text:
MO diagram of N2(2-)
In the MO diagram of $\ce{N2^2-}$, does s-p mixing happen to a significant degree to change the ordering of orbitals - $\pi_\mathrm{2p}$ orbital below $\sigma_\mathrm{2p}$? As it is isoelectronic with oxygen, I think that the s-p mixing shouldn't change the ordering.

In order to test whether there may be $\ce{s}$-$\ce{p}$ mixing and also to check whether there may be actually a bound state, I optimised this molecule (at PBE0-D3/def2-TZVPD) using NWChem 6.6 and analysed it using natural bonding orbitals (NBO 5.9).
I found following minimum geometry:

Figure 1: Minimum geometry found for $\ce{N2^{-2}}$ (distance in angstrom)
So there seems to be a bound state.
Natural charge on each $\ce{N}$ is -1.0 as would be expected (natural electron configuration was found to be $\ce{1s^{2.00} 2s^{1.65} 2p^{4.17} 3s^{0.06} 3p^{0.05} 3d^{0.07} 4d^{0.01}}$).
Wiberg bond order in the NAO basis for the $\ce{N-N}$ bond was found to be 2.2511.
Two bonding NBOs were found:
    (Occupancy)   Bond orbital/ Coefficients/ Hybrids
-------------------------------------------------------------------------------
  1. (2.00000) BD ( 1) N  1- N  2
              ( 50.00%)   0.7071* N  1 s(  0.00%)p 1.00( 99.74%)d 0.00(  0.24%)
                                                 f 0.00(  0.02%)
              ( 50.00%)   0.7071* N  2 s(  0.00%)p 1.00( 99.74%)d 0.00(  0.24%)
                                                 f 0.00(  0.02%)
  2. (2.00000) BD ( 2) N  1- N  2
              ( 50.00%)   0.7071* N  1 s( 36.78%)p 1.70( 62.66%)d 0.01(  0.51%)
                                                 f 0.00(  0.06%)
              ( 50.00%)   0.7071* N  2 s( 36.78%)p 1.70( 62.66%)d 0.01(  0.51%)
                                                 f 0.00(  0.06%)

That is,
$$\pi_{\ce{N1-N2}} = 0.7071 \ce{p}_{\ce{N1}} + 0.7071 \ce{p}_{\ce{N2}}$$

Figure 2: $\pi_{\ce{N1-N2}}$ bond of $\ce{N2^{-2}}$
$$\sigma_{\ce{N1-N2}} = 0.7071 \ce{sp^{1.70}}_{\ce{N1}} + 0.7071 \ce{sp^{1.70}}_{\ce{N2}}$$

Figure 3: $\sigma_{\ce{N1-N2}}$ bond of $\ce{N2^{-2}}$
Furthermore,
NATURAL BOND ORBITALS (Summary):

                                                    Principal Delocalizations
          NBO                 Occupancy    Energy   (geminal,vicinal,remote)
===============================================================================
Molecular unit  1  (N2)
  1. BD ( 1) N  1- N  2        2.00000     0.18488
  2. BD ( 2) N  1- N  2        2.00000    -0.43475  10(g),42(g)

Thus, according to the simple Lewis picture provided by NBO,

$\ce{s}$-$\ce{p}$ mixing ($\ce{sp^{1.70}-sp^{1.70}}$) happens for the formation of $\sigma_{\ce{N1-N2}}$, as expected;
the formation of $\pi_{\ce{N1-N2}}$ happens through pure $p$ hybrids; and
$\pi_{\ce{N1-N2}}$ lies in energy above $\sigma_{\ce{N1-N2}}$ by 0.61963 a.u..

According to "Experimental Search for the Smallest Stable Multiply Charged Anions in the Gas Phase" Physical Review Letters vol. 83, page 3402: "It has
also been concluded that no diatomic or triatomic dianions
are stable" http://casey.brown.edu/chemistry/research/LSWang/publications/84.pdf

And Physical Review A vol. 60, page 3515 says "Theoretical
attempts to understand the stability of the dianions already
discovered and directed towards finding ever smaller molecules
capable of supporting two extra electrons soon followed
[18–30]. A summary of the status of these calculations
is given in Scheller, Compton, and Cederbaum [31]. To
date, the smallest molecular dianions calculated to be stable
are alkali-metal halides of the form MX3
-2 (M=Li, Na, or
K; X=F or Cl)" http://journals.aps.org/pra/pdf/10.1103/PhysRevA.60.3515

You are completely right, this is highly unstable, as thoroughly supported by the articles provided and further citations therein.
In fact it has been stated, "a molecular dianion (such as a diatomic) must be stable not only with respect to the ejection of an extra electron but also to fragmentation" (ref. 31 of your second citation).

Thus, either $\ce{[N2]^{-2} -> [N2]^{-1} + e^-}$ (and maybe even $\ce{[N2]^{-1} -> N2 + e^-}$, I'm not sure) or $\ce{[N2]^{-2} -> 2 N^{-1}}$ (or $\ce{[N2]^{-2} -> N^{-2} + N}$?) is a favourable, spontaneous process.
This, in turn, is due to very low transition states (fast processes) and/or very stable products.
But I have not claimed stability, but the existence of a bound state.
At the level of theory applied, I found one bound state and analysed its wavefunction. No more, no less.

The singly-negative charged $\ce{N2-}$  has been calculated as bound between lengths 1.4 and 2.5 Angstroms http://aip.scitation.org/doi/pdf/10.1063/1.478408 , so I'm surprised that adding another negative charge would make the bond length shorter and still be bound.

Also N2(2-) has a triplet GS, while the answer seems to imply a singlet one.

Well, s-p mixing only occurs when the s and p atomic orbitals are close in energy (it's a necessary but not sufficient condition). Even though $\rm N_2^{2-}$ is isoelectronic with $\rm O_2$, the lower effective nuclear charge on nitrogen should make its s orbitals a little closer to the energies of the p orbitals than they would be in oxygen. I think you still would have s-p mixing.
Clarification: I mean to cast doubt on the conjecture that there's no s-p mixing here. As you point out in the comments, electron repulsion is a consideration too. We can't definitively say which effect is more important without looking at the actual orbital energies.

But the two added electrons would cause big repulsion forces, which would increase the difference in energy between s and p, right?

There would be some additional repulsion which would increase the difference, but those two added electrons go in different orbitals, so it wouldn't be that big a difference. Really we'd have to look at the actual orbital energies to answer your question definitively.

I would be delighted if I could get an accurate answer.

@FredSenese  Fred wrote, "Well, s-p mixing only occurs when the s and p atomic orbitals are close in energy."  It doesn't always occur even then.  The lone pairs in water are an example.  One lone pair is basically in a p orbital and the other in an sp orbital.  That's very different from what was (still is) taught.  S-P mixing is difficult (impossible?) to predict without experimental data such as molecular shape, ESCA (for lone pairs), etc.  In addition to the water example, s-p mixing in NH3 and PH3 is different, how would you predict this difference without experimental data?

@ron, in your NH3/PH3 example, what role would the different effective nuclear charges play on the difference in energy between s and p orbitals? Wouldn't that predict the difference?

@FredSenese  It doesn't seem to be taught that way, many people learn that PH3 is tetrahedral like NH3.  But following up on your point, due to more nuclear screening in P than N, wouldn't the 3s and 3p orbitals be expected to be closer in energy and therefore more likely to mix (hybridize)?  Just the opposite of what is observed.

Well, I don't teach that PH3 is tetrahedral---it's not. And the effective nuclear charges for 2s/2p in nitrogen are 3.847/3.834, and for 3s/3p in phosphorus they're 5.642/4.886 (I'm using Clementi's numbers here). So wouldn't the situation be a bigger separation in P, and in accord with what's observed?

@FredSenese Yes, based on those numbers.  Water is still a problem.  Why do the 2 OH bonds and one of the lone pairs hybridize and the other lone pair doesn't.  I'm still skeptical that s-p mixing can be predicted by simple methodologies.

We agree on that, actually; I just wanted to provide a counter argument for Marko's assertion that there wasn't any s-p mixing. Electron correlation needs to be accurately included in the calculation to see and understand effects like that...  When I'm back on campus I'll do a quick search for some $\rm N_2^{2-}$ calculations that include electron correlation; if I find anything I'll update my answer.

It don't think that $\ce{N2^{2-}}$ can exist as a free species.
$\ce{N-}$ is unstable.  In other words, the electron affinity of nitrogen is negative.
To create $\ce{N2^{2-}}$ you would need to bring together 2 negatively charged $\ce{N-}$ which are each already unstable due to their negative charge.
According to Experimental Search for the Smallest Stable Multiply Charged Anions in the Gas Phase Physical Review Letters vol. 83, page 3402:

It has also been concluded that no diatomic or triatomic dianions are stable

And how is this relevant to the question?

@user34388 If there is no bound state of the ion, how can it be modeled using linear combinations of bound atomic orbitals such as s and p?

Fair enough....


text:
What material is synthesized in the rubber egg demo?
One of the popular do-at-home chemistry demonstrations is placing an egg in a container of vinegar for two days. After that period most if not all of the eggshell has dissolved and the outer part of the egg has congealed giving the egg a 'rubber' like appearance and feel.
So chemically the vinegar (acetic acid) reacts with the egg protein (albumen).
Chemically speaking, what is this product?

So far I've reached the fact that the protein has these aminoacids: Glycine, tryptophan, tyrosine, and cysteine.

The shell dissolution reaction is the calcium carbonate in the egg being dissolved by the acetic acid, resulting in release of $\ce{CO2}$ and formation of calcium acetate.
$$\ce{CaCO3 + 2 CH3COOH -> Ca(CH3COO)2 + CO2 + H2O}$$
The rubberization of the egg white is primarily a protein denaturation reaction.  Albumin in egg whites is a globular protein, meaning each molecule is a single chain that is tightly bundled into a ball-like ("globular") form.  In this form the molecules can stay nicely solvated and the egg white is liquid.  However, the low pH from the acid (and also possibly the high concentration of calcium acetate) causes each chain to unfurl and then to bind with other neighboring chains, forming a complicated cross-linked tangle of molecules.  This tangle is no longer a liquid but is a rubbery solid-like material.
So the products are primarily calcium acetate, $\ce{CO2}$, and denatured albumin.

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text:
Does adsorption violate thermodynamics?
My textbook reads as follows:

When a gas is adsorbed, the freedom of movement of its molecules become restricted. This amounts to decrease in the entropy of the gas after adsorption, i.e. Entropy change is negative.

But now, I am unable to understand why it isn't a violation of the second law of thermodynamics, which states that the entropy change of a system can never be negative. Please guide.

Oh, no not again that... Entropy decreases locally by but raises globally - If this wouldn't be possible you wouldn't exist

This I think is the most commonly asked question about entropy! (Not the exact question, but the concept)

The second law of thermodynamics states that the entropy of the universe always increases.
$$\mathrm{d}S > 0$$
In the case of adsorption the entropy of the system; the gas being adsorbed; decreases but the entropy of the surroundings;the rest of the gas and the surface (and everything else in the universe); increases and this outweighs the decrease in entropy of the system.
$$\Delta S_\mathrm{sys} < 0 \qquad \Delta S_\mathrm{surr} > 0$$
$$|{\Delta S_\mathrm{surr}}| > |{\Delta S_\mathrm{sys}}|$$
This is because adsorption is an exothermic process and so the surroundings are heated up and therefore increase in entropy. If you consider the Gibbs energy change for adsorption it will be negative because the negative $\Delta H$ term is larger than the positive $-T\Delta S$ term. This chem.uic.edu page goes through the maths of this very nicely.

The original URL got broken, and I replaced it with WBM version. However, it looks like there is some font issues with the PDF making it a questionable source for learning, so you might want to either write down the "maths" here.

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text:
Why does stannous chloride occur despite the octet rule?
Shouldn't reaching an octet be any atom's "goal"?
However, I've recently learned about cases that are either expanding octets, or have lesser than "enough" electrons for an octet abiding. e.g.:
S in sulfur hexafluoride (Expanding octet into $\text{3d}$)

Source
B in boron trifluoride (It's "hextet", instead of octet)

Source
However, both $\ce{SnCl2}$ and $\ce{SnCl4}$ are existent. The latter is explainable with octet, but not the former. Surprisingly, $\ce{SnCl2}$ is more stable!

How?
Why is the phenomenon happening?

Dave pointed out that the octet nonabidingness is happening in the vapor phase. So I redirect the question to ask about the vapor phase; since that's what I'm looking for.

Well, it's inly a "rule" of octet, not a law :)

Rule, law, whatever! They look the same to a nonnative English speaker. :)

related: http://chemistry.stackexchange.com/questions/5484/what-elements-dont-observe-the-octet-rule, http://chemistry.stackexchange.com/questions/13949/how-can-sulphur-and-phosphorus-expand-their-octet, http://chemistry.stackexchange.com/questions/19433/how-can-iodine-bond-with-5-fluorines-in-iodine-pentafluoride?lq=1

$\ce{SF6}$ does not expand it octet by way of using 3d orbitals.  It is often taught that way, particularly in introductory classes.  A better description involves treating it as a hypercoordinated molecule.  See these earlier answers for an explanation of how hypercoordination can be applied.  When finished you should be able to apply the concept to $\ce{SF6}$.

reference 1
reference 2

$\ce{BF3}$ is electron deficient and reacts rapidly with molecules containing lone pairs of electrons (amines, oxygen containing compounds, etc.) to stabilize the octet configuration.  It has also been observed that the $\ce{B-F}$ bond length is somewhat shorter than expected.  This has been explained by invoking resonance structures such as

and

To whatever degree these types of resonance structures contribute to the true description of $\ce{BF3}$ they help explain the shortened bond length and also achieve an octet around the central boron.
A similar explanation can be applied to $\ce{SnCl2}$, drawing a resonance structure with a $\ce{Sn=Cl}$ double bond creates an octet around the central tin atom.  To whatever degree such resonance structures contribute to the real description of $\ce{SnCl2}$, they will help satisfy the octet electronic configuration around the central tin atom.

ron, so is it not because of the inert pair effect? I have an up-to-a-point understanding of both concepts.

I don't think the inert pair effect plays a role in explaining the non-octet description for $\ce{SnCl2}$.  It may be involved in explaining the molecule's bond angles.

According to Theoretical and experimental studies of the electronic structure
of SnCl2 by photoionization with synchrotron radiation AIP Conf. Proc. 258, 60 (1992)
$\ce{SnCl2}$ has a molecular orbital structure as below (HOMO at the top)
Sigma-antibonding: 36% Sn s, 22% Sn p, 41% Cl s
Nonbonding 99% Cl p
Nonbonding 99% Cl p
Nonbonding 7% Sn p , 92% Cl p
Nonbonding 3% Sn s, 6% Sn p , 90% Cl p
Sigma bonding 14% Sn p, 83% Cl p
Sigma bonding 48% Sn s, 6% Cl s 45% Cl p
Nonbonding 3% Sn p, 96% Cl s
Nonbonding 5% Sn s, 2% Sn p, 92% Cl s
So overall it seems:
There is a total bond order of only 1.
Much of the Sn p and some Sn s electron density has been donated to Cl p orbitals.
There is little mixing of Sn s and p orbitals expect in the antibonding orbital.
additional references:
A Symmetry Rule for Predicting Molecular Structures (This reference explains how, based upon second order Jahn Teller effect, the symmetry of molecular orbitals predicts geometry, for example bent vs linear, and says that the symmetry predicts bent for SnCl2)
and Metal oxidation state effect in photoionization of gas‐phase metal halides (which further discusses the molecular orbitals of SnCl2 discussed above)

There is another rather more simple answer to the question:
$\ce{SnCl2}$ can be thought of as a salt to a much greater extent than $\ce{SnCl4}$. In fact, tin is so often thought of as a metal, that it takes some people time to understand that $\ce{SnCl4}$ is in fact not a salt but much more a molecule.
Then, when looking at $\ce{Sn(II)}$, it becomes clear that the unaccompanied ion would have the electron configuration $\ce{[Kr] 4d^{10} 5s^2}$ — doesn’t that look stable to you? Add two chloride ions that took the electrons originally having been in the $\ce{5p}$-orbitals and voilà.

I know that this answer simplifies a lot of things that maybe shouldn’t be simplified in all cases. But to the best of my knowledge, it explains the compound and its stability well, and should therefore be considered. Once the question turns into ‘why is $\ce{Sn(II)}$ ionic, while $\ce{Sn(IV)}$ better described as molecular?’, it of course becomes invalid.

in a simple way since tin belong to group 14 the stability of lower oxidation state i.e +2,is comparitively more stable than +4 oxidation state due to inert pair effect (reluctance of s orbital electron to take part in bond formation due to poor shielding ability of d orbital which causes the outer s orbital electron to be attracted towards nucleus.)thus tin loses two p electrons only.the outer s orbital electron does not take part in bond formation might.i hope i am right.

Hi and welcome to chemistry.stackexchange.com. Please take care to use correct orthography (that includes capitalisation) and punctuation. Also please note that ‘Google X’ is usually not considered a good answer on this site; ideally you would explain the phenomenon directly and link to sources directly.

sorry got that.

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text:
Why is NH2 negatively charged?
Whenever I look up the Lewis structure for $\ce{NH2^-}$, it shows 8 electrons. Now I understand the octet rule, but I still don't see how an electron can just appear from nowhere. There are 5 valence electrons with Nitrogen, and 2 more, one from each hydrogen. I know that formal charge is calculated by subtracting the nonbonding electrons and number of bonds (or number of electrons in bonds divided by 2), but why is the formal charge of $\ce{NH2^-}$ "-1", and why the extra electron out of nowhere? Because 5 - 3-2 is still just 5

For atoms in the first 2 rows of the periodic table the octet rule arises as a consequence of filling orbitals four orbitals which each hold 2 electrons - each of which appears as a pair of dots in your Lewis structure.
Were you to construct the Lewis structure for neutral $\ce{NH2}$ you would indeed have 5 electrons from the nitrogen, and two from hydrogen, 7 i.e. 3 pairs and 1 extra electron on its own in a orbital.  Electrons in unpaired orbitals are however very reactive, the pairing of electrons lowers the total energy of the system.  As such, it is more favourable for the $\ce{NH2}$ molecule to remove an electron from some other molecule or atom near by and form ions e.g. $\ce{Na}$ to create $\ce{Na+NH2-}$.
Also by creating the charges, you create an electrostatic attraction between the ions which also lowers the energy and makes the ion more stable than the neutral molecule.

+1 for the explaination that $NH_2$ doesn't automatically have a negative charge, but rather it steals a charge from another atom/molecule to become $NH_2^-$.  It just happens so mindbogglingly fast that we gloss over that detail and go straight to the ion.

Good answer, but you might as well touch the notion of a radical, which neutral $\ce{NH2}$ is, by actually drawing its Lewis structure you just mentioned. And you can then show in comparison the Lewis structure of negatively charged $\ce{NH2-}$ which is the result of neutral $\ce{NH2}$ "stealing" an electron from some other molecular entity.

$\ce{NH2-}$ doesn't actually exist on its own, it's only shown that way for parsing a reaction,  it is transferred at the same time the group that is giving it away gains something else, for example sodium amide is Na+NH2- so in order to be separated the NH2- needs somewhere to go as well as the Na+, always wondered why they bother teaching confusing nonexistent intermediates anyway, ions only exist in the non time they are exchanged (or electrocuted).

Um...sodium amide. $\ce{NH2-}$ definitely does exist.

This post also does not answer the question about why $\ce{NH2-}$ (extant or otherwise) is an anion. If you are concerned about the existence of the amide anion, feel free to ask a question about it.

Were you thinking strictly organic? Don't forget about inorganic chemistry and don't forget that even though ammonia is only a very weak acid there exists such reactive elements as the alkali metals.

This is actually quite good point. Every single anion may be "free" only in special situation, otherwise it's bound. Every single "ionic" bond has some covalent character.


text:
Why is electronegativity increasing with oxidation state increase?
I may be a rudimentary fact or something really "not-that-hard-to-figure-out", but

In inorganic chemistry it is common to consider a single value of the electronegativity to be valid for most "normal" situations. While this approach has the advantage of simplicity, it is clear that the electronegativity of an element is not an invariable atomic property and, in particular, increases with the oxidation state of the element. Wikipedia

Really? That means for example $\ce{C}$ in $\ce{CO3^{2-}}$ is more electronegative than a "bare" C atom? Why is that so?
Interestingly, I'm not able to find relevant info in the Net (and the wikipedia citation that should have hopefully got me results is behind a big bad paywall), so am leaving "reference-request" for the case of additional study.

Yes absolutely, electronegativities are hardly static values when you start combining elements and forming ions or molecules.
Electronegativity is simply a measure of an ability of an atom to attract electrons in a covalent framework.
So keeping in mind that like charges repel (electrons repel electrons) ... it follows that elements in a high oxidation state are more electronegative - i.e. more able to attract and stabilize electrons. High oxidation state elements have essentially been "robbed" of their electrons and likely bear some degree of partial positive charge ... this is "good" for attracting and stabilizing nearby electrons.
Conversely consider a lone pair - which is defined to have an electronegativity of 0. Lone pairs cannot stabilize other electrons and don't attract other electrons to themselves ... the electron-electron repulsion would be too great.

Thanks an amount of Avogadro's number, and would you mind adding something (paper, page or whatever) relevant as for a more thorough studying?

https://books.google.com/books?id=g5dYyJMBhCoC&pg=PA12&lpg=PA12&dq=electronegativity+of+sp2+hybridized+carbon&source=bl&ots=vrqWWrRVDH&sig=aozEVO52wDL4t0Ri64iRawCtsI8&hl=en&sa=X&ei=8OzTVMfMGJDBgwTHx4CgAg&ved=0CDkQ6AEwBw#v=onepage&q=electronegativity%20of%20sp2%20hybridized%20carbon&f=false


text:
Mass of "orange" in 25% orange juice
Hypothetical question:
Knowing that orange juice has 25% concentration of orange. And knowing mass of juice which is 200g. What is mass of oranges included in juice?
Assuming that orange is a molecule, am I right that mass in 25% juice would be 25% x 200g = 50gram?
I am thinking about simple formulas, without mols on basic level.
If I know mass of entire liquid and know that 25% of acid is a concentration, so do I have to simply computer percentage, or rather computing for acid and rest differently?

No. Molecular mass doesn't change with concentration alteration. (Your question is sounding a little bit wierd to me)

Are you asking how many grams of a molecule in orange juice would be expected from 25% juice? Perhaps with the assumption that the juice is made up of the same compound?

@Melanie Shebel, it sounded like that was what was being asked at first. That's not what's being asked now.

If the mass of the juice is 200g, then the mass of orange in the juice would be 50 grams.
You just need to rephrase your question in terms of "mass" instead of "molecular mass":

Knowing that orange juice has 25% concentration of orange, and knowing mass of juice is 200g, what is mass of orange included in juice?
Assuming that orange is a molecule, am I right that mass of orange in the juice would be 25% x 200g = 50 gram?

Then it makes sense.

Yes, as Fred says, sometimes it's better to edit the question.


text:
Redox reaction with tin and sulfuric acid solution
I need to write the redox reaction from tin and a sulfuric acid solution. Here is the solution given:

The first half-reaction: $\ce{Sn->Sn^2+\+2e^-}$
  The second half-reaction: $\ce{2H+\+2e^-->H2}$
  Adding both reactions will give you: $\ce{Sn\+2H+->Sn^2+\+H2}$

Why is the sulfuric acid solution replaced with $\ce{H^+}$?

The sulfuric acid dissociates in water: $$\ce{H2SO4 -> H+ + HSO4-}$$$$\ce{HSO4- -> H+ + SO4^{2-}}$$
When writing redox equations the spectator ions are left out because they do not participate in the redox reaction.
In this case it is the hydrogen ions which are oxidising the tin and the sulfate and hydrogensulfate ions play no part in the reaction so they are left out. Including them in the hydrogen half equation would give (with sulfate ions): $$\ce{2H+ + SO4^{2-} +2e- -> H2 + SO4^{2-}}$$
Clearly the sulfate ions are unnecessary and so they are omitted from the equation.

I doubt the answer because I'm sure you'll get $\ce{Sn_{(s)} + 2H2SO4_{(aq)} → SnSO4_{(aq)} + 2H2O_{(l)} + SO2_{(g)}}$ as your Products
anyways , The Arrhenius concept of acids and bases is as follows: an acid is a substance that when dissolved in water increases the concentration of the hydrogen ion,  $$\ce{H+}$$
When Sulphuric acid is dissolved in water it gives you $$\ce{H+}$$and $$\ce{SO4^{2-}}$$ ions.
Sulphuric acid is a VERY Strong Oxidising agent, that means that it will oxidise  your Tin
therefore we get  $\text{Sn}\rightarrow \text{Sn}^{2+}+2\text{ e}^-$
in the second part I think you have reacted your 2 electrons formed in the previous reaction with Hydrogen ion, But I doubt this part


text:
How to calculate bond enthalpies from first principles?
Is it possible to calculate bond enthalpies using the periodic table and the molecule's geometric structure? I am interested in a way of calculating this without getting the answer from measurements or pre-existing bond enthalpies.

Firstly the following basic facts can be helpful :

Ionic bond is stronger than covalent bond in most cases .
A metal non-metal bond is more ionic.
Non-metal non-metal bond is covalent in nature.

By considering Valence Bond Theory , we can draw the fact that overlapping of larger orbitals ( usually of large atoms ) with smaller orbitals is less efficient and will result in a weaker bond .
For example ,
HF>HCl>HBr>HI is the order of their bond strengths .
This is because of the fact that as we go down the group , atomic size increases and so does its p-orbital which takes part in the overlapping with small s-orbital of H , making it a weaker bond .
Electronic configurations of various elements ( which is derived from the atomic number provided in the periodic table ) can be used to make the Lewis structures of most molecules and then the number of covalent bonds can tell comparitive enthalpies. ( more the bonds , more is the enthalpy )
Also , some Periodic tables try to be more informative and provide Pauling EN alongside the atomic number and mass . In that case , bond polarity can be measured by calculating the EN difference . Usually , more the polarity of the covalent bonds, more they act like an ionic bond and are more stable .
Though these tips can be of help in most comparisons , relying on actual values cover anomalous behaviors and exceptions which bump along our ways .


text:
Is there a clear distinction between the various intra- and intermolecular interactions?
Wikipedia lists dipole-dipole, hydrogen bonding and Van der Waals under intermolecular interactions, and hydrophobic, ionic and covalent under intramolecular interactions.

Is there a clear distinction between each and every one of these interactions? For example: isn't hydrogen bond a private case of VdW? aren't ionic bonds and hydrophobic interactions private cases of dipole-dipole interaction?
Is there one single chart that differentiates all of these types of interaction?

No, there aren't clear distinctions, particularly between intramolecular and intermolecular interactions.
First, a caveat that we generally divide molecules along covalent bonds. So those are strictly intramolecular, in my opinion. (Network solids are a bit tricky to classify.)
Let's go through the rest one at a time.

Dipole-dipole: These are usually intermolecular, but certainly in larger, more flexible molecules, intramolecular polar-polar interactions can change the molecular shape and preferred conformation. (Think about tying together two bar magnets on one piece of string. Certainly they'll interact.)
Hydrogen bonds: Yes, many of these are intermolecular. However, there definitely are intramolecular hydrogen bonds:

Van der Waals: Again, these can be intermolecular. In even a moderate molecule, intramolecular non-bonded van der Waals interactions can change conformations and geometries. Consider axial/equitorial preferences in cyclohexanes or other types of "steric clashes" because of big bulky groups. These are the effects of van der Waals interactions.

OK, so all the intermolecular interactions are also intramolecular ones too.

Hydrophobic interactions: I usually think of these as intermolecular but similar considerations make sense as intramolecular for the reasons I outline under van der Waals interactions above.
Ionic: I'll just think of these as different partial charges in a molecule. Clearly electrostatic interactions occur both inside a molecule (e.g., electrostatic non-bonded interactions changing conformations and geometries) and as intermolecular forces, e.g., water-cation interactions.

So besides covalent interactions, which we use to divide "inside" and "outside" a molecule, all of these are both intermolecular and intramolecular interactions.
I'm not sure I've seen a great chart. Maybe someone else can comment.

I've read that it's not vdW repulsion in ethane that causes the molecule to favor a staggered conformation but rather hyperconjugation ...your thoughts?

http://www.nature.com/nature/journal/v411/n6837/full/411565a0.html

Definitely hyperconjugation and other effects can cause conformational preferences. But there are definitely longer range VdW steric effects, eg substitutions around the inter-ring bond in biphenyls.

I came across this chart, which seems to differentiate firstly by "basis of attraction" and secondly by bond energy, and it seems that anything above 600kJ/mol is intramolecular, and anything below 75kJ/mol is intermolecular. What do you think?

While these trends may be largely true, for the reasons I give above, all the interactions can be both intermolecular and intramolecular.

so one final question (rephrasing #1 in the original question above): how do I know to name a given interaction, if for example it can be at the same time ionic, hydrophobic and dipole-dipole?

Well, those three are quite different. Hydrophobic interactions are generally non-charged components. Ionic are fully charged. Dipole are polar (e.g., not charged, but clear partial charges, e.g C-F). So for that, you just look at the molecules themselves.


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Why is FORTRAN so commonly used in computational chemistry?
I've been using Ruby to write scripts for research, but I want to get into some heavier stuff that Ruby is just too slow for. I noticed there are a few things written in C and C++, but there is an oddly large proportion of software used in computational chemistry that is written in FORTRAN (in which I have zero experience.)
Why is FORTRAN used in computational chemistry? From what I understand, FORTRAN is kind of on the ancient side (“punchcard” old.) I was a bit shocked to find fairly recently written tutorials for FORTRAN.
Is it a sort of, "this is how we've always done it," thing or is there an efficiency aspect to FORTRAN I'm overlooking?
Note: I may have FORTRAN confused with later programming languages with similar names.

Because people started writing code a long time ago, and Fortran was (and still is) heavily optimized for formula translation (e.g. Calculations).

Keep in mind that modern Fortran (95, 2003, 2008) is very different from FORTRAN 66 or 77.

Yes, you overlook something. 1) Fortran has slighly different semantics. It does not have true pointers, so compiler can be a bit more liberal in optimizations. 2) there is a lot of legacy fortran code 3) fortran is hell lot simpler, than C and especially 'object-oriented' languages.

For certain array computations, Fortran is potentially faster than C/C++ due to its assumption that pointers never alias one another.

"not quite punchcard"? https://upload.wikimedia.org/wikipedia/commons/5/58/FortranCardPROJ039.agr.jpg

This particular language war (http://en.wikipedia.org/wiki/Software_wars) appeared quite couple of times on Stackoverflow http://stackoverflow.com/questions/13078736/fortran-vs-c-does-fortran-still-hold-any-advantage-in-numerical-analysis-thes?lq=1  http://stackoverflow.com/questions/31672/learning-fortran-in-the-modern-era/168565#168565  http://stackoverflow.com/questions/953869/is-the-return-worth-the-investment-in-learning-fortran?lq=1  http://stackoverflow.com/questions/146159/is-fortran-faster-than-c  Plagued by people who are 30 years back and think it is still written FORTRAN as here.

FORTRAN was one of the few options available to develop calculation specific programs. As to punch cards the column format (which is part of the language specification) was designed with punch cards in mind.

@DavidZhang That's easily fixed with use of the restrict keyword, which was added a whopping 16 years ago.

Let's not forget that many scientific programmers are actually terrible programmers, and it is much easier to write something reasonable in Fortran than in C++.

https://www.quora.com/What-is-the-most-popular-programming-language-used-in-computational-chemistry?share=1

LOL - Wrote my first program in 1971 in FORTRAN using punch cards. Wonder how many folks know how to setup the control card for a keypunch. Another lost sign of being a nerd was knowing how to wind a paper tape for a teletype punch reader.

I don't think that's really true anymore.
Some Fortran use is historical (i.e., early codes were developed in FORTRAN because that was the best programming language for number crunching in the 70s and 80s). Heck, the name stands for "formula translation."
Some Fortran use is because of performance. The language was designed to be:

especially suited to numeric computation and scientific computing.

Many times, I find chemistry coders sticking to Fortran because they know it and have existing highly optimized numeric code-bases.
I think the performance side isn't necessarily true anymore when using modern, highly optimizing C and C++ compilers.
I write a lot of code in C and C++ for performance and glue a lot of things with Python. I know some quantum programs are written exclusively or primarily in C++. Here are a few open source examples:

Psi4 - Written in C++ and Python
MPQC - Written in C++
LibInt - Written in C++ for efficient quantum integrals.
LibXC - Written in C with Fortran "bindings" for DFT exchange-correlation functionals

This is my opinion, but my recommendation for faster performance in chemistry would be Python with some C or C++ mixed in.
I find I'm more efficient coding in Python, partly because of the language, partly because of the many packages, partly since I don't have to compile, and that's all important.
Also, you can run Python scripts and functions in parallel, on the GPU, and even compile them, e.g. with Numba. As I said, if I think performance is crucial, I'll write pieces in C or usually C++ and link to Python as needed.

Based on sheer usage, I highly recommend Python for scientific programming right now. Some pieces for a recent project were already available as efficient NumPy or SciPy functions. :-)

I'm seriously sold on Python. I haven't written much in Python, but I have found a lot of really helpful scripts written in it.
When it comes to C or C++, which do you recommend? I've only ever really used Ruby and Perl (and a tiny bit of Python) and I really didn't care for Perl.

My suggestion for learning languages is to have a project where you need it. I taught myself Python recently simply to use NumPy image processing. I'd suggest using Python for a while, then if there's a performance-critical portion, try rewriting in C or C++. Many "learn language X" guides exist online and in book form.

@GeoffHutchison, I would never recommend something just on its sheer usage.

Most wide-spread QC packages are written in fortran, though.

@Wildcat There are advantages in terms of faster prototyping if you do not need to re-invent the wheel.

@permeakra I believe that's largely due to history. Gaussian was started by Pople in the 70s. GAMESS was similarly started well before C gained much acceptance. More modern projects are largely mixed or C++.

@permeakra For what it's worth, I know that Pople taught himself some C++ when he started Q-Chem. Some of the template meta-programming in projects like libint would be hard or impossible to replicate in Fortran.

@GeoffHutchison C/C++ has some quircks in semantics (specifically, indirect access using pointers and array name = pointer), that restricts reordering of memory reads and writes. For this reasons, fortran code can be optimized a lot better. For this reason, Fortran is the best language for number crunching from all wide-spread languages. In modern times of opencl the host language becomes less important, though, so other options become available.

@permeakra I agree, but I don't want to get into a debate about which language is best. Certainly both of them are used to make highly performant chemistry codes, and OpenCL and other features make language choice less important.

Fortran has, for ages, had native multidimensional arrays with compiler and optimizer support. C / C++, on the other hand, simulates that with vectors of vectors. Lots of people kept using Fortran for chemistry and other scientific purposes for precisely that reason.

Actually Python is an interpreted scripting language, which means your code is translated by a interpreter at runtime. So there is a tradeoff between development time and speed. For mathematical/chemical calculations no programming language out will beat c/c++. If you are looking to run code in parallel you should learn to make your programs multi-threaded

Although I should mention that there is a learning curve for efficient and effective c/c++ but I challenge you to demonstrate otherwise.

@AlanZ2223 If the new programm is intended to run on opencl-enabled software, there is no much difference from what language to utilize opencl. And C++ is one of the worst existing languages ever.

I think it does make sense to provide a somewhat alternative view and to clarify the matter.
FORTRAN vs. Fortran
First off, one has to distinguish the old FORTRAN from the new Fortran, where, by convention, the name of the old language is written usually in all caps. The old FORTRAN (all the way up to FORTRAN 77) is indeed still used because of tons of legacy code, but the new Fortran (starting from Fortran 90) is used mainly because it is a very elegant and simple yet powerful and efficiently implemented language for number crunching.
DSL vs. GPL in general
Note specifically, that even modern Fortran, in my opinion, is domain-specific language (DSL) and herein lies its relative weakness comparing to some general-purpose languages (GPL) like the mentioned C++: Fortran is specialised for a particular task (number crunching) and might be not so suitable for some related tasks (say, automated analysis of the final results, their graphical representation, etc.)
General-purpose languages, such as C++, give you more flexibility (in language features, in 3rd party libraries, etc.) so that you can solve not just the primary task (number crunching) but also the related task using the same language. If, however, you choose Fortran for number crunching, you often have to use one more language (e.g. Python) for these related tasks. Think about it as of using two different DSLs: one for the primary task, another one for related tasks. Of course, you could also use Python together with C++, but an experienced C++ developer (which you are supposed to be if you choose to do some number crunching in it) would not necessarily take advantage of using another language instead of his C++ beast.
Fortran vs. C++ specifically
All this is somewhat subjective, but anyway, here are my 5 cents. Overall, Fortran is simpler than C++, but (and because) C++ is more feature-rich. Basically, it because Fortran is DSL, while C++ is GPL. And as I said, this is subjective to some degree, and besides, the complexity is one of the most complex things in the universe, so we could start a debate about it, but hey, just one word: templates and the discussion is in principle over. Templates make C++ the beast, but everything comes for its price.
Note that I didn't say that C++ is more powerful, since, in my opinion, "more feature-rich" does not always mean "more powerful". It depends first of all on do you actually need these additional features in the first place or not? Seriously, do you need the level of generosity (and metaprogramming) C++ templates provides for number crunching? Not necessarily. And if so, C++ would not be more powerful than Fortran, although, it will still be more feature-rich.
Templates are not the only feature of C++ that Fortran does not have. Exceptions and Standard Library - are two other noticeable features that Fortran does not have. Again, it is not so likely that you will greatly benefit from these features for number crunching. But some other tasks they might be very helpful, so C++ as GPL includes them while Fortran as DSL for number crunching does not. Think about Fortran as of a "suitcase language" just for number crunching and about C++ as of a "trunk language" for everything.
There is also one great feature of the Fortran language which C++ still could not catch up with: modules. I mean, real modules, and not ancient preprocessor machinery which irritates me more than everything else.

C++ has real modules; if you don't like the implementation details, then just use #pragma once.

FORTRAN used to have an edge when it came to speed due to having much better optimizing compilers, in part due to its relative simplicity. Now that C and C++ compilers are nearly on par (and occasionally better), other factors such as programmability come into play. However, there is a lot of legacy code out there as well.
I work in quantum chemistry, and many of the programs that I work with have a mix but are starting to incorporate more C++. The PSI4 package even mixes Python and C++ to gain the benefits of speed and usability of each.
Here is a list of various programs and the language(s) they use.

ORCA - C++
MPQC - C++
PSI4 - C++, Python
PySCF - Python
Q-Chem - FORTRAN, C++
CFour/ACES - FORTRAN, recently some C++
NWChem - FORTRAN, C
GAMESS(US) and GAMESS(UK) - FORTRAN
Gaussian - FORTRAN
Molpro - FORTRAN
Dalton - FORTRAN, some C
MRCC - FORTRAN
MOLCAS/OpenMOLCAS - FORTRAN
DIRAC - FORTRAN
ADF - FORTRAN
CASINO - FORTRAN
COLUMBUS - FORTRAN
CP2K - FORTRAN
TURBOMOL - FORTRAN

Quantum chemistry packages

C/C++ code in Computer Language Benchmarks Game is written in a very "hacky" way. Nobody will do so "in production" since such code is almost unmaintainable. Maintainable C/C++ would not beat the Fortran one.

Sadly that is sometimes true. However, since C and C++ optimizing compilers have been catching up, there are other considerations such as usability.

@Wildcat I disagree; libraries such as Blitz++ using expression templates are neither hacky nor unmaintainable; they are very easy to use and maintain, and regularly beat FORTRAN in linear algebra.

@Alice, first, I did not say any word on complexity of C++ libraries. And secondly, they are indeed very complex. If you have 10 years of C++ development, then you possibly could understand expression templates, but look where you are: it is chemistry site out there.

@Alice: I have never heard of Blitz++. Maybe it's good, but large user base is also important.

@Alice: Can you provide an example where Blitz++ beats FORTRAN?

FORTRAN is a less expressive language than C++, and that allowed older compilers to optimize much more efficiently.  In modern compilers, there is little to no difference in performance.  There are just a few places where FORTRAN's weaker safety guarantees allow more extreme optimizations.  You likely won't notice the difference unless you spend 6-10 years developing FORTRAN.
Generally speaking, it is easier to do development in the newer languages because they better take advantage of modern computing capabilities.  FORTRAN is very strict in its format because it had to be compiled by computers of the mid-1970's.  You tell the computer exactly what to do in its language, rather than telling it what you want to do in a more human readable language.
FORTRAN is still used for two major reasons that I can uncover:

Some problems are simple enough where the expressiveness of C++ is wasted, and you can actually make a clearer program with the simpler FORTRAN.
Many programs are made by the previous generation.  FORTRAN was, in the past, simply better for computational work.

As a matter of pure opinion, as a developer who has developed in perhaps a dozen languages, there is no reason I would recommend FORTRAN for any developer for any task unless there is a legacy FORTRAN codebase to interface with.  People are simply not learning the language anymore, meaning you will have to maintain any code you write forever.  It will be hard to hand off to anybody.

@CortAmmon This is so far from true that it is intellectually dishonest. Modern FORTRAN is in use in CERN on the LHC for gods sakes, and not new code using a legacy base either; many things have been extensively rewritten or written from scratch. FORTRAN is not even close to less expressive than C++, as shown by the myriad of libraries and tasks it has been put to use on. Furthermore, tons of people are still learning it; it's a required programming language in some environments (mostly scientific).

This got an upvote, which drew my attention back to it.  I felt it would be useful to post a link to the StackExchange programmer survey from 2019.  To Alice's credit, there may be small close-knit populations which develop FORTRAN, and in those populations, it can be a maintainable language.  But in the larger developer population, it's so unpopular that it didn't even make the list of languages people develop with (the lowest on the list was WebAssembly, at 1.1%)

I stumbled on an article in J. Appl. Cryst. that partially is related to the original question.  Comparing Fortran to other programming languages in the context of crystallographic algorithms in a reusable software framework yields this condensed diagram:

(doi 10.1107/S0021889801017824)
It should be added that the article relates to software packages written in the old Fortran 77 dialect.

While languages like Python are not especially fast (it's basically interpreted not compiled) this is misleading. Python makes it easy to access highly optimised compiled libraries for specific computational tasks. So you can combine the ease of programming and still achieve high speed number crunching. That is one reason for its growing popularity.

Is this saying that C has better runtime performance and programming efficiency?

I believe there are two main reasons for the wide use of Fortran in the computational chemistry community. The first was because of its efficiency: FORTRAN means Formula Translator and the scientific/numeric calculation was the main reason for developing this code.
Nowadays, Fortran is still very efficient, however, some other languages (C++ for example) are as efficient as Fortran. But so why do we still code in Fortran?? Basically, the majority of Electronic structure software is written in Fortran and no one is so crazy to translate millions of lines. Therefore we keep it writing in Fortran when adding blocks of code!
On the other hand, if a new software had to be written, there would be no practical reasons to choose the Fortran language. For example, the recent software MIDAS (Aarhus University) for obtaining nuclear wavefunctions has been completely written in C++. Personally, I use Fortran90 as language and it is not as rigid as the old FORTRAN77... and I like it! :)
I actually asked myself the same question you posted when I started my Ph.D.

authors: Vladimir F Героям слава

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text:
Reason behind Opaque translucent transparent
I would like to know the reason that some materials allow light to pass (transparent), blocks it a little (translucent), or completely reflects or absorbs it (opaque.)
I'm don't have a PhD, so answer this in a way that is easy to understand.

A physics questions- its no doubt a duplicate of a question in that StackExchange.

Neither I am an active user of physics nor I have copied from it. Thanks for helping me @user2617804

Metals are opaque by virtue of a 'sea' of electrons, which reflect the light.

Opacity or transparence is a matter of light reflection.

The way light is reflected or absorbed depends on severals parameters, first the composition of your material. If it contains molecules or crystals absorbing the visible light, you'll see it black, or colored with the color corresponding to the wavelenght the less absorbed.
There is also crystallographic considerations to take care of. If your material structure is well ordered, dimension of your latice will be set so to reflect the light, then you'll see it transluscent or opaque. For something without ordered structure, you'll probably see it transparent if the visible light isn't absorbed. We'll speak about "amorphous" material then like glass for instance.
edit : if you feel easy with those concepts, I can go further in the explainations, but since you asked for a really vulgarized answer I haven't developed more.


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How to know when electrons will be added onto Lewis Structure from surrounding atoms to become ion?
How do I know when I should add an electron to a single electron? Do I just always do this? I've came across this in the textbook in the form of the amide ion $\ce{NH2}$, among others. It was shown to have 8 electrons, despite nitrogen only having 5 valence electrons, and each hydrogen only having 1. The last came from some surrounding atom to complete the outer shell and thus become more stable. So, do I always do that when there is a single electron not part of a pair?

@Tlwolf I think you're not really asking about lewis structures, but about anions, radicals and cations.

Tlwolf your title gave me the impression that you are mixing up ionic and covalent bonds, though I later observed you don't.

I was under the impression ionic bonds are between a metal and nonmetal, such as sodium chloride, right? Due to the large difference in electro negativity? But I do believe @Mithoron may be right in that I am talking about cations and anions, I am just unsure about when I show the structure as having a charge by adding an electron (or maybe taking one away, in the case of cations?) I don't know if I just always do this for a compound that has a left over electron not in a pair, or in certain instances

It all depends what you want to have written - without this 8-th electron you'd have a radical - in this case it would be unstable (cation even more) but sometimes they aren't - chemistry is complicated.

Lol yes it is, that's why I'm trying to get a head start before my freshman year even starts, and this site seems full of chemistry enthusiasts who are knowledgable

http://chemistry.stackexchange.com/questions/24673/why-is-nh2-negatively-charged

I think I know where your confusion comes from. Firstly, I would want to note an important point:

Covalent bonds consist of pairs of electrons shared by two atoms, and bind the atoms in a fixed orientation. Chemwiki.UCDavis

This is the characteristic that makes them different from ionic bonds. The surrounding atom will not become an ion in its full sense.
I would want to describe a way of drawing Lewis electron dot structures that will avoid any confusions.
$\ce{H2O}$:
We have two Hydrogen atoms and an Oxygen atom to begin with. They have this lewis structure:

Source 1
Follow these instructions for "normal" molecules (the ones that abide octet in a perfect sense):

Compute all of the electrons that are in the valence shells of separate atoms. For instance, an O atom has 6 es in the valence shell and each of H atoms have one. So you get: $$6+(2*1)=8$$
I prefer to call this $\text{n}_v$. V represents valence shell.
Compute all of the electrons the atoms will have after the compound has been formed. That is, when all of them abide octet. In the instance of water, O will have eight (just like other octet abiding atoms) and H atoms, as they are exceptional, will have two. So you get:
$$8+2*2=12$$
I prefer to call this $\text{n}_f$. F represents "final". You can name it whatever you want.
If you want to know how many electrons are bonding and are shared between atoms, subtract $\text{n}_v$ from $\text{n}_f$:
$$\text{n}_f - \text{n}_v \rightarrow 12-8=4$$
For an ease in calculation, I call this $\text{n}_b$. As you might have just figured out,
If you want to know how many es are nonbonding and "belong" to a single atom, subtract $\text{n}_b$ from $\text{n}_v$ to get:
$$\text{n}_v - \text{n}_b \rightarrow 8-4=4$$
Call this whatever you want. I rather calling it $\text{n}_{nb}$.

With the information you just gathered you can draw the lewis structure for most of the covalent compounds. In the case of water, 4 bonding es mean two covalent bonds, right? Plus, H has no other electrons than it's only one. So O has 4 nonbonding electrons. Thus, you may come to something like this:

Source 2
$\ce{NH4+}$
In ions, we have to take the charge of the ion into an account. Positive charge means lesser electrons than a neutral species, and negative charge means more es in a comparison with a neutral covalent compound. Thus, we should add to or subtract from $\text{n}_v$.
So let's get on with this thing's lewis structure:

$\text{n}_v = 5+(4*1) -1=8$ (Notice (-1) that was because of a 1+ ion charge.)
$\text{n}_f = 8 + (4*2)=16$
$\text{n}_b = 16-8=8$
$\text{n}_{nb} = 8-8=0$

There is an important thing to note here. Nitrogen has 5 es in valence shell, but ends up having four bonds. This is because an H atom had lost it's electron to become n ion AND, with an empty orbital. So, N shares two electrons instead of one which is described as dative covalent bonding. We get this structure in accordance to the calculations and thoughts:

Source 4
Finally, $\ce{NH2-}$
Now this case is easily solvable through the explanations for ammonium.

$\text{n}_v = 5 + 2 +1=8$ The +1 is because of the charge of the ion.
$\text{n}_f = 8 +2*2=12$
$\text{n}_b = 12-8=4$
$\text{n}_{nb} = 8-4=4$

Put some thinking, and you should get this (ignore the other ones if you like):

Source 5

Those formulas are helpful, thank you. I'm just still unsure if there's some basic rule to know when a single electron needs to have another added to it (like NH2 -1) It's just I thought I had the concepts down pretty well, until I ran into this exception where an extra electron was present that I couldn't find a way to account for

I'm attempting to expand my answer a little bit. I see it incomplete. And, I will include what you mean.

Ok, I think that pretty much clears it up for me, thanks. So it's just a matter of following the octet rule. So then I'll assume I pretty much always take this into account except for some select exceptions to the octet rule then?

This is a really nice rule for most of the compounds, but it may result in confusion in the case of resonance structures and some of the exceptions to the octet and such. If you see yourself compromised in those cases, it's better to make a though-based approach, rather than calculations. The only way to deal with them is practicing dealing with them!

Alright, I'll keep practicing in that case aha. But thanks for your help


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Solubility of gold salt solution?
I was wondering what salts of gold are soluble and which are insoluble. From what I understand, silver is soluble as silver nitrate but insoluble as most other common salts like chlorides and sulphates. How does this work with gold? I was also wondering if there are any rules for solubility of salts, or whether there is some explanation for the solubility that would help in guessing the solubility of a salt?
I started thinking about this because of a question I came across in which there was a mixture of gold, silver and copper to which hot concentrated nitric acid was added, and an undissolved residue of given mass remained. Then $\ce{HCl}$ was added and more precipitate was formed. The question asked you to find the percentage by mass of gold, silver and copper. Judging by the general solubility trend of these metals, I'm guessing the first precipitate was gold and the second silver, but I was also wondering if there are some solubility rules...

As a side point the second precipitate would have been $\ce{AgCl}$ not $\ce{Ag}$

Just a note that silver chloride is more soluble in solutions with chloride and/or ammonium ions, such as salt water or ammonium chloride. See http://www.saltlakemetals.com/Solubility_Of_Silver_Chloride.htm.

I know this is old but check out http://www.ausetute.com.au/solrules.html as well as https://www.google.com/search?q=solubility+of+salts+table&tbm=isch. There's lots of tables and handy rules, and the answer to this question is definitely there.

Old question, but I thought I'd add to it:
In your example, the first undissolved solid is indeed gold. It's not easily dissolved by $\ce{HNO3}$ alone, but requires a mixture of $\ce{HNO3}$ and $\ce{HCl}$ (aqua regia). $\ce{Ag}$ and $\ce{Cu}$ would dissolve by hot $\ce{HNO3}$. The precipitate formed by addition of $\ce{HCl}$ is indeed $\ce{AgCl}$, so you're on the right track.
A brief list from experience:
[At room temperature in aqueous solution]
$\bullet \ce{Au2S}$: insoluble (quite sure this is the precipitate formed in basic solution when sodium metabisulfite is added to $\ce{HAuCl4}$. Either that or $\ce{Au2S3}$).
$\bullet \ce{AuCl3}$: highly soluble (can be formed by dehydrating and evaporating $\ce{HCl}$ from $\ce{HAuCl4}$, which is the product of $\ce{Au}$ dissolved in aqua regia).
$\bullet \ce{KAu(CN)4}$: soluble
$\bullet \ce{KAu(CN)2}$: soluble
$\bullet \ce{AgCl}$: insoluble
$\bullet \ce{CuCl2}$: highly soluble
$\bullet \ce{CuCl}$: soluble, but harder to synthesize than $\ce{CuCl2}$
$\bullet \ce{(Au/Ag/Cu)(NO3)_x}$: soluble ($x = 1,2$)
I think the overall result here is that gold, silver, and copper salts generally follow one of the primary trends of solubility: that is, that $\ce{NO3-}$ salts are always soluble regardless of cation, and $\ce{Cl-}$ are usually soluble except in cases $\ce{Ag+}$, $\ce{Pb^{2+}}$ and $\ce{Hg+}$.

Here is What I know:
Gold(III) chloride is Highly Hygroscopic [1] and is highly soluble in water,
I don't know which salts are not soluble never researched it,
maybe Gold-Mercury amalgam is not soluble in water but its not a salt..
[1] = absorbing or attracting moisture from the air.

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text:
How does the production of pure, crystalline silicon happen via the Czochralsky process?
I have a few questions regarding the production of pure, crystalline silicon:
As far as I know Silicon usually appears in nature in the form of sand or $\ce{SiO2}$ and then is split apart and brought into the form $ \ce{Si + 2CO} $ ;
I am imagining now a hot glowing molten mass.
However then I don't understand the following methods to separate the Si from the elements you don't want to have, like iron, carbon etc.
How do you remove the impurities, why do those methods work?
For example you use a few other chemical reaction to bring Si into several other forms like $ \ce{HSiCl_3} $ and split them apart later on.
I don't understand at the moment, how this helps to separate those materials as in: There is still this big mass with lots of different chemical connection you don't want to have.
I also lack an actual understanding of the Czochralski process to further purify your mass of silicon. I guess want has to accept that silicon likes to join pure crystalline silicon and adapt to its structure. However I don't see, why one is not using this method beforehand for example, if one already has raw silicon you get from splitting up $ \ce{SiO_2} $ if it works so greatly. And if it not does work so greatly, why don't you have other impurities ?

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The Czochralski process, unlike the float zone growth method, does not purify the silicon.  In addition, carbon is intentionally added the the melt to improve the mechanical properties of the final silicon wafers.  Oxygen gets in as well (not desired).  And then, of course, Boron is added as a dopant.

How does trichlorosilane $\ce{HSiCl3}$ help in the purification of silicon?
Trichlorosilane is a highly volatile compound with a boiling point of 32 °C. This means that it can be destilled off the raw material and separated from impurities.
The purified trichlorosilane can either be decomposed thermally
$$\ce{4 HSiCl3 ->[\Delta] Si + 3 SiCl4 + 2 H2}$$ or reduced with hydrogen
$$\ce{HSiCl3 + H2-> Si + 3 HCl}$$
This does give you rather pure silicon, but it is, at best, polycrystalline.
What's the Czochralski process good for?
Note that this in't just a melting/cooling cycle!
The crucial point in the method is the use of a rotating, monocrystalline seeding crystal, dipped into (previously purified) molten silicon. The goal is to obtain huge, monocrystalline silicon rods of defined diameter, that can be used in the production of wafers.


text:
Is it possible to identify the source of a protein hydrolysate
I have a powder which only declares to be "Protein Hydrolysate"
Meaning that the all the protein has been broken down.
We've tried to identify the source of the powder with DNA research. But no usable DNA (parts) where found.
Is there a way to identify the source of this powder? (ie. chicken, pork, cow, shark, dolphins or unicorns)
For clarification, we failed to find any DNA in the powder. so we're looking for alternative ways to analyse it.

Unless I misunderstood your question: Protein != DNA

DNA research? Unless there's a total misconception in this question it's very ambiguous.

I've added a clarification in the question. DNA is protein, but since we're talking about a hydrolysate all protein and therefor DNA is broken down to it's amino acids (or small pieces of protein). Meaning that DNA analysis gives zero result. So we're looking for alternative methods to analyse the powder

Proteins are made of 22 common types of natural aminoacids. The combination of those results in a sea of different proteins. Furthermore, you can have a protein with for example 300 cysteines, 250 glycines and 50 tryptophans, while another one contains 300 glycines, 250 tryptophans and 50 cysteines. They will vary very very very deeply.

DNA is not a protein, as @MARamezani states. See http://www.answers.com/Q/Is_DNA_a_protein.
There is a prevalence (or lack) of some amino acids in certain sources. For example, maize ("corn" in America) is deficient in tryptophan and lysine, so if your sample is short on those two, it is more likely that it is from corn. A list of some foods and amino acid content is at http://www.ljhs.sandi.net/faculty/aquesnell/Biology%20Notes/Chapter%204/4.8-4.13/Essential%20Amino%20Acids.pdf [there appear to be some errors, though]. According to that source, yeast may be deficient in phenylalanine, so that is another indicator.

You would need an analysis of the relative abundance of each amino acid, some reference showing their abundance in various sources, and additional analysis to calculate relative abundances in a mixture, e.g. corn supplemented with yeast hydrolyzate.


text:
How do I excel in Chemistry?
I wanted to ask an online group of experts/community about how I could improve in chemistry. I study in 10th grade currently and don't get very good grades. Most of the time its a B or C.
I have to select science stream, in which physics, chemistry and maths are compulsory. I am good in physics and maths, but what should I do in chemistry?
I have a hard time grasping concepts like Ionisation Energy, structure of benzene etc. and these are only going to get 100 times tougher next year!

Chemistry needs a devoting reader. You should try to love chemistry as much as you want! That'll be the ignition for lots of reading and stuff. My pointers:

Love chemistry.
Learn in an organized matter. (For example, do not jump from studying about covalent bonds' basics to reaction mechanism.
Take notes of the most important points of what you learn.
Look for open-source online tutorials and not only the books.
Some chemistry concepts are related to each other. Learn the first one's basics and then read the other one's. This will build up your understanding, literally. For example, first learn about electronegativity and then study the covalent bonds' basics. This is the exception for rule 2.
Try to differentiate between what's taught in schools to be true and what really is. For example, look for exceptions of the periodic trends that are taught in schools.
Be imaginatively practical. Don't just say enough for the lewis structure of water, but do it about $\ce{SF6}$.
Always try to go relatively higher than what your textbook has written. We have a proverb in my mother language meaning "If you have 100, you do have 90."
Ask what misunderstandings you have in here!
Never look to doing a research as a waste of time.
When approaching to solve a chemical problem, do it step by step.
When approaching to solve a problem, don't forget to write units of measurement.
When you reached an answer to the problem, ask yourself: "Is this possible?" If you've calculated 5 grams of water to be $\frac{18}{5}$ mol, and you know that a mole of water is 18 grams, shouldn't something be wrong here?
For any definitions, uses, notions, and practical experiments you encounter, ask yourself: "What is this for, either in real life, experimental sciences or theoretical chemistry?"
Look for online exercises, free of charge, but usually important.
Follow the latest news that are being discussed everywhere in the world of chemistry.
For specific concepts in chemistry, e.g.: Periodic table, doing a research about the history is very helpful.

This is as far as I can count. You can ask "why"s in comments.

Get a chemistry set and do experiments and have fun.
When I was a kid, much younger than you, I asked my mom for 75 cents to buy a chemistry set at a yard sale in our neighborhood, but she wouldn't give it to me.
But a friend that was a couple years older bought it and we would use it at his house. Eventually I got my own chemistry set and would constantly be looking for ways to expand it.
So by the time I got to high school chemistry I had intuition about chemistry.
Once my high school teacher said, I think quoting some else, that students today ('today' being a long time ago now) know all about atomic orbitals but don't know what color permanganate is.
So given that you are good at math and physics, you could only be missing an intuitive aspect of chemistry that you get from practical experiments.

Read a lot.
Do exercises.
Read stackexchangechemistry :P
Try this kind of website also maybe : Massive Online Open Courses
Alos you can find plenty of nice second hand chemistry book on this website : used books

Thank you! I do read , but not that much .do abebooks ship in INDIA?

Of course they ship everywhere. Some sellers are also located in India as well.

I would actually start with a general rule of thumb for any topic: Don't strive to learn the content, strive to want to learn the content.  The learning will often happen on its own.
Google and Wikipedia can give you all the Chemistry facts you could ever need.  In fact, more than you could ever need.  If all you do is learn facts, you'll eventually be outstripped by those who played on the playground, and then asked Google.
The true learning comes from being able to integrate those facts with real life, and that is a skill you develop for yourself.  There's no textbook to teach it.  Once you find uses for the chemistry facts you learn, you'll suddenly find they're a WHOLE lot easier to learn.  Those who played on the playground will suddenly be awed by how easily you digest information and how quickly you turn it into application.
The path is different for every person.  However, I want to give a shout out to Dave PhD's answer of "Get a chemistry set and have fun with it."  If there is one "best path" to learn Chemistry, that's it.  Chemistry is so much more interesting when you come across a problem in life such as "how do I get gum off my shoe," and find that not only do you know enough about the chemical properties of the latex in the gum to understand why it is so {bleepity-bleep} sticky, but you happen to have the chemicals on hand to break it down because you were using them in a cute color-changing experiment the other night!
In an ideal world, one day you'd realize you need a chemical to solve some cool problem, but you don't have it.  In a flash of inspiration, you realize that you can do a chemical reaction with stuff you do have to make the chemical you don't.  All you need is some starch to complete the reaction.  Mom doesn't have starch on hand?  Saltine crackers will do nicely, I just have to wash the salt off of it so that the salt doesn't interact.  Once everything connects in your mind, you'll find memorizing ionization energies is a breeze because you know that, one day, you'll use that information!

Start from bit to bit. When you don't understand what it means, stop for a while and then read the thing again

I can only speak for mathematics, though I guess, it is similar: In fact I can only stress, what Babounet said and say, that you have to really work hard... really hard until your hands hurt from writing and read until you fall asleep.
You should also work on your mindset. It is not a bad thing to not know something, however one really has to want to understand something and if necessary take every help he/she can get. Thus I think Stackexchange is a good place for you.

I guess you haven't started with organic....after that you will love it. Try to develop a logic behind everything you read. Think about the concepts. Make the reaction that you were cramming "obvious" for yourself. It might become your strong subject, perhaps more stronger than physics and maths!

Speaking as a software engineer who took chemistry courses, I can tell you that I found organic chemistry more challenging than a lot of my other courses.  In retrospect, I think this had to do with my style of learning/retention, which is primarily auditory.  Organic chemistry involves a lot of spatial reasoning and structural diagrams, which are often much more difficult to describe in words than to draw.  It's worth learning these skills, of course.  Just something to consider when considering why a subject seems difficult or easy.

Well, I'm in class 12th and in my previous year I just sat in front of my book waiting for a miracle to happen. But when organic started, I started liking chemistry and also started liking inorganic a bit. Organic actually clarifies the behavior of elements with carbon and that is really very helpful even in inorganic, since now we have a rough idea as to how compounds react with each other. I even had a hard time at first, but now I don't hesitate to prove my teacher wrong :)

I am really dismayed that none of the above answers even so much as mention laboratory work. Get thee into a laboratory and get as much hands on experience with chemistry as you can. It will give you reasons and motivations to learn the theory. Also, you may find that you have a special talent--a "green thumb" for making things work in the lab, even if book learning is not your strong suit. And--it is fun!

True, but working in lab makes you wanting to learn the theory, and to learn the theory, you have to follow some of my advices. And I doubt he can actually work in a lab since he would asked directly his colleague or supervisors ...

@Babounet OP says she is in 10th grade, so about 15 years old

So I was right assuming she can't actually perform lab work.

15 used to be the age when you took high school chemistry, and once upon a time that meant doing hands in experiments. Why are you saying she can't do lab work?

You can watch videos related to chemistry as a starter. Later you will developed an interest in it naturally.


text:
How to rationalise the protonation state of amino acids
I am learning about amino acids in various $\ce{pH}$ environments and how their groups are protonated or deprotonated in these environments.
I don't need to mention a specific acid for my question, I will just use the generic form of one:
$$\ce{H2N-CH(R)-COOH}$$
The $\ce{H2N}$ group has a $\mathrm{p}K_a$ of $8$, I believe, and the acid has a $\mathrm{p}K_a$ of $3.1$.
If I drop these into a solution with $\ce{pH} = 12$, the $\ce{COOH}$ will donate it's hydrogen ion, but the $\ce{H2N}$ will not. It will only add another hydrogen ion if it is basic compared to the solution to become $\ce{H3N+}$. Why is it, that it won't donate its hydrogen ion like $\ce{COOH}$, when it is acidic compared to the solution, to become $\ce{HN-}$?

You can visit this page to find out how to make your future posts better.

Oh, nothing wrong. I just wanted to note that \ce{} is used to write chemical formulas and equations. You would be free of lots of formatting with it as it automatically renders numbers and ion charges in favorable behavior. (No, not really :)

@BjarniJóhannsson Meta contains much more information than the [help]. Please have a look here and here. Even if it is tempting, please do not use markup in the title field, see here for details.

Let's assume you drop the amino acid in water at $\ce{pH} = 7$.
You'll have $$\ce{ H3N^+{}-CH(R)-COO^-}.$$
Then you can go to $\ce{pH}= 12$ and the amino acid loose the proton.
You don't get rid of the second proton because the $\mathrm{p}K_a$ is defined as $$\mathrm{p}k_{a}=-\lg\frac{[\ce{NH2}]\cdot[\ce{H+}]}{[\ce{NH3+}]}.$$
This means $\ce{NH3+ -> NH2}$.
You could define another $\ce{p}K_a$ for $\ce{NH2 -> NH^{-}}$.
The $\mathrm{p}K_a$ for this would be much higher than 12 - at least around 35.
You may look at this site: How to use a $\mathrm{p}K_a$ table
You can compare the amine $\ce{NH3}$ (without the rest of the amino acid) with your $\ce{NH2}$ group of the amino acid.


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Memorizing the periodic table
Several people have said that the key to understanding chemistry is through memorizing the periodic table.
I want to ask if there is a simple technique to learn it, or if I just have to remember every element as it is?

What would a high school student need to do with Lutetium?!

Just as an additional information

Anyone who says that the key to chemistry is memorizing the periodic table is either messing with you very confused. I've never written a chem exam where the periodic table was not given, even if it was not needed (read biochem). Having said that, if you want to take the time to learn the periodic table by all means go ahead, but it's really not necessary to set time aside for it. You'll learn it over time anyways.

IMO, they are wrong, and all those pictures of the table are for sth, aren't they?

I teach high school chemistry, and a number of my students have this idea for some reason; I'm not sure where it comes from, honestly.  Learning the name and symbol of common elements (through krypton, let's say, and a few of the better known ones like silver, gold, mercury, lead, and such beyond that) is useful, but it isn't necessary for understanding chemistry.  Understanding periodicity and the reason why the periodic table is the way it is is useful.  Memorizing The Elements is fun...

That is a terrible advice. Learn how to read it, that is all you need.

(H)ai Li_Na (K)i Rb (Cs)e (Fr)iyaad... I hope you get this mnemonic by Indian chemistry teachers ^_^

You got it all backwards. In fact, it is the other way around: understanding chemistry is the key to memorizing the periodic table.

I was expected to know it at least to plutonium for my University final examination (Oxford, a few decades back). Once you've got to that level it's pretty easy to work it out from knowing the chemistry of the elements and from years of experience except for the Lanthanides which, Ce, Eu, and Yb excepted I had to memorize. At school I can see no point in this exercise.

@IanBush You may or may not be pleased to hear that a few decades later, Oxford is still pretty much the only place that doesn't give people the Periodic Table for exams. :) I can relate to your comment very well. Of the lanthanides, I only remember La, Ce, Eu, Yb; I devised mnemonics for the rest of the f-block (including actinides).

@orthocresol Thanks for the update, Oxford changes as rapidly as usual! But the early actinides are fairly straightforward, just remember the maximum valence and that takes you to Np, which was near as damn it good enough for my porpoises.

I accidentally learned it.  I thought that I should learn the s and p blocks as they contain many important elements.  Then I thought that I should also learn the first two rows of the d block.  Then I got carried away.  I learned the rest of the d block, the f block up to plutonium, and finally even the transuranic elements.  I am trying to resist learning the recent additions.

@IanBush Personal experience: chemistry school in France with the periodic table only in the teaching lab mineral / analytical chemistry, but none in the lecture halls, nor at hand or sight during the exams.  "You had grammar school and your two years in classes préparatoires, so you are expected to know this detail by heart, at least up to U."

You can memorize the periodic table in one night, simply by emulating best-practice memorization techniques and doing what memory experts do. Common sense, right?
Memory experts and world champion memory ‘athletes’ activate the enormous natural power of their visual memory by using visualization and association mnemonic techniques.
That’s a fancy way of saying they create mental pictures and link them together in their mind. It’s incredibly simple but amazingly fast and effective.
Watch YouTube’s #1 “How to Memorize” video and you’ll probably amaze yourself with how easily you can remember and recall 15 random words in order, using one of these techniques.
The foundation technique most memory experts use is the Method of Loci (or Memory Palace or Journey Method). Think of a particular journey you take every day, and picture certain locations along the way.
For example, imagine leaving home in the morning and travelling to work or school. You might walk out your front door, through the front gate, and get on a bus.
At each location you visualize an object that represents what you’re trying to remember. Because the chemical elements themselves can be difficult to visualize, you substitute them with an object that you will naturally associate or link to the element itself.
For example, ‘hydrogen’ sounds similar to ‘hydrant’, so when you visualize a hydrant sitting at your front door, you’ll be prompted to remember ‘hydrogen’. When you picture a large helium balloon tied to your front gate, you’ll remember helium. And when your bus begins talking with a ‘lithp’ (how people with a lisp pronounce ‘lisp’), you’ll be prompted to recall lithium.
These established memory techniques have been proven by over 50 years of academic research in fields like cognitive psychology. Google ‘memory palace’ or ‘world memory champion’ and you’ll discover they’re the fastest and most effective methods to memorize a deck of playing cards and a lot of other geeky things.
The method used in the video above is called the Link and Story Method, and is based on the same principles of visualization and association. The weakness of this method compared to the Memory Palace is the amount of time it takes to create the (intentionally) bizarre and crazy story to link all the words (or chemical elements) together.
But all the work has already been done for you at How to Memorize the Periodic Table. This animated video course is the fastest way to memorize the periodic table because it uses best-practice visual memory techniques.
All the mental images and association links described above have already been created, and transformed into engaging whiteboard animation videos. You just need to sit back and watch, and let the amazing natural power of your visual memory do its thing.
What about other techniques?
Most other methods people suggest to memorize the periodic table rely on verbal memory, but don’t activate the enormous power of your visual memory.
Flashcards or equivalent apps are convenient but don’t provide an association or link between chemical element names, meaning they rely on rote memorization. Repetition by itself is not meaningful, takes an unnecessarily long time and effective retention is low.
Acronyms and acrostics are ‘first letter mnemonics’. You could use the acronym HHeLiBeBCNOF (pronounced ‘heeliebeb kernoff’) to remember the first nine elements. It's a nonsense word, but it condenses nine names into one mental prompt or cue.
Or the acrostic “Here He Lies Beneath Bed Clothes, Nothing On, Feeling Nervous” would equate to H He Li Be B C N O F Ne.
Acronyms chunk words together, which is good (even if they are nonsense) and acrostics use more meaning, but either way you'll only have the first letter or two to remind you of each element's full name. That's tough for 118 elements!
The first letter cues don’t prompt you enough to recall the complete element name, so acronyms and acrostics can be great for the first 20 elements, but not for all 118.
Songs are also popular, whether you’re a fan of Tom Lehrer or ASAP Science. A catchy tune gives better association and meaning than acronyms and acrostics, but you still have to rely on bucketloads of repetition. They’re a great way to make repetition fun, but songs only tap into your verbal memory, not your powerful visual memory.
Bottom line, you should play to your natural strengths. Your brain loves pictures. And that makes visual memory techniques 10X more powerful than verbal memory techniques (like songs and acronyms) to memorize the periodic table.

There isn't really much sense in memorizing the periodic table. The elements you often use you will know them by heart after a while. And you can always use a table when you need it for the others.
That being said if you really want to do it, mnemonics are probably the best solution to memorizing the whole table. You can find some here :
https://www.mnemonic-device.com/chemistry/periodic-table/mnemonic-device-for-the-periodic-table/
or here :
http://en.wikipedia.org/wiki/List_of_chemistry_mnemonics#Periodic_table

Tom Lehrer had a pretty catchy song...

@JonCuster Yes, but they aren't in sequence.

Trying to learn chemistry by memorizing the periodic table is pretty superficial.  I would recommend learning chemistry by doing chemical reactions.  Make plastics, mix acids and bases (correctly please), ask your self why milk curdles, try to light water on fire and then ask yourself why it doesn't work even though it is made of hydrogen and oxygen.
Look at calcite and notice that it splits text on a page into two.  Why is ruby red?  Why is the sky blue?  Can you make blue smoke the same way?  Everything that grabs your attention is valid, and pay enough attention to safety so that you don't blow up/poison yourself before you know enough chemistry to do things safely naturally.
Curiosity is the only way to learn chemistry, so enjoy it.
One day you will wake up and realize you know most of the periodic table.  Then you will probably memorize the rest, or the useful pieces because you can't resist it, and that is totally fine.  As long as it doesn't take you away from melting plastic with acetone (under a fume hood.)

There are a couple of elements you'll want to really get to know, but other than that there's no real reason to memorize the periodic table. To me, memorizing the periodic table sounds like a quick way to hate chemistry.
I recommend getting to know a few things about

Hydrogen
Carbon
Nitrogen
Oxygen
Fluorine
Sulfur
Chlorine
Sodium
Potassium
Bromine

Really the best thing to learn about those is how many bonds they make and what they tend to bond with. These are the elements that I most often see on exams and in examples since they're used in a lot of textbook reactions.
For the other elements, just get a feel for where they are on the periodic table so you can quickly look them up. This is especially helpful during exams. As you're taking chemistry courses, you'll notice you'll REALLY learn how to find your way around the periodic table and start memorizing the molar masses of various elements.
I highly recommend just looking up each element (this will take a lot of time) on Wikipedia or checking out the book, "The Elements." That way, you can kind of get to know elements and if you're really wanting to nerd out, you might even find that you have favorites.

Go to this website
and click on "Sitemap" in the almost left upper corner, then select "Spiele" (which means "Games") and then "Tetris". Have fun playing Tetris with the periodic table!
The key to understand chemistry is not by memorising the periodic table. It is more like the language of chemists, but not its understanding. The key to understand chemistry is logic and the realisation that chemistry is not black and white but an interplay of many factors.

I think those people meant the important ones. Try to remember the first 20 elements and some useful that you often study in school like iron, silver, bromine, etc...
But since you really want to remember the whole periodic table of elements, try searching for the "asapscience new periodic" song on youtube. I found it useful for remembering the elements in order.

You could print out a blank periodic tables and fill them in one column (group) at a time. You will be surprised at how quickly you can memorize the whole periodic table. By initially filling the periodic table in by group you will learn to associate the elements of each group together. The "periodic" part of the periodic table is the fact that the elements of a given group tend to have similar chemical properties. So the reason it might be helpful to memorize the periodic table is that it helps you quickly associate any element with its group and therefor its chemical behavior.

I read all the answers by experts, but I'm also a student and have the same interest as you.
I would suggest going on YouTube and using it to discover your own rhyme or trick. Be patient, look for patterns in their names, and find a meaningful rhyme that will be easy for you to remember.
Doing this on your own has the benefit of you remembering it. You did all the work in creating your mnemonic. The work you put forth in doing this makes it a much better method than learning a trick from others.


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Aqua regia and nascent chlorine
Aqua regia has the famous property of being able to dissolve noble metals, like gold and silver. I read that this has to do with the generation of nascent chlorine. How does this happen and how does it help to dissolve such metals?

Aqua regia is a concentrated mixture of nitric and hydrochloric acids. The two acids react to generate soluble nitrosyl chloride, water and nascent chlorine: $$\ce{HNO3 + 3HCl -> NOCl + 2H2O + 2Cl}$$
Secondly, the gold is oxidised by nitric acid to gold(III) ions: $$\ce{Au <=> Au^{3+} + 3e-}$$
This reaction is normally highly unfavourable and so very few gold(III) ions are formed if you put gold into only concentrated nitric acid. However, the nascent chlorine is highly reactive as it has an unpaired electron in a high energy orbital and it is able to react with the gold(III) ions to form the soluble tetrachloroaurate(III) complex. This removes the gold(III) ions from the equilibrium, driving the reaction to completion.
Overall the reaction is: $$\ce{Au + 3HNO3 + 4HCl -> [AuCl4]- + 3NO2 + H3O+ + 2H2O}$$

For the most part you are right, but just a minor detail... Chlorine in [AuCl4]- is obviously Cl(-1), so the atomic chlorine(0) cannot react with the gold(III) ions to form [AuCl4-] because there is no reducing agent. I would put it one of two ways: either atomic Cl(0) oxidises the gold to Au(III), itself being reduced to Cl(-1); or HNO3 oxidises Au(0) to Au(III) and the Cl- ions from HCl complex with the gold atoms to form [AuCl4]-.

@orthocresol could you move this into the answer, and delete the comments? Thanks!

Chlorine gas can react with gold powder to generate $\ce{AuCl3}$. In an aqua regia solution, the driving force is the creation of the tetrachloroaurate complex, $\ce{[AuCl4]-}$.

The nascent part describes how the gas chlorine is formed in aqua regia and how it is different from standard chlorine gas. The idea is that the chloride ions from hydrochloric acid don’t all have to reach the oxidising $\ce{HNO3}$ species at the same time but do so one after the other. So that instead of immediately forming a chlorine molecule $\ce{Cl2}$, chlorine radicals $\ce{Cl^.}$ or something close to that are first formed. These then have a much greater reactivity than gaseous chlorine since they have an inherently less stable electronic structure.
A similar phenomenon is observed with e.g. zinc in hydrochloric acid. The reaction is pretty simple and generates hydrogen gas:
$$\ce{Zn + 2 H+ -> Zn^2+ + H2}$$
However, you can think of this as two single electron transfers (each hydrogen ion only gets one electron) with two hydrogen radicals having to meet up to generate the gas; this is why $\ce{Zn + HCl}$ has a better reducing ability than gaseous hydrogen.
As for how aqua regia is able to dissolve otherwise inert metals like gold: it all boils down again to nitrosyl chloride and nascent hydrogen being stronger oxidising agents than simple $\ce{HNO3}$.


text:
Why and when is lysine charged
I'm reading about Lysine and doing some calculations on it's charge under various $\ce{pH}$.
Depending on the source, Lysine is usually stated as a charged amino acid with a $\ce{NH3+}$ group. However the video I'm watching is calculating the charge at 5 pH and it writes Lysine out with $\ce{NH2}$ and goes on to say that because the solution is pH 5, then $\ce{NH2 -> NH3+}$ and arguments that THIS is why Lysine is charged.
Does the literature state Lysine as a charged acid, assuming it is at pH 7? Or is the video I am watching wrong and it should not gain a charge as $\ce{NH3}$ will not accept a hydrogen?
Bonus: Why are only SOME groups given a Pka value?
http://www.geneinfinity.org/images/aapks.jpg
http://www.cem.msu.edu/~cem252/sp97/ch24/ch24aa.html
How would I calculate the charge for Isoleucin for example?

The side chain of lysine is charged at all pH lower than 10.53 (according to the table you linked), and so it would be charged at pH 7. What happens is $\ce{NH2}$ is more or less stable, but once you start to have any decent amount of hydrogen in your solution it will donate its lone pair to become $\ce{NH3+}$.
Bonus answer: only the amino acids with ionizable side chains are given pKa values, because, well, they're the only ones that are ionizable. In other words, they're the only groups that will accept or donate a proton at any reasonable pH (you could theoretically ionize glutamine, but the pH would be ridiculously low, and well outside of the world of biochemists who are only really concerned with pH levels inside the body).

Thank you. I wish I could select more "best answers" but unfortunately only one can be selected and this one was most fitting for my level.
I also found a video that helped me understand the topic better if anyone finds this thread in the future: https://www.youtube.com/watch?v=WxNYfc98Iqw

At pH 7 lysine has a net charge of very close to +1.
The carboxylic acid group is fully deprotonated (-1 charge).
The alpha amino group is about 99% protonated (+1 charge).
The side chain amino group is fully protonated (+1 charge).
You need to look at the pK of each ionizable group, consider what fraction of the group is in each ionization state (for example using the Henderson Hasselbach equation), and add the charges of each group to get the net charge.
For isoleucin, at pH 7, the net charge will be almost neutral, because 7 is well above the carboxylic acid pK of 2.32 and significantly below the amino pK of 9.76.  There will be a small fractional negative charge because 7 is closer to the pK of 9.76 than to the pK of 2.32.  Note in the first linked table in the question, there is a column for "pI".  The pI will be the only pH where the amino acid has a net neutral charge.

isoelectric point of Lysine is ~ pH 9.7; That clearly means at physiological pH 7.4, Lysine is acidic (N-terminus is deprotonated) with +1 charge?

@bonCodigo The second pKa of lysine is 8.95.  At pH 7.4 the ratio of protonated to deprotonated will be 10^(8.95-7.4) = 35.   So about +0.97.

LOL @DavePhd, that's a precise calculation. I meant to confirm my understanding, that Lysine is in the acidic range in this case. Still, Isoelectric point is calculated using Ka1 and Ka2. So why do we care about Ka2 again here?

In aqueous solutions amino acids are usually considered as zwitterions.
Even as solid salt they are charged.
I can't really say, why usually it is drawn uncharged. It may confuse but it isn't important when you want to know which charge it really has. Just take a look at the pH and the pka values.
In a solution the charge depends on the pH of the solution. If there are additional groups which can protonate or deprotonate you have to take a look at their pka.
Since Lysin has to amino groups and one carboxylic acid group.
The pka of the both aminogroups is much higher than your pH of 5, so both are protonated. You have two positive charges.
The carboxylic acid is deprotonated at pH 5 because its pka is 2.18. So one negative charge here.
That means Lysin has in total one positive charges at pH 5.
In water you have always charges because you have a acid and base in one molecule, so they can react with each other. COOH protonates the $NH_2$ and $NH_2$ deprotonates the COOH.
The charge of Isoleucin is an easy example because you have no side chains, just COOH and $NH_2$. You can take a given pH (maybe 5) and then take a look at the pka of both functional groups as written above. Then add the charges.
Or you can take a look at the pI. This works even with side chains. If the pH = pI you have a zwitterion with on positive and one negative charge, so the whole molecule is neutral. If your pH is higher than the pI it is mor basic and something will deprotonate so you have a negative charge.
If pH < pI something is protonated und you have a positive charge.
pI refers to that point there half of all
BONUS:
Just take a close look at the functional groups of the side chains. Since most times pka is refered to aqueous solutions only those groups are given a pka value which can be protonated or deprotonated in water. A Methyl group for example $R-CH_3$ can't be deprotoned in water. so why should they give a pka value? It wouldn't be helpful.


text:
Ion concentration in acid and base
Hello here is a question that is a bit confusing I hope you can help me solve it and understand it:

Determine the hydronium and hydroxide ion concentrations in a solution
  that is $10^{-4} M \ce{Ca(OH)2} \text{ AND } 10^{-4} M \ce{HCl}$

Note: the "and" is actually capitalised in my book..
So what I understood from this question is that they are both in one solution. Now I thought of two ways to solve it:
A) the concentrations are equal and they are both strong which means they we're neutralise each other completely and end up being a neutral solution and both $\ce{H3O+}$ and $\ce{OH-}$ will be $10^{-7}$.
B) the calcium base has 2 $\ce{OH-}$ while HCl has 1 H+ so the base concentration is double that of the acid which means that the solution is going to be basic but I don't know how can I solve this to find the actual concentrations.
Which way is right and when does an acid and base really react to become neutral?

Your second (B) way of thinking is correct.
In the solution there are twice as many OH- ions as H+ ions.
All the H+ from HCL will react with half the OH- from calcium hydroxide.
Half the OH- will remain ($10^{-4} OH-).
Then use $K_w$ to determine the final concentration of H+.

Ok, I solved it and got for the H3O+ 10^10. And for pH =10.... But I have another question why is there H3O+ left if they all reacted with the OH- ?

because aqueous solutions must obey the equillibrium Kw=[OH-][H+] = 10^-14.  Even pure water dissociates to provide H+ and OH-.

Virtually all of the H3O+ ions reacted, but still very little amounts left because of water protolysis.

authors: مارنس السويدي