text:
Residual hydroxyl groups in silicone resins
Silicone resins are solutions of branched-chain siloxanes containing residual hydroxyl groups. What is meant by "residual hydroxyl groups"?

Silicones are produced by hydrolyzing chlorosilanes( in which chlorine atoms are replaced the hydroxyl groups). These hydroxyl groups undergo condensation polymerization with removal of water molecules to form polysiloxanes or silicones.
Now, residual hydroxyl groups in your question means that some hydroxyl groups has not undergone condensation yet. Solvent is evaporated and resin is heating resulting condensation of residual hydroxyl groups to form cured resinous film.

It means there are hydroxyl groups that have not bound to (think: intermolecular forces) or reacted with something else that are still present on the siloxane chains.


text:
How to remove hydrate from Fe(NO3)3 . 9H2O?
I am currently in a dilemma where I need to have pure Iron (III) nitrate, but I have Iron (III) nitrate 9 hydrate in order to do a lab experiment where I investigate voltaic cell. I am pretty sure the hydrate will mess up the results, so I want to remove the hydrate to give $\ce{Fe(NO3)3}$. Is there any way for me to do this?

What kind of voltaic cell would require an anhydrous salt? I thought they are mostly about water solutions.

Out of curiosity, is your voltaic cell going to be a wet cell? If so, account for the water. Even many "dry" cells have an electrolyte paste.

@BenNorris It is! What do you mean by accounting the water? Should I take note of the change in water levels?

You just need to heat the sample.Water of Crystallization
For the ferric nitrate I recommend adjusting the temperate at 110$^\circ$C see wikipedia page

I can't find anything on that Wikipedia page indicating that you could obtain the anhydrous material by heating to 110C. I did see that the nonahydrate boils at 125C.

True, boiling of nonahydrate means that it is losing water, but it is never stated that it loses all water, or that it doesn't hydrolyze in the process. More importantly, when the OP is going to shoot himself in the leg, it is a good practice not to help them with that, even if that's what the question literally asks for.

In the Water of Crystallization page it states that it loses water.
I suggest that you do a TGA or DSC beforehand. This will identify how much of the water is lost during heating and at what temperature

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text:
Why do seemingly reliable sources give NFPA 704 Health ratings ranging from 1-4 for elemental mercury?
Recently a user posted the question Do Mercury have a NFPA 704 classification?.
Why the present question is not a duplicate:
Although there is no accepted answer to the above question, I think it's a fairly obvious "yes", and that my question rather is regarding how there could be such a diversity of classifications arising from this single, widely used standard.
Background:
According to the US National Fire Protection Association (NFPA), NFPA 704: Standard System for the Identification of the Hazards of Materials for Emergency Response:

This standard shall address the health, flammability, instability, and related hazards that are presented by short-term, acute exposure to a material under conditions of fire, spill, or similar emergencies.

These are typically displayed in the widely recognizable "fire diamond" (generic image from Wikipedia, not for mercury):

Here, blue indicates health hazard, red indicates flammability, and yellow indicates chemical reactivity.
Data:
The following are some values given from different sources for the "health" rating (scale of 0-4, where 0 is the most benign and 4 the most hazardous) for elemental mercury:
The National Institute for Occupational Safety and Health (NIOSH): Health = 1
MSDS from NIST: Health = 2
MSDS from Fischer Scientific: Health = 3
MSDS from Sigma-Aldrich: Health = 4 (link from comment on first link above by MaxW)
SDS from Sigma-Aldrich: Health =2
Question:
My question is simply: why is there such disparity in the NFPA health ratings given for elemental mercury?

I believe that the radical rage of possible hazard ratings for elemental mercury arise from different assumptions about how the mercury is handled. I can't prove this, but it makes sense when you consider the possible hazards arising from mercury use.
Properly handled liquid mercury is not particularly harmful. There are no acute risks and a competent person could drink it and handle it with little risk. In a chemical laboratory, for example, it is common to use mercury manometers to provide limited gas overpressure for supplies of nitrogen or argon. This involves bubbling gas through mercury. As long as the vapour is safely vented to a scrubber, this is not a hazard to people in the lab.
Equally, chemists can use mercury safely with few precautions as long as they clean up any spills thoroughly. The hazard primarily comes from vapour buildup which, if it is allowed to happen, will create serious long term chronic problems.
One example of potential danger is if non-experts come into contact with mercury. Some old gas-flow meters for domestic gas supplies used mercury which could be discharged from the meter in certain rare situations, leaving droplets of mercury around the floor near the meter. This can (and has) created a great deal of hazard in a domestic environment. This hazard is magnified enormously, for example, if the householder tries to clean the mercury spill by vacuuming it up. The result is a fine spray of mercury droplets inside a warm vacuum cleaner, which vastly increases the vapour emission and may spread fine droplets widely through the house. If badly ventilated, this can be a serious chronic issue.
The two situations span a wide range of hazard. But the difference is the context: in experienced hands mercury is not very hazardous; in other hands it is potentially very dangerous.
In short, the different rating may reflect different assumptions about who is using the mercury: someone who knows what they re doing or someone who does not.

Thanks for the answer, and what you are saying may well be the case. Trying to think of a comparable situation (another chemical that is safe in experienced hands, dangerous in others) I looked up NFPA 704 Health values for isopropanol, acetone and ethanol. I consistently found 1,1 and 2, respectively. Given that ethanol is safe to ingest in small amounts and lethal in slightly larger amounts, it seems by your argument that there would be some variation in health rating. And since the other 2 are actually far more toxic when ingested, I would expect some sources --continued--

to have higher health ratings for acetone and isopropanol than for ethanol. My comparison of the health ratings for these compounds compared to mercury isn't perfect, and what you are saying may be correct. I just wish there were some more concrete answer citing specific, confirmed criteria for this disparity.

Mercury vapor built up over time causes a huge host of problems.  Metallic mercury at room temp can be played with as long as it's cleaned up properly.    Boiling mercury is obviously worse.  I agree with NIOSH, metallic mercury has been incredibly low on my list of scary substances for some time.  I'd much rather handle it bare-handed than lead, for example.

While this attempts to answer the question, it does so in a terrible way.

Sorry, had a glass of Hg for breakfast and I'm feeling a bit funny.  Point taken though.

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text:
What happens to potassium iodide in the "elephant Toothpaste" reaction?
General Questions on "Elephant Toothpaste"
Equations:
1)$\ce{~2H2O2(aq)->2H2O(l) + O2(g)}$
Context:
I have a few questions on the reaction commonly referred to as "Elephant Toothpaste". You see, I am doing a basic demonstration of chemistry to little kids. I certainly don't think I am qualified to perform this, (Well, I can, but I want to know the details behind this, which is why I am asking for help.) but I have no choice. So Here we go!
Procedure:

Pour ~$\pu{50 mL}$ of hydrogen peroxide1 into graduated cylinder.
A "Squirt" (Not my wording) of Dish Soap into the graduated cylinder.
~$\pu{10 mL}$ of potassium iodide ($\ce{KI}$) Solution into Graduated Cylinder.

The result is...Foam, and lots of it.
130% concentration
So, here is my question:

What happens to the potassium iodide?

@Riggycat: What is the concentration of potassium iodide in your experiment?

https://en.m.wikipedia.org/wiki/Elephant%27s_toothpaste. Potassium iodide just act as catalyst.

the iodine turns into gas but the heat burns it away

The potassium iodide serves as a catalyst for the decomposition of the hydrogen peroxide by facilitating oxygen removal from the solution so the reaction is carried to completion. The rapid decomposition allows you generate a lot of gas quickly and the long cylindrical shape of the graduated cylinder directs the momentum of the gas in a vertical direction to achieve the gushing effect. Since potassium iodide is a catalyst in the reaction, it is not consumed in the process of making the foam, but exists as the ions that come from dissolving the salt in the hydrogen peroxide solution.
Reaction mechanism on wikipedia
P.S. This reaction is exothermic, so if you go really large scale, expect your vessels and the foam to be hot to the touch.


text:
Gaussian Error: Zero Point Energy Calculation on Fluoroform fails
I have been performing isodesmic reaction calculations and one of my molecules is Fluoroform. It has already been optimized and its frequencies checked at the HF level of theory. However, when I went to calculate the energy at  B3LYP/6‐311+G(3df,2p) I received an error. I have isolated my error to the route section.
Below is my input file:
#T B3LYP/6‐311+G(3df,2p) Temperature =0

Optimization of Fluoroform

0 1
C
F     1    B1
F     1    B2   2    A1
F     1    B3   2    A2   3    D1
H     1    B4   2    A3   3    D2

B1  1.490000000
B2  1.490000000
B3  1.490000000
B4  1.090000000
A1  109.4712206
A2  109.4712206
A3  109.4712206
D1 -120.0000000
D2  120.0000000

Additionally, here is the error output Gaussian prints on termination:
QPErr --- A syntax error was detected in the input line.
 #T B3LYP/6â€311+G(3df,2p) Temperature =
           '
 Last state= "GCL"
 TCursr=      3523 LCursr=        10
 Error termination via Lnk1e in /cluster/software/gaussian/gaussian-09-D.01/g09/l1.exe

As you can see in the error message
#T B3LYP/6â€311+G(3df,2p)

there is some Unicode garbage in your input file; the dash between the 6 and 3 isn't an ASCII dash. Deleting and replacing it should fix the problem.


text:
Why do species not absorb an electron in mass spectrometry?
The following is a direct quote from my high-school chemistry text-book regarding mass spectrometry:

The ionization process in the mass spectrometer involves an electron from an electron gun hitting the incident species and removing an electron. The collision can be so energetic that it causes the molecule to break up into different fragments.

This fails to make sense to me for two reasons:

Why wouldn't the majority of these electrons simply pass through a given substance, like the alpha particles in Rutherford's Gold Foil Experiment?
Why do positive ions/functional groups, such as ammonia, not absorb these electrons, instead of allowing them to break the species apart (and make them more positive in the process)?

Any justifications for these discrepancies would be greatly appreciated.

Partial answer to 2. "In electron impact (E.I.) mass spectrometry the molecule is bombarded with highly energetic electrons that knock a weakly bound electron out of the molecule. If you think this is strange, think of throwing bricks at a brick wall: the bricks do not stick to the wall but knock loose bricks off the top of the wall." (from Clayden's Organic Chemistry). The electrons have too much energy to be absorbed.

The 'duplicate question' is fine in so far as it goes but it does not  answer part (1) or part (2) of this question   but states that energetic electrons can ionise molecules. (Neither does it give any details such as cross section for ionisation vs. electron impact energy etc.) You should certainly ask part(1) again as this has been closed off here.


text:
Would a quarternary ammonium electride conduct electricity?
If suppose, I crystallised a quarternary ammonium electride could this conduct electricity similar to a metallic conductor? Metals have lattice atoms as charged metal ions are a sea of electrons. This leads me to believe a quartnernary ammonium electrides are quite similar, with lattice points being quartnernary ammonium ions and a sea of electrons.
Is this true? Can I consider quarternary ammonium electrides as metallic conductors?

https://en.wikipedia.org/wiki/Electride describes it...

You can't conclude that by thinking alone; some experiments are necessary.
True, an electron in an electride is kinda "free", in that it isn't connected to any particular atom. But that doesn't mean it can roam the entire structure, free as the wind. In fact, it sits in a potential well formed by the neighboring ions and molecules, only the well is so wide and shallow that it looks more like a puddle. To decide the conductivity question, you want to know how one puddle relates to the others, and the answer may be any of the following:

connected to form a continuous sea, much like in a metal;
connected to form a network of channels;
isolated but still close enough for electrons to tunnel between them;
isolated and spaced wide apart.

All in all, theoretically an electride can be anything from a metallic conductor to an insulator, depending on its composition and structure.

I don't understand what you mean by 'potential well'. Can you please elaborate a bit on that?

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

I see, so your saying the electron simply stays within the low potential region and just oscillating?

That depends on the shape of the low potential regions and how far apart they are placed, that's what I said. It can be a continuous sea, or it can be a multitude of isolated puddles.

In that case, is it possible to find out whether the electron will be delocalized or stuck in a well? I think higher temperatures can delocalize the electron.

Sure, just measure the conductivity and you'll see. You may do so at different temperatures, too.

Alright then, thank you. That has clarified me.


text:
If a bond dissociation energy is listed, does it refer to homolytic or heterolytic fission?
I am perfectly sure than heterolytic fission of a bond needs more energy than homolytic ones, since it requires opposite charge separation. When the bond dissociation energy is listed, which one are they talking about?
Ex: Suppose bond energy of A-B is 500kJ/mol. Is the energy to make A radical + B radical, or A cation + B anion, or A anion + B cation?

Suppose you read in a book that bond energy is 260kJ/mol, only they don't say which bond was that. You'll get confused and angry at the author, and rightly so. What did they think they were doing? This number is useless unless we know what it refers to! Now, this is the answer to your own question as well.

@Ivan Neretin Exactly!

As a 'rule of thumb' the radicals are usually found at a lower energy than the ions, e.g.  NaI, which means that there is more than one dissociation energy depending on what the products are. As @Ivan Neretin writes these should be stated in any table/book.

This Wikipedia article summarizes the definition of bond-dissociation well:

Bond-dissociation energy (BDE or D0) is one measure of the strength of a chemical bond. It can be defined as the standard enthalpy change when a bond is cleaved by homolysis, with reactants and products of the homolysis reaction at $\pu{0 K}$ (absolute zero). For instance, the bond-dissociation energy for one of the $\ce{C–H}$ bonds in ethane ($\ce{C2H6}$) is defined by the process: $\ce{CH3CH2–H → CH3CH2• + H•}$

Thus the definition of bond-dissociation refers to homolytic fission, and in your example the bond energy of $\ce{A-B}$ is the energy required to make "A radical + B radical".


text:
Hybridisation of complex compound having octahedral geometry
$$\ce{K[Co(NH3)2Cl4]}$$
I have problem finding the hybridisation of this compound. The strong field ligand $\ce{NH3}$ forms the minority among the ligands, so do I consider the hybridisation to be $\ce{sp^3d^2}$ ? Next, in this compound, $\ce{[Co(NH3)Cl3]}$, the number of strong field ligands is equal to the number of weak field ones. Can someone please explain the basis on which I can find out the hybridisation using crystal field splitting.

If you calculate the oxidation state of the central metal atom in both of these cases, you'll find it to be $+3$. With $\ce{Co^3+}$, all ligands behave as strong field ligands except in the cases of $\ce{[CoF6]^3-}$ and $\ce{[Co(H2O)3F3]}$. Thus, in both of the cases you've mentioned, the ligands will cause pairing of electrons, the complex will be of a diamagnetic nature and the hybridization will be $\ce{d^2sp^3}$.

So would  [Co(NH3)Cl5] also have d2sp3 hybridisation?

@Pewpaled If you meant $\ce{[Co(NH3)Cl5]^2-}$, then my answer is  yes, $\ce{Co^2+}$ also acts as a central metal atom so be sure to calculate the oxidation number of the complex you want to deal with.

How do I find out the hybridisation for other transition metal ions then?

@Pewpaled Excluding the exceptional cases of cobalt (+3), metal ions with a $d^8$ configuration, $d^9$ configuration with a strong field ligand etc. the spectrochemical series of CFT will give a correct answer. You can even VBT if you you're aware of addtional properties, for instance the magnetic behavior of your complex. Afaik the configuration is experimentally determined, undertaking the task theoretically will be a mere speculation.

I also had a confusion regarding  finding hybridisation of four coordination complex compounds. How do we know if a complex compound is square planar or tetrahedral given the compound.  How does it depend upon the spectrochemical series. Uhhhh..The actual confusion is till which ligand in the spectrochemical series do I consider them to be strong and does it depend on the number of strong ligands. If yes, then how. I certainly need your help because I'm facing difficulties solving questions based on finding hybridisation of complex compounds

@Pewpaled The ligands that lie above water are considered to be strong field ligands in the spectrochemical series. Here's the one I use: $\ce{I- < Br- < S^2- < SCN- < Cl- < NO3- < N3- < F- < OH- < ox^2- < H2O < NCS- < CH3NC < pyridine < NH3 < en < NO2- < PPh3 < CN- < CO < NO/NO+}$

Let us continue this discussion in chat.

Thank you so much... that helped more than I can ever think. That is the ultimate answer to my doubts...The chat room was not working though...I'm logged in but it says to log in and then it fails logging me in.. Nevermind ..tysm

Here are the revelant messages from the chat room: As for the ligands if the complex has a CN=6 with 3SFL sand 3WFLs then consider all the ligands to be SFLs, if the complex has a CN=4 with 2SFL and 2WFL, take all the ligands to be WFL. If the complex is sp3 hybrid, it's tetrahedral and if it's dsp2 hybrid, it's square planar, the hybridization is solely dependent on the ability of the ligands to force the electrons to get paired up. I'm glad I could help you, if you get stuck, feel free to ask your queries, I'll be more than happy to help. :)

It read it but the problem was I couldn't chat back...Thank you again

If in four coordination no. Complex compound if the no.of SFL are equal to 2 and WFL =2 for 3d elements should I consider it to be dsp2 or sp3?

That depends, if there's a vacant d-orbital in the penultimate shell, the hybridization can be $dsp^2$, else it will be $sp^3$.

So what would be the hybridisation of [Ni(NH3)2Cl2] and K[Ni(NH3)Cl3]

Would it be sp3 for both?

Yes, you're correct.

And would it be dsp2 for [Ni(NH3)3Cl]+?

Exactly, $dsp^2$, $NH_3$ is a strong field ligand.

Then would that mean four coordination complex compounds having SFL more than or equal to 3 would be square planar for 3d elements and always square planar for 4d and 5d elements irrespective of the number of strong ligands?

https://chemistry.stackexchange.com/a/72729/43942 this might help in case of $d^8$ configuration. See, it depends on the configuration, you first write down the configuration of the CMA by using Hund's, Pauli's and Aufbau's rules. Then if it's a SFL, pair up the electrons in the $t_2g$ orbitals first and then proceed to fill $e_g$ orbitals in case of octahedral complexes. You might want to look that up if you're unsure what it means.

That helped a lot

https://chemistry.stackexchange.com/questions/76726/why-is-it-wrong-to-use-the-concept-of-hybridization-while-studying-complexes


text:
How to make a sephadex slurry
I want to purify crude alpha-amylase with column chromatography. I am using a spehadex 75, but for some reason I can't find any information on how to make the slurry.
I can quickly find a lot of information on how to pack and store a column, but what I really want to know is how much powder and buffer do I mix together?

Sephadex resins swell as they absorb buffer, so the amount of buffer you need to add is dependent on the amount that the specific resin you are using is expected to swell. It is also dependent on the final thickness of slurry that you want to achieve. Too thick - the resin will not be suspended well enough and will pack poorly. Too thin - the resin at the bottom will settle before enough buffer has flowed through to allow you to add more, and again you will not get good packing (and it will take a lot longer to pour).
Sigma Aldrich provides technical documents online with recommendations for preparing Sephadex slurries from dry powder, which you can find here.
Here are some the the key points. The expected swelling of the Sigma Aldrich resins is provided in units of mL per 1 g dry resin:
Sephadex G-10 Medium: 2-3 mL/g
Sephadex G-25 (all grades): 4-6 mL/g
Sephadex G-50 Fine: 9-11 mL/g
In all cases, add buffer and allow the resin to swell (1-3 hours at room temperature), then top off the buffer so that the settled swollen resin is ~75% of the total volume. That will give you a slurry of appropriate thickness when stirred.

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text:
Why do metals appear lustrous?
I came across a question asking me the reason for the lustrous appearance of many metals. The answer stated that it was due to the presence of free electrons in the metal. But I don't understand how this works. How do the free electrons affect the reflection of light from the metallic surface?
Thanks.

Free electrons have continuous energy spectrum, so they can respond to electromagnetic radiation of any wavelength. Also, related: https://chemistry.stackexchange.com/questions/72818/is-chromium-the-shiniest-metal

see also the answers to this question https://chemistry.stackexchange.com/questions/16633/why-is-gold-golden

Light is an electromagnetic wave. A metal has a large cloud of relatively free electrons (electrons that are loosely bound to the metal surface). When a beam of light is incident on a metal surface, it polarizes the electron cloud, i.e. some regions on the metal become relative more "positive" while some regions relatively become more "negative". Thus, this induces a field which makes the electrons to start oscillating. This oscillation generates another electromagnetic wave which opposes the incident radiation, (An ideal metal will completely oppose the incident light radiations), and hence our incident light rays get reflected.


text:
Cleaning with sodium bicarbonate
I have a small metal, most likely silver, cross. It is an old object and has turned dark. In order to clean it, my mother told me to boil it for a few minutes in water containing sodium bicarbonate.
Could someone explain what kind of reaction(s) occur(s)?

If originally made of silver, the dark layer on your jewlery is silver sulfide, $\ce{Ag2S}$.  The removal suggested by your mother is an electrochemical one, that will reduce the silver back into metallic, shiny $\ce{Ag}$, if in addition of hot water and bicarbonate, salt and an other metal acting as reducing component is in electrical contact with the silver piece are present.  Here, some foil of aluminum will serve well as sacrifical anode.
To quote UMass Amherst, presenting it as a lecture demonstration, the chemistry may be summed up to this equation
$$\ce{3 Ag2S(s) + 2 Al(s) + 3 H2O (l) –> 6 Ag(s) + 3 H2S (aq) + Al2O3(s)   }$$
The addition of salt (which is absent in the equation) serves to increase the electric conductivity of the solution, hence increasing the rate of this process.  Immerse the jewlery in the hot solution (it needn't be boiling), for example wrapped in the creased foil for some minutes; then rinse and wipe it with a cloth.

Thank you. Any idea how the silver sulfide was produced in the first place?

@ΚελεσίδηςΑντώνιος Silver sulfide is the product of the reaction of silver and hydrogen sulfide.  Hydrogen sulfide is a colourless gas, with the smell of "rotten eggs".  It is, for example, result of the digestion of organic matter (especially in absence of oxygen, as in landfills, wastewaters) already containing sulfur -- cysteine and methionine, for example, are two amino acids containing it. In addition, small amounts of hydrogen sulfide are produced by humans as signaling molecule, too.  It is found in small quantities in the air; sufficient -- over some time -- to cause the reaction obser

authors: Αντώνιος Κελεσίδης


text:
Kitchen Counter Endothermy?
I am not a chemist.  But I am curious...
This evening I was making ingredients for a "lemonade bar", where people mix various flavors into a lemonade base to create custom flavors (mint + cucumber is a favorite, strawberry + ginger is another).
I wanted to offer "really sour" as one of the mix-ins, so I put about an ounce of food-grade powdered citric acid in a jar and added enough apple juice to make it dissolve [see note].  I shook it vigorously, and as the citric acid dissolved, I noticed that the jar got fairly cold.
That's neat! So what's the process that makes apple juice + citric acid get cold?
(For extra credit: are there other kitchen counter endo- or exo-thermic reactions I should try with my six year old son?)
update
I just repeated the experiment mixing tap plain water and citric acid. It was also endothermic, dropping the water temperature from 21.5 to 11.5 celsius.
[note] And yessir, it's really sour!  :)

Breaking up the lattice of the solid citric acid costs energy. That energy is taken from the solvent. Same reason that water gets colder if you add salt

Dissolving ammonium nitrate is quite endothermic, and is used in cold packs, e.g. for athletic injuries.


text:
Effect of Temperature on Chemical Reaction Equilibrium
here is the question (part I):

For part (i), I need some assistance, I cannot figure out how to do the question. I know eventually what to do, it's just working out the equilibrium constants I'm having trouble with. So to start, I want to work out the K values at $\pu{600K}$ and $\pu{800K}$; I do this by considering the total pressure:
$$
\text{Total Pressure} = 1 \text{ bar}
$$
Initially:
$$
P_\ce{O2} + P_\ce{HCl} = \pu{1bar}
$$
Since there are $4$ moles of $\ce{HCl}$ per $\ce{O2}$ we can write:
$$
5P_\ce{O2} = \pu{1bar}  \to P_{\ce{O2}} = \pu{0.2bar}
$$
$85\%$ conversion means that in the end, $85\%$ of the oxygen has reacted; therefore, there is $\pu{0.2bar}*0.15 = \pu{0.03bar}$ oxygen remaining, therefore $\pu{0.12bar}~\ce{HCl}$ remaining. In addition to this, each mole of oxygen reacted results in $4$ new moles being created, and therefore there are $(0.85*0.2*2)\,\mathrm{bar}$ of water (and also chlorine) produced. With this information, I can calculate my equilibrium constant, right?

What is the equation for determining the temperature dependence of the equilibrium constant for a reaction?

@ChesterMiller -G = RTlnK?

@ChesterMiller Basically, I know if I assume the enthalpy of reaction is constant I can write: $ln\frac{K}{K_1} = \frac{\Delta H^o_T}{R} - (\frac{1}{T}- \frac{1}{T_1})$, so I can find the enthalpy of reaction. But I do not feel my method is right for getting the K values, does the total pressure have to remain constant at 1 bar? To me that does not make sense

No.  The total pressure does not have to remain constant at 1 bar.  What you need to do is review the derivation of the equilibrium constant from $\Delta G^0$

@ChesterMiller Hmm, I know you start at dG = -sdT + vdP + sum(chemical potentials), and I know the subsequent steps, I don't see how it can help though :/

Are you familiar with $\Delta G=\Delta G^0+RT\ln{Q}$?

@ChesterMiller Yes sir, and at equilibrium , $\Delta G = 0 \implies - \Delta G^o_T = RTlnK$

So, when you use the equation $K=\exp({-\Delta G^0/RT})$, does the total pressure have to be 1 bar?

No, there are no restrictions, @ChesterMiller

Then I can't understand why you feel that you cannot use the equation involving $\Delta H^0$ to determine the value of K at 600K.

@ChesterMiller My confusion stems from the question not specifying whether the pressure is maintained at 1 bar or not, if it is then my above working is wrong as the total pressures do not add up to 1 but if the pressure is not constant then my working works, I'm just confused with the system really.

This all goes back to the definition of $\Delta G^0$.  Please state your understanding of $\Delta G^0$

@ChesterMiller Ok, so if we consider the reaction, A + B --> C, then: $\Delta G^o = \mu^0_{C} - \mu^0_{B} - \mu^0_{A}$, basically the difference in chemical potential for the pure components at T.

You omitted that the pressure of each pure component is 1 bar.  So, in state 1, you have pure reactants (in separate containers) at temperature T and 1 bar.  And, in state 2, you have pure products (in separate containers) at temperature T and 1 bar.  Now, if you change the temperature, holding the pressure of each component constant at 1 bar, you have for each component $dG_A=S_AdT$

@ChesterMiller Right, so the equilibrium constant is independent of pressure,  so whatever K I get it should be independent of the total pressure in the system, be it 1 bar or something else. But when I am working out K by considering that K = $\frac{P_c}{P_aP_b}$ where those are the partial pressures, suppose I start with 0.2 bar partial pressure of A, if 85% reacts, is the final partial pressure of A 0.2*0.15, even if the pressure of the system changes?

ah okay, I get is now @ChesterMiller, thank you for your help. Changing the pressure doesn't affect the equilibrium constant but it does have an affect on the position of equilibrium. In my case, all I need to do is set up the question in terms of extent of reaction (which I know is 0.85), the pressure is 1 bar for the reaction, with this information I can calculate the equilibrium constant. Then do what I said earlier. Thanks a lot.

Nicely analyzed!!!!


text:
Why is oxygen so special?
Why do reactions with oxygen produce flame (i.e. light), heat and hence a lot of energy? I mean, why only oxygen, why not any other element like nitrogen? What is special about the electronic configuration/arrangement or molecular geometry that confers it with this property?
Maybe you
will cite examples of other reactions involving a lot of energy and producing flame, but I am sure there is nothing as familiar to mankind (in terms of the frequency in their occurance) as the reactions involving this compound. Reactions with oxygen are so ubiquitous and vital to life and thus they were classified under a separate category, i.e. combustion.

Many other materials will undergo rapid oxidation to the point of flaming in halogen atmospheres.

Your question is quite broad, so I will tackle it in parts.
Element abundance

there is nothing as familiar to mankind (in terms of the frequency in
  their occurrence) as the reactions involving this compound

This is because frequency depends on abundance. If you have a look to the element abundance on the earth oxygen is the first:
Element Amount
O   44.8
Mg  22.
Si  21.5
Fe  5.8
Ca  2.3
Al  2.2
Na  0.3
K   0.03
Sum 99.7

We have then to take care of the "kind" of oxygen you are referring actually to $\ce{O2} $ that is mainly produced by photosynthesis so linked to the evolutionary history of our planet. If you have a look to the atmosphere these are the ratio.
Element        Formula    %
Nitrogen        N2      78.084
Oxygen          O2      20.946
Argon           Ar      0.9340
Carbon dioxide  CO2     0.04
Neon            Ne      0.001818
Helium          He      0.000524
Methane         CH4     0.000179

Electronegativity

why not any other element like nitrogen?

This is due mainly to the electronegativity, actually, there are other molecules that have greater electronegativity and hence they are good oxidizing agents but because of their abundance, they are not so common.

Dioxygen, $\ce{O2}$ is a very special molecule. A good majority of organisms on earth use dioxygen to breathe and survive. Oxygen is also the second most abundant element in the sky, as well as the most abundant elements in the earth's crust. What makes it so special? Its because of its abundance, which is obvious, and also because of it's strong oxidizing power.
If you are a student below 11th grade, you might be taught that oxidation is addition of oxygen or the removal of hydrogen. I will now, introduce a new definition of oxidation.
Oxidation: The loss of electrons from a substance is called oxidation.
The reagents that cause substances to become oxidized are called oxidizing agents. $\ce{O2}$ is an example of an oxidizing agent. A popular oxidizing agent in analytical chemistry is Potassium Permanganate ($\ce{KMnO4}$ ).
Many times, an actual loss of electron doesn't take place, rather a shift of electrons from one element to another within the same molecule occurs, which is also considered as oxidation. To explain such stuff, we introduce the concept of oxidation numbers. An easy way to find oxidation number of an element in a compound is:
Ox.num=(Num. of bonds formed with electronegative atoms)-(Num. of bonds with electropositive atoms)
For example, take carbon dioxide. It's structure is:

Note that bonds made by an atom with a similar atom are not considered (like $\ce{O-O}$ or $\ce{S-S}$).
Carbon makes four bonds with more electronegative atoms (oxygen). So it's oxidation number is +4.
If oxidation number of any atom of a molecule is increased through a reaction, we say that oxidation has taken place. Note that as this happens, the oxidation number of the oxidizing agent will decrease.
Now, finally (with a deep breath), lets come to your question.
Speciality of Oxygen:
Oxygen in nature, as dioxygen has a 0 oxidation number. Since it's so electronegative, it almost always forms bonds with electropositive atoms, and so its oxidation number in compounds is always negative. Oxygen itself has such a high tendency to rip off electrons from other molecules. This is what makes oxygen very special.
There are so many examples of oxidation reactions with oxygen:
Combustion: Burning stuff itself is an oxidation reaction. If you burn coal(carbon), its oxidation number  goes from 0 (in pure carbon) to +4 (in carbon dioxide).
Respiration: Even our own bodies have to do an oxidation reaction to keep us with energy to do our daily work, like asking and answering questions on StackExchange. Glucose, which has a oxidation number of 0 (equal number of bonds with oxygen and carbon in its formula of $\ce{C6H12O6}$ ) is oxidized by the oxygen we breathe to carbon dioxide.
Much of the interesting properties of oxygen is due  to its oxidizing power. While many oxidizers are known, like fluorine for example; oxygen is, and will always, be the most well known one!

First of all, good answer. Also notice that I edited you question using the convenient and powerful \ce (for chemical equation) notation. It makes formatting chemical equations easier and formats them correctly according to the meta discussions here and here. Just FYI. Most importantly, keep up the good questions and answers!

Fluorine is even more oxidizing than oxygen, but we don't see biological reactions centered around fluorination. Might have to do with fluorination being too fast, and oxidation being in the sweet spot of kinetic constants.

@svavil you're absolutely right, fluorine is more oxidizing than oxygen, but oxygen is more commonly around. The reason why you don't observe fluorine oxidations much is because it isn't around much!


text:
Does Victor Meyer test distinguish alcohols and alkyl halides?
I read from a source that alkyl halides and alcohols cannot be differentiated by the Victor Meyer test. It says that both react similarly with the reagents and form the same products. Is this correct, and if so what are the reactions for alkyl halides.

What do you mean by reactions for alkyl halides? Afaik victor Meyer test is primilarily used to differentiate between 1°, 2° and 3° alcohols and not to differentiate alcohols from alkyl halides. (It's even not an effective test for 3° alcohols because they along with phenols and enols, do not respond positively to this test).

Well, as of what I have read, alkyl halides and alcohols are said to react in the same way with the reagents used for Victor Meyer test. That's what I've asked.

@AaronJohnSabu Just "as far as I have read" doesn't cut it. We invite you to logically decide whether its really apt or not. If you're confused, come over here and ask, mention exactly why you're conflicted.

I wanted to know if it was really apt or not. I am conflicted because I want to know whether alkyl halides undergo the same reactions. I want to confirm it.

Aaron,
You could be correct if your alkyl halides to be ditinguished are 1,2,3 degree alkyl IODIDES.
Because we ultimately are converting 1,2,3 degree alcohols into IODIDES in step-I; followed by reactions with NaNO2/HCl and then NaOH.
@Pritt Balagopal
He had a valid point, when he said "alkyl halides and alcohols are said to react in the same way with the reagents used for Victor Meyer test"


text:
What are the uses of ethyl phthalate
The last time I went to my junior college I remember seeing huge blue colored drums labeled ethyl phthalate. So just out of curiosity I decided to check upon this compound. Wikipedia entry containes some information about it's toxicity and usage, but I couldn't understand that part clearly. Could someone help me by listing the possible uses and properties of ethyl phthalate?

It's commonly used as an organic solvent is fairly benign as such things go. Was this in a chemistry department storage facility of similar?

Its main use, the reason for barrels, was a heat exchange medium in liquid heating baths. It has been replaced by heating mantles and silicone oils.

Wikipedia — Diethylphthalat (German) contains more use-cases for ethyl phthalate:

Formulation auxiliaries for plant protection products;
Fixer for fragrances;
In the process of ethanol denaturing;
In manufacturing of cellulose esters as a very well-gelling and light-resistant plasticizer;
A plasticizer for cellulose-based varnish binders and as enteric coating for medicaments;
A part of cosmetic products due to its film-forming, softening and hair-conditioning properties;
Explosives phlegmatizer;
A carrier in the dyeing process of cellulose acetate or triacetate fibers.


text:
Iodoform reaction of alkyl halide?
I read from a source that 2-iodopropane undergoes iodoform reaction. What reaction mechanism is undergone here, and why does this happen?

Mention the source.

Well, it was an online test for JEE Adv Prep

Yes, Aaron.
It will.
Because Iodide is very good leaving group. In iodoform test, we use base... which will convert 2-iodopropane into propan-2-ol. The latter will give haloform the same way as ethyl alcohol and any other methyl ketone does!

Oxidation of isopropanol to acetone using halide? This isn't a common reaction

Okay. How then would ethanol be giving haloform reaction...
This is oxidation by halogen [NOT BY HALIDE]. Halogens are good oxidizing agents!

Are you able to give a mechanism for the reaction? Or a source in the original literature oxidising isopropanol with just a halide/halogen?

https://en.wikipedia.org/wiki/Haloform_reaction
Go to mechanism section
Reaction sequence 2

In that case, will any CH3CHX- group give the iodoform reaction? (X = halogen)

Yes!
Except that of CH3CHF-R


text:
How does acidity affect electrode potentials?
What is the role of acidity in the decrease of formal potential of $\ce{Fe^{3+}$/$Fe^{2+}}$ in a sulfuric acid and phosphoric acid mixture?
I've observed that the formal potential of the $\ce{Fe^{3+}$/$Fe^{2+}}$ electrode in a mixture of sulfuric acid and phosphoric acid, $E_0 = \pu{+0.61V}$ is much lower than the standard electrode potential, $E_0 = \pu{+0.76V}$.
I came up with a plausible explanation that phosphoric acid would cause $\ce{Fe^{3+}}$ to get precipitated as  $\ce{FePO4}$ and the presence of excess acid will help in the formation of a complex $\ce{[FeHPO4]+}$. Am I correct in my observation? How would the situation change if the medium was basic?

Mamun, you must point out where youre stuck. Suppose, I was wondering which α-carbon of a ketone is more acidic. I would ask the question, and also add how I try to find out. I would also include something like "I considered both resonance and inductive effects, but the former increases stability, while the latter decreases it".

For your situation, you wonder how acidity affects the electrode potential. Try to solve your situation. Add some points as to why acidity might affect the potential, or some points saying why acidity doesnt affect.

what i know is phosphoric acid precipitates Fe3+ as FePo4 ,,then excess phosphoric acid complexes with it to form FeHPO4+ in 'acidic medium'    now i dont understand what is the problem if this complex is in basic medium.?

related https://chemistry.stackexchange.com/questions/13963/why-is-the-water-reduction-of-oxygen-favored-in-acidic-conditions https://chemistry.stackexchange.com/questions/14111/why-do-acids-favor-the-oxidizer-and-bases-the-reducer

omg! how did my question change by itself!!

@Mamun We (myself and berryholmes) changed it for you.

There are numerous reactions that involve iron and many involve either $\ce{H+}$ and $\ce{OH-}$. Take a look at this table and search for "$\ce{Fe}$". The non-standard potential for each one depends on the concentration of the species involved, and for the ones with hydrogen or hydroxide ions this means it includes the pH. This is described by the Nernst Equation. The potential varies as the natural log of the concentration, so you would add or subtract the pH or pOH, depending on the half reaction in question.


text:
Slight FTIR band intensity changes across several different copper salt spectra: relevant for characterization?
I have several spectra of different samples which are probably all copper (neutral and/or basic) acetate(s) and which are very similar. I present them overlayed and (some of them) stacked:

My question is: can I assume all the samples are qualitatively identical or do subtle/slight intensity differences in the same bands mean a qualitative difference?
These are all spectra made with a pellet of KBr and I intend a qualitative characterisation to compare with literature, not a quantitative one.

Chemists prefer IR spectra with (remnant) %transmittance along the ordinate axis, and with one (or even two) brakes in the abscissa at about 2200 1/cm (or even a twice at about 1000 1/cm) instead of a straight linear axis; it renders the spectrum more legible.  In most cases, the software of the spectrometer offers this and it is only a few clicks away.  Is it a transmittance spectrum with a pellet of KBr, or a neat film/ATR?  Do you intend a qualitative characterisation to compare with literature, or a quantitative one? Do the traces represent several batches of one (what?) intended product?

(Answered after your edit of your question:)
Some variation in the location of the absorption bands (vs. 1/cm) and their intensities (vs. remnant transmittance) is unavoidable due to the interaction of your analyte (here: your copper salt/salts) with the matrix (here: pellet of KBr) compared to recording a neat film of your sample.  The spectra depicted by the trace in magenta and blue seem to be a little bit different from the others, perhaps due to some humidity (KCl exposed to air, or a sample not completely dried, or both, ...).  On first view of the current representation, the other look very similar, compared with each other.

As your data represent infrared spectroscopy, figure out how to change the representation such that your ordinate is in remnant %transmittance instead of absorption.  Let the spectrometer's software do this for you.  For a visual inspection of the data, a split in the abscissa, too, for example around 2000 1/cm eases the visual inspection a lot -- equally something typically offered by the software of the spectrometer. (On good instruments, both changes do not alter the raw data collected by the spectrometer.) For example:

(source, the picture was edited by addition of horizontal lines)

Determine the barycentre of your absorption bands. Likely, as you start with IR spectroscopy, a deviation of about 1/cm is not an issue  (ask your instructor).  In other words, if the peak pic by the software suggests 1837.54 1/cm, a report like 1838 1/cm likely is good enough (ask your instructor).
For each important absorption band, have a look about the shape, note if it is a particular broad absorption band -- in your example, around 3350 1/cm, for example.
Beside location, an other important information provided by IR spectroscopy is the intensity of the absorption band.  In the representation of %T, imagine three equal parts between the baseline on top, and the most intense absorption band on the bottom.  A bit like the three horizontal blue dotted lines added to the picture above. The absorbtion bands in the first strip are then (semiquantitatively) classified as of weak, the ones reaching till the second strip of medium, and the ones reaching down to the third strip of strong intensity.  (As an example, the edited spectrum above contains an absorption band of medium intensity at about 3500 1/cm, and two strong ones at about 1700 and 1500 1/cm.)
Compare localisation and absorption band intensity of the spectral data recorded by you, with data found in a reference (textbook, journal article, data base).  Are all important absorption bands found in your spectrum mentioned in the literature reference?  Do you find all important absorption bands described in the literature reference in your spectrum, too?  Are there some important bands in addition or missing?
Of course you may compare sample #1 with sample #2, with sample #3, etc.,
too.

Only by growing experience you will learn what important absorption bands in infrared spectroscopy are.  Because -- in contrast to other spectroscopic techniques -- it is rather the exception than the general case that you may correlate each absorption band with a structural element in the structure of your compound.
For future reference I suggest to record the IR data either already in the %T mode, or (second best) to convert them into this form.  This allows you to check rapidly if even the most intense absorption band does not hit a transmittance of 0% (such an occurence were a rapid indication of a sample with to high concentration of X in your pellet of KBr.)


text:
Products of ozonolysis of 3-methyl-3-pentene?

Can someone please help me explain the steps on how to predict the products?

2-butanone and butyraldehyde, assuming you have drawn the correct substrate and made an error naming it.
Ozonolysis with DMS workup splits the double bond to give two carbonyl groups.

You shouldn't give final answer for such homework questions.


text:
Shape of a wavefunction
I am trying to plotting the angular parts of a wavefunction ($Y_{(m,l)}(\theta, \phi)$) with Wolfram Mathematica. this is the table of spherical harmonics.
I think the shape of these function should looks round each part of p and d orbitals is a ellipsoid looks like an egg, like this figure:

I have 2 question with my result:
1. All my figure of p orbitals only have half part the orbital above(examples:
$p_0$ orbital and $p_y$ orbital). And all the d orbitals looks longer than that above (example: $d_{xz}$).

2. When I square the spherical haromics ($[Y_{(m,l)}(\theta, \phi)]^2$), The p orbitals looks the same as the top figure (example: $(p_y)^2$) ,but the p orbitals getting far longer and thinner than that before squared, also the figure in the top (example:$(d_{xz})^2$)

What's wrong with my plot or the shape is not looks like eggs? Do I miss something that make the figure only have half part?

You are plotting different functions. The figures have no reason to be more than vaguely reminiscent of each other.
Now, those nice puffy things above are the isosurfaces of $\psi$ function. This is a function of three variables, and it just so happens that it can be represented as a product:
$$\psi(r,\varphi,\vartheta)=R(r)\cdot Y_{\ell,m}(\varphi,\vartheta)$$
You fix a value, find all points in 3D where the function takes that value, and plot them. That's what an isosurface is.
(By now, you might have noticed that the radial part $R(r)$ never appears at all in your attempts, so they got to be producing something different. And so they do.)
OK, then what are your plots? These are the spherical graphs of the angular component alone. You stay at the origin, point at the direction $(\varphi,\vartheta)$ with a stick having length $Y_{\ell,m}(\varphi,\vartheta)$, and leave a dot there. You don't care at all about the value of $\psi$ at that location. Besides, when $Y_{\ell,m}$ is negative (which it is half of the times), you point in the opposite direction. This, BTW, is the reason for having only one lobe in your p-orbital-like plots. No wonder you've got both lobes after you've squared $Y_{\ell,m}$ and thus made it always positive.
It could have been worse, you know. Think of a cone with its tip at the origin. Is it an isosurface of any function? Of course it is. What function? For example, $\cos\vartheta$ would do. What is the spherical graph of $\cos\vartheta$? A sphere, as you already know. Does a sphere look like a cone? Er, well...


text:
How can I increase the current on a voltaic cell for high school lab?
I'm creating a voltaic cell project for my high school chemistry students. This is a design challenge for them to use electronegativity differences to produce enough voltage and current to power a small DIY motor and fan. So far, I have been able to generate about 1.7V but only 0.4mA using lots of spirals of Mg ribbon along with a Cu electrode using porous cups (and not a salt bridge). I've increased the concentration of Copper nitrate. What methods might increase the current? We're have enough voltage to run 1.5V DIY motors but they won't run due to lack of current.
We're doing this: http://static.nsta.org/files/tst1409_30.pdf

Increase the area of the electrodes (e.g more ribbon).
Decrease the resistance of the electrolyte (e.g. reduce the spacing between electrodes, increase concentration of salts).
Reduce polarization of electrodes (bubbles on surface, etc.). Scrape or lightly sandpaper the electrodes and test immediately after immersion, before bubbles form.

Thank you for the ideas. I put quite a lot of Mg ribbon spirals. I'll try all of these in combination. A friend said maybe increasing the contact of the electrode and the wires might help. Would that help?

It could if the electrode is quite dirty or oxidized, but normally the resistance of the electrolyte is far greater than contact resistance of wire and electrode.

That helps. These are new electrodes so shouldn't be a problem. I'll keep working on it. thanks


text:
Salt hydrolysis or buffer solution
Whenever there a question comes to find the pH of the given solution, I always gets confused to identify the solution if it is a buffer solution or whether it will undergo salt hydrolysis .
Can anybody explain this to me?

This would be much easier to answer if you could [edit] in a couple examples.

When you have just a salt left with either [or even both] of its ions coming from a weak acid/base, it is a case of SALT HYDROLYSIS. I mean to say no free acid or base is left.
e.g,
100 mL 0.1 N CH3COOH + 100 mL 0.1 N NaOH ---> 100 mL 0.05 N CH3COONa
This need to be solved by SALT HYDROLYSIS
While if you have a salt(from the same weak acid/base) + a weak acid or base present, i.e., both are present, it would be a case of BUFFER.
200 mL 0.1 N CH3COOH + 100 mL 0.1 N NaOH ---> 100 mL 0.033 N CH3COONa + 100 mL 0.033 N CH3COOH (left unreacted)
This need to be solved by BUFFER

Avoid answering questions with low votes.


text:
Why does a titration have to be precise?
When we do titrations (w/ phenolphthalein) in lab, I know it's fun to see how close we can get the solution to clear, but why does my teacher place so much emphasis on accuracy and criticize overtitrating by just one drop? In our calculations, we end up just rounding volume of titrant used to the nearest tenth of a mL.

Well you really shouldn't be rounding to the nearest ml... If you do that then of course there's no point in being precise.

I've edited: I think I meant to say nearest tenth, which is what most burets are calibrated to; I thought the implication was still clear. I still don't think I drop will affect calculations that much?

Yes, one drop might not affect your calculations that much.
However, at the end of the day, it is simply good laboratory technique to be as accurate as possible. If you only care about your titrations in the context of a lab grade, or the concentration of some solution that does not affect your life, then yeah, it's not a big deal. Nobody really cares whether your concentration is 1 M or 1.005 M.
That said, in general, you should always aim to carry out experiments to the best of your ability (and not just for "fun"). If you can do something well and get accurate results, then it's simply not good practice to settle for less.
Fundamentally, chemistry is an experimental science, and we can only talk about all the things we do now because of people who were very careful with their experiments and measurements.
Whether your teacher is being too strict or not is not a question for us to answer. Every individual will have their own approach to teaching the subject. However, what I can see is that he/she clearly wants you to take the lab seriously. Their insistence on your accuracy is almost certainly because of this, rather than any technical reason why one extra drop might lead to disastrous consequences.

This makes sense--not sure why my post got so much hate, but it's a bit clearer now.

Because, near end point in titration pH shoots very fast with every single drop. To get the exact value of the concentration of your titrant, it is a MUST.
2 drops will be ~ 0.1 mL
I belive you ought to be rounding it off to the first decimal place, and not to the nearest mL. If you do so there is no point of skipping the end point by 20 DROPS!!!

You are correct in saying that near the end-point, the pH change will be very large even for a small volume of titrant. However, the change in pH is not relevant if the sole purpose is to determine the concentration. If the correct value is $\pu{20 cm^3}$, and you measure $\pu{20.05 cm^3}$, the final calculated concentration will be off by 0.25%; it does not matter whether the extra drop leads to a pH change of 0.01 or 1.

This was my point, thanks. I'm trying to clarify the difference between change in pH (certainly not a linear relationship with respect to volume) and what you're trying to find--typically something related to number of moles of base added, which IS linear with respect with number of MILLILITERS of base added. So being off by a drop shouldn't matter much. So I guess my question now becomes--is the purpose of being really accurate just for when you're recording pH rather than recording volume of titrant?

I remember my professor back then saying the in the case of overtitration always back tirate and adjust accordingly. And I get what your teacher is trying to do, since you're still at the university (?) level or still in school, they're supposed to guide you in applying accurate laboratory techniques. Though when you're doing routinary work in industry, you get so used to doing titrations (at least for a few people I know) that you start to get a feel of how to adjust in such a way that your results are still pretty spot on. At the school level, you're going to have to live with the horror that comes with overtitration.


text:
Is it possible to extract mass concentration from DLS measurement?
DLS (dynamic light scattering) gives the particle size distribution (number percent vs. size) of particles in suspension. It also gives the count rate (cps, counts per second).
Knowing this and the physical properties of the dispersing medium and particles, is it possible to extract the mass concentration of the particles?


text:
Overall effective activation energy for parallel reactions
Today, I was solving a question from Chemical Kinetics.
Consider the following parallel, first order reactions:
$$\begin{align}
\ce{A -> P} \tag{1} \\
\ce{A -> Q} \tag{2}
\end{align}$$
with rate constants $k_1$ and $k_2$, and activation energies $E_1$ and $E_2$ respectively.
I was asked to calculate the overall activation energy $(E)$ for the decay of A.
In my book a formula for $E$ is given to be :
$$E = \frac{E_1k_1+E_2k_2}{k_1+k_2}$$
This formula works well and provides correct answer, but I was trying to derive this expression using the Arrhenius equation. But I didn't succeed. I could deduce that,
Since:
$$ k_1=A_1 \mathrm e^{-E_1/RT}~; ~ k_2=A_2 \mathrm e^{-E_2/RT}$$
and
$$k_{\text{eff.}}= k_1+k_2 = A_{\text{eff.}}\mathrm e^{-E/RT}; $$
Using these both:
$$k_1+k_2= A_1 \mathrm e^{-E_1/RT} + A_2 \mathrm e^{-E_2/RT}$$
Now, I have no clue what to do.

What chemical kinetics book are you using?

The activation energy $E_\mathrm a$ should, in general, be viewed as an empirical quantity that characterises the sensitivity of the rate to temperature. From this perspective, it may be obtained via the equation
$$E_\mathrm a = RT^2 \left(\frac{\mathrm d \ln k}{\mathrm dT}\right) \tag{1}$$
This relation can be "derived" from the Arrhenius equation. (Quotation marks because the Arrhenius equation itself is only an empirical relation.)
$$\begin{align}
k &= A\exp\left(-\frac{E_\mathrm a}{RT}\right) \\
\ln k &= \ln A - \frac{E_\mathrm a}{RT} \\
\frac{\mathrm d \ln k}{\mathrm dT} &= \frac{E_\mathrm a}{RT^2}
\end{align}$$
If a reaction rate has Arrhenius-type behaviour, i.e. $k = A\exp(-E_\mathrm a/RT)$, then equation $(1)$ simply returns the usual interpretation of the activation energy. However, equation $(1)$ is useful in that it allows us to define an "activation energy" in the case where the rate constant does not have the mathematical form $A\exp(-E/RT)$. The parallel reactions that you have described fall exactly into this category. The sum of two exponential functions generally does not return an exponential function, so the overall rate constant $k = k_1 + k_2$ should not be expected to conform to an Arrhenius description.
If we substitute
$$k = k_1 + k_2 = A_1\exp\left(-\frac{E_1}{RT}\right) + A_2\exp\left(-\frac{E_2}{RT}\right)$$
into equation $(1)$, then we get:
$$\begin{align}
E_\mathrm a &= RT^2 \cdot \frac{\mathrm d}{\mathrm dT}\left\{ \ln\left[A_1\exp\left(-\frac{E_1}{RT}\right) + A_2\exp\left(-\frac{E_2}{RT}\right)\right] \right\} \\[8pt]
&= RT^2 \cdot \frac{A_1(E_1/RT^2)\exp(-E_1/RT) + A_2(E_2/RT^2)\exp(-E_2/RT)}{A_1\exp(-E_1/RT) + A_2\exp(-E_2/RT)} \\[8pt]
&= \frac{A_1E_1\exp(-E_1/RT) + A_2E_2\exp(-E_2/RT)}{A_1\exp(-E_1/RT) + A_2\exp(-E_2/RT)} \\[8pt]
&= \frac{E_1k_1 + E_2k_2}{k_1 + k_2}
\end{align}$$
as desired.

The effective activation energy you quote is the weighted sum of the two individual ones to P and Q. The fraction of molecules going to P is $k_1/(k_1+k_2)$ and so $k_2/(k_1+k_2)$ go to Q. Thus the effective activation energy is the sum of these two fractional values, i.e.
$$E_{eff}=\frac{k_1}{k_1+k_2}E_1 + \frac{k_2}{k_1+k_2}E_2 $$
which is your equation.
(Note that the rate constant for the decay of A and rise-time of P and Q is the same and is $k_1+k_2$; P and Q are only kinetically distinguished by the fraction of A that ends up as P and Q not by how quickly they are produced).


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What is created when "Iron-Out®" is mixed with household bleach?
I accidentally mixed bleach in a toilet bowl that had Iron-Out in it. The gas almost overwhelmed me, and left me gasping for breath for a good 15 minutes. What in the heck did I create with those two chemicals?

This is potentially quite serious. Some people can have severe idiosyncratic reactions to sulfur dioxide. Go to ER right now. If you start to have breathing difficulties again, call an ambulance.

Good that you left for fresh air.
Assuming it is the same product I know, it calls here for caution:

The below mentioned data safety sheet available (SDS) here mentions for the very product as composition

and recommends as first aid (same source):

Upon the action of bleach, you likely generated sulfur dioxide, gas with pungent smell. Refraining from a remote diagnosis, but if you accident happened in a less ventilated space and you inhaled a lot, consultation of a physician may be adequate as sulfur dioxide ($\ce{SO2}$) is harmful and maybe you equally inhaled vapours of bleach, too.  Likely, mixing "iron out" and bleach together yielded a considerable heat, too.


text:
How well will molten iron chrysolite (fayalite) react with hydrogen?
The equation balances nicely, but would the reaction tend to proceed this way? What temperature and time would be needed?
$\ce{Fe2SiO4 + 2H2 -> 2H2O + 2Fe + SiO2}$
There is no problem with supplying the energy to maintain a molten mix under the envisioned scenario, even up to pretty high temperatures. It's a solar furnace on the Moon. It can be left to boil as long as there is sunlight, which in some polar locations would actually be for months, elsewhere for 2 weeks at a time.
I have been trying to narrow the questions I ask here down as I seek an approach to bulk glass production in an advanced lunar colony. In consideration of that, it is important to bear in mind that the entire process is at high temperature and there is no exposure to the atmosphere, other than the gases that come from the mix - the furnace is outside, in the hard vacuum. The steam is pumped off and condensed, and if it is advantageous to maintain low pressure the pumping can achieve that.

It was helpfully pointed out to me that fayalite reacts with oxygen to produce silica and magnetite. The oxygen is consumed but more can be obtained, although it is an energy intensive process. So, that is another line of investigation. Their opinion was that a question on that regard likely fits better on Earth Science. I'm still researching it before asking something.

I assume water is fairly precious also. Fayalite can react with water to form magnetite, silica, and molecular hydrogen. The context is geologic formations, so I think the paper I saw is talking about pretty high pressures and temperatures for this reaction.


text:
Nernst equation calculations
I am stuck on 2 problems and would be grateful if anyone decided to give me a proper explanation as to how to solve them. I am not sure how to use the given information to correctly solve what is being asked. I have included the answers so its not a matter of academic integrity, I am genuinely lost.

Given the equation:
  $$\ce{2X^3+ + 3 Y -> 3 Y^2+ + 2X}$$
  The standard reduction potential for the X cation is $\pu{0.498 V}$.
  The standard reduction potential for the Y cation is $\pu{0.187 V}$.
  Using the Nernst equation, calculate the $\log$ of the equilibrium constant $K$ for the above reaction.

The answer is $\log K = 31.52$.
Using the answer, I attempted to work backwards:
$$0.685 = \frac{0.0592}{n} \times 31.52 \implies n = 2.724 ,$$
which I don't understand since $n$ is supposed to be the number of electrons.

Given the equation:
  $$\ce{2 X+ + Y -> Y^2+ + 2 X}$$
  The equilibrium constant for the reaction is $5.912 \cdot 10^{13}$.
  You measure the concentration of $\ce{Y^2+}$ to be $\pu{0.0674 M}$.
  Calculate molar concentration of $\ce{X^+}$.

The answer is $\pu{33.8 nM}$. This one I'm not even sure.

In (1) are you sure you have calculated the change in potential correctly?

For (1), I think you made the mistake of using a different value for overall cell potential when you worked backwards. Because if you use the correct overall cell potential, you get $K = 31.52$, your $\log K$ would be different if you used $0.685$ (which would make $\log K = 69.43$). Using the correct $E_\mathrm{cell}$ will get you the correct value of $n$ that you're looking for.
To solve for (2), think of what $K$ stands for. It's the ratio of something. Rearrange the ratio to solve for the unknown. It might seem confusing since you might think of an electrochemical solution, but that isn't always the case.

Thank you for your help! I was able to solve both of them.

@user44270 You can express your gratitude by upvoting and accepting the answer; also, you can post your own answer.

Your cell potential is wrong. You've added up both the reduction potentials instead of adding up one reduction and one oxidation potential. The oxidation potential of $\ce{Y}$ would be the negative of its reduction potential, so on adding you get 0.311, which, when substituted in the Nernst equation, would give you the right answer. Here $n = 6$ because there are 2 $\ce{X}$ atoms losing 3 electrons each.


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Why is a solution of PCl3 more acidic than MgCl2?
Why is a solution of $\ce{PCl3}$ more acidic than $\ce{MgCl2}$? I thought that both $\ce{MgCl2}$ and $\ce{PCl3}$ would react with water to form $\ce{HCl}$ so both solutions should have a very low pH. But does $\ce{MgCl2}$ actually form $\ce{HCl}$ in water or it simply dissolves and forms $\ce{[Mg(H2O)6]^2+}$, but if it only forms this, how is it slightly acidic?

I thought $\ce{MgCl2}$ would form a pretty neutral solution. The only reason I can think of for the formation of an acidic solution is the low solubility of magnesium hydroxide could result in precipitation of some of the hydroxide, so the solution could become slightly acidic. Compared to the situation with $\ce{PCl3}$, however, I'd say a solution of $\ce{MgCl2}$ should be pretty neutral.

Hydrolysis does not have to result in precipitation of hydroxide.

There is no such thing as a water solution of $\ce{PCl3}$. It reacts with water, and quite vigorously. Once the reaction is over, there is no more $\ce{PCl3}$ in the solution. Arguably, there is no $\ce{MgCl2}$ in the other solution either, but it would form (as a hexahydrate) if you evaporate the solution; $\ce{PCl3}$ would not.
The reason for this is that $\ce{Mg}$ is a metal and $\ce{P}$ is a nonmetal. They are different elements, about as different as two elements can be. They have nothing in common, except for a few sparse random traits, like being able to burn brightly in air when ignited. No wonder they are different in this regard as well. $\ce{MgCl2}$ hydrolyzes only very slightly (say, one part in a million). $\ce{PCl3}$ hydrolyzes completely.


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Does IUPAC nomenclature have the ability to name all organic compounds?
We all know about IUPAC nomenclature. There are rules to name straight chain compounds, cyclic ones, polycyclic ones, ones with functional groups, and so on. But I have come across several examples of compounds which I simply cannot name, most of them being pharmaceutical drugs. Take the heme group complex in hemoglobin:

I'm not sure, but systematic nomenclature appears not suitable to name this structure. A popular antacid, Zantac, also known as ranitidine has this structure:

Can compounds like these even be named systematically using IUPAC rules?

Definitely not.
You got yourself in trouble specifying all organic compounds, because there is a truly immense, mind-boggling number of possible compounds. No one even knows how to accurately determine such a quantity. A very rough estimate, making some incredible simplifications such as the use of only carbon, hydrogen, oxygen, nitrogen and sulfur atoms, is that there are some $10^{63}$ "reasonable" unique structures for compounds with a molecular weight below $\rm{500\ g\  mol^{-1}}$. Thanks to combinatorics, chemical space is enormous. We (or all sapient species in our observable volume, for that matter) will never come close to scratching its surface.
Furthermore, IUPAC nomenclature is largely created a posteriori. That is, though there are many rules trying to cover as many bases as possible (in the process becoming quite unwieldy at times), eventually some unexpected compound with unusual connectivity is discovered and becomes of wide interest. Thus standardising its nomenclature and that for closely related structures becomes fundamental to allow communication between scientists. A recent example of this occurred with fullerenes, which quickly jumped to prominence after 1985. IUPAC just had to create an entire new section of nomenclature for this class of compounds, which is not at all uncommon.
The closest thing to an absolute method of describing a compound's structure is to have a table of positional data (XYZ coordinates) giving the relative positions between the atoms determined from X-ray/neutron diffraction. Any attempt at simplifying this data will be lossy, whether the structures are drawn (not too lossy) or named (very lossy).
The structures you show have comparatively simple IUPAC names, in fact. Heme is a type of porphyrin, which is a widely occurring framework in biomolecules. The central framework can have its positions numbered and the substituent in each one read off separately. Regarding Zantac, the Wikipedia page for the compound states its IUPAC name in the "Identifier" section in the box at the right, namely N-(2-[(5-[(dimethylamino)methyl]furan-2-yl)methylthio]ethyl)-N'-methyl-2-nitroethene-1,1-diamine.
As some interesting examples of the relations between available nomenclature and chemical space, consider the following:

1,1,1,2,2,2-Hexaphenylethane : A molecule with a simple structure and a simple IUPAC name which likely cannot exist in reasonable conditions.
Maitotoxin : an awe-inspiring biomolecule with a rather large structure but containing fairly simple connectivity between atoms, whose IUPAC exists but is quite complex - disodium (2S,3R,4R,4aS,5aR,6aS,7aS,8R,9R,10R,11aR,12R,12aR,13aS,14aR)-10-[(2R,3R,4R,4aS,6S,7R,8aS)-6-[(1R,3R)-4-[(2S,3R,4R,4aS,6R,7R,8aS)-6-[(1R,3S,5R,7S,9R,10R,12R,13S,14S,16R,19S,21R,23S,25S,28R,30S)-25-[(1S,3R,5S,7R,9S,11S,14R,16S,18R,20S,21Z,24R,26S,28R,30S,32R,34R,35R,37S,39R,42S,44R)-11-[(1S,2R,4R,5S)-1,2-dihydroxy-4,5-dimethyloct-7-en-1-yl]-35-hydroxy-14,16,18,32,34,39,42,44-octamethyl-2,6,10,15,19,25,29,33,38,43-decaoxadecacyclo[22.21.0.0³,²⁰.0⁵,¹⁸.0⁷,¹⁶.0⁹,¹⁴.0²⁶,⁴⁴.0²⁸,⁴².0³⁰,³⁹.0³²,³⁷]pentatetracont-21-en-34-yl]-9,13-dihydroxy-3,7,14,19,30-pentamethyl-2,6,11,15,20,24,29-heptaoxaheptacyclo[17.12.0.0³,¹⁶.0⁵,¹⁴.0⁷,¹².0²¹,³⁰.0²³,²⁸]hentriacontan-10-yl]-3,4,7-trihydroxy-octahydropyrano[3,2-b]pyran-2-yl]-1,3-dihydroxybutyl]-3,4,7-trihydroxy-octahydropyrano[3,2-b]pyran-2-yl]-2-[(2S,3R)-2,3-dihydroxy-3-[(1S,3R,5S,6S,7R,8S,10R,11R,13S,15R,17S,19R,21R,22S,24S,25S,26R)-6,7,11,21,25-pentahydroxy-13,17-dimethyl-8-[(2R,3R,4R,7S,8R,9R,11R,13E)-3,8,11,15-tetrahydroxy-4,9,13-trimethyl-12-methylidene-7-(sulfonatooxy)pentadec-13-en-2-yl]-4,9,14,18,23,27-hexaoxahexacyclo[13.12.0.0³,¹³.0⁵,¹⁰.0¹⁷,²⁶.0¹⁹,²⁴]heptacosan-22-yl]propyl]-4,8,9,12-tetrahydroxy-hexadecahydro-2H-1,5,7,11,13-pentaoxapentacen-3-yl sulfate
This hydrocarbon : A molecule which likely exists, with a seemingly very simple structure, but with slightly quirky connectivity which makes naming it a challenge. Add a few more bridges and I'm sure you can break any existent nomenclature rules.

What do you mean by 'lossy'? How does structure drawing cause 'loss'?

I mean "lossy" in the sense of data compression. There are lossless techniques for compressing data, where the exact original data can still be recovered, and there are lossy techniques, where some of the original data is discarded. Lossless methods are usually much more complex, so in many cases we opt to use lossy ones that do the job well enough. Another way to look at it is if you're translating between different languages; some things just can't be described properly without the original terms.

I'm not so sure about data in the context of computers, but writing a structure, how do we lose data or information?

When you write down a structure, you necessarily build on a set of rules (e.g., H means hydrogen, a single line between two atoms means a covalent single bond, "ethyl" means a $\ce{C2H5}$ chain, etc.), and if there is nothing in your set of rules capable of describing a certain situation, then either you can't write it down, or you do so at the expense of discarding some subtle characteristic of the compound. In the latter case, if someone else later only sees what you have written down, rather than the actual thing, they will not be able to recover the structure in full detail.

I think you came to wrong conclusion, and that it is possible to name any possible compound using some existing IUPAC nomenclature system. If I got the question right, it does not ask "Is is possible to generate list of systematic names of all possible chemical structures", or "Is it possible to generate unique preferred systematic name for any existing compound?", but "Is it possible to generate (some valid, not necessary preferred) IUPAC systematic name for any existing or possible chemical compound?"

@mykhal Though you are certainly more knowledgeable about nomenclature than I am, I think you're being overconfident. Even though IUPAC does a great job and it takes effort to break existing rules, there are always fringe cases, and almost all of chemical space is made of fringes, even though we turn our eyes away from it and stay in the simpler regions for our own sanity. The fact that IUPAC constantly plays a game of catch-up with new sections of nomenclature should be a very strong indication that what we have now is not enough.

@NicolauSakerNeto, you are probably right, I have now remembered some very complex supramolecular assemblies, which would be probably impossible to name, definitely their exact stereochemistry.

@NicolauSakerNeto I just reread your answer again, and it's still very interesting to read :) . I found out a simple compound bromochlorofluoroiodomethane is also hypothetical despite it's simple structure. I don't like to edit answers being the asker, but you could add it if you think it's a good one.

@PrittBalagopal Thank you for your approval! Indeed, CIBrClF has not been synthesized even though it is conceptually very simple, but in this case it is likely an experimental limitation rather than a physical one. The compound is most likely stable in isolation, we just haven't found a reliable synthetic route to make macroscopic amounts of it. I wouldn't be surprised if some molecules of it have already been synthesized accidentally in minute amounts during the synthesis of other freon-type compounds.

(cont.) This is different from, for example, hexaphenylethane, which is predicted to spontaneously split into two triphenylmethyl radicals with a very low (if any) activation energy barrier

SMILES seems to do fine with naming everything.

Interesting question. Let's start with your examples.
You probably meant real heme, fixed:

"Heme" is recognized name in IUPAC tetrapyrrole nomenclature.
But it can also be named as (protoporphyrinato)iron(II).
Protoporphyrin can be named as
3,7,12,17-tetramethyl-8,13-divinyl-2,18-porphinedipropionic acid, or
2,18-bis(2-carboxyethyl)-3,7,12,17-tetramethyl-8,13-divinyl-2,18-porphine.
Porphine can be named as 21,22-dihydroporphyrin.
Porphyrin could be then named systematically as some ugly a,b,c,d-tetraazapentacyclo[X.Y.Z.UV,W.RS,T.OP,Q]tetracosane-i,j,k,...-undecaene (italic letters would be replaced with correct numbers).
How do we put these names together with iron? By using organometallic coordination compound nomenclature. But let us name that organic ligand with somewhat more recent phane nomenclature, that will produce less ugly name, as it is a powerful tool able to make abstractions from the local structures to "superatoms" in a simplified skeleton (can be used nicely to name the local interesting looking organic compound).

So the final name could be
[54,73-bis(2-carboxyethyl)-14,33,53,74-tetramethyl-13,34-divinyl-11H,51H-1,3,5,7(2,5)-tetrapyrrolacyclooctaphane-2(32),4(52),55(6),75(8)-tetraene-N,N′′-diyl-κ4N,N′,N′′,N′′′]iron(II).
I likely have some mistake there, but I can construct the structure according to the rules and check if it is correct, and if not, fix it.
The second example is named by some software in a fraction of a second as (Z)-N1-{2-[({5-[(dimethylamino)methyl]furan-2-yl}methyl)sulfanyl]ethyl}-N′1-methyl-2-nitroethene-1,1-diamine.
So yes, IUPAC nomenclature is constructed for chemists to be able to name pretty much anything. Some strange functional group is present? You can decompose it to building blocks (ad absurdum, you can name -COOH group as hydroxycarbonyl).
However, these rules are constructed by humans, and as far as I know, there is no formal proof that it can name any possible structure. It almost certainly is not (yet) describe e.g. stereochemistry of highly complex molecules with certain topological isomerism, e.g. complicated knots. Is "pyrrol" component name from the heme systematic name too trivial? It is already correct IUPAC name. But you can name it "more" systematically as 1-azacyclopenta-2,4-diene (if I'm not wrong with numbering). On the lowest level, the raw weighted graph of element nodes, the systematic nomenclature probably cannot operate in a practical way. The structural formula is worth a thousand IUPAC words.


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Examples of non-volatile, but flammable liquids?
As far as I know, most flammable liquids are volatile. Are there flammable liquids that are non-volatile?

The element cesium melts at 28.4°C which is slightly above "room temperature" but below human body temperature, so I think we can consider it a liquid in "normal" conditions. It is so flammable it readily self-ignites in the air, but if you put it in inert atmosphere you will see that it isn't particularly volatile.

Flammability and volatility of a liquid are related.
According to the Globally Harmonized System of Classification and Labelling of Chemicals (GHS), the term “flammable liquid” is defined as follows:

Flammable liquid means a liquid having a flash point of not more than 60 °C.

The European Agreement concerning the International Carriage of Dangerous Goods by Road (ADR) uses a very similar definition.

(…) have a flash-point of not more than 60 °C (…).

(Note that the definitions in some older national regulations and standards may refer to different temperatures.)
The flash point is the lowest temperature of the liquid at which application of an ignition source causes the vapour of the substance to ignite. (The flash point is an experimental value and not a constant physical-chemical property of the liquid. The exact definition and the resulting value depend on the used experimental method.) Thus, actually the vapour above the liquid (i.e. the mixture of vapour and air) and not the liquid itself is ignited. Therefore, for the liquid to be considered flammable, there must be a significant concentration of vapour in the air above the liquid at the given temperature.
If the volatility of the liquid is too low, it cannot be a flammable liquid. For example, biodiesel (flash point: 180 °C) and typical vegetable oils are not flammable liquids although these liquids can be burned.

This addresses the technical definition of the word flammable but not the question which but its very nature is not using the technical definition.

Most flammable liquids are flammable precisely because they are volatile (and combustible at that), so that they easily form vapors which easily catch fire. That being said, various oils are flammable (albeit less so than gasoline) and at the same time not very volatile.

for example vegetable cooking oil is pretty non-volatile and burns with a flame ok, although it hard to get it to ignite (at least on my gas cooker).

But there are plenty of volatile liquids that are not flammable, for example dichloromethane which is a volatile as diethyl ether but won't burn at all. Combustibility (which is what the question seems to be asking about rather than the the more technical "flammability") has little relationship to volatility.

Methylene chloride will burn and explode upper explosive limit23% lower limit 13%. Even trichloroethylene will explode

combustibility and volatility are unrelated
Whether things burn or not (i.e. whether they are flammable) is unrelated to whether they are volatile. Volatility only determines whether things are easy to set alight.
Consider coal, for example. This is a non volatile solid but it is flammable though hard to set on fire. Long chain extracts from crude oil (tars and bitumen) are (technically) liquids but not at all volatile. They burn, though not easily (which is why some motorway accidents involving fires take so much time to repair). A range of other hydrocarbon fractions from crude oil are more liquid but not very volatile as are many vegetable oils (used in cooking because they are not volatile even at 200 °C but clearly flammable as you can see if splashed into a flame).
And there are plenty more examples. The confusion arises because many very volatile liquids are volatile enough that their vapours are flammable at room temperature. There is a good reason why you don't use bunsen burners anywhere near ether (bp ~40 °C): the vapour will ignite and flash back to the vessel even if the flame is far away. But many non-volatile things will also burn.

This answer is in direct contradiction to Loong's answer. I agree with Loong's definition of flammable as it is used this way in the scientific and industrial communities. What you are talking about is combustible. A substance can be "combustible" without being "flammable." It is unclear whether the OP's understanding of the word "flammable" is closer to yours or Loong's definition. This may explain why you have downvotes (I have abstained from voting either way.)

@LevelRiverSt You are right if the definition of flammable requested was the narrow technical one. But, if that is what was wanted, the question is tautological. So, in context, I addressed the informal version which creates a better grasp of the the key concepts.

Every answer that equates volatility with flammability is wrong, this one is correct.

To tell the truth,
"Flammable" means substance with easy ignitable vapors.
But topicstater asked about "easy ignitable"- wider set of substances.
Among weakly volatile ones I can add:
1)potassium-sodium eutectic alloy,liquid at ambient temperature.
2)white phosphorus - melts at ~40C,selfignites in the air.
3)thermally unstable chemicals like peroxides/nitrocompounds/some metal organic substances.
Di-tert butyl peroxide (CH3)3COOC(CH3)3.
tert butyl hydroperoxide (CH3)3COOH
( https://pubchem.ncbi.nlm.nih.gov/compound/tert-butyl_hydroperoxide#section=Computed-Properties )
They easily react with oxygen of air (with or without heating) with subsequent selfheating to ignition. Or violently decompose, selfheat and finally burn.

Cooking oil is flammable, but doesn't evaporate or evaporates very slowly over months or years. http://www.oliveoilsource.com/asktheexpert/does-cooking-oil-evaporate It's flammable because if you heat it up enough, it will start to burn. Once it starts to burn, the heat will increase causing it to burn faster.

I am sorry to have entered this discussion so late.
Flammable means how easy it is to set on fire.
Volatile means how quickly it evaporates.
It is easy to think of liquids that are both flammable and volatile: ethyl ether, pentane, gasoline, etc.
Some liquids are volatile but NOT flammable. They include tetrachloroethylene, chloroform, hexafluoroisopropanol and water.

The question was whether there were liquids that are flammable and non-volatile.


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Titration with EDTA and Eriochrome black T
I have a titration of Calcium Ions with EDTA. When I reach a volume of 9.6mL I the liquid turns slightly blue and after that the more I add the bluer it becomes. Now at what point is my titration finished? When I notice the first change or as soon as adding more EDTA doesnt change the color?

Suggestion: Prepare a stem solution (i.e. of known $c(\ce{Ca^{2+}}$).  Perform the titration several times.  Once "as if you were unaware about $c(\ce{Ca^{2+}}$" till the point all Calcium is chelated (which may be, taking known concentration of your stem solution and its volume, calculated in advance).  Keep a small amount of this solution, for example in a test tube.  Repeat the titration, intentionally adding more EDTA solution, than necessary; equally keep some of this solution as visual reference in a test tube. Now perform a titration with your unknown, with the visual aides beside.

Assuming you've used Erichrome Black-T as an indicator, the second your whole solution changes from wine red to blue, as in the moment you realise there is a change in colour, however light it is, that's your end point and you should stop.
Make sure you use a white tile under your conical flask and remember to pause the titration and swirl the flask little bit every now and then as you near the end point. That should even out the colour and make the change noticeable before it's too late.

I see, thanks. Could you also explain why this is so?
As far as I know ErichromeBlack-T is pink as long as there are metall ions that it can bind to, so while there are pink and blue particles (first change) there whould still be some Calcium not bound to EDTA but to ErioT

I'm a bit unsure about what your question means but I'll try to answer.Erichrome Black-T is actually blue in colour in its protonated form but turns wine red on complexing. That is why it is imperative to swirl the the contents of the flask, to distribute the reagent evenly so that any colour change is uniform. It's only when the whole solution uniformly turns the lightest blue possible that you must stop. There should be no trace of pink of course, but just like it's unadvisable to have a deep phenolpthalein pink while titrating, it's the same here. Does this answer your question?


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What is the longest systematic name used for a compound that actually exists?
The answer to this question (Does IUPAC nomenclature have the ability to name all organic compounds?) which references the complex biomolecule Maitotoxin which has a ridiculously long systematic name.
I doubt that this is the longest systematic name ever used for a synthesised and fully characterised compound. What is?
PS I'm sure we can name many proteins using stupidly long names, but I suspect that is cheating. Let's stick to lab-made compounds and the published literature.

This question can never have a confident single answer, I'm afraid. Voting to close as too broad.

@hBy2Py On the contrary, it can have a confident answer if we restrict it to names used in the literature. That answer might change over time, but, in principle, we can always have a single answer.

In principle, yes, there is always only one winner in this race. But in practice, we never will have a single answer due to the epistemological problem: the 'world record name' as tracked here will just keep getting longer as people on the site happen to discover new chemicals with longer names, or as people who know of long names happen to come across this question. I stand by my vote.

@matt_black Used in literature is not a really good criterion. Since several software is capable to generate names, it is by chance if someone put a full name somewhere in his paper or supplementary information, while mostly calling it compound 3 in the manuscript.

@hBy2Py Flagging and Voting are meant to be anonymous. I strongly advise you to keep it that way.

@PrittBalagopal Welcome to Chem.SE. They are anonymous if the voter/flagger wishes them to be. Revealing one's votes and/or flags cast is entirely acceptable and common.

@PrittBalgopal If anything, I would much prefer if people told me why they downvoted/flagged my post rather than just receiving a bunch of downvotes. I believe the site actually suggests that you explain in the comments why are downvoting something and how it can potentially be improved.


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Mechanism for Amino-Imidazole Ring Fission
A reaction I've tried several times in the past is the hydrogenation of nitroimidazole derivative shown below. The amino product is extremely unstable and rapidly decomposes.
This instability is noted in may pieces of literature and the explanation tends to only go far as saying 'ring fission' occurs.
Question: What is the mechanism of this ring fission? To be clear, I am not interested in getting this reaction to work.... It is no longer a target of mine.
Perhaps the entire mechanism is not determinable, but I presume the amine of one molecule attacks another molecule's imidazole ring? Where would this be? How is fission occuring?
EDIT: Possibly something to do with the apparently favoured tautomerism between these two forms:
https://i.stack.imgur.com/AhNsE.jpg

I think this product should be stable, but would react with substrate in this case. IMO condensation of amino and nitro group could occur.

@Mithoron I've found evidence in some old textbooks that the decomposition takes place via a Van Slyke determination mechanism? Mechanism is below. I don't see how it applies given I have no acid in my mixture.
http://imgur.com/a/GbyKc

Maybe -HCN after lone pair donation from amine. This may cyclicize with the phenyl. Look at 123 triazole decomposition it will probably be similar.

I'm afraid the answer to how this ring fission occurs is... it doesn't, you just made some mistake.
Here and here the authors used the same reaction as you and got the product in good yields. I think you might not realized that the product is a very hydrophilic amine (or salt, depending on workup). If you did an extraction of your reaction mixture then the product surely ended up in the aqueous phase.

Can anyone explain why this answer gets downvoted? I provided the publications that back my claim and it is an explanation of what the question author observed.

I'm sure you already had an idea, but it's most likely the language. We're not trying to be prudes, but it unnecessarily detracts from the point you are trying to make.

OK, I'm sorry. I'll remember it and not use such languange in future.

I surmise that if you edit it the downvotes may be retracted, since you have a quite valid point.

@JanRzymkowski  You are wrong. I quote from the very first paper: "Palladium-mediated hydrogenation... afforded the
corresponding amino derivatives as free bases, which were immediately
converted into hydrochloride salts, as the compounds
are unstable as free bases..." Down voted.

@JanRzymkowski From the second paper: "The reduction of the nitro group, relatively easy in benzene derivatives, has proven challenging
in azole systems due to metastability of the
resulting amines".

@JanRzymkowski In my case, I wanted to amide couple the reduced amino imidazole with an acid which is impossible as my amine degraded before I could isolate it as the hydrochloride salt (strategy followed by paper 1 and others). Additionally, no extraction was carried out/necessary as following Pd hydrogenation you usually filter over celite and evap. your solvent to obtain the compound.

@orthocresol He does not have a valid point at all. Ring fission of amino imidazoles is well established and reported in the very same papers he has cited. To question my synthetic capabilities in the lab is one thing, but to misrepresent literature is just not right, hence, why I feel the need to set the record straight.

@Hazinga perhaps a poor choice of word on my part, I only meant to say that from a moderation POV the answer itself should be allowed to stand (whether it is true or not, I do not know. I did not fact check). I have had similar difficulties isolating 1,2-diaminobenzenes (after hydrogenation of o-nitroanilines) so I am not surprised that aminoimidazoles are problematic, but I don't know anything about the exact nature of the problem.

@orthocresol absolutely and no worries about the wording.

This is just a guess, but it could be worth considering carbanions.  The nitrogen atom containing the phenyl group, when donating its electrons into the ring (in the image at the top, the donation I am talking about is to the left of the donating nitrogen atom) the nitro group has enough electron withdrawing power to accept the donation from the adjacent bond.  Thel amine formed from the reduction of this nitro group, may be basic enough to deprotonate any remaining nitro compound (imidazoles are amphoretic).  If this is true, when the electron donation occurs, a carbanion may form.  With a carbanion could accelerate decomposition.

One explanation could be that the imadazole is hydrogenated as well, leading to a decomposition of the starting material to phenylamine, ethylamine and ethane:

Or something similar, e.g. formation of phenylamine and 2-aminopentane.

I see you are getting downvoted.  Can you supply empirical or literature evidence to back the claim?

Well, its just an idea. If you think its wrong, then explain why/provide an alternative.

While over reduction of the pyrrole ring is conceivable, reduction of the pyrrolidine ring is without merit.


text:
Does H2 + Cl2 produce 2HCl a redox reaction
Is $\ce{H2 + Cl2}$ produces $\ce{2HCl}$ a redox reaction?

Anything that involves change of oxidation states of reactants in the products or exchange of electrons can be considered a redox reaction.
Here, $\ce{H2}$ (oxidation state = 0) changes to $\ce{H+}$ (oxidation state = +1) and $\ce{Cl2}$ (oxidation state = 0) changes to $\ce{Cl-}$ (oxidation state = -1)
So, $\ce{H2}$ has been oxidised and $\ce{Cl2}$ has been reduced, making this a redox reaction.

While I agree about the formal change of oxidation states let us be clear; Hydrogen Chloride is not an ionic substance, it is covalently bonded

My mistake. Of course, I shouldn't have called them ions. I'll amend that in a second.


text:
Synthesize coumarin from 4-allyl-2-methoxyphenol (eugenol)
I'm trying to figure out how to get the following reaction: (coumarin from eugenol)

My immediate thoughts are

$\ce{OH}$ and $\ce{OCH3}$are bad leaving groups.
Could I just add $\ce{HI}$ to the $\ce{C10H12O2}$ to convert the methoxy ($\ce{OCH3}$) group to an $\ce{OH}$group? Then, I could add $\ce{Zn}$ to remove the $\ce{OH}$groups?
Or, is there a way to make the methoxy ($\ce{OCH3}$) group into methanol, which would be a better leaving group?
Once the methoxy and hydroxy groups are removed, I'm left with the benzene ring and the attached butene. I would need to add the $\ce{CO2}$ to form the second benzene ring, but I'm not sure how this would be done.
I might be thinking about this all the wrong way. I was thinking that I need to remove the methoxy and hydroxy groups since they're on a benzene ring, and my product's benzene ring doesn't have these groups. Maybe I don't need to get rid of the methoxy and hydroxy groups at all. Maybe I just need to shift the bonds around?

Look up the Claisen rearrangement. Boron tribromide  is the standard reagent for removing Me from phenol

Could you give any contex? The product is a very common, cheap chemical with no obvious route from the starting material - is this homework question? Sadly, textbook problems are aften very far from real syntheses.

Since no-one has offered an answer I'm going to try but as I have no database access some of this is a bit flaky.

Heat (maybe microwave) to give multiple Claisen rearrangement products then separate to isolate 2-allyl, 6-methoxyphenol. This is the flakiest step.
Hydroborate with oxidative work-up to take the allyl to propionic acid.
Ring close (PTSA, Toluene)
Reflux over activated MnO2 to introduce double bond.
Treat with BBr3 to remove Me from MeO group
Zn reduction of OH to H

I am in agreement with the comments of @jerepierre and @Jan Rzymkowski. The deoxygenation of phenols by zinc dust is used on occasion on SE but I have been unable to find it in the primary literature. Jan is correct. Who would try to convert eugenol (1) to coumarin (10) when salicylaldehyde is the logical precursor? Nonetheless, I'll offer a solution that is somewhat similar to the one provided by @Waylander.
The base t-BuOK in DMSO isomerizes eugenol (1) to isoeugenol. Protection of the phenolic group of isoeugenol as the acetate permits conversion of the carbon chain to a carboxyl group via vanillic acid (3). The decarboxylation of vanillic acid is modeled after a literature report for the conversion of benzoic acid 11 to propofol (12). [Cucumber juice is claimed to effect this reaction via a biotransformation.] Phenol 4 is allylated and subjected to Claisen rearrangement to afford allylphenol 6. The allyl chain can be parlayed into an aldehyde group by the methods employed earlier. Salicylaldehyde itself has been converted into coumarin using the Perkin condensation. Waylander's use of boron tribromide for demethylation is an excellent choice. If indeed zinc dust works, reduction of the double bond may be problematic. Reinstallation of the double bond lacks elegance but it can be accomplished [a) LDA/THF, PhSSPh; b) H2O2; c) heat.]