text:
Anion formation in an alpha particle-neutral atom interaction?
How can an alpha particle/beta particle/gamma rays interacting with a neutral atom lead to an anion forming? Because i can see how they cause loss of electrons leading to cations forming, but how can anions form?
cheers

When high energy particles rip through matter a great portion of the interaction is with electrons. The electrons which are ionized from atoms just don't vanish. So for every cation of some sort that is created there is an anion of some sort created.

If all you have is an atom and an alpha particle whizzing by to kick an electron out of the atom then yes. But how often do you see an atom sitting out on it's own?
If everything happens as before except that now there are other atoms around, when that electron is knocked loose of the first atom, it will likely come to rest in some other atom thereby creating an anion.


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Single step transformation of a carboxylic acid into an aldehyde
What are the different ways of transforming a carboxylic acid into an aldehyde in a single step ? (as opposed as for instance first transforming it to an ester then to an aldehyde)
The bottom of this page doesn't indicate any such way.
I am not looking for a "big-list", so the most common ways would be enough.

DIBAL is the best way I know of.

@DGS The link in my question says that DIBAlH reduces it to alcohols. Hence the 'need' to transform it into an ester before using DIBAlH

You're right. There are instances of success, but it's not as good as I thought.

$\ce{Li, RNH2 (R=Me, Et)/NH4Cl/H3O+}$

Burgstahler, A. W.; Worden, L. R.; Lewis, T. B. Direct Reduction of Carboxylic Acids to Aldehydes by Lithium in Ethylamine. J. Org. Chem. 1963, 28 (10), 2918–2919.
Bedenbaugh, A. O.; Bedenbaugh, J. H.; Bergin, W. A.; Adkins, J. D. Synthesis of aldehydes and secondary amines from carboxylic acids via imines. J. Am. Chem. Soc. 1970, 92 (19), 5774–5775.

$\ce{BH3.SMe2/PCC}$

Brown, H. C.; Rao, C. G.; Kulkarni, S. U. A Convenient Conversion of Carboxylic Acids into Aldehydes. Synthesis 1979, 1979 (9), 704–705.

$\ce{(CH3)2CHC(CH3)2BH2}$

Brown, H. C.; Heim, P.; Yoon, N. M. Selective reductions. XVII. Reaction of thexylborane in tetrahydrofuran with selected organic compounds containing representative functional groups. Comparison of the reducing characteristics of diborane and its alkyl derivatives. J. Org. Chem. 1972, 37 (19), 2942–2950.

$\ce{(CH3)2CHC(CH3)2BHCl.SMe2}$

Brown, H. C.; Cha, J. S.; Nazer, B.; Yoon, N. M. Exceptionally facile reduction of acyclic and alicyclic carboxylic acids to aldehydes by thexylchloroborane-dimethyl sulfide. J. Am. Chem. Soc. 1984, 106 (25), 8001–8002.
Brown, H. C.; Nazer, B.; Cha, J. S.; Sikorski, J. A. Selective reductions. 38. Reaction of thexylchloroborane-methyl sulfide complex in methylene chloride with selected organic compounds containing representative functional groups. Comparison of the reducing characteristics of thexylchloroborane, thexylborane, and diborane. J. Org. Chem. 1986, 51 (26), 5264–5270.
Brown, H. C.; Cha, J. S.; Yoon, N. M.; Nazer, B. Selective reductions. 39. Partial reduction of carboxylic acids with thexylchloroborane-methyl sulfide. A direct and simple aldehyde synthesis. J. Org. Chem. 1987, 52 (24), 5400–5406.

$\ce{(CH3)2CHC(CH3)2BHBr.SMe2}$

Cha, J. S.; Kim, J. E.; Lee, K. W. Facile reduction of saturated and unsaturated carboxylic acids and their salts to aldehydes by thexylbromoborane-dimethyl sulfide. J. Org. Chem. 1987, 52 (22), 5030–5032.
Cha, J. S.; Kim, J. E.; Oh, S. Y.; Lee, J. C.; Lee, K. W. Direct conversion of saturated and unsaturated carboxylic acids into aldehydes by thexylbromoborane-dimethyl sulfide. Tetrahedron Letters 1987, 28 (21), 2389–2392.

9-BBN/Li 9-BBN

Cha, J. S.; Kim, J. E.; Oh, S. Y.; Kim, J. D. One-pot conversion of carboxylic acids to aldehydes through treatment of acyloxy-9-borabicyclo[3.3.1]nonanes with lithium 9-boratabicyclo[3.3.1]nonane. Tetrahedron Letters 1987, 28 (39), 4575–4578.

9-BBN/t-BuLi/9-BBN

Cha, J. S.; Kim, J. E.; Kim, Y. S. One-pot transformation of carboxylic acids into aldehydes through stepwise treatment of acycloxy-9-borabicyclo[3.3.1]nonanes with Full-size image (<1 K)-butyllithium and 9-borabicyclo [3.3.1]nonane. Tetrahedron Letters 1987, 28 (49), 6231–6234.

DIBAL-H

Zakharkin, L. I.; Khorlina, I. M. Zh. Obshch. Khim. 1964 34, 1029; J. Gen. Chem. USSR 1964 34, 1021.

bis(N-methylpiperazinyl)aluminum hydride

Muraki, M.; Mukaiyama, T. Aminoaluminum hydride as new reducing agents. I. Selective reduction of carboxylic acids to aldehydes. Chemistry Letters 1974, 3 (12), 1447–1450.
Hubert, T. D.; Eyman, D. P.; Wiemer, D. F. A convenient synthesis of bis(N-methylpiperazinyl)aluminum hydride: a reagent for the reduction of carboxylic acids to aldehydes. J. Org. Chem. 1984, 49 (12), 2279–2281.

activated silyl carboxylates

Corriu, R. J. P.; Lanneau G. F.; Perrot M. The one-pot conversion of carboxylic acids to aldehydes via activated silyl carboxylates. Tetrahedron Letters 1987, 28 (34), 3941–3944.

titanium-catalyzed Grignard

Sato, F.; Jinbo, T.; Sato, M. The Reduction of Carboxylic Acids to Aldehydes by Dichlorobis[π-cyclopentadienyl]titanium-Catalyzed Grignard Reactions. Synthesis 1981, 1981 (11), 871

N,N-dimethylchloromethyleniminium chloride and lithium tri-t-butoxyaluminum hydride

Fujisawa, T.; Mori, T.; Tsuge, S.; Sato, T. Direct and chemoselective conversion of carboxylic acids into aldehydes. Tetrahedron Letters 1983, 24 (14), 1543–1546.

$o\text{-}\ce{HSC6H4OH, POCl3, HClO4/LiAlH4/H2O, HgCl2}$

Costa, L.; Degani, I.; Fochi, R.; Tundo, P. Pentaatomie heteroaromatic cations. Note III. A Convenient Synthesis of Aldehydes from Carboxylic Acids via 2-Substituted 1,3-Benzoxathiolium Perchlorates. J. Heterocyclic Chem. 1974, 11, 943–948.

$o\text{-}\ce{(NH2)C6H4, PPA/NaOet/MeI/NaBH4}$ or $\ce{LiAlH4/H3O+}$

Craig, J. C.; Ekwurire, N. N.; Fu, C. C.; Walker, K. A. M. Conversion of Carboxylic Acids into Aldehydes and their C-1 or C-2 Deuteriated Derivatives. Synthesis 1981, 1981 (4), 303–305.

But @Hipp didn't ask about big list ;)

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How to decide from the solubility product constants which hydroxide precipitates first?
Salts of $\ce{Fe^3+}$ and $\ce{Al^3+}$ are in a beaker of water. If NaOH is added slowly, what compound would precipitate first?
For me, I think it's $\ce{Fe^3+}$ because of its lower $K_\mathrm{sp}$ value when it reacts with $\ce{NaOH}$. The two equations I wrote were

$\displaystyle\ce{Fe^3+(aq) +3NaOH(aq) -> 3Na+(aq) +Fe(OH)3(s)}$
$\displaystyle\ce{Al^3+(aq) + 3NaOH(aq) -> 3Na+(aq) + Al(OH)3(s)}$

The lower the $K_\mathrm{sp}$ value, the more insoluble the substance is (the precipitate is insoluble). So, $\ce{Fe}$ would precipitate first.
Is this correct?

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

I think this deserves reopening. That being said, neither of those reactions will happen as drawn, but your reasoning is sound. Just make sure you don't have redox reactions occuring. I think you just left out + charges, though, but I want to make sure you don't think $\ce{Na^0}$ is forming.

Additonally, bear in mind that precipitation of both of these hydroxides would be sensitive to pH. For instance, at higher pH (more basic environment), you aluminium hydroxide precipitate would disappear due to the formation of a hydroxocomplex; you need "moderately" basic conditions..not sure of the exact values.

If the question is about what compound would precipitate first, then there can be at most one answer.
Given equal stoichiometry  for the precipitation of $\ce{Fe(OH)3}$ and $\ce{Al(OH)3}$ and solubility product constants of $6 \cdot 10^{38}$ and $4 \cdot 10^{15}$ respectively, $\ce{Fe(OH)3}$ will be the correct answer.
$$\ce{Fe^{3+}_{(aq)} + 3 OH^{-}_{(aq)} -> Fe(OH)3_{(s)}}$$
However, in practice the precipitate will probably contain less water than stoichiometric $\ce{Fe(OH)3}$, and also contain small amounts of $\ce{Al(OH)3}$

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Did my teacher specialist about sulfur compounds made an error in part of Ramberg-Bäcklund rearragement mechanism?
I have a lot of troubles with what I circled in red with the big question mark near. I am not speaking about stereochemistry but about conservation of number of hydrogen atoms, and also there is actually no reason for the electrons on the carbon near $R_1$ to attack the other carbon. In my lesson, the teacher didn't show how the extra hydrogen was removed.
Can you explain what my teacher didn't show or forgot to include?

Thank you in advance!

Yes there is a problem with the left-hand side. There is no leaving group and you would end up with a pentavalent carbon.

You're right. That one on the left is definitely not right. I've never seen anything like that as a mechanism. Maybe he left a step out where a carbene forms and inserts into one of the C-H bonds?

@DGS he didn't speak about carbene at this time of the lesson. :/

It's also the Ramberg-Bäcklun>d< reaction.

I wouldn't expect him to do so, since that isn't how the reaction is thought to work, but that's the only way I could see that anion reacting.

https://commons.wikimedia.org/wiki/File:Ramberg_B%C3%A4cklund_Reaktion_Mechanismus1.svg

@Loong this mechanism is in the right part of my sheet.

Do you want a proper mechanism for the formation of the (E)-product?

@DGS oh yes I should be glad !!! :D

As discussed above, the pathway shown on the left is incorrect because it essentially shows insertion into a $\ce{C-H}$ bond by an anion with loss of $\ce{X^-}$, which is not the easiest path for this reaction to take by a long shot.
The actual mechanism for this reaction is as shown below.

First, the non-halogenated alpha-carbon is deprotonated.
The carbanion formed arranges itself such that the it can attack the halogenated carbon and in such a way that the two R-groups are cis to each other.
Attack occurs, with loss of $\ce{X^-}$. Product is cis.
In water, due to lack of solubility, decomposition is faster than redeprotonation and the (Z)-alkene forms.
In organic solvent, base deprotonates an alpha-carbon, which can then scramble.
The main reprotonation product is the trans product.
The trans product decomposes, and the (E)-alkene is the major product.

A few notes:

The red path makes almost completely (Z) product. The blue path makes a mix of the two with the (E) product as the main product.
Purple star: the reason the anion attacks to form the cis ring is unknown, but there are several theories, apparently.
Green stars: cheletropic ring decomposition is symmetry forbidden, so this is theorized to be a two-step decomposition through either diradical bond fissions or a dipolar mechanism.

DGS, could you try to rebuild your picture wirh some program like ChemDraw, Chemsketch or Marvin Sketch? Though white board self drawn graphics are very fancy, they are not as easy to read as a proper digital version.

@pH13 I can and will, but at this point I haven't had the chance.


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The diference between covalent molecule lattice and covalent network lattice
Can you tell me the difference between covalent molecular lattice and covalent network lattice?
Thanks for your help.

Do you know an example of each, to begin with?

water and diamond

Good. Well, that pretty much sums it up: nearly all covalent crystals are harder and have higher melting points than molecular crystals.

Are you asking how their bonding is different (the chemistry) or how their physical properties are different?

Covalent molecular compounds are actual molecules. Examples include $\ce{H_2O, NH_3, CO_2, N_2, I_2, SO_3, N_2O_4}$ etc. They are made of molecules which are made of covalently bonded atoms. Covalent network compounds form a network of bonds, and the whole object can almost be considered a "molecule," such as diamond (C network), silicon (Si network), graphene (C network), and black phosphorous (P network).
It's not a coincidence that all but one of the covalent molecules that I listed are gasses at room tenperature and pressure. The intermolcular bonds between molecules are relatively weak. On the other hand, every molecule in a network solid is surrounded in a network of strong covalent bonds.


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Is HCl(l) acidic?
If in a reaction mixture HCl is given in liquid phase, will it behave as an acid ?
I know that an aqueous solution of HCl is acidic but am confused about the liquid phase.

It would act as an acid if there were even slightly basic atoms available to protonate. Pure liquid HCl probably would probably have a pretty high $\text{p} (\ce{H2Cl^+})$, though, due to the very low basicity of the chloride ion. In other words, it would have a very low autoionization constant, similar or less than that of $\ce{H2SO4}$ (which has a higher pKa pure than it does in water), based on their similar aqueous pKa values.
I say "less than" because stabilizing ions would be harder due to the smaller size and higher concentration of charge of the  $\ce{H2Cl^+}$ and $\ce{Cl^-}$ ions relative to the more diffuse $\ce{H3SO4^+}$ and $\ce{HSO4^-}$ ions.

In fact, autoionization constant of $\ce{H2SO4}$ is not that low; it is way higher than that of water. Not sure about liquid $\ce{HCl}$, though.

@IvanNeretin Indeed, $\ce{H2SO4}$ is has a much higher autoionization constant than water, but as far as mineral acids go, I think it's fairly low. Interestingly, sulfuric acid autoionizes to water and hydrogen pyrosulfate. It's autoionization constant is $7x10^-5$. Nitric acid, however, has an autoionization constant of $2x10^-2$. I haven't yet found one for $\ce{HCl}$, and I'm still looking, but I believe it would be much less than either of these.


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Why is boric acid soapy to touch?
Bases are soapy to touch because of soap formation on contact with lipids present on our skin. If this is the case then why and how is boric acid soapy to touch?

@DGS            Yeah , but how H3BO3 is self lubricating?

From what I've been able to find, it is similar to how graphite acts as a lubricant. Boric acid is flat and it hydrogen bonds to form layers of flat networks. These slide on each other like layers of graphite do, reducing friction.


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How to apply the Heisenberg Uncertainty Principle in calculations on the movement of bacteria?
In this problem, I know the formula, but I get confused with all the steps and how to use what I got, to get to what I need.

A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of $m = \pu{0.100 fg}$ (where a femtogram, $\pu{fg}$, is $\pu{10^{−15} g}$) and is swimming at a velocity of $v = \pu{9.00\mu m/s}$, with an uncertainty in the velocity of  $9.00\%$. E. coli bacterial cells are around $\pu{1 \mu m}$ ($\pu{10^{−6} m}$) in length. What is the uncertainty of the position of the bacterium?

I used the formula $$\Delta x \Delta p \geq \hslash/2$$
I plugged in these numbers:

Mass of bacteria $(m) = \pu{0.100 fg} = \pu{0.1 \times 10^{-15} g} = \pu{10^{-16} g}$
Velocity of bacteria $(v) = \pu{2 m/s}$
Uncertainty in velocity = 9.00

I am getting this answer $1.67\times10^{-9}$, and it is wrong. I don't know what I am doing wrong.

Well why don't you tell us what you know and then we might be able to help.

$\newcommand{\upmu}{{\large\unicode[Times]{x3BC}}}%Remove this line when upright greek characters are implemented by mathjax$
(Hover over the boxes to see the results.)
The velocity of the  bacteria is $\mathbf{v} = 9~\mathrm{\upmu m/s} = 9\times10^{-6}~\mathrm{m/s}$, with an uncertainty of $9.00\%$ you can calculate the absolute uncertainty in the velocity $\Delta \mathbf{v}$ as the difference of the upper bound $(\mathbf{v} + 0.09\mathbf{v})$ and the lower bound $(\mathbf{v} - 0.09\mathbf{v})$.

 $$\Delta \mathbf{v} = (\mathbf{v} + 0.09\mathbf{v}) - (\mathbf{v} - 0.09\mathbf{v}) = 0.18\mathbf{v} = 1.62\times10^{-6}~\mathrm{m\,s^{-1}}$$

The momentum is defined as
$$\mathbf{p} = m\mathbf{v},$$
hence the absolute uncertainty in the momentum is
$$\Delta\mathbf{p} = m\Delta\mathbf{v}.$$
The mass is $m = 0.100~\mathrm{fg} = 1.00\times10^{-19}~\mathrm{kg}$.

 $$\Delta\mathbf{p} = m\Delta\mathbf{v} = 1.62\times10^{-25}~\mathrm{kg\,m\,s^{-1}}$$

Now you can use this to calculate the uncertainty in position $\Delta\mathbf{x}$.
$$\Delta\mathbf{x}\Delta\mathbf{p} \geq \frac{h}{2}$$
Planck's constant is $h = 6.62607004\times 10^{-34}~\mathrm{m^2\,kg\,s^{-1}}$.

 $$\Delta\mathbf{x} \geq \frac{h}{2\Delta\mathbf{p}}\\\Delta\mathbf{x} \geq 2.07\times10^{-9}~\mathrm{m}$$

Since you have not included any of the steps how you arrived at your result, I cannot help you finding your mistake.

I think we can argue about the fact that the uncertainty is on the magnitude of the velocity or on the velocity itself. If the result is still different from OP's corrections, the uncertainty can be divided by 2 and the result should correspond to the solutions.

@R.M. I cannot make sense of what you have written. Also what corrections are you talking about?

The OP states that its result is wrong, so I am assuming that he is comparing it with some solutions. Concerning you answer, I interpreted the statement "with an uncertainty in the velocity of 9.00%" as $\Delta v = 0.09 |\vec{v}|$, which is half of your uncertainty. I whink that the two interpretations are equally correct.

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Factors affecting mass of a gas-evolving reaction mixture over time

Excess magnesium was added to a beaker of aqueous hydrochloric acid on a balance. A graph of the mass of the beaker and contents was plotted against time (line 1).
What change in the experiment could give line 2?

The same mass of magnesium but in smaller pieces
The same volume of a more concentrated solution of hydrochloric acid
A lower temperature

Original question:

The answer is just II. However, I thought that both I and II would answer the question. Why is it just II?

$I$ is definitely not the case. They would have converged by the time they flattened out if the same amount of magnesium was used. See the addition to Hippalectryon's answer below and the note I made there. The only way the final mass would change (out of the three given) would be if more $\ce{H2}$ formed, so more $\ce{HC}l$ would have to be used.

Related: Question regarding nitric acid, magnesium powder and initial reaction rate

A screenshot or picture of an exercise is not searchable. Please consider rewriting it (which I did already - please consider this for future reference), so that it can be of help for future visitors.

The reaction that occurs here is
$$\ce{Mg (s) + 2 HCl (aq) -> MgCl2 (aq) + H_2 (g)^}.$$
As the reaction progresses, $\ce{H_2(g)}$ is released and hence the mass decreases.
I believe you are right, both I and II could be correct.
II, because a greater quantity of $\ce{HCl}$ will lead to more $\ce{H2}$ leaving, hence a greater mass decrease.
I, because putting the magnesium in smaller pieces is likely to increase the speed of the reaction, and hence increase the mass of $\ce{H2}$ that has been created in a given time. However, since the end of both functions is "flat", one could assume that the reactions won't evolve any more, thus invalidating this answer (if one waits long enough for the reaction to not evolve any more, the speed of the reaction will in most cases not affect the end result).

The problem with $I$ is that while, yes, the rate would be slower, the final mass would be the same. Here, at time $t=\infty$ $m_{II}\ne m_I$.

@DGS I didn't really consider that the end of the given timeline was an equilibrium state... but now that you're pointing that out, it indeed may be what the graph meant to convey. Amended.

Well, the reaction won't be in equilibrium due to evolution of $\ce{H2}$ (though, I don't think it would be in equilibrium either way). Since both lines go horizontal, both would be at completion, so they should have converged by end of the graph.

The first word of the question is "Excess", which means that all of the dissolved HCl will be converted into aqueous magnesium chloride solution. Starting with the same amount of magnesium (and acid) the final mass must be equal. | @DGS wouldn't the rate be faster with smaller sized grains of magnesium?

@Martin-マーチン Of course it would, but the final masses would have to be the same. Same amount or hydrogen ions to make hydrogen means same amount of gas evolved. These have gone to completion, or at least you can see the asymptote they're each trending toward. Clearly different. But if higher concentration acid was used, but same volume, that means more magnesium can react and more hydrogen gas can evolve, thereby ending with a lower final mass.

Or were you not disagreeing with that? At first I thought you agreed. But your question to me threw me off.

@DGS Of course I agree, my question referred to your first comment, where you said "the rate would be slower". Btw. I'd have liked your comment (to the question) as an answer.

@Martin-マーチン Oh, whoops. I don't know why I said "slower" — I definitely meant "faster." =/ My bad.

I. False – because the magnesium has a higher surface area in this reaction, the rate will proceed faster, producing a steeper slope in the graph; however, the final mass of the solution will be the same in both experiments. We would note that the lines converge when each system reaches equilibrium.
II. True – the higher concentration of $\ce{HCl}$ means more $\ce{H2}$ gas will be lost from the system at equilibrium. Thus, the line would flatten beneath line 1.
III. False – a lower temperature would mean the average kinetic energy of molecules in the system would be lower; therefore, a smaller fraction of the molecules will have sufficient energy to react with others. In this case, we would note an elongated curve that converges with the first.

Please don't use MathJax as a styling element and also please don't use sizing commands where not necessary.


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What happens to particulate matter after collection via electrostatic precipitator?
What I understand about electrostatic precipitators is that the particles become ionized due to corona discharge and thus are attracted to the oppositely charge plate. This would work as long as the precipitator is turned on and plates maintain a charge.
However afterwards, when the precipitator is turned off, wouldn't the collected particulate matter tend to regain their charge neutrality? What if it goes back to its original state? Wouldn't the whole action of the precipitator then become moot?
How is the precipitated pollutant stored and prevented from escaping in an electrostatic precipitator?

Could you provide some background for your question? I can figure out from your question what an electrostatic precipitator likely is, but I am not familiar with them and there is not enough context in your question for me to know HOW they are used. What application is this for? That would likely influence the answer. Also, I am not certain the [tag:electrochemistry] tag is appropriate, since electrochemistry is more about using redox reactions in 1) analytical situations and 2) electrolysis. This application seems more about electrostatic attraction and not redox chemistry

An an electrostatic precipitator does need to be cleaned periodically. Typically you turn off the electricity, remove the filters, and either blow them out or wash them out somehow. https://www.youtube.com/watch?v=Y-CXjdMx8HM

@BenNorris it's an air filtration device. You can find more about it on wikipedia - Electrostatic Precipitator. I don't think it involves chemical redox reactions. Though it does involve ionization of particles.

@MaxW do they stay on the filters because of electrostatic attraction (particles are negatively charged, plate is positively charged) or do they regain charge neutrality and settle on that filter?

When the filter is charged the stuff in the air is  electrostatically  attracted. When the charge is turned off the gunk is mechanically stuck on the filter till it is cleaned off.


text:
why the buffer capacity depends on the ratio of salt and acid in acidic buffer?
According to the Henderson–Hasselbatch equation  when  pH =  Pka for  any  buffer solution then  why  we  say  that  this is  the  best  buffer?

Related: Why is Buffer capacity maximum when the ratio of its components is 1?

When we say it's the best buffer, we mean that the solution can resist pH change in either direction, and this occurs when $\ce{pH=pK_a}$. In the Henderson-Hasselbatch equation, $\ce{pH = pK_a + log\frac{[base]}{[acid]}}$, this is achieved as long as [base] = [acid]. The actual amounts of the acid and base forms, however, determines how far it can resist pH change.


text:
Why don't the solute molecules in a volatile solvent also escape to contribute to vapor pressure?
I'm currently studying relative lowering of vapor pressure with the addition of a  solid non volatile solute.
The book explains it with the notion that some of the non volatile solute also appears at the surface between the liquid solution and the vapor. However, i don't understand why don't the molecules of the solute at the surface just escape? Obviously, a solid can't vaporize (Unless you raise the temperature, (And maybe reducing the pressure?)) however the molecules of the solute that are at the surface between the solution and the vapor and should not be held by the strong intermolecular bonds of the solid.
Hence, if i understand this correctly, they, just like the water (Or other solvent) molecules should be constantly moving around in the solution and sometimes 'bump' into other molecules to gain enough velocity to escape the solution. The few molecules that are are the surface between the solution and the air should be getting frequently hit with high energy molecules from below, which should constantly knock such molecules into the vapor. This would mean two things:

The solute also has a small partial vapor pressure
The vapor pressure should still lower, as once solute particles are knocked out into vapor, a few of the solvent particles should escape through the 'opening' before another solute can take it's place. Rinse and repeat, like a valve being opened and closed. Hence, while there would still be a lowering of vapor pressure, there shouldn't be any permanent layer being formed between the solution and the vapor.

But both of these are in contradiction to what's in the book. So where did i go wrong in my reasoning?

Many solids have vapor pressures. Imagine any solid you can smell. 2. I don't think that explanation is giving the hard physics behind vapor pressure reduction due to solute (it's apparently entropy-based). It's a massively oversimplified representation.

It is true that "non-volatile" is a relative term and that virtually all compounds have some, possibly infinitesimal, vapor pressure, but the idea is that this is negligible. And you are also correct that "...the molecules of the solute that are at the surface between the solution and the vapor and should not be held by the strong intermolecular bonds of the solid."
Concerning that latter statement however, there are many reasons that a compound might maintain very low volatility as a dilute solute, even though the intermolecular forces within the bulk solid are not at play. For example, compounds having a large molecular weight tend to be non-volatile, and these will still require much more kinetic energy to be "bumped out" of the solvent, meaning this will be a statistically rare event compared to the transfer of solvent molecules to the vapor phase.
Ionic solids are another type of solute that are typically quite non-volatile. In order for an ionic species occupying a solute surface site to escape into the gas phase, in the case of an aqueous solvent for example, a rather large amount of energy would be required to dissociate the ion from it's solvation sphere and kick it out into the gas phase as a free ion.
I hope this answers your question "where did I go wrong in my reasoning". Don't hesitate to ask for clarification in the comments if I missed something.


text:
Type of adsorption in catalysis
why type of adsorption takes place during catalysis? It can't be physisorption because it involves formation of chemical bond. It shouldn't be chemisorption too because it is characterised by desorption, where as desorption takes place in chemisorption. So basically what kind of adsorption is involved in catalysis?

What type of catalysis?

@DGS            In general, i.e. both hetero and homogeneous catalysis.

But like transition metal catalysis? There are just so many ways catalysis can occur. Oh, I just saw. Heterogeneous catalysis is mentioned on the chemisorption Wikipedia page. However, I don't think homogeneous catalysis works through adsorption.

@DGS            Yeah sure but, chemisorption do not involve desorption but in heterogenous catalysis desorption takes,place.

What do you mean "chemisorption does not involve desorption"? I think this is may be a misunderstanding. You can have hydrogen adding and eliminating from a platinum surface, but because the H-H bond breaks and the Pt-H bond forms, it is chemisorption, even though the reaction is in equilibrium. It is still a chemical reaction and chemical change.

@DGS             Yeah, that was my misunderstanding, thank you , should I delete this question?

Nope. Someone might have the same question some day. You could always provide a formal answer yourself and probably get some additional reputation in the process.


text:
Why wouldn't solvents be interchanged in isotonic solutions?
While studying osmosis, my book remarks that

Two solutions having same osmotic pressure at a given temperature are called isotonic solutions. When such solutions are separated by semipermeable membrane no osmosis occurs between them. For example, the osmotic pressure associated with the fluid inside the blood cell is equivalent to that of 0.9% (mass/volume) sodium chloride solution, called normal saline solution and it is safe to inject intravenously.
On the other hand, if we place the cells in a solution containing more
than 0.9% (mass/volume) sodium chloride, water will flow out of the
cells and they would shrink. Such a solution is called hypertonic. If
the salt concentration is less than 0.9% (mass/volume), the solution
is said to be hypotonic. In this case, water will flow into the cells
if placed in this solution and they would swell.

But why would the osmosis simply stop? If I'm understanding osmosis correctly, then it is simply due to the fact that there is an SPM separating the two solutions, and hence the solvent (Or rather, the one with the size smaller than what the SPM allows, although I doubt it'd make a very big difference for a dilute solution) flows from the lower to higher concentration and eventually, equilibrium is attained.
In this case, even if you do inject somebody with a saline solution with $0.9%$ mass/volume sodium chloride solution, the SPM of the blood cell should still trade the solvent from the injection with the material inside (Since equilibrium does not translate to no flow, as confirmed by experiments with heavy and light hydrogen in the formation of NH3) it and the blood cell should still swell up. Obviously, that doesn't happen, though, or else I would have quite a different memory of getting a shot! Where did I go wrong in my reasoning?
Note: In the above, I'm assuming that the fluid inside the blood cells is something other than water. Please correct me if I'm wrong.
Edit: I realized the error. It says in the above paragraph itself that the fluid is water.

So did you figure out the answer to your question? If so, could you please create an Answer below?


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Error propagation
I have a question regarding finding the error associated with the $r_\mathrm e$ term in the equation
$$r_\mathrm e=\sqrt{\frac h{8\tilde B_\mathrm e\pi^2\tilde c\mu}}$$
Is there anyway I can find the uncertainty of $r_\mathrm e$ using any formula. $h$, $\pi$ and $\tilde c$ and $\mu$ all are constant. Only $\tilde B_\mathrm e$ have an uncertainty. Any readings that I could tap into?

Upto how much decimal you want accuracy?

just 1 significant figure

I just want to know what is the general equation that I can tackle this type of question, there should be a formula that i could apply to determine the uncertainty

When two quantities are multiplied or divided, the relative error in the result is the sum of the relative errors in the multipliers.

If the individual uncertainties of $h$, $\tilde B_\mathrm e$, $\pi$, $\tilde c$, and $\mu$ are not correlated, the total uncertainty of $r_\mathrm e$ may be estimated using the general formula:
$$u(r_\mathrm e) = \sqrt{\left(\frac{\partial r_\mathrm e}{\partial h}\right)^2 u^2(h)+\left(\frac{\partial r_\mathrm e}{\partial \tilde B_\mathrm e}\right)^2 u^2(\tilde B_\mathrm e)+\left(\frac{\partial r_\mathrm e}{\partial \pi}\right)^2 u^2(\pi)+\left(\frac{\partial r_\mathrm e}{\partial \tilde c}\right)^2 u^2(\tilde c)+\left(\frac{\partial r_\mathrm e}{\partial \mu}\right)^2 u^2(\mu)}$$
If only $\tilde B_\mathrm e$ has a relevant uncertainty, the formula can be simplified to:
$$u(r_\mathrm e) = \sqrt{\left(\frac{\partial r_\mathrm e}{\partial \tilde B_\mathrm e}\right)^2 u^2(\tilde B_\mathrm e)}$$
where the partial derivative of $r_\mathrm e$ with respect to $\tilde B_\mathrm e$ is
$$\frac{\partial r_\mathrm e}{\partial \tilde B_\mathrm e}= - \frac{1}{8}\frac{{\sqrt 2 h}}{{\sqrt {\frac{h}{{\tilde B_\mathrm e{\pi ^2}\tilde c\mu }}} {\tilde B_{\mathrm e}^2}{\pi ^2}\tilde c\mu }}$$


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Is everything in organic chemistry arbitrary?
I am a student who is trying to study organic chemistry. I'm using McMurry's book for it. As a beginner in chemistry, I have a fundamental question about chemistry itself. Are most of the chemistry's reaction rules, proved in a way of experiment and probability? Especially for organic chemistry. I really can't get why the reactions occur and there seem to be too many exceptions. When I ask teachers, they say it is because they just found them by experiment. But it seems to be too much for to just say that all the knowledge is just from experiences. There must be some fundamental rules.
So my question is

How do we find organic chemistry reactions?
Where do we use the reactions that we find in books?
Are there other important rules in a molecule or the electrons that are in the molecule besides the induction and resonance effect?
Can't organic chemistry be more rigid (I can't find a good word for this)? Studying McMurry's book, the products of the reactions seems to be determined arbitrarily.

Read Clayden and throw McMurry away...

I might not go as far as @orthocresol suggests, but you're asking good questions and I think going beyond McMurry might be a good idea.

:D I cannot conscientiously recommend that anyone use McMurry though. The book is organised by functional group (a very poor organisation imo) and reads much more like a list of reactions. There's far too little conceptual reinforcement. Lots of useless figures (in particular, the repetitive electron density plots irritated me). Trust me - many people I know picked up Clayden and never looked back. I can't say it has all your answers but it is a huge improvement.

However, Anslyn and Dougherty is great way to understand just how arbitrary organic chemistry isn't, when you want to make the jump into hardcore physical organic chemistry.

BTW, I am going to VTC as too broad. If you split them up into multiple questions, I would love to write some answers. But as it stands, each bullet point warrants an entire essay.

Functional group makes little sense, but that's how the old school guys do it.

I throw in a bit here. First organic chemistry is really about trying to make some particular compound. What do you have to start from and what reactions to get there... // Studying in a book is good, but a lab to go with the book learning is really helpful. Products are rarely 100% yield. Incomplete reactions, and side reactions are the norm. Purifying a compound made is also sometimes quite a challenge. The hands-on effort is really helpful.

@MaxW, as an organic chemist "raised" as a physical organic chemist now working in a natural product synthesis lab, I don't know whether to begrudgingly agree or to argue vehemently. There's two sides to that coin. This isn't wrong, but I also think it's incomplete.

@DGS - I agree that the answer is dumb downed. The point that I was trying to make is that in general you don't want just to make "some" product. The art in organic chemistry is getting a good yield of a high purity product. Thus some "textbook reactions" wouldn't really work in an industrial plant.


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Mechanism for the CBS reduction?
I don't get the CBS reduction of ketones in organic chemistry. I can draw two ''chair'' structures for the reduction of phenyl,methyl ketone and each gives a different enantiomer. Both have the phenyl equatorial, I don't know what I'm doing wrong. Please, help.

You don't really make the stereochemistry of the oxazaborolidine obvious in either case here. I'm guessing you probably would have to use the opposite enantiomer between to two to get these results. I think the top structure would have the hydrogen dashed back and the bottom would have the H wedged forward.

Naah, the hydrogen must be cys to boron, but that's how it always is, since when you add BH3 it can only coordinate the nitrogen from above (the bicyclic compund is folded like a book), otherwise it would be too hindered. But I think I found the answer.  I should have used the boat conformation :D Thank you for your answer.

That may be true, but you didn't add stereochemistry for anything.

Okay, so in the bottom one, you have the bulky group on the niteogen axial, and in the top one the bulky group is equatorial. So the top answer makes more sense to me, but do you actually know the answer? If so, you might want to answer your own question.


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Lanthanides react with mineral oil?
I have an element collection and noticed that the lanthanide (rare earth) element samples seem to be reacting with the mineral oil I've used to help prevent oxidation.
When I originally placed the samples I researched the lanthanides and found that the lower atomic numbers indeed tend to oxidize in air. Even though I've stored them in jars, I also filled the jars with mineral oil, similar to how I've stored the alkali metals. And at least for the alkali metals, the mineral oil has served as a good barrier to keep out oxygen.
So why would the lanthanides, which are a lot less reactive to oxidation, corrode so quickly? Is there another, perhaps organic reaction going on, between the lanthanides and the mineral oil?

Someone ... posted another answer to this question, and I voted it up. Today I caught a glimpse of a message I believe from that someone: "Why did you delete my post ...?"  Now I can find no trace, no evidence of the post or the comment. I assure you - I did not delete the post, and it's a mystery to me who would/could delete it.

David Hamric did a four year Rare-Earth Metal Long Term Air Exposure Test
and has a wonderful set of time-lapsed photographs on his site.
Europium is the most reactive to oxidation.
Several sources, such as Avalon Advanced Materials: Europium make statements like:

Samples of the metal, even when coated with a protective layer of mineral oil, are rarely shiny and it readily ignites in air above 150°C.

So rather than reacting with the mineral oil, I would say that europium and possibly others are reacting with oxygen despite being covered with the mineral oil.
THE SOLUBILITY OF O2, CO, AND N2 IN MINERAL OIL AND THE TRANSFER OF CARBON DIOXIDE FROM OIL TO AIR. The Journal of Biological Chemistry (1927)
 vol. 72,  pages 545-548 gives quantitative values for the solubility of oxygen in mineral oil, but I can't understand what units are being used. It says solubility is 0.134 +/- 0.004 at 24 degrees C.  I think this means 0.134 volumes of gas phase oxygen at standard temperature and pressure per volume of oil. Whatever the units are, it does show that significant oxygen can be present in mineral oil.
Also, I'll quote your own answer to another question:

Even when these [alkali] metals are stored under mineral oil, oxygen can dissolve in the mineral oil and react.

Original Source: Management of time-sensitive
chemicals (II): Their
identification, chemistry
and management

Your answer - almost complete. It's clear the lanthanides indeed oxidize. I knew that. It's a fact that mineral oil is not a perfect barrier to O2; I knew that. What I don't understand is why given that for example potassium and europium, each under mineral oil, the europium appears to oxidize at a much greater rate. I've not run the experiment, but expect in open air, the opposite would be true. Is it that the potassium forms a somewhat protective skin, but europium does not?

@docscience Sorry if my answer wasn't helpful. Because the question said "reacting with the mineral oil" it made me think you believed the metals were reacting with the mineral oil rather than oxygen. Maybe the oxides that form on the europium are more oil-soluble or more easily penetrated by the oil than those of potassium.


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Why is there no Pi-backbonding in BeCl2?
The $\ce{Cl}$ atoms in $\ce{BeCl2}$ donate a lone pair to the $\ce{Be}$ atom of another molecule so that Beryllium can complete its octet.

What I don't understand is why don't the $\ce{Cl}$ atoms donate the lone pair to the $\ce{Be}$ of the same molecule and form $\pi-$backbonds?

So your asking why BeCl2 isn't monomeric in solid phase? Related http://chemistry.stackexchange.com/questions/21764/which-lewis-structure-for-becl2-is-more-commonly-seen


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Le Chatelier's principle with pure solids and liquids
In the following reaction:
$$\ce{A(s) <=> B(g) + C(g)}$$
since adding a pure solid is not supposed to shift the equilibrium in any way, does that mean that I can essentially remove all of the solid except for 1 molecule, and the equilibrium will not shift at all since the concentration of the solid stays the same? Similarly, does this mean that I can essentially fill up the container with the solid, and no extra B and C will be produced?

Your first statement is right.
You just need a certain amount of solid or pure liquid for the reaction to occur, just like how putting in more solid in a saturated solution doesn't increase dissolution, while you need solid in there to call it an equilibrium between aqueous and solid phases.
However, the amount of solid or liquid does change the kinetics of the reaction: the rate at which it proceeds to equilibrium.
But if you put in some solid A, the Q is 0, so some B and C will form. It's the same reason why putting in a sugar cube in water increases aqueous sugar concentration: there is an equilibrium between solid and aqueous components, it's just that the constant means the maximum amount of aqueous concentration.


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Entropy and enthalpy of hydrogen ion?
The absolute entropy of $\ce{H^+ (g)}$  is  $\ce{108 kJ/mol.K}$. However, the absolute entropy of $\ce{H^+ (aq)}$ is exactly 0. Why is that? Also, the standard enthalpy of $\ce{H^+ (aq)}$ is 0, so what is it the standard state of?

For H+, it is simply conventionally defined as such. It is not the same as the value for H2(g), which is 0 because H2(g) is the standard state of hydrogen.

Its entropy is not zero, but you can define (arbitrary) a reference state.

The reference state can not be the two states at the same time. You should give more details about your problem.


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Can anyone identify this glassware?
Hoping for a little help identifying this lab glass , I don't know anything about lab glass , this was in a storage unit I'm purchased at auction,,, there is what I think is called a condense & 2 round beakers or flasks,, one is 50ml & the other is 200ml ,,, all the male & female connector are marked SJ GSA CO 28/15 ...PLEASE is there anything you can tell me about what there are called & what they would be used for.

The glassware isn't particularly valuable. Finding someone who wants it will be the problem. If you were going to buy it new, I'm thinking maybe $150.
It is indeed a distillation setup. The large flask is the pot in which you'd put the stuff that you wanted to boil. The small flask would collect the stuff that you boil off. The nipples on the condenser are for rubber hoses to connect to a water supply so that the condenser jacket can be water cooled. Water goes in on the bottom and out on the top. In order to use this you're missing some parts.

You need a couple of ringstands and clamps to hold the whole thing together.
Two thermometers would go into the top of the condenser. One to monitor the pot temp, and the other to monitor the condensate temp.
You need some metal clamps to hold the flasks to the condenser.
You need something to heat the pot, a Bunsen burner would do.

Not looking to use it , As I said it was in a storage unit
I purchased at auction , I buy to sell everything in the unit. Was just interested in finding out what it is so I know how what to call it when I try to sell it

Careful - You are missing a distillation adapter between the condenser and the receiving flask. Without the adapter, the receiving flask should simply be placed under the end of the condenser to receive the distillate, rather than attached to it. Otherwise you would be heating a closed system, possibly leading to a catastrophic failure.

Good point that I failed to mention. It looked to me as if the thermometer holes are not ground glass. I assumed that there was a pressure exit there, but didn't mention that.

The joints are reminiscent of the Buchi ones for the receiver flask on a rotary evaporator under vacuum.  Could this be for vacuum distillation?

Not looking to use it , As I said it was in a storage unit
I purchased at auction , I buy to sell everything in the unit. Was just interested in finding out what it is so I know how what to call it when I try to sell it


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Why aqueous hydroxide is slippery (beyond just saponification)
A recent question mentions:

Bases are soapy to touch because of soap formation on contact with lipids present on our skin.

This is always the reasoning I hear, as well, and searching the internet gave the same results. However, when working with nitrile gloves, glass with aqueous sodium or potassium hydroxide on the outside always feels slippery. I believe that even nitrile-on-nitrile (rubbing the fingers together) gives the same slippery feeling. This effect is even stronger for the liquid in the base bath, but I assume that's because alkoxides would be detergents similarly to soap molecules in the absence of water.
Point is, if saponification is the culprit, why does the exterior of the glove experience a similar soapy effect to that my fingers feel? Could it be negative charges on the exterior of the glass? Is it possible for hydroxide to react with nitrile gloves to cause some similar kind of electrostatic repulsion, maybe by deprotonation or nucleophilic attack of the nitrile? This seems like a reasonable idea as far as glass goes, but I don't think that reasoning crosses over to gloves. Of course, I am not sure it really happens with glove-on-glove contact. I will safely test that theory again and edit if it's incorrect.

Hmmm... This PDF says that nitrile isn't a good glove for use with sodium hydroxide. This one, however, says the opposite. I was thinking, and I hope I don't havs to worry about this: $$\ce{OH^- + RCN -> ROH + CN^-}$$

I know this is a little late, but couldnt it just be because liquids make things slippery? You can slide your hand/glove around a glass object easily when its covered by water/other non-oils.

@ChildishJack It's true that liquids can make things more slippery, but there is a definite difference between glass with water on it and glass with aqueous hydroxide on it. The basic solution feels kind of oily on the surface (but rinses off with water), but plain water doesn't have the same level of effect.

Glass itself is soluble in strong alkalis, forming sodium silicate solution, which is a syrupy, somewhat slippery-feeling substance. You can use sodium silicate to make a fire-resistant coating or high-temperature cement.
As you state, it's also possible the glove itself is being attacked. If the glass feels slippery in $\ce{NaOH}$ and $\ce{KOH}$ solutions, but not in $\ce{NH4OH}$, then it is most likely a silicate forming, though.


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Stabilization of ipso and para position in Birch reduction by EWG of aromatic ring
I'm trying to grasp the Birch reduction. In the lecture about it, it's said that the EWG group stabilizes ipso and para position through resonance. I tried to understand that by drawing resonances without success. What is wrong in it?

You seem to have an extra dot (electron) in the first structure, which is muching things up all over the place.  What is going on there?

sorry;; that extra electron is from Na

So this is sodium donating an electron to an aromatic compound with an electron withdrawing group?

TL;DR: This question may not be readily answered with Lewis structures.
The problem is the electron is delocalized into an antibonding orbital.  By analogy, we can examine this concept with molecular oxygen.  The Lewis structure for molecular oxygen:

This clearly is a double bond, with all spins paired.  Now observe the molecular orbital diagram below:

The top energy levels clearly show oxygen with two unpaired electrons, which contradicts the Lewis structure completely.  There is no real way to simultaneously convey these two situations.
Similar diagrams exist and may be used for the aromatic system with an electron-withdrawing group exist, but they are rather complicated.

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


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Reagent for identification of Water Hemlock (Cicuta) or Cicutoxin?
Chemical reagents are used for macroscopic identification of some wild mushrooms. Are there similar reagents used to identify Water Hemlock (or potentially cicutoxin) as a supplement to visual identification of this incredibly poisonous plant?


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Why are 7-membered rings less likely to form than 5- and 6- membered rings?
I see 5- and 6-membered ring products as the major products even when 7-membered rings are possible. Why is that 5- and 6-membered rings are more likely to form than 7-membered rings? Is it even true?
My textbook says a 5-membered ring is the major product and the 7-membered ring is the minor product in intra-molecular aldol condensation reaction of 6-oxoheptanal.
Major Product: 1-(cyclopent-1-en-1-yl)propan-2-one
Minor Product: cyclohept-2-ene-1-carbaldehyde and 2-methylcyclopent-1-ene-1-carbaldehyde
This is not the only reaction where I found a 5-membered ring as a major product when a 7-membered ring was possible.

In this particular case, the product formation is thermodynamically controlled and five-membered rings are more stable than seven-membered (less ring strain), so that's the end of the story. In general, though, for kinetically controlled reactions, ron's answer is important.

Related: https://chemistry.stackexchange.com/questions/131572/what-does-the-intramolecular-aldol-condensation-of-6-oxoheptanal-form

There are two factors that need to be considered to answer your question - entropy and enthalpy.
Entropy favors the formation of smaller rings. If we consider a hydrocarbon chain, the number of possible conformations increases dramatically as we go from 3 carbons to, for example, 10 carbons.  Only a few of these conformations will be correctly arranged so that the two ends of the chain are close to each other, which is necessary if we are to form a bond. So, entropically speaking, it will be much more probable to form a 3-membered ring than a 10-membered ring. This is shown in the following graph where 3-, 4- and 5-membered ring formation is entropically most favored.

Note that a less negative $\Delta S$ favors the reaction since $\Delta G=\Delta H - T \Delta S$.
Enthalpy comes into play when we consider the strain in the transition state leading to the possible products. Here is a chart showing the relative ring strain as a function of ring size.

As expected, strain is the highest in 3- and 4-membered rings.
If we overlay (add) these entropic and enthalpic effects we would get a graph similar to the one pictured below.

In general, we find that 5-membered ring formation is usually the favored pathway.  This web article (J. Am. Chem. Soc. 1984, 106, 1051-1056.) is a nice review of this topic that may be useful for a more in-depth reading on the subject. The above web article is also the source of the above three images.


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Microscale Experimentation vs. Preparative Synthesis during drug development
I'm trying to understand the drug development stages, techniques and procedures. My understanding on the subject:
Microscale experimentation

Is used on early stages to find compounds reacting with some target
proteins/whatever
Is done on a very small scale (< 1mg) since most of
the compounds are not going to react with the target
Techniques: Combinatorial Chemistry

Preparative Synthesis

Is done on later stages when we found a number of potentially useful compounds
Is used to synthesise compounds in larger quantities (5-200mg) for further testing and possible conversions to other compounds with better characteristics (e.g. lower toxicity or better distribution). We need larger quantities because we will conduct a lot of experiments with these compounds down the road (including microscale experimentation?)
Techniques: High-Throughput Chemistry

Could someone please correct and possibly extend my explanation?

What exactly do you want clarification about? You seem pretty clear on the subject to me.

I didn't know if I was understanding all the details correctly, wanted to confirm my vision. If you're saying that everything is okay, then I'm happy :)


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Finding the energy given out when 2 moles of reactants are mixed and ignited from a balanced equation
The question I am trying to answer is this:

I only know about Hess' law, bond enthalpies and enthalpies of combustion and have tried to use all concepts in order to somehow get an answer but none of them have worked. I first tried to write a balanced equation for what the reaction of 2 moles of reactants would look like but I don't believe it could be balanced. Then, I tried using bond enthalpies but I couldn't have done that either since the equation couldn't be balanced. Next, I tried using enthalpies of combustion, and the same thing happened. So, I'm wondering if there's anyway to answer this question, preferably with the concepts mentioned above.

The N2O4 is going to define the maximum amount that will be used up in the reaction, since it has the higher stoichiometric number of the reagents. That is, the second term will always be less and the amount consumed will be defined by the N2O4 because that two moles will limit the N2O4 and in doing so reduce the amount of the second term consumed- there will be excess second compound after the N2O4 has been consumed.
For example, the ratio of first to second term is 5:4, so if we consume all of the second term (2 mols) this requires 2.5 mols of N2O4, which we don't have. So we are limited with the N2O4.
So, knowing that, with two mols of N2O4 the amount is reduced to 2/5 of the enthalpic calculation. That becomes 5116*2/5 = 2046, or answer B.
Apologies for awkwardness of this answer, on a mobile device and have to keep swapping from question to answer.


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Naming unsaturated hydrocarbons
$\ce{CH2=CH-CH=CH-CH2-CH3}$
Is this hexa-1,3-diene or hex-1,3-diene?
And why?
When do you add an extra a and when do you not?

Ignoring the numbers for a moment.
This will probably help you in most exams: You generally never see 2 vowels together. When 2 vowels come together, remove the first one. e.g.: methane + al = methanal (notice that I removed 'e' from the combined word), butane + one = butanone, but, in case of dione because the suffix doesn't start with a vowel, one has to just join the two: butane + dione = butanedione.
Thus, in your case, we just have to join the two: hexadiene, there is no need of removing a vowel.

According to the general methodology described in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a double bond is expressed by changing the ending ‘ane’ of the name of the corresponding saturated parent structure to ‘ene’.

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. (…)

For example, the name ‘hexane’ is changed to ‘hex-1-ene’.

If the compound contains multiple double bonds, a multiplicative prefix is placed before the ending ‘ene’ and the letter ‘a’ is inserted between the root of the name of the parent structure and the multiplicative prefix.

P-31.1.1.2 The multiplying prefixes ‘di’, ‘tri’, etc., are placed before endings denoting unsaturation to indicate the number of multiple bonds of each kind, as required, for example, ‘diene’ and ‘triyne’. For euphonic reasons, when the endings ‘ene’ and ‘yne’ are preceded by a multiplying prefix and a locant the letter ‘a’ is inserted. (…)

See also:

P-16.8.2 For euphonic reasons, the letter ‘a’ is inserted between the root of the name for polyenes, polyynes, and polyenynes, and the numerical multiplicative prefix ‘di’, ‘tri’, etc., preceding the ending ‘ene’ or ‘yne’.

Therefore, the correct name for the compound given in the question is hexa-1,3-diene (not hex-1,3-diene).


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Is ammonia an acid or a base?
Is $\ce{NH_3}$ an acid or a base? N has a higher electro-negativity (3.04) than hydrogen (2.20), therefore nitrogen attracts the bond's electrons easily. The reaction should produce $\ce{H^+}$ because the electrons of hydrogen are attracted by nitrogen.
According to my Chemistry book, my assumption seems to be wrong because $\ce{NH_3}$ is considered a base. Why am I wrong?

Oxygen has a higher electronegativity than nitrogen. Therefore, which reaction should be more likely: $$\ce{NH3 + H2O <=> NH4^+ + OH^-}$$ $$\ce{NH3 + H2O <=> H3O^+ + NH2^-}$$ This example only applies in water, of course, but $\ce{NH3}$ will act as a base in many other instances, also.

Indeed, $\ce{NH3}$ is an acid and a base, like water and many other compounds.

Ammonia is a base because it's lone pair on nitrogen accepta protons. Ammonia could act as a really weak acid.

The difference isn't that stark. However, you have to bear in mind the Configuration of electrons on the central nitrogen. We often will draw ammonia with the electron pair atop the nitrogen. That lone pair will readily accept a proton, making it a base.
To put it another way, ammonia is far more ready to accept a proton than to release one thanks to the lone pair.
Of course, it also depends on the environment it's in. But in general, you'll usually see ammonia (or just nitrogens in compounds) referred to as basic


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How to prove at what pH the color of an indicator will change based on its pKa?

The equivalence point of an acid-base titration can be determined by choosing the appropriate colored indicator. This indicator is a weak acid $\ce{HA}$. The undissociated acid has a different color than it's ion $\ce{A-}$. The $\mathrm{p}K_\mathrm{a}$ value of a colored indicator is given to be $\mathrm{p}K_\mathrm{a}=6$. At what value of the $\mathrm{pH}$  will the color change? Please explain your answer. (Assume that both colors are equally well recognizable).

I've had this question on an exam and I can't seem to figure out what exactly the answer could be.
I would intuitively say that the color change happens at $\mathrm{pH}=6$. If the $\mathrm{pH}$ is lower than 6 then there are less $\ce{A-}$ ions and more $\ce{HA}$ (acid-indicator) in the solution. Hence the solution will take on the color of the undissociated acid-indicator. If the pH is higher than 6 then the $\ce{HA}$ starts to dissociate more to $\ce{H3O+}$ hence there being a higher $\ce{A-}$ concentration and the solution takes the color of the indicator's conjugate base. But I'm not sure if my thought process is right or how to 'prove' it's right.

What ideas do you have? This is a homework (http://meta.chemistry.stackexchange.com/a/142/6998 ) question, so you should present your current train of thoughts

Thanks for the answer!
Well, I would intuitively say that the color change happens at pH=6.
If the pH is lower than 6 then there are less A(-) ions and more HA(acid-indicator) in the solution. Hence the solution will take on the color of the undissociated   acid-indicator.
If the pH is higher than 6 then the HA starts to dissociate more to H3O+ hence there being a higher A(-) concentration and the solution takes the color of the indicator's conjugate base.
But I'm not sure if my thought process is right or how to 'prove' it's right.
Thanks again!

@Michael please [edit] your question and include your thoughts. Voting to leave open

I second @IͶΔ, but I just did that for you. Unfortunately I have no time to answer this, so I give you some hints. You are on the right track. Assume you need about a 1:10 ratio of colour1 to colour2 to actually see a change. Try using the Henderson-Hasselbalch equation to explain it. The answers differ when you increase or decrease the pH.

I'll throw this in. Usually the indicator is easier to monitor one way than the other. For example with phenolphthalein it is easier to watch the pink color appear than disappear. It certainly is impossible to judge "maximum pinkness." So you generally titrate with a base when using phenolphthalein as the indicator.

Thanks for the support guys! I think I found the right answer.

Thank you guys for your answers! I am fairly confident I have found the correct answer, or at least one of the possible correct answers. Here's it goes:
The indicator reacts with water according to the following reaction equation:
$$\ce{HA + H2O <=> H3O+ + A-}$$
Then using the Henderson-Hasselbach equation, I arrive at the following point:
 $$ K_s = \frac{[\ce{H3O+}]*[\ce{A-}]}{[\ce{HA}]} $$
Now, at the equivalence point, the concentration of the indicator an it's conjugate base are equal, hence:
 $$\require{cancel} K_s = \frac{[\ce{H3O+}]*\cancel{[\ce{A-}]}}{\cancel{[\ce{HA}]}} $$
$K_s = [\ce{H3O+}]$. After taking the negative log of both sides I arrive at the equation: $pK_s = pH$ (which was given to be 6).
Now, when the solution is acidic(pH lower than 6 and high [$\ce{H3O+}$] concentration), Le Chatelier sais then the balance will shift to the left of the top most equation. Favoring the indicator $\ce{HA}$ as opposed to it's conjugate base. Naturally, if the solution contains more indicator molecules, then it will also take on the indicators color.

 Analogue to the previous reasoning, when the pH reaches values above those of 6, Le Chatleier sais that the reaction will favor the formation of the indicators conjugate base [$\ce{A-}$](the right side) and the solution takes on the color accordingly.

No... you didn't take Martin's hint. Either [HA] = 10*[A-] --or-- 10*[HA] = [A-]. If you've every done a titration with phenolphthalein, then the end point is when the first tinge of pink appears. [HA] = 10*[A-] which is on the acid side of the pKa.


text:
lassaigne's nitrogen detection test
I was testing an organic dye for nitrogen today in lab.
First I added powdered ferrous  sulfate (.2 g) and the mixture gave prussian blue precipitate. Then I boiled and cooled it. When I acidified the mixture using diluted sulfuric acid, the mixture cleared and only a few small blue crystals remained, when there should've been a bluish-green precipitate, ensuring the presence of nitrogen(organic dyes should contain nitrogen).
Where did it go wrong?

Did you google "Lassaigne's nitrogen" ? The test seems quite different than what you did.

Not all organic dies include nitrogen. It's helpful, for sure, but not a rule. It sounds like this one does, though. However, MaxW is right.

thanks MaxW for your reply.I did look 'lassaigne's nitrogen test' and it does seem like I was doing this test differently. but that's how we were taught to do the nitrogen detection test.I did the same test on several other compounds containing nitrogen and it gave positive results. I never added ferrous chloride in any of those test yet it showed positive results.that's why I can't be sure if I was doing the test wrong


text:
Angular Momentum Of S-subshell of an atom?
The angular momentum of every S-subshell of an atom is 0 by Azimuthal Quantum No. Relation. But if angular momentum of S-subshell is zero. Then by,
Angular Momentum =Mass×Velocity×Radius;
Radius of S-subshell shoul be 0 which is not the the case. Then why do we consider Angular Momentum of S-subshell zero.

Because electrons do not behave classically. $\vec{l} = \vec{r} \times \vec{p}$ is a classical relationship. This is one of the shortcomings of the Bohr model.

@orthocresol I thought the relation $\vec{l} = \vec{r} \times \vec{p}$, where $\times$ denotes the cross product, carries over from classical to quantum mechanics simply by replacing the vectors with their respective quantum mechanical operators? I've always taken that for granted. Are there any exceptions I should know about?

@Philipp Yeah, the ang mom relations $\hat{l}_x = \hat{y}\hat{p}_z - \hat{z}\hat{p}_y$ etc. are derived from the cross product. I think the problem with OP's question is that he's implicitly using the Bohr model by talking about the "radius" of the s orbital - the electron isn't going in uniform circular motion around the nucleus with a defined value of $r$ and so it's not appropriate to use the classical relation here (which applies to the movement of a particle). I'd say $l = 0$ for an s orbital simply because the s orbital is an eigenfunction of $\hat{l^2}$ with eigenvalue 0

Your confusion arises from the wrong choice of equation for the angular momentum. Your equation
\begin{align}
\lVert \vec{l} \rVert = m \lVert \vec{v} \rVert \lVert \vec{r} \rVert \ ,
\end{align}
where $\vec{l}$ denotes the angular momentum vector ($\lVert \vec{l} \rVert$ is its length, i.e. the angular momentum), $\vec{v}$ the velocity vector, $\vec{r}$ the radius vector  and $m$ the mass, does already entail some assumptions that are not necessarily valid. It derives from the more general equation
\begin{align}
\vec{l} = \vec{r} \times  \vec{p} ,
\end{align}
where $\vec{p}$ is the momentum vector which under certain circumstances can indeed be written as $\vec{p} = m \vec{v}$ (there are a few subtleties that I don't want to touch here, keyword: canonical momentum, since they don't relate to the problem at hand) and $\times$ denotes the cross product between two vectors. If you now take the absolute value of the angular momentum vector and use the properties of the cross product you arrive at
\begin{align}
\lVert \vec{l} \rVert &= \lVert \vec{r} \times  \vec{p} \rVert \\
&= \lVert \vec{r} \rVert \lVert \vec{p} \rVert \sin \theta \ ,
\end{align}
where $\theta$ is the angle between the vectors $\vec{r}$ and $\vec{p}$. So, the factor that is missing in your equation is the $\sin \theta$ term, whose absence means that your equation implicitly assumes that $\theta = 90°$ since $\sin 90° = 1$. And indeed if this would be the case, then for $ \lVert \vec{l} \rVert = 0$ to be true either $ \lVert \vec{r} \rVert$ or $ \lVert \vec{p} \rVert$ had to be equal to zero. However, since this assumption isn't made anywhere in the quantum mechanical description of an atom the $\sin \theta$ term has to be kept and there is another way to achieve $\lVert \vec{l} \rVert = 0$, namely $\vec{r}$ and $\vec{p}$ being either parallel ($\theta = 0°$) or antiparallel ($\theta = 180°$) to each other because then the $\sin \theta$ term is equal to zero.
Of course, for the quantum mechanical description you have to replace the vectors by the quantum mechanical operators and the absolute values of the vectors by the expectation values of the operators.


text:
Can't you form a ring from almost any number of sides from carbon?
Recently, we have been learning Chemistry in my physics class at school. We learned that Carbon is one of the elements that has the most variety in the compounds it can formed, being able to form four covalent bonds. This got me thinking- I know you can form Carbon into hexagonal rings- graphite does this- but shouldn't you be able to form almost any kind of polygon from carbon, with each one double bonded to the two next to it?

Yes you theoretically could but it would be very unstable. http://chemistry.stackexchange.com/questions/33497/a-monocyclic-6-carbon-ring-with-6-double-bonds

You'll do better to do 1.5 average bond order. Examples

Or just "1.5 bond order," I guess.

Carbon bonds have certain preferred angles. For example, the tetrahedrally symmetrical methane, which has sp3 hybridization,, $\ce{CH4}$, has an angle of 109.5 0. In benzene, where there is sp2 hybridization, the angle is 120 0.
These are the preferred angles for bonds. In some molecules, the geometry forces them to odd angles, such as cyclopropene, with a 60 0 bend. This imposes stress, much like making Tinker toy models where you have to bend the sticks a bit. Overdo this and the stick (bond) snaps.
So, here are some cyclic (ring) compounds with the following number of carbons, and at least one alkene bond:
3: cyclopropene
4: cyclobutene
5: cyclopentene
6: cyclohexene
7: cycloheptene

cyclooctene

Notice in the drawings of the molecular structure that larger rings are increasingly asymmetrical, degenerating to a twisted mess. So yes, very large carbon "rings" are possible, but eventually cannot hold their shapes. Some compounds contain more than on C=C bond, such as 1,3-cyclohexadiene, but I know of no cyclic compounds where all carbons contain a double (alkene) bond.
You can extend this, though, to build up three-dimensional polyhedra, such as cubene or the elegant (though not double-bonded) buckminsterfullerene.

The OP asked about cycloalkene rings (name?) not cycloalkane rings.


text:
Is peroxynitrous acid an isomer of nitric acid?
Structure of nitric acid:

Structure of peroxynitrous acid:

Nitric acid and peroxynitrous acid have the same structural formula, $\ce{HNO3}$ and in both the compounds, O.N. of nitrogen is +5. So, is peroxynitrous acid an isomer of nitric acid?
It seems that peroxynitrous acid is a very, very unstable compound and I don't think it has been synthesised in pure, stable form. This video shows the preparation of peroxynitrous acid but it readily converts into nitric acid and thus the peroxynitrous acid yield is very poor.
Keeping aside its stability and considering only their structure and structural formula, is peroxynitrous acid an isomer of nitric acid?

Yes http://goldbook.iupac.org/I03289.html

Per the definition of "isomer", it is. Why would you think it wouldn't be?

@IͶΔ perhaps I was considering its stability. It is a very, very unstable compound and I hardly doubt it would exist.

Note that "I hardly doubt that it has been synthesised" means "I am pretty sure it has been synthesized"...

@BuckThorn You are being pedantic, aren't you? ;). Anyways, I was so lazy at that time that I forgot to check the Wikipedia page. On the last line, it clearly says "It is an isomer of nitric acid" . Sigh, the question loses its purpose though I must appreciate hBy2Py's answer. He summarized his Ph.D career in his answer.

I was just a little confused by the statement and thought it worth cleaning up. And, I've seen much worse questions ;-) .... both question and answer are interesting.

Note that the N oxidation number is formally +3, as oxidation number of oxygen atoms in peroxy group is -1.

(@BuckThorn) Nilay, if only: despite spending a fair bit of time working on the peroxynitrite synthesis and related experiments, I never actually got any of that work to the point of publication -- all of my journal articles were from other chemical systems!

Keeping aside its stability and considering only their structure and structural formula, is peroxynitrous acid an isomer of nitric acid?

Yes.

It seems that peroxynitrous acid is a very, very unstable compound and I hardly doubt that it has been synthesised in pure, stable form. This video shows the preparation of peroxynitrous acid but it readily converts into nitric acid and so the peroxynitrous acid yield is very poor.

Indeed, it is impossible to isolate peroxynitrous acid $\left(\ce{ONOOH}\right)$ in aqueous solution, due to the rapid decomposition to nitric acid. The yield is not just poor, it is zero.
However, the conjugate base of peroxynitrous acid, peroxynitrite $\left(\ce{ONOO^-}\right)$, happens to be extremely stable. (Apropos the question, peroxynitrite is an isomer of nitrate.) So, as long as a synthetic approach is devised that keeps the product strictly in its anionic/alkaline form, one can actually produce peroxynitrite solutions and salts and work with them in experiments.  In fact, I synthesized sodium peroxynitrite solutions myself as part of my Ph.D. research.  You can read about the procedure in Section 4.2.4 of my thesis (free for download).  A short literature review of various $\ce{ONOO-}$ synthesis methods is also included there.
Briefly, I loaded aqueous solutions of (1) $\ce{NaNO2}$, (2) $\ce{H2O2}$ plus $\ce{HCl}$, and (3) $\ce{NaOH}$ into syringes fitted onto a syringe pump.  I then connected the syringes to a tubing network that first mixed the acidified $\ce{H2O2}$ with the $\ce{NaNO2}$, forming $\ce{ONOOH}$, and then quickly mixed the product solution with the $\ce{NaOH}$.  The alkali "caught" the $\ce{ONOOH}$ while a significant concentration still remained, providing a bright yellow solution of alkaline peroxynitrite.  After decomposing residual $\ce{H2O2}$ by treatment with $\ce{MnO2}$, these solutions held their concentrations for a couple of years when stored at $-80\,^\circ\mathrm C$.

Can't not prattle on about my Ph.D. work any time I get the chance...!


text:
Does a frying pan use radiation?
I have a project to prove that the colour black absorbs more light than the colour white in a creative way. Our first idea was to time a white frying pan and black pan and see which would melt the ice cube first, but is this proving radiation? Because I looked at saw pieces of metal above where the flame would go, making me guess pans use conduction. Will a flame ever give off enough radiation to be noticeable?

When cooking using a pan over a flame, most of the heat doesn't come from the light of the flame (radiation), but rather the hot gases created by combustion in the flame.

True, most of the heat is carried by convection. That being said, flame does produce a good deal of thermal radiation, more than enough to be noticeable. You have to devise some ingenious way to separate that from convection.

A light mill proves that black absorbs more light than the color white.

Isn't the heat the pan is putting into the food almost purely conduction? The heat gets to the pan through all three modes of heat transfer, but once it's there, I don't think you would get much cooking going if you suspended a steak an inch above the pan. Sounds like a good way to warm some milk before bed... if you want to wait an hour.

You may estimate the contribution of thermal radiation by using the Stefan–Boltzmann law, which describes the power $P$ radiated from a black body in terms of its temperature $T$:
$$P=\sigma\cdot A\cdot T^4$$
where $\sigma=5.670\,367(13)\times10^{-8}\ \mathrm{W\ m^{-2}\ K^{-4}}$ is the Stefan–Boltzmann constant[reference] and $A$ is the surface area.
The formula can be corrected for real materials by introducing the emissivity $\varepsilon$:
$$P=\varepsilon\cdot\sigma\cdot A\cdot T^4$$
For an ideal black body, the emissivity is $\varepsilon=1$. For polished metal surfaces, the values are significantly lower.
For example, for a hot black frying pan with $\varepsilon\approx1$, $A\approx600\ \mathrm{cm^2}=0.06\ \mathrm{m^2}$, and $T\approx230\ \mathrm{^\circ C}\approx500\ \mathrm K$, the power may be estimated as
$$\begin{align}
P&=\varepsilon\cdot\sigma\cdot A\cdot T^4\\
&=1\times5.670\,367\times10^{-8}\ \mathrm{W\ m^{-2}\ K^{-4}}\times0.06\ \mathrm{m^2}\times\left(500\ \mathrm K\right)^4\\
&\approx200\ \mathrm W
\end{align}$$
However, in addition to thermal radiation, heat transfer also occurs via thermal convection and thermal conduction. You could exclude these modes by placing the frying pan in a vacuum.
You could also estimate the total heat transfer due to all three modes from the electrical power of the stove that is required to keep the frying pan at a constant temperature.


text:
How to calculate the rate of reaction of Sodium and Chlorine
$$\ce{2Na(s) + Cl2(g)   ->    2NaCl(s)}$$
How exactly do you calculate the reaction rate of the following equation?
Since reaction rate is usually calculated in moles per $\rm dm^3$ per second (M/s) and the reactants here are solid. I want to have a equation of $$R = K \times\ce{[Na]}^x\ce{[Cl]}^y$$

Reaction rate constant completely depends upon the reaction order (i.e. the stoichiometry of the slow step). What do you think the slow step is in this case? Also, there are plenty of first-order reactions that require one species to be excited or activated first (as well as other common reaction orders), so saying they're usually in units of $\frac{mol}{L\cdot s}$ is... I don't know. Well, a mistake, as I'm pretty sure you will find out on this question.

Nah, nevermind. This one comes out.

Rate laws are determined via experimental data. Reactant concentrations are held constant while others are modified in order to determine the effect of each reactant's concentration on a reaction. For this reason, one cannot determine the rate law from just a stoichiometric equation.
$$\ce{2Na + Cl2 -> 2NaCl}$$
We can propose reaction mechanisms to make an educated guess of the rate law:
\begin{align}
\ce{Na + Cl2 &-> NaCl + Cl}&&\text{(slow step)}\\
\ce{Na + Cl &-> NaCl}&&\text{(fast step)}\\
\end{align}
Using Hess' law, we can combine these to make sure it equals the original equation.
$$\require{cancel}
\begin{align}
\ce{  && Na + Cl2 &-> NaCl + \cancel{Cl}}\\
\ce{+ && Na + \cancel{Cl} &-> NaCl} \\\hline
\ce{  && 2Na + Cl2 &-> 2NaCl} \\
\end{align}$$
The coefficients of the slowest elementary step in reaction mechanism determine the exponents of the rate law. Hence:
$$\text{Rate}=k[\ce{Na}]^{1}[\ce{Cl2}]^{1}$$
Or just
$$\text{Rate}=k[\ce{Na}][\ce{Cl2}].$$

What about that part where sodium is a solid?


text:
Chemical measurement questions
I have some basic chemical measurement questions. I have to mix 3% of potassium sulfate with calcium sulfate as an accelerant. How do I calculate the 3%? Base it off 100%? Does this question make sense?
Also, I need to measure out 5cc amounts of powder from a large bucket of powder. Is there a measuring cup for cubic centimeters?

Who has you measuring chemical powders by volume? The amount of chemical you'd get would be highly dependent on how much settling of the powders there is. I'd highly recommend looking at the source material for your protocol, and seeing if there's a weight equivalent for the 5cc listed anywhere. Measuring powders by volume is only suitable for low-accuracy tasks like cooking.

Thank you for taking the time to answer. Appreciate the insight

Generally a specification like 3% of potassium sulfate is wt%. Such a solution would have 30 g in a about a liter of water (weighing 1 kilogram). $$30/1030 = 2.91 \%$$ which should be close enough. You could use 970 g of water to be more exact. But just filling a beaker to the 1 liter mark would work fine.
I never though of cooking in Europe, but I'd guess that they do have measuring spoons in some $\mathrm{cm}^3$ (mL) units. Again good enough would almost certainly work. $$ 1 \text{ teaspoon} = 4.92892 \text{ mL}$$

Thank you for taking the time to answer. Appreciate the insight


text:
Exothermic liquid-liquid mixing
Does anyone know of a good example of two liquids mixing very exothermically? As a reference, I'm thinking more like NaOH and water boiling when mixed at room temperature than ethanol and water are exothermic but you'll never notice making a gin and tonic.

like sulfuric acid and water?

Not that I would do sulfuric acid and water except under very controlled situations.

I have to do the safety shtick on this. Always add sulfuric acid to water. The reaction generates a lot of heat. It is not unsafe for a chemist, but caution and respect are required. Safety glasses should be worn, you should have a lab coat or apron, and rubber gloves would be a good idea. You should have access to a safety shower and an eyewash - just in case... //  If you drop concentrated sulfuric acid on the floor be careful. It is very slippery.

Yeah, like sulfuric acid and water! Thanks. As for safety, I don't want to do it, just looking for an upper bound for what reasonable enthalpies of mixing are. Loong, if you want to post an answer I'll accept it.

Adding water to sulfuric acid generates so much heat that the diluted solution boils and splatters sulfuric acid. To make matters worse sulfuric acid is dense, so the diluted solution "floats" on the surface where it can splatter the most. very very nasty... It is kind of like dumping some water on a really hot skillet. It is definitely not the kind of mistake that anyone would make twice.

Water & ethanol have a (small) positive mixing enthalpy? I'd doubt that very much.


text:
Counterintuitive Reaction of Aluminium On Steel Balls
There is a certain video which I just watched about 2 large cast iron balls. If the balls are struck together, no reaction happens, but if one is covered in aluminium foil and then struck, a large quantity of sparks are produced. This flies in the face of my metalworking experiences with these metals. When cut, steel reacts with oxygen in the atmosphere producing large quantities of sparks, but aluminium will not produce any sparks, despite being cut with an abrasive wheel or grinded with an abrasive wheel.
So why does the aluminium spark when the balls are grinded together? Could the temperature be higher forming a reaction with oxygen? Is there pure aluminium in the center of the foil which does not have the typical Alumina composition on the exterior and is being exposed to the atmosphere and is being oxidized?
Video: (CrazyRussianHacker): https://www.youtube.com/watch?v=w7yWoO7vqvI

Physics comment: thermal conductivity combined with strength - Al is softer and has a much higher TC than steel. On Al you tend to rip bits out quite easily while on steel, frictional heating has a much greater effect. So the steel particles are much hotter.  Whether they get red-hot from this alone is another matter, which a quick google couldn't answer.

So are they steel or are they cast iron?

@MasonWheeler I believe that cast iron is a type of carbon steel.

Nope.  Cast iron is an iron/carbon alloy, but its carbon content is too high to be steel.  Steel is silvery in color and malleable; cast iron is black in color and brittle.

@MasonWheeler Yeah, I just read the Wikipedia page on it, and apparently it contains $Si$ as well.

connected with http://meta.chemistry.stackexchange.com/questions/3184/unresolved-question-from-physics-se and http://physics.stackexchange.com/questions/106121/why-does-grinding-steel-create-sparks-but-aluminium-doesnt

This is actually related to the thermite reaction, where the rust on the balls reacts with the pure aluminum in the foil. Here is a pdf detailing the experiment.
The activation energy for the reaction is reached through the friction and pressure, and the rust on the balls is actually very important, since it is the exchange of oxygen between the iron oxide and aluminum that releases the heat as iron and aluminum oxide form.

Oh my god, that's so dumb of me, I forgot the balls were rusted and in my head I thought that they were just regular steel balls.

While the thermite reaction between the iron oxide of the rusted balls and the aluminium foil is certainly the major factor in the resulting lightshow, even unoxidized steel balls could potentially ignite a small amount of the aluminium foil, in a similar fashion as how the piece of paper is burnt in this video.

Here's what happens without the rust: https://youtu.be/JTcE9wbbgZ8?t=323


text:
pH of a buffer made with boric acid and potassium dihydrogen borate
The Problem

Calculate the pH of a buffer that is $\pu{0.200M}$ $\ce{H3BO3}$ and $\pu{0.122M}$ $\ce{KH2BO3}$. The $K_\mathrm{a}$ for $\ce{H3BO3}$ is $7.3\times10^{-10}$

What I Tried
For each of these methods, I used $\pu{0.200M}$ as the concentration for the acid, $\ce{H3BO3}$, and $\pu{0.122M}$ as the concentration of its conjugate base, $\ce{H2BO3-}$.

Method one – Henderson–Hasselbalch equation
$\mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a}) = 9.13668$
${\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{0.122}{0.200}\right) = 9.13668 + (-0.21467) = 8.92201}$
Method two – Rice Table
$$
\begin{array}{c|lcr}
 & \text{$\ce{H3BO3}$} & \text{$\ce{H3O+}$} & \text{$\ce{H2BO3-}$} \\
\hline
\text{I} & 0.200 & 0 & 0.122 \\
\text{C} & -x & +x & +x \\
\text{E} & 0.200-x & x & 0.122+x
\end{array}
$$
use x is small approximation
$K_\mathrm{a} = \ce{\frac{[\ce{H3O}][\ce{A-}]}{[\ce{HA}]}}$
$\ce{K_\mathrm{a} = \frac{0.122x}{0.200}}$
$x = 0.200 \times \frac{K_\mathrm{a}}{0.122}$
$x = 1.19672*10^{-9}$ – This equals the number of moles $\ce{H3O+}$
$\mathrm{pH} = -\log([\ce{H3O+}]) = 8.92201$

Check $x$ is small approximation: $\frac{x}{0.122} = 9.081\times10^{-9}$ which is less than $5\ \%$, so the approximation is valid.
My question
The answer I got using both of these methods, $\mathrm{pH} = 8.92$, is not correct. The only conclusion I can come to is that I have missed some nuance of this question or I am making a careless mistake. Can somebody please tell me what I'm overlooking here?

How do you know that the answer of 8.92 is wrong? What is the answer supposed to be?

@pH13 boric acid does not give a proton but rather accepts an $\ce{OH-}$ to form $\ce{[B(OH)4]}$. So isn't it $\ce{K[B(OH)4]}$ instead of $\ce{KH2BO3}$?

Boric acid is quite complex, so I don't really start without knowing where to go.

I don't know what the answer is supposed to be; it is a multiple choice question on Mastering Chemistry, and when I picked 8.92 it said I was wrong. I didn't list the other choices because I didn't want anyone to think I was trying to get someone to answer the problem. Really I'm just looking for some insight as to what I could be missing.

From where did you get the acidity constant? Was it given or did you search it yourself?

It was given. Everything listed under "The Problem" was copied verbatim from the question.

Then actually, I'd be interested in what the multiple choices are.

They were:  3.52,   10.77,   9.87,  9.46  and 8.93

@JennyAnn There's nothing wrong with what you did, you just need more confidence in knowing that you are right.  Check your work in a problem like this intuitively saying to yourself, "there is a little more acid than conjugate base, so the pH will be a little below the pK".

Depending on the level of rigor required, the solution to this problem could range from honors high school level chemistry to upper level graduate school inorganic chemistry.  The complexity of the chemistry of aqueous boric acid is well described in this reference:
    [http://www.crscientific.com/experiment4.html][1]

The author cites one source in which it is stated that there are 10 different equilibrium systems in boric acid solutions!
If you are in honors or AP high school chemistry or even in freshman inorganic chemistry, I suspect that you would be expected to solve the problem exactly as you did.  Perhaps the test writers wanted an interesting weak acid and conjugate base for the problem and either did not worry about the true complexity of the boric acid chemistry or were not aware of it.  In addition, the author(s) of the questions may not be the ones who provided the solutions to the problems.  Now here are the twists.
There is some debate about whether or not boric acid behaves as a triprotic acid with three successive hydrogen ion transfers to finally produce BO3- + 3 H3O+. (https://en.wikipedia.org/wiki/Boric_acid#Properties)  In accordance with the triprotic model three separate university websites (North Carolina State University, University of California Santa Barbara, and the University of Washington) cited the Ka1 of boric acid to be 5.8 x 10-10.
The competing boric acid dissociation model is well described in the crscientific source above and, in summary, begins with B(OH)3 (another way to write boric acid) acting as a Lewis acid:
    B(OH)3 (aq)  +  H2O  ⇄   B(OH)4- (aq) +  H+ (aq)

Further reactions involving B(OH)4- (aq) introduce species such as H2B4O7, HB4O7- and B4O72-.  What is very curious is that the K of the reaction of B(OH)3 (aq) and H2O is said to equal 7.3 x 10-10 in a Wikipedia article
 (https://en.wikipedia.org/wiki/Boric_acid#Properties). This is the value of Ka1 you were given in the problem, but the problem is based on the triprotic boric acid model!  I encourage you to rework the problem with a value of 5.8 x 10-10 for the Ka1 just to see if the answer for this calculation equals one of the other choices in the problem.

LOL - I got hooked on this. Using the Henderson-Hasselback Equation as @JennyAnn did you get
$${pKa = -log(K_a) = 9.23657}$$
and
$${pH = pK_a + log (\frac{0.122}{0.200}) = 9.23657 + (-0.21467) = 9.02190}$$ I don't see how that helps...

@Steve Lantz, I did rework it with that value for the Ka and got the same answer as MaxW. I am in taking a general chemistry course in college, so I think the method is correct. The system must be wrong.
Completely aside from the answer to the problem, your post was very interesting. I am taking an intro level general chemistry course, and I had no idea that there were still debates about the way molecules behave; I'd just assumed we knew everything already. Thank you for taking the time to answer my question

Update: Per the advice of @DavePhD , I decided to trust my work and I approached my teacher with the problem. He just emailed me to say there must have been some error in the online system; 8.92 is the correct answer.
Thanks so much to everyone who took the time to help me with this!

Thanks for letting us know. My perception was that when you really beat on a problem like this you learn something. // If the problem was supposed to be analyzed at a more complex level then you would have needed a whole lot more equilibrium constants.

Hi Jenny Ann.  I agree with MaxW about how much you learn from really going after a problem!  It throws all of us into a tizzy when there is an error in a test question, which is what I suspected from the beginning.  @DavePhD was right when he suggested that you should trust your work.  Still, the chemistry adventure we embarked on was cool.  I see the possibility of a chem major in your future ...


text:
Wattage to produce 6L of air from Water in less than a second?
So I understand the question is probably vague to a scientist, but this is all the specifics I could mustard up with my non-scientific background.
So I understand the basics of the process and have used it occationally in a garage environment to fiddle around, but never really thought about the applications until a particular Indiegogo campaign claimed to have created artificial gills small enough to hold in your mouth and generating 45 minutes of breathable air.
The campaign have gotten a wide spread of media attention and it got me thinking, how much wattage would be required to produce 6L of air over the course of a second which happens to be a rough average of a adults lung capacity.
I understand that submarines use this process to create breathable air, but as someone mentioned they have a nuclear reactor onboard and "it would not be wise to attach that to your face".
Would shortening a Lithium Ion/Polymer battery be sufficient to create such a quick reaction? Or are we talking megawatts in order to produce such a volume of air in that time span?
Update to the numbers used:
Pointed out by @DavePhD, you don't use 6L per breath, rather 6L on average per minute.
Taking into account that this would be a physical exercise under water, you'd normally use anything between 14-17L per minute, on average 16L.
Also noted is that I should have used P instead of the term wattage, thanks @Loong

Please note that the quantity electric power shall not be called “wattage”.

@Loong What's it called? Sorry for being ignorant and novice but would Joule be better? Oh wait, it's probably not as easy as that heh. Man I'm out on thin ice here concerning my knowledge in the area and should probably not keep on guessing!

Now you are mixing up quantities and units. The name of the quantity is “electrical power” (symbol: $P$). The name of the unit in this case is “watt” (symbol: $\mathrm W$); however, you could use any other unit of power. Since quantities are themselves always independent of the unit in which they are expressed, a quantity name shall not reflect the name of any corresponding unit. Therefore, you should avoid the name “wattage”.

Instead of 6L per second, people breath about 6L per minute.  6L is the lung capacity, but a typical breath is much less and the interval between breathes is about 4s.

@DavePhD Makes sense, since it takes roughly 1-2 seconds to breathe out the carbon dioxide. And you're right, come to think of my average air consumption while diving back in the day was 16L per minute (remember, it's a physical exercise). I'm glad you corrected me on this one!

I guess it would have to be the amount energy released by the reaction $$\ce{2H2 + O2 -> 2H2O}$$ where you have four liters of hydrogen gas and two liters of oxygen gas, all divided by one second (or less, increasing the wattage as you decrease the time).
You asked about "air" — but you can't get normal air from electrolysis of water.
So, the heat of combustion for hydrogen is $286\ \mathrm{kJ/mol}$. $$286\ \frac{\mathrm{kJ}}{\mathrm{mol}} \times \frac{4\ \mathrm L}{22.4\ \frac{\mathrm L}{\mathrm{mol}}} = 51.1\ \mathrm{kJ}$$ and that is assuming atmospheric pressure which is certainly low, but how low depends on your depth below the surface of the water. So with the "barely under the surface" energy of $51.1\ \mathrm{kJ}$, we can then divide that by seconds: $$\frac{51.1\ \mathrm{kJ}}{1\ \mathrm s} = 51.1\ \mathrm{kW}$$ so you would need a bare minimum of $51.1\ \mathrm{kW}$ of power to convert water into a 67:33 mixture of $\ce{H2:O2}$. That seems like a lot of energy in a small amount of time to me: think about a six-liter balloon with the perfect combustion ratio of hydrogen and oxygen. That reaction is a lot faster than one second, but it's still a lot of energy. Tough to get into a small, portable battery and have it fill a capacitor with that amount of e energy every few seconds.
I'll be honest, I'm just doing these calculations on-the-fly, and I may have made a mistake or five, but it seems reasonable to me. If anyone takes any issue with my answer, please, feel free to edit/give suggestions.
Edit: This Wikipedia page has some information on hydrogen-oxygen diving air. They use a 96:4 ratio. Not sure why, but if that's the best ratio, this would take significantly more energy per breath, since a lot of oxygen would just get dumped.

Based on this information I've calculated that you would need roughly 5 LiPo 12S batteries of 60C with  5Ah capacity (which produce 300A over 44.4V each continiously) with a potential burst of up to 600A each for shorter durations. The size of these batteries would be roughly 10cm wide and 26cm heigh and 32cm long, It would be a odd formfactor to carry around and by the design schematics of the Indiegogo device I'd say impossible. But Plausable in say a back pack. Drawing that much current would not last 45 min either from this setup.

@Torxed I understood very little of that, but I can agree that carrying around with those dimentions would be awkward. You might have an easier time getting dissolved gasses out of solution from the water... Like fish.

Sorry for the code language, basically "5000mAh 60C" Lithium Polymer battery, 60C standing for the discharge rate the material can handle ((60*5000)/1000) is the calculation for the ammount of Amps that particular battery can dish out). Hehe indeed it would be an aswkard thing to carry around on your face. It would however be less clumbsy than a regular diving tube that you would spend hours to pressurize. I'm surprised that no one has done this yet if this all adds upp.

"gills" means taking dissolved oxygen out of water, not performing electrolysis.  The OP is referring to this concept: http://www.techinsider.io/can-triton-device-indiegogo-pull-oxygen-water-2016-3

@DavePhD I'm aware of that. But he also talks about submarines and making breathable air from water. I answered that part, and it's more feasable to make work than the dissolved oxygen thing, apparently, it's just still terrible.

@DGS Just to be sure I didn't screw up, in enlight of new information (that it's 6L per minute, and not per second), but during phsyical exercise I would need say 12L per minute. Would the end result per second be 1.7kW? Just making sure my calculations add up in regards to your example where 6L of "air" per second came out as 51.1kW. That formulate to me added up to 1.7kW/second if i based it on 12L/per minute. Sorry for a very fuzzy question, my way of explaining formulas and stuff is that of a 4 year old.

That does sound more doable. But then realize that the air they use that is hydrogen and oxygen throws away 7/8 of the oxygen that would be produced, so it might have to work eight times as hard unless it uses that oxygen to reenergize itself (without burning the hydrogen -- that might be terribly hard, I think) and finding some fuel that doesn't weigh you down wouldn't be easy, either... Well, maybe a methane balloon, huh? But then you have the opposite issue. Well, I guess hydrocarbons would actually make you more buoyant. Just cumbersome.

It's worth noting that even assuming you can bring all this electricity to bear, if you apply it to seawater, you'll mostly get chlorine instead of oxygen IIRC.

@Aesin Oh... Yeah. Well, that's a good way no to drown, too.