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Diatomic Molecules/Elements
What is the difference between diatomic molecules and elements?
Also, If oxygen is element of a single atom, then what makes it a diatomic element? But, then I read that a single oxygen atom can bind with another element to form a compound, but it's a diatomic element with two atoms.
Please can someone enlighten some clarity on this subject. Also, it would help to see some examples to see the differences and context of how these diatomic molecules and elements interact.

How these elements interact? That's a little bit too broad. Like one's going to write more than thousands of books to cover it. Oxygen molecule is diatomic. That's all, and that means the most stable form of O in nature is $\ce{O2}$.

Anyways, welcome to chemistry.SE!

But, why is a single oxygen atom not an element, or is it? So, are you saying that a single oxygen atom will easily bind with another atom of a different element to become stable? And that two oxygen atoms are stable and not easily bind with other atoms of different elements?  -LAG

$\ce{O2}$ is thermodynamically way more stable than a single O atom. A single O is considered a radical. It's very reactive and will react as soon as it's "allowed" to. :)

So, all diatomic elements have the same characteristics as you described for oxygen. I am new to chemistry, and I don't want to assume. So, that's why all of the questions and restatements. Thanks.

Well, I don't wanna cause confusions. Take $\ce{N}$ for example. Nitrogen molecule is diatomic, and a free Nitrogen atom is very reactive. If you mean that by "characteristics" then yes.

But Nitrogen molecule is very different from Oxygen molecule. They have very different properties.

Diatomuc molecules means constituted by two atoms. The latter can be the same (element) or different. In cases,the diatomic molecule of an element can be its stable form in standard or common conditions

From the IUPAC Gold book ( https://goldbook.iupac.org/terms/view/C01022 )
Chemical Element
(1) A species of atoms; all atoms with the same number of protons in the atomic nucleus.
(2) A pure chemical substance composed of atoms with the same number of protons in the atomic nucleus. Sometimes this concept is called the elementary substance as distinct from the chemical element as defined under ($1$), but mostly the term chemical element is used for both concepts.

The form an element usually has is distinct from the fact it is an element
The key for a substance to be described as an element is that it contains only one type of atom. But that leaves a variety of possible structures even for one element.
Some elements–noble gases–consist of isolated atoms, all others consist of molecules or structures containing multiple atoms of the element. Oxygen mostly appears as a diatomic gas under normal conditions but sometimes appears a a triatomic gas (ozone). Some elements are usually seen as more complex molecules. Sulfur has a variety of allotropes (that's what different structures of pure elements are called) many containing rings of 8 sulfur atoms but others are known. Carbon has an interesting range of structures. One is a ball-shaped molecule of 60 carbons, graphite is a soft compound consisting of planes of hexagons weakly held together, diamond is a network of interlinked carbon tetrahedra and is very strong.
Elements, in other words, can be found as a variety of structures and many single elements can be seen with more than one structure. Some usually appear as single atoms, some as small molecules, some as large molecules and some as network solids. Some elements can be seen in multiple forms. All those forms are still elements. Don't confuse the form with the fact the substance is an element.

An element is the smallest entity that has it's own characteristics in Chemistry. Any two different elements differ fundamentally in respect to the number of protons in their nuclei. An element is always of single atom. It's the characteristic properties of that element that would decide in what form or kind of molecule it will be found in nature.
Diatomic element is a kind of molecule in which two atoms of the same or different elements have combined chemically to form a molecule. If we restrict ourselves to the context of elements, then diatomic element is one among various allotropes of the element.
Oxygen is found in nature as diatomic and triatomic (ozone). The reason we don't find monoatomic oxygen often around is that an atom of oxygen is highly reactive. You may have heard of nascent oxygen but it exist for very short time in some chemical reactions.  O, O2 and O3 are different allotropes of the same element oxygen. Besides these, there are some other allotropes of oxygen that are listed in the given reference.

You do not distinguish properly between a chemical element and it's atoms.

A chemical element is a substance consisting of atoms with the same particular proton number as its smallest particles. Even better is the Porphyrin's answer.


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Writing pressure equilibrium constant expression for a reaction with a single solid product
How would you write the $K_\mathrm p$ expression for the reaction: $$\ce{4 Ag(s) + O2(g) \rightarrow 2 Ag2O(s)}$$
I know normally it would be products over reactants, but there is a solid on the product side and they aren't included, so I don't know how you would write it?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website. Personally, I don't think you haven't tried solving this, so I don't think this question will be closed.

The expression $\mathrm{K_p}$ is an expression of the equilibrium of partial pressures. So, it just includes the the partial pressure of the oxygen because that's the only gas.
$$\mathrm{K_P} = \frac{1}{\ce{P_{O2}}}$$
More information about the equilibrium constant $\mathrm{K_p}$ can be found here.

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Are acids and bases always aqueous?
I am tying to determine if an acid is a liquid or aqueous solution? Please can some one enlighten me?

There are varying definitions of"acid" (or "base") depending on solvent system; some do not involve water at all. For example, there are Brønsted–Lowry acids (proton donors) and Lewis acids (electron acceptors), though the Arrhenius definition is based on aqueous solutions dissociating into $\ce{H+}$ (acid) or $\ce{OH-}$ (base) ions.
Of course, many acids are not liquids, such as solid citric acid, which may be anhydrous or the monohdrate.

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How to get a protein's dihedral angles ordered by variance?
The setting: I want to simulate protein docking and let some dihedral angles vary, but in order to keep it low-dimensional I have to select those which are most likely to change.
What I've thought of so far:

The easiest solution would be some tool/database that simply returns a list of dihedral angles ordered by their variance (something like http://www.ifm.liu.se/compchem/msi/doc/life/cerius46/qsar/theory_descriptors.html#515786, but that only returns the number of "meaningful" rotamers)
Take a set of different conformations and use principal component analysis on the phase space consisting of the dihedral angles.
Take Ramachandran plots and try to combine all those conditional variances (if φ is …, then ψ has some amount of freedom) into a more "holistic" statement.

The problems I encountered:

I couldn't find such a list.
Searching rcsb.org, I only found about ~100 conformations (for albumin), but if I'm not mistaken the format isn't the exactly the same, so it'd be kinda hard to compare. Besides, ~100 samples won't yield good results for >100 dimensions (=angles).
The only approach for which data seems to be available, but it just looks like a real pain.

If I understand you, I can't imagine how to solve the your problem with this approach. Also I don't get the physical meaning of the proposed solution.
If you have to deal with a molecule with more than 100 rotatable bonds, you are just in problems.
You could find the rotatable bonds with some tools (see for example OpenBabel). But the problem will just start.
At first, you can not just try one by one the $N$ angles ($\{\theta _i \}$) (by changing the angle value $n$ times and trying each of them), because for exploring all the conformational space you will just have $n^N$ points. And you just can't know a priori which are the most important angles (in case (I think not) that effectively exist the most important angles).
Also note that not every point ($\theta _1, \theta _2, \cdots,\theta _N)$ is acceptable, just because parts of the molecule can superpose with other parts.

"Take a set of different conformations and use principal component analysis on the phase space consisting of the dihedral angles."

I think that this is not possible. The values of the rotatable angles will be just chosen by you.
Even more, suppose that you have two rotatable bonds very close each other, in a linear part of the molecule. If not a large superposition ocurrs, you will have no diference in the efect of them. But the effect of them will be very correlated. So, you can get useful info. with PCA.
A possible approach
As I know very few of biology, I would make a research about if there are prefered configuration, that are much more common than others (because of biological facts).
If you are forced to work with the entire configurational space, the computational approach that comes to my mind is:

Use very fast method for energy computations that allow you set the connectivity of the atoms. This will prevent changes of the covalent bonds due to superpositions.
Use the Monte Carlo method, this should explore the configurational space giving you a reasonable idea of this space.

Final note
I find extrange doing docking with a large protein and ¿without a ligand? If there is a ligand for binding to the protein that we are talking about, I won't take too much care of the protein. The scoring functions or energies are few accurate, I find hard to believe that just one found configuration would represent accurately the problem (considering the large degeneracies involved)

First of all, thank you very much for the extensive answer!
—3rd paragraph: That's what I meant with "conditional probabilities" for the 3rd point, like "assuming all angles except $\Theta_i$ fixed, $\Theta_i$ can take $x$% of all possible angles without creating some overlapping" (this can be read off Ramachandran plots)—however, the task of combining all those conditional probabilities into a single one looks pretty tedious
—PCA: I wanted to feed it with real world data, not randomly sampled conformations which may be invalid. I'm not sure why there's a problem with correlated atoms.

—possible approach: Do you happen to know where I can find a list of preferred configurations or should I try to construct them myself by ranking Monte Carlo generated conformations?
—final note: To be honest, I have no idea about the chemistry; I'm doing this for a mathematics seminar—I thought protein-protein docking would be the way to go, so that's why I want to let at least one protein stay somewhat flexible.

a) "assuming all angles except..." . As the overlapping depends on many angles, you should do, for each $\theta _i$, $n^{N-1}$ configurations.
b) I bet that the data is not available. About correlated bonds: Suppose that here is a contiguous rotatable bonds. A rotation about one of them will do an almost equal effect on the "surface" of the molecule (where the dispersion forces are important for binding). So, I will show a similar importance in PCA (so, even if it is would be possible, PCA can't give you useful info.). Now, imagine that there are 2 rotatable bonds not too close each other ...

For each of them, you could get $n^{N-1}$ values.  A distribution of values for each bond, that will then to be very close to the other one. Can PCA detect this? I bet no, just because both angles are almost equally important.
c) "or should I try to construct them myself by ranking Monte Carlo generated conformations?" Yes!
d) "I thought protein-protein docking would be the way to go" I would be difficult to predict protein-protein binding omiting solvent (water) effect, for the energy variations and specially for entropy changes. Ligand protein would be much more simple and physical meaning

I would be interesting to hear a biochemistry (or similar) opinon

Ok, I guess I will limit the project's scope to protein-ligand docking then. Thank you very much for taking your time!


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Is Bromide anion acting as a base in reaction with ethylamine and methylbromide?

What is happening in Step 2 and would you please give the reaction mechanism for it as well?

The bromide anion is acting as a counter anion to balance out the ammonium charge in step 1.
Given sufficient heat, step 2 can occur, with bromide acting as a weak base.  This is likely to produce gaseous produces which would drive the equilibrium as the gasses escape.
Mechanistically, this is possible, but synthetically, it is more likely another base would be added in a typical synthesis.


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Are these reaction mechanisms correct?

Are these reaction mechanisms correct for Ammonia and Bromide anions acting as a base respectively?

Can't say if its correct, but it sounds ok

I haven't seen the 'dative' arrow used as you have to show a bond between nitrogen and hydrogen (top right) or nitrogen and methyl (bottom right). A regular line without the lone pair on nitrogen is more conventional.

If you use dative bond arrows, which are a representation of an ionic lewis structure, the charge should be located at the hydrogen in the first reaction and at the carbon in the second reaction. In the second reaction, the product should not have a dative bond any more, as the molecule changed with the abstraction of the proton.

Except for the missing equilibrium arrow (these reaction are acid-base equilibria), they are fine. Or do you have any objections :) (what makes you doubt) ?
Also take @jerepierre's comment in account. This would probably be considered wrong if at an exam.

Well i wasnt sure if this is the reaction mechanism of a lewis base

How do you check if mechanism is correct/ realistic? Just adding up? Isn't it the less reactant the more likely it's gonna happen?

Could you please elaborate more?

authors: أميره الزهراني


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Are molecules chiral if a rotation leads back to the same compound?
I understand that if you rotate a molecule and it has the same configuration as its enantiomer, then it is indeed the same compound as the enantiomer (meso), and hence is achiral (despite having chirality centers).
If compound is rotated and the resulting Fisher projection has exactly the same configuration after its rotation, what does that mean in terms of chirality? (I'm not talking about enantiomers here,) It will have a different enantiomer (and hence not be meso), and will be polarizable, but the compound itself has exactly the same perspective if you rotate it. Does this hold any signifance chemically or stereochemically, or is it meaningless?
Take for example D-idaric acid (2S,3R,4R,5S)-2,3,4,5-tetrahydroxyhexanedioic acid).
In the below example of Idaric acid, the enantiomer appears to be a different compound. When the compound is rotated 180 degrees, it looks exactly the same.

In the below example of galactaric acid, when the compound is rotated 180 degrees, it is a different image. This is despite it being meso.

Obviously, being meso has a lot of chemical significance (no polarization of light). But does a compound possessing the property of being rotated 180 degrees, and having the same image (in Fischer form) present any chemical significance compared to a compound that doesn't do this?

Be careful when manipulating Fischer projections. You are allowed to rotate around an axis perpendicular to the page (described as 'rotate 180 deg), but you can't rotate around an axis in the plane of the page (described as 'vertical flip'). I believe what you're doing is really drawing the mirror image, which is not the same thing. Remember that the Fischer projection implies that all of the horizontal bonds are coming out of the plane and the vertical bonds are going into the plane. The vertical flip would place the horizontal bonds into the plane.

I didn't think of that jerepierre, good point, thanks.

Related: What is the perfect definition for chirality?

The property that you are describing for D-idaric acid is called C2 symmetry. This means that when you rotate a molecule (or any object) by 180 degrees, it appears identically to the original orientation. In D-idaric acid, the C2 symmetry axis is halfway along the bond between C3 and C4.
Having a C2 symmetry axis doesn't tell us if the compound is chiral or not. Achiral compounds have an (internal) plane of symmetry or a point of symmetry. D-Idaric acid does not meet either of these requirements and is a chiral molecule.
Galactaric acid does not have a C2 symmetry axis but does have a plane of symmetry. Galactaric acid is achiral. Since galactaric acid has stereocenters but is not chiral, we describe it as meso.

I'll add a couple of examples of the chemical utility of C2 compounds when I get a chance later.

Just the answer I'm looking for thank you. When you mention both D-Idaric acid and Galactaric acid having C2 symmetry, I notice Galactaric acid isn't the same when rotated 180 degrees?

@user4779 Sorry, that wasn't correct. Galactaric acid is not C2 symmetric.

You can have "symmetric" chiral molecules like you describe.
This is a subtle point and your particular choice isn't chiral.
In typical organic chemistry, instructors mention about mirror images, but leave some of the subtle details to later courses.
There are many types of chiral molecules and some of them look quite symmetric. Consider a pinwheel or propeller. You can rotate along the axis of the propeller and it will be symmetric, but there's a tilt/twist to the blades:

So these two images are mirror, but there's a chiral axis and the direction of rotation is reversed. They are enantiomers. This case has three-way symmetry, but it's possible to have a "two leaf" pinwheel that's still chiral. Proper rotations don't make mirror images.
Your case, D-idaric acid, is C2 symmetric.


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How to find density of element metals
I was wondering, how do I determine what metal (element) has the highest density by using the periodic table? Is it possible?

You look it up. Chemistry is empirical. Theory often fails. That's why periodic tables often have the relevant numbers on the table.

One way you can do this is by looking at the packing structure of the metal.
As an example, if you look at Wikipedia, you see that Tungsten has a body-centered cubic crystal structure. This means that in each unit cell, there are going to be two Tungsten atoms. We can then predict the density of a perfect Tungsten crystal lattice using some geometry and unit conversion.
First off, I'll give you an equation which you can prove to yourself quite easily so I won't go into that. Density of a crystal is: $$\rho=\frac{n*M}{N_A*V}$$
Where, $n$ is the number of atoms in the unit cell, $M$ is the molar mass of the atom, $N_A$ is Avogadro's number, $V$ is the volume of the unit cell.
So, for Tungsten this comes out to be $$\rho=\frac{2*183.83 g*mol^{-1}}{6.022*10^{23}*(\frac{4*139*10^{-10}cm}{\sqrt{3}})^3}=18.45\frac{g}{cm^3}$$
The experimental density of Tungsten is $19.33 \frac{g}{cm^3}$.
The answer is usually a little better than that, but still quite close.
The only information you need to do this calculation which isn't on a periodic table is the packing structure and the atomic radius.
Something which is noteworthy is the atomic-packing-factor, $APF$, which comes from finding the ratio of volume of the atoms to volume of the unit cell and represents how much space the atoms fill in the cube, or how efficient the structure is at packing.
For the body-centered cubic (BCC), $$APF=\frac{2*\frac{4}{3}\pi r^3}{(\frac{4r}{\sqrt{3}})^3}=0.68$$
That means BCC, takes op 68% of the total available space per unit cell for equal sized spheres.
Check out this link if you want more info on that.
So, to answer the actual question which how do we find a trend with all this, we now know that density depends on radius, which we already have a trend for, molar mass, which also has a very simple trend, and packing structure, which is the real unknown.
There is this from this page,

In the resonating valence bond theory, the factors that determine the choice of one from among alternative crystal structures of a metal or intermetallic compound revolve around the energy of resonance of bonds among interatomic positions. It is clear that some modes of resonance would make larger contributions (be more mechanically stable than others), and that in particular a simple ratio of number of bonds to number of positions would be exceptional. The resulting principle is that a special stability is associated with the simplest ratios or "bond numbers": 1/2, 1/3, 2/3, 1/4, 3/4, etc. The choice of structure and the value of the axial ratio (which determines the relative bond lengths) are thus a result of the effort of an atom to use its valency in the formation of stable bonds with simple fractional bond numbers.
  which I don't actually understand but seems to explain why certain lattices are chosen.

Basically, using the fact that radius decreases going right and molecular weight increases going right, we would predict that density would increase uniformly across the periodic table for elemental metals, except that various metals pack in different ways. Hexagonal Close Packed is the most efficient packing system, so I wouldn't be surprised to find that being associated with many high density metals.
I hope that gives a good idea of how there is sort of a trend, but also why no trend is truly there.
EDIT:
To figure out which has the highest density, I would start by figuring out which pack in a Hexagonal Close-Packed structure as that is the most efficient packing structure with an $APF$=.74

There are two most efficient packing structures: HCP and FCC (face-centered cubic). They have identical packing factor.


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Relating change in internal energy and to that in enthalpy
The question is

For which exothermic reaction is $\Delta\,E$ more negative than
  $\Delta\,H$?
(A) $\ce{Br2(l) <=> Br2(g)}$
(B) $\ce{2C(s) + O2(g) r 2CO(g)}$
(C) $\ce{H2(g) + F2(g) -> 2HF(g)}$
(D) $\ce{2SO2(g) + O2(g) r 2SO2(g)}$

r is supposed to be reaction? Either a one headed or equilibrium arrows. The "correct" answer is supposed to be D.
However, using the formula $H = U + PV$ to $\Delta H = \Delta U + \Delta PV$, I found that $\Delta U$ would be less if $\Delta V$ was positive.
Thus, more gaseous products should be produced. As a result, I chose B.
Note, A is not exothermic, so would not be considered.
Could anyone confirm my answer or offer an explanation for my answer being wrong?

The last reaction should have $\ce {SO3}$ as the product

You are right. B is the correct choice.
$$\Delta H = \Delta E + \Delta PV$$
Since $\Delta E $ is more negative than $\Delta H$, $$\Delta H > \Delta E$$.
Hence $$\Delta PV>0$$
$\Delta E \space $has to be negative. So option A is ruled out.
Among the other choices, B is the only one where number of gaseous moles are increasing from left to right, leading to a positive value of $\Delta V$ and hence $\Delta PV$.


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why does phosphorus makes 5 bonds with oxygen in P4O10 whereas 3 bonds in P4O6?
how phosphorus forms 2 types of bonds with oxygen i.e. 5 in $\ce{P4O10}$ and 3 in $\ce{P4O6}$ when only three are needed to complete its octet?

Did you consider the different oxidation states of phosphorus in both oxides?

Welcome to chemistry.SE! I wouldn't rely too much on octet if I were you, especially when it's about phosphorus. Your question won't make sense if we not rely on octet, which is the case about group 15.

@Ϻ.Λ.Ʀ.  -- too bad there's no such simple reply for Cu(OH)2. I mean, as simple as Chinmay's.

Phosphorus has electronic configuration of [Ne]3S2 3P3 As it is a 3rd period element it also have a vacant 3D orbital.
In case of P4O6 only the 3P3 electrons take part. Resulting in formation of 3 bonds with O. But in case of P4O10 one of the electron of 3S is excited to vacant 3D orbital. Now P has 5 unpaired electron and it can form 5 bonds. Therefore P forms 5 in P4O10.
Now, about the octet. Not every compound follow the octet rule. Just to understand The whole chemistry depends upon STABILITY i.e lesser energy of molecule. Therefore, if by forming more bonds (and does not obeying Octet rule) the molecule is gaining more stability, it will do so.

The shape of the 4 Os around P is tetrahedral . Also, according to JD Lee's Concise Inorganic Chemistry, the double bond P=O in P4O10 contains p(pi)-d(pi) backbonding, where a full p-orbital on oxygen overlaps with an empty d-orbital on P ... this seems to point towards an sp3 arrangement with one electron pair from P being "given" to an empty p orbital on O, otherwise how will O attain it's octet ..?


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How to remove nitrate from water sample?
Are there any simple methods to completely remove nitrate ions from few ml of impure water sample?
One method I am curious about is the following process:

Add one or more reagents to water in test tube, which would crystallize nitrate ions into some solids
Then filter out the solids from water such that the water is now nitrate free.

Is the above process possible? If so, what are some reagents that can be used?

Precipitation won't work
Nitrate removal is a very tough problem in wastewater treatment.  One of the reasons is that nitrate salts of nearly every common cation are highly water soluble.  Unfortunately, I can't think of a way to precipitate nitrate from your water sample.
How commercial nitrate removal processes work
Nitrate is removed from wastewater through a microbiological process called denitrification.  In this process, organic carbon substrates are used as food by special anaerobic bacteria.  In the absence of oxygen, the bacteria use nitrate as an electron acceptor so they can oxidize their food and obtain energy via respiration.  As an example organic food molecule, consider hexose sugar.  Then the reaction could be written as:
$$\ce{5 C6H12O6 + 24NO3- -> 30CO2 + 24 OH- + 18H2O + 12N2} $$
It will be difficult to purify a few mL of water in this way unless you know exactly the BOD of the water and are an experienced microbiologist with access to cultures of the appropriate bacteria.
Another not very good option
Another possibility would be electrochemical removal of the nitrate.  That is a topic which has been researched but which is not practices commercially to any significant extent.  Here is one paper where you can read about the technique.  I don't think it would be very easy for a home chemist either.
The tried and true
That leaves more tried and true methods.  You've simply got to distill your water!  Boil it and collect and condense the steam and you'll have substantially purified water to behold and enjoy!

Thank for the info. Looks like nitrate detection and removal are quite complex tasks for someone like me, without the relevant technical expertise.  I feel i should be looking for commercial solutions.  If possible can you please suggest where i can get info about the available commercial solutions.  Should i  post a new question on this forum, with my exact requirements? Is there anyother better place on StackExchange for that question ?   Thank you for your time and effort.


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How do tie-line slopes affect raffinate/extract envelope in ternary phase diagrams?
I am struggling to understand ternary phase diagrams for reactor engineering. The course slides offer little to no help so I'm hoping to find an answer to this question

How does the slope of the tie-lines affect on which side of the
  equilibrium line the extract/raffinate is?

From what I understand is that if the lines slope down, the extract is on the right and the raffinate is on the left. If they slope up, the sides are flipped. Is this correct?
Sidenote: does anyone know a good resource where ternary diagrams for chemical/reactor engineering are well explained?

I can't really comment on the extract/raffinate aspects - that is more chemical engineering that I never did.  However, in metallurgical phase diagrams (binary, ternary, ...) the only point is that the line (plane in ternary) touches the axes at the free energy value that corresponds to the equilibrium value.  BUT - you are free to reference that free energy to anything you want to. Using absolute (SGTE), relative phases, or whatever, you will end up with the exact same diagram, but different tie line slopes.

It has been 7 years for this question, but it may help others with the same doubt.
First of all, ternary phase diagrams are not related to reactor design, they are related to the field of liquid-liquid extraction. This is, we have a liquid flow of two components, but only one is of interest, so we try to obtain an exiting flow that is enriched in the component of interest. We do this by adding a third component, the solvent, with the hope that the resulting liquid-liquid equilibria behaviour will help us.
The tie-line is the line (it may be a curve) that represents the equilibrium compositions of the two liquid phases, that given enough time, will be formed by an an initial solution that lies along the tie line. The nature of the tie-line solely depends on the behaviour of the equilibrium established by the three componenents. Its "slope" is a result of this behaviour, and it has nothing to do with respect to on which side you are going to end.
Lets give an example. Say we have a mixture of components $A$ and $C$, but we want $A$. Mixing both with $B$, the solvent, results in the emergence of liquid-liquid equilibria, delimitated by the following dome:

If we mix an adequate molar flow of $F$ (containing $A$ and $C$) and of $S$ (containing the three components, but very rich in the extracting solvent $B$), the result is the mixture $M_1$. Given enough time, we will end up with two molar flows, in equilibrium with each other and of the following characteristics:

An extract $E_1$, that we generally don't care, with a high composition of $B$ and then of $C$.
A raffinate $R_1$, with a high composition of $A$, that we really do care.

The slope of the tie-line depends, as I said, in the equilibrium established. I will on purpose increase the slope of that tie-line:

All the tie-lines should change in reality, but for illustrating purposes, I just changed the one that we are interested in. As we can see, the equilibrium is much more favorable to us now. The raffinate $ R_1 $ has a much more pure composition in $A$ than before.  Did the slope send $ A $ to the right and $ B$ to the left? No, they stay where they were, they don't get flipped.
For more details in LL extraction, you can check Equilibrium-Stage Separation Operations in Chemical Engineer by Henley & Seader. However, Mass Transfer by Treybal is also fine, and a classic.

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Can I make a pH 7 buffer from phosphoric acid and sodium hydroxide?
I am an electronics guy with not so much knowledge of chemistry.
I would like to prepare a pH 7 buffer to test pH meters.
All web references I have found so far describing how to make a phosphate buffer use
disodium phosphate (Na2HPO4) and Monosodium phosphate (NaH2PO4) as ingredients.
I have a good supply of 85% phosphoric acid and solid sodium hydroxide. Can I make a pH 7 buffer from these ingredients instead or is there a catch?

Generally you'd get a standard solution to calibrate your meter.

You should be able to do this with little trouble.  The only caveats:

Make a stock solution of phosphoric acid and also of NaOH.  When you add them to each other to make your buffer, heating could be a problem, much moreso than if you were using the mono- and di-basic sodium phosphate salts.  So you may want to cool on ice when mixing these compounds.
You can calculate approximately how much NaOH and phosphoric acid to mix in order to get a pH 7 result, but it's best to use a pH meter to verify that you've reached pH 7.


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Trouble understanding 5 orientations of d orbitals
I understand $d_{xy}$ and $d_{yz}$ and those orientations but ones like $d_{z^2}$ are very strange to me.
Can someone please explain that to me?

Does this help?

Welcome to chemistry.SE! Yes. I wanted to add @Klaus's link too. Does that solve your problem?

Generally speaking, and very informally (I'd get shredded if I put this an answer), you want all 3 planes and all 3 axes to have a double dumbbell in them.  This is not possible with 5 orbitals,  3 orbitals each put two dumbbells in the planes of coordinate system (xy,yz,xz). The remaining two have 4 dumbbells occupy 3 axes.  So 1 of the dumbbells 'folds up' into the donut shape.  This is just how the math works out.

And even shorter than Lighthart’s description: This is just how the maths works out.


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Why are enynes named as -en-yne and not -yn-ene?

The preferred IUPAC name for the compound above is hept-1-en-6-yne. I understand that the numbers are chosen in order to give the alkene the lower locant, as explained in Should double or triple bonds have preference for low locants in IUPAC nomenclature?.
However, I'm wondering why the name has the form -en-yne instead of -yn-ene. In other words, why is it not "hept-6-yn-1-ene" instead?
Furthermore, if this is an IUPAC specification, is there any rationale for why en-yne should be preferred to yn-ene?

Anecdotally, some nomenclature decisions were made to make names easier to pronounce. enyne is easier than ynene.

The name is given according to alphabetical order like in the cases of prefixes, bromo preceds chloro, similarly, -ene preceds -yne. (E>Y) .

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.
  (…)

Since, in names, the ending ‘ene’ always precedes ‘yne’ (with elision of the final letter ‘e’ in ‘ene’), the given compound is a ‘hept-x-en-y-yne’.
Since the lowest sets of locants in ‘hept-1-en-6-yne’ and ‘hept-6-en-1-yne’ are equivalent (1, 6), preference for low locants is given to the double bond. Therefore, locant 1 is assigned to the double bond and locant 6 is assigned to the triple bond.
Thus, the correct name is hept-1-en-6-yne.

thank's for your thoughtful and helpful answer. Certainly, I understand that this is the IUPAC Blue Book states that this would be the correct ordering, but I'm mostly interested in why that is the case. Specifically, you've highlighted a line "...The ending 'ene' always precedes 'yne'..." but why is this the case?

There has to be a rule and this happens to be it.  Some choices are necessary but arbitrary: e.g. which side of the road to drive on.  All countries have made a choice but it is a pity that they did not all make the same one.


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Analogies of azide?
Azide is a fairly common ion, but why aren’t there similar versions such as $\ce{P3-}$ or $\ce{B3+}$? Do they exist, is there something wrong with these ions as to make them more likely to just break apart rather than for ionic compounds.

Compare the covalent radii of $\ce{N}$ ($75\,\mathrm{pm}$) and $\ce{P}$ ($106\,\mathrm{pm}$).
Phosphorus is a bit on the chubby side. Larger atom-to-atom distance means less overlap of p-orbitals in multiple bonds: $\ce{P=P}$ and $\ce{P#P}$ bonds are much weaker than similar bonds between nitrogen atoms.

Yet boron is still fairly small, barely 7 pm larger, or is this significant in atoms?

Boron is electron deficient so while p orbital overlap is okay it would rather form clusters than linear catenated cations.You can make molecular compounds with boron-boron double and triple bonds but they are generally really unstable.

Nitrogen is special in that its pi bonds are stronger than its sigma bonds. This explains why organic compounds containing nitrogen connected with only single bonds or non-resonant double bonds (especially -N=N-) tend to be unstable.
Almost every other element has stronger sigma than pi bonds so they prefer single bonds. Azide analogues will most likely dissociate and/or polymerise.

How could one determine that this statement: 'Nitrogen is special in that its pi bonds are stronger than its sigma bonds. '  is true?  Can one make a nitrogen-nitrogen pi-bond in the absence of a nitrogen sigma bond and then do a thermochemical measurement?


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Amorphous to crystalline: Opposite trend?
Well my book says that on heating- amorphous solids crystalline at some temperature. As far as I know , amorphous solids have a very disordered structure compared to crystalline solids and on further heating the level of disorder should increase. So why do they become more ordered or crystalline at high temperatures or am I missing something fundamental in here?

What they say in your book is right. Take the case of polyethylene terephthalate (PET) which is a kind of polyester used in water bottles for example. It's an amorphous polymer (remember the bottles are transparent). When it is heated, the strained chains have the possibility to move and reorder themselves and the polymer becomes partially crystalline. This fact can readily be proved experimentally by heating a piece of PET (over its glass transition temperature, around $74 ^0\mathrm{C}$), then quenching the sample. It becomes white and not transparent which means it's partially crystalline.
We can monitor this transformation using a specific instrument called differential scanning calorimeter, where we can notice the formation of an exothermic peak corresponding to crystallization.

But why does this happen? Isn't it violating thermodynamic principles?

Of course not. You can imagine the polymer chains entangled and frozen in an amorphous state. The role of heating is to liberate these entangled chains, so some  parts of chains are arranged in regular order, these regions are called crystalline regions. In between these ordered regions chains are randomly oriented. I agree this is weird. But this  property is specific of long chains polymers


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Predicting orbital angular momentum effects on magnetic moments
For metal complexes with $A_{2}$ or $E$ ground state terms there is angular momentum contribution to the magnetic moment which is generally positive for more than half-filled subshells and negative otherwise, and related to the spin-only magnetic moment by
$$m_\mathrm{eff} = m_\mathrm{SO}\left(1-\frac{\alpha\lambda}{\Delta}\right)$$
where $\Delta$ is the ligand field splitting parameter, $\alpha$ is the orbital angular momentum degeneracy of the state and $\lambda$ is a constant.
My questions:

When do exceptions to this behaviour occur ? For example I have found that $\ce{Co(bipy)_{3}(ClO4)}$ has a lower experimental magnetic moment than the spin only formula predicts despite being a $\ce{Co(I)}$ (d8) compound with a $^{3}\!A_{2g}$ ground state, but I do not understand why.
Is the deviation from the spin only formula always positive for $T$ states ? If not can we predict its sign similarly ?
EDIT: a simple way to do this is via Hund's third rule. If the d-orbitals are half or more full, the highest J state will be lowest in energy, and then the deviation for the T state will be positive ! Exactly why is probably rather subtle.

Just one little thing: your experimental data may or may not be good. It happens with published, cited data, too. There are several factors (impurities, anisotropy, mainly) that can influence measure values. Anisotropy is especially important in materials with large angular momentum...

When do exceptions to this behaviour occur ? For example I have found that $\ce{Co(bipy)_{3}(ClO4)}$ has a lower experimental magnetic moment than the spin only formula predicts despite being a $\ce{Co(I)}$ (d8) compound with a $^{3}\!A_{2g}$ ground state, but I do not understand why.

According to ADVANCES IN INORGANIC CHEMISTRY AND RADIOCHEMISTRY, Volume 12, page 193 "The tris(bipyridal) cobalt(I) cation has a magnetic momment of 3.2-3.4 B.M. which is independent of temperature from 70K to 340K".
Since there are two unpaired electrons, the spin only value is $\sqrt{2(2+2)} =2.83$.  So your experimental data are inconsistent with the literature, as 3.2-3.4 is greater than the spin only value, 2.83.

Is the deviation from the spin only formula always positive for $T$ states ? If not can we predict its sign similarly ?

No, the deviation from the spin only value is not always positive for complexes with T ground states. The main characteristic of complexes with T ground states is that the magnetic moment is temperature dependent, because the spin-orbital coupling splits the T state into states that differ in energy by an amount close to kT.
For example $\ce{Cs2VCl6}$ has a magnetic moment of 1.4 at 80K and 1.8 at 300K, with the spin only formula predicting 1.73.  So deviation can be negative, and isn't necessarily positive or negative even for the same compound, depending upon temperature.
See section 4.3 of Dr. Mark D. Spicer's magnetism lecture for more information.

My value for the magnetic moment is 2.05 BM (source unpublished, 2005) - is there a modern source corroborating your higher value ?

@J.LS Not exactly, J. Chem. Soc., Dalton Trans., 1997, Pages 519–530 cites to 2 references that have values of 2.53 and 2.89, but these are not really more modern, even though the citing article is.

Why might you get a value lower than the spin only magnetic moment in this case ? Is this a TIP effect ?

@J.LS I don't really have a good answer for you.  The value 2.05 is significantly lower than the 3 indepedent published values (2.53, 2.89 and 3.2-3.4) as well as the value of 2.88 for the isoelectronic Ni(II) complex.  There is discussion in the Dalton Trans. reference of whether the complex is really Co(II) with a bipy being reduced, but the magnetic moment values are said to rule this out.  With a different magnetic moment value, perhaps a different conclusion would be reached.


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Relative stability of keto- enol form of acetyl chloride
Why is the keto form of acetyl chloride more stable even though the enol form is stabilized by the hydrogen bonding?

The enol form of acetyl chloride isn't stabilized by hydrogen bonding. Can you include a picture of what you mean?

The enol form is stabilised more than or comparable to ketones in phenol and 1,3 or 2,4 type diketones.

The stability of keto-enol compounds depends on the solution they are present in.
Keto compounds are stabilized due to inter-molecular hydrogen bonding when present in a polar solvent (for example water).
Enol compounds are stable when present in non-polar solvents, in which intra-molecular hydrogen bonding stabilizes them.


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Equilibrium Constant and Surface Area
At equilibrium, the rates of the forward and backward reaction become equal. In a reaction
A(s) giving B(g) + C(g)
Increasing the surface area of A increases the rate of the forward reaction, but the backward reaction is between two gases and is unaffected. By the time the rate of the backward reaction becomes equal to that of the forward reaction, more of A must have converted into B as compared to how much it did when there was less surface area. Since the activity of a pure solid is unity, it doesn't appear in the expression for K. So K must have increased. The equilibrium constant on the other hand depends on the ∆G standard for a reaction. The standard ∆G is unchanged, so K must be unchanged. How does one resolve these two?

In a reaction A(s) giving B(g) + C(g) Increasing the surface area of A increases the rate of the forward reaction, but the backward reaction is between two gases and is unaffected.

This is an assumption that you're making and it is thermodynamically and geometrically questionable.  If increasing the surface area of $\ce{A}$ increases the rate at which it sublimes/gasifies to form $\ce{B + C}$, it means that the reactive molecules of $\ce{A}$ are only those at the surface, in which case it is misleading to denote the reactant as $\ce{A(s)}$.  It is really $\ce{A(surface)}$ which is reacting.  Similarly, if the rate of the reverse reaction is not really dependent on the surface area and surface geometry of $\ce{A(s)}$, then it implies that very quickly molecules of $\ce{B}$ and $\ce{C}$ would condense to form a nanocrystallite of $\ce{A}$ that is one molecule big.  These nanocrystallites of $\ce{A}$ would have, collectively, nearly infinite surface area.  So the backward reaction creates surface area under the assumptions you have made.  You have to account for that created area.
Solid-phase reactions are complicated by phenomena such as crystal nucleation and other mechanisms of crystal growth.  Treating solids as uniformly reactive materials with unit activity is inappropriate for understanding such phenomena.  Another way to say it is that the assumption that solids have unit activity and don't appear in expressions for $K$ or $\Delta G$ is an assumption, one that is inapplicable when dealing with reactions whose rates depend on surface area.

I always thought solids reacted at surfaces - isn't that true?

It is mostly true.  But to account for that fact kinetically, you have to explicitly model surface sites.  Instead of A(s) -> B(g) + C(g), you could write something like A(surf) -> B(g) + C(g) + A*, where A* is an empty site on the surface of A.  Now, the back reaction rate depends on the "amount" of A*, i.e., on the availability of those surface sites.


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Order of lewis basic strength
My teacher told me that order of lewis basic strength is inversely proportional to electro negativity. Comparing between iodide ion and chloride ion, chloride ion is more electronegative than iodide ion therefore iodide ion should be more basic than chloride ion but the order is reverse that is chloride ion is more basic than iodide ion. Why this is so? Is it related to localised and delocalize lone pair?

My teacher told me that order of lewis basic strength is inversely proportional to electro negativity.

I agree with your teacher's guideline. Now, note that this cannot be your sole guideline. Lewis basicity (and acidicty) depend on a number of factors including:

Strength of the corresponding Lewis acid or Lewis base and the strength of the resulting bonds in the Lewis adduct.
Solvent effects
Geometry of the interacting species
Electronegativity

Comparing between iodide ion and chloride ion, chloride ion is more electronegative than iodide ion therefore iodide ion should be more basic than chloride ion but the order is reverse that is chloride ion is more basic than iodide ion.

I think you are confusing Bronsted basicity with Lewis basicity. It is true that in water (a polar, protic solvent) chloride ion is a stronger Bronsted base than iodide ion.
In terms of Lewis basicity, however, things become a lot more complicated. Iodide ion has the more readily available lone pair electrons for donation since iodide ion is less electronegative than chloride ion.
On the other hand, the strength of bonds to iodine are generally not as strong as bonds with chloride ion. This is a result of poor orbital overlap between iodide ion - which is a big atom - much bigger than many Lewis acids.

I asked my teacher today the same question and he told that since both iodide and chloride ion have delocalize lone pair both iodide and chloride ion have delocalize lone pair in their own vacant orbital therefore the electron donating tendency decreases. here we can't compare using electro negativity. And comparing them since iodide ion has more big size than chloride ion therefore the delocalization tendency is better in iodine therefore chloride ion is better lewis base. Any comments?

I mostly agree; both have electrons in their orbitals but I'm not a fan of the term "delocalization" - that implies that the electrons are spread across multiple nuclei - but in chloride ion and iodide ion there's only one nucleus.
Again, there are multiple factors to consider - not just electronegativity.

Thanks for answer. I asked my teacher today the same question and he told that since both iodide and chloride ion have delocalize lone pair in their own vacant orbital therefore the electron donating tendency decreases. here we can't compare using electro negativity. And comparing them since iodide ion has more big size than chloride ion therefore the delocalization tendency is better in iodine therefore chloride ion is better lewis base. Any comments?

@Vaibhav - I've edited my answer to address your concerns. I've also reconsidered my position on which ion is more basic in the Lewis sense ... so please consider re-reading the entire answer. I believe my reasoning was mistaken earlier.


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What are all the structural isomers of a tertiary amine with the sum formula C6H15N?
According to the solutions there exist 7 different structural isomers 'containing 3 degree amines' for the molecular formula  $\ce{C_6H_{15}N}$
I tried and made a few, here are those

I have been trying since an hour and all I made were the repetitions of these 5 only. Could anyone suggest me the remaining 2?

There is a methyl ethyl propyl amine and methyl ethyl isopropyl amine .

Welcome to chemistry.se! Find out about the site by taking the [tour] and about formatting in the [help]. If you get helpful answers, please consider accepting one.

Please note that this is just a visualization of the comment given by LDC3!
Here are the two missing isomers:

The drawings were made using BKChem under Linux. The program is in the repositories of all major Linux distributions, There also seem to be Windows and OSX ports at BKChem which I haven't tried. The program was initially written by Beda Kosata and is currently maintained by Reinis Danne. Sources are available from her repository on gitorious.


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Do sugar alcohols have the potential to get a person drunk?
From what I understand is ingesting high amounts of sugar alcohols can lead to various effects such as abdominal pain and diarrhea, my question is, is there a certain amount of sugar alcohol if ingested cause drunkenness (even if it causes a person to be in a critical state or death) ?

What do you mean by "sugar alcohol"?

Welcome to chemistry.SE! It's high time you and @matt_black take a look at the almighty Wikipedia. :P Oh, and a cross-site duplicate: http://biology.stackexchange.com/questions/13547/can-sugar-alcohols-and-glycerol-in-particular-cause-intoxication

@matt_black Here is a list of sugar alcohols: http://en.wikipedia.org/wiki/Sugar_alcohol#Common_sugar_alcohols

Sugar alcohols are only partially absorbed. It is very unlikely anyone will become intoxicated with sugar alcohols. Also, their effects are less likely to cause intoxications.


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How to determine limiting conductivity of specific ions?
Given a table with equivalent conductivity of certain electrolytes, how can I determine the conductivity of separate ions? I know that you can find the difference between for example $\ce{K+}$ and $\ce{Na+}$ if you know the limiting conductivity of $\ce{KOH}$ and $\ce{NaOH}$, but how do you calculate the actual value for those two ions?
EDIT:
I've found a different source for the table, my apologies. Let me rephrase my question. Many texts on electrolytic conductivity refer to a table containing the molar conductivity of individual ions, (2). What I'm wondering is, how are these values obtained? Can they be calculated with data from (1)?

Apologies, I picked that table from the first image I could find on google. I've updated my question now.

The Colorado State one was fine, it was just the "homework as a service" one that was a bit suspect.  Sometimes people spam sites like that on SE to get search engine recognition, so I tend to err on the side of caution.  Thanks for your attention to the question!


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Difference between intermediates and transition states
Can someone please explain what exactly the difference between an intermediate and a transition state is?
I understand that they are formed as part of the process in converting the reactants to products in a chemical reaction. Other than that, all I know is that usually a transition state exists very briefly (if at all) whereas an intermediate can be isolated. But why?

An intermediate is a short-lived unstable molecule in a reaction which is formed inbetween the reaction when reactants change into products.
Whereas, transition state is just the state before formation of new molecule(involves breaking of bonds of reactants and formation of new ones)
An intermediate differs from a transition state in that the intermediate has a discrete lifetime (be it a few nanoseconds or many days), whereas a transition state lasts for just one bond vibration cycle. Intermediates may be unstable molecules (in which case they are called reactive intermediates) or highly stable molecules.
The difference between them can be better described through the energy profile diagram.
Transition states are local energy maximums and have partial bonds. This might be one of the reasons why they cant be isolated as intermediates.

Throughout, I will be using definitions from a Journal of Chemical Education article$^1$ by Keith Laidler.
I'll start with describing an intermediate, as I think it is an easier concept to grasp. An intermediate, according to Laidler, is a short-lived compound formed during a composite reaction. This intermediate is a product of one elementary reaction and a reactant for another. An intermediate must be a minimum on the potential energy surface; this means it will have some finite lifetime (dependent on how deep this minimum is) and could potentially be isolated under the right conditions.
A transition state is a saddle point on the potential energy surface, meaning it is a maximum along the reaction coordinate and a minimum along all other coordinates (this is a slight simplification as it is possible, though rare, to have higher order saddle points where it is a maximum along more than one coordinate). Note that this definition only applies to an elementary reaction; a composite reaction doesn't have one overarching transition state. Laidler defines the broader category of transition species to refer to any structure that occurs while moving along the reaction coordinate between reactants and products. This is an important distinction; note that while a transition state is impossible to detect (it exists only for a very small range of atomic configurations and thus its concentration is always vanishingly small) transition species as whole can be detected as they exist, however fleetingly, in much higher concentration due to being defined for a wider range of atomic configurations. Laidler gives as an example the reaction $$\ce{F + Na2 ->F...Na...Na->NaF +Na^*}$$ where certain emissions surrounding the sodium D-line have been attributed to these transition species.
As some what of an aside, it is worth mention how potential energy extrema and saddle points are determined from a computational perspective. First, we find extrema and saddle points by determining regions where the force vector (first derivative of the energy with respect to all coordinates) is zero. Given the Hessian matrix for a molecule, which gives the second derivative of the energy with respect to all pairs of the 3N coordinates of the molecule (these might be normal modes or cartesian coordinates), one can determine whether the compound is an min/saddle point by examining the eigenvalues of the matrix. If all the eigenvalues are positive, this means the compound is a minimum; if exactly one eigenvalue is negative, it is a saddle point; if $n$ eigenvalues are negative, it is an $n^{th}$ order saddle point; if all eigenvalues are negative, it is a maximum (this is hardly ever the case for chemical systems).

Laidler, K. J. Just what is a transition state? J. Chem. Educ. 1988, 65 (6), 540. DOI: 10.1021/ed065p540.

A transition  state  cannot  be  isolated while  an  intermediate  can   be  isolated. A transition  state  is  a  chemical  species which  has only  fleeting  existence and represents an energy  maxima  on  reaction  coordination  diagram . While  an  intermediate  lies in  depression  on potential  energy  curve . Therefore  actual  lifetime  of  an intermediate  depends on the depth  of  the  depression. A shallow  depression  implies  a low  activation  energy for  subsequent  steps  and  therefore  short  lifetime.   While deeper  the  depression  longer is the  lifetime of intermediate.


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Why do Allylic Halides prefer SN2 reaction over SN1?
As explained by my instructor,

Allylic Halides form highly stable Transition States. They are stable because they are:

$sp^2$ hybridized
possess a conjugated system of pi bond and empty p orbital

However, if we consider the $S_N1$ mechanism, a carbocation intermediate is stabilized by resonance. Why is the formation of the transition state more favorable than the formation of the resonance stabilized intermediate?

You and your instructor are both correct, it is your premise that's wrong.
Your instructor's explanation about the allyl double bond stabilizing the $\mathrm{S_{N}2}$ transition state is correct. The following figure shows the stabilizing interaction.

image source
If we remove the attacking nucleophile and the leaving group and place a positive charge in the system, then we have the $\mathrm{S_{N}1}$ allyl intermediate.  As you point out, this intermediate is resonance stabilized, and therefore, the transition state leading to it is also stabilized.
Your premise,

Why do Allylic Halides prefer SN2 reaction over Sn1?

is not true.  There are many cases where allylic halides react preferentially by an $\mathrm{S_{N}1}$ process.
Since both the allylic $\mathrm{S_{N}1}$ and $\mathrm{S_{N}2}$ reactions are stabilized, there is a delicate balance between the two pathways.  We can shift from one mechanism to the other by changing reaction conditions.
For example, allyl chloride reacts by an $\mathrm{S_{N}1}$ mechanism to produce allyl alcohol when we place it in a 50:50 mix of $\ce{H2O:EtOH}$ at 45°C.  If instead we run the reaction in the presence of ethoxide ion in ethanol at 45°C, the corresponding ethyl ether is formed by an $\mathrm{S_{N}2}$ process.

Can you provide the condition when SN2 dominates SN1 among allyl halides?

@ZOZ See the last paragraph of the answer for an example.

im asking for condition...not example

solvent and temperature (the conditions) are included.

I think you did'nt get me.I saw the reaction conditions you gave for those specific reactions.Is there any check list (sort of) which you use to decide SN1 or SN2 in such situations (allylic halides) ?

It's hard to say, but generally, lower temperatures, less polar solvent and stronger nucleophile will favor SN2

Thank you!BTW if you have some time do see if you can answer this question (or leave a comment) http://chemistry.stackexchange.com/questions/59268/can-ces-n1-reaction-occur-in-compounds-with-leaving-group-attached-to-1-deg

What if the conditions include a lewis acid e.g. AlCl3 and in a water solvent? Will this favour an sn1 reaction by abstracting the halide before (poor)nucleophile attacks?


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Why tautomers are considered to be the same chemical compound?

Because of the rapid interconversion, tautomers are generally
  considered to be the same chemical compound. (Wikipedia)

But looking at the examples provided on that same Wikipedia article, it is obvious that different chemical reactions would take place in each case (for example enol vs. keto).

(source)

You're right, that's why I don't consider tautomers to be the same chemical compound.  When I buy a bottle of a acetone, I know both the keto and enol forms are in the bottle.  Maybe Wikipedia was saying that when I use acetone, even though I just call it acetone, both the keto and enol forms are present.

Hydronium hydroxide is far more reactive and polar than water, too.  Why do some people consider it the "same compound" as water?  No matter how much you purify water, you just can't seem to be rid of the hydronium hydroxide.

There are a number of types of rapid chemical equilibria that basically cannot be turned off. Tautomers are probably the best known examples, although fluxional behavior is observed in many inorganic and organometallic systems.
The origin for most tautomers is that protons are really quantum-mechanical. They can tunnel just like electrons can delocalize. Obviously different tautomers will have different stability, much like some electric resonance structures are clearly more stable.
The comments to your question above indicate that most chemists would just call it "acetone" but of course there's some prop-1-en-2-ol present too.
Yes, the reactivities might be different.
It's a partly a matter of the balance of the equilibrium (i.e., that for many tautomers, only one form predominates). There are exceptions, e.g.
phenols (from Wikipedia):

In certain aromatic compounds such as phenol, the enol is important due to the aromatic character of the enol but not the keto form. Melting the naphthalene derivative naphthalene-1,4-diol 2 at 200 °C results in a 2:1 mixture with the keto form.

It's also a matter of the time-scale. For example, at room temperature if I take an NMR, do I see two sets of peaks? What about IR or other characterization techniques? If I can't tell the difference because there are one set of peaks, I think many chemists will consider there to be "one compound" even if multiple tautomers are present.
For an example of this (from the cheminformatics perspective), Marc Nicklaus of the NIH/National Cancer Institute gave a great talk at the 247th ACS national meeting "The NCI/CADD Group's InChI Usage and Analysis of Tautomerism for InChI V2".
The talk focuses on tautomers and what software is trying to do to derive simple rules (as opposed to lots of quantum calculations or experiments) to determine one standard tautomer. But they also measured several tautomers by NMR:
These would obviously be chemically quite different, but one form clearly predominates.

I'll point out that there are plenty of cases where multiple tautomers exist and are observable by characterization techniques. But I think your question falls into what I describe - many chemists think of tautomers as the same compound because there's often one predominant form, you can't usually separate them (e.g., enol-keto) or because you can't observe the two forms via NMR or other characterization technique.


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Why interconversion in tautomers is so rapid?
Looking at the examples in Tautomer on Wikipedia, it seems that the hydrogen needs to travel a significant distance (~20Å) for the interconversion to take place. What lets it happen so fast? Why interconversion is preferred over maintaining its position?

(source)

I think this is too similar to your other question to deserve its own entry. You can just edit the other question adding a small sentence asking about tautomerisation kinetics.

tautomerisation isn't quick 2)where did you found sth like 20A ?

Tautomerisation doesn't proceed by an intramolecular mechanism generally, and it is also rarely fast.

@J.LS, on Wikipedia it says "Because of the rapid interconversion, tautomers are generally considered to be the same chemical compound.", but I don't really know what "rapid" actually means in this context.

I think that refers to the fact that interconversion is fast enough that you get equilibrium population of tautomers.

If you are using a protic solvent (the most common situation when you're dealing with polar functional groups), then you are "cheating" because the solvent can generate its own proton transfers, making the reaction look rapid.


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What is the reason for strong acids/bases dissociating in water?
I have read that the actual definition of a strong acid is one which will fully dissociate in water, so please do not turn the problem on its head when answering. It is also my understanding that every acid reaction is in fact reversible (just as with any other reaction) and yet the thing that is different about strong acid forward reactions is that they are shifted so far forward (the equilibrium constant is very high) to the extent that hardly any of the reactants remain after reaction. The acid reactions follow the laws of chemical reaction kinetics and the reason for this is that the Gibbs free energy decrease of the acid reaction is fairly high and so according to a Maxwell-Boltzmann Curve of energy distributions, only a very small fraction of the molecules will have enough energy to actually revert back to their undissociated form.
My real question is what is it about the electronic (or otherwise) structure of these compounds, for example $\ce{HCl}$, as opposed to weak acids, e.g. $\ce{CH3COOH}$, that makes them more easily dissociated in water. It might also be helpful to learn the same about strong and weak bases. I think it might have something to do with the electronegativity of the ions involved.
If you can help answer the above, please could you also explain the weak/strong conjugate rule (that if a weak/strong acid/base reacts it will produce a strong/weak conjugate base/acid respectively).

The easy answer:
It comes down to the strength of the bonds. Or that covalent bond strength is directly related to the bond dissociation energy and thus dissociation.

A little Physical Chemistry:
(The following is how Pauling explained this.)
In physical chemistry, they have something called bond dissociation ($D_o$).
From Atkins (9th Ed 10.4): "bond dissociation energy, $D_o$, the energy required to separate the atoms to infinity or by the well depth $D_e$." - What is meant here by well is a potential well – for now just think of that as the 2D box talked about in physics.
Now lets shift gears. You may be familiar with bond enthalpy (sometimes just called bond energies). A common formula used in general chemistry is $\Delta H_\mathrm{rxn} = n(\text{bonds broken}) - n(\text{bonds formed})$
It turns out that $\text{bonds broken}$ or $\text{bonds formed}$ is actually referring to the bond energy or bond enthalpy.
It actually works out that there is a direct relationship to bond dissociation energy and bond enthalpy; which makes sense. It is elegantly related by:
$\Delta H^o_\mathrm{bond} = D_o + \frac32 RT $

I guess the natural question to ask now is why do some bonds have a greater enthalpy of formation?
This was not well understood for a long time. In reality there is a lot of factors that can come into play. The way that Pauling initially came up with dissociation energy, $D_o$, in The Nature of the Chemical Bond followed this protocol:

Spectroscopy, (I did this with FTIR for example, but there are several choices) to get a molecular potential energy curve.
Extract the well depth, $D_e$
Find the dissociation Energy using $D_o = D_e - \frac12 h\nu$

In modern thermo:
Because of the relationship with enthalpy, we can measure enthalpy with calorimetry and back calculate.
But the best description uses statistical thermodynamics … Which I will not get into.

It's not what OP asked - he asked why some of those bonds are weaker... Also your answer looks rather impolite.

Was not my intention to be impolite; my apologies. Will edit this up.

@PhysicalChemist Hi, thanks for your reply! Firstly, just wanted to ask what you were referring to by the '2D box in physics'? Also, am I right in saying that when you refer to '3/2RT' this is equivalent to the kinetic energy of ideal gaseous molecules - this would therefore mean that the equation you have stated means the bond enthalpy is equal to the energy needed to separate the atoms in a stationary state accounting only for electrostatic forces + the energy needed to overcome their ideal kinetic energy? Just need some back-up on my deductions from what you wrote in that sense?

The rule is not  "weak/strong conjugate rule" .
The strength of an acid is expressed by its acidity constant, Ka. The higher the value of the constant, the stronger the acid. Likewise, the strength of a base is expressed by its basicity constant, Kb. For a conjugate acid-base pair, in water, it is Ka . Kb = Kw = 10*-14 (25 ºC). So, the rule is "the stronger the acid, the weaker the conjugate base and vice-versa", which is not exactly the same thing. Just try with "weak" acetic acid (Ka=10*-5); its conjugate base, the acetate ion has a basicity constant Kb= 10*-9. Which is the weak and which is the strong?


text:
Finding is the concentration of H⁺ in a benzoate buffer solution after adding HCl

Q: A buffer with a $\mathrm{pH}$ of 4.44 contains $\pu{0.21 M}$ of sodium benzoate and $\pu{0.12 M}$ of benzoic acid. What is the concentration of $\ce{[H+]}$ in the solution after the addition of $\pu{0.050 mol}$ of $\ce{HCl}$ to a final volume of $\pu{1.3 L}$?

A: So these are the steps that I have taken for this:
The reaction is:
$$\ce{C6H5COOH +H2O -> C6H5COO- + H+}$$
The $K_ \mathrm{a}$ of benzoic acid is $6.3 × 10^{-5}$
But, $K_\mathrm{a}=\ce{\frac{[C6H5COO-][H+]}{[C6H5COOH]}}$
We don't know the concentrations of $\ce{[H+]}$ so we solve for that. We then assume at equilibrium $\ce{[C6H5COO-]}$ is equal to it.
Then: $6.3×10^{-5}=\ce{\frac{[C6H5COO- ][H+]}{[C6H5COOH]}}$ and
$\ce{[H+]}=0.00272$
But this is wrong. I think I'm doing something wrong in my problem, but really can't figure out what.

Welcome to chemistry.SE! Please have a look at here and fix your question's formatting to make it more readable. And well done for asking a homework Q appropriately!

Much appreciated @MARamezani

You missed the information you wrote above. The pH is 4.44, which means the $[H^+]$ is $3.63\times 10^{-5} M$. Also, it is rare that the 2 concentration are equal at equilibrium.

Initially, it's true that you have the following equilibrium between sodium benzoate and benzoic acid:
$$\ce{C6H5COOH + H2O <=> C6H5COO- + H3O+_\mathrm{(aq)}}$$
When you add  $\ce{HCl}$ to your buffer solution, sodium benzoate reacts with the acid, and the aforementioned equilibrium is displaced to the left:
$$\ce{C6H5COO- + H3O+_\mathrm{(aq)}  <=>  C6H5COOH + H2O }$$
Now, the equilibrium constant of this reaction is:
$$K=\frac{1}{K_ a}=0.16 \times 10^5>10^3$$
The reaction is quantitative (total). So, we can rewrite it:
$$\ce{C6H5COO- + H3O+_\mathrm{(aq)} ->  C6H5COOH + H2O }$$
Now, we have to construct an ICE table:
Assuming that we didn't dilute the solution when we added $\ce{HCl}$ (this point is not clear in your question): $$\ce{[C6H5COOH]_0} = \pu{0.12 M}$$
$$\ce{[C6H5COO^-]_0} =  \pu{0.21 M}$$
$$\ce{[H^+]_0} = (\frac{0.050}{1.3} + 10^{-4.4}) \mathrm{M}\approx 0.0385 \mathrm{M}$$
As you can clearly seen $\ce{[H^+]_0}$ is the lowest, and so ion hydronium is the limiting reactant.
At the end of the reaction, we have:
$$\ce{[C6H5COO^-]_\mathrm{(eq)}} = (0.21-0.0385) = \pu{0.1715 M}$$
$$\ce{[C6H5COOH]_\mathrm{(eq)}} = (0.12+0.0385) = \pu{0.1585 M}$$
$$\ce{[H^+]_\mathrm{(eq)}} = \epsilon$$
Substituting all these numbers in the expression of $K$
$$K = \frac{ 0.1585}{0.1715\epsilon}=0.16 \times 10^5$$
By solving the above equation, we find:
$$\ce{[H^+]} = \epsilon = \pu{5.82 \times 10^{-5} M}$$
$$\mathrm{pH} = 4.23$$

Here is a different approach:
Since we know the initial pH, as well as, the concentration of the benzoic acid "a" and its corresponding base "b", we can calculate the pka for benzoic acid:
pH = pka + log Cb/Ca => 4.44 = pka + log (0.21/0.12) => pka = 4.20.
Adding the strong acid HCl to the buffer, will convert the corresponding base "b" into benzoic acid "a". Suppose we start with 1 liter of the buffer. This solution will contain 0.12 mole of "a" and 0.21 mole of "b". At equilibrium we get: Ca = (0.12 + 0.050)/1.3 = 0.1307 (M) and
Cb = (0.21 - 0.050)/1.3 = 0.1230 (M).
The pH at equilibrium will be: pH = 4.20 + log(0.1230/0.1307).
=> pH = 4.17

The given solution is a buffer of the equilibrium:
$$\ce{C6H5COOH + H2O <=> C6H5COO- + H3O+} \tag1$$
The buffer question can be easily solved using Henderson-Hasselbalch equation. In case you don't know how to derive it, we'll do it from the scratch. Following expression can be written from the equation $(1)$:
$$K_\mathrm{a}=\ce{\frac{[C6H5COO-][H+]}{[C6H5COOH]}}$$
Take $\log$ on both side:
$$\log K_\mathrm{a}=\log\left(\ce{\frac{[C6H5COO-][H+]}{[C6H5COOH]}}\right) = \log [\ce{H+}] +\log\left(\ce{\frac{[C6H5COO-]}{[C6H5COOH]}}\right)$$
Multiply both side by negtive sign:
$$-\log K_\mathrm{a}= -\log [\ce{H+}] -\log\left(\ce{\frac{[C6H5COO-]}{[C6H5COOH]}}\right) \; \Rightarrow \; \mathrm{p} K_\mathrm{a}= \mathrm{pH} -\log\left(\ce{\frac{[C6H5COO-]}{[C6H5COOH]}}\right)$$
After rearranging the last equation, you got:
$$\mathrm{pH} = \mathrm{p} K_\mathrm{a} + \log\left(\ce{\frac{[C6H5COO-]}{[C6H5COOH]}}\right) \tag2$$
The equation $(2)$ represent the Henderson-Hasselbalch equation of the equilibrium reaction $(1)$. Accordingly, any changes to $\ce{[C6H5COO-]}$ and $\ce{[C6H5COOH]}$ changes the $\mathrm{pH}$ of the solution. For example, let's check the $\mathrm{pH}$ of original conditions ($\mathrm{p} K_\mathrm{a} = -\log (6.3 \times 10^{-5}) = 4.201$):
$$\mathrm{pH} = \mathrm{p} K_\mathrm{a} + \log\left(\ce{\frac{[C6H5COO-]}{[C6H5COOH]}}\right) = 4.201 +  \log\left(\ce{\frac{0.21}{0.12}}\right) = 4.444$$
Let's look at what happens when a strong acid such as $\ce{HCl}$ is added to the system. The following reaction would happen (See the $ICE$ chart):
$$
\begin{array}{lccc}
\ce{&C6H5COO- &+ & HCl &->  & C6H5COOH & + & Cl-} \\
\text{I} & 0.21 \times 1.3 && 0.050 && 0.21 \times 1.3  \\
\text{C} & - 0.050 && - 0.050 && + 0.050 \\
\text{E} & 0.21 \times 1.3 - 0.050 && 0.0 && 0.21 \times 1.3 + 0.050 \\
\end{array}
$$
After addition of $\ce{HCl}$ assuming volume didn't change,
$\ce{[C6H5COO-]} = \frac{0.21 \times 1.3 - 0.050}{1.3} = \pu{0.1715 M}$, and $\ce{[C6H5COO-]} = \frac{0.21 \times 1.3 + 0.050}{1.3} = \pu{0.1585 M}$. Now apply these values to the equation $(2)$:
$$\mathrm{pH} = \mathrm{p} K_\mathrm{a} + \log\left(\ce{\frac{[C6H5COO-]}{[C6H5COOH]}}\right) = 4.201 + \log\left(\ce{\frac{0.1715}{0.1585}}\right) = 4.235 $$
Hence $[\ce{H+}] = 10^{-4.235} = 5.82 \times 10^{-5}$.


text:
Is the momentary dipole in between a cell membrane due to the van-der-Waals forces?
There are van-der-Waals forces within non-polar molecules. I read it that is mainly due to the instantaneous movements of the electron cloud's distribution. When looking at symmetrical molecules like $\ce{CH4}$, the concept isn't hard to understand. But I have a question regarding the bioelectricity and action potential of human cell membrane.
e.g. Outside the cell membrane the ion concentration is high and inside it's low, thus it creates an instantaneous dipole effect between the cell wall. Can this be due to van-der-Waals forces?

Looks your rather confused - it's simply difference in charge - basic physics not VdW or sth.


text:
What is degree of dissociation for polyprotic acids?
My textbook says that degree of dissociation= [HA]dissociated/[HA]initial * 100%.
However, it also says that citric acid has a greater degree of dissociation than acetic acid because citric is triprotic whereas acetic is monoprotic.
This reasoning doesn't fit with the top definition because, more protons donated doesn't necessarily mean more acid molecules dissociated (triprotic acid vs monoprotic acid for example; for every acid molecule dissociated, triprotic acid donates more protons).
So my question is, how does one define degree of dissociation for polyprotic acids?

I think you missed the word greater in the second line after "citric acid has a ..."

I think this resembles the question Confused about strong/weak dibasic and tribasic acids?
Saying that the degree of dissociation $(\alpha)$ of citric acid is greater because its triprotic is not correct. The degree of dissociation should be calculated for each dissociation separately. The degree of dissociation of citric acid's first dissociation $\ce{C6H8O7 <=> C6H7O7- + H+}$ is greater than that of acetic acid, because its $\mathrm{p}K_\mathrm{a}$ is smaller (3.13 vs 4.76)$^1$, meaning it's a stronger acid.
The second value of $\alpha$ for citric acid, however, will be very to close to acetic acid's, because the second dissociation has approximately the same $\mathrm{p}K_\mathrm{a}$ (4.76)$^1$. The third $\alpha$ will be smaller, since the $\mathrm{p}K_\mathrm{a}$ of the third dissociation is greater (6.40)$^1$.
This can be seen in the common polyprotic acids dissociation graphs, like this one (source):

Note that $\alpha$ is represented on the $y$-axis in a 0-1 scale. For each of the four states of the molecule, there is a different $\alpha$, and different values of $\alpha$ happen simultaneously.
$^1$ $\mathrm{p}K_\mathrm{a}$ values taken from Wikipedia.


text:
Ignoring spin, consider an electron in a hydrogen 2p orbital, what is its orbital magnetic moment?
I know that a magnetic dipole moment is given by $$\mu=\frac {-e}{2m}I$$ and that the z component of angular momentum is $$m_j\hbar.$$ However, I have also seen that angular momentum $I$ is given by $$I=\hbar\sqrt{l(l+1)}.$$ Is this also the component in the z direction or is it different? Are both expressions for angular momentum right? Which one do I use to answer the question?
(This question was asked in a problem sheet that I have been asked to do over the vacation. See below for context of question).

The first equation you have is for spin. Let's look at it in a operator form
To properly treat the upcoming discussion, quantum electrodynamics is introduced to describe the eigenstates of spin. The eigenstates of spin has a direct correlation with spin. Formally we depict it as the following
$$ μ_B =-g \frac {eℏ}{2m}  \frac { \boldsymbol {S}}{ℏ} $$
where μ is the magnetic moment operator, g is the g factor, ℏ is plank’s constant over 2π, m is the mass of the particle and S is the spin operator.

For orbital magnetic moment (your question) you use:
$$\mu =\hbar\sqrt{l(l+1)} \boldsymbol {\mu_B} $$
Recall:
s: l=0
p: l=1
d: l=2
f: l=3
So plugging in:
$$\mu =\hbar\sqrt{1(1+1)} \boldsymbol {\mu_B} $$
$$\mu =\hbar\sqrt{2} \boldsymbol {\mu_B}$$
(Units are $ \boldsymbol {\mu_B}$)

How are you defining $\mu_b$? Because, I understand it to be $\frac{e\hbar}{2m}$. If that is the case, shouldn't $\mu=\sqrt 2 \mu_b$, rather than $\mu=\hbar \sqrt 2 \mu_b$


text:
What is the E/Z configuration of cyclic compounds?
What are the configurations of 1-bromo-3-methylcyclopent-1-ene and 1-nitro-2-(propan-2-yl)benzene? Are they E or Z? Are they cis or trans?

What are the rules for these types of isomerism?

It's not necessary to include the E/Z descriptors for these compounds, because the alternative geometries would be far too strained to be stable molecules. However, there is no reason that we can't apply the rules to these compounds and assign the E/Z descriptor.
In each case, compare at the substituents on each side of the pi-bond. For the top compound, one side of the alkene has Br compared to a carbon group, and Br is higher priority. On the other side, hydrogen is lower priority than a carbon group. The higher priority groups are trans, which makes the geometrical stereochemistry E.
For the bottom compound, we must compare two carbon groups, and the alkene containing one is higher priority. On the other side, nitro is higher priority than the carbon group. Again, the higher priority groups are trans, which means that the stereochemistry is E.

In general, it is unnecessary to define the E/Z stereochemistry until the ring size is 8.

What if it's a bicyclo compound.. Will it show cis - trans

@harambe it depends on the strain for the specific isomers you're thinking about. Feel free to open another question if you have an example in mind.

What is the meaning of cis and trans?
cis means on the same side (or face), trans means on the opposite side (or face).
Can I use cis and trans for cylic compounds?
Yes, you can describe the relative orientation of substituents on a ring using cis and trans, such as in the case of 1,3-dibromcyclohexane.

Can I use cis and trans for alkenes?
Yes, in priciple. But you should rather use E and Z. Note that these descriptors are taken from German.

E means entgegen and refers to opposite directions (or sites), while Z means zusammen, which is together, that is the same direction (or site).
What about the compounds in my question?
E/Z or cis/trans are not applicable here.
In the first case, the $\ce{Br}$ atom is in the ring plane, the methyl group can be above or below it. You could describe the absolute stereochemistry at the centre bearing the $\ce{CH3}$ substituent using R or S.
In the second case, there's no alkene at all but a disubstituted benzene. The relative orientation of the substituents on these systems can be described by assigning the proper numbers or by using the ortho- meta- or para-prefix. For this compounds, 2-isopropylnitrobenzene or o-isopropylnitrobenzene would be ok.


text:
Allotropy and polymorphism
Polymorphism is defined as the ability of a solid material to exist in more than one form or crystal structure, whereas allotropy is defined as the property of some chemical elements to exist in two or more different forms, in the same physical state, known as allotropes of these elements. (From wikipedia).
In the case of crystal solids, it seems clear that allotropy is a particular case of polymorphism... but this is always true? That is, both terms, polymorphism and allotropy refer always to crystal structures? Polymorphism definition mentions "crystal structures" but "crystal" does not appears in the definition of allotropy...
Thanks in advance.
UPDATE
Thank you for your answer Then, in some cases allotropy is more general than polymorphism. For instance, dioxygen and ozone are allotropes but not polymorphisms. But... my question arises from my TRUE/FALSE set of questions which I have to answer, and one of them is:

In a non-crystal material the concept of allotropy or polymorphism does not exist.

What would you answer to that? My conclusion is that the question is not clear... For polymorphism is true whereas for allotropy it is not.

I took the liberty to merge your comment that you added in an answer back into your question. You can always extend your own question or add comments to it. I agree that the question is unclear. Btw, unless I'm wrong, greetings to Catalunya.

In my own work, I've never used the term polymorph for anything but molecular crystals in a highly but differently ordered solid state.

In the case of crystal solids, it seems clear that allotropy is a particular case of polymorphism.

Yes, allotropy is the polymorphism of elements in the same state. This does not mean that each solid allotrope has to be highly ordered: amorphous forms of carbon, phosphorus or selenium are considered as allotropes too.
The definition of allotropy isn't confined to the solid state: ozone is an allotrope of dioxygen.


text:
How was it determined that the electron observed in the cathode ray experiments was the same particle that gave an atom its balancing negative charge?
How was it determined that the electron observed in the cathode ray experiments was the same particle that gave an atom its balancing negative charge?
Couldn't there have been an entirely different negatively charged particle? What justified that assumption?

Welcome to chemistry.SE! It's always better to have the main question in the body too. Alright, now I have to go check back my textbook... :)

The idea that the atom contains small indivisible quantities of electric charge was developed in the 19th century by various contributors (e.g. Faraday, Stoney, Laming, Weber, and Helmholtz).
When Thomson (Nobel Prize in physics 1906) discovered the electron in 1897 and estimated its mass and charge, he used the free electrons of cathode rays.
The important experiment that showed that atoms contain electrons was the discovery of the Zeeman effect in 1896, i.e. the splitting of spectral lines in the presence of a magnetic field. Lorentz explained the phenomena with his electron theory. In 1902, Lorentz and Zeeman jointly received the Nobel Prize in physics.


text:
Galvanic Electrochemical Cell Without half cells
Is it possible to create a galvanic cell without a half cell?
In the book I was reading, Illustrated Guide to Home Chemistry Experiments, the following lab (Laboratory 16.4) was suggested:
add 2 M HCl and add copper metal to one end of the beaker, with magnesium at the other end. Then, add patch cables to connect the metal plate to an LED to have physical evidence of success.
However, I don't understand how useful work could be generated without half cells.
To my understanding, we add HCl to react with the magnesium and copper, to produce ions ($\ce{Cu+}$ and $\ce{Mg^2+}$). Then, they can spontaneously produce work due to their eV differences.
What I am missing is how the work can be useful, as opposed to just generating heat in the solution.
I would believe that, rather than traveling though the patch cables, electrons can travel directly through the solution and produce heat.
Thanks for reading and I really appreciate any help!

HCl won't react with Cu.

Please consider giving the green checkmark (or upvote or both) to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this!

This differs from the lemon galvanic cell (aka, 'battery') in several important ways: 1) Mg is used as the anode, rather than Zn, 2) there is no salt bridge, unlike in the the lemon cell, where the lemon serves as a crude 'salt bridge', and 3) the electrotyle is 2 M HCl rather than citric acid (and other stuff) in lemon juice. An open circuit voltage will be produced, easily measured with a digital multimeter in voltage input mode. A short circuit current will be produced, easily measured with a digital multimeter in current input mode. It may be possible to light a red LED, since they have low threshold voltage and low current requirements. I suspect it would light the red LED, though feebly.
But regardless of whether or not there is an external connection of the Mg and Cu electrodes, there will be a rapid redox reaction at the Mg electrode: $$\ce{Mg(s) + 2 H^+ <=> Mg^{2+} + H_2 (g)}$$
This unwanted competing redox reaction is the same as the net reaction obtained from the two relevant half cell reactions and it justs wastes some of the available energy as heat in the cell's vessel. So some electrical power is available by connecting an electrical load (e.g., red LED or resistor) between the two electrodes. But the parasitic oxidation of the Mg anode will continue until the electrode is consumed or the HCl is almost entirely consumed. (Magnesium will very slowly react with boiling water.)
Notes: 1. Clean copper, i.e., without surface compounds or so-called 'patina', does not react with 2 M HCl at room temperature. So the 2 M HCl does not contain any $\ce{Cu^+}$ or $\ce{Cu^{2+}}$ ions. 2. The reaction of Zn with 2 M HCl is considerably slower than is the reaction of Mg with 2 M HCl.
For more information, and at a much deeper level, see K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", Journal of Chemical Education, 95 (2018) 1801-1810.

Electrons cannot move through the solution. Only ions can. Therefore the circuit will still complete properly and work will be done.
Similar idea to the lemon battery. Two dissimilar metals will have different potentials and things will work out.


text:
Effect of dilution on pH for HCl
Is it necessary to consider common ion effect when suppose a strong acid like HCl is added to water?
For example in this question: 10 mL of $10^{-6 } \ce{mol/L \ HCl}$ solution is mixed with 90 mL of $\ce{H2O}$. pH will change approximately?
Normally by using $SV=constant$ I get pH will change by 1 unit (without taking common ion effect).
How will common ion effect affect the answer? Help.

Yes, you will definitely need to account for common ion effect.
However, use Kw to compute the pH. Assume that there is a complete dissociation by $\ce{HCl}$. However, you would need to solve the according quadratic for Kw.
as such, $\ce{H+= concentration~~ H+ of~~ HCl + x ~~and ~~OH- = x}$.
You should get a result around 6.79=pH.


text:
pKaH of pyridine versus other imines
Why is pyridine so much more acidic than other similar imines (5.5 versus ~9, see picture)? Wouldn't delocalization of the positive charge into the aromatic system stabilize it and hence decrease acidity? Also the lone pair of pyridine is not delocalized into the perpendicular $\pi$-system and hence not any less available than the one of the projected imine.

(Source: Organic Chemistry by Clayden et al. p. 1149. )

For future visitors, the corresponding passage in the 2nd edition is at the top of p 726.

This is not a definitive answer to the question but seems to at least give the insight that the premise might not be true (i.e. pKa of typical 6-membered cyclic imine is ~9). I decided to e-mailed the main author of the book that made the above claim, prof. Jonathan Clayden. He kindly replied (if you read, many thanks!) the following:

I have been trying to find the original data for the statement that
  the cyclic imine (or rather its conjugate acid) has a pKa of around 9.
  Typically, imines have a pKa between 5 and 7, so this seems unusually
  high for a start. If this number is wrong then it means that pyridine
  may have quite a typical pKa for a cyclic imine of this type. I've not
  had time for a detailed search but I suspect this number may not be
  fully accurate. Imine pKas are notoriously hard to determine because
  of their hydrolytic instability.

If any additional information comes up, I will edit this answer accordingly.

+1 for going to source for more information! Hopefully that might cause  errata and an update in a future edition.

Why is pyridine so much more acidic than other similar imines?

When we discuss the acidity of a compound, we are really discussing where the equilibrium between the acid and its conjugate base lies.  Therefore, when we compare acidities we are really comparing equilibria, the relative stabilities of the reactant (acid) and product (conjugate base).
The following diagram illustrates the equilibrium between the iminium ion and the corresponding imine, it's conjugate base.

We can't draw any good resonance structures for the imine, but we can draw 2 resonance structures to describe the iminium ion, A and B in the figure. In the pyridinium ion - pyridine case we can draw 2 resonance structures for pyridine (just like in benzene), but we can't draw any resonance structures for the pyridinium ion that delocalize the positive charge without destroying the aromaticity of the ring.
So, in the iminium ion - imine case

we can stabilize the iminium ion through resonance - this stabilizes the iminium ion and shifts the equilibrium to the left making the iminium ion less acidic.

Whereas in the pyridinium ion - pyridine case

we can't draw good resonance structures that stabilize the positive charge in the pyridinium ion - this makes pyridinium ion more acidic.

Both factors operate in the same direction and tend to make the pyridinium ion ($\mathrm{p}K_\mathrm{a}=5.5$) more acidic than the iminium ion ($\mathrm{p}K_\mathrm{a}$ $\sim$ 9).

(Just a minor detail.) Regarding your opening paragraph, the authors make it very clear early on in the book that pKa refers to the pKa of the conjugate acid, so there is a bit of context missing in the picture above.

One interpretation of acidity is the ability to stabilize a negative charge (since the electron is generally kept by the acid, and the proton harbors the positive charge, if we go by bronsted lowry)
Thus, the negative charge is delocalized and thus stabilized. Its impact is not felt as greatly on one atom, and therefore it is more stable, not seeking a proton as much.
Also, the negative charge is better for nitrogen, an electronegative atom which can better stabilize it (but this doesn't explain the difference. I think the main difference is resonance)
This website may help:http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/heterocyclen/arom_gessaet_heterocyclen.vlu/Page/vsc/en/ch/12/oc/heterocyclen/sechsringaromat/sechsring_aromat.vscml.html

@Jori is talking about acidity of cation (which he should mention specifically).


text:
Meaning of prefix p,p′-?
I recently stumbled across the substance "p,p′-terphenyldicarboxylic acid" in a scientific journal when recently reading about metal–organic frameworks.
Could someone explain the prefix p,p'- to me?

Related to this is the $p,p'$ prefix for DDT, where we have two para chloro substituents on two different phenyl rings.

I am guessing that this is the compound you are interested in:

Sigma labels this compound as p-Terphenyl-4,4′′-dicarboxylic acid (or 1,4-Bis(4-carboxyphenyl)benzene).
In your case, the p,p' indicates that the 2 carboxylic acids are in the para positions on 2 of the phenyl rings.

As LDC3 has pointed out, the two p stand for para. But taking into account that there are three different terphenyls, p,p'-terphenyldicarboxylic acid is ambiguous.

An unambiguous name would denote both the terphenyl core, as well as the positions of the $\ce{COOH}$ substituents on the outer rings, such as in p-terphenyl-4,4'-dicarboxylic acid.


text:
How can I determine the quantum number of a electron lost during ion formation?
How do quantum numbers change for ions?

A sample question: Give the set of four quantum numbers that could represent the electron lost to form the Rb ION from the Rb atom.

The answer given is $n=5; l=0; m_l=0; m_s= +\frac12$. I chose the answer: $n=4; l=1; m_l=1; m_s= -\frac12$.
I am looking for an explanation. I am confused because I thought the question asked me to describe $\ce{Rb+}$ and the answer given is the same that I would choose for $\ce{Rb}$ at ground state. If it is asking me to describe the electron, how is that technically possible if I don't know where the electron went? Sounds silly and I know I'm over thinking it. Some insight would be appreciated.

It seems you have the correct mindset and required knowledge to solve this problem, but just made a tiny slip-up in interpretation. You were asked to describe the electron lost from the rubidium atom. Write out the electronic configurations for the neutral rubidium atom and the rubidium cation, and compare. Is it clear now?

Yes, it is clear now. Representing the electron lost from the Rb atom as n=4; l=1; ml=1; ms= -1/2 was really describing Kr. Thanks for helping me reach my 'aha' moment.

If you feel up for it, we encourage users to answer their own questions when possible! Describing what was on your mind and how you came to understand the correct solution may help future readers.

The correct answer has been given in the comments section of this post. I am providing the answer here in accordance with a meta post discussion.
The question you were asked was in reference to the quantum numbers that correspond to the electron lost when $\ce{Rb}$ goes to $\ce{Rb+}$. The electronic configuration of $\ce{Rb}$ is $[\ce{Kr}]\ 5\mathrm{s}^{1}$; thus, you need the set of quantum numbers for a $5\mathrm{s}^{1}$ electron. That set of quantum numbers is:
$$n = 5;\ l= 0;\ m_{l} = 0;\ m_{s} = +{1\over2}$$


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How do you know what inter-molecular interactions are present?
I'm studying for a chem exam and there are often questions of this form: What kind of inter-molecular interactions are important in (insert compound here). Then the possible answers are any combination of dipole-dipole, hydrogen bonding, and dispersion force.
My question is: When ever a molecule has hydrogen bonding does that mean there is also dipole-dipole?
How do you tell if a molecule has a dipole, for instance CH3COOH,  has one but how do you know?

Look at the electronegativities of the neighbouring atoms. If an atom has a stronger electronegativity than a neighbouring one, it will draw some of its charge, and there will exist a dipole moment. Be careful however in molecules where the charge may be drawn equally, such as in symmetric molecules. These molecules won't necessarily have a net dipole moment.


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Hydrogen and Helium ideal gas behavior in room temperature
Which one of hydrogen and helium behaves more like ideal gas in room temperature...
I thought hydrogen gas would show more ideal behavior in room temperature because it has lighter mass (less no of electron) although it's diatomic, so there would be less attraction.
However, I saw one question on the book and they said that the helium gas is more likely to show the ideal gas behavior in room temperature, not the hydrogen gas..
So, can someone please give me an idea how my thinking is wrong?.
Thanks for advance

For inert gas molecules, intermolecular interactions would be by Van der Waals forces, which are based on the total volume and polarizability of the gas molecule.
The diatomic hydrogen is larger and more polarizable of the two.

He is more nearer to ideal behaviour than hydrogen because ideal behaviour also depends on the charge spread over area or size of molecule. So the charge density of hydrogen is expanded increases the size of hydrogen due to which interactions between hydrogen molecules increases and hence show less ideal character as compared to helium and helium charge density is bound to single nucleus. Therefore the low charge density of hydrogen show little

You do not complete the last sentence. Please check and edit.