text:
Why is an ionic bond a chemical and not a physical bond?
Ionic bonds seem to be intermolecular but are classified as chemical bonds.

"Ionic bonding is a type of chemical bond that involves the electrostatic attraction between oppositely charged ions." -  Wikipedia, definition of an ionic bond
"The physical force of attraction which holds atoms and molecules in a matter is called a physical bond. van der Waals' forces and coulombic forces are physical forces." - Chemistry Stack Exchange, definition of an intermolecular force
"A chemical bond is an attraction between atoms that allows the formation of chemical substances that contain two or more atoms. The bond is caused by the electrostatic force of attraction between opposite charges, either between electrons and nuclei or as the result of a dipole attraction." - Wikipedia, definition of a chemical bond

Before an ionic bond is formed, oxidation and reduction must occur. For example, let's use sodium chloride. A chlorine atom will steal the outer electron from a sodium atom. The chlorine is gaining an electron, i.e. being reduced. The sodium is losing an electron, i.e. being oxidized. I agree that this is a purely chemical process.
Now that the atoms are oppositely-charged ions, they can attract and form an ionic bond:
$$\ce{Na+ + Cl- -> NaCl}$$
My question is this: How is this bond defined as a chemical bond?
Let's take water; $\ce{H2O}$. The oxygen forms two polar-covalent bonds with two hydrogen atoms. Each hydrogen shares two electrons with the oxygen. The electrons are being shared, so this bond an intermolecular force. The atoms stay together because they both depend on the electrons they share.
Now let's take more than one water molecule. These molecules will form hydrogen bonds. The hydrogens on the water molecules do not often have an electron around them because the oxygen has a greater attraction with them. This leaves the lonely proton of the hydrogen to make that end of the water positive while the oxygen is negative with its electrons. The negative end of the water (oxygen) will become attracted to the positive end of another water molecule (hydrogen), forming our anticipated intermolecular force. Does this sound familiar?
I hope it does. According to my knowledge, an ionic bond is that, but instead of a dipole-dipole interaction, it's a monopole-monopole interaction. If you look at the bond itself and not the reaction forming it, it acts just like an intermolecular force.

Geez, chemistry is physics applied to molecules. I think you got it from http://chemistry.stackexchange.com/questions/230/what-is-the-difference-between-physical-and-chemical-bonds - I don't know why this guy even asked about "physical bond" as normally no bond is called like that.

No, this is something I've been wondering for a long time. I asked my chemistry teacher but to no prevail. Also, I hear "physical bond" all the time. It's generally a hydrogen bond, a dipole-dipole bond, or a London Dispersion Force bond.

Calling intermolecular forces "physical" is very bad idea as all interactions in whole universe are physical - it's misleading.

The problem stems from an insistence on categorizing properties and changes as distinctly nuclear, chemical, or physical.  Many properties and changes fit these categories nicely, but there are others (dissolution of an ionic compound, for instance) that do not.

Thanks everybody for trying to help. I know that it's not technically a physical bond but an intermolecular force, but it didn't answer my question.

Well, although this went into semantics not serious ion-dipol analogy don't lose hope yet - you may get another answers and your question is already quite upvoted, which raises the chances :)

Ionic bond is pretty much physical.

It doesn't make much sense to create such categories for bonds. Sometimes categories help to explain differences between systems, but when they live out their usefulness one doesn't need to take them too seriously... I think the concept is similar to surface absorption where we like to distinguish physical and chemical ones, based on the charachter of the bonds to surface.

Short version:
We don't call bonds "physical", there are chemical bonds and other types of interactions between particles. The chemical bonds are classified this way because they make up molecules, salts, polymers and such, which are the materials chemists are interested in studying, and not particularly because of their sub-atomic/electrostatic characteristcs.

Your question relates to two frequent sources of confusion:

How do we define a chemical bond? What makes a bond between two atoms ionic or covalent?
Where does chemistry (hence chemical interactions) ends, and consequentially where does physics (and hence physical interactions) begins?

Both questions have been pretty much answered here on Chemistry SE and also on Physics SE. I'll give some links to related questions and also a shrot summary.
1. How do we define a chemical bond? What makes a bond between two atoms ionic or covalent?
Your question didn't go much into covalent bonds, but anyone trying to understand anything past the surfaces of chemical bonds should know that the definition of a bond as covalent or ionic is simply an easier way for us to classify things, because when the bonds are categorized in groups we can say that one group behaves in a way, and the other behaves in a different way. The truth is, all bonds have some ionic and some covalent character to them.
Good answers here on Chemistry SE about that can be found here:

Can 100% covalent bonds exist?
What happens if the electronegativity difference is exactly 2.1?

2. Where does chemistry (hence chemical interactions) ends, and consequentially where does physics (and hence physical interactions) begins?
For a Physics pont of view, see: What distinguishes between physics and chemistry?
For a Chemistry point of view, see: Differences between chemical physics and physical chemistry?
Basically, it's clear that both sciences have a lot of overlapping, and perhaps the main difference is the approach taken and how the results will be used. While chemists usually focuses on understanding the structure and interactions between substances and materials, in order to be able to improve them or develop new ones, physicists are mostly interested in understanding the exact laws of nature and representing them with mathematical equations that can be used to explain and predict phenomena. This definitions do not seem so different, and that's not supposed to be wrong: We do have the Physical Chemistry and Chemical Physics fields.
Ok, but Why is an ionic bond a chemical and not a physical bond?
Two important things were mentioned by @Mithoron in the comments. The term "Physical bond" is not frequently used by chemists, (or anyone I suppose), pretty much because all interactions in whole universe are physical. This single quote by @Mithoron is related to the two points I highlighted above. Both "types" bonds "follow" the same laws, and there is no hard line between them.
It's also worth noting that the definition you gave for Physical Bond was taken from an answer without a single upvote, which is an indicator that it isn't what the Chemistry SE community thinks (despite the fact that Google liked it very much and uses it in it's short-answer for the query). Another indicator that "Physical Bond" isn't really a type of bond is that there aren't definitions of it out there (Yahoo answers doesn't count), not even a Wikipedia article.
So why do we call them chemical bonds?
That's pretty much because the bonds we call chemical are those which make up molecules, salts and other materials which are mainly studied by chemistry. The definition of chemical bond varies a lot (as explained here), but they are the bonds which create what chemists study.
Other types of bonds which aren't called chemical are the intermolecular interactions, which are obviously of interest to chemistry, but also the Strong Force, which is the force that binds protons and neutrons (nucleons) together to form the nucleus of an atom. (I hope it's okay to call Strong Force a bond, I'm not sure to be honest).

If this answered my question, it was buried in there somewhere. I guess what I'm truly asking is the similarities between an ionic bond and a dipole-dipole movement, along with the similarities between an ionic bond and a covalent bond. It seems like the former has more similarities and thus should be considered an intermolecular force.
P.S., I apologize for saying "physical bond." That's what I was taught in  my AP chemistry class, but I now agree that it is wrong.

@DylanSpano No need to apologize. You're right, I focused my answer primary on the basic concepts you mentioned and not on the specific bond properties you talked about. A dipole-dipole interaction happens because of a permanent dipole created by a covalent bond between atoms with very different electronegativities, but this dipole is not the same, or as strong, as a formal charge which the atoms have when they lose or receive an extra electron. This is why dipoles are represented with $\delta+$ or $\delta-$  while formal charges are simply + and -.

@DylanSpano I think that you'll get a more detailed answer if you ask a new question specifically about the differences and similarities of ionic bonds and dipole-dipole interactions, remarking that you're interested in their electrostatic properties and not simply on when they happen. Without the "physical bond" confusion your question will be clearer.

I did so here


text:
Why is HNO3 a stronger oxidising agent than H3PO4?
Why is $\ce{HNO3}$ a stronger oxidising agent than $\ce{H3PO4}$? $\ce{N}$ and $\ce{P}$ have the same oxidation number. Is it the electronegativity difference between $\ce{N}$ and $\ce{P}$?

Do you think the electronegativity of N helps? N: 3.04 while P: 2.19.

Yes, may be. N atoms accept electrons more easily than P atoms, and this can affect his reduction potential. But I'm not sure, the difference in electronegativity is not that big to explain the huge difference in reduction potentials.

related http://chemistry.stackexchange.com/questions/30863/why-is-no-bromine-liberated-in-the-reaction-of-potassium-bromide-and-concentrate

Let me create background first. Oxidizing agents are the chemical that helps something else oxidize and itself gets reduced.(reduction in charge)
$\ce{N}$ in $\ce{HNO3}$ is in the +5 oxidation state - how do we know that? $\ce{H}$ is +1, $\ce{O}$ is -2 and the overall $\ce{HNO3}$ has a zero net charge. The same goes for $\ce{H3PO4}$ resulting in +5 oxidation state of $\ce{P}$.
That +5 is the highest common oxidation state for $\ce{N}$. If $\ce{N}$ is reduced to say +4 as in $\ce{NO2}$ or +2 in $\ce{NO}$, it's charge is reduced and $\ce{N}$ is therefore an oxidizing agent for that particular reaction
So, $\ce{HNO3}$ is an strong oxidizing agent because the $\ce{N}$ can be readily reduced.
But $\ce{H3PO4}$ is a poor oxidizing agent. Here are two basic reasons:

Nitrogen does not possess $d$-orbitals in valence shell and so its covalency is limited to 4. $\ce{N}$ can however achieve a formal oxidation state of +5 as  in the $\ce{NO3-}$ ion. The inability of $\ce{N}$ to unpair and promote its $2s$ electron results in that $\ce{N(+5)}$ is less stable than $\ce{N(+3)}$. However, $\ce{P(+5)}$ is more stable than $\ce{P(+3)}$ and phosphoric acid shows less oxidising properties.
The affinity of phosphorus for oxygen is greater than that of nitrogen; as a result, phosphonic acid ($\ce{H3PO3}$) is good reducing agent.
$\ce{H3PO4 + 2H+ + 2e-> H3PO3 + H2O}$; $E=-0.276 \ce{V}$

Thank you so much, now I get it. I didn't think of the contribution of d-orbitals in P.

No, there is no need to bring hypothetical d-orbitals of phosphorus into the game. Electronegativity of the two elements, compared with that of oxygen is enough.

Frost Diagrams explain it very nicely as well. Look at the G/F = zE/V, very positive for HNO3, which makes it a potent oxidizing agent (much less stable). H3PO4 is basically the complete opposite.

The simplest explanation lies in electronegativity. N is more electronegative than P. And a general trend is that the more electronegative an element is, the more unstable  it is in high oxidation states. And since strong oxidizing agents typically are unstable, one can suspect that nitric acid is a stronger oxidizing agent than phosphoric acid.
However, to truly answer your question, one would need to consider the relative stability of the oxidized state and reduced state of the agent. And then lots of different considerations may come into play. For instance, nitric acid develops NO gas, giving an entropic contribution that phosphoric acid does not have.
Even though arguing with electronegativity is simple, it can be useful in a lot of different cases. For instance, high voltage cathode materials for Li-ion batteries depend on electronegative d-metals like Ni and Co. This is because a good Li-ion cathode needs to be a strong oxidizer.

To put it simply,  in the lewis structure of nitric acid the nitrogen atom has a formal charge of +1 whereas, in the case of orthophosphoric acid,  phosphorus bears no formal charge.
Nitrogen being an electronegative element  couldn't bear the positive charge and thus attracts electron from the other substance with which it reacts.
Also notice that ,nitrogen of nitric acid is sp2 hybridized and phosphorous of ortho phosphoric acid is sp3 hybridized  meaning that  nitrogen  is more electronegative   ( electron -pulling charge density increases with  more s orbital involvement)  thus, due to positive charge & sp2 hybrid nature, HNO3 is stronger  oxidizing agent than H3PO4 despite  the fact that, they both have same oxidation number .in my personal opinion, if one just goes about saying that HNO3 is stronger oxidizing agent, just because Nitrogen is more electronegative than phosphorous-  it is not a very good answer  on grounds that,  nitrous acid (HNO2) is not a stronger oxidizing agent than H3PO4 -  despite the fact that Nitrogen is more electronegative than  phosphorous ;  here, taking the lewis structure of HNO2 into account ,we find that N bears no formal charge with sp2 hybridization   more over, it shows smaller oxidation number than P of H3PO4.  also the  presence of  lone pair of electrons
makes it  a sort of reducing agent ( to a  little  extent) these all facts are drawn out to say HNO2 is weaker oxidizing agent than H3PO4;  in a nutshell, oxidization and electro negativity are bulk properties - one can't easily predict without giving thoughts to the bulk.

To put it simple is to put it wrong, at least in this case.

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text:
Finding equivalent weight of acid given its reaction to form a salt

Find the equivalent weight of an acid if it is given that $\pu{1.321 g}$ acid reacts with magnesium to give $\pu{1.72 g}$ salt. Options:

$49$
$36.5$
$45$
none of above.

The second choice is given as the answer.
I tried to solve the question. I used the formula
$$\frac{\text{wt. of acid}}{\text{equivalent weight of hydrogen}  \text{ eq. wt. of anion say 'x'}} = \frac{\text{wt. of salt}}{\text{eq. wt. of magnesium   eq. wt. of anion}}$$
And after solving I got $x=35.41$ taking equivalent weight of hydrogen as $1$, and $x=35.38$ taking equivalent weight of H as $1.008$. The answer is different in each case. Is this correct? But the correct answer given is $36.5$

If known salt is given I can find the solution. But I have no idea about magnesium. We can equate no. of gm. eq. wt. of metal = no. of gm. eq. wt. of salt.

I mean we have weight of metal divided by equivalent weight of metal = wt. of salt divided by equivalent weight of salt. If known salt is given (e.g. MgCl or MgSO4) we can find equivalent weight of salt. And hence by solving the above equation we know the ew of metal. But in this question ew of salt is not given and I dont kmow that how Mg helps to find ew of metal. We also dont have weight of Mg. I need weight of salt or Mg. I am preparing for entrance exam and this is an objective question. 4 options are given: 1)49 2) 36.6 3)45 4) none of above. In answer option b is given.

The trick is to recognize that the acid and the salt share an anion $\ce{A}$, for example:
$$\ce{2HA + Mg -> MgA2 + H2}$$
Thus the equivalent weights of the acid and the salt are functions of the only one unknown, the equivalent weight of $\ce{A}$.

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text:
How will the equilibrium partial pressure of carbon dioxide be affected?
\begin{equation}\ce{CaCO3(s) -> CaO(s) + CO2(g)}\quad\Delta H^\circ = +556\ \mathrm{kJ/mol} \tag{endothermic}\end{equation}
How will the equilibrium partial pressure of  $\ce{CO2}$ be affected if the volume is decreased?
So, I thought of course the moles will increase when you decrease the volume hence $pV = nRT$, thus $p = \frac{nRT}{V}$. Therefore, $p$ is inversely proportional to $V$. But, the right answer is NO CHANGE, can anyone help me to understand why?

The equilibrium constant for the reaction is
$$K=\frac{P_{\ce{CO2}}}{P^\circ}=e^{-\Delta G^\circ/RT}$$
Changing the volume changes neither $\Delta G^\circ$ or $T$ (at least this is implicitly implied), so $K$ and, hence, $P_{\ce{CO2}}$ does not change.
For an ideal gas
$$P_{\ce{CO2}}=\frac{n_{\ce{CO2}}RT}{V}$$
So if $V$ decreases then $n_{\ce{CO2}}$ must also decrease in order for $K$ to remain constant. So, when the equilibrium is perturbed by the volume-decrease the equilibrium shifts to the left to re-establish equilibrium, in accord with Le Chatelier's principle.

Okay let's recall what the Le Châtelier's principle is all about: (in short) whenever a system in equilibrium is disturbed the system will adjust itself in such a way that the effect of the change will be nullified.
Now let's look at your problem:
$\ce{CaCO3(s)->CaO(s) +CO2(g)}$  $\Delta H* = +556 \ce{KJ/mol}$ (endothermic)
Your ideas are correct, if we reduce the volume we are in fact increasing the pressure. (and in a way, this displeases our system)
Our system, naturally, reacts by shifting to a side which exerts a lower pressure or has fewer moles of gas (the left in this case).
Our system nullifies the effect of the stress applied to it.
Now because this shift occurs, we observe no change in the equilibrium partial pressure of carbon dioxide.
I hope this makes sense? Maybe, someone smarter than me can offer a clearer explanation.

Ok, we if we do know that the rxn goes to the left wouldnt the pressure of CO2 then decrease? i still don't get how it can remain the same and shift to the left at the same time as it kinda contradicts all the le chatelier problems i did before

Okay. Think of it this way..you reduce the volume (i.e increase the pressure). Now the reaction shifts, pressure dec. (i.e original pressure is restored)?


text:
Corrosion of a galvanic couple made with silver and gold
If a bangle is made out of gold and silver, connected with each-other would there be corrosion happening?
If so, can it be explained using the galvanic series?
Also do those metals undergo oxidation under normal conditions?

A very interesting question - I wonder if the carat (varying ratio of Au and Ag) value affects the rate of corrosion (if any are present)

As you state, though silver is not very reactive, silver jewelry would corrode more rapidly when in contact with gold and a bit of sweat as electrolyte. Though the difference in electronegativity of pure gold and pure silver is 0.15V, and is considered acceptably corrosion resistant, commercial jewelry contains copper and other metals in the alloys. There is clearly evident corrosion in photos of copper/silver jewelry. According to the Victoria and Albert Museum staff, "All metals, with the exception of 24 carat gold, suffer the effects of corrosion."
However, in a bangle, physical abrasion is likely to cause more damage and loss of material than galvanic corrosion. Wear the jewelry and enjoy it, but it will not last forever.


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Triflate use in metathesis reactions
Why are triflates so widely used in coupling reactions, but not tosylates, mesylates nor nosylates?

As a result of the -I effect of the trifluoromethyl group, triflate is a better nucleofuge than mesylate and tosylate.
As compared to tosylate and nosylate, triflate is rather small. This might be an advantage in reactions at "crowded" centres.
Triflates do not undergo side reactions, such as hydrogen abstraction reactions, etc. A triflate, once introduced, can easily survive several steps in a synthesis.


text:
Why are pi bonds only formed when sigma bonds are formed?
While studying about bonding there was one statement that "pi bonds can only be formed only with sigma bonds" as we know that in double bond there is 1 sigma bond and 1 pi bond but then one question arises: Why is a pi bond formed only when a sigma bond is formed? Is it possible to form a pi bond without any formation of a sigma bond?

There is reported som few examples of pi-bonds without sigma-bonds outside transition metals. See link
However, I haven't read the original research article myself, so I can't say whether absence of a sigma-bond is only when you calculate the bond"order" by one model only; or whether more models or experimental data beyond molecule structure supports this claim.

A $\pi$ bond has a plane of symmetry along the bond axis. It cannot be formed by s-orbitals; it needs at least p-orbitals to be created. $90\,\%$ of all bonds described some time or another are somehow involving carbon, nitrogen or oxygen. (In fact, I probably underestimated). But these elements can only use p-orbitals to create $\pi$ bonds. To do that, one needs a p-orbital that is ortohogonal to the bond axis. So you run into the problem that you have an orbital pointing in one direction, but want to bond into another direction — hardly optimal, especially since there likely is already another orbital pointing in the direction you need to give a $\sigma$ bond.
Transition metals can use d-orbitals for $\pi$ bonding. They can actually point towards the atom they want to bond with so there is a greater chance of using them due to higher overlap. However, there will usually also be a different orbital pointing directly in the bonding direction which again will bond earlier and would give a $\sigma$ bond.

So, can we never have a π bond without a σ bond? And is the σ and π bond representation valid for all compounds. I still have a lot to learn (I am in Grade 12), but I think these models would not be true for all systems...much like most of what I have learnt till now.

@AbsoluteZero There is nothing philosophical preventing a compound from having a π but not a σ bond between two atoms, it’s just that usually orbitals are such that a σ bond will be formed before a π one. The distinction between σ and π (and others) is valid for all compounds much like all atoms have occupied s orbitals and all atoms after beryllium have occupied p orbitals in their ground state.

$\sigma$-bonding MOs tend to have lower energy than $\pi$-bonding MOs so they will be formed first.  One explanation is that the $\sigma$-bonding MOs have a lot of $s$-AO character and $s$-AOs have lower energy than $p$-AOs.
I don't know of any double bond that is purely $\pi$-bonding. Basic MO theory suggest that $\ce{C2}$ should have a double bond made of two $\pi$-bonding MOs because the $\sigma$-anti-bonding MO is lower in the energy than the $\pi$-bonding MOs.  However, experiment tells us that $\ce{C2}$ has two unpaired electrons so this simple picture cannot be right.  This is explained in greater detail here and here.

In transition metal complexes with 2 metal centers that are bound to each other you can have the situation that the $\pi$-MOs (and sometimes also the $\delta$-MOs) lie energetically below the bonding $\sigma$-MO (all formed from d-orbitals).

I have a problem with your statement "One explanation is that the $\sigma$-bonding MOs have a lot of $s$-AO character and $s$-AOs have lower energy than $p$-AOs." While this is true for some cases it is not the main reason for $\sigma$ orbitals being lower in energy than their $\pi$ counterparts (sometimes, like in $\ce{C2}$, it is even the other way round: sp-mixing raises the energy of the higher lying $\sigma$ orbitals so that they lie above the $\pi$ orbitals). The more important reason is the better overlap in $\sigma$-type interactions compared to $\pi$-type interactions.

related http://chemistry.stackexchange.com/questions/32407/why-can-there-not-be-more-than-one-sigma-bond-in-a-set-of-bonds?rq=1, http://chemistry.stackexchange.com/questions/594/bonding-in-c%e2%82%82-a-carbon-carbon-quadruple-bond

All above answers are correct but I want to add some more details.
According to 'molecular orbital theory' molecular orbital diagram of $\ce{B2}$ is

Clearly $\ce{B2}$ forms 1 $\pi$-bond without any $\sigma$ bond.
But as stated in comments
,"Just because there is also an antibonding orbital doesn't make the bonding orbital not existing. Bonding in diatomic molecules is incredibly hard to explain and even then there's room for interpretation."
Source

It is more complicated than that. There is obviously a sigma bond, it is the lowest orbital in your scheme. Related: Why is diboron (B2) paramagnetic?

@Martin-マーチン do you mean $\sigma_1$ as lowest orbital? Your diagram in indicated question is similar to mine. I can't get you where you founded a sigma bond or it cannot be explained by MOT?

I think $\ce{C2}$ might work for what you want to show.

@NisargBhavsar if you get time just add it as your answer else I will edit my answer when I will get time thanks for the thought I forgot it.

@NisargBhavsar there is definitely a sigma bond in C2! https://chemistry.stackexchange.com/q/594/4945

Just because there is also an antibonding orbital doesn't make the bonding orbital not existing. Bonding in diatomic molecules is incredibly hard to explain. And even then there's room for interpretation. Please consider reading Jan Jensen's answer.


text:
How to get a precise reading from pipette?
While doing titrations, we are instructed to use pipette and they demonstrated how to use a pipette. But still after all this, we face difficulties of how to take precise reading from pipette.
Example-

Due to the concave water meniscus, until where to take the reading ?
Is it really important to take every single drop of solution from pipette ?
as while putting solution in beaker we see little bit drop left inside pipette.
And what's the best way to hold pipette to avoid breaking it and easy to see the reading which we are taking ?

The bottom of the meniscus (in the middle) is your measurement point and should lie on the line that you are trying to read.
This is completely dependent on the type of pipette that you are using. Some pipettes are designed so that you just leave the last drop and some you are supposed to just touch the tip of the pipette to the surface of what you've dispensed already to get the last bit. TD pipettes stand for "to deliver" (leave the last bit). TC pipettes stand for "to contain" (take the last bit out).
Again, this is dependent on the pipette. If you have a classic bulb pipette then you shouldn't hold it by the bulb as it is the weakest part and that's how you end up with glass in your hand.

Also if you have to take the last bit of liquid out you don't have to blow but just close off the top end of the pipette with your finger and the gently hold the bulb with your palm and let it heat up with the warmth of your hand. This will let the last drop fall out.

@Papul That was something new which I just got to know. And its really helpful. :D

@ShashankSharma Yeah I saw this on YouTube somewhere. Felt like an idiot blowing the pipette all these years.


text:
Is there any way of representing a dative/coordinate covalent bond?
I was wondering if there was any way of representing a dative/coordinate covalent bond on a chemical diagram.
A single bond can easily be represented as a single line, a double bond as a double line and a triple bond as a triple line.
Take the below structure of ethene, a compound with single and double non-coordinate covalent bonds: each bond is easily represented by a single and double line.

However, the ammonium ion does have a coordinate bond between the lone pair of electrons on the nitrogen and the H+ ion from whichever compound donated it (e.g. hydrochloric acid). Despite this, I am unable to see how a coordinate bond could be represented, as, to me, it just looks like 4 non-dative covalent bonds.
Is there any way of representing a coordinate/dative covalent bond on diagrams like these?

I have usually seen a coordinate bond represented by an arrow.  The arrow points from the atom donating the lone pair to the atom accepting it

I suggest doing a Google image search on "dative bonding"

Generally, you wouldn't describe the bonds in an ammonium ion as dative bonds. They are usually considered to be proper covalent bonds. The term "dative bond" is typically used for bonds in (transition) metal complexes between ligands and a metal center. But for those complexes Lewis structures have lots of weaknesses and are rarely a good representation of the actual bonding. Yet they are used for the sake of convenience. Often an arrow from the ligand to the metal center is used to indicate a dative bond.

(source: uni-erlangen.de)
But just as often you see normal "covalent bonds" (straight lines) like in traditional Lewis structures being used.
Often both kinds of representations are even intermixed.
For a more detailed view on the subject of bonding in coordination complexes and its representation have a look at this answer of mine.

According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), an ion may be best represented by its molecular formula when exact connectivity is not important. The formula should be placed in square brackets, and the charge should be written as a superscript following the rightmost bracket.
$$\ce{[CuCl4]^2-}$$
If it is important to depict atoms and bonds explicitly, the ion should be drawn in full. That structure should be enclosed in large square brackets, and the charge should be reported as a superscript following the rightmost bracket. (For single fragments, it is permissible to depict simply a corner at the top right of the diagram, with the charge as a superscript following the corner.)
Bonds representing coordination from one atom to a single other atom should be represented as normal plain single bonds.

The use of dashed bonds or dative bonds to represent coordination is not acceptable.

Plain bonds should be used for nonstereogenic atoms, even if the substituents are not physically coplanar. If a coordination bond is attached to an atom that does have a specified stereochemical configuration, then a hashed or wedged bond depiction should be used instead.
The structure should be drawn without concern for delocalized charges. Other structural depictions that involve localized charges should be avoided, even if that produces a diagram where some atoms appear to have non-standard valences. It is not acceptable to add charges that depict formal zwitterions.


text:
How to analyze a acid and salt mixture using titrimetry?
Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, $0.1\:\mathrm{M}$ $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.
\begin{align}
\ce{ M(OH)3 &~<=> M(OH)^{+}_{2} + OH^{-} } &
\mathrm{p}K_{\mathrm{b}1} &= 0.5 \\
\ce{ M(OH)^{+}_{2} &~<=> M(OH)^{2+} + OH^{-} } &
 \mathrm{p}K_{\mathrm{b}2} &= 0.7 \\
\ce{ M(OH)^{2+} &~<=> M^{3+} + OH^{-} } &
 \mathrm{p}K_{\mathrm{b}3} &= 9.0
\end{align}
All of the species in above equilibria are water soluble.
Progress so far:

$\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

\begin{align}
\ce{ M^{3+} + {H2O} &~<=> M(OH)^{2+} + H+ } &
\mathrm{p}K_{\mathrm{a}1} &= 14.0 - 9.0 = 5.0 \tag{a} \\
\ce{ M(OH)^{2+} + {H2O} &~<=> M(OH)^{+}_{2} + H+ } &
\mathrm{p}K_{\mathrm{a}2} &= 14.0 - 0.7 = 13.3 \tag{b}\\
\ce{ M(OH)^{+}_{2} + {H2O} &~<=> M(OH)_3 + H+ } &
\mathrm{p}K_{\mathrm{a}3} &= 14.0 - 0.5 = 13.5 \tag{c}
\end{align}

Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
Looking at the $\mathrm{p}K_\mathrm{a}$ values of $\mathrm{(b)}$ and $\mathrm{(c)}$ it can be assumed that reaction $\mathrm{(a)}$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.
Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

You mention both the titration curve and indicators. Are you planning to monitor the reaction with a pH meter? If so, an indicator might not be necessary.

@Molx No. No pH meters. Just want to sketch the curve before carrying out the titration.

What is the approximate concentration of the metal oxide? and of the hydrochloric acid?

@LDC3 No other information is given.

When you use the align environment, you can use & as an anchor: use odd appearances to align at that point, use even appearances  to separate equations in the same line. The \tag{...} will produce whatever you give as argument in text mode just left of the right margin.

Prologue
The following will show you not only how to sketch a titration curve but how to produce an analytical form of a titration curve. So this might not be an easy solution even if I try to keep it as easy as possible. There is a wonderful yet german book that has (to my knowledge) never been translated to english which describes everything of the following in detail:

Bliefert, Claus; Linek, Alfred; Morawietz, Gerd (1978): PH-Wert-Berechnungen. Weinheim: Verlag Chemie.

There might also be some imprecise terms because I don't know many of the english counterparts to the german terms I am familiar with.
Answer
Let's think a bit about your problem first to simplify it from scratch:

As you mentioned, reaction (a) will take place right after all protons from hydrochloric acid have reacted and before reaction (b) and (c) would start.
The pK$_a$ of (b) and (c) are so close to another that they are unable to be separeted by titration and
the pK$_a$ of (b) and (c) are so high that they are unable to be determined by titration in diluted aqueous solutions.

If we take a look at the titration curve of phosphoric acid, we can see what is meant by "too high to be determined". You can clearly identify the two inflection points for the first two protons but there is no third inflection point that could be determined. What is likewise to be seen, is that one can neglect this third deprotonation step in titration curves of diluted solutions (yellow vs. blue titration curve).

Right now we can reduce your problem to a mixture of hydrochloric acid and M$^{3+}$. To continue we need to establish a proton condition (Protonen-Bedingung):
$$c(H_3O^+) = c(OH^-) + c(A^-) + c(X^-) - c(HB^+) \tag{1}$$
It means that protons come from the autoprotolysis of water (c(OH$^-$)), from hydrochloric acid (c(A$^-$)), from your metal ion (c(X$^-$)) and that the protons react with the used base (c(HB$^+$)).
To define an equation for the acid residue anions a small equation system (where small c's denote actual concentrations in solution and the big C denotes the origin concentration)
$$\begin{align}c(HA) \cdot k_a - c(H_3O^+) \cdot c(A^-) =&~0\\c(HA)+c(A^-) =&~C(HA)\end{align} \tag{2.1}$$
needs to be solved to give
$$c(A^-) = \frac{C(HA) \cdot k_a}{c(H_3O^+)+k_a}. \tag{2.2}$$
This can be done analogously for the other acid and also for the base.
What remains is the concentration of the hydroxide ions from the autoprotolysis of water that is simply determined by rearranging the equation for the autoprotolysis itself:
$$c(H_3O^+) \cdot c(OH^-) = k_W \Leftrightarrow c(OH^-) = \frac{k_W}{c(H_3O^+)} \tag{3}$$
Combining everything so far we get the following proton condition:
$$x = \frac{k_W}{x} + \frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}} - \frac{x \cdot C(B) \cdot k_b}{x \cdot k_b + k_W} \tag{4}$$
where $c(H_3O^+)$ has been replaced by $x$ for the sake of simplicity.
The plot of titration curves usually shows the pH over the base volume (see eq. 8) or the degree of neutralisation ($\tau$ or $\gamma$) that is defined as $$\tau = \frac{c(B)}{C(HA)} \tag{5.1}$$
In the case of your mixture of two acids - and as always neglecting the increase in volume - it can be written as
$$\tau = \frac{c(B)}{C(HA)+C(HX)} \tag{5.2}$$
To sketch the titration curve let's say that we know the two concentrations of the acids. To calculate (5.2) we also need to know c(B) that can easily be obtained by rearranging (4) to get:
$$c(B) = \left(1+\frac{k_w}{x \cdot k_b}\right)\left(\frac{k_w}{x}-x+\frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}}\right) \tag{6}$$
If one decides to include the increasing volume the concentrations of the acids and the base have to be extended to give
$$\begin{align}C(HA) \rightarrow&~\frac{C(HA) \cdot V(HA)}{V(HA)+V(B)}\\C(B) \rightarrow&~\frac{C(B) \cdot V(B)}{V(HA)+V(B)}\end{align} \tag{7}$$
Inserting (7) in (4) and rearranging for V(B) we finally get a neat equation
$$V(B)=\frac{V(HA)~\left(\frac{k_w}{x}-x+\frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}}\right)}{\left(x-\frac{k_w}{x}+\frac{C(B) \cdot k_b}{\frac{k_w}{x}+k_b}\right)} \tag{8}$$
As the pH is defined as $pH = -log_{10}(c(H_3O^+))$ and $c(H_3O^+)=x$ you can replace every $x$ in equation 8 with $10^{-x}$.
Now you are able to calculate with e.g. Excel enough (V(B),pH)-pairs to draw a titration curve and get some information about your system of two acids during the titration with your desired base.
The following figure shows some examples of your mixture using equation 8.

Please let me know if someone is interested in the full equation including also the second and third deprotonation step with pK$_a$ values 13.3 and 13.5 and I will add it here.

(+1) This is great! Glad I finally found it and it confirms the equation I have been using (your equation 8) for a long time.

authors: Диана Бауыржанкызы


text:
Why are hydrogen ions always associated with another molecule?
I was reading N.C.E.R.T. class XI part 2 book and chapter Hydrogen,when I came across this statement:

Loss of the electron from hydrogen atom
  results in nucleus ($\ce{H+}$) of ~1.510–3 pm size.
  This is extremely small as compared to normal
  atomic and ionic sizes of 50 to 200pm. As a
  consequence, H+ does not exist freely and is
  always associated with other atoms or
  molecules.

So a question arose in my mind that why it is a consequence of small size of hydrogen nucleus that $\ce{H+}$ does not exist freely and is always associated with other atoms and molecules?

Think of what you are referring to when you say H+ (or a hydrogen ion)

i am referring to  a proton.

http://chemistry.stackexchange.com/questions/22193/is-it-possible-for-hydrogen-to-lose-its-electron

The way I understand it is (and my understanding is by no means perfect, or complete), as you pointed out correctly: a hydrogen ion is in fact a proton. The proton is a "bare charge" and as you rightly said, "tiny". a
This makes it extremely reactive (in a sense), and thus in a chemical system of any sort would immediately seek out and associate with the electron clouds of a surrounding molecule.
For instance, in an aqueous solution it forms $\ce{H_3O+}$ (hydronium ion, which are further solvated by water molecules and form a manner of aggregates).
An extremely interesting way of looking at it was pointed out in the comments (just putting it here, so no one misses it)

An interesting way to thinking about the reactivity of a bare proton is to calculate its charge density and compare with other monocations, such as Li+. This will make it clear why free protons will abstract electrons from just about anything. –  Nicolau Saker Neto

Do note (as far as I know, someone might correct me if I am wrong) free protons are fairly stable (by free I mean not bound to electrons or other nucleons) and usually don't decay into other particles. They exist in several naturally occurring situations, high energy plasmas come to mind.

Free protons are extremely if not indefinitely stable to nuclear decay, with a half-life of at least about $10^{35}$ years. An interesting way to thinking about the reactivity of a bare proton is to calculate its charge density and compare with other monocations, such as $\ce{Li+}$. This will make it clear why free protons will abstract electrons from just about anything.

Nicolau, thank you for your comment, and I do prefer your way of looking at the problem, hence I have edited and added it to my answer. hope you don't mind

Sure, no problem at all. I actually wound up discussing it a long time ago, in this answer.


text:
Significance of single point energy when calculating interaction energies
I am currently investigating about the interaction behavior of a few atoms in certain conditions.

Is it possible to use the concept of single point energy to represent the atomic interaction energies or I have to go other way around?
What is the basic difference between potential energy and single point energy?

The very language you use is a bit strange, so I'm not sure we are on the same page. How do you actually investigate "the interaction behavior of few atoms"?

@Wildcat sorry for that. I'm basically using DFT/mp2 to obtain the interaction energy(potential energy or single point energy)  between atomic pairs like (O...H) or (N...H) or (F...H).

What you can obtain using electronic structure calculations is the single point energy. If you do a number of calculations for few different nuclear configurations you than single point energies will define the potential energy for nuclear motion. But it is not the interaction energy. For details, see my updated answer.

Single point energy arises in the framework of the Born–Oppenheimer approximation and corresponds to just one point on the potential energy surface. Physically it is the total energy of the molecular system with its nuclei beeing fixed (or clamped) at some particular locations in space. In other words, it is total energy of the molecular system within the so-called clamped nuclei approximation.
Mathematically, if you develop the Born–Oppenheimer approximation step-by-step you can easily see that single point energy it is the sum of the electronic energy and nuclear repulsion potential energy,
$$
U = E_{\mathrm{e}} + V_{\mathrm{nn}} \, ,
$$
where the electronic energy $E_{\mathrm{e}}$ is the solution of the electronic Schrödinger equation,
$$
    \hat{H}_{\mathrm{e}} \psi_{\mathrm{e}}(\vec{r}_{\mathrm{e}}) = E_{\mathrm{e}} \psi_{\mathrm{e}}(\vec{r}_{\mathrm{e}}) \, .
$$
The fact that at this point we use the symbol $U$ which (alongside with $V$) is usually used for potential energy to mean the single point energy is justified a little later. Namely, when we introduce the Born-Oppenheimer approximation which give rise to the nuclear Schrödinger equation,
$$
    \Big( \hat{T}_{\mathrm{n}} + U(\vec{r}_{\mathrm{n}}) \Big) \psi_{\mathrm{n}}(\vec{r}_{\mathrm{n}}) = E \psi_{\mathrm{n}}(\vec{r}_{\mathrm{n}}) \, ,
$$
it is easy to recognize that the values of the single point energy $U$ for all possible nuclear configurations define the potential energy for nuclear motion. So, it is in this sense that the single point energy is related to the potential energy.

Update: it became clear that OP misunderstood the notion of the single point energy $U$. Indeed, once we do few single point calculations for different nuclear configurations, the resulting $U(\vec{r}_{\mathrm{n}})$ is the potential energy for nuclear motion. However, it is not the interaction energy between some fragments, though, the interaction energy contributes to it. So if one wants to obtain the interaction energy one has to decompose the potential energy $U(\vec{r}_{\mathrm{n}})$ into its parts.
There different ways to perform the energy decomposition, just to name a few without any particular order:

SAPT (Symmetry-Adapted Perturbation Theory) a separate program (few of them, to be more precise) which can be interfaced with different quantum chemistry codes.
NEDA (Natural Energy Decomposition Analysis) which is available as a part of NBO package.
Morokuma decomposition already available in some quantum chemistry codes (GAMESS-US, for instance).
LMO-EDA (Localised Molecular Orbital Energy Decomposition Analysis) also available in some quantum chemistry codes (GAMESS-US, for instance).

Ok, so what I'm getting is single point energy from DFT/mp2. If I get these single point energies for multiple atomic-configurations(as possible) in a particular distance-range then am I to get the potential energy surface (applicable to those atomic configurations and limited to that distance-range) ?!

Yes, you need to scan the PES (as we call it) at least in the direction of one coordinate, say, by elongating the bond. You will end up with the curve which is basically a slice of the PES. But interaction energy is just a part of the potential energy, so you have to decompose the potential energy than.

can you refer some literature for me on this matter. Thank you.

Single point energy is a point on the potential energy surface.
Suppose you have a single atom which has both potential energy (Due to interaction between electrons and protons and since there is no other atom or external force, all the potential interactions come from interaction between it's own electrons and protons.) and kinetic energy due to electronic motion. So we can write a hamiltonian (PE+KE) here. Here if you want to optimize this geometry you just need to optimize electonic hamiltonian because there is only one nucleus. It can be called single point energy as well as optimized energy. Now for two H atom, each have KE and also PE due to previously discussed interaction plus interaction between electrons and nucleus of each other. For example equlibrium H-H distance is 0.745 Angstrom (B3LYP theory). So if you put two H in a distance of 0.75 or greater, they will attract each other and if you put them in a distance less than equilibrium distance they will repel each other. And eventually they will come to equilibrium distance. So your potential energy curve will look like morse potential curve. This is optimization and the lowest point you find is optimized distance. Now, you want to find the force or you just want to know the energy when they are 0.8 Angstrom apart. What you can do is you can freeze their nucleus at those position and solve the schrodinger equation. Here since there is no nuclear degree of freedom, it will only optimize electronic hamiltonian to remove any kind of electronic overlap or to make the orbital orthogonal. This is called the single point energy. If you find all the single point energy for all distance, you can plot the potential energy surface and you can also find the lowest point on the PES, which is optimized geometry or you can find the maximum energy, which would be transition energy.

thanks for elaborate explanation. But how can I be able to explain the greater pot. energy of [C-C-C-O  H-N-C-C] compared to [C-O  H-N]. While all that matters as per my interest is the pot. energy between O and H atoms.

Your question is not clear. Can you please make it clear? I think you are having problem with the term used in computational chemistry because your question was also ambiguous.

I need to get the potential energy between O and H atom(given a particular cartesian geometry) using DFT/mp2. Its all going fine. But if I add some adjacent atoms too with this calculation (viz. C-C-C-O....H-N-C-C) then the potential energy value of interaction between O and H is getting higher. May be this is because dispersion issue. But how can I be able to explain it more scientifically .!?


text:
Reaction kinetics, relating half life to reaction rate

How do I relate the half life to the overall rate of reaction?
I argued that from the data, doubling the partial pressure of either reactant, keeping the other constant, will half the half life.
So try t1/2 =
$\frac{\ce{[1]}}{\ce{[k][PA_0][PB_0]}}$ and since pressure of A is always greater than B, so t1/2 = $\frac{\ce{[1]}}{\ce{[k'][PB_0]}}$ ?
I then wrote:
$$\frac{-\mathrm{d}[PB]}{\mathrm{d}t} = k [P_B]^m $$, where k = k[PA]

Recall the meaning of the half life $t_{1/2}$: At $t_{1/2}$, a concentration (or partial pressure) is decreased to the half of its initial value.
It is crucial to realize that this represents exactly the $\mathbf{-\frac{dp_A}{dt}}$ in your rate equation!
With other words, $-\frac{dp_A}{dt} = k\cdot p_A\cdot p_B^2$ becomes $-\frac{54}{2\cdot100} = k\cdot54\cdot(1.0)^2$ for your first set of data.
Solve for $k$ and do the same for the other data sets. What is the result?

I'm trying to understand how you can say that the concentration at half life can be equal to the rate of reaction - it looks like you are not considering the differential? It may be that it works out for this problem, but it seems like the first step should be to integrate the proposed rate equation, then make the substitutions.


text:
Is an ionic bond more like a covalent bond or an intermolecular force?
I have asked a question loosely asking this, where I confused terms and did not specify what I wanted to know here, so I formed a new question.
What are the differences and similarities between ionic bonds and covalent bonds and ionic bonds and dipole-dipole interactions?
It seems to be that an ionic bond is nothing but a stronger dipole-dipole interaction. The only difference is where the electrons are.
For example, let's take a sodium cation and chloride anion bond. The sodium had its electron taken away from it. It is now positive. Chloride had an electron added to it and it's feeling negative. Now the two will attract because of these charges.
Now let's take two water molecules. The hydrogen of one will be positive because it is less electronegative and thus the electron density will be greater around the oxygen, which is more electronegative, making it negative. Now the hydrogen of one and the oxygen of another will attract due to the charges.
It seems like the only difference is that if you split the sodium chloride and the water's hydrogens and oxygen, the sodium and chloride will still have a charge, but the hydrogens and oxygen will not.
By textbook, a dipole-dipole interaction is an intermolecular force while a covalent bond, as well as an ionic bond, is a chemical bond. It seems to me that, yes, the forming of ions is a chemical reaction, but the ionic bond is an intermolecular force.

Covalent bonds are directional but ionic bonds are not. A cation can be surrounded by a number of anions and vice versa and they will have ionic bonds equally strong.

That sounds to me like the ionic bond is even more like a dipole-dipole interaction.

An ionic bond could maybe be described as an inter-ionic force. All electron interactions are most accurately described by wavefunctions and quantum mechanics, but in practice we use successively more detailed approximations for convenience, stopping at the lowest level of detail that suits our needs at the time.
At the lowest-detail end of the spectrum, you have ionic bonds - we usually model these as simple Coulombic (a.k.a. charge-charge or electrostatic) interactions. This is good enough to describe bonding in most cases where the bond is between a metal and a non-metal. We model the bond as a complete electron transfer from one atom to another, which results in an electrostatic charge. The Coulomb force between the charges holds the ion together, and so we call it a molecule. What we really mean is "these atoms are now stuck together."
As we move along the periodic table and the metal atoms become less metallic and the non-metals become more metallic, at some rather arbitrary point we say "the ionic approximation is no longer good enough" and start seeing the bond as covalent. What we really mean is that the electrons are not fully transferred between atoms, and so charge-charge interactions don't explain bonding well enough. At that point we can no longer ignore quantum mechanics, and have to acknowledge that the electrons exist in molecular orbitals, not atomic orbitals.
That is the main difference between ionic and covalent bonds. As for dipole-dipole interactions, it again goes back to what the electrons are doing. In a molecule with a dipole moment, what this means is that within the molecule there are polar covalent bonds, and the geometry is such that the electron density on the surface of the molecule is lopsided - one side has more electrons than the other. The dipoles can align and will be attracted to each other via the Coulomb force, but this is not the same as how we model an ionic bond - in that case, one atom gives up one or more electrons completely. In a molecule with a dipole moment, one (or more) atoms just gets "more" of the electron density. Then the two separate molecules, each with their own dipole, are attracted to each other.
We call ionic bonds "intramolecular" forces because they are what hold molecules together - without ionic bonds, you couldn't have a salt crystal, for example. When you break apart sodium chloride, you get sodium ions and chloride ions.
We call dipole interactions "intermolecular" because they are what make separate molecules stick together. When you pull apart two water molecules, you still have two water molecules.
Thinking about it this way, an easy way to distinguish between intermolecular and intramolecular forces is to ask yourself whether you still have the same substance after pulling things apart. If you do, it must have been an intermolecular force. If you don't, it must have been a bond or an intramolecular force.
note - For very large molecules like proteins it gets a little tricky - you can have forces between sites on the same molecule that act like what we normally describe as intermolecular forces.

Thank you for the reply. I respect and understand the answer, but I'm not sure. Using your example of intramolecular forces: if you take just one NaCl molecule, it is just that. Just a NaCl molecule. Together, they make a NaCl crystal lattice. However, this is the same with solid H2O; ice. A single H2O molecule is just that. However, if you put more and more together, it, too, forms a crystal lattice. This being said, ionic bonds and intermolecular forces seem to make up the structure of molecules while covalent bonds form the structure of just one molecule (except covalent-network solids)

Adding onto that, since I ran out of space, it seems ionic bonds can determine the structure of one molecule and the structure of the molecules together. Meanwhile, dipole-dipole does the structure of the whole molecule, and covalent bonds determine the structure of a single molecule, discluding covalent networks. From this perspective, an ionic bond is both an intermolecular force and a chemical bond.

Ions don't form molecules in the sense you are thinking of, they form what are called "formula units" - the lowest whole number ratio of elements in the ionic solid.
I think you understand the physics: dipole interactions and ionic interactions are both electrostatic in nature. Where you are getting hung up is on the nomenclature. A molecule is two or more atoms joined by a covalent bond. By definition, ionic bonds can't be intermolecular forces.

I don't want to hear that something cannot happen "by definition." What I am trying to say is that the definition doesn't seem right.

Well, good luck with that approach!


text:
Respect to other chemicals in reduction/oxidation potentials
From what I understand, the standard reduction/oxidation potential table is based on the hydrogen half reaction which makes a reduction and oxidation potential of 0 respectfully.
In a research internship I'm in now, we are looking for more efficient batteries, they are basing potential candidates in respect with magnesium.
My problem is seeing how the 'in respect' part works and how it affects the system. Maybe it's just hard for me to relate to it in my mind so maybe a more common analogy would help?

Just to clarify, are you looking for the equivalent oxidation/reduction potentials with respect to other specific metals instead of hydrogen?

You cannot define ‘vacuum’ oxidation potentials, or otherwise derive them de novo without having some zero-point. There is, technically, no self-suggesting zero-point around — unlike for example temperature, where $0\,\mathrm{K}$ is equivalent to ‘molecular movement frozen’.
Part of the reason is, that the electron(s) lost or gained during the redox reaction must go or come from somewhere. Therefore, one has chosen the hydrogen cell as a centre of the scale. Why?

Hydrogen is the easiest element with the least electrons.
It’s simple to build.
It actually also comes with a good further distinction: The one between noble and non-noble metals.

You can actually easily recalculate the standard potentials with respect to another half-cell like magnesium: Just add/subtract that cell’s potential with respect to hydrogen to all other potentials. It’s a bit like moving the temperature scale’s zero point from $0\,\mathrm{K}$ to $273\,\mathrm{K} = 0\,\mathrm{°C}$.

This isn't strictly true. There have been a few definitions of absolute electrode potentials, but they don't really add anything useful outside of specialized applications and are difficult to measure, so they're not generally used.


text:
Explanation of the law of equivalence
According to the law of equivalence, the number of equivalents of each reactant and product in a reaction are equal. Why is this the case?

equivalents are defined that way, btw they aren't really used in practice now

http://chemistry.stackexchange.com/questions/18693/confusion-with-law-of-equivalence

Aren't they used for titration ? btw I am preparing for JEE (Engineering entrance exam in India) and its syllabus includes equivalents.

https://en.wikipedia.org/wiki/Equivalent_%28chemistry%29

Many students are still taught about it but in practice generally moles are used.

Because of the law of conservation of mass?


text:
In a nucleophilic substitution reaction, does the other atom to which the nucleophile is bonded matter?
For example,
1-bromopropane + $\ce{NaSH ->}$ Propanethiol + $\ce{NaBr}$
vs.
1-bromopropane + $\ce{H2S ->}$ Propanethiol + $\ce{HBr}$
Which is more valid, and why? Or are they equally valid?

When undergoing an SN rxn, does the other atom to which the
  nucleophile is ionically bonded matter?

It can matter.  If the counterion shifts the covalent-ionic nature of the bond connecting it to the nucleophile, then the nucleophilicity of the nucleophile can change (increase or decrease).  However in your example, the counterions don't make a significant difference.

Which is more valid, and why?

On paper, they are both valid but
$\ce{H2S}$ is very toxic and corrosive, so the first method using sodium hydrosulfide would likely be preferred.  However, in industry, where raw material cost is very important, both of these methods are successfully used.
However both methods suffer the same drawback - once some of the desired thiol is prepared, it can react further with another molecule of alkyl bromide to produce a thioether; usually a mixture of thiol and thioether is obtained.
In the lab, neither method is used
In the chem lab, the preferred method to prepare simple thiols is through the use of thiourea.  Note the resonance structures for thiourea (just disregard the lower half of the diagram) that place negative charge on the sulfur, making it a good nucleophile.

(image source)
In the first step the thiourea reacts with the alkyl halide via an $\mathrm{S_{N}2}$ reaction to yield an isothiouronium salt, which is subsequently hydrolyzed by sodium hydroxide to generate the thiol. This synthesis can be carried as a one pot, two step reaction.

(image source)


text:
Do carboxyl groups behave like bases and amino groups behave like acids when peptide bonds are created?
Peptide bonds are synthesized when the carboxyl group of one amino acid molecule reacts with the amino group of the other amino acid molecule, causing the release of a molecule of water.
Water molecules consist of a hydroxyl from a carboxyl group and a hydrogen from an amino group.
And that seems a bit strange to me, since carboxyl is an acid and should give a hydrogen (instead of hydroxyl) and amino groups are bases and should get proton (and not give a hydrogen).
So for me it looks like in creating this bond, the carboxyl group behaves like a base and amino group behaves like an acid. Is that correct?

To paraphrase what Jan said, you should really look at the reaction mechanism to understand why things happen the way you described.

This paper examines the reaction mechanism for Gly peptide bond formation in the gas phase: http://pubs.acs.org/doi/abs/10.1021/jp504924c

Yes, an amino acid consists of a carboxyl group, commonly called carboxylic acid, plus the amino group. And yes, a carboxyl group can act as an acid, i.e. release its proton to give a carboxylate. Conversely, an amino group can act as base, its free electron pair accepting a hydrogen to give an ammonium derivate. That is in fact what happens in aquaeous media or in the solid state of amino acids: the carboxylate protonates the amino group creating a zwitterion.
But these are only one type of reactions both groups are capable of performing. In the case of the amino group, as long as it is not protonated (i.e. still has an unbonded electron pair) it can attack nucleophilicly. Maybe you have heard of $\mathrm{S_N}2$ type reactions, that are rather similar.
The nucleophilic attack of the nitrogen could happen on the carboxylate’s proton — but then we wouldn’t be gaining anything, since it would just be the same acid-base reaction. (And this wouldn’t be considered a nucleophilic attack, usually.) Instead, the nitrogen needs to go to the carbonyl group — which happens to be the most positive carbon in an amino acid, so should be a great target for a nucleophile.
The initial result, as I tried to show in the scheme below, is a tetrahedral carbon with a negative charge on the ex-carbonyl oxygen and a positive charge on the nitrogen. If one of the nitrogen’s protons jumps across to the ex-hydroxy oxygen, that one gets positively charged, and then electron pairs move, we release water and gain our peptide bond.

(The picture is really only the basic mock-up and no real effort put into it. It neither exactly matches the biological peptide bond mechanism, nor does it match the mechanism that a chemist would use in the lab. But it conveys the basic picture.)
So while you did technically use the basicity (or postivie-charge loving) of the nitrogen, you see that the acidity of the carboxyl group plays no role in the reaction itself.


text:
Bending of water and methanol towards a charged objects
I just had a question about why does water, methanol or any other polar substances tend to bend towards the charged object.
I think it is because of a slight dipole dipole movement between the charged object and the polar substance, but not sure if that's the reason.

Here's a bunch of videos you may find interesting: 1, 1 cont., 2, and 3. If you have access to articles, this one seems like an important read

As you state, in an electric field, a polar molecule such as methanol or water, having an uneven charge distribution, is attracted (or repelled) by an electric field. This orients the molecule, but is not primarily responsible for the attraction of the bulk substance.
An electric field causes a force on even a nonpolar molecule such as methane or sulfur, though. You can pick up powdered sulfur with a charged piece of plastic because of electrostatic induction, which moves electrons in the bulk material.
This induced charge is similar to the effect of moon (and sun) on tides, pulling the water underneath the moon more strongly than it pulls on the more distant core of the earth, and on the core more strongly than on the ocean on the opposite side of the planet, causing uneven distribution of the water. In a similar way, an external field causes an uneven distribution of charge.


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Vinegar CH2O ? COH2
we must write empirical formula of $\ce{CH3COOH}$ as $\ce{CH2O}$ or $\ce{COH2}$ ? Alike ? Any nomenclature?
Somebody told me it was because of bonding, different way of writing the chemical formula mean different way or bonding, but I'm not sure if it is correct or not,

The reason that acetic acid's chemical formula is often written as $\ce{CH3COOH}$ is that the $\ce{COOH}$ portion of the formula indicates that the compound is a carboxylic acid.  You'll also see it written as $\ce{HC2H3O2}$ sometimes, again to indicate an acid, this time through the leading $\ce{H}$.
An empirical formula is only intended to give the relative number of atoms in a compound; writing the formula to indicate the type of compound is not usually done.  I would typically follow something like the Hill System for this purpose: carbon, hydrogen, then other elements in alphabetical order, so $\ce{CH2O}$ here.  Empirical formulas are never named.

Even though your answer is right, I find the idea of using the empirical formula rather ugly. The normal thing to do for an organic compound would be to express the molecular formula in the Hill System: C2H4O2. There is much to be lost and little to be gained from dividing everything by 2 to get the empirical formula CH2O,  though this was indeed what the OP asked for. On the other hand, for inorganic substances it's more normal to see, for example, the empirical formula (P2O5 or S) than the molecular formula (P4O10 or S8.)

@steveverrill I agree that finding an empirical formula from a molecular formula is very nearly pointless, particularly for organic compounds.  The empirical formula's usefulness is primarily in identifying compounds, for example from combustion analysis.  Working from empirical to molecular makes sense, but I know from experience that many chemistry instructors do not emphasize that aspect of it and instead teach it as an analogue of reducing fractions to simplest form.  I considered addressing that in my answer, but the question really doesn't ask about why the empirical formula exists.


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Why do some solvents bubble more than others when boiling?
This question about boiling milk reminded me of my everyday lab work at the rotavap: Some solvents and solutes are feared for the intense bubbling they create while the solution is boiling — be it at ambient pressure like the milk or at reduced pressure in the rotavap. For most of the infamous solvents, the bubbles will even create a somewhat stable foam (that usually ends up on the wrong side of the rotavap).
Well known in my lab are dissolved triphenylphosphane and the solvent toluene while isohexane (‘mixture of hexane isomers including methylcyclopentane’) is known to hardly ever give bubbles.
Why is this so? Why are bubbles often so much more stable even for just a solvent (toluene)?

This is such an annoyance in the lab! Anyone care to share techniques to avoid/minimize it, especially in rotavaps?

@Nicolau you might ask a separate question about it.

Boiling depends on certain properties of a solvent:

Tendency to form an emulsion of bubbles due to low surface tension and separation of solvent phases, as in soap bubbles. Try boiling some dish-washing detergent in water.
Tendency to superheat, i.e. to go well above the boiling point and then to suddenly form bubbles, which is more pronounced at lower pressure. This can cause violent bumping, in addition to foaming. The cleaner the apparatus, the more likely, so boiling stones (chips) are added to nucleate more gentle boiling. A glass mug of water heated in a microwave can bump violently enough to blow open the door (and blow the fuse).


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Identifying Alkyl Ligands as Pi Donors or Pi Acceptors in Transition Metal Complexes
Just wondering if there was a systematic way to identify an alkyl ligand as a $\pi$ acceptor or $\pi$ donor in a transition metal complex?
Although various sources indicate that weak field ligands are typically $\pi$ donors and strong field ligands are $\pi$ acceptors, I'm having trouble with the more uncommon ligands not listed in the spectrochemical species, namely $\ce{CH2}$ groups directly double-bonded to metals, or $\ce{CH2-tBu}$ groups double bonded to ligands.


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Why must NaOH be kept dry during storage?
Why should the Erlenmeyer flask used to store $\ce{NaOH}$ be dry? Why should it be kept covered at all times?

Without a proper closed container, sodium hydroxide readily attracts water and become a viscous solution (concentrated sodium hydroxide). This solution readily attracts carbon dioxide in air and reacts with it.
$$\ce{CO2(g) + H2O(l)<=>H2CO3(aq)}$$
$$\ce{2OH^{-}(aq) + H2CO3(aq) -> 2H2O(l) + CO3^{2-}(aq)}$$
(The $\ce{OH^{-}}$ in the second equation comes from $\ce{NaOH}$.)
This reaction decrease the amount of $\ce{NaOH}$ and thus the titration result will contains error.

Actually, the first equation can happen even without sodium hydroxide. Carbon dioxide in air is known to spontaneously dissolved into water and become carbonic acid. This is why the distilled/deionized water which in contact with air will have pH of around 6 after some time. With the presence of sodium hydroxide in water, the second equation greatly drives the equilibrium of the first solution to the right. Concentrated sodium hydroxide is used in some experiment as a carbon dioxide absorber.

Another possible problem that might occur is that concentrated sodium hydroxide can react with glass. Althought this is a very slow reaction, it might causes some unexpected error or ruin the glassware.

The reaction with glass is not really a slow reaction, especially when it was a highly concentrated solution.


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Diagonal relationships in the Periodic Table
I have often wondered about diagonal relationships between elements on the periodic table, and the most often cited explanations revolve around charge-density considerations.
But other than that, what other factors could possibly contribute to this phenomenon?
EDIT: What I mean by "diagonal relationships" is that there are certain similarities in chemical properties that have been observed in diagonally adjacent neighbors in the 2nd and 3rd period example: Li-Mg, Be-Al, and B-Si.
Link to a wikipedia article (not much to be found here though)
The explanation that I personally have most often encountered is that the  charge-to volume ratios (charge density), say for Li-Mg cations is roughly the same, and hence could account for some of the similarities in behaviors.
What I am looking for is what can be some other contributing factors (if any) to this phenomenon. I am not expecting very concrete answers because I believe this phenomenon is not very well understood.

What do you mean by the "phenomenon" of "diagonal relationships"?  How are "charge-density" considerations involved?  All elements have at least one diagonal neighbor.

I looked up the term on Wikipedia: http://en.wikipedia.org/wiki/Diagonal_relationship  You might want to link to or refer to this restricted definition in the Q.

What I remember from studying C-P 'diagonal relationships' (for multiple bonds e.g. carbene vs phosphinidene, alkyne vs phosphaalkyne) is that similar electronegativity also played a role, also by affecting the valence orbitals.
However, perhaps it is better viewed from a different vantage point:
the main group elements behave rather similarly from 3rd row onwards, but there is a distinct difference between the 2nd and 3rd row elements and the explanation is pretty straightforward:
The 2s and 2p orbitals are roughly similar in 'size' since the 2p orbitals are the first p-orbitals and are pretty compact. Therefore 2s and 2p orbitals mix happily.
Going to the third row, the 3p orbitals need to be orthogonal to the 2p orbitals and as a consequence they are much more diffuse than the 3s orbitals, resulting in less efficient s-p hybridization. This also manifests itself in the so-called 'inert pair' effect whereby e.g. Si will resist hybridization in favor of a lone pair. There is some beautiful work by Knutzelnigg from the 80s (?) describing these effects.
This 2nd-3rd row discrepancy is somewhat counterbalanced if you step to right, increasing Z and thereby contracting the orbitals giving rise to this apparent diagonal relationship.
Similar arguments make that for the 3d TMs the hybridization is more efficient for 4s-3d than for the 4d/5d TMs. Consequently, 3d TMs tend to behave differently from their heavier sisters, which in turn are closer together (although for the 5d's the relativistic effects start to kick in more heavily...).

The diagonal relationship is not a very good way to group elements, it makes sense only if we consider highly polar or ionic compounds. when you take compounds of 2nd and 3rd period following diagonal relationships where they are having their max oxidations state, the charge to volume ratio of the atoms/ions become more or less similar. Also when there is lot of positive charge on an atom say al3+, its orbitals become shorter and hence the bonding orbitals overlap in a similar ways as it does in the diagonally related 2nd period element(be2+). Hence the nature of compounds and the way the diagonally related elements react is similar.
EDIT: by Be2+ and Al3+ i am trying to say that Be is in +2 oxidation state and Al is in 3+ state


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Reliable automatic addition of hydrogens and charge assignment to small molecules at given pH?
I have some PDB structures of small molecules with missing hydrogens. Is there a software available that places missing hydrogens and assigns a charge to the molecule that is reliable? I am looking for an automatic tool that returns a PDB with added Hydrogens and the overall charge (I can do some programming if I need to combine output from multiple programs to accomplish this or something.)
It is enough for me if you suggest a couple of FREE tools (I don't have resources to buy a commercial license, but I am academic, in case they offer free academic licenses), with associated papers that evaluate their performance that I can cite. For example, a free tool that calculates pKa values of ionizable groups would solve my problem, if you can also suggest a tool that automatically detects ionizable groups in molecules.
I have used Avogadro to add hydrogens (though I don't know how to compute the charge with it), but I've found that some hydrogens aren't well placed. Is there a better tool available for this task?

Could you give an example of Avogadro producing mis-assigned hydrogens, along with what you would expect to find? Different chemical toolkits may give different answers for some calculations depending on what algorithm they have implemented - but it's hard to recommend one that does it 'right' if we don't know what 'right' is here.

Possible duplicate: http://chemistry.stackexchange.com/q/4105/5017

I'm not sure I understand what you mean by "compute the charge?" Do you mean that you want software that will automatically assign protonation (and charges) for arbitrary molecules at arbitrary pH? No program is a substitute for understanding protonation rules yourself.

@GeoffHutchison A program is a good substitute when you have to evaluate the protonation state of thousands of molecules. On the other hand, I don't have a solid foundation in chemistry and understanding these rules would help me anyway. Can you point me to some references? Thanks.

@GeoffHutchison By "compute the charge" I mean that it will place the hydrogens and also print somewhere the total charge of the protonated molecule.

If I understand you question correctly, there isn't a tool which can take as input a chemical formula and then output accurate pK values for all the "active" groups in the molecule. You'd need to add the pK values of the active groups to the database for each compound, and then a program could predict how the molecules would behave as a function of pH. Take an amino acid as an example. The acid end has a pKa value and the amine group has a pKb value. With the two the molecule's protonation can be accurately modeled in solution as a function of pH.

If you're looking for comprehensive protonation and evaluation of $pK_a$, I suggest my answer to a previous question. There is specialized software for evaluating protonation and $pK_a$ based on large databases.
Avogadro has some simple rules, mostly for amino acid residues, but is not intended for automated evaluation of thousands of molecules.

I'm just looking for a reliable, and quotable method to add hydrogens to small molecules. I wan't to be able to cite a paper where the method is evaluated. The method should allow me to determine where the Hydrogens are located in my small molecule. Can you suggest a tool to do this?

There is currently no such method as others have commented here. You could, in principal, compute all tautomers and protonation states exhaustively for a small molecule, compute energies using a semiempirical method (e.g., PM7 with MOPAC) and pick the most stable one.

If the molecule is a protein you can use the PDB2PQR webserver.  It uses PROPKA to predict $pK_a$ values and optimizes the hydrogen bonding network

My molecules are small ligands. Sorry I didn't specify that in the original question.


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Are vitamins in tablet are more efficient than the vitamins in their natural source?
In medical shop I have seen that there are loads of vitamins tablets available but I was wondering like does vitamins in tablet are more efficient than the vitamins which are available in their natural source like in fruits ?

The big question: How do you define more efficient?

Well, there are reasons why dietitians tell people to eat fruits not tablets.

This is perhaps slightly tangential, but it should be noted that there is a lot of debate whether vitamin/mineral supplements are actually useful to most people without a specific requirement, and they can even be toxic. Results in support of nutrient supplementation seem to be inconclusive at best for the general population to which it is marketed (those actually rich enough to purchase them), though of course in people with proven deficiencies it can be a life-saver.

For some vitamins, such as Vitamin C, which are always found in the form of a unique chemical substance, there is no chemical difference between the vitamin in a tablet--synthetic or otherwise--and the same vitamin as found in a foodstuff. Still, how well you absorb a vitamin might be affected by what else you consume at the time - for example a given amount of Vitamin C might be absorbed more or less efficiently as an ingredient in orange juice than as a pill, even though it is the same substance in both cases. But these differences might not necessarily be significant.
Other vitamins, such as Vitamin A or Vitamin D, are found in multiple closely related chemical forms, any of which the body can use to fulfill its need for the vitamin. In that case, the body may absorb or utilize the vitamin more efficiently when it is in one form relative to another. Fats in the diet can further influence your absorption of a fatty vitamin like Vitamin D.
Minerals, such as iron, are notorious for being made easier or harder to absorb depending on their chemical form and how they are bound to other components of the food. So, for example, iron in greens is known to be slower/more inefficient for the body to extract than iron in liver.

I'd add that iron in vitamin pills or many breakfast cereals (among other places where it's added) is in little bits of iron metal, not $\ce{Fe2+}$ in any form, and as you note, the ligand can influence iron absorption.

For most vitamins, it's much better to consume them in dissolved form (as they would be in most foods). You can test how much dissolves by putting a vitamin tablet into a cup of warm-to-hot (around body temp) vinegar and seeing how much of the pill dissolves. If it doesn't dissolve, it's exiting your body without any chance to use it.
So, for most vitamins, no.

This does not answer the question.

This is even too far too away to be a comment. But anyways, welcome to the site! You might take a look at the help center to find out what's a good answer in our community. This answer of yours doesn't count as one, unfortunately.


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Why are H2O asymmetric while CO2 is not?
What is the reason for water molecules $\ce{H_2O}$ being asymmetric while other molecules like $\ce{CO_2}$ being completely symmetric? See the pictures below:

In both case there is an electronegative difference between the atoms in the molecule. I have read that the angle in the water molecule is always a specific value. Is this a special thing for water or is there a common mechanism I should know?

That's because of the lone pairs of oxygen. We have the notorious VSEPR, the mathematical hybridization, and the ultimate molecular orbital theory to explain this. Have you studied any of them?

Related: Why is H₂O a dipole?


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Least likely product of a hydrocarbon cracking reaction?

The hydrocarbon $\ce{C17H36}$ can be cracked. Which compound is the least likely to be produced in this reaction?
A: $\ce{C3H8}$ B: $\ce{C4H8}$ C: $\ce{C8H16}$ D: $\ce{C16H34}$

Is it D because cracking's purpose is the breaking of long hydrocarbon chains into smaller ones?

Yeah well, a secondary carbocation is more stable and is formed way faster, so hexadecane must be the most obscure compound.

There are two ways to look at this question.  One is simple carbon atom counts.  Another is the saturation of the products.
Carbon atom counts.  The question doesn't specify which of the 24894 isomers of heptadecane is to be cracked.   Suppose it is "normal" or n-heptadecane.  Cracking usually begins with the homolytic cleavage of a carbon-carbon bond.  To make a $\ce{C16}$ product from n-heptadecane by cleaving carbon-carbon bonds, there are only two potential bonds to cleave, one at one end of the molecule, another at the other end.  In either case, to obtain a $\ce{C16}$ product, only a single cleavage would be allowed, because further reactions would break more C-C bonds and change the product.  Cracking is a high temperature reaction that is entropically driven.  The system is increasing the disorder.  Imagine rolling four 6-sided dice, and the total comes up 17.  Then you take away either one or two of the dice (whether its one or two is random).  What is the probability that the remaining dice sum to 16, compared to say 8 or 4?
Saturation.  Choices B and C are unsaturated hydrocarbons.  Choices A and D are saturated hydrocarbons.  Both saturated and unsaturated compounds can be formed by cracking, but most modern cracking processes tend to focus on making unsaturated hydrocarbons.
Thus, primarily because of simple statistical unlikelihood, but also because of saturation considerations, D is the correct answer.
Wikipedia's article on cracking is very helpful.  See here.$%edit$

I am sorry but I don't understand what you're trying to explain with the dice example. Are you trying to say that more than just one bond of $\ce{C17H36}$ will be cleaved? If so, how exactly does the dice example illustrate this?


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Phosphate buffer from phosphoric acid and K2HPO4?
Is it possible to make a solution of ph 7 phosphate buffer solution using phosphoric acid and $\ce{K2HPO4}$ ? How can I calculate the amount of $\ce{K2HPO4}$ needed for 1L of phosphoric acid ?
Edit:
I have 50 mL of 0.2M $\ce{H3PO4}$ solution. How can I convert this solution into 50 mL of pH 7 buffer solution by adding (only) $\ce{K2HPO4}$ ? How can I calculate the weight of $\ce{K2HPO4}$ considering all the equilibria present in the $\ce{H3PO4}$ solution and by the application of Henderson-Hasselbalch equation ?

What concentration do you want? 50 mM or 1.0 M?

@LDC3 Say 1.0 M.

I suggest you first consider the following reaction:
$$\ce{H3PO4 + 3K2HPO4 -> 2HPO4^{2-} + 2H2PO4- + 6K+}$$
It appears, that transforming all $\ce{H3PO4}$ to equal amounts of $\ce{HPO2-}$ and $\ce{H2PO4-}$
requires 3 mole equivalents of $\ce{K2HPO4}$. If you add 3 mole equivalents of $\ce{K2HPO4}$ you will end up in a situation where the concentration of $\ce{[HPO2^{-}] = [H2PO4^{-}]}$, i.e. at the $\ce{pH} = pK_{a2} = 7.21$. Then, I suppose you use the $\ce{HH}$-equation to figure out the rest.
At pH = pka2 = 7.21 the concentration of [H2PO4(-)] = [HPO4(2-)] = 0.40 M. This is because we have added 3 mole equivalents of K2HPO4 to 50*0.2 = 10 mmole of phosphoric acid, i.e. we have reached a total concentration of phosphoric acid protolytes of (3*50*0.2 + 50*0.2)/50 = 0.80 M.
Now, since we wanted to reach pH = 7.0, we have theoretically added too much of K2HPO4. We suppose the excess amount is equal to x.
At pH = 7.0: [HPO4(2-)] < [H2PO4(-)].   At  this pH, only HPO4(2-) and H2PO4(-) are present in significant amounts in the solution. We can then calculate the following:
7.00  = 7.21 + log ([HPO4(2-)] - x/[H2PO4(-)]) = 7.21 + log (0.4 - x)/0.4) => x = 0,1533.
Now, initially we had 50*0.2 mmole of phosphoric acid. To reach pH = 7.0 we should then add 3*50*0.2 - 0.1533*50 mmole =  30 - 7,66(5) = 22,34 mmole of K2HPO4 or 3.8(9) gram.
Please check the calculations.

The edit of my answer does not look good. It should read HPO4(2-)!

Thanks for the reply. Can you please explain how that reaction happens ? I mean what about $\ce{H3PO4 + K2HPO4 -> 2 H2PO4^- + 2K+} $ ? And I want the pH to be 7.0 not 7.21.

@Bive I think thats the correct equation now isn't it?

You now tell us that the final concentration should be 1,0 M.
This cannot be right. Initially, you had 50 ml 0,2 M H3PO4, i.e. 10 mmole.
If you add K2HPO4 to reach a final concentration of 1,0 M, the pH of
the final solution will have a pH much higher than 7,0. If we approximate the volume of the solution to be constant, you have to add 5 mole equivalents of K2HPO4 to achieve 1, 0 M.
Initial: 50 ml*0,2 M = 10 mmole => Final: 50 ml * 1,0 M = 50 mmole?
I think you should stick to your original presented problem, which is interesting, since the problem does not state the final concentration.


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Does the temperature of water affect its ability to extinguish fires?
Since water can be either hot or cold, does its temperature affect its ability to extinguish fires?

A fire needs 3 things as shown in the image:

Water reduces the heat as it turns into steam; and hot water turns into steam faster. But the most important thing water does, is to prevent oxygen getting to the fuel. So as long as there is liquid water, it does a great job.

This really depends on the fuel, don't you think? Water may not be the most efficient retardant for a grease fire.

@ringo The OP wants to know about water, so I limited my answer to that.

To turn to steam, you need to (1) raise the liquid to the boiling point, and (2) add the enthalpy of the phase transition.  So, yes, boiling water will be less effective than cold water, but the enthalpy is still a big term (vague memory, about half as effective). Coupled with the smothering noted by @LDC3, use whatever water is available and don't wait for the tap to cool down.  (and don't pour water on a grease fire).


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Are there any free phosphorus-31 NMR spectral libraries?
I know that there are some sites like Sigma Aldrich and AIST for $\ce{^1H}$ NMR and $\ce{^13C}$ NMR, but was wondering if there were any free sites to compare $\ce{^31P}$ NMR spectra?

There are quite a few it seems, a couple of examples are below:
University of Wisconsin Phosphorus NMR data, starts with a shifts overview.
Stanford's Magnetic Resonance Laboratory site P NMR at the SMRL, provides comparisons and a protocol for testing.
A general NMR site is NMR Wiki - has many links to many sources.


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What are the possible changes that might occur if I add an inert gas to a chamber containing another few different gases?
As an example, let's assume that we have a gas $\ce{P(g)}$ inside a chamber and when the temperature inside it increases to a certain value there is an equilibrium that prevails inside the system.
So for the sake of argument, let us assume that this equilibrium is:
$$\ce{P(g) <=> Q(g)  + R (g)},$$
And let's assume the figure below shows that chamber when equilibrium is established:

What I want to know is, what are the possible changes that might occur if I add an inert gas $\ce{I(g)}$ to this chamber?

Quoting from the Wikipedia page on Le Chatelier's principle, we have to distinguish two cases:

The total volume is constant:

Adding an inert gas into a gas-phase equilibrium  does not result in a shift of the equilibrium in this case. This is due to the fact that adding an inert gas does not change the partial pressures of the other gases in the container. While it is true that the total pressure of the system increases, the total pressure does not have any effect on the equilibrium constant; rather, it is a change in partial pressures that will cause a shift in the equilibrium.

The total volume is not constant:

If, however, the volume is increased, the partial pressures of all gases will decrease which results in a shift of the equilibrium towards the side with the greater number of moles of gas.

Could you explain why the partial pressure will not change upon the addition of an inert gas in  when the volume is constant

The only additional pressure in the container will come from the inert gas. The pressures exerted by the reactants/products have no way to get larger or smaller (To calculate the partial pressures, use P = nRT/V, using the molar amount of the individual gas. You'll see that because your volume is constant, as is your temperature, the partial pressure must remain equal).

The partial pressure of a gas  is the hypothetical pressure of that gas if it alone occupied the volume of the mixture at the same temperature. So, if the volume is constant in your problem,  the partial pressure  $p_i=n_iRT/V$ is constant.

I'm pretty sure this answer is strictly correct only for ideal gases.  With strongly interacting molecules, at sufficiently high pressures, or at sufficiently low temperatures, I would think adding moles to the system would start to directly affect the equilibrium constant.

Of course! you're right. The answer is valid for ideal gases.

Okay,thanks a lot , for all of you . Now I have got the point


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How to separate a 5 atom thick layer of Cu from Au
I've become very curious about (electrochemical) atomic layer deposition (ALD) via surface limited redox replacement.
Since the technique allows for deposition of metals in atom thick layers it's possible to create films of nearly exact atomic thickness. However, I haven't found any reference that describes how one can remove the film from the (usually Au) substrate that it's deposited on.
Hypothetically speaking, if one used ALD to generate a film of $\ce{Cu}$ that is 5 atoms thick (arbitrary length and width), how would one go about removing the film in-tact? Is it even possible?

Very carefully with a pair of small tweezers.

In most cases I've seen, ALD is performed on the intended substrate - that is, there's no need to separate it. In other cases, you use a sacrificial substrate that you can remove easily.
If you wanted to separate the film from the substrate, there are various techniques, but nothing comes to mind for separating Au from Cu. I find it incredibly unlikely in the specific case of a 5-atom thick layer of Cu, which would likely not survive typical techniques (vs. bending, mushing, forming a larger packing, etc.)
Some typical techniques for separating two films:

Adhesive tape (as mentioned in a comment) - works for many films, e.g., graphene, other layered materials, where you physically pull apart layers
Swelling (e.g., separating Au from mica) - including various intercalation methods used in graphenes and graphene oxides and other layered materials, so that the layers separate more easily
Dissolving, etching, or eroding the substrate (e.g., back-etching a Si wafer to expose an $\ce{SiO2}$ or thin metal film deposited on the wafer)

In my group, for example, we make "template-stripped gold" where Au is deposited on cleaved mica substrates. Then the thin gold layer is glued to glass substrates and the mica is separated from the Au by swelling / degredation, leaving a very smooth Au surface.
This technique can work with fairly thin layers of gold (~10-30 nm) but there's a large difference between Au and mica, making separation possible. Not true with Au/Cu.

I'll comment that I would love to see some technique capable of separating 5-atom thick metal films.

+1 for dissolving or etching the substrate. It would be hard to selectively etch gold underneath copper, but with a proper choice of substrate, I would say that is the way to go.

@BurakUlgut I'm just coming up empty with ways to etch out the gold and leave the copper, especially in this scenario. I guess if you timed it extremely carefully, you'd remove all the gold and leave copper.

I'd suggest electrostatic repulsion. Charge both layers with electrons and they'd separate easily. Then use a electron microscope to manipulate the layer.

Adhesive tape would work for some combinations (used for graphene, for example), but since gold and copper alloy so well, I'd assume the copper would have interpenetrated the gold surface layer, to some extent.

I agree with the comment above (i.e., some amount of alloying in such a thin film) but even if Au/Cu was fully segregated, I don't see how charging the film will separate the layers. The film will be highly conductive, so I don't see how electrostatic repulsion will occur - added electrons will be highly delocalized.

This could have been written as a comment.


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Anti elimination from vinylic halides
An $\mathrm{E2}$ (or $\mathrm{E1cb}$) mechanism requires antiperiplanar configuration. In case of restricted rotations, like in cycles, where rotation is not possible to achieve the antiperiplanar configuration, elimination occurs to favour the Hofmann product instead of the Zaitsev, if antiperiplanar configuration is possible at the less substituted β C.

In menthyl chloride, antiperiplanar configuration is only seen with the red $\ce{H}$, so 2 menthene is the only product, although it is not the most substituted.

In the molecule above, due to restricted rotation, the leaving group $\ce{Cl}$ and the base attacking the $\ce{H}$ will be on the same side. So will elimination be possible in this case?

An antiperiplanar while being the most efficient is not the only configuration from which elimination occurs. A synperiplanar configuration can give you an efficient elimination as well especially in the case of a double bond which prevents rotation and holds the eliminated H and leaving group fixed at the perfect positions for a syn elimination. It is all a question of orbital overlap: It is best in the case of an antiperiplanar configuration since the orbitals are perfectly aligned in this case, but in the synperiplanar configuration they are at least partly aligned and good overlap occurs.

Elimination can occur from vinylic halides.  This related question may be helpful.

But if in an E2 mechanism, there is the possibility of both antiperiplanar and synperiplanar configurations, the anti is always preferred - despite the fact that it may be the less substituted product (in reference to the example of menthyl chloride in the question)?

An E2 reaction will not occur with this reagent.
An E1 reaction, on the other hand, could do so (though quite slowly: E1 is an all around pathetic reaction mechanism, and this molecule does nothing special to stabilize it). The chlorine will leave and there will be a carbocation on the end of the alkene, whereupon the proton from the other side of the double can be extracted with a base in a second step, leaving you with an alkyne.
That said, if you want the alkyne, it would be infinitely infinitely easier to start with the cis isomer of the molecule you have and just run it through very strong base. E2 is far far better than E1 in terms of reaction rate and yield.