text:
Does rate of Evaporation change with surface area?
Two vessels of the same volume, in the
same location and open to the atmosphere. One is of a small diameter, the
other is of a larger diameter.
 The question is will the vessel with the
larger surface area evaporate the liquid
faster than the smaller one?
If not! I need answer to other question ,if i am not wrong.
Why does vapour pressure decreases with decrease in surface area?

this question was discussed there:
http://physics.stackexchange.com/questions/55205/whats-the-surface-area-of-a-liquid-how-does-evaporation-increase-if-the-surfac

Why does vapour pressure decreases with decrease in surface area.

Well, it does not. Vapour pressure is a function of temperature only. It is in fact the equilibrium constant for the reaction $\ce{H2O_{(l)}->H2O_{(g)}}$, and like all other equilibrium constants, is independent of surface area, amount of the species but depends only on the temperature. At a given T, only for one specific value of the partial pressure is water and its vapour in equilibrium.
That said, water is not actually in equilibrium when in an open vessel evaporating into the atmosphere. The vapour pressure above the water filled vessel is far less than what is required for equilibrium. Water constantly evaporates as a desperate attempt to increase the vapour pressure to its equilibrium value, but since the vessel is open and the vapours not confined, the equilibrium value is seldom reached. If the surface area is larger, it means that the liquid phase can evaporate a larger amount of water while attempting to attain equilibrium. Therefore, larger surface area increases the amount of water that is lost as vapour while attaining equilibrium, since it is that part of the liquid exposed to the vapour phase which tries to maintain equilibrium by evaporating or condensing. Evaporation is essentially a surface phenomenon, as opposed to boiling which becomes considerably a bulk phenomenon.
Since it is practically the surface layer which participates actively in the phase equilibrium, larger the surface, greater the attempt to attain equilibrium and quicker is the evaporation.

Nicely done! Specifically, I liked the visual part of this answer- water is trying hard to equate its fugacity, but it's always coming short. That's why after few days we will see empty vessel.

It depends what you mean by "rate". If the question is in which vessel does the liquid disappear fastest, then the answer is yes. But some people might assume that rate means rate per unit area in which case the answer would be no as that just depends on the vapour pressure.
It is pretty easy to observe the effect with any volatile liquid.

Evaporation means the change from liquid to gas. That's why water in the larger surface will evaporate more than smaller surface. The water surface in a larger vessel will have more direct contact with the air. Another thing is that evaporation will be influenced by pressure and temperature and both will influence the boiling point. In this point the liquid evaporation will occur.
The pressure inside the vessel which different surface, should be different and the pressure difference also found in between inside and outside the vessel.


text:
Photosynthetic rate calculation

What are the steps in calculating the photosynthetic rate ($\pu{\mu mole}~ \ce{O2} \pu{m2/ min}$) for a leaf that has an area of $\pu{0.0025 m2}$ and a $\ce{O2}$ evolution
rate of $0.096~\ce{O2}/\pu{min}$?

To do so , the first step is to convert it to $\mu L$ i.e. $\pu{960 \mu L} ~\ce{O2}/\pu{min}$.
The second step should be to convert it to $\pu{\mu moles}$. So, how do we do that ?
EDIT : In my textbook, the following calculation has been done but I am not able to understand it...

$\pu{960 \mu L}~\ce{O2}~\pu{/L} / [(273+23)/273) \times 22.423]$

I know that $22.423$ is the volume of a mole of gas and $\pu{273K}$ is $\pu{0^\circ C}$ but I am not able to understand how that fits here.

Using the general case equation
$$\frac{P_1V_1 }{T_1} = \frac{P_2 V_2} {T_2}$$
and given that pressure is constant at $\pu{1  atm}$ in your experiment, the equation simplifies to
$$\frac{V_1 }{T_1} = \frac{ V_2} {T_2}$$
We know $V_1=\pu{960 \mu L}$ and it appears that the experiment is being run at $\pu{(273+23)K}$, so we need to calculate what the volume ($V_2$) would be at standard T ($\pu{273 K}$).  This is given by
$$V_2 = \frac{V_1 T_2}{T_1}$$
if we next divide $V_2$ by $\pu{22.423E6 \mu~L}$, we now know how many moles of $\ce{O2}$ we have.

You did beat me by seconds. Sometimes, mathjaxing the ideal gas law is a burden :D

It's too bad that SE doesn't permit the OP to select more than 1 correct answer.

I think that upvoting several answers is possible, but only one can be accepted. You were faster and explained in more detail than I would have done!

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text:
Defining and testing custom made DFT functionals?
What is a good and free software for this endeavor? Any recommended articles on the subject?
I want to design dft functionals, trying out different parametrizations and optimizing mixing constants for exchange and correlation contributions for specific data sets. Even add PT2 perturbation energy into the mix if possible.
I am aware of Gaussian's user-defined functional options, but would like to have a bit more control in the parameters and contributions, instead of just 6 parameters.

You may want to read up on Don Truhlar's work. Also note some modern variations on the theme.

Depending on what you're after, you might want to look for programs using libxc, so you can completely define your own functionals.
From the libxc section of the Octopus wiki, these include:

Abinit – plane-wave code
APE – an atomic code
Atomistix ToolKit – numerical orbitals code
AtomPAW – projector augmented wave functions generator
BigDFT – wavelet code
CP2K -  A program to perform atomistic and molecular simulations of solid state, liquid, molecular, and biological systems.
DP – Dielectric Properties, a linear response TDDFT code
Elk – FP-LAPW code
ERKALE – a DFT/HF molecular electronic structure code based on Gaussian orbitals
exciting – FP-LAPW code
GPAW – grid-based projector-augmented wave method
JDFTx – plane-wave code designed for Joint Density Functional Theory
MOLGW - a small, but accurate MBPT code for molecules
octopus – real-space (TD)DFT code
Yambo – solid state and molecular physics many-body calculations code

On a more personal note, and this is just my opinion, if you're just arbitrarily mixing functionals, you should really make sure you have a scientific justification for doing this, otherwise you're just generating numbers without basis.

Thanks Aesin, theses resources were what I was looking for. I have used Octopus before and it works really good. I completely agree with your stance on getting "numbers without basis".

@Aesin  Even if you don't mix-a-match functionals, no matter what you do, the DFT method should always always ALWAYS be calibrated before running with it.  Of course not everyone can implement rigorous wave function methods (e.g. CCSD(T)) but un-calibrated DFT is just bad science in general.

@LordStryker: I disagree somewhat: if you understand what the origins of the functions and parameters used in your functional are, how they relate to your simulated system, and are aware of obvious possible sources of limitations, I think calibration isn't always necessary. Helpful, yes, and a good idea, but not necessary. I guess they're two ways of approaching the functional verification: empirical vs theoretical matching.

@Aesin It depends, I suppose, on what property you're trying to analyze.  In my work, DFT has done well with qualitatively reproducing energetics of non-covalent interactions between weakly-bound clusters (especially with the recent dispersion corrected schemes of Grimme).  However, there is severe disagreement with the Hessians produced by (most) of the DFT methods implemented and Hessians from wave function theory. In every case, perfect agreement was not observed.

@LordStryker: Hah, yes, in this case I'd say "obvious possible sources of limitations" -> "Typical forms of KS-DFT are really bad at this".

If you're trying to modify the source code and run DFT jobs, then you basically have only three options:

PSI4 (very powerful, written in both C and Fortran)
NWChem (very powerful and fast)
GAMESS (kind of fast, fully written in FORTRAN)

If you would like to play with codes and tweak it as you wish, not worrying about the speed and optimization level, then you must try PyQuante. They recently added DFT and MP2 codes in it, so you it will be easy for you to play with. Needless to say, it runs slowly because the top layer is written in Python, while the inner layers are written in C and FORTRAN. I guess you will love playing with it!

Since you mentioned "free", you maybe want to have a look at GAMESS from the Gordon group.
I have really no idea whether it's fit for the task, but last time I checked it was free for academic use.

There's of course Turbomole, but according to an old thread in a forum from 2007, they weren't fond of custom functionals either at that time. The statement of one of the developers might be worth reading anyway.

Thanks! I use GAMESS extensively, but they have no easy way of using custom made DFT functionals. Their  help file only describes already-made functionals.

Maybe, Molpro has one of the easiest ways to do it
http://www.molpro.net/info/2012.1/doc/manual/node194.html
though, not free

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text:
Can polyatomic ions (CO₃, PO₄, SO₄, NO₃) be considered conjugated systems?
From my perspective these resonance structures allow these specific polyatomic ions to act as donor-acceptor molecules. Many donor-acceptor molecules also tend to be conjugated systems because they have chains of alternating conjugated π orbitals.
So does this imply that these ions (due to their resonance structures) can act as though they had a conjugated system or am I making too big of an assumption?
Curious on others perspectives/if I'm totally of my rocker.

I think you're a bit off ;)
To my understanding, the IUPAC Gold Book restricts the term conjugation to organic chemistry. I'm highly biased and too much an organic chemist to disagree :D

@Klaus Warzecha: If the OP substituted "delocalized" in place of "conjugated", would that make for a better statement?

I appreciate the thoughts guys! I think what I'm hunting for is what lies in the stricter definitions of what makes something conjugated vs what makes something resonant. Though in both cases I would consider the electrons delocalized as @ron mentioned. Thanks guys!

"Conjugated" implies a 1,3-shift to move an electron or hole.  It is a remnant of LCAO modeling that is obviously wrong but fantastically useful short of the Woodward-Hoffmann rules.  I'm not sure inorganikers would like the name as such. MO modeling is accurate but unwieldly.  "Delocalized" or "resonance hybrid" is good, certainly for inorganic systems that may have (virtual) d-orbital participation.
Consider nitrate.  Inorganikers would say $\ce{N^{5+}}$, $\ce{^{-}O-N(=O)2}$, with five bonds to the nitrogen.  The negative charge 1,3-shifts around all three oxygens.  Organikers would see it as $\ce{N^{3+}}$, $\ce{[^{-}O{}-]_2N^{+}=O}%edit$, with four bonds to the nitrogen.  1,3 shifts, etc.  Is nitrate ever a bidentate ligand?  YES!  But it is just another resonance structure.
Inorg. Chem. 35(24) 6964 (1996)
http://pubs.acs.org/doi/abs/10.1021/ic960587b

Few people would really assign five bonds to nitrogen in nitrate ion.  We can explain its characteristics with only four bonds.  Yes the oxidation state is +5 -- consisting of +4 for four bonds to more electronegative atoms, and +1 more by sharing one more electron pair than the normal valence of three.

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text:
Unfavorable reactions
How can you make an unfavorable reaction go backwards?
From my understanding if we wanted a reaction at equilibrium to go backwards (i.e. form reactants from products), we would add more product into the solution.
But I'm not sure how the reaction being an unfavorable one changes the above statement.
After making the reaction go backwards, we have changed the reaction rate going backwards correct?

If the process we're discussing is an equilibrium, and if the forward reaction is "unfavorable" (I take that to mean that the products are of higher energy then the reactants), then there are still ways to shift the equilibrium in whichever direction you want.  For example
A <-> B + C
using LeChatelier's principle, if we increase the pressure in our vessel, we will push the equilibrium back towards A, the opposite will occur if we decrease the pressure.  Temperature, catalysts, reagent concentrations, etc. can all affect the position of an equilibrium.


text:
Equilibrium rate constants forward and backwards
Correct me If I am wrong, given a reaction at equilibrium, if we say add more concentration of products to the reaction, the rate going backwards would increase to try to get back to equilibrium.
I was wondering if the rate forward stays the same or slows down?

In simple cases, the rate is equal to the rate constant "k" times the concentration.  The forward rate constant and the backward rate constant don't change, but the rate of change (rate constant times concentration) will change as the concentration of a reactant or product changes.

I see, so there is a distinction between rate constant denoted by a K and rate. So in my above example the Rate backward is slowing down after being initially sped up, and the rate forward is being increased because reactants are being formed from products?

yes, also a capital "K" typically refers to an equilibrium constant and lower case "k" refers to a rate constant.  Keq = kforward/kbackward

Yes, if we add products, then the reverse reaction will try to make more reactants. If we take away products, then the forward reaction will try to make more products. Not on topic: If you increase pressure, the side with less moles will be favoured. And more things. This is all Le Chatelier's Principle.


text:
Does HCl kill algae?
I live in the tropics, and have a water tank that sits in the sun, forming algae. Every year, I need to scrub the thing out. The plumber I’ve hired for the job wants to flush my water pipes with HCl to kill any residual algae. The pipes are mostly PVC, with braided metal hose and metal fittings here and there. Should I let him? If not, why not?

Are they metal pipes?  If you want to kill algae as such, flush with a diluted quaternary ammononium disinfectant, let sit a day, then flush everything very well indeed, and again.  If you have rubber or polymer fittings, don't do that, either, for the stuff diffuses in.

From wiki: Quaternary ammonium compounds ("quats"), such as benzalkonium chloride ... that looks like exactly what I need, thanks :)

Clean the tank with bleach and flush it several times.  Once clean, put a piece of copper pipe in the tank and that should help it stay clean.  Don't use HCl, too strong for the job.

The wiki under Uncle Al's post also says copper is great at destroying microorganisms. Tossing copper pipe in the tank is excellent advice.

I didn't read this but it seems relevant:  https://homeguides.sfgate.com/copper-control-algae-97682.html


text:
How does iron rust in basic conditions?
I performed an experiment about the corrosion of iron. I used acid and base in the experiment.
I understand that the acid speeds up the corrosion of iron. But I don't understand why the base also corrodes iron. I can't find out the answer even after I've googled it.
In my experiment I mix an ulcer drug into a glass of water, then I put in an iron nail. After I let it sit for a few days, the iron nail becomes rusty. I don't know of any reaction between a base and iron that gives rust.

Ulcer drugs are neutral to very weakly basic. That means there is no sense in your question.

Iron can oxidize to "rust" (which is actually a mixture of hydrated iron(iii) oxide $\ce{Fe2O3*nH2O}$ and iron(iii) oxide hydroxide $\ce{FeO(OH)}/\ce{Fe(OH)3}$). The iron hydroxide species look like they could come from basic solutions.
Even if consider only the formation of $\ce{Fe2O3}$, this can occur by several pathways.
Using Wikipedia's Standard electrode potential data page as a source of known redox half reactions, the process may be something like this:
In acidic medium
The net reaction in acid is between the iron and the protons in the acid to produce $\ce{Fe^{3+}}$, which reacts with water in a redox neutral (but generates more acid) process to form the insoluble oxides/hydroxides. The initial redox reaction is spontaneous. The water is the net source of oxygen.
$$\ce{2Fe(s) + 6H+(aq) - 2Fe^{3+}(aq) + 3H2(g)} \ \ E^\circ=+0.04\ \text{V}$$
$$\ce{Fe^{3+}(aq) +3H2O -> Fe(OH)3(s) + 3H+(aq)}$$
$$\ce{Fe(OH)3(s) -> FeO(OH)(s) + H2O}$$
$$\ce{2FeO(OH) -> Fe2O3 + H2O}$$
In basic medium
The net reaction in base is between oxygen (in the air, which is not in contact with your submerged nail) and the iron in the nail. The iron initially reacts with the hydroxide ions, which are regenerated by the reduction of oxygen at the surface of the water.
$$\begin{align}
\ce{4Fe(s) + 8OH- &-> 4Fe(OH)2 + 8e-} \ \ \ &&E^\circ=\pu{+0.89 V}\\
\ce{4Fe(OH)2 + 4OH- &-> 2Fe2O3 +4e-}\ \ \ &&E^\circ=\pu{+0.86 V}\\
\ce{3O2 + 6H2O + 12e- & -> 12OH-}\ \ \ &&E^\circ=\pu{+0.40 V}\\ \hline
\ce{4Fe(s) +3O2(g)& -> 2Fe2O3(s)}\ \ \ &&E^\circ=\pu{+1.95 V}\\
\end{align}$$

Where would hydrogen gas come from for the first reaction?

The iron metal reduces protons ($\ce{H+}$) to hydrogen ($\ce{H2}$).

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text:
How does a space group determine crystal structure?
If I have an intermetallic and want to know the specific arrangement of atoms in the lattice, how can I determine that? What I mean is the Cartesian (xyz) position, or the (abc) positions in the lattice. In my case, I am working with Ni3Sn4, which has a space group C2/m. The space group diagram can be found here.

To build lattice you need the space group, the three unit cell dimensions, the three unit cell angles, each atom's fractional coordinates (e.g., a *.cif file overall),
Acta Cryst. B38, 598 (1982)  doi:10.1107/S056774088200346X
Ni3Sn4, C2/m, a=12.214, b=4.060, c=5.219 A, beta=105.0 degrees (others 90 degrees by default), V=250 A^3 (verify), Z=2 (formula units/unit cell)
plus
http://www.chem.gla.ac.uk/~louis/software/ortep3/ or
http://www.shapesoftware.com/00_Website_Homepage/ or
http://www.iucr.org/resources/other-directories/software?result_42405_result_page=A
the big kahuna.
For 3D visualization, if you are not practiced,
http://www.berezin.com/3d/3dprism.htm


text:
What is the implication of not being able to balance the complete combustion reaction equation of methanol?
This question has been intriguing me since 10th or 11th grade. The teacher just told us about it but didn't get into the details of the why. Recently I asked a biochemist but couldn't get an answer. So I bring it here.
$$\ce{CH3OH + O2 -> CO2 + H2O}$$
is the complete combustion equation of methanol. But the problem is it can't be balanced. What is the practical implication, and significance of this fact?
Side question:
What is the mathematical reasoning behind the inability to balance the equation?

Wait, what do you mean it cannot be balanced?

Maybe it was some other equation. Just can't recall it, I'll update the question.

Please do. I'm very curious now.

I'm pretty sure every single valid chemical equation balances, as long of course as you know all the products. You could conceivably use a reaction as an analog computer (a very cumbersome one!) to solve a linear system of equations. If nature does it, then math must do it.

@NicolauSakerNeto Back then not being able to balance that equation was bugbear. Now not being able to recall is one. Argh.

This question has awoken the fire of linear algebra in me! I believe it is possible to prove that any system of equations resulting from a valid reaction has $n$ variables and $n-1$ equations, i.e. the system is always underdetermined by one equation and therefore has infinite solutions, all of which lie in a line in $\mathbb{R}^n$ sand are multiples of eachother by some arbitrary real number. Let me try formalizing the argument. Unfortunately I'm kind of tired, so I can't guarantee I'll be able to wade through the notation properly.

After playing with the question in my mind, it turns out to be quite interesting. I cannot prove that all valid chemical equations can be balanced, because it's actually the opposite cause and effect; all equations that can be balanced are valid chemical equations. Unfortunately I cannot provide as formal a proof as I thought, but I may write something up later.

@NicolauSakerNeto You are quite right but I'm a TA for a gen. chem course and nobody seems to get that! "But sir, why are the coefficients 2,3,2,4 instead of 4,6,4,8?"

@CHM I haven't had the opportunity to teach class yet, but I would retort with something like "They can be. They can also be 20-30-20-40, and they can also be 634-951-634-1268, and even $\frac{1}{12}$-$\frac{1}{8}$-$\frac{1}{12}$-$\frac{1}{6}$. But it turns out that using the smallest numbers possible while avoiding fractions makes calculations simpler! Do you want to work more for no reason?"

Here is the balanced eqn...
$$\ce{2 CH3OH + 3 O2 → 2 CO2 + 4 H2O}$$

While i was typing the the mathjaxed equation, you were faster submitting. So I beautify yours :-D

@KlausWarzecha Can you beautify me next??

@LordStryker I might be good, but not THAT good :-P

beatify or beautify?

@ron Blessed are the cheesemakers.

Mass is conserved, charge is conserved, ignore spectator ions until the end.  If it is a redox reaction, begin by listing each active atom's reduction or oxidation, e.g.,
Mn(+7) + 5e to Mn(+2)
Cr(+3) - 3e to Cr(+6)
and find the least common denominator (here, 15).  That gives you the primary stoichiometry, here 5Cr(+3) (going to chromate) with 3Mn(+7) (coming from permanganate).  Add the other atoms, slop in H2O, OH-, or H+, add back the spectator ions. Confirm the same number of each kind of atom on both sides.  Try it out,
CuSCN + KIO3 + HCl to CuSO4 + KCl + HCN + ICl + H2O
[Cr(N2H4CO)6]4[Cr(CN)6]3 + KMnO4 + H2SO4 to K2Cr2O7 + MnSO4 +CO2 + KNO3 + K2SO4+ H2O


text:
How does molecular structure affect physical properties?
The question sounds vague and non-specific because my background in chemistry is limited to undergrad college courses.  However, I am interested in learning the correlations between the atom-by-atom structure of molecules (simple, complex...) and the resultant physical properties thereof.
So far, it seems most the material online directly referencing the phrases "molecular structure" and "physical properties" is in the form of highly specialized research publications.  Surely, there is a Basics to the Relationships between Molecular Structure and Physical Properties 101 out there, somewhere.
Ultimately, my questions sum up to the following:

Where may I go to learn these things?
What material may I read?
Who might I ask?
Where do I begin?

Thank you very much.

Predicting (approximate) physical properties starting with molecules is complicated, below.  Nobody has ever produced a way to accurately predict how molecules will pack in a crystal structure.  Big Pharma is desperate to predict crystal polymorphs and their properties. Slightly vast amounts of grant funding pursued predicted super-explosive octanitrocubane.  When it was finally synthesized it crystallized in a less dense crystal structure than expected, deeply disappointing the DoD.
http://www.chemspider.com/ACDLabs.aspx
http://pubs.acs.org/doi/abs/10.1021/ed064p575
http://molsoft.com/mprop/
If you have a way to predict crystal structures given complex formula units, you're a big winner.

Until then, Accelrys and others make a fortune with their software ;)

So simple crystal formation prediction is a good place to start, structure-wise.  Thank you.

Uncle Al is definitively right, especially on polymorphism.
But maybe you just want to start on a more basic level.
Check the web for some introductory information on

boiling points of linear hydrocarbons of different length
boiling points of alcohols as compared to their parent hydrocarbons
melting points of saturated vs unsaturated fatty acids

to get a feeling how structure and different interactions between molecules influence the physical properties.

Melting and boiling points, then...  That sounds like a good start.  Already, what I've found on listings of intermolecular forces makes most reference to those properties.  Thank you.


text:
What happens to the hydroxide anion during the electroreduction of water?
The reduction of water at a cathode is represented by the following equation:
Cathode (reduction): $\ce{2 H2O_{(l)} + 2e^{−} → H2_{(g)} + 2 OH^{-}_{(aq)}}$
If this is done in a divided cell, what would happen to the $\ce{OH-}$ anions? The hydrogen gas would accumulate and bubble would form. Would the $\ce{OH-}$ anions just build up in the solution? Do they interact with each other naturally forming $\ce{H2}$ and $\ce{O2}$?

Will you observe evolution of hydrogen at all if your half cells are separated by a impermeable wall?

Would you please post species involved in both the half cells.

You only show one a half-reaction, reduction at the cathode.  Write the other half-reaction, oxidation at the anode to give oxygen.  A real world water electrolysis cell requires a salt electrolyte for conductivity.  $\ce{NaCl}$ could be electrolyzed to $\ce{Cl2}$ and $\ce{Na}$, the latter reacting with water to give $\ce{H2}$ and $\ce{NaOH}$.
http://www.ineris.fr/ippc/sites/default/interactive/brefca/image16.gif
If you want $\ce{H2}$ and $\ce{O2}$, make the electrolyte $\ce{NaOH}$.  $\ce{Na2CO3}$ can be interesting.


text:
Any references for adsorption of ions to metal surfaces?
I'm specifically interested in ions that would adsorb to gold surfaces, however I would be incredibly appreciative if someone could direct me to a textbook reference or something more expansive, that i could use to reference in the future.
I've been doing some reading and found the following regarding Silver:
Silver Adsorbing
Chloride, Bromide, Iodide, Urea, thiocyanate
Silver Nonadsorbing
Fluoride, Perchlorate, Sulfate
I know Gold leeches with cyanide and I think bisulfide.
 I actually found a good set of lecture slides that discusses chemisorption an Oxygen atom across a series of Transition metals. Also I'm also looking for something about physiorption not chemi...
 The slides reference a book called "Surface Science:Foundations of Catalysis and Nanoscience" by Kurt Kolasinski"...I guess I'll start there, but would love to hear about other references anyone else may have.

Almost all late transition metals adsorb $CO$ and isoelectronic particles, like organic isocyanides. Some of them (including gold) also adsorb sulfides and derivitives of S(-2), like alkylsulfides and ditiocarbamates, but the latter bonds very strongly (many catalysts are poisoned by tiny amount of sulfur in the feed). I tried to find a good book in the past, but found that there is not comprehensive systematic book, since the field only recently gained tools to be studied effectively. Your best bet is google scholar with subsecuent begging your scientific friends for help to get the articles.

Wish this was an answer so I could throw some points your way. I really appreciate it man! My position has moved me out of the ChemE field, so my current peers lack any chemistry at all. I'll have to go rub some old contacts shoulders. Thank you so much! And I'll let ya know if that book is actually any good. It's funny because the papers I have seen published in the 70-80's state that "X" adsorbs to Silver, but they never have any references. They all state whether these ions adsorb "specifically", regularly, or if they're non-adsorbing ions.

@permeakra You could flesh out your comment some more and it would be a very good answer!


text:
Heat produced when magnesium is reacted with acid

Calculate the heat produced when a strip of $\ce{Mg}$ metal with a mass of $\pu{0.0801 g}$ is reacted with $\pu{50.0 ml}$ of $\pu{1.0 M}$ $\ce{HCl}$ to raise the temperature of the water by $\pu{7.6 °C}$ (change in temperature is $\pu{7.6 °C})$ Calculate the heat produced when a mole of this metal is used.

I have this so far:
$$q = mc\,\Delta T$$
$m =\pu{0.0801 g}$ and the change in temperature is
$$\Delta T = T_2 - T_1 = \pu{7.6 °C} - \pu{4.184 K} = \pu{3.416 K}.$$
Is that right? I have no idea what the specific heat $c$ is. Is it the specific heat of $\ce{Mg}?$
What do I do with the $\pu{50.0 ml}$ of $\pu{1.0 M}$ $\ce{HCl}?$

It can help to imagine the experimental setup.
Imagine a thermally insulated reaction container, let's call it calorimeter, that is filled with 50 ml 1 M $\ce{HCl}$ ( = a solution of hydrogen chloride gas in water).
We have to measure a temperature change ($\Delta{} T$), that comes from an exothermic reaction. Where would we best measure this temperature change? Our reaction medium apparently is a good choice. Therefore, there's most likely a thermometer (insulated from the outside) in good thermal contact with the water.
Now we let the magnesium strip fall into the reaction medium (using some clever mechanic) while our calorimeter is closed), let it react and measure the temperature change until the temperature reaches some final (maximum) value.
As you wrote, the temperature change ($ \Delta{} T$) is +7 °C.
I don't see that an initial temperature was given, but that's not important. We have the $ \Delta{} T$.
Now we need a specific factor that correlates the heat liberated in the reaction with the temperature change in a specific medium.
Here, the specific heat capacity of water comes into play.
There you go.

As user user2617804 noted the problem doesn't appear to be well thought out.
The major problem here is that converting mass of magnesium to amount of substance gives
$$n(\ce{Mg}) = \frac{\pu{8.01E-2 g}}{\pu{24.3 g mol^-1}} = \pu{3.30E-3 mol}$$
So, $\pu{50 mL}$ of $1.00$ molar acid only has $\pu{5.0E-2 mol}.$ That little acid can't react with $\pu{1.00 mol}$ of magnesium, so there will be $0.95$ excess moles of magnesium.
Since the magnesium is in excess, Using proportions if $\pu{3.30E-3 mol}$ of acid raises the temperature $7.6$ degrees, then $\pu{5.0E-2 mol}$ of acid should raise the temperature by
$$\Delta T = \frac{\pu{5.0E-2 mol}}{\pu{3.30E-3 mol}}\times \pu{7.6 °C} = \pu{115 °C}$$
So even if the water started at $0\text{ }^\circ\text{C}$ that much heat would boil some of the water off.
The other problem here is that the specific heat of the $\pu{0.95 mol}$ $(\pu{23.1 g})$ of magnesium which would be left is not the same as the specific heat of water.
To really solve this problem, we require

initial temperature of water;
initial temperature of magnesium metal;
specific heat of water;
specific heat of magnesium.


text:
Are all PP(Polypropylene), i.e. resin code 5 plastics, safe for microwave and steaming?
If a plastic is marked PP as in Polypropylene (i.e. resin code 5), is it automatically safe for steaming and microwaving, such that it wouldn't leach chemicals into the food(besides not melting)?

No.
Just because the polymer is of one or the other kind does not let you say what industrial process actually synthesized it. Subsequently it is impossible to know what trace chemicals are left in the polymer packaging.
If if isn't labelled as microwave-safe, it isn't!
They test the package for it's microwave compatibility, not one type of polymer each! It seems that resin code 5 polymers are often used for microwave-proof packaging, but again: You don't know how it was made, you have no idea what's in there, don't microwave it (together with food or something else that you then ingest).

+1 for the caution but maybe it is not so dangerous I've found only this study but seems no traces were found...

@GM  In that study https://web.archive.org/web/20120105064009/http://www.mcpherson.edu/science/cantaurus/Vol7/99-Torrison.pdf   They were testing particular brands of "microwaveable plastic dishware" and found they didn't leach a hazardous amount. But I wasn't asking about particular brands or if things labelled as microwave safe are microwave safe.   tschoppi addressed what I asked though.

Consumer polyolefin plastics contain trace mold releases that adversely affect biological studies.  For human food use, one presumes Generally Recognized as Safe as long as the contents stay water wet, below 100 C.  Burning on residues (cheese) allows the temperature to skyrocket.
Polyoelfin containers are penetrated by hot lipid and especially hydrocarbons (red lycopene in tomato products).  Do NOT use dishwasher "bleaches" to remove the internal stain.  The active ingredient is dispersed dibenzoyl peroxide (Bondo cure).  Sure, it diffuses in and free-radical breaks up the molecule's conjugation, rendering then colorless.  It also free-radical scissions all plastics in the dishwasher.  Things go brittle and fail.

Basically any plastic used in a microwave is problematic.   I have seen several containers at work marked "microwave safe" that clearly reacted into the food when used in the microwave.


text:
Why is the maximum % for a gold alloy 92%?
The assignment question is asking us to explain why you can't get more than 92% gold in a gold alloy, and I can't seem to find out why it's specifically 92% in either the textbook or online.
Any help is appreciated. Thank you.

I would expect this sort of question to come with an associated phase diagram. Is there one nearby?

It's not that you can't, it's that you usually don't.
The majority of Gold goes into jewellery with some in coinage (more in the past, we make gold coins now but they don't circulate much). But pure gold is too soft to be useful in most applications (coins have to last, rings have to not scratch too much and they aren't much use if they bend too easily). It is also expensive. So the most used alloys have to make a compromise between usability and cost. There are also some that try to achieve a particular aesthetic effect like a different colour.
This imposes limits of practicality. At the cheap end alloys that look like gold might be mostly other metals. But at the high end the issue is to get the correct aesthetic effect while retaining the appeal and value of gold without having something  too soft or weak to be practical. Many coloured golds are only about 75% gold (18 karat) because that gives a wide range of colours. But top end "pure" gold objects can be made at 22 karat (92% isn). We could make richer alloys, but they would be too soft.
So, mostly we don't. We can make many alloys with more gold, including many that would be useless as jewellery like the AuBe alloy mentioned in another answer (it is far too brittle), but you will rarely see them as they are not useful for anything.


text:
What is the mechanism for the oxidation of benzene to maleic anhydride using vanadium pentoxide?
What is the mechanism for the oxidation of benzene to maleic anhydride using vanadium pentoxide, $\ce{V2O5}$?

A talk given by Ken Waugh named The Detailed Kinetics of the Partial Oxidation Reactions: Benzene/Butane to Maleic Anhydride, Ethylene to Ethylene Oxide,[1] proposes the following mechanism on page 4:

[1]: Acessible as pdf via rsc.org


text:
Reason for vibration of molecules
Why do molecules of a body exhibit vibrational motion? Is it due to interaction between various molecules of the body, or interatomic interactions between atoms in the same molecule? If it is because of any of them, how can these interaction cause the molecules to vibrate?

Atoms and molecules have energy even at 0 K.  This energy causes them to vibrate (among other things).  Interactions between various components of the molecules (nuclear/nuclear repulsion, electron/nuclear attraction, electron/electron repulsion, etc.) determine their motion (rotational, vibrational, translational).  Each component has a finite amount of energy associated with it.

Geeze all the answers are too complicated. In the simplest sense a molecule in free space can "vibrate" because the atoms which make up the molecule can move in different directions relative to the position of the whole molecule. See images within https://en.wikipedia.org/wiki/Infrared_spectroscopy#Theory

Matter is energy and the zero point vibrational mode is part of matter and is not caused by energy; it is energy.

Here is a picture and an equation that may add to the comment LordStryker made.  Molecules vibrate because they have energy. First the equation, as $T \to 0$ (absolute zero), the first term will vanish, but the molecule will still have an energy given by $h\nu/2$.  So even at absolute zero a molecule will have some vibrational energy, this is called the "zero point energy".  The parabola (representing the hydrogen well, for example, that two hydrogen atoms will "fall into" when they bond and become a stable hydrogen molecule) gives a pictorial representation of why this must be - energy levels are quantized.  Even the lowest level is above the bottom of the parabola, so even in the lowest energy level at absolute zero, the atom or molecule will still have a non-zero energy which will cause it to vibrate.
$\displaystyle\varepsilon = \frac{h\nu}{\mathrm e^{h\nu/kT} - 1} + \frac{h\nu}{2}$

Can we tap that for free, infinite energy?

@LordStryker  Lots of people have tried / are trying, but no success to date, that I am aware of (and I'm sure it would be big news!).  Google "zero-point energy" and you'll see some links.

Molecules have zero-point energy as a consequence of the Heisenberg Uncertainty Principle, which directs that it is not possible to simultaneously know the position and momentum of a particle, in this case a vibrational quantum. The zero point energy, $h\nu/2$, is that minimum energy needed to satisfy uncertainty.
In contrast rotational motion can have zero energy, i.e. the molecule does not rotate, but it is then not possible to know in which direction the molecule is pointing. As soon as it starts rotating its spatial orientation is limited to only certain values; again uncertainty limits this.

The uncertainty principle explains not causes.  Vibration or rotation are the energy in the molecule and are a result of the QM properties of matter and energy. They are not caused by the "presence" of energy. Changes in their levels are quantized and are the reason for the larger heat capacities of molecules over atoms. There is a zero rotational energy and most likely a zero translational energy. Can ZPE be harvested? QM energy can be; think lasers, microwaves and photosynthesis. ZPE requires more effort; It must wax and wane in nuclear fission, fusion and other nuclear reactions.

Heat
Molecules vibrate because of heat. Or we could say this the other way round: heat is because molecules vibrate. If we ignore quantum stuff (molecules vibrate even at zero kelvin) then the major reason for vibration is because  things are warm. And the reason why any particular molecule vibrates is because is is constantly bumping into all the other molecules around it. The more vibration, the hotter.
In a molecular gas, the molecules hit each other a lot and this ensures that any energy they have is well distributed among the molecules (in a gas much of the thermal energy is tied up in the kinetic motion of the molecules or atoms, but for molecules, some of it will be distributed in vibrations and the constant bumping into each other will ensure that the energy is well distributed).
Solid or liquids are similar but much of the energy is distributed to those vibrational modes. Solids, where the atoms or molecules in them don't have kinetic energy by definition (that is, they are locked in relative place, which is more or less the definition of a solid), most of the energy is tied up in vibrations either of atoms/molecules vibrating around their location in the solid or by stretching vibrations of the bonds inside the molecules. Molecules in the solid interaction with each other (if a neighbour molecule is vibrating strongly, you will pick up some of that energy because you are close to and interacting with that neighbour).
For example, heat one end of a bar of metal and the atoms will vibrate more strongly. But those vibrations excite the atoms near the hot end and they start to vibrate more strongly as well. They in turn do the same until the whole bar is warmer and all the atoms are vibrating more vigorously.

I think "atoms or molecules in them don't have kinetic energy by definition" is misleading. The translational KE of molecules in a solid correlates with temperature just as in a liquid or gas. The difference is that the translational motion in a solid appears as a very small amplitude vibration. The difference between these translational vibrations and a true vibrational mode is that the center of mass is moving. The vibrations tend to couple on a large scale in the form of phonons.

Taking a stab at theoretical physics here.
As long as there is some energy, the electrons and other sub atomic bits will move randomly. This randomness will result in tiny fluctuations in the distribution of energy /mass inside the atom resulting in vibrations that can be passed on to adjacent atoms.
The fluctuations will also result in fluctuations in charge distribution inside the atom, that can affect the state of adjacent atoms, again causing vibrations and rotations.
Before long, a feedback loop could in theory arise between adjacent atoms whereby the motions of a neighborhood of atoms may appear random to the macro observer, but be quite deterministic at the sub particle level.
I have no proof of this from an experimental standpoint, but from my understanding of the topic as an energy engineer, this makes sense to me. Cheers


text:
How to achieve clean dishes and glasses?
I heard that there is a standard about cleaned glass, namely

Chemically clean glass supports a uniform film of water, with no hanging droplets visible.

I tried to clean my dishes at home using detergent. After several efforts, the ceramic-made dish looks like a uniform film of water just after finishing the cleaning. If I hold the dish for a while, the water starts to concentrate at regions, forming a droplet.
In addition, in this video
http://www.youtube.com/watch?v=m7liHlmv5Kc
at 2 min, there is water droplet in the flask.
My questions are:
(i) Does the clean standard have any time range? Does it apply to arbitary time scale?
(ii) If the answer to question (i) is yes, how should I do to make the dishes so clean that there is absolutely no water concentrating at certain regions? I do not have a dishwasher.
P.S. Video demonstration is highly appreciated.

DO you want an answer for chemical glassware or for home dishes and glasses?

for home dishes and glasses

There are several laboratory techniques to rigorously clean glass. Here are a few of the more common ones, to add to the other answers. There are probably more procedures, and it's also common to perform more than one treatment to maximize efficiency.
Practically none of these have any use at home, because they are too dangerous, too expensive, and just outright unnecessary. A soapy scrub will do just fine cleaning anything in 99% of cases, even if water still beads.
1 - Ultrasonication in pure solvents or detergent solutions.
Often, the main impurities on the surface of glass are organic substances which are present at all times in air outside of strictly clean environments, and which strongly adhere to the glass when they come into contact. By dipping them into a solution with some affinity for the organic substances, it is possible to strip them off. Common options are regular dishwashing liquid (usually sulphonic acids with large alkyl chains), solutions of quaternary ammonium salts, aqueous ethanolanime, or pure deionized water, ethanol, isopropanol, acetone, among many others.
It is much more effective, however, to couple the aforementioned compounds with ultrasonication rather than simply dipping or scrubbing. In an ultrasonicator, the liquid is vibrated at frequencies higher than $10\ kHz$. Though macroscopically the liquid doesn't seem to change much, on the microscopic level there is chaos. The high frequency vibrations locally produce very fast solvent jets and cavitation bubbles, which effectively sandblast anything closeby. Glass manages to stay mostly unaffected due to its stiffness and strong bonding, but anything not very firmly attached to its surface will be flung away.
2 - Prolonged exposure to high temperatures
Common glass is thermally stable and can be heated strongly before it softens, so it can take temperatures at which many other substance decompose, evaporate or burn (if exposed to oxygen). Different types of glass have different maximum temperatures depending on their composition. Also, the thermal expansion while heating and contraction while cooling can be problematic for delicate glassware.
3 - Soak in base bath
Strong bases react with many substances and help dissolve them off the surface of the glass. However, while very inert, glass itself will also slowly react with concentrated basic solutions at room temperature. This also helps clean because if very persistent stains manage to survive the basic conditions, the glass surrounding them will begin dissolving slightly, and after a while the stain ends up detaching from the substrate as the glass behind it will have dissolved away. The regular options are concentrated $\ce{NaOH}$ or $\ce{KOH}$ in water or ethanol.
4 - Soak in highly oxidizing solutions
Glass is composted mostly of silicon already in its most oxidized form ($\ce{SiO2}$). Hence exposure to very oxidizing conditions will not damage the surface of the glass, while anything else on it is likely to dissolve, be it organic or inorganic matter. Classic options for dipping the glass are acidic piranha solutions (a mixture of pure $\ce{H2SO4}$ and concentrated $\ce{H2O2}$), basic piranha solutions (concentrated aqueous $\ce{NH4OH}$ and concentrated $\ce{H2O2}$), sulphocromic solution ($\ce{H2SO4}$ + $\ce{H2Cr2O7}$, in disuse due to the toxicity of $\ce{Cr^{6+}}$) and sulphonitric solution ($\ce{H2SO4}$ + $\ce{HNO3}$). These are all pretty heavy-duty mixtures. For example, here's why piranha solutions earn their name.
5 - Treatment with ozone/oxygen plasma
This is a step up from even the piranha solutions. If all you want to do is oxidize organic matter on the surface of the glass, then why not cut out the middleman and directly bathe the glass in reactive oxygen? The glass sample is first put into a closed chamber containing oxygen (air works, but pure oxygen is better). If ozone is being used, then the chamber is lit from the inside by a mercury lamp, which produces lots of ultraviolet light. The light is capable of dissociating oxygen molecules and producing ozone, which then is directly exposed to whatever is on the surface of the glass. In the case of oxygen plasma, the chamber is pumped down to a low pressure of pure oxygen gas before a coil produces strongly varying electric and magnetic fields inside the chamber, generating all sorts of oxygen species such as singlet oxygen (an electronically excited form of standard, triplet oxygen), free oxygen atoms, and several ions and radicals. Here's a nice video on how this process works in practice.
After all the work you put into cleaning a glass surface, it will immediately start adsorbing impurities if outside a vacuum, and it may take as little as a few minutes before it is too dirty to use in very sensitive and specialized applications. I have found that water already beads significantly again less than 10 minutes after the final treatment.

I am not that familiar with cleaning procedures, but the 'rule' to support a uniform film of water without droplets seems to be an easy proxy for checking whether there are still contaminated spots on the surface. In that case, a phenomenon called contact line pinning will occur, which results in droplets. The uniform film will only be possible if there are no contaminated spots.
The fact that there are droplets however doesn't necessarily indicate that the glass is not clean, because it can also be a result of just too much water being present. The maximum thickness of the uniform water film is governed by a balance between gravity and surface tension as $\sqrt{\frac{\gamma}{\rho g}}$ and for water is approximately 2.7 mm (see e.g. capillary length or the great book by de Gennes on capillarity and wetting phenomena).
If the water film becomes thicker then the 2.7 mm it will start to form droplets as well. So the 'cleaning rule' is only useful if you have removed sufficient water from the surface. In the video I'm quite sure that they have not removed sufficient water yet, so the rule cannot be applied.
In your case you are holding the glassware probably such that water starts to drain from 1 side to the other (due to gravity), at the moment that the layer at the bottom exceed the capillary length you will start to see droplets.
An additional point to consider: drying a water-wet piece of glassware in an oven only yields a clean surface when you have used deionized water (as they do in the video). If you use normal tap water then you will have some calcification due to the salts in the water that end up on your glass surface.
So to try and answer your questions based on these considerations:
i) Yes, there is a time range if you have gravitational drainage of the film. For anything that is not completely horizontal you will see this.
ii) Water concentrating is not the issue as long as it is deionized. If you really want spotlessly clean glasses: use deionized water after the whole cleaning procedure and just dry the glass in a warm place. If you don't have deionized water: you can make reasonably well deionized water by boiling tap water and catching the steam in a clean cup.

Thanks for your answer. If too much water caused the droplet, why in that youtube video, the droplet did not form in the bottom of the flask?

@user26143 You are right that there is a droplet at the side which suggests pinning thus a non clean surface. I do think that this droplet is on the outside of the flask, not on the inside, but that is hard to see.

"Clean" depends on what you consider to be dirt.  detergent leaves a spotty adsorbed layer of detergent.  "Clean" glass by removing its surface. Begin by having a lab coat, chemical gloves, and a full face shield.  If it gets on you it will dig holes.  Instant blindness in an eye.
5-gallon plastic (not vinyl!) bucket with a cover, 4 gallons of biological denatured ethanol, pound jar of of potassium hydroxide pastilles.  Shove crushed ice into the jar with mixing to form a viscous slurry (exotherm!).  Pour with mixing into the alcohol (probably have to do this twice to get all the KOH dissolved).  This is a KOH bath.  (Standard denatured alcohol will react to go red and stinky, but still work.)
Soak Pyrex in a KOH bath overnight to remove everything except hydrocarbon grease and metal oxides.  Soak soda lime glass for an hour or three, for it slowly dissolves.  Rinse with DI water, soak in 0.1 M HCl for an hour [#SiO(-) K(+) to #SiOH], rinse well with DI water, store totally submerged in DI water.  Water will perfectly sheet.  If the surface is allowed to dry, start again.  Never put your hand in there.  Wear long heavy gloves and positive eye protection.  Immediately, thoroughly wash off skin contact.
NEVER allow glass flats to stack flat under water.  They will bond (a micron of water layer is the hydrogen bonding adhesive).  Try sliding them apart without applying orthogonal pressure.
Chromate/conc. sulfuric acid) cleaning solution cover the glass with a layer of chromium(VI) species.  Things like NoChromix work, but you still have the conc. sulfuric acid.
Note added in proof: NEVER put graduated glassware in a base bath, for they lose calibration.  Frosted syringe barrels are iffy. Aluminum will fizz hydrogen, corrode, and disappear.

And just another reason why I'm glad I'm a computational chemist.


text:
How to calculate the frequency of cyclohexane conformational interconversion?
My book mentioned that the energy barrier for cyclohexane to reach the half-chair conformation is $50.6\ \mathrm{kJ\ mol^{-1}}$. It says that from this value, it was calculated that cyclohexane undergoes $10^5$ flips per second.
How did they calculate this? We can assume that all the conformational states possible will be of the same energy as the chair conformation (so they all encounter the same energy barrier).
I'm more interested in the principles behind the calculation – why the equation works – than just seeing the calculation done.

Look up application of the rate of reaction equation given barrier height and energy available at room temp, $kT$ (Boltzmann's constant).
$k = A \mathrm e^{ - \frac{E_\mathrm a}{RT} }$
http://research.cm.utexas.edu/nbauld/teach/cyhexconf.jpeg
Twist-boat leads to mathematically perfect geometric chirality,
Symmetry: Culture and Science 19(4) 307-316 (2008).


text:
Where does the 9th electron go in a N=O bond?

In the first resonating structure you can see 5 unpaired electrons and 4 shared electrons on nitrogen, then isn't this a extended octet? If it is so, then in which orbital that 9th electron is revolving?
In the second resonating structure you can see that oxygen has 5 unpaired electrons on nitrogen and 4 shared electrons, then isn't this a extended octet? If it is so, then in which orbital that 9th electron is revolving?
Can anyone explain completely about the above questions? If this question is already been asked please comment me the link of that question. Thank you in advance.
Please explain with boxes if you can.

Nitric oxide is a tough molecule to represent as a Lewis structure. However, you have made one common mistake in your structures. I also want to clear up a misconception about resonance that's present in your post. Finally, I will need to introduce a different bonding picture of $\ce{NO}$ to answer your question about which orbital the unpaired electron is in.

The Lewis Structure of $\ce{NO}$

The mistake you made in drawing those Lewis Structures is that you used all of the electrons in the nitrogen and oxygen atoms and not just the valence or outer-shell electrons.
A Nitrogen atom has 7 total electrons in the following configuration:
$$1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$$
Of these seven electrons, only the $5$ in the second shell (energy level) are available for bonding. These are the valence electrons. The $1s$ electrons are core electrons. They are not available for bonding because their probability density function maxes at less than the atomic radius.
Similarly, an oxygen atom can only use the six electrons in its second shell for bonding: $$2s^2 2p_x^2 2p_y^1 2p_z^1$$
The bonding in $\ce{NO}$ thus only has 11 electrons and not 13 (like you have). With two fewer electrons, it's easy to place the unpaired electron in a way that does not violate the octet rule.
$$\begin{aligned}
&.. \ \ \  .. &&\  &&&\ ..\ \ \ ..\ \\
\ce{.} &\ce{N=O:} &&\ce{<->} &&&\ce{:N=O.}
\end{aligned}$$

Resonance

The most common misconception related to resonance is unfortunately suggested by the name of the phenomenon. The two resonance structures drawn do not exist as separate structures inter-converting. They are attempts to represent a compound using Lewis structures that has bonding that is not localized. The true structure of nitric oxide is some hybrid of these resonance contributors.

Where is the unpaired electron?

To answer this question, we need molecular orbital theory. A molecular orbital diagram describes the bonding in terms of constructive and destructive overlap of atomic orbitals. A MO diagram dispenses with the need for resonance - all electrons are clearly shown in their orbitals: $\sigma$ or $\pi$, bonding or anti-bonding. Since this answer is running long already, I won't go into how to construct such a diagram.
Here is the MO diagram for nitric oxide:

The unpaired electron (highlighted in blue)is in a $\pi$ anti-bonding orbital that is formed from the destructive overlap of out of phase $p$ orbitals, one from nitrogen and one from oxygen.

Excellent answer as always, Ben. I have a slight nitpick about the molecular orbital diagram you found though; most references seem to have the relative energies of the $\pi_{2p}$ and $\sigma_{2p}$ switched. I found this excerpt of a book, which on page 379 has an excellent discussion of the electronic structure of $\ce{NO}$ using molecular orbital theory.

The creator of this image must have thought $\ce{NO}$ to be more similar to $\ce{N2}$ than to $\ce{O2}$. The increasing degree of $s-p$ interaction causes the relative position of the $\sigma_{2p}$ and $\pi_{2p}$ to switch between $\ce{N2}$ and $\ce{O2}$.

Don't forget to start at a lower energy for the oxygen atom orbitals, because of the higher electronegativity (whatever that is, right...). Because of that, the single electron is energetically closer to N than O.

what will be the hybridisation of NO molecule, is hybridisation is defined For NO like in ClO2???

Your Lewis structures contain too many electrons. A molecule of $\ce{NO}$ has 11 valence electrons (5 from nitrogen, 6 from oxygen). Your proposed Lewis structures have 13 electrons, so both of them wound up having one atom with 9 electrons to compensate.  With simple valence bond theory, the most accurate structure has a bond order of 2.5 resulting from the combination of two canonical structures, one with a double bond and another with a triple bond, and where both structures have the odd electron on the nitrogen atom.
Edit: As Ben Norris mentions in his answer, there actually is no satisfactory Lewis structure for $\ce{NO}$, which actually has a bond order of 2.5. The closest representation is a doubly-bonded molecule, and it is not possible to write a canonical structure with triple bond without going over the octet in one of the atoms. This happens because naive valence bond theory lacks the concept of antibonding bonds.


text:
Z:#Neutrons Ratio
Is there a fixed ratio that can be used to calculate the number of neutrons in the nucleus of the most stable isotope of an element by knowing the atomic number?
Side question: if Z is the symbol for atomic number, are there symbols for the number of neutrons and/or atomic mass?

The symbols $Z$, $N$, and $A$ are typically used to describe a nucleus:
$Z$ : atomic number (= number of protons)
$N$ : neutron number
$A$ : mass number ($A = Z + N$)
There is no fixed $Z:N$ ratio throughout the whole periodic table for stable isotopes, but a belt of stability in the Chart of Nuclides.


text:
Would 2 C₂ be 2 moles of diatomic carbon, or 4 moles of carbon?
Would molecules like $\ce{C2, H2}$, etc., be considered 1 mole of diatomic [element] or 2 moles of [element]?

You can have 1 mole of $\ce{C_2}$ which contains 2 mole of $\ce{C}$ just as you can have for example 1 mole of $\ce{C_2H_5OH}$ which contains 2 mole of $\ce{C}$, 6 mole of $\ce{H}$ and 1 mole of $\ce{O}$.
The point is that you always have to mention whether you are indicating moles of molecule or moles of atoms. That is why a statement like: "I have 1 mole of oxygen" is ambiguous, because it could be referring to oxygen gas (i.e. $\ce{O_2}$) or to atomic oxygen (i.e. $\ce{O}$).

C2 is available from your regular supplier, but it certainly exists, has been monitored spectroscopically and used in reactions, see my comment to this question

Of course I meant that it is NOT commercially available ;)

@KlausWarzecha good to know! Never heard about it.

$\ce{C_2}$ is the blue-green Swan band emission of comets' ion tails and the blue in gas flames. The big question is its electronic structure (presumably not $\ce{•C#C•}$).  Its paper synthesis in cryogenic gas matrix (UV cleavage) or in vacuum (flash pyrolysis or UV) seems to be easy, from
ROO-C(=O)-C#C-C(=O)-OOR, the peresters of acetylenedicarboxylic acid with R = $\ce{-CF_3}$, $\ce{-C(CH_3)_3}$, or $\ce{-Si(CH_3)_3}$.
Strings of alpha-omega diradical polycarbyne, $\ce{(-C#C-)_n}$ or $\ce{(=C=C=)_n}$ (your choice in LCAO, same in MO), in carbon arcs are suggested to pericyclicly close to fullerenes.

I'm not quite sure of how this answers my question. Could you explain a little further and use words a high school student could understand?

$\ce{C_2}$ is one mole of $\ce{C_2}$ and two moles of carbon atoms.  A box of eggs is one dozen eggs and 12 eggs simultaneously.  The fun is in the footnotes.  How can you make it on a bench top, what can you do with it?  The cold synthesis byproduct is squaric acid.  It's a hoot of a compound, including derived squarylium dyes,  http://link.springer.com/static-content/images/967/chp%253A10.1007%252F7081_2008_117/MediaObjects/978-3-540-79064-8_117_Fig1_HTML.gif

What an eccentric performance.

The diradical recombines with the two liberated alkoxy radicals to an acetylene diether. That antarafacially 2+2 dimerizes to the squaric acid precursor. Aqueous acid triiodide to oxidize and hydrolyze.  God save us from the congenitally inconsequential.


text:
Nearest branch numbering
If the longest carbon chain is symmetric numbering-wise (meaning the numbering schemes are identical both ways) and has all alkyl substituents, should one prioritize the largest alkyl group?

I am not sure what is your concern here, but if you are asking about naming a compound like $\ce{CH3CH2CH(CH3)(CH2)_4CH(C2H5)CH2CH3}$, I believe the priority is alphabetical (i.e ethyl>methyl), and hence this compound will be 3-ethyl 8-methyl decane.

Thanks - that's what I what I was looking for. I was confused about if one should take into consideration the size of the substituent.

If numbering from each end of the parent chain gives the same locants, the next criterion stipulates that lowest locants are assigned to the substituent that is cited first as a prefix in the name. This means in effect that alphanumerical order is used to establish the numbering when a choice is possible since alphanumerical order is used to establish the order of citation of substituent prefixes in the name.
On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g. If no decision can be attained by application of P-14.4(f), the lowest set of locants is selected according to P-14.4(g) in their order of their citation in the name.
For example, a correct systematic name is 4-ethyl-5-methyloctane (not 5-ethyl-4-methyloctane) since ethyl is cited first as a prefix in the name.

Another correct systematic name is 4-ethyl-5-propyloctane (not 5-ethyl-4-propyloctane). Again, the reason is that ethyl is cited first as a prefix in the name. The size of the substituent is not relevant in this step.


text:
Comparison of boiling points
I was comparing boiling points of $\ce{CHCl_3}$ and $\ce{CH_3Cl}$. According to me B.P. of $\ce{CHCl_3}$ should be higher due to it's higher molecular mass than $\ce{CH_3Cl}$, but the answer is opposite ,ie. B.P. of $\ce{CH_3Cl}$ $\gt$ B.P. of $\ce{CHCl_3}$.Why?

Molecular mass has NO pertinence to BP, water and ammonia.  Look up the BP of perfluorocyclohexane, hexafluorobenzene, and the hydrocarbons.  If I want to volatize iron, ferrocene.  If I want to volatize iron more, $\ce{Fe(hfac)_3}$ where the ligand is the bidentate enolate anion of 1,1,1,5,5,5-hexafluoro-2,4-pentanedione.  Big MW.  What are the MW and BP of $\ce{UF_6}$?  $\ce{-Si(CH3)_3}$ also confers volatility.
Boiling points are more about polarizability, intermolecular van der Waals and such. Bigger molecules are overall electronically floppy.  Benzene, toluene, xylenes – nice BP steps, but not because of MW. Look up $\ce{SF_6}$ bp.  $\ce{I(CF3)_7}$ boils near 0 °C.  Calculate that MW, and its vapor’s mass/liter at STP compared to dry air at 29 g/liter.

but the answer is opposite ,ie. B.P. of $\ce{CH3Cl}$ > B.P. of $\ce{CHCl3}$.

Really?
$\ce{CH3Cl}$ has a boiling point of −24 °C. It is a gas.
$\ce{CHCl3}$ has a boiling point around 61 °C.

Edit: Your argumentation is valid, there was probably just some confusion in the notes.


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How to determine Delta H from the heat of combustion?
The heat of combustion of 2-propanol at 298.15 K, determined in a bomb calorimeter, is -33.41 kJ/g. For the combustion of one mole of 2-propanol, determine (a) $\Delta U$ and (b) $\Delta H$
So I managed to determine $\Delta U$ by first finding the amount grams in one of 2-propanol and then using that number with 33.41 kJ/g...
I'm not able to determine Delta H as I'm not sure what to do... Please help and guide me.
Thanks!

Are you supposed to determine $\Delta_rH$ for the reaction (which is $\Delta_cH$ for the combustion of 2-propanol) or are you supposed to determine $\Delta_fH$ for the formation of 2-propanol?

If the first case, it might be hard.

The change in internal energy $U$ is equal to the sum of the heat transferred $q$ and the work done $w$. If we assume that only pressure volume work is done, then:
$$\Delta U=q+w = q-p\Delta V $$
The negative sign in front of $p\Delta V$ is to ensure that a decrease in volume is work being done on the system (system gains energy).
As a thermodynamic potential, internal energy is only easy to use if the pressure is constant (and $\Delta V$ is measurable), which is not true in a bomb calorimeter, or the volume is constant (so $\Delta V =0$), which is true in a bomb calorimeter.
At constant pressure: $\Delta U=q$.
Enthalpy is defined such that
$$\Delta H = \Delta U + \Delta(pV) = q - p\Delta V + p\Delta V + V\Delta p $$
$$\Delta H = q + V\Delta p$$
Thus, enthalpy change is only easy to use if the volume is constant (true in a bomb calorimeter) and the pressure change is knowable (you were not given that data) or if the pressure does not change (so that $\Delta p=0$).

The second case is simpler.

If you were asked to determine $\Delta_f H$ for the formation of 2-propanol, you probably are encouraged to assume that for the combustion reaction $\Delta_rH\approx \Delta_cU$ and go from there. Then, you will need a balanced equation (2-propanol + oxygen forms what?) and some standard enthalpies of formation for the products.
$$\Delta_rH \approx \sum_{products}{\Delta_fH}-\sum_{reactants}{\Delta_fH}$$


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Calculating approximate pH of polyprotic acids
When I took up ionic equilibria and titrations after a long break, I found it hard to solve the questions regarding pH calculations of polyprotic acds. Consider these two questions as examples:-

Find the approximate pH of a solution with $0.4M\text{ }\ce{HCO3-}$ and $0.2M\text{ }\ce{CO3^2-}$$K_{a1}=4\times 10^{-7}$,
  $K_{a2}=4\times 10^{-11}$
Find the approximate pH of the resulting solution formed after mixing $20ml$ of $0.1M\text{ }\ce{H3PO4}$ and $20ml$ of $0.1M\text{ }\ce{Na3PO4}$. Express in terms of $pK_{a1},pK_{a2},pK_{a3}$.

I have mentioned 2 questions here because I want to know the method to solve such type of questions and not these two questions here in particular. The problem arises when there is a polyprotic acid with hydrolysable salts. For the second case, the ongoing equilibria are the dissociation of the three acids and the hydrolysis of their respective salts. For the first one, there are just two dissociation equilibria(and the dissociation of water). Even by writing down the equilibrium expressions of these reactions, I am unable to reach the final expression with pH or pOH. And, all the solutions have to be approximate (using approximations like the negligible contribution of hydrogen from the dissociation of water, low value of the successive ionization constants) and not at all accurate.
I would appreciate if someone explained how to do such numericals, taking one of the above as an example (I already have the answers for the aforementioned questions. I just want to use them to illustrate the general principle behind these questions).
Let me know if my question looks too homework-like and off-topic.

You can always solve problems such as these by writing down the charge and mass balance equations, and the definitions for every acid/base/self dissociation constant for every relevant species in the medium. It takes a while but it is exact. Or do you want to use approximate techniques precisely so that you don't have to rely on the more lengthy procedure?

@NicolauSakerNeto The thing is, these have to be solved in less than a minute. The options given are also very approximate. Therefore, I need to use a lot of approximations. And I am not able to figure out equation-simplifying assumptions. I seem to have forgotten how I used to solve these a long time ago.

I would be hard-pressed to solve the second question in a minute while having to take all three acidic equilibria into account. A very coarse approximation is to consider that the first proton of $\ce{H3PO4}$ acts as a strong acid, quantitatively going to $\ce{H2PO4^{-}}$, and that  $\ce{PO4^{3-}}$ acts as a strong base, quantitatively protonating to $\ce{HPO4^{2-}}$. Now you only have to consider the equilibrium for $K_{a_2}$, and by the Henderson-Hasselbach equation you get $pH=pK_{a_2}$. You can then attempt to modify this equation to insert $K_{a_1}$ and $K_{a_3}$.

I think I got it. Write the acid dissociation equilibrium for $K_{a_1}$, then replace the term $[H_2PO_4^-]$ with an expression in terms of $K_{a_2}$, now replace the term in $[HPO_4^{2-}]$ for an expression in $K_{a_3}$. You get a "higher order" Henderson-Hasselbach equation: $pH=(pK_{a_1}+pK_{a_2}+pK_{a_3}-log\frac{[PO_4^{3-}]}{[H_3PO_4]}) / 3$. Assuming the ion concentrations are approximately equal, then pH is merely the arithmetic mean of the pKa's. Each order of magnitude of difference between the ions concentrations only creates a 0.33 error in the approximation.

@NicolauSakerNeto Although the second method of averaging all the three $pK_a$ is more satisfying, the first one seems to be the method which yields the answer in this case and in a similar one. I do not know why $\ce{PO4^3-=H3PO4}$ is not a good approximation. Thanks for the help.

For the first scenario, the Henderson-Hasselbalch equation is probably good enough. In this case, we assume the major species to be $\ce{HCO3-}$ and $\ce{CO3^{2-}}$ and we will use $K_2$ in the equation.
$$\text{pH}=\text{p}K_2 + \log_{10}{\frac{[\ce{CO3^{2-}}]}{[\ce{HCO3-}]}}$$
I was working on a more general approach using systematic treatment of equilibrium, and it was driving me crazy. I will try to come back to it for the second part of you question.

That's perfectly fine.I posted the question after I was driven crazy by trying to find a exact solution and then substituting the appropriate assumptions. I believe, we need to make the assumptions first and then formulate the equations.

I Think that is correct. For example, in the triprotic acid case, you end up with a term like $[\ce{H+}]^3 +K_1[\ce{H+}]^2 +K_1K_2[\ce{H+}]+K_1K_2K_3$ in the denominator, and for most acids $K_1K_2K_3$ is fairly small and can be ignored.

For the first question, Ben Norris has already provided a way to solve it.
Coming to the second part of the question, I'll suggest you to take a look at the following graph for maleic acid.
$\ce{H2M}$ denotes maleic acid, $\ce{HM-}$ denotes hydrogen maleate and $\ce{M^2-}$ denotes maleate.

You can clearly see that $\ce{H2M}$ and $\ce{M^{2-}}$ can never be simultaneously dominant in the solution.
Same is the case here.
$\ce{Na3PO4}$ and $\ce{H3PO4}$ react to form $\ce{Na2HPO4}$ and $\ce{NaH2PO4}$.
$$\ce{Na3PO4 + H3PO4 -> Na2HPO4 + NaH2PO4}$$
Now, you can calculate the $pH$ of this solution with the help of Henderson-Hasselbalch equation.
$$\mathrm{pH = pK_{a2} + log\frac{[\ce{Na2HPO4}]}{[\ce{NaH2PO4}]}}$$
Here, ${[\ce{Na2HPO4}] = [\ce{NaH2PO4}] = \pu{0.05 M} }$ in your case.
So, the pH of the resulting mixture would be $pK_{a2}$.
Image source:
Fundamentals of Analytical Chemistry by Douglas Skoog, Donald West, F. Holler, Stanley Crouch.


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Reaction of alkenes with bromine in presence of sodium chloride

Consider the following reaction
           $$\ce{CH3\bond{-}CH\bond{=}CH2 ->T[$\ce{Br2}$/$\ce{NaCl}$]?}$$

What would be the product(s) for this reaction? I am confused because there are two nucleophiles here, $\ce{Br-}$ and $\ce{Cl-}$. So, there are two alternatives for the second step. Let me explain what I am saying.
First of all, the $\pi$-electrons will attack an electrophile which in this case is a $\ce{Br+}$ which is obtained form polarisation of the $\ce{Br\bond{-}Br}$ and it is added to the less substituted carbon, which complies with Markovnikov's rule, forming the carbocation below:
 $$\ce{CH3\bond{-}CH+\bond{-}CH2\bond{-}Br}$$
Now the carbocation will be attacked by a nucleophile. There are 2 nucleophiles in this problem, as I have stated earlier.
So, my question is which one of them will attack the carbocation and subsequently what will be the product? Will it be only 1,2-dibromopropane or only 1-bromo-2-chloropropane or a mixture of both? Please give the reasons.

The first step is $\ce{Br-Br}$ approaching the pi-cloud, being polarized, loosing a $\ce{Br^{-}}$ to form a 3-cyclic bromonium cation.  Given solubility and reactively inert solvent, $\ce{Br^{-}}$ (large poor nucleophile) competes with $\ce{Cl^{-}}$ (smaller, better nucleophile) for ring-opening capture.  Now factor in relative concentrations.  Now factor in thermodynamics (reversibility, $\ce{Cl^{-}}$ wins for the stronger $\ce{C-Cl}$ bond) versus kinetics (smaller $\ce{Cl^{-}}$ vs. initial tight ion pair rapidly collapsing to product).  Temperature probably makes a product ratio difference, too.
If your linear ion intermediate is controlling, the $\ce{Cl}$ comes in at the 2-position by default.  If my cyclic ion intermediate is controlling, the $\ce{Cl}$ comes in at the 1-position by steric hindrance. And if you do it in water, you get halohydrin byproducts.
Keep the reaction way from light or you will get radical abstraction and allylic substitution.

Which product/reaction pathway is major? Bromine or chlorine addition?


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Unexpected reaction during acid hydrolysis of cellulose
Recently I tried to hydrolyze some cellulose:
I added some acid to it ($\ce{H2SO4}$) then noticed nothing. Even after a few days there was no visible change and after doing some tests for glucose, those were negative too.
So at the end - expecting nothing would happen - I added some sodium hydroxide ($\ce{NaOH}$) to the solution to neutralize the acid and then the most unexpected thing happened, the solution began to warm up and there was an immediate color change (turning from white (from the cellulose) to yellow/brownish) I suspect that color came from the dehydration of the glucose  that was produced at that moment due to instantaneous hydrolysis of cellulose. I finally diluted the solution with more water and the reaction stopped
So somehow there was no interaction between the sulfuric acid and the cellulose, but when sodium hydroxide was added, the solution did more than just neutralize the acid!
Does anyone have any idea why this is?

We miss some data to make a good hypothesis.

The concentration of the $\ce{H_2SO_4}$ solution.
The temperatures.

These are the main two factors that affect cellulose hydrolysis. Usually are used temperature above 200 °C and pressure above 25 MPa. For me its strange that the exothermic reaction between SA and $\ce{NaOH}$ could lead to an increase of temperature such that.  However some studies show that the reaction can occur at lower temperature with concentrated acid.
Another possible solution is the formation of alkali cellulose.

I used 50 grams of cellulose, then I added 55 ml H2SO4 solution (99%) and it was conducted at normal room temperature and normal air pressure And in the end I added about 10 sodium hydroxide pellets (1 pellet being around 1cm long)

@LievenB was very concentrate maybe you have done a partial hydrolysis using the heat provided by the reaction. Maybe the crystalline structure was broken by the neutralization reaction. I don't know...

and so the yellowish/brown colour is then because of further hydrolysis (by H2SO4) of the produced glucose?

@LievenB yes this for me is quite probable, search in youtube glucose and sulphuric acid and you will see a lot of yellow and black. However I never read about hydrolysis with 99% concentrated acid, you should use a more diluted solution in my opinion...

ok - thanks! I'm planning on doing some more experiments with the hydrolysis of cellulose by using sulfuric acid so I'll definitely use your advice!

Take all the precautions! Never add water to sulphuric acid but viceversa and avoid strong exothermic reaction without the right protection! Good luck!

Chromophoric bi-product of alkali hydrolysis on cellulose is called oxycellulose. It is this which will have produced the colour - like you get on old clothes. This acid will have made the cotton stiffer/brittle, but the colour will come from the alkali.

From the comments:

I used 50 grams of cellulose, then I added 55 ml $\ce{H2SO4}$ solution (99%) and it was conducted at normal room temperature and normal air pressure, and in the end I added about 10 sodium hydroxide pellets (1 pellet being around 1cm long)

So unless I am missing something, in an attempt to hydrolyze cellulose, no water was added to cellulose powder.  The exo hydrolysis of cellulose is
$$\ce{(C6H10O5)_{n} + H2O -> C6H12O6 + (C6H10O5)_{n-1}}$$
Since water is a reactant, adding it is probably required to make the reaction work!
These are some hypotheses to consider:

Mostly anhydrous sulfuric acid and cellulose probably did not react, although perhaps some solvolysis to form sulfated glucose monomers may have happened.
When NaOH was added, water was formed nearly instantly by reaction with the acid, and the liberated heat strongly heated the reaction.
It doesn't sound like you added enough NaOH to neutralize the acid completely.    The reactions you observed probably happened in acidic conditions.

I think the strong heat and water formation that happened led to simultaneous hydrolysis and dehydration of the cellulose, leading to a complex mixture of hydroxymethylfurfural and further polymerized/carbonized derivatives.
I'm a little surprised that the 99% sulfuric acid you used did not carbonize the cellulose eventually even without water.  It's a very powerful dehydrating agent.$%edit$

Aside from the exotherm... Repeat the experiment with warming.  Glucose can air-oxidize, removing the aldehyde with which glucose tests react,  In fact, you can get the peracid from aldehyde free radical air oxidation, ruining reductive glucose assays toward any ring-opened glucose present.  Glucose slow hydrolytic release in a vessel exposed to air can be tested by doing it in seltzer as the reaction medium, then close the jar or cap cap after fizzing subsides, thus purging solution and containment with carbon dioxide.

I don't understand quite well what you are saying, could you please clarify? (I haven't learned yet about peracids,....)

R-C(=O)-H + •O-O• = R-C(=O)• + •O-O-H = R-C(=O)-OOH Oxygen is a triplet diradical.  It abstracts the aldehyde hydrogen, flips around, and bonds.  A peracid has a carbonyl bonded to -OOH not to -OH.  Peresters are with peroxides R-00H not with alcohols R-OOH.  The Alpha effect makes peroxide anions strong nucleophiles.   Absence of steric hindrance promotes fast kinetics.


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My sample of gallium does not melt
I held gallium under a heat lamp. It didn't melt. I didn't even find a thin film of liquid gallium. Can anyone explain why? Does gallium rust?

Where did you get it? Are you sure it's gallium? Gallium does rust slowly in air I think, forming an oxide which doesn't melt nearly as easily, but I don't think that would be the case. Try dropping it in some almost boiling water to be sure it doesn't melt.

It takes quite some time to get gallium to melt. Even my small sample (in a plastic bottle) needs to be kept on a warm (>50 celcius) household radiator for 15-30mins to show much melting.

Having recently had an opportunity to play around with some gallium of my own, I have a few thoughts.
It's possible you have a different metal than what you thought you had, but frankly, I don't think that's very likely.
It takes a surprising amount of absorbed heat to melt gallium. Putting a sizable chunk of gallium in $35~^\circ\mathrm C$ water doesn't cause it to melt immediately. Even boiling water doesn't melt it instantly. Because gallium is a metal, it conducts heat fairly well, so when heated gently, it won't melt until the temperature of the whole piece gets close to the melting point.
As mentioned by user5134, a shiny metal like gallium isn't going to pick up heat very quickly from a radiative source like a heat lamp. If you wait long enough, it'll definitely melt. If you want to melt the gallium relatively quickly, then you'll want to put it into physical contact with something warm. The standard method for melting solid gallium in a container is to just immerse the closed container in hot water (not boiling water, especially if the container is glass).

Gallium metal may be obtained in the elemental form.  A classic demonstration is to hold gallium in your hand and watch it melt.
http://astounde.com/wp-content/uploads/2012/06/gallium.jpg
The energy from the heat lamp must be absorbed by the gallium, but if the IR light is reflected, then it may not be getting hot enough to melt.
It's also possible that you have an alloy of gallium and some other metal and that the melting point is higher than for pure gallium.

Maybe it is another metal.
If you are sure it is gallium try melting it with your hand, smaller pieces melt faster.
Gallium oxide layer is too small to make a difference, it takes hundreds of years for it to completely rust.
It is very important to note that storing it in a metal container will ruin your sample and that storing the gallium while it is a liquid in a glass container is dangerous because it expands when solidifying.
You can also measure gallium' density around 6 g/cm3 with a water cup placing the gallium and seeing how much volume it increased and weighting it (remember that gallium amalgamates metals so measure them with not metal tools.
Divide the weight in grams with the volume and you should get a number lower than 6 and higher than 5.7.

Elemental gallium does not occur in free form in nature, but as the gallium(III) compounds. So it mustn't be gallium but something else.

The asker may have attempted to purchase metallic gallium online. Whether they received what was expected is another matter, though.

Likewise aluminum, iron, tin, zinc, sodium, magnesium, chromium, etc. Yet these metals can be obtained in pure form from their compounds.


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Are some alkali metals better conductors than others?
I want to know if there is any difference at all between the alkali metals as far as conducting electricity goes.

The answers show that the resistivities/conductivities don't vary monotonically. I wonder if anyone can give a good explanation why.

Li
Conductivity 1.08×10^7 S/m
Resistivity  9.26×10^(-8) mΩ
Na
Conductivity 2.18×10^7 S/m
Resistivity  4.59×10^(-8) mΩ
K
Conductivity 1.43×10^7 S/m
Resistivity  7.0×10^(-8) mΩ
Rb
Conductivity 8.0×10^6 S/m
Resistivity  1.25×10^(-7) mΩ
Cs
Conductivity 5.3×10^6 S/m
Resistivity  1.9×10^(-7) mΩ
Fr
Conductivity 1.4×10^6 S/m
Resistivity  7.1×10^(-7) mΩ

Here is the total resistivity ρ (in $10^{-6}\ \Omega{}\,\textrm{m}$ at 273.16 K) from Electrical resistivity of alkaline earth elements,
T. C. Chi, J. Phys. Chem. Ref. Data, 1979, 8, 439-497:
Li  8.53
Na  4.33
K   6.49
Rb 11.54
Cs 18.75
Fr 28.00


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Are ionic bonds stronger than covalent bonds?
A covalent bond involves overlapping of orbitals while an Ionic bond involves charge separation.
Why are bonds formed by the overlapping of orbitals weaker than charge separation; why is an ionic bond stronger than a covalent bond?

This isn't always true - for example the $\ce{H-H}$ bond has an enthalpy of 436 kJ/mol whereas $\ce{NaCl}$ has a 'bond energy' (lattice enthalpy divided by six $\ce{Na-Cl}$ 'bonds' for each Na or Cl atom) of 131 kJ/mol.

In a fully covalent bond, you only have the mixing of the orbitals, as you said. But it's not only that, you also have coulomb interactions between the nuclei, that slightly raise the energy of the molecular orbitals, which results in asymmetric energies.
In a fully ionic bond you also have a coulomb interaction, but this time with the different sign, because you have a positive and a negative charge. These want to be as close together as possible, resulting in a stronger bond.
Generally speaking: Coulomb interactions have a greater impact than molecular orbitals on the energy, and thus the strength, of a bond.
It's almost like saying that electromagnetic forces are stronger than gravitational forces.

Thank you, that was so well explained, would you suggest me a good book for chemistry, I have a very bad chemistry teacher...

There are many books around, I suggest browsing the science section at the next university bookstore and finding out which type of book you want to study things with. I use this one, it covers most of the stuff that I learned in my first year at university.

If ionic bonds are stronger, then why in a salt solution, it's the salt's ionic bond that separates, but water retains most of its bonds despite having covalent bonds?

Two atoms will always (with a few corner-case exceptions) form the strongest bond between them that they can.
If they form a covalent bond then that is because the covalent bond is stronger than the alternative ionic bond (at least after taking into account the ionisation energies). This is either because the covalent bond is strong (good orbital overlap) or the ionisation energies are so large that they would outweigh the ionic lattice enthalpy.
If they form an ionic bond then that is because the ionic bond is stronger than the alternative covalent bond. This is either because the covalent bond is weak (poor orbital overlap / mismatched orbital energies) or the ionisation energies are relatively small compared to the lattice enthalpy.

This is interesting, but does not explain whether covalent bonds or ionic bonds are in general the strongest.


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cooling/refrigeration essentiality for carbonated softdrinks
How does cooling/refrigeration help in maintaining the taste and shelf life of food products, especially CSD (Carbonated Soft Drinks)? For example, chilled drinks have a nicer taste than a normal drink and may have longer shelf life. Why is that?

Have a look at this, this and possibly this. Then come back and specify your question a little more, as it is too broad at the moment.

Welcome to Chemistry.SE. I incorporated info from your comment into the question and cleaned it up a little. This question may still be too broad.

reasons for the cooling/refrigeration is to -ensure that the carbonic acid in the drink hasn't all decomposed (=no bubbles). Keeping it cold slows down the bacteria reproduction (=longer shelf life) and as for the nicer taste - it tastes more fresh when its cooled

It has a longer shelf life because the bacteria, (which is the reason it will go bad), cannot reproduce as fast in cold temperatures.
As far as taste goes with carbonated soft-drinks, I think it is mostly just about preference.

Sodium benzoate (which is in a lot of drinks), or other preservatives, prevent bacteria growth, so the drinks don't need to be chilled.

Refrigeration condenses the CO2 and allows for a "softer" bubble feel in the mouth.
As for shelf life of soda, it should not effect a "canned" soda one way or another.


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Why does the position of equilibrium depend only on the relative Gibbs energies of the products and the reactants?
The position of equilibrium, I read, is dependent only on the relative Gibbs energies of the products and the reactants. Why is this so? Surely one must take into account the entropy of mixing of the products and the reactants, which wouldn't be dependent on the Gibbs energies, to work out the lowest point (energy wise) and find the position of equilibrium. Why, therefore, is the position of equilibrium between products and reactants only dependent on the relative gibbs energies of the products and reactants.
Thanks!

The entropy of mixing of what exactly?

If $\Delta G = \Delta H - T\Delta S$, then there is a Gibbs energy of mixing that could in theory be factored into the maths. However, see tschoppi's answer.

I quote Wikipedia:

In thermodynamics the entropy of mixing is the increase in the total entropy when several initially separate systems of different composition, each in a thermodynamic state of internal equilibrium, are mixed without chemical reaction by the thermodynamic operation of removal of impermeable partition(s) between them, followed by a time for establishment of a new thermodynamic state of internal equilibrium in the new unpartitioned closed system.

It seems to me that the scenario which you describe is an already mixed solution of reactants. So you have already maximized the entropy, because for every molecule that changes according to the reaction, you get the new molecule at the same place. The overall entropy with respect to the mixing doesn't change.
Note that the position of the equilibrium is only then dependent on the relative gibbs free energies when the reaction is not kinetically hindered (when the activation energy barrier is low enough to be overcome).

If I have reactant A and product B. At 100% A the Gibbs Energy is the gibbs energy of A. If at 100% B gibbs Energy of B. But in the middle, we have entropy of mixing A and B, causing more "disorderliness", and therefore more entropy. Also the relative enthalpies of each substance has to be taken into account. Why, therefore, does it depend only on the relative Gibbs Energies, when the entropy of mixing is not dependent on them?? Thanks for the additional detail at the bottom! :)

The last paragraph does not make sense to me. When the reaction is kinetically hindered, it does not attain equilibrium, but the position of equilibrium is still defined by $\Delta_\mathrm{r} G = 0$. If you can find the appropriate catalyst, the reaction will proceed to equilibrium.

In this answer, there is a good explanation how the entropy of mixing relates to reactions not going to completion.

Why does the position of equilibrium depend only on the relative Gibbs energies of the products and the reactants?

The short answer is that the approximate entropy of mixing is a simple function of the reaction quotient Q, so it is implicitly considered when calculating the equilibrium constant from $\Delta_\mathrm{r} G$.

Surely one must take into account the entropy of mixing of the products and the reactants, which wouldn't be dependent on the Gibbs energies, to work out the lowest point (energy wise) and find the position of equilibrium.

Yes, the entropy of mixing is important in describing an equilibrium and figuring out the equilibrium constant. Without the entropy of mixing, all reactions would go to completion, see Why does equilibrium exist? and Why can't a reaction go to completion?.
The Gibbs energy of reaction has an enthalpy and an entropy component:
$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} H - T \Delta_\mathrm{r} S$$
The enthalpy does not depend on concentrations (or reaction quotient Q), but the entropy does:
$$\Delta_\mathrm{r} S = \Delta_\mathrm{r} S^\circ - R \ln Q$$
The term $R \ln Q$ is the entropy of mixing (relative to everything being at standard state, i.e. $Q = 1$). Having both reactants and products present results in a higher entropy than having just reactants or just products.
Where is the mixing?
There is no actual mixing, just making reactants or products, but the expression is the same. You could do a thought experiment where the reaction proceeds via an enzyme that takes up reactants from one container and puts them into another (so reactants and products would never be in the same container). In a second step, you would mix the two containers so that reactants and products are in the same container.
Equilibrium and the entropy of mixing
As a result of the concentration-dependent entropy term, the Gibbs energy of reaction changes (increases) as the reaction goes towards equilibrium. When it reaches equilibrium, it is zero. The expression for $\Delta_\mathrm{r} G$ is:
$$\tag{1}\Delta_\mathrm{r} G = \Delta_\mathrm{r} H - T \Delta_\mathrm{r} S$$ $$ = \Delta_\mathrm{r} H^\circ - T (\Delta_\mathrm{r} S^\circ - R \ln Q)$$ $$ = \Delta_\mathrm{r} G^\circ + R T \ln Q$$
At equilibrium, $Q = K$ (that's the special name we give to $Q$ for a set of concentrations that is at equilibrium) and $\Delta_\mathrm{r} G = 0$. If we make these substitutions in equation (1), we get:
$$\tag{2}0 = \Delta_\mathrm{r} G^\circ + R T \ln K$$
Solving for $\Delta_\mathrm{r} G^\circ$, we get the well-known relationship between standard Gibbs energy of reaction and the equilibrium constant:
$$\Delta_\mathrm{r} G^\circ = - R T \ln K$$
The left-hand side does not make any reference to concentration dependence (it refers to a single state, the standard state), whereas the right-hand side contains the concentration-dependent term (entropy of mixing times temperature, again compared to the standard state where Q is one).


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Predicting bond-strength of metal carbonyls
The metal carbonyls (and similar organometallic compounds) involve a combination of sigma bond, a pi bond and backbonding. The bond strengths under consideration are the metal-carbon bond and the carbon oxygen bond. Here is the original question:-

Predict the order of $\ce{C-O}$ bond strength in the following:-
I) $[\ce{Mn(CO)_6}]^+$
II) $[\ce{Cr(CO)_6}]$
III) $[\ce{V(CO)_6}]^-$

All of these are isoelectronic, and hence whatever the difference of bond strengths, must arise from the difference in the resultant nuclear force of attraction. Since $\ce{Mn+}$ has the highest charge density, I would assume that the $\ce{M-C}$ bond strength of the manganese compound is the strongest (vanadium being the weakest) and therefore, the $\ce{C-O}$ bond strength must follow the opposite order, as the strengthening of the metal carbonyl bond should weaken the carbon oxygen bond.
But the answer given is III<II<I, exactly the opposite of my prediction. Why is it so? A relevant factor which I didn't consider is the back bonding, but I am unsure of how to integrate that into my prediction. Also, how would we make a prediction if the given compounds were not isoelectronic?

CO bond strength is reciprocal to the extent of back donation involved in the synergic bonding. The correct answer to the question is II<I<III.

As the π-antibonding orbital of CO is filled by electrons from the metal, they weaken the C-O bond compared with free CO.  The metal-carbon bond is strengthened in kind. That's what the books says.

Considering that, how can we predict the order of bond-strength?

It is due to negative charge on vanadium. Vanadium in this comples ion is negatively charged and thus has some extra electrons readily available for donation to the $\pi$-anti bonding molecular orbitals of carbon in $\ce{CO}$ molecule. Thus $\ce{V-C}$ bond in $\ce{[V(CO)5]-}$  is much stronger as compared to other two and hence $\ce{C-O}$ bond order is least in this among the three. If this might help. This what my teacher taught me.

The structures are isoelectronic so the negative charge does not signal that "extra" electrons are available.


text:
Salinic bridges in voltaic cells
I'm doing a study a study on voltaic cells.
I've been working with them for a while and one of the questions that I've had is how does the salt bridge affect the cell. I mean, I know it allows ions to move and that it'a a basic component of the pill. But, does its' concentration affect the voltage of the cell itself?
Thank you.

A n ideal salt bridge just eliminates liquid-liquid junction potential without affecting the equilibrium of the cell. A non ideal salt bridge changes the observed voltage depending on the relative mobilities of the ions in the salt bridge solution.

A cell has an internal ohmic-equivalent resistance.  The salt bridge is a resistive path for the flow of ions.  Ions, plus their hydration shells and possible coordination shells, have low drift velocities through a medium.  The salt bridge can constrain the ion current of the cell and its material equilibrium rate.  A poor salt bridge by length, diameter, concentration, and permeability will increasingly adversely affect a cell's current and voltage as discharge rate increases.

I don't think it affects the voltage of the cell as :
E°(cell)= E°(red,cathode) - E°(red,anode)
so it doesn't depend on the salt bridge !

Theoretically it does not, although many factors ensure the observed voltage is lower than the theoretical voltage including the resistance of the salt bridge.