text:
Assumptions made in calculating melting point from thermodynamic data
I'm fine with doing these questions, such as:
For ammonia, enthalpy of fusion is 5.65 kJ/mol, and entropy of fusion is 28.9 J/K.mol
Hence find an approximate melting point:
Well $\Delta G = \Delta H - T\Delta S = 5.65*10^3 - T(28.9)=0$ Which yields $T=196K$. This is quite close to Google's value.
What I'm thinking, though, is that we shouldn't be able (in real life) to determine a melting point so easily, because $\Delta_r H$ and $\Delta_r S$ both vary with temperature, so in reality this isn't as simple as solving a linear equation for T?
I'd just like some confirmation here, as my textbook isn't giving much away on this point :)

Enthalpy and entropy of fusion cannot vary with temperature really. They can vary with pressure, in which case the melting point changes. ;)

In real life the melting point is determined by measuring the temperature. The entropy of fusion is usually determined by dividing the enthalpy of fusion by the melting temperature.
$$\Delta S_f = \Delta H_f / T_f$$
Your text book calculation just solves this equation for $T_f$ to recover the temperature that probably had been measured before.
Note that for constant pressure the temperature is constant $T_f$ as long as the phase transition takes place.

Ok, I think there's something I left out from my question: I don't know what conditions the thermodynamic data is quoted at. For instance, the entropy and enthalpy of fusion could have been at 298K, quite different from reality.
If they were quoted at melting point, then I'd have no problem with solving for T. But what if they aren't?

By 'enthalpy of fusion at 298K', I mean hypothetical values for 298K (although the process can't occur at 298K) - values achieved by applying Hess's law

What the respective free energies of the solid and liquid are doing away from the melting point is irrelevant - you know they aren't equal. At the melting point, the two must be equal, so the enthalpy and entropy of fusion must balance each other out in the manner you describe. There is nothing magic about it. Now, determining them experimentally is the hard part, but once determined the melting point is implicitly defined by them.


text:
Effects of Groups on the Nucleophilcity of Nitrogen

The reactivity of nitrogen mustards increases with increasing
  nucleophilicity of the central nitrogen atom. Select the most and least reactive from each of the following groups of nitrogen mustards.

My Attempt
I know that II will be the most reactive since the methyl group will donate electron density to the nitrogen atom due to inductive effects, making it more nucleophilic and hence more reactive.
I also thought that III will be the least reactive since the ester group pulls more electron density away from the nitrogen atom compared to the carbonyl group. I also thought in general that esters are stronger deactivators than carbonyl groups. However the answer is I.
After searching on the internet, it turns out that esters and carbonyl groups are roughly equally strong deactivators. So how am I suppose to determine which one will be less nucleophilic?

Your analysis of "pulls more electron density away" is a bit too simplistic. Both of them are approximately equally electron-withdrawing via the inductive effect, but in terms of resonance, which is the main factor here, they are quite different.
Compound I is an amide. Amides are non-nucleophilic on nitrogen because the N lone pair is delocalised into the C=O π* orbital. In fact, amides act as nucleophiles via the carbonyl oxygen and not the nitrogen.
In compound III, there is also the same delocalisation, but a way of looking at it is that the other oxygen (on the -OMe group) also wants to donate its lone pair into the C=O π*. In a sense this means that there is some degree of competition between the nitrogen and the oxygen, and the nitrogen lone pair is no longer delocalised to such a great extent.
Formally speaking, I think the best way of putting it is that the HOMO of compound III has a greater coefficient on nitrogen than the HOMO of compound I. You can also draw resonance forms to show that there is a smaller contribution from the resonance form with positively charged nitrogen.

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text:
Determine if an acid base reaction will occur
I'm wondering why some acid base reactions occur, let's say for example:
$$\ce{CH3CH2OH + H2O <<=> CH3CH2O- + H3O+}$$
Why does this reaction occur, because the alkoxide ion is a really strong base right?
Why should its conjugate acid give off a $\ce{H+}$ to water, if it was a strong base let's say $\ce{NaOH}$ it would make sense to me, because the $\ce{OH-}$ is probably a stronger base then the alkoxide ion.
Question:
Why does the above reaction take place, considering the fact that water is a really weak base?

Certainly some of the amine will be protonated. The extent to which that happens depends on the equilibrium constant.

so an acid base reaction will Always occur? @MaxW

$$\ce{CH3CH2NHCH3 + H2O <=> NH2(CH3)(CH2CH3)^+ + OH^- }$$

If you're asking generally then it's much to broad.

I narrowed my question down, hopfully you can help me now @Mithoron

You seem to have understood it, given that you have shown in the size of your arrows that the equilibrium lies mostly to the left hand side. What exactly are you confused about?

I would think the reaction won't happen at all, because the alkoxide ion is a really strong base so why would it's conjugated acid (CH3CH2OH) donate it's H+ to H2O at all? @bon

@RickBeeloo Every reaction is an equilibrium. Its just that for some reactions, the equilibrium lies far to the products or reactants side. Here, the equilibrium will lie far to the reactants side.

@RickBeeloo Don't edit your questions to make them into different questions. Ask new questions instead.

Yeah I want to but stackexchange says I can't because these 2 haven't been answered yet @bon

@RickBeeloo I'll give you an answer to this one later when I'm free.

OH- is not a stronger base than the alkoxide ion.

Ow ok, but if the alkoxide ion is so strong, why should it's conjugated acid undergo a reaction'? Because it don't want to lose it's proton? @orthocresol

It does but only to a very small extent, as bon already indicated in the answer.

The first thing you need to realise is that every reaction is in equilibrium, but some reactions have an equilibrium so far to one side that they are effectively complete (or they don't go at all). The position of equilibrium is determined by the relative stability of the products and the reactants. $$\ce{CH3CH2OH + H2O <<=> CH3CH2O- + H3O+}$$
The strength of acids can be compared by looking at $\mathrm{p}K_\mathrm{a}$ values. The $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in water is -1.7 and the $\mathrm{p}K_\mathrm{a}$ of ethanol in water is 16. So you are correct that hydronium is a much stronger acid than ethanol - or equivalently, ethoxide is a much stronger base than water. However, some of the water will deprotonate the ethanol and so the reaction will go to a small extent. This is why the equilibrium arrows are written with a big arrow on the reverse reaction and a small arrow on the forward reaction.
The equilibrium constant for the reaction is related to the standard Gibbs energy change for the reaction by: $$\Delta_\mathrm{r} G^\circ = -RT\ln K$$

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text:
Collective term for acid/base; hydronium/hydroxide
When writing about acids and bases it would seem very natural for me to abstract the set of both acids and bases and have a single noun for them, instead of cumbersomly listing them both each time. Compare anions and cations, where one can simply say 'ions' if one means both kinds.
Is there a established collective term? If not, what is the elegant way to speak about them?
About the same thing applies to hydronium/hydroxide. Is there a collective term? What about 'water ions'?

I've wondered the same thing, too, before...only I never thought of asking.

"Electrolytes" spring to mind, but are not exactly what you're looking for. "water ions" sounds good to me.

"Ionizable groups/compounds" could work

Wouldn't "ionizable groups/compounds" include salts in addition to acids and bases?


text:
Does clay absorb water because of polarity?
From what I have recently learned about ceramics, clay is essentially sheets of silicon and oxygen, bonded by covalent bonds. Between them, water keeps clay from forming a full lattice of $\ce{SiO4^{2-}}$. What I am wondering is, does clay absorb the water because of the water's polarity, i.e. are water molecules bonding with hydrogen bonds to the negatively charged oxygen atoms of the ceramic sheets? Or is water there because pressure from the environment forces it in between the sheets (which I know would then be a physics question)?

I am no expert regarding clay (which is a surprisingly varied field) and will therefore give a more general answer. I suspect that there is also quite complex interaction with ions on the surfaces of the layers and such things.
Materials can be generally categorized on a rather continuous scale, the extremes of which are called hydrophilic and hydrophobic (or lipophobic/lipophilic which are synonyms).
If something is hydrophilic, it "likes" to be wet, i.e. it is energetically favourable for the surface to be wet because this minimized surface energy of the system. The opposite applies to hydrophobic substances.
As you have observed yourself, clay is quite hydrophilic.
One can indeed estimate whether a material is hydrophilic based on its chemistry: water likes to form hydrogen bonds and water complexes (e.g. around Na+). Therefore things are more hydrophilic the more of said things they have. A good example is that sugars and carbohydrates are very soluble in water despite being large, whereas even small alkanes are barely soluble.
The explanation involving air pressure makes no sense, because to move matter there has to be a pressure difference. But if you imagine clay as a porous substance, the pores might as well be filled with air and there is an equilibrium. In fact (as easily observed when burning earthware) clay gets more voluminous with higher water content.

I was thinking in hydrophilic terms, yes. Your points are exactly why I was thinking that this could be the reason. By pressure, however, I never meant AIR pressure; clay forms (as I understand it) in water, and the thought was that the pressure from water (oceans, groundwater, etc.) pressed itself inside the sheets and simply stayed there when they were exhumed for use.

There still is no pressure difference; I either don't understand you or you have an erroneous understanding about pressure.

I assume there is no pressure issue, then. The thought was that the pressure of deep waters above could press water through the seabed or nearby layers, similar to the pressure of groundwater. I guess that is not the case.

Somehow you still sound confused, but I guess you have to add more information if that is the case. I did never figure out what exactly confused you :(

I perfectly understand your explanation, it's very well explained. But it is the basic premise. I just need to know if that premise applies to clay or not.


text:
Will 2-methylbutanal undergo aldol condensation
Will 2-methylbutanal undergo aldol condensation?
We are taught that it will not because the +I effect makes the carbanion unstable, but if we make an enolate ion, isn't it resonance stabilized by the C=O group?

There is no reason why the aldehyde cannot undergo an aldol reaction with itself. Enolates are not rendered unstable by the presence of additional alkyl substituents; in fact they are considered to be more stable. (See: formation of silyl enol ethers with $\ce{Et3N}$/$\ce{Me3SiCl}$.)
However, there is no way it can eliminate $\ce{H2O}$ to form the α,β-unsaturated ketone, since the aldol product does not have any α-hydrogens. Strictly you cannot call it an aldol condensation, although often no distinction is made in the literature between an aldol addition and a condensation.

The formation of the β-hydroxyketone above was reported as long ago as 1924, the reagent being potassium carbonate. More recently, in an attempt to make a 1,3-dioxolane from 2-methylbutanal with potassium t-butoxide and 2-chloroethanol, it was found that the enolisable aldehyde underwent "self-condensation" instead: Org. Lett., 2006, 8, 3745.


text:
Can heteroatoms with lone pairs be chiral centres?
If a compound has a carbon atom with four different groups covalently bonded to it, it is called asymmetric and enantiomers of the compound can exist.
But imagine if one has a different central atom, such as a nitrogen or a sulfur where one of the four "groups" is an electron pair (see examples below).

Would such a thing still behave like a "normal" asymmetric carbon? If not, how else does it behave, i.e. are there enantiomers? Is my assumption that those will still have $\mathrm{sp^3}$ hybrid orbitals wrong? I would also enjoy pointers to literature, I was unable to find any.

related http://chemistry.stackexchange.com/questions/16190/chirality-of-heteroatoms http://chemistry.stackexchange.com/questions/464/why-dont-trigonal-s-and-p-compounds-undergo-inversion-at-room-temperature

Generally, amine nitrogens will not behave like a normal asymmetric carbon. Simple amines are roughly $\mathrm{sp^3}$ hybridiized and the molecules you use as examples do have 4 (we include the lone pair of electrons as a substituent) different substituents around the central nitrogen atom. So in principle me might consider it asymmetric or chiral.  But simple amines can undergo a process called nitrogen inversion, which essentially coverts one enantiomer into the other.

(image source)
However, if one can find a way to slow down or eliminate the nitrogen inversion process, then chiral amines can be isolated. One way to achieve this is to incorporate the amine nitrogen into a 3-membered ring (an aziridine). Now to achieve the planar state necessary for nitrogen inversion requires the bond angle in the 3-membered ring to open from 60° to 120°, an impossible task. Aziridines containing a chiral nitrogen atom have been isolated and characterized.

The source for your image covers the exact system I was curious about. But it says it doesn't undergo nitrogen inversion. Is that true, or is the inversion simply of on consequence because the carbons are locked?

Actually you mentioned already that the planar state is the issue, so it probably doesn't invert. Not because the final energy of the inversion is quite high, but because the transition state energy is extremely high.

Thats's correct, the quinuclidine nitrogen being constrained in a rather rigid ring system also does not invert.

Strictly speaking, an aziridine can invert without opening the bond angle in the ring to $120°$.  The ligand attached to nitrogen could do all the moving.  But then you would have to distort the bond angles to the ligand.

Two cases are possible:

The nitrogen case with inversion frequently happening. Here, the compounds inverts via a planar transition state going from $\mathrm{sp^3}$ hybridisation to $\mathrm{sp^2+p}$ and back to $\mathrm{sp^3}$. Because of this, asymmetric nitrogen atoms that are not forced into place due to steric strain (i.e. as bridgehead atoms in 1,4-diazabicyclo[2.2.2]octane) are impossible.
The sulfoxide case. Sulfoxides $\ce{{R^1}-SO-R^2}$ have a free electron pair and a trigonal pyramidal configuration on the sulphur atom. The lone pair is in an orbital with a high s-contribution meaning that the nitrogen inversion highlighted above is rather complicated. Thus, sulfoxides can be assumed to be chiral on sulphur (although in practice they are often generated racemicly).

The case I called sulfoxide case (because that is generally the most well-known example) generally applies to all atoms with a single lone pair period three and up as the s-contribution to bonding orbitals decreases rapidly. Another common example of a heteroatom which can be a chiral centre is phosphorus, e.g. in phosphines $\ce{PHR^1R^2}$. Thus, the nitrogen inversion is actually more of a special case than a general rule.

In the case of N, the compound is chiral but not isolable as it inverts. (compare quinuclidine). The diastereomors are not resolvable.
In the case of sulfoxides, these are isolable and chiral.
In the case of Carbon.Carbanion intermediates are nonresolvable but neutral ground state carbon does not invert spontaneously.

There are some exceptions, but simple carbanions generally do undergo rapid pyramidal inversion leading to racemisation.

i wrote that in haste


text:
How is mean square displacement defined in Brownian motion?
I am trying to calculate mean square displacement (MSD) of water in a simulation to calculate its diffusion coefficient. However I am little confused on its exact form. Can anyone please explain it in little detail?
My confusion is: from wikipedia first paragraph on MSD page
$$MSD = 1/T * \sum_\text{t=1}^\text{T} (x(t)-x_\text{0})^2$$
My understanding of it was that:
$$MSD =(x(0)-x_\text{0})^2/1 + \{(x(0)-x_\text{0})^2 + (x(1)-x_\text{0})^2\}/2 + \{(x(0)-x_\text{0})^2+(x(1)-x_\text{0})^2+(x(2)-x_\text{0})^2\}/3$$
... and so on
That is that each term in MSD is cumulative sum of all previous terms devided by time. However most of online programms that i saw (for example matlab programme at the last page of this link)
tries to calculate MSD as  just simple square of distance between initial position and position at time T. Can some one please explain if it is a cumulative sum of previous displacements or not.

Try at physics.SE, it belongs there more.

I think you misunderstand the $\sum$ symbol and the fact that the term $1/T$ multiply the whole sum.
If you want to sum the integers between 1 and 3 and multiply this sum by 5
$$
S = 5 * (1+2+3)
$$
you can write
$$
S = 5 * \sum_{n=1}^3 n.
$$
If you now look at your espression to compute the MSD after a total simulation time $T$, you have almost the same expression. You sum the square distances after each time step $t$ and then you divide by the total time T the whole sum. For example for $T=3$ you have
$$
MSD(3) = \frac{1}{3}[ (x(1)-x_0)^2 + (x(2)-x_0)^2 + (x(3)-x_0)^2].
$$
In general this becomes
$$
MSD(T) = \frac{1}{T} \sum_{t=1}^T (x(t)-x_0)^2.
$$

The mean squared distance is a measure of deviation: how far away two things are.
Let's say we want to find the MSD of a particle from it's origin. What we do is find the square distance at specific time intervals, say every second. We then make a table of data:
$$\begin{array}{c|c|}
 \text{time} & \text{square distance} \\ \hline
 1 & 3.6 \\ \hline
  2 & 3.8 \\ \hline
 3 & 3.4 \\ \hline
 4 & 3.2 \\ \hline
\end{array}$$
The formula for the average is adding everything up and dividing by the number of things, so in this example:
$$MSD = \frac{3.6+3.8+3.4+3.2}{4}$$
What I've done above is stated more generally, and much more elegantly, as the formula you stated,
$$MSD = 1/T * \sum_\text{t=1}^\text{T} (x(t)-x_\text{0})^2$$
Your cumulative sum formula doesn't make intuitive sense, it instead finds the sum of the mean square distance at every time increment (there is no reason why you'd want to do that)

Thank you for your clear answer. Forgive me if my question seems too general or silly; but I want to know now that we have the MSD of "one single" particle in a random-walk environment, what is it useful for?

At least in the software I use, MSD is based only on the displacement vector from the initial position. It's computationally cheap and very good at differentiating between diffusion and Brownian motion. In the former you'll get almost always a smooth positive gradient whereas the latter would give random gradients.


text:
How to filter out magnesium acetate (a soluble salt) quickly and cheaply from water
I am conducting an experiment that involves reacting a weak acid and a weak base to form water and a salt with this overall equation
$$\ce{Mg(OH)2(s) + 2CH3COOH(l) <=> 2H2O(l) + Mg(CH3COO)2(aq)}$$
Currently, I am trying to filter out the salt to purify the water; however, there are specific restrictions.
The filtration process must be able to:

be done quickly. I can't wait a long time for the water to be distilled
be done at home. This filtration needs to be something that the average person can  could do with household ingredients or ingredients they can buy at a drug store (Assume that the $\ce{Mg(CH3COO)2}$ solution is given to them pre-made)

So far, I have tried to use Gelatin as a clarifying agent, which is typically used to clear beer. However, I couldn't find much info on using gelatin with salt water. Does anyone have any insight about using gelatin as a clarifying agent in this solution? Is there a better way to do such a thing that still meets the requirements?

I'm not sure what you're trying to accomplish, so I'm not sure if this will help, but have you considered ion-exchange? You can't filter something soluble, but something like ion-exchange resin is pretty easy to come by and is very fast and effective at removing soluble salts.

it seems like ion-exchange resin beads would work. Where could someone buy some of these beads? What type of resin would I use to remove this specific salt? Will it remove the Magnesium and Acetate ions completely or replace them with Sodium ions?

Actually, I found this pdf showing that I can use a cation and anion resin beads. All that remains is where I could find such a thing

Ion-exchange resin can be made with $\ce{H+}$ cations on one type of resin and $\ce{OH-}$ anions on another type of resin. These would leave the water without the any of the magnesium or acetate ions effectively without replacing the ions with other ions. This is how we deionize for general chemical use. This happens according to the reactions below:
$$\ce{2HSO3R + Mg^{2+} + 2C2H3O2- -> Mg(SO3R)2 + 2HC2H3O2}$$
$$\ce{R4N+OH- + HC2H3O2 -> [R4N+ +\ C2H3O2^{-}] + H2O}$$
In these examples, the sulfonates/sulfonic acids and ammonium cations are tethered to the resin so that they can't dissolve.
The downside to this is that if you are trying to measure something else in the system, if it is ionic, it will be removed as well.
I've never had to source ion-exchange resin myself, so I'm not sure of the easiest way to go about it. There's pleanty of deionization resin on Amazon. The mixed stuff is simple to use, just mix it with your water, agitate for a while and then filter it out and you'll have removed all the ions. That is expensive, though, because it's single use.
You can also buy regeneratable ion exchange resins. You don't want to mix the regeneratable stuff, and you have to do two separate ion-exchange steps, one with the cation-exchange and one with the anion-exchange. It's pretty easy to regenerate, though, as far as I know. You just use strong acid with the cation-exchange to redissolve the soluble cations and use a hydroxide solution to regenerate the anion exchange resin. It can't be done forever because there are some cations and anions that just won't want to redissolve, but if you're doing it a lot, might be worth it. I can't seem to find the anion stuff except in bulk.
I'm guessing the mixed type would be more appropriate for most applications. If you really need it quickly, I think you can buy the filters at Lowes/Home Depot and just run it through those or open them up and get the resin out.
Update: These filters are the only ones I saw that said they actually have the "dual ion-exchange system." Their website actually says it gets down to "000 total dissolved solids," which is pretty impressive relative to other water filters from what I've seen (depending on what "000 TDS" actually means).

Thanks a lot, that solves the finding the resin part. Also, at first your equation seemed confusing to me compared to what i found here, but I then realized that the water from the second equation comes from the hydroxide and hydrogen ions from the resin beads. I'm still confused on which type of cation and anion resin beads would be best to use, but i think It would be better to use one that doesn't need filtering, since the pitcher won't work well since every person who wants to replicate this has to spend $20 extra

If the pitcher is the only possible and/or most efficient way, it might work, but is there anyway to use the resins without filtration? Could I just use beads that exchange the ions (if they're not too hard to come by) and then filter them out with filter paper?

@Abob78 That's exactly what my students do, all they do is put the resin in a vial with the water, shake it up and decant it off. Works fine. The beads aren't like grains of sand or anything, they're pretty easy to get out. The "mixed bed" resin I linked above should work, but I just don't think its as readily available at a local store. Could be wrong, but I didn't find anything online. If I were in a pinch, I'd use the filter pitcher or, because I'm destructive, I'd gut the filter, see how much was in one and decide if it was practical to do that way.

Central Pennsylvania water is really hard, some of the hardest water in the country, apparently. These ion exchange beads soften the water to the point that we can't even get a reading for calcium or magnesium ions in the water by atomic absorption.

do you happen to know where I could buy ion exchange resin beads containing OH- and H+ in stores? I've found some online but they go for hundreds of dollars because they come in bulk amounts (more than 1 lb).

@Abob78 I can't find any in stores. This stuff was the best I could find. In-stores is tough, that's why I said about the filters and how I would just dismantle one if I needed it on short notice.

I specifically could buy them online, I'm talking about if a student wanted to do this at home. By in store, I meant something readily available. Most of what I've seen goes for hundreds of dollars in bulk meant for industrial use. If there's something cheaper and possible smaller scale, even if it's online, it would be great. They don't even have to be beads or reusable, I just thought that beads would just be preferred.

@Abob78 As I said, I can't find them in stores beyond what I've already told you.

Okay, then how much of the disposable mixed stuff would i need to get rid of, let's say, 14.2g (.1 moles) of Mg(CH3COO)2 in 360 mL [or grams] (.2 moles) of water?


text:
Why do ionic compounds have such high melting points and boiling points? Do they have IMAFs?
Molecules have IMAFs that determine the melting and boiling points. What about with ionic substances? What keeps the different compounds together? Do they experience IMAFs? I've heard that it's the ionic bond itself, but doesn't the bond remain after melting, and even if it doesn't then how come covalent bonds don't cause the boiling and melting points in molecules?
Sorry for all the questions, main ones are in the title.

What are IMAFs?

Do you know what an ionic compound looks like? Do you think, for example, NaCl is just a collection of NaCl molecules floating around held together by so-called "intermolecular" forces?

Obviously ionic compounds aren't even molecules so they can't experience IMAFs, the question was getting at what holds the particles together so that they're a lattice that doesn't want them to be liquids or gases?

The particles are positively and negatively charged. Positive attracts negative.

The total binding energy in $\ce{NaCl}$ is more than the sum of the binding energy of all the "pairs" of ions. It is because all the ions are interacting. See the wikipedia article on the Madelung constant. https://en.wikipedia.org/wiki/Madelung_constant

Molecules have IMAFs that determine the melting and boiling points. What about with ionic substances? What keeps the different compounds together? Do they experience IMAFs?

The problem is that you are trying to apply a paradigm that was developed for covalent compounds (the distinction between covalent bonds and intermolecular forces) to ionic compounds. They don't play by the same rules. In an ionic compound, all the ions are held together by electrostatic forces, which mean exactly the same thing as ionic bonds. There is no discrete molecular unit in an ionic compound. The word "molecule" does not apply.
Consequently, there is no such thing as an intermolecular force in an ionic compound. There are only ionic bonds.

I've heard that it's the ionic bond itself, but doesn't the bond remain after melting

Loosely speaking, it does not remain after melting. The reason why they have such high boiling points is therefore because of the fact that you have to break the ionic bonds for it to begin melting.

and even if it doesn't then how come covalent bonds don't cause the boiling and melting points in molecules?

Covalent bonds don't break during melting or boiling, with some exceptions (see: giant covalent compounds).

You say that in an ionic compound all the ions are held together by electrostatic forces. So in an NaCl lattice for example, every Na+ experiencse an Coulombic attraction towards every Cl- and this is the push against melting? What is the push against boiling then since the bonds don't remain in the liquid? Thanks a lot for all the time you've given me so far.

That's why I said loosely speaking. The Coulombic attraction remains in the liquid phase (it will always remain, even in the gas phase). You could think of it as: in the solid phase, the ions don't have enough energy so they remain stuck to each other. In the liquid phase, the ions have enough energy to "escape" each other momentarily, but they can't get sufficiently far away. In the gas phase, you've given them so much energy that they just fly everywhere, disregarding the attractions. Even this is kinda simplified though. http://en.wikipedia.org/wiki/Ionic_compound#Melting_and_boiling_points


text:
How is Al2Cl6 covalent and Al2O3 ionic?
I encountered a question where the type of bonding for various oxides are prompted. Two of which are $\ce{Al2Cl6}$ and $\ce{Al2O3}$.
I thought that chlorine atoms are more electronegative than oxygen atoms, so if one of them is ionic and the other is covalent, surely chlorine would be ionic, because it may be able to "tear off" the electrons from aluminium atoms.
This is evidently not the case according to the answer key of this question. I would like to know an explanation for the ionic/covalent properties of these bonds.

You thought wrong. Oxygen is more electronegative.

short answer: both have comparable ionic character, but chloride ions are large in comparison with oxygen, so ionic lattice with high coordination numbers cannot  form.

You are right that the difference in electronegativity is responsible for the ionic bonds in aluminum and oxygen, and normally that would be the case with chlorine too, but $\ce{Al2Cl6}$ is a special molecule.
The structure of $\ce{Al2Cl6}$ can be viewed as two aluminum atoms covalently bonded to four chlorine atoms each. Two of the chlorine atoms bridge between the two aluminum atoms. The following is a possible representation:

A special type of bonding called 3 center 4 electron bonding is present in the structure. These bonds result from a combination of a filled p orbital and two half filled p orbitals. This causes there to be a filled bonding and a filled non-bonding orbital. The bond orders between each bridging chlorine and aluminum atom are 0.5; the structure of 3 center 4 electron bonds makes it so that the bonding orbital is delocalized over both the bonds. $\ce{Al2Cl6}$ has two of these bonds and they are responsible for the covalent nature of the molecule.

The 3c-4e bond demands a linear geometry about the central atom. Because the bond arises from the head-on combinations of 3 p orbitals. Thus, I can't understand how it is the case here.

Note this is the structure of AlCl3 in the liquid and low temperature gas phases. In the solid it is a layered structure with octahedrally coordinated aluminium, see e.g. https://www.cs.mcgill.ca/~rwest/wikispeedia/wpcd/wp/a/Aluminium_chloride.htm . In Al203 aluminium is also octahedrally coordinated. Maybe they aren't so different ...

Sir, are these 3 centre 4 electron bonds stronger than the terminal bonds? (Couldn't find on the internet)

$\ce{Al2Cl6}$ is covalent as the electronegativity difference between Al and Cl is 1.5 which is less than 1.7.
$\ce{Al2O3}$ is ionic as the electronegativity difference between Al and O is  2.0 which is greater than 1.7.

$\ce{Al2O3}$ is ionic due to relative size of oxygen and aluminium and polarizing power of Al,
 (since we know that aluminium has a charge of +3, provides three electrons) in case  of $\ce{Al2Cl6}$ & $\ce{AlCl3}$ , it seems to be covalent due to similarities like banana bonding & bigger radius of Cl (in compression to oxygen). Banana bonding is not possible in $\ce{Al2Cl6}$ due to size of Cl atom . Radius of Cl even gets bigger when it forms an anion by receiving the electron from aluminium , The aluminium's cation being smaller than its original atom has a high polarizing power which attracts and distorts the electron cloud of the chloride ion (has a high polarizability) and forms a covalent bond  in case  of $\ce{Al2Cl6}$ & $\ce{AlCl3}$ ($\ce{AlCl3}$ form ionic for very shot time) . So is not in the case of $\ce{Al2O3}$, as the oxide ion's atomic radius is not large enough for the aluminium cation to distort it thus it stays in ionic bonding.

Firstly let's take $\ce{AlCl3}$, since we know that aluminium has a charge of +3, provides chlorine with three electrons (1 to each chlorine). The chlorine being in the second period has a larger atomic radius (in comparison to oxygen). This even gets bigger when it forms an anion by receiving the electron from aluminium. The aluminium's cation being smaller than its original atom has a high polarizing power which attracts and distorts the electron cloud of the chloride ion (has a high polarizability) and forms a covalent bond.
So is not in the case of $\ce{Al2O3}$, as the oxide ion's atomic radius is not large enough for the aluminium cation to distort it thus it stays in ionic bonding.
In short, $\ce{AlCl3}$ does form an ionic bond at first but it's a really small transition state for like nanoseconds, so it quickly forms a covalent bond by the polarizing process.

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


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How does pH level change the concentration of dissolved oxygen (DO) in sea water?
I am planning to investigate the relationship between $\mathrm{pH}$ and dissolved oxygen.
I would like to have a reason without taking in account the plants, animals and decomposer roles.
I want to know if $\mathrm{pH}$ affects the concentration of dissolved oxygen. For example:
If I added $\ce{HCl}$ to water, would the concentration of DO change?
In same way, if I add $\ce{NaOH}$, would the concentration of DO change?

In short: no, it does not, at least until you reach the truly extreme concentrations.

The equilibrium of dissolved-to-gas-phase oxygen is not affected by the aqueous hydronium ion concentration ($\ce{H3O+}$), and thus the pH of water does not affect dissolved oxygen concetrations of sea water.


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Why does NaOH catalyze H2O2 decomposition?
I had a lab practice in which I had to test the effectiveness of different catalysts for the decomposition of $\ce{H2O2}$. Among them, there was: $\ce{KI(aq)}$, $\ce{FeCl3(aq)}$, $\ce{MnO(s)}$, $\ce{NaOH(aq)}$, $\ce{HCl(aq)}$ and $\ce{I2(s)}$.
In different test tubes, I added some $\ce{H2O2}$ and some catalyst. After a while, I added $\ce{NaOH}$ as well. I observed an increase in the effervescence for $\ce{KI}$, $\ce{FeCl3}$, $\ce{I2}$ and $\ce{MnO}$. However, when the mixture was only $\ce{H2O2}$ + $\ce{NaOH}$, nothing happened. I've made some research, but can't find an explanation for this...
I believe $\ce{NaOH}$ increases the rate of the reaction of the other catalysts, but why?

According to THE HOMOGENEOUS BASE-CATALYZED DECOMPOSITION OF HYDROGEN PEROXIDE J. Phys. Chem., 1961, vol. 65 , pages 304–306:
$\ce{H2O2 + OH- <=> HO2- + H2O}$
$\ce{H2O2 + HO2- ->}$ [6-membered ring with 2 "H"s bridging 2 "O-O"s, with another H bonded to one of the "O"s] $\ce{-> H2O + O2 + OH-}$

It is clear that the rate will go through a maximum
  when the peroxide is 50% ionized, that is
  $[\ce{H2O2}] = [\ce{HO2-}]$.

So OH- catalyzes decomposition of peroxide, but a maximum is reacted when the amount of hydroxide added is half the amount of peroxide.  If more hydroxide than peroxide is added, the catalytic effect is lost.


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Sodium azide to Nitrogen ratio in airbags
I've been given this question and I don't know how to solve it.

Airbags are designed to protect the occupants of a car during a collision. They use the decomposition of sodium azide, $\ce{NaN3}$. Three chemical reactions occur rapidly after impact and produce nitrogen gas ($\ce{N2}$) to fill the airbag.
The equations summarise the overall process that takes place.

$\ce{2 NaN3 -> 2 Na + 3 N_2}$
$\ce{10 Na + 2 KNO3 -> K2O + 5 Na2O + N2}$
$\ce{K2O + Na2O + 2 SiO2 -> K2O3Si + Na2O3Si}$

What is the ratio between the number of sodium azide ($\ce{NaN3}$) molecules used and the number of nitrogen ($\ce{N2}$) molecules produced in the overall process?

I initially just used the first equation to get a ratio of 2:3, but obviously that's not right. I have no clue where to start. How do you do this? (How this can be considered a Year 10 question baffles me)

It baffles me also. You need also to consider the second equation, which produces N2 gas as well. The first equation generates Na, and this is consumed in the second equation.

Assuming you have an excess of $\ce{KNO3}$, the ratio of $\ce{NaN3}$ consumed to $\ce{N2}$ produced would be 5:8.
Let's renormalize the first two equations, dividing the first through by 2, and the second through by 10.  By doing this we can see, directly, that in the overall reaction, 3/2 + 1/10 = 8/5 = 1.6 molecules of $\ce{N_2}$ are produced for every molecule of  $\ce{NaN3}$ consumed.  If you prefer an integer ratio, it's 1:(8/5) = 5:8.

$\ce{NaN3 -> Na + 3/2 N_2}$
$\ce{Na + 1/5 KNO3 -> 1/10K2O + 1/2 Na2O + 1/10 N2}$

The initial ratio 2:3 is correct. $\ce{2NaN3 = 2 Na}$ molecules as well as $\ce{6 N}$ molecules. The products of the reaction, $\ce{2Na}$ is equal to two $\ce{Na}$ molecules and $\ce{3N2}$ is equal to $\ce{6 N}$ molecules. So as you can see the equation is perfectly balanced. Hope this helps.


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How can eosin be acidic and negative at the same time?
The broadest acid-base definition is done by Lewis and he states that a lewis acid is:
"a compound or ionic species which can accept an electron pair from a donor compound."
And all the sources say that eosin is an acid and a negatively charged molecule.
I think eosin cannot accept a pair of electrons if it's already negatively charged, thus, it should be considered as a base.
A clarification on this subject would be perfect.

There are many examples of negatively charged acids. Look at polyprotic acids. As examples with biological relevance, $\ce{H3PO4}$ is an acid, but $\ce{H2PO4^{-}}$ and $\ce{HPO4^{2-}}$ can also act as acids. Similarly, $\ce{H2CO3}$ is an acid, but $\ce{HCO3^{-}}$ can also act as an acid.

There is nothing unusual in being an acid and a base at the same time. Many organic molecules (for example, pretty much any peptide) are like that.

@NicolauSakerNeto That is true, but what proton do you see on that molecule that is acidic when the molecule is negatively charged? These things look more like Lewis acids to me (Michael acceptors, maybe, to be specific).

@SendersReagent do you mean the molecular structures of eosin? (at https://en.m.wikipedia.org/wiki/Eosin)

There are a lot of hydrogen atoms on aromatic rings. Can't they be ionized?

@m1clee Indeed I do. Aromatic protons are usually very acidic. I don't really see any in there that I would describe as acidic.

Aren't "Aromatic protons" the hydrogen atoms on ch=ch-ch=ch-ch=ch  benzene ring for instance? So there are aromatic protons? And you say that they are very acidic.  So I can't see any problem here.

By the way I am not a chemist and right now I am just trying to understand your question. So if you think I don't have some basic  knowledge based on my last two replies, I'd be very happy if you could adress them.

Geeez, there's only an bad wording on Wikipedia - the picture shows anions of the acid as they are important - it's an acid so it is present in dissociated form in diluted solutions.

And no those anions aren't acidic in water solution. I guess @Senders wanted to say "aren't usually". Also edited this article you're referring to. Check it as it is now.

@Mithoron If that's true biologists misuse the term "acid" all the  time. For instance they call dna, "nucleic acid" whereas it is always in an aquaus solution with tons of negative po3(-2) in it which are conjugate bases. So it must be nucleic base, if we use the terms in a correct, unconventional way.

@Mithoron Aren't usually* correct.

DNA is nucleic acid and caring only about it's conjugate base doesn't change that. Still it's usually more or less neutralized with various counterions, so one could say it's in form of a salt or tight ionic pair, like bonded eosin.


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How to analysis the amount of reactive oxygen atom in specific molecules
From a Phd dissertation here, I read about the material follows:

"Odd oxygen is a measure of oxidative capacity of the atmosphere. It contains species with an oxygen atom avaliable to serve as an oxidizing agent."

The author summed the species as $O_3$, $O$, $O^1D$, $NO_2$ and $2\ NO_3$, etc as the whole odd oxygen.
The species which has more than one reactive oxygen atoms are counted multiple times to represent he total number of reactive oxygen atoms.
How to analysis the amout of reactive oxygen atom of certain species like  $NO_2$ and $2\ NO_3$


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Calculating the electrode potential from thermochemical data
How can I calculate the electrode potential from thermochemical data?
My attempt: I know that the electrode potential is dependent on the enthalpies of: atomisation, first ionisation, second ionisation and hydration. So I summed up all to get $\Delta G$ and used the formula $\Delta G=-nFE$,but I am not getting the same value as given in the table.
Does the electrode potential depends on factors other than those given in the table, or there is something more to it than just addition?

You are summing $\Delta H$ values, how do you expect to get $\Delta G$?

@orthocresol     But delta S is not mentioned in the table, how could I find it?

You can't find it from the table. So the table isn't enough.

As stated above by @orthocresol, these values of $\Delta H$ alone are not sufficient to find the electrode potential. To do so, you would need $\Delta G$ for each reaction, so you would either need to find a table of $\Delta G$ for these reactions, or you would need a table of $\Delta S$ along with the table given.
But your question is why do we care about those particular values to begin with. The reason for this is that those reactions summed together give the reduction reaction for that metal. For example, I will look at the reactions of iron.
The reduction reaction for iron: $$\ce{Fe(s) \to Fe^2+(aq) + 2e^-}$$
All the other reactions in the order they are listed in the original post:
\begin{align}
\ce{Fe(s) &\to Fe(g)} \\
\ce{Fe(g) &\to Fe^+(g) + e^-} \\
\ce{Fe^+(g) &\to Fe^2+(g) + e^-} \\
\ce{Fe^2+(g) &\to Fe^2+(aq)}
\end{align}
If you add together these reactions, you should obtain the reduction reaction. Consequently, by Hess's law for Gibb's free energy, the sum of the $\Delta G$'s for these reactions will give you the $ΔG$ for the reduction reaction.
Then, it is just a matter of converting the $\Delta G$ to a reduction potential using the equation given in the original question:$$\Delta G^{\circ}=-nFE^\circ$$ where $n$ is the moles of electrons, $F$ is Faraday's constant, and $E^\circ$ is the standard reduction potential.


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Concentration and volumes
In the reaction:
$$\ce{H2SO4 +CaCO3 \to CO2 + H2O + CaSO4}$$
Does altering the concentration of the sulfuric acid change the volume of the carbon dioxide produced [the volume of the sulfuric acid is constant]

Do you know about Le Chatelier's principle

The concentration is defined as the amount of particles per unit volume. If you have a greater concentration of sulfuric acid, there are more sulfuric acid 'particles' ready to collide and react with the calcium carbonate particles. This is a simple result from the collision theory of chemical reactions.
So, increasing the concentration of the sulfuric acid should increase the amount of carbon dioxide (and other products). However, this is not always the case.
Consider the case where calcium carbonate is the limiting reagent/reactant. This basically means that all of it gets used up before the other reactant gets used up. In a case like this, increasing the sulfuric acid concentration will not affect the amount of products, simply because there's not enough calcium carbonate.
It's like asking if having more flour is going to help you bake more cake. Often, it does. But if you don't have enough of the other ingredients, it doesn't matter how much flour you have! Obviously, this is an incomplete analogy because you could probably still make a cake, albeit a nasty one.


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Which species gives electron in the following reaction?
In a given reaction like this:
$$\ce{FeO4^2-  + 3e^-  + 8H+ <=> Fe^3+ + 4H2O}$$
From where does the electron come  in the acidic medium, as there is no other species to get oxidised in the medium? Also if this were in basic medium then why there is no oxidation of any other species in the medium?

This is not a reaction, but a half-reaction. The electron comes from the other half-reaction.

Expanding on Ivan's comment, what you have here is a reduction half-equation (electrons are being accepted). As the name suggests, this only shows half of the actual reaction. You also need to have the oxidation half-equation (electrons being donated) and add them together to get the entire equation for the reaction. When you add these two half-equations together, the electrons should cancel out.

What you have here is a reduction half-equation as the $\ce{Fe(IV)}$ is accepting electrons to get reduced to $\ce{Fe(III)}$. As the name suggests, this only shows half of the actual reaction as if something is getting reduced, something else must be getting oxidised. So what you need is the oxidation half-equation which shows the species that is getting oxidised and donating electrons. When you add these two half equations the electrons will cancel each other out and you will get your equation for the reaction.
For example, one possible oxidation half-equation is the following:
$$\ce{Mg -> Mg^{2+} + 2e-}$$
Now if you times the co-efficients of the reduction half-equation by 2 and times the co-efficients of the oxidation half-equation by 3 and then add them together you will get the following equation:
$$\ce{2FeO4^{2-} 3Mg + 16H+ -> 2Fe^{3+} 3Mg^{2+} + 8H2O}$$
As you can see all the electrons balance out and that is your complete equation for the reaction.
Sorry, I don't think I really understand your second question about basic medium. Could you please expand on it to clarify what you mean.


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Facial selectivity of iodonium formation in an iodolactonisation
I am trying to understand the stereochemistry of iodolactonisation of this cyclohexene. I can't work out if the iodonium ion would form on top or the bottom.

There is $\ce{CH2COO-}$ and $\ce{Me}$ pointing up, and $\ce{CH2OTBS}$ and $\ce{Me}$ pointing down. My guess would be iodine attacks from the top because there is a bulky $\ce{TBS}$ group pointing down as well as a methyl, however there is a semi-big $\ce{CH2COO-}$ group pointing up right next to the double bond. Would that have more of an effect as it is nearer to the double bond?

Here, it does not matter which face is preferred. The formation of the iodonium ion is reversible and it can be formed on both faces. However, the iodonium ion can only be attacked by the carboxylate if it is formed on the opposite face from the carboxylate, which is the bottom face in your diagram. Attack on the top face is said to be unproductive because it cannot lead to iodolactonisation.


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Mechanism for the Cleavage of Diborane?
The cleavage of diborane was presented in a recent lecture, and it was said that the borane could be cleaved unsymmetrically if sterics allowed, or symmetrically if they did not, as shown in the image below:

However, no mechanism for this cleavage was provided, nor can it be found in Housecroft and Sharpe's "Inorganic Chemistry", our course textbook, nor can I find much about it online.
I have postulated such a mechanism:

Is it known how this cleavage occurs, and if it does indeed follow this mechanism? Is my attempt a reasonable proposition?

I cannot discern the failure to account for charges in your mechanism.

@Lighthart, neither can I now actually; I guess I was having a bit of a mind-blank as I posted this; sorry about that, I'll edit the question appropriately. I'd still like to know if this is a reasonable mechanism though, or whether the correct one is known.

Inorg. Chem., 1987, 26, 43 may or may not be relevant - I don't have the time to read in detail now. It does mention the word "intermediates" and is indexed under "inorganic reaction mechanisms" on SciFinder. Feel free to steal it to write an answer.

Related, unanswered: http://chemistry.stackexchange.com/questions/27826/why-is-diborane-cleaved-symmetrically-by-trimethylamine-but-unsymmetrically-by-a

Is it known how this cleavage occurs?

The mechanism proposed for the cleavage reaction involves an initial attack by the donor on one boron atom in diborane, leading to cleavage of one $\ce{B-H-B}$ bridge.

This is followed by the attack of a second donor molecule, cleaving the remaining $\ce{B-H-B}$ bridge. If the boron atom in the pendant $\ce{BH3}$ group is attacked, two moles of adduct result, whereas attack on the atom already carrying a donor produces cation and borohydride:

Support for this view comes from the fact that the intermediate ($\text{I}$) is detectable by tensimetric titration of amine-borane with diborane and by $\ce{^{11}_{}B}$ nmr.
Source: Boron Hydride Chemistry, Earl Muetterties


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Bond length increase when replacing nitrogen with phosphorus for NH3 and NF3
There is an increase in bond length of about $40\rm~pm$ from $\ce{NH3}$ to $\ce{PH3}$, but only by about $20\rm~pm$ from $\ce{NF3}$ to $\ce{PF3}$. I understand that the bond length will increase with the larger $\ce{P}$ atom, but I cannot see why the changes are so different when the atom is changing in the same way.

I'm sorry I don't have time for a fuller answer,  but the simplest explanation is that both nuclei in a bond affect the bond length, and it is roughly related to the nuclear charge.  Since there are two factor interactions here, the change from H->F cannot be isolated from the change of N->P when trying to figure on trends.

Related question with very related answer. Another related answer.

This answer of mine basically leads the way to answering your question but it is focussed on bond angles, not lengths. So let me rewrite it to focus on lengths.
All four compounds would ideally want a set of $90^\circ$ angles which nitrogen cannot achieve due to steric stress; instead, the angles are expanded to $102^\circ$ ($\ce{NF3}$) and $107^\circ$ ($\ce{NH3}$). This seems couner-intuitive, because the smaller atoms require a larger bond angle. But looking at it closely, we realise that the $\ce{N-H}$ bond is simply shorter so the hydrogens are closer together hence more strained and hence further apart. Hydrogen being a very small atom the bond lengths cannot be extended well.
The fluorine atom on the other hand is much larger, allowing not only for a greater standard bond length but also for greater bond length variations. It can extend the bonds when compared to $\ce{NH3}$ to allow for a smaller bonding angle and in turn allow for a larger s-contribution to nitrogen’s lone pair. Lengthening (that is, weakening) the bonds slightly allows for a greater stabilisation on the central atom and thus a more stable overall molecule.
Another important issue is the fact that a smaller bond length in $\ce{NF3}$ would bring the fluorine atoms closer together causing electrostatic repulsion of their lone pairs (although that is what is commonly meant with steric stress). Hydrogen, having no lone pairs, does not have this issue.
In phosphorus’ case, both bond angles are almost perfectly $90^\circ$, so we can expect an almost unstrained molecule. The smaller hydrogen still calls for an equally short bond but the larger fluorine can now come closer making the bonds stronger, because the angles are still near-perfect and phosphorus’ lone pair has a very high s-contribution. In fact, the $\ce{P-F}$ bonds are rather strong; much stronger than their $\ce{N-F}$ counterparts.
Rationalising the entire thing we come up with:

The difference between $\ce{NH3}$ and $\ce{PH3}$ ($40~\mathrm{pm}$) is approximately equal to the larger phosphorus atom’s size. (The difference between the empirical atomic radii is $35~\mathrm{pm}$.)
In $\ce{NF3}$’s case, the bonds are lengthened by some $15$ or $20~\mathrm{pm}$ to accomodate for steric stress both. This weakens the bonds but stabilises the atoms, especially nitrogen. Also, steric stress between the fluorine atoms is reduced.
$\ce{PF3}$ has the bond lengths one would expect.

I saw your other answer but could not relate it to bond lengths... but this makes sense now! Thank you.


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Finding concentration and moles given final and initial pH
Questions:

How many moles of $\ce{NaOH}$ are needed to change $500~\mathrm{L}$ of solution with $\mathrm{pH}\ 2$ to $\mathrm{pH}\ 11$?
A swimming pool contains 2 million litres of water at $\mathrm{pH}\ 7.80$ and the $\mathrm{pH}$ needs to be changed to $6.80$. How many moles of $\ce{HCl}$ are needed?

Not quite sure how to tackle these types of questions.
Would I have to find the moles of hydroxide/hydrogen ions to neutralise the solution first (ie. moles of hydroxide ions needed to reach $\mathrm{pH}\ 7$) and then from there calculate the amount of hydrogen/hydroxide ions needed to get to final $\mathrm{pH}$?
Or do I just calculate the difference between the moles of ions in final $\mathrm{pH}$ with moles of ions in initial $\mathrm{pH}$?
I have tried both methods but I don't have the answer to the questions so I'm just trying to understand the concept.

Both the cases are similar, so I'll show you the calculation for the fist case only:
The $[\ce{H+}]$ ion concentration changes from $\pu{10^-2~M}$ to $\pu{10^-11~M}$. Assuming water at $\pu{25^\circ C}$, this implies the change in $[\ce{OH-}]$ ion  concentration from $\pu{10^-12~M}$ to $\pu{10^-3~M}$. Since the volume of the solution is fixed at $\pu{500L}$, the amount of $\ce{OH-}$ ions has changed from $500\times\pu{10^-12 moles}$ to $500\times\pu{10^-3 moles}$. This change is exactly the number of moles of $\ce{NaOH}$ that need to be added to this solution, since $\ce{NaOH}$ is a strong base and dissociates with $100\%$ yield.


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Why does a silanized glass surface become hydrophobic?
What is the physical reason that a glass surface becomes hydrophobic by silanization? (e.g. is it due to increase in surface roughness due to the long chains which is created on the glass surface? or ...)

Because long alkyl chains are hydrophobic...

Water has many, rather strong intramolecular bonds called hydrogen bonds, which influences many of its properties, for example being fluid even at high temperatures.
Wheter some thing is hydrophilic/hydrophobic is influenced mostly by the availability of places where hydrogen bonds can happen, these are either
1. bonds between hydrogen and an element with high electronegativity (e.g. O; N)
2. an orbital already filled with 2 electrons.
Because water molecules are also polar, charged or polar particles are also more hydrophilic. (e.g. salts)
Hydrophobic substances simply lack those features.
A good example illuminating this is that even small alkanes are quite insoluble in water whereas starch, a large molecule with many hydroxy groups is still soluble.
Of course water "tries" to minimize  its  surface energy for example by forming spherical (least surface are) drops. If a surface has many possibilities for  bonds on the other hand, water wets it.
A silanized surface has many long alkane chains, and therefore few possibilities to form bond, ergo it is quite hydrophobic. (In contrary to the normal glass, which  has many atoms with (formal) charge.)


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Monitoring H -> D exchange
I was wondering how the exchange of hydrogen by deuterium, like on arenes like benzene

could be monitored for a reaction on a scale larger than say 10 g of product.  While I know that C6D6 is commercially available, used for NMR spectroscopy, I recognize

the observable signal of the protons bound to carbon is attenuated by the advancement of the reaction, and
the very deuterium installed equally serves as lock signal for the 1H-NMR spectrometer (similar to the deuterium in CDCl3).  So there might be an experimental problem, from the spectrometer, too.

But what is a typical standard compound may serve as an added inner standard in such a reaction, provided protocols accessed so far mention extended heating / autoclave or/and presence some hydrogen gas to initiate the reaction as reaction conditions?  Do commercial suppliers carry out the exchange hydrogen by deuterium, and (perhaps by virtue of even larger batches) obtain the 99%+ isotopically enriched product by sophisticated distillation / zone melting, recycling the partially deuterated product?

While this side-steps several of your questions, I suspect in the real world, most people would monitor this reaction using mass spectrometry rather than NMR.  It's more sensitive and you would directly quantify the relative amounts of each class of isomers (C6H6 vs C6H6D vs C6H4D2, etc.) with essentially no sample workup required.  I love NMR (I run an NMR facility), but for many situations, the best answer is to use MS (faster, more sensitive, etc.)

@S.Burt Some experiments later & aiming for the exhaustive H-D-exchange, 1H-NMR was used for an estimate of "non-deuterated starting material" by addition of an internal standard, added after the reaction.  Rather a conservative one, because poly-substituted arenes' patterns in 1H-NMR may be more complex anyway.  Probably yes, a (tandem?) MS might be more suitable to quantify the products, esp. if there are multiple protons to be substituted (like in doi 10.1002/jlcr.1818) and a simple GC-MS is unable to separate them.  Zooming into the large TIC-peak however showed different isotope patterns.

You've managed to squeeze a few questions into one here, but all nice questions, so let's have a go at answering them.

What standard could be used for monitoring this type of reaction?
As you indicate, any internal standard needs to be inert to any reaction process, and many industrial processes for deuteration are fairly extreme, so it is unlikely that a reliable standard could be found. Instead, what is commonly done is to add an internal standard post-reaction. That is, take an accurately measured aliquot of the reaction mixture and add a known amount of standard. This is commonly done for quantitative analysis, and is easy and accurate to do. Take 0.5mL of your reaction mixture and add 0.1ml of a stock solution of your standard. Common standards include TMS, TSP, acetonitrile, methylsulfone, benzene, 1,4-dioxane and dichloromethane and of course the one to use wil depend on the expected peak ranges of your material. Wise spectroscopists use a stock solution containing two standards.
Many modern spectrometers can use an electronic form of providing an internal integration standard. The ERETIC method (search NMR ERETIC) introduces a highly accurate and reproducible concentration measurements by applying a synthesised electronic reference, and can be postioned anywhere in the spectrum out of the way of other peaks. This method is used extensively in qNMR methodology.
A precaution that you need to take care with to ensure accurate results is in ensuring no loss of solvent volume (ie changes to concentration) over time of your reaction mixture.
Interference of 2H signal in experiment. It is entirely possible to run an experiment without an internal lock. You can use an external lock, or simply turn of field sweep for short durations (up to a few hours). Alternatively, it is possible to use 19F as a spectrometer lock on spectrometers where the hardware allows. In the event of requiring an internal lock, this too would be no problem, providing the signal was well away from your peaks of interest. Similarly, large amounts of 1H signal from other reactants could easily be eliminated by solvent suppression methods.
Commercial deuteration of common solvents for NMR typically use processes involving D2O. D2O is manufactured by a number of means, such as the Girdler sulfide process, and is pretty cheap to manufacture. Other solvents will use multi step processes to drive the exchange. DMSO, for example, is heated over CaO to induce partial exchange, and the process repeated several times before (almost) complete conversion.

+1 (beside acceptance), both for the markers and their directions provided; and esp. the hint not to overlook the addition of a standard /after/ the reaction occurred  -- recognized as useful not only for NMR spectroscopy.


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What properties of later elements are predicted by scientists?
Though scientists have predicted up to element 172 (unseptbium) what properties do they predict any elements later on the extended periodic table may have?

Later then 172?

related http://chemistry.stackexchange.com/questions/8073/the-last-elements-atomic-number

Though elements past 92 are quite unstable, so that chemical properties are difficult to determine, heavier elements behave more like metals, since the outer electrons are well shielded from the attraction of the positive nucleus and will therefore be easily lost.
Another predictable property is specific heat, which decreases with increased atomic mass.


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What makes carbon special?
We are home schooling my son, and right now we are covering chemistry- ionic bonds, covalent bonds, etc.  We've talked about how carbon is "special" in that it can form very complex molecules (organic chemistry), and I have done some hand waving by saying that it is its ability to make four bonds that causes that specialness.
He, quite reasonably, asked why nitrogen, with its ability to make three covalent bonds, couldn't be almost as special as carbon.  I told him that it probably can make pretty complicated molecules, but that only being able to make three bonds is a limiting factor.  That didn't seem to satisfy him, and to be honest, it doesn't totally satisfy me.
So why is carbon so special?  Is it just the four bonds vs. three, or is it something else?

Catenation (this is mainly what allows those complicated molecules)

Maybe this earlier answer will help.

I think that the link that Ron mentioned answered the question adequately, although let me add my own take.  (What makes carbon special and versatile?)
I heard that the reason carbon is able to make such stable multiple bonds is that its orbitals are close enough to its nucleus that they form extremely effective overlap when bonding.  It is therefore the best atom because it is the smallest atom that can form four bonds.  When you have 5 valence electrons already and can only make 3 bonds with 1 lone pair, that usually leads to weird bent shapes (although nitrogen is also pretty common in organic compounds).
It's kind of like, which hotel is more versatile in booking large families—a hotel with four open rooms right next to a breakfast bar (carbon), a hotel with three open rooms and a full room right next to a breakfast bar (nitrogen), or a hotel with any amount of open rooms, except you need to walk a mile for breakfast (any other element)?

What is also important for carbon is that its bond to hydrogen is relatively strong and almost nonpolar.  Therefore C-H bonds only react homolytically with high activation energies unless specifically activated in a molecule. The idea that C forms 4 reasonably strong bonds in a neutral molecule is a pretty close, good reason.


text:
Bonding in Square Antiprismatic Compounds
Both $\ce{IF_8^-}$ and $\ce{XeF_8^{2-}}$ have (distorted) square antiprismatic geometries.

I've been wondering how to explain the bonding in such compounds.  Hypercoordinate bonding (as used to explain the bonding in $\ce{PCl_5}$, for example) doesn't seem to work in these cases.  Nor does the resonance explanation used in the case of $\ce{IF7^-}$ (which would involve something like $\ce{XeF_8^{2-} <-> XeF6 + 2F-}$ in this case) seem very strong.  Can the bonding and geometry in the above octacoordinated compounds be explained using  s and p orbitals?

As a quick note: in addition to MOs when considering this compound you should keep in mind that the structures contain extremely polarised bonds, and thus elesctrostatic interactions play important role.

From what research I've done, there isn't a definitive answer, though people have ideas about what causes this phenomenon. Both sources here cited note the lack of adherence to the VSEPR model, and hypothesize that the steric effects of the eight fluorine atoms prevent the central unbonded lone pair from existing in a definite space, thereby distorting the antiprismatic shape.
To paraphrase an inorganic chemistry textbook:

$\ce{[XeF8]^{2-}}$ is difficult to interpret using VSEPR theory. There is a lone pair on the central xenon atom, which in the VSEPR model would be predicted to occupy a definite position about the central atom as do lone pairs in $\ce{NH3}$, $\ce{SF4}$, and $\ce{IF5}$. In $\ce{[XeF8]^{2-}}$, however, there is no definite location for this central lone pair.
One explanation is based on the degree of crowding around xenon—with eight fluorines attached, repulsions between the xenon-fluorine bonds are too strong to enable a lone pair to occupy a well-defined position. This gives $\ce{[XeF8]^{2-}}$ a very slightly distorted square anti prism structure, with one face being slightly larger than the opposite face. Although this distortion may be a result of the way in which these ions pack in a crystal structure, it is also possible that the distortion is caused by a lone pair exerting some influence on the size of the larger face.

From a piece of literature specifically analyzing the antiprismatic structure of xenon(VI):

We believe that the larger $\ce{Xe-F}$ separations are due to a weak interaction between some fluorine atoms and the neighboring cations, thus causing a slight but significant distortion of the antiprism. However, the distortion might also be explained as a manifestation of some steric activity of the xenon lone pair. Since the eightfold xenon coordination observed here provides no clearly defined ninth coordination position for the lone pair, it appears to offer little support for electron repulsion models which have been used with considerable success in rationalizing the structures of other noble gas compounds.


text:
Understanding effect of ph on solubility of salt
In the following picture why Ksp is not simply S2- ? Why is f included with one species only? Please explain the last part.

What does f represent and why it is used in molar solubility?

I am also very confused on what is happening here. I have no idea what f is suppose to be and I don;t know how they got that expression for Ksp. I will will write up an answer on how I would do this which is hopefully much clearer

$f$ is defined in the question as the fraction of $[\ce{X}]{tot}$ that is ionized.  It's just a shorthand abbreviation for $\frac{[\ce{X-}]}{[\ce{X-}] + [\ce{HX}]}$.  By charge conservation, the concentration of metal $\ce{M+}$ is equal to $f S$, so $K{sp}$ can be written as $[S] * f[S] $.

Let assume we have the following equilibria:
$$\ce{MX \rightleftharpoons M+ + X-}$$
$$\ce{HX \rightleftharpoons H+ + X-}$$
The expression for their respective equilibrium constants are the following:
$$\mathrm{K_{sp} = [M^+][X-]}$$
$$\mathrm{K_a = \frac{[H^+][X^-]}{[HX]}}$$
We also know the following (let S be solubility):
$$\mathrm{S = [M^+] = [X^-]_{tot}}$$
Now we just need one more equation before we can start solving for S which is a mass balance. Since all the $\mathrm{X^-}$ that is produced either reacts with water to form $\mathrm{HX}$ or remains as $\mathrm{X^-}$, we get the following equation :
$$\mathrm{[X^-]_{tot} = [X^-] + [HX]}$$
Now lets put $\mathrm{[X^-]}$ and $\mathrm{[HX]}$ in terms of $\mathrm{[M^+]}$:
$$\mathrm{[X^-] = \frac{K_{sp}}{[M^+]}}$$
$$\mathrm{[HX] = \frac{[H^+][X^-]}{K_a} = \frac{K_{sp}[H^+]}{K_a[M+]}}$$
Now let plugs those values into our mass balance and replace $\mathrm{[X^-]_{tot}}$ and $\mathrm{[M^+]}$ with S:
$$\mathrm{[X^-]_{tot} = \frac{K_{sp}}{[M^+]} + \frac{K_{sp}[H^+]}{K_a[M+]}}$$
$$\mathrm{S = \frac{K_{sp}}{S} + \frac{K_{sp}[H^+]}{K_a\times S}}$$
$$\mathrm{S^2 = K_{sp} + \frac{K_{sp}[H^+]}{K_a}}$$
$$\mathrm{S = \sqrt{K_{sp} + \frac{K_{sp}[H^+]}{K_a}}}$$
This gives you the exact same expression as the one on the sheet but hopefully this is much clearer for you.

Could you please explain why X- tot=X- + HX ?

@JM97 That is because the total X- that is produced either reacts with water to form HX or it just remains as X-. Therefore X-tot = X- + HX


text:
Safely storing hydrogen
I have built a solar panel which is used to add electrodes to $\ce{H2O}$ and $\ce{NaCl}$, evolving $\ce{H2}$ and $\ce{O2}$.
How can I safely store the hydrogen in the long-term?

Hydrogen can't and won't explode as long as it is kept separate from oxygen, unless you're storing it under high pressure.  So as long as your hydrogen container can't break or leak under all sorts of environmental hazards (heat, fire, dropping, collision, degradation, etc.), you should be fine.  Essentially, this question is more about risk management to consider all the possible scenarios of a potential leak.
Ultimately it depends on how and where you will store your hydrogen container, and the potential dangers of that environment.  So storing hydrogen under low pressure within an argon-filled safe would be a lot safer than storing it under high pressure in a mobile oxygen tent.

So you are saying hydrogen becomes flammable when coming in contact with oxygen? meaning if it comes in contact with H2O the excess Hydrogen will ignite? THAT IS EXTREMELY USEFUL TO KNOW! I am looking into argon now.

Obviously you misunderstood.  I wasn't talking about the Water substance being ignited. I was more over talking about https://www.youtube.com/watch?v=jdxi-PvNYS8 again, my question was a safe way on storing hydrogen without the risk of ignition, ie. Oxygen, tempature ect..

The video is fake, notice the cut after taking the batteries out, and don't believe everything you see on YouTube. Pay attention to the comments and downvotes.

@Kelvin While you certainly have a point, we all would appreciate if you could be nice.

@Kelvin i only google and posted a random video. My point is hydrogen builds up when being split from oxygen from a negative electrode terminal and becomes flammable. ie. Hydrogen gas. i did not watch the video, i have tested this myself, but the results were only a sudden ignition with a loud pop.

Ok, then how do you separate the hydrogen from the oxygen gas from electrolysis?  In any case you need to remove all oxygen from the hydrogen, and then keep oxygen from mixing back in.  As long as the hydrogen has no oxygen, it cannot ignite - as per my original answer.


text:
Relationship between ionization energy and orbitals
My chemistry book says that the ionization energy depends on the orbital an electron is taken from?
I'm not sure I understand what it means. Could someone please explain this?

I don't understand it either. Are they talking about one element or different elements, to begin with?

http://chemistry.stackexchange.com/questions/16186/whats-the-difference-between-ionization-energy-and-orbital-energy


text:
What is hybridisation of XeF6 in solid state?
According to the following formula:
Type of hybridisation/steric no. = (no. of sigma bonds + no. of lone pairs)
it should be $\ce{sp^3d^3}$, however according to my textbook it is $\ce{sp^3d^2}$ in solid state. That is how it was written (in my textbook), emphasising on the fact that "in solid state it is $\ce{sp^3d^2}$". How does solid state of a compound change its hybridisation?

Forget about applying hybridization outside the second row, especially in 'hypervalent' compounds. I know, that it is common to use and sometimes works, but it is incorrect.
The $\ce{XeF6}$ molecule is a hard spot. While, indeed, experimental data suggest that it adopts distorted octahedral geometry in the gas phase, there is evidence that the minimum is very shallow.
The octahedral structure of $\ce{XeF6}$ (which is probably a local minimum, or at least found to be one in more than one calculation) can be pretty easily described in terms of three-center four-electron bonds.
There is no good agreement on the nature of the stereo-active electron pair in $\ce{XeF6}$. It seems, that repulsion with core d-shells is important, but the recent articles I could get my hands on has little to no rationalization on the fact. Still, it must be noted that the minimum is very shallow, meaning that rough qualitative theories like VSEPR and MO LCAO are too rough for a good rationalization anyway.
The solid state typically involves a lot more interaction, and, as result, compounds may adopt very different structures. The simplest one example I can think of, I could point to $\ce{SbF5}$ for example. A monomer adopts trigonal-bipyramidal structure. However, in the solid state, a tetramer with octahedrally coordinated antimony and four bridging fluorines is formed. In terms of VSEPR, it is a move from $\mathrm{sp^3d}$ to $\mathrm{sp^3d^2}$. Similarly, $\ce{XeF6}$ has a crystal phase (one of six) that involve bridging atoms.

Solid state does have an effect on the hybridization. Inorganic compounds generally split into cationic and anionic parts to form a lattice and the lattice energy compensates for the energy lost. For example
$$\ce{2PCl5 -> [PCl4]+ + [PCl6]-}$$
Such is the case with $\ce{XeF6}$. In the solid state, this occurs:
$$\ce{XeF6 -> [XeF5]+ + [F]-}$$

Are you sure $\ce{XeF6(s)}$ exists as $\ce{[XeF5+][XeF7-]}$? Because that's what your last sentence implies.

@orthocresol Actually, this is one of the early proposed crystal compositions. Later was considered that depending on modification $n$-meric $\ce{(XeF5^+F^-)_n}$ ($n=2..6$) units are present. So it makes sense to review  square pyramidal $\ce{XeF5^+}$ ions instead of $\ce{XeF6}$(Hoyer, S.; Emmler, T.; Seppelt, K. Journal of Fluorine Chemistry 2006, 127 (10), 1415–1422. DOI  10.1016/j.jfluchem.2006.04.014).

You should edit to make that clear. I downvoted because it looks to me like you are proposing $\ce{[XeF5]^+[XeF7]^-}$.

Oh, I agree that $\ce{[XeF5+][F-]}$ (or polymers thereof) is OK. I just don't ever recall seeing $\ce{[XeF5+][XeF7-]}$ and the way I'm reading it, that's what your last sentence implies, by drawing the analogy to $\ce{PCl5}$. (cc @andselisk - maybe you know something I don't)


text:
Reaction of an alkyl halide with silver nitrite
I want to know that whether both alkyl nitrite and nito alkane are produced or if nitroalkane is the only product. There are many different (contradicting) things written in different books that is confusing me.

Reaction of alkyl with KCN is different from reaction of alkyl with AgCN

On treating ethanolic solution of haloalkane with silver nitrite ($\ce{Ag-O-N=O}$), nitroalkane is formed because since the bond between $\ce{Ag-O}$ is covalent, the lone pair on nitrogen acts as an attacking site for nucleophilic substitution.
$$\ce{R-X + AgNO2 -> R-NO2 + AgX}$$
But on the other hand if haloalkane is treated with potassium nitrite ($\ce{KNO2}$), alkyl nitrite is formed as major product because since the bond between $\ce{K-O}$ is ionic in nature, the negative charge on oxygen serves as an attacking site.
$$\ce{R-X + KNO2 -> R-O-N=O + KX}$$

But my question is , Is it the only product or the product with higher yield

@BEWARB It is the only product.

Can $AgNO_2$ not exist as nitrite-N?

@Internet Guy no, silver nitrite is nitrite-O as the silver is bonded to oxygen atom. See this: https://chemistry.stackexchange.com/questions/55300/difference-between-the-structure-of-nitrito-o-and-nitrito-n/55336#55336

@Nilay Ghosh but the other O has 2 lone pairs, why dosen't it attack the methyl group ...?


text:
Can triethylamine/pyridine be used in the synthesis of Labetalol?
As part of my homework, I had to plan the synthesis of labetalol, a blood pressure drug, from its component parts. One of the steps involved was the following SN2 reaction:

The solutions manual just shows the two being added together, but my mind harkened back to previous reactions I had learned in which triethylamine and pyridine were used when $\ce{HCl}$ was being eliminated. Would that be a reasonable thing to do in this instance, or would these amines compete with 4-phenylbutan-2-amine?

I'm more worried about multiple alkylations of the nitrogen.

If the solution manual shows these two added together, then I wouldn't worry about multiple alkylations in this problem (IRL is a different matter.) That chloride is going to be extra electrophilic because of the adjacent carbonyl. I'd say this reaction would probably be fine without any base, so triethylamine or pyridine should be fine to do the final deprotonation.

I would actually be curious to see what the "component parts" are for your retrosynthesis...

I disagree with IT Tsoi's answer.  Triethylamine absolutely will compete, and form ammonium salts.  Those salts may react with 4-phenylbutan-2-amine to lead to product, but more likely they will precipitate.  And be general gunk in the reaction flask.  Pyridine will probably be less of an issue.
In general direct alkylation of amines is a pretty ugly reaction with overalkylation products likely.  A better reaction in this vein would be to make the acetamide and alkylate that with the chloroacetophenone, and then deacetylate, or similar strategy.
If you must alyklate the amine, use Hunig's base (N,N-diisopropylethylamine) which is too sterically hindered to alyklate.  Also, add the electrophile to excess nucleophile, if the reaction is faster this will decrease polylaklyation probabilistically, and the carbonyl alpha to the amine may have enough electron withdrawing character to slow down the increase in nucleophiliciity that you get from a more substituted amine.
Further mucking up this route is the phenol with two carbonyls on the ring, that is probably quite acidic as well, and may be deprotonated with the base scavengers.  Use a phosgene protecting group to cover up both the phenol and the aramide, and this problem will be side-stepped.

I don't understand how the ammonium salts will react with the amine. Triethylamine will not react with the chloride.  I don't think overalkylation is an issue in general with haloacetates. Also we shouldn't worry about phenol (and the amide) deprotonation since triethylamine or the amine reagent used are not basic enough

Triethylamine and Pyridine will not compete with the primary amine - their main purpose would simply be to capture the acid. Neither of them have a proton they can readily lose, and compared to the primary amine they are sterically hindered so their rate of attack (if ever) will be slower too.
It is a reasonable thing to do (adding base) because acidification will lead to the protonation of the amine rendering it unable to attack so your reaction won't run to completion or be slower than it should be.

Afaik, pyridine is actually sterically less hindered. Otherwise the rest you say holds true.


text:
Second law of thermodynamics: How do entropy changes in the system and the environment count in?
$$\frac{q_{rev}}{T} = \Delta S$$
Here, what does $\Delta S$ signify? Does it mean the change in entropy of the system or the surroundings?
How are the entropies of the system, the surroundings, and the universe related to each other, and which entropy is used in the Gibbs free energy equation?
consider this problem
H2O(l) -> H2O(g) at 373 k and 1 atm pressure then which of the following is correct?
a) ΔSsystem >0 ΔSsurrounding > 0
b) ΔSsystem <0 and ΔSsurrounding < 0
c) ΔSsystem>0 and ΔSsurrounding < 0
d) ΔSsystem <0 and ΔSsurrounding > 0

The $\Delta \mathrm{S}$ corresponds to what $\mathrm{q_{rev}}$ you are using. For example if plug in the $\mathrm{q_{rev}}$ of the system, then you get the $\Delta \mathrm{S}$ of the system. The same goes for the Gibbs Free energy equation. I don't understand what you mean by how is entropy of system,surrounding and universe related to each other. Could you please expand on what you mean.

@ Nanoputian sir thank you very much for replying,please see this problem

@Nanoputian thank you very much for replying sir, please see this problem
H2O(l) -> H2O(g) at 373 k and 1 atm pressure then which of the following is correct?
a) ΔSsystem>0 ΔSsurrounding > 0
b) ΔSsystem<0 and ΔSsurrounding < 0
c) ΔSsystem>0 and ΔSsurrounding < 0
d) ΔSsystem <0 and ΔSsurrounding > 0
i know that ΔSsystem is positive but how to determine ΔSsurrounding?

I may be wrong since I didn thermodynamics a while back, but I pretty sure the answer is C. Firstly I am assuming the water is heat reversibily, then $\mathrm{q_{sys} = q_{sur}}$. Therefore, $\mathrm{\Delta S_{sys} = -\Delta S_{sur}}$. Since you are convert a substance from the liquid phase to the gas phase, the entropy of the system must be increasing and hence >0. Therefore the answer is C

I also recommend that you edit your original question to include this problem.

The equation given in the question is an incorrect statement of the definition of entropy and is only applicable at constant temperature. Entropy is properly defined as $\mathrm{d}S = đq_{\mathrm{rev}}/T$ (integration holding $T$ constant yields the equation in the question).

@Nanoputian
sir but at 373 k and 1 atm pressure won't the system be at equilbrium since that is just boiling point.and won't the entropy of universe be zero at equilbrium? this is my doubt. and order decreases but the system is in equilbrium so we can't say which direction reaction proceeds,so how can we say entropy of system increases?

@Nanoputian when you see something that shouldn't be in the comments, you shouldn't engage. Comments aren't discussion boards.

@PhMgBr sorry about that. I will keep that in mind in the future.

I think your first question (regarding the meaning of $\Delta S$) has been adequately covered in the comments. I hope the following will clarify your other concerns, though it isn't entirely clear what exactly you don't know.
The Second Law tells you that the total entropy of the universe will always increase, i.e $\Delta S_{universe} > 0$.
Logically, we can split the 'universe' into a system and surroundings, so:
$$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings}$$
From the second law, this obviously implies that
$$\Delta S_{system} + \Delta S_{surroundings} > 0$$
Regarding the multiple choice question you asked in the comments, this means that we can immediately eliminate option B. If both $\Delta S_{system}$ and $\Delta S_{surroundings}$ are negative, their sum cannot be positive.
Clearly, within the system, order decreases as liquid is turned to gas, so entropy increases. So we know that $\Delta S_{system} > 0$.
In order to make the water boil, heat is transferred away from the surroundings, so $\Delta S_{surroundings} < 0$. If you knew how much water you were boiling, you could even figure out exactly how much the entropy decreases, because you know that the energy required to boil a mass of water is given by $Q=mL$, where $L$ is the latent heat of vaporisation.

sir but at 373 k and 1 atm pressure won't the system be at equilbrium since that is just boiling point.and won't the entropy of universe be zero at equilbrium? this is my doubt. and order decreases but the system is in equilbrium so we can't say which direction reaction proceeds,so how can we say entropy of system increases?


text:
How to prepare solutions of different concentrations of potassium from stock solutions with different concentartions?
I need to prepare different concentrations of $\ce{K+}$ ($10$, $20$, $30$, $40$, $\pu{50 mM}$) by diluting a solution of $\pu{100 mM}$ $\ce{K+}$ with an isotonic buffer of $\pu{5 mM}$ $\ce{K+}$.
I need $\pu{4 mL}$ of each concentration.
I really don't remember how to calculate this. Because I need to consider the $\ce{K+}$ within the isotonic buffer right, because they will also add amount of $\ce{K+}$? If I would have diluted the $\pu{100 mM}$ $\ce{K+}$ with just water, then I understand how to do.

The only thing you need to remember in chemistry is conservation of matter. "Just water" is a solution where $\ce{K^+}$ concentration is simply zero.

Thanks for answearing, but I still dont understand how to prepare the 4 concentrations :/

If you mix two solutions of different concentrations with each other you can easily determine the resulting concentration from the general formula for it:
$$c = \frac{n}{V}\tag1\label{concentration-general}$$
You have one solution with $c_1 = \pu{100 mM}$ and one with $c_2 = \pu{5 mM}$, therefore with $\eqref{concentration-general}$ you can determine the amount of substance for each, and the total amount of substance:
$$n_\mathrm{tot} = c_1\cdot V_1 + c_2\cdot V_2\tag2\label{amount-of-substance-total}$$
Assuming there is not volume contraction, we can simply add them to know the total volume:
$$V_\mathrm{total} = V_1 + V_2 \tag3\label{volume-total}$$
The total concentration therefore is:
$$c_\mathrm{total} = \frac{n_\mathrm{total}}{V_\mathrm{total}}
  = \frac{c_1 V_1 + c_2 V_2}{V_1 + V_2}\tag4\label{concentration-total}$$
For your dilution series you will need to modify it, a $\eqref{volume-total}$ is fixed.
$$V_\mathrm{total} = V_1 + V_2 = \text{const.}$$
Therefore we modify $\eqref{concentration-total}$:
\begin{align}
  c_\mathrm{total} &=
    \frac{c_1 V_1 + c_2 (V_\mathrm{total} - V_1)}{V_\mathrm{total}}
    \tag{4a}\label{concentration-total-mod-a}\\
  c_\mathrm{total} &=
    \frac{c_1 (V_\mathrm{total} - V_2) + c_2 V_2}{V_\mathrm{total}}
    \tag{4b}\label{concentration-total-mod-b}\\
\end{align}
Now you have one remaining unknown, solve for it:
\begin{align}
  V_1 &= \frac{c_\mathrm{total} - c_2}{c_1 -c_2} V_\mathrm{total}\\
  V_2 &= \frac{c_\mathrm{total} - c_1}{c_2 -c_1} V_\mathrm{total}\\
\end{align}
Unless you have very accurate equipment, however, I do suggest using larger quantities than $\pu{4 mL}$.
\begin{array}{rrrrrr}
c_\mathrm{total}/\pu{mM} &
V_\mathrm{total}/\pu{mL} &
c_1/\pu{mM} &
c_2/\pu{mM} &
V_1/\pu{mL} &
V_2/\pu{mL} \\\hline
10 & 4 & 100 & 5 & 0.21 & 3.79 \\
20 & 4 & 100 & 5 & 0.63 & 3.37 \\
30 & 4 & 100 & 5 & 1.05 & 2.95 \\
40 & 4 & 100 & 5 & 1.47 & 2.53 \\
50 & 4 & 100 & 5 & 1.89 & 2.11 \\\hline
\end{array}


text:
Determination of structure by 13C and 1H NMR spectroscopy

I'm trying to interpret a $\ce{^{13}C}$-NMR and $\ce{^1H}$-NMR spectrum. So far, I've decided that $\ce{CH3-C}$, $\ce{CH2-CH3}$ and $\ce{CH-CH}$ exist in the structure. I'm not sure but I guess there should be different seven carbons too.
I've come up with the structure below but I feel like there is not enough information to be sure. What should I need to consider to go further at this point?

Hi and welcome  to Chemistry SE. The structure you are proposing is not correct. You should be able to work back through your proposed structure and see evidence for each part of the molecule. For instance, your molecule has 4 non-equivalent methyl groups in it, but the 1H spectrum only shows 2 distinct methyl groups.

As a clue to get you thinking in the right direction (we don't want to just give you the answer here) count up the total integral of signals in your 1H spectrum - I count 10. Your molecular formula indicates there are 20. This would suggest that each 1H group is duplicated in the molecule.

This is a tricky little problem, given there is an element of symmetry to consider. As mentioned in the comments, once you come up with a proposed structure, and your one is not a bad first attempt, you should go through and test the spectrum against what you would predict from your structure. Unfortunately, in your suggestion, there are 4 non-equivalent $\ce{-CH3}$ environments, and the spectrum only shows two, so we can eliminate that structure as a possibility. But don't be disheartened — that is the process of structure elucidation! Propose a stucture, test it, refine or eliminate it. And then start again.
Looking at the $\ce{^1H}$ spectrum, we can see:

$\ce{-CH3}$
$\ce{-CH2}$
$\ce{-CH3}$
$\ce{-CH}$
$\ce{-CH}$

We can construct the $\ce{-CH2CH3}$ spin system based on coupling, and the chemical shifts of the two downfield protons indicate a double bond (and, using some old tricks of the trade, looks to be cis — the splitting here is much less than about $14~\mathrm{Hz}$, which would be the expected splitting between the outside lines of the methyl triplet). The $\ce{CH3}$ at about $3.3$ is in the general ball park for a methyl ether, I'd say.
That gives us 10 protons, but the molecular formula says $\ce{H20}$ … so we must have symmetry in there somewhere which gives us:

2 $\ce{-CH2CH3}$
2 $\ce{-CH3}$ (probably $\ce{-OCH3}$)
2 $\ce{-CH=CH-{}}$

So, looking at the carbon spectrum:
I count 7 carbon peaks.
From the proton information, we have 5 $\ce{C}$ environments, each appearing twice, which gives a total of 10 carbons. The other 2 carbon environments occur once only in the molecule.
From DEPT experiments (which nobody in the real world should ever run when they can instead run a phase-edited HSQC instead) we can see we have

$\ce{CH3}$ at $9.9$ and $51.4$
$\ce{CH2}$ at $38.0$
$\ce{CH}$ $132.8$ and $133.2$
$\ce{C}$ at $47$ and $110$.

We know (or should know) what the non-quaternary carbon cantres all are. We worked that out from the $\ce{^1H}$ spectrum. So let's concentrate on the quaternaries. The peak at $110$ is interesting, and is typical of a carbon centre attached to 2 oxygens. Many sugars have carbons that come around this area. The other quaternary at about $47$ is typical of an aliphatic quaternary; not overly exciting.
So somehow we need to construct a molecule that has 2 non-equivalent quaternary centres that have a plane of symmetry through them, with the $\ce{-CH2CH3, -OCH3}$ and $\ce{-CH=CH-{}}$ attached. Each of these three spin systems must be isolated, otherwise we would see coupling to the vinylic protons.
As this is clearly a homework question, I'll leave it to you to piece together, and ask further questions if you need, but my answer gives the exact same predicted spectra in ChemDraw as your spectra.


text:
Stereospecific in SN2 reaction
(S)-1-bromo-1-fluoroethane reacts with sodium methoxide to give pure (S)-1-fluoro-1-methoxyethane.
This is an SN2 reaction so inversion should take place , hence the product should be R but it is not so . Why?

Think of this : how is absolute configuration (R, S)  assigned, according to the Cahn-Ingold-Prelog rules ?

@julien but reactant is (S) while product is also forming (S) configuration

@julien now I have edited the question

You're right in that the SN2 Reaction Mechanism will invert the stereocenter. However, the naming convention is based on relative priority and the mechanism by which the product is formed has little to do in how it is named. This appears to be where your confusion lies since you're associating the mechanism and the named configuration when they are not related at all
In the case of an SN2 reaction:
If the attacking nucleophile had the same relative priority as the leaving group (in this case anything with an atomic number > F, since Br is the heaviest among H, F, and C), then you're right, S would turn into R.
However, in the case you quote the nucleophile is MeO-, and O has an atomic number lower than F. Following the CIP rules, we would have to assign the absolute configuration of 'S' even though the stereocenter has been inverted relative to the original compound.


text:
How does changing the concentrations of intermediate products/reactants change the rate law?

A = CH2(CN)2
B = CH(CN)2-
C = BrCH(CN)2
The rate law above is derived from applying the steady state approximation.
I am trying to work out how to modify the rate law to accomodate the following conditions:
1) [Br2] is low relative to [H+]
2) [Br2] is high relative to [H+]
And in the 2nd case how this affects the RDS and order of reaction.
It seems to me like these conditions would just affect the rate coefficient, not actually change the rate law.
EDIT:
$$
\frac{d [CH(CN)_{2}^{-}]}{dt} = k_{1}[CH_{2}(CN)_{2}] - k2[CH(CN)_{2}^{-}][Br_{2}] - K_{-1}[CH(CN)_{2}^{-}][H^{+}]
$$
$$
Applying\ Steady\ State\ Approximation:
$$
$$
k_{1}[CH_{2}(CN)_{2}] - k2[CH(CN)_{2}^{-}][Br_{2}] - K_{-1}[CH(CN)_{2}^{-}][H^{+}] = 0
$$
$$
k2[CH(CN)_{2}^{-}][Br_{2}] + K_{-1}[CH(CN)_{2}^{-}][H^{+}] = k_{1}[CH_{2}(CN)_{2}]
$$
$$
Dividing\ by\ [CH(CN)_{2}^{-}]:
$$
$$
k2[Br_{2}] + K_{-1}[H^{+}] = \frac{k_{1}[CH_{2}(CN)_{2}]}{[CH(CN)_{2}^{-}]}
$$
$$
[CH(CN)_{2}^{-}](k2[Br_{2}] + K_{-1}[H^{+}]) = k_{1}[CH_{2}(CN)_{2}]
$$
$$
[CH(CN)_{2}^{-}] = \frac{k_{1}[CH_{2}(CN)_{2}]}{(k2[Br_{2}] + K_{-1}[H^{+}])}
$$
$$
\frac{d [BrCH(CN)_{2}]}{dt} = {k_{2}[Br_{2}][CH(CN)_{2}^{-}]}{}
$$
$$Rate\ Equation:$$
$$
\frac{d [BrCH(CN)_{2}]}{dt} = \frac{k_{2}[Br_{2}]k_{1}[CH_{2}(CN)_{2}]}{{k_{2}[Br_{2}]}+k_{-1}[H^{+}]}
$$
So if $ [Br_{2}]$ is low compared to $ [H^{+}]$ then I would remove $ [Br_{2}]$ from the rate equation giving:
$$
\frac{d [BrCH(CN)_{2}]}{dt} = \frac{k_{1}[CH_{2}(CN)_{2}]}{k_{-1}[H^{+}]}
$$
If $ [H^{+}]$ is low compared to $ [Br_{2}]$ then I would remove $ [H^{+}]$ from the rate equation giving:
$$
\frac{d [BrCH(CN)_{2}]}{dt} = k_{1}[CH_{2}(CN)_{2}]
$$
So in this case the reaction is 1st order and the RDS no longer proceeds backwards with rate coefficient $k_{-1}$.
Is this right or have I made a mistake somewhere?

How do Br an H interact?  Are C, B or A = Br?  What does Br2 have to to with this.  This question is pretty mangled right now.

My apologies I've fixed the error and added identities of A, B and C.

Rate law above is not correct. Try to derive it yourself via steady state approximation and you will quickly figure out some answers.

The steady state approximation allows us to assume that the concentration of $\ce{B}$ does not change.
$$\dfrac{d[\ce{B}]}{dt}=0$$
Thus the rates of production and consumption of $\ce{CH(CN)2-}$ must be equal.

$\ce{CH(CN)2-}$ can be produced by one pathway:

$$\ce{CH2(CN)2 ->[k_1] CH(CN)2- + H+} \\ \mathrm{rate}=k_1[\ce{A}]$$

$\ce{CH(CN)2-}$ can be consumed by two pathways:

$$\ce{CH(CN)2- + Br2 ->[k_2] BrCH(CN)2 + Br-} \\
\ce{CH(CN)2- + H+ ->[k_{-1}] CH2(CN)2} \\
\mathrm{rate}=k_2 [\ce{CH(CN)2-}][\ce{Br2}]+K_{-1}[\ce{CH(CN)2-}][\ce{H+}]$$

Thus:

$$\dfrac{d[\ce{CH(CN)2-}]}{dt}=k_1[\ce{CH2(CN)2}] -\left( k_2 [\ce{CH(CN)2-}][\ce{Br2}]+K_{-1}[\ce{CH(CN)2-}][\ce{H+}] \right)= 0$$
Now you can can solve for $[\ce{CH(CN)2-}]$ and substitute into the rate law equation for the second step:
$$\dfrac{d[\ce{BrCH(CN)2}]}{dt}=k_2[\ce{CH(CN)2-}][\ce{Br2}]$$
You should get a final rate law that contains $[\ce{CH2(CN)2}],\ [\ce{Br2}],\ \&\ [\ce{H+}]$, and then you can answer your questions.

Update based on revised question.

Your new rate late is correct:
$$\dfrac{d[\ce{BrCH(CN)2}]}{dt}=\dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_2[\ce{Br2}]+k_{-1}[\ce{H+}]}$$
Your interpretation of the simplification you can do is correct for one case and almost correct for the other.

$[\ce{H+}] \ll [\ce{Br}]$

In this case, we assume that $k_{-1}[\ce{H+}]\ll k_2[\ce{Br}]$ so
$$k_2[\ce{Br2}]+k_{-1}[\ce{H+}] \approx k_2[\ce{Br2}] \\
\therefore \
\dfrac{d[\ce{BrCH(CN)2}]}{dt}\approx \dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_2[\ce{Br2}]}\approx k_1[\ce{CH2(CN)2}$$
which is as you figured.

$[\ce{H+}] \gg [\ce{Br}]$

In this case, we assume that $k_{-1}[\ce{H+}]\gg k_2[\ce{Br}]$ so
$$k_2[\ce{Br2}]+k_{-1}[\ce{H+}] \approx k_{-1}[\ce{H+}] \\
\therefore \
\dfrac{d[\ce{BrCH(CN)2}]}{dt}\approx \dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_{-1}[\ce{H+}]}$$
Note that this approximation does not remove the $k_2[\ce{Br2}]$ term in the numerator.

I think I've got the correct rate law now and I've added it to the original post with my answers to the questions. Have I done it right?