text:
contradiction between Gibbs energy variation and redox potential
Let's consider the following two redox reactions;

$\ce{C_2H_4O_2 + 2\;O_2 -> 2\;HCO_3^- + 2\;H^+}$
$\ce{C_2H_4O_2 + 1.6\;NO_3^- -> 2\;HCO_3^- + 0.8\;N_2 + 0.8\;H_2O + 0.4\;H^+}$

Considering that the standard Gibbs energies of formation of the involved chemical species are the following;

$\ce{C_2H_4O_2}$: -393.8 kJ.mol-1
$\ce{HCO_3^-}$: -586.9 kJ.mol-1
$\ce{NO_3^-}$: -111.3 kJ.mol-1
$\ce{H_2O}$: -237.2 kJ.mol-1
$\ce{O_2$, $N_2$, $H^+}$: 0 kJ.mol-1

Then the standard Gibbs energy variation associated to those reactions are respectively -804.4 kJ.mol-1 for the oxidation of acetic acid by oxygen and -816.0 kJ.mol-1 for the oxidation of acetic acid by nitrate.
It appears that the oxidation by nitrate is more exergonic that the oxidation by oxygen in standard conditions, when computed this way.
However, the redox potential of $\ce{0.25\;O_2 + H^+ + e^- -> 0.5\;H_2O}$ is +0.811 V, while the redox potential of $\ce{1/5\;NO_3^- + 6/5\;H^+ + e^- -> 1/10\;N_2 + 3/5\;H_2O}$ is 0.746 V. I haven't found any data for the redox potential of acetic acid, however the redox potential of its electron acceptors indicates that an electron transfer from acetic acid to oxygen should imply a larger Gibbs energy variation than an electron transfer from acetic acid to nitrate (according to the $\Delta G = nFE$ formula), which contradicts what has been previously computed from Gibbs energies of formation.
I am new to electrochemistry, so what did I misunderstand here?

Assuming that you have made no maths errors in calculating $\Delta G$ from the Gibbs energies of formation, I think the first thing I would check is your reduction potential for $\ce{NO3-}/\ce{N2}$. That half-equation you have written is not balanced and I couldn't find the value for the reduction potential (I did a brief search in Atkins as well as Wikipedia).

Apart from that, I can't see anything wrong with your line of reasoning. There are a couple of typos ($\ce{NO3+}$, and $\Delta G = -nFE$ (minus sign is missing)) but they don't affect the logic. Perhaps I am being dense too, I don't know.

Yes you are right, there were some typos in the $NO3^-/N_2$ half-reaction. However this doesn't affect the conclusion. For the record, those redox potentials come from Stumm and Morgan, 1996

Looking up the redox potentials in tables vs. Hydrogen as standard at zero volts the redox for Oxygen is
$$\ce{O2 +  4H^+ + 4e <=> 2H2O} \qquad \qquad E^\circ = +1.229~\mathrm{V}$$
so the tables you use may be vs. calomel electrode as standard.
The redox potential for the nitrate reaction has to be constructed from others as
$$\begin{align}
\ce{NO3^- + 3H^+ + 4e}\, &\ce{<=> HNO2 + H2O} & E_1^\circ &= +0.934~\mathrm{V} \\
\ce{2HNO2 + 4H^+ + 4e}\, &\ce{<=> N2O + 3H2O} & E_2^\circ &= +1.297~\mathrm{V} \\
\ce{N2O + 2H^+ + 2e}\, &\ce{<=> N2 + H2O} & E_3^\circ &= +1.766~\mathrm{V}
\end{align}$$
The overall half-reaction, after multiplying the nitrate reaction by 2 (to balance) and adding is
$$\ce{2NO_3^- + 12H^+ + 10 e <=> N2 + 6H2O} \tag{1}$$
The redox potential for the reaction $(1)$ is calculated from $n_\mathrm{r}E_\mathrm{r}^\circ = n_1E_1^\circ + n_2E_2^\circ +....$ where the n's are the number of electrons in each half reaction and $n_\mathrm{r}$ for the new reaction.
The $\ce{NO_3^- to N2}$ redox potential is therefore $1.246~\mathrm{V}$, calculated as
$$E_\mathrm{r}^\circ = \frac{(4 \times 0.934) + (4 \times 1.297) + (2 \times 1.766)}{10}$$
The two values, $1.229~\mathrm{V}$ and $1.246~\mathrm{V}$, would now seem to agree with your $\Delta G$ values. They may be the same within error as may your $\Delta G$ values.
(Redox Values are from the CRC Handbook of Chemistry and Physics, 75 ed.)

Wonderful, I will openly admit I was too lazy to fact check :)

"so the tables you use may be vs. calomel electrode as standard" no, it is that the OP is using redox potentials at pH 7

I biochemistry it is normal for reactions involving protons to use pH 7  as a standard state thus $\ce{O2 + 4H^+  + 4e <=> 2H2O}$ has a potential of $ E^{0'} = 0.816 $ V, but this does not matter if all potentials are on the same standard state. We only need the difference in values.

@porphyrin it matters because the number of moles of  H+ is different in the two equations, so changing 1M to 10^-7M affects them differently.  The difference is different at pH 7 vs pH 0.  I will post data later.

@porphyrin so at pH 7, according to table 11-1 here https://books.google.com/books?id=6TMvdZQiySoC&pg=PA207&dq=nitrate+redox+potential+Stumm+and+Morgan+1981&hl=en&sa=X&ved=0ahUKEwjthbWz3bTPAhUCLSYKHduZAhQQ6AEIHDAA#v=onepage&q=nitrate%20redox%20potential%20Stumm%20and%20Morgan%201981&f=false the redox potential are as OP is saying.  And the reason the differences are different is as explained here http://www1.lsbu.ac.uk/water/water_redox.html .

@DavePhD Yes you are correct, correcting for the pH is the proper thing to do and  I was not precise enough in my comment. However, in my answer I used tabulated standard redox values not ones based on pH 7, which makes for a  simpler calculation.

The standard states of the Gibbs energies of formation (usually 1M concentration for ions) must match the standard states for the redox potentials.
The redox potentials you used are for pH=7, which means [H+]= 10^-7 M.
See this link for sample calculations of the effect of [H+] on redox potentials.


text:
Solvents for tartrates / tannins that won't attack glass
At the bottom of my wine decanter (which due to its shape is fantastically hard to clean mechanically) a layer of reddish brown precipitate forms. I believe this is probably potassium bitartrate crystals, which are coloured reddish brown by the presence of tannins, which in essence is a polyphenol.
I would like to know whether there is a chemical that I could use to clean the decanter with minimum mechanical action. I have tried water (with and without soap), alcohol (in the form of vodka), isopropanol without success. I have yet to try acetone. Preferably the solvent should not be greatly toxic (at least I'd like to be able to wash it out with water and reuse the decanter), and not react with high lead-content glass (which is what decanters are made out of) - hydrofluoric acid is thus to be avoided (as usual). It is unsurprising alcohol doesn't work as the reason tartrates precipitate is apparently because they are not soluble in alcohol. It would also be useful if I could purchase the solvent outside a lab environment.
My understanding is that potassium bitartrate is more soluble in acid solutions. I've tried citric acid (made up from powder) and vinegar, but the effect if any is modest.
I don't much care if the method used is to react with (rather than dissolve) the precipitate.
(Note this question is in part a thought experiment. I realise that in fact the best way to clean decanters is generally mechanical despite the difficulties, and I'm familiar with the copper beads method and the egg-shell method, as well as no end of sponges on bendy sticks. However, I'd like to know whether there is a chemical way. Also note that I added an 'equipment' tag as it's possible that some know-how from cleaning chemistry glassware might be useful here)

Strong cleaning solutions of an acid and an oxidizer which would be used in a chemistry lab are not suitable for home use. It wouldn't be safe for you or your drains.

Do you have a water pick for dental cleaning? I'd think that a well aim blast would dislodge the crud.

@MaxW No I don't. In this particular decanter the neck is narrow, and the crud sticks in a place where there is no straight line exit via the neck - not sure a pick would reach it. In any case I'm trying to find a chemical solution (preferably one safe for me and my drains too).

Did you try a dishwasher soap in warm water?

@MaxW I've tried hot soapy water (obviously) and washing-up liquid. Dishwasher soap (i.e. for machines) is not recommended for decanters of this sort as it is too abrasive and can scratch the glass and make it opaque. The best current method I have involves sponges on bendy sticks, soapy water and washing up liquid, but to repeat this isn't a 'how do I clean my decanter' question, it's a 'what chemicals could I use to attack/dissolve the precipitate' question.

Have a look at Bar Keepers Friend, which may or may not be available where you live. It's essentially oxalic acid with some abrasives in it (see the FAQ section of the linked site for more information). You might try swirling around some of the stuff in solution in the decanter and not using mechanical force, thus preventing scratching the glass. In any event, it's great for removing stains and, as the name suggests, works well on things that might stain and are kept around the bar such as wine. Good luck!

Try saturated (!) calcined soda solution. It does dissolve tanines, though slowly. carbonate is a relatively strong base, so it may deprotonate bitartrate, but the shift unlikely to be big.

In a lab, I would use something from this list http://chemistry.stackexchange.com/a/1357/485

I'd recommend a consumer product that was much more acidic than acetic or citric acids.   You can get products such as drain cleaners (I think) that contain sulfamic acid or hydrochloric acid.   If that didn't work, then I'd go the other route and recommend ammonia, especially mixtures of ammonia and rubbing alcohol.  (Rinse everything with tons of water between attempting an acid wash and then a basic wash, obviously.)

The CRC Handbook indicates that potassium d-tartrate and dl-tartrate are soluble to the extent of 100-150 grams per 100 mL H2O vs about 0.5 grams for the bitartrate. Rather than using acid to clean the decanter, alkali would neutralize the bitartrate to the much more soluble tartrate . Washing soda (Na2CO3) should not be too alkaline, but even sodium bicarbonate (NaHCO3) might work. Neither is expected to be harmful at the level that might exist after rinsing.
The tannins or other organic might slow things down. Bleach (NaClO) is alkaline, and although bleach is toxic if ingested, it should present no problem if the decanter is rinsed. Water could be flushed around inside the decanter, then tested with litmus paper. Or the decanter could be rinsed with vinegar and a little honey to remove the bleach: the vinegar makes the liquid acidic (HClO) and the honey gets oxidized to gluconic acid, consuming all the bleach - leaving a sweet, tart taste.

Really interesting. I'd tried baking soda (did not seem to help much), but not washing soda. Slightly fearful of bleach in case it leaches (somehow) into the glass or the scratches in it (in seems quite soft), but I can't immediately think how bleach would do that. Perhaps I should pluck up the courage to try this.

source: Stack Exchange

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

sort: votes


text:
Is the rate of an SN1 reaction really independent of the type of nucleophile?
Suppose we take two reactions
$\ce{(CH3)3CCl + H_2O \rightarrow (CH3)3COH +HCl}$
$\ce{(CH3)3CCl + CH3OH \rightarrow (CH3)3COCH3 +HCl}$
I know in $\mathrm{S_N1}$ reactions the RDS (Rate Determining Step) is the formation of a carbocation intermediate.
But I am quite certain that methanol is a more bulky and weaker nucleophile, as compared to water. Should that make the rate of the second reaction lower than the first reaction (after the RDS is crossed)?

The key concept here is that both steps of the reaction are happening simultaneously, not sequentially. The steady state theory says that there will always be a pretty small amount of carbocation in the reaction mixture that will be stable over time. The faster it is generated, the faster it will be consumed. Therefore the reaction rate is only dependent of the carbocation generation rate, as long as this step is significatly slower than the formation of the final product, enough to make the time of this second process negligible.

Nope, the rates wouldn't be affected, even if the nucleophiles used are different (though if you're dealing with highly concentrated reactants, it might be a different story...).
Judging from the way you framed the question, you probably already know a fair deal about both the Unimolecular and Bimolecular Nucleophilic Substitution (SN) reactions. So I won't start from scratch.
In the SN1 case, as the name would suggest, the rate depends on only one concentration factor, namely the substrate. You already (correctly) mentioned that a carbocation intermediate is formed. At this stage, its just a matter of whether the nucleophile performs a 'frontal attack' or a 'rear attack', and doesn't really involve the nucleophilicity or the size (at least not much in your example) of the nucleophile. This is different from the SN2 reaction, which requires species with high nucleophilicity  and appropriate size to react (rear attack) with the substrate, leading to an unstable transition state.
How about an analogy?
Suppose you've got a lot of magnets stuck on your fridge door (at least I sure do..). Now suppose you're standing at a distance from the fridge, and you want to place another magnet on the fridge by throwing it. Since you've already got a lot of magnets stuck there, its going to be really hard to make your magnet land there. The weaker your magnet, the lower the chances of you landing it on the fridge door. However, if you have this really strong neodymium magnet, this wouldn't be much of a task; one throw, and it'll land so hard, it'll send the other magnets flying. This is a lot like the SN2 scenario. The attacking group must be a stronger nucleophile that the leaving group.
Now suppose your fridge door is devoid of magnets [which is precisely what my mother requires of our fridge.... :( ]. Then clearly throwing a magnet and hoping it'll land is a very, very good possibility. In this case it doesn't really matter whether the magnet is square or circular, large or small, strong or weak (not too weak). Apart from how large your fridge door is, there's no other factor that would keep the magnet from sticking to the fridge (albeit you aim). This is a lot like your SN1 scenario. The nucleophile's characteristics don't really matter (at least not in most typical cases). As long as there's space on the door, the magnet will stick.

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


text:
Logic behind this statement (On Transcription)
There's this little paragraph in my textbook, parts of which I find redundant. My teacher is absolutely fine with this ("Aaron, you've got to write down what's in the textbook if you want to score in exams, nothing extra...."), but all the same.....
It's regarding 'Why does transcription occur at one DNA strand at a time?'

According to their first point, forming two complementary RNA strands would result in the production of two different proteins for the same character.
Now according to their second point, formation of two complementary strands of RNA side-by-side would just result in them pairing to form dsRNA, which would be transcriptionally inactive and therefore wouldn't take part in the translation process.
The thing is, they've mentioned the second point after the first point, which I find confusing. For if the second point were correct, the first point wouldn't even exist.
Is there a sound reason for the points being stated this way? Am I missing something here?

This should better be posted in biology.SE

This textbook would never have made it into my classroom. The second "answer" is an arguably correct concept but is specious as a reason. The first "point" has a silly justification.
The real reason is an evolutionary one. In polypeptides, particular sequences of amino acids fold in ways that provide function. Tiny variations in this sequence can render the resultant polypeptide non-functional. With this sort of sensitivity what are the odds that the non-coding strand, whose sequence is dictated by the coding strand, could actually possess a viable gene? Furthermore, genes evolve over time. Any changes in the sequence of one would necessitate a change in the other. What are the odds of that being efficatious?
That said, I believe there are some cases where a gene on one strand has some overlap with a gene on the sister strand. I doubt these are lengthy or common, possibly occuring with the help of introns, portions of the DNA message that are "spliced out" of the final RNA transcript.


text:
What is the difference in the reducing power of various ionic hydrides?
By definition "Ionic or saline hydrides are composed of hydride bound to an electropositive metal, generally an alkali metal or alkaline earth metal."

What is the relative order of reactivities of all the ionic hydrides of alkali and alkaline earth metals?Is there any general trend?
Is the reactivity of alkali/alkaline-earth metal hydrides greater than $\ce{LiAlH4}$ ?
Is the reactivity of alkali/alkaline-earth metal hydrides greater than $\ce{NaBH4}$ ?

I know reducing nature of $\ce{LiAlH4}$>$\ce{NaBH4}$.

Can't NaH reduce carbonyls? @orthocresol and cant NaBH4 deprotonate compounds?

Can you explain why NaBH4 or LiAlH4 is slower than the s block hydrides ?

@orthocresol Hey,sorry for the late follow-up question.But can you clarify something?Why is breaking B-H bond tougher than breaking Na-H bond?Aren't ionic bonds stronger than covalent bonds?Or is it because NaH is ionic it already dissociates in water?

@Aaron Abraham has written a nice answer to the original question.However I would like to combine his arguments along with the points made by @orthocresol.

In case of covalent compounds larger and weaker bonds in hydrides facilitates donation of hydrogen for reduction reactions.
S-block elements too have some covalent nature.So the previous fact can be applied for them too using Fajan's rules.
In general the top most element of both the groups in s-block has maximum covalent nature.Covalent bonds are generally weaker than ionic bonds (with some exceptions).So the reducing power is maximum at top and minimum at bottom.
Alkaline earth metal hydrides tend to be more covalent than alkali metals hydrides due to higher charge density on alkaline earth metals.So Alkaline earth metal hydrides have more reducing power compared to its counterpart.
S-block hydrides are usually never used in reduction of carbonyls and similar compunds.However they are used for deprotonation.
Even $\ce{NaBH_4}$ can be used to deprotonate.However they are much weaker and slower than s-block hydrides as for them to react the $\ce{B-H}$ bond has to be broken which is difficult and slow.

Al-H bond is really weak though, LiAlH4 deprotonates stuff really fast.

But slower than s-block hydrides?Right ? @orthocresol

No idea actually.

@orthocresol Are any values available on the net to comapare the rates?Well even I have not much idea...

I don't know. I don't think that such a thing would exist, deprotonations/protonations occur at a very fast rate, and for the most part we are more concerned about the extent of de/protonation (a thermodynamic question) rather than the rate (a kinetic question).

This is a very generalized answer but I'll go for it anyhow, who knows how it'll turn out to help (if it does add to what you already know, I suspect you may already know about what I'm going to type....)
The metrics you're considering are 'Reactivity' and (since you brought up LiAlH) 'Reducing character'. Clump these together, and it turns out what you're looking for is the ability for these hydrides to donate/release their hydrogen.
Now I've reasons to believe you already know the 'reactivity' trend for p-block hydrides (correct me if I'm wrong). Figuring out the reducing character of the s-block ionic hydrides involves similar logic.
In the p-block hydrides, you've probably learn't that as you move down the group, the E-H bond length increases (I use 'E' to denote p-block elements). If the bond length increases, the easier it is to cleave, therefore the hydrogen is released more readily. Here's an analogy if that bit seems a bit hard to grasp. Think of the E-H bond as a pencil. The longer it is, the easier it is to break; the shorter it gets the more difficult it is to break (try it......P.S-this credits for this analogy goes to my Chem. Teacher).
In a nutshell, for the p-block hydrides, as you move down the group the tendency to donate hydrogen increases.
Now back to the s-block. You can't apply the logic you used for p-block hydrides here (at least not directly ;) ).
At first glance, the main impediment you'd see is "s-block hydrides are ionic but p-block hydrides are covalent, so I can't figure out the trend the same way...". You'd be correct, but if you trying extrapolating (in reverse) the logic you use for p-block hydrides, see what happens.....
Since they're all ionic hydrides, and since you want the reactivity trend, what you'd want to do is compare the covalent nature of each s-block hydride.
As you'd already know, an ionic bond isn't 100% ionic, and the same can be said for covalent bonds. You call them ionic or covalent, if the ionic or covalent nature of a bond is relatively significant.
So by using Fajan's rules (assuming you know those as well) you can figure out that LiH is the most 'covalent' of the ionic s-block hydrides, and therefore more prone to donating hydrogen thereby acting as a reducing agent. This Wikipedia page just confirms that...
As you move down Groups 1 and 2, the reducing character of the s-block hydrides decreases.
Also LiAlH and NaBH are much, much stronger reducing agents than plain ole' s-block hydrides.

Can you give reasons for your last sentence?

$\ce{BeH2}$ is the most covalent hydride among s-block hydrides.


text:
Is ozone really a chemical element?
In our chemistry book is written that ozone ($\ce{O3}$) is a chemical element. Also our chemistry book gives a definition that agrees with that. But our teacher doesn't agree.
His reasons:

If ozone is a chemical element why it isn't in the periodic table?
It is not true with the things that Robert Boyle has written in his book.

Then what is the best definition for chemical element and is ozone a chemical element?

http://goldbook.iupac.org/C01022.html: an element is a "species of atoms with the same number of protons in the nucleus". Ozone isn't an atom. Ozone is an allotrope of oxygen. Oxygen is the element.

You might want to read up on this Wikipedia article
My answer is a bit oversimplified, but I think it would serve your purpose.
When you refer to a chemical element, as it is, you'd be referring to one of the 118 elements in the Periodic Table. Something's a chemical element if it can't be broken down into more atoms that can be located on the periodic table. Ozone is not an element for the same reason dioxygen is not an element.....it can be broken down into three identical constituent particles, which you can locate on the Periodic Table (oxygen).
As @orthocresol pointed out, ozone is an allotrope of elemental oxygen.

$\ce{O3}$ is not a chemical element, but is a molecular form or allotrope of an element. Sulfur, for example, exists in different allotropes: puckered rings of $\ce{S8}$, a red triatomic ($\ce{S3}$) gaseous form, similar to ozone, etc.
Oxygen can exist temporarily in monatomic form, as ordinary diatomic oxygen in air, as triatomic ozone, even as $\ce{O8}$, a red solid. All are the same element, just different allotropes.
That said, it was difficult for the early chemists to prove that different allotropes were really the same element. Antoine Lavoisier had to burn diamonds to show that they were made of carbon, like graphite... and that experiment rather annoyed the new egalitarian French Republic. It goes to show that chemistry can be dangerous.


text:
How would you separate sugar from water?
I'm a student and I've been getting interested in chemistry. I really like experiments and was brainstorming of possible things to try out, one of them was separating a homogeneous mixture. I thought of sugar water and tried looking it up online but couldn't find anything like I wanted. Everything says to use crystallization but I don't understand why you can't just boil away water and be left with sugar like you can with salt water. Anyways, what methods/ways can I separate sugar from water?

You are more or less right about boiling water away. Crystallization per se is not a method (rather, it is what happens when you use some actual methods). You can't point a wand at your solution and say "Crystallize!"

@IvanNeretin Nah, you have to say something like "expecto crystallize". I am not sure, never quite got my letter from hogwarts.

If you start with a known 'pure' mixture of sugar and water, then what you propose is fine, to an extent, if you are just after separating the two ingredients.
Crystallisation is a purification method for extracting a solid from solution; it will allow you to separate pure solute from the solvent. The remaining solution would still contain a small amount of dissolved material, and hopefully all of your impurities.
If you wanted to isolate pure solvent from the solution, you would need to use distillation. After collecting the distillate, you would be left with a gummy residue of semi-decomposed sugar (and your impurities).
Sugar is hygroscopic, and starts decomposing at just over 180°C (~365°F). The problem that you would have with your proposed method is that in order to remove the final remnants of the water, you will need to heat well above the boiling point of water. Google 'making caramel' if you can't think of what will happen. Even a very small amount of decomposition of your sugar will mean you are left with an impure solid. Back to square one. The reason that you can simply boil the water off salt is that it is far more stable and can be heated safely to much higher temperatures.
So, there are a few ways to separate these two ingredients, depending on your end goals:

To collect just solute (sugar): heat over very low heat to evaporate as much water as possible to give yourself a saturated solution. Then crystallise sugar and filter (in the lab you'd wash with a suitable solvent and dry under vacuum)
To collect just the solvent (water): distill over moderate heat to collect water. To collect as much as possible, the heat used to boil off the solvent needs to be slowly increased, especially as you approach the final portion.
To collect both solute (sugar) and solvent (water): You would remove solid first, then distil solvent. Unless you are starting with a saturated solution, you would need to do a partial distillation to remove some solvent, then crystallise your sugar, then complete the distillation. You'd still be left with a horrible brown gummy residue.

Sounds like a great home project, though. Just be sure to have done some suitable research to allow you to make use of your materials if it doesn't quite go to plan :-)

Amazing answer! Just what I was looking for


text:
How to raise pH without generating CO2
What buffering chemical/supplement will reduce lactic acid for runners but not generate $\ce{CO2}$?  Sodium bicarbonate does so but causes gastric upset.

Trying to reduce lactic acid build up in the muscles is really messing with the body's biochemistry and feed-back loops.

I don't think this question has an answer.  First, muscle soreness is not caused by lactic acid.   Second, distance runners run aerobically, meaning very little lactic acid would be generated by running in the first place.  (Sprinters are the exception here.)  Third, even if lactic acid buildup was a problem (which it isn't), both the cellular cytoplasm and human bloodstream are very well-buffered.  The cytoplasm is primarily buffered by potassium glutamate and the circulation by bicarbonate (created from respiration).  Production of "lactic acid" would really be production of lactate salts.

Thank You , Curt . Would you have any suggestions of a supplement to help the "burn" felt in the muscles when a runner is tiring ?


text:
Conductive liquid, that after evaporation will allow me to use my gloves with a smartphone
I'm looking for a conductive liquid that, after it evaporates, leaves the material that has been exposed to it conductive.
I'd like to use my smartphone with gloves. I do have a pair of professional, expensive gloves, with "smartphone friendly material" on the index finger-pad, but it works dreadfully. I'd like to use the tip of my finger, and not the finger-pad, to interact with the smartphone surface.
So I thought, let's wet the material in some conductive liquid, let it dry, and done!
I did try this approach with colloidal silver and it does not seem to work.
The next on my list is a solution of water with iron rust ( I am unsure about the effect, but I'd give it a try).
Any other ideas? It must be human-safe. Colloidal silver would be perfect (antibacterial properties) if it could be made to work.

When conductive liquids go dry, the remaining solid residue is rarely conductive. Try stitching a thin copper wire into the fingertip.

Possible solution could be achieved with conductive thread/strand - but it requires interaction with the gloves :/

Personally, I'd suggest something like silver paint. I wouldn't use iron oxides because they're not very conductive.

This sounds alot like the X Y problem. Please look at this post and make sure you aren't making the X Y mistake in asking a question.


text:
Can I submerge a polyester shoe underwater to wash it, and for how long?
I'm wondering if this will damage the shoe somehow. The shoe doesn't include washing instructions. My shoe actually has four materials listed, they are:
-Polyester
-Polyuretan
-Natural rubber
-Nylon
I don't know any chemistry, I'm ignorant. I think this substances can come in many forms, so there is a lot of uncertainty. But if you answer for at least what the title asks, I'm fine, I'll make more questions for the rest of the materials.

Have you thought of wiping with a damp cloth?

@long Yes, but I don't want to do that, thanks. I want to explore this  different method in particular, also if the shoe has sweat and dirt inside that won't take it off.

It should do no harm, other than possibly removing some dye or sizing (which makes fabric hold its shape), provided that the shoes are dried thoroughly afterwards to prevent fungus and bacteria from growing in the wet material and  particularly in the foam insole. Don't use too much heat (much above body temperature), since heat will damage the glues and melt polyurethane.


text:
How many empty orbitals?
I'm really confused about orbitals.
I know there's 1s, 2s, 2p, 3s, 3p, 4s, 3d, etc.  From my understanding, each of those is an orbital.  So for example, carbon has 3 orbitals (1s-2, 2s-2, 3p-2).  Is that right?
If it is, now I have a question about EMPTY orbitals- I don't understand this at all.  For example, take carbon.
Well carbon has 6 electrons and it would be: 1s-2, 2s-2, 2p-2. The p ones have 3 suborbitals, so carbon has one empty SUB-orbital??  But how does any element have an empty orbital?  Isn't their outermost orbital the one that has at least one valence electron in it (i.e it's not empty)?

zillions of empty orbitals. Look at the spectral series for hydrogen. https://en.wikipedia.org/wiki/Hydrogen_spectral_series

So the orbital is not empty, but one of its suborbitals is empty. What is your question?

I don't think this question deserves the downvotes. The person simply had an issue of nomenclature [suborbitals is the incorrect term (or atleast quite uncommon) while orbitals is the correct term for what he had in mind].

There is 1  ns orbital, n=1,2,..
There are 3 np orbitals, n=2,3,...
There are 5 nd orbitals, n=3,4,...
There are 7 nf orbitals, n=4,5,...
Electrons in multiple p,d,f, orbitals differe in their orbital angular momentum.

The orbitals are:
$$\large 1s,\; 2s,\; 2p_x,\; 2p_y,\; 2p_z,\; 3s,\; 3p_x,\; 3p_y,\; 3p_z,\; 3d_{xy},\; 3d_{yz},\; 3d_{zx},\; 3d_{x^2-y^2},\; 3d_{z^2}\; ...$$
In your example, the electronic configuration of carbon is:
$$ \ce{_6^{12}C}=1s^2 \; 2s^2 \; 2p_x^1 \; 2p_y^1 \; 2p_z^0 $$
My point is, the classification is like this: there are four (commonly occuring) types of orbitals, the s-orbitals, the p-orbitals, the d-orbitals and the f-orbitals. The first shell consists of only the $1s$ orbital. The second shell consists of the $2s$ orbital and the three $2p$ orbitals: $2p_x, 2p_y, 2p_z$. Similarly $3d$ is just the blanket name for the five different $3d$ orbitals as mentioned above.
Thus in carbon, the $2p_z$ orbital is empty, and the $2p_x$ and $2p_y$ orbitals are half filled (holding one electron). Actually, you could equivalently left the $2p_x$ or $2p_y$ orbital empty, it doesn't matter.
You may be confused with the naming system of the separate orbitals ("what is this $2p_x$ and $3d_{xy}$ and $3d_{x^2-y^2}$ this guy is talking about?"). Well, they are just names given to the separate $p$ and $d$ orbitals, based on their mathematical nature. (See atomic orbitals on Wikipedia).

Thank you.  What then is a suborbital? I thought the 2px, 2py, 2pz were the 3 suborbitals of the p orbital.

I am hearing the word suborbitals for the first time really...it is not in common use! Ideally you should use the word orbitals (specifically, atomic orbitals) as it is referred to in the Wikipedia page, unless your syllabus demands that specific word.

An orbital is a region of space where an electron may occupy, usually we draw a hard boundary at where the electron exist 90% of the time but it actually extends to infinity (ie. an electron can be anywhere in the universe).
Naming. In an atom, there’s principal quantum shell, sub-shell, then orbitals. “Suborbitals” do not exist. For example, in the second principal quantum shell $n = 2$, there are one $2s$ and one $2p$ sub-shell. The $2p$ subshell is made out of 3 $p$ orbitals: the $2p_x$, $2p_y$ and $2p_z$ orbitals, where each orbital can hold a maximum of 2 electrons.
In carbon, one of the $p$ orbitals is empty. What does it mean? It means that this orbital is the region of space an electron will occupy if there is an electron. It’s just a region of space.

I feel compelled to mention that for the advanced reader, even this is not quite correct: an orbital is not a region of space, but rather a single-electron wavefunction, which forms part of the overall, multi-electron, wavefunction (to a certain approximation). This makes it difficult to ascribe physical meaning to an empty orbital as you do in the final paragraph: if you were to add an electron to a carbon atom, the Hamiltonian would be different, and consequently the extra orbital required will not be exactly the same as the vacant orbital in the carbon atom.

... and I do not mean to say that this is a bad answer in any respect: it is just that this description of an orbital will only work up to a point in one's chemistry education, hence my addition of a disclaimer.

I agree! Thanks for the clarification

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


text:
How are Kleenex Tissues Infused
Kleenex tissues are infused with long lasting scents - their scent is released when crushed. How does this work? I's like to know how to reproduce similar effect for another purpose. Who would be the person who could develop this for me? A Chemist - A Pharmacist?

This FAQ from the manufacturer will get you somewhere, but I suggest contacting them directly to resolve your question.

I think that the company might be not so eager to share their trade secrets. I guess that they use microencapsulation similar to the 'scratch and sniff' stuff. A chemist with the experience in the pharmaceutical industry may be the right person to develop your product.


text:
Isotope pairs in different states at room temperature?
Does any pair of isotopes exist which are in different states at room temperature—either in isolation or as part of an otherwise-identical compound?

Please Elaborate.

I'll try ... two isotopes of some element that exist in different states of matter from each other at room temperature--all other variables being equal as well?

Typically no, but processes such as the Gridler-Sulfide or COLEX process can greatly enrich an isotope.

Ordinary water and heavy water differ in boiling point by $1.4^\circ,\rm C$. Guess $\ce{HF}$ and $\ce{DF}$ might be the very example you are after (not that it would be of any use for anything, though).

@Ivan Neretin Would $HF$ and $DF$ have different state? I am sorry I cant find any data about b.p. or m.p. of $DF$.

I don't know these either. Like I said, it might be.

I'll assume that H2, HD and D2 have very different properties regarding their spin isomers.

It’s not exactly room temperature but it’s probably the closest you can get to two isotopes having different states so that they can be separated by physical means:
At very low temperatures (below $1~\mathrm{K}$) liquefied helium will separate into a superfluidic helium-4 phase and a normal liquid phase containing mostly helium-3. The superfluid and the normal liquid can be considered different phases and are immiscible.

Isotopes differ in the number of neutrons in their nuclei.  For very light elements like H the differences in atomic weight are relatively significant.  So if transfer of a hydride or proton is a rate-limiting step in some kinetic process, then the mass of the nucleus can have a measurable effect.  This 'kinetic isotope effect' can be used to tease out mechanisms of some chemical reactions, esp. the mechanisms of enzyme catalysis.
    In the realm of physics, the neutron/proton ratio determines the net spin of a nucleus, with 2 consequences.  First, nuclear spin determines if that nucleus is a fermion or boson, with consequences esp. at low temperatures (eg. He-3 vs He-4).  Another significant effect is seen in the presence of a strong magnetic field.  In this case, nuclei with different spin states possess different energies.  Absorption of EM radiation (in the RF-uWave frequencies) causes the nuclei in the lower energy state to 'flip' to the higher state.  They will spontaneously return to the original state, emitting the rf energy they previously absorbed.  Different isotopes can differ significantly in the way their spins couple with EM fields, an effect that NMR spectroscopy, MRI imaging and similar technics make use of.


text:
Why cotton and wool clothes shrink after washing them?
I want to take measures to avoid my clothes shrinking in the washing machine and in manual washing. I read here if the water is hot that contributes more the shrinking.
The Wikipedia article is kind of as they say in Wikipedia, a "stub" (too short) and doesn't explain the reasons very scientifically.

Wool, in particular, has very "scaly" fibers that mat together forming a dense felt. Natural oils prevent this (no shrunken sheep), but cleaned wool, agitated or shaken, shrinks due to the fibers gripping each other.
"Cotton shrinks because of the tension that is applied to its yarn and fabrics during the construction of the clothing. The tension is released by heat from the washer, dryer, steam and even sunlight, which causes the fabric to be reduced to its natural size." However, mercerization, formaldehyde and other reagents can "pre-shrink" cotton and prevent subsequent shrinkage in washing.


text:
What is the simplest formula for the mineral beryl which contains 5.03% Be, 10.4% Al, 35.5% Si and 53.58% O?

Beryl is a mineral which contains 5.03% $\ce{Be}$, 10.4% $\ce{Al}$, 35.5% $\ce{Si}$ and 53.58% $\ce{O}$. What is the simplest formula for beryl?

So I tried to follow a 3 step rule to solving this equation:

Divide each % by it's own atomic mass
Divide each of those by whichever is smallest
Find the lowest whole #

Amount of $\ce{Be}$ = 5.03% / 9 = .558 moles
Amount of $\ce{Al}$ = 10.04 / 26.98 = .372 moles (smallest number)
Amount of $\ce{Si}$ = 31.35% / 32 = .979 moles
Amount of $\ce{O}$ = 54.58% / 16 = 3.3 moles
Proportionate amount of $\ce{Be}$ = .558 / .372 = 1.5
Proportionate amount of $\ce{Al}$ = .372 / .372 = 1
Proportionate amount of $\ce{Si}$ = .979 / .372 = 2.63
Proportionate amount of $\ce{O}$ = 53.58 / 16 = 3.3
From here I don't know what I have done wrong, but these numbers are all whack silly and I can't see to get a number that leads me to the answer.
The correct answer is $\ce{Be4Al(SiO3)8}$

You have the wrong atomic mass for Si. I'd use 4 significant figures for all of the atomic masses also.

Good catch! I didn't even spot that. 32 is sulfur, not silicon.

If you "cheat" (or want to use this as a check), you can render oxidation states which, in this setting, will be Be=+2, Al=+3, Si=+4, O=-2.  No other oxidation states of these elements are likely to be found in naturally occurring minerals.  Whatever formula you get should have these oxidation states, each multiplied by the number of atoms for that element, add up to zero.  The answer in the back does not, the correct formula given in a correct answer below does.

Firstly the "correct" answer seems to be wrong, which is not an uncommon occurrence, so please do not always take answers at the back of books to be correct. Wikipedia lists the formula of beryl as $\ce{Be3Al2(SiO3)6}$ which is much more in accordance with what you have done.
That aside, there are two things you need to correct.

The percentages that you have in your first step are different from those quoted in the title. For example, for silicon, you divided 31.35 by 32, but the title says that there is 35.5% silicon. Please, be more careful when you are pressing numbers into a calculator or the computer. The same issue exists for aluminium and oxygen.
In your second step, oxygen, you seem to have done a calculation pertaining to the first step? You meant to do 3.3/0.558, correct?

Fix those and you should obtain the correct answer - and not the one in the back of whatever text you are using.

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


text:
What is the reason of explosion of batteries?
In social sites we see many pictures showing awareness of not using smartphone while charging because of the reason that battery explode while charging, is really batteries explode and if yes than what is the reason behind it.

Welcome to the site!  Your question sounds interesting but is a bit tough to understand.  Can you give an example of an article or posting that refers to batteries exploding curing charging?  Are you asking about any types of phone in particular?

http://www.androidauthority.com/busting-the-myth-yes-cell-phones-can-explode-42582/ go through it

So is bursting the same as exploding? http://nypost.com/2016/09/11/samsungs-new-smartphone-may-catch-fire-us-officials/

Which type of batteries? That is an important question!

The most common cause for a battery to explode is the contents of it heating to a point where it exerts a force on the battery too large for it to handle. Basically, if the battery feels very hot, it could be under large pressure and the object that it is powering should be powered off if possible. The recent reports of smartphones catching fire are due to a manufacturing error by Samsung on the Galaxy Note 7. A portion of these phones were not insulated correctly and it caused a short circuit that ended up burning holes through the phones. This was not an explosion, but the phones did catch on fire. In conclusion, you can use your phone while charging if it's not a Galaxy Note 7 (which has been recalled).

It would be useful to know how to put out such a fire.

Like iPhone more?


text:
Intercalation and Ionic Bonds
What I know:
Coming from a physics background and a frequenter of more glib stack exchange sites, I only recently found out that intercalation is a thing when studying Lithium-Ion Batteries.
As I understand it, intercalation happens when a material, such as graphite, has "layers" which have a small charge. Oppositely charged particles become attracted to these layers and diffuse their way between the sheets. This is common in graphite, such as in this image from Wikipedia:

My understanding is that coulomb forces are responsible for forming this structure. The purple ions will not move out because they experience a coulomb attraction to the black sheets of atoms shown above.
I also remember from undergraduate and grade school classes that ionic bonds are also a thing. Once again, wikipedia has a great graphic showing what is going on in ionic bonding:

Some chemical species looses a charge, one gains a charge, and the bond is formed by coulomb forces between the two.
Careful reading of my reasoning notes that these two situations arise because of coulomb forces shoving things together. This has lead me to a simple question: what is the difference between the attraction in intercalation and ionic bonding? I've asked around, and nothing seems to settle well with me.
The Explanations So Far
I have asked some people more knowledgeable in chemistry than me what the difference is, and they've not satisfied this question, as shown by the following italicized text. These explanations are:

Ionic bonds require the transfer of electrons, whereas intercalation does not. This explanation implies that two ions who became so in another way do not form ionic bonds with each other, in spite of their clear attraction to one another.
Ionic bonds are "evenly matched," whereas intercalated ions and their sheets are not. My problem with this is that the same mechanism still attracts them, so why call them something different? What about Tin(IV)Chloride, which matches 1 tin with 4 chlorines and intercalation of, say, $\text{KC}_8$?
The difference in the strength/length of the bonds is the difference. We physicists tend not to have different names simply because of magnitude; gravity makes things fall to each other, no matter the magnitude!
It's a "paper difference" only; intercalated cations are just as bonded to their negatively charged sheets as any other ion in other bonds. Paper differences are terrible, for many, many reasons!
Intercalation is a subset of ionic bonds. I'm actually okay with this, except it should be mentioned more clearly.

For the sake of clarity, my questions are: What is the difference between ionic bonding and the attraction in intercalation? If I've already hit a correct/best explanation, what about my reservations?

I'd also create the "intercalation" tag, but that's above my current rep here.

"Ionic bonds require the transfer of electrons" - nope... not quite. Ionic bonds can and will exist between any pair of oppositely charged species, and it doesn't matter how they came about.

And, very simplistically speaking, an ionic bond is a force of attraction between these cations and anions. Intercalation refers to the act of inserting a certain species into a compound that already exists. For example, you can intercalate $\ce{Li+}$ ions into $\ce{TiS2}$. Once you have shoved these $\ce{Li+}$ ions in (never mind how you get them in), then they may experience ionic bonds of their own, with $\ce{S^2-}$ ions.

To start, the animated graphic for NaF doesn't really show what is happening in a crystal of NaF. In the solid salt a particular $\ce{Na^+}$ cation is not associated with a particular $\ce{F^-}$ anion. Rather there is a regular 3D structure. So the lattice energy has to be summed over all ions in the ccrystal. The summation is given by the Madelung constant.
The image shown in the intercalation would belie the actual arrangement. Think of the graphite units as having various shapes/sizes. So a stack won't always have say 80 carbon atoms in each layer but may have 40 carbon atoms in one layer, 120 in the next and 80 in the third.
So an intercalation material won't have the nice 3D structure that an ionic bonded salt would have, even though there might be some stoichiometry for the intercalation material.


text:
Thermodynamic vs kinetic reaction control with radical substitution
Doing a revision for a chemistry class test tomorrow, I stumbled upon thermodynamic and kinetic reaction control in the context of radical substitution.
Here, my chemistry teacher said, that thermodynamic reaction control would basically be about favoring stabilized (originally instable) transition states, e.g. by increasing temperatures. According to him, this would "thermodynamically prefer" the substitution of tertiary hydrogen atoms relative to primary ones.
However, he also told us that kinetic reaction control would, instead, prefer the substitution of primary hydrogen atoms. More precisely, the fastest reaction (which would result in the substitution of primary hydrogen atoms due to less steric hindrance) would be favored under certain conditions.
The question is, under which conditions a certain type of reaction control would be prefered.

I would certainly say that thermodynamic control is associated with higher temperatures, not lower. This is to allow reversibility of the kinetic pathway (abstraction of primary hydrogens) and supply sufficient energy to overcome the larger activation energy (since the TS is more unstable).

An example I can remember would be that a split (tertiary) hydrogen atom would lead to a charge at the carbon atom. In return, this could, be stabilized better by inductive effects (from neighbouring alkyl groups).

The carbon isn't positively charged. One could perhaps say it is electron-poor, but a radical is certainly not a carbocation. 2) I must confess I have never actually heard of any real examples of td vs kin control in radical halogenation. March 6ed and Clayden 2ed also have nothing on this. The classic example of td/kin control occurs with the addition of bromine to conjugated dienes, or addition of HX to α,β-unsaturated carbonyls. I am not really convinced that the concept is applicable here, although I may very well be wrong.

I realised, perhaps, a better explanation of why I am skeptical. Since a tertiary radical is most stable than a primary radical, the transition state leading to abstraction of a tertiary H is, by Hammond's postulate, more stable than that leading to abstraction of a primary H. This is solely based on electronic factors, and while steric factors certainly exist, I am not convinced that they are significant enough to overturn the relative energy of the transition states that we established earlier. Personally, I have not read anything that indicates so.

Thermodynamic control typically implies reversibility of all steps. Say you had three reactants $\ce{A, B}$ and $\ce{C}$ which could react either to product $\ce{A-B-C}$ or to product $\ce{A-C-B}$ (equation $(1)$). Then thermodynamic control would assume that the ratio of products depends only on the difference in Gibbs free energy between them, signified by the equilibrium constant (equation $(2)$).
$$\ce{A-C-B <=>[$\Delta G_1$] A + B + C <=>[$\Delta G_2$] A-B-C}\tag{1}$$
$$\frac{\ce{[A-C-B]}}{\ce{[A-B-C]}} = K = - \frac{\Delta\left(\Delta G\right )}{RT}\tag{2}$$
On the other hand, kinetic control implies that the reactions are basically irreversible and that it depends on the transitions states and their energies which product will be formed preferentially. This is often expressed as a difference in rate constants $k$ as can be seen in equation $(4)$ for the reaction $(3)$.
$$\ce{A-C-B <-[$k_1$] A + B + C ->[$k_2$] A-B-C}\tag{3}$$
$$\frac{\ce{[A-C-B]}}{\ce{[A-B-C]}} = \frac{k_1}{k_2}\tag{4}$$

One (simple) method of achieving thermodynamic control is to heat the reaction to oblivion, until one can assume that all transition states are sufficiently populated and the energy differences of products can take effect. Kinetic control, on the other hand, is often achieved by cooling down and higher dilution, since both allow the difference of transition states to take greater effect.
To apply this to the radical halogenation of hydrocarbons, Hammond’s postulate must be invoked, which basically states that the lower-energy intermediate (or product) will be reached by a lower-energy transition state. Since radicals are electron-deficient species, a tertiary radical is more stable than a secondary one, which in turn is more stable than a primary one. Thus, kinetic control implies that the lower-lying transition state determines the dominating product which would be the one leading to a tertiary radical.
A secondary factor under kinetic control is the number of hydrogens that can react. Of course, given a molecule such as isobutane, there are much more primary hydrogens ($9$) than there are tertiary ones ($1$). Yet, the reaction rate of the tertiary hydrogen is still much faster. Kinetic control still favours the tertiary halide.
Under thermodynamic control, the heats of formation of the different products must be examied. But conincidentally, these lead to the same conclusion. Since the $\ce{C-Br}$ bond is polarised towards bromine, the molecule is more stable if the ipso-carbon is tertiary. However, since we are no longer comparing rate constants but Gibbs free energy values, the ratio will be a different one.
Thus, both control types would favour the same product — a tertiary halide — albeit to different extents. This is the standard case! Only rarely, such as in certain Diels-Alder reactions does the main product differ when changing from thermodynamic to kinetic control.

As you said, kinetic control compares the TS leading to the alkyl radical. Does TD control here compare the intermediate (tertiary vs primary radical), or the final product (tertiary vs primary alkyl halide)? I'm actually not sure. (It doesn't affect your argument in any way, since you explored both - just wondering.)

@orthocresol I’m going to say the final product, although I can’t back it with examples. (The stereotypical Diels-Alder example doesn’t work since we’re comparing transition states there.)

Basically any reaction will proceed according to the conditions available. While predicting the product of a chemical reaction we generally tend to guess the most stable product we can think of. But the important thing we forget are the condition needed for it. If a reaction occurs in an environment where the temperature is very low, in some cases the a certain product may be stable but the chemical factors required for the proper collision at that position to give the desired product would be difficult. Thus the reaction will form some other product as a major product by making the reactant molecules collide at a position where there may be less steric crowding or something else on the same lines. This may alter the transition states, thus altering the product. But at high temperatures due to increase in the velocity of molecules more collisions happen and we may get our desired stable product. Thus the product on the first case is kinetically favoured and the product in the second case is thermodynamically favoured.
And to be case specific as I said before kinetically favoured product can be one which is less steric ally hindered so in your case primary carbons are less sterically hinderedd,thus when high temperature isnt available , it will collide with the primary carbon.
Hope it helps


text:
At what temperature does charcoal burn?
According to an old Wikipedia revision, charcoal burns at temperatures as high as 2600 °C. Other sources which seem less reliable give a figure of around 1000 °C. The Wikipedia article has since been updated to state a temperature of 1100 °C.
I know that it depends on the airflow, but how high of a temperature can normal charcoal (bought for BBQs) burn? Can it be used to melt metals with a melting point such as iron or steel?

Melting of steel is possible (how do you think steel was invented in the first place?), but requires special setup. 2600 °C seems unrealistic.

The high value of 2600 °C probably corresponds to the adiabatic flame temperature of a certain charcoal in air (i.e. a theoretical temperature based on the heat of combustion of charcoal and the heat capacity of the involved gases, without any heat losses from the flame).

In open air at atmospheric pressure (BBQ) expect about 1000 °C or lower. It's enough to reduce iron, but not enough to melt it. That's why steel was first invented as porous substance, which must undergo thorough casting to form compact metal. In oxygen and specially constructed furnace about 1500 °C is expected.

Perhaps even 2000 in oxygen. But I doubt about any useful setup to obtain 2600 from burning charcoal.

The problem with the question is that there is no unique temperature at which charcoal burns. There could easily be a 1,000 degree difference between the temperature in an open barbeque and a specially constructed furnace built with insulating walls and forced air flow. Iron can be melted but furnace construction is a skilled job and the limiting factor in the temperature reached.

This question has been unanswered for a while, but there are some great comments which I will use as part of the answer, as well as some answers to other questions here and on the Physics site.
The temperature is dependent on the location
In well-controlled flames such as a candle or a Bunsen burner, you can measure the temperatures in different locations of the flame and get a wide range of temperatures. Here is the example of the candle, copied from this Physics StackExchange answer:

In the question, the OP asks:

[...] how high of a temperature can normal charcoal (bought for BBQs) burn? Can it be used to melt metals with a melting point such as iron or steel?

You would need the hottest part, not some average temperature. So if you want to melt iron or steel samples, you would have to have some part of your setup get sufficiently hot to melt the sample, and this space would have to have sufficient volume so that heat conduction through the sample does not lower the temperature below what you need.
The temperature depends on the design of the furnace

[OP] I know that it depends on the airflow [...]

It depends on the airflow, the temperature of the supplied oxidant, the amount of heat transferred to the product gases (carbon dioxide and perhaps others). It also depends on the design and dimensions of the setup:

[Matt Black in comment to question] The problem with the question is that there is no unique temperature at which charcoal burns. There could easily be a 1,000 degree difference between the temperature in an open barbeque and a specially constructed furnace built with insulating walls and forced air flow. Iron can be melted but furnace construction is a skilled job and the limiting factor in the temperature reached

So there is no typical temperature associated with a given fuel. However, you can ask something specific, like about the temperature on the surface of the charcoal when they have burnt for a while. However, there is the concept of adiabatic flame temperature:

[user7951 in comments] The high value of 2600 °C probably corresponds to the adiabatic flame temperature of a certain charcoal in air (i.e. a theoretical temperature based on the heat of combustion of charcoal and the heat capacity of the involved gases, without any heat losses from the flame).

Ways to measure the temperature
You can either measure the temperature by inserting a probe into the flame, or by observing the radiation that is emitted (and deduce the temperature based on theory and some calibration data). Some methods works better for sooty flames whereas other work better for soot-free flames (see introduction to this manuscript).
Color due to black body radiation
If there is a solid in the fire (e.g. the charcoal, glowing embers or smoke, i.e. small solid particles), they will emit black body radiation. The surface of the sun, at first approximation, is a black body that emits white light, corresponding to a surface temperature of about 5,600 degrees Celsius. The color of light bulbs is described by a temperature, even though modern light bulbs don't use black body radiation, whereas incandescent light bulbs did. Red-glowing char coal corresponds to a temperature of about 700 degrees Celsius.
If you want to measure temperatures of the gases, you can lower a wire into the flame and check how it changes color.
Not all colors you observe are from black body radiation. For example, gases can emit light from electronic transitions (e.g. sodium vapor lamp) and vibrational transitions (e.g. Swan bands).
In the example below, a metal inoculation loop (probably nichrome) is placed into a flame for sterilization. The orange color of the loop is due to black body radiation, the blue flame color is due to Swan bands, and the orange color of the gas above the loop is due to sodium electronic transitions (sodium is on the loop before sterilization):

Source: Wikimedia
Infrared spectrum
You can measure the temperature of a gas from the vibrational states the particles are in. The infrared spectrum of carbon dioxide, for example, depends on the temperature. Temperature "guns" use the infrared emission to estimate the temperature of a sample, but they work best with surfaces.
Thermocouple
You would not be able to use a conventional alcohol or mercury thermometer in a charcoal pit. However, there are thermocouples that have a wide temperature range. One technical difficulty of using them in a flame is the formation of soot on the thermocouple, making the measurement less accurate.
Safety considerations
If you have a very hot fire that can melt metal, you have to make sure your grill doesn't melt (or deforms and collapses). Furnaces are made from material with higher temperature stability than steel, and very hot flames (such as an acetylene burner) are designed in a way that the hottest part of the flame points away from the metal parts.
You can always make a fire hotter by making it bigger as long as the air supply is ensured (camping fire vs. bonfire vs. forest fire), but that makes it more difficult to keep under control.

You are confounding temperature with heat evolution in your last paragraph Bigger is not necessarily hotter; There is more heat evolved.

@jimchmst The ratio of surface to volume gets smaller, so less heat escapes.


text:
What is the empirical formula of the polyethylene?
Polyethylene is a polymer of only carbon and hydrogen.  If 2.300 g of the polymer is burned in the oxygen it produces 2.955 g H2O and 7.217 g CO2.  What is the empirical formula of the polyethylene?
So my method for this problem is
1) find the mass of carbon from CO2
2) find mass of H from H20
3) Find mass of O by subtracting
4) cont. answer with formula from % composition
Therefore:
(7.217 g x 1mole x 1 mole Carbon x 12 grams) /  44g CO2 = 1.96 g
(2.955 g x 1mole x 2 mole Hydrogen x 1g)     /  18g H2O = .328 g
1.96g / 12g = 0.163
.328  / 16g = .0205 (smallest #)
.163 / .0205 = 8 Carbon
.0205 / .0205 = 1 Hydrogen
The answer is CH2 :/, but I am having a hard time finding out where I slipped :(

.328g is hydrogen, so you shouldn't divide by 16g

$1 \dfrac{\rm{mole(C)}}{\rm{mole(\ce{CO2})}}*\dfrac{7.217 ~\rm{g(CO2)}}{44.01 ~\rm g(\ce{CO2})/\rm{mole(\ce{CO2})}} = 0.1640 ~\rm{mole(C)}$
$2 \dfrac{\rm{mole(H)}}{\rm{mole(H2O)}}*\dfrac{2.955 ~\rm{g(H2O)}}{18.015 ~\rm g(\ce{H2O})/\rm{mole(\ce{H2O})}} = 0.3281~\rm{mole(H)}$
Since fewer moles of C, normalize on C
$\rm C = 1$
$\rm H = \dfrac{0.3281}{0.1640} = 2.000 $
So the empirical formula is $\ce{CH2}$

Just some comments: First, it is poor practice to use constants with so few significant figures that the constants contribute to the error of the results. H has a mol mass of 1.00797, C 12.011 and O 15.9994 not using 1.008, 12.001, and 15.999 is sloppy/lazy. I got a mass of C as 1.969686 (ignoring significant figures "inside" a calculation is better than trying to manage them; apply significance to the results after you're done number crunching. You got a mass of 1.96. One of us is very wrong. A 2% error 1.96 vs 1.97 FROM CALCULATION is really inexcusable (unless you're doing it in your head, or need it to keep your spaceship from crashing in the next 2 minutes and are doing it on a wall with your own blood...). Next, your "calculation":(7.217 g x 1mole x 1 mole Carbon x 12 grams) / 44g CO2 is RUBBISH. The use of the unnecessary parentheses suggest you don't understand when to use them, but that is a minor issue. Dimensional analysis (working with the units of measure) of that calculation results in your 'final' units to be (grams² x moles x moles C ÷ grams CO2 )
This indicates a profound misunderstanding of what you are doing.
In a mole of CO2 there is 1 mole of C and two moles of O. You probably know this. That means there are 12.011 grams of C and 31.998 grams of O in one mole (44.009 grams) of CO2. The calculation is 12.011 (grams C)/(12.011 +31.998)(grams total) = 0.2729 (g C/g total) and there's nothing wrong with claiming this as 0.2729 without any units (g/g "cancel") and the ratio can be used without having to continue to manage its units making further calculations easier. It is true that 7 g of CO2 can be multiplied by 0.2729 to get 1.97 g of C. (note that if we carry through the units of the ratio 0.27 gC/gTot that the dimensional analysis gives us gTot (same as gCO2) x gC/gTot which gives us units of gC → showing that we've done it right (in other words, it can be useful to carry the units through calculations). You should NOTE that "moles" don't come into this, although arguably you could claim that it isn't 12.011 gC; it's 12.011 g C per mole (or per mole C). NONE of this justifies what you did in multiplying (not dividing!) by moles! Wrong, wrong, wrong. Doing the  calculation for the H, it is 0.33068 grams of H another HUGE mistake you made (0.3307 g of H is 0.3281 moles of H, which is, I suppose where you screwed up).
Think of dimensional analysis as a check that your calculation makes sense. What you did makes no sense at all, imho. But don't get discouraged; the process of understanding what you're doing, rather than throwing things at the wall takes practice time.

Did you hear about sth called paragraphs? Also formatting?

Thanks Li Zhi - I know the work I've shown doesn't really mirror the work I've done on paper as I can't find the right formatting here on my computer (I wish I was a boss like MaxW hehe).  But thanks for your comments and feedback.  I see the error of my ways #fml

@LiZhi This is not the appropriate tone to take if you're trying to provide constructive feedback.


text:
Is it possible to catch a cold by inhaling an inorganic salt?
My instructor once while teaching qualitative analysis told us that there's a salt in laboratory which, if inhaled, causes cold and cough. Is that true? Because today I was identifying a salt in the lab which was $\ce{AlCl3}$ and now I can feel that I've cold and cough. Is it true? Could there be any connection?

How is this related to qualitative analysis?

Qualitative analysis is related to finding salts.

I don't advise you finding salts by exploring their effect on your body.

But that was purely accidental. Our sir told us that it can happen. So today it did. I want to if there's a connection. If the teacher was right or not

Welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] to learn more about how it works.

A "cold" is viral infection of the respiratory tract. That is: it is caused by a virus (and the human body's inability to protect itself from it) and affects the nose, throat and/or lungs. I hope you are not asking whether a chemical compound (a salt) is a virus. Part of our bodies immune response is the mucus that coats our nasal passages and throat, the lungs have a similar protective layer. Inhalation of a wide variety of chemical compounds can damage the mucal layer (reduce our immune response) allowing a virus to gain a foothold. Inhalation of a variety of chemicals (most chemicals) is TOXIC and will damage or kill healthy tissue. You are sadly mistaken if you think your question has anything to do with qualitative analysis. It is a question about lab safety, or if your knowledge of human health is limited, as it seems it is, a question of medicine and health. AlCl3 shouldn't be inhaled. It reacts with water to form hydrochloric acid. You should have learned this, if you are working with it. If your teacher is so incompetent that he didn't teach you this, your parents should report him to his superiors. A fume of it can kill you (it would take a lot, not likely in qualitative analysis lab), and lesser concentrations will ...surprise, surprise, damage your lungs.

You're mostly right, but there is a proper way to do a smell test to confirm certain compounds. There are also chemicals for which doing a smell test would be really stupid. The object in a proper smell test is to get a whiff of the airborne smell, not snort the solid. // One use for a smell test is that a mercaptan is added to the natural gas piped into homes. The smell is explicitly used to alert users of a gas leak.

Breaking the different thoughts in the answer into separate paragraphs would help too.


text:
How to make soap without sodium hydroxide as the starting material?
These days I'm working on a school project and I need to make soap without using sodium hydroxide.
If I really need to use it and there is no alternative, how can I obtain it from chemical reactions?

Use potassium hydroxide?

If you can't buy it from a chemical vendor I don't know if you'd be any better off making it yourself. Do you happen to have a lot of sodium metal laying around? If not do you have the means to make a molten salt electrolytic cell? And all this just for some soap?

I think that i can make a molten dalt electrolytic cell...and it's not question of just some soap but question of something that i really want to do...i love challenges

Many hardware stores sell it as drain cleaner. I know ACE does. However, beware: NaOH is a nasty chemical and on top of that there is no guarantee of purity on something sold for drain cleaning, so it could contain even worse. You might be able to find higher quality lye at craft stores. It'll still melt your skin off, though.

http://www.wikihow.com/Make-Your-Own-Soap I would advise you not to try to make NaOH.It can be extremely dangerous.Use the link I gave.

In short: Yes, you can make soap without buying sodium hydroxide and the method adds a historical touch to the school project.
Back in the days when food was cooked on open fireplaces or in ovens fired with wood, the remaining ashes (of the burnt wood) were collected for good reason.
Leach the ashes with water and filter out the insoluble residue. Measure the pH of the solution and you'll recognize that it is alkaline. This is due to the formation of soluble carbonates when the wood was burned.
You can now either try to use the solution for the saponification of vegetable oils or animal fat, or evaporate the solution to dryness in order to obtain a solid material which will mostly consist of potassium carbonate.

You can heat a small amount of baking soda ($\ce{NaHCO3}$) to ~9000C (a reddish-orange glow) to decompose to first washing soda ($\ce{Na2CO3}$) and then to sodium oxide ($\ce{Na2O}$). When it cools, slowly and carefully add the $\ce{Na2O}$ to water to get $\ce{NaOH}$ solution.
N.B. $\ce{NaOH}$ is caustic: it will damage skin and eyes, and can even dissolve a glass container.


text:
will mercury vaporize in boiling water in a flask?
Will mercury vaporize with the water vapours if boiled in a flask full of water?
Also if we store mercury under water in the beaker will it still give vapours?
If some mercury is bound to vaporize , will there be any mercuric oxide formation or not ??

Well, some of it will. Related: http://chemistry.stackexchange.com/questions/51346/vapor-pressure-of-immiscible-liquids/51348#51348

Mercury has a boiling point of 356.73 °C. Could @DrMoishe Pippik please provide a reference for his claim of vaporization at room temperature?

@PRIVATEPRIVATE - Please note your remark is more suitable as a comment. I have edited it to alert DrMoishe Pippik (added @ sign, removed double quotes) that their input is requested. Finally,  your comment should be made under the relevant answer you are having an issue with, so I will leave it up to you to move it and delete it from this thread.

Mercury evaporates to some extent even at room temperature: Vapor pressure: 0.002 mm Hg at 250C (~300 K). "Although the vapor pressure of elemental mercury is low [at room temperature], an atmosphere that is fully saturated with mercury vapor contains approximately 18 mg/m . The levels attainable in
indoor airs at room temperature can therefore greatly exceed
safe levels and result in poisoning." Mercury metal stored under water in an open container will still release Hg vapor into the air, as it is insoluble in water. Since mercury vapor, and worse yet, mercury compounds that might be formed are so toxic, I would not suggest experimenting with it except with proper guidance and with proper equipment, such as a negative-pressure hood feeding into a scrubber and a containment area for possible spills.
At one time, mercury was used to treat STD's and was used by hat-makers to preserve furs, with serious neurological damage to the users. It accumulates through the food chain to cause some fish to be toxic. If feasible, avoid using mercury in the lab and substitute a somewhat safer chemical.

The go into a bit more detail, the expression "mad as a hatter" originated from hat-makers being poisoned by mercury.

It is much harder than you think to reach a dangerous level of mercury vapour in the atmosphere. First you need to have poor ventilation (no airflow or atmospheric turnover in the room), then you need either lots of time or a mechanism to heat or disperse the mercury to enhance the rate of vaporisation. This is rare even when mercury is spilled (last time I know of involved trying to clean a spill with a domestic vacuum cleaner: a bad idea as it both heats and disperses the droplets).

The problem is, as you state, a spill will result in droplets going into inaccessible nooks and crannies. This leads not to acute toxicity, but little by little, long term (chronic) toxic effects. And, if bacteria metabolize the mercury to an organometllic compound such as methylmercury, the effects are far worse. See https://en.wikipedia.org/wiki/Methylmercury.

Mercury vaporization visualized: https://www.youtube.com/watch?v=JABbofwD3MI

Mercury metal exposed to the atmosphere will (dependent on the specific location) soon gather dust particles upon it's surface if left undisturbed. In this state and at 23C the mercury will not evaporate into the atmosphere as the combination of dust and temperature will work against this. I'm talking about a very small spill, that one would have to get down upon hands and knees to visualize. Droplets in the order of less than 1mm.
In a previous job (9.5 years experience-senior technician) on extremely rare occasions  had to either deal with small spills (onto the floor) or a yearly cleanup of old Mercury diffusion vacuum pumps (very hazardous because it involves actually heating at high temperature the Hg to help clean the pump- all conducted under stringent safety procedures and apparatus, to prevent any mercury from entering the atmosphere. When we tried to detect small mercury spills that had been in cracks in the flooring, it was not possible with a very sensitive mercury detector. However, if one jumped or caused large vibrations near the spill, it could be detected but only extremely close to the source. This is because a 'fresh surface" of mercury will have been exposed due to the vibrations moving the settled dust to different locations on the surface of the droplets.
So to answer your questions, will depend upon the conditions the mercury and water is subjected too. Boiling the water will definitely liberate Hg vapour! It is not that simple an answer. As stated above unless you understand the physical and chemical properties of Hg and it's interaction with the particular environment, better not to "experiment" with it. I suggest reading this: https://journals.aps.org/pr/abstract/10.1103/PhysRev.26.859


text:
How to set the memory for the Gaussian 09 cubegen utility?
I am connected to a supercomputer via SSH and my cluster uses Linux Scientific, I am using the Bash (Unix Shell), I load the modules of Gaussian 09 and Gaussview. I am trying to print orbital densities for a very large tetramer system, but when I try to generate the orbitals on Gaussview, the system crashes because of insufficient memory. On the internet, I found that using the "cubegen" keyword I can generate the cube files of the orbitals. The keyword gives me the power to modify the amount of memory but I do not know how. I go to interactive mode so I can use computational resources directly and on the command shell I do the following:
 (I try changing Gaussian's memory's environmental variable directly at the folder I am at)
GAUSS_MEMDEF=16gb

then,
cubegen 16000000 MO=HOMO xxx.fchk xxx.cub 0 h

But it still crashes with the following message,
Out-of-memory error in routine DecCub-RdMO (IEnd=      34279201 MxCore=      33546152)
Use %mem=33MW to provide the minimum amount of memory required to complete this step.
Error termination via Lnk1e at Wed Sep 28 16:41:41 2016.
Segmentation fault (core dumped)

How can I solve this problem? How can I allocate memory to Gaussian from the command window (I do not know where Gaussian folders are located since I just load it as a module, given that the software is already installed on the system)

First, you need to know which shell is used. You can type echo $0 to see which shell you are using.

Anyway, I guess (from the error message), that GAUSS_MEMDEF=33000000 cubegen bla-bla should work fine, at least in Bash shell. IIRC, GAUSS_MEMDEF is set in words, so since the error message says 33MW are needed, use the value 33000000, or a bit more, say, 35000000.

Note that cubegen nprocs kind fchkfile cubefile npts format cubefile2, so the second number should be the number of processes you'd like to use not the memory.

Btw. GAUSS_MEMDEF=16gb cubgen [...] should also work fine. It must be set immediately before the command though. As a over the thumb rule you can use 1MW are about 8MB for a 64 bit machine.

For cubegen you want to set GAUSS_UDEF: ```
export GAUSS_UDEF=4GB

By reading the information from Gaussian about cubegen, you would find out, that use of cubegen is like the following:

cubegen nprocs kind fchkfile cubefile npts format cubefile2

They further elaborate about nprocs:

nprocs
Number of shared memory processors used for electrostatic potential calculations. A value of 0 is equivalent to 1 (it is the default). Note that this parameter must be included if other parameters are specified. Previously this parameter was used to specify the amount of memory to allocate. The GAUSS_MEMDEF environment variable should be used instead.

You can read about setting GAUSS_MEMDEF on the site about the default.route file.

Environment Variables
All of these directives can also be set via environment variables or UNIX command line arguments. The environment variable GAUSS_XDEF provides a line equivalent to -X- in the Default.Route file. Similarly, the command line argument below specifies the same setting:
$ g09 -x="value"
Similarly, all of the following have the equivalent effect:
Default.Route file: -M- 4GB
bash shell: export GAUSS_MDEF=4GB
tcsh shell: setenv GAUSS_MDEF 4GB
command line: g09 -m="4GB" …

Based on that, you could try the way by exporting GAUSS_MDEF to the amount that you want, or by using the commandline. But I have no idea if the latter works with cubegen instead of g09. Let me know. :)

The latter will not work with cubegen, I tried. Setting (export) the environment variable works very well (it affects the whole session though). It also works immediately specifying it before the command GAUSS_MEMDEF=X cubegen ....


text:
Why is the helium wavefunction a linear combination?
the general solution for the wave function of Helium is a linear combination of the two possible states:
$$
\text{state 1:} \quad \psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2}) \\
\text{state 2:} \quad \psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1})
$$
(where $\vec{r}_{1}/\vec{r}_{2}$ denotes electron 1/2.)
But aren't the products of these wavefunction the same? Why then make a linear combination; it would be just like same function +/- same function?
$$
\psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2})
=
\psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1}) \, ?
$$

"the general solution for the wave function". It is not correct. The wavefunction is the solution of the Schrödinger equation. Also, that combination is not the general solution of the T.I.S.E.

First of all, note that
$$
\psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2})
\neq
\psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1}) \, ,
$$
since same functions in these products ($\psi_a$ and $\psi_b$) have different arguments ($\vec{r}_{1}$ and $\vec{r}_{2}$) to the left and to the right of the inequality sign.

Now, to the question, why do we have a linear combination. For a system of two indistinguishable particles (say, electrons) the obvious requirement is that there should be no observable difference between the system in state $\psi(1, 2)$ and the system in state $\psi(2, 1)$,
\begin{equation}
    |\psi(1, 2)|^{2} = |\psi(2, 1)|^{2} \, ,
\end{equation}
where $1$ and $2$ stand for coordinates of particles.
This implies that
\begin{equation*}
    \psi(1, 2) = \pm \psi(2, 1) \, .
\end{equation*}

And in order to satisfy this equality in what is often called the orbital approximation, where the state of a many electron system $\psi$ is represented as a product of a single-particle states $\psi_i$, the wave function $\psi(1, 2)$ should be actually written as not just a product of $\psi_1$ and $\psi_2$ (none of which, neither $\psi_1(1) \psi_2(2)$ nor $\psi_2(1) \psi_1(2)$, is symmetric or antisymmetric), but rather as the following linear combination of these products,
\begin{equation*}
    \psi(1, 2) = \frac{1}{\sqrt{2}} \Big( \psi_1(1) \psi_2(2) \pm \psi_1(2) \psi_2(1) \Big) \, ,
\end{equation*}
where $1 / \sqrt{2}$ is the normalization factor.

The choice of the sign in the expressions above is not arbitrary, but rather determined by the particles spin $s$, namely,
\begin{equation*}
    \psi(1, 2) = (-1)^{2s} \psi(2, 1).
\end{equation*}
Particles with integer spins, called bosons (photons, alpha particles), are described by a symmetric wave function
\begin{equation*}
    \psi_{\mathrm{s}}(1, 2)
    =
    \frac{1}{\sqrt{2}}
    \Big(
        \psi_1(1) \psi_2(2)
        +
        \psi_1(2) \psi_2(1)
    \Big) \, ,
\end{equation*}
while the particles with half-integer spins, called fermions (electrons, protons, neutrons), are described by the anti-symmetric wave function
\begin{equation*}
    \psi_{\mathrm{a}}(1, 2)
    =
    \frac{1}{\sqrt{2}}
    \Big(
        \psi_1(1) \psi_2(2)
        -
        \psi_1(2) \psi_2(1)
    \Big) \, .
\end{equation*}

thanks for the great answer; but why aren't those products the same, the terms of the functions depend on n,l,m, but are those quantum numbers independent of the position vector of the electrons, or has a distinct vector a given set of n,l,m ?

@hanna, well, physically, the $\psi_a(\vec{r}{1}) , \psi_b(\vec{r}{2})$ and $\psi_a(\vec{r}{2}) , \psi_b(\vec{r}{1})$ are the same in a sense that they describe the exact same state of a 2-electron system, namely, a state in which one lectron occupies $\psi_a$ spin orbital, while another occupies $\psi_b$. And, of course, since electrons are indistinguishable it does not matter which electron occupies which orbital: the state described by $\psi_a(\vec{r}{1}) , \psi_b(\vec{r}{2})$ is equal to one described $\psi_a(\vec{r}{2}) , \psi_b(\vec{r}{1})$ in every observable way.

@hanna, mathematically though, it is quite apparent that the $\psi_a(\vec{r}{1}) , \psi_b(\vec{r}{2})$ product differs from the $\psi_a(\vec{r}{2}) , \psi_b(\vec{r}{1})$ one, which is odd, since we have then two apparently different mathematical description of the same state! This already suggests that there is something wrong with describing the state by these individual products, they both should be included in the description.


text:
Vacuum at different altitude
I have tried to find this searching the internet and I am not sure if I understand it. I want to turn water from solid to gas without going into the liquid phase - Sublimation.
The altitude above sea level is 1500 meter about 5000 feet. I have small 1 stage vacuum pump and a cheap vacuum gauge. The pump is rated to 5Pa (at sea level). Atmospheric pressure is around 85kPa. The vacuum gauge reads 20 inHg but what is that in Pascal, if I convert it using Google or the calculator it give me a value of 67kPa.
Two questions:
1. Can sublimation occur at an altitude of 1500 meter above sea level and atmospheric pressure of around 85kPa

How does one convert vacuum inHg into pascal

I am not chemist nor a chemist student and just like experimenting with stuff

You do not say what 20 in Hg is, I assume  inches ?  More often we use mm Hg ( or 1 torr) as a non SI  measure of pressure. The conversion is 1 mm Hg is equivalent to 133.322 pascal.

@porphyrin yesterday. Yes, inHg, I meant inches

Because you are manipulating the pressure using a vacuum pump, the altitude change should not matter. If you look at the phase diagram for water in the attached link, you can see that if you keep your system below 0C, say at -10C, than you will cross the solid-vapor equilibrium line and your ice will start to vaporize when the vacuum pump decreases the pressure to ~200 pa.
https://en.wikipedia.org/wiki/Phase_diagram#/media/File:Phase_diagram_of_water.svg
To convert from pa to in Hg, I suggest searching for the conversion factor. From google: 1 in Hg = 3386.39 pa. So if you have a reading of say, 2 in Hg on your gauge then: $ 2 inHg*(3386.9 pa)/(1 in Hg) = 6772.78 pa $


text:
Yellowish tinge during titration of oxalic acid with potassium permanganate
One of the our chemistry practicals involve finding the molarity of an oxalic acid solution by titrating it with standard $\ce{KMnO4}$ solution.
The oxalic acid solution is heated and little sulfuric acid is added to titration flask before starting the titration.
It was my personal observation that adding the $\ce{KMnO4}$ solution rapidly causes a temporary appearance of a light yellow colour. Slow and gradual addition of $\ce{KMnO4}$ does not lead to this observation.
If sulfuric acid is not added to the titration flask, then also a yellowish tinge appears as we get closer to the end point.
What is the for this yellowish tinge? Is there any intermediate species formed (maybe a complex one) that leads to this observation?

marginally related http://chemistry.stackexchange.com/questions/4310/what-is-the-reaction-between-oxalic-acid-and-potassium-permanganate

Without knowing the concentrations of the involved substances and the values of some other parameters, we can only guess what has happened during your experiment.
The intended analytical reaction is the reduction of purple permanganate to colourless $\ce{Mn^2+}$:
$$\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O}$$
If you add the permanganate solution too quickly, the reduction may be incomplete because not enough reducing agent (here: oxalic acid) or acid $(\ce{H+})$ is available in the reaction zone:
$$\ce{MnO4- + 4H+ + 3e- <=> MnO2 + 2H2O}$$
The observed colour may be caused by a cloud of finely dispersed particles of dark brown manganese(IV) oxide $(\ce{MnO2})$, which dissolves when the solution is well mixed again so that the intended reaction continues.
The same reaction may occur when you don’t add enough sulfuric acid, since $\ce{H+}$ is consumed during the reaction (8 mol $\ce{H+}$ per 1 mol $\ce{MnO4-}$).

thanks for the explanation. If you need the concentrations, they are as follows: a)20ml of 0.025M Oxalic acid in titration flask(determined by the experiment). b)1 full test tube of 2M sulfuric acid(approx 15ml). c)0.01M KMnO4 in the burette.


text:
Gauche effect and bond length

Which of the following compounds has the longest bond length?
(A) $\ce{H3C-CF3}$
(B) $\ce{FCH2-CH2F}$
(C) $\ce{F2CH-CHF2}$
(D) $\ce{F3C-CF3}$

My logic was based on the relative electronegativities of the two carbons. The more fluorines there are on one carbon, the more polarised the C-C bond will be, and the longer the C-C bond length will be (since the bond is weakened). Thus, I think that the answer is A.
Is this right?

Actually, the more the polarization of a bond, the stronger it is, since both the atoms are drawn close together due to electrostatic attraction, decreasing the bond length and increasing the bond stability (think about ionic bonds. Your reasoning would dictate them to be very weak, while we know how strong they are!). Actually, I would predict $\ce{CH3-CF3}$ to have a really small bond length (the bond is probably the strongest, would someone confirm this please?), since the fluorinated carbon has quite a positive charge, while the other one is relatively negatively charged (not negative in the absolute, just relative to the other carbon).
All the remaining options have equal number of fluorines on either carbon, differing only in their number. These three are tricky. Consider $\ce{CF3-CF3}$. Both of its carbons are severely deficient in electrons. Each badly needs electrons, and since it can't argue with fluorine (fluorine is an unreasonable bully), so both carbons try to drag the shared pair of electrons towards itself. This ultimately results in them both coming closer to each other. As we reduce the number of fluorines on each carbon, their desparation is reduced, thus increasing the bond length. Thus the answer is (B).
Edit: If @pentavalentcarbon's calculations of bond lengths are correct, then my reasoning was too simplistic, and indeed it seems so now. So if we dig deeper, $\ce{CH2F-CH2F}$ could have the second smallest bond length and $\ce{CF3-CF3}$. Why? Well, it is because of hyperconjugation, similar to the gauche effect (read it up to get better understanding of the gauche effect, although I explain it in short here also).

So, what happens is that in the case of $\ce{CH2F-CH2F}$, the bonding C-H molecular orbital (MO) overlaps with the antibonding C-F MO, transferring some of its electron density to the C-F MO (see diagram). This interaction is stabilizing (since delocalization of electrons stabilizes the molecule), and also introduces slight double bond character in the C-C sigma bond, thus serving to shorten it.
Now in $\ce{CH2F-CH2F}$, there will be two such interactions, but then there are also two such interactions in $\ce{CHF2-CHF2}$! There is a difference though. In the latter case, there are two fluorines on each carbon, which heavily draw electron density away from the carbon through the sigma bonds. This decreases the sigma bond density in the C-H bond, which being poor in electrons, gives relatively low stabilization when overlapping with the antibonding C-F MO. So $\ce{CHF2-CHF2}$ has a slightly higher bond length than $\ce{CH2F-CH2F}$.
Using this same reasoning, we can see why $\ce{CH3-CF3}$ has the smallest bond length (three favourable hyperconjugational interactions!) and $\ce{CF3-CF3}$ has the longest bond length (no hyperconjugation).

I calculated the bond lengths to be (in angstrom) A: 1.511 B: 1.526 C: 1.544 D: 1.573.

@pentavalentcarbon Your comment suggests that my answer and the textbook's answer is wrong. Either I am missing something or there is quantum mechanics involved. I am also curious, so could you add an answer of your own too, please? I would love to understand the reasons behind it!

I agree with the newer edit. Penta's calculations make a lot of sense too. I expect the gauche effect to shorten the bond length, in accordance with the resonance form that Jan drew here involving a C=C double bond. Ah, you mentioned the slight double bond character too! Very good. I would, however, say that the gauche effect is a specific example of hyperconjugation (which is a very general term). +1 anyway.

So to summarise things as I understand it so far: hyperconjugation (shortens bond) competes with electron withdrawal by F atoms (repulsion lengthens). The effect is best seen compared to ethane which has a bond length 153.5. Therefore structures (A) and (B) are shorter, just, due to hyperconjugation being dominant, (C) the two effects just about balance and (D) repulsion of C-C nuclei make bond stronger due to electron withdrawal by F atoms. Do the experimental bond lengths support this?

@porphyrin You could say so, but instead of treating it as C-C nucleic repulsion, I feel that it should be treated as which molecule shows more hyperconjugation effect! I didn't find any good sources for experimental bond lengths for these molecules.

@orthocresol Hey, the link in your comment has a really nice explanation! Thanks for sharing it!

Many of these cases hinge on some weird effect. Evidently the gauche form of 1,2-Difluoroethane is more stable than the anti form which is sort of weird.
This wikipedia article on the Gauche Effect list CC bond of gauche 1,2-Difluoroethane as 150 pm and anti form as 151.4 pm.
This reference says 151.2 pm for C-C bond in 1,1,2,2-tetrafluoroethane. (High-Resolution Microwave and Infrared Molecular-Beam Studies of
the Conformers of 1,1,2,2-Tetrafluoroethane, Stephen C. Stone et al., Journal  of  Molecular Spectroscopy 192, 75–85 (1998).)
This reference says the C-C bond length is shorter for hexafluoroethane than then the corresponding hydrocarbon. (Fluorocarbon and Related Chemistry
By R. E. Banks, M. G. Barlow,  ISBN-13: 978-0851865140, page 10.)
This answer on this website says ethane C-C bond is 153.51 pm.
So to summarize:
$$ \small
\begin{array}{lcc}
\hline
\text{Compound} & \text{Bond Length (pm)}\\
\hline
\text{gauche 1,2-difluoroethane} & 150 \\
\text{anti 1,2-difluoroethane} & 151.4 \\
\text{1,1,2,2-tetrafluoroethane} & 151.2 \\
\text{Hexafluoroethane} & \text{<ethane C-C} \\
\text{Ethane} & 153.51 \\
\hline
\end{array}
$$

Hexafluoroethane vs ethane can easily be explained with Bent's rule: http://chemistry.stackexchange.com/q/65195/4945


text:
Computational heat of reation vs. using literature values
My question is I don't understand how to explain why or what I did wrong to have such large differences between my computational heat of reaction and heat of reaction using enthalpies of formation from literature, and my experimental heat of reaction.
I performed a lab in which three methods were used to compute the heat of reaction for the reaction:
$$
\ce{Na2CO3 (s) + 2 HCl (aq) -> 2 NaCl (aq) + H2O (l) + CO2 (g)}
$$
First method I used was using experimental data from solution calorimetry and from my results I got that the average heat of reaction was $\textbf{-89.835 kJ/mol}$.
For my second method I found the heat of reaction computationally with Orca. I was instructed to use this equation instead of the original equation to calculate heat of reaction for the above reaction.
$$
\Delta U = U \left[ \ce{CO2 (g)} \right] + 3U \left[ \ce{H2O (l)} \right] - 2U \left[ \ce{H3O+ (aq)} \right] - U \left[ \ce{CO3^{2-} (aq)} \right]
$$
So basically I used Orca to find the energy for each of the molecules, and I plugged those values into their proper U(molecule) in the above formula and I calculated delta U and then I used the approximation of $\Delta H = \Delta U + R \Delta T$ to transform it into heat. My overall answer ended up being $\textbf{-378 kJ/mol}$. My work (data from Orca optimization is listed first):
\begin{align*}
\ce{CO3^{2-} (aq)} &= -264.0667 \, \text{Ha} \\
    \ce{H3O+ (aq)} &= -76.8491 \, \text{Ha} \\
      \ce{H2O (l)} &= -76.4409 \, \text{Ha} \\
      \ce{CO2 (g)} &= -188.5873 \, \text{Ha} \\
    1 \, \text{Ha} &= 627.503 \, \text{kcal/mol} \\
          \Delta H &= \Delta U + R \Delta T \\
          \Delta U &= \left[ -188.5873 + (3*-76.4409) \right] - \left[ (2*-76.8491) + -264.0667 \right] = -0.1451 \, \text{Ha} \\
-0.1451 \, \text{Ha} &\rightarrow 627.503 \, \frac{\text{kcal/mol}}{\text{Ha}} * -0.1451 \, \text{Ha} = -91.0507 \, \frac{\text{kcal}}{\text{mol}} \\
-91.0507 \, \frac{\text{kcal}}{\text{mol}} &* 1000 \, \frac{\text{cal}}{\text{kcal}} * 4.184 \, \frac{\text{J}}{\text{cal}} * \frac{1}{1000} \frac{\text{kJ}}{\text{J}} = -380.956 \, \frac{\text{kcal}}{\text{mol}} \\
\Delta H_{\text{rxn}} &= -380.956 \frac{\text{kJ}}{\text{mol}} + \left[ 8.314 \frac{\text{J}}{\text{mol K}} * (273.15 + 25.27) \, \text{K} * \frac{1}{1000} \frac{\text{kJ}}{\text{J}} \right] = -378.475 \, \text{kJ/mol}
\end{align*}
My third method I used enthalpies of formation (values for looked up from various sources) and my work is shown below.
\begin{align*}
\ce{Na2CO3 (s)} &= -1130.7 \, \text{kJ/mol} \\
       \ce{HCl (aq)} &= -167.2 \, \text{kJ/mol} \\
 \ce{NaCl (aq)} &= -407.27 \, \text{kJ/mol} \\
   \ce{H2O (l)} &= -285.8 \, \text{kJ/mol} \\
   \ce{CO2 (g)} &= -393.5 \, \text{kJ/mol} \\
\Delta H_{\text{rxn}} &= \left[ (2*-407.27) + (-285.8) + (-393.5) \right] - \left[ (2*-167.2) + (-1130.7) \right] = \mathbf{-28.74 \, \textbf{kJ/mol}}
\end{align*}
I'm not really sure how to say why the computational method did so poorly, I would have thought the computer would have done a decent job. Please let me know if something looks wrong in my calculations! Thanks for all the help! Even just a little advice will help this non-chemistry nerd!

Can you give more details about how you performed the ORCA calculation? The settings line (everything after ! near the top) should suffice.

For the molecules that were in aqueous state I used ! B3LYP def2-TZVPP COSMO(Water)         !OPT        For molecules that were not in water I used   ! B3LYP def2-TZVPP ! OPT

I don't understand your approximation $\Delta H = \Delta U + R \Delta T$.  It would seem to make sense only for an ideal gas.  Also you plug in ~300 K as a temperature; was there really a temperature change of 300 K during your experiment or simulation?

It's also unclear why in the second method you assume the carbonate is dissolved (you speak of $\ce{CO3^2-}$ ions) while in the other examples you assume it is a solid.

I was told to use the delta H = U +Rt equation from my professor. The t in the equation is not a delta t, it is just t, and since everything is in kJ/mol whatever t is used doesn't make much of a difference, so I just used the temp. That my reactions took place in which was about 25C. Before the reaction took place sodium carbonate was in solid form and it was reacted with .1M HCl. I was given the second equation with all the U(molecules) from my professor and was told that finding the energy of this equation and converting it to heat would somehow relate to the original reaction.

I think you're comparing apples and oranges. Any thermodynamics textbook will give H=U+PV. You need to consider which thermodynamic ensembles you are using under which boundary conditions and then compute the right quantities. I suggest looking at the Gibbs free energy. Then calculate the right properties with ORCA. For your calorimetry, I suggest you consider whether you are actually performing the experiment for the described equation. I think your third method is correct for the described reaction equation.