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Are there any reactions with no activation energies?
Are there any reactions with no activation energies? Our professor just told us there are no chemical cliffs. Is this true?
I read something about nuclear decay as being a reaction with no activation energies. Why is this so, and how does this work?

Bond formation between atoms has no activation energy, at least for simple cases. Consider the reaction between two hydrogen atoms to form hydrogen gas. The potential energy curve only slopes downwards from infinity to some critical value, approximately equal to the bond length. Electron affinities (when exothermic) also likely have no activation energy.

@NicolauSakerNeto Very good points. If you find the time it would make a good answer.

@Philipp I'm rather busy lately, so I should probably avoid getting into more stuff. If anyone wants to build on my comment, feel free to do so!

As an aside, if you had a "cliff" could you even purify the high energy compound? Seems like it would just "roll off" and spontaneously react. Having activation energies creates a valley that traps molecules in a stable state.

@NicolauSakerNeto I understand that this is indeed case. But wouldn't there be some repulsion between the hydrogen atoms when they approach each other?

@TanYongBoon If you squeeze the atoms together too far, yes, they will repel each other. However, if you set two atoms (with adequate spin states) at rest, far apart in a perfect vacuum, they will mutually attract via the electromagnetic force (as well as the gravitational force, though that effect is tremendously weaker).

Marcus-Hush theory describes electron/charge transfer rates using conventional transition state theory, e.g.:
$\ce{A+ + B -> A + B+}$
Now one might naively assume that there's no activation barrier in a simple electron transfer, but that's not true. Since the geometry of $\ce{A+}$ and $\ce{A}$ as well as $\ce{B}$ and  $\ce{B+}$ are not identical, there is a reorganization energy of both the molecular species and the solvent or environment. This reorganization energy ($\lambda$) serves as an activation barrier for the charge transfer.
Now the convenient thing about charge transfer reactions is that you can tune the $\Delta G$ of the reactants.
Marcus predicted that the maximum rate would occur when the $\Delta G^0 = -\lambda$ and thus the $\Delta G^\ddagger = 0$. If the reaction becomes more thermodynamically favorable, you enter the inverted region and the activation barrier goes up and the reaction rate goes down.
This is a long-winded answer to say there are some chemical reactions with no activation barrier, in addition to the other cases here. Usually the key phrase describing these in papers is "barrier-less reactions"

I'd give a +10 if possible for bringing up electron transfer processes! Btw, Rudolpg Marcus's Nobel lecture can be found here.

have a look at the data in this reply to another question on electron transfer where you can see where the reaction has no barrier.  http://chemistry.stackexchange.com/questions/63545/how-does-the-inverted-marcus-region-explain-chemiluminescence/63546#63546

So of course it is hard to tell the difference between the case where there is no activation energy and the case where there is a very small activation energy.  Given that...
Many reactions, such as simple acid-base reactions are diffusion controlled.  This means that reaction occurs at every encounter, and that this is not an activated process.  Another example involves benzene.  Say we generated the molecule 1,3,5-cyclohexatriene, it would immediately isomerize to benzene (unequal bond lengths, localized pi electrons -> equal bond lengths, delocalized pi electrons).  If we put this on a reaction coordinate it might look something like this,

I've always thought of this as a process with a negative activation energy, but maybe zero would be a more accurate description (can you have a negative barrier?).  I suspect it applies to many reactions where there is a large thermodynamic difference in stability between product and reactant favoring product and where there is minimal structural reorganization.

There isn't such a thing as a truly negative barrier, since if the path looked something like your diagram above, it would classify as either a product or a stable intermediate. That said, this diagram gives an example of a barrier-less reaction.

ron, could you please edit your answer to account for the concerns Geoff raised? Thank you!

The issue with some of the proposed reactions of the above posts is activation energy, if its measured in Gibbs free energy, should also subject to entropy, since ΔG=ΔH-TΔS. However the wikipedia article on activation energy (as of this writing) suggests when reaction rates decrease with activation energy, because the molecules were prepared to react at room temperature. The only way for the activation energy equation to work would be through would be for negative activation energy. Reactions like this are simple: i.e. two things smash together and will combine if the potential energy is larger than their momentum. But say for instance you heat this reaction up. You've accelerated these particles (that is, heated them, since heat can be considered kinetic energy) the reactants are so fast and have so much momentum they simply bounce off each other, and so the  becomes meaningless.
As an aside, you can have 'cheating' occur in normal reactions with hydrogen which exhibits quantum tunneling (because its so tiny). So you can have product form from a reaction while the reactants are way below the activation energy expected. So a smartass rejoinder to your professor's comment would be "sure, cliffs may not exist, but slippery slopes sure do!"

Welcome to Chemistry.SE! As an activation energy is a barrier to reaction progress (e.g., ron's answer) you cannot define a negative activation energy. That's a meaningless statement. If there's no barrier (i.e., it's completely downhill from reactants to products) it's just barrier-less. See my comments about Marcus theory for an example of decreasing reaction rates - with no need for negative activation barriers.

The reaction between two ions to form a molecule has no activation energy as they are attracted to each other and therefore no force, and no energy is required to bring them together. e.g
$\ce{H+ + H- -> H2}$

I seem to remember (haven't got the reference on hand) that singlet methylene (|CH2) inserts into the C-H bond with zero activation enthalpy.  It's certainly an enormously reactive and high-energy species.

I read something about nuclear decay as being a reaction with no activation energies. Why is this so, and how does this work?

You're asking too much. I can answer you in short:
Nuclear Decay Reactions occur when the nuclear force$^1$ between nucleons is not able to hold the nucleons together.
Now as it can't hold large nucleons, the nucleus starts emitting particles or following special processes  to reduce mass or nucleons and reach a stable zone, the choice of process depends on where the nucleon is as per a graph $^3$.

$^1$.one of the four fundamental forces of nature. If there was no nuclear force, everything would repel apart. The reason being, if there was only electrostatic force between protons, the whole nucleus$^{1a}$  would repel away. So another "nuclear force", acts into play, which is blind to the charges, and acts according to another intrinsic property$^{1b}$ called color$^{1c}$ . So now when nucleons are at the order of 1 fermi$^{1d}$ , the force is attractive and stops acting very (very) quickly (dying).
$^{1a}$ actually, only protons, to be specific
$^{1b}$ like mass, charge, etc.
$^{1c}$ and involves variety of particles called gluons, mesons, hyper-mesons, quarks, etc.
$^{1d}$ named after great scientist, $1$ fm$=10^{-15}$ m
$^{2}$$\beta^-,\beta^+,\ce{_2^4He}(\alpha)$, K-capture etc.
$^3$ which predicts that is no. of protons = no. of neutrons, then nucleus is stable ,generally, only for lighter nucleus(es) and for heavier ones it bends a little upward like a parabola ($y=x^2$)


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Is there a heat capacity of an object when it is melting?
Is there a heat capacity of an object when it is melting?
I’m questioning it since the temperature doesn’t change during phase transitions and the unit of heat capacity is joule per kelvin.

Yes there is a heat capacity even at the melting point.  The reason the temperature doesn't change is that you need to add the enthalpy of melting to convert solid to liquid in equilibrium.  However, under the right conditions you can supercool the liquid under the melting point, and the temperature will behave as you would expect, needing to pull a certain amount of energy out to supercool by a degree.  But this merely delays having to deal with the enthalpy of melting until solid uncleared - it is a kinetics issue, not a thermodynamics issue.  So, it is clear that the heat capacity of a phase is continuous through the phase transition (and beyond).

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Why does the photoelectric effect depend on the frequency of the light?
In experiments to detect the photoelectric effect, a clean metal was irradiated by monochromatic light and electrons were emitted.
Why was monochromatic light used in the experiment, and why does the frequency of the light have to be above a threshold frequency?

The photoelectric effect is described by the following equation
$$E_\mathrm{max} = h\nu - \mathrm{WF_M}$$
where $E_\mathrm{max}$ is the maximum kinetic energy of the electron escaping from the metal surface, $\nu$ is the frequency of the incoming photon and $\mathrm{WF_M}$ is the workfunction for the particular metal. The kinetic energies of all electrons emitted are distributed from $0$ to $E_\mathrm{max}$.

Why was monochromatic light used in the experiment? Why is a proper
  frequency used and not any other frequency?

The experiment is typically performed by scanning through a continuous range of monochromatic wavelengths from lower to higher energy.  At some specific wavelength, the observer will notice that electrons start to be emitted (the threshold).  As the scan continues to wavelengths with even more energy, the emitted electrons will increase in kinetic energy.  From the threshold energy and the above equation, the experimenter can determine the workfunction of the metal.  If the experiment had been run with light containing many different wavelengths (non-monochromatic light), electrons would still be ejected, but you wouldn't know what the threshold wavelength was and you wouldn't be able to determine the workfunction of the metal being studied.
If you'd like more information on the photoelectric effect, here's a good, concise reference.

The physical background of the experiment is that inside the metal the electrons can occupy states up till a point. These states are more or less continuous, however the electrons fill these states till an energy, called Fermi level. This energy is lower than the energy of electron in vacuum, so the electron needs an extra energy to jump. If the Fermi energy was not lower than vacuum, the electrons would just leave the metal, and fly away. When PE happens and a photon is absorbed, and it gives enough energy, and electron can jump out.
The important point here is that a single photon gives all the energy to a single electron. In a classical theory, light is just wave, it has no reason to excite a specific electron. Also, if it is continuous wave, electrons could collect enough energy just "waiting longer" if the frequency of light is lower. But PE works this way, because light is quantized: one photon is not the same as two photons with half energy. To jump, the electron needs enough energy from a single photon, because the chance that it is hit by two photons is practically zero, and there may not be empty states along the way and so cannot just add up energy piece by piece.
Monochromatic light
You don't need monochromatic light. If you use monochromatic light, than all the photons correspond to a given energy, therefore easier to interpret the data, but it is just a practical requirement. If you use normal light, all kind of photos will come, some bellow threshold, some above, so all you see is the mixture of zillion energies.
Threshold energy
As I said, this is the whole point. The electron cannot just absorb 3 and a half photons to get enough energy. If you have enough energy from a photon, you see PE, if not, there is no PE. The threshold energy which tells you if the energy is enough.

The wave theory fails to explain the observations
Now, one might try to explain this photoelectric effect with waves of
  light: light waves impinge on the source plate. They give energy to
  electrons on the plate. The electrons fly off the source plate to the
  receptor. The wave theory predicts several things:
The more intense the light, the more energy the electrons will have
  when the fly off the plate. If the light is very feeble, one may have
  to expose the source plate for several seconds or minutes until enough
  waves strike it to knock electrons loose. Waves of any frequency ought
  to knock electrons free. Careful measurements in the lab, however,
  showed that these predictions were wrong, wrong, wrong.
The energy of the electrons does NOT depend on the intensity of the
  light. The electrons always appear AS SOON AS the light reaches the
  plate (though a feeble light produces only a few). NO electrons are
  produced if the frequency of the light waves is below a critical
  value.

Albert Einstien while Explaining Photo electric Effect assumed the following things.

Light is made up of photons.
One photon can only knock out 1 electron

So As per Planks Equation  E = h(nu) and $E= mv^2/2 $
so a specific frequency till which electrons are not ejected is called thresh hold frequency.
$$ W=hf $$ where W is the work function, h is Planck's constant, and f is the threshold frequency.
After that, the energy supplied to the electrons is then converted to kinetic energy
Thus Total Energy $$ E = W + KE $$
where KE is just the kinetic energy of electron ejected.
Which means "The higher The frequency More will be the speed of ejected electron"
PS: - Monochromatic light is used so that there is no intermixing of frequencies!

The term monochromatic comes from the Greek words mono, meaning
  single, and chroma, meaning colour. So monochromatic light literally
  means light of one colour. In scientific terms, it means light of a
  single wavelength or frequency. Light is a term for the visible and near visible
  portions of electromagnetic radiation

ALSO,
ALBERT EINSTIEN RECEIVED NOBEL PRIZE FOR THIS EXPLANATION NOT FOR RELATIVITY!

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What does "a metallic oxide has 40 % oxygen" mean?

Question
  If a metallic oxide has $40\,\%$ oxygen, find the equivalent weight of
  the metal.

This amounts to finding the atomic weight of the metal and the charge on the cation (in effect identifying the metal itself).
I didn't know how to do this, so I looked up the solution. It goes as follows.

Solution
Assume the oxide has the formula $\ce{MO}$. Let the atomic weight of $\ce{M}$ be $x$
  Then
  $$\frac{16}{x + 16} = \frac{40}{100}$$
  Simplifying and solving for $x$ yields
  $$x = 24$$
  Thus the metal is magnesium and the equivalent weight is $12$.

My problem is, how can we assume that the metallic oxide has the formula $\ce{MO}$? For all we know, the metallic oxide may be $\ce{M2O}$ or $\ce{M2O3}$. In fact, if we assume the latter formula we get
$$\frac{48}{2x + 48} = \frac{40}{100}$$
giving
$$x = 72$$
which I believe is Hafnium.

Since you are not using units, you made a mistake in your calculation. X is the atomic mass of the metal. In the first calculation; it is 24 amu, in the second calculation, it is 72 amu. Since there are 2 metal atoms, each atom has a mass of 36 amu. The closest atom with a mass of 36 amu is chlorine, but that is not a metal; therefore $\ce {M2O3}$ is not the formula.

But the question is still valid. Can we assume that the metallic oxide is $MO$ without testing the other possibilities?

In real world problems you may well have other constraints such as knowing stoichiometry. However, if you truly know only what is stated in the problem, you'd be wise to consider other stoichiometries.  In any case it certainly isn't wrong to consider more possibilities and rule them out ... so long as you get the numbers right.

Your question is a good one, and no, you can't assume that the metal is $\ce{MO}$ necessarily.  However, you can find the equivalent weight of the metal without any assumptions about its formula.
The equivalent weight of a metal combining with oxygen is the mass that reacts with 8g of oxygen.  We know that the compound contains 40% oxygen, so it must contain 60% metal.  Suppose we had 100g of this compound; that would be 40g O and 60g M.
$40g\ \ce{O}:60g\ \ce{M}$ (Divide by 5)
$8 g\ \ce{O}:12g\ \ce{M}$
The equivalent weight of the metal must be 12g.
If you had been asked to identify the metal you would have needed the formula (or would have needed to test the likely formulas to see if they produced reasonable results.)
Given an equivalent weight for M of 12g:
$\ce{M2O-> 12g/mol, carbon}$ (not a metal)
$\ce{MO-> 24g/mol, magnesium}$
$\ce{M2O3-> 36g/mol, chlorine}$ (not a metal)
$\ce{MO2->48g/mol, titanium}$
Both magnesium and titanium fit the information that was given, $\ce{MgO}$ and $\ce{TiO2}$ are common oxides of those metals.

40% O2
So, 40/16 = 2.5
60% M
So, 60/24 (given for metal M)
    =2.5
Therefore, metal M: oxygen= 2.5: 2.5
                                                 =1:1
So, the formula is MO

For homework questions we have a policy where you do not simply post the answer without explanation but rather that you give hints that lead the asker towards their own solution. We are not a homework completion service.


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Why are all the orbitals that have the same principal number in Hydrogen degenerate?
In hydrogen, all orbitals with the same principal quantum number 'n' (1,2,3...) are degenerate, regardless of the orbital angular momentum quantum number'l' (0,1...n-1 or s,p,d..). However, in atoms with more than one electron, orbitals with different values of l for a given value for n are not degenerate. Why is this? Surely the radial distribution functions are similar for hydrogen (in that there's still penetration of orbitals and so on). Or is it that orbitals with different values of l are degenerate for a value of n greater than would be occupied in that particular atom's ground state?

This excellent answer by user Philipp seems to answer your question precisely.

The answer is right there in your question. It's only the interaction of multiple electrons in an atom like He, Li, Be, etc. that makes the different angular momenta wave functions differ in energy.
Consider this.. For the one electron system, why should a $p$ or $d$ orbital differ in energy from an $s$. What makes them differ?
In the multi-electron case, the $p$ orbitals have different spatial extent, different angular components, so the electron density caused by an electron in that orbital will interact differently with the other electrons. In other words, you need to have more than one electron for the "shape" of the $p$ and $d$ and $f$ orbitals to matter to the other electrons.
In the H atom, there's only one electron, so there's no electron-electron repulsion to differentiate the $s$, $p$, and $d$ orbitals.

I'm sorry to say that I dont find this answer at all satisfactory.  For example why is degeneracy lost in the case of the He atom.  If we are to discuss it in terms of orbitals that are occupied then we are only dealing with the 1s orbital.  (not the p and d that the above answer appeals to).

@Dr.EricScerri - no problem - you can of course write your own answer if you wish. (I expect that you've thought about how to explain this far more than I have.) That's how StackExchange works.

Actually I have always had great difficulty in connecting removal of degeneracy with the mere presence of more than one electron.  This is why I have been searching for answers.  Let me raise a further question in the hope of getting clear on this issue.  He of course has 2 electrons.  Will the energies of transitions from 1s to 2s be the same as between 1s and 2p?  I think the answer is YES.  But this is a case in which the relative penetration of electrons in orbitals fails to give an answer as I see it.  www.ericscerri.com

@Dr.EricScerri - No, for Helium, the transitions from 1s to 2s and 1s to 2p have different energies. (e.g., NIST tables. That's probably worth a separate question.

This was asked a long time ago, but I think I'll add an answer.
The answer is penetration and shielding. For a hydrogen atom, if its electron is promoted to 2s or 2p, the only thing that determines its energy is its radial distance, determined by quantum number n. There is no electron-electron repulsion, obviously.
For Helium, however, if one of its electrons is promoted to 2s or 2p, there's still an electron remaining in the core 1s shell, so the other electron in the 2s or 2p orbitals will be shielded because of electron-electron repulsion. Since 2s is shielded less by the core than 2p, it experiences a higher Zeff and is able to penetrate the nucleus more and hence has a lower potential than 2p, which is shielded more by the core and has a lower Zeff than 2s. Therefore, 2s is lower in energy than 2p because of penetration and shielding.

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Reductive amination of (protected) 3-Oxopropanenitrile: Will azetidin-2-ol be formed via intramolecular ring closure?
What happens during this reductive amination?
I got the first product; the lithium aluminium hydride will reduce the nitrile to an amine. But what about the reductive animation step? As far as I can see the C=O will be protonated by the strong tosic acid, and then there will be a competition between the methanol and the internal amine to attack C=O. Actually, the amine is able to attack the C=O without the C=O being protonated, but still, this intramolecular seem too favorable given that we are forming a 4-membered ring. So will the methanol outcompete the amine? In that case what's up with the reductive animation step?

This is what I think happens after the second step ... I am going through the intramolecular route ... since that's the only way we get an iminium ion, on which the sodium cyanoborohydride can act.

In the condensation to make the iminium intermediate, the hydroxyl should have been eliminated.

Just curious, did the instructor ever say what he thought "product 2" was?

@ron I'll let you know when I find out

Your first product looks alright, $\ce{LiAlH4}$ will reduce the nitrile group to an amino group.

But I don't think that product 1 will do an intramolecular reaction to form an azetidin-2-ol. The reason is that such a reaction, being a 4-endo-trig reaction, doesn't comply with Baldwin's rules and is thus disfavoured for kinetical reasons. If you look at the orbitals that would have to interact for the 4-ring to form you see why this is so:

The "tether" by which the amino group is attached to the carbonyl group is simply too short for the molecule to get comfortably into a conformation that allows the amino group's lone pair orbital to efficiently overlap with the carbonyl group's $\pi^{*}$ orbital.
In my opinion you will get the following reactions instead (the $\ce{NaBH3CN}$ is a typical reagent for selectively reducing an iminium ion):

But closing 8-rings is often a tricky thing. Since the chain is long the activation entropy is rather high, which makes the ring-closing reaction slow (compared to 3- or 5-rings) and intermolecular reactions with other amines in the solution might compete with the intramolecular reaction I proposed.

I agree for product 1. For product 2, if you go for the intramolecular route then the product of a reductive amination should be azetidine. The 2nd step of your mechanism in the bottom picture is not correct (give it another check). However I am not convinced it will go intramolecularly because of the ring strain of the intermediate. Maybe at best you can get a mixture of the azetidine with the open chain product and possibly oligo/polymerisation products, a mess. You need to check the literature for a precedent.
In respect to the question about MeOH: Methanol should not compete with the amine, as the amine is a lot more nucleophilic. In fact MeOH is a common solvent or cosolvent for reductive aminations (or alkylations).


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Devitrification vs glass transition temperature

This graph comes from a so-called "differential thermal analysis". It is studying the effects of temperature on glass.
I don't understand what the difference between "glass transition" and "devitrification" is, and why one would be endothermic and the other would be exothermic.
I'm having trouble analysing these types of graphs. Does anyone know of a simple source online that I could check out?

This graph comes from a so-called "differential thermal analysis". It is studying the effects of temperature on glass.

Differential thermal analysis (DTA) is a method that heats two sample pans, one with a sample and one empty as a reference so that the effects of the sample pan can be negated. The heating is performed at a constant heat flow and the temperature of the sample is compared to the reference. Any exo- or endothermic process will add or subtract additional heat to the material and cause a spike in temperature upward or downward.

I dont understand what the difference between "glass transition" and "devitrification" is, and why one would be endothermic and the other would be exothermic.

Glass transition is the temperature where bonds of a glass can relax and stress can be anneal out of a sample. It is neither endothermic or exothermic but the specific heat capacity increases as the atoms now have additional degrees of freedom.
Devitrification is where enough is put into the bonds that the atoms may rearrange and form crystalline structures from their vitreous state (devitrification). Crystallizing is an exothermic process because the atoms become more ordered and minimally space compared to a vitreous state, this reduces the strain in the bonds, releasing energy.

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Volume of different gas in a container
I know that technically two or more gases in one container should have different volumes. But when you imagine the gas molecules being mixed together, it look sorta like the two gases have the exact same volume, that is the volume of the container.
Also, in one of the questions I did on gas laws, wet hydrogen and hydrogen are shown to have the same volume in a container. Given the barometric pressure, I subs tract the vapor pressure at the certain temperature from it. Then I sub the pressure of dry hydrogen gas into the ideal gas law along with the volume of the container to find the number of mole of hydrogen gas.

This is the part in my textbook that I am confused about. It shows how partial pressure fraction is equal to the corresponding mole fraction. But this is all under the assumption that the volume and other things are constant. I am lost right here, because Avogadro's theory states the volume and mole number of a gas are directly proportional.
( wait...wouldn't this mean that pressure, mole and volume are all directly proportional to one another???)
Can anyone give a answer, along with a proof/ a source?

Your second idea, that they have the same volume, is correct.  Each gas in a mixture of gases stored in the same container has the volume of that container.  Remember that gases are mostly empty space, so it is not a problem that their volumes overlap.
As is implied by your comment about subtracting the vapor pressure of water from the pressure of wet hydrogen, the gases in a mixture can have different partial pressures. Perhaps that is your point of confusion.
The only case where they would have different volumes would be in a more accurate gas model than ideal gases, something like the Van der Waals model, where gas volume is corrected by a small amount to account for the portion of space that is taken up by the gas particles.
In short, each component of a mixture of gases stored in a given container shares the same volume and temperature and may (or may not) have different partial pressures and number of moles.

Wait even partial pressure of ideal gas depends on their mole ratio which is volume ratio. So you mean they all got the same pressures?

@Doeser No, that's not what I meant.  If I take 2 moles of oxygen and 2 moles of nitrogen and put them in the same container, they will have the same pressure and volume.  However, if I have 1 mole of oxygen and 2 moles of nitrogen and I put them in the same container, the gases will have different partial pressures but will have the same volume (the volume of the container.)  Mole ratios for gases aren't measured by volume, they're measured by pressure.  Check out Dalton's Law of Partial Pressures for more information.

But they are proportional according to Avogadro

@Doeser Avogadro's Law only applies if pressure and temperature are kept constant.  That is not the case here, where volume and temperature are constant; that's why they could be canceled in the image from your book. A real life example: I fill a tire with nitrogen until it is at 100kPa pressure. If I then put in an equal number of moles of oxygen, the tire's volume won't increase (much), but the pressure inside the tire will double. The oxygen is filling the same space as the nitrogen. They share a volume, and each has the same partial pressure, 100kPa, and the total gas pressure is 200kPa.

Sorry, after thinking about it all I still felt there are two important question unanswered:                           according to your tire analogy, the pressure is not constant because of the later added oxygen, but how do you keep the pressure constant, by putting in nothing?  What about releasing half the nitrogen and putting in half the oxygen that you are about to put in? Now, do they still share the same volume? Can you also give a example of a vander wal model?

I was giving a real world model that compared to this situation, that is, where volume was a constant, not one where Avogadro's law applied.  In such a situation pressure and moles are directly proportional rather than volume and moles.  In my tire example, all of the scenarios share the same volume.  Tires are built to keep a nearly constant volume across a wide range of pressures.  Remove half of the nitrogen, same V, smaller n and P.  Add it back in using oxygen, same V, original n and P.  Double it, same V,  greater n and P.  All four variables, n,V,P, and T, are interrelated.

Van der Waals made a model that involved gases that were slightly more real.  His model accounted for the volume of the gas particles and the attraction between them.  That's all there is to it.  Any real gas is a van der Waals system.


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Do acids and bases have the same coefficients if they neutralize one another?
The questions states: Ammonia has the formula $\ce{NH3}$. Household ammonia is a dilute aqueous solution of $\ce{NH3}$. Aqueous ammonia is a base that can neutralize acids.
If 18.2 mL of 0.800 M $\ce{HCl}$ solution are needed to neutralize 5.00 mL of a household ammonia solution, what is the molar concentration of the ammonia?
Express your answer with the appropriate units.
I used the values I was given; converted the $\ce{HCl}$ to moles; divided the moles of $\ce{HCl}$ by the liters of ammonia solution. However, I do not understand why this is correct.
$\ce{HCl}$ is an acid and $\ce{NH3}$ is a base. How does finding the molar mass of $\ce{HCl}$ help one calculate the molarity of $\ce{NH3}$?

Look at the chemical equation. One mole of $\ce{HCl}$ will neutralize one mole of $\ce{NH3}$. $\ce{HCl}$ is an acid so it loses a hydrogen ion and ammonia is a base so it gains a hydrogen ion.
$\ce{HCl + NH3 -> NH4Cl}$
So, you found the number of moles $\ce{HCl}$ added, and this equals the number of moles in the 5.00 mL sample of $\ce{NH3}$. Divide that by the volume and you get concentration, which is moles per liter. You don't need to find the molar mass of $\ce{HCl}$ and you didn't find it as described in your calculations above.

I guess I'm trying to have a deeper understanding of ratios when it comes to moles. Since the above reaction is 1:1 ratio I will assume that means that no matter if the elements were. let say, Chromium or Chlorine; these both would have the same amount of moles if they were part of a product of a reaction. For example If I was seeking the amount of moles of Chromium given that I had 100 grams of Chlorine in a reaction of CrCl. 100 grams of Cl equals 2.8206 moles Cl, thus since the ration is 1:1 Cr also equals 2.8206 moles. Is that correct?

@user137452 Yes. Your comment above is spot-on. If it's a balanced equation (i.e., the chemical equation has the lowest common integers like above), then there's a 1:1 ratio or 1:2 ratio or whatever is indicated in the equation. See more about limiting reagents which might help further your understanding.


text:
Net charge of Peptides at pH1 and pH0
Taking Glycine and aspartic acid. As both the amino acids are fully protonated at pH 0 ad pH 1, how does the peptide bond form? Is it a two step mechanism where the OH from an alpha carboxyl group bonds with a hydrogen from the alpha amino group of the other amino acid to form water then the H+ from the amino group being released to form a peptide bond?

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

First, note that the protonated amino group of an amino acid like glycine or aspartic acid has a pKa of about 9.6 - the number for glycine. At pH 1 it is indeed "fully protonated", but more correctly about 1 molecule in 10^(8.6) will have a free amino group at any given point in time (always in flux due to the rapidity of proton transfer). This amino group can react with a carboxyl group (from another amino acid) at the carbonyl carbon to give a tetrahedral intermediate, which in turn can eliminate water to give the peptide bond.
Under equilibrium at pH 0 or 1, the energetics typically favor the reverse of this reaction - hydrolysis of the peptide bonds to give the amino acids - this is what goes on in the stomach when you digest a protein - but it is possible that these conditions could produce a small amount of material from the reverse reaction - the amide-bond forming process.


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Using knowledge of radial distribution functions, how is it possible to explain the different electron configurations of V3+ and Calcium
I recognize that the 3d orbital decreases in energy to lower than the 4s once it becomes occupied (even if I don't completely understand why?!). However, how is the difference in electron configuration of Ca and V^3+ explained by the radial distribution functions? This is a question from a problem sheet at my University and I can't seem to find an answer relating to RDF's anywhere.

It would be helpful if you included everything you mention into your question, e.g. pictures of the radial distribution functions and the electron configurations of $\ce{Ca}$ and $\ce{V^{3+}}$. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

Please see this question for a better phrased version of this one: http://chemistry.stackexchange.com/questions/18988/how-does-the-radial-distribution-function-of-vanadium-differ-from-that-of-calciu

Please don't double-post questions. Instead of posting another question with improved phrasing it would have been better if you edited this question of yours.

sorry but do you know the answer?

I'm desperate. I have two problem sheets in for Monday and a mountain of work ontop.


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Smallest whole number coefficient?
What is the smallest whole number coefficent of $\ce {Cr^{2+}}$ when the equation is balanced?
$$\ce {Cr^{2+}_{(aq)} + Al_{(s)} -> Cr_{(s)} + Al^{3+}_{(aq)}}$$
I don't understand, how can this be balanced? And why would 3 be the smallest coefficient?

Currently, the charges are not balanced. You need to balance the charges to have a balanced equation.

You need to balance the reaction, that's it. But you should know step wise method, because if you use try and error method you might waste your lot of time!

You need to factor in the electrons that are being moved around
Chromium is doing this:
$\ce {Cr^{2+}_{(aq)} + 2 e- ->  Cr_{(s)}}$
Aluminum is doing the reverse of this:
$\ce {Al^{3+}_{(aq)} + 3 e- -> Al_{(s)}}$
Now, like a math equation flip around the aluminum reaction.
And balance out the electrons
$\ce {[Cr^{2+} + 2 e- ->   Cr_{(s)}] \times 3}$
$\ce {[Al_{(s)} -> Al^{3+} + 3 e- ] \times 2}$

$\ce {3Cr^{2+} + 6 e- + 2 Al_{(s)} -> 3Cr_{(s)} + 2 Al^{3+} + 6 e-}$
Hope this helps


text:
Validity of Trouton's Rule
I would like to know under what circumstances is Trouton's rule obeyed by liquids and why certain systems i.e. substances may deviate from Trouton's rule.
By the way I know what Trouton's rule is and I have a vague idea that sometimes Trouton's rule is not obeyed due to the presence of hydrogen bonding in liquids such as water but I would like to know more about other reasons why certain situations the rule is invalid.

you may find my answer here useful? The GSE is derived by assuming that Trouton's rule is valid. The GSE shows best results for small organic molecules termed "druglike". http://chemistry.stackexchange.com/questions/31020/why-does-solubility-decrease-with-temperature-increase/31022#31022

The rule says that for many liquids the entropy of vaporization is almost the same, around ${\rm 85\,J K^{-1}} $.
The (partial) success of the rule is due to the fact that the entropy of a gas is so much larger than that of any liquid. Therefore, the entropy of the initial state (e.g. the liquid) usually makes no big difference.  However, if the liquid presents hydrogen bonding or any other kind of high ordered structure, its entropy will be particularly low and the entropy gain during vaporization will be larger than usual, too.

It's true that the rule states that entropy of vapourization is $85 – 88~\mathrm{J/K}$.However, the rule hardly works for high ordered substances exhibiting hydrogen bonding.
Other factors like the enthalpy of vaporization for a long chained organic molecule {strength of Van der Waals forces} may also play some significance role.

Welcome to Chemistry! Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Please take a read them before posting further, and use correct punctuation and capitalization.


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Why does a glass tube show white bands after being heated with a torch?
When I use a torch to split a glass tube, there's always a hazy white band which appears. Does anyone know what the cause of this is?
It was a hydrogen-oxygen torch. The glass is fused quartz glass, which should be pure silica.

It is sublimed silica according to the following reference at page 8.
http://www.public.asu.edu/~aomdw/GLASS/book.pdf


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What causes Starch's lump creation when solvent (butter/water/oil) is added and how to prevent them
When liquid(water/butter) is added to starch lumps of starch are formed. Which are completely dry from inside. Adding more solvent doesn't seem to help therefore it is not a case of saturated solution. So what does cause the lump formation ? and how can it be prevented ?

The trick here is that you're not actually dissolving the starch. Starch is made up of small semi-crystalline granules of amylose and amylopectin which are actually pretty much insoluble at room temperature. Some water can get in as amylose breaks up the crystallinity, but it doesn't really dissolve. When you drop it into water, it hydrates a bit on the outside, holding it all together, but water doesn't make it into the inside. If you stir or shake it, you can break the clumps up into smaller pieces, eventually getting at most individual hydrated granules.
It's not until you heat the mixture that the granules really break apart and allow water in, in a process called gelatinization. This is what makes starch useful in cooking as the starch can form a more diffuse network, thickening the liquid.

Have you ever made gravy? If the gravy starts out lumpy it stays lumpy. Boiling helps but doesn't get rid of the lumps. You must mechanically break the large lumps when cold, or by adding fat to the starch before adding any water. You add just a little water at a time breaking up lumps until you thin out the mixture. Then you boil to get the gravy to thicken.

Put the starch in a jar with a lid.
Add a little cold water.
Put on lid and shake well.
Use starch mixture as desired.

But from chemistry's point of view how does lump form? Why doesn't solute dissolves in solvent completely given that solution is not saturated.


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What is the type of dependence of the rate of reaction on the amount of catalyst?
I have performed an experiment where I had some test tubes with the reagents and different amounts of catalyst. I started the reactions and measured the time they took.
Then I calculated the rate of reaction and plotted it in function of the amount of catalyst (in mL) in each test tube.
However I can't decide between a logarithmic fit line and a linear one (though the logarithmic one has a higher $R^{2}$). The data points are below.
If it helps, the reaction we used was the one of potassium permanganate with oxalic acid. We also added sulfuric acid and the catalyst used was manganese sulphate.
What would make more sense?

0   4e-05
0.2 7.22e-05
0.4 0.000136
0.6 0.000159
0.8    0.000183
1    0.0002

Neither of the choices makes sense without a mechanistic reason to believe that. What I mean is that it is very tricky to fit experimental data to infer the order of the reaction (in this case wrt the catalyst). If on the other hand you have a reaction mechanism that predicts a first order dependence, then it makes absolute sense tot fit a linear curve

One other thing: do NOT use $R^2$ to decide whether one fit is better than another. It is not meant for that. The only thing $R^2$ tests is how well the current fit (linear, logarithmic, whatever) does as compared to just averaging the values (i.e. assuming a constant value).


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Why is NaCl insoluble in CCl4?
Now I know it's something related to the fact that $\ce{CCl4}$ does not have a net dipole moment but I can't connect the dots.

In order for a substance to dissolve, the free energy change for the dissolution must be negative.  For the enthalpy change to be negative the heat of solvation (the solvent solvating the solute) must be greater in magnitude than the heat needed to break up the forces holding the solid solute together (in the case of NaCl lattice energy).  The entropy change of the solution process is accounted for by the gain in entropy from breaking up the solid and the loss of entropy from organizing solvent around the solute.  In forming most solutions, the entropy change is positive.
In this case the ability of carbon tet to solvate sodium and chloride ions is too low to make up for the lattice energy.  Carbon tet is nonpolar.  A good solvent for NaCl would be polar and the negative end of the solvent would surround sodium ions and the positive end would surround chloride ions.
http://www.biology.arizona.edu/biochemistry/tutorials/chemistry/graphics/nacl2.gif


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Absorbtion of light by a coordination compound
A $\ce{[M(H2O)6]^2+}$ complex typically absorbs at around $600\ \mathrm{nm}$. It is allowed to form a new complex $\ce{[M(NH3)6]^2+}$ that should have absorbtion at?
The absorbtion is because of transition of $\mathrm{e_2g}$ electron to $\mathrm{t_{2g}}$ and vice versa which depend on crystal field stabilization energy as:
$$\Delta = \frac{hc}{\lambda=600\ \mathrm{nm}}\approx2.0\ \mathrm{eV}$$
But I don't know how to convert it in terms of $\Delta_\text o$ so that I can know the actual configuration.
In $\ce{H2O}$ there will be no pairing of electrons as it is weak and reverse with $\ce{NH3}$.
This looks useful but isn't without the metal configuration:
$$\frac{\Delta_{\text o,\ce{H2O}}}{\Delta_{\text o,\ce{NH3}}}=\frac{\lambda_{\ce{NH3}}}{\lambda_{\ce{H2O}}}$$

The given wavelength of $\lambda=600\ \mathrm{nm}$ may be directly converted to various other quantities, namely wavenumber $\tilde\nu$, frequency $\nu$, photon energy $E_\text{p}$, and molar energy $E_\text{m}$:
$$\tilde\nu=\frac{1}{\lambda}=16\,700\ \mathrm{cm^{-1}}$$
$$\nu=\tilde\nu c=\frac{c}{\lambda}=5.00\times 10^{14}\ \mathrm{s^{-1}}=500\ \mathrm{THz}$$
$$E_\text{p}=h\nu=\frac{hc}{\lambda}=3.31\times 10^{-19}\ \mathrm{J}=2.07\ \mathrm{eV}$$
$$E_\text{m}=N_\text{A}h\nu=\frac{N_\text{A}hc}{\lambda}=199\ \mathrm{kJ\ mol^{-1}}$$
With regard to electronic spectra of coordination compounds, usually wavenumbers $\tilde\nu$ expressed in $\mathrm{cm^{-1}}$ are used.
The electronic configuration of $\ce{M^2+}$ in the given complex $\ce{[M(H2O)6]^2+}$ is unknown. Nevertheless, we may assume that the complex has octahedral geometry and that the observed absorption maximum probably corresponds to promoting one electron from the $\mathrm{t_{2g}}$ orbitals ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, and $\mathrm{d}_{yz}$) to an $\mathrm{e_g}$ orbital ($\mathrm{d}_{x^2{-}y^2}$ or $\mathrm{d}_{z^2}$). Thus, it directly gives an estimate for $\Delta_\text{o}=10\ \mathrm{Dq}$ as $16\,700\ \mathrm{cm^{-1}}$.
The energy splitting of the spectral terms depends on the ligand field strength. If the $\ce{H2O}$ ligands are replaced by the strong-field ligand $\ce{NH3}$, the value of $\Delta_\text{o}=10\ \mathrm{Dq}$ and the wavenumber of the corresponding absorption maximum are increased.
An empirical equation for estimating the value of $\Delta_\text{o}$ for any pair of metal and ligands is $$\Delta_\text{o}=f \cdot g$$ where $f$ is the ligand parameter and $g$ the contribution from the central atom. The parameter $f$ describes the ligand strength relative to water; i.e. the value for $\ce{H2O}$ is $f=1.00$. The empirical value for $\ce{NH3}$ is $f=1.25$.
Since the ligand field splitting for $\ce{[M(H2O)6]^2+}$ is $\Delta_\text{o} = 16\,700\ \mathrm{cm^{-1}}$, the corresponding ligand field splitting for $\ce{[M(NH3)6]^2+}$ may be estimated as $\Delta_\text{o} = 1.25 \times 16\,700\ \mathrm{cm^{-1}}= 20\,900\ \mathrm{cm^{-1}}$.
The resulting wavenumber $\tilde\nu=20\,900\ \mathrm{cm^{-1}}$ may be converted to a wavelength of approximately $\lambda=480\ \mathrm{nm}$.
Since $\lambda=\frac{1}{\tilde\nu}$, the new wavelength could have been directly calculated from the given original value of $600\ \mathrm{nm}$ using the parameter $f=1.25$ for $\ce{NH3}$:
$$\lambda=\frac{600\ \mathrm{nm}}{1.25}=480\ \mathrm{nm}$$


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Buffer Power of the amphiprotic substance
The mass balance for a generic amphiprotic substance $[\ce{AH^-}]$ this:
$$
[\ce{HA^-}]=C \cdot \dfrac{K_{a1} [\ce{H_3O^+}]}{K_{a1}\cdot [\ce{H_3O^+}] + [\ce{H_3O^+}]^2 + K_{a1} \cdot K_{a2}} \quad Eq.1
$$
where:
 C = initial concentration of the amphitrofic substance
The Eq.1 derivative  is:
$$
\dfrac{d[\ce{HA^-}]}{d[\ce{H_3O^+}]} = -\dfrac{C \cdot [\ce{H_3O^+}] \cdot K_{a1} \cdot (2 [\ce{H_3O^+}] + K_{a1}}{(K_{a1}\cdot [\ce{H_3O^+}] + [H_3O^+]^2 + K_{a1} \cdot K_{a2})^2} + \dfrac{C \cdot \cdot K_{a1}}{K_{a1}\cdot [\ce{H_3O^+}] + [\ce{H_3O^+}]^2 + K_{a1} \cdot K_{a2}} \quad Eq.2
$$
The zero of this derivative is found at $[\ce{H_3O^+}] = \sqrt{K_{a1} \cdot K_{a2}}$, where is the maximum of $[\ce{AH^-}]$.
The buffer power (BP) can be defined $BP = \dfrac{dC_B}{dp \ce{H}} = -\dfrac{dC_A}{dp \ce{H}} = - 2.3 \cdot [\ce{H_3O^+}] \cdot \dfrac{d[\ce{HA^-}]}{d[\ce{H_3O^+}]} \quad Eq.3$
For example:
If $K_{a1} = 6.2  \times 10^{-8}$ and $K_{a2} = 4.8 \times 10^{-13}$ and C=0.1, the maximum oncentration of the specie $\ce{AH^-}$ is found at $[\ce{H_3O^+}] = \sqrt{6.2  \times 10^{-8} \cdot 4.8 \times 10^{-13}}  = 1.725 \times 10^{-10} \quad pH=9.76$
The curve that represents the buffer power, plotting the Eq.3, is:

In this way I can say that at maximum concentration of the amphiprotic substance the buffer power is zero. Is this statement true?

As the curve that represents the buffer capacity does not simply rely on one fraction of the whole system but on all species of the acid-base system it is not enough to only consider the derivative of $c(\ce{HA-})$.
You also have to take care of the other deprotonated species in your system. For the case of a diprotic acid this means that $c(\ce{A^2-})$ is also important albeit not much as you can see in the following figure.

Figure 1: Buffer capacities for a diprotic acid with different amounts of assumptions taken into account – denoted by their plot labels. The lower graphs are zoomed versions of the upper graphs.
As you have come to the mass balance for $c(\ce{HA-})$ I don't have to show you how to come to the mass balance for $c(\ce{A^2-})$:
$$c(\ce{A^2-}) = \frac{c_0(\ce{H2A})~k_{a1}~k_{a2}}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+k_{a1}~k_{a2}}$$
This means that you have to derive the following equation:
$$C_B=\frac{c_0(\ce{H2A})~\left(c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}\right)}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}}$$
or if you want to include he self-ionization of water it has to be
$$C_B=\frac{k_W}{c(\ce{H3O+})}-c(\ce{H3O+})+\frac{c_0(\ce{H2A})~\left(c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}\right)}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}}$$
Let's derive (with the help of D. D. van Slyke, J. Biol. Chem. 1922, (52), 525-570 and Mathematica (or by hand if you've got time for that))
$$\beta=-\frac{\mathrm{dC_B}}{\mathrm{dpH}}=-\ln(10)~c(\ce{H3O+})~\left(\frac{\mathrm{d C_B}}{\mathrm{d}c(\ce{H3O+})}\right)$$
$$\frac{\mathrm{d C_B}}{\mathrm{d}c(\ce{H3O+})}=-\frac{k_W}{x}-1-\frac{c_0(\ce{H2A})~k_{a1}~c(\ce{H3O+})~(2~c(\ce{H3O+})+k_{a1})}{(c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+k_{a1}~k_{a2})^2}+\frac{c_0(\ce{H2A})~k_{a1}}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+k_{a1}~k_{a2}}$$
Using this final equation you should come up with plots like mine and see that the buffer capacity (for $c_0(\ce{H2A})=0.1~\ce{mol/L}$) at $\ce{pH}=\frac{1}{2}\left(\ce{pK_{a1}}+\ce{pK_{a2}}\right)$ is not zero albeit not being very high.

complete: $\beta=0.0014~\ce{mol/L}$
complete w/o water: $\beta=0.0013~\ce{mol/L}$
w/o $c(\ce{A^2-})$: $\beta=0.0001~\ce{mol/L}$
w/o $c(\ce{A^2-})$ and water: $\beta=1.1 \cdot 10^{-10}~\ce{mol/L}$

So ... is it correct to say that at maximum concentration of the amphiprotic substance the buffer capacity is zero? No but for your to acidic constants it's nearly zero.
But is it always zero at $c_\text{max}(\ce{HA^-})$?
Let's look at those systems a little bit closer. Imagine a few systems and their corresponding buffer capacity curves with $pK_{a1}=4$ and $pK_{a2}$ values between 4 and 10 like in the following figure.

If you create a plot with the buffer capacity over $\frac{1}{2}\left(pK_{a1}+pK_{a2}\right)$ you would end up with something like this

What you can see now is that the buffer capacity at $c_\text{max}(\ce{HA-})$ depends clearly on the distance between both $pK_a$ values. If they are close together the buffer capacity at $c_\text{max}(\ce{HA-})$ is high and if $\Delta pK_a \approx 3$ the buffer capacity at $c_\text{max}(\ce{HA-})$ is nearly zero but never exactly zero.


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Is an ester or a ketone more acidic?
Two organic chemistry textbooks I've read (by Bruice and by Tandon) say:

Esters are more acidic than ketones, because the resonance between the two oxygen atoms gives less opportunity for the delocalization of the electron pair on the alpha carbon in esters contrary to what happens in ketones.

But, in esters, when the resonance happens between the two oxygen atoms, the oxygen atom gains a positive charge which is highly unstable, given that oxygen is highly electronegative. So, shouldn’t the order of acidity be the other way round?

So, shouldn’t the order be the other way round?

Yes, it is the other way around.  Ketones (pKa ~ 20) are more acidic than esters (pKa ~ 25).  The figure below shows the resonance structures for the enolate anion, a conjugate base of a ketone or ester.  Resonance structures II and IV stabilize the enolate anion in the ketone and ester respectively.  In the ester, there is an additional resonance structure, V. Through resonance structure V, the alkoxy group in the ester donates electrons to the carbonyl.  This movement of electron density towards the carbonyl will tend to destabilize the formation of any negative charge on the carbon attached on the other side of the carbonyl (electrostatic repulsion).  This reduces the likelihood of the formation of an enolate anion on that carbon. Therefore, in the ester, to whatever extent structure V contributes, structure IV will contribute somewhat less.  Hence, the ester enolate anion resonance structure IV plays a smaller role and the enolate anion is less stabilized in the ester compared to the ketone.  Consequently, the ester is less acidic than the ketone.

Excellent answer (as usual). Just a small addendum: Both cases can be considered a Y aromatic system if you include hyperconjugation from the R groups. Then it is obvious, that in the ester case the $\pi$-system is better delocalised (larger), hence leads to a more stable configuration.

So this is an example of the exception of the rule that the greater the number of resonance structures the greater the stability. We should take account the stability (energy) of each resonance structure.

Is it really an exception? The ester is indeed more delocalised, but not by much. So when one has to deal with similar conjugated systems, analysing the resonance structures and comparing them face-to-face will reveal the answer.


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How does the radial distribution function of Vanadium differ from that of Calcium and how does this affect the ionic electron configurations?
When Vanadium is ionised it loses the 4s electron first, meaning that it's 3+ ion has a different electron configuration to Calcium despite it being isoelectronic. Can it be explained in terms of radial distribution functions? Maybe the maxima of the 3d and 4s become further apart as the effective nuclear charge increases?

Here is a related question

I think @NicolauSakerNeto mentioned a number of excellent points in his linked answer. Frankly, I think he deserves a bounty.
There are several effects occurring, which give different electron configurations to the transition metals. A very nice review article is available: "The Full Story of the Electron Configurations of the Transition Elements" J. Chem. Educ., 2010, 87 (4), pp 444–448

"d orbital collapse": Because of the difference in angular-momentum "centrifugal force" in quantum mechanics, incomplete shielding of core electrons, and nuclear attraction, the $Z_{eff}$ - effective nuclear charge - for 3d and 4s electrons are different. As a rule of thumb, electrons are typically lost from the 4s first,
electrostatic repulsion in 4s vs. 3d. It's interesting you mention V, since $\ce{V^0}$ is $[3d^3 4s^2]$ and $\ce{V+}$ is $[3d^4]$. The usual explanation is that while 3d is closer to the nucleus and energetically favorable, once you start filling the 3d electrons, the electrostatic repulsion is high, and for the neutral atoms, it's favorable to have one or two 4s electrons, which are more diffuse.

There are other things going on, particularly later in the periodic table, where relativistic and spin-orbit coupling effects occur. Also, many times the electron configuration can change in a free atom or ion compared to a complex.


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Does bleach turn a substance white or colorless?
From what I understand, bleach works to weaken the ability of a substance to absorb light. Thus this seems to be "whitening" i.e. reflecting all light. In contrast, colorlessness (according to Wikipedia) implies that light can pass through the object without scattering.
So is 'colorless' a misnomer when used to describe bleach?
(Also, to double check when you bleach something, is it still there? Only that it is now white?)

It bleaches the dye. So, if the matrix that contained the dye is see-through, then it is colorless. If not, it will appear white (like tissue for example). White is a phenomenon of small particles that disperse light.

First, let's think about what makes a substance coloured: a substance will appear to be coloured if it absorbs light in the visible spectrum. Beta carotene:
absorbs blue/green light (400–500 nm), so it looks red-orange when white light is shined on it as the blue/green light is subtracted from the white and what remains is what reaches our eyes. The details aren't dreadfully important, but you'll notice that the molecule has a lot of conjugated double bonds. This is what allows the molecule to absorb visible light and this conjugation is what produces the colour in many things that might stain clothing or the like. The size of this conjugated system affects how it absorbs light. In general, the more conjugation, the more absorption occurs and at longer wavelengths (towards red).
Substances used as bleaches tend to be oxidizing agents. Again, the details aren't too important, but one thing oxidation can do to these conjugated molecules is break them apart or convert double bonds to other functionalities. By disrupting the conjugated system, one changes the absorption properties of the molecule. Breaking the molecule apart or removing double bonds shrinks the extent of conjugation, reducing absorption and shifting it to shorter wavelengths. If the wavelength is short enough, the absorption is the in the UV range, which is invisible to the human eye, making it colourless.
Now, to your questions:

From what I understand, bleach works to weaken the ability of a substance to absorb light.

This is as described above.

Thus this seems to be "whitening" i.e. reflecting all light.

Reflective/scattering properties are distinct from absorption. e.g. if I add bleach to an aqueous solution of a dye, it may go from a coloured clear solution to a colourless clear solution, but if I bleach a shirt stained with a dye, the shirt doesn't become transparent—the reflective properties are determined by the fabric, not the dye. Bleach simply changes the structure and thus, absorption of the dye, it doesn't render them reflective.

So is 'colorless' a misnomer when used to describe bleach?

Colourless is the correct term to describe the effects of bleach—it makes coloured substances colourless.

Also, to double check when you bleach something, is it still there? Only that it is now white??

The coloured molecules undergo a chemical change so they aren't the same substances with which you started. Whether the products of the reaction are still there depends on what they are. It may very well be that the oxidation products of a dye are more soluble and easier to remove when you wash a piece of clothing. The bleaching itself implies a chemical reaction that alters the absorption of coloured substances, not their removal.

So colorless and opaque appears white. Colorless and transparent looks like air and clear water?...

Right. Coloured or colourless refer to absorption while opaque and clear/transparent refer to reflection and they're more or less independent: one can have coloured and clear/colourless and clear/coloured and opaque/colourless and opaque. Things get a bit more complicated if you bring scattering into it, of course.

I disagree with this answer. Bleach, by definition, is something that makes white. If you go from a color to a less opaque state, it isn't bleaching. After bleaching a clear red liquid, one should have a pink or whitened liquid, at least as turbid as the red. If it becomes transparent you just neutralized the dye, not bleached. Edit:this was supposed to be a comment...


text:
Melting ice off of trucks fast
When a truck comes into the shop with extreme ice that is rock hard, is there a way to remove it fast? Any tricks of the trade?

There are plenty of ways of doing it, most of which are not practical. This being a chemistry site, I don't think that chemistry will be of much use in removing pre-exitant ice. I would just use hot water via hose and sprayer. However, chemistry may be able to provide an anti-icing coating that helps prevent the buildup of ice before it reaches your shop. Is that something you are interested in?

Are there any restrictions? I have suggestions for how to remove the ice in a zippy instant...but it would probably not be very good for anyone in the vicinity.

Water has a high heat capacity, so I agree with JoeHobbit that a spray of hot water via hose would probably work best. A power washer that is capable of heating the water might be worth getting. A secondary advantage of doing this is that the water spray would remove accumulated road salt, which accelerates corrosion. Short of that, a space heater that blows hot air on the truck would be a less effective but also less messy solution. A moving stream of hot water or air should be far more effective than static immersion.
Small amounts of ice deposits can be removed with an aerosol deicer spray.
See, there is actually a lot of chemistry going on here.


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How to determine the composition for a sucrose/ zinc nitrate solution from its freezing point?

A $4~\mathrm{g}$ mixture of sucrose ($\ce{C12H22O11}$) and zinc nitrate ($\ce{Zn(NO3)2}$) is dissolved in $150~\mathrm{g}$ of water. If the resulting solution freezes at $-0.768~^\circ\mathrm{C}$, what is the mass percent of sucrose in the mixture? Assume that zinc nitrate dissociates completely in water, $K_\mathrm f = 1.86$. (Answer: $2.8\%$)

First find the molality of solute particles.  Assume i = 1; we're just counting how many things are dissolved and it doesn't matter whether they are ions or molecules at this point.  I'm using m for mass and b for molality throughout to avoid confusion.
$\Delta T = ibk_f\implies\frac{\Delta T}{k}=b$
$b=\frac{0.768}{1.86}=\pu{0.413 molal}$
Next find how many moles of particles were dissolved.
$b = \frac{n_{\text{solute}}}{m_{\text{solvent}}}$
$n_{\text{solute}}=b\times m_{\text{solvent}}=0.413\times 0.150 = 0.0619~\text{moles of particles}$
Now determine the mass of each species based on this total number of moles.
$n_{\text{particles}}=3n_{\ce{Zn(NO3)2}}+n_{\ce{C12H22O11}}$
$n_{\text{particles}}=3\frac{m_{\ce{Zn(NO3)2}}}{M_{\ce{Zn(NO3)2}}}+\frac{m_{\ce{C12H22O11}}}{M_{\ce{C12H22O11}}}$
You know the total number of moles of particles, 0.0619, and you know the molar masses of both zinc nitrate and sucrose, but you don't know either the actual mass of zinc nitrate or of sucrose.  However, you do have one other piece of information that allows you to set up a system of equations.
$m_{\text{solute}}=m_{\ce{Zn(NO3)2}}+m_{\ce{C12H22O11}}=\pu{4g}$
Solving this system of equations should give you the masses of both solutes.

Hint:
$$i\neq3\text{, but }\underbrace{\frac x{342}}_{\text{sucrose}}+\underbrace{\frac{4-x}{189}\times3}_{\text{zinc-nitrate}}$$


text:
Efforts made to reduce the amount of CFC
Till now I have heard $\ce{CO2}$ is the main reason of global warming. I have studied the remedies to control $\ce{CO2}$ many times through books, awarness programs, etc.
But today I learned about Chlorofluorocarbon ($\ce{CFC}$). It have 38 million times more "Global Warming Potential" than $\ce{CO2}$    according to UN.
In my book they have mention the sources of $\ce{CFC}$ like aerosols, refrigerators, air-conditioners, foams in fire extinguisher, etc.
But it does not tell what effort we should make or scientist are doing in order to reduce harmful of $\ce{CFC}$.
I have heard that $\ce{CFC}$ is replaced by 1,1,2,2-Tetrachloroethane, nothing else.
So, I want to know what else we are trying to reduce harmful effect of  $\ce{CFC}$?

I don't know this topic well enough to provide a proper answer, but I believe that the problem with CFC's lies mostly in their ability to form chlorine radicals at high altitude.  These act as catalysts to destroy ozone.  The volume of CFC released each year is nowhere near as much as CO2.  They've been banned in the US for about 2 decades outside of medical and research purposes.  Elsewhere in the world, I don't know.  The 38 million figure seems high.  Global Warming Potential has a list of chemicals ranked in this way.

@JasonPatterson About 38 million i don't know, but this what my textbook says. And it is compared to CO2.

After learning more about environmental chemistry I got this.
On 16 september 1987, 95 countries accepted not to use ODS(Ozone Depleting Substances)  after signing Mortal Protocol. Later other countries also accepted it.
After that United Nation have decided to celebrate 16 september as "Ozone protection day" at international level to spread awareness.


text:
Why is Electrostatic Potential Energy positive when the charges are like?
We know that in an atom charge of an electron at infinity is zero. As it approaches the nucleus it become more and more negatively charged.
We also know that E.P.E is positive when charges are like and negative when charges are opposite.
Well does it mean that as an electron approaches nucleus there is repulsion which causes the approaching electron to go away. As the electron go away it is getting closer to to infinity point because of which electron become less and less negatively charged. So the E.P.E is positive .Nucleus of atom is like earth and the electron are objects.The farther the object move from earth it has more Potential energy.

The charge of an electron doesn't go to zero when it approaches infinite separation from a nucleus.  An electron has an inherent charge regardless of this separation.  However, the interaction energy between a negatively charged electron and a positively charged nucleus goes to zero as the separation goes to infinity.

There is no way to know the actual potential energy of a system, we can only determine it relative to some predefined zero value.  In the case of electrostatic potential, zero potential energy is defined as the energy between charges when their separation is infinite.
The charge of the electron doesn't change in this scenario; it is always $1.6\times 10^{-19}\ C$.
When oppositely charged particles near one another, they lose potential energy (their energy becomes more negative.)  When particles with the same charge near one another, they gain potential energy.  The attractive force between oppositely charged particles does work as their separation decreases, and the repulsive force between similarly charged particles requires work to be done on them in order to move closer to one another (so potential energy goes up.)


text:
Why is active mass of a pure solid or liquid always taken as unity?

Active mass is defined as the molar concentration ie. number of Gram-moles per litre.

My book then wrote

Active mass of pure solid/liquid is always 1 .

The book reasoned that

Molar concentration is directly proportional to density. Since density of solid or liquid always remains constant, the active mass is taken as 1 .

Really? Doesn't the density of the solid reactant decrease when the forward reaction is about to reach the equilibrium??
In a word, I am greatly confused. What is the reason for taking the active mass 1 ? Please help.

Does the activity of a solid or liquid change over the course of a reaction?
The density of a solid or liquid reactant doesn't change over the course of a reaction. The mass and volume do as it is consumed, but the ratio of the two is constant. If the reaction causes a temperature change then there are small changes in density, but that would also alter the equilibrium constant and the concentrations of the species involved.
The equilibrium constant is based on activities, which you can think of as the ratio of the concentration (molar density) at standard state to the concentration in the reaction. Since solid/liquid density is constant at a given temperature, activity is too.  It takes quite a large change of temperature or pressure to make a substantial change in the density of a substance, assuming no phase changes occur.
What about gases, why don't they show the same phenomenon and have fixed activities? (Asked in comments.)
For solids or liquids, as mass decreases the volume also decreases by a proportional amount. For example, $\mathrm{100mL}$ $\ce{H2O}$ $\mathrm{= 100g}$ and half as much, $\mathrm{50mL}$ $\ce{H2O}$ $\mathrm{= 50g}$. Both have molar densities of $\mathrm{55.5 mol/L}$.) Gases keep the volume of their container, so if you remove some via a reaction, the mass of the gas decreases, but its volume remains the same, and its concentration decreases as a result.
Why is active mass given a value of 1 then?  Is this simply a convention?  Shouldn't its value depend on the compound? (Asked in comments.)
No, it's not a convention. Activities, active mass or any other activities, are really ratios. The ratio here is a comparison between the activity under experimental conditions and the activity at standard state under standard conditions.  For gases that is 1 bar of pressure at 25°C, for solutions that is 1 molar at 25°C and 1 bar, and for solids and liquids that is the molar density of the substance in its most stable phase and allotrope at 25°C and 1 bar.  The activities of real substances require some additional work and there are particular compounds that do not behave quite as nicely as expected, but the basic idea of activity being the ratio between experimental state and standard state is sound.
With that in mind, consider the activity of solids and liquids.  Their standard state is their density at standard conditions.  Their experimental state is their density at experimental conditions.  It takes significant pressure and temperature changes to cause substantial changes in the density of a solid or liquid, so their experimental state is approximately equal to their standard state, and the ratio of these two states is therefore quite close to 1.

can you please provide some relevant text/reference to read more on the same topic.

@Md_Asher I'm sorry, I don't have a good current reference, I just remember this from my schooling and later use as a chemistry teacher.  As is often the case, the Wikipedia article on the topic of Thermodynamic Activity is well written.  A good college level Physical Chemistry text should include a discussion of activity.

The activity of a liquid can change over the course of a reaction, for instance, for a liquid-liquid equilibrium, where both liquids are miscible.

See also https://chemistry.stackexchange.com/a/59331/104336

Density is an intensive property that is:

An intensive property is a bulk property, meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system. Examples of intensive properties include temperature, refractive index, density, and hardness of an object.

Probably that answers your question, for more detail see this

This one as well.  It's quoted fine, but please cite the source.

Does the activity of a solid or liquid change over the course of a reaction?
The density of a solid or liquid reactant doesn't change over the course of a reaction. The mass and volume do as it is consumed, but the ratio of the two is constant. If the reaction causes a temperature change then there are small changes in density, but that would also alter the equilibrium constant and the concentrations of the species involved.
The equilibrium constant is based on activities, which you can think of as the ratio of the concentration (molar density) at standard state to the concentration in the reaction. Since solid/liquid density is constant at a given temperature, activity is too. It takes quite a large change of temperature or pressure to make a substantial change in the density of a substance, assuming no phase changes occur.
What about gases, why don't they show the same phenomenon and have fixed activities? (Asked in comments.)
For solids or liquids, as mass decreases the volume also decreases by a proportional amount. For example, 100mL H2O =100g and half as much, 50mL H2O =50g. Both have molar densities of 55.5mol/L.) Gases keep the volume of their container, so if you remove some via a reaction, the mass of the gas decreases, but its volume remains the same, and its concentration decreases as a result.
Why is active mass given a value of 1 then? Is this simply a convention? Shouldn't its value depend on the compound? (Asked in comments.)
No, it's not a convention. Activities, active mass or any other activities, are really ratios. The ratio here is a comparison between the activity under experimental conditions and the activity at standard state under standard conditions. For gases that is 1 bar of pressure at 25°C, for solutions that is 1 molar at 25°C and 1 bar, and for solids and liquids that is the molar density of the substance in its most stable phase and allotrope at 25°C and 1 bar. The activities of real substances require some additional work and there are particular compounds that do not behave quite as nicely as expected, but the basic idea of activity being the ratio between experimental state and standard state is sound.
With that in mind, consider the activity of solids and liquids. Their standard state is their density at standard conditions. Their experimental state is their density at experimental conditions. It takes significant pressure and temperature changes to cause substantial changes in the density of a solid or liquid, so their experimental state is approximately equal to their standard state, and the ratio of these two states is therefore quite close to 1.


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Why does my coffee creamer have titanium dioxide in it and is it safe?
From the manufacturers website of a coconut french vanilla coffee creamer:

Ingredients: ORGANIC COCONUT MILK (WATER, ORGANIC COCONUT CREAM), DRIED CANE SYRUP, NATURAL FLAVORS, COLORED WITH TITANIUM DIOXIDE, DIPOTASSIUM PHOSPHATE, CARRAGEENAN, GUAR GUM.

My guess, from looking at the abstract of the report (which is behind a paywall), "Titanium Dioxide Nanoparticles in Food and Personal Care Products" is that the the metal is simply used for the (white) coloring. Is this all it is used for?
From the paper abstract:

... human exposure analysis to $\ce{TiO2}$ through foods identified children as having the highest exposures because $\ce{TiO2}$ content of sweets is higher than other food products, and that a typical exposure for a US adult may be on the order of 1 mg Ti per kilogram body weight per day.

Is there any potential risk (toxicity) to consumption of $\ce{TiO2}$ at these levels?

$\ce{TiO2}$ is is the most widely used white pigment out there, and it is considered to be non toxic. You'll find it used as food coloring commonly in candies, but in this instance  it's in your coconut sweetener. Aside from food, it's used as a primary pigment in paint (the more $\ce{TiO2}$ generally the higher the quality of paint, because it results in higher opacity), solar cells, toothpaste, and as you've mentioned, sunscreen.
The only health concerns I found were from inhalation

Titanium dioxide dust, when inhaled, has been classified by the International Agency for Research on Cancer (IARC) as an IARC Group 2B carcinogen, meaning it is possibly carcinogenic to humans. The findings of the IARC are based on the discovery that high concentrations of pigment-grade (powdered) and ultrafine titanium dioxide dust caused respiratory tract cancer in rats exposed by inhalation and intratracheal instillation. The series of biological events or steps that produce the rat lung cancers have also been seen in people working in dusty environments. Therefore, the observations of cancer in animals were considered, by IARC, as relevant to people doing jobs with exposures to titanium dioxide dust.

If I recall correctly, the exact same mechanism is why sand is labeled "potentially carcinogenic" by some MSDSses. As long as you're not huffing large amounts of the stuff, the health risk is basically nonexistent.

I just read an article* which says that it depends on whether it contains nanoparticles or not. It seems to affect the microvilli:

chronic exposure diminished the absorptive projections on the
  surface of intestinal cells called microvilli. With fewer microvilli,
  the intestinal barrier was weakened, metabolism slowed and some
  nutrients—iron, zinc, and fatty acids, specifically—were more
  difficult to absorb. Enzyme functions were negatively affected, while
  inflammation signals increased.

Reference
*https://phys.org/news/2017-02-food-additive-candy-gum-digestive.html

Colloidal particles would be fine for any food additive and such particles are tremendously larger than nanoparticles.


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CO2 removal from mushroom growing tent
I'm trying to grow oyster mushrooms that like high temperatures. They look great in an indoor terrarium, and can eventually yield several tasty meals :)
However oyster mushrooms are hard to grow indoors, because they need low $\ce{CO2}$ levels (<1000 ppm according to Paul Stamets), AND high humidity (>95%).
Warm climate oysters tend to grow fast, and can generate quite a lot of $\ce{CO2}$ fast.
If I vent the air in tent about 10 times a day, humidity will suffer (and I can't do that all the time).
If I close the tent, mushrooms will be stunted and club-like, and yields will be next to nothing.
I'm thinking about putting a tray with $\ce{NaOH}$ crytals on it, probably with a fan to slightly move the humid air around a bit.
An other method would be to use an aquarium pump to bubble air through NaOH solution.
My questions are:

Does this look like something feasible? Can some $\ce{NaOH}$ remove enough $\ce{CO2}$ from the air to keep it under 1000ppm?
Which of the above methods seem to be the better one given the aquarium pump can bubble about 100-200 litres of air in every hour and the tent is about 500 L ?

Removing CO2 from gasses is a process called sweetening in industry. I'm really not sure how feasible this would be to implement on a tiny scale like this but maybe something worth researching!  http://en.wikipedia.org/wiki/Amine_gas_treating

I would not use $\ce{NaOH}$ as it is potentially dangerous and unnecessary costly. Soda $\ce{Na2CO3}$ solutions can accept quite an amount of $\ce{CO2}$. In case of large use, regenerable adsorber could be beneficial, preferably diethanolamine as having highte boiling point.

Could you grow a plant along with the mushrooms to use the CO2?  Something like a spider plant grows fast and doesn't take much light.  Vegetables would be a better solution, but they're going to require stronger lighting.  Another idea is to just pump fresh air into the tent, through a container of water to maintain humidity, and continuously displace the excess CO2.

The plant would require additional lighting, and the heat from the light course would also have to be removed. That is too complex and frankly I have even more doubt in the efficiency as the NaOH solution.
Since the mushroom blocks don't generate much $CO_2$, the costs of the NaOH is not a problem.
I would need to quesstimate how fast a NaOH solution with given surface area can absorb $CO_2$ from the air at atmospheric pressure given the concentration and temperature.

...maybe algae could remove $\ce{CO2}$, but that's a different story. At least they would be easier to arrange in the terrarium than plants with leaves. :)

How about circulating cool water through the mushroom habitat to absorb CO2 and then routing it outside the area with the mushrooms where it is heated to reduce the solubility of the CO2?  This way the same water can be recycled and the humidity kept high.

@brinnb this route is possible, but energetically extensive.

@netom I still would not use $\ce{NaOH}$ as long as I have a choice. The solid can inflict nasty chemical burns and drops of the liquid can damage eyes. So, it is recommended to avoid $\ce{NaOH}$ as far as it is possible.

What if the heating is solar?

The setup is already given, the tent stands in the corner of a poorly lit living room. No solar heating is possible. I'm avare of the safety problems with $\ce{NaOH}$, and I can work with it. $\ce{Ca(OH)2}$ is also a possiblity.

http://www.instructables.com/id/Simple-Algae-Home-CO2-Scrubber-Part-1/

There's little light in the tent, so algae and photosynthesis in general is out of question. I'm looking for something more passive, low maintenance. I'll probably take a chance with $\ce{Ca(OH)2}$

Getting ideal humidity levels is much less important than CO2 levels. Take instruction from the kitchen kit manufacturers and just mist to drenching twice a day. Mushrooms that don't get enough oxygen have long stems and undeveloped caps, baby mushrooms (primordia) fail to develop.

Thanks Doug, this information opens up a lot of possibilities.

Please stay away from sodium hydroxide, it's way too nasty to use in non-lab setup. You can handle it safely in dry conditions, but in contact with water, it becomes sticky and causes bad burns.
As a substitute, I suggest $\ce{Ca(OH)_2}$, readily available as slaked lime. You can structure it into relatively high-surface sculptures and it solidifies into common plaster upon binding enough $\ce{CO_2}$. It is often used for this purpose due to its low price, low toxicity and ease of handling.
In your situation, I would just put some vertically fixed mesh soaked in very viscous suspension of calcium hydroxide. In the high humidity conditions you'd have no problems with normal drying, so once it solidifies it has done its job and you can just hit the mesh against the ground to crumble the plaster away and put on new layer. It is advisable to use gloves and goggles, but it is not as necessary as it would be in the case of sodium hydroxide (look at all the masons around).

Thank you for the advices. I already tried $\ce{Ca(OH)2}$ with limited success. The vertical mesh idea is very good, i'll try it thank you!

Actually, the vertical mesh setup doubles as a humidifier! I tried to use a large piece of textile soaked in thick calcium hydroxide slurry. If it was kept humid, no other means of humidification were needed. No ventillation was used beside accidental, and the oysters looked much better. Since the system is pretty much closed, only a little humidity escapes. (The reaction itself produces a small amount of water too, although probably it does not matter much.)  Thank you for the suggestion!

If your chamber was small enough, you could try having racks of germinating seeds. As they sprout they would be absorbing a minuscule amount of co2. They wouldn’t need light. If you’ve ever made sprouts in your kitchen cupboard this can be easily achieved. I am going to try this. Sunflower would be ideal as they germinate easily, cheap and easy to come by.

Stratification
Best of all, use no chemicals.  If you have an air circulating fan in your chamber, use it intermittently.  When not running, ambient air enters small hole(s) at the top of your chamber, and carbon dioxide will escape out small hole(s) at the bottom of your chamber.  CO2 at standard temp/press is invisible, but it definitely will sink beneath the unmoving oxygen and nitrogen of ambient air.

You should probably expand on this to address the OP's query. As it stands, it's borderline NAA.


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Is it possible to normalize the concentration equilibrium constant between different reactions for comparison?
A follow up question from this answer:

Compare:
  $\ce{A(aq) +B(aq)->AB(aq)}$
  $\ce{2A(aq) +B(aq)->A2B(aq)}$
We have two similar reactions but the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants.  Similarly, with a constant solid/liquid concentration (activity = 1) the equilibrium constant effectively ignores that product or reactant.

Is there a way to "normalize" the $K_c$ of reactions like that shown above, such that a fair comparison can be made, regarding whether one reaction is more product favoured than the other?

Since the equilibrium constant depends on the concentration of the species involved, there is no way to normalize it in the way I think you imagine it. A fair comparison of equilibrium constants is already given. The equilibrium constant is, as its name gives away already, constant. As such there is no normalization to be made.
Also, the quote is slightly off the point where it says that

the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants

because the equilibrium constant might actually be the same (except for the unit, of course). The way you calculate the constant changes, but inferring from that that the value changes drastically is simply overreaching.
If $K > 1$ then the equilibrium lies on the side of the products, if $K<1$ then the equilibrium lies on the side of the reactants.

Is there any scenario where chemist would even compare equilibrium constants of different reactions like I described, or is K simply used to infer the product-reactant relationship for some single reaction? I have a feeling that my question may not even be practically valid.

I have never compared the equilibrium constants of different reactions. Now rate constants are quite the different matter. $K$ here is just used to calculate amounts of substance at equilibrium.

The equilibrium constant is also the ratio of rate constants, when using the standard definitions (activity, etc.) it is possible to compare these. However it is much easier to compare reaction energies directly to compare reactions of different types.


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"Asymmetric molecules are necessarily polar"
Is a molecule with no symmetry necessarily polar?
Can a polar molecule still have some elements of symmetry (i.e. mirror image is the same as itself)? I think so ... because we can have achiral tetrahedral carbon molecules that are polar - i.e. methylene chloride.
So, is it proper to say that a molecule that lacks symmetry is necessarily polar?
Or can a more nuanced statement based on symmetry be made? Is there a "minimum" level of symmetry needed to achieve non-polarity?

There appears to be a lot of negative inversions in this post.  All asymmetric molecules are polar to some degree.  Some polar molecules with have symmetry elements.

@Lighthart can you expand this into an answer?

The answer to this question depends on what your threshold is for polar or what context you mean. For example, diethyl ether is more polar than hexane, but I wouldn't really consider either polar. If you mean polar as in having a non-zero dipole moment, then you're probably correct, but that's nearly every single compound.

I mean non dipole moment @jerepierre

No two atoms have exactly the same electronegativity.  Consequently any molecule that is not symmetric has a nonsymmetric distribution of electron density, and as such is formally polar.  Others have suggested that there is a threshold of polarity, underneath which the nonsymmetric distribution of electron density is not sufficiently nonsymmetric so as to create a dipole of meaningful magnitude.  Compare t-butane to  t-butylchloride, for example.

Let's start by defining a polar molecule as a molecule with a dipole moment.

Is a molecule with no symmetry necessarily polar?

Yes, a molecule with no symmetry belongs to point group $\ce{C_1}$ (e.g. the only symmetry element it contains is a $\ce{C_1}$ axis, rotation of the molecule 360/1 degrees converts the molecule back into itself).  The molecule is asymmetric and polar.

Can a polar molecule still have some elements of symmetry (i.e. mirror
  image is the same as itself)? I think so ... because we can have
  achiral tetrahedral carbon molecules that are polar - i.e. methylene
  chloride.

Yes, methylene chloride, vinyl chloride, methanol, chloroform, etc. all contain planes of symmetry, but are still polar.

So, is it proper to say that a molecule that lacks symmetry is
  necessarily polar?

If I understand correctly, this is a repeat of your first question.

Or can a more nuanced statement based on symmetry be made? Is there a
  "minimum" level of symmetry needed to achieve non-polarity?

Yes

All asymmetric molecules are polar.  Such molecules belong to point group $\ce{C_1}$
Molecules belonging to point group $\ce{C_{n}}$ with n>1 are polar. 1,3-dimethylallene is an example (it has $\ce{C_2}$ symmetry, rotate it by 360/2 degrees about its $\ce{C_{2}}$ axis and it converts back to its original representation)
Molecules belonging to point group $\ce{C_{s}}$ are polar.  A sigma plane either contains or bisects molecules in this point group, and this plane is the only symmetry element.  Methanol, bromocyclopropane, vinyl chloride, etc. belong to this point group and are polar.
Molecules belonging to point group ${C_{nv}}$ are polar.  Formaldehyde and methylene chloride (both ${C_{2v}}$), chloroform (${C_{3v}}$) and hydrogen chloride (${C_{\infty v}}$) are examples.  This point group contains a ${C_{n}}$ axis and n sigma planes containing the axis.

If the molecule is not in one of these point groups, then it is not polar and does not have a dipole moment (or said differently, molecules in all other point groups [${D_n, S_n, C_{nh}, D_{nd}, D_{nh}, T_d, O_h}$ and higher] are not polar and do not have dipole moments).

"All asymmetric molecules are polar." but I thought earlier in your post you said that some polar molecules contain elements of symmetry

Asymmetric molecules belong only to point group ${C_1}$, the only symmetry element they possess is a ${C_1}$ axis.  They are all chiral, polar and have dipole moments.  As described above, other non-asymmetric point groups that contain other symmetry elements may also be polar.

To be polar the molecule must not have a centre of inversion. This is the only condition and this restricts the point groups that it can belong to.  The common point groups with a centre of inversion are $C_i, D_{2h} ,D_{4h}, D_{6h}, D_{8h}, D_{3d}, D_{5d}, S_6, T_h, O_h, I_h$ & $D_{\infty h} $.

This is perhaps a pedantic point, but an object lacking inversion symmetry can still “accidentally” have zero dipole moment. Presence of inversion symmetry implies zero dipole moment, but absence of inversion symmetry does not (strictly speaking) imply non-zero dipole moment.

Consider the following (absolutely imaginary) molecule: $\ce{CX^1X^2X^3X^4}$ where the $\ce{X}$ atoms are all of a different kind but have the same electronegativity.
This molecule is asymmetrical but has no dipole moment.
While an imaginary example is nice and well, I think that in reality, electronegativities will always differ, if even ever so slightly. As a result, there will be a total dipole moment on a molecule with no symmetry. The question you really have to ask now is: What is the threshold for a molecular dipole moment to consider the molecule polar?

what's the geometry of this 5-atomed molecule?

The first atom is a carbon, so I'd say tetrahedral ;)


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Dissolving silver braze from tungsten carbide
I'm working out the best process to chemically strip silver brazing from tungsten carbide (WC) bits. My main objective is to recover the silver as a relatively pure metal. When I use $\ce{HNO3}$ at anywhere from 20-70%, it dissolves the solder and I end up with a red solution.  When I try to use cementation (putting copper metal in the solution) to precipitate the $\ce{Ag}$ as sponge, it isn't working.
Can any one help me figure out what's going on here?  The brazing being used should be 50-60% $\ce{Ag}$ by weight. I'm not sure yet if there is $\ce{Pd}$ or not.

Well, to start, if you're trying to recover silver for its value, it's probably not economical unless you have to strip hundreds of bits—silver's not worth a whole lot. Having said that, what did you actually observe? The reaction is pretty slow; it could take several hours. Also, if the silver concentration is low and you use a lot of copper, the reduced silver deposited could be quite thin.

Also, keep in mind that copper can itself be oxidized by concentrated nitric acid so you might be better off concentrating the solution to get silver nitrate crystals, then dissolving those to use with copper.

I add 35% HNO3 to fresh WC bits and I end up with a reddish/magenta solution.  If I boil off all the liquid, I get a dark purple (like potassium permanganate) thick liquid before it finally hardens into a black mass.  When I add dist. H2O to the black mass and put clear copper metal, the solver does NOT precipitate.  When I do this same process on a smaller scale I can get the silver to stick to the copper as sponge.

You don't really want to boil all the liquid down, just remove enough water to start precipitating silver nitrate. I'm not quite sure what the issue could be, but I found this: http://www.crscientific.com/article-silver.html which seems reasonable.

I think the red color is coming from the cobalt binder in the cemented WC (it's 6-9% Co).  As the HNO3 is consumed, the Co more readily goes into solution. If I use >50% HNO3, the solution stays clear or brown until the HNO3 concentration is lower and it can more readily attack the Co.  The silver braze itself has Ag (50-60%), Cu (15-33%), Sn, Zn, and may be Cd, Ni, and Mn.  There should be a number of nitrate salts in solution.  There is something preventing the AgNO3 from precipitating as a metal when I put copper metal in the solution.  Any ideas? Thanks for all the info so far!!!

If the problem is Co, you should be able to separate out the silver as AgCl by adding HCl, which is very very insoluble, though I can't think of any reason cobalt should make the difference. If anything, cobalt should be much easier to dissolve than silver. The other thing is if you're using too much acid, your final solution may still have enough nitric acid to directly oxidize the copper, preventing the reduction of silver. It seems like adding acid in small increments just until everything is dissolved and diluting afterwards might be prudent.

Most common silver brazes are 50 to 90 % silver with copper or maybe zinc.  A complication may be the WC; it is sintered with 2 to 20 % cobalt ( usually).

No expert here, but it seems that if the copper added becomes coated with a monolayer of silver from your very complicated solution, more silver may not deposit as it does from pure solutions of silver nitrate.
One way to keep silver depositing would be to use two copper electrodes and electrolyze your solution with a relatively low voltage. The most inactive metals (Ag, Pd) will plate out first onto the negative electrode, and the positive electrode will lose an equivalent amount of copper. Finally, your solution will have been depleted of silver and enriched in copper, and the silver will be on the negative electrode. If you start with a silver wire, you will have a purer result, but that may be unimportant. If you electrolyze too fast, you will plate out copper with the silver.


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Can imines be catalytically hydrogenated?
Can imines be catalytically hydrogenated? According to the DEA this appears to be possible. However I cannot find any other references to catalytic hydrogenation of imines anywhere else.

http://thedea.org/synthesis.html

Interesting to know why they use catalytic hydrog that requires some more expertise for the ambitious Walter White wannabe (handling gas cylinders, getting hold of one, getting hold of the catalyst) when simply $\ce {NaBH_4}$ should do the trick

@K_P Once you move from reaction with 1-10 mmol of costly source to reactions of 1-10 mol of cheap resource the use of costly reducers, like $\ce{NaBH4}$ becomes questionable at best. Despite common misconception, such works still happen.

@permeakra ... why not use Na(CN)BH4?

@Dissenter Because it costs much more than hydrogen gas. It is easier to use, but at larger scale the cost becomes significant enough to move to cheaper alternatives.

@permeakra Sure hydrogen is cost effective but I was not talking about professional labs. I assumed that for an amateur (the page linked looks like it is targeting this kind of group) it would be easier to find and handle sodium borohydride than a hydrogen cylinder and catalyst. I'm sure they can recover their shortfall selling this kind of product

Yes, imines can be catalytically hydrogenated. Chapter 8 of Nishimura's The Handbook of Heterogeneous Catalytic Hydrogenation for Organic Syntheses is "Hydrogenation of Imines, Oximes, and Related Compounds" and gives several examples to work with.
The most analogous example in the chapter to your pictured reaction is Archer et al.'s preparation of 3α-(2-diethylaminoethyl)-aminotropane, wherein tropinone was condensed with N,N-diethylethylenediamine to give the imine, which was then catalytically hydrogenated at between room temperature and 50 °C over platinum oxide or Adams catalyst with 60-70% yield.
Archer, et al. J. Am. Chem. Soc. 1957, 79, 4194

Several methods on the reduction of imines are outlined on Wikipedia. Another method using catalytic hydrogenation is given here.

Can you summarize what's at the links in the body of the answer, please.

@jonsca I thought somebody would mention that. There are more than 10 reactions on the web page and since Wikipedia is unlikely to disappear (although the page might get moved), there is enough information (i.e. reduction of imines) for anyone to search Wikipedia and find it.

I would agree that Wikipedia won't likely move anytime soon, but you never know.  Anyway, just adding something, even if you put, for example, "including palladium and ruthenium catalysis methods" or something to that effect makes it more self-contained and gives someone some keywords to search on to find a new link if that one went bad.  I realize it's too time and space consuming to list out all the information, so I'm certainly not expecting a whole laundry list.