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Stability of Sulfides - backbonding?

A stability model premised on Lewis acid-base covalent bonding can be proposed to rationalize the stability shown by sulfides such as CuS and Ag2S. For instance, although the oxide anion is a stronger Bronsted base than sulfide anion, sulfide anion can be (and usually is) a stronger Lewis base than oxide anion since S has a lower EN than O. Furthermore, recall that Lewis base strength depends on the Lewis acid with which the Lewis base interacts. For the sulfide anion, since sulfide anion possess empty valence 3d orbitals, Lewis acid-base interactions are often very strong if the Lewis acid is a low oxidation state metal ion with a fairly large number of valence d electrons. for such cases, sulfide-to-cation sigma coordinate covalent bond formation couples with pi-type coordinate covalent bond formation because metal ion d-electron density "drifts" back to empty valence 3d orbitals on sulfide anion. This results in covalent bond interactions that have substantial, strong multiple ond character and formation of a metal-sulfide lattice which is rather macromolecular in character. In sum, such as lattice must be very stable ...

1) Is the above excerpt describing pi-backbonding?
2) Is pi-backbonding more favorable for metals in a low rather than a high oxidation state because such metals still have substantial electron density that needs to be stabilized - and preferably stabilized by something more electronegative than a metal?
3) Metal d-orbitals overlap with p or d-orbitals of the non-metal. How does this work? I always see diagrams such as these:

Is the overlap between the metal d-orbital and non-metal p-orbital as poor as depicted? Are the d-orbitals really at a 45 degree angle relative to the p-orbital?
4) How well is the above excerpt written? I feel that two improvements could be made:
A) The passage seems to imply that Bronsted acidity and basicity do not depend on the corresponding base or acid. Wrong implication. HCl in HBr solvent won't be a strong acid.
B) Why electron density might just "drift" to the non-metal could be explained explicitly. I.e. electronegativity.

Silver oxide and silver sulfide have roughly similar energy of formation. Given higher energy of atomization per atom for $\ce{O2}$ molecule in comparison to $\ce{S8}$, this suggests stronger bonding in $\ce{Ag2O}$ rather than in $\ce{Ag2S}$.

@permeakra my professor makes the argument that Ag2S is MUCH less soluble than Ag2O in water (Ksp of Ag2S is ~10^-51 while Ksp of Ag2O does not even approach that by a few trillion trillion). Does his argument hold water? I was just thinking that this only suggests that one is less polar than the other (like dissolves like).

How Ksp of Ag2O can be measured, if $\ce{O^{2-}}$ does not exist in water solutions?

@permeakra Ksp of Ag2O is not necessarily defined as [Ag+]^2[O^2-] ... http://goldbook.iupac.org/S05742.html

1) Yes and no ... the d-d orbital overlap will almost certainly have pi-orbital character (angular node along bond axis), and it could reasonably be considered "backbonding" because the electron density is going in the canonically "unexpected" direction (i.e. away from positively charged metal center).  On the other hand, chemists (particularly inorganic chemists) usually restrict the phrase "pi-backbonding" to refer to back donation from the d-orbitals of a neutral or ionic metal center into anti-bonding orbitals of a ligand, typically an olefin or a carbonyl ligand.
2) Yes .. that's essentially correct ... d-orbitals tend to be rather diffuse, and so can overlap with lobes of diffuse orbitals on ligands (or counter-ions, as with the metal sulfides).  In traditional metal-ligand pi-backbonding, lobes the d-orbitals overlap with lobes of an unoccupied anti-bonding orbital of the ligand.  In the example given above, they overlap with lobes of unoccupied d-orbitals on the sulfur.  For a given system, unoccupied, or virtual, orbitals tend to be much more diffuse than occupied orbitals.
3) Those diagrams are just intended to capture the qualitative aspects of the bonding interaction, such as the symmetries of the orbitals involved.  Remember, the orbitals are 3-dimensional shapes that describe the probability of finding an electron with a particular set of quantum numbers in a given region of space.  Depictions like the one in your post are helping to visualize the orbital by assuming a fairly tight "tolerance" for the probability. For example, those images might depict the surfaces that enclose 75% of the total probability density.  If you expand that to 90%, or even 99%, the lobes maintain the same basic shapes but get much larger, which might make the overlap seem more significant, but could also be confusing for other reasons.
4) I think it is quite well-written, but I suppose it would depend on the broader context in which it was being presented.  I guess it is presented at an inappropriately high level for an introductory chemistry textbook, for example.

"I guess it is presented at an inappropriately high level for an introductory chemistry textbook." Curiously enough, this is an excerpt from a freshman chemistry text. I like it, however ;). Even if it only took me a year or so to actually understand what the text was trying to say!

Also doesn't the d in d orbital stand for diffuse?

Actually, yes .. it's part of an old convention for grouping of spectral lines .. "sharp", "principal", "diffuse" and "fundamental" .. as it became known that these grouping arose from atomic levels with different angular momentum quantum numbers, the numbers themselves took on those designations: $s$ for $l=0$, $p$ for $l=1$, $d$ for $l=2$ and so on ... but it's just coincidental that "diffuse" is also qualitatively descriptive for the nature of the d-orbitals in my answer to your question.

one last question; what do you think this line means? "This results in covalent bond interactions that have substantial, strong multiple ond character and formation of a metal-sulfide lattice which is rather macromolecular in character." What does it mean by macromolecular in character? What's micromolecular?


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Why does Cobalt have different coordination numbers in its complexes with water and thiocyanate?
I'd like to know why does cobalt(II),  $\ce{Co^2+}$, form $\ce{[Co(H2O)6]^2+}$ complex with water while  it forms $\ce{[Co(SCN)4]^2-}$ with the thiocyanate? Why is there $\ce{6H2O}$ but only $\ce{4 SCN-}$? And is the cobalt's  coordination number 6 in these cases?
Thanks.

Welcome to chemistry.SE! Please take some of your time for a visit at the help center or take a look at this page to enhance the formatting in your future posts.

the coordination number is the number of coordinate bonds to the metal ion: six for the hexaaqua complex and four for the tetrathiocyanate complex. as for why I too am dying to know as I haven't been able to find any good explanations on the internet

Water, being neutral, has a stabilizing effect on positive ions, such as $H^+$. Thiocyanate has a negative charge. I think that after 4 ions have clustered around the cobalt, that the charge repel more thiocyanates from bonding to the cobalt.

That's a good thought @LDC3, but if that were all there was to it, wouldn't you expect the thiocyanate complex to have two water ligands as well?  However, when you add $SCN^-$ to a solution of hexaaquo cobalt (II), you get $[Co(SCN)_4]^2-$, not $[Co(SCN)_4(H_2O)_2]^2-$.  I am still scratching my head on this one .. water and thiocyanate are both weak field ligands, so I would expect the same coordination geometry for both of them.  I must be missing something.

Maybe you need to take into account the geometry in addition of charge stabilisation. $\ce {H2O}$ is bended while $\ce {SCN-}$ is linear. I'm not sayin' that's the answer but maybe it should be taken into account :x

It seems coordination number changes because $\ce{SCN-}$ i weak ligand in spectrochemical series and water isn't, but I don't understand this.

Firstly, in water Co(II) is not exclusively $\ce{[Co(H2O)6]^2+}$, there is also a small amount of tetrahedral $\ce{[Co(H2O)4]^2+}$, 0.08% at room temperature increasing substantially at high temperature. Canadian Journal of Chemistry, 1980, 58(14): 1418-1426.
According to Greenwood's "Chemistry of the elements" 2nd edition page 1131, the more polarizable the ligand, the more the tetrahedral form of Cobalt (II) is preferred since fewer ligands are required to neutralize the metal ion's positive charge.

I don't know whether it's correct or not but what I think is due to steric factors .
Given the SCN- is pretty large ligand as compare to H2O .
Also there might be some additional stabilisation in water as ligand due to intra atomic hydrogen bonding.

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sort: votes


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Is there a reaction happening between Coconut Oil & Hydrogen Peroxide?
On a whim, I mixed together some coconut oil and drug store (3%) hydrogen peroxide and washed my hands with it. It definitely felt less greasy than pure coconut oil, left my hands feeling clean and soft :)
Did any chemical reaction occur between the coconut oil and $\ce{H2O2}$ to make a new compound? My guess would be, since Coconut Oil is mainly lauric acid, small quantities of peroxylauric acid may have been produced?
I only had high school chemistry so I could be wrong. Just my guess. Anyone know for sure?

You might have oxidized the fats in the oil to some extent.

Did any chemical reaction occur between the coconut oil and H2O2 to make a new compound?

Most likely, no chemical reaction occurred.
Whether any reaction occurred depends on the conditions you used when mixing the oil and hydrogen peroxide. How long did they mix for? What temperature was the mixture at? Did you add anything else to the mixture?
Hydrogen peroxide is used in the production of epoxidized vegetable oils. However, the reaction is performed for prolonged times at elevated temperatures with concentrated hydrogen peroxide in the presence of a catalyst.
I am guessing that you just poured the two together at room temp, then soon after used the mixture to wash your hands. Compared to the commercial reaction, you are missing time, heat, and a catalyst, so it is very unlikely that any reaction occurred.

My guess would be, since Coconut Oil is mainly lauric acid, small quantities of peroxylauric acid may have been produced?

Oils are composed of triglyceride esters of fatty acids, not fatty acids themselves. So you would not see any peroxylauric acid.

It definitely felt less greasy than pure coconut oil, left my hands feeling clean and soft :)

It felt less greasy because it was less greasy: you added water to it (3% hydrogen peroxide is $\ce{H2O2}$ in water). You probably made an emulsion of oil and water and used that to wash your hands (kind of like using soap).

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Which of the following atoms has the largest first ionization energy?

Of the following atoms, which has the largest first ionization energy?

$\ce{Br}$
$\ce{O}$
$\ce{C}$
$\ce{P}$
$\ce{I}$

I got confused between $\ce{O}$ and $\ce{Br}$. The answer given is $\ce{O}$ although in the periodic table $\ce{Br}$ is closer to Group 18 (noble gases). Why is the answer $\ce{O}$ then?

I actually think this is an unfair question, unless you were given some additional information, or you were specifically instructed to memorize specific ionization energies, rather than learning the trends.  The trend for ionization energies is that they increase from left to right across rows, and from bottom to top in columns.  Based on only that information, I can't see how one should distinguish between $O$ and $Br$ ... yes, it happens to be true that $O$ has a higher IP than $Br$, but how are you supposed to tell? Suppose the example had used $S$ instead of $O$, and $I$ instead of $Br$?  Those elements have the same relative positions in the table, but in that case, it would be $I$ that has the larger ionization energy.
So, in this case I think your confusion was quite understandable .. I would never construct a test question like this unless the student had some additional information.  It's not even true that you can always use the rule about them increasing from left to right along rows .. in the first row, $N$ actually has a slightly higher IP than $O$, because there is a penalty for disrupting the half-filled p-shell in $N$, which is anomalously stable.  You can see the periodic trends here.

The larger that atoms get, the outer electrons experience whats called a shielding effect.
Electrons that are closer to the nucleus have a stronger attraction than the electrons further away. Because the electrons in oxygen experience less shielding, they have a stronger affinity.
Oxygen's stronger electron affinity contributes to the observed electronegativity of the atom.

On the most basic level, electronegativity is determined by factors like the nuclear charge (the more protons an atom has, the more "pull" it will have on negative electrons) and the number/location of other electrons present in the atomic shells (the more electrons an atom has, the farther from the nucleus the valence electrons will be, and as a result the less positive charge they will experience—both because of their increased distance from the nucleus, and because the other electrons in the lower energy core orbitals will act to shield the valence electrons from the positively charged nucleus).

This answer is sort of right, because the trend in shielding ..usually expressed as effective nuclear charge.. is the underlying reason for the trend in ionization energies.  It does also happen to be the underlying reason for the trend in electronegativities, but electronegativity actually describes a different phenomenon.  The electronegativity of an atom describes its tendency to polarize electrons in covalent bonds.

Ok so I thought about the question to and I thought that Bromine should have a higher first ionization energy but then if you really think about it, the answer would be oxygen. Look at it this way. Oxygen desperately needs two electrons. And Bromine needs only one more. So if you look at it, it would be harder to lose another electron when you already lost two and it would be easier to lose one more if you are Bromine. And also if you look at a basic ionization energy trend, it goes to the top right corner.
Imagine you were broke, would you be willing to pay to get something even though you know that you need money? No like Oxygen.
Think of Bromine as an investor who needs a little bit more $ for an achievement, Bromine would be more likely to sacrifice its electron or to invest in something like an investor.
Really bad analogies but hopefully you get the idea.

Almost everything about this answer is incorrect or misguided ... except the part about IE's increase toward the top right corner in general.


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Beta Decay - Doesn't Add Up
The definition of beta $(\beta^-)$ decay is that an electron is produced.
For example,
$$\ce{^131_53I -> ^131_54Xe + e^-}$$
So iodine forms Xe and releases one electron with a -1 charge only and no weight.
My question is that if an electron is released, then there is still a proton left so shouldn't it still be iodine? Then you should gain an extra proton. I personally feel like something is missing here, Also about Positron Emission or Positron Decay.
Anyone know why or what?

Edit: The "duplicate" question picked by many does not fully answer the question as well as bon did here.

This is really a physics question but I will answer it anyway.
I'm not sure how much subatomic physics you know so I will give two different versions.
Simply put, a neutron decays to form a proton and electron.
$$\ce{n ->  p+ + e-}$$
This explains why the proton number increases by one to form Xenon.
More properly a down quark inside a neutron decays to form an up quark, an electron and an electron antineutrino as a result of weak nuclear force interactions.
$$\ce{d -> u + e- + \overline{\nu}_{e}}$$
This causes the neutron ($\ce{udd}$) to turn into a proton ($\ce{uud}$) leading to the increase in atomic number.
This type of beta decay is know as $\beta^{-}$ decay. There is also $\beta^{+}$ decay which is a similar process where a proton becomes a neutron, positron and an electron neutrino:
$$\ce{^23_12Mg -> ^23_11Na + e+ + \nu_{e}}$$
$$\ce{u -> d + e+ + \nu_{e}}$$
http://en.wikipedia.org/wiki/Beta_decay $%edit$

I'll just stick with the textbook and not ask any questions. Haha. Don't have time to comprehend when exam is in a couple of days.

Unless you have a particular need or desire to understand particle physics the first equation should allow you to work out whats going on

Then why do they write the electron after the arrow, isn't it just with the Xe. I don't see the need to write the electron. Just say that Xe forms.

Could you give me an example of the $\ce{B^+}$decay. Like you explained with $\ce{B^-}$.

The electrons are ejected from the nucleus at very high energies, and therefore velocities, and so are not captured by the atom. This is why we usually write them seperately.

What is the e in the Positron Emission? Shouldn't that be just a regular -e.

$e^{+}$ is a positron which is the antiparticle of an electron. It has to be $e^{+}$ not $e^{-}$ in order to conserve charge

But when a proton becomes a neutron, what about the electron that is associated with that proton? So shouldn't the electron get ejected if a proton turns to a neutron?

Let us continue this discussion in chat.

when writing nuclear decay equations we are only interested in nuclei and the decay products. the electron in an orbital that i think you are referring to remains there.
the positron (and the electron in beta minus decay) comes from the nucleus and has nothing to do with the electrons in orbitals around the atom.

When we talk about an unstable atom that will decay by emitting $\alpha$, $\beta$ or $\gamma$ radiation, the nucleus is the thing that's unstable, right?
As Wikipedia puts it,

An unstable atomic nucleus with an excess of neutrons may undergo β− decay, where a neutron is converted into a proton, an electron and an electron antineutrino (the antiparticle of the neutrino)

You can forget the anti-matter part. Just remember than in $\beta$ decay there is a neutron from the nucleus that is being "devolved" into eminently a proton and an electron. The electron leaves the atom with high velocity and energy, but the proton remains. So, for each atom that a $\beta$ particle is released from, we have:
$$\ce{^A_zM -> ^A_{z+1}M+}$$

Essentially, a neutron is converted to a proton, releasing an electron in the form of beta radiation. That is why the mass stays constant but the number of protons increases by 1.

I don't quite understand that, any clarification please.

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What will eventually happen to uncovered vegetable (seed) oil exposed to the air?
If I leave some olive oil uncovered and fully exposed to indoors air for enough time (though away from UV exposure or any other environmental factors), what will happen to the oil?
I read here that some of its constituents ('lighter' molecules) will start to evaporate but that heavier ones will be left behind, which will 'not quite' evaporate.
If that is true (it may not be), would the non-evaporative part of the oil eventually harden and assume a cake-like consistency, almost like wax or hardened animal lard stored at refridgerator temperatures?

Some of the lighter constituents could evaporate. There are a lot of minor components in olive oil, and they are largely responsible for the fragrance and flavor. As these evaporate, the fragrance and flavor would change. It is unlikely that they would all evaporate though, as many of these components likely have high boiling points/low vapor pressures.
The oil itself would not evaporate, but it would not solidify either. At most you might notice a thickening of the oil.
Over long periods of time the oil may react with oxygen to form various oxidation products. This process is so slow, however, that it is definitely not a concern for olive oil you might leave out in your kitchen. The oil will start tasting and smelling stale long before significant oxidation occurs.

Just to be clear, the oil will taste and smell stale (then rancid) due to oxidation.  I realize that you were referring to a more thorough version of the process.  :)

The funny thing is, I was wondering if it may actually be useful for something (as a material) - rather than worrying about it - and upon googling 'epoxidized' I discover: 'Epoxidized soybean oil, better known by its acronym, ESBO, is a plasticizer used in polyvinyl chloride (PVC) plastics' - so there you go...

epoxidized olive oil || that's fantasy. Oxidation does happen, but via radical route, leading to hydroperoxides and products of their decay.

@permeakra, you are correct. I think I just got distracted by the prevalence ESBO and didn't think through the actual mechanisms of oxidation applicable here.

The key to the answer is what kind of oil do you have, there are differences. In general, oils with big amount of double bonds MAY oxidize by oxygen from air. The process is radical, UV-induced, and very slow. Some other oils, with no or too little double bonds do not oxidize. and remain mostly same.
The process of oxidation (drying) initially produces hydroperoxides, later able to serve as a source of more free radicals, leading eventually to inter-molecular linkages. I found contradicting  notions for what these linkages consist of (it is either oxygen bridges or direct links similar to polymers), but the end process if well known and used in varnishes. However, varnishes has extra components: they are usually diluted by some solvent to reduce viscosity and always contain a metal catalyst to accelerate drying of the oil, that also is usually heated for prolonged period of time, again, to increase the speed of drying.
The considerations above ignore obvious process of evaporation of light impurities and possibility of bio-degradation (I'm not aware of bio-degradation processes for pure fatty oils, but there are bacteria that live in almost undiluted toluene and there are bacteria that thrive on petrol, so I'm open to this idea).

Vegetable oils do lose some lighter constituents through evaporation, particularly substances responsible for aroma, but the primary degradation pathways, making it rancid, are through oxidation, hydrolization and microorganism metabolism; see http://www.ivannikolov.com/how-cooking-oil-goes-rancid/ and http://en.wikipedia.org/wiki/Rancidification.
Some tree-nut oil, in particular, quickly develops a noticeable off-taste. BHT, BHA and other additives (often applied to packaging rather than to the food itself) help decrease oxidative rancidity.
That said, vegetable oil used for lubrication in old machine parts certainly does thicken to the consistency of wax, though I do not know how much of that is due to evaporation vs. degradation.


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Is it possible to make broad spectrum light absorbing paint?
Is there such a paint / pigment that absorbs most wavelengths of light (beyond the visible light range) that can be used for solar heat generating applications?  If not, would it be possible to make it?  Ideally, this paint would absorb light from the sun in an efficient manner and become heated quickly by solar light.  It should probably at least absorb infrared, and any other wavelengths of light that produce significant heat when absorbed.  This should be far more effective than normal black paint as black paint only "for sure" absorbs light in the visible spectrum, and this desired paint would absorb other types of light from the sun.

Carbon nanotube black is highly absorbent in and outside the visible spectrum, see http://www.extremetech.com/extreme/186229-its-like-staring-into-a-black-hole-worlds-darkest-material-will-be-used-to-make-very-stealthy-aircraft-better-telescopes.
Of course, that does not make it ideal for energy conversion, as it would also need at least the following qualities to be practical.

A good conductor of heat to the heat-exchanger surface below
An enduring finish, resistant to abrasion, heat and environmental degradation
Economical, both in material cost and application
Safety in use and in manufacture


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Whats the difference between entropy and the (dis)order of a system?
Entropy is often verbally described as the order/disorder of the thermodynamic system. However, I've been told that this description is a vague "hand-waving" attempt at describing what entropy is. For example, a messy bedroom doesn't have greater entropy than a tidy room
My question is why is this the case? Also, what would better describe entropy verbally?

Essential: http://ariehbennaim.com/books/entropyd.html.

Briefly, spontaneous processes tend to proceed from states of low probability to states of higher probability. The higher-probability states tend to be those that can be realized in many different ways.
Entropy is a measure of the number of different ways a state with a particular energy can be realized. Specifically, $$S=k\ln W$$ where $k$ is Boltzmann's constant and $W$ is the number of equivalent ways to distribute energy in the system. If there are many ways to realize a state with a given energy, we say it has high entropy. Often the many ways to realize a high entropy state might be described as "disorder", but the lack of order is beside the point; the state has high entropy because it can be realized in many different ways, not because it's "messy".
Here's an analogy: if energy were money, entropy would be related to the number of different ways of counting it out. For example, there, there are only two ways of counting out two dollars with American paper money (2 1-dollar bills, or 1 two-dollar bill). But there are five ways of counting out two dollars using 50-cent or 25-cent coins (4 50-cent pieces, 3 50-cent pieces and 2 quarters, and so on). You could say that the "entropy" of a system that dealt in coins was higher than that of a system that dealt only in paper money.
Let's look the change in entropy for a reaction $\rm A\rightarrow B$, where A molecules can take on energies that are multiples of 10 energy units, and B molecules can take on energies that are multiples of 5 units.
Suppose that the total energy of the reacting mixture is 20 units. If we have 3 molecules of A, there are 2 ways to distribute our 20 units among energy levels with 0, 10, and 20 units:

If we have 3 molecules of B, there are 4 ways to distribute 20 units among energy levels with 0, 5, 10, 15, and 20 units:

The entropy of B is higher than the entropy of A because there are more ways to distribute the same amount of energy in B than in A. Therefore, $\Delta S$ for the reaction $\rm A\rightarrow B$ will be positive.

what would better describe entropy verbally?

We can also use "measurement of randomness" or "amount of chaos" or " energy dispersion"

Initially in 1862,  Rudolf Clausius asserted that  thermodynamic process always "admits to being reduced to the alteration in some way or another of the arrangement of the constituent parts of the working body" and that internal work associated with these alterations is quantified energetically by a measure of entropy change.
But later after few years Ludwig Boltzmann translated word alteration from Rudolf Clausius' assertion to order and disorder in gas phase molecular systems.
But if you see latest books they use concept of energy dispersion instead of order or disorder to explain entropy.

Source: Wikipedia
Also have look at physics S.E.

measurement of randomness??

@RutvikSutaria I have added link have a look there. :)

I will take a crack at this although I admit that this topic sometimes confuses me as well. Here is how I like to think about entropy. Consider a box containing equal amounts of gases A and B. So, we have a fixed volume and fixed number of molecules. Let us also isolate the box from its surroundings so that it has a fixed energy as well. We have just created a microcanonical ensemble (constant NVE). The ensemble consists of every possible configuration of molecules having the same NVE. Now, if we were able to step back and take a broad view of the ensemble, we would observe that the vast majority of boxes contain a rather bland homogeneous mixture of gases A and B. In fact, they would be indistinguishable for all practical purposes. Let us count the number of boxes in this state and call the number $W_1$. Continuing with the box counting, we find that there are only a handful of distinguishable boxes left, but they are quite interesting! One box, for instance, might have all of the A molecules crowded together in one corner and all of the B molecules in another corner! This one box can be labelled $W_2$. We start to understand that some configurations of molecules will be extremely improbable because they represent such a small fraction of the total number of possible configurations. Boltzmann quantified this relationship as $S=k \log W$. If we plug $W_1$ into this equation we get a high value for the entropy, $S$, because $W_1$ is so large. On the other hand, if we plug in $W_2$ the entropy we get is very low.
So, to sum up, entropy is really the measure of how likely a given system configuration is when compared to all of the possible configurations. In this sense, I would argue that you could say a messy room has a higher entropy than a clean room because there are so many more ways a room could be considered "messy" than "clean".

Can you make it bit more readable. This seems too boring! :'(

That's probably about the best I can do. I am a rather boring person.

Breaking down the wall of text might be helpful.


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Can we really see the bonds?
I was thinking is there really bond present at microscopic level or atoms/molecules are just nearby and are connected with force which is not visible(like gravitational force between earth and sun) and we make bonds just for understanding.

electrons exist

All credit to Zhang et al. "Real-Space Identification of Intermolecular Bonding with Atomic Force Microscopy" Science Vol. 342  no. 6158  pp. 611-614.
Yes, direct images of bonds, not only covalent bonds but also intermolecular hydrogen bonds have been recorded.
It is the electron density that is being observed, covalent and hydrogen bonds involving high electron density between the atoms.
Scanning Tunneling Microscopy can also be utilized to directly observe the electron density of bonds.

ionic bond implies that the electron density stays with the individual ions, and there is electrostatic attraction between the ions, but nothing has purely ionic bonding, there is always some covalent character

Interesting to note that since we can actually observe electron density with nc-AFM, the actual bond-order of said bonds can also be determined in some cases, as shown in "Bond-Order Discrimination by Atomic Force Microscopy" by Gross et al.

I'd be really careful about interpretation of bond order. Consider that even in "simple" molecules like CO, it's hard to define a real bond order. Certainly the work by @DavePhD and cartographer suggest there's increased interatomic electron density.

Does this work for non planar compounds?

No I don't think that is possible.
It's all about the attractions (the wanting of a full octet) between the atoms, to be honest we don't even have the technology to view the actual atoms itself (if we do then we are at the initial stages). So these "bonds" that you are talking about are just imaginary connections between atoms so people don't get confused or overwhelmed.
Take an ionic bond, one of the strongest bonds out there, we know it is caused by strong electrostatic attractions between the atoms in which one completely takes an electron or gives it up. If there was like a connection between them it would look almost surreal. It would make more sense if they are beside each other.
Now for the covalent bond, in a covalent bond atoms are sharing electrons, in this case the electron clouds merge, this maybe hard to grasp but that is just how this works. That is why for most covalent bonds we look at the Lewis Structures.
Now take a look at an ionic "bond".

Take a look at the covalent "bond".

Check out the double and triple "bond". - Bad Image

By looking at the above picture you can see that a triple "bond" is just two more electrons.
So in reality to make these drawings of atoms easy, we have came up with a bond. There might be "bonds" but one has to note that they are attractions instead. Here I gave you a couple examples, mostly 2D but if you want to really get the idea you should take a look at the 3D structures that include everything. But you have to also note that we don't know how far our current depictions are true, I just gave you the most plausible ones, so don't believe everything as a fact.
To sum it up: You Will Not See The Bonds (Not the sticks and ball kind). Though you may see the attractions.

Bonds of the stick and ball kind are an extreme simplification, the "real thing" are molecular orbitals, and those can be seen in STM microscopy & co.


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In order to be a gas does the pressure exerted by the gas has to be more than the atmospheric pressure?
In order to be a gas does the pressure exerted by the gas has to be more than the atmospheric pressure?
Really simple, I think yes to this question but teacher says no. So what is your view into this?

Is the question inspired by boiling of gas

@DSinghvi I don't think you can boil a gas, and no.

@AwesomeFlame123 It's a joke - if you want to boil sth it's vapor pressure should be at least as big as the pressure of the atmosphere - it doesn't say anything about gas pressure generally.

It would be nice if you could give your questions more meaningful titles. It would be even nicer if you could include your reasoning, why you think your answer is correct and the other one is false.

Your teacher is right. For example, at room temperature, the pressure of water vapor over a pond or lake might be around 25 torr; that's less than atmospheric pressure, but the water vapor is undeniably a gas.
Maybe your confusion stems from a belief that gases must be produced by boiling liquids (since a liquid must have a vapor pressure equal to the external pressure before it begins to boil). But a liquid can also become a gas through evaporation, no matter what the external pressure is.
In evaporation, molecules escape from the liquid's surface and become a gas.
Molecules in the liquid are in constant motion, and they frequently collide with each other. Every now and then one will get knocked off the surface of the liquid by the others. This can happen at any temperature or pressure the liquid exists at.
In boiling, bubbles of gas form throughout the liquid. This can only happen when the temperature is high enough to make the vapor pressure equal to the external pressure.

But doesn't a gas have to exceed atmospheric pressure to become a gas? Then how can it be a gas.

Certainly a liquid boils when its vapor pressure reaches atmospheric pressure... but a liquid can become a gas at temperatures and pressures lower than its boiling point by evaporation.

That is what that has always confused me, how can something be a gas at lower temperatures and pressures but it boils when its vapor pressure reaches atmospheric pressure. How can this be? You said evaporation but can you be more specific.

I've edited the difference between boiling and evaporation into the answer.

How can they escape liquid's surface? Just like that?

Molecules in the liquid are in constant motion, and they frequently collide with each other. Every now and then one will get knocked off the surface of the liquid by the others.

Oh, Only If I knew this earlier then I wouldn't be so confused. So just to be clear, a substance has to exert more pressure above the atmospheric pressure? For it to be a gas, not taking evaporation into account.

Think about what the molecules are doing to keep it clear. In a liquid, they're churning around but sticking together; in a gas, they're moving like rifle bullets, and not sticking together much at all. The higher the temperature, the faster the molecules move. In boiling, the liquid molecules are moving so fast that they break away from each other, even when they're buried inside the liquid. The bubbles have to form with enough pressure to push back the atmosphere on top of the liquid, or they'll be squashed. In evaporation, molecules are being knocked off the liquid surface by collisions.

Ok then, I got your point, now lets look at evaporation, once something is evaporated and is a gas can it exist as a gas forever even if the pressure and temperature is much higher?

No. Suppose we had a container containing nothing but water.  Suppose it is only half full, so there is empty space. There is no  pressure from other gases. If we heat this container enough, the water will boil and eventually turn completely into a gas.
In short, clearly other things influence the state of matter in everyday life besides pressure from atmospheric gases. In this example, I illustrate how water evaporating in a vacuum is determined by temperature.  I think  this is a good example because it involves just one substance and doesn't bring into question how to think about mixed gases and partial pressures.
There is an extremely useful equation that connects these relationships beautifully:
$PV = nRT$
P stands for pressure, V for volume, T for temperature, n for the number of molecules, and R is a constant.
This equation basically says that
any change in pressure or volume to a closed system of gas will result in a corresponding change to the temperature of the system.
Thus, as long as $P_1V_1\neq P_2V_2$, you can always think about a change in volume or pressure as a change in temperature for a closed system.

Your italicized sentence is incorrect; 1) a closed system has a constant number of molecules and 2) you can easily change volume and pressure without changing moles of gas or temperature, e. g., half the pressure and double the volume. Also, the ideal gas equation has nothing to do with why substances change state; an ideal gas never changes state.

@FredSenese Thank you for your corrections. I have edited the answer.

If you know Dalton's law of partial pressure that "in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases"
And since pressure is non-negative quantity then:
P(total)=P1 + P2 + ... + Pn
SO the pressure of gas in atmosphere in fact, cannot be greater than atmospheric pressures .
http://en.wikipedia.org/wiki/File:Dalton%27s_law_of_partial_pressures.png
And also I am not saying that gas stored in container cannot have pressure greater than atmosphere . The best example is your LPG cylinder.

Ok, so you are saying that a gas cannot exceed atmospheric pressure, then how is something going to be a gas if it cannot exceed atmospheric pressure?

Is it the substances intermolecular forces that determine if it can be a gas or not at a specific temperature.

@AwesomeFlame123 Who said gas cannot exist above atmospheric pressure. I just said that the gases in ATMOSPHERE which surrounds us, we breathe in cannot exceed atmospheric pressure by definition


text:
Why is the pKa (COOH) of alanine and glycine similar but the pKa (NH3) different?
Alanine is heavier with a ($\ce{-CH3}$) than glycine which has an ($\ce{-H}$). What makes them have different $\mathrm{p}K_\text{a}(\ce{-NH3+})$ $(\mathrm{p}K_\text{a}: \text{Ala}=9.69, \text{Gly}=9.60)$ but similar $\mathrm{p}K_\text{a}(\ce{-COOH})$ $(\mathrm{p}K_\text{a}=2.34)$ when the rest of the amino acid is similar? Why does changing a hydrogen for a methyl group affect the $\ce{-NH3+}$ but not the $\ce{-COOH}$?

Also there are three amino acids (glycine, leucine, aspartine) which have same $\mathrm{p}K_\text{a}(\ce{-NH3+})$ but different $\mathrm{p}K_\text{a}(\ce{-COOH})$.

Can you add the pKas for both amino acids?

I found the following pKa values for glycine and alanine from the CRC Handbook of Chemistry and Physics via Sigma-Aldrich:
Glycine $\ce{COOH}$: 2.34
Alanine $\ce{COOH}$: 2.34
Glycine $\ce{NH3+}$: 9.60
Alanine $\ce{NH3+}$: 9.69
First, although the values for the carboxylic acid are identical as reported, you would see a difference if you found a value reported with enough precision. The carboxylic acid is not immune to changes in the structure of the rest of the molecule.
Second, the difference between 9.60 and 9.69 is pretty small. In most practical situations, the two amino acids would behave identically with regard to the $\ce{NH3+}$ pKa.
The fact that the magnitude of the difference is larger for the $\ce{NH3+}$ pKa values can likely be attributed to proximity. In this case, the relevant proton is 2 bonds away from the $\alpha$ carbon, whereas the proton in the $\ce{COOH}$ group is 3 bonds away.


text:
Cell overpotential diagrams
Is there anyone who knows a book where I can find a detailed explanation of cell overpotential diagrams, like the ones shown here?
Thanks in advance!

Not sure what kind of explanation you are looking for, but
Electrode Dynamics, A. C. Fisher, Oxford Chemistry Press, 1996, ISBN: 019855690X
is a great concise explanation of overpotentials among other things.

I have always thought that corrosion textbooks give some of the most concise and enlightening overviews of electrochemistry. One of my favorites is:

Corrosion and Surface Chemistry of Metals, by D. Landolt, EPFL Press, 2007, ISBN:  0849382335

This book will explain to you how to interpret polarization diagrams.  The emphasis, of course, is on corrosion phenomena, but the same type of reasoning applies to any electrochemical system.  The author does a great job of breaking down all the sources of overpotential at an electrode and shows many polarization diagrams along with the discussion.  It really is a great book and I highly recommend it!


text:
Significant Figures: To expand or not to expand?
4 * (2 + 2) = ?
4 * 4 = 20  (rounded to 1 significant figure)
OR
4 * 2 + 4 * 2 = 16

2.0 * (0.5 + 0.5) = ?
2.0 * 1.0 = 2.0
OR
2.0 * 0.5 + 2.0 * 0.5 = 2

When I multiply 4 by 4, I get 16, not 20.

4 has one significant figure.  When you multiply two numbers with one significant figure, you are supposed to get a new number with one significant figure.  16 rounds to 20.

What the heck this has to do with chemistry ?!

There wasn't a laboratory science category so since most of the other significant figures questions are placed in the chemistry category, I decided to put my significant figures question in the chemistry category too.

Significant figures are math lessons used in chemistry, not chemistry. This Q should get removed. Your explanation puts this in chemistry category; and I'm not meaning to be insulting or something, but with this explanation we should merge math into chemistry!

This question appears to be off-topic because it is about significant figures outside of a chemistry context.

In the first example, the 4(2+2), if you were ABSOLUTELY going to adhere to significant figure rules, then 20 would be the correct answer, though I don't believe many people would fault you for saying 16. In your second case, 2.0(0.5+0.5), the appropriate answer here would indeed be 2.0.
In any case, significant figures are not really valuable when referring to purely mathematical equations like you did. We use them in order to avoid making errors when it comes to the accuracy of a result, in other words, significant figures are there so that your answer is not more specific than your measurement was. In mathematics, this is not an issue, as you are not making any measurements. So, while you COULD use significant figures in a situation like this, it really is not quite the appropriate situation in which to do so.


text:
Which one of the compounds listed would we use in order to get the highest concentration of sulphate ion in a saturated solution?

Solutions are prepared by dissolving a sulphate salt in water. Which one of the compounds listed would we use in order to get the highest concentration of sulphate ion, SO42-, in a saturated solution? What would be the concentration of sulphate ion in this case?
A. $\ce{CaSO4}$ $K_{sp}$ = 7.1E-5
B. $\ce{SrSO4}$ $K_{sp}$ = 3.4E-7
C. $\ce{BaSO4}$ $K_{sp}$ = 1.1E-10
D. $\ce{RaSO4}$ $K_{sp}$ = 4.0E-11

My first instinct was to simply look at the $K_{sp}$ values.. Since $\ce{CaSO4}$ has the largest $K_{sp}$ value, it should be the most soluble, right?
But then I looked at a data page referring to the solubility of common compounds in water (Please see image. According to the table, $\ce{SO4^2-}$ has low solubility with $Ca$, $Sr$, and $Ba$.
My question is- does the solubility depend on the $K_{sp}$ value (was my first instinct right), or is the table correct?
Any ideas would  be appreciated.

The $K_{sp}$ values are the way to go .. the other information you mentioned are just some convenient rules of thumb for predicting relative solubilities of compounds when you don't know the $K_{sp}$ values.  Also, the 4th choice in your list .. radium .. probably also belongs on that list, since it's the next alkali earth element after barium, but it's included because it's a radioactive element, and people don't commonly use or encounter it.


text:
Are the values for enthalpy, internal energy and Gibbs free energy the same for a particular process?
Internal energy, enthalpy and Gibbs free energy are all units of energy. So, for any particular process where energy is lost from the system the same amount is given off to the surroundings. Am I right in thinking that a change in internal energy/free energy/enthalpy is all a measure of how the energy of the system changes in a process and as a result should be equal to each other for the same process, albeit under different conditions?
Let me expand slightly:
$$dU<TdS-pdV$$ So for a spontaneous change to occur, the internal energy must decrease provided that the entropy and the volume are kept constant. The internal energy will continue to fall until equilibrium is reached. The exact amount of energy lost by the system is gained by the surroundings. As a side question: how exactly is the energy lost to the surroundings?
$$dH<TdS+VdP$$ The enthalpy of the system decreases for a spontaneous change, provided that the entropy and pressure are kept constant. Energy lost by system = energy gained by surroundings.
$$dG<VdP-SdT$$ The Gibbs free energy of the  system decreases for a spontaneous change, provided that the temperature and pressure are kept constant. Energy lost by system = energy gained by surroundings.
For a given process, surely the change in internal energy (at constant S,V) be equal to the change in enthalpy (at constant P,S) and Gibbs free energy (at costant P,V), right? The above argument seems to make sense to me but enthalpy can be defined as $H=U+PV$ which would suggest that the two values would be different wouldn't it?

While enthalpy, internal energy, and Gibbs's energy are all types of energy, they are not all measuring the same quantities. Internal energy is the energy of the molecules in your system. Enthalpy takes into account the internal energy plus whatever work is being done by or on your system. And Gibbs's energy is the maximum amount of non-PV work that you can get from your system. While these three quantities are all in units of energy, they are not just different terms for the total energy. Since enthalpy and Gibbs's energy involve work, and hence exchange of energy with the surroundings, these quantities will be different than just the internal energy of the system alone.


text:
For a system to undergo a spontaneous change at constant temperature and pressure the Gibbs free energy must decrease, but how?
$$\mathrm dG < V\,\mathrm dp - S\,\mathrm dT$$
For a process at constant pressure and temperature ($\mathrm dp = \mathrm dT = 0$), and in the absence of any non-pV work, $\mathrm dG < 0$. Since energy cannot be created or destroyed (and Gibbs free energy is a measure of energy) where does the energy go when this spontaneous change occurs? Likewise for internal energy: $\mathrm dU = T\,\mathrm dS - p\,\mathrm dV$?
Also, why is this even useful? I understand that it's easy to hold temperature and pressure constant in a lab but we only seem to measure enthalpy anyway using $\mathrm dH = C_p\,\mathrm dT$. If it's so easy to keep temperature and pressure constant yet so hard to keep entropy constant then why do we measure enthalpy (which decreases for a spontaneous change when entropy and pressure are held constant) rather than Gibbs free energy. I suspect I am barking up the wrong tree and I am hoping that you'll be able to put me right; the only thought that I had was that the fact that Gibbs free energy muct decrease for a spontaneous change would help to predict whether a spontaneous change would occur when done in a lab - given that $p$ and $T$ will be constant; is this right?

The inequality holds for a spontaneous or irreversible change; it'll be an equality for equilibrium or reversible change.

@A.K., at risk of sounding like a pedant, I just want to point out that variables such as $G$, $V$, $p$, $T$, $S$ are supposed to be italicised. The only thing in the question that should be upright is the differential symbol $\mathrm d$. To put everything in roman (\mathrm or rm) is technically incorrect: there is a guide on meta, see e.g. https://chemistry.meta.stackexchange.com/q/443/16683

If you want to track where the energy is going, do a first-law analysis (compute heat and work for a given change in internal energy or enthalpy).
If you want to predict whether a process occurs spontaneously or not, or you want to study an equilibrium, the second law is the tool you need. It's convenient to apply the second law by looking at free energy changes, since they let you focus on the system rather than the entire universe.
When we're interested in experimental internal energies or enthalpies we're tracking energy transfers. When we're interested in the Gibbs free energy we're usually answering questions about process spontaneity, or equilibrium conditions, or maybe we want to compute non-PV work. Different tools for different jobs.

Let me edit some of my comments below into my answer, because the comments are scrolling off the page and they make it clearer why you can't really track energy transfers just by looking at the Gibbs free energy:
The Gibbs free energy is an energy in the sense that it is the part of the energy that is "free" to do useful work. It does indeed have units of energy, and $\Delta G$ can be used to compute the maximum  possible nonexpansion work.
It is not an energy in the sense that it is not conserved in a process, the way internal energy and enthalpy are. $G$ is derived partly from entropy. It's quite possible that $G_\text{system} + G_\text{surroundings}$ isn't constant in a process.
If you have an isothermal process, $\Delta G= \Delta H−T\Delta S$. You can see from that equation that it's possible to decrease the Gibbs free energy simply by increasing the entropy, without any energy transfer at all. DavePhD's answer gives a concrete example of that.

Gibbs free energy is still a measure of energy though isn't it? (it's units are Joules). So if it decreases, surely the energy of the system decreases doesn't it?

If you have an isothermal process, $\Delta G = \Delta H - T\Delta S$. You can see from that equation that it's possible to decrease the Gibbs free energy simply by increasing the entropy, without any energy transfer at all. DavePhD is giving you a concrete example of that.

So, is it reasonable to say that the Gibbs free energy isn't really energy?

Gibbs free energy isn't an energy in the sense that it's conserved. It's derived from entropy. It's quite possible that $G_{system} + G_{surroundings}$ aren't constant in a process.

...That's the essence of my answer, really, if you want to track energy changes, look at energies (or enthalpies); if you want to look at spontaneity or equilibrium, look at Gibbs free energies.

Thanks. I just find it confusing given that it has energy in its name and has units of energy.

Think of free energy like potential energy. You can convert between kinetic and potential energy, and total energy is conserved. Free energy is a form of energy - it is a thermodynamic potential energy, and it is the amount of energy in a "useful" form. When the free energy of a system decreases, it means that the energy has transferred from a useful form into a non-useful form (via entropy increase)

You aren't alone---in fact, the IUPAC Gold Book says calling it "free energy" is deprecated. They want to call it the Gibbs energy, or (better, I think) the Gibbs function.

Imagine you have a divided insulated container.  On one side is helium gas and on the other side is neon gas.  Both sides are uniformly at standard temperature and pressure.
You remove the divider.
What happens?
The gases mix, because there are greatly more microstates we would consider "mixed" compare to not mixed, statisitically.
Entropy (S) increased, and in the ideal gas approximation there is no change in temperature or pressure (but there is a significant change in entropy of the system).
Gibbs so-called "free energy" decreased solely because entropy increased.
No energy went anywhere.

Very interesting. It'd be nice to see similar examples involving just vibrational, rotational or electronic entropy. Do you know any website or book?

I think the other answers here are pretty good, but it seems to me that the main point of confusion is over what it means for energy to be conserved.
You know that total energy is conserved, but you also know that it can take different forms. Gibbs Free energy is just one form of energy. When Gibbs free energy decreases, it goes somewhere - either as heat released by the system, or into increasing the entropy of the system. Those are really the only two options, because this is how Gibbs energy is defined.
For your other questions:

why is this even useful?

Your thoughts at the end are correct - it is useful because it lets us determine spontaneity, and also the amount of energy available to do "useful" work - useful in the sense that we can control its transformation to other forms of energy.

If it's so easy to keep temperature and pressure constant yet so hard to keep entropy constant then why do we measure enthalpy (which decreases for a spontaneous change when entropy and pressure are held constant)

We track enthalpy even though it is the thermodynamic potential for constant P & S systems because at constant T & P, in the absence of non-PV work, it is also equal to the heat produced or absorbed by the system.
In the lab that might not seem like a big deal, but if you are designing a steam engine, air conditioning system, or a chemical reactor, being able to predict the amount of heat that is absorbed or lost during a transition is very useful!

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text:
How could you dissolve a coating of CaCO3(s) from a metal plate?

A metal plate has an unwanted coating of $\ce{CaCO3(s)}$. You must find a way to clean this metal plate without causing any scratches or abrasion. How might you dissolve the coating?

I know I should find a substance that will dissolve the coating, and then soak the plate in it. How do I determine an appropriate substance?

Acetic acid, aka vinegar

It might take some time, but I would use vinegar.
The acid in vinegar is strong enough to dissolve the calcium carbonate, but it is not strong enough to dissolve metals (unless the metal reacts with water).

Do you mind explaining the chemistry behind that?

@McB I think it's because if a metal doesn't have a negative redox potential, it won't spontaneously oxidize (my redox chem is extremely rusty). I'm not sure what the difference between acetic acid or (what I imagine would work better) hydrochloric acid would be though.

@NickT Depending on what the metal is, hydrochloric acid could dissolve the metal (especially zinc and copper).

@LDC3 acetic acid will happily attack zinc, and HCl anecdotally won't (e.g. the "hollow penny" demo); what determines it though? More concentrated HCl might, as it would shift whatever equilibrium and thus (I think) the redox potential.

I recommend sulfamic acid with formula $\ce{H3NSO3}$. The reaction is irreversible, no dangerous compound:
$$\ce{2 H3NSO3 + CaCO3 -> Ca(H2NSO3)2 + CO2 + H2O}$$


text:
why do we consider CuO as a base?
According to Arrhenius acid/base theory, a base is a substance which when dissolved in water furnishes $\ce{OH-}$.
But $\ce{CuO}$ is insoluble in water and does not produce $\ce{OH-}$, so why do we consider it a base?

There are two other widely used acid/base theories that apply here: Brønsted–Lowry theory and Lewis theory.
A Brønsted base is defined as a proton acceptor. Technically, $\ce{CuO}$ is capable of reacting with $\ce{H+}$ to form hydroxide ions and eventually $\ce{H2O}$. Although $\ce{CuO}$ is insoluble in water (as you said), this reaction can occur in other solvents or in a biphasic system (where the proton is in solution and the $\ce{CuO}$ is present as an undissolved solid).
A Lewis base is defined as an electron donor. Generally, this means it has a lone pair of electrons that is available for reactions. The oxygen atom in $\ce{CuO}$ has lone pairs of electrons available.
$\ce{CuO}$ is considered a base because it fits these two descriptions.

Though in practice making bulk $\ce{CuO}$ act as a base is not an easy feat.

An additional way to consider it: $\ce{CuO}$ is defined as a base in the Lux-Flood theory because it is an oxide ion donor; it donates oxide to oxide acceptors (Lux-Flood acids)
An example of a Lux-Flood acid-base reaction involving $\ce{CuO}$ is the high pressure reaction with carbon dioxide to give $\ce{CuCO_3}$   (source).

$\ce{CuO}$ is a basic oxide and when it reacts with water it gives $\ce{Cu(OH)2}$ which is a base as it gives $\ce{Cu^2+}$ and $\ce{OH-}$ ions when dissolved in water.
So technically $\ce{CuO}$ is a base rather a basic oxide.
Generally all metallic oxides and hydroxides are bases.

I know you use the word "generally", but not all metal oxides act (mostly) as bases - CrO3 for instance is best considered acidic. And pedant alert, you mean metal oxide, VO is a metallic oxide, Al2O3 isn't.


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In the elementary steps of a reaction, why do we assume that the coefficients of a reactant correspond to its order?
Recently, I was taught that the coefficients of the reactants in an elementary step of a reaction indicates that reactant's order. Why is this the case? I thought that the order of a reactant was not related to its coefficient.
For example, in the following termolecular reaction mechanism
$$\ce{A + 2B -> C + 2D}$$
the rate law is $k[\ce{A}][\ce{B}]^2$. Why is this the case?
EDIT: Also, for equilibrium reactions, do we always assume the coefficients of the reactants to provide the order for those reactants? What about for equilibrium reactions involving multiple steps?

"I thought that the order of a reactant" - do you mean order of the reactant or order of the reaction?

I mean the order of the reactant. Is this an incorrect assumption? From the ChemWiki, it states that for chemical reactions involving multiple elementary steps, the orders of the reactants are not necessarily the coefficients of the reactants.

That's for reactions involving multiple elementary steps.

For reactions involving multiple steps, do we use the coefficients of the slowest step of that reaction to write the rate law? Or is the order of each reactant (in multiple step rxns) unrelated to its coefficient? Also, does this multiple step rule apply to equilibrium reactions as well, or do the coefficients always provide the order for equilibrium reactions?

I'll have to leave that question for someone else; it's late and I can't think it through atm. Good question though.

If a reaction has just one step, it's named an elementary reaction. In this reaction all of reactant directly participate on reaction and effectively collide with each other. So, the order of reaction corresponds to its coefficient, but if a reaction is not a elementary reaction, there isn't any reason for colliding all reactants with each other. For example, consider the following overall reaction:
$$\ce{A + B + D -> F}$$
Assume the corresponding mechanism is constituted of two elementary reactions:
$$\ce{A + B -> C}$$
$$\ce{C + D -> F}$$
As shown above, there is no need to collide A with D or A with C ... Then, in these type of reactions coefficients are not corresponded to order of reaction .

In these multiple step type reactions, would the coefficient of the slowest step be used to determine the order of the reactants?

yes ,it is correct in a reaction with multiple steps,the coefficient of the slowest step would be used to determine the order of the reactants.

Is it right to say if the order for the reaction is seen to be the coefficients of the reaction then it must be elementary? Is it right?


text:
Calculating the volume of 1 mole of liquid water
I am doing some rough calculations to test my understanding of elementary chemistry. In what follows, I calculate the volume of one mole of liquid water at about $\pu{4 ^\circ C}$ and $\pu{1 atm}$:
Method 1: Using the molar mass and density of water
The molar mass of water is $\pu{18 \times 10^-3 kg mol-1}$, and the density is $\pu{1 \times 10^3 kg m-3}$. Hence, the volume should be
$$V = \frac{(\pu{1 mol})(\pu{18 \times 10^-3 kg mol-1})}{\pu{1 \times 10^3 kg m-3}} = \pu{1.8 \times 10^-5 m3}.$$
Method 2: Using the size of an individual water molecule
If we model the mole of water as a cube, since there are $6.022 \times 10^{23}$ water molecules, each side of the cube would have $8.44 \times 10^{7}$ molecules.
Using the water data page on Wikipedia, I used trigonometry to calculate the H–H distance as being approximately $\pu{150 \times 10^-12 m}$. I would think the dimensions do not change too much when there are many molecules.
From this, I find that the volume is
$$V = [(8.44 \times 10^7) \times (\pu{150 \times 10^-12 m})]^3 = \pu{2.05 \times 10^-6 m3}$$
which is an order of magnitude smaller than Method 1.
What is the reason for this discrepancy?

The molecule length value you are using is the problem. It is more like 300 pm, which is the van der Vaals diameter and includes regions where electron density is significant. You have calculated an internuclear distance, but the nuclei of one molecule won't come close to those of another molecule because of electron-electron repulsion.
This will change your answer by a factor of 8.
See Finney, J. L. The water molecule and its interactions: the interaction between theory, modelling, and experiment. J. Mol. Liq. 2001, 90 (1-3), 303–312. DOI: 10.1016/S0167-7322(01)00134-9.


text:
What value of temperature does T represent in Gibbs Free Energy Equation?
Temperature before, or after reaction? Temperature change? Or what is it?

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Wow, Graham, this is turning out to be a long debate! Would you mind showing us that what you meant by "Gibbs Free Energy Equation"? Graham: Of course not. :)

We can not calculate the exact amount of G (Gibbs free energy), only we can calculate change of that,so we must put temperature change on this equation:
$$\Delta G = \Delta H - \Delta TS $$

Temperature can be constant or not,if it is constant we can change the equation to deltaG = deltaH - TdeltaS

Yes it can. But if it would be inconstant then the equation needs expansion. By the way, visit this page to make better posts.

@MARamezani The answer seems almost correct because the equation deals with partial derivatives. Your answer is just a simplified equation.

We've got:$$G_{(p,T)}= U+pV-TS$$
Where G is Gibbs free energy, U is the thermodynamic system's internal energy, p is pressure, V is volume, T is temperature and finally, S in entropy. In a chemical reaction, we can change the equation above to have:$$\Delta{G}_{(p,T)}= \Delta{U} +p\Delta{V}- T\Delta{S}$$
Which is then "interpreted" to:$$\Delta{G}_{(p,T)}= \Delta{H} - T\Delta{S}$$
This is what learners like me learn during the high school. Note that Gibbs free energy (measured in here) works well when deciding about the spontaneity of processes that occur in constant temperatures and pressures.
So if you're to use the equation, know that you oblige the poor poor system to be in a constant temperature! Think about it. If you're being asked a question from things you've studied, you will definitely run into phrases like STP or RTP in the question content. There is, however, a "formula expansion" to where we have variable temperatures; but of course, that wouldn't use the equations above.
If it's still vague for you, you can comment me.

I think the question is really clear. It seems you're just answering a high school question. It's a conceptual question not a high school exam.

Nah @Mohebifar, I just wanted to, or maybe tried to, look at the Q from his/her point of view. I just could have answered "the T of system which is constant."; instead, I took an approach a learner with some research on the topic would take, a position s/he is in. Plus, the "stp or rtp" question example is just an evidence for the fact, nothing more.

It is definitely false and that's why I am telling you that. The equation you wrote, is just an special case, not a universal equation. This is what you can find in high school books.

That's exactly what I say! Read the whole answer. The equation only works in constant p and T, if this is the especial case you're talking about. Graham hasn't provided the equation in his mind; but since the Net is full of this equation, and since this question can rise when talking about this equation, I assumed this was what that was in his mind.

Sorry I should have clarified, yes this is a question for AP chem so basically high school, and the equation I had in mind is dG=dH-TdS (d meaning delta)

The temperature must remain constant so it is the value both at the start and the end. I will try to explain this simply as, from your question, I suspect that you have just been introduced to $G$ with the equation $G=H-TS$.
The Gibbs free energy helps to indicate whether a process/reaction will be spontaneous at conditions where the pressure and temperature remain constant. This is useful because these are the conditions that occur in lab when we do an experiment. As I am sure you know the Gibbs free energy must decrease for a spontaneous process; in other words $\Delta G<0$.
Notice that the equation $G=H-TS$ does not involve $\Delta G$ is only gives a relation between Gibbs free energy and enthalpy and entropy. It is clearly useful to consider changes in $G$ so lets say $G$ changes by an infinitesimally small amount: $\mathrm{d}G$ (this basically means the same as $\Delta G$ but the change is much smaller) we can change the expression so that $\mathrm{d}G$ is written in terms of infinitesimal changes in entropy temperature and enthalpy:
$$(G+\mathrm{d}G)=(H+\mathrm{d}H)-(T+\mathrm{d}T)(S+\mathrm{d}S)$$
Expanding the brackets in the last expression:
$$G+\mathrm{d}G=H+\mathrm{d}H-TS+S\mathrm{d}T+T\mathrm{d}S+\mathrm{d}S\mathrm{d}T$$
I will now rearrange to make the following step clear:
$$\mathrm{d}G+[G=H-TS]+\mathrm{d}H+S\mathrm{d}T+T\mathrm{d}S+\mathrm{d}S\mathrm{d}T$$
Notice the $G=H-TS$ in the equation (highlighted in square brackets) – this can now cancel because it’s the same as having $G$ on each side. This leaves:
$$\mathrm{d}G=\mathrm{d}H+S\mathrm{d}T+T\mathrm{d}S+\mathrm{d}S\mathrm{d}T$$
The last term is the product of two infinitesimally small changes (very very small numbers) so is basically zero and can be ignored. This leaves:
$$\mathrm{d}G=\mathrm{d}H+S\mathrm{d}T+T\mathrm{d}S$$
Now, remember towards the start of my answer I said that this is useful at a constant temperature and pressure. Well, if temperature is constant $\mathrm{d}T=0$ so the $S\mathrm{d}T$ term vanishes leaving:
$$\mathrm{d}G=\mathrm{d}H-T\mathrm{d}S$$ and for larger changes: $$\Delta G =\Delta H -T\Delta S$$
It is crucial that you know, and hopefully now understand after my explanation, that the above equation is only valid at a constant pressure and temperature.
So in short $T$ is the constant value for temperature.


text:
Drying Clothes Outside Below Freezing
It is commonly accepted that you can hang your wet clothes outside to dry even when the winter temperature is below freezing. It is mentioned here
http://grist.org/living/ask-umbra-can-i-line-dry-my-clothes-in-the-winter-too/
that water in the clothes will freeze to ice, then sublimate into water vapor, making the clothes dry.
Question: Why does the ice in the clothing sublimate, while the ice on the roads don't sublimate? Why is there a difference between what happens to ice on the clothing versus ice on the roads?

Aside from the question.. I'd be a bit hesitant to hang my clothes outside when the water will freeze. The clothing might eventually get dry, but the expansion of the water into ice certainly might damage the fabric. I'm happier drying inside in the winter.

Good point @GeoffHutchison. Perhaps, then, this will be useful for my clothes that have become too small over the years - worth a shot, right?

As long as there is less than 100% relative humidity, the ice will sublime to some extent.
Just in the past week it has been in the -12 °C to -5 °C range in my area, and the snow and ice on the ground did sublime, on the order of 1 cm per day, more in the sun less in the shade.
For quantitative information see "Sublimation from a seasonal snowpack at a continental,
mid-latitude alpine site" Hydrol. Process. 13, 1781-1797 (1999).

Just a guess, but I assume the process would be quicker in hanging clothes than on the ground because it would access more moving air, which would move any humidity created from the sublimation away and drier air in.

On cold days, the wet clothes dry due to the water in the clothes that freezes, then sublimates, to create water vapour that escapes the clothes, therefore drying it.


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What happens to energy changes in non-reversible processes?
Take internal energy for example:
$$dU\leq TdS-PdV$$
Firstly, I would like to make clear that the following question/argument relies on the following statement being correct: the left hand side of the above equation represents the change in internal energy of the system and the right hand side represents the work done on the system by the surroundings. Please clarify if that is correct.
The two equations, above, are in fact inequalities where it only becomes an equality when the system is at equilibrium (when changes are reversible). However, when  the system is not in equilibrium the inequality implies that the work done on the system by the surroundings is greater than the internal energy gained by the system; how can this be? Where does the energy difference between the energy put in and the energy gained by the system go?
Secondly, at a constant $S$ and $V$, spontaneous changes occur only when $dU \leq 0$. What's going on here? Is it as simple as: the system must lose internal energy in a spontaneous change or does it mean that work must be done on/by the system in a different way (eg. electrical work)?

$\Delta U = q + w$ and $dU = dq + dw$
This is the first law of thermodynamics.  It is not an inequality.
$dU\leq TdS-PdV$ is correct when there is only PV work. It is incorrect when there is other work such as electrical or gravitational.

the right hand side represents the work done on the system by the surroundings. Please clarify if that is correct.

The PdV term alone represents the reversible work, the TdS term represents reversible heat

Secondly, at a constant $S$ and $V$, spontaneous changes occur only when $dU \leq 0$. What's going on here? Is it as simple as: the system must lose internal energy in a spontaneous change or does it mean that work must be done on/by the system in a different way (eg. electrical work)?

the system must lose internal energy in a spontaneous change

The equality assumes equilibrium  though doesn't it? This is certainly the case for G: $dG\leq VdP-SdT$. What happens in that instance?

$\Delta U = q + w$ and $dU = dq + dw$ do not assume equilibrium.

If you write $dU\leq TdS-PdV$ as an equality, that assumes quasistatic (near equilibrium) and only PV work.

Sorry, I see now I think. However, $dU \leq TdS-PdV$ is a valid inequality isn't it? Assuming equality at equilibrium?

I have edited my question now - I see my error.

as an inequality, $dU\leq TdS-PdV$ is valid even without assuming equilibrium, but assuming only PV work.  As an equality, it assumes quasistatic (near equilibrium) and only PV work.

I understand that but the right hand side still represents energy put into the system and the left is the change in energy of the system right? So where does the energy go if he two sides of the equation are not equal (i.e. when the process is irreversible)

Perhaps my question is better put in another way. When at a constant P and T $dG\leq w_e$ indicating that the change in gibbs free energy at a constant P and T is the electrical work done. In a non-reversible process, where the loss of Gibbs free energy is greater in magnitude (the electrical work is also negative) then where does the rest of the energy go?

(response to second to last comment) the right hand side no longer represents energy put into the system, because TdS no longer represents q

Why? It has units of energy.

when the process in irreversible, S can increase without any change in U.

Okay. May you please address the second point in my question.

yes, I added to my answer

Thank you. How does the system lose the energy? As heat?


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Why is pyridine a monodentate ligand, but pyrrole a tridentate ligand?
Is it correct to say pyridine’s non-bonding electron pair does not participate in electron resonance while pyrrole’s non-bond electron participates in electron resonance? Would therefore pyridine’s resonance not be disturbed so act the molecule can act as a monodentate ligand?

I don't understand how pyrrole can be a tridentate ligand, can you give an example where it acts as a tridentate ligand?

I think he says about coordination using whole ring, like with cyclopentadienyl

Still, I've  never seen pyrrole coordinated like that - examples welcome!

From what I gathered in the comments you confused a few terms. Both compounds are monodentate, i.e. they can only coordinate one metal at a time. However, pyridine usually donates one electron pair, while pyrrole can donate three (to one metal) — I haven’t seen a structure where pyrrole does that, though.
The reason is pretty simple: Pyrrole is much like the cyclopentadienyl anion except that one carbon atom is replaced by nitrogen. If it wants to coordinate in any way, it will have to do that like a slice of bread from its side, forming a $\eta ^5$ complex. Much like ferrocene in the picture below.

Pyridine, on the other hand, has no hydrogen on the nitrogen atom; instead it has a lone pair pointing outwards, but still in the plane of the ring. This allows pyridine to donate two electrons in a $\sigma$-like fashion to a metal centre, $\eta ^1$-style. Of course, pyridine could also perform a sandwich-like coordination much like benzene sometimes does. However, the direct two-electron $\sigma$ donation will be much stronger and therefore largely outweigh a sandwich-type donation in most compounds.

I have seen $\eta^5$ coordination of pyrrole, e.g. here. Presumably, that means a 6-electron interaction.

There are also examples of $\eta^6$ coordination of pyridine.

@GeoffHutchison I didn’t say ‘no-one has seen it’, I said ‘I haven’t seen it’ which held true until now. (But I wouldn’t have gone looking for it, I would have assumed it to exist but being less common.) Thanks for the nice examples!

Fair enough. Organometallic chemists seem to relish the idea of making "less common" compounds.


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How does this method of calculating the water equivalent of a calorimeter work?
I am doing a AP Chemistry lab in which we determine the heat capacity of a calorimeter (a concept that seems like it should depend on a lot more than just the calorimeter.) We have to calorimeters with water of different temperatures (which is tracked the whole time) but equal mass. We then mix them. We then track temperature again. We then extrapolate to the time of mixing (with a linear regression for some reason.) We look at how much energy was lost by the hot water and gained by the cold water based on all 3 lines (each calorimeter before, and the mixed calorimeter after.) Next we take the difference of the energy gained and energy lost at a single instant divide by the temperature change of the cold calorimeter (I have no idea why the cold one) and this is the calorimeters heat capacity.
I am having trouble understanding this, since at a single instance, the energy gained by the hot water and lost by the cold should be equal, and any difference would be error. It also seems completely random to divide by the temperature of the cold water and not the hot. What is going on here?
P.S. Later in the lab, extrapolating to the time of mixing is used to find the energy change due to a reaction, which makes more sense.
P.S.S One Calorimeter was fully mixed into the other (I forget which one, but either way does not make sense, and calorimeters themselves (like the material it was made from) had negligible heat capacity.))
P.S.S.S It is the teacher's first year doing this lab I think.

Maybe it's the cold one because it can't be the hot one. The hot one loses heat in many ways, and that makes the calculations more "approximate".

@PyRulez "One Calorimeter was fully mixed into the other (I forget which one, but either way does not make sense, and calorimeters themselves (like the material it was made from) had negligible heat capacity.))" What do you mean by it ?You are finding water equivalent and saying calorimeter heat capacity is negligible.It's so contradicting.

So you're mixing hot and cold water in a calorimeter, and from the temperature change read from a time versus temperature curve, you're finding the heat capacity of the calorimeter. You're not understanding why the heat capacity is computed the way it is.
First, identify all of your heat flows. You've got heat coming from the hot water, heat absorbed by the cold water, and heat absorbed by the calorimeter. If energy is conserved, $$q_{hot} + q_{cold} + q_{calorimeter} = 0$$
One thing that's confusing you is that last term, $q_{calorimeter}$. It isn't zero. It's very small, since your calorimeter is Styrofoam, but it will be measureable. Measuring it is a central goal of this experiment!
Now, you have a time versus temperature curve for the contents of the calorimeter. You're extrapolating the temperature plateaus before and after mixing to the time of mixing to figure out the initial and final temperatures. Since you had the cold water sitting in the calorimeter initially, and you dumped the hot water into it, the final temperature minus the initial temperature gives you the temperature change for the cold water. It also gives the approximate temperature change for the calorimeter if you can assume that the temperature of the calorimeter is the same as the temperature of its contents.
The heat capacity of the calorimeter is $$C = \frac{q_{calorimeter}}{\Delta T_{calorimeter}}$$
You know $\Delta T_{calorimeter}$ from your time vs temperature graph. You know $q_{calorimeter}$ from the energy conservation equation above; $q_{calorimeter} = -q_{hot}-q_{cold}$. Therefore you know $C$. You should know how to get $q_{hot}$ and $q_{cold}$ from the masses and initial and final temperatures of the hot and cold water (and the specific heat of water). Can you take it from there?

The point is that heat is not just stored in the water, but also in the calorimeter sturcture.  Heat that is unaccounted for just considering the temperature and amount of water must have been taken up by the calorimeter itself.

This should really be a comment. Any way, it has now been addressed in the question.

@PyRulez Old men like me say, "you can lead them to water but you can't make them drink".  When you say "calorimeters themselves...had neglibile heat capacity", you are contradicting your earlier statement "we determine the heat capacity of the calorimeter".   The point is that heat is not just stored in the water, but also in the calorimeter sturcture. Heat that is unaccounted for just considering the temperature and amount of water must have been taken up (or released) by the calorimeter itself.

When one talks about the heat equivalent of a calorimeter, you also take heat lost intro account. These calirimeters mostly leak their heat, and does not contain it (Styrofoam.)

@PyRulez The phrase "heat capacity of the calorimeter" refers only to the heat stored in the material of the calorimeter per degree. It does not include heat transfer in or out of the system over time.  See http://www.chm.davidson.edu/vce/calorimetry/heatcapacityofcalorimeter.html  For someone to give a more helpful answer, I think you need to give more information in question.  As far as heat leaking into or out of the calorimeter, the rate is proportional to the difference between the inside and outside temperatures, in accordance with Newtons law of cooling.

@PyRulez, it sounds like you're accounting for heat lost into the surroundings with your time vs. temperature curve. By extrapolating the initial and final temperatures at the time of mixing, you are compensating for temperature drift.


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Why does the Born equation give the Gibbs free energy of solvation rather than enthalpy of solvation?
The Born equation gives the difference in energy required to charge a particle in a vacuum and in solution which results in the work required to transfer an ion from a vacuum into solution. It is derived thus:
$$w=\int_0^Q\frac{Q}{4\pi\epsilon r}dQ-\int_0^Q\frac{Q}{4\pi\epsilon_0 r}dQ$$
Where $\epsilon=\epsilon_r\epsilon_0$ ($\epsilon_r$ is the relative permittivity of the medium/solvent - a dimensionless quantity). $w$ is work done.
$$w=\frac{1}{4\pi\epsilon r}\int_0^QQdQ-\frac{1}{4\pi\epsilon_0 r}\int_0^QQdQ$$
$$w=\frac{Q^2}{8\pi\epsilon r}-\frac{Q^2}{8\pi\epsilon_0 r}$$
$$w=\frac{z^2e^2}{8\pi\epsilon r}-\frac{z^2e^2}{8\pi\epsilon_0 r}$$
So for a mole of ions the work done is equal to:
$$\frac{N_Az^2e^2}{8\pi\epsilon r}-\frac{N_Az^2e^2}{8\pi\epsilon_0 r}=\triangle_{solv}G^{\theta}$$
My textbook seems to suggest that the last step makes it Gibbs free energy of solvation rather than just work done. I have two issues with this. Firstly, why is it Gibbs free energy and not, say, enthalpy of solvation or internal energy of solvation? Secondly, why would scaling the number of ions to one mole make it into Gibbs free energy? Isn't Gibbs free energy an extensive property with units of Joules? I know that the Gibbs free energy (and enthalpy) are often given with units of $Jmol^{-1}$ but isn't this strictly molar gibbs free energy/chemical potential?

1) Because $-\Delta G$ is the maximum possible nonexpansion work done by a system in a constant temperature and pressure process. You can write $$\Delta G \le w_{non PV}\quad (\text{constant T and P, closed system})$$
where the equality sign holds for reversible changes.
Here's proof:
$$\begin{align*}
H &= U + P V&\text{(definition of enthalpy)}\\
dH &= dq + dw + d(PV) &\text{(because dU = dq + dw)}\\
dG &= dH - T dS - S dT &\text{(because G = H - TS)}\\
&=dq + dw + d(PV) - T dS -S dT &\\
&= dq + dw + d(PV) - T dS & \text{(when the change is isothermal)}\\
&= dw_{rev} + d(PV) &\text{(if reversible,}\ dq_{rev} = T dS\text{)}\\
&= dw_{electrical,rev} + dw_{expansion,rev} + P dV + V dP & \text{(partitioning the work)}\\
&= dw_{electrical,rev} + V dP & (dw_{expansion,rev} = - P dV)\\
&= dw_{electrical,rev} & \text{(if pressure is constant)}\\
\end{align*}$$
2) It isn't the scaling to one mole that makes the result a Gibbs free energy (see 1); that just makes it a molar Gibbs free energy. You can use either extensive or intensive Gibbs free energies in calculations; whatever is most convenient. Just make it clear (either by context or notation) which one you're using.

Would you please elaborate on why $-\triangle G$ is the maximum electrical work?

Also, why is it an inequality? Surely a loss of free energy would always be equal to the energy gained by the surroundings by non-pv work if pv-work is not available?

Thank you very much for the proof. When the change is not reversible  the change is gibbs free energy is different to the electrical work done. How is this possible given that energy cannot be created nor destroyed?

OK, I added the proof to my answer... the inequality applies to either irreversible or reversible processes. For an irreversible process you'll have $\Delta G < w_{nonPV}$. Watch the signs, $w_{nonPV}$ is minus the non-PV work done by the system, so you could write $w_{by, nonPV} < -\Delta G$ for an irreversible, closed system, constant T,P process.

Thanks, I just don't get how there can be an inequality. dG is negative and it's more negative than the electrical work done on the system which is also negative. Put more simply, the gibbs free energy lost by the system is less than the electrical work done by the system - where does the rest of the work/energy go?

"When the change is not reversible the change is gibbs free energy is different to the electrical work done. How is this possible given that energy cannot be created nor destroyed?" Look at the line in the proof where I say "if reversible, $dq_{rev} = TdS$". To see where the inequality for an irreversible change comes in, you can change that step to "if irreversible, $dq_{irrev} \lt T dS$" and take it from there.

I understand where it comes in but how can energy lost by the system not equal the work done by the system?

Please see my last comment. I am not sure that my issue is with the clausius inequality

That statement ($dq_{irrev} < T dS$) is one statement of the Clausius inequality---that's the central concept you want to clear up. That means $dq_{irrev} - T dS < 0$  in my proof, and that term will remain in the final result if you work it through for an irreversible process. There is no energy being lost; there will be some heat dissipated  in the irreversible process.

Isn't heat also non-pv work though?

No; heat and nonexpansion work are fundamentally different; work is an energy transfer that occurs when you have organized motion against an opposing force. Heat is energy transfer via random molecular motion.


text:
Reviving a 6 Volt lead-sulfuric acid battery
I have a 6 Volt sulfuric acid battery used to power a golf cart that has all the acid drained out, and the usual solution was to add more water to the battery however I am afraid that would not work this time since there is no acid left. Could I solve it by adding more acid from other batteries? or do I need to be accurate in the amount of acid added? or could I just solve it by adding water? If you could also give me an explanation as to how these batteries work and why adding water has any benefit I would appreciate it.

Depending on how long the battery has been sitting empty even adding electrolyte might not help.
Your best bet is to fill the battery with distilled water and put it on a charge, charge it for a couple of days if possible.
Get yourself a hydrometer (used for checking the specific gravity of your electrolyte) and check the water's gravities. After you charge the battery for awhile check them again. If the gravities start to rise, and the liquid is reasonable clear stop charging the battery and let it sit 12-24 hours. (If however the water turns cloudy/grey the battery is scrap get a new one).
After the battery has sat and cooled down, empty the water into a container, and neutralize with soda ash, or baking soda. BE CAREFUL even though what you put it was water, what will come out might/will be corrosive. Dispose of properly.
Now go to your local auto parts store and buy a box of battery electrolyte its usually sold as 1.28 spg, fill the battery with electrolyte re-charge until your gravities come up, and voila, use just re-conditioned a battery.
Be sure to wear rubber gloves / goggles or face shield and any other appropriate safety gear. Remember that when a battery charges it releases hydrogen sulfide gas which is explosive so no open flames sparks etc....
Of course the battery may only last your a couple of months or another year or so, and it may not even come back at all. But I have used this procedure with great success to revive old batteries....and squeeze a bit more life out of them.
Jeff
Industrial battery specialist for 30 years.

Obviously handling sulfuric acid is hazardous.  If all the acid leaked out, I would get a new battery, otherwise the acid will leak out again and possibly hurt someone or damage the vehicle.
If there is no acid, certainly adding water will not help.  If you do add acid, the concentration of acid needs to be correct.
Lead-acid batteries do not contain pure sulphuric acid, but acid dilute with water.  The concentration of acid can increase over time due to electrolysis of the water to hydrogen and oxygen gases.  If the concentration of acid is too high (solution density above 1.19 g/ml), adding water to dilute the acid is beneficial.


text:
It is said that atoms cannot be created. If so, then how did atoms get created after the Big Bang?
I read somewhere that atoms cannot be created. If this is true, then how did the atoms form after the Big Bang? Also, does this mean that the number of atoms in our universe has remained the same since then? Also, what happens as the universe expands? Does the atoms just scatter or are new atoms being created?

It looks like your question is rather abot physics and astronomy than chemistry.

Perhaps what you read is not true.  A simple counterexample would be alpha decay, which turns one atom into two. Of course, fusion turns two atoms into one...

Simply put atoms are not created or destroyed in chemical reactions, but in nuclear they certainly are.

Hey, just that there's the tag atoms doesn't mean this is a chemistry question!:D Please take a look at how these concepts are categorized at the reliable sites in the Net. This is astrophysics.

This question appears to be off-topic because it is about astrophysics more than chemistry. It surely can be dealt with better at physics.SE or astronomy.SE.

Historically, in chemistry texts it was common for there to be statements like "All the results of chemistry prove that the ultimate atoms of bodies are unchangeable and imperishable...by no natural process can material atoms be either created or destroyed" Textbook of Chemistry by Draper 1846.  So historically the princple that atoms are not created or destroyed has been the foundation of chemistry and the question is relavent to chemistry in that sense, but the details of the answer are nuclear physics and astrophysics.

@DavePhD, the question title and contents are astrophysics, the concepts are astrophysics, its answer is astrophysics, its parents are astrophysics... :D You be the judge. The almost-a-little-higher-than-trivial fact that chemists at "Dalton"'s age believed the atom to be non-breakable, non-producible and non-destructible doesn't really make it a chemistry Q rather than a physics Q.

"I have been engaged in experiments which suggest that the atom can be artificially disintegrated. If it is true, it is of far greater importance than a war!" Rutherford, 1917.  The belief prevailed long after Dalton.  I suppose the relavence to chemistry depends upon how important one considers the history of science to science education.

Atoms can be created and destroyed.  For example, in nuclear fusion two or more atoms make one atom.  In nuclear fission, one atom makes two or more atoms.

how did the atoms form after the Big Bang?

The theory of this process in known as Big Bang Nucleosynthesis

Starting from protons and neutrons, light elements formed.  Later, in stars, nuclei up to iron formed through nuclear fusion.  Nuclei heavier than iron formed from supernovae by the r-process and asymptotic branch giant stars by the s-process.

Also, does this mean that the number of atoms in our universe has remained the same since then?

No.  The number constantly changes due to fusion, fission, supernovae, creation of neutron stars, black holes and possibily quark stars.

Also, what happens as the universe expands? Does the atoms just scatter or are new atoms being created?

Currently, this process does not create or destroy atoms.  However, according to the theory of accelerating expansion, there is a senario referred to as the "big rip" in which all atoms would be destroyed.


text:
Second electron affinity of sulfur and oxygen
Why is the second electron affinity of oxygen greater than that of sulfur?
I think it should be like- second electron gain enthalpy of oxygen should be greater than the sulfur but electron affinity should be less than sulfur.

Electron gain enthalpy= - electron affinity

To RobChem..may u please tell that what's exactly greater for oxgen-electron affinity or electron gain enthalpy?

They have the same magnitude but different sign. For oxygen, the process where one electron is added to the neutral atom is exothermic so the electron gain enthalpy is negative but the electron affinity is positive. Look up the work affinity and it might help.

I was exactly asking for the SECOND electron affinity and electron gain enthalpy between oxygen and sulphur. .. I know that second electron gain enthalpy of oxygen is greater than sulphur (with +ve sign) ..so it should have lesser electron affinity than that of sulphur (with -ve sign) ...no?

Actually I was solving some questions and the answer given there was rather odd.. then I turned to Google but got messed up with them.. anyways ur reply is awaited and thanks for paying ur attention. .. m I right what I said just now?:)

I know how you feel. Maybe it's better to first check easier resources. Did you read what was written at wikipedia?

@Ϻ.Λ.Ʀ. wikipedia dont have any article about electron gain enthalpy...but have one about about electron affinity

The electron affinity of an atom or molecule is defined as the amount of energy released when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion.
It is the tendency of an atom to gain electron. It is numerically equal to the negative of electron gain enthalpy.
Although fluorine has the highest electronegativity, chlorine has the highest electron affinity and this is because the considerable repulsion in the tightly packed $2p$ subshell of fluorine.
While electron gain enthalpy is enthalpy change when an isolated gaseous atom attain an electron to form a monovalent gaseous anion.
Chlorine ($\ce{Cl}$) has more negative electron gain enthalpy than fluorine the reason is same  as above.

Source:
Wikipedia and chemistry reader


text:
Is oil shinier than water? Why?
The physics part: The part that talks about being shiny, I agree, is mainly physics. But proceed in my question, and you'll see how it's a chemistry one.
We know things are shiny if they either produce light, or deflect light well. Producing is not our topic in here, but deflecting is. Metals are considered shiny objects.
Mad scientist (the site, not our fellow friend!) has provided a nice explanation for why metals are shiny: In short, because free electrons (electrons are partially "free" in metals, due to the especial way of bonding) deflect more photons than usual and do not let the inner layers of atoms in the metal absorb them.!
(These three differ in shine because of impurity; not related)
Metals aren't the only materials that are exceptionally "shiny". We are aware of many crystals that are either well-cut or well-formed, which result in total internal refraction and that makes them very shiny too. There's a related question in physics.SE wherein the diamond's shiny properties are questioned.
The organic-chemistry part:
Saying the word "oil" talks about a vast variety of chemicals; therefore, in an attempt to narrow down the question, I only ask about liquid-at-RTP triacylglycerol (the famous triglycerides which exist usually in plants or fish). The chemical structure is as following:

My question is: Organic molecules like these aren't metals, are they?! So since a thorough research (maybe seemingly) didn't get me answers about why these are good light reflectors I'm asking for a way that will allow me to compare the reflectivity of these liquids, primarily from their structural formula. e.g.: Are these liquids "shinier" than water? Why? About water, I would say that for example, Hydrogen atoms in correspondence to the Oxygen atom, make water have hydrogen bonding, which is the main cause of surface tension. Surface tension causes the surface of water to be "smoother" and thus the reflectivity increases. On the other hand, it can't be only about Hydrogen bonding, can it?
(If the answer might be too broad, you may guide me to any resource helping me to find an answer, and that's why I add the tag: reference-request.)
Image sources: http://www.luxrender.net/wiki/images/2/2f/Shiny_yellow_specular_series.jpg ; http://library.med.utah.edu/NetBiochem/FattyAcids/7_2.html ; http://guweb2.gonzaga.edu/faculty/cronk/biochem/images/triglyceride.gif

The Fresnel equations, http://en.wikipedia.org/wiki/Fresnel_equations, are used to calculate reflectivity of transparent substances based on the index of refraction of the liquid vs. that of air or other medium. There are complications introduced by conductivity, and solids may display anisotropic reflectivity, see http://www.hindawi.com/journals/isrn/2012/732973/.
Dispersion staining is a technique for staining microscopic specimens depending on the change in the index of refraction with the color of light (dispersion), see http://en.wikipedia.org/wiki/Dispersion_staining. This uses the shininess of substances in a liquid, often an organic immersion oil, to help distinguish or identify them.
Note that metallic reflection is based on a "sea" of electrons bouncing back an inverted wave, and differs from refraction. Liquid metals such as mercury are shiny for that reason, but I do not know if there are sufficiently conductive organic liquids that would have metallic reflection.