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Absorbance vs titrant added in acid-base titration
I was working on this question:

Nitrophenol is a colorless weak monoprotic acid
  ($\mathrm{p}K_\mathrm{a} = 7.2$) whose conjugate base is bright yellow. To
  $2.00\ \mathrm{mL}$ of a solution of $\mathrm{0.0100\ M}$ nitrophenol is added
  $\mathrm{1.00\ M} ~\ce{NaOH}$ in $\mathrm{0.001\ mL}$ portions, and the absorbance
  of the solution at $\mathrm{485\ nm}$ is monitored. What does the
  graph of $A_\mathrm{{485}}$ as a function of added volume of $\ce{NaOH}$
  look like?

The answers given were:

My reasoning for this question was that the absorbance is proportional to the concentration of conjugate base present in the solution. As the titration progressed towards the equivalence point ($\mathrm{0.02\ mL}~\ce{NaOH}$ added), the concentration of base increased. Then, as more base is added beyond the equivalence point, the concentration of base decreased, since no more base is being produced but the solution is being diluted. Therefore, I thought the correct graph was (A). However, the correct answer ended up being (B). So why would the concentration of base stay constant past the equivalence point?

Once all the conjugate base is formed in the solution it doesn't get any more yellow.  The twist here is that the small extra volume of titrant has a negligible effect on the overall volume of solution.

@MaxW Ohh the solution originally had a volume of 2 mL so adding the 0.02 mL of titrant doesn't change the concentration much. Thanks.


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How many hydrogen bonds in these compounds?
We usually get some questions about how many hydrogen bonds these compounds can create, we use it to differentiate between the boiling point of each one of them so it would be great if you could help me.
So what I know that $\ce{H2O}$ has 4 hydrogen bonds those are 2 from hydrogen atoms and 2 from other molecules that has 2 hydrogen atoms connected to the 2 lone pair of electron that the oxygen has.
In $\ce{HF}$, It has 2 hydrogen bonds. The same explanation implies here.
In $\ce{NH3}$, 2 hydrogen bonds because $\ce{N}$ doesn't have the required amount of lone pairs to make more hydrogen bonds.
But what I really want to know is about alcohols. In the way that I am explaining it, it should have 2-3 hydrogen bonds but when I check some books and articles they say that it only creates 1.
So if you can explain how many each of them creates and what it depends on that would be awesome.

Interesting question. I suspect the answer is more complicated than you might expect.

Like I am pretty sure there are some other factors that play here. That why I am asking about the alcohols. But the way of me thinking seems to work with everything except alcohols.

Just summing up the number of hydrogen bonds will not do, accepting and donating are two different things.

Various papers suggest that the average number of hydrogen bonds per molecule in water is less than 4 and possibly as low as 2 and that it varies depending on temperature. I suspect a similar effect will be seen in alcohols but I can't find any specific references to it.

This a good question for discussion to clear concept on H-bonding in alcohol.
Hydrogen bonding occurs between molecules where you have a hydrogen atom attached to one of the very electronegative elements - fluorine, oxygen or nitrogen.
In the case of alcohols, there are hydrogen bonds set up between the slightly positive hydrogen atoms and lone pairs on oxygens in other molecules.
The hydrogen atoms are slightly positive because the bonding electrons are pulled away from them towards the very electronegative oxygen atoms.

Note:  If you want to be fussy, the diagram is slightly misleading in that it suggests that all of the lone pairs on the oxygen atoms are forming hydrogen bonds. In an alcohol that can't happen. Taking the alcohol as a whole, there are only half as many slightly positive hydrogen atoms as there are lone pairs. At any one time, half of the lone pairs in the total liquid alcohol won't have hydrogen bonds from them because there aren't enough slightly positive hydrogens to go around.
In the diagram, to show the 3-dimensional arrangement, the wedge-shaped lines show bonds coming out of the screen or paper towards you. The dotted bonds (other than the hydrogen bonds) show bonds going back into the screen or paper away from you.

So as a result from what the picture says, That it can create up to 3 bonds but if that happen then another molecule have to create 1 bond hydrogen bond. So In average alcohol can create 2 bonds for each molecule right?

yeah. 2 H-bonds per molecule of alcohol

There are also such things as bifurcated hydrogen bonds and there is the rest of the molecule to consider and the steric effects that it has. These complicate matters and I would be interested to see a higher level answer than this.

Hydrogen bonds require a $\ce{\delta{+}}$ hydrogen atom capable of forming a hydrogen-bond, so $\ce{H2O}$ makes 2 H-bonds, and $\ce{NH3}$ has 3 H$\ce{\delta{+}}$ atoms capable of forming these bonds.
However hydrogen bonds also require an "active" lone pair capable of accepting a hydrogen-bond, where $\ce{H2O}$ can do 2 of these again, but $\ce{NH3}$ can only accept 1.
So, does water make 4 bonds or 2? Does ammonia make 1 bond or 2?  The answer is both, but it depends on how you count it.  If you look at an individual molecule in a cluster of water molecules it may appear to be surrounded and held in place by four H-bonds; it makes two with the hydrogen and accepts two with the lone pairs.  That molecule is involved in 4 hydrogen bonds.
But if you take 100 water molecules and count how many hydrogen bonds there are between them, the answer will be about 200 because each molecules makes 2 bonds.  If you consider each molecule making 4 bonds then you are double counting each bond being made and accepted.  The same is true for ammonia, each molecule makes one hydrogen-bond, it also accepts one.  You could say that an individual molecule is involved in two bonds, but if you count them all up each molecule only makes one each.
Because of these two different viewpoints, some texts will say water makes 4 and alcohols make 2 bonds, and some will say water makes 2 and alcohols make 1 bond.  They are both right, but right in different ways. Whichever way you count them, you must be consistent. If this is related to an exam syllabus, the correct way is defined by that.

Well, It doesnt make sense. Ammonia will create 1 h bonds through it's hydrogen atom but it will also have another hydrogen atom that another molecule provides a H atom and the Nitrogen in the first molecules provides  its pair of lone electrons. So as in total it has 2 of hydrogen bonds."
Same applies to water.

@Biker that other hydrogen bond in created by another ammonia molecule. I'll try an edit to clarify.

never thought about these different perspectives, Great! @Spontification

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Side products during the electrolysis of Copper Sulfate?
Using good quality reagents, during the electrolysis of $\ce{CuSO4}$ (at $\mathrm{15V ~ \& ~ 2A}$), I have noticed a significant amount of Hydrogen generated at the negative electrode in addition to copper deposition.
This generation occurred at all voltages, which surprised me. I expected a only the reduction of copper in the reaction. The solution was blue, so I know that copper sulfate was still in solution. At first, I thought that the copper might be reacting to the sulfuric acid being generated, but the copper that has fallen off the electrode is not bubbling.
I am using a platinum and copper electrodes.
Is $\ce{H2}$ generation happening as a result of the weak sulfuric acid in solution?
Can you calculate the products of simultaneous electrochemistry reactions?

which electrodes are used. either platinum/ inert or copper one.

Hydrogen is not released. at cathode - copper is deposited and at anode - O2 gas released.

Hydrogen evolution is common at high current densities. The reason is that hydrogen cation and to slightly lower extent hydroxyde anion are highly mobile in water solutions, so hydrogen transport (and hydrolysis of water) is kinetically favoured over salt ions transport (and salt electrolysis). This is a constant nuisance in electronchemical processes and can be avoided only by use of voltages below water electrolysis voltage or use of special cathodes (such as carbon), as hydrogen evolution is kinetically hindered on some cathodes.

What books would you recommend for me to learn about this process, permeakra7? -Honestly, I am more interested in understanding the mathematical relationship that predicts the mechanics of this hydrogen transport behavior. All my books predict only Cu and O2 as products. I tried lower voltages and it did seem to diminish the Hydrogen production, but not remove it entirely. I am going to try some different electrodes in addition to my copper/platinum pair tomorrow and see what that does.

@Mcruggs: If you're interested in a good fundamental understanding, I would recommend Electrochemical Methods by Bard and Faulkner. There's a whole chapter on electrochemical kinetics (chapter 3 I think) that derives general rate expressions which take mass transport into account.

There are a few key factors that I expect are contributing here:

$\ce{CuSO4}$ concentration – If your $\left[\ce{Cu^2+}\!\right]$ in solution is low, then that will disfavor deposition, as permeakra noted in a comment. You want at least $25$ to $100~\mathrm{g\over L}$ of copper in the electrolyte.
Cell voltage – $15~\mathrm V$ is an exorbitantly high voltage for copper electrodeposition. In order to avoid gassing at the cathode, you'll want to operate down in the $1.5$ to $2.0~\mathrm V$ range, tops.
Electrolyte conductivity – you don't mention anywhere that you've added other constituents to try to increase the conductivity.

If this is a home experiment, I would try dissolving the $\ce{CuSO4}$ in table or pickling vinegar instead of plain water. That will give you more charge carriers in solution, and also decrease the $\mathrm{pH}$, both of which should help.
If this is a school or work experiment with access to proper lab equipment and chemicals disposal services, I would use $1\%$ to $10\%~\ce{H2SO4}$ as the base electrolyte. That will really get your conductivity up, and your cell voltage down.

Finally, on the methods side, I would recommend choosing your voltage in order to initially target a current density in the vicinity of $\sim\!10~\mathrm{mA\over cm^2}$, as a value where I would expect to see little to no $\ce{H2}$ evolution. You can always nudge the voltage up from there as needed.

Is $\ce{H2}$ generation happening as a result of the weak sulfuric acid in solution?

Almost certainly not.

Can you calculate the products of simultaneous electrochemistry reactions?

Not with much certainty. Most of the time the best one can do is make qualitative arguments about what reactions might be happening.  Detailed modeling of all reactions occurring at an electrode is still prohibitively difficult, as far as I know.

Hydrogen evolution will be favored by increased hydrogen concentration (lower pH). As the copper plates out, the Cu++ ion is replaced by H+. The very first plating out should be with the least hydrogen, and then hydrogen gas evolved should increase as the electrolysis continues. How can we reduce the pH without precipitating the copper?
I have not done the experiment, but I think that adding a bit of ammonia would convert the H+ to NH4+ and raise the pH enough to reduce hydrogen evolution without precipitating copper (Cu++ --> soluble Cu(NH3)4++). You might get more creative and use other amines.
Cell voltage is important also, as suggested in the comments. High current forces something to deposit; copper is preferred, but if the current demands, hydrogen will deposit also.
If you want to get really inventive, you might consider a cathodic poison that is not incorporated into the copper deposit. Mercury would work, but is a no-no for many reasons; sulfur and arsenic compounds also probably not good; maybe a long chain water-soluble polymer, or maybe even a sugar would coat the copper cathode and interfere with hydrogen deposition without becoming incorporated. Good luck!

Platinum is expensive. How about using a copper anode? No H2SO4 generated - only more CuSO4!

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What are point groups and the benefits of using them?
I have heard vague information about point groups and symmetry before but I am unsure how they actually help me predicting physical properties of molecules. How do I assign point groups and what good does it do me?
Please answer at a low level.

Point groups are a very valuable tool for analysing molecules without knowing much about them. They rely on analysing a molecule’s symmetry and deducing both physical parameters and the shape of molecular orbitals from the information gathered. The downside is that one must know a molecule’s geometry before one can apply group theory to it, but for most molecules the geometry is known anyway.
Picture a molecule, for example ammonia $\ce{NH3}$. Remember that it has a trigonal pyramidal shape with $\angle(\ce{H-N-H}) \approx 107^\circ$ Consider all atoms of the same type (only applicable to hydrogen here) identical. Look at this molecule naively and think what kind of symmetry it has.

Figure 1: Ammonia (image taken from Wikipedia).
One of the easiest symmetry elements that you can find is probably a mirror plane that transects the molecule along one $\ce{N-H}$ bond and places the image of the second hydrogen onto the third hydrogen. They are shown in figure two by the blue-ish ovals. A plane of symmetry is typically called $\sigma$.

Figure 2: Planes of symmetry in ammonia (image taken from Researchgate)
You may also see, that if you stick an axis into the nitrogen atom through its lone pair (the thick, blue axis in figure 2), you can rotate the molecule $120^\circ$ around this axis and it will again look the same. Three rotations along this axis make $360^\circ$, so due to some formalism (which is really helpful if you think about it), the rotation is called $C_3$.
All we have gotten so far is a simple set of symmetry operations but they don’t help much. We need to put them together. We can define the connection of two symmetry operations by saying ‘we’ll perform one first, the other after it’; since neither change the molecule the final molecule will remain unchanged. Now to prove that the symmetry elements present in a molecule constitute a mathematical group, we need to prove the group axioms:

Closure The connection $A \circ B$ of two elements of a group must always be an element of the group.
I just proved this before. A symmetry operation will transform all hydrogens onto a hydrogen position, no two hydrogens on the same and nitrogen will always remain unchanged. We will always gain the same molecule. We just need to make sure we find all possibilities correctly when searching.

Associativity $(A \circ B) \circ C = A \circ (B \circ C)$.
If you think about this practically, it makes sense. Why should it matter whether I rotate first and apply a mirror plane later or the other way around?

Existence of an identity element One element must exist so that $E \circ A = A$ and $A \circ E = A$.
Well, this is easily done; we just include the identity, i.e. no transformation as one of our symmetry operations. It is labelled $E$.

Inverse elements For each element $A$ there must exist an element $A^{-1}$ so that $A \circ A^{-1} = E$.
This requires a bit of working but in the end we can prove this. For $\sigma$, the same $\sigma$ is always inverse element. For $C_3$, $C_3^2$, the rotation around $240^\circ = -120^\circ$ is the inverse element.

The good thing is that if we didn’t identify all the possible operations initially, we can use the connection $\circ$ to indentify the ones we missed.
Now because there is only a certain number of inherently different groups around, they have been given names for easier identification. Our ammonia example would be the point group $\bf\it C_\mathrm{3v}$. $C$, because we have a rotation axis, $_3$ because rotation is $\frac{1}{3}$ of the full circle and $_\mathrm{v}$ because the planes of symmetry are vertical to the rotation axis (rather than horizontal).
(Naming point groups is rather simple. Find the most important axis — the rotation with the highest index. If that is the only rotation, it is a $C$ group. If there are rotations perpendicular to that one, the group is called $D$. If there is an improper rotation, it is labelled $S$. The highest index of rotation is also the group’s index. $_\mathrm{v}, _\mathrm{h}$ and $_\mathrm{d}$ denote planes of symmetry vertical, horizontal or diagonal to this axis. Finally, $C_s$ is the group that contains only one plane of symmetry, $C_i$ contins only the inversion and finally, tetrahedra ($T_\mathrm{d}$), octahedra ($O_\mathrm{h}$) and icosahedra ($I_\mathrm{h}$) have their own names due to their high symmetry.)
Part of the beauty of point groups is already apparant here. If we know what symmetry operations transform the atoms onto each other, this must also be true for some physical properties — those that can be expressed as vectors. Most notably, a dipole moment must transform onto itself during any symmetry operation, so ammonia’s dipole moment must be along the rotation axis. Molecules that display inversion symmetry cannot be a dipole. Molecules can only be chiral (but are always chiral) if their point groups do not contain inversion, planes of symmetry or improper rotation (a rotation followed by a mirror image perpendicular to the rotation axis). There are more examples, but these are probably the most important.
Finally, for each point group a character table exists. This groups sets of similar symmetry operations together according to a set of specific rules (e.g. the two rotations $C_3$ and $C_3^2$ will be grouped together) whose details I won’t go into, and also includes irreducible representations. Any spec of the molecule (and all its symmetry-equivalent specs) must transform as one of these irreducible representations. The $\mathrm{p}_z$ orbital of nitrogen in ammonia (the one along the $C_3$ axis) transforms according to the irreducible representation $\mathrm{a_1}$ — essentially, $\mathrm{a_1}$ means that something will always be the same. The $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals together will transform as $\mathrm{e}$. Knowing the irreducible representations of atomic orbitals helps to predict molecular orbitals, because only those of the same irreducible representation will mix and combine (for others, the constructive overlap is always $0$).
You can google character tables to find lists of them, and thereby also which point groups actually exist.

Point groups are even more powerful than shown here; but I tried to keep the application parts low-level. There are a number of questions here on Stack Exchange tagged group-theory that go more into depth.


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Absorption Spectrum Using a Spectrophotometer
After using a spectrophotometer, I have obtained absorbance values. To create an absorption spectrum, should I convert these values to percentage absorbance? (I've seen it in textbooks as such)
If I were to convert absorbance to percentage absorbance, would I simply multiply all values by 100%?

There is no such value as "percentage absorbance."

This is simple mathematical conversion. Firstly, you find the fraction of each absorbance value.
Let, the absorption spectrum has a three absorbance values say $A_1, A_2$ and $A_3$.
Now, find the fraction of each absorbance as per following:
fraction of $A_1 = A_1/(A_1+A_2+A_3)$
fraction of $A_2 = A_2/(A_1+A_2+A_3)$
fraction of $A_3 = A_3/(A_1+A_2+A_3)$
Now convert all the respective fraction values into percentage as per:
percentage of $A_1 = (\text{fraction of }A_1 \cdot 100)\%$
percentage of $A_2 = (\text{fraction of }A_2 \cdot 100)\%$
percentage of $A_3 = (\text{fraction of }A_3 \cdot 100)\%$


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Beta lactone query
While studying lactones I read that usually gamma and delta lactones are formed. Why aren't beta lactones formed. The first thought that came into my mind was angle strain. But then I thought of epoxides which have more angle strain and still are more stable. Please help.

Now similarly in the above reaction why isn't beta lactone formed?

Two things. 1) the lactone has an additional sp2 carbons, whereas epoxides do not have sp2 carbons, so the analogy is not accurate. 2) 3-membered rings are not more thermodynamically stable than 4-membered rings - see the sections on ring strain in any (proper) org chem textbook e.g. Clayden or March. However,  their formation is kinetically favoured - that means they are easy to form.

@orthocresol could u explain y 3 mannered ring is thermodynamically stable?

I said it was not thermodynamically stable. That's mostly because of angle and torsional strain.

@AdityaKumar. Which book is this? Please, tell me book name and its writer.

$\beta$-Lactones can be prepared by cycloaddition of ketenes with aldehydes

@K_P can u put a pic of the mechanism?

@kiransolanki "Solomons and Frhyle's Organic Chemistry" it is for graduation.

The "textbook" reaction is the reaction of ketene with itself but here is one method I got from google that can yield mixed $\beta$-lactones from a ketene and an aldehyde: http://organicreactions.org/index.php/Catalytic_asymmetric_ketene_cycloadditions

It is all about angle strain. You may have heard about the Baeyer strain that consideres six-membered rings to be unstrained and all smaller ring sizes to have angle strain, getting more as the ring gets smaller. This is also relevant for the creation of rings — such as but not limited to lactones — five- and six-membered rings are kinetically possible, smaller rings are generally not formed by intramolecular substitution reactions. (Note the emphasis, of course the synthesis of smaller rings is possible due to other mechanisms, such as in the Prilezhaev reaction to synthesise epoxides.)
You also seem to have an incorrect assumption about the stability of epoxides. Epoxides are generally not stable with respect to nucleophilic subsitution. In fact, the Baeyer strain is so high that epoxides are commonly taught as the only aliphatic oxygen that can be displaced in an $\mathrm{S_N2}$ reaction to give the alcoholate anion and which does not need to be protonated first.

But as far as I know there exists more epoxides than lactones. Am I right? If yes then why doesn't angle strain affect them? In fact I've rarely come across a 4 membered ring with an oxygen.

@AdityaKumar Well, it is very easy to synthesise epoxides and much like primary halides (especially bromides and iodides) while they are generally susceptible for nucleophilic displacement a satisfactory nucleophile must actually be present for the substitution to happen. Oftentimes you are lucky and halides/epoxides survive a reaction. The oxetane rings do exist but they are less common because they don’t have as large a synthetic value in either way.


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Reason for the differences in third ionisation energy between elements (magnesium and fluorine)
The question is "Why is the third ionisation energy of magnesium higher than that of fluorine?"
Although I understand that differences in ionisation energy have several factors governing it, e.g. the shielding effect, distance from the nucleus, whether the electron is in its own orbital or paired with another electron etc.
The answer to this question is that "Even though the valence electrons of both elements are removed from the same shell, Mg has greater nuclear charge."
I am confused what the answer means about nuclear charge. Doesn't the third ionisation energy of Mg and F have the same nuclear charge as they are both 3+?

The ionisation energy, as a rough guide, can be related to the effective nuclear charge and the principal quantum number:
$$\mathrm{IE} \propto \left(\frac{Z_\mathrm{eff}}{n}\right)^2$$
In the ionisation of $\ce{Mg^2+}$ and $\ce{F^2+}$, both electrons are being ionised from the second principal quantum shell, so $n = 2$ in both cases. However, the main reason why the 3rd IE of magnesium is larger is because the nuclear charge $Z$ is larger, and not because of any so-called stability of half-filled or fully filled shells.
Contrary to what you thought, the nuclear charge is not the charge on the species. It is the charge of the nucleus - hence "nuclear charge". The $\ce{Mg}$ nucleus has 12 protons, and the $\ce{F}$ nucleus has 9 protons. So, there is a greater electrostatic force pulling the electron towards the nucleus in $\ce{Mg^2+}$ than in $\ce{F^2+}$, and consequently it is harder to remove the electron from $\ce{Mg^2+}$.
In truth, the raw nuclear charge must be modified by the shielding constant $\sigma$ to obtain the effective nuclear charge, $Z_\mathrm{eff}$. However, the variation in $\sigma$ between the two compounds is really negligible compared to the variation in the actual nuclear charge.


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Why are thiolates more nucleophilic than alkoxides?
Why is $\ce{RS-}$ (a thiolate) more nucleophilic than $\ce{RO-}$(an alkoxide)?
According to my logic, the larger size of the sulfur atom will allow the negative charge to be dispersed more as compared to the negative charge on the oxygen atom. Thus, shouldn't $\ce{RS-}$ be less nucleophilic as compared to $\ce{RO-}$ ion?

$\ce{S-}$ is quite bigger in size hence it shows lesser hydrogen bonding in protic solvents hence they are better nucleophiles in protic solvents. But when we take polar aprotic solvents, no hydrogen bonding is possible so, nucleophilicity $\propto$ basicity and your argument becomes valid.

It depends entirely on the solvent that the reaction is being conducted in. If the solvent is polar protic, the thiolate anion is more nucleophilic because of polarizability factors. Whereas if the solvent is polar aprotic basicity takes precedence as the gauge for nucleophilicity.

It's true that thiolate ion is more nucleophilic than alkoxide ion. You are going in a right way. The order of nucleophilities is according to polarisability of atom. As we know, the size of atoms goes on increasing down the group.
atomic size: $\ce{S>O}$
Due to increase in size of atoms, the electrons are more loosely bound to nucleus. Hence, the atoms are now easy polarise. As a result, electrons are easily polarisable down a group.
Thus, $\ce{RS-}$ of thiolate ion is more polarisable than $\ce{RO-}$ of alkoxide ion. Therefore, thiolate ion is more nucleophilic than alkoxide ion.
The polarisability of nucleophile is more important in case of nucleophilic substitution reaction where the nucleophile electrons are polarised by the positive charge of electrophile and bond is formed during the reaction.
As nucleophilicity is a kinetic factor. More stronger the nucleophile, greater is its rate constant. The following is relative rate constant of both nucleophile on reaction with iodomethane by using methanol as reference whose rate constant is taken as unity:
$\ce{CH3-O-};~k  =  2\cdot 10^6$
$\ce{CH3-S-};~k  =  1\cdot 10^9$

Hmmm.... does that mean thiolate ions are more basic than alkoxide ions?

no.. there  is not always that basicity parallels nucleophilicity. Though Here, alkoxide ion is more basic than thiolate ion.

And why is that? I don't get the logic behind it. :(


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Cubane systematic name
Why is cubane called pentacyclo[$4$.$2$.$0$.$0^{2,5}$.$0^{3,8}$.$0^{4,7}$]octane?
I am especially intrigued by the pentacyclo part.

Related: How to name tricyclo carbon compounds?

Apparently because it has five cycles.

How was it determined (five cycles)?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), polycyclic hydrocarbons having an independent secondary bridge are named on the basis of a bicyclic system as follows.
The main ring of a polycyclic hydrocarbon ring system is selected so as to include as many skeletal atoms of the structure as possible (here: 8 atoms).

The main bridge is the bridge that includes as many of the atoms as possible that are not included in the main ring (here: 0 atoms).

Bridges other than the main bridge are called secondary bridges:

P-23.2.6.1 Naming polycyclic alicyclic hydrocarbons
Rings not designated by the bicyclic system described above (…) are defined by citing the number of atoms in each secondary bridge as an arabic number. The locants of the two attachment points of each secondary bridge to the main ring are cited as a pair of superscript arabic numbers (lower first) separated by a comma. The numbers indicating independent secondary bridges (bridges that connect atoms of the bicyclic system) are cited
  in decreasing order. (…)
P-23.2.6.1.1 The prefixes ‘tricyclo’, ‘tetracyclo’, etc., in place of ‘bicyclo’, indicate the number of rings in the polyalicyclic system. The number of rings is equal to the number of bond cuts necessary to transform the polycyclic system into an acyclic skeleton, unbranched or branched.

Therefore, the correct systematic name is ‘pentacyclo[4.2.0.02,5.03,8.04,7]octane’. However, the retained name ‘cubane’ is used in general nomenclature and as preferred IUPAC name (PIN).

This is a revelation for me: The number of rings is equal to the number of bond cuts necessary to transform the polycyclic system into an acyclic skeleton, unbranched or branched.

Loong has already taken care of how the IUPAC system names cubane. I only wish to give the mathematical (more precisely, graph-theoretic) view on why cubane is considered pentacyclic.
Consider the following two graphs:

You can check that you can assign labels to the edges and vertices of both graphs such that there is a one-to-one correspondence between them. In graph theory parlance, these two graphs are isomorphic.
Thus, the cubic graph can be represented by either graph. The second "planarized" graph, in particular, clearly displays the five "rings" in cubane. Mathematically, we have Euler's formula $\text{vertices} - \text{edges} +\text{faces}=2$. Since one face has been "projected to infinity" by the flattening, the formula for counting rings now becomes something like $\text{rings}=\text{bonds}-\text{atoms in ring system}+1$. For cubane, we thus have $12-8+1=5$ rings.

I'm not really sure that this answers the question. It doesn't have anything to do with naming but it is somewhat relevant to the last part so I won't flag for deletion for now (others may disagree though).

@bon OP: "I am especially intrigued by the pentacyclo part."

@bon I agree. It's always interesting to think about flattening your because the molecular bond graph tells you the topological index.


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Hyperconjugation vs. steric hindrance: which is stronger?
If we talk about a series of molecules: ethene, prop-1-ene, but-2-ene, etc., we see both hyperconjugation (which increases stability) and steric hindrance (which decreases stability). So which one is more powerful i.e. which of these molecules will be most stable?

Stable with respect to what?

@NotNicolaou, probably thermodynamic stability with respect to each other. In which case ethene would be the most stable because of most no. of alpha-H and least steric hindrance.

@NotNicolaou As in my textbook, more number of alkyl groups attached to double-bonded carbon cause increase in steric hindrance which make the molecule unstable (simply unstable to exist).

Can you cite the textbook so I can read the whole thing?

@Tamoghna my teacher told that more number of alkyl groups would provide more alpha-H and thus increase hyperconjugation and stability, though i doubt whether those hydrogens will actually be alpha-H.

@NotNicolaou actually this one just came to my mind and i discussed it with my teacher, but was not satisfied with the answer. The point that "more number of alkyl groups attached to double-bonded carbon cause increase in steric hindrance which make the molecule unstable" was told by my teacher.

You're right, those will become beta-H and won't contribute to the stability.

It's not just double-bonded C, but rather any C atom attached to a functional group.

So, simply, would more alkyl groups increase stability? (i didn't get what you just posted)

It's a contradict to explain the stability of substituted alkenes by using hyperconjugation and steric hinderance both. I think, it is more better to consider heat of hydrogenation factor in case for stability of substituted alkenes. The stability of substituted alkenes is explained by Hyperconjugation factor while the stability of geometrical alkenes (cis-trans) is explained by steric hinderance.
The stability of alkene can be determined by measuring the amount of energy associated with the hydrogenation of the molecule.  Since the double bond is breaking in this reaction, the energy released in hydrogenation is proportional to the energy in the double bond of the molecule. This is a useful tool because heats of hydrogenation can be measured very accurately. The ΔHo is usually around -30 kcal/mol for alkenes. Stability is simply a measure of energy.  Lower energy molecules are more stable than higher energy molecules.  More substituted alkenes are more stable than less substituted ones due to hyperconjugation. They have a lower heat of hydrogenation.  The following illustrates stability of alkenes with various substituents:

In disubstituted alkenes, trans isomers are more stable than cis isomers due to less steric hindrance because in trans bulky groups are in opposite sides so less repulsion between them as there is more repulsion in case of cis due same side of bulky groups. Also, internal alkenes are more stable than terminal ones. See the following isomers of butene:

Overall stability : trans-but-2-ene > cis-but-2-ene > propene > ethene

I still have a doubt. In your first diagram, if we look at the third molecule (let it be the trans isomer) and fifth molecule, how will you explain the stability as fifth molecule has more steric hindrance than third one. I cannot understand "The stability of substituted alkenes is explained by Hyperconjugation factor while the stability of geometrical alkenes (cis-trans) is explained by steric hinderance".

It is simple. here is a trick. when u compare all substituted alkene which have different molar mass with each other (not for geometric isomers i.e. cis-trans), take hyperconjugation factor. In case of alkenes of same molar mass i.e. with geometrical isomers, take steric hinderance factor.

So, let your 'R' in third and fifth molecule have same molar mass. Now you can apply both steric hindrance and hyperconjugation because they are not geometrical isomers. So which one is more stable?

Actually my question is that which one of hyperconjugation and steric hindrance is more effective in increasing stability of the molecule?

Trans alkenes are only more stable than cis alkenes in the case of acyclic systems. Trans-cycloalkenes smaller than about 12 carbons are unstable due to ring strain. See here for details.

@kiransolanki I get your point now of not comparing compounds of different molar masses by steric hindrance. Then why did my teacher tell me hex-3-ene, pent-2-ene and but-2-ene for comparing steric hindrance? (This is what made me confuse)

@aayushsrivastav. don't know. what is going on ur teacher mind that time. but, here all three hex-3-ene, pent-2-ene and but-2-ene can be compare by hyperconjugation.


text:
Why are geminal diols unstable?
The most stable of:
(a) $\ce{(CH3)2C(OH)2}$
(b) $\ce{CH3 C Cl (OH)2}$
(c) $\ce{CH3CN}$
is $\ce{CH3CN}$, because the other two are geminal diols. How can we tell that geminal diols are unstable?

See What makes C=O more stable that C(OH)₂

the question is on stability of different gem-diols not for comparison of stability of C=O and diols.

@kiransolanki That's also discussed in the answer.

I'm voting to close this question as off-topic because the answer is addressed in the link provided by Ron.

also related http://chemistry.stackexchange.com/questions/42756/formation-of-a-stable-hydrate

Formation Of Gemdiol
Carbonyl compounds react with water to give gem diols. This reaction is catalysed by acid. The reaction is reversible reaction.

Gem diols are highly unstable compounds hence equilibrium favours the backward direction. The extent to which an aldehyde or ketone is hydrated depends on the stability of gem diol.
Stability of gem diols depend on the following factors:
(i) Steric hindrance by +I group around α-carbon decreases the stability of gem diols. +I group decreases stability of gem diol and hence decreases extent of hydration.
(ii) Stability of gem diols mainly depends on the presence of –I group on α-carbon. More is the –I power of the group more will be stability of gem diols.
(iii) Intramolecular hydrogen bonding increases stability of gem diols. –I groups present on carbon having gem diol group increases strength of hydrogen bond.
More is the strength of hydrogen bond more will be the stability of gem diol.
Overall stability : (CH3)2C(OH)2 < CH3CCl(OH)2 < CH3CN

Your stability order seems incorrect. I also suggest to add at least one (and better three or more) examples of stable gem-diols.

explain ur saying.

Ask me tomorrow again if you can't guess yourself.

If you still didn't guess, elimination of HCl is much more preferable than of water.

Geminal diols are unstable due to repulsion between the two oxygens present on the same carbon.If a compound is able to disperse the electron pair of oxygens by phenomenon like hydrogen bonding,resonance,-H effect(hyperconjugation) etc,their stability will increase.

Simply not true.

Well this is what i was taught in school, and it does help to solve questions based on the stability of diols.I must be wrong then,could anyone please tell me what's the correct concept?

Earlier answer is better


text:
Is it correct to write Na2B2O4 as 2NaBO2?
My teacher wrote the reaction for heating of borax like this:
$$\ce{Na2B4(OH)4 .8H2O->Na2B4O7 + 10H2O}$$
$$\ce{Na2B4O7->B2O3  +Na2B2O4}$$
And told us that $\ce{Na2B2O4}$ can be written as $\ce{2NaBO2}$,is this correct?
If yes then why?

No, $\ce{Na2B2O4}$ can not be written as $\ce{2NaBO2}$. $\ce{Na2B2O4}$ means 1 mole of $\ce{Na2B2O4}$ and $\ce{2NaBO2}$ means there are 2 moles of $\ce{NaBO2}$. Both are different compounds and have different molar mass.
Actually in real $\ce{2NaBO2}$ should be written as $\ce{(NaBO2)2}$. Now, both $\ce{Na2B2O4}$ and $\ce{(NaBO2)2}$ are same.


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Does SiO compound exist?
If carbon monoxide can exist then why can't we have silicon monoxide?
Both carbon and silicon belong to group 14 and have similar chemical properties.

Short answer: It does, but unstable and hard to detect reliably..
Long answer.
First of all, we should look at structure of carbon dioxide and silicon dioxide. It is pretty easy to see that silicon forms polymeric dioxide with no double bonds, while carbon forms molecules with two double bonds. It is not an anomaly, obtaining silicon compounds with double bonds $\ce{Si-E}$ is tricky at best. This is attributed to larger atom size and more diffuse $p$-orbitals for silicon, making formation of double bonds unfavoured in most circumstances.
Now, let's look at carbon monooxide. It is a molecule with bonding very similar to that in dinitrogen molecule. The carbon atom has formal negative charge and oxygen has formal positive charge. However, because of high electronegativity of oxygen, electron density is pulled back towards oxygen, so the resulting dipole moment is diminutive. The molecules contains not double, but triple bond, making formation of similar molecule for silicon highly unfavorable. Still, silicon monooxide can be detected in gas phase.
Now, why solid phase with composition $\ce{SiO}$ is not formed is a more interesting question. Unfortunately, such questions not always have an easy answer, as rules guarding stoichiometry of solid phases is often base on arcane concepts of atom packings and electron count. It is entirely possible, that such a phase would be obtained one day in some exotic conditions. For clarity, I would use sterical considerations. $\ce{Si-O-Si}$ fragment is roughly linear, while $\ce{Si}$ atom favors tetrahedral coordination. Going from here, only two simple phases with no dangling bonds can be formed, based on diamond structure. It is structures with either $\ce{Si}$ or $\ce{SiO4}$ units in the structure nodes. It IS an oversimplification, as silicon dioxide forms many rather esoteric crystal structures, because $\ce{Si-O-Si}$ fragment isn't perfectly linear, but is is close enough in my opinion.
Do not rely on oxidation states in prediction of existence of covalent compounds,   +2 oxidation state for silicon is a thing. It exists in $\ce{Si6Cl12}$, adopting structure similar to cyclohexane.

A solid with the composition does form according to https://en.wikipedia.org/wiki/Silicon_monoxide.  It is, however, polymeric and not molecular SiO.

Since your question is a school-level question, I think the answer is as follows. Though carbon and silicon have similar electron configurations in their valence orbitals, silicon is one shell higher than carbon. That makes the bonding energies different. $\ce{Si}$ only has an oxidation number of $\ce{+4}$, so it must have $\ce{2}$ oxygens to form $\ce{SiO2}$. Carbon, however, can have an oxidation number of $\ce{+4}$, $\ce{-4}$ and $\ce{2}$, so it can form $\ce{CO}$.
In reality, $\ce{SiO}$ exists in nature. Silicon monoxide is the chemical compound with the formula $\ce{SiO}$ where silicon is present in the oxidation state $\ce{+2}$. In the vapour phase it is a diatomic molecule. When $\ce{SiO}$ gas is cooled rapidly, it condenses to form a brown/black polymeric glassy material, $\ce{(SiO)_{n}}$, which is available commercially and used to deposit films of $\ce{SiO}$.
Silica itself, or refractory materials containing $\ce{SiO2}$, can be reduced with $\ce{H2}$ or $\ce{CO}$ at high temperatures,
$$\ce{SiO2(s) + H2(g) ⇌ SiO(g) + H2O(g)}$$
As the $\ce{SiO}$ product is collected and removed, the equilibrium shifts to the right, resulting in the continued consumption of $\ce{SiO2}$.

Whether the glassy material is actually silicon monoxide is a widely disputed issue. See this paper to get an insight.

The gist I think is that SiO is a very reactive species and only really exists in a gaseous state under "exotic" laboratory conditions on earth.

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


text:
Does the change in Gibbs free energy depend on initial and final concentrations?
Suppose there is a decomposition reaction, and by varying different factors, the equilibrium can be disturbed to alter the final concentration of compounds. So, does the value of ∆G here depend upon the initial and final concentration? The basic rule of ∆G = G(products) - G(reactants) seems unsatisfactory, because, it seems pretty absurd to me, that the value of ∆G is the same when 0.01% of a gas decomposes, and when 85% of it decomposes.

You're mistaking molar and total enthalpy

You may got confused: you assume an equilibrium, yet your intuition seems to be about a non equilibrium decomposition.

$$\mathrm dG = -S\mathrm dT+ V\mathrm dP + \sum \mu_i \mathrm dN_i$$
If (products) and (reactants) are at the same temperature and pressure then $\mathrm dT$ and $\mathrm dP$ are both $0$, and $\mathrm dG = \sum \mu_i\mathrm dN_i$. Integration gives $\Delta G = \sum \mu_i\Delta N_i$. $\mu$ is the chemical potential which is just the specific molar Gibbs free energy. For the decomposition of $\ce{AB -> B + A}$ this is $$(\mu_\ce{A}\Delta N_\ce{A} + \mu_\ce{B}\Delta N_\ce{B}) - \mu_\ce{AB}\Delta N_\ce{AB}$$

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Please use MathJax for chemical and mathematical expressions; I have edited your post to do so. Check out the [help], this meta-post or this one.

Where does that formula come from? All I've ever studies is that ∆G = ∆H - T∆S and that ∆G = G(products) - G(reactants)...

The equation $\Delta G = G(products) - G(reactants)$ refers to a change of state in which you start out with pure reactants and end up with pure products.   Of course, if you start out in thermodynamic equilibrium state A (containing an equilibrium mixture of reactants and products) and you make a change so that you end up in thermodynamic equilibrium state B (containing a different equilibrium mixture of reactants and products), the change in free energy is not given by that equation.  However, you should be able to calculate what that change in free energy actually is (between states A and B).

And, how would that be? Also, if once I completely decompose 1 mol of a substance, and after that 2 mol of it, is the ∆G same? Does the quantity of matter not influence it in any way??

I can show you how to calculate this, but there would be some "arithmetic" involved.  Are you sure you feel up to it?  Are you familiar with working with  "chemical potentials?"  With regard to your second question, if I understand you correctly, the $\Delta G$ would not be the same.  Since free energy is an extensive property, the quantity of matter does influence it.


text:
Are there atoms with different numbers of protons, but the same number of electrons, with different orbital occupations?
Imagine there is an electron-stripped nucleus with $Z$ protons, in high vacuum and isolated. Then you add electrons, one by one, and check which orbitals are filled. Will this depend on $Z$ at all, or does it only depend on the number of electrons that are added?
In other words, will an ion have different orbits filled than the corresponding neutral atom with the same number of electrons?
In other words: are there atoms with numbers of protons $Z_1$ and $Z_2$, both with $n\leq\operatorname{min}(Z_1, Z_2)$ electrons that have different orbital fillings?

Configuration does depend on nucleus, see for example http://chemistry.stackexchange.com/questions/2469/how-can-one-explain-niobium-s-weird-electronic-configuration

Yes, the number of protons in the nucleus does affect how the electrons are filled into orbitals. There are a number of real-world examples that usually stem from transition metals since they do not only have a rather rich oxidation state chemistry but also two energy levels (3d and 4s) which are rather close by each other.
To give you a concrete example, take $\ce{Fe^2+}$ and $\ce{Cr^{\pm 0}}$. They both have the same number of electrons. However, chromium’s ground state electronic configuration is given as $\mathrm{3d^5 4s^1}$, while iron(II) typically has $\mathrm{3d^6 4s^0}$.
The general explanation is that the higher nuclear charge in the iron nucleus attracts the electrons better than the lower nuclear charge in chromium. This difference causes the orbitals to contract. The d-orbitals which are closer to the nucleus, are contracted more and thus experience more stabilisation than the s-orbitals of the next shell.

Since electrons prefer to maintain their lowest energy states, electrons orbiting a stripped nucleus will occupy the lowest available energy levels before filling higher levels (1s before 3p, for example). This is called the Aufbau Principle.
Your question about a striped nucleus is a little confusing. Obviously an electron filling a hydrogen ion will have different energy states compared to an electron rushing toward a helium ion.
Check out https://en.wikipedia.org/wiki/Aufbau_principle as a starting point to your research.
Hope this helps!

My next few minutes will be absorbed by analyzing the link given by Mithoron. I have learned about the Aufbau Principle quite recently, and that is exactly what made me ask this question: is the Aufbau Principle influenced or independent of the number of protons in the nucleus? Will the sequence in which new electrons are "assigned" to the shells be independent of Z?

@ Ruggiero, I don't see any wrong in your answer, and I thank you for trying to help. It was not me that downvoted it and I really can't see any reason for anyone to do so.


text:
Jahn-Teller Distortions in Square Planar Complexes?
A Jahn-Teller distortion is predicted whenever a non-linear symmetric molecule has degenerate orbitals and has unequal electron occupation in those degenerate orbitals. Of course, this most often is discussed in the context of octahedral complexes, but it also occurs in other species - it's a general effect.
When I was teaching inorganic chemistry this term, my students asked if this could occur in square planar complexes. My immediate answer was "yes because, of course, there are degenerate MOs:"

Later, I was trying to figure out what the exact geometric distortion would be. As Wikipedia notes:

Jahn–Teller theorem does not predict the direction of the distortion, only the presence of an unstable geometry.

I consulted the Walsh diagram posted above from Orbital Interactions in Chemistry by Albright, Burdett, and Whangbo.
My initial inclination would be a "scissor" distortion towards a tetrahedral shape since I know $\ce{T_d}$ complexes undergo similar Jahn-Teller distortions.
Based on the Walsh diagram, depending on the electron count (e.g., $\ce{d^3}$ as suggested by the students) it seems as if you would always have a Jahn-Teller distortion predicted, because you would have at least 1 electron in the $\ce{2e}$ orbitals (the structure would be $\ce{D_{2d}}$ as indicated in the figure.
I understand based on LFSE, such complexes wouldn't be stable, but the question remains: Wouldn't it drop down to $\ce{C_{2V}}$ to break degeneracy? If so, how?
What is the specific geometric Jahn-Teller distortion for unstable (e.g., $\ce{d^3}$ square planar complexes?

Introduction
Thank you for prompting me to look for square planar complexes that are not $\mathrm{d^8}$; I learnt some valuable stuff while researching the answer to this question! Do take this answer with a grain of salt, because I cannot access the paper I’m referencing here on holidays (I could when I’m back at work) but the supplementaries confirm what I’m trying to prove.
Dataset
Scott et al. reported the crystal structure of a square planar vanadium(II) complex in 1990. The central vanadium ion is surrounded by the oxygens of four substituted phenolates. A set of two oxygens is also coordinating a lithium ion to give an approximate trigonal planar geometry on the oxygens, and said lithium ions are further coordinated by a THF solvent molecule (again giving a trigonal planar geometry around lithium) such that there is an approximately straight line going through the following atoms: THF-oxygen, lithium, vanadium, lithium, THF-oxygen. I shall define this as the $x$ axis for further discussion; the square planar geometry being in the $x,y$ plane and $z$ being vertical.
Experimental section
I took the atom coordinates of vanadium, the six oxygens, the four oxygen-bound phenol carbons and the two lithii from the supplementary information. My first step was to move the coordinate system’s origin into the vanadium ion. Then I transformed into spherical coordinates and adjusted the axes so that when retransforming back into cartesian coordinates my definitions from above are valid.
Results
The vanadium-coordinating oxygens are slightly displaced from the $x,y$ plane — in sets of two such that one is about as much above the plane as the other is below the plane. This distortion carries through onto their corresponding carbons. The lithium ions are then again almost spot on $z=0$ — also coordinated by one oxygen with positive $z$ and one with negative $z$.
The oxygens do not form a perfect square. Two oxygens are slightly displaced outwards in $x$ direction; two in $y$ direction. The $y$ displacement is almost twice as strong as the $x$ displacement.
Discussion
The distortion of the square planar geometry resembles the one you predicted: Two opposite oxygens above the plane and two below the plane. This is coherent with a reported $\angle(\ce{O-V-O}) = 162^\circ$ (diagonally across the square).
The authors also reported a large $\angle(\ce{O-V-O}) = 103^\circ$ and a small $\angle(\ce{O-V-O}) = 79^\circ$ all between two neighbouring oxygens and a set of longer and shorter $\ce{V-O}$ bonds differing by $2~\mathrm{pm}$ all in accordance with the slight displacement in the crystal coordinates.
Seen as a whole, the entire thing is indeed en route towards tetrahedral, but also features additional slight distortion that further reduces symmetry from a $D$ point group to a $C$ point group.

Conclusion
The $\mathrm{d^3}$ ion vanadium(II) features a distorted square planar geometry. The distortion seems strong enough to consider a point group of $C_\mathrm{2v}$ around vanadium rather than $D_\mathrm{2d}$.
Reference
M. J. Scott, W. C. A. Willish, W. H. Armstrong, J. Am. Chem. Soc. 1990, 112, 2429. DOI: 10.1021/ja00162a059.

You can start from either $D_{4h}$ or $T_d$ and use group theory to determine the symmetry of the Jahn-Teller active modes by determining the direct product of the degenerate state term symbol with itself.  Only the symmetric portion of that the direct product less any totally symmetric irrep can serve as J-T active mode(s). For $T_d$, the J-T active modes would be $e \text{ and } t_2$ (since $t_2 \otimes t_2 = a_1 \oplus e \oplus [t_1] \oplus t_2$).  One component of the e-mode is the "scissoring" mentioned above.  Depending on your perspective, this mode is also a distortion toward $D_{2d}$ and then further to $D_{2h}$ symmetry.


text:
Can gases be "immiscible"?
Anyone know of any examples of gas mixtures that phase separate, analogously to the way immiscible liquids do?
They would have to be extremely non-ideal. I might guess one might have to be much higher molecular weight than the other and at least one if not both would have to have significant intermolecular interactions (although of different kinds).

Can an "extremely non-ideal gas" still be called a gas? One could argue that a liquid is a type of extremely non-ideal gas. Depending on how much you relax that condition, something might eventually work, but it might not resemble a gas much by then.

I'd guess you could get phase separation as you approach the critical point of one (or both) of the fluids as you suggest. This is obviously a bit of a fluffy question but I'm curious about things that a kid would identify as a gas.

That's about what I thought you were trying to go for with your question. Unfortunately in that case I'm pretty sure the answer is that there is no simple way for gasses to spontaneously phase separate, so you'll probably have to bring in some rather heavy considerations (e.g. a gas of photons, a plasma, some external potential, intrinsically quantum mechanical effects, etc) which aren't exactly intuitive.

Side question, you assumed that the molecular weight would have to be imbalanced between the gases. If I was looking at liquids to guess if the were miscible, I would be looking at the surface tension. Are there 2 liquids with similar surface tensions that will not mix? Are there liquids with similar weights that do mix, don't? Ahh, whoops, excellent question mate. +1

Can liquids be immiscible? Sure they can. Now what is the difference between a gas and a liquid, if you can smoothly walk from one to another without any transition - that is, if you walk around the critical point?

This question provides an interesting thought experiment. The rub comes with the "analogously" part of "...analogously to the way immiscible liquids do."
You could ask, for example, are dogs analogous to jellyfish...but the answer would be "yes and no."
Gases do "settle out" on planetary scales. Helium can (and will) escape the earth's atmosphere. So if you made a bottle big enough to hold the earth inside, the helium would be the "top layer."
However, when we think of the miscibility of liquids, we are often thinking about the interactions between molecules that force congregation or separation (polar non-polar interactions, for example) and not gravity related thermodynamic-style considerations. If you compress a gas out of a mixture of gases, then you made a liquid, or tricked a gas into separating using Graham's law.
If we mean, miscibility as the property of substances to mix in all proportions. Then yes, we can mix gases in a bottle at standard state. Dalton's law would hold.
If we mean miscibility as a strict homogeneity only seen under conditions of rest then nothing is absolutely miscible, really.
But...I love your question and it got me researching some fun articles.
Check out atmospheric concentration and vertical structure, it is fascinating!
http://ruc.noaa.gov/AMB_Publications_bj/2009%20Schlatter_Atmospheric%20Composition%20and%20Vertical%20Structure_eae319MS-1.pdf
It might also interest you to look up some articles on mixture settling. http://webserver.dmt.upm.es/~isidoro/bk3/c07/Mixture%20settling.pdf
Hope this is useful to you!

Random baloney, this. Gases mix perfectly, under practically all conditions except in ultracentrifuges and >100km. As do miscible fluids.

Good question but there is no such thing as immiscible gases. Immiscibility is caused by surface energy/tension.  Gases do not have surfaces and therefore lack surface tension.

Yes.  Not at standard conditions, but at elevated temperatures and pressures there can be gas-gas immiscibility.  See Liquids and Liquid Mixtures by Rowlinson for more details.  Some examples include argon+ammonia, methane+ammonia, ethane+water, carbon dioxide+water, and benzene+water.
At the pressures where these occur the densities are comparable to liquids, but the temperature is above the critical temperature of both components, so gas-gas equilibrium is an appropriate term.

The composition of a mixture of gases is due to gravity, which separates, competing with thermal motion which mixes. Gases are totally miscible but that does not mean that the concentration has to be uniform throughout a mixture. Both gases have a lower concentration lower down, but the gradient is less for the lighter gas, so hydrogen would be almost uniformly distributed in a room, but chlorine will be far more concentrated near to the floor but don't try hydrogen and chlorine as they explode in strong light!

This answer is not completely my own and uses some ideas from Mcruggs' answer. I think that at 21°C and 1 atmosphere, the answer is no but it might be possible when either the pressure is 800 atmospheres or the temperature is 3 K. Even if a gas at this temperature and pressure is exactly at its boiling point, its atoms will still be widely spaced apart because according to Nelson chemistry 11, its volume for a given amount and pressure varies linearly with its Kelvin temperature and not with the amount above its boiling point. You can have 2 gases that can't mix in every proportion because they react with each other on mixing but no two gases have a tendency to have a sharp surface between them and be in solubility equilibrium with each other. Since the molecules are so widely spaced apart, their attractions to each other are negligible so when 2 gases that don't react with each other are put together a container, the tendency for molecules to diffuse greatly exceeds the tendency for them to them to molecules to attract molecules of the same type even if the gases are the polar HCl and the nonpolar He gas. According to Bubble's answer at https://physics.stackexchange.com/questions/76842/the-statistical-nature-of-the-2nd-law-of-thermodynamics, the second law of thermodynamics is probably not an absolute law. The statement that all gases that don't react with each other or condense upon mixing in a closed container at this temperature will almost evenly mix which can be deduced from the second law of thermodynamics turns out to be true anyway. From the presence of gravity, the proportion of each gas might vary continuously with height at a really low rate. If you start with a heterogeneous mixture of gases of different density, the denser gas might sink to the bottom in the near future but they're eventually uniformly diffuse into each other just like in my video at https://www.youtube.com/watch?v=ay0mJHPvj88, tea can sink to the bottom of a cup of hot water after a short time but will eventually uniformly diffuse into the water.


text:
Desorption instead of adsorption from solution using activated charcoal
I was just trying to see the effect of concentration on adsorption in solutions, i.e. I was trying to verify the Freundlich isotherm. I used oxalic acid and a batch of homemade activated charcoal (using this site).
First I prepared solutions of different concentrations. Then I added the activated charcoal, stirred it for 5 minutes and filtered the solution. When I titrated the filtrate using KMnO4, after calculating the concentration, I found that the concentration of the solution had been increased!
I tried the same process with different concentrations of oxalic acid solution and the result was the same. Can anyone tell me where did I go wrong?
Note: I used the same batch of activated charcoal in all the experiments. (Even then, how did the concentration of the solution increase? Shouldn't it be the same or lower than the initial concentration?)
For titrating oxalic acid, I heated the solution to about 60 degrees C.( After filtering the solution)

In such experiments you typically run a blank. So add the homemade activated charcoal to just some water, filter, then titrate that filtrate as a blank.
There are two possibilities for the increase.

(1) Your batch of homemade activated charcoal was contaminated with who knows what soluble compound. (This seems the most likely...)
(2) Quite possible that some of the charcoal particulates are small enough to not be trapped by the filter.

Another way to run the experiment would be to add the homemade activated charcoal to just some water, filter, then wash the wet charcoal into the oxalic acid solutions (this would remove soluble compounds from the charcoal). Then dilute all the solutions to some co0nstant volume, filter, and titrate the filtrates. (You should still run a blank.)

then how should i carry out the experiment?

Edited answer which hopefully answers your question.

authors: 金匙室內設計工程有限公司


text:
How does polydimethylsiloxane stick to glass?
Polydimethylsiloxane (PDMS) is the main ingredient in Rain-X, a treatment that adheres to windshields.

How does PDMS adhere to class? It can't form covalent Si-O bonds with glass, according to one source. Makes sense, as the Si is already coordinated with four groups in PDMS and none of them seem to be easily removable.
A similar windshield treatment chemical, FAS17, was noted by the same source as being able to form covalent bonds with glass in the presence of an acid to cleave the ether linkages.

The source, however, didn't detail how exactly PDMS would adhere to glass. So what major forces would be involved in the adhesion of PDMS to glass?
Source:
http://hydrophobicsam.blogspot.com/2012/04/science-of-self-assembling-monolayers.html

If I had to guess, I'd go for Van der Waals interactions, which are the usual culprits when something is adsorbed but not forming covalent or ionic bonds with the surface.

This paper https://www.researchgate.net/profile/Jordan_Berg/publication/3329826_Studies_on_Surface_Wettability_of_PolyDimethyl_Siloxane_PDMS_and_Glass_Under_Oxygen-Plasma_Treatment_and_Correlation_With_Bond_Strength/links/0a85e53a032f8ec175000000.pdf also says Van der Waals forces.  In order to get chemical bonding to occur a plasma treatment is required.

@user1945827 - I would suggest writing that up as a more thorough answer, including quoting relevant passages from the link, since links can break. (You may also want to use the DOI rather than a link to a particular website.)

From what I've read, it does bond to glass in the same way as FAS.  However, since the addition of acid doesn't break down the polymers of PDMS into monomers, dimers, and trimers as it does for FAS, the bonds formed to the glass are weak, which explains why most customer reviews of Rain-X complain that it doesn't last very long.

If you ask a chemist about PDMS and its adhesion to glass, she will immediately think of the work done in constructing micro-reactors.  This is because one method of manufacturing micro-reactors is to stick a patterned piece of PDMS to a piece of glass; the PDMS is patterned such that it has reactor channels that are sealed when it is stuck to the glass.  The glass is usually a micro-scope slide or something similar.
With the above in mind, in order to answer your question I began to search through the literature which describes the preparation of micro-reactors from glass and PDMS.  What they were saying was that in order to form a bond between the molecules within the glass and the molecules within the polymer (PDMS) the glass had to be pre-treated,

"most of the research papers mention the
  use of oxygen plasma for developing chemical (siloxane) bonds be-
  tween the participating surfaces"

from "Studies on Surface Wettability of Poly(Dimethyl)Siloxane (PDMS) and Glass Under Oxygen-Plasma Treatment and Correlation With Bond Strength", DOI: 10.1109/JMEMS.2005.844746.
Since the manner in which your material is applied does not have any sophisticated pre-treatment then the only alternative method of adhesion (as pointed out earlier in this thread) is Van der Waals forces.
If this isn't quite the answer you expected, at least it is a good start.

One reason for surface cleaning is that glass and most other surfaces accumulate "gunk" (various organic molecules) over time. A clean surface, naturally will produce better adhesion.

As the above comments mention, van der Waals interactions are the primary types of intermolecular forces forces involved in the adsorption of any molecule onto a surface. Given that PDMS does not react with the glass surface chemically, it is pretty much certain that it is adhering through the combination of many weaker intermolecular forces (dipole/dipole, dipole/induced-dipole, and induced-dipole/induced-dipole interactions). You do not need covalent bonds for a molecule to adhere to the surface, given that there are enough weaker intermolecular forces holding it in place (covalent bond dissociation energies are on the order of hundreds of kcal/mol, whereas van der Waals interactions are an order to two orders of magnitude weaker). PDMS is a polymer, so given its length it should not be a problem to have many van der Waals interactions with the surface.
Indeed, if we consider the molecular structures of PDMS and glass (which is made from silica), we see that the Si-O bonds in PDMS and glass are permanent dipoles (oxygen is more electronegative than silicon). Therefore, dipole/dipole interactions are certainly at play. Given the rest of the structure, it is also safe to assume that dipole/induced-dipole interactions will be present, as well as induced-dipole/induced-dipole interactions.
However, from a quick search I learned that glass is not only composed of silicon and oxygen. Glass can also contain additional species such as calcium and sodium ions which are the result of treating silica with sodium carbonate and/or calcium carbonate during the manufacturing process. Compounds such as these are typically added to lower the melting temperature of silica as well as to keep the finished glass from deteriorating (source: http://www.cmog.org/article/what-is-glass). Given that there are these additional ions (and perhaps others) present in  glass, it follows that ion/dipole and ion/induced-dipole interactions are also contributing to the adsorption of PDMS onto the glass surface, in addition to van der Waals forces.
Because these forces are directional, I suspect that PDMS adsorbs onto the glass such that the Si-O bonds are facing the glass surface. This way, it may maximize its interaction with the Si-O dipoles in the glass, as well as the calcium and sodium cations. As a result, the less polar methyl groups would be left pointing outwards from the surface, giving the coated surface its hydrophobic properties.

As someone who works in the silane chemical manufacturing business, I can tell you most of the answers here are a little short sighted. Yes plasma is used all the time - for more difficult substrates. They're not using plasma pretreatment on windshield glass.
It may rely on van der Waals interactions, but that would likely make for a mess of a coating.
I would guess Rain-X is getting away with just not saying how their PDMS is functionalized. It likely has silanol or alkoxy functionality that can hydrolyze and form a covalent bond with the glass. Because there is so little functionality, relative to siloxane chain, they don't have to call out exactly what it is in the SDS, and can get away with saying "oh yeah, it's PDMS".

The simple answer to the question is hydrogen bonding to the glass.  Glass has silanol (>Si-OH) groups on the surface. by rinsing with an acid you can ensure that the proton is on the glass rather than other net positive species.  In addition, perhaps some of the glass is etched or dissolved slightly, thereby cleaning the surface. If you look at the PDMS structure you can see that the oxygens are positioned with their lone electron pairs available to hydrogen bond with the glass.  This allows the hydrophilic side of the PDMS linear molecule to associate with the glass while also directing the hydrophilic portion of the molecule outward.  The result is a glass surface that appears to be very hydrophobic.  The surface tension of the water is so strong that the water rolls off the windshield easily - the water beads on the low energy or hydrophobic surface.

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


text:
Questions related to distinguishing dimagnetism and paramagnetism and when is there magnetic properties
If a transition metal ion has both paired and unpaired electrons, would it be considered as showcasing paramagnetism or dimagnetism? I know that Fe2+, which has an abbreviated electronic configuration of [Ar]3d6 shows paramagnetism even though 1 of the d-orbitals have paired electrons. How about a transition metal ion with 7 / 8 electrons in d orbitals? Do we determine paramagnetism / dimagnetism based on whether there is greater number of paired / unpaired electrons or if a transition metal ion has unpaired electrons, it must be paramagnetic? I was thinking that there are many other orbitals in Fe2+ like the 3s, 3p orbitals and all of which contain paired electrons, so won't that outweigh the unpaired electrons in 3d orbitals?
Also, the teacher taught the idea paramagnetism and diamagnetism in the context of transition metals and said that transition metals show magnetic properties. Why don't other species show magnetic properties cause they can also have paired / unpaired electrons?

Do we determine paramagnetism / diamagnetism based on whether there is greater number of paired / unpaired electrons or if a transition metal ion has unpaired electrons, it must be paramagnetic?

HyperPhysics writes:
"Diamagnetism is a property of all materials and opposes applied magnetic fields, but is very weak. Paramagnetism, when present, is stronger than diamagnetism...
That means the diamagnetism from any paired electrons is almost negligible compared to the paramagnetism from any unpaired electrons present - even if it's just one unpaired electron. So, as long as there are unpaired electrons, the material will be observed to be paramagnetic.
Watch out with your $\ce{Fe^2+}$ example. You are right in saying it is a $\mathrm{d^6}$ ion, but the ligands attached to it will determine whether it is diamagnetic (the low-spin case, e.g. with $\ce{CN-}$ ligands) or paramagnetic (the high-spin case, e.g. with $\ce{H2O}$ ligands). This phenomenon is explained by crystal field theory.

Why don't other species show magnetic properties cause they can also have paired / unpaired electrons?

All species do show magnetic properties. The most famous example is diatomic oxygen, which is paramagnetic because of its unpaired electrons. Water, which has all its electrons paired, can be shown to be diamagnetic (again, just search Google - there are even YouTube videos on it).


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Phosphorus standard state
Why is white phosphorus considered to be the standard state of phosphorus although it is only metastable? Red phosphorus is thermodynamically more stable than white phosphorus, yet it is not the standard state.

It is correct that white phosphorus is usually used as the standard state for tabulating thermodynamic data, in particular by the Committee on Data for Science and Technology (CODATA).
A comparison of thermodynamic data for phosphorus and some of its compounds can be found in Rard, J. A.; Wolery; T. J. The Standard Chemical-Thermodynamic Properties of Phosphorus and Some of its Key Compounds and Aqueous Species: An Evaluation of Differences between the Previous Recommendations of NBS/NIST and CODATA. J. Solution Chem. 2007, 36 (11), 1585–1599.. This publication also includes an explanation for the use of white phosphorus as the standard state:

Although white phosphorus is not the thermodynamically stable allotrope, the red and black forms are difficult to prepare in pure form, which makes them less suitable for quantitative thermodynamic measurements.

Similarly, from the other side of the Atlantic, A.T. Dinsdales' SGTE Data for Pure Elements, originally published as CALPHAD 15(4) 317-425 (1991), says:

[The] reference phase is normally the phase stable at 10$^{5}$ Pa and 298.15 K. The exception to this rule is phosphorus for which, by convention, the white form is chosen as the reference phase because the more stable red form is difficult to characterise.


text:
How long the block starting with element 121 will be?
I remember from my chemistry classes that (after the initial irregularities) a new block of elements starts every two periods. After the initial s-block and p-block following it shortly, we have d-block starting at period IV, and f-block starting at period VI.
Now that Element 118 has been discovered, we're about to open period VIII and we're two elements short of a new block.
What block will it be? How many groups, what name etc?

As you noted, this is a very appropriate question in light of the IUPAC announcement that we have just finished filling Period 7!
The names of the subshells s, p, d, and f are named after the old spectroscopic terms sharp, principal, diffuse, and fundamental. We ran out of fancy names after that, so the subsequent subshells are named in alphabetical order - g, h, and so on - which means that after the 8s block is filled, we would theoretically have a 5g block.
The orbitals in the g subshell would be labelled with the quantum number $l = 4$, so $m_l$ would take integer values between $-4$ and $4$ (inclusive) giving a total of nine g-orbitals. Each g-orbital could hold two electrons with opposite spins, so the g-block would have $18$ electrons.
However, it is worth noting that the electronic configurations may or may not obey the aufbau principle fully. Whether the 5g orbitals will actually be filled or not will certainly not be easy to determine, considering how short the half-lives of those elements are likely to be.
Wikipedia has an article which talks about it.


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Breakdown of butyl rubber
I'm in my high school's theatre crew and of course we build a lot of props and such. However I just opened (yesterday) an old drawer of screwdrivers and they smelled like puke! I know that the chemical responsible for that smell is butyric acid, so the rubber must have been a butyl rubber. Could someone point out the chemical equation that shows the reaction for this?

If ever, it would be rather isobutyric or isovaleric acid, as butyl rubber is copolymer of isobutene with few percents of isoprene.

One really does not know much until a chemical analysis is done.


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Is it possible, given a target temperature, to create an alloy that melts at precisely that temperature?
Assuming the target melting points are between the melting points of mercury and tungsten, is it possible, given a target temperature, to create an alloy (or compound) composition that has a direct (say eutectic) transformation to liquid at +/- one degree of the target temperature?
If yes, is there a method for identifying the elements to use based on the target temperature?

That is actually a hard question to answer properly. Do you mean an alloy (or compound) composition that has a direct (say eutectic) transformation to the liquid?

The answer hinges on how you define "precisely". Melt within +/- 5 degrees yes. All melt within +/- 0.01 degree no. So you'd have to make a 80/20 by weight mixture of the metals which by the phase diagram should melt at temperature T. You then melt the metals and quickly quench the melt. The composition will be almost homogeneous, but not absolutely.

Thanks for the feedback, I've updated the question based on your comments

Probably no. Alloys melt at precise temperature in two cases: eutectics and intermetallics.  Given number of metals it is reasonable to expect that some eutectic or intermetallic would melt quite close to required temperature. But identifying such composition is another matter entirely.

You can identify materials (alloys and many others) with certain physical characteristics (such as melting point within a temperature range) and see what emerges from a database - I use Matweb for this sort of thing on occasion. I entered the temperature range of -38.83 $^{\circ}$C to 3422 $^{\circ}$C and got 619 hits under the "Solder/Braze Alloy" category. I now know that 97% lead, 2.5% tin, and 0.5% silver solder melts between 303-310 $^{\circ}$C, with the lower number corresponding to the solidus and the higher one to the liquidus. Might be of help to you...

The other facet to this is do you want the metal to melt/solidify at that temperature or just melt once at that temperature. // !@#$%^& theoretical rabbit holes.

Adding to the above - from one eutectic point one may be able to add other elements to move the eutectic point around. This, of course, rapidly becomes quite complex. (As I type this I'm also calculating the Al-La phase diagram. So the thermodynamics of phase transitions are kind of up my alley.) I'll also add that many intermetallics do transform directly to the liquid.

What do you mean by create? If experiments allowed, then sure, yes.

thanks all! If anyone wants to venture an answer I'd accept it. Based on your comments it sounds like it's possible but there's no straight line from target temp to composition parts.

If you open the question up to more than just metals the answer is likely yes, otherwise, no


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Interchange of axial and equatorial groups in Me2NPF4

The $\ce{^{31}P}$ NMR spectrum of $\ce{Me2NPF4}$ when recorded at room temperature consists of a quintet with intensities $1:4:6:4:1$. However, at $-100~\mathrm{^\circ C}$, the spectrum is as shown below:

Propose a structure for $\ce{Me2NPF4}$ that is consistent with these observations.

The phosphorus has five groups surrounding it so it must have some sort of five-coordinate geometry. The low temperature spectrum shows a triplet of triplets with similar coupling constants, which is what you would expect from a trigonal bipyramidal structure with two axial and two equatorial fluorines. On this basis I assigned the structure as below:

However, I don't understand why all four fluorines are equivalent at higher temperatures. Does this mean that they are constantly interchanging and if so, what is the mechanism for this?

https://en.wikipedia.org/wiki/Berry_mechanism

See "Suggest possible structures for the compound PF3(Ph)(NMe2) using the NMR data" for some background and follow the link to Berry pseudorotation.

What's really bothering me about this NMR spectrum is the lack of $\ce{H-P}$ coupling. I made a number of phosphonate compounds of the general type $\ce{RP(O)(OCH3)2}$ for my PhD, and the methoxy groups showed  $^3 J_{\ce{H-P}} \approx 10-20\ \text{Hz}$. In just about any structure I could propose for $\ce{Me2NPF4}$, there should be some $\ce{H-P}$ coupling.

@BenNorris The lack of $\ce{H-P}$ coupling is due to quadrupolar relaxation by the intervening nitrogen.  See my comment to Jan's answer for more detail.

Obviously, in the structure you drew you have two sets of two fluorines which are non-identical: The axial ones that form a 4-electron-3-centre bond with phosphorus and the equatorial ones that are bound by the classical 2-electron-2-centre bond. And also obviously, at higher temperatures this difference vanishes so the fluorines must somehow interconvert into each other (that is the easier explanation when compared to a different structure and is thus preferred by Occam’s razor) rapidly on the NMR-timescale. You noticed this already.
The mechanism for this is called Berry pseudorotation, and it involes a square pyramidal transition state as shown in this image taken from the Wikipedia page:

(The image is drawn with pentacarbonyliron but just exchange iron with phosphorus, four carbonyl groups with fluorine and the fifth with dimethylamine and voilà.)
Since this mechanism involves a significant change in the binding situation (we no longer can write two clearly 2-e-2-c and one 4-e-3-c bond(s)), we need a certain activation energy — that is usually supplied thermally. At $-100~\mathrm{^\circ C}$, there is not enough thermal energy to induce this change; at room temperature there is.
Ben Norris mentioned another important observation in a comment: There is not $^3J_\ce{PH}$ coupling visible. I suspect that this is due to the spectrum actually being $\ce{^{31}P\{^1H\}}$ — a proton-decoupled phosphorus spectrum.

The absence of $^3J_\ce{PH}$ coupling is due to the fact that the coupling path involves the nitrogen and $\ce{^14N}$ is a quadrupolar nucleus - quadrupolar relaxation effectively washes out the coupling. Jan, I suspect you understand this, but for others who might like to read a bit more about quadrupolar relaxation and when it will and will not effectively decouple nuclei, here are links to two earlier answers on the subject (ref_1,

ref_2)

@ron Actually, I didn’t know about that. I hope I remember to read your references soon (kinda bad time atm) and then hopefully correct my answer accordingly. Thanks for pointing out!


text:
Why is conjugation not possible if there is not present alternate $\pi$ molecules according to MO theory?
I was reading this page at Chemwiki; there they were discussing conjugated-$\pi$ bond.
In order to show that conjugation requires  alternate double bonding, they cited the example of 1,3,6-heptatriene:

To be considered conjugated, two or more π bonds must be separated by only one single bond – in other words, there cannot be an intervening sp$^3$-hybridized carbon, because this would break up the overlapping system of parallel 2p$_z$ orbitals. In the compound below, for example,  the C1-C2 and C3-C4 double bonds are conjugated, while the C6-C7 double bond is considered to be isolated, due to the effect of the sp$^3$-hybridized C5.

I've a problem with their explanation; firstly are they following MO theory or VB theory? While describing molecules with MO, they, suddenly from nowhere, used the concept of hybridisation. Why?
Can anyone explain only with he help of MO theory why, there needs to be alternate $\pi$ bonding in the molecule for conjugation to happen without use of hybeidisation?

I have a feeling that you don't exactly understand what MO theory is so you may don't know what your asking for - there was no MO stuff in what you mention.

I'll try and answer when I have access to my Mac.  But in the meantime, your exaxt question is answered in organic chemistry by Warren, Clayden and Greeves...

I think some help will be to read "Orbitals are not real". Orbitals are mathematical constructs. What's more important is that MO and VB theory are not competing, but complimentary, approaches to modeling multi-electron wavefuntions. They emphasize different aspects of electron behavior and depending on what you want to know, the maths may be simpler for one over the other (or whatever approximations were needed).

And to second NotNicolaou, just about any organic chemistry textbook will answer this question using MO theory only. It has to do with accessible orbitals having appropriate symmetry.

@Mithoron: Are you saying MO is not required for conjugation?


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Diffusion into PMMA
My thesis is about effect of cigarette smoke to PMMA polymer, apparently cigarette smoke have some amount of eugenol which is mildly acidic and they will evaporate when the cigarette is burned.
In the normal situation PMMA is subject to constant hydrolysis but the reaction is very slow, when there is an acid present the acid become a catalyst for hydrolysis reaction.
my hypothesis is the PMMA will be weakened because it absorbs some water from humidity and when there is eugenol vapor in air , Eugenol will diffusing to the water in PMMA polymer. now I need to know how to calculate how much eugenol concentration present in water inside the polymer.

I have found that my specimen at least contain 14722.5 µg water.

"concentration present in water inside the polymer"? Your polymer is solvent not water

@Mithoron I dont understand , why my solvent would be the polymer not the water inside the polymer ?

In a solution, the solvent is the component present in the highest amount. Not all solvents are liquids.

@BenNorris now I get it :D , but if the eugenol is dissolved in PMMA polymer rather in water inside the polymer how could it make it the environment acidic ? isn't acid only present in the form of Aqueous ?


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Why we use -adien- / -atrien- infixes if we need to indicate double bond positions with numbers?
For exemple: instead of Buta-1,3-diene why we don't use but-1,3-ene. In my mind, there would be no confusion when using the nomenclature that way because we use one number to indicate the position of each bond. Is there any particular reason for the use of these infixes?


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Why is Borax used as a primary standard?
It's given in my book that borax is a useful primary standard for titrating against acids. I have heard that primary standard is a pure reagent and does not gain any weight by absorbing moisture. (Hence any error during weight measurement is prevented).
I got this from a website that came first on searching "primary standard"

Primary standard is a reagent that is extremely pure, stable, has no waters of hydration, and has a high molecular weight .

But borax (represented as $\ce{Na2B4O7.10H2O}$) has water of hydration.

Yes borax is a hydrate, but the reason it is favored for titration is that it gives consistent results, which to most scientists, is far more important than accuracy.
The $\ce{ .\!10H2O}$ hydration number will remain constant and will not fluctuate to $\mathrm{11}$ or $\mathrm{9}$ within normal temperature and humidity ranges. This means that per mass the same amount of sodium ions are dispensed every time and gives the same results time after time and experiment after experiment even with different batches of powder. Also the tetraborate conjugate base is very weak, which is ideal for a titration agent as well.


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Why does copper react only with nitric acid?
When we are entering the chapter for acid and base, one of the reactions is about reacting metal with acid to form a salt and hydrogen gas. My teacher also informed us that a metal which is at a lower position than hydrogen in the electrochemical series, such as copper, cannot react with acid. It is because copper cannot displace the hydrogen atom from the acid.
However, I have seen several videos from YouTube which show copper can react with nitric acid as follows:
$$\ce{Cu + 4HNO_3 -> Cu(NO_3)_2 + 2H_2O + 2NO_2}$$
My question does why nitric acid react with copper, unlike the other acids? And are there any other acids which can react with copper?

Copper is an unreactive metal and doesn’t react in normal circumstances with dilute acids. However, it does react with nitric acid. Why is this?
Nitric acid is an oxidizing agent and the reaction is not the usual acid + metal reaction. The products are oxides of nitrogen instead of hydrogen. The actual nitrogen oxide formed depends on the concentration and temperature of the acid.
There are actually two equations for the reaction of copper with nitric acid. It depends on whether the nitric acid is concentrated or not. If it is concentrated and in excess then the ratio is 1:4 copper to nitric acid. If it is dilute then the ratio is 3:8.
$$
\begin{align}
\ce{Cu + 4HNO3 &–> Cu(NO3)2 + 2NO2 + 2H2O}\\
\ce{3Cu + 8HNO3 &–> 3Cu(NO3)2 + 2NO + 4H2O}
\end{align}
$$
Copper also reacts with hot conc. $\ce{H2SO4}$ but, here $\ce{H2}$ gas is released as follows :
$$
\begin{align}
\ce{Cu(s) + H2SO4(aq,conc) &-> Cu^2+(aq) + SO4^2-(aq) + H2(g)}\\
\ce{Cu(s) + H2SO4(aq, dil) &-> no ~reaction}\\
\ce{Cu(s) + HCl &-> no ~reaction}
\end{align}$$

Not $SO_4^{2-}$ in concentrated sulfuric acid.  The anion is bisulfate, $HSO_4^-$, instead.


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Solubility of Silver Nanoparticles at different pH
Consider the following exercise:

(i) What happens when you put $\ce{Ag10}$ and $\ce{Ag5}$ clusters into an aqueous solution of $\mathrm{pH} = 13$?
(ii) What happens when you put $\ce{Ag10}$ and $\ce{Ag5}$ clusters into an aqueous solution of $\mathrm{pH} = 5$?

In the previous questions they have asked to calculate the solubility product of $\ce{AgCl}$ and the standard potentials of $\ce{Ag5}$ and $\ce{Ag10}$, which I calculated them to be $\pu{1.4E-10 V}$, $\pu{-0.5184 V}$, $\pu{0.0816 V}$, respectively. The question  has given no other data.
I have no idea how these 3 values that I have calculate has anything to do with the solubility of $\ce{Ag5}$ and $\ce{Ag10}$, let alone how I am suppose to relate it to pH.
Could anyone point me into the right direction.

I don't think the $K_{sp}$ of AgCl has anything to do with it, but if you gave me this question without further info, my first thought would be "surely one species must be favored over the other under different conditions." If you assume there's excess silver ions in the solution, could you maybe use the Nernst Equation to figure out where the equilibrium lies?

@chipbuster how could I figure out where the equilibrium lies from the nernst equation? Also, how can I relate this to pH?

Let's say you have hydrogen ions in solution, along with some silver ions. There will be some half reactions that turn silver clusters into silver ions, and vice versa. The Nernst Equation will tell you what the potentials are away from equilibrium (e.g. if you add acid, you move the equilibrium--the Nernst eqn tell you how much). You can use the pH to find the concentration of H+ ions that you're actually dealing with. Note that this is just a sketch of an idea---I don't know if you can actually make the calculation work out (you may be missing some info you'll need along the way).

This is no way an answer but it is only a vague guess.  I am guessing both Ag10 and Ag5 has surface charges(+ve, -ve or Zero). In low  Ph there will be lots of H+ ions which will react with negatively charged silver cluster and make it neutral and at high Ph -OH ions will react with positively charged Ag cluster


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Color differences between the different physical states of oxygen
Why is oxygen gas colourless whereas in liquid and solid states oxygen is blue in colour ? Also why is the blue colour more intense in the solid state than in the liquid state ?


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Hexagonal unit cell vs crystal system
Is the hexagonal unit cell (resulting from hexagonal closed packing) same as the unit cell of hexagonal crystal system?
The unit cell of hexagonal crystal system is described by the following sides and angles: $a=b\neq c$, $\alpha=\beta=90^\circ$, $\gamma=120^\circ$,
where $a,b,c$ are the sides of unit cell and $\alpha,\beta,\gamma$ are the angles.

This is the unit cell of hcp

In the unit cell of hcp there are 17 atoms whereas in unit cell of hexagonal crystal system there are 8. Is this true?

I this, you are totally confuse with meaning hexagonal unit cell and hexagonal crystal system. First of all, I make you to clear about these two terms.
Hexagonal unit cell is single entity of hexagonal crystal system. It means there are a no. of repetitive hexagonal unit cells inside a hexagonal crystal system.
Yes, the diagram given in the question is of hcp i.e. hexagonal unit cell.
But, your next statement : In the unit cell of hcp there are 17 atoms whereas in unit cell of hexagonal crystal system there are 8.Is this true?
is incorrect. One thing I want to clear you that as per saying there are 17 atoms in unit cell of hcp is wrong. We can make the statement after correction is : To make a unit cell of hcp, we need 17 atoms. It means that there are 17 atoms present in unit cell hcp but to make a unit cell of hcp we require 17 atoms.
Now, we calculate the no. of atoms present inside the unit cell of hcp or we can say this as no. of effective atoms of unit cell of hcp.
Calculation of no. of effective atoms of unit cell of hcp :
No. of effective atoms of unit cell of hcp = 3 (there are a total of three atoms present inside the unit cell of hcp) * 1 (as all three atoms are in main unit cell of hcp, no part of atom is outside the unit cell of hcp) + 2 (there are two atoms present a centre of each base of hexagonal unit cell) * 1/2 ( half part of each atom is inside the main hexagonal unit cell) + 12 (there are a total of 12 atoms present at 12 corners of each base of hexagonal unit cell) * 1/6 ( one-sixth part of each atom is inside the main hexagonal unit cell) = 6 (not 8)

Ohkay i got your point i meant 17 atoms are used to make a hcp unit cell whereas we use 8 atoms to make a hexagonal crystal unit cell.I thought that hexagonal crystal unit cell is used to make hcp but after seeing the difference in their unit cells I am not sure whether  hexagonal crystal system unit cell and hcp unit cell are linked or not.

So basically I want to know the difference and similarities between hcp unit cell amd hexagonal crystal system unit cell amd how they are interlinked or different.

"we use 8 atoms to make a hexagonal crystal unit cell". This is again wrong. It is 6 not 8. Yes, both are interlinked with each other. Let's I clear u by simple example. There is a big cube in which a no. of small but of same sizes cubes are placed inside the given big cube with no vacant spaces between the small cubes. Now, a single small cube is called hexagonal unit cell and big cube is called hexagonal crystal system. I think u got my point.

I am not talking about effective atoms in a unit cell,I am just talking about number of atoms present in a unit cell.And I know the difference between hexagonal crystal system lattice and hexagonal crystal system unit cell.I want to know the difference between hexagonal closed packing unit cell and hexagonal crystal system unit cell.There are 8 atoms in hexagonal crystal unit cell and 17 in hexagonal closed packing unit cell,by saying the 8 and 17 atoms I am not mentioning the effective atoms present but the number of atoms present in each unit cell.I think you have misunderstood my question.

First can u tell where is written "There are 8 atoms in hexagonal crystal unit cell and 17 in hexagonal closed packing unit cell". Now, I already provide u the difference between the two. If u did not get my point, u tell me what is your thinking or opinion about their difference. So, on that I tell u my opinion.

Only pink portion of blue hexagonal prism is the hexagonal unit cell that is 1/3 rd of the hexagonal prism is the unit cell. This picture probably clears your doubt.