text:
Dichloromethane solubility in water
I heard dichloromethane is soluble in water 1:50, but I also heard it is not because it can't form hydrogen bonds. I'm confused. Is there a way to extract it? Would putting the water to $40~\mathrm{^\circ C}$ make it evaporate trough water?

1:50 is misleading. Wikipedia gives dichloromethane’s solubility in water as $20,\mathrm{\frac{g}{l}}$ which is approximately 1:50 w/w, but ratios are usually understood v/v. (weight/weight versus volume/volume)

@Jan That's a good start for an answer...

"it is not because it can't form hydrogen bonds" - Just bear in mind that solubility isn't a black and white thing. Dichloromethane is pretty insoluble in water (as opposed to something like ethanol) because it doesn't H-bond with water, but that doesn't mean it is entirely insoluble.

The solubility of dichloromethane in water (as well as the solubility of water in dichloromethane) depends on temperature.
At a temperature of $T=20\ \mathrm{^\circ C}$, the solubility amounts to $17.6\ \mathrm g$ dichloromethane per $1\ \mathrm{kg}$ water; i.e. the mass fraction of dichloromethane is $w_{\ce{C2H2Cl2}}=0.0173$ or $w_{\ce{C2H2Cl2}}=1.73\,\%$.
[source: “Aqueous Solubility and Henry’s Law Constants of Organic Compounds”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.]
(See here for values at other temperatures.)
You cannot simply evaporate pure dichloromethane (boiling point: $40\ \mathrm{^\circ C}$) from such mixtures since dichloromethane and water form an azeotrope, which boils at $38.8\ \mathrm{^\circ C}$. The mass fraction of water in this azeotrope is about $w_{\ce{H2O}}=1\,\%$.

You could fractionally distil the mixture until the distillate no longer contains dichloromethane. The remaining water in the distillation flask would then be free of dichloromethane.

Is this mean at 40C i will evaporate the azeotrope wich have 1% water in it ? I have no interest in pure DCM, I want DCM-Free water.

@simon - yes. Even if your lose some water in getting the DCM off, you'll be left with what you want.


text:
Does the electrochemical series reverse in electrolysis?
Are these two statements true?

In a galvanic cell at standard lab conditions (SLC), the strongest reductant is $\ce{Li (s)}$ and the strongest oxidant is $\ce{F2 (g)}$.
In an electrolytic cell at SLC, the strongest oxidant is $\ce{Li+ (aq)}$ and the strongest reductant is $\ce{F- (aq)}$.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

I thought so but thanks I'll keep that in mind

No, only the first is. The oxidising or reducing property of a species is tied to the species, not to the process in which the species is used. Therefore regardless of whether you are running a galvanic cell or an electrolytic cell, $\ce{F2}$ will always be a strong oxidant and $\ce{Li}$ a strong reductant.
The only difference about the two types of cells is that in the galvanic cell, you are letting the spontaneous cell reaction happen, which produces energy. In the electrolytic cell, you are using energy to force the non-spontaneous cell reaction to happen. The spontaneous cell reaction refers to the direction of the reaction with $\Delta G < 0$ (or equivalently, $E > 0$).
By this, I mean: you can use a galvanic cell to carry out the reaction $\ce{2 Li + F2 -> 2Li+ + 2F-}$. Since $\ce{Li}$ is a strong reductant and $\ce{F2}$ is a strong oxidant, this reaction will generate a lot of energy.
You can use an electrolytic cell to reverse that reaction (although you'd have to use molten and not aqueous $\ce{LiF}$). But since $\ce{Li+}$ is a weak oxidant and $\ce{F-}$ a weak reductant, you're going to need lots of energy to carry that out. Which makes perfect sense!

SLC is an abbreviation for Standard Lab Conditions; i.e. 25 ºC  temperature and 1 atm pressure.  From what I've read, when you electrolyse molten NaCl, the oxidant is Na$^+$ and the reductant is Cl$^-$. Isn't that consistent with my second statement?

@inspd I see, the more frequently used abbreviation is STP (standard temperature and pressure). You're right about the identities of the oxidant and reductant, but I think we might have different ideas about what the word "strong" means. If you refer to this page, "strong oxidant" means the LHS species for a half-reaction with a large positive $E^\circ$. "Strong reductant" means the RHS species for a half-reaction with a large negative $E^\circ$.

then it should be "best in the LiF liquid" not "strongest in the chemical series" am I  right ?

@inspd Essentially that means a strong oxidant is a species that is easily reduced, and a strong reductant is a species that is easily oxidised. The fact that you would require a lot of energy for the electrolysis of NaCl means that the participating species (Na+ or Cl-) are hard to reduce/oxidise - which makes them weak oxidants/reductants.

so when we electrolyse LiF(l), the best (but weakest) oxidant is:    Li$^+$ and the best (but weakest) reductant is:    F$^-$

I would rather not use the terms in electrolytic cell at all, for they are misleading. There are no oxidants and reductants here, really. It is electric current that oxidizes and reduces.

@IvanNeretin So if it is the electric current that oxidises and reduces, then Li isn't an oxidant? You're saying that Li would be described as an oxidant (maybe to make things conceptually easier), but it isn't really oxidising anything, right ?

Right, so let's stop calling it an oxidant (though nominally it is).

Haha you can still say that the oxidising / reducing properties of the species (ie whether they are strong/weak oxidants/reductants) affects the energy required for the electrolysis, even though they do not behave as true oxidants/reductants. Anyway just to sum it up my point is that the direction of the reaction does not affect the order of the electrochemical series. Thanks @IvanNeretin for the explanation

Makes sense. Also I'll use the terms "best oxidant" and "best reductant" instead (in comparison to other compounds, e.g. $\ce{H2O(l)}$ is the best oxidant in $\ce{NaCl(aq)}$).  Much appreciated @orthocresol and IvanNeretin.

@inspd You know, that's exactly what he was trying to stop you from doing! Because H2O behaves neither as an oxidant (which is a species that oxidises something else) or a reductant (a species that reduces something else) in the electrolysis. H2O is oxidised and reduced in the electrolysis, but technically it does not carry out the process. So I think he (and I) would prefer if you write "most easily reduced species" and "most easily oxidised species". :)

Yeh that's what I meant, I gotta watch the semantics.. I'll use what you said, thank you

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text:
How to identify the reduction in a given redox equation of iron chloride and chlorine?

The equation shows a redox reaction between iron(II) chloride and chlorine gas.
  $$\ce{2FeCl2 + Cl2 -> 2FeCl3}$$
  Which equation describes the reduction process in this reaction?
  \begin{align}
\ce{2Cl- &-> Cl2 + 2e-}\tag{A}\\
\ce{Cl2 + 2e- &-> 2Cl- }\tag{B}\\
\ce{Fe^2+ &-> Fe^3+ + e-}\tag{C}\\
\ce{Fe^3+ + e- &-> Fe^2+}\tag{D}
\end{align}

I couldn't understand why the answers are not C instead of B.
The correct answer is B. A and D cannot be the answer.

Because what is reduction, really? That's when you add electrons to something.

A redox reaction consists of 2 parts, oxidation and reduction These two parts can be written up as half-equations where one half-equation shows oxidation and the other shows reduction.
It seems that you have mixed up the oxidation and reduction half-equations. Oxidation is losing electrons while reduction is gaining electrons. This can be remembered by the mnemonic:

The question is asking for the reduction half-equation. To do this, you must first recognise what is getting reduced (in other words, what is the oxidant). This can be done by looking at the oxidation states for each element before and after the reaction. Since it is getting reduced, it is gaining electrons and therefore its oxidation state should decrease after the reaction. So by looking at the oxidation states, it should be clear that chlorine is getting reduced as its oxidation state goes from $0$ to $-1$.
Therefore the answer is B
Note: The answer definitely can't be a or c since they are oxidation half-equations

Another mnemonic I've found useful is "LEO the plussy cat says GER".  Lose electrons for oxidation, (often producing) cations with a plus charge, and gain of electrons is reduction.

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Predict flame test color?
If you know the composition of any sample of compound, is it possible to predict it's flame test color?
For eg, if you can given Sodium Chloride, then without actually doing the flame test, can you mathematically (and using physics and chemistry obviously) come up with it flame color, ie yellow-orange?
Do the flame test colors follow any pattern for different metals? (for eg, does it follow the spectrum?)
Does the group (for eg alkali metals or transition metals) affect the flame color?

Same question intrigued the brightest minds of early 20th century and as an answer, the quantum theory came to the light. See e.g. Schroedinger equation.

Great. Than learn a little bit about it, write answer to your own question and aim for a Self Learner badge!

Atomic and ionic emission spectra are empirically determined. Some emission spectra, e.g., for iron, thorium, uranium, are ferociously complicated. There are extensive tables of emission line wavelengths and relative intensities available from NIST, etc. But no one can yet predict the thousands of iron emission lines. Even just for the alkali metals, there is no obvious pattern: Li gives crimson flame color, Na gives the famous yellow D lines, K gives violet, Rb gives red (as per the element name) and Cs gives sky blue (as per its element name).

https://chemistry.stackexchange.com/questions/143663/why-flame-testing-different-ions-give-different-colors
https://chemistry.stackexchange.com/questions/147950/flame-test-of-alkali-metals-and-alkaline-earth-metals
https://chemistry.stackexchange.com/questions/31476/why-do-s-block-elements-impart-color-to-a-flame
https://chemistry.stackexchange.com/questions/75303/why-flame-color-of-salts-is-determined-by-metal

https://chemistry.stackexchange.com/questions/150824/reason-for-sodium-chloride-to-show-colour-in-flame-test
https://chemistry.stackexchange.com/questions/149459/color-of-flame-test-for-alkali-metals
https://chemistry.stackexchange.com/questions/20516/why-they-have-different-flame-colour

https://chemistry.stackexchange.com/questions/92011/is-it-possible-to-predict-the-color-of-the-flame-test-result-based-on-the-elemen
https://chemistry.stackexchange.com/questions/141382/wavelengths-of-colors-emitted-by-alkali-metals-during-flame-test
https://chemistry.stackexchange.com/questions/64785/spectrometry-in-chemistry-flame-test-lab

If you know the composition of any sample of compound, is it possible
to predict it's flame test color?

By current standards, unfortunately not. The calculations will be extremely complicated. If you recall, there are > 92 natural elements. The flame test works only for a few elements, so basically it is a useless test today. The Bunsen burner flame is a low temperature flame. It cannot break any sample or molecules into atoms. You need a very high temperature flame.
The keypoint to understand is that most of the elements in the periodic table in a  strongly emit light in the ultraviolet region in a high temperature flame. So you will not be see the true flame color anyway, even if it were predictable by calculations 100 years later.

I wouldn't say flame tests are useless as they are widely used for simple samples. And many analytical instruments, in effect, rely on them for identifying elements in samples.

Agree about the flame test being useless now, except for lecture demo purposes and maybe some quick and dirty niche testing. It would be interesting to know if flame photometry is still used much anymore, given far more powerful (albeit much more expensive) alternatives like ICP-OES, etc. And, of course, techniques like AAS have nothing to do with flame emission.

@EdV, Flame photometry is still useful in clinical chemistry for Li, Na, K determination in blood or urine.

The concept of flame test is limited to a simple Bunsen burner and introduction of a solid /solution with the help of a metallic loop and watching the color of the flame. Good for a demo as EdV stated. It is useless for real samples because Na is everywhere. Flame photometry, and other atomic absorption/emission techniques using flames are still widely used although plasma emission techniques have been more popular in pharma.

Prediction of color is different thing to prediction of emission spectrum, as the former is subjective.
Even if we could accurately predict the radiometric emission spectrum,  color evaluation, based on summary evaluation of photometric spectrum would be on experimental subjective assesment with multiple conclusions.
Especially if women are involved. We men with 16 basic colors can agree more easily.

Wait, where are the other 9 colors? ;-)

@EdV It would be too many for men, too little for women. ;-)

In principle, yes. In practice, it is much more complex (and you don't need to do it as others have already made the observations)
The colour of a flame test is a consequence of the combination emission lines from electronic transitions in the flame. Most useful lines are narrow and involve transitions in isolated atoms between different electron orbitals. The perceived colour is derived from the strong lines in the visible spectrum (eg the well known sodium lines around 589 nM which look orange and dominate the visible spectrum in flames containing sodium).
The specific position of the lines is far more diagnostic of the substance than the colour which will be a combination of all the strong visible lines and there may be more than one strong line. Some common compounds have fairly distinct colours, though, which is why it is easy to distinguish, for example, lithium (red), sodium (yellow) and potassium (violet/turquoise). This is why even simple flame tests can be useful.
And you don't need to calculate anything to identify the emitting substance in the flame: others have observed the emissions and classified what all the lines represent in terms of the electronic transitions that cause them. This is why techniques such as atomic absorption or emission spectroscopy are often very useful at characterising samples.
As for the original question about whether these lines could be predicted computationally from first principles the answer is, in principle, yes. In practice things are a lot harder because quantum mechanics is hard.
A full picture of all the possible lines could, in principle, be derived from a model of the isolated atom's possible electron energy levels. A very thorough treatment could give a theoretical picture of those levels and all the possible transitions between them. Even the transition probabilities which partly determine the brightness of the lines. This picture when filtered to eliminate the lines in the IR and UV could give a good idea of the colour of the emission spectrum.
But those calculations are hard as an exact value for the electron energies is impossible to do analytically for anything more complex than an isolated hydrogen atom. The amount of computation required to account for all the possible interactions that determine the energy levels is huge and rarely worth doing when it is easier to understand the observed spectrum than it is to calculate it from first principles.
It is also worth noting that the complexity of the possible spectra for even the simplest atom, hydrogen, explains why there are no simple relationships between colour and the individual lines. There are more than 30 relatively simple lines in the hydrogen spectrum that are easily observed and they range from the hard UV to the far IR in wavelength (in principle, there are an infinite number as every possible orbital can contain an electron and every pair of orbitals can generate a unique transition but many of the higher orbitals will rarely contain electrons and will get closer and less distinguishable). Some transitions will only be observed in extreme conditions unlikely to occur in a simple flame. More complex atoms will display far more complex patterns and the number of interactions between orbitals that result in splitting simple energy levels will increase rapidly.
But the biggest reason why simple relationships between colour and the emission lines doesn't follow simple patterns is that visible light only occupies about one octave of the possible electromagnetic spectrum. Actual emissions span wavelengths from the UV to the IR of dozens of octaves. So the wavelengths that determine colour are a small selection from a much larger range of wavelengths. This makes the pattern of colour much more arbitrary than any consistent patterns in the full range of wavelengths. And this view ignores the even harder problem of estimating the relative brightness of the lines which requires a full understanding of the quantum transition probabilities and the likely population of different energy levels at a given temperature.
So, in principle, you could calculate the full range of emission spectra for different atoms and filter it for lines in the visible spectrum and derive some idea of colour. But this is very much harder than observing the emissions from a pure sample and using them to identify an unknown.

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text:
Combustion properties of Dimethyl Ether
I have a C++ code which calculates the combustion thermodynamic variables (temperature, density) and the mass fractions of Dimethyl Ether for different equivalence ratio. I would like to check my results but I couldn't find anything on google. Most of the papers about DME that I've found show the laminar flame speed.
Please refer me to some paper or book if you know any. A graph showing the adiabatic flame temperature vs equivalence ratio would be ideal.

Please refer me to some paper or book if you know any. A graph showing the adiabatic flame temperature vs equivalence ratio would be ideal.

In 2014 it appears that Glaude, Sirjean, Fournet et al. published a paper that gives exactly the data you are looking for.  See Figure 4 from their paper, which I have also included below.

Possibly relevant text from the paper:

the
  maximum flame temperature is obtained for slightly rich
  mixtures, at an equivalence ratio around 1.05-1.1, because of
  the lower impact of dilution by nitrogen when the amount of
  air decreases. At the strict stoichiometric conditions, the rate
  of the endothermal molecular dissociations is also at its
  maximum which tends to depress the temperature.

Also this next bit:

The results show that alcohols have flame temperatures
  comparable to those of conventional fuels, contrary to DME
  that develops significantly higher temperatures. This is
  consistent with the respective values of the heating values of
  these fuels and thermochemical equilibrium data.
I couldn't find anything on google.

I found the paper by Googling dme adiabatic flame temperature.

I never expected to get such a complete answer! Thank you for that. Also, I googled dme adiabatic flame temperature again just to be sure and did not find that paper. I guess google knows you better than me!


text:
Double bond formation in alkenes during cracking
In thermal or catalytic cracking, energy is used to break a long chain alkane into smaller alkanes/alkenes. Since the energy goes into breaking the bonds, how is the C=C bond in alkenes formed?

Bond formation is exothermic and so favourable. Can you explain exactly what is confusing you?

Bond formation occurs as a result of the cracking - which is endothermic and requires heat. I understand that the input heat energy goes to breaking the bonds, but how is the double bond then formed?

Do you want the mechanism for cracking? Bond breaking is endothermic, bond formation is exothermic.

Alright, thank you, not fully clear, but I get the general drift. :)

I don't know if I have understood well your question.
During the process of cracking, homolysis of long chains occurs. Radical intermediates result from breaking the bonds. These intermediates are highly unstable and undergo rearrangements. As a consequence of the rearrangements,  smaller alkanes and alkenes are formed. We have to point out that, due the the high reactivity of radicals, the control of radical reactions is very difficult and we can not readily predict what happens inside the chemical reactor.
I hope I've answered your question.

To add a small point to Yomen's excellent answer:
Alkyl radicals can be turned into alkenes through loss of a hydrogen radical.  For example suppose during ethane cracking, an ethyl radical is produced.  It can lose a hydrogen radical to form ethylene.
$$\ce{^{\cdot}CH2-CH3 <=> CH2=CH2 + H^{\cdot}}$$
Alkyl radicals and H radicals are both relatively unstable, so this reaction is an equilibrium.  Many other reactions that produce, interconvert, and consume radicals also happen during cracking.  Ultimately the driving force is the high temperature, which favors products which more entropy.  The entropy of many small molecules is much higher than a few larger molecules because of the additional translational degrees of freedom of the small molecules.  That is what favors production of smaller molecules such as ethylene and hydrogen, as compared to larger molecules such as ethane.


text:
Why can't element with paired electron have colour
I know that elements with unpaired electrons are coloured especially the d-block element which can be explained by crystal field theory. But, why can't element with paired electrons give colour, is it because the energy gap is too large ?

It's not about whether the electrons are paired or not. It has to do with whether the electronic transitions in the molecule lie within the visible spectrum

to give you counterexamples, Ni(II) in its dimethylglyoxime complex has 8 d electrons and all of them are paired but the complex is red. On the other hand $\ce{O2}$ gas has 2 unpaired electrons but it's colourless.

@orthocresol actually, liquid oxygen has a blueish colour, or so I heard.

@Jan I've seen that before in a demonstration... which is why I said O2 gas :D Anyway I went to read up. Turns out that the blue colour is because of two O2 molecules being excited simultaneously, which has a very low probability of occurring in the gas phase (hence colourless). Probably wasn't the best example to use but I guess it's passable.

@orthocresol Disregard me, I overread the 'gas' in there ^^' But on that note: chlorine gas is greenish, bromine is brown, iodine is purple, liquid chlorine is yellow.

So what should we really look for

To write up a proper answer, are you talking about elemental compounds, i.e. those only made up of atoms of a single element such as $\ce{O2}$, or are you talking about compounds in general where the chromophore (the structural entity that creates the colour) is a single (metal) atom?

Of single element

Your assumption is wrong. There are elements with paired electrons which are coloured.
Colour is a product of electronic transitions in atoms or molecules. If those transitions match the energy of a photon of visible light, then the substance will be coloured (though there are some other rules about how intense the colour will actually be).
Some elements do show colour. Halogens, for example, exist as diatomic molecules in their native state and are all coloured. Chlorine is a greenish gas and bromine is an intensely red liquid. Sulphur is usually found as a yellow solid. All have paired electrons.


text:
Is it possible for Cu to reduce Cu2+?
I tried a slightly different set up of a Galvanic cell by inserting the $\ce{Zn}$ electrode into the $\ce{CuSO4}$ solution and the $\ce{Cu}$ electrode into the $\ce{ZnSO4}$ solution. This gave a voltage reading of 0.85 V. I also observed that the $\ce{Zn}$ electrode was getting coated with copper as the reaction progressed.
I had two hypotheses in mind -

The reaction is limited to only one half cell: the $\ce{Zn}$ electrode is oxidised and the $\ce{Cu^2+}$ in the solution is reduced. This explains the coating on the $\ce{Zn}$ electrode, but it does not explain the voltage. Doesn't the voltage mean that there is some interaction between the two cells?
This is my main question: Is it possible that the $\ce{Cu}$ atoms on the electrode are losing electrons that then travel through the circuit and reduce the $\ce{Cu^2+}$ ions in the second half cell? A test for $\ce{Cu^2+}$ ions in the $\ce{ZnSO4}$ solution after the reaction had progressed was inconclusive.

What exactly is happening in this set up?

Did you also measure the current?

No, just the voltage. Why current?

If there was both voltage and significant current it is direct evidence that electrochemical phenomena were happening.  With only voltage, it is possible that essentially no current was flowing, and that the voltage could be coming from capacitive or other non-electrochemical effects.  Do you know the overall cell resistance of your setup in Ohms?

No resistance either, sorry. Only took voltage readings.  But will take your point into consideration and do another experiment taking the current readings as well. But isn't the voltage quite high to be coming from something else? Your normal Galvanic cell voltage is around 1.1V, not very far from 0.85V.

Zinc is higher in the reactivity series than copper, so simply by placing a piece of zinc metal into a copper sulphate solution, you will get a displacement reaction where the zinc metal loses electrons and becomes $\ce{Zn^2+}$ ions, and $\ce{Cu^2+}$ ions gain those two electrons to become copper metal.  This will happen until the zinc electrode is coated with copper and no more zinc ions can get into the solution.
The standard half-cell potentials are measured using standard 1M solutions, and deviations from this concentration cause the equilibria to shift according to Le Chatelier's principle:
$\ce{ Cu^{2+}_{(aq)} + 2e^{-}<=> Cu_{(s)}} \quad  E^{o}=+0.337V$
Since the $\ce{[Cu^{2+}_{(aq)}]}$ is zero for the copper electrode, this system is far from equilibrium and the half-cell potential will be closer to zero.  In the other half of the system, the $\ce{[Cu^{2+}_{(aq)}]}$ is less than 1M due to the reaction with the zinc electrode, but it is still significantly greater than zero.  The result is a slight potential for copper metal to become ions and for the electrons to travel through the circuit to reduce some of of the copper(II) ions in the other half cell.  Using the Nernst equation for this isn't valid since concentrations close to 1M are too high for accurate results on one hand (concentrations greater than $10^{-3}$M deviate from experimental values), and the use of the equation is on the assumption that the ion is in equilibrium.  If we use it naively we do get something:
$E_{cell} = E^{o}_{cell} - (\frac{0.059}{2})log(Q) \approx 0-0.0295log(\frac{\ce{[0.0000000001]}}{\ce{[0.5]}}) = 0.37V$
As the approximation of the copper(II) ion concentration approaches zero the potential difference for the cell approaches infinity, which goes some way to demonstrate this is only valid within limits since you don't get a 100,000 Volts electric shock when you put a piece of metal into pure water.  Nevertheless it describes a trend and supports the Le Chatelier interpretation.
However, if the circuit is connected using a voltmeter this has a sufficiently high resistance to prevent current from flowing, so although there is potential for the copper to reduce itself through the cell, this can only happen if electrons can flow.  Even without the high resistance the reaction of the copper metal forming ions may be so slow, that very little current flows and few ions are formed which may explain the inconclusive test result.


text:
Orbital mixing in square planar D4h complexes
Consider a square planar $\ce{ML4}$ complex with $D_\mathrm{4h}$ symmetry. Which orbital should the $\mathrm{a_{1g}}$ SALC of the ligands mix with, $\mathrm{3d}_{z^2}$ or $\mathrm{4s}$? Both of those AOs transform as  $\mathrm{A_{1g}}$.

Why can't all three orbitals mix?

Could you refresh my memory in the meaning of SALC?

Symmetry adapted linear combinations.

How should I draw the MO diagram if all three mix together?

All orbitals of the same symmetry will interact (mix) to some degree.
So the ligand SALC will interact with both $\mathrm{a_{1g}}$ orbitals on the metal. The degree of interaction will depend on how close in energy the ligand $\mathrm{a_{1g}}$ is to the $\mathrm{4s}$ or $\mathrm{3d}_{z^2}$ orbitals on the metal.
Generally, the ligand orbitals are closer in energy to the $\mathrm{3d}_{z^2}$ orbital than to the $\mathrm{4s}$.


text:
Why would phosphorus pentoxide be called 'phosphorus(v) oxide'?
There's a question in the chemistry textbook I use with my GCSE class which asks pupils to identify which of various compounds feature ionic bonding. One of the compounds is phosphorus(V) oxide, which I gather is an accepted name of phosphorus pentoxide (or tetraphosphorus decoxide, if you prefer). But why? One of my pupils reasoned that it must have ionic bonding, because I'd just explained to them how Roman numerals in brackets indicate the number of electrons lost. This is evidently an exception, but none of the guides to chemical naming I've found explain why!
I figure the explanation must lie in subtleties of the oxidation state that I haven't yet got my head round - Wikipedia's page on oxidation state lists phosphorus pentoxide as having oxidation state +5, and phosphorus trioxide as +3, but I'm not entirely following why.
Any clarification much appreciated!

$\ce{P4O10}$: assign oxidation states. Result: $\ce{O^{-II}}$ and $\ce{P^{V}}$. Anything unclear?

"Roman numerals in brackets indicate the number of electrons lost" - They indicate the oxidation state. This only truly corresponds to the number of electrons lost when one talks about pure ionic bonding. Phosphorus in $\ce{P4O10}$ is in oxidation state +5 but it does not mean it exists in the form of $\ce{P^5+}$ ions. It means that, if you cleaved all P-O bonds heterolytically and assigned all the electrons to the more electronegative element, oxygen, then the phosphorus would exist as $\ce{P^5+}$. Until you do that, P is happily sharing electrons with O in a covalent bond.

This previous question (and answer) about the assignment of oxidation states may help clarify: http://chemistry.stackexchange.com/q/10944/16683

Thanks @orthocresol, that helps a lot. Having never studied (or taught) oxidation state in any detail, I've always been a little hazy on how it works in covalent compounds, but it's starting to make sense.


text:
Tele-substitution mechanism
According to IUPAC goldbook:

A tele-substituion is a substitution reaction in which the entering
  group takes up a position more than one atom away from the atom to
  which the leaving group was attached:

And that's the only example given. What is the mechanism of this reaction?

Have you drawn it correctly?  If you got this compound, NC1=NC=CN2C1=NN=C2 I could perhaps see a Dimroth rearrangement.

My smiles should be NC1=CN=CN2C1=NN=C2 (the N and C got transposed) but if this is the product, I can see a Dimroth.

Please elaborate. The image was taken from the IUPAC gooldbook.

Hi @Marko I'm wondering whether or not there is a mistake.  I've checked out the Gold book which references Pure and Applied Chemistry, 1994, 66, 1077 but this doesn't go any further.  Reaxys reaction query as starting material doesn't give the reaction; instead the literature search gives the following paper, Tet Lett 1974, 4539 http://www.sciencedirect.com/science/article/pii/S0040403901922130 which is very close to giving an explanation.  I'm guessing that that paper will provide the answer and show that there's an error in the Gold book.

Even if it is close to an explanation, I can't access it.

I can only access the first page.  Nevertheless, I'll contact IUPAC and tell them that they should double check it.

http://web4.uwindsor.ca/users/j/jlichaa/reference.nsf/0/10ff8b04ff3a317885256d88005720f6/$FILE/ejic-2002-1269-arene-Cr-rose.pdf - but not about azacyclic

this is similar http://pubs.acs.org/doi/pdf/10.1021/jo00862a005


text:
Looking for an isomerization reaction that involves a color change for classroom demonstration purposes
I am in the process of designing a chemical kinetics course for undergraduates. I would like to demonstrate isomerization equilibria, and therefore am looking for a reaction A <-> B fulfilling the following criteria:

At least one of the reactants has a color;
The equilibrium is reached in a few minutes so that the color change is easy to follow (not too fast or slow);
The reaction is safe and could be demonstrated in a beaker on the table;
The reaction is reversible and e.g. by gentle heating the equilibrium could be shifted easily.

Titration of some weak acids/bases with suitable indicators might also be good, but of course these are not isomerisations and are a bit more complex than a simple A <-> B reaction.
Any creative ideas are welcome. Thanks!

If you can use gases, how about dimerisation of $\ce{NO2}$ to $\ce{N2O4}$ (decreasing temperature).

The photoisomerization / thermoisomerization of various azobenzenes dyes seems like a place worth investigating.  It could be tough to find a combination of "long" (several minute" equilibration times, aqueous solubility, and a molecule you can buy easily, but there is probably one somewhere.

@Jan Good idea with NO2 but there are safety concerns. We had a setup at the university: two sealed glass bulbs with NO2 in them, connected by a glass tube. Warm one of the bulbs in a water bath, put the other bulb in an ice bath... looked nice.

@CurtF: thanks for the tip, I'll be looking into it and report back.

You might use thermochromic cobalt chloride in a water/ethanol (or isopropanol) mixture... though this color change is due to complex formation and change in the amount of water of hydration. See http://chemistry.about.com/od/colorchemistryprojects/a/Color-Change-Liquid-Thermometer.htm.


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Do coefficients matter in rate law?
For reaction $\ce{2A→B}$ (elementary step), according to the rate law, rate $= k [A]^2$.
In some calculations, we use $k[A]^2$ as the production rate of B. Why isn't it ${1 \over 2} k[A]^2$?
In this case, is the consumption rate of A also $k[A]^2$? If so, why are they the same when in fact two molecules of A produce one molecule of B?
I'm really confused about this part. Can anyone explain why coefficient doesn't play a role in this calculation? Thanks!

This is similar but probably not a duplicate of http://chemistry.stackexchange.com/questions/32304/law-of-mass-action

This confusion pops up often around here.
The problem usually comes down to insufficient consideration of units.
When you say "rate", what do you mean?  It has units of mol per volume per second, e.g. $\mathrm{\frac{mol}{L\cdot s}}$.  But that isn't enough information.  You need to know moles of what.
First example: if the rate is defined as loss of A
For example, if you say that the "rate" means the disappearance of A, then the units of the rate are really $\mathrm{\frac{mol~of~A}{L\cdot s}}$.  If so, then as long as:

The reaction is $\ce{2 A->B}$; and
The rate as you have defined it can be expressed by $r_A = k_1~[\ce{A}]^2$ ...

Then the rate of appearance of B will be $r_B = \frac{1}{2} k_1~[\ce{A}]^2$
Second example: if the "rate" is defined as appearance of B
If conversely you suppose that the "rate" refers to the appearance of B, then the units of rate are really $\mathrm{\frac{mol~of~B}{L\cdot s}}$, and if so, then as long as

The reaction is $\ce{2 A->B}$; and
The rate as you have defined it can be expressed by $r_B = k_2~[\ce{A}]^2$ ...

Then the rate of disappearance of A will be $r_A = 2 k_2~[\ce{A}]^2$
Third example: a totally separate definition of rate
You say its an elementary reaction.  Let's say that we didn't know that and wrote the reaction as $\ce{200 A -> 100 B}$.  The rate, according to this new example definition, is the rate of the reaction.  The units of the rate are $\mathrm{\frac{mol~of~"reaction"}{L\cdot s}}$. If the reaction is still 2nd-order then the rate of the reaction will be $$r_r = k_3~[\ce{A}]^2$$
According to our reaction, 200 moles of A disappear for each mole of reaction, so $$r_A = 200 k_3~[\ce{A}]^2$$  and 100 moles of B appear, so $$r_B = 100 k_3~[\ce{A}]^2$$
Final thoughts
You will notice that I used different subscripts to distinguish between the $k$ values for the different assumptions.  That's because they aren't equal to each other.  The value of the "rate constant" $k$ depends on how you define the rate.
No matter how we define "the rate" and what we define as the stoichiometry (e.g. $\ce{2A -> B}$ or $\ce{A -> 1/2B}$ or $\ce{200 A -> 100 B}$, the rates must be equal.  Therefore:
$$r_A = 200 k_3~[\ce{A}]^2 = 2 k_2~[\ce{A}]^2 = k_1~[\ce{A}]^2$$
That is,
$$ 200 k_3 = 2 k_2 = k_1$$
But no matter which way we decide to write it, it is still a second-order reaction because "the rate", no matter how we define it, is proportional to $[\ce{A}]^2$.

So it's just depends on how I define rate and the k I mentioned actually has different values, right? Clear now. Thank you very much!

With respect to what mention in the above "Final thoughts" :
IF we define the stoichiometry ( e.g.4A⟶2B), what is the rate of disappearance of A in our reaction? is it equal 4K3[A]^2 which is wrong or how we can define it according this new stoichiometry /

You can't express a reaction as $\ce{4A -> 2B}$ because you can divide by 2 to get $\ce{2A -> B}$. If you could multiple by any integer then you'd have an infinite number of rate equations.

NO    + 1/2Cl2   -----> NOCl    how to  relate the rate of the reaction with rate of disappearance NO/

I'm confused by all the new comments on this old answer of mine.  First, I think my answer does answer the question.  Second, you absolutely can define the reaction as $\ce{4A -> 2}$ B, in which case the rate of disappearance of A will be $-4 k_4\ce{[A]}^2$.  There are in fact an infinite number of possible rate equations and an infinite number of corresponding rate constants.  You just have to make sure your reaction definition is consistent with your rate equations.

Just a doubt ..In many textbooks they dont explicitly mention whether the rate in the rate law expression is the rate of disappearance of reactant A or appearance product B..Do we by default assume its the rate of disappearance of A..Like in the OP's question what could his rate expression probably refer to?

@CurtF. here you said" No matter how we define "the rate" and what we define as the stoichiometry (e.g. 2A⟶B or A⟶12B or 200A⟶100B, the rates must be equal. Therefore:
rA=200k3 [A]2=2k2 [A]2=k1 [A]2     " this seems true that rate of consumption of reactant must not change as reaction doesn't change . But what about the net rate of reaction ? (which is defined as rate of consumption of reactants divided by their stoichiometric coefficients and rate of production of products divided by their stoichiometric coefficients ) .That will change here then .But it shouldnt as reaction remains same

@CurtF. (continuation to prev comment ) reaction remains same since only stoichiometric coefficients are changed .  So shouldnt both the overall rate and the rate of production and consumption be same ?

@CurtF. What does it mean "each mole of reaction"?

@RaghavMadan: it will change because when you change "the reaction", you are changing what the "net rate of reaction" means.  For 200 A -> 100 B, if the rate of "the reaction" is 1 mol / second then the rate of disappearance of A is 200 mol / s.  If you define the reaction as 2A -> B, then if the rate of this very different "the reaction" is 1 mol / second, then the rate of disappearance of A is 2 mol / s.

@Apurvium: If "the reaction" is 200 A -> 100 B, then each mole "of reaction" refers to 200 mol of A and 100 mol of B. If the reaction is 0.5 A -> B, then each mole of "reaction" refers to 0.5 mol of A and 1 mol of B.

@CurtF. ok , so net rate of reaction changes on changing to stoichiometric coefficient .But rate of consumption and production must remain same as the reaction is same .Rate of consumption /production is a more important and usefull quantity than overall rate of reaction . To keep one of them constant the other has to change .So we prefer to keep rate of consumption and production constant instead of overall rate .Am i right here ?

The rate of production of $\ce{B}$ is given by:
$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = k[\ce{A}]^2$$
The rate of consumption of $\ce{A}$ is given by:
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -2k[\ce{A}]^2$$
The reason for having the coefficients this way is because of the way the rate of the reaction, $r$, is defined (source: IUPAC Gold Book):
$$r = -\frac{1}{a}\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = \frac{1}{b}\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}$$
where $a$ and $b$ indicate the stoichiometric coefficients in the reaction $a \ce{A ->} b \ce{B}$ (2 for A, 1 for B). If you assert that the rate of the reaction is equal to$k[\ce{A}]^2$, then you will obtain the first two equations above. You may choose to define it otherwise - read Curt's answer. But ultimately the most important thing is that you respect the fact that
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -2 \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}$$
This equation says that for every molecule of $\ce{B}$ that is produced, two molecules of $\ce{A}$ are consumed. If you went against this, your reaction wouldn't make any sense at all. (Note that this, in general, only applies to a reaction that does not involve significant buildup of any intermediates. For a single-step reaction such as the one you proposed, this holds true as there are no intermediates.)
Do see this previous question for more discussion about the reaction you have mentioned.

Nice IUPAC reference!  I should have known IUPAC had an official declaration on the appropriate way to represent stoichiometry.

@orthocresol Which $k$ to be used in the equation, $kt=\ln \frac{[A]_0}{[A]_t}$ for first order reaction, $\ce{2A -> B}$? All are constant but $k$.

@Apurvium You need to be careful here. The integrated rate law you quoted assumes that the stoichiometric coefficient of A is 1. If you take the correct definition of k as shown in the second equation in my answer, and integrate that, you should find the integrated rate law gets an extra factor of 2.

@orthocresol As $k$ changes, rate of reaction and half-life of reaction also changes. For radioactive disintegration, the stoichiometry is always one and half-life is constant for a particular nuclear reaction. It does not make sense to have different half-lives here. But what does it mean if we multiply any nuclear reaction by some factor?

@Apurvium exactly the same as your previous question. The equation t_1/2 = ln 2 / k is only valid for A -> B. If you have 2A -> 2B then there is an extra factor of 2 in that equation. Again this is derived from the integrated rate law. I suggest going through its derivation again.

@orthocresol I got it but my question was about radioactive disintegration for which half-life is always constant for a radioactive species.

@Apurvium I don't get your question, then. The half-life is constant, regardless of what stoichiometric coefficient you use. That was precisely my point: $k$ can change (by a factor of $n$) but $t_{1/2}$ doesn't change (because to get it you would divide by that same factor of $n$).

@orthocresol Finally I derived it! But I was trying to reason out like this: For the reaction $\ce{2A->2B}$, the rate of reaction becomes half but rate of disappearance of A remains the same as for $\ce{A->B}$. If we start with equal number of moles of A, say $a$ moles, then the time taken to reduce the number of moles of A to $a/2$ will remain same in both the cases. Am I correct?

@Apurvium Yes, and this is entirely natural: because the physical description of the system can't change, regardless of how you (an external observer) choose to describe the system.

@orthocresol Does any book explain these confusions in detail?

@Apurvium Not that I know of, but it's not that hard to figure it out for yourself, to be honest. You must think carefully about which quantities are human-defined and are thus arbitrary (e.g. the rate constant $k$: you can't measure $k$, you can only measure something else from which you calculate $k$ — but how you do this depends on your definition, i.e. the stoichiometric coefficients) and what quantities are directly observable and thus cannot be changed (e.g. half-life, or in general, any time $t$ taken for $\ce{[A]}$ to decrease by a given factor).


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Why do self-replicating molecules dissociate?
I am new to chemistry (have only done a GCSE) so I apologize for my lack of knowledge. But I am confused about why self-replicating molecules dissociate.
It is my understanding that a self-complimentary molecule can gather the parts needed through bonding to create another copy of itself on its recognition surface. Once the new molecule is constructed I am unsure why it would then depart from the original one during this dissociation phase. Why would all of these bonds suddenly break apart? What exact mechanisms are involved in this process?

Well, allow me to answer since I worked on self-replication during my master’s thesis.
The thing about all these associationen is that there is always an equilibrium to be considered. That is also true for chemical reactions: Technically, every reaction is at an equilibrium of some kind.
$$\ce{A + B <=> C + D}$$
or in the special case of associations:
$$\ce{A + B + AB <=> A-B-AB}$$
For nomal bond-creating and bond-breaking reactions, the energy involved in the whole process is relatively large — breaking a carbon-carbon bond requires approximately $400\,\mathrm{\frac{kJ}{mol}}$! However, if we are merely talking about association as in complexation, the energies involved (i.e. gained or lost) are rather small. This es especially also true for the activation energy, meaning that forming and dissociation of such complexes are rather fast and at equilibrium the whole time. It is important to realise that the bonds formed for the self-replicating complex are all intermolecular and only the bond between the fragments is a covalent, intramolecular bond.
Let’s take a look at an exemplary self-replicating complex:[1]

On the right half of this picture we have the two fragments, on the left we see the template. All solid lines are actual (intramolecular) bonds while all dashed lines are intermolecular associations — in this case hydrogen bonds. The intermolecular dashed interactions build up and break down rapidly with low energy barriers. The new intramolecular bond is formed between carbon and nitrogen on the aromatic rings (not formed on the right, formed on the left) and represents the energy liberated in the reaction. This one is only formed, not broken.
To form the bond, the two fragments need to linger on the template long enough. That is generally no problem, despite rapid association and dissociation.
In fact, form too stable of a complex (i.e. one that won’t dissociate quickly) and you have effectively eliminated the complex’ ability to self-replicate.

[1] A. Terford, G. von Kiedrowski, Angew. Chem. 1992, 104, 626. DOI: 10.1002/ange.19921040523
In English: Angew. Chem. Int. Ed. 1992, 31, 654. DOI: 10.1002/anie.199206541

Great! So the constituent fragments of a self-complimentary molecule are bonded together inTRAmolecularly as they are gathered on the recognition surface (presumably before fragments have a chance to dissociate individually). These bonds require lots of energy to break (and so won't at equilibrium). The gathering process is done easily/quickly through association. Then at some point the formed molecule will then dissociate, which like association requires little energy, is based on intERmolecular forces, and can happen at equilibrium. Is that right?

@Jonathan I added an image and a reference; maybe that can clear the confusion that arose with intramolecular and intermolecular bonds.

Jan, I respectfully disagree with your statement: "Technically, every reaction is at an equilibrium of some kind." Many reactions, such as the burning of hydrogen in air, are not in equilibrium--there is no "reverse burning" going on that dissociates water back into hydrogen and oxygen. I do agree that in considering any reaction, it is worth asking whether there is an equilibrium involved.

@iad22agp I respectfully disagree with your respectful disagreement. Reversablility of all reactions is a fundamental thermodynamic property. Only that in the case of combustion (no matter which one) the equilibrium is 99.999999… % towards the product side.

@Jan, I think you are confusing the term equilibrium with the term reversible. Take a simple system, where two chemicals A and B can interconvert, but have a high kinetic barrier (activation energy) for interconversion. At a sufficiently high temperature, they do interconvert, and an equilibrium is established, where the A:B ratio reflects their relative stabilities. However, at a low temperature, they do not interconvert, so you can add more A or more B and obtain a non-equilibrium mixture that remains stable at low temperature. (contd.)

(contd.) That system - at that temperature - is not in a state of thermodynamic equilibrium. It might eventually reach equilibrium if you waited long enough, however.

@iad22agp Let us continue this discussion in chat.


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Why does ammonium carbonate smell?
Why is it that $\ce{(NH4)2CO3}$ smells while other salts with ammonium like $\ce{NH4Cl}$ does not smell? Does it have something to do with the equilibrium reaction? I know it has nothing to do with the coefficient of $\ce{NH4}$ in ammonium carbonate. I am pretty sure the water in our nose causes it to dissolve. I do note that $\ce{CO3^2-}$ is basic along with ammonium is acidic while the chloride ion is neutral. So does it have to do with the fact that $\ce{(NH4)2CO3}$ is more neutral than $\ce{NH4Cl}$?

Yes it does, though people usually put it in different words, like "Ammonium carbonate, being a salt of a weak base and a weak acid, is prone to..."

@IvanNeretin can you elaborate, i dont quite understand what you are trying to say

See, what happens when you heat ammonium salt (any of them)?

@IvanNeretin i see, but why does (nh4)2co3 smell stronger than nh4cl?

Basically, because $\ce{H2CO3}$ is a weaker acid than $\ce{HCl}$.

@IvanNeretin how does tha matter? also, can you please expand into an answer?

There are two reasons.

As mentioned by Ivan Neretin in the comments, carbonate is the conjugate base of a much weaker acid than chloride is the conjugate base of.  In other words, carbonic acid is less acidic than hydrochloric acid, and carbonate is more basic than chloride.  This means that carbonate solutions will have higher pHs than chloride solutions, in general.  At higher pH, the equilibrium between $\ce{NH4^{+}}$ and $\ce{NH3}$ is shifted more towards $\ce{NH3}$, which is volatile and smelly.
Carbonate is also potentially volatile.  If you leave ammonium carbonate, even in solution, around forever in an open container, it will eventually completely disappear.
$$\ce{(NH4)2CO3(s) -> H2O(g) + 2 NH3(g) + 2CO2(g) }$$This also favors more vaporization of ammonia.  In contrast, in ammonium chloride, chloride is essentially completely non-volatile.  Thus ammonium could not evaporate without a buildup of negative charge in the solution, highly unfavorable.

Actually, I am curious as to whether there is any appreciable formation of urea via $\ce{(NH4)2CO3 -> CO(NH2)2 + 2H2O}$. Ordinarily I would try to search myself but I'm on mobile now, might get round to it later.

....Urea itself has no smell, so just disregard me.

https://en.wikipedia.org/wiki/W%C3%B6hler_synthesis

@Jan yep that's the experiment that got me thinking, it was ammonium cyanate though. Was just wondering if there might have been something similar with the carbonate


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Negative charge of orbit
Why is the energy of orbit (of electron) negative. Explain.
Electrons are negatively charged and electrons revolve round the nucleus in the orbit hence the orbit is negatively charged but what are the proves for electron's negative charge.

The negative energy of orbitals has nothing to do with the negative charge of electrons. You could have an antimatter world in which a positron orbits an antiproton and the energy of the positron would still be negative. It has to do with the solution of the Schrodinger equation which shows that normalisable eigenfunctions of the Hamiltonian correspond to negative energy eigenvalues (specifically, orbitals are bound states in a Coulomb potential which goes to 0 at infinity, so their energies are negative).

Negative energy of a system means lower energy than a set zero energy level. We set the zero level for an electron-nucleus pair to the energy of independent particles (electron and nucleus are infinite distance). It is just a convention that makes calculation easy, but you could choose the zero anywhere.
If the energy is lower than that (i.e. negative), the particles are interacting with each other and they don't have enough kinetic energy to split apart. If they have positive energy, it means the electron has enough kinetic energy to just fly away from the nucleus wherever it want. When it flies away, it is not an atom anymore. This happens in plasma or when you hit the atoms with some high energy (ionizing) radiation.


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Coordination compounds with alkene ligands
I came across this compound: $\ce{[Pd(η^2-C2H4)Cl2(H2O)]}$
What does that '$\eta^2$' sign mean? I am guessing that it means that the ligand is bidentate because if it wasn't, I don't think it is possible (well it would be rare) for a metal to be only co-ordinates by 3 ligands.
If this is the case, how can the carbon atom in ethene bond to Pd as it doesn't have a lone pair? Does the double break and hence the 2 carbons bond to the metal?

I took the liberty of changing your title. See, it's not really about nomenclature, it's more about the nature of this compound.

‘because if it wasn't, I don't think it is possible (well it would be rare) for a metal to be only co-ordinates by 3 ligands.’ There are two chlorines, water and ethylene. Assuming ethylene be bidentate, that would be five ligands; otherwise four. Perfectly normal for palladium. Also, complexes such as $\ce{[Ag(CN)2]-}$ (two ligands) exist.

Ethylene is a monodentate ligand and there are 4 ligands total: ethylene,  two chlorides, and water.
Regarding hapticity, you can find a great answer here: What does the eta notation mean in the naming of a transition metal complex?
While a chapter can be dedicated to the nature of organometallic bonding, I think it is practical to say that it is the two electrons in the pi-bond that are involved in bonding to the metal center.

It is not carbon that is donating a pair - instead, it is a $\pi$ bond as a whole. Its two electrons, which were shared between two carbons, are now shared between them and palladium in a kind of multicentered bond.
There is a huge class of compounds where organic molecules are attached to a metal in the same way. With larger conjugated $\pi$-systems, we may have ligands of $\eta^5$ coordination (look up ferrocene) and more, all the way to $\eta^8$ and maybe beyond that.

I would +1 this, but the OP obviously does not know about hapticity, so I feel you should include a note on what the notation actually means.

That's covered by the other answer already, so why bother.


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How fast does MgSO4 absorb water from the air?
I had to do a lab to figure out the water of hydration of magnesium sulfate. However, after the heatings, I had a loss 0.45 g of water resulting in 0.63 g of anhydrous magnesium sulfate. This gives 0.02497 mol of water and 0.005237 mol of anhydrous magnesium sulfate. Thus, this would imply that 0.02497/0.005237 = 4.77 = 5 is the water of hydration.
However, my chemistry teacher specifically said that the actual number for the MgSO4 he gave us was 7. I couldn't figure out exactly where I went wrong, but my only guess would be that the wait of 5 minutes for cooling the substance down. Some research shows that MgSO4 is a great drying agent, and the humidity during the time I carried out the experiment was 75% humidity.
Doing the calculations, during the five minutes of cooling, it must have absorbed 0.01169 moles of water in order for 7 to drop to 4.77. Is this plausible given 75% humidity? I don't know anything about Chemical kinetics yet.
Edit: Here are the procedures.
1. Measure the mass of the empty crucible with the lid.

Add some amount of the (powdered) hydrated salt in the crucible, and measure it's total mass with the lid.
Heat the crucible for 20 minutes over a bunsen burner, with lid slightly open.
Wait for 5 minutes, and then measure the amount of mass after heating.

The relevant data is given above.

This is a great question, and unfortunately, I don't have an answer for you.
My first instinct would be to question how well you heated the material and under what conditions. Insufficient heating would mean that you might still have water remaining in the material.
Anhydrous, powdered magnesium sulfate will absorb water relatively quickly at ambient conditions, though it will depend on the temperature of the solid, how anhydrous it is and the humidity of the air. Something more to consider is that the crystalline form will absorb water at a slower rate. Were you dealing with powder or crystals?
Unfortunately, it isn't possible to judge the kinetics without more information. Perhaps you can provide your procedure?

I added the procedures and also I was dealing with powder.

Great, thanks. The following paper should help to answer your question.  Their procedure, by coincidence, is actually quite similar to the procedure you performed. Figure 2 gives a nice idea of what happens to nH2O in humid environment with respect to time. Granted, they are using a powdered form of the monohydrate salt, but I think it is fair to say the results would be similar.  http://www.sciencedirect.com/science/article/pii/S1876610214003087


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pH calculation - adding distilled water after titration
Will someone help me figure out how can I solve the following question?
We are given 110 ml of acetic acid (pKa=4.76), with concentration of 1M, to which we add 10 ml of NaOH with concentration of 1M.
Afterwards, we add the solution to 880 ml of distilled water.
What would be the pH of the solution, after the addition of water?
Will you please help me figure this out?
I obviously need to use Henderson-Hasselbalch here, but how does the addition of water change things?
Thanks a lot in advance

So I presume you can work out the initial H-H equation. I got pH 5.8? might be worth checking.
Then use the equation $$c_1 \cdot V_1 = c_2 \cdot V_2$$
Where $c_1 = [\ce{H+}]$
$[\ce{H+}] = 10^{-\mathrm{pH}}$
so I got $[\ce{H+}] =10^{-5.8} = 1.58489 \times 10^{-6}$
Now we can rearrange our original equation: $(c_1 \cdot V_1)/V_2 = c_2$
$c_1=1.585\times10^{-6}$
$V_1=120\ \mathrm{ml}=0.12\ \mathrm L$
$V_2=(120\ \mathrm{ml}+880\ \mathrm{ml})= 1\ \mathrm L$
$c_2=1.902\times10^{-7}$
$\mathrm{pH}=-\log[\ce{H+}]$
$\mathrm{pH}= -\log(1.902\times10^{-7})$
$\mathrm{pH}= 6.72$
Hope this helps, Its been a long time since I did anything like this so let me know how it goes and if this is correct. cheers.

Perhaps I can help guide you through this problem.
First, what is the reaction that occurs between acetic acid and NaOH? Write it down and check that it is balanced. What are the reactants and what are the products.
Next, using the stoichiometry of the reaction figure out how much reactants and products you have at the end of the reaction (is the limiting reactant used to completion?)
Then you can figure out the molarity of your reactants and products (keep in mind the addition of water will change your final volume)
Finally, enter the molarity of your acid and conjugate base into the H-H equation and solve for pH.
Happy to help you work through it if you get stuck.


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Why does acid react with skin
Why does acid reacts with skin??
I can understand the acid would react with blood as it is alkaline in nature. But when i googled pH of skin, its around 4-5, "it is estimated that the 'natural' skin surface pH is on average 4.7, i.e. below 5"(Reference) The thing that i could guess is that only H+ are the ions that would react with our skin since only H+ are common in every acid. But since our skin is also acidic how will an acid react with another acidic material

The shortest answer I could give is that not every molecule in a skin cell is acidic. Many biomolecules are also prone to hydrolysis under strongly acidic conditions

but what about dil. sulfuric acid or dil. acetic acid. acetic acid is even less acidic but there is still burning sensation with a bit if conc. acetic acid

1 molar (that is: dilute) sulphuric acid still has a pH of approximately zero, so for sure it will react somehow. Acids can react with much more than ‘just’ alkaline solutions.

isn't the question narrow enough to answer??? it is as simple that "why does acid reacts with our skin to give burns?" :(

Acids first participate proteins in the skin to form a barrier.  Alkali compounds tend to digest tissue and penetrate deeper into the skin/body.


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How could I find the solubility of hydrocarbons such as iso- and n-Butane in liquid Methane?
I am trying to figure out the dissolution capacity of iso and normal butane in methane at very low temperatures (110K). This would allow me to make compositions that prevent freeze out of these hydrocarbons at these very low temperatures. For example I would like to prepare a 5% mix of iso-butane in methane and make some measurements at 110K. However, if the solubility of methane is only 3%, I will have a 2% bias in my final measurement.
Any ideas would be much appreciated.
Thanks.


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Why is the combination of hydrated aluminium oxide and magnesium hydroxide used to treat heartburn?

Why is the combination of hydrated aluminium oxide and magnesium hydroxide used to treat heartburn?

I get why $\ce{Mg(OH)2}$ would be used since its a strong base and is badly soluble in water. Thus it won't do harm to the oesophagus in comparison to something like $\ce{NaOH}$. Aluminium oxide is also poorly soluble, however it is a weak base so what is the point of combining this substance with magnesium hydroxide instead of just increasing the dose of magnesium hydroxide?

Different antacids (used to treat heartburn) have different chemical constitutions - see this table on Wikipedia. As to why someone would choose one over the other, I admittedly have no idea

Hydrated $\ce{Mg(OH)2}$, aka milk of magnesia, is an osmotic (saline) laxative, as well as antacid.
$\ce{Al(OH)3}$ is also an antacid, but has the opposite effect on the bowels: it's costive, i.e. a binding agent. Aluminum has also been linked (with much uncertainty!) with Alzheimer's disease, so it has become less popular as an antacid.
The combined effects of $\ce{Mg(OH)2}$ as laxative and $\ce{Al(OH)3}$ as a costive might tend to counteract each other, so as to avoid interfering with colonic function.
However, it would be far simpler to use $\ce{CaCO3}$ as an antacid, unless the release of $\ce{CO2}$ bubbles or excess calcium in the diet is an issue.


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Why do transition metals form particularly stable complexes?
The pre-transition metal ions can form complexes but the ligands are weakly bound unless they are multidentate and thus benefit from the chelate effect. For example, crown ethers and cryptands.
However, transition metal ions can form a vast array of stable of stable complexes with monodentate and multidentate ligands. Why is there such a difference between the pre transition elements and the transition metals in this respect.
My initial thoughts are that the transition metal ions will have a higher charge and smaller radius so the electrostatic attraction is larger. I also thought about the d-orbitals but the pre transition elements have d orbitals available, they're just not filled. As a result, is increased stability related to exchange energy or perhaps the lower energy d orbitals? Are there any more factors? Are my suggestions valid?

To illustrate my answer, I am going to use a modified version of the picture of a simplified molecular orbitals scheme Professor Klüfers (LMU Munich) uses in his internet scriptum to the coordination chemistry I course.

On the left you can see the metal’s orbitals, on the right the group orbitals of six $\unicode[Times]{x3C3}$ donors (assuming no $\unicode[Times]{x3c0}$ interactions of any kind). $\mathrm{e_{g}}$ and $\mathrm{t_{2g}}$ correspond to the metal’s d-orbitals, $\mathrm{a_{1g}}$ is the 4s orbital and $\mathrm{t_{1u}}$ are the 4p-orbitals.
You can clearly see that the 3d-orbitals are similar in energy to the ligand group orbitals, and that they have the correct symmetry to mix. The greatest stabilisation comes from the $\mathrm{e_g}$ orbitals.
Empty d-orbitals can thus be thought to help with coordination by stabilising the coordination compound. Completely filled d-orbitals (e.g. $\ce{Zn^2+}$) are contraproductive and therefore zinc’s coordination chemistry is much like sodium’s.


text:
Original concentration calculation homework
A student performs the following gravimetric analysis of iron ions in a water system
$$\ce{Fe^2+  (aq) + CO3^2-  (aq) <=> FeCO3 (s)}$$
If this student used excess carbonate and obtained the following data, what was the original concentration in moles/L) of $\ce{Fe^2+}$ in the water sample?
Total Volume of Solution: 100.00 mL
Mass of $\ce{FeCO3}$ collected: 23.758 grams.
Change 100 ml to 0.1 liters $x + 115.583 = 23.758 \times 0.1 = 0.02~\mathrm{l}$
- would this be how you solve this problem?

Hi, just a heads up: in order to make a new paragraph, you have to make two line breaks after a paragraph. Regarding your solution itself, the first thing I noticed is that you converted mL to L wrongly. Apart from that I'm afraid you'll have to make it a little clearer what you are trying to say. What does $x$ represent? Where does 11.583 come from?

Trying to find the original concentration of the equation I used M1V1 = M2V2

liters should be .1 liters   M1 is the molar mass of FeCO3 = 115.853g  X V1 = 23.758g X .1 litter = .02 liters - would this be the original concentration

The mass of $\ce{FeCO3}$ is given as $m_{\ce{FeCO3}}=23.758\ \mathrm g$.
You already found that the molar mass of $\ce{FeCO3}$ is $M_{\ce{FeCO3}}=115.85\ \mathrm{g\ mol^{-1}}$.
For a pure sample, molar mass $M$ is defined as
$$M=\frac mn$$
where $m$ is mass and $n$ is amount of substance.
Thus, the amount of $\ce{FeCO3}$ can be calculated as follows:
$$\begin{align}n_{\ce{FeCO3}}&=\frac {m_{\ce{FeCO3}}}{M_{\ce{FeCO3}}}\\[6pt]
&=\frac {23.758\ \mathrm g}{115.85\ \mathrm{g\ mol^{-1}}}
\end{align}$$
The chemical equation
$$\ce{Fe^2+ + CO3^2- <=> FeCO3}$$
indicates that $1\ \mathrm{mol}$ of $\ce{Fe^2+}$ yields $1\ \mathrm{mol}$ of $\ce{FeCO3}$. Therefore,
$$n_{\ce{Fe^2+}}=n_{\ce{FeCO3}}$$
The volume of the solution is given as $V=100.00\ \mathrm{ml}=0.10000\ \mathrm l$.
The concentration $c$ is defined as
$$c=\frac nV$$
where $n$ is amount of substance and $V$ is the volume of the solution.
Thus, the concentration of $\ce{Fe^2+}$ can be calculated as follows:
$$\begin{align}c_{\ce{Fe^2+}}&=\frac {n_{\ce{Fe^2+}}}V\\[6pt]
&=\frac {n_{\ce{Fe^2+}}}{0.10000\ \mathrm l}\end{align}$$

I consider this to be the blueprint of how to answer [tag:homework] questions!

authors: Tàkëìtì Bõårï


text:
Zwitterion (amino acid) equilibrium and the equilibrium constant
I just had my biology test paper back from my teacher, and my teacher insisted that when functioning as a buffer in an alkaline solution, the amino acid deprotonates by the ammonium group instead of the carboxyl group, because amino acids in solution are mostly zwitterionic. I thought zwitterions only consist a small portion of the mixture, but checking Wikipedia only further confirms what my teacher said.
In an attempt to find mathematical proof, I tried to derive the equilibrium constant between zwitterionic and completely neutral forms of amino acids, with the assumption that the solution is pure and only constitutes the said acid(and its zwitterionic isomer) and water. This was my working:
Using $K_\mathrm{a}$ for carboxyl group and $K_\mathrm{b}$ for amine group.
$$\ce{H2N-CHR-COOH <=> H3N+ -CHR-COO-}$$
\begin{align}
[\ce{H2N-CHR-COOH}]   &=  [\ce{H2N\bond{-}}] = [\ce{-COOH}]\\
[\ce{H3N+ -CHR-COO-}] &=  [\ce{H3N+\bond{-}}] = [\ce{-COO^-}]
\end{align}
\begin{align}
K_c &= \frac{[\ce{H3N+ -CHR-COO-}]}{[\ce{H2N-CHR-COOH}]}\\
    &=\sqrt{\frac{[\ce{H3N+-CHR-COO^-}][\ce{H3N+-CHR-COO^-}]}{[\ce{H2N-CHR-COOH}][\ce{H2N-CHR-COOH}]}}\\
    &=\sqrt{\frac{[\ce{H3N+\bond{-}}][\ce{-COO^-}]}{[\ce{H2N\bond{-}}] [\ce{-COOH}]}}\\
    &=\sqrt{\frac{[\ce{H3N+\bond{-}}]}{[\ce{H2N-}][\ce{H+}]} \times
      \frac{[\ce{-COO^-}][\ce{H+}]}{[\ce{-COOH}]}}\\
    &=\sqrt{\frac{[\ce{H3N+\bond{-}}][\ce{OH-}]}{[\ce{H2N\bond{-}}][\ce{H+}][\ce{OH-}]} \times \frac{[\ce{-COO^-}][\ce{H+}]}{[\ce{-COOH}]}}\\
&\text{(now assuming $[\ce{H2N\bond{-}}] = [\ce{H2O.H2N\bond{-}}]$)}\\
    &=\sqrt{\frac{[\ce{H3N+\bond{-}}][\ce{OH-}]}{[\ce{H2O.H2N\bond{-}}][\ce{H+}][\ce{OH-}]} \times \frac{[\ce{-COO^-}][\ce{H+}]}{[\ce{-COOH}]}}\\
    &=\sqrt{\frac{K_\mathrm{b}}{K_\mathrm{w}} \times K_\mathrm{a}}\\
    &=\sqrt{\frac{K_\mathrm{a} K_\mathrm{b}}{K_\mathrm{w}}}
\end{align}
With $K_\mathrm{w}$ being the denominator, it is easy to see that $K_c$ would be sufficiently large that nearly all amino acids in aqueous solution exists in the zwitterionic form, that is contrary to my initial belief; bringing in the $K_\mathrm{a}$ and $K_\mathrm{b}$ values further confirms that.
However, I am not content with just that. I'd like to ask:

What are the faults (calculation errors, false/inappropriate assumptions, etc) in my working, if any?
What is the main reaction mechanism that mediates this equilibrium? Is it an intramolecular transition or is it brought about by the solvent, or something else?
Can this be considered as a form of tautomerization? If so, is it prototropy? Hence, is the $K_c$ above the so-called tautomeric ratio?

Let’s start by saying that there are two $\mathrm{p}K_\mathrm{a}$ values: one corresponding closely to that of a carboxylic acid, one corresponding close to that of an amino group. Also, since protonation and reprotonation of both groups is completely independent you should have an easier way of arriving at this solution, too. Maybe I’ll write something up later.

@Jan the pKa above was that of the carboxyl group, and pKb that of the amino group. Or are you saying that I need the amino pKa too, though IMHO that seems very insignificant.

No, that was meant along the lines of ‘each amino acid can be deprotonated (at least) twice. The first deprotonation chemically corresponds nicely to a carboxylic acid, the second to an amino group’. I didn’t read through your argument well enough (yet) to be sure if it’s good or bad ^^'

@Jan Well that is what I was referring to in the 2nd sentence in my last comment. IIRC the pKa of the amine group was around 9 so I guess I could omit that, but maybe I'm wrong.

If you assume that the groups react independently, you have two equilibria with equilibrium constants $K_a(\ce{-NH3+})$ and $K_a(\ce{-COOH})$:
$$\ce{^+(H3N)-R <=> (H2N)-R + H+}$$
and
$$\ce{R-COOH <=> R-COO- + H+}$$
You are interested in the overall reaction
$$\ce{^+(H3N)-CHR-COO- <=> (H2N)-CHR-COOH}.$$
You can take a two step route to achieve this:
$$\ce{^+(H3N)-CHR-COO- <=> (H2N)-CHR-COO- <=> (H2N)-CHR-COOH}$$
The equilibrium constant for the first step is $K_a(\ce{-NH3+})$, and for the second step is $1/K_a(\ce{-COOH})$. Combined, you get a result of
$$ K_\text{combined} = \frac{K_a(\ce{-NH3+})}{K_a(\ce{-COOH})}$$
Because the $K_a$ values of ammonium and carboxylic acid are different by about four orders of magnitude, the uncharged species occurs about ten thousand times less often than the zwitter ionic form in aqueous solution. In general, comparing p$K_a$ values of two groups competing for a proton tells you which group will usually have the proton (the one with the larger p$K_a$) and how unlikely a proton transfer to the other group would be ($10^{\Delta pK_a}$).

What are the faults (calculation errors, false/inappropriate assumptions, etc) in my working, if any?

I don't understand line 2 and 3 of your argument, and how you get from line 5 to 6. The result you get differs from mine in a square root.

I thought zwitterions only consist a small portion of the mixture

This depends on the solvent. In aqueous solution, zwitterions are the common species, whereas in organic solvent, the uncharged species is the common one.

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


text:
what is the reaction between copper and an acid
what happens when copper is reacted with hydrochloric acid may be dilute or concentrated.Should it displace hydrogen?if not what will happen if yes why?i think it will not react because copper is placed below hydrogen

Well, you are pretty much correct, but if you are interested you may want to read about $E^\circ$ values to find out why copper is "below hydrogen".

you can react it if you use hydrogen peroxide. I believe you can etch copper on PCBs this way


text:
Can nitric acid be used to create an acidic medium for redox reactions?
While reading about redox reactions, every time I found that $\ce{H2SO4}$ is used for creating an acidic medium.
My question is - can I use nitric acid to create an acidic medium?

One reason why nitric acid isn't often used is because it is also a rather strong oxidising agent. That leads to undesired reactions

Yes, you can.
However, nitric acid is an oxidizer, so it can interfere in some cases. If a strong reducer (like magnesium) is involved, nitric acid may act as an oxidizer even in diluted solutions. Obiously, it acts as an oxidizer in concentrated solutions, producing rather nasty nitrogen oxides, and if used without prior knowledge, it may produce unstable explosives.
For this reason, one generally use hydrochloric acid or sulfuric acid. Mind you, in case oxidation is not a problem, nitric acid may be used with no problems. Another reason to use sulfuric acid is that it is dirt cheap. It is the cheapest alternative.
Still, in concentrated solutions sulfuric acid does show oxidating properties, so in some cases sulfuric acid is not suitable either. Unfortunately, there is no common and reasonable alternative in the case, except significanlty weaker phosphoric acid.

If magnesium is involved, any acid will definitely act as an oxidizer even in diluted solutions.

@IvanNeretin chuckle Most acids will symply produce hydrogen though, not an unholy mixture of various toxic products.

That's true, $\ce{HNO3}$ is quite special.


text:
Structure of Benzyne
I have seen in a lot of pictures that benzyne looks like:

But aren't $\ce{C_6}$ and $\ce{C_5}$ $-$ $\ce{sp}$ hybridized?
And hence bond angle should be $180^{\circ}$ or at least near about it?
Something like :

Or,

?

Please don’t ask two unrelated questions at once. I feel that the cycloallene question could stand by its own right. (But please ask ‘do they exist’ rather than ‘why do they not exist’. If need be, add that you never heard of them.)

Benzyne is but a short-lived reaction intermediate. I'm not sure its geometry is known with any certainty.

Wikipedia has a good summary of the situation here. Remeber that geometry determines hybridisation, not the other way round - i.e just because a triple bond is depicted doesn't mean there is $sp$ hybridisation.

related http://chemistry.stackexchange.com/questions/215/why-is-benzyne-an-intermediate

To address an implicit question, it is standard practice to draw line structures with simple angles and bond lengths and leave it to the reader to interpret what the real structure might actually look like based on their knowledge and intuition.

There's been an interesting recent development on the nature of arynes such as benzyne, by direct observation of individual molecules through microscopy. It seems that the molecular geometry is more consistent with a ring containing four double bonds (three of them consecutive) rather than two double bonds and a triple bond.

Yes, but.
Yes, in the conventional low-level models, one would consider the two carbons in benzyne you mentioned as sp-hybridised. And that does mean that their orbitals seem to be pointing the wrong way.

However, a better picture would be to use an orbital which is a lot closer to sp²-hybridisation, and an even better picture would calculate orbitals computationally (I can’t do that, but Orthocresol can).
The thing is, all these pictures show, that the triple bond in benzyne is rather different from those in linear molecules or large cycloalkynes such as cyclodecyne ($\ce{C10H16}$). It contains a large angular strain, not unlike very small saturated rings such as cyclopropane. The strain causes the triple bond to be rather weak and easily cleaved/attacked — which paves the way for all reactions in which benzyne is an intermediate.
Larger cycloalkynes with eight or nine carbons have been isolated, but for benzyne only in situ studies have been made according to my quick check on SciFinder two seconds ago. In layman’s terms, you can think of the angle strain by saying ‘I am a triple bond, I want a $180°$ angle, but all I can have is $120°$ so I’m rather unhappy!’ A straight-up $180°$ angle is not possible, however, because of the other two atoms in the benzyne ring.
Kukolich, McCarthy and Thaddeus published two papers (one together with Tanjaroon) on microwave-spectrum measurements to determine the structure and compare it to in silico structures. Their results give a $126°$ angle along $\ce{C^1-C^6-C^5}$ (your numbering scheme). This is more obtuse than in benzene ($120°$) but still far away from $180°$. However, the major constraint is the angle $\ce{C^6-C^1-C^2}$, which is compressed to a mere $111°$.[1,2]
Before the objection is raised and as ron pointed out in the comments: The triple bond is indeed a bond and not a diradical as can be derived from its behaviour. E.g, it concertedly and stereospecificly adds to dienes.

References
[1] S. G. Kukolich, C. Tanjaroon, M. C. McCarthy, P. Thaddeus, J. Chem. Phys. 2003, 119, 4353. DOI: 10.1063/1.1593015
[2] S. G. Kukolich, C. Tanjaroon, M. C. McCarthy, P. Thaddeus, J. Phys. Chem. A 2004, 108, 2645. DOI: 10.1021/jp031344p

Nice answer.  I would just add that although the aryne bond is strained, there is still enough overlap between the two orbitals for it to be considered a bond rather than a biradical.  For example, o-benzyne reacts in a concerted, stereospecific manner with dienes.

@ron Added a line to implement that.

If you are interested in the computational bits: https://chemistry.stackexchange.com/a/114929/16683

@orthocresol I’m always interested in the computational bits; thanks for linking =3


text:
Is there any explosion hazard with an enclosed graphite crucible melting silver?
the crucible volume is about 2 ounces.   The final design of the crucible will be enclosed tightly except for a small .5mm diameter hole at the bottom and a 1mm diameter hole at the top through which silver solder will feed.
I was wondering.  In my experiments should I worry about the thing exploding?

you should worry about your graphite slowly oxidizing. Slowly.

So producing CO2?

Well, if both holes get clogged somehow, and then you seal it shut and start heating... though the volume is too small for a decent explosion, this does not sound good.

What type of shield do you recommend?  Will some acryllic panels work?  The size of the crucible internally is now 6mm ID x 30mm Depth.

Silver will absorb oxygen readily from air while molten and spatter as it cools. This is hardly an explosion, but some care should be taken to avoid being spattered with hot particles ejected from the crucible or plugging the holes.


text:
Is it possible to make a fuel cell that runs on triglyceride?
I apologize if this is a silly question. The reason I ask is I'm wondering if it would be possible to build a fuel cell that runs on body fat. I know ones for butane exist, could they work for this? If not, is there any good reason it isn't possible?
Basically I'm imagining some kind of flexible inductive charger in your hand, or wrist for a smart watch, and connected to a fuel cell that's connected to the circulatory system and runs off body fat.
It just seems silly to lug external batteries along on long on bike rides when the whole point of the ride is to get rid of excess chemical energy anyway.
I obviously don't have the skills to pursue this, but I just want to know if the dream is silly. :)
Thanks!

It is certainly possible to oxidize fats electrochemically, but there are a few problems with the idea that I can think of:

Triglycerides do not flow freely through the blood as they are too hydrophobic and are instead carried by lipoproteins that would have to be removed. Similarly, free fatty acids are transported bound to serum albumin.
The blood concentration of triglycerides and fatty acids is relatively low most of the time in healthy individuals.
Existing fuel cells that work on fatty acids seem to be mostly microbial fuel cells, which are quite slow.

I think it would be more practical to use glucose from blood instead, using an enzymatic fuel cell, as blood glucose levels are relatively high and the oxidation is easier to accomplish (though there are still a lot of technical challenges to be resolved—biofouling, efficiency, device lifetime, etc.).

Thanks a glucose based fuel cell looks like something that is in active development. Very exciting!

Glycerol might make a good fuel for fuel cells. There is currently a glut of it on the market.


text:
How to round significant figures correctly?
Calculate how many grams of p-nitrophenol are required to prepare $\pu{250 mL}$ of a $\pu{11 mM}$ solution. (Answer to 3 significant figures.)
I worked it out as
\begin{align}
\frac{250}{1000}\,\mathrm{L} \times (\pu{11E-3 M}) \times \pu{139.11 g//mol}
  &= \pu{2.75E-3 mol} \times \pu{139.11 g//mol} \\
  &= \pu{0.3825525 g}
\end{align}
Is that $\pu{0.383 g}$ or $\pu{0.382 g}$ when approximated to 3 significant figures?
I think it is $\pu{0.383 g}$, but that's only going by the general rule of "up" approximating a five.

I'll point out that you made a mistake when multiplying 250 ml times 11. The result is 2.750 not 2.75. 250 ml implies +/- 0.5 ml.  If it was +/- 10 ml then it should have be written as $2.5 * 10^2$ ml.

The 11 is the "limiting" precision. It is 11 +/- 0.5.

In rounding with significant figures, you want to avoid undue precision, yes, but it's also important to minimize error. It's this error minimization which is the primary goal of all "round to nearest" approaches.
So in short:
\begin{align}
0.383 - 0.3825525 &= 0.0004475\\
0.3825525 - 0.382 &= 0.0005525
\end{align}
$0.0004475$ is less than $0.0005525$, so you should round up to $0.383$, as that's the nearest value, and will minimize any error you introduce by rounding.
Even & Odd?
So what's this about the even/odd rule? Is there any validity there?
This technique is called "Banker's rounding" and is also attempting to minimize error.
When a number falls exactly in the middle of the two values to be rounded to (that is, when the digit to round is "5" exactly), each of the two difference is equally far away. So to a first order, both will result in an equal amount of error.
But picking a consistent direction (e.g. "always round up") introduces a subtle bias in the sums over the long run. This is because the errors are biased.
.     |  Always  |  Up   | Banker's | Banker's |
Value | Round Up | Error | Rounding | Error |
------+----------+-------+---------+------+
2.0   |     2    |  0.0  |    2    |  0.0  |
2.1   |     2    | -0.1  |    2    | -0.1  |
2.2   |     2    | -0.2  |    2    | -0.2  |
2.3   |     2    | -0.3  |    2    | -0.3  |
2.4   |     2    | -0.4  |    2    | -0.4  |
2.5   |     3    | +0.5  |    2    | -0.5  |   <<<<<
2.6   |     3    | +0.4  |    3    | +0.4  |
2.7   |     3    | +0.3  |    3    | +0.3  |
2.8   |     3    | +0.2  |    3    | +0.2  |
2.9   |     3    | +0.1  |    3    | +0.1  |
3.0   |     3    |  0.0  |    3    |  0.0  |
3.1   |     3    | -0.1  |    3    | -0.1  |
3.2   |     3    | -0.2  |    3    | -0.2  |
3.3   |     3    | -0.3  |    3    | -0.3  |
3.4   |     3    | -0.4  |    3    | -0.4  |
3.5   |     4    | +0.5  |    4    | +0.5  |   <<<<<
3.6   |     4    | +0.4  |    4    | +0.4  |
3.7   |     4    | +0.3  |    4    | +0.3  |
3.8   |     4    | +0.2  |    4    | +0.2  |
3.9   |     4    | +0.1  |    4    | +0.1  |
4.0   |     4    |  0.0  |    4    |  0.0  |
------+----------+-------+---------+-------+
63.0  |    64    | +1.0  |    63   |  0.0  |   - Totals

With "always round up" you get a consistent $+0.5$ bias per decade, and this is consistent even if you extend the range to more decimal places. You can balance the errors for $0.00...01$ to $0.49...99$ with those for $0.99...99$ to $0.50...01$, but there's that dead center $0.50...00$ (aka $0.5$) which is the odd man out - consistent rounding up (or down) will introduce a bias because of this entry.
So what are you to do? Well, introduce a rule which rounds that entry up half the time and rounds down the other half. Bankers' rounding is the most common such rule. This rule says that if the digits to be rounded off are $5$ exactly, then round such that the last digit of the number after rounding is even. (Note that this is equivalent to a rule to round up if the digit to the left of the $5$ is odd, but round down if the digit to the left is even - two different ways of saying the same thing.)
0.255 -> 0.26 ... because 6 is even (or round up because 5 is odd)
0.265 -> 0.26 ... because 6 is even (or round down because 6 is even)
0.275 -> 0.28 ... because 8 is even (or round up because 7 is odd)
0.285 -> 0.28 ... because 8 is even (or round down because 8 is even)

But again, this is the rule only if the number to be rounded falls exactly between the two possibilities. If the number is closer to one or the other, then round to the nearest.
0.2453   -> 0.25 (Round to nearest)
0.2553   -> 0.26 (Round to nearest)
0.2651   -> 0.27 (Round to nearest)
0.275000 -> 0.28 (Falls exactly between 0.27 and 0.28, apply even/odd rules)
0.275378 -> 0.28 (Round to nearest)
0.285000 -> 0.28 (Falls exactly between 0.28 and 0.29, apply even/odd rules)
0.285023 -> 0.29 (Round to nearest)

Note also that you should only round once, and don't cascade your rounding:
0.14555 -> 0.1    (Round to nearest)
0.14555 -> 0.1456 -> 0.146 -> 0.15 -> 0.2  (WRONG!)

I agree with your objection but you have the wrong examples. The first table should be for 0.5, 1.5, 2.5 and so on. Over 10 possibilities rounding with even odd "balances." //The second table is wrong. You compare to left digit. // The last table should compare rounding a 4 digit number to 3 digits or to 2 digits. So 53 always rounds up since it is more than 50. So for 0.2550  you'd have to apply the odd/even rounding rule when rounding to two digits, but you would not need the odd/even rule to round 0.2553 to two digits.

@MaxW The first table is just to demonstrate that there's a bias, not how Bankers' rounding fixes it - I'll extend it to help clarify. On the second table, I think you're using the same rule, but with different phrasing - I'll clarify. On the third I'll add additional examples.

In second table you have 0.2X5. Rounding up/down depends on if the X is even or odd. // In third table 2.553 rounds to 2.55 if we rounding to 2 decimals and 2.6 if we're rounding to one decimal. But if we rounding to two decimals then 53 is bigger than 50 and it is always up. // Think of it this way. 2.549 rounds to two decimals as 2.5. You don't round to 2.55 then to 2.6.

In your bankers example 2.5 round to 2, so the "error" averages out. That is why you don't have to flip a coin at 5 (which is wrong in the accepted answer).

Would it be ok for me just to edit your answer? If you don't like my edits you can flip it back.

I agree; when there are numbers to the right of 5, it does not stay just 5, it becomes a quantity bigger than 5, thus it makes sense to round 5 up in this case.

My teacher gave a rationale for rounding to the even number, even if the ending is not EXACTLY 5.  For example 0.2533 would round to 0.2.  His rationale was: due to imprecision the digits after the 5 must be ignored because they are insignificant.  Could you explain why this is wrong?

Random rounding rules do, in general, assume that digits are largely independent. However, for example, in the realm of commerce and price marketing, prices are not likely composed of random digits. So, depending on how the underlying numbers were created, perhaps one should think twice before assuming on aggregation some perceived benefit.


text:
Do hydrogen peroxide stabilizers affect oxidizing ability
I plan to use hydrogen peroxide for a redox reaction in oxidizing $\ce{I-}$ to $\ce{I}$.  I found 20 volume hydrogen peroxide as a hair developer, but it contains stabilizer.  I imagined these would have little effect on its oxidizing potential, but I wanted to make sure.  Will any of the common stabilizers used for hydrogen peroxide inhibit its oxidizing ability.

Take a tiny droplet of each solution and check if they will react. Chemistry is an experimental science, after all.

I imagine the purpose of the "stabiliser" is there to inhibit the disproportionation to H2O and O2, not to stop it from oxidising other stuff. Although there might very well be some effect since disproportionation is essentially H2O2 oxidising itself... Ivan is right, just test it yourself.

Shouldn't hinder your oxidation much, just keep in mind they will still likely be around when your reaction is finished.


text:
Molar concentration calculation with absolute uncertainty
A solution of $\ce{F-}$ is prepared by dissolving $(0.0933\pm0.0004)\ \mathrm g$ of $\ce{NaF}$ $(M= (41.989\pm0.001)\ \mathrm {g/mol})$ in $156.00\pm\ \mathrm{mL}$ of water. Calculate the concentration of $\ce{F-}$ in the solution and its absolute uncertainty.
First of all, uncertainty aside, I got $0.0141$ for the concentration of $\ce{F-}$. However, this is not the correct answer (according to Sapling Learning).
Furthermore, what technique should I use to find the absolute uncertainty?

The concentration is correct but you might have had a rounding error introduced since I got the value 0.0142 mol/L. See this answer where I explain the rounding error. Bear in mind the textbook may be wrong - it happens, not infrequently. 2) What is the uncertainty in the volume of water? 3) Google for "error propagation" and try to see if you can apply those formulas - there's a handy table here

Generally, it is preferable to write down the complete quantity equation before plugging in the numbers. If necessary, perform all mathematic operations on quantities symbolically. Do not plug in the numbers until you have only one equation for the desired result. The same applies to the equation for the uncertainty of the desired result.
In this case, the concentration $c$ shall be calculated. The mass $m=0.0933\ \mathrm g$, the molar mass $M=41.989\ \mathrm{g\ mol^{-1}}$, and the volume $V=156.00\ \mathrm{ml}$ are given. The uncertainty of the mass $m$ and the molar mass $M$ are given as $u(m)=0.0004\ \mathrm g$ and $u(M)=0.001\ \mathrm{g\ mol^{-1}}$, respectively. The uncertainty of the volume $V$ is not given; however, when a number is given without any further information, it is generally interpreted so that the last digit is rounded. Therefore, the given value $V=156.00\ \mathrm{ml}$ is assumed to represent a value between $155.995\ \mathrm{ml}$ and $156.005\ \mathrm{ml}$, or $V=\left(156.000\pm0.005\right)\ \mathrm{ml}$. Thus, the assumed uncertainty of $V$ is $u(V)=0.005\ \mathrm{ml}$.
The concentration $c$ is defined as
$$c=\frac nV\tag1$$
where $n$ is amount of substance, which is unknown.
Since the molar mass $M$ is defined as
$$M=\frac mn\tag2$$
the concentration $c$ can be calculated according to
$$\begin{align}c&=\frac m{M\cdot V}\tag3\\[6pt]
&=\frac {0.0933\ \mathrm g}{41.989\ \mathrm{g\ mol^{-1}}\times 156.00\ \mathrm{ml}}\\[6pt]
&=1.42437\times10^{-5}\ \mathrm{mol\ ml^{-1}}\\[6pt]
&=0.0142437\ \mathrm{mol\ l^{-1}}
\end{align}$$
In this case, we may assume that the uncertainties of $m$, $M$ and $V$ are not correlated. Therefore, the relative uncertainty of the concentration $c$ may be estimated as
$$\begin{align}
\left(\frac{u(c)}{c}\right)^2 &= \left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2\tag4 \\[6pt]
\frac{u(c)}{c} &= \sqrt{\left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2}\tag5
\end{align}$$
Thus, the absolute uncertainty of the concentration $c$ is
$$\begin{align}
u(c) &= \sqrt{\left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2} \cdot c\tag6 \\[6pt]
&= \sqrt{\left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2} \cdot \frac m{M\cdot V}\tag7 \\[6pt]
&= \sqrt{\left(\frac{0.0004\ \mathrm g}{0.0933\ \mathrm g}\right)^2 + \left(\frac{0.001\ \mathrm{g\ mol^{-1}}}{41.989\ \mathrm{g\ mol^{-1}}}\right)^2 + \left(\frac{0.005\ \mathrm{ml}}{156.00\ \mathrm{ml}}\right)^2} \cdot \frac {0.0933\ \mathrm g}{41.989\ \mathrm{g\ mol^{-1}}\times 156.00\ \mathrm{ml}} \\[6pt]
&=6.10687\times10^{-8}\ \mathrm{mol\ ml^{-1}}\\[6pt]
&\approx0.00006\ \mathrm{mol\ l^{-1}}
\end{align}$$
Note that the same result may be obtained using the general formula
$$\begin{align}
u(c) &= \sqrt{\left(\frac{\partial c}{\partial m}\right)^2\cdot u^2(m)+\left(\frac{\partial c}{\partial M}\right)^2\cdot u^2(M)+\left(\frac{\partial c}{\partial V}\right)^2\cdot u^2(V)}\tag8 \\[6pt]
&= \sqrt{\left(\frac1{M\cdot V}\right)^2\cdot u^2(m)+\left(-\frac m{M^2\cdot V}\right)^2\cdot u^2(M)+\left(-\frac m{M\cdot V^2}\right)^2\cdot u^2(V)}\tag9 \\[6pt]
&= \sqrt{\left(\frac1{41.989\ \mathrm{g\ mol^{-1}}\times 156.00\ \mathrm{ml}}\right)^2\times \left(0.0004\ \mathrm g\right)^2\\
+\left(-\frac {0.0933\ \mathrm g}{\left(41.989\ \mathrm{g\ mol^{-1}}\right)^2\times 156.00\ \mathrm{ml}}\right)^2\times \left(0.001\ \mathrm{g\ mol^{-1}}\right)^2\\
+\left(-\frac {0.0933\ \mathrm g}{41.989\ \mathrm{g\ mol^{-1}}\times \left(156.00\ \mathrm{ml}\right)^2}\right)^2\times \left(0.005\ \mathrm{ml}\right)^2}\\[6pt]
&=6.10687\times10^{-8}\ \mathrm{mol\ ml^{-1}}\\[6pt]
&\approx0.00006\ \mathrm{mol\ l^{-1}}
\end{align}$$
The numerical values of the concentration $c$ and its uncertainty $u(c)$ should not be given with an excessive number of digits. In this case, it suffices to quote $u(c)=0.00006\ \mathrm{mol\ l^{-1}}$ and accordingly $c=0.01424\ \mathrm{mol\ l^{-1}}$; i.e. $c=\left(0.01424\pm0.00006\right)\ \mathrm{mol\ l^{-1}}$ or $c=0.01424(6)\ \mathrm{mol\ l^{-1}}$. In some cases, however, it may be necessary to retain additional digits to avoid round-off errors in subsequent calculations.

Note
The large equations for the uncertainty $u(c)$ may be confusing and susceptible to errors.
To verify the result, you should check whether the calculated quantity and its absolute uncertainty have the same unit (here: $\left[c\right]=\left[u(c)\right]=\mathrm{mol\ ml^{-1}}$).
Furthermore, you should compare the relative uncertainty of the result with the relative uncertainty of the input data. Usually, the relative uncertainty of the result is greater than or equal to the relative uncertainty of the input data. In this case:
$$\frac{u(m)}{m}=\frac{0.0004\ \mathrm g}{0.0933\ \mathrm g}=0.0042872\approx0.43\ \%$$
$$\frac{u(M)}{M}=\frac{0.001\ \mathrm{g\ mol^{-1}}}{41.989\ \mathrm{g\ mol^{-1}}}=2.38158\times10^{-5}\approx0.0024\ \%$$
$$\frac{u(V)}{V}=\frac{0.005\ \mathrm{ml}}{156.00\ \mathrm{ml}}=3.20513\times10^{-5}\approx0.0032\ \%$$
$$\frac{u(c)}{c}=\frac{6.10687\times10^{-8}\ \mathrm{mol\ ml^{-1}}}{1.42437\times10^{-5}\ \mathrm{mol\ ml^{-1}}}=0.0042874\approx0.43\ \%$$
We find that, in this case, the uncertainty of the concentration $c$ is mainly caused by the uncertainty of the mass $m$.