red

title(PlotTitle)
minor.tick(nx=5, ny=5, tick.ratio=0.3)

dev.off()

And the results:

R is a great suggestion, though I hardly believe this code will be useful to anyone else the way it is. With this amount of detail for limits, data transformation and such, it's probably better to plot the bare chromatogram x/y values with lines and let people do the adjustments themselves.

@Molx Yes, people who want to plot their own would need to adjust the limits of course - I would have thought that is obvious? Maybe one could write an automatic function - however if this is possible I wouldn't know how to do this.


text:
Hybridization of oxygen in Nitrate ion and the location/bond of nitrogen's lone pair
I have tried to apply the rules and basics that I learnt so far. But I am confused about the hybridization of oxygen atoms which are making the single bond in nitrate ion.
Following are the steps I used. The average bond order of 4/3 makes the molecule unstable.

Based on the steps, the single bonds are made of $\ce{sp^3}$ of oxygen and $\ce{sp^2}$ of nitrogen, the bonds are sigma bonds. However that doesn't seem like the case. If I look at $\ce{HNO3}$ dissociation equation in the water, lone electron pair of nitrogen is entirely shared with (co-ordinative covalent bond) one of the oxygens making a single bond.
$\ce{ HNO3(aq) -> H+(aq) + NO3^{-}(aq) }$
But considering $\ce{NO3}$ ion individually, without thinking about nitric acid, I can't find the long pair to be in the same type of co-ordinative bond. Isn't there an unbonding sp2 electron with nitrogen (probably one from the lone pair)?
The other issue, why isn't oxygen hybridizing to be $\ce{sp^3}$ here? What am I doing wrong here?

You indeed can think that these two $\ce{O-}$ are sp3-hybdiridzed. But I would rather explicitly call them $\ce{O-}$ rather then just $\ce{O}$.

nintrate-ion is isoelectronic to carbonate ion. The bonding is very similar except that in nitrate due to positive charge on nitrogen, the bonds are stronger.

The actual structure of the nitrate ion is symmetrical. It forms a trigonal planar molecule.
This means that the bonds between oxygen and nitrogen are all the same. The resonance structures are a simple way to explain that structure using two-electron bonds. In reality the "extra" electron is delocalised across all three bonds (this will involve the nitrogen lone pair and nitrogen will be sp2 with no free lone pair).
The average bond order will be, as you calculate 4/3 but this doesn't, as you claim, make the molecule unstable.


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Pyramid Structure Name
I would like to know what the name is for the structure drawn below. This looks like a pyramid with 4 atoms at each vertex. Does the name of the structure depend on whether the atoms are all the same or if they are a combination of different ones?

Trigonal pyramid; btw you mean atoms in corners not molecules/edge?

Yes they can be atoms. Is that name, just Trigonal Pyramid? Not any more or special name?

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

Somewhat related http://chemistry.stackexchange.com/q/32325/15489

If the four faces of this shape are equilateral triangles, that would make this shape a tetrahedron (literally meaning four faces). This is the molecular shape associated with the $\mathrm{sp^3}$ hybridization. The shape is independent of what atoms constitute the corners.

Tetrahedra have central atoms.

@Mithoron: no, they don't.

Gerhard is right.  As evidence, https://en.wikipedia.org/wiki/Tetrahedrane

The argument about hybridisation of course only applies to (ideal) tetrahedra that have a central atom and to this central atom alone.

A tetrahedron is an abstract geometrical shape, just as a cube, a sphere or a dodecahedron. They only describe the shape of the body, not the content.  They do not have anything to do with atoms, but are of course often used to describe the outer appearance of an arrangement of atoms in chemistry.

The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) and the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) include polyhedra and corresponding polyhedral symbols that describe the geometrical configuration.
When an atom is attached to three atoms or groups (e.g. coordination compounds with a coordination number of three), the geometry may be trigonal planar or trigonal pyramidal. For the structure given in the question, the geometry is obviously trigonal pyramidal, not planar. Thus, the coordination polyhedron is a trigonal pyramid; the corresponding polyhedral symbol is TPY-3.

This coordination polyhedron is not to be confused with a tetrahedron (polyhedral symbol: T-4), which applies to a coordination number of four, or an atom that is attached to four atoms or groups.


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Why 4s lose electron to 3d?
In chromium and copper, the are the configuration fill up like
$$\mathrm{3d^5  4s^1}$$
but not $\mathrm{4s}$ first as it has lower energy level. and why only these 2 elements.

Copper doesn't have $3d^5$ in its base state, but a completely filled $d$ orbital.

If you've just only been recently introduced to the electron configurations, then an appropriate answer would be: "the half-filled 3d orbital in Cr and the completely filled 3d orbital in Cu are particularly stable". For a more detailed but also potentially very technical answer, look up exchange energy.

This is a very popular question in the chemistry world, and it has been asked and answered before: http://chemistry.stackexchange.com/questions/151/why-do-elements-in-columns-6-and-11-assume-abnormal-electron-configurations

In any element after Argon [1s22s22p63s23p6],
the repulsion force between the elctrons increase at a great extent and also the atom becomes very stable. So the elctrons tend to move in far 4s orbital first and after filling it completely , fill the 3d orbital. About chromium and copper, they are exceptions. Any half filled or fully-filled orbitals are more stable then rest of the configurations, so in Chromium one electron of 4s comes to 3d orbital and a stable 3d5 is obtained, similarly in Copper one electron from 4s comes to 3d orbital and forms stable 3d10.
I hope this answers your question,
If you have any further doubts you can ask me :)


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Sodium Polysulfide production from NaOH and Sulfur
Does the sulfur also oxidize the $\ce{OH-}$ radicals after the $\ce{Na2S4}$ chains form? Something like this:
$$\ce{9S + 4NaOH -> 2Na2S4 + SO3 + H2O}$$

If so then I can expect a mixed solution of Sodium Sulfate and Sodium Polysulfide(s) after boiling of the Sodium Hydroxide/Sulfur mixture.

If you calculate the oxidation numbers for everything before and after, you will see that the oxidation number of the oxygen atoms will always be $-2$. However, sulfur $\ce{S},\ (0)$ is being both reduced to $\ce{S4^2-},\ (\frac{1}{2})$ and oxidized to $\ce{SO3},\ (+6)$. Sulfur is both the oxidant and reductant in what might be called a disproportionation. Sodium hydroxide is thus acting as both a source of oxygen atoms in the formation of $\ce{SO3}$ and probably as the electrolyte if this is occurring in aqueous solution.

Oxidation numbers:

Element    Reactant     Product
S           0            -1/2 in polysulfide
S           0            +6 in sulfur trioxide
Na         +1            +1
O          -2            -2 in sulfur trioxide and water
H          +1            +1

The half reactions (unbalanced) are:

Oxidation: $\ce{S -> SO3}$
Reduction: $\ce{S -> S4^2-}$

You might get some sodium hydrogen sulfate forming through the followin;
$$\ce{NaOH + SO3 <=> NaHSO4}$$

Now to the real challenge, extracting the sulfides from the rest of the products/leftover reactants. The leftovers should be minimal as a sulfur excess existed after solution cool down to ambient. The liquid portions were drained off and the remnants disposed of. Currently I am involved with reducing 1M  H2SO4 in hopes of using its desiccant behavior to carbonize sugar impregnated cotton cloth for use in the electrodes. It would seem that the MathJax does not work in the comments.

MathJax still works in comments $\ce{H2SO4}$ is produced by $\ce{H2SO4}$ just like normal.

I left out the $ signs

Note that $\ce{SO3}$ reacts strongly and rapidly with $\ce{H2O}$ to give $\ce{H2SO4}$.  I'd expect this to consume two more equivalents of $\ce{OH-}$ (giving $\ce{SO4^2- + 2H2O)}$, and that those reactions would be more rapid than the disproportionation reactions.
Dr. Nathan E. Stott's proposed reactions which involve sulfur oxidizing hydroxide to give oxygen aren't favoured energetically, under normal conditions.  Oxygen's a much stronger oxidant than sulfur.

This seems to be rather a comment than an answer.

In short, the answer to the original question is "no, the proposed reaction wouldn't come out like that".

Here are a list of equations that could potentially occur for this reaction depending on the amounts of sulfur and sodium hydroxide used as reactants:
4 NaOH + 2 S8 $\rightarrow$ 2 Na2S8 + 2 H2O + O2

4 NaOH + S8 $\rightarrow$ Na2S3 + Na2S4 + 2 H2O + SO2
4 NaOH + S8 $\rightarrow$ Na2S2 + Na2S5 + 2 H2O + SO2
4 NaOH + S8 $\rightarrow$ Na2S3 + Na2S5 + 2 H2O + O2
4 NaOH + S8 $\rightarrow$ Na2S2 + Na2S6 + 2 H2O + O2
4 NaOH + S8 $\rightarrow$ 2 Na2S4 + 2 H2O + O2

6 NaOH + S8 $\rightarrow$ 2 Na2S3 + Na2S2O3 + 3 H2O

12 NaOH + S8 $\rightarrow$ 4 Na2S + 2 Na2S2O3 + 6 H2O

Displacing oxygen? Really?


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How are the chemical reactions of the table of standard reduction potentials experimentally determined?
My background: I've read several books and articles on electrochemistry and even do photoelectrochemical research in a lab, so I'm somewhat familiar with the concepts and techniques.
In texts on electrochemistry, the standard reduction potential and its meaning is usually axiomatically presented early on and everything follows from there. What I would like to see in a treatment of the subject is an explanation of how the chemical equations of the reduction half reactions are actually determined. For example, let's take the equation $\ce{Cu^{2+} + 2e- <-> Cu}$. How did they figure out that it is in fact two electrons reducing the $\ce{Cu^{2+}}$ ion?
One could do open circuit potential measurements to determine the standard reduction potential, but it would seem to me that you would have to know the ionic state of the element a priori to use this technique. One could also apply a voltage until the reaction "turns on" then approximate that onset voltage as the standard reduction potential, but this wouldn't be strictly true as it would be a non-equilibrium measurement and would also be subject to the kinetics of the surface. Does anyone know of a book or other treatment showing how the chemical equations are determined or better yet has experience doing this?

The standard reduction potentials were mostly discovered by carefully measuring the reduction and oxidation of solid metals (or dissolved ions) in electrochemical cells, and monitoring the electrical current through the wire connecting the two electrodes. (The image below should have an ammeter instead of of a voltmeter, but you get the idea)
Take, for example, an electrochemical cell of zinc and copper.

At the anode, the solid zinc will dissolve to become Zn2+. At the cathode, the Cu2+ ions will be reduced, and become solid copper. During the redox reaction, the scientist measures the current across the two electrodes.
After some time, the scientist could pull the two electrodes out of solution, dry them off and mass them. They could then calculate the moles of zinc that dissolved from the anode, and the moles of copper added to the cathode. In addition, by measuring the current over the period of the experiment, they would also be able to calculate the number of electrons that passed from the zinc anode to the copper cathode.
In this instance, they would calculate that for every mole of zinc that dissolved, 2 moles of electrons were produced. Concurrently, 2 electrons were consumed for each mole of copper reduced! These measurements, over time, made it possible to determine the oxidation states of many metals and soluble ions.
(image from moodle2.rockyview.ab.ca/)

That seems pretty reasonable and probably historically accurate. Would you mind giving a citation supporting why you believe this is true?

No problem - the modern table of standard reduction potentials was first developed early on using Michael Faraday's research (published in early 1800's). His works culminated in a set of rules called Faraday's laws of electrolysis. There's a nice treatment in the J. of Chem. Ed. (https://dx.doi.org/10.1021%2Fed031p226), and a quick search for laws of electrolysis produce a summary in wikipedia.

Re-read this answer after four months and realized that it doesn't fully answer the question. First of all, if you had a voltmeter in series with two half cells as drawn above, there would be negligible electrolysis due to its high resistance. Did you mean to put it in parallel? In any case, even if it were in parallel if you had the reaction running I don't think you'd quite measure the equilibrium voltage (i.e. the standard reduction potential). How would this be corrected for?

Yeah, it should be an ammeter to measure current instead of a voltmeter. I didn't create the image, but should have made that significant change more clear in the answer (edited for clarity)


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Cyclopentyl to cyclohexyl carbocation rearrangement

Here is my problem. The mechanism seems pretty straightforward, but I am having trouble converting the cyclopentane to a cyclohexane. During a synthesis problem, does the starting material add the extra C onto its chain during carbocation rearrangement or a methyl/hydride shift?
As for the problem at hand, I believe the first step is for the pi bond to attack the $\ce{H+}$, and that will leave me with a carbocation at the more substituted carbon. From there, I know $\ce{HOCH3}$ will attack, but somehow, the molecule needs to rearrange its format in order for the $\ce{HOCH3}$ to come in. What needs to happen here?

I figure there's an alkyl shift after the initial protonation.

Your first step is correct. The reactive exomethylene group attacks the proton leaving a secondary carbocation (remember Markovnikovs rule). Now the five-membered ring can rearrange into a more stable six-membered ring by releasing ring strain. This is a fast intramolecular step. Finally methanol attacks the rearranged carbocation releasing a proton which is used catalytically in the next reaction.
Remember though that you can often not be completely sure about a mechanism before it is verified in practice (i.e. in the lab through an experiment). Sometimes true mechanism can be quite different from what appears to be most reasonable on paper!


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What is the difference between electric current and electric charge?
Are electric current and electric charge the same thing? I'm just getting back into relearning chemistry, so a more in-depth answer about how electricity works in general would be wonderful.
Edit: Thank you all for clarifying that this belongs to the Physics SE. That being said, I guess this question doesn't have much relevance to me anymore. I'm honestly not sure if I should delete the question or not, so I'll leave that up to the mods to decide. Thanks again.

I think this should go on physics SE. The question itself seems too broad imo (basically you're asking about the whole of electrostatics). You might find the MIT 8.02 lectures by Walter Lewin good, just watch as much or as little as you need. The materials are not hosted on MIT any more since Lewin got sacked for online sexual harassment last December, but they are available on other places online (just Google for it), and regardless of what he did to his students his lectures were very, very good.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I think this question is indeed very broad and it will most likely be too broad for physics as well.

I'm voting to close this question as off-topic because it belongs to Physics.SE.

I don't think it should be migrated if it's gonna be closed as too broad there too. Since the author isn't interested in their question, I dunno what else we could do. But migrating isn't an option.

Electric charge is a bus. Electric current is a bus that is driving somewhere.

Your question is very broad. The first thing. No they are not the same. Simple explanation for these terms can be:
Charge: Its the "amount" of unbalanced electricity in a body.  Its usually measured in the form of electrons (either excess or deficiency). Its units is coulomb.
Current: It is the "flow" of electrons from higher electric potential towards the lower. Its unit is ampere.
To visualize how they are not the same is by considering two situations (without going into any calculations). Two bodies (A & B) are connected via some copper wire. A has 1 million Coulombs charge & the second one has 1million & 1 coulombs charge. Now individually these charges are enormous, but as the difference between the two is only 1 coulomb, its a very low difference in charge, so a little current will flow between them.
Now if you change the bodies' charge. A has 100 Coulombs & B has 10 .. much more current will flow between the bodies, because there is a way bigger potential difference.
ps. and yes its a physics questions.

edit: error, sorry. wrong post.


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Can ingredients in Tiger Balm cause any harmful effects on PVC figures?
My roommate often uses Tiger Balm and sometimes browses through my collection of vintage Star Wars figures. Can the residue on his fingers cause harm to the PVC figures or cause the paint to crack or chip?

According to this Wikipedia entry on Tiger Balm, there is camphor (among other chemicals) in the formulation of this product. This chart of chemical resistance recommendations for PVC indicates that it is not recommended that PVC should be exposed to camphor. Therefore, it is probably best to reduce or eliminate contact of Tiger Balm with the figures.
As for the paint cracking or chipping as a result of exposure to Tiger Balm, that might depend in part on the composition of the paint, and that cannot be addressed unless you can provide that information.


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Why are porphyrins colored when porphyrinogens are colorless?
In heme synthesis, porphyrinogens are colorless but upon their oxidation (from removing 6 hydrogen atoms), they become colored porphyrins. Why does the change in color occur? Is red the only color it can turn into or are there others (if so, what does that depend on)?

Another color is green: chlorophyll. A synthetic molecule is tetraphenylporphyrin, which is blue.

The precursor porphyrinogen does not have a conjugated pi system that spans the macrocyclic ring. On oxidation to porphyrin, however, the macrocyclic ring becomes fully "conjugated", i.e. connected by sp2 hybridized atoms. The HOMO/LUMO spacing in such systems is much narrower than in non-conjugated systems, which means the photon energy required to excite electrons into the LUMO is lowered from "deep" UV to the visible region, resulting in strong absorbance of light in the visual frequency range.


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Belousov-Zhabotinsky reaction: questions about rate determining step, k and activation energy
In one of our lab courses, we performed a variation of the BZ reaction, the experimental procedure is outlined below:
Experimental Procedure

10 mL of 3.0 M $\ce{H2SO4}$ was added to test tube T1.
Test tube T1 was placed on a stand and immersed in the water bath maintained at 35 °C
5.0 mL of 0.25 M $\ce{KBrO3}$ was added to test tube T2.
2.5 mL 0.02 M $\ce{Ce(NH4)4(SO4)4}$ in a 1.0 M $\ce{H2SO4}$ aqueous solution was added to test tube T3.
5.0 mL 1.0 M $\ce{CH2(COOH)2}$ was added to test tube T4.
Test tube T2 and T3 were added to test tube T1 and then the contents were swirled.
Test tube T1 and T4 were put into the water bath for 10 minutes.
While test tube T1 was still in the water bath, the contents of test tube T4 was put into test tube T1 and the timer was started.
Changes were observed while the tube T1 was vigorously shaken in the water bath.
The length of the induction period was recorded.
The reaction was allowed to oscillated 10 times, and the length of the period of oscillation was recorded.
The experiment was repeated at $\pu{40 ^\circ C},~\pu{45 ^\circ C},~ \text{and}~\pu{50 ^\circ C}$.

Also, here's the data set that I obtained:

Now, based on the instructions in our lab manual I plotted the following curves:
Now, my questions:

The manual instructs us to calculate activation energy using the period of induction and the period of oscillation from the Arrhenius plots above, which I calculate as $\pu{69.3 kJ/mol}$ (oscillation period) and $\pu{51.4 kJ/mol}$ (induction period).
But in doing so, we utilise the fact the $k = \frac{1}{\mathrm{period}}$ (i.e., $\ln(K)$ corresponds to $\ln(\frac{1}{\mathrm{period}})$. We were just given this information with no explanation. I can intuitively guess that for a first order reaction $k$ has the dimension $\mathrm{s}^{-1}$, but is there a stronger reason for making this selection?
I would guess the rate-determining step occurs during the induction period (since it is the slowest), but then the activation energy is lower. I would think that the energy barrier of the rate-determining step should be higher. Am I seeing a relation where none exists?
I noticed that the period of the reaction changed when the reaction mix was stirred more vigorously (temperature  and starting concentration being the same). Is that possible? If that is the case, then does that mean that the period isn't a reliable measure?

BZ reaction is very complicated, see https://en.wikipedia.org/wiki/Oregonator for simplified model.

I realise your homework was probably due a LONG time ago but I just stumbled across this.
In answer to (3) yes, the reaction is notoriously sensitive to a wide range of conditions and stirring is one of them...when the system is stirred faster (probably generating a vortex) oxygen is more readily available to react with radical species in the reaction - effectively these will react with oxygen rather than the bromide. As such the bromide in the system is above the threshold level for oscillations to occur - if you were to slow this down again then you'd see the oscillations resume.
Also I don't get why they'd make you use cerium sulfate...everyone knows ferroin is better.


text:
Why soak filter paper in K2SO4?
From a description of my experiment:

Potassium iodide crystals are place in the middle of filter paper soaked with potassium sulphate and phenolphthalein. Two electrodes are placed on the filter on two opposite sides. The electrodes are connected to a battery. The objective of this experiment is to see the movement of ions.

Why soak the filter paper in potassium sulphate and not just soak it in potassium iodide?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

For this experiment, you'll be seeing an interesting color change when Iodine is oxidized at the anode to form the brownish $\ce{I2}$ on the electrode and yellow $\ce{I3-}$ in solution.
The reduction of $\ce{H+}$ to form $\ce{H2~(g)}$ and $\ce{OH-}$ will turn the indicator pink at the cathode.
The cool thing about your experiment is that you will not see any redox reaction at all (at either electrode) until iodide ions diffuse to the anode! That is, unless you use a high voltage battery, $\gtrapprox2.0~\mathrm{V}$, you will not have sufficient energy to oxidize $\ce{SO4^2-}$ to $\ce{S2O8^2-}$.
This makes the experiment an exercise in patience, and then sudden color display! Measuring the time it takes can yield data that can help you calculate the diffusion coefficient of $\ce{I-}$ if you wanted to take it another step!
In summary: Soaking the filter paper with KI would result in the electrodes producing color changes immediately. I suspect the sulfate was used in this experiment to illustrate the diffusion of dissolved ions, because the redox reaction would not take place until the iodide ions reached the anode.

What is S2O4?  Presumably you mean oxidation of sulfate to persulfate or something, but could you correct the formulae?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. I agree with Curt, that S2O4 does not make much sense, maybe you mean $\ce{2SO4^2- <=> S2O8^2- + 2e-}$?

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details. States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

@Martin-マーチン Thanks! Yes, I meant to write the sulfate ion, not S2O4. Sorry for not adhering to recommendations - just started on here! I appreciate your patience.

Whilst this is a good answer I don't quite see how it answers the question. Perhaps you could be more specific.

@bon - Sorry for not being clear! The poster asks why the filter paper was soaked in potassium sulfate and not potassium iodide. Soaking the filter paper with KI would result in the electrodes producing color changes immediately. I suspect the sulfate was used in this experiment to illustrate the diffusion of dissolved ions, because the redox reaction would not take place until the iodide ions reached the anode. Please let me know if that clears up my answer!

@Dan Burden You should edit this into your answer.


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Why does solvation release energy?
Whether dissolving acids and salts is exothermic or endothermic, depends on the lattice and solvation energy. Breaking bonds in the lattice requires energy and is endothermic - that is clear. However, I do not see how solvation would release energy. I do understand that solvation is a type of creating new bonds (and creation of bonds generally releases energy), but how is it that the, eg., Al(H2O)6 3+ is at a lower energy level than Al 3+?

Can you be more specific about why you think they shouldn't be exothermic ? Does @Mithoron's link answer your question?

So the answer is entropy change? The system became more orderly so it compensated with releasing energy?


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What are the units of 85000 Molecular Weight of a polymer
I have a PVA bottle, which says that the molecular weight (Mw) of PVA is between 85000 - 124000.
I know that this means that there are repeated chains of the PVA monomer, but what are the units to the molecular weight (of PVA). Is it $\frac{\mathrm{g}}{\mathrm{mol}}$? I am having trouble grasping this concept. Or does 85000 - 124000 mean the number of monomers in the chain of this PVA?

Yes, most likely it is g/mol. Or Da, if you prefer.

Relative molecule mass is often referred to as molecular weight and is dimensionless. http://goldbook.iupac.org/R05271.html

Related: Quick and simple explanation of molar mass, molecular mass and atomic mass

First of all $M_w$ is not the number of monomers in a chain of PVA. The number of monomers in a chain is called the degree of polymerization.
$\overline {M_w}$ is the weight average molecular weight. It is a way of determining the molecular mass of a polymer. As polymer molecules have different chain lengths, the average molecular mass will depend on the method of averaging. For example, we can use the number average molecular weight which is based on different type of averaging.
As for the unit, it's as you mentioned: g/mol. In biochemistry, they use the dalton (Da) as a mass unit for biopolymers (1 $N_a.$Da= g/mol, where $N_a.$ is the Avogadro number). So, when we say a protein has a molecular weight of 64000 g·mol$^{−1}$, it has  also a mass of 64 kDa.
P.S. To be rigorous, we have to say ** molecular mass** instead of ** molecular weight**. But this was an old way of naming things.

So is the molecular weight in case of PVA between 85000 - 124000 g/mol?

@http://chemistry.stackexchange.com/users/17750/user5139637 Yes, it is.

$1\ \mathrm{Da} = 1\ \mathrm{u} = 1.660,538,921(73) \times 10^{-27}\ \mathrm{kg}$

@Loong you are right. But when we say a protein with a molecular weight of 64000 g·mol−1, it has  also a mass of 64 kDa.


text:
Correct Electron Configuration of Ni-28
I have seen two electron configurations for $\ce{_{28}Ni}$ in my chemistry textbook:
$$[\ce{Ar}]~3d^8~4s^2$$
$$[\ce{Ar}]~3d^9~4s^1$$
Which of those is actually the correct one (in reality)? Why exactly?

NIST suggests here and here (search for Ni) that $\mathrm{[Ar] (3d)^8 (4s)^2}$, specifically in the state $\mathrm{^3 F_4}$, is the fundamental electronic state for neutral nickel atoms in empty space. However, as Wikipedia describes, if you don't specify the term symbol for the configuration, then the average energy of all terms for the $\mathrm{[Ar] (3d)^9 (4s)^1}$ configuration is slightly lower than the average energy of all terms in the $\mathrm{[Ar] (3d)^8 (4s)^2}$ configuration.
Now, why one configuration has a lower energy than another is a much harder question to answer in close cases such as this, as it involves a precise quantitative analysis of electronic energies for every possible electronic configuration of a nickel atom (a multielectronic atom). The general idea is that even though, for $Z=28$, the configuration $\mathrm{[Ar] (3d)^8 (4s)^2}$ is higher in energy than $\mathrm{[Ar] (3d)^9 (4s)^1}$ if you look at the sum of occupied orbital energies, the latter case has more interelectronic repulsions, which causes its total energy to be slightly higher. A lot of information on this can be found in this previous Chem.SE question and the several nested questions within.
Interestingly, the gas-phase fundamental configuration of the monocation $\ce{Ni^+}$ is $\mathrm{[Ar] (3d)^9}$, which further shows how nickel's electronic configuration is teetering on the boundary between the options.