text:
What is the dipole moment direction of an aliphatic amine, i.e., R-NH2?
Since the electron donating property of $\ce{N}$ and $\ce{C}$ is different, there is partial charge transfer between $\ce{N}$ and $\ce{C}$ atoms in aliphatic amines, namely $\ce{R-NH2}$. What is the charge transfer direction? Can the following be true? And how large is this dipole moment?

More generally, how can one determine both the direction and quantity of the charge transfer dipole moment given two common organic groups? Or what is the order of electron donating property of common groups?

I really do not see an easy and straight forward way of determining the dipole moment's direction and quantity. It is very much dependent on how the molecule is currently shaped, i.e. rotations of the alkyl chain and also the electron cloud. Of course you could go ahead and calculate the dipole moment from tabulated values of certain bonds (for example), but this can be really tedious and it might now be always right.
You can probably find tabulated values for most common compounds online. Try http://cccbdb.nist.gov/ or maybe just google will do. If you are lucky enough to have access, the handbook of chemistry and physics has a large number of values. There is an online verison, but you can often find it in libraries, too.
For the meantime, I will try to illustrate something. In the series $\ce{R-NH2}$ the simplest one is $\ce{R~=~H}$, ammonia. Here the dipole moment is straight forward isn't it? I believe it is a little more complicated than that. Unfortunately you cannot just take the charges of the atoms in the molecule, since they are not point charges. Have a look at the total electron density, you will notice, that the isosurface is more extended on top of the Nitrogen, which is expectable, since it is the area where the lone pair is supposed to be. This also coincides with the highest occupied molecular orbital (HOMO). In this special case, the dipole moment is almost a direct correlation of that. (If you would add up all dipole moments of all occupied molecular orbitals, you will also obtain the total dipole moment.)
Below you see an isosurface of the total electron density in blue on the left, the isosurface of the homo in the middle and natural charges with the resulting dipole moment on the right. The values are taken from a DF-BP86/def2-TZVPP calculation with Gaussian09.

What we can learn from this is, that the lone pair has a huge effect on the dipole moment.
Let's go a bit further and look at $\ce{R~=~CH3}$, methyl amine. It can be expected, that the dipole moment is somewhat smaller than in ammonia, due to the methyl group. What you can also observe, that it is dependent on how the methyl group is rotated. Of course in room temperature, the group is freely rotatable and it will result in a mean (pulsating) value for the dipole moment. However, what you can observe here is, that the direction of the dipole moment is aligned with the direction of the lone pair. As you go further down the series this generally will hold, However, the magnitude will be decreasing since your "quite unpolar" moiety is becoming bigger.

For ethyl amine, $\ce{R~=~C2H5}$, the trend holds. But to not extend this answer too much, I only pick one representative amongst all possible conformations.

And for propyl amine, $\ce{R~=~(CH2)2CH3}$, we can observe it analogously, with decreasing magnitude of the dipole moment.

As the last example I have chosen heptanyl amine, $\ce{R~=~(CH2)6CH3}$, to example the trend. I removed some of the charges to not overcrowd the image. If you have a close look at the HOMO, you will notice, that it only significantly extends to the first carbon. The situation for any aliphatic amine can be expected to be very similar.

(I have no idea, why the vector for the dipole got bigger and bigger, while the magnitude was decreasing.)
TL;DR:
The dipole moment in aliphatic amines is about 1 Debye and the direction is through the nitrogen directed to the lone pair.

Does the computer program consider tunneling inversion of ammomia? Nobel prize winner Anderson said the dipole moment is zero in "More Is Different" Science, Vol. 177, (1972), pp. 393-396 http://robotics.cs.tamu.edu/dshell/cs689/papers/anderson72more_is_different.pdf.  NIST says dipole moment is 1.470 Debye http://cccbdb.nist.gov/exp2.asp?casno=7664417#NSRDS-NBS10 Dulal Ghosh seems to argue for Anderson's "dipole is zero" in "Nanoscience and Advancing Computational Methods in Chemistry" 2012.

@DavePhD No it calculates the electronic structure at the given arrangement and from that the dipole moment. Therefore it is only a snapshot. Inversion is fast in ammonia, with the transition state having no dipole moment, hence it is well possible, that you are observing zero debye, but I thought that would be off the point for this answer.

Ok, it's a difficult topic for me to understand.  I guess classically we say the electons are moving fast and the inversion is fast, but the wavefunction description has the electon probability density distributed and the hydrogens (and lone pair) partially on both sides of the nitrogen in a probability distribution.  The program considers inversion classically and the electrons with QM.  The inversion barrier is about the same for methylamine and ammonia, maybe all primary amines too. It's difficult for me to relate all this to an experimental measurment.  Maybe I should ask a new question.

@DavePhD Actually the QM programs do not treat inversion at all, as they just solve the electronic Schrödinger equation for one set of coordinates at the time. So you only see one point and its resulting dipole moment. But I guess dwelling into that would be an awesome question for this site.

Since the nitrogen atom is more electronegative than the carbon atom it is bonded to, at least in the case of a simple aliphatic amine, I would venture to say the direction of the dipole will be with a slight negative charge toward nitrogen. The amine nitrogen has nonbonding electrons on it, but the adjacent carbon is saturated and cannot accept the lone pair electrons on the nitrogen, and thus nitrogen in this example will not have a partial positive charge. However, if R is aromatic, the situation will be different.


text:
What is the direction of charge transfer between Aluminum and aliphatic amines physisorbed on top?
Is there dipole formation upon physisorption of an aliphatic amine to an aluminum substrate? What is the charge transfer direction? What if the aluminum substrate is clean, surface oxidized or covered with -OH groups?
What if aluminum is replaced by other types of metals, such as Au and Ag?

Amines could bond to the aluminum surface via a Lewis acid-base reaction between the electron rich amine nitrogen and the electron poor Al atom on the surface (perhaps formed by loss of a surface hydroxyl group as hydroxide ion). The direction of charge transfer would be from the amine nitrogen to an aluminum atom. The resulting "Lewis salt" would have a partial positive charge on nitrogen and a partial negative charge on aluminum (like an aluminate ion or an amine-aluminum chloride adduct).

source: Stack Exchange

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

sort: votes


text:
Are all solutions of weak acid/bases buffers?
I am having a difficult time understanding what makes a buffer a buffer.
Buffers in my textbook are defined as a solution of a weak acid or base and their conjugate acid/base. So if I were to just dissolve acetic acid (a weak acid) in water why would this not be defined as a buffer. Acetic acid incompletely dissociates into an acetate ion (the conjugate base) and a Hydrogen ion. Why is it that for a base to be made the acetic acid would have to be mixed with something like Sodium acetate if acetic acid incompletely dissociates into a conjugate base in the first place? Could someone please help me understand why this is the case?

The term "buffer" implies a purpose of resisting changes in pH when small amounts of acid or base are added.  The buffer will be most effective when the pH equals the pKa of the acid.  This is when there are equal amounts of acid and conjugate base. For each pH unit the system deviates from pH=pKa, the buffer is about a factor 10 less effective.
If you just dissolve acetic acid in water, yes it is true that there will be a small fraction that ionizes to yield the conjugate base, acetate.  However, the pH will be far from the pKa of acetic acid, and the system will not be very effective as a buffer.
In summary, "buffer" is not a very precise term, but it implies that there are significant amounts of both the acid and the conjugate base.

I totally agree with @DavePhD and I want to show you how the pH dependence of the degree of acid-base ionisation provides the background of the buffering mechanism. Titration curves effectively illustrate the influence of the pH on the ionisation of weak acids and bases. The Figure below shows the change in the pH of a 0.1 M acetic acid solution on its titration with a strong base. If you just dissolve acetic acid in water, yes it is true that there will be a small fraction that ionizes to yield the conjugate base, acetate. However if you add the slightest amount of a base, for example, it will result into a steep increase in the pH (look at the beginning of the titration curve) and the solution is not a buffer.
That initial steep increase in the pH slows down quickly, and a pH plateau appears in the range of 4.0-5.8. As in this range $[\ce{CH3COOH}]$ is comparable to $[\ce{CH3COO-}]$ (i.e. their ratio is within a tenfold range), we will have a two-component system containing both a weak acid and a weak base—i.e. a buffer. The centre of the plateau is at 50 % titration where $[\ce{CH3COOH}]=[\ce{CH3COO-}]$. Therefore, according to the equation $\mathrm{pH}=\mathrm{p}K_\mathrm{a}+\log{\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}}$, the $\mathrm{pH}$ will just equal the $\mathrm{p}K_\mathrm{a}$. Following the plateau (upon the exhaustion of the buffer capacity), the pH will again increase steeply, and the top of the titration step will be reached, i.e. the point of neutralization at which the degree of titration of the acetic acid will be 100 %

From the figure above, the region marked inside the rectangle is a buffer. Because it can provide the most efficient pH stabilization against both acidic and basic shifts. In this region, the ratio of the two components (acid and conjugated base) is one or close to one. Accordingly, a buffer is effective in the $\mathrm{pH}$ range of its $\mathrm{p}K_\mathrm{a} \pm 1$ $\mathrm{pH}$ unit

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


text:
Hardy Schulze law
I have trouble understanding this particular law.
It states that:

Greater the valency of the flocculating ion, greater will be its coagulation power.

Why does this happen? What is the intuition behind this?

Coagulation is a process which involves coming together of colloidal particles so as to change into large sized particles which ultimately settle as a precipitate or float on the surface.
Coagulation is generally brought about by the addition of electrolytes. When an electrolyte is added to a colloidal solution, the particles of the sol take up the ions which are oppositely charged and thus get neutralized. The neutral particles then start accumulating to form particles of a larger size which settle down.
The quantity of the electrolyte which is required to coagulate a definite amount of a colloidal solution depends upon the valency of the ion having a charge opposite to that of the colloidal particles. This observation of Hardy and Schulze are known as Hardy Schulze law.
It can be defined as:
Greater is the valency of the oppositely charged ion of the electrolyte being added, the faster is the coagulation.
The idea behind this is the attractive electrostatic forces between ions of opposite charges. The greater the charge of the ions, the greater is  the attractive electrostatic forces between ions:
$f=-\frac{qq'}{4\pi\epsilon_0r^2}$, the faster is the coagulation.
Hence, for the coagulation of negatively charged arsenious sulphide sol., trivalent cations are far more effective than divalent cations which in turn are more effective than monovalent cations. Similarly for coagulation of positively charged ferric hydroxide sol, tetravalent anions are more effective than trivalent anions which are more effective than divalent anions which in turn are more effective than monovalent anions.
The minimum amount of an electrolyte that must be added to one litre of a colloidal solution so as to bring about complete coagulation or flocculation is called the coagulation or flocculation value of the electrolyte. Thus smaller is the flocculation value of an electrolyte; greater is its coagulating or precipitating power.

What's the effect of pH?

In order to explain the Schuze-Hardy rule, you have to take into account that you are working with a colloidal dispersions stabilized electrostatically.
The stability of this dispersions is well described by the DLVO theory. According to this theory, the particles interaction potential can be expressed as the sum of two contributions, $$V_T = V_A + V_R$$
the attractive interaction $V_A$ (due to VdW forces) and a repulsive electrostatic interaction $V_R$ (due to the overlap of the diffuse double layer around these particles). The particle is sytable if the interaction potential result in a barrier that avoids the particle getting close enough to reach the strong interaction region. The following diagram illustrates a repulsive potentical.

A simple model to describe the repulsive potential is provided by the Gouy-Chapman model. For an spherical particle this potential has the form
$$V_R = V(r0) e^{-\kappa r}$$
where $\kappa^{-1}$ is known as the Debey length, and provides a measurement of the extent of the diffuses double layer. $\kappa$ depends on the ionic-strength ($I$),
$$ \kappa \propto \sqrt{I} $$
$$ I = \frac{1}{2}\sum_{i} c_i z_i^2 $$
Thus, an increase in the ionic strength is results in a smaller Debye length, more compacted double layer and a smaller barrier to aggregation. Notice that for the same concentrations of ions, the ionic strength increases with the square of the ion charge. The Schulze-Hardy rule is an empirical realization of these property.
I enclosed some images that illustrate some predictions for spherical gold nanoparticles. Unfortunately, I lost the reference where I found them and I am not able to find it again.

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


text:
What are examples of group 14 elements forming dπ - pπ bonding?
While studying the Group 14 elements, I came across the anomalous behaviour of carbon. In that there was a point:

Carbon forms only p$\pi$-p$\pi$ type bonding while other elements in the group form p$\pi$-p$\pi$, as well as d$\pi$-p$\pi$ type bonding.

I am searching for molecules in which  group 14 elements form d$\pi$-p$\pi$ bonds.

http://en.wikipedia.org/wiki/Transition_metal_carbene_complex || However, I would like to see the context. It is common to older textbooks to mention d-orbital involvement for 3+ rows for p-block, while modern quantum chemistry calculations show that this involvements is limited even if present.

For $d\pi-p\pi$ bonding we need to consider other elements of group 14 except $\ce{C}$. (i.e, $\ce{Si, Ge ,Sn}$ etc.) which have got a vacant $d$ orbital of proper symmetry and energy to come into play bonding with attached group  having lone pair of electron in p orbital of comparable energy.
    For example $\ce{SiF4}$. $\ce{GeCl4}$ etc.
In tetrahedral geometry the orientation of vacant  d orbital of the central atom (dxy. dyz,dxz) may overlap with filled p orbital of the ligating atom  and hence $d\pi -p\pi$ bonding is possible.

$\ce{SiF4}$ is tetrahedral, there's no $\pi$ overlap between Si 3d and F 2p? correct me if I'm wrong. Imo a better example is $\ce{N(SiH3)3}$.


text:
Equivalent points for 3 fold and 6 fold symmetry
In a crystal Lattice how to find the equivalent points for 3 fold and 6 fold symmetry?

Can you be a bit more precise about what you want to know? My initial response would be "find the rotation axis and you'll see the equivalent atoms". If you can give us a little more on what you find confusing (e.g., how you find a rotation axis in the first place) it would really help. Thanks!

Okay..So you have to fix the Z axis and rotate and then find the equivalent points...Let's Say we have (X,Y,Z) as our initial points..so we keep the Z axis fixed and do a rotation...so this is the info..I can give u at most I mean that is what the question..I need help


text:
Oxide surfaces develop a charge in H2O. Do we have any, even vague, handle on how long does an indivdual charge last on avg. before being neutralised?
We often know, from e.g. titration experiments, what the average surface charge density of a surface is.
How do these charges vary over both time and space (i.e. dynamics)? I have been unable to find much (well, any really!) literature on this.
Example
Taking the example of oxide surfaces, you have the following processes going on:
$\ce{S-O- + H3O+<=>[\text{K1}]S-OH +H2O}$ #Protonation/Autoionisation
$\ce{S-OH + OH- <=>[\text{K2}]S-O- +H2O}$ #Deprotonation/Protonation
$\ce{S-OH + H3O+ <=>[\text{K3}]S-OH2+ +H2O}$ #Protonation/Deprotonation
If you know (from experimental/theoetrical calculation of the Equilibrium Constants K1, K2, K3), that you have 100 negative charges per micrometer squared (i.e we are at high pH). Is it a more realistic picture on a picosecond/nanosecond timescale to consider it being something like 10,000 charges per micrometer squared which are being regularly neutralised over time or simply 100 charges per micrometer which are fixed?

Proton transfer reactions in water such as the ones you wrote down are diffusion controlled under ambient conditions (rate constant about 1e+12 per sec.) A snapshot in time will have a certain distribution of charges on the surface, but these are continually in flux, so any one charged site will have an exceedingly short lifetime. You could reasonably represent that system with a fixed set of charges on the surface, for some types of models (e.g., states at equilibrium).


text:
n-factor for ammonium nitrite?
The given reaction is $$\ce{NH4NO2 \rightarrow N2  +  2H2O}$$
The way I did it was
$$\ce{NH4+ \rightarrow \frac{1}{2} N2 + 4H+ + 3e-}$$
and
$$\ce{NO2- +3e- +4H+ \rightarrow \frac{1}{2} N2 + 2H2O}$$
Therefore n factor is 3
But the book I am referring to says 6. Can someone please help?

"n factor" of oxidizing or reducing agent = total number of electrons transferred per a mole of the reactant.
When more than one element in a substance is oxidized/reduced, the n-factor of the substance is taken as the sum of the n-factors of the individual elements of the substance.
This is the case here.You have  ammonium ion that undergoes oxidation reaction (n-factor= 3) and  nitrite ion that undergoes reduction reaction (n-factor= 3).
So, the n-factor of ammonium nitrite is 6.

Even though one is getting reduced and the other oxidised? Because if we usually add half reactions, the net e- transfer is not the sum. Am i right?

It's a matter of definitions of terms. Let's take the case of separate oxidizing and reducing agents. Permanganate ions reactiong with fer(II) ions.  "n factor" of KMnO4 when it's reduced to ion Mn2+ is five. Of course, when permangante ion is reduced, there is another chemical substance,  ion Fe (II) that is oxidized to ion Fe(III), and the n factor of it is 1.

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


text:
Hydrolysing large molecules to reveal amino acids
Consider Oxytocin:

I am asked to hydrolyze Oxytocin and reveal 5 amino acids which are the result of this hydrolysis.
I must admit I have no idea where to begin. I know that generally amino acids are on the form of $\ce{NH_2-CH(R)-COOH}$, so perhaps looking at these double bonds to oxygen may be a starting point, as I assume they will become the acid group? But I really don't know how to move forward from there!

It's highly irrelevant, but it would be nice of you to put the source of the images. Some of them may even be subjects to copyright.

It's from a local newspaper in Denmark. I think it's a public image and therefore ok to use.

Oh, just wanted to remind. In this case, I don't think it's necessary.

Just to be sure: Do you ask for a real experimental procedure or just for a hint on how to identify (visualize) the different amino acids on paper?

looks like more than 5 amino acids to me

I am only to identify 5. I really don't know how to take it apart and identify these. All I know is:
I need to add water to certain parts of the molecule to make new connections exposing amino acids. A likely sort of identifier could be the double bonded oxygens.

You are on the right track.  $\ce{NH_2-C(H)R-COOH}$ is the formula for an alpha amino acid, one where the amino group of the molecule is attached to a carbon atom one carbon atom away from the carboxylate carbon.  This is the most common type of amino acid found in biochemistry.  (A beta amino acid would be $\ce{NH_2-C(H)R^2-C(H)R^1-COOH}$).
When amino acids condense into a polymer or oligomer they do so by forming peptide bonds, also called amide bonds.
$\ce{R^1COOH + NH2R^2 -> R^1CONH2R^2 + H2O}$
So to find the places in oxytocin that would be hyrolyzed into amino acids, look for individual carbon atoms that are attached to both (i) an oxygen atom via a double bond and (ii) a nitrogen atom while (iii) not being attached to any other heteroatoms.  (Item (iii) is generally necessary to exclude other functional groups such as carbamides or carbamates, but these groups are not found in oxytocin.)  For oxytocin, your impulse to look just at carbon double-bonded to oxygen (item (i) only) is OK, but be warned that this procedure is not precise enough in general, because molecules with e.g. ketone moieties will have C=O bonds but will not be hydrolyzed.
The red-circled amino group is a free amino group:  the carbon attached to it is not bound to a carbonyl oxygen, so it is not a place where hydrolysis will happen.  (However, look one carbon atom to the left of C atom attached to the circled N....)
@DavePhD is right; there are more than five amide bonds in oxytocin.  I think there are a total of ten.  Eight are the result of amide bond formation between alpha amino acids and two are the result of amide bond formation between an acid side chain and unsubstituted ammonia.

Thanks, I have identified the carbon atom to the left of the one attached to the free amino group. I see that it is double bonded to an oxygen but I don't see how it is bonded to a nitrogen?
Also is it not bound to the CCHNH group straight to its left?
I'm also looking at various double bound oxygens and the carbons attached to them seem all to be attached to either a hydrogen or a carbon, therefore heterocarbons?

Hi Bjarni, I worry you are being mislead by structural abbreviations.  HNHC is a weird shorthand for H2N-C.  Maybe it would be easiest for you to find a better picture/structure of oxytocin to use for your problem.

authors: Šhřëýă Ķãśhýâp


text:
Which orbitals of the hydrogen atom are degenerate for n=3?

Which of the following orbitals are degenerate in the hydrogen atom with $n = 3$?

A. II and III only;
B. I and IV only;
C. I, II, and IV only;
D. II, III, and IV only;
E. all.

The answer says its E. All of them.
First of all isn't there only 1 electron in hydrogen? And how could the $\mathrm{s}$ orbital be degenerate? Doesn't degenerate mean there are multiple places pairs of orbitals can be?

First of all isn't there only 1 electron in hydrogen?

yes

And how could the s orbital be degenerate? Doesn't degenerate mean there are multiple places pairs of orbitals can be?

"degenerate" means having the same energy.  "Degenerate" refers to a set of orbitals.  It doesn't make sense to say one orbital is degenerate.
Solving the non-relativistic Schrodinger equation, all the orbitals for a given "n" are degenerate.  Energy only depends upon n.
More complete consideration including relativity, spin and quantum electrodynamics shows that they are not all degenerate however.

What I meant to say was "Doesn't degenerate mean there are multiple places pairs of orbitals can be within the same subshell?" I'm still confused why the answer is all of them. I thought the s orbitals were never degenerate because there is only one place and like you said it doesn't make sense to say one orbital is degenerate. So why is it degenerate? what orbital with n = 3 has the same energy as it? And how do you know?

If n=3, then l could be l=0 (s), l=1 (p) or l=2 (d).  So when n=3, the degenerate orbitals (according to the non-relativistic Schrodinger equation) are 3s, the three 3p orbitals, and the five 3d orbitals.

I'm not thrilled about this question, since II and III could easily be a 2p and 1s orbital, respectively. There's nothing to indicate that they're really a 3p and 3s orbital. So my first instinct was choice "B" because I identified the "p" and "s" of a different n than the clearly 3d orbitals.

I don't like it either, on the one hand the question does  specifically says "n=3", but II looks like 2p, not 3p because 3p would have a radial node http://winter.group.shef.ac.uk/orbitron/AOs/3p/

Same thing with a 3s.. there would be two radial nodes in a 3s.

But for 3s, you wouldn't be able to see the nodes in such drawings, in the sense that there is still an outer spherical equal-probability surface.

Its not an ideal question, I agree, and it is a pity that external examiners did not point this out (assuming its a year 1 exam question),  but presumably at this level the examiners thought that generic orbital diagrams are acceptable.

is this only valid for Hydrogen? Or for any element is it valid?

@paracetamol only for hydrogen

The one electron of hydrogen, when excited, reaches 3rd energy level and degeneracy is determined by $(n + l)$, where $n$ - energy level, $l = 0$ for $\mathrm{s}$, $1$ for $\mathrm{p}$, $2$ for $\mathrm{d}$, $3$ for $\mathrm{f}$.

First of all yes there is only 1 e- present in atom of H. But here the atom is "hydrogen like" which can be He+, Li+2 Be3+, etc. The question was twisted at the point n=3, when it said n=3 then the atom can be "hydrogen like" as I mentioned above.
Now n=3 makes the above orbitals with 3s, 3p, 3d and 3f. Now (l) quantum number for 3s=0 for 3p=1, for 3d=2 and for 3f=3.
Now, electron can jump in case of 3s=0 to 1, 3p=0 to 3, 3d= 0 to 5 and 3d= 0 to 7
Thus, we can say that from above orbitals, there will be degeneration in all of them.

I can make sense of your last two sentences. Are you saying that all orbitals in the third layer are degenerate?

There are 3f orbitals? That's kind of new to me.

@Mithoron Just for your information, your edit (i -> I) dequeued the item from the VLQ queue, and the "recommend deletion" never got executed. That's a bummer, isn't it ;)

Ok so the best way to think of this is to compare two atoms. Take for example {Ne} which has 10 electrons so this atom would have a number of n=3. The problem asks what orbitals would be degenerate for n=3 and we know that H has a number of n=1 so this would be just 1s^1 the question is a trick question to make you think that all orbitals above the n=1 orbital are degenerate so really even though it gives you hydrogen its just to throw you off. Picture a n=3 atom for this problem that has no element attached to it in that case all orbitals would be equal energy therefore it would be option E because all orbitals s,p,and d would be degenerate. Hope that helps took my a while to conceptualize this.

Welcome to chemistry. Feel free to take a [tour]. Visit the [help] for any unanswered questions about the site. I don’t get your answer. But neon’s configuration is $\mathrm{1s^2\ 2s^2\ 2p^8}$, so there is no $n = 3$ in there …


text:
Why does the potential difference of a current drop, when silver nitrate is added to deionised water?
Semi-related to Why do some salt ions in solution conduct electricity better than others?
After conducting more experiments, using deionised water as an analyte only, I found the potential difference decreases as more silver nitrate is added. This goes against what I was lead to believe that it is actually disruption in forms of "pure" water that actually conduct electricity (i.e. pure water should not technically conduct electricity at all).

This graph shows how the potential difference drops, from my findings. The last data points plotted are at 1cm3 of silver nitrate added; the interval is 0.05cm3.
So, why is it that I see a massive drop in potential when even the slightest (0.50cm3) amount of silver nitrate is added?

I think your image link is broken... Unless we're only meant to see 3 different x-values.

This is correct. That large drop is the only point of concern, otherwise the graph just plateaus out.

Ah ok. I'd still include the full graph just to avoid confusion, but that's just me.

Also, what are you measuring the voltage drop across?

The potential difference is measured across deionised water (obviously liquid) with a silver bar electrode and a graphite reference electrode.

The potential of an electrode in the absence or paucity of redox reactions at the interface is not predictable using the Nernst equation. This situation is known as an unpoised electrode and the electrode's measured potential is related to the formation of an electrical double layer at the interface, which behaves very much like a capacitor.
If you try to measure the potential of an electrode that won't easily oxidize or reduce in deionized water, due to the very low concentration of ions, establishing the double layer is slow and easily disrupted. You can think of it as having a large solution resistance in series with the double layer capacitance. When there are few ions, the resistance is very high. On that same note, how good the voltage measurement instrument is becomes important. Any current that a voltmeter draws will induce a large voltage at the electrode (as there is a large resistance to bringing ions to the interface, excess charge at the electrode is not easily balanced)—a voltmeter with a 10 nA input bias current may induce 100–200 mV of error in a very resistive solution, so it takes very good instruments to deal with these types of measurements.
This error is proportional to the solution resistance and even adding a tiny amount of ions to deionized water will dramatically reduce the resistance. I'm guessing that repeating the experiment with a supporting electrolyte will prevent this effect.
The other trouble here is that neither of your electrodes can be assumed to maintain a fixed potential in your solution. The potential of the silver electrode should vary according to the Nernst equation once you have an appreciable quantity of silver ions in solution, but the graphite electrode is not in equilibrium with any components of your solution and its potential is free to fluctuate as the solution composition changes. Without a supporting electrolyte, adding the silver nitrate may have a large effect on the potential of the graphite electrode.


text:
Gas-Phase Functionalization of Carbon Nanotubes
I'm new to the subject of carbon nanotubes; However, I recently studied a little about them. But I failed to understand the gas phase functionalization.
As I understand, this technique is used to improve the solubility of the nanotubes.
However, I failed to understand the chemistry and physics of these processes so resumidad, considering that there have been many experiments in this field, but I failed to understand the basis of this process.
Someone could explain?
Another question: What causes that the nanotubes are hydrophobic?


text:
Why do copper (II) complexes contain so many valence electrons?
Consider tetraaminecopper (II) ion. $\ce{[Cu(NH3)4]^2+}$ or $\ce{[Cu(NH3)4(H2O)2]^2+}$.
The copper(II) ion has the electron configuration $\mathrm{[Ar]\ 3d^9}$. How do the 4 electron pairs from the ammonia ligands form the coordination bond with the copper ion, if the 3d subshell is almost entirely occupied?

I'll address $\ce{[Cu(NH3)4(H2O)2]++}$, but the reason is similar for $\ce{[Cu(NH3)4]++}$
The 9 d-electrons in the complex are in non-bonding and antibonding orbitals.
See pages 6-9 of Ligand Field and Molecular Orbital Theory.

the ligand σ orbitals are derived from AOs with energies much lower than those of the metal d-orbitals. Thus, the six bonding MOs of the complex are mostly ligand-orbital in character. These six bonding orbitals can accommodate the 12 electrons provided by the six ligand lone pairs. Therefore, the electrons provided by the ligands are largely confined to the ligands in the complex.

So, to a large degree, the 12 electrons from the ligands are in bonding orbitals, but the 9 d-electrons are in nonbonding or antibonding orbitals.

The d electrons of the metal centre...occupy the lower energy, non-bonding metal $t_{2g}$ set and the higher energy, antibonding $e_g$ set. Therefore, the metal based d electrons tend to reside largely on the metal atom. In summary, the frontier orbitals of the complex are the non-bonding, entirely metal based $t_{2g}$ orbitals and the antibonding $e_g$ orbitals, which are mainly metal in character anyway.

‘But the reason is similar for $\ce{[Cu(NH3)4]^2+}$’ — Do you have a recent source for tetraamminecopper(II)? In liquid ammonia, the hexaamminecomplex is formed to the best of my knowledge. Also, is that structure tetrahedral?

@Jan "Thus, for example, addition of NH3 to [Cu(H2O)4]2+ in Equation(15.1) has four possible positions available for substitution, whereas addition of NH3 to [Cu(NH3)3(H2O)]2+ in Equation(15.4) has only one possible position available for ..." https://books.google.com/books?id=-twqBwAAQBAJ&pg=PA490&dq=cu(nh3)4+2%2B&hl=en&sa=X&ved=0ahUKEwjqqbargr_QAhWDSyYKHZESBE4Q6AEISjAI#v=onepage&q=cu(nh3)4%202%2B&f=false

@Jan "The crystal structure consists of cations Na+ and [Cu(II)(NH3)4]2+ and of catena-anions, [Cu(I)n(S2O3)2n]3-. The coordination around Cu(II) is purely square planar, no other atom completing the coordination polyhedron to give any pyramidal ..." https://books.google.com/books?id=IEpGAAAAYAAJ&q=cu(nh3)4+2%2B+crystal+structure+square+planar&dq=cu(nh3)4+2%2B+crystal+structure+square+planar&hl=en&sa=X&ved=0ahUKEwjXxMfYg7_QAhVL7CYKHZLVADgQ6AEIMDAE

This is best explained using a molecular orbital scheme of a typical octahedral complex. Now, this is assuming $O_\mathrm{h}$ symmetry which is not correct for $\ce{[Cu(NH3)4(H2O)2]}$; the actual symmetry of the complex is $D_\mathrm{4h}$. However, it is close enough for discussion of why the complex can form, and we can even use a gedankenexperiment to find out the MO scheme with the correct $D_\mathrm{4h}$ symmetry. The scheme also ignores π interactions which are present but not important in ammine complexes.

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Metal orbitals on the left-hand side; ligand orbitals on the right-hand side. Molecular orbitals in the centre. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.
The first thing we notice is that the metal’s $\mathrm{t_{2g}}$ orbitals (that would be $\mathrm{d}_{xy}, \mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$) do not take part in metal-ligand interactions; they are nonbonding. This explains why complexes with populated $\mathrm{t_{2g}}$ orbitals are often stable and favourable. Next, we see that the six σ-symmetric ligand group orbitals transform as $\mathrm{a_{1g} + e_g + t_{1u}}$ which is exactly the transformation of the metal’s remaining d-orbitals, its 4s and three 4p-orbitals. Thus, these are all destabilised while the ligands’ orbitals are lowered. This is the net gain of ligand bonding.
Still, the metal’s $\mathrm{e_g}$ orbitals are still antibonding, i.e. $\mathrm{e_g^*}$. Electrons in antibonding orbitals are generally unfavourable. But if we take a $\mathrm{d^9}$ case and calculate the bond order:
$$\begin{multline}\mathrm{B.O.} = \frac{n(\text{bonding electrons}) - n(\text{antibonding electrons})}{2 \times n(\text{bonds})}=\\
\frac{\mathrm{(a_{1g} + e_g + t_{1u}) - e_g^*}}{2\times 6}
= \frac{12 - 3}{2\times 6} = \frac{9}{12} = \frac34
\end{multline}$$
A bond order of $\frac34$ is larger than $0$ and thus favourable. Thus, having the bond is better than not having it. The MO explanation is that the destabilisation of the populated orbitals is less than the stabilisation of populated orbitals: most of the stabilisation is offset by destabilising unpopulated orbitals, not populated ones.
And what changes if we remove symmetry by going from octahedral $O_\mathrm{h}$ to Jahn-Teller distorted $D_\mathrm{4h}$? Well, what happens is that ligands are removed on the $z$-axis, so simply spoken anything containing $z$ contribution will be stabilised. That includes one of the two $\mathrm{e_g^*}$ orbitals as well as two of the three $\mathrm{t_{2g}}$ orbitals. Similarly, the ligand orbitals that used to be $\mathrm{t_{1u}}$ are split up into a set of two and a single one, depending on their principal direction. These will no longer be degenerate. The same applies to the metal’s p-orbitals, which are also $\mathrm{t_{1u}^*}$. However, the basic picture remains the same; stabilised orbitals are more populated than destabilised ones for a net gain in the overall system.


text:
On the stability of deuterium
The isotope of hydrogen, tritium, is indeed radioactive resulting in $\ce{^3He}$ through beta decay.
But what of the radioactively 'stable' isotope deuterium?
Are all deuterium atoms destined to be deuterium for eternity or is there any process other than radioactive decay that could lead to transition such as say $\ce{^1H}$?

No, deuterium is completely stable. I found the answer at Hyperphysics, and it has to do with the mass energies of the products and reactants of this hypothetical reaction.
The decay of deuterium would be
$$\ce{D -> P + N + e + \bar{\nu}}_e$$
where $\ce{D}$ is deuterium, $\ce{P}$ is a proton, $\ce{N}$ is a neutron, $\ce{e}$ is an electron and $\ce{\bar{\nu}}_e$ is an electron antineutrino. The combined mass energies total to $1877.05\,\mathrm{MeV}$ (mega electron volts). But the mass of $\ce{D}$ is $1875.6\,\mathrm{MeV}$, so the decay cannot happen because conservation of mass/energy would be broken. There's no way to get around this, not even imbuing the initial deuterium atom with extra energy of some other sort. Deuterium will never decay unless, for some reason, the proton turns out to be unstable.
If, for some reason, deuterium did decay, it could go to $\ce{^2He}$ … which would then decay back to deuterium.

Perhaps it is interesting to also point out that while deuterium is stable, it is relatively weakly bound. One consequence is that it undergoes fusion comparatively easily, at temperatures as low as ~$8 \times 10^5 K$. Almost all the deuterium present in the Universe was created during the big bang nucleosynthesis, and since then it has been the first nuclear fuel to ignite in any object with mass above approximately 13 times that of Jupiter.

@NicolauSakerNeto Interesting. I knew about fusion and nucleosynthesis, but I didn't know that it's relatively weakly bound. Is the fact that it's weakly bound unusual for stable isotopes?

As can be seen here, deuterium is the composite nucleus with the smallest binding energy per nucleon of any naturally found isotope; it would seem there are just not enough nucleons to allow for lots of binding via the strong force. Not only that, its composition is very close to a much more stable nucleus, helium-4. The combination makes deuterium quite special in astrophysics and nuclear physics. Precisely why it should be that way is a rather difficult question to approach quantitatively.

Nice answer, +1.  This isn't my area, but is deuterium "completely stable" if it turns out protons are unstable?  Some physics theories predict proton decay and the experimental lower bound on proton half-life is "only" 10^34 years.

Well, if protons are unstable, the entire idea of stable vs radioactive nuclei kind of goes out the window - nothing would be completely stable.

@JonCuster I know; I wanted to cover all the possibilities.

@CurtF. No, it would be unstable (over very long scales). Did I say that explicitly, or did I forget?

Helium-2 doesn't convert to D but simply breaks apart.

In the Standard Model of physics (plus some assumptions on gravitons), baryon number can be violated in multiples of three, as long as the difference between baryons and leptons remains constant. The process is called sphaleron transitions.
Theoretically we could have
$$ D \to \bar p + 2e^+ + \bar \nu_e$$
The problem is that baryon number violation has never been measured (it would have a very, very small probability) and in any case it would not belong to chemistry.
Note that in any case, protons would still be stable in this context.

If we are looking for a "decay" mode in deuterium, try fusion to helium-4, especially in "brown" dwarfs (which are actually reddish in color).


text:
How many stereogenic centres are there in discodermolide?

It is generally said that carbon-carbon double bonds cannot behave as stereogenic centres but the above molecule contains three such centres. Can somebody please point out all the stereogenic centres (including the double bonded carbons) and explain how and why this is so?

I'm confused.  Who is saying the C=C double bounds are stereogenic?  Various materials I find, including (this one)[http://www.scs.illinois.edu/denmark/presentations/2004/gm-2004-07_06.pdf], say it has 13 stereocenters.  That is the same as the number of carbons connected to "stereo" bold or hashed bonds.  None of those carbons have double bonds.  Why do you think it contains C=C stereocenters?  Please consider editing your question to explain in more detail.

@CurtF. I am confused, too. Who is saying, that carbon-carbon double bonds are not stereogenic? Interchanging the substitutes at these centres would lead to a different stereoisomer (as long as the substituents are different of course). Unfortunately the definition is not given in the goldbook and hence not straight forward.

Can you give your definition of stereogenic centre. Are you referring to chiral centres?

Strange it seems there's cis/trans and R/S confusion, but they're totally different isomerisms.

OK, I just looked up "stereocenter" and "stereogenic" and to my surprise these terms supposedly include the cis-trans isomerism of C=C bonds as well as chiral isomerism.  In that case, however, many of the popular descriptions of the discodermolide are wrong, because in many places the molecule is referred to as having 13 stereocenters.  But if you include the C=C bond "stereocenters", the total would go up to three more.

The IUPAC Gold Book does include E/Z double bond isomers in its definition of 'stereogenic', which makes sense. I'm surprised that there's no definition of 'stereocenter'. I would NOT include double bonds in counting stereocenters because the E/Z designation is defined by the substitution at two atoms (all carbon in this case). In my mind, a stereocenter is limited to the classic 'four different things attached to one atom (carbon)'. I would say that discodermolide has 13 stereocenters and 16 stereogenic units. http://goldbook.iupac.org/S05980.html

There are 3 E/Z olefins and 16 asymmetric carbons total. The answer is 16. The entire olefin (both carbons) are a stereogenic center. It is the involvement of both carbons of the double bond (their juxtaposition) that makes the molecule otherwise than it would be. These designations are binary...R/S, E/Z, there is no other way to have the generation of stereoisomers (configurational isomers) arise. Consider how many maximum possible stereoisomers you can have. Now do you really think it matters if semantics calls it a 'unit' or a 'center'?

Summary
The confusion probably arises because of incompletely standardized definitions of at least some subset of the words "stereocenter", "stereogenic", "stereogenic center", "stereogenic unit".
IUPAC definitions
The place to turn for authoritative definitions of these terms is IUPAC's Gold Book, as noted by @jerepierre in the comments.  Here is what it has to say:

stereogenic unit (stereogen/stereoelement)
A grouping within a molecular entity that may be considered a focus of stereoisomerism.
At least one of these must be present in every enantiomer (though the
  presence of stereogenic units does not conversely require the
  corresponding chemical species to be chiral). Three basic types are
  recognized for molecular entities involving atoms having not more than
  four substituents:

A grouping of atoms consisting of a central atom and distinguishable
  ligands, such that the interchange of any two of the substituents
  leads to a stereoisomer. An asymmetric atom (chirality centre) is
  the traditional example of this stereogenic unit.
A chain of four non-coplanar atoms (or rigid groups) in a stable conformation, such that an imaginary or real (restricted) rotation
  (with a change of sign of the torsion angle) about the central bond
  leads to a stereoisomer.
A grouping of atoms consisting of a double bond with substituents which give rise to cis-trans isomerism.

Thus, the unequally substituted C=C moieties in discodermolide are "stereogenic units" according to IUPAC.
Stereocenter
Unfortunately, this term is not defined in IUPAC's Gold Book.
Stereogenic center
Unfortunately, this term is not defined in IUPAC's Gold Book.
Wikipedia
Wikipedia is not as authoritative as IUPAC but it is likely that chem.se readers will encounter wikipedia entries for these terms far more often than IUPAC's in the course of regular web searching.
Stereogenic unit
No Wikipedia entry
Stereogenic center
Redirects to "stereocenter", see below.
Stereocenter

A stereocenter or stereogenic center is an *atom* bearing groups such that an interchanging of any two groups leads to a stereoisomer.

This definition does not apply to C=C double bonds, because only an atom can be a stereocenter, not a group of atoms.  Supporting this interpretation is a complete lack of reference to cis-trans isomerism on the rest of the page.
Usage by chemists
Discodermolide is referred to in many sources as having 13 "stereocenters" or "stereogenic" centers.  This conforms with the Wikipedia, atom-centric definition of stereocenter.  Here are some example sources:

Wikipedia's discodermolide article: (+)-discodermolide has
...13 stereogenic centers...
A presentation from 2004:
molecule possesses: 13 stereocenters
A paper from 2003: Paper title: Regulated-Stereoselective Construction of Thirteen
Stereogenic Centers...

All these uses comport with both Wikipedia definitions of "stereocenter"/"stereogenic center".  Chemists do not seem to use the IUPAC term "stereogenic unit" very much in actual practice.   I could not find any relevant pages using both the term "discodermolide" and the phrase "stereogenic unit", for example.
Bottom line
It is just possible to pass a thread of consistency through the eye of this needle of confusing terminology.

Discodermolide has thirteen (13) stereogenic centers.
Each of those stereogenic centers is by definition also a stereocenter, and is also a "stereogenic unit".
In addition to those 13 stereocenters, there are three additional "stereogenic units" in discodermolide corresponding to the unequally substituted C=C bonds.

I've probably heard 'stereogenic element' or 'stereoelement' more often than 'stereogenic unit' in practice. Now that I've seen 'stereogenic unit' in the Gold Book, I'll probably go with that to stay away from 'element' which obviously has a unique meaning to chemistry.


text:
What is hard and soft water and how does the process of dissolution work?
This is a very basic question about solubility, especially solution of table salt in water and solution of sugar in water. I have heard about terms like hard water, soft water, filtering of water to remove impurities etc., but I don't understand these well enough. My questions are:

If table salt is dissolved in water, why do we call it a solution? Don't say that a chemical reaction has taken place and some new product say $\ce{NaH2OCl}$ etc. is formed.
By mere filtering using a cloth or a machine, how can we get rid of minute impurities, (which make water hard) which are not even visible to naked eyes?

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

There is quite a good wikipedia article about solution 2. You can't.

Hard water and soft water differ only in the presence or absence of certain impurities:

Hard water: Water with impurities in the form of minerals containing calcium and magnesium added in.
Soft water: Water with these minerals removed.

One thing to note is that hard water is not the slightest bit dangerous. Studies have been conducted by the World Health Organization and the National Academy of Sciences, and no evidence was found linking hard water to and adverse health effects - or any effects at all!

By mere filtering using a cloth or a machine, how can we get rid of minute impurities, (which make water hard) which are not even visible to naked eyes?

This is impossible. Many devices have been proposed, but none have been shown to work. Quoting from this article,

There are a variety of devices that claim to manage hard water scale using primarily magnetic or electrical technology. Manufacturers generally claim the devices utilize energy to alter the behavior of compounds or elements within the water. They do not claim water chemistry is altered. In fact, the hardness of the water before and after treatment is not changed. This complicates assessment of their performance. While protocol has been established to assess the effectiveness of ion exchange water softeners (NSF/ANSI 44), at this time, no recognized agency in the U.S. has established protocol to assess the effectiveness of physical water treatment devices. Therefore, questions remain as to their effectiveness.

Chemicals are the only way to go, unfortunately.
More information on hard water can be found here and here.

If table salt is dissolved in water, why do we call it a solution? Don't say that a chemical reaction has taken place and some new product say $\ce{NaH2OCl}$ etc. is formed.

In case you're interested, the reaction is
$$\ce{NaCl + H2O \to Na+ + Cl- +H2O}$$
It fits the very definition of a solution:

a homogeneous mixture composed of only one phase.

$\ce{NaCl}$ is separated into ions that separate uniformly throughout the water.

The other answer wasn't really answering the question completely, and I was feeling too lazy to add a better one!

Note that:

A chemical reaction is a process that leads to the transformation of one set of chemical substances to another.

Salt when put into water, dissolves into ions as $\ce{Na+}$ and $\ce{Cl-}$.The positive and negative part of the ions are attracted by the oppositely charged water molecules. The ions are stable and do not undergo further reactions.
For the second part of your question, external impurities can be removed by filtering if they are of considerable size. Dissolved impurities like phosphates, sulphates cannot be removed by a cloth. The hard water has to be then 'treated' to make it soft again.


text:
How does zinc become oxidized when it's not in direct contact with copper ions in a galvanic cell?
Whenever a galvanic cell is explained in textbooks/videos, it usually starts by showing that if you put a piece of $\ce{Zn}$ in a $\ce{Cu^{2+}}$ ion solution, the $\ce{Zn}$ will become oxidized and transfer $\ce{2e-}$ to the $\ce{Cu^{2+}}$ ion, so the $\ce{Cu^{2+}}$ becomes a neutral $\ce{Cu}$ metal.
I understand this, but in a galvanic cell, the $\ce{Zn}$ never actually touches the $\ce{Cu^{2+}}$ ions, the $\ce{Zn}$ and $\ce{Cu}$ are only connected with wires. I don't understand how the $\ce{Zn}$ can be oxidized and transfer ions through the current.

Zinc electrode is anode and copper electrode is cathode. The two electrodes are connected by a electrically conducting wire. At anode, $\ce{Zn}$ atoms are oxidised to $\ce{Zn^2+}$ which go into the solution, and the electrons are left on the metal strip.
The electrons produced in the oxidation half-reaction at anode are transferred through the wire to $\ce{Cu}$ electrode where they are consumed. The $\ce{Cu^2+}$ ions are reduced to metallic $\ce{Cu}$ which coats on the $\ce{Cu}$ strip.
Thus, oxidation and reduction half reactions occur at separate electrodes and the current flows through the wire. A salt bridge maintains neutrality in both the compartments by a flow of ions.
The overall cell reaction can be given as:
$$\ce{Zn (s) + Cu^2+ -> Zn^2+ + Cu (s)}$$

How can the Zink atoms be oxidized in the first place? How can some Zink that is just minding its own business,placed in a solution of Zinksulfat(away from the Copper ion solution), suddenly lose electrons?

Because the dissociation of zinc into ions leads to decrease of its overall energy. The free energy change for this reaction is around $-213  kJ$ and as the entropy change isn't significant, the overall reaction is spontaneous.
Also,  It works in that direction and not the reverse (at standard conditions) because copper atoms are more closely and strongly bonded in the solid metal than zinc atoms.

So the zinc reacts with the zink ions in the zinksulfat solution and thus releases electrons? I thought that in order for an element to send electrons to another, the first element has to be lower in the galvanic series.

No, the zinc does not react with the zinc ions. It dissociates into electrons as it is energetically favourable for it to dissociate rather than staying as a metal.

Could you explain why it is energetically favorable for zink to dissociate rather than stay as metal? I could understand it if zink was put in a copper ion solution(since Copper ions are positively charged and zink is lower than copper on the galvanic series), but in this case i dont see what the oxidation agent is. Thanks.

This process is thermodynamically governed by the Gibbs free energy equation. The equation is $\Delta{G} = \Delta{H} - T\Delta{S}$. For dissociation of zinc, the free energy is about $-213kJ$, thereby making it spontaneous. Read more here


text:
Mimic a volcano eruption with household chemicals?
How can one mimic a volcano eruption using household chemicals safely?
We've tried it in a testing tube with baking soda and vinegar. This wasn't quite as impressive as expected.
On the web we've found recipes that add food coloring (red) and dish washing liquid to the mix. I guess the latter is for some bubbling effect.
What we have:

A volcano made from paper-mâché (classic volcano cone form, perhaps a bit steeper than natural)
Center hole big enough to fit a 1.5 liter water bottle

Note: this is supposed to be done with kids, so please don't propose any method that would generate poisonous fumes or similar.

Meanwhile tried a soda ($\ce{NaCO_3}$, i.e. not baking soda $\ce{NaHCO_3}$) solution poured into a dish washing solution, then adding a citric acid solution. It was pretty good, but a bit too quick. Vinegar seems to work better than citric acid, too. Also I used warm water to dissolve the reagents.

Not taking excessive precaution is somehow contradicting "eruption". If you want there to be a more impressive reaction, you need to be more meticulous. I would suggest potassium permanganate and glycerin for a "start". Though that is more impressive than safe.

Cool question! Vinegar or citric acid on the one side, baking soda on the other is a safe and reasonable setup! Red or orange food dyes are good for the lava colour. I'm wondering if adding some corn starch to make the mixture somewhat thicker will add to the effect. Kids usually like it when it gets more messy :D. On a side note, back in the days school vulcanoes were created by heating a cone of ammonium dichromate with a burner. The effect is dramatic (fire at the top, the cone grows and changes the colour to gray-green) but the dichromate is cangerogenic!

Starch is an interesting idea. KMnO4 is definitely out, as the volcano is meant to be reused. From experience I'd expect the color of all lava-flooded "terrain" to get a brownish to black tone after the treatment with KMnO4. Also, I am not able to get this much, as the pharmacists are really rather paranoid these days. When one pointed to the potential explosive traits of it, I could barely keep myself to suggesting silver nitrate in its stead, as it would probably be more reactive and you'd get even less "bang for the buck" ;) ...

Mentos and some sort of red soda?

Take a warm bottle of champagne.

The 'elephant's toothpaste' experiment with hydrogen peroxide decomposition works well to produce large amounts of foam and might be suitable for what you are looking for.
Low concentration hydrogen peroxide solutions are readily available and not dangerous provided you don't get it in your eyes or drink it or anything silly.
Fill the bottle with some hydrogen peroxide solution, add a few drops of orange food colouring and some dish washing liquid and then add something to catalyse the decomposition. You could use something like potassium iodide or manganese dioxide to give you a very rapid decomposition but if you want a slightly less violent and more sustained reaction then apparently dried yeast dissolved in warm water works quite well (yeast contains good amounts of catalase, an enzyme which catalyses the decomposition of hydrogen peroxide).
Having never done this experiment, I can't tell you exactly what quantities to use so you should probably test it before doing it as a demonstration. However, there are some descriptions of the experiment on the internet to give you a good starting point.
http://www.sciencebob.com/experiments/toothpaste.php
http://www.stevespanglerscience.com/lab/experiments/elephants-toothpaste

Safety is a critical issue here, and it is very possible to simulate eruption safely, with a simple combination of household items.
A couple that I have seen and tried with some of my younger relatives - quite surprising some of them:
Ketchup and Baking Soda Volcano, this takes a bit of trial and error to get the consistency right. Mix some detergent and water with the ketchup - then react it with baking soda. (Be sure to read the ingredients list on the ketchup first). The reason the eruption occurs is due to:

The ketchup contains vinegar, which is dilute acetic acid. The acetic acid reacts with the baking soda to produce carbon dioxide gas. The gas bubbles expand and rise through the liquid, bubbling out the ketchup.

We also tried adding vinegar (acetic acid) to ketchup for an eruption. Once again, trial and error is needed to get the best consistency required.
The classic Mentos in Coca Cola is a good one as well, but can be too fast and messy (use diet coke to avoid sticky residue). This may or may not be viable due to its lack of viscosity.


text:
MO scheme for coordination complexes (tetrahedral) - Heteroligand vs. homoligand
How is the MO diagram drawn for a tetrahedral complex with heteroligands (e.g. $\ce{[NiCl2(PPh3)2]}$) different from a tetrahedral complex with the same type of ligand? Do we consider the same "a1 and t2" SALCs as in the case of same ligand type tetrahedral?

Bis(triphenylphosphine)nickel chloride can have be tetrahedral or square planar depending on conditions. In any case the SALCs for a tetrahedral complex with heteroligands will depend on the point group of the complex. For [NiCl2(PPh3)2], the point gorup is C2v and not Td like for same ligand tetrahedral. You need to check the character tables

Why does [NiCl2(PPh3)2] have a C2v point group? Wouldn't it be more logical if it have Td point group? If so, did you not include the two Cl ligand in determining the point group because I think the only way the complex would be a C2v is that when its geometry is similar to water.


text:
Reaction between elements in a periodic table
Can we know the reaction between any two elements in the periodic table? If yes then can we know the reaction between any three or more elements in the periodic table?

Welcome to chemistry.SE! Um, I see problems in your question: 1. I really don't get what causes your question. What is confusing you? 2. If you're asking about "teach me chemical reactions" you must be asking something like "teach me all of chemistry."

Can you say more precisely, what are you asking? What do you mean by can we know? Reactivity of all relative stable elements is known.

We don't even know completely the reaction of one element with itself.
For example, consider carbon.  There are constantly new forms of carbon being discovered.


text:
Do molecules with axial chirality have "stereogenic units"?
This question is a follow-up to a recent question on discodermolide stereochemistry that I answered here.  In the course of reading IUPAC's definition of "stereogenic center", I wondered if axially chiral molecules such as BINOL have a "stereogenic unit" according to IUPAC.
The part of IUPAC's definition that may (or may not) apply to BINOL is:

A chain of four non-coplanar atoms (or rigid groups) in a stable conformation, such that an imaginary or real (restricted) rotation (with a change of sign of the torsion angle) about the central bond leads to a stereoisomer.

Here is the structure of (S)-BINOL:

Does the four-atom chain consisting of the O-bond carbons plus the transannular C-C-bound carbons qualify?  I can't quite work out what is meant by "with a change of sign of the torsion angle" and "stable conformation".  If it does qualify, suppose temperature is increased to the point where spontaneous racemization happens (and suppose that any side reactions or decomposition can be ignored).  Is that chain no longer a stereogenic unit?

Yes, axially chiral compounds have a stereogenic unit. Biphenyls (such as BINOL) and allenes are common types of compounds that exhibit axial chirality. As you describe, for BINOL the stereogenic unit is the set of atoms including the carbons joining the aryl groups and one carbon further from the biaryl junction. IUPAC's description is simply meant to define a torsional angle, or thought of another way, define two planes around an axis.

If it does qualify, suppose temperature is increased to the point where spontaneous racemization happens (and suppose that any side reactions or decomposition can be ignored). Is that chain no longer a stereogenic unit?

This is a great question that forces us to think about what we mean by 'stable' and under what conditions is something stable.
Reference to 'stable conformation' means that the two conformations cannot interconvert at ambient temperature. In BINOL, there is restricted rotation around the biaryl bond, which means that the two stereoisomers cannot interconvert under ambient conditions.
Let's think though about a much simpler compound with four carbons in a row that can be used to define a torsional angle: butane. The gauche conformation actually meets the definition of chiral: it's mirror image is not superimposable on itself. However we don't consider butane to have a stereogenic unit (or consider butane to be a chiral compound) because those two conformations can interconvert readily at room temperature.

So those are two ends of the spectrum if we consider room temperature: BINOL has a stereogenic element but butane does not.
Let's say we heat up (S)-BINOL to a temperature that allows for rotation around the biaryl bond. As you say, the sample will racemize. At that temperature, the compound no longer has a stereogenic unit because rotation is no longer restricted. Allow the sample to cool, and the stereogenic unit returns, although the sample will be racemic.

What about hexahelicenes?  There aren't four atoms to point to.  But is that when "rigid groups" parenthetical exception to needing four atoms applies?

@CurtF. You can pick out a set of four atoms to define the torsional angle. Multiple sets actually.

If you can pick multiple sets, doesn't it mean that there isn't one well-defined "unit" of stereogenicity?

@CurtF. I don't think so. In BINOL, we could choose different sets of four atoms to define the torsional angle, but it's still a single stereogenic unit.


text:
Which chemical reactions occur between my tongue and the food I eat?
There are various chemical reactions that can occur between materials.

Which chemical reactions occur with the tongue?
How come all reactions are reversible (nothing sticks to my tongue)?
What's special about the tongue-water interaction that makes it tasteless?

There shouldn't necessarily be sticking to determine a chemical reaction. 2. All reactions are reversible. Some just don't get reversed cuz it's not favorable for them (i.e.: Thermodynamically or such) 3. What has your first question to do with the second?! (Separate questions should be asked in separate posts.) Sorry, I don't mean to be harsh or picking on you or something. All I intend to do is help improve the quality of the question.

@MARamezani, by "reversible" I meant easily removed by mechanical means; I think the nature of the chemical reaction gives the food its taste, hence the question about water: what's so special about it that it's tasteless..?

About water, it's simply because our tongue buds do not sense any taste. Water isn't the only thing that "our tongue considers tasteless". About the other one, I'm not ready to answer yet.

1) I would guess that all of the different interactions are possible except for ionic and covalent because those are much stronger than the rest and it would be difficult for the food molecules to disassociate from the taste receptors to be used again. Enzymes in saliva break down complex food molecules to smaller and simpler ones.
2) Started explaining in 1, molecules are not usually permanently bound to the taste receptor in order for the taste receptor to be used again (another molecule being able to bind to it). Some toxins form covalent bonds to receptors, such as curare. I'm not sure what kinds of receptors taste ones are but toxins target different specific receptors. Curare binds to nicotinic acetylcholine receptors, found in muscles.
3) We don't taste things unless a molecule binds to a taste receptor.


text:
Why did they provide the value for the enthalpy change of fusion?

The metal gallium melts when held in the hand; its melting point is $\pu{29.76 ^\circ C}$. How much energy as heat is removed from the hand when $\pu{5.00 grams}$ of gallium initially at $\pu{20.0^\circ C}$ melts? The value of $\Delta H_{fusion}$ is $5.576~\mathrm{kJ~mol^{-1}}$ and the specific heat of gallium is $0.374~\mathrm{J~g^{-1}K^{-1}}$. Take the final temperature to be $\pu{29.76 ^\circ C}$.

I got:
$$q = (\pu{5.00 g})(0.374~\mathrm{J~g^{-1}K^{-1}}) (\pu{9.76^\circ C})$$
$$q=\pu{18.3 J}$$
So I reason that since $\pu{18.3 J}$ of energy in the form of heat is required to melt gallium from at an initial temperature of $\pu{20.0 ^\circ C}$ to a final of $\pu{29.76 ^\circ C}$ then this is all the energy required to do this but why did they give $\Delta H_\text{fusion}$?

You calculated the energy required to change solid gallium at 20°C to solid gallium at 29.76°C.  Once solid gallium reaches 29.76°C, additional energy is required to melt it.  The energy required is given by the $\Delta H_{\text{fusion}}$.  So to complete the question, transform the $\Delta H$ units to a grams basis instead of a mole basis, multiply it by the weight of gallium in the problem, and then add that result to the one you already calculated.  That will be the energy required to change solid gallium at 20°C to liquid gallium at 29.76°C.


text:
Reason for inconsistency in specific gravity of calcium choride added to water
I am seeing a significant change in specific gravity of products upon addition of $\ce{CaCl2}$ to water. The mass and volume of what goes in doesn't match my total specific gravity of what comes out. I am curious as to why that is the case?

When you say, "of products", do you really mean "of the solution"? If so:
The solvent mass and the solute mass add up to the mass of the solution.
But solvent volume and solute volume often don't add up to the volume of the solution. This is because the forces between molecules in the solution are different from the forces between molecules in the pure solute and the pure solvent.
In your solution, the calcium ions and chloride ions attract water molecules around them. The water molecules are packed more tightly than they would be if the ions weren't there. That can increase the density of the solution relative to the total volume of the unmixed solvent and solute. Another effect is that the ions in solution are packed less tightly than they are in solid $\rm CaCl_2$.
You can see which effect wins by dumping some calcium chloride into a volumetric flask. Fill the flask up to the mark with water, and then mix until all the calcium chloride dissolves. The solution probably doesn't quite come up to the mark anymore, because the calcium and chloride ions have drawn the water molecules in around themselves, making the solution denser than it would be otherwise.
More detail than you may want below:
The volume of the solution $V$ is given by the following equation:
$$V = n_1 \overline{V}_1 + n_1 \overline{V}_2$$
where $n_1$ and $n_2$ are moles of calcium chloride and moles of water in the solution, and $\overline{V}_1$ and $\overline{V}_2$ are the partial molar volumes of the calcium chloride and the water.
These partial molar volumes are actually derivatives of the volume of the solution with respect to moles of calcium chloride ($\overline{V}_1 = \partial V/\partial n_1$) and with respect to water ($\overline{V}_2 = \partial V/\partial n_2$). They are related to each other (when one rises, the other falls) and they vary with the composition and temperature of the solution in a rather complicated way.

This ushered me forth in a direction that finally made sense. I ended up finding this website, http://www.chem.latech.edu/~ramu/cmen522/homework/hw06_04a.pdf, that further explained the dynamic as well. I gathered what was necessary to be done, but wasn't quite sure how to apply it. I am currently a chemist for the oil field and am trying to find an accurate representation of percent water and CaCl2, as well as the correct specific gravity of the complete solution. Given a 10.5 Mud weight (10.5 lb/gal), I have struggled to find volumes necessary to add and partial volume actually present.

Further explained - I applied the formula designated V /cm3 = 1001.97 + 30.2207m1/2 − 36.7889m + 37.0838m3/2 − 5.7447m2 for the corrected solution volume and was really close to the correct "loss" of volume from the packing you described. The two following formulas for partial volumes −36.7889 + 15.1104m−1/2 + 55.6257m1/2 − 11.4894m (for CaCl2) and 18.048 + 0.2722m1/2 − 0.3341m3/2 + 0.1035m2 (for water), don't add up to the total solution loss. I am a bit baffled as to why this is the case. Granted my physical chemistry knowledge is a bit dated, I would assume my excel document is correct.


text:
How to calculate a rough estimate for the volume of a nitric oxide molecule?
I was wondering what would be the best way to calculate a rough estimate for size of a nitric oxide molecule to use in comparison to a box.
I was thinking volume would be the best method, but this gave me an answer in dm3/mol which I can't use for comparing against the volume of a box.

The size of what? A molecule? A certain amount of molecules?

Though this seems very clear to you, it's ambiguous for us. Details need be added. But anyways, welcome to chemistry.SE!

What details should I include? I'm trying to calculate a rough estimate for the size (volume?) of a Nitric Oxide molecule.

NO is in equilibrium with weak dimer form. May be important.

Take the volume per mole value (b) from a van der waals constants table (0.0289 L/mol) and divide by $6.022 \times 10^{23}$ molecules per mole.
This will give you a volume that you can compare to the box.

This gives such a small number,10^-26. I was thinking it would be around 10^-15 or larger. Does this look correct to you?

for one molecule, 10^-26 liters sounds much more correct than 10^-15 liters.  The distance from the center of the oxygen atom to the center of the nitrogen atom is 115 picometers.

On the MP2/def2-QZVPP level of theory the $\ce{NO}$ bond length is $\pu{113.5 pm}$ and the molar volume is estimated based on $\pu{0.001 e/bohr3}$ density envelope as $\pu{241.539 bohr3/mol}$ or $\pu{21.555 cm3/mol}$. The molecular volume therefore is about $\pu{3.58\times10^-29 m3}$.


text:
What's the pH of vinegar containg 5% acetic acid?

Vinegar generally contains 5% acetic acid. We would expect the pH of vinegar to be approximately:
a. 0
b. 3
c. 7
d. 9
e. 12

I don't have the key for this question, so I just want to make sure that I got it right and what I got doesn't seem to be one of the answer choices.
My answer was 1.3, so I rounded it to 1 and there wasn't an answer choice. Here is what I did:
$$\log(.05/1) = -1.3010$$
How else would you do it because you were already given the final concentration from the original concentration?
Am I missing something here?
Edit: The professor said to use this formula for calculating it:
$$K_\mathrm{a} = \frac{(.05x)^2}{x}$$
and got 3, which made no sense, any way this is possible?

What does 5% acetic acid mean, i.e. 5% of what? What quantity do you need to calculate the pH? What property of acetic acid to you need to calculate the pH?

Well we know that vinegar is mostly composed of water and acetic acid so I guess they are asking for the pH of that 5%?

What's the formula for pH?

I thought it was the - of the log of the molarity of the Hydronium ion concentration. Why are you asking me :)?

Right, but you don't know the molarity of the hydrogen ion concentration. You have the concentration of acetic acid as a percentage...

So what you are saying is that use the .05 and solve for the Hydronium concentration and calculate the pH from that? And then I can get the Ka values from a chart too.

Let us continue this discussion in chat.

Did the question say to round it?

@L.B. No but I think it is implied.

The key word there is "approximately". Anyway,it usually is a knee-jerk response for students to calculate. With approximation, yes, the question is usually conceptual. Look at the choices first before computing. You might already find the correct answer.

LOL - yes round it. As in "The pH is 'round 3."

PS - the professors solution is totally wrong.

You are not given a concentration, but a percentage value. Assuming its mass percent (wt%), you first have to calculate the actual concentration of the solution like this:
5% Acetic acid means in a $\pu{1 kg}$ solution, you can find $0.05 \times \pu{1000 g} = \pu{50 g}$ acetic acid. With a molar mass of $\pu{60.05 g/mol}$, this is similar to
$$n = \frac{\pu{50 g}}{\pu{60.05 g/mol}} = \pu{0.83 mol}$$
The concentration $C = n/V$, therefore is
$$C = \frac{\pu{0.83 mol}}{\pu{0.95 L}} = \pu{0.874 mol/L}$$
Acetic acid is a weak acid. This means it does not dissociate completely and $\mathrm{p}K_\mathrm{a}$ has to be taken into account to calculate the pH. The $\mathrm{p}K_\mathrm{a}$ of acetic acid is about $4.76$. The pH is then defined as:
$$\mathrm{pH} = 0.5 (\mathrm{p}K_\mathrm{a} - \log C_\mathrm{a}) = 0.5 (4.76 - \log 0.874) = 2.4$$

Which isn't one of the choices either unless you round up and most people don't round up until 2.5.

True but If I didnt make a mistake, this is the correct calculation. The closest answer is a pH of 3 which is also what google says about the pH of vinegar.
The answer options just suck.

I see what you're getting at there. I was just slightly concerned since the person asking wasn't able/willing to round 1.3 and get a close answer.

also, I just found this: https://answers.yahoo.com/question/index?qid=20110109075809AAGGAwF Its the exact same solution.

Well there 'ya have it :)

Consider the choices. Since it's acetic acid, you can rule out any pH values that indicate a basic or neutral solution. Knowing that acetic acid is a weak acid and that vinegar is a fairly dilute solution, you can also rule out any pH values that would indicate either a strong acid or a highly concentrated solution. Eliminating those impossible values leaves only one choice as the answer.
The trick with a question like this is to realize that it's multiple choice for a reason. It's intended to be a conceptual problem, not a calculation problem. In general, a good rule for multiple-choice questions like this is to first see if you can solve it using general concepts; only if you can't should you reach for you calculator.

I know you are very right, but what if you were to show your work? I know that without work 3 is the most plausible answer but with work I can't mathematically support my assumptions.

To show work (or to get an exact answer in general), I'd use basically the same method that danijoo did. The wording of the question is clear that the desired answer is only approximate, so calculating that the actual pH is about 2.4 is good enough. You're not rounding to the nearest integer; you're rounding to the nearest answer choice. That said, the only assumptions required for the method I gave are that the desired answer is one of the choices given and that work isn't required. The rest follows logically from general principles of acids, bases, and pH.

You don't need a calculation to know the answer. You aren't likely to use a pH 0 acid to make a salad :-). 7 is neutral and therefore not acid. 9 and 12 are bases. The only answer left is 3.

A rehash of j_foster’s answer

I know this question was over 2 years ago, but I've just joined and wanted to see if I could improve on the responses already given and practice the formatting of text while also perhaps helping someone else who might come along this type of problem.
Acetic acid is a weak acid, as others have mentioned. Acetic acid does not fully ionize in water. $\mathrm{p}K_\mathrm{a}$ becomes necessary here.
The $\mathrm{p}K_\mathrm{a}$ of acetic acid is $4.76$. You should have been permitted to locate this value in a table or otherwise been provided it on the test form, unless you were required to memorize acid dissociation constants, in which case I apologize for what must have been a course destined to forever char (oxidation!) your memory and appreciation for the magical world of chemistry.
From the definition of $\mathrm{p}K_\mathrm{a}$ we see that $$\mathrm{p}K_\mathrm{a} = -\lg K_\mathrm{a}$$ which implies that the $K_\mathrm{a}$ of acetic acid is $10^{-4.76}$, which is $1.74 \times 10^{-5}$.
Now, consider that,
$$K_\mathrm{a} = \frac{[\ce{H^+}][\ce{A^-}]}{[\ce{HA}]}$$
We are asked for $\mathrm{pH}$ in the problem, and we know that:
$$\mathrm{pH} = -\lg [\ce{H^+}]$$
So let's work on isolating $[\ce{H^+}]$ so that we can get the $\mathrm{pH}$ for our answer.
Let's consider it a given that our acetic acid is in pure water which means, when it reaches equilibrium, it does so to equimolar portions of protons and acetate anions, viz.,
$$\ce{HA<<=>H^+ + A^-}$$ where the $K_\mathrm{a}$ of this equilibrium we determined above to be $1.74 \times 10^{-5}$.
As we mentioned above, $$K_\mathrm{a} = \frac{[\ce{H^+}][\ce{A^-}]}{[\ce{HA}]}$$
We talked about the very reasonable assumption that $[\ce{H^+}]$ and $[\ce{A^-}]$ are in equimolar proportions on dissociation of $[\ce{HA}]$. If we let $$x = [\ce{H^+}] = [\ce{A^-}]$$ then it follows that:
$$K_\mathrm{a} = \frac{x^2}{[\ce{HA}]}$$
Since the $K_\mathrm{a}$ of acetic acid is $1.74 \times 10^{-5}$, what are we left with? What do we know? That $5~\%$ is coming into play now I think. We know that per cent is "per 100," but "per 100" of what? In chemistry, in the absence of specific details otherwise, my assumption is molarity (moles/litre). Molality is used at times (mass fraction) as may be volume and weight percentages, mole fractions, etc. Given just the "$5~\%$" statement in the problem, I would take that as molarity of the acid ($[\ce{HA}]$).
Molarity is concentration! So the equations above, particularly the one that equates $K_\mathrm{a}$ to concentrations of products/reactants should work!
If $1.74 \times 10^{-5} = K_\mathrm{a}$ and $0.05 = [\ce{HA}]$, then it follows that
$$x = [\ce{H^+}] = \sqrt{1.74 \times 10^{-5} \times 0.05}$$
On my calculator, I get $x = 0.000933 = [\ce{H^+}]$.
So, the $\mathrm{pH} = -\lg([\ce{H^+}]) = -\lg(0.000933) = 3.03$
Looks pretty close to "B" to me!

Percentage concentrations are actually well defined as $% = 100 \times m(\text{solute}) /m(\text{solution})$. So a $5~%$ solution of acetic acid (we assume: in water) means that $\pu{5g}$ of acetic acid are dissolved in $\pu{95g}$ of water. This is not molarity. However, please take an upvote for the rest of the answer.

+1 Yeah while the use of % to indicate a percentage of 1M is innovative (read unintuitive) the fact that the answer comes out so close to the multiple choice answer that is the only possible option is kind of scary.  It is possible that the person preparing the question made this same assumption.

The percent could either be percent by mass, by volume, or by moles. Definitely not a quantity with dimensions such as molarity.

“EXPECT” is the key word here. It has nothing to do with an equation. The PH values are quite funny 0? 7? 9? 12? Really??? I haven’t been in a chemistry class in almost forty years…. For vinegar to be shelf stable it needs an acetic acid concentration around 5% and a Ph under 4 (mine have run anywhere from 2.4 to 3.6). I’m sure the equations listed here work for acetic acid solution of 5%. However vinegar’s PH is not created by its acetic acid percentage (exclusively I should say). Type of fruit, sugars when harvested, which yeasts were used etc etc

The point of the problem is to train chemists on the use of equations to compute properties such as pH of solutions. I think you are not addressing the problem.


text:
What is the difference between the Joule and Joule–Thomson coefficients?
Is there a difference between the Joule coefficient and the Joule–Thomson coefficient? Also, I am having a hard time understanding what they mean.

The Joule coefficient is $$\left(\frac{\partial T}{\partial V}\right)_U$$
Experimentally, Joule attempted to measure this value by expanding gas into an evacuated, insulated container, thus ensuring $U$ is constant.
The Joule–Thomson coefficient is $$\left(\frac{\partial T}{\partial P}\right)_H$$
Experimentally, this is realized by expanding a flow of gas in an insulted pipe from a high pressure upstream region to a lower pressure downstream region, the two regions being seperated by a porous frit.
For more information see THE JOULE AND JOULE-THOMSON EXPERIMENTS by Dr. J. B. Tatum, University of Victoria.

So, the Joule coefficient is a ratio of the change in temperature to the change in volume with constant internal energy, and the Joule-Thomson coefficient is a ratio of the change in temperature to the change in pressure under constant enthalpy? In the Joule experiment, the gas does no work, and the temperature doesn't change?

This almost correct, only your statement "the temperature doesn't change" is incorrect.  Joule didn't detect a temperature change, because his equipment wasn't good enough, but for a real gas (not an ideal gas) there is a temperature change.


text:
Helium gas expand isobaric isothermal
If $4\ \mathrm{mol}$ helium gas expand against constant pressure in isothermal process, how can calculate $W$?
Is it $W=-nRT\ln\frac{V_2}{V_1}$ as defined by isothermal or is it $W=-p\left(V_2-V_1\right)$ as define from isobaric?
Also, is enthalpy equal to $0$ in such process?

Since the process is isothermal ($q=0$) and isobaric, then we can say that $\Delta H = 0$ since at constant pressure $\Delta H = q$.

i also thought ike that but then i statred to consider that atually isothermal mean dT=0 and not Dq/

Ran, you are right that isothermal means that dT = 0.  Heat transfer generally occurs in isothermal processes, so Q is not zero.

The "constant pressure" in the question is probably meant to refer to the atmosphere, not necessarily the helium.  If the helium were at constant temperature, constant number of moles, and constant pressure, how would it be expanding???

actually it mention constant external pressure of 2 atm and isothermal expand under 320k...

If the helium is expanding isothermally against a constant external pressure $P_{ext}$, the work is calculated as
$$w = -P_{ext}(V_2 - V_1)$$
This process is irreversible. The first expression you gave for work is (sort of) the expression for work for a reversible isothermal ideal gas expansion.
If helium is assumed to be an ideal gas, then $\Delta U = 0$ because the process is isothermal. You have $q = -w$ (and not zero) for this process. $\Delta H$ will be zero too, since $$\Delta H = \Delta U + \Delta(PV) = \Delta U + nR\Delta T = 0$$.

thanks. if helium no consider ideal gas i assume cannot calculate dH. correct?

Both ΔU and ΔH will be small but nonzero if the gas isn't ideal; you can compute them if you use the equation of state for the gas.


text:
Why don't ions have one equilibrium potential? (Nernst equation)
I know that equilibrium potentials are dependent on the ratio of ion concentrations inside and outside of the cell and temperature. I also know that the equilibrium potential is reached when there is no net movement of ions across the membrane, i.e. the chemical and electrical gradients are equal. Why don't the ions move until they reach the same ratio of inside and outside concentration every time? For example, why doesn't ion "x" always diffuse until the ratio of inside to outside concentration is 15 to 1? Thanks in advance!

Good question.  The answer (as I understand it) is that if there were only a single type of permeable ion, then there would be only one equilibrium ratio, and it would depend on the transmembrane potential.
However, in real life, there (a) is usually more than one membrane permeable ion, and more importantly (b) the transmembrane potential is not a fully independent variable but is affected by the transmembrane permeation of all those ions.
Imagine sovling a single instance of the Nernst equation when you know the electrical potential $\Delta \psi$:

$\frac{\Delta G}{F}=\text{ion motive force}=0=-m \Delta \psi+ RT\ln{\frac{[X_{out}]}{[X_{in}]}}$

If you know $\Delta \psi$ (and $R$ and $T$) and you are at equilibrium (so $\Delta G$ is 0), then the ion concentration ratio will always go to the same value, as you suggested in your question.
But now suppose there are many equilibrating ions, say sodium, potassium, protons, and chloride.  And suppose you don't know $\Delta \psi$.  Now, there is a Nernst equation for each species, as well as Gauss's law that $\nabla \centerdot \bf{E} = \frac{\rho}{\epsilon}$.  And probably also total material balances like $V_{out}[X_{out}] + V_{in}[X_{in}] = N_{tot}$ for each ion.   To fully solve the system to determine equilibrium, all equations have to be solved at once!  I hope you can see that (a) doing this is complicated and (b) if parameters like the volume of the cell $V_{in}$ or even more likely the total amount $N_{tot}$ of ions in the system changes for one ion, it could completely change the ratio $\frac{[X_{out}]}{[X_{in}]}$ for other ions.  That's why in general, the ratio is not the same every time.

Even when there is only one permeable ion, my professors have led me to believe that the equilibrium potential for that one ion is not a constant. Also, the Goldman equation is used when there are multiple ions that are permeable.

The Goldman equation is a more general case that can be used when ion fluxes are not zero, i.e. the membrane is not at equilibrium.  For equilibrium calculations, the Nernst equation applies.

Looking at the wiki page for the Goldman equation, I now see that the membrane reaches its reversal potential (net current from all ions is 0) but doesn't necessarily mean that the ions themselves are at equilibrium. Can you see any way for there to be multiple equilibrium potentials for a specific ion when only that ion is permeable?


text:
Can artificial sea water be prepared by mixing sea salt with water?
I am looking for ways to easily prepare artificial sea water. The chemical composition doesn't have to exactly match that of natural sea water. On-line I can find recipes to prepare sea water. But I was wondering, can I just buy natural sea salt and dissolve it in fresh water (Until I get the required salinity ).
The commercial shrimp farms in my region, that use natural sea water cause enormous environmental damage. That is the primary motivation behind my interest in artificial sea water. I want to explore if artificial sea water can be a eco-friendly sustainable alternative.
My intention is not to breed shrimp, but just to grow the shrimp that I get from hatcheries. And yes, I am planning to get the help of people with relevant expertise in this. I just want to gain some prior knowledge and experience before I go for their help.
Some of my present concerns are similar to those of oceanarium though on a much much smaller scale :-) Like them I also want to prepare sea water, which I can control to my exact needs. Locally we get these "raw sea salt" used for industrial purposes, which are claimed to be very similar to natural salts found in sea. They do not contain iodine, caking agents or any other additives etc. I am guessing probably the sea salts that oceanarium imports might be same as these (probably of much better quality).

Well it seems you're not entirely unaware of the issues, so that's good, but I don't have enough knowledge to advise you on much more. All I can say is that the younger the shrimp start off, the harder it is to cultivate them properly, because different phases in their life require specialized conditions, including hormonal dosage. Shrimp farming is a fairly popular industry where I am, and from what I've been told it's no walk in the park. Try in small scales and work your way up, but be prepared to spend a lot of effort on getting it to work.

@Jason If you preserve the cookies in your browser, you can ask questions as an unregistered user.  If you are uncomfortable with your Google "real name" being shown, you can use one of the other OpenID providers, which don't have that restriction.  That's a Google requirement, not one from Stack Exchange.

If shrimp farms cause environmental damage they do so because they return water to the sea. Making your own water won't help that: not returning to to the sea will. So don't try to make you own sea water, use the abundant source of sea water (the sea) and fix the problem with discharging back into the sea.

There is sea salt and then there is sea salt. What is bought as sea salt in a grocery store is almost entirely sodium chloride. So potassium and calcium are almost none existent in such salt.

Yes, it is possible though not easy to make artificial seawater by mixing sea salt and fresh water. An interesting example of this happens at the oceanarium in Lisbon, Portugal, which is one of the largest in Europe. After considerable investigation, it was found that the seawater present in massive quantities just outside the building could not be economically filtered, purified and adjusted to attend the surprisingly demanding and delicate needs of the marine organisms to be displayed. The chemical composition of seawater is actually quite complex, varying significantly with the location and time of collection. As with all life, marine life can be very sensitive to small changes in concentration of compounds, not to mention possible biological incompatibilities between the micro- and macrobiota present in seawater from different regions. The solution chosen for the Lisbon Oceanarium was to import some 16.5 metric tons of sea salt monthly from the Red Sea in Israel and mix it with local fresh water in order to replace the tank water. [1] [2] (Couldn't find a reference in English)
Having said that, you really should consult people experienced in setting up seawater aquariums before trying anything yourself. Seawater aquariums have significantly tougher upkeep and maintenance requirements compared to fresh water aquariums; you can't just mix any old cooking sea salt into water, it has to be in the right concentration, and replacement of the tank water has to be more gradual and frequent, not to mention that the salty water will corrode the actual aquarium more quickly. A good aquarium supplies store should be able to better inform you.
Edit: Now I realize you may actually be interested in cultivating shrimp rather than putting them on display. That's a whole different matter, many many times more difficult. You really need to contact some experts in marine cultivation.

Yes. If you do however, that would likely unsaturate the solution. I don't know if certain animals need certain amounts of salt, but doing this without adding salt might kill them.