text:
Knowing about Sensor Film Kit
I am very new to this site and this is my first ever question. I want to know what is this black material in this sensor film kit. I am creating a weight sensor by using this very sensor kit. Now I am short of this black film. I want to purchase it now locally but do not know the composition of this material.
This material behaves as a resistance between two aluminium foil, just to inform.
If any one can help me in this.

This really should be on the electrical engineering section. // I don't know what specific material this is, but Google "Conductive foam" and you should find some. The resistance changes as the foam is squeezed.

@Max In this case I agree that it would probably better fit another site. But I am only one. Before asking to migrate something it would be preferable if the question is closed by the community. Then I will research if it is a good fit for the target site. Still, thank you for your effort.

It's called a force sensing resistor. It's basically a film of small conductive particles mixed into a non-conductive polymer matrix on top of a conductive substrate. The more force on the film, the more conductive particles can touch the layer underneath, reducing the resistance. Sensitronics owns the original patent for the technology and their site shows how they're constructed, though you'll probably have to dig up the patent to find out exactly what's in the force sensing layer.


text:
Mechanism for reaction of ketene with water
In the Arndt-Eistert synthesis, a ketene is the intermediate. The ketene is then hydrolysed in the presence of $\ce{Ag2O}$. What is the mechanism for the hydrolysis of the ketene?
Is an oxonium ion formed or will there be nucleophilic attack of $\ce{H2O}$ on the carbonyl carbon?

Don't capitalise common nouns like ketene and oxonium.

The silver salt, which can also be $\ce{AgCF3COO}$, is added to the mixture to cleave off $\ce{N2}$, creating a singlet carbene which induces the 1,2-shift to generate the ketene. From then on, it has no further purpose.
Our solvent is water, which could technically act as an acid, a base or a nucleophile. Reacting as a base does not forward our mechanism, so we need to consider whether water is stronger as an acid (left-hand mechanism in scheme 1) or as a nucleophile (right-hand mechanism).

Scheme 1: Two possible mechanisms for ketene hydrolysis.
A few thoughts on the two mechanisms, first. In whichever way water reacts, it will react with the most reactive site, first. In case of a nucleophilic attack, the most electrophilic carbon is clearly the ketene carbon. Water has the possibility of cleaving either the $\ce{C=C}$ double bond or the $\ce{C=O}$ double bond. The $\ce{C=O}$ bond is preferred because it usually represents the molecule’s LUMO and the resulting carbanion would be rather unfavourable and probably immediately flip to an oxyanion. The final [1,3]-proton shift cannot occur in a concerted manner for electronic reasons. It is also highly unlikely that a single water molecule is both acceptor and donor as I drew it for brevity. I would assume a pair of molecules acting from both sides in a stepwise manner.
In case of the water-acidic mechanism, the most basic position in the molecule would be the oxygen atom so one would expect protonation there first. Again, I would assume the [1,3]-proton shift to be non-concerted which could provide major flaws to the mechanism. I also do not see the carbon position having sufficient basicity to be protonated in any noteworthy manner in rather neutral water. But let’s just assume it could happen for the sake of the following argument.
When it comes to rating water’s properties, we should note that it’s a mediocre base, a rather bad nucleophile but an even worse acid. So while one can save the nucleophilic mechanism by saying ‘yeah, water’s not a good nucleophile, but it’s the best we have floating around and the ketene position is rather reactive’ we cannot save the acidic mechanism in any similar way. Ketones are, in good approximation, unprotonated in aquaeous solution, so protonating an even worse position is all the more unlikely.
Of course, it is conceivable that both mechanisms happen to some extent. By playing around with the solution’s pH, you could probably favour either: an acidic solution will possibly facilitate the acidic mechanism while a basic solution will definitely facilitate the nucleophilic one — hydroxide is a so much better base than water.

Keep in mind that this analysis is just for the very first attack of water on the ketene. After the carboxylic acid is produced, an acid catalyzed mechanism surely is operating.

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text:
Is there a universal minimum energy state?
I'm imagining this: if everything was separated (even on the nuclear level) and then the parts were recombined so that they were in the lowest possible energy state, do we have any way of knowing what that might be? Particularly, would it be something simple like a helium atom or could it be something very complex and unusual? And I mean that you might grab a few parts and make a helium atom or grab a whole bunch of parts and make whatever else.
I guess one problem is we need a way of totaling how much the energy is decreasing per some amount of something else. Similar to how nuclear binding energy is often expressed on a "per nucleon" basis.
I'm interested to see what people think.

That's pretty much what happens in the stars, so yes, the end result is more or less well known.

This is an interesting and actually surprisingly hard question, which reaches the edge of our understanding of physics, so really a physicist would be more equipped to answer this.
I'm not sure if there is rigorous mathematical/physical proof that energy must have a minimum. Energy is a continuous, real-valued, state function, which as far as I know makes it mathematically very well-behaved, so it's unlikely that it can display weird behaviour. The extreme value theorem suggests that taking any finite collection of states into account, at least one of them must be a global energy minimum. However, I'm not sure if the Universe has a finite amount of states it can assume (probably not, if it is infinite in size). All that said, even if strictly speaking there is no global minimum, it's likely there is some asymptotically low value. So what is it?
What you're trying to imagine, essentially, is a Universe where there is no kinetics, only thermodynamics. Equivalently, you're thinking about the state of the Universe at $\mathrm{t→+ \infty}$. All finite kinetic barriers are surpassed and all finite distances are traversed. With this much time, a lot of unintuitive events take place. Two very interesting sources to read are "Time without end: Physics and biology in an open universe",  (1979) Rev. Mod. Phys., 51, 447, and "A dying universe: the long-term fate and evolution of astrophysical objects", (1997) Rev. Mod. Phys., 69, 337. There are many processes, varying from almost certain to extremely speculative, which can happen in overlapping time scales, so it is hard to determine exactly which route the Universe (or a toy model of a universe) might take. I'll present several events which may happen as the Universe marches on towards $\mathrm{t→+ \infty}$, and thus which reach closer to the minimum energy state of the Universe, but be aware that there are many uncertainties, especially near the deep end.
Starting with chemistry, allowing systems to evolve for very long periods would have some interesting consequences. Only the most stable compounds could survive, so for example, all nitrogen atoms would eventually recombine to form gaseous $\ce{N2}$. Other very stable molecules would be $\ce{H2O}$, $\ce{CO}$, and $\ce{H2}$. The Universe would turn into a mixture of small molecules. This would take at least something of the order of $10^{65}$ years. For comparison, the Universe is currently $1.38\times 10^{10}$ years old.
But why stop there? Chemical energy is only one of many factors to be taken into account in the total energy, and really it's one of the smallest. For example, nuclear energy scales completely dwarf chemical energy scales, so forget about optimising molecular abundances to minimise energy, because optimising atomic abundances will reach much lower energies. What would really happen is that, given tremendous amounts of time, light/heavy nuclei would overcome the barrier to fusion/fission, even at very low temperature and pressure. As you mention, the ultimate effect would be that all of the atoms in the Universe would fuse/fission into $\ce{^{62}Ni}$, the nucleus with the highest nuclear binding energy per nucleon. This would take on the order of $10^{1500}$ years.
This still isn't the end, though. Given a ridiculous amount of time, much of the nickel could still coalesce and compress into degenerate neutron matter (like neutron stars), releasing even more energy. The kinetic barrier for this is truly colossal, though, so it would take some $10^{10^{76}}$ years.
It's possible to keep going. Sufficiently large chunks of matter could compress further into black holes, releasing even more energy. Depending on the mass of the lightest possible black hole, this process may take "only" $10^{10^{26}}$ years, or up to a scale comparable to neutron matter formation. If the lightest possible black hole has a non-zero mass, then objects lighter than that threshold may be stable indefinitely. The black holes that did form would evaporate almost instantly relative to this time scale, producing an abundance of subatomic particles which would eventually decay into photons and neutrinos.
There's another speculative process which may severely undercut the previous possibilities, however. Currently many types of string theory posit the existence of proton decay. If it does happen, it could happen as quickly as $10^{40}$ years. This would be the end of Chemistry in the Universe; all atoms would eventually decay into photons, neutrinos, electrons and positrons. Black holes would survive this, but they would evaporate by $10^{100}$ years at the latest, quickly resulting in a Universe sparsely filled only with photons and neutrinos.
Last but definitely not least, it may be that even completely empty vacuum is not the minimum energy state of the Universe. It's conceivable that the vacuum we know is less stable than another type of vacuum, separated by an enormous kinetic barrier. If this is the case, then given enough time there could be a vacuum metastability event, a tremendously energetic process in which the Universe as we know it would cease to exist, being replaced by another with entirely different physics.

Wow. Thank you for that incredibly detailed answer. I'll have to ponder some of that. I'll take a look at those papers too. Thanks.

And I mean that you might grab a few parts and make a helium atom or
  grab a whole bunch of parts and make whatever else.

Why helium if hydrogen is simpler? The universe is indeed dominated by hydrogen: monatomic hydrogen is the most abundant chemical substance in the Universe, constituting roughly 75% of all baryonic mass.1
According to the widely accepted Big Bang theory, once the universe became cool enough (at about 379,000 years) the electrons and nuclei combined into atoms, and the atoms were mostly hydrogen atoms because previously despite the presence of neutrons most protons remained uncombined as hydrogen nuclei. The reason is that when the universe was hot protons and neutrons transformed back and forth into each other, and their ratio was determined solely by their relative masses: it was about 1 neutron to 7 protons.
A bit later, once it was cool enough, neutrons combined with protons to form deuterium and helium nuclei by a process called Big Bang nucleosynthesis. Eventually these nuclei gave rise to deuterium and helium atoms. The theoretical estimates of the end result of Big Bang nucleosynthesis are (in mass abundances): about 75% of hydrogen-1, about 25% helium-4, about 0.01% of deuterium and helium-3, trace amounts (on the order of 10−10) of lithium, and negligible heavier elements.
The bottleneck of to the absence of a stable nucleus with 8 or 5 nucleons was passed only much later in evolving and exploding stars by stellar nucleosynthesis, so that the elements heavier than lithium and beryllium were formed. The triple-alpha process should be mentioned specifically in that respect in which three helium-4 nuclei are transformed through the unstable beryllium-8 nuclei into a carbon-12 nuclei which is finally stable. Note that the triple-alpha process is very unlikely and requires very high temperatures (about $10^8$ K and higher) for nuclear fusion, and thus, it needs these special conditions and a lot of time to produce much carbon.

1) "Baryonic" is important here, because most of the mass-energy in the universe is not even in the form of baryons or chemical elements: the universe is dominated by the dark matter and dark energy which collectively amount for about 95% of the total mass–energy content of the universe.

On a per nucleon basis, Ni-62, Fe-58, and Fe-56 are all about 930 MeV.
However, as explained in Strange Quark Matter and Compact Stars

Theoretically the energy per baryon of strange quark matter may be below 930 MeV, which would render such matter more stable than nuclear matter.

Thanks for that great reference. Whenever I get time to try to comprehend 62 pages of dense physics I'll take a look at that :)

@jheindel section 3.3 is the part most relevant to your quesiton

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Why does the Temperature Decrease?

This is from a calorimetry lab. $50\ \mathrm{ml}$ of acid and base were mixed. Why does the temperature keep decreasing after 1 minute and why isn't at a constant rate?
Is it because heat is lost to the surroundings?

Source: https://fronter.com/uio/links/files.phtml/1520337012$477794732$/Lab/Laboppgaver+-++Prelab++-+Rapportmal/Oppgave+2+all.pdf
This for a Lab Report Hand-IN at my University. Blindern University of Oslo.

Heat loss to surroundings is probably it.

The link is not visible to the general public. This is unlikely due to heat loss, as that wouldn't be linear. Especially not if it wasn't well-mixed, which it probably isn't. More information is going to be needed.


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How to recycle olive oil manufacturing wastes?
The fact that only 20% of the olive used to make olive oil becomes olive oil and the other 80% is spread between hard wastes and liquid water wastes  made me want to ask this question. Isn't there a way to recycle the other 80% and use it in an effective way either in other industries (energy, agriculture, or cosmetics)?
I'd like to know if there is a simple and low-cost way people can do so.

The components of the 80% waste material could definitely be recycled using common procedures ( eg distillation, solvent extraction ) and could be used for many purposes like you mentioned or by converting the waste material to a more useful product.
I'm not sure what they make from the waste from olive oil production exactly but a good example of similar biotechnology is the use of waste cooking oils to synthesize biodiesel.

Are there some already established means of turning the waste into cooking oil ? if the answer is yes, how do they proceed about it ? I'd like to make a prototype for the machine or the mechanism that does that !!

It depends how much fatty acids are left in the waste product . They definetly could be extracted and refined to a pure product, whether this is practical to do so I'm not sure hence why it's the waste product. Other compounds present could be converted to fatty acids but I'm not sure if they would be ideal for cooking. There could be many different mechanisms and machines that could be used to do this and many variables it depends on what the impurities present are and what you want to make exactly then determine the appropriate procedures and equipment to use.


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A substance that is immisiscible with water and that can be made cloudy
I need to identify a substance that is immiscible with water that is or can be made cloudy.
At the moment I'm using glycerol, but I don't know how to make glycerol cloudy nor is it perfectly immiscible with water.
Clarity:
I need to make a bottle containing two liquids. One part being water mixed with rhodamine and the other part another liquid that is immiscible with the rhodamine water mix that is cloudy. The other chemical can either be cloudy already or be mixed with another substance so that it can become cloudy.

Made cloudy? What do you mean?

It smells but I think chloroxylenol 48 mg in 250 ml ethanol mixed with water would be what your looking for, I'm not sure what it's called where you are but it's a common antiseptic sold everywhere, in Australia the most common brand is Dettol.
Also , if something is completely immiscible with water it will form two phases , both relatively clear, eg water/xylene.
If you mix the solution it will go cloudy but then seperate, the longer it takes to do this depends on how much it was mixed, an emulsion can be formed but that's not what your looking for.
You want something that's only very slightly or sparingly soluble but doesn't seperate into a different phases once it's mixed.
Dettol came straight into my mind but I'm sure there's many more. Arrowroot or cornflour mixed with water would produce a milky/cloudy solution.


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How does one calculate chemical shift in a gauge independent manner?
The literature on ab initio calculations of chemical shift in NMR experiments usually provide Lamb's and Ramsey's formulae as the solution. Yet the expressions explicitly depend on vector potential gauge choise. Some authors claim that the theory is intrinsically gauge invariant, giving no proper proof yet. Can the solution be expressed in an explicitly gauge invariant form? Some literature on this.
NOTE: I know about London orbitals theory, yet it's still not fully understood by me, how it can deliver a type of expression I want. Some suggestion on where this is clarified are very welcome.

Would you be satisfied with a worked-out set of equations using London/gauge-including orbitals?

Actually I've already found the answer, that satisfies me

The references below describe the answer.
Prog Nucl Magn Reson Spectrosc. Author manuscript; available in PMC 2012 May 1.
Published in final edited form as:
Prog Nucl Magn Reson Spectrosc. 2011 May; 58(3-4): 176–201.
Published online 2010 Dec 15. doi: 10.1016/j.pnmrs.2010.10.003
PMCID: PMC3058154
NIHMSID: NIHMS248338
Chemical shift tensors: Theory and application to molecular structural problems
Julio C. Facelli
Nucleic Acids Res. 2014 Dec 16; 42(22): e173.
Published online 2014 Nov 17. doi: 10.1093/nar/gku1006
PMCID: PMC4267612
Accurate ab initio prediction of NMR chemical shifts of nucleic acids and nucleic acids/protein complexes
Andrea Victora,1 Heiko M. Möller,2 and Thomas E. Exner1,3,*
A comparison of models for calculating nuclear magnetic resonance
shielding tensors
James R. Cheeseman, Gary W. Trucks, Todd A. Keith, and Michael J. Frisch
J. Chem. Phys. 104 (14), 8 April 1996

Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the references.


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How do you make napalm?
Not a recipe from movies, but actual napalm. I need it for a college experiment. I need the good kind preferably, but any recipe will suffice.

I'm voting to close this question as off-topic because the intention behind it seems rather dubious.

The original meaning of napalm referred only to the thickener, not the flammable substance which it thickens, and not the mixture of the two.

The aluminum soaps of this mixture of acids was christened "napalm" from "nap" for napthenic acids and "palm" for the coconut fatty acids.

Quoting from the 1951 article They Don't Like "Hell Bombs"; Napalm + Gasoline = "Hell Bombs", Armed Forces Chemical Journal, Volume 4, page 27
Coconut fatty acids include palmitic acid.
Also, from THE SPECIFIC HEAT OF NAPALM–GASOLINE GELS Canadian Journal of Research, 1949, 27a(4): 151-155, footnote 3:

Napalm is a commercial product consisting of a basic aluminum soap of naphthenic, oleic, and palmitic acids in the approximate ratio of 1 : 1 : 2.

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Which alcohol will most easily react with HCl to form an alkyl halide?
Which alcohol will most easily react with HCl to form an alkyl halide?
A. Primary alcohol
B. Secondary alcohol
C. Tertiary alcohol
D. All alcohols react equally to form alkyl halides

This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Tertiary alcohols have greater reactivity with hydrogen halides than secondary alcohols, which in turn have comparatively more reactivity than primary alcohols. For tertiary alcohols, the reaction proceeds by an SN1 mechanism which prefers a more substituted alcohol.
Therefore answer is C

Add explanations, please.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. While we appreciate your contribution to the site, it would be preferable to not answer questions, that do not comply with our homework policy.


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Relation between ionisation enthalpy and nature of bonds in boron
My notes given by institute states that:

Due to small size of boron, the sum of its first three ionization enthalpies is very high. This prevents it to form $+3$ ions and forces it to form only covalent compounds.

How is the covalent nature of bond of boron related to the sum of its first three ionization enthalpies?
EDIT :- I am directly or indirectly asking to prove fajan's rule for boron.

At first, let me give a picture of what ‘relatively high’ actually means. I present the sum of the first three ionisation enthalpies of some elements typically found as metal(III) ions:

Boron $6887.4~\mathrm{\frac{kJ}{mol}}$
Aluminium $5139~\mathrm{\frac{kJ}{mol}}$
Scandium $4256.7~\mathrm{\frac{kJ}{mol}}$
Iron $5281.4~\mathrm{\frac{kJ}{mol}}$
Cobalt $5640.4~\mathrm{\frac{kJ}{mol}}$
Ytterbium $4195.2~\mathrm{\frac{kJ}{mol}}$

We see that boron’s ionisation energy sum is $1.2~\mathrm{\frac{MJ}{mol}}$ larger than that of the next largest in my sample. Also keep in mind that boron is rather small per se so a $\ce{B^3+}$ ion must be even smaller.
But what does that actually mean? To generate an ionic compound from its elements, one usually considers an overall formation enthalpy composed of bringing both elements into the gas phase, turning them into ions and letting the crystal lattice form — this last step liberates most energy and is usually the driving force for salt creation. In boron’s case, the cations of such a salt would be $41~\mathrm{pm}$ — exceptionally small; way smaller than $\ce{Al^3+}$ ($68~\mathrm{pm}$) which is already tiny and smaller than $\ce{Be^2+}$ whose $59~\mathrm{pm}$ are typically cited as the smallest ionic radius known ($\ce{H+}$ doesn’t count).[1]
The structure of $\ce{BeF2}$ can already be considered as the greatest possible compromise in keeping the cations and — more importantly — the anions apart from themselves; hence its spacious structure. These problems would be multiplied if an even smaller cation were used here: It would be extremely hard to generate a crystal lattice that would allow for sufficient separation of the anions from each other since the cation is so tiny (and we need lots of anions: three negative charges per boron!).
So all things considered, ionic boron salts would have a very unstable crystal structure and would hence be hard to generate, especially since the lattice enthalpy cannot counteract the large ionisation enthalpy required for boron adequately. Luckily, covalent compounds are always a possibility and boron’s atomic orbitals are well shaped and well placed in energy to allow covalent bonding.
Note that boron’s electronegativity of $2.0$ on the Pauling scale also shows how badly boron would form ionic bonds: The tiny sizes of ions would draw electrons in in an even stronger way, potentially reducing boron before it would reach the ionic state.

The Polish chemist Kazimierz Fajans simplified this in his set of Fajans’ rules:

A compound will be covalent, if:

there is a high positive charge;
the anion is large; and/or
the cation is small.

A compound will be ionic, if:

there is a low positive charge;
the anion is small; and/or
the cation is large.

Note that in ambiguous cases such as $\ce{AlF3}$, (small cation, high positive charge but small anion) the rules fail to predict a mainly ionic compound, but correctly make us assume a rather high covalent character of said ionic bonds.

[1] Even beryllium forms rather covalent compounds most of the time. Due to the electronegativity difference of $2.5$ on the Pauling scale, we may just be able to consider solid $\ce{BeF2}$ as an ionic compound. All of these are borderline cases in one way or the other, though.

Alright, i have understood this: small size of boron -> high energy needed to make +3 ion  ->  very unstable in ionic form and therefore in lattice form ->  makes covalent bond. am i right?

@manshu Yeah, more or less.

an hour ago, i came across something called Fajan's rule which says 'the smaller the ion the greater the tendency to covalency'. I am just just asking you if i can add this term in the question so that in the future, the help seekers find this answer helpful. so should i add this term or is it something different?

@manshu Thanks for pointing the rule out to me; I learnt something new again ^^

aye aye captain.


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Why does melting point depend on external pressure?
The melting point of a substance is the temperature at which the solid form's vapor pressure equals the liquid form's vapor pressure. I read that the vapor pressure of something depends ONLY on the temperature. So as the external pressure changes, the substance's vapor pressure should stay the same if temperature is constant. In that case, the melting point should not be affected by external pressure.
I understand that intuitively, melting point increasing with pressure makes sense, as more energy is needed to overcome larger external pressure in order to liquefy. But how can this be explained with the vapor pressure definition?

This is a duplicate. All 1st order phase changes have a volume change, hence there will be a PV term in the free energy. Thus, the pressure matters.

Please go back and check your source for the opening statement. Melting point is a transition between the liquid and solid phases of a substance and not its vapor phase.

@iad22agp I know that melting point is the transition between liquid and solid phases. What I was saying was that it is defined as the point where the liquid and solid forms have identical vapor pressures. For reference, this is from pg. 827 of Chemical Principles 7th ed. by Zumdahl/Decoste.

@swenger Thanks for the reference. I can see how at the melting point, liquid and solid are in equilibrium, so maybe it follows that they would have to have identical vapor pressures at that temperature. But at a lower temperature, you only have the solid phase present, so how can you measure the vp of the liquid form?

@iad22agp I think you can measure the vp of the liquid below its normal melting point by changing the external pressure. Here's a diagram showing the vp of ice and water as the temperature approaches 273K. link


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Reaction of 1-(1-chloroethyl)-1-methylcyclobutane under E1 Conditions

What is the major product of the following reaction?
$\hspace{40 mm}$
$\hspace{20 mm}$

It looks to me that the $\ce{Cl}$ group leaves then there's an alkyl shift to where the $\ce{Cl}$ group leaves in order to form a tertiary carbocation. Then $\ce{EtOH}$ attacks the beta hydrogen and forms a double bond to give product B:
$\hspace{35 mm}$
However, the answer I came up with was incorrect. Any ideas why?

It's a good attempt, but as I said there are quite a number of alkyl groups you could migrate - you might want to try to list all of them and see why each of them may or may not be so favoured.

Before solving this problem, you should know the migrating aptitude of groups. https://en.wikipedia.org/wiki/Migratory_aptitude

I've thought of another one I guess. I'm pretty confident nobody actually mentioned it as something that happens in my ochem class but could the ring expand with an alkyl shift?
So it would create product A. Problem is why would A be favored over B? Both involve a tertiary carbocation... is the carbocation in a ring more stable or something?

I would say it is because there is less ring strain in A than in B.

You are correct that an alkyl shift will take place but there are two possible options for which group moves. Your option is possible but there is another option which involves a ring expansion to generate product A.

This mechanism is favoured because it relieves the ring strain in the four membered ring. Since the reaction is run under reversible conditions, the thermodynamic product (with the least ring strain) will dominate.

The migration of the primary alkyl group is also favoured over the methyl group since it can stabilise the positive charge in the TS better.


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Odd Acidity Trend in Hexaaqua Metal Complexes
In the following link (three quarters down the page) it states that theoretically there should be a proportional relationship between the ionic radius and pKa of a hexaaqua metal complex (provided that the charge of each metal is the same).
The website's reasoning for this is that due to the smaller ionic radius, there is a smaller volume meaning that there is a greater charge density. Therefore this will cause a greater distorting effect on the electrons in the $\ce{O-H}$ bond. Hence this will enable the $\ce{H}$ to be deprotonated more easily, resulting in a lower pKa value.
However it later says that experimental data don't fit this relationship shown by one of the graphs in the website:

Obviously $\ce{Cu^2+, Cr^2+}$ and $\ce{V^2+}$ don't follow this relationship at all. The author of the website states that he has no idea why this happens.
I was wondering if anyone here could provide a more in depth explanation on how ionic radius effect acidity which perhaps can explain why these metals don't follow this linear relationship.

To start off with, there are two graphs on the site you linked, the second one being this one:

This, as the author of your site also states, is a much nicer graph in that it explains all but two ions ($\ce{Cr^{II}}$ and $\ce{V^{II}}$) well.
I also went around and checked different crystal structures of hexaaquametal(II) complexes for their $\ce{M-O}$ bond length to see if that may have any influence. The results are:

Vanadium: $212~\mathrm{pm}$ (octahedral)
Chromium $203~\mathrm{pm}$ (axial) and $245~\mathrm{pm}$ (equatorial; Jahn–Teller distorted octahedron)
Manganese: $214$ to $217~\mathrm{pm}$
Iron: $211~\mathrm{pm}$ (almost octahedral)
Cobalt: $208$ to $213~\mathrm{pm}$ (distorted octahedron)
Nickel: Did not find data
Copper: $200~\mathrm{pm}$ (axial) $220~\mathrm{pm}$ (equatorial)

On the lower end of this spectrum we find copper and chromium which indeed have rather low $\mathrm{p}K_\mathrm{a}$ values. This could be due to a shorter $\ce{M-O}$ bond meaning that the water molecules experience the charge stronger. (This is also why I favour the graph I put in this answer: It puts copper closer to ‘where it should be’ in ionic radii terms.)
A second thing to consider is the ligand field stabilisation enthalpy LFSE. This is strongest ($-12~\mathrm{Dq}$) for vanadium since that ion only has three electrons which all occupy the lower energy levels. This would mean that ligand exchange for $\ce{[V(H2O)6]^2+}$ is kinetically strongly inhibited especially for vanadium, since any loss of a ligand would need to overcome the stabilisation of $-12~\mathrm{Dq}$. Hence, the water molecules are strongly bound to vanadium and experience its charge all the time. On the other hand, copper has rather weakly bound ligands exemplified in how simple it is to create $\ce{[Cu(NH3)4(H2O)2]^2+}$, the deep blue tetraammincopper(II) complex.
(Note: I did not find any papers confirming or disputing this hypothesis, but I only spent a short time looking.)
Right after vanadium we encounter chromium. It only has an LFSE of $-6~\mathrm{Dq}$ — which is still pretty good, considering it all derives from unpaired electrons. Overall, I think this ion can be considered to be almost like vanadium(II) albeit a little smaller, with an extra electron and a shorter metal-water bond length which would explain its position.
Next up is manganese which looks as if it were the exception. But it actually isn’t. First off, its LFSE is $0$, thus water molecules can diffuse on and off as they see fit, therefore overall weakening the effect the ion’s charge has on them.
From iron onwards, we can again consider the water ligands bound slightly more tightly since there is again LFSE to be gained. However, remember that while the $\mathrm{t_{2g}}$ orbitals are non-bonding with respect to metal–ligand σ interactions, the $\mathrm{e_g}$ level is antibonding. We have a half-filled $\mathrm{e_g}$ level which will weaken the overall $\ce{M-O}$ bonds. This could not only explain the larger $\ce{M-O}$ distances of manganese and iron when compared to chromium but also their weaker acidity. (A weaker bond would also mean weaker charge transfer.) I suspect that this does not yet really matter for chromium due to its Jahn–Teller distortion being somehow favourable overall, and that chromium’s acidity stems from the closer water molecules.
From then on, the explanation seems rather simple, since the ion radii and the $\mathrm{p}K_\mathrm{a}$ are almost in a linear correlation. The question is how the ionic radii were deduced (which is none-trivial) and why they differ so strongly with the two sources the author of your linked site used.

I'll be honest. I don't completely know. So this is pure speculation.
My initial instinct was be that $\ce{[Cr(H2O)6]^{2+}}$ and $\ce{[Cu(H2O)6]^{2+}}$ have Jahn-Teller distortions. Obviously some of the other complexes do too, so that may be only a small part of the explanation.
More importantly, I suspect, is that $\ce{V^{2+}}$ and $\ce{Cr^{2+}}$ have very high enthalpies of hydration relative to their expected ligand-field stabilization energies. That is, the "expected enthalpies" (without LFSE) is nearly linear from $\ce{Ca^{2+}}$ to $\ce{Zn^{2+}}$, but when you subtract the LFSE, you see V, Cr, and Cu as "unexpectedly stable" (from Johnson and Nelson, Inorg. Chem., 1995, 34 (22), pp 5666–5671):

So if I were picking three transition metals that have high enthalpies of hydration relative to their LFSE, it's those three. In short, the water complexes are fairly stable, influencing the ability to pull a proton off the aqua ligands.


text:
Deciding E1/E2/SN1/SN2 for cyanide ion
The $K_\mathrm{a}$ for $\ce{HCN}$ is $6.17 \cdot 10^{-10}$ with a $\mathrm{p}K_\mathrm{a}$ of $9.21$
Thus, the cyanide ion is a strong base.
Also, the cyanide ion is a good nucleophile.
So in the reaction of alkyl halides with $\ce{KCN}$, a mixture of products must be formed depending on the solvent and alkyl group.
However my text suggests that the reaction  proceeds only via $\mathrm{S_N2}$.
Shouldn't the product formed depend on the alkyl group, say for a tertiary alkyl group, the reaction can go via $\mathrm{S_N1}$ mechanism for polar protic solvents and via $\mathrm{E2}$ for polar aprotic solvents and as the cyanide ion is strong base, the reaction is unlikely to go via $\mathrm{E1}$ and due to steric factors $\mathrm{S_N2}$ is also ruled out.
Also is there any  difference if we use $\ce{KCN}$ or alcoholic $\ce{KCN}$?

"Cyanide ion is a strong base" - are you really sure about that? What is the definition of a strong base?

@orthocresol HCN is a weak acid.So its conjugate base cyanide ion should be a good base.But i think you are pointing to the resonance between N and C that lowers its basicity.

"The conjugate base of a weak acid is a strong base." No matter what your teacher or textbook says that is unequivocally wrong. Get it out of your head.

Yasir sometimes you have to accept that what you're being taught is what you should stick to. That oversimplification is another example. Cyanide ion isn't a strong base, period.

So what about the statement that "The conjugate base of a weak acid is strong base"?

It's wrong.

@yasir Usually, a single counterexample is enough to invalidate a claim. Counterexample one was $\ce{CN-}$, which is a weak base. Counterexample two would be $\ce{NH3}$ which is a weak base and whose conjugate acid, $\ce{NH4+}$ is a weak acid.

what about the leaving group nature for cyanide ion? Now since you clarified it's a weak base, it should be a good leaving group, as the charge would be dispersed.

@yasir Not all weak bases have dispersed charges, and nucleofugicity depends on other factors, too. If cyanide were a leaving group, it would cleave a carbon-carbon bond which is a bad thing to do, energetically.

@Jan , I don't agree. Please see the last step of Benzoin condensation.The reason why Cyanide ion is used in this reaction is because it's a good nucleophile and leaving group.
See Here

@yasir This is the special case of addition/elimination from $\mathrm{sp^2}$ carbons where generally slightly different rules apply. Nucleophilic displacement of $\ce{CN-}$ via an $\mathrm{S_N2}$ mechanism is considerably harder.

A general method for estimating $\mathrm{p}K_\mathrm{b}$ values is to use the equation $\mathrm{p}K_\mathrm{b} + \mathrm{p}K_\mathrm{a} = 14$ where $\mathrm{p}K_\mathrm{b}$ is that of the conjugate acid; in this case, applying it gives us $\mathrm{p}K_\mathrm{b}(\ce{CN-}) \approx 4.79$. We can generally assume that ‘strong’ and ‘weak’ in $\mathrm{p}K_\mathrm{b}$ terms are similarly to be defined as for $\mathrm{p}K_\mathrm{a}$ values. Therefore, cyanide is a weak base and the premise of this question is incorrect. This observation is consistent with the fact that neutral cyanide solutions are relatively safe as long as no acid is introduced to protonate $\ce{CN-}$.
Note that the general statement ‘the conjugate base of a weak acid is a strong base’ is incorrect. The opposite is often true, namely:

The conjugate base of a strong acid is generally a weak base.

And vice-versa if you swap all occurences of acid and base. The word generally is required here, since compounds such as $\ce{H2Cl+}$ are very strong acids whose conjugate bases are still very strong — but these compounds are not typically easily accessable and certainly not in aquaeous solution. A weak acid needs to have $\mathrm{p}K_\mathrm{a} > 15$ for the conjugate base to be a strong one but those are seldomly still considered ‘acidic’. An example would be water ($\mathrm{p}K_\mathrm{a} = 15.7$ therefore $\mathrm{p}K_\mathrm{b}(\ce{OH-}) \approx -1.7$) which is a weak acid with a strong conjugate base.
Since it is such a weak base, it has no access to $\mathrm{E2}$ pathways and $\mathrm{E1}$ pathways are disfavoured against $\mathrm{S_N1}$ if either is applicable. We don’t even need to consider $\mathrm{E1_{cb}}$. Thus, cyanide will generally react via an $\mathrm{S_N2}$ pathway unless the conditions for $\mathrm{S_N1}$ are much better.

Can you address the last part about KCN and alcoholic KCN? Do they alter the likelihood of an SN2 reaction?

@yasir Refer to ringo’s answer for the alcoholic bit. ^^

@Jan Will substitution take place or elimination in reaction of NaCN with t butyl bromide. Some sites say it will be E1 whereas according to you SN1 will be major product

By definition, a weak base or weak acid is something that does not completely ionize in solution. Consider acetic acid, which is weak acid, and has a $\mathrm{p}K_\mathrm{a}(\ce{CH3COOH}) \approx 4.756$. Now consider that $\mathrm{p}K_\mathrm{b}(\ce{CN-}) \approx 4.79$. This means that acetic acid ionizes even more than the cyanide ion does, though it is still a weak acid. Clearly then, the cyanide ion is a weak base, and cannot react with haloalkanes via $\mathrm{E_1}$ or $\mathrm{E_2}$ pathways.
What I assume you mean when you ask about the differences between $\ce{KCN}$ and alcoholic $\ce{KCN}$ is the use of alcohol as a solvent. A polar aprotic solvent (like DMSO or acetone) will greatly increase the the nucleophilicity of $\ce{CN-}$, because they effectively solvate $\ce{K+}$. They do not solvate anions like $\ce{CN-}$ as well though, meaning it will be relatively free to react with the haloalkane. A polar protic solvent (like an alcohol or water) will decrease the nucleophilicity of $\ce{CN-}$ because it solvates it, and thereby reduces its ability to participate in a reaction.
For your reference, the relative rates of reaction of 1-bromobutane with azide happens 1300 times faster in DMSO than in methanol$^1$. Although azide is not the same molecule as cyanide and this is only data for a primary haloalkane, the trend should be clear.
1 - Brown, W., & Iverson, B. (2014). Organic chemistry (7th ed., p. 359)


text:
Separate Dextrose & Caffeine in water?
Can a dextrose / caffeine powder be (mostly) separated by dissolving in water then filtering the water?
Dextrose is readily soluble in water (1-10 parts solvent to 1 part solute).
Caffeine anhydrous is sparingly soluble in water (30-100 water : 1 caffeine).
I'm assuming water would dissolve nearly all the dextrose (and a bit of caffeine), and upon reaching saturation, the caffeine will remain out of solution in solid form and easily filtered.
Repeating the process would refine (increase) the quantity of caffeine vs. dextrose in the remaining filtered solids.
Thoughts?
References:
Caffeine Anhydrous - BASF
Dextrose - NBS

Yes, exploiting differences in solubility is indeed one of the easiest ways to separate compounds - however in practice I suspect a chemist would use more "sophisticated" methods (not sure exactly what, maybe column chromatography) since dissolving it in water leads to quite a bit of product loss.

Non - polar solvent extraction  would be a good method.

@Joel any household ingredients that could do that?

@Baker , I don't know what you have at your house! I suggest to study the chemistry involved in a non-polar extraction of alkaloids then you should be able to sort it out...


text:
Is heating oil identical to fuel oil?
I don't quite understand the categorization of middle distillate petroleum products.
Is heating oil the same thing as fuel oil which is the same thing as diesel, or not? Can someone explain the differences (if any) between the common middle distillates?

Diesel is identical to the stuff sold in Germany (and probably others) as ‘heating oil’ but they are coloured differently for tax reasons. (You’re not allowed to drive on heating oil because the country wants its fuel taxes.)

In the USA, it is divided into 3 categories by sulfur content, <15ppm, 15-500ppm and >500ppm, with newer vehicles being required to use <15ppm and home heating oil permitted to have the highest sulfur content (though states can further restrict)

Home heating oil and motor diesel are very similar in terms of density and chain length (basically same stuff) with minor differences in sulfur content (at different eras in different countries) and with colorant for tax purposes.  [Similar to comment above.]
In general the middle distillates would have a bit more sulfur in them if just based on distillation cut.  In order to meet lower sulfur requirements for diesel, additional processing is needed in the refinery to get the sulfur out.  As the US (and other countries) started putting in stricter and stricter sulfur requirements for diesel (for pollution reasons), it has had the impact of driving diesel prices up to be more than gasoline--traditionally it was cheaper.
[Note also that, at least in the US, it is possible to buy untaxed, undyed diesel for use in farm machinery.]
Note that "RFO" or residual fuel oil is a very different beast from home heating oil.  It is much heavier and higher boiling point fraction in the column.  It also typically has MUCH more sulfur in it (from concentration, the sulfur containing compounds tend to be heavier).  RFO is used in large onshore power plants (declining in the US but still some remain and many overseas) and also in ship power plants ("bunker fuel").  Ship use is both in steam boilers and in "marine diesels".  Larger shipboard diesels have evolved to become larger and larger, competing with steam turbines, and they are designed specially so that they can use a more crappy fuel (don't need motor diesel). RFO is also often burned (for heat exchangers and/or cogen) in island refineries (that lack access to natural gas).
Note that shipboard use of RFO is expected to decline after 2020 because of new restrictions on open ocean fuels (the "IMO law").  RFO burning is already prohibited in close proximity to North American and European coastal waters. The issue is sulfur content.  Ships can continue to burn the RFO if they install onboard sulfur scrubbers, but current indications are most will just use a lower sulfur fuel.  The easiest way to do this is to use something close to a motor diesel (rather than desulfurize RFO itself).  Will have a little more sulfur than most land diesels, but still somewhat desulfurized...and just because it is a middle distillate "cut" inherently has much less sulfur than RFO (which collects the nasties when distilling).  Also there are still questions about enforcement (i.e. cheating) or even legality of regulating open ocean operations.  But overall impact will be a big drop in prices for RFO as market goes away and more pressure on middle distillate prices.  [Ships can also shift to LNG but this is even trickier than scrubbers because, in addition to engine modifications, the infrastructure to bunker LNG does not exist in most ports.  But perhaps over time, we will see migration to LNG, a zero sulfur fuel.]
A decent reference is here:  https://www.amazon.com/Petroleum-Refining-Nontechnical-Language-William/dp/1593701586 It is a few years old so won't have the IMO discussion but will give you a good overview on the whole situation and is VERY friendly in terms of plain language.  I bought it before going into the industry and worked through the whole book in a few days...totally helps with some initial grounding in things.  After that you can fill in the blanks with whatever you are doing (marketing, buying, refining) by just asking for specs.
See also: https://www.eia.gov/tools/glossary/index.php?id=petroleum EIA glossary--you may also find some other info if you scratch around that website...a lot of pages and FAQs are a little buried.

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


text:
Calculate the radius of an Ag atom
The question I have been given is:

Silver atoms in a metallic lattice only fill up $88\,\%$ of the space ($12\,\%$ is empty). The density of silver is $10.5\ \mathrm{g\cdot cm^{-3}}$. Assuming that silver atoms are hard spheres ($V=\tfrac43\cdot\pi\cdot r^3$, when $r$ is atomic radius), what is the radius of a silver atom? Give the answer in units of $10^{-12}$ meters.

The atomic mass of $\ce{Ag}$ is 107.8682.
My solution:
$$V=0.88\times V$$
$$V=\frac{0.88\times10.5\times6.022\times10^{23}}{107.8682}=5.158\times10^{22}\ \mathrm{cm^3}$$
$$V=\frac43\cdot\pi\cdot r^3 \Rightarrow r=\left(\frac34\cdot\frac V\pi\right)^{1/3}$$
Then I switched to the $10^{12}$ meters, the result was $4.953\times10^{17}$ and it is not correct. What am I doing wrong?

I've added the information about the atomic mass of $\ce{Ag}$ in an effort to clarify for you and others what information you'll need in order to do the problem.

actually Ag crystallizes in FCC and the spheres fill up $$\dfrac{\pi}{3\sqrt{2}} \approx 0.74048$$

If you had included the units in your calculation, you would have noticed why your equation is not correct.
Molar mass $M$ is defined as
$$M=\frac mn\tag1$$
where $m$ is mass and $n$ is amount of substance.
Since the Avogadro constant $N_\mathrm A$ is
$$N_\mathrm A=\frac Nn\tag2$$
where $N$ is the number of particles, the mass $m$ of one atom $(N=1)$ is
$$m=\frac M{N_\mathrm A}\tag3$$
Density $\rho$ is defined as
$$\rho=\frac mV\tag4$$
where $V$ is volume.
Thus, the volume of a sample is
$$V=\frac m\rho\tag5$$
Using Equation $\text{(3)}$, the volume $V$ can be calculated for a single atom:
$$V=\frac M{N_\mathrm A\cdot\rho}\tag6$$
Assuming that a fraction of $88\,\%$ of the volume $V$ is filled with a hard sphere, the volume $V_\text{sphere}$ of the sphere is
$$\begin{align}
V_\text{sphere}&=0.88\times V\tag7\\[6pt]
&=0.88\times\frac M{N_\mathrm A\cdot\rho}\tag8
\end{align}$$
Since the volume of a sphere is
$$V_\text{sphere}=\frac43\pi r^3\tag9$$
where $r$ is the radius of the sphere, the radius $r$ is
$$\begin{align}
r&=\sqrt[3]{\frac{3V_\text{sphere}}{4\pi}}\tag{10}\\[6pt]
&=\sqrt[3]{\frac{3\times0.88\times M}{4\pi\cdot N_\mathrm A\cdot\rho}}\tag{11}\\[6pt]
&=\sqrt[3]{\frac{3\times0.88\times 107.86820\ \mathrm{g\ mol^{-1}}}{4\pi\times
6.02214076\times10^{23}\ \mathrm{mol^{-1}} \times 10.5\ \mathrm{g\ cm^{-3}}}}\\[6pt]
&=1.53\times10^{-8}\ \mathrm{cm}\\[6pt]
&=1.53\times10^{-10}\ \mathrm m\\[6pt]
&=153\times10^{-12}\ \mathrm m\\
\end{align}$$


text:
Gibbs free energy in chemical reaction
I know this kind of question has been asked many times but I think this is a bit different.
We have $G = H - TS$. Here I am talking about chemical reactions at constant pressure and temperature. We can write $\Delta G = Q - Q_\mathrm{rev} = \Delta H - T \Delta S$.
For an irreversible chemical reaction, for example an exothermic one, if $q = -10~\mathrm{kJ~mol^{-1}}$ (exothermic), then $Q_\mathrm{rev}$ could be $-9$, $-8$, $-7$, etc.  The difference $(Q – Q_\mathrm{rev})$ would be negative ($-1$, $-2$, $-3$, etc.).  This represents the portion of $Q$ that can be converted into non-PV work.  The remainder of the heat is dispersed into the surrounding and is wasted. Right?
So here $Q$ as absolute value is more than $Q_\mathrm{rev}$, so we have more heat generated than a reversible chemical reaction that can be used as work ($Q = 8~\mathrm{kJ~mol^{-1}}$ and $Q_\mathrm{rev} = 2~\mathrm{kJ~mol^{-1}}$). So why is it said that a reversible reaction produces more useful heat? Again, I am talking about chemical reaction and not mechanical process.

Can you cite a reference for the equation $\Delta G = Q - Q_\mathrm{rev} $?

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwiI2K7HzJ_JAhXL1hoKHTkmANUQFggiMAE&url=http%3A%2F%2Fwww.usna.edu%2FUsers%2Fchemistry%2Fmorse%2F_files%2Fdocuments%2FSC112-Chapter18%2Foptional18.doc&usg=AFQjCNHwJqyIG_IR8g7YSAphCqQxhqFYQQ&sig2=8n2Uv6uPvkD3kkVBsSlSzw&bvm=bv.108194040,d.d2s
Actually I found it on google and it is from www.usna.edu (a doc file). It is logical that ΔG=Q−Qrev since ΔS = Qrev/T

Can you explain why you find it logical, given that $\Delta G$ and $\Delta S$ are functions of state, while Q is not a function of state?  Can you also provide the exact web site address?

Hello, it is a doc file inside the website which is downloaded automatically when you press on the link. The original website is www.usna.edu (US navy academy)

You can search it on google type: optional18 + usna.edu

In my judgement, the development in the usna document that led to the relationship  you cited is fallacious at many levels.  This is not a peer reviewed document, and I don't think you can find this equation in any peer reviewed literature.  I invite you to search the peer reviewed  literature to see if you can.


text:
Transition Metal Compounds
So I have a few dumb questions. I'm still in high school and am doing something on magnetic properties of transition metals. I'm facing a few roadblocks.
1) What're compounds like $\ce{CoBr2, CoCl2, AgI, AgBr, AgCl, AgF, MnBr, Co(CN)2}$, supposed to be? Are they the same as the coordinate complexes in something like $\ce{[Fe(H2O)6]}$?
2) If they're not, how do they bond with halogens like $\ce{Cl}$ or ligands like $\ce{OH/CN}$? Is it similar to ligand interaction and d-orbital splitting?
3) If they're neither complex ions nor have split d-orbitals, what determines their magnetic properties?
I'm having a bit of trouble finding the answers. I'm very new to this whole transition metal thing. On some websites compounds like ferrous bromide is considered a and on another it's b. I'd appreciate any help!

What’s your education level?

Sadly still in highschool. We have yet to take this and I'm sorry if I asked something embarrassingly dumb. But my teacher's of very little help and I'm in a bit of a bind.

In that case, I totally understand your confusion. You might want to add that to the question; there's an [edit] button on the bottom left below the tags.

It would have to be something about their electron configurations having two singly-occupied orbitals. I'm not too familiar with MO diagrams for these compounds though. But yeah, (para)magnetism generally has to do with having singly-occupied orbitals, if I remember correctly from my days in I-Chem.

"On some websites compounds like Ferrous Bromide is considered a and on another it's b." What is a and b?

What're compounds like $\ce{CoBr2,CoCl2,AgI,AgBr,AgCl,AgF,MnBr,Co(CN)2}$, supposed to be?

That depends.
In a lot of cases, especially if we are talking about large anions with low charge density, such as nitrate and perchlorate, and moderately small cations, like $\ce{Co(II)}$, obtaining an anhydrous salt is a long, tiring quest, if possible at all. For example, anhydrous metal nitrates for small cations in general are obtained in anhydrous solvents only.
In case we are talking about medium anion with moderate charge density, like $\ce{Cl-}$ and the cation binds with its neighborhood strongly, like $\ce{Cr^{3+}}$, a lot of structures with different hydration degree may be obtained, such as
$\ce{[Cr(H2O)3Cl3], [Cr(H2O)4Cl2]Cl, [Cr(H2O)5Cl]Cl2, [Cr(H2O)6]Cl3}$ in addition to anhydrous form $\ce{CrCl3}$ (and again, obtaining the anhydrous salt from water solutions is virtually impossible)
Sometimes a salt is unsolvable in water. This typically means that it is anhydrous, like silver halides and many sulfides, or at least low-hydrated, like $\ce{Cu(CH3COO)2\cdot H2O}$
In some cases a salt may have even more weird structure. For example, $\ce{MoCl2}$ actually contains $\ce{[Mo6Cl8]^{4+}}$ unit with $\ce{Mo}$ atoms forming an octanedron and $\ce{Cl}$ atoms occupying positions on top of each of the octahedron's faces. This unit is rather stable and the remaining chlorine my be exchanged with the unit intact.
Similarly, a series of oxoacetates is known, built around $\ce{M3O}$ unit. And so on....
Unfortunately, most transition metals are rather unique.  There is a strong vertical similarity between second and third rows of d-elements, but even it is not perfect.

If they're not, how do they bond with halogens like Cl or ligands like OH/CN? Is it similar to ligand interaction and d-orbital splitting?

Typically yes.

If they're neither complex ions nor have split d-orbitals, what determines their magnetic properties?

d-orbital splitting happens for any atom with said orbtials in neighborhood with symmetry below spherical, i.e. in all chemically viable cases.

Whenever you are looking at a substance that contains a transition metal you are always looking at some kind of coordination compound and the d-orbitals will split in some way. Generally, all chemistry with transition metals should be discussed with d-orbitals first and other things brought into consideration later. (The same is true for non-transition metals except for the d-orbitals part: They, too, will form complexes of some sort to stabilise their negative charges. However, their ions are usually if not always diamagnetic so do not contribute to magnetism.)
Finding out what an exact case is, is non-trivial. Let’s take chromium(II) acetate monohydrate $\ce{Cr(CH3COO)2 . H2O}$. We first need to find out how many d-electrons are present altogether. Since we have chromium(II) and chromium is in the 6th group, we arrive at $\mathrm{3d^4, 4s^0}$. (General rule: Empty the higher s-orbital immediately, if you haven’t yet reached the charge, remove from the d-orbitals.) $\mathrm{d^4}$ means that all four electrons could pair up for certain d-orbital splittings but they also could form a high-spin complex with four unpaired electrons or a low-spin one with two (see figure 2 below). We next need to know about the structure. Searching on Wikipedia, we find the following image (figure 1):

Figure 1: Dimeric structure of $\ce{Cr(CH3COO)2 . H2O}$
Chromium (the silverish/grey balls) obviously has an octahedral environment. For octahedral environments, as you may know, we get a certain d-orbital splitting which can lead us to one of the following pictures (figure 2):

Figure 2: d-Orbital splittings in vacuum, high-spin and low-spin environments.
Our ligands (acetate and water) are low-field ligands, thus the picture in the complex is better described by the middle, high-spin splitting than by the right-hand, low-spin one. We could stop here and assume chromium(II) acetate to be paramagnetic and thus, maybe magnetic, depending on whether it exhibits ferromagnetic, ferrimagnetic or antiferromagnetic coupling. But we shouldn’t. Because if we measure this complex we find it to be diamagnetic. Erm, what happened? Well, you see a chromium-chromium bond between the two metals in figure 1. In the end, it turns out, that the unpaired electrons make up a quadruple bond. If we assume the $z$-axis to be along the chromium-chromium axis we get:

A σ bond between the two $\mathrm{d}_{z^2}$ orbitals;
two π bonds between the $\mathrm{d}_{xz}$ and the $\mathrm{d}_{yz}$ orbitals; and
one δ bond between either the $\mathrm{d}_{x^2 - y^2}$ or the $\mathrm{d}_{xy}$ orbitals (physically equivalent).

Every bond pairs up a single electron of the original paramagnetic chromium(II) to give an overall diamagnetic state.
You realised how complex the analysis is for a seemingly simple compound. Most importantly, it is very hard if not impossible to deduce whether a compound will have magnetic property $x$ a priori — unless you arrive at the conclusion that there are no d-electrons and the metal must thus be diamagnetic.

In a nutshell, the answer to your questions are:

Always. But it is non-trivial to distinguish what kind of complexes they would be.
Well yes, but that is already given by a yes to question 1.
See above.

Unfortunately, there is no simple answer.


text:
Specific Heat Capacity of Heavy Water
Water has an unusually high specific heat capacity due to it's hydrogen bonds.
Why is it that the change in isotope causes a 10% difference in the heat capacity? How does the added neutron in the nucleus make any difference to the bonds?

Googling 'heat capacity of heavy water' yields various scholarly reports dating back to the 1930s.

True, but the ones I've looked at are extremely long with lots of equations I can't make sense of, so an understandable answer would be nicer if possible... Should I edit the question?

The Wikipedia page (3rd hit for me from Google) on 'Orders_of_magnitude_(specific_heat_capacity)' lists Water as 4186 J/kg/K and Heavy water as 3767 J/kg/K - so it makes about a 10% difference.

Thank you! I just realised I messed up and the edit I put in after your first comment didn't show up. I've improved the question now, I hope it's better!

The added mass of a deuteron relative to a proton reduces the O-D stretch vibration frequency relative to O-H. This reduces the energy that can go in to that degree of freedom, reducing the heat capacity.

I agree with Jon Custer. A far I know the added neutron will not change the Schrodinger equation. So electronic contribution is same for them but mass has effect on vibrational frequency. You can also look for KIE (kinetic isotope effect) in the standard computational chemistry literature.

Specific heat capacity // H2O: 74.539 J mol-1 K-1 // D2O: 84.42 J mol-1 K-1 // http://www1.lsbu.ac.uk/water/water_properties.html // So "heavy water" takes more energy to heat than "normal" water.

@JonCuster I may be wrong, but around room temperature, do vibrational degrees of freedom really contribute to heat capacity? Additionally,  I think the increased mass of the deuterium atoms would increase the energy stored in the rotational degrees of freedom, which would actually increase the heat capacity of $\ce{D2O}$, as suggested by MaxW's link (which I trust more, given the background of the site).

The values on Wikipedia seem to be the same on a per mol basis, which leads me to believe someone just took the same molar heat capacity for both as an approximation, and converted to the specific heat capacities according to their respective molar masses masses

The heat capacity per mole is higher for heavy water. The hydrogen bonding is stronger.

I posted an answer here the other day which was completely wrong! I apologise and have left the incorrect text below for posterity to record how wrong people who should know better can be!
Looking on the NIST fluid properties database I see that the specific heat capacity (i.e. the heat capacity per unit mass) of D2O is just 1.5% higher than the specific heat capacity of H2O. So at 298 K this makes the molar heat capacity of D2O (84.963 J/K/mol) an astonishing 12.7% higher than H2O (75.38 J/K/mol). I am astounded and have no explanation.
Everything below this line is WRONG!
The molar heat capacity is almost exactly the same. i.e. it takes the same energy to heat the same number of molecules by one degree Celsius.
The difference in the *specific* heat is just because of the difference in density. D2O molecules have a relative molar mass of about 20 compared with 18 for H2O. Since the bond lengths are similar, the density is higher by a factor 20/18 - about 11% higher. So a given mass of D2O has about 11% fewer molecules in it.

According to this page http://www1.lsbu.ac.uk/water/water_properties.html the molar heat capacity of D2O is more than 10% higher than H2O. Are these values wrong then?

I will check with some independent sources and get back to you. The 'internet' seems to have specific heat capacities i.e. per unit mass that are within 1.5% of the light water value (e.g. http://hedhme.com/content_map/?link_id=22146) . That is the level of change I would have expected for the molar heat capacity. It seems suspicious to me that the molar heat capacity should change by almost the factor as the density. As it happends, this answer really matters to me (!) but I think I need more reliable sources than are openly available on line.

Hi. I was completely wrong. I apologise. I have changed the original post to reflect that.


text:
Does Beta-ionone posses the same anti-ageing properties as Retinoids if applied topically to the skin?
I read a scholarly article on oxidisation and collagen structure in the dermis and they mentioned the use of Retinoids topically. so I looked into retinoids (admittedly on Wikipedia) for an overview (Vitamin A and Retinoic acid, retinol etc) in the possible use on anti-ageing, which happened to state that carotenes are apparently "metabolized to β-ionone, and thus have vitamin A activity because they can be converted by plant-eating animals to retinol and retinal. Carotenoids that do not contain the β-ionone moiety cannot be converted to retinol, and thus have no vitamin A activity."
I cannot find anywhere by googling this, and as I am an amateur perfumer, not a chemist, with a genuine curiosity for an innocuous bottle of Beta Ionone that I use for fragrancing. I would be really grateful if someone could point out in layman's terms what is meant by "vitamin A activity" and could explain-
If using Beta Ionone on the skin it can possess the same qualities of the retinoids that are sold as anti-ageing products?
P.S I realise that beta-ionone is a possible cause for irritation on skin in certain amounts so this is not the question of safe amounts, it is the reaction in terms of the aforementioned that I am interested in.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] to learn more about it.


text:
Do filter-specific dyes exist? Equivalently, do dye-specific filters exist?
Apologies if this isn't chemistry enough for chemistry.se. I'm happy to go elsewhere if there are suggestions.
Blue + Red = Purple is arbitrary here. If any different colors might make this more feasible, please say so (preferably with some explanation as to why).
I'm interested in having a jug of purple liquid.
When I turn it upside-down, it will empty through two bottlenecks into two smaller jugs. I would like for one of the smaller jugs to now contain red liquid, and the other to contain blue. Eg, I would like for one bottlenecks to actively filter out the red/blue portions of the original purple liquid.
This process has to be repeatable.
That is, I should be able to turn this contraption back over on itself and the blue/red will recombine to be purple, and then for the separation of colors to happen again when it's flipped a third time. It's not very suitable for me if this system only works a few times before losing its punch, but it may be acceptable if I had to replace or clean filters once every 50 or 100 filtrations.
Looking forward to some affirmative answers, or even some suggestions on how to better search for this.

Can you give any context as to why you'd want to do this? As a classroom demonstration of actual science or for a different purpose?

I'd like to build one of these, but with color separation. https://www.youtube.com/watch?v=CAkMUdeB06o

Color has no direct importance in filtration and this question doesn't seem properly researched.

@Mithoron How about with the "illusion" of color filtration such as in the example of the YouTube video where those containers are clear mixed with blue liquid, you could have those be colored and the liquid be colored to just give this illusion rather than actually filtering out the chemicals, etc. within the liquid. In other words, a  "magic" trick, spooky... Boo!!!! Wahaa ha ha! Do some research on a magic trick of this nature and post back with an updated question with more detail. . . David Copperfield may be lurking around chimistry s.e. to provider further guidance. Hocus Pocus, kazam!!


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Nomenclature of Organic Compound from Condensed Formula
I am just unable to name $\ce{C6H5CH2C6H5CH2C6H5}$ using IUPAC nomenclature.
I tried searching online however I couldn't find a name for it.

The formula you have given is incorrect - the middle phenyl group should be C6H4. Furthermore, you didn't give any information about the substitution pattern on the middle ring. The current answer is just one of the possible isomers.

Related: What is the correct nomenclature for diphenyl methane?

Are you looking for this? 1,4-dibenzylbenzene
http://www.chemspider.com/Chemical-Structure.205991.html?rid=80e8a67d-0782-4794-bf27-db26ccd1ce39&page_num=0
What I did was go to their Structure tab and input what I thought to be the compound you typed. If this isn't it, you can probably draw what you think it is.
Hope this helped!


text:
Irreversible chemical reaction in a battery
"In reversible reactions, as the reactants react with other reactants to form products, the products are reacting with other products to form reactants."
http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Chemical_Equilibria/Reversible_vs._Irreversible_Reactions
Inside an ideal battery, the chemical reaction is reversible (at equilibrium), thus the oxydation-reduction reaction Zn + Cu2+ = Zn2+ + Cu is happening in both direction at the same rate right?
Now, for a real battery, there's resistivity inside it due to many factors, thus the chemical reaction is irreversible. My question is, during the irreversible reaction, is the reaction happening in both directions but at different rate due to resistivity inside the battery or it is just like the combustion reaction happening in one direction?


text:
Determine the concentrations of the three ionic forms of glycine present if 1.0 mol of glycine is used to make 1.00 L of aqueous solution

Determine the concentrations of the three ionic forms of glycine present if 1.0 mol of glycine is used to make 1.00 L of aqueous solution. $\mathrm{p}K_1 = 2.34, \mathrm{p}K_2 = 9.60$. Do you need to make any other assumptions to simplify the calculation?

Source: David Ball's Physical Chemistry.
I tried to solve this problem but I think it's quite flaw and naive. So I want to put it here if anybody have an improvement.
The question asked us to determine the concentrations of 3 ionic forms of glycine in aqueous solution. Let me call

$\ce{Gly}$ = $\ce{NH2-CH2-COO^-}$ at high alkalinity
$\ce{GlyH}$ = $\ce{NH3^+-CH2-COO^-}$ at neutral
$\ce{GlyH2}$ = $\ce{NH3^+-CH2-COOH}$ at high acidity

I can calculate the isoelectric point $\mathrm{pI} = (\mathrm{p}K_1 + \mathrm{p}K_2)/2 = 5.97$
If I assume "used to make $\pu{1.00L}$ of aqueous solution" means putting glycine into water, $T = \pu{25^\circ C}$, thus $\mathrm{pH}$ must be 7.00. However, the isoelectric point is 5.97. That means the solution cannot contain completely GlyH but also an $x$ amount of Gly.
Therefore, I use the Henderson-Hasselbalch equation to determine the amount of Gly and GlyH in the solution
$$
\begin{align}
\mathrm{pH} &= \mathrm{p}K_2 + \log\frac{[\text{Gly}]}{[\text{GlyH}]}\\
7 &= 9.6 + \log \frac{x}{1-x}\\
x &= 0.0025 M = [\text{Gly}] (\text{rounded})\\
[\ce{GlyH}] &= \pu{0.9975 M}\\
\text{Obviously } [\ce{GlyH2}] &= \mathrm{0 M}
\end{align}
$$
I think the wrong with my solution is to assume a fixed value of water pH. In common sense I think when glycine is put into water, it would drag the pH of the solution down a little bit, but not over pI, it means $\mathrm{p}K_1 < \mathrm{pI} < \mathrm{pH} < 7 < \mathrm{p}K_2$. But I really don't know how to find the value. Please guide me because I tried my best but really don't know how to solve this.

"[GlyH2] obviously is 0 M" - Why?

Because pH is higher than the pI. I don't think GlyH2 would be presented above pI.


text:
Why can mercury oxide be decomposed into mercury and oxygen upon heating?
As I am a grade 9 student, please use simple English that does not involve many technical words in Chemistry. Thank you.
I have learnt from my teachers' handout that physical methods of separation include heating, and that a compound can only be broken down into elements by chemical methods such as electrolysis.
But why can mercury oxide (a compound) be decomposed into mercury and oxygen upon heating (physical method)?
Or is it simply because heating is a chemical method? (that my teacher probably was mistaken)

Heating can be a chemical or a physical method.

In simple terms, heating is indeed a physical method. You might have heard of heat being, in fact, molecules/ions moving faster. There’s nothing in there per se that would make it a chemical method.
But. The atomic particles that are moving quicker can bump into each other with more energy (because they are faster — think of a car crash which is worse at high speeds); this can be understood as activation energy, i.e. energy that allows reactions to proceed.
So by adding enough heat energy, you are inducing a chemical change which leads to the (thermal) decomposition of mercury oxide into its elements.


text:
Irreversible chemical reaction in a battery
Inside an ideal battery, the chemical reaction is reversible (at equilibrium), thus the oxidation-reduction reaction $\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}$ is happening in both ways at the same rate right?
Now, for a real battery, there's resistivity inside it due to many factors, thus the chemical reaction is irreversible. My question is, during the irreversible reaction, is the reaction happening in both ways but at different rate due to resistivity inside the battery or it is just like the combustion reaction happening in one way?

The original poster has a serious misconception by equating a reversible chemical reaction with a reaction that is in equilibrium. The electrochemical reaction that allows the battery to deliver energy to the load (such as the copper/zinc reaction mentioned) is always reversible in theory by reversing the current flow. However in addition to the electrochemical reaction, there are other chemical reactions going on inside the battery which are not reversible as well as geometrical changes that may not allow the reactants to be close enough for the electrochemical reaction to run in reverse. In batteries designed to be recharged, these geometrical and irreversible chemical changes are minimized to allow as many recharge cycles as possible. Batteries not designed to be recharged therefore usually can't be recharged fully and have very few recharge cycles not to mention a high probability of gaseous or corrosive liquid leakage.
At equilibrium, both forward and reverse reactions are going at the same rate, but this is only discernable microscopically. At a macroscopic level you see nothing happening. So a battery is never at equilibrium while you are using it since at that point the battery is completely dead. Then (or preferably sooner) it is time to recharge the battery or throw it away.

Hello, Than for an ideal battery how can you explain that the potential difference remains constant? And for a real battery, during discharge does the reaction happen in both ways with the first way having higher rate? or only one way?

"In reversible reactions, as the reactants react with other reactants to form products, the products are reacting with other products to form reactants."
http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Chemical_Equilibria/Reversible_vs._Irreversible_Reactions

http://physics.stackexchange.com/a/126977/72921 See the answer here

The reaction you describe is what known in chemistry as an equilibrium reaction, symbolised by the double sided arrow:
$$\mathrm{Zn}+\mathrm{Cu^{2+}}\leftrightarrow \mathrm{Zn^{2+}}+\mathrm{Cu}$$
The equilibrium constant of the reaction, $K$, which is the ratio of the products of the activities of the species on the right and the products of the activities of the species on the left is fairly high which means the reaction tends to run more or less to completion from left to right. In a cell (or battery: multiple cells in series) that results in a flow of electrons from the Zn metal to the Cu electrode, resulting in a net positive EMF.
But by reversing the current by applying the reverse voltage over the electrodes the reaction can also be made to run from right to left. The original amounts of reactants and reaction products can thus be exactly restored, assuming the total amount of charge (current times time) is the same as the charge that was discharged during 'normal' use of the cell.
Internal resistance simply arises from resistance of flow to the charge carriers (electrons and solvated ions, in this case).

When current flows it carries ions away from the battery (electrolyte) interface where the reaction can take place. That is why the reactions then take place at a different rate in the forward and reverse directions: the concentrations of ions changes.

this is due to resistivity?

Not so much "due to" as "gives rise to" I would say. But you might get better answers on the chemistry stack.


text:
Formation of ferrous chloride or ferric chloride

Question
When $\ce{HCl}$ reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride. Why?
Answer. Its reaction with iron produces $\ce{H2}$.
$\ce{Fe + 2HCl -> FeCl2 + H2}$
Liberation of hydrogen prevents the formation of ferric chloride.

Now my question is why couldn't it be: $\ce{2Fe + 6HCl -> 2FeCl3 + 3H2}$ ?

Acids can oxidize iron since the redox potentials $E$ for $\mathrm{pH} = 0$ show that $\ce{H+}$ can oxidize $\ce{Fe}$ to $\ce{Fe^2+}$; but non-oxidizing acids cannot further oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$:
$$\begin{alignat}{2}
\ce{2H+ + 2e- \;&<=> H2}\quad &&E^\circ = +0.000\ \mathrm{V}\\
\ce{[Fe(H2O)6]^2+ + 2e- \;&<=> Fe}\quad &&E^\circ = -0.440\ \mathrm{V}\\
\ce{[Fe(H2O)6]^3+ + e- \;&<=> [Fe(H2O)6]^2+}\quad &&E^\circ = +0.771\ \mathrm{V}\\
\end{alignat}$$
However, oxygen can oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$:
$$\begin{alignat}{2}
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad &&E^\circ = +1.229\ \mathrm{V}\\
\ce{[Fe(H2O)6]^3+ + e- \;&<=> [Fe(H2O)6]^2+}\quad &&E^\circ = +0.771\ \mathrm{V}
\end{alignat}$$
Therefore, iron is oxidized to $\ce{Fe^3+}$ in dilute hydrochloric acid when in contact with air.

Raising the pH to 6.5-8 greatly increases the speed of oxidation by oxygen (see https://www.researchgate.net/publication/6438169_Morgan_B_Lahav_O_The_effect_of_pH_on_the_kinetics_of_spontaneous_FeII_oxidation_by_O-2_in_aqueous_solution-basic_principles_and_a_simple_heuristic_description_Chemosphere_68_2080-2084)


text:
Why does the mixing of sigma 2s and 2p orbitals lower the energy of the sigma 2s orbital and raise the energy of the sigma 2p orbital?
My general chemistry textbook (General Chemistry Principles and Modern Applications, Tenth Edition) says that for bonding in diatomic molecules with Z ≤ 7, orbital mixing occurs between the $\sigma_{2s}$ and $\sigma_{2p}$ molecular orbitals because they are close enough in energy. As a result, the energy of the $\sigma_{2s}$ molecular orbital is lowered and the energy of the $\sigma_{2p}$ molecular orbital is raised.
My question is why: if the $\sigma_{2s}$ orbital mixes with the $\sigma_{2p}$ orbital, wouldn't the modified $\sigma_{2s}$ orbital be higher in energy, as a result of mixing with an orbital slightly higher in energy?
Likewise, wouldn't the modified $\sigma_{2p}$ orbital be lower in energy as a result of mixing with an orbital slightly lower in energy? Why is it the other way around?

See also http://chemistry.stackexchange.com/q/14417/ and http://chemistry.stackexchange.com/questions/27287/

This is a frequent source of confusion in my experience - you'll see there are at least two other very similar questions here already.
Let's take the first point: why do the $\sigma_{2s}$ orbitals get lower in energy when "mixing" with the $\sigma_{2p}$?
I'm going to ignore "mixing" and use the word "interact" instead. If I consider the 1s orbitals in $\ce{H2}$, I take two 1s atomic orbitals and they (of course) interact. So one orbital goes down in energy, and one orbital goes up in energy.

This is a common theme - if there's any degree of overlap between two orbitals, one will become more stable and one will become less stable. In the diagram, we claim "bonding" and "anti bonding" but in more complex molecules, these terms become hard to identify (e.g. benzene).
The theme remains: if orbitals interact (i.e., have the same symmetry), some will go down in energy, some will go up in energy.
That's the answer to your first question. Since the $\sigma_{2s}$ orbitals do interact with $\sigma_{2p}$ the 2s will go down in energy.
Your second question is about the $\sigma_2p$ going up in energy.
I could answer with a formula from quantum mechanics, but let's argue qualitatively. Here's a diagram for two interacting orbitals:

This is much like our case with the 2s and 2p orbitals. We see that the result is a system that's most likely more stable - the orbital lower in energy is even lower because of the interaction.
If we consider the reverse case - that the 2s goes up in energy, and the 2p goes down in energy, we would have a system that's less stable. In theory, this could happen - if the interaction is negative. In other words, the system would be more stable without the interaction than with.
In short, the "mixing" or interaction between 2s and 2p orbitals occurs because the species becomes more stable.

So, the reason the σ2s orbital goes down in energy and the σ2p orbital goes up in energy has to do with the fact that 2s orbitals are more inclined to be lower in energy and 2p orbitals are inclined to be higher in energy?

I'm not sure I'd use "inclined," but yes - the 2s is lower in energy to begin with, so it gets even more stable because of the favorable interaction with the 2p.

But doesn't the fact that the 2p orbital increases in energy make it somewhat less favorable?

The question is whether the total energy is favorable.


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Heating paper to a yellowish smoke
I put some papers into a biscuit tin and put it over fire and baked it. Then I made a small hole at the top and saw yellowish smoke coming out. Any idea what the smoke could be?

Hi and welcome to chemistry.stackexchange.com. Since you already took the [tour] all that is left to point you towards the [help] if you want to learn more about how the site works. Your question is rather hard to answer, because there is a lot that can happen with paper in the heat. It may well get closed for being too broad. Please don’t let that discourage you from contributing more in the future.

What you've performed is, technically, "thermal decomposition of cellulose", or pyrolysis. During heating, there are large number of changes producing many different chemicals, so the smoke you saw was a mixture of different chemicals.
Pyrolysis is used to produce many industrial chemical, such as "wood alcohol" (methanol), and during gasoline shortages, to run internal combustion engines.

As an addition, because the image of the answer above may (wrongly) lead to the perception wood gas generation were old and thus an obsolete technology in the automotive industry, see post WWII examples on https://en.wikipedia.org/wiki/Wood_gas_generator.


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Would a redox reaction still happen between the two electrodes only connected by a wire?
The galvanic cell I'm studying has Pb as reducing agent and Cu as oxidizing agent, and as I understand it the reason that the electron flows from the Pb electrode to the Cu electrode without anyone doing anything is because it's thermodynamically favorable, ie a spontaneous reaction. Would this flow of electron still happen if I only have two electrodes only connected by a wire? This doesn't seem possible, but I'm not sure.

You need some kind of charge neutraliser, i.e. a salt bridge (through which ions can travel in the opposite direction). Otherwise you would end up with a positively and a negatively charged macroscopic area which isn’t possible.

Also you would need an electrolyte for the Pb and Cu electrodes to be immersed in.


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Does Chemical Formula of Matter Influence on how would it be Charged?
Firstly: Suppose we rub glass ($\ce{SiO2}$) against paper (which is made from cellulose $\ce{(C6H10O5)_{$n$}}$). We receive positively charged glass and negatively charged paper. Is that because of total ionization energy of molecule stronger than glass or crystal system too?
Secondly: When we rub atoms against other atoms, do electrons accelerate each other by the same electrostatic charge and change their orbitals to higher orbits due to synchrotron radiation they emit. Or is it because of the fact that protons are more close are ionizing electrons by electrostatic attraction? Am I correct?

Crossposted from http://physics.stackexchange.com/q/219862/2451

Firstly: Suppose we rub glass ($\ce{SiO2}$) against paper (which is made from cellulose $\ce{(C6H10O5)_{$n$}}$). We receive positively charged glass and negatively charged paper. Is that because of total ionization energy of molecule stronger than glass or crystal system too?

Although atoms and molecules like to be uncharged there is a difference between the electron affinity and the ionization energy. Electron affinity is a measure of willing a substance is to accept an extra electron. Ionization energy is a measure of how willing a substance is to donate electrons.
When the paper is rubbed against the glass the surfaces of the two substances come into physical contact. So the electron affinity of the paper is greater than the ionization energy of the glass rod, so electrons are stripped from the glass and are transferred to the paper. Enough electrons are transferred to balance the affinity/ionization relationship. This is known as the Triboelectric effect.
To be totally honest the Triboelectric effect is not well understood. There are a lot of enigmatic interactions in how the surfaces contact, how many times the surfaces contact when rubbing, how specific contact points my heat up, or how some physical transfer may take place. Read the Wikipedia article for more details on these points.

Secondly: When we rub atoms against other atoms, do electrons accelerate each other by the same electrostatic charge and change their orbitals to higher orbits due to synchrotron radiation they emit. Or is it because of the fact that protons are more close are ionizing electrons by electrostatic attraction? Am I correct?

There are a number of erroneous concepts in the setup/question. The first explains the phenomena.
Synchrotron radiation is electromagnetic radiation and has absolutely nothing to do with the electrostatic experiment that you are discussing.
The electrostatic experiment that you are discussing does not "accelerate electrons." So rubbing a glass rod with paper doesn't make a particle accelerator.
Neither the paper nor the glass has bare protons floating around.


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Proton-decoupled phosphorus spectrum of a ruthenium complex

Regarding the complex $\ce{[RuH2(CO)(PPh3)3]}$, there are three possible octahedral stereochemical configurations,
a) Draw these isomers indicating which, if any, are chiral and identify the spin systems in each case
b) Sketch the expected appearance of the $\ce{^{31}P\{^1H\}}$ ($\ce{^1H}$ decoupled) spectra
c) Sketch the expected appearance of the $\ce{^1H}$ spectra

My attempt:
I think I have completed (a) but would appreciate on any advice for (b) and (c)

For the isomer on the far left I think the decoupled phosphorus NMR would show a doublet and a triplet.

The middle one I think it might too be a doublet and a triplet.

The isomer on the right I'm leaning towards doublet of doublets but I'm unsure.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I improved the formatting of your post with MathJax; for more information on how to do so yourself, check out the [help], this meta-post or this one. Also note that your question qualifies as a [tag:homework] question as per our homework policy. Therefore, please show your work, e.g. the results you have for a) and how you would approach b) and c).

(cont’d.) Otherwise your question may get closed.

Let’s call the isomers mer-trans, mer-cis and fac from left to right for easier addressing.
What you didn’t indicate yet is which symmetry elements the different ones have and if they are chiral; this is helpful in determining the look of the NMR spectra because it helps us identify enantiotopic and homotopic atoms which give identical spins. The mer-trans isomer is clearly the one with the highest degree of symmetry: I judge its point group to be $C_\mathrm{2v}$. The other two should be $C_\mathrm{s}$, if I did the looking correctly.
Once we have identified the symmetry, we need to find out which atoms are chemically and/or magnetically equivalent. For the two $C_\mathrm{s}$ compounds it is pretty easy: If the plane of symmetry transforms one atom onto the other, they are chemically equivalent but enantiotopic (so magnetically non-equivalent). If the plane of symmetry contains the atoms in question, they must be chemically non-equivalent, thus have different spins. For the mer-trans isomer it may look slightly more tricky, but note that spinning around the $C_2$ axis will always place one atom of the same kind in question on top of the other thus again making them chemically equivalent and homotopic. (Also note that the three phosphorus atoms can never all be chemically equivalent to each other since one always occupies a special position by symmetry.)
With our sets of chemically equivalent or non-equivalent atoms in hand — to check: I arrived at sets of 1, 1 and 2 hydrogens and always two phosphorus; in one phosphorus set we have a pair chemically equivalent atoms, the other two contain a pair of enantiotopic ones — we can predict the look of the spectra and what will couple to what. Note that the $\ce{^{31}P\{^1H\}}$ spectrum will not contain any couplings to hydrogen atoms.

For mer-trans, your reasoning is sound and I also expect the result to be a doublet and a triplet; the coupling constants must be identical. If I had to guess, I would say that those closer to $\ce{CO}$ should have a higher chemical shift due to $\ce{CO}$’s electron withdrawing nature; but it could be due to a push-pull mechanism that its influence is greater in the para-position.
mer-cis should also be a doublet and a triplet for more or less the same reasons. The chemical shifts must be non-identical but could be closer to each other due to every phosphorus being a neighbour of $\ce{CO}$ (which is still the single most differentiating species in the compound).
fac cannot be a doublet of doublets since that would be a single signal for all three; the three are, however, chemically non-equivalent. The single special phosphorus will again display a triplet because it is coupling to two distinct phosphorus atoms. The other two should show up as a doublet. I’ll let you do the reasoning.

When proceeding to the hydrogen spectra, remember that $\ce{^{31}P}$ is also a magnetically active, spin ½, $100~\%$ abundance nucleus. Thus, you must consider the coupling of the protons with each other (if applicable) plus the coupling of the protons with the phosphorus atoms (the magnetically non-equivalent ones). In no particular order you should arrive at:

One spectrum containing two distinct proton shifts, each with two doublet couplings and a triplet coupling. I cannot say a priori which one should be larger or smaller.
One spectrum containing a single proton shift which is split up into a doublet and a triplet. Again, I cannot say which coupling is stronger.
One spectrum containing a single proton shift, split into a doublet of doublets of doublets (ddd). At least here I don’t have to say which coupling is stronger because they are all distinct.

Retroactively assinging proton spin systems, I get $\ce{AX, A2}$ and $\ce{AA'}$; for phosphorus I arrive at $\ce{AX2, AA'X}$ and $\ce{AX2}$, again in no particular order for all six spin systems (i.e. you would still need to assign them to the correct structures). If you wish to combine the spin systems into one entire one per structure, use $\ce{M}$ rather than $\ce{A}$ in the hydrogen ones (or $\ce{K}$ and $\ce{R}$ instead of the hydrogen $\ce{AX}$). Note however that I am terrible at spin systems and that you should take those with a grain of salt.

I hope that someone good at spin systems can confirm that part of my answer to be correct.


text:
How does hyperconjugation affect bond lengths in alkenes?
How does the number of hyperconjugates affect the $\ce{C=C}$ bond length in the compounds propene and 3,3-dimethylbut-1-ene?

In general, is bond length directly or inversely proportional to the number of hyperconjugates?

Have a look here for a detailed explanation on how hyperconjugation works on a frontier-orbital level. In propene and 3,3-dimethylbut-1-ene you have 3 bonds that can take part in hyperconjugation: in propene it's the 3 $\ce{C-H}$ bonds and in the other compound it's the 3 $\ce{C-C}$ bonds. So, the premise of your question seems to be wrong: The number of "hyperconjugates" is the same, only their types differ. Concentrating on the type, I would say the $\ce{C-C}$ bonds should be a bit more effective at hyperconjugation than the $\ce{C-H}$ bonds.

The higher the number of alpha hydrogens present in a molecule, the more resonance there will be which will decrease the bond order which was earlier 2, thus decreasing its strength and increasing its length.

The answer is correct but could be explained a lot better.

Because of hyperconjugation, C2-C3 single bond in propene acquires some double bond character ans hence is little shorter (1.49 Å) than the normal C-C single bond length (1.54 Å).
However in 3,3-dimethyl-but-1-ene there are no alpha hydrogens to operate the hyperconjugation effect and hence the C2-C3 single bond is not affected.

It directly depends on the bond order.
Because bond order=(number of covalent bonds in that molecule)/(number of resonance structures)
Hence ,more the hyperconjugative structures less will be its bond order.
As bond order is inversely proportional to bond length,less the bond order more the bond length.
Therefore more hyperconjugative structures is directly proportional to bond length.
Hence with greater no of alpha hydrogen they will have greater bond length.
Hope this helps..


text:
Analysis of mixture by precipitation and complexometric titrations
I've performed an experiment about chloride mixtures. First using $\ce{AgNO3}$ as titrant, I've titrated the mixture to determine the amount of total chloride. Then by direct titration with $\ce{EDTA}$, I wanted to determine the amount of $\ce{Mg}$ ions.
I'm confused about the amount of total chloride ions as there are two sources of chloride ions.
I'd really appreciate if someone can help me here.
Thank you so much.

You should specify more details - what is in the mixture, etc. In any case, why should the chloride ions from the two different sources be any different? There's no label attached to the $\ce{Cl-}$ that tells the $\ce{Ag+}$ "this ion came from the first component". As such, $\ce{Ag+}$ is going to precipitate every single chloride ion inside that mixture.

Safe to calculate like what? Regarding your "another question", there's no way to calculate the amount of $\ce{Mg^2+}$ without the EDTA titration. You will have one equation in two unknowns. Why don't you try setting up a system of equations? Let $x$ be the number of moles of $\ce{MgCl2}$ and $y$ be the number of moles of $\ce{NaCl}$. What is the total number of moles of $\ce{Cl-}$? What is the number of moles of $\ce{Mg^2+}$?

There is no guarantee that the amount of $\ce{Cl-}$ from $\ce{NaCl}$ is equal to the amount of  $\ce{Cl-}$ from $\ce{MgCl2}$ - so how can you add them up to give $2n(\ce{Cl-})?$ They are independent of each other. If you use $x$ and $y$ as I suggested earlier, you might understand it better.

Let $x$ be the amount of $\ce{MgCl2}$ in the original mixture, and $y$ be the amount of $\ce{NaCl}$.
Both of these salts dissociate completely in aqueous solution to give their constituent ions:
$$\begin{align}
\ce{MgCl2 (s) -> Mg^2+ (aq) + 2 Cl- (aq)} \\
\ce{NaCl (s) -> Na+ (aq) + Cl- (aq)}
\end{align}$$
Therefore, the total number of moles of chloride ions in the mixture is:
$$\begin{align}
\eta_{\ce{Cl-},\text{total}} &= \eta_{\ce{Cl-}\text{(from }\ce{MgCl2})} + \eta_{\ce{Cl-}\text{(from }\ce{NaCl})} \\
&= 2\eta_{\ce{MgCl2}} + \eta_{\ce{NaCl}} \\
&= 2x + y
\end{align}$$
$\ce{Ag+}$ ions react with $\ce{Cl-}$ ions in a $1:1$ stoichiometric ratio:
$$\ce{Ag+ (aq) + Cl- (aq) -> AgCl (s)}$$
Therefore, we have $\eta_{\ce{Ag+}} = \eta_{\ce{Cl-},\text{total}} = 2x + y$.

$\ce{EDTA}$ only complexes magnesium, and again in a $1:1$ stoichiometric ratio:
$$\ce{Mg^2+ (aq) + edta^4- (aq) -> [Mg(edta)]^2- (aq)}$$
Therefore, $\eta_{\ce{EDTA}} = \eta_{\ce{Mg^2+}} = \eta_{\ce{MgCl2}} = x$.

From your titration, you would have determined both values $\eta_{\ce{Ag+}}$ and $\eta_{\ce{EDTA}}$. Therefore, you have a system of two simultaneous equations in two unknowns, which is extremely simple to solve:
$$\begin{align}
\eta_{\ce{Ag+}} &= 2x + y \\
\eta_{\ce{EDTA}} &= x
\end{align}$$

@Mel - the answer key gives the Mg+2 as mmol/(100 ml of solution). Without knowing how large a sample was titrated you can't determine that part of the experiment. The answer key to me indicates that two different samples were taken from the solution in question. A 10 ml sample was titrated with the Ag solution and a 25 ml sample was titrated with EDTA.


text:
Why do BF3 and NH3 react when B has an expanded octet in the adduct?
I know that some elements like sulfur are able to have an expanded octet due to vacant $\mathrm{d}$ orbitals. However, boron and nitrogen do not have any such vacant place so how is it that $\ce{NH3}$ with its extra electron pair is able to share it with $\ce{BF3}$ and form a compound?
This is an example of a Lewis acid-base reaction that I read in my textbook.

Draw the Lewis structures for $\ce{BF3}$, $\ce{NH3}$, and the adduct $\ce{F3B.NH3}$. Are you really sure boron has an expanded octet in the adduct? Are you sure that boron even has a full octet in $\ce{BF3}$?