text:
Why do we use dG < 0 to describe a spontaneous process?
With something like a reaction or phase change, all sources use the criterion that $dG < 0$ for the reaction to be spontaneous and then substitute an appropriate expression for $dG$ to the specific application. For phase change my book stats $(\mu'-\mu'')dn' < 0$ where $'$ is phase A and $''$ is phase B, $\mu$ is the chemical potential and $dn' > 0$ means gaining of molecules in phase A.
However, reactions require activation energy and phase change require nucleation (I think this is kind of like an activation energy); both of these processes require an increase in $G$ before a larger decrease. Why do we not use $\Delta G$ (between initial and final equilibrium states) instead?
Edit:
I am confused by the use of dG (a rate differential) as opposed to $\Delta G$ (a net change between end states) because I'm not sure how one would integrate this for say, a phase change, since it must pass through a potential barrier where G must increase. Is the integral just $G_2-G_1$ between the end states and thermodynamically the process ignores the barrier (since thermodynamics deals with equilibrium only)? A typical thermodynamic process I'm imagining is quasistatic compression, where we can get work from the integral of PdV and P is defined as it goes through an infinite number of equilibrium states; I'm not sure if it's possible to draw a parallel here for free energy and phase change/reacting systems.
Edit 2:
From this graph it looks like G increases then decreases to stable equilibrium after phase change. If we find dG/dr from this graph and integrate dG(r), should we get the same $\Delta G$ as from thermodynamics (before and after phase equilibrium)? If so, this seems to me like finding work of a rapid piston-cylinder compression of an insulated system; PdV is undefined since we are not passing through a set of equilibrium states but we can still find work from $\Delta U$. Difference is that in this case, our initial state is technically not in equilibrium/stable since the system favours a phase change, so we are beginning from a state that technically does not exist on the phase diagram (like if we have a supercooled/superheated liquid).

Activation barriers are kinetic phenomena. In pure thermodynamic considerations they are not included. There only the free energy of the initial and the final state are considered. The way which your system takes to get from initial to final is of no concern.

definitely +1 to Philipp here ... I would say though you can compare thermodynamically the transition complex .. but as I said in the other question it is a mix up between Kinetics and thermodynamics :)

@Philipp I suspected that I was mixing up concepts. My confusion is that thermodynamics deals with equilibrium initial/final states (compressing piston from two equilibrium states etc.). However, phase transitions for instance only happen when the system is dis-equilibrium (i.e. supercooling liquid), how do we define an initial equilibrium state for such a case?

@Yandle When judging the spontaneity of a process from the thermodynamics perspective you compare $G$ for an initial equilibrium state to $G$ for a final equilibrium state. "Usual" thermodynamics is not able to describe non-equilibrium states .

@Philipp I am confused by the fact that dG = 0 when a system is in equilibrium, but dG is not 0 when the system say, has too much of phase A than phase B (chemical potentials do not balance). In that sense, I am confused about how the initial state is in equilibrium.

@Yandle What you would do in your phase example is the following: You calculate $G$ for the given conditions $p, T$ for phase A. This gives you the equilibrium value of $G$ as if there was no phase B and no opportunity for a phase change. Then you calculate $G$ for the given conditions $p, T$ for phase B. This gives you the equilibrium value of $G$ as if there was no phase A. If the value for phase B is lower than the value for phase B than the system will spontaneously engage in a phase change reaction.

@Philipp I  am imagining equilibrium state as like the starting point of a piston-cylinder compression, it won't change unless disturbed. If given Phase A at T and p and if G for phase B is lower than phase A, then how come phase A is considered to be an equilibrium state as it will spontaneously change?

@Yandle It is more a hypothetical equilibrium for phase A. It would be realized only if there was no opportunity for a phase change to B.

@Philipp How do we differentiate hypothetical equilibrium and any point during the nucleation/transition to Phase B, since neither one is in reality stable?

Maybe its best to see how we generally derive the Gibbs energy... If you think about this then the answer should be clear ;)
The Gibbs potential is a function of $P$, $T$ and $N$. That is to say:
\begin{equation}
G=G(T,P,N)
\end{equation}
Where $T$ is the temperature, $P$ is the pressure and $N$ is the number of particles.
\begin{equation}
dG=\frac{\partial G}{\partial P}\bigg|_{T,N}dP+\frac{\partial P}{\partial T}\bigg|_{P,N}dT+\sum _{i=1}^{I}\frac{\partial G}{\partial N_i}\bigg|_{P,T,N_j\neq i}dN_i
\end{equation}
Use some Maxwell relations (Look these up):
\begin{equation}
dG=VdP-SdT+\sum ^{I}_{i=1}\mu _i dN_i
\end{equation}
By definition we state the following formula at constant temperature and pressure:
\begin{equation}
dG=\sum _{i=1}^{I}\mu _idN_i
\end{equation}
Therefore you are actually just expressing the Gibbs energy in a different way when your book quotes the chemical potential. In fact the very definition of the chemical potential is:
\begin{equation}
\mu _i=\bigg(\frac{\partial G}{\partial N_i}\bigg)_{P,T,N_{j\neq i}}
\end{equation}
I don't know your mathematical ability but if it helps...its pretty clear that this is just a simple Legendre transform of the internal energy to switch between $U$ and $G$.
(As an aside), with a little product ruling we get:
\begin{equation}
dG=\sum _{i=1}^{I}\big(\mu _idN_i+N_id\mu _i\big)
\end{equation}
This leaves us with the Gibbs-Duhem relation:
\begin{equation}
\sum ^{I}_{i=1}N_id\mu_i=-SdT+VdP
\end{equation}
Which is a handy result.
Also just to add on the end here about nucleation points... lets imagine a precipitation in solution....As the size of the nucleus grows (as more bits "clump" together) the Gibbs energy of the particle will rise ... this is due to two competing effects:
1) the interface interaction (increasing energy)
2) the volume free energy (decreasing energy)
The total energy is just the resultant of these processes ... at small size the surface interactions outcompete the volume energy liberation .... as the radius of the particle increases we see a switch.

I follow the math, but it's the logic I'm having trouble with. I edited my original question to make it more clear.

Is this first or second year university chemistry? (I read about chemistry in my spare time and don't think I've ever come across this so I'm a little curious).

@SherlockHolmes This is usually introduced in 1st year and expanded upon in second year...at least in the UK.

A really good book on classical thermodynamics is by Finn ... I read that before I tacked statistical mechanics ... It helped a lot ..

I thought about this a bit more and added a second edit.

@Yandle give me a couple of days but I'll get back to you.

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text:
Ionic bonds- are electrons transferred?
In an ionic bond, are electrons actually transferred from one atom to the other to form a positive and a negative ion OR are ionic bonds just extremely polar covalent bonds? Sorry if this seems like a simple question but I've heard this somewhere so just wanted to make sure.

Have a look at this answer of mine. It is quite related to your question.

So are you saying the electrons in say Cl in NaCl are just very far from the nucleus (so it isn't in Na's shells)?

I wouldn't put it that way. I think you termed it quite good with "extremely polar covalent bonds". You usually find some covalent contribution left in every bond. It might be very, very weak but it should be there. However, describing some bonds such as that in $\ce{NaCl}$ as purely ionic has advantages in most situations because it makes the treatment easier while giving quite accurate results.

This is not a simple question at all, in fact, it is an interesting insight in chemical bonding.  You make a valid assertion with your question:

are ionic bonds just extremely polar covalent bonds?

This answer is a simplistic way of looking at it.
According to ChemGuide page Electronegativity, there is no real clear-cut 'boundary' between covalent and ionic bonding. The website describes that there is a 'spectrum of bonding' based on the electronegativity (chart shown below) of the constituent elements.

Image source: UC Davis ChemWiki
So, the greater the electronegativity, the greater the polarity or "pulling power" on the bonding electrons towards elements with a higher electrongativity.
In the following diagrams (from the ChemGuide site linked before), A and B represent arbitrary elements, and the green and red dots represent bonding electrons. Starting from a situation where the polarity is zero as the electronegativity of the constituent elements are the same (e.g. diatomic molecules $\ce{O2}$, $\ce{Cl2}$ etc), as in diagram I.
As the polarity increases, the bonding electrons move towards the more electronegative member, causing the bond to become increasingly polar (II), then eventually to the point where the difference is enough (III) for:

To all intents and purposes, A has lost control of its electron, and B has complete control over both electrons. Ions have been formed.

I bolded the 'to all intents and purposes' to emphasise that this is a simplistic generalisation
(I)

(II)

(III)

As the lowest electronegativity is not zero (it is Francium at 0.7), there will always be an influence on the bonding electrons. So, the reality is that A does not completely lose its influence on the bonding electrons.
For example, if we were to look at a compound combining the most and least* electrongative elements (as appear in the diagram above): francium fluoride $\ce{FrF}$, even though fluorine has a far stronger electronegativity than francium, the electronegativity of the latter is not zero, so, the influence of the francium is very small, but not zero - indicating that purely ionic bonds don't exist.

As noted in the comments, cesium is likely to be less electronegative than francium, however the point made above still stands.

This is nicely summarised by an article from the University of Malta:

Pure ionic bonding is not known to exist. All ionic compounds have a degree of
  covalent bonding. The larger the difference in electronegativity between two
  atoms, the more ionic the bond.

Francium is probably more electronegative than caesium due to relativistic effects.

True, but it still makes a good example. Asides, I was just using a simplistic example - I'll make that more explicit.

As a side question: has FrF ever been synthesized, as far as you know?

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text:
Regarding the free energy requirements for reacting gases
Continued from my previous question here.

With regards to the Boltzmann factor, I am confused on what type of terms we can substitute for E. I know kinetic energy is one, free energy as well, but can we also do internal energy, enthalpy etc.?
From this link it looks like internal energy and molecular kinetic energy are related and for real gases the internal energy term contains other factors. Since U is a part of G, is it correct to say that kinetic energy in Maxwell Boltzmann Distributions make up a part of free energy?
I am still unclear as to whether I can interpret activation energy as a difference in free energy between average reactant energy and activated complex like diagram 1?
Is it correct to say that, reaction cannot occur unless the reactant have free energy above that of the activated complex, but even if this was satisfied other factors such as minimum collision kinetic energy, orientation etc. must also be met for reaction to occur?

Addendum for Q3 and Previous Post:
My confusion is with regards to the relationship between Maxwell Boltzmann Distributions like diagrams 1 and 2, if there even is one.
The M-B distribution implies to me that

activation energy $E_a$ (which I assume to be that in Arrhenius equation) is absolute given it is marked on an absolute energy scale
activation energy refers to a minimum absolute kinetic energy
activation energy is constant with temperature.

My confusion:

My impression has always been that activation energy (in the Arrhenius equation) refers to the difference in free energy between the transition state and reactant and that $E_a = \Delta G^{++}$ as noted in the latter two graphs, which appear to contradict the M-B Distribution. I'm not sure if there's a formal definition for activation energy that defines what "energy" it is.
The relationship between KE and activation energy (if any)
The effect temperature has on reactant and transition state free/kinetic energies. I believe that higher temperature lowers activation energy if it is the relative energy between the two.

Diagram 1:

Diagram 2:

I think what you are fundamentally saying is you are trying to relate classical thermodynamics to the molecular properties. It is best to have some firm familiarity with classical thermodynamics before attempting this. Classical thermodynamics does not require a molecular interpretation at all! (It helps a lot but is by no means necessary)!. I personally feel the answer I am about to write will help you more than trying to wade through terms and ideas/principles.... I hope so at least. If you don't like it then maybe someone else can provide a more quantitative solution.
1) The whole point of statistical thermodynamics is to provide a molecular interpretation. To do this we have to divide the system up into smaller sub-systems as I described earlier. The key to it is that there can be many "configurations" that give rise to the same macroscopic properties. The aim is to find the most probable one ... which is where your distributions come in.
It is the energy of the microstates that is what you put in so that you can calculate the various macroscopic properties you quote above. This is the best way to think about it. When you get to advanced theories this definition may change slightly. For now just accept that it is the energy of the micro state you are putting in.
In my honest opinion I don't know why courses teach the Maxwell-Boltzmann velocity distribution in a statistical thermodynamics course. I feel while a nice introduction to dealing with ensembles ... it leads to confusion (when I was learning this I had very similar questions with regard to what can be related to what, it only comes with time and exposure to the subject. So I know exactly where you are coming from!).
2) Whenever you see kinetic energy you should be thinking temperature rather than trying to get the individual molecular speeds... what does knowing the instantaneous position/velocity of a mole of gas particles at one instant in time really lead to? (not much)... I haven't really seen anyone combine the thermodynamic potentials with something like velocity. If anyone reading this has then I would love to hear what you have to say!! When I was first learning thermodynamics I clung on to what I was familiar with ... potential / kinetic energies ... now I don't really think about them when you say thermodynamics at all ... I don't think about $G$ as being a "part" of $U$ ... I just treat them as if they were separate (hmm how to phrase this) "properties" of a given system under certain conditions.
In fact I would go further to say that I view them as mathematical tools ... functions of variables of the system that I can manipulate to express molecular properties in terms of measurable quantities ....
3) Yes. You can indeed As is done by most theoretical chemists use Gibbs energies to calculate the thermodynamic activation barrier in the starting material and transition state.... The problem, which admittedly I may have confused you earlier about, is that this is done via the Arrhenius equation A kinetics equation. When in early undergraduate years you don't mix kinetics with thermodynamics. Later on it becomes (bar a factor of $2\pi$ or so ) acceptable to do so ... but you should not attempt this yet ....
A crucial point is that the activation energy graph's x-axis is the reaction coordinate. The reaction coordinate is a generalised coordinate which can be used (whatever it is) to describe the progression of the reaction.
Normally we deal very nicely in cartesian $x,y,z$ coordinates ... the idea of a generalised coordinate is that it can be used to reduce the dimensionality of the mathematical problem with some nifty rethinking of how we choose to describe the system .... for instance when talking about a simple harmonic diatomic molecule .. instead of choosing the location of both atoms in space (requiring 6 position coordinates) to fully describe the system as a function of time (as it oscillates back and forth ) i could use the relative separation (the internuclear distance) relate to the equilibrium bond length.... in this way it becomes much easier to describe the system ....
That was just one example of a generalised coordinate.... something that makes the description of the problem al little easier.... (especially if the system is subject to constraints, the generalised coordinates remove the constraint equation... anyway where were we.)...
Bottom line: ** there is a big difference between reaction coordinate and the Maxwell-boltzmann graph you linked.
4) This question in itself has hit on a very big topic. This is the realm of classical thermodynamics. What you need to understand is the difference between the standard Gibbs energy and the Gibbs energy .... For the following just imagine that $A$→$B$.
In a very brief overview it all boils down to this:
The Gibbs free energy is a mathematical function... don't try to pin down WHAT it is... its just another energy we can describe the system in .... It is a function of $P$, $T$ and $N$. Lets make it easier and say that the number of particles does not change. i.e $dN=0$ ...
Since the Gibbs energy is a function of $T$ and $P$ if we change either of these variables ... $G$ will change too .... The starting materials will have a certain $G$ and the products will have a certain $G$ ... The difference between these under standard conditions is fixed by definition that we are controlling the variables. This is like the thermodynamic drive for the process if it were to occur at our specified standard conditions .... Over the course of the reaction this value does not change since as I said it is the difference between the Starting Material and the Products under given conditions. .... This is also for pure values of the starting material and the product...
So far so good right?
As your reaction proceeds the composition varies i.e $dN\neq 0$. As I pointed out in another answer from the definition of the Gibbs energy this means that the value Gibbs function now changes... if you want a molecular picture it is due to the entropy of mixing adding in another factor to the "disorder" that has so far not been accounted for...
For an illustration please follow this link:
http://www.public.asu.edu/~laserweb/woodbury/classes/chm341/lecture_set5/lecture_set5.html
There are therefore two measures of the Gibbs energy. The one under standard conditions... unchanging for a given reaction... it is this one that has the $\Delta G^{\small o}$.. it is this one you get your electrode potentials off... it is this one that leads to the the equilibrium constant....
The other one is ever changing as the reaction proceeds since the composition of the reaction vessel is ever changing as $A$ goes to $B$. At equilibrium the net rate of change of A to B and B to A is now zero right... this in fact defines an equilibrium point.... shifting it slightly in one way or the other will cause the system to respond to re-establish equilibrium .... Most importantly we define equilibrium as the point where $dN=0$ .... for both types (in this case one would go up as the other went down ... ) but at equilibrium $dN=0$ .... This means that $dG=0$ too from the mathematics I showed you before ....
Therefore equilibrium is the point where the change in the Gibbs energies of two substances in a mixture is zero.... The reaction has no more drive if you like .... You can equivalently say their chemical potentials are equal too ... but leave that for now.....
So thats it ... Im sure if you ask question 4 to someone else they will give you a more complete answer.... but since its getting on for 2:00 am here I think I'm done. Hope that helps and again if I have made any errors ... please edit away :)
edit
What you are describing on the end of question 4 is know as the steric factor... you basically compare the frequency of successful collisions in the transition state geometry "linear" for example with respect to a full sphere of reactivity .... this gives you your factor :)
Edit
So the problem is your definition of the energy. I dot know if you have seen this before but perhaps this will help,...
Yes you can indeed use statistical mechanics to obtain a value of $G$ ... The best way to do it is to get an expression for the Helmholtz free energy and then just Legendre transform it to $G$ ... $G=A+pv$.
Whilst yes at a simple level the internal energy does indeed "contain" the kinetic energy ... when you learn a bit more about quantum mechanics .. we never talk about kinetic energy ... I think it would help to actually leave the classical results behind and focus on the actual version of events...
With reference to the uncertainty relation .. how do you think kinetic energy is something we can talk about at a molecular level? The answer is you don't. Once QM comes in it is not useful to talk about velocity.
That being said how do we describe motion ? after all the Schrödinger equation is just an equation of motion. All the Schrödinger equation describes is the time evolution of a vector state in a certain vector space - Hilbert space, Which is the space in which we describe quantum mechanics...
With this in mind ... all motion of the system is quantised ... can only have discrete values .... non-continuous values ... etc etc...
Translational motion is quantised... so is vibrational ... so is rotational ..... each degree of freedom has its own quantised set of discrete energy levels ....
Equipartition theorem uses classical ideas and distributions to distribute all the available energy across these states ... roughly it says that each has energy of $\frac 1 2 KT$....
To say $U$ includes kinetic energy is a primitive way of saying the internal energy is a function that describes the energy of all the modes in the system... (you would need a distribution to see how they are populated mind) ....
Equipartition theorem doesn't get it quite right when you consider the low temperature theoretical and experimental heat capacity of a substance..... Even at low temperature you are still expected to populate all modes with $\frac 12 KT$ ... but this doesn't happen since $KT$ does not have enough energy to populate said modes ... i.e it doesn't reach the quantised energy level requirement .... therefore fewer modes are available at low temperature...
Bottom line is: just try and separate the different preconceived competing ideas you have when it comes to statistical mechanics .... have you had a lecture course on this yet?
it is certainly not about having "enough Gibbs energy to react"... The Gibbs energy between a reactant and transition state $\Delta G ^{\ddagger}$ is not a variable thing as you are saying ... (obviously for a given set of chemicals),.... If it changed all the time then why would we write them down to compare them? The Gibbs energy in your picture is not the same Gibbs energy that goes to zero at equilibrium ...
\begin{equation}
\bigg(\frac{dG}{dq}\bigg)_{T,P}=\Delta _r G^{\small o}+RT\ln Q
\end{equation}
at equilibrium the left hand side .... the gradient of the graph in that link i sent you before ... is zero ...
The rate of change of the Gibbs energy with respect to the extent of the reaction is by definition zero at eqm.
When equilibrium is reached $Q=K$ that is to say the reaction quotient becomes the equilibrium constant ....

From here, I assume a relation between microstate and G can also be made? 2. The reason I say G includes molecular KE is that G=U+PV-TS and U includes KE. 3/4 My questions arise from graphs like this I don't understand what "energy" is on the x-axis and why one can mark a point called activation energy when it's a relative value not an absolute state. At a high level, reactants need to meet min. G (of transition state) and steric factor to react?

Ill edit my response to try to address these. :)

I added an addendum my OP for Q3 and my previous post to outline the source of my confusion/misconceptions so they are easier to address. I want to get a grasp of classical concepts before thinking about QM so for now, is it ok think the primitive way and view molecular KE as "velocity" that is part of macroscopic U?

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text:
What is the molecular formula of the nitrate of a metal Q that forms a carbonate of the form QCO3?

A metal $\ce{Q}$ forms a carbonate and a nitrate. The molecular formula of the carbonate of $\ce{Q}$ is $\ce{QCO3}$.  What is the molecular formula of the nitrate of $\ce{Q}$?

The answer says it is $\ce{Q(NO2)3}$ which I think is incorrect; I think the answer should be $\ce{Q(NO3)2}$. Am I right?

Yes, you were right.

OK, this takes my back to my high school days when I have to deal with my chemistry/gym teacher, which I previously mentioned on this forum (surely to the disbelief of many), which may be a barometer on the quality of science education in some locales. The obvious correct answer is, as you have argued because you have some elemental foundation in chemistry. The question most likely suffers from a typo. Those unsure should not be your first source on chemistry. – AJKOER 3 mins ago

The carbonate anion is $\ce{CO3^{2-}}$. This tells us that the metal cation is doubly positively charged:
$$ \ce{Q^{2+} + CO3^{2-} -> QCO3} $$
The other two anions mentioned in your question are

the nitrite anion $\ce{NO2^-}$
the nitrate anion $\ce{NO3^-}$

Since the question you're concerned with asks for the formula of the nitrate of $\ce{Q}$, the correct answer would be $\ce{Q(NO3)2}$.
However, the compound $\ce{Q(NO2)2}$ is, at least from a charge perspective, also correct. But that is the nitrite of $\ce{Q}$.

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text:
Understanding a concentration term
I am having a problem wrapping my mind around the idea that in $230\:\mathrm{mL}$ of $0.615\:\mathrm{M}$ $\ce{KCl}$ there is $0.141\:\mathrm{mol}$ of $\ce{Cl^-}$ ions. I guess I still do not understand moles.
Is this calculation saying that there $0.141\:\mathrm{mol}$ of $\ce{K}$ and $\ce{Cl}$ in one mole of $\ce{KCl}$?
I know this may be a silly question, but I just want to make sure I am following the correct way of thinking when it comes to this concept.

Your concentration is 0.615 M. The M stands for molar concentration, which is the number of moles per liter of solution. Thus, your concentration of 0.615 M (mol/L) means that in one liter of solution there will be 0.615 moles of solute (KCl). Similarly, a solution with concentration of 0.615 g/L would have 0.615 grams of solute per 1 liter of solution.
A mole is just like any other unit for amount, except that it is built on a very large number. Since, for example, there at $6.02\times 10^{23}$ atoms of chlorine in 35.45 grams of chlorine, we generally do not discuss numbers of atoms (or molecules) as a useful measurable amount. How much would a dozen atoms of chlorine weigh? Much less than any of our balances or scales could measure!
One mole of stuff is an amount of stuff equal to the number of carbon atoms in 12 grams of $\ce{^{12}C}$, most common isotope of carbon. This number happens to be Avogadro's Number - $6.02\times 10^{23}$, which is really large! This is also the number of atoms or molecules present in a number of grams equal to one formula weight of that substance, so it is a convenient number. We know that when we weigh out 12.01 grams of carbon, 1.008 grams of hydrogen, 15.99 grams of oxygen, etc., we know that we always have the same number of atoms in each sample, because each is one mole. We tend not to worry too much about the large number hiding in the definition of one mole.
Now, if you have one mole of a compound, like $\ce{KCl}$, then you know that there is one mole of the $\ce{KCl}$ formula unit in that sample. Since each $\ce{KCl}$ formula unit has one $\ce{K+}$ and one $\ce{Cl-}$, then each mole of $\ce{KCl}$ has one mole of $\ce{K+}$ ions and one mole of $\ce{Cl-}$ ions.
Now, you have your 0.615 M solution, and you only have 230 mL of it. How much $\ce{KCl}$, $\ce{K+}$, and $\ce{Cl-}$ is in that sample? Well, you would have 0.615 moles of each in 1 liter, but you do not have one liter. You have 230 mL. Since there are 1000 mL in a liter, we can do a little maths:
$$230\text{ mL}\times \dfrac{1 \text{ L}}{1000\text{ mL}}\times\dfrac{0.615\text{ mol KCl}}{1\text{ L}}=0.141 \text{ mol KCl}\equiv 0.141 \text{ mol}\,\ce{K+}\equiv 0.141 \text{ mol}\,\ce{Cl-}$$

One thing that may help is to keep in mind that a "mole" is simply a name for a number, like "dozen" (12) or "gross" (144) or "score" (20).
The difference is that a "mole" is a very large number (approximately 6.02x1023) - one that is typically only reached when you're dealing with things the size of atoms.
So 0.141 moles of KCl breaks into 0.141 moles of K+ and 0.141 moles of Cl-, just as 2 dozen eggs can be broken down into 2 dozen yolks and 2 dozen shells (and two dozen egg whites).
"Molar" simply means "moles per liter", and can be treated similarly to eggs per carton. If there are two dozen eggs per carton and you have three quarters of a carton left, you have 2 dozen eggs/carton * 0.75 carton = 1.5 dozen eggs left. You also have 1.5 dozen yolks and 1.5 dozen eggshells, because each egg has both a yolk and a shell. Likewise, having 750 mL of a 2 Molar solution of KCl means that you have 2 moles KCl/liter * 0.75 liters = 1.5 moles KCl. You also have 1.5 moles of K+ and 1.5 moles of Cl-
Keep in mind that just like "dozen" and "score", you have to keep track of what you're measuring. A dozen cars and a dozen wheels are two different things. Two boxes with 1.5 dozen toy cars per box have 3 dozen cars total but 12 dozen wheels, because there are 4 dozen wheels per 1 dozen cars (which is entirely equivalent to saying there are 4 wheels per 1 car). Likewise, for every mole of CaCl2 there are 2 moles of Cl- (which is equivalent to saying there are two chlorines for each "molecule" of CaCl2), so you'll have to work that factor of two into your calcuations if you're going from quantities of CaCl2 to quantities of chloride ions.


text:
What are some of the simplest chiral molecules in biological systems
I have understood generally for a while there is a significant preference in "life" for so-called "left-handed" (chiral) molecules, though I'm reading things now that say "life" sugars are right-handed, while "life" amino-acids are left-handed.
I'm wondering: how simple do these molecules get (particularly the molecules that are in intelligent life, if there happen to be any that are exclusive to intelligent life)?

Don't get hung up on the "left-handed" and "right-handed" designations for amino acids and sugars. Those are just conventions for distinguishing one mirror image (enantiomer) of a chiral molecule from another. It just so happens that based on a particular way of drawing (Fisher projection) amino acids and sugars that it appears that the configurations are "left-handed" or "right-handed."
As for how simple chiral biological molecules get, the simplest ones that I can think of are alanine or lactic acid. I'm not sure what you consider "intelligent", but these are likely to be found in all life (as we know it).

I'm not trying to get hung up on it, as I understand it has to do with how it bends light (at least in one classification system), but that's probably good advice for others with a similar question. I'm only looking for directing further research as far as needed for science fiction (a story I have in mind that probably will never get written), and Google wasn't really able to "understand" my question. These two look like a good enough starting point - Thanks!

as far as "intelligent", I was going for anywhere from things that have a "central" nervous system to things that we consider self-aware - pretty loose definition, really


text:
Will I get bromine by electrolysis of the brine from the Dead Sea?
I know that electrolysis of a saturated solution of sodium chloride with an inert anode (eg., graphite) is a handy way to obtain pure chlorine. I have a sample of saturated salt solution from the Dead Sea which is reported to have up to 5% bromine concentration, the domination anion being $\ce{Cl}^{-}$, of course.
Will I get bromine on the anode by electrolysis of Dead Sea water?
In general, if I have a mixture of chlorides and bromides, what determines the ratio of electrolytically produced pure $\ce{Cl}_2$ and $\ce{Br}_2$?

Is your sample from the northern Dead Sea itself, the southern ponds where the hotels are, or the southern ponds where the factory mines the brine? Br and Cl contents differ across these areas.

It is from Ein Gedi beach, about 15m from shoreline, filled close beneath surface. Cooling down to room temperature caused precipitation of a small amount of clear crystals.

The short answer is probably a bit, depending on the potential you apply.
There are at least a few factors involved:

How easily are $\ce{Cl-}$ and $\ce{Br-}$ oxidized?
$$\begin{align*}\ce{2Cl^{-}_{(aq)} -> Cl_{2(g)} +2e-}\; &E°_{\mathrm{ox}}=-1.36\ V\\\ce{2Br^{-}_{(aq)} -> Br_{2(aq)} +2e-}\; &E°_{\mathrm{ox}}=-1.09\ V\end{align*}$$
At standard activities, the two ions are pretty similar in oxidation potentials, so there is probably a range of potentials where both reactions occur to a similar degree or it might be possible to achieve some partial selectivity for bromine formation. The Nernst equation can be used to calculate equilibrium concentrations for a given potential, but so long as the potential applied is well over the standard oxidation potential, the exact values don't really matter.
What difference do the concentrations make?
In most cases, what determines how much of either reaction occurs is mass transport. For a redox reaction to occur, the reacting molecule must be very close to the electrode surface (<10 nm). For a planar electrode at a single potential in a standing solution, mass transport occurs by diffusion and the current that flows should be directly proportional to concentration in the bulk. In principle, the Cl to Br ratio should be the same as their activities in solution, though in some situations, other factors like the speed of the electron transfer can be limiting, so this may not be the case, especially since there is so much chloride compared to bromide.
What about the electrodes?
As electrolysis is a surface phenomenon, the electrode and its interactions with the reactants can be important. Assuming you pick an electrode which doesn't itself react with the chloride or bromide, the exact material can affect the behaviour of the cell. Even though the Nernst equation may predict a certain oxidation potential, interactions of the reactants and the electrode may result in varying amounts of additional potential (overpotential) needed to cause the reaction.
What happens to the bromine?
A certain amount of bromine can dissolve in water, but depending on the pH, it can rapidly disproportionate back to bromate and hypobromate: $\ce{Br2 + 2OH- → Br- + OBr- + H2O}$ Also, the chlorine gas can react with the bromide to make bromine and bromine is also known to react with metals (use carbon electrodes?). In short, you can probably create an electrolytic cell that produces some bromine, but it's difficult to say in what ratio with chlorine and what its exact fate would be.


text:
Mechanism for hydroxyethylation on nucleobases (e.g. Thymine)
I'm trying to figure out the following reaction mechanism
There's a reference paper by Yanagi and Akiyoshi(1959) showing the hydroxyethylation of imides resulting in the following

I have two questions :
(1) Why would the hydroxylethylation occur in the 1 position instead of the imide at the 3 position? I would imagine that this reaction occurs for the more acidic amine, but I have no idea why there doesn't seem to be any literature about that reaction.
(2) Why would this reaction produce a HydroxyEthyl and not a urethane, as shown here :
Any insights would be appreciated, thanks!
Yanagi and Akiyoshi : http://pubs.acs.org/doi/abs/10.1021/jo01090a601
Reaction in red (from a patent application): http://www.huntsman.com/performance_products/Media%20Library/a_MC348531CFA3EA9A2E040EBCD2B6B7B06/Products_MC348531D0B9FA9A2E040EBCD2B6B7B06/Carbonates_MC348531D1109A9A2E040EBCD2B6B7B06/files/jeffsol_alkylene_carbonates_synthesis_of_hydroxy_alkyl_urethanes.pdf

The second part of the question is easy to answer: the imide attacks at C3 of propylene carbonate instead of C1 (the carbonyl carbon) because carbonates are stabilized (double ester-resonance), whereas carboxylate is a so-so leaving group.
The initial product from nuclephilic attack at C3 is a carbonate half-ester, which subsequently loses CO2 to form the observed N-hydroxyethyl imide.

I would expect that the two N-H groups would be similar acidity. The conjugate base from deprotonation at N-1 is resonance stabilized through both carbonyls, through the carbonyl at C-2 directly and through the carbonyl at C-4 via the C-5/6 double bond.


text:
HNMR Assignment
I have a molecule

and i need to assign the hnmr. I have a printed spectrum

So far my answer is:
chemical shift_______splitting pattern_______coupling constant
 2.32                quartet            6.6 Hz & 13.2 Hz

 2.66                singlet                   -

 3.66                triplet                6.5 Hz

 5.11         triplet of doublets         1.5 Hz & 16.8 Hz

 5.81   doublet of doublet of triplets   6.9 Hz & 10.2 Hz & 13.8 Hz & 17.2 Hz

I am having trouble assigning them to the hydrogens. I think the H on the OH is a singlet and the H on the right hand side of the double bond is a ddt as it has cis (13.8 Hz) and trans (17.2 Hz). Then on the left of the double bond, the cis H would be the quartet as it has cis coupling (13.2) and the trans H would be the td as it has trans coupling (16.8 Hz).
But if these are correct then does that mean the four hydrogens on carbons 2 and three are the triplet group?

You haven't quite rationailsed the spectrum correctly; If you look at your molecule you will see that you have 6 distinct 1H species, but you have only described 5. Take particular note of the signals that you have described as a td at 5.11ppm. This should be symmetrical about the centre peak if it was a true td, and this signal clearly is not. So, it actually comes from two overlapping signals; one with a large coupling (~17Hz) and the other a medium coupling (~10Hz). Also, a quartet will only have one coupling constant - this will be the separation of any of two adjacent peaks in the quartet, and so is 6.6Hz, as you have described. So perhaps this is not a true quartet.
When assigning your peaks, you should first look at your chemical shifts;

Vinylic protons typically come in the range 3.5-8.5, but commonly 5-7ppm
alkyl protons alpha to a heteroatom 3-4.5ppm
alkyl protons alpha to vinyl group 1.5-2.5ppm
hydroxyl protons can come just about anywhere, but true alkyl-OH 1H will typically come 1-5 ppm

So from this information, you should be able to broadly assign the 6 protons in your molecule.
You then need to look at coupling constants to further refine your assignments. For vinylic groups, you can rely on the following general coupling constants:

trans coupling Jbc 14-19Hz
cis coupling Jac 4-12 Hz
geminal coupling Jab -4-4 Hz
vicinal coupling Jcd 4-10 Hz

This should allow you to complete the assignment of this spectrum. See how you go from here.
A useful approach to these type of assignment problems, especially if you know the structure, is to look at the molecule and try to predict the spectrum you'd expect, rather than looking at the spectrum and working out what's what.

Looking at your molecule, I'd expect:

Ha to come in the vinylic region, and appear as a dd (cis and geminal coupling to Hb and Hc)
Hb to come in the vinylic region, and appear as a dd (trans and geminal coupling to Ha and Hc)
Hc to come in the vinylic region, and appear as a ddt (cis, trans and vicinal coupling to Ha, Hb and Hd)
Hd to come in the alkyl region alpha to a double bond, and appear as a dt (coupling to Hc and He)
He to come in the alkyl region alpha to a heteroatom, and appear as a t (coupling to Hd)
Hf to come somewhere and appear most likely as an exchange broadened singlet

==============
Edits to answer questions raised in comment to the original answer.
Below are two splitting diagrams to help you rationalise the splitting for the peaks centred around 2.3 ppm and 5.10 ppm.

Let's first look at your upfield peaks (2.10ppm), which is the splitting diagram on the left. This comes from signal Hd, and should show coupling to Hc (which will give rise to a doublet) and also He (giving a triplet). So this signal is actually a doublet of triplets. Normally if the doublet coupling and triplet coupling are quite different, then you observe two clear sets of triplet peaks. However, in this case, Jdc is approximately the same as Jde (within the natural linewisth of the peaks), and so you get overlap of some of peaks, creating what looks like a quartet. It is still a dt, with 6 lines, but you can't resolve them all. So, you still have two coupling constants, but both couplings are about 6.6Hz.
The second diagram (on the right) is for your sets of signals around 5.10 ppm. This is actually the overlap of two independent signals. Ha will occur as a dd, with cis coupling to Hc (~14Hz), and geminal coupling to Hb (~2Hz, so very small, and realy only broadens the peak in this case.) This is the signal on the right.
Hb will also occur as a dd, with trans coupling to Hc (~17Hz), and again, geminal coupling to Ha. This is the signal on the left. It just so happens (life is like that sometimes), that one set of lines for each signal overlap, to create what looks like an aysmmetrical triplet. So, here are your two signals, Ha and Hb. Integrating the area under the peaks serves as a valuable tool to tell you how many signals each peak represents, and in this case you would have clearly seen that the signals at 5.10ppm come from 2 protons.
As a note, geminal couplings in vinylic systems is usually quite small, and often is of similar magnitude to allylic coupling (4J couplings, such as Jad, and this is why the peaks for protons a and b actually look a little more complicated than just doublets of doublets. The main take home message about double bond couplings at an introductory level really comes from the cis and trans couplings.

I dont understand how you tell which hydrogen doesn't have a peak. So you are saying that the td at 5.11ppm is two peaks, how do you know what it really is and how would you write this down as I have only been given 5 peaks with 6 distinct hydrogen environments. Also, for the quartet I got a coupling constant of 6.6 and 13.2Hz so do i take an average or just ignore the 13.2Hz one.


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Is cyclobutadiene antiaromatic?
Cyclobutadiene is very unstable. But, some sources claim that this instability can be attributed to other factors such as ring and angle strain rather than antiaromaticity.
According to some, cyclobutadiene is simply non-aromatic (as opposed to antiaromatic) because it doesn't even have a fully conjugated pi system.
What is/are the real reason(s) for its instability?

I always stumble at the words anti-aromatic. Is this a well defined concept? I believe not. From my point of view, your whole question is a little bit philosophical, as we dive in the realm of interpretation only. Also one should not confuse instability with high reactivity...

If we go back to your earlier question on Frost diagrams (I've reproduced the key figure below), we see why simple molecular orbital theory or the "$4n+2$" rule suggests that benzene is aromatic while cyclobutadiene is antiaromatic.

Forming a planar, conjugated, 6-membered ring and placing 6 π-electrons in it creates a π-system that is energetically more stable than 3 conjugated C=C bonds (e.g. in hexa-1,3,5-triene). On the other hand, forming a planar, conjugated, 4-membered ring and placing 4 π-electrons in it creates a π-system that is energetically less stable than 2 conjugated C=C bonds (e.g. in butadiene).  For these reasons we say that the first system, benzene, is "aromatic", while the second system, cyclobutadiene, is "antiaromatic". Other measurements and physical phenomenon such as reactivity, bond lengths, ring currents, etc. support these conclusions.
Free cyclobutadiene has been observed as a transient intermediate.  Further theoretical analysis suggests that due to its lack of aromaticity it will distort from a square structure to a rectangular one with alternating single and double bonds (Jahn–Teller effect).

sources claim that this instability can be attributed to other factors such as ring and angle strain rather than antiaromaticity

It turns out that other systems involving the cyclobutadiene skeleton have been prepared and studied.  Derivatives of both the cyclobutadiene dication[1] ($4n+2$, $n=0$) and the cyclobutadiene dianion[2] ($4n+2$, $n=1$) have been prepared and studied.  Their stability is much less than that of benzene, but much more than that of cyclobutadiene. Placing two charges in these small rings results in extremely high coulombic repulsions in both the dianion and dication, and this may be a significant part of the explanation as to  why they are less stable than say, benzene.  The fact that they exist as planar structures, with appropriate ring currents and are more stable than cyclobutadiene suggests that the instability observed with cyclobutadiene is not due to ring strain and angle strain alone.  Rather the aromatic stabilization - antiaromatic destabilization suggested by simple MO theory seems to be involved.

References

Olah, G. A.; Staral, J. S. Novel aromatic systems. 4. cyclobutadiene dications. J. Am. Chem. Soc. 1976, 98 (20), 6290–6304. DOI: 10.1021/ja00436a037.
Takanashi, K.; Inatomi, A.; Lee, V. Y.; Nakamoto, M.; Ichinohe, M.; Sekiguchi, A. Tetrakis(trimethylsilyl)cyclobutadiene dianion alkaline earth metal salts: new members of the 6π-electron aromatics family. Eur. J. Inorg. Chem. 2008, 2008 (11), 1752–1755. DOI: 10.1002/ejic.200800066.

I saw mention that dianion is non-planar according to one study.

It was computational paper for unsubstituted dianion, not derivtives though.

Strictly cyclobutadiene is non aromatic. Still it answers.

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When mixing lye and citric acid, which citrate will form?
I'm trying to combine precise amounts of citric acid and lye to get sodium citrate with minimal amount of reacted acid/base. I'm not so sure how stochiometry works, but I could sure learn that from some web article.
What's the real problem is, that I don't know which of three sodium citrates will form. Depending on the number of $\ce{Na}$ molecules, the amounts of $\ce{NaOH}$ greatly differs.
Which forms and why?

Citric acid has three pKs: 3.1, 4.8 and 5.4.  If you add 3 equivalents of lye, you will get essentially only trisodium citrate.  If you add a lesser amount, you will have a mixture of the various citrates (at least from microscopic point of view), as well as some remaining citric acid if you add less than one equivalent.
One "equivalent" being one mole of lye per mole of citric acid.

What is the pH?  You can find the pKa's of the three acidic protons in citric acid and use the Hendersson Hasselbalch equation to find what percent of each proton is ionized at that pH.
http://www.chemteam.info/AcidBase/HH-Equation.html

Depending on the citrate form, there will also be some ammount of unreacted acid.

Yes, the concentration of all the possible citrates will be determined by the pH.

Too bad I'm using Earl Grey tea to determine pH.

Make an Arduino pH meter.  You have to buy the electrode.  Also, universal paper is cheap.

I really wish to make arduino pH meter. I can't afford arduino however, and I have horrible problems when programing Atmel chips - both Atmega 328 and AtTiny.


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Why it is not yet possible to produce ethanol selectively in the Fischer Tropsch reaction?
The Fischer Tropsch reaction between $\ce{CO}$ and $\ce{H2}$ has the potential to give a wide range of products. However by judicious choice of catalytic metal it is possible to produce $\ce{CH4}$, $\ce{CH3OH}$, and other alkanes selectively. $\ce{Fe}$ and $\ce{Co}$ are most commonly used catalysts where $\ce{Fe}$ produces less $\ce{CH4}$, more oxygenated products and more alkenes and $\ce{Co}$ produces more $\ce{CH4}$, less oxygenated products and more alkanes. There are a number of other catalysts that can be used, for example $\ce{Rh}$ which can produce $\ce{C2}$ and higher oxygenates and hydrocarbons. This includes the production of $\ce{CH3OH}$. However, it is not selectively produced but is produced in a mixture of other products and in small amounts. My question is why can $\ce{CH3OH}$ not be produced selectively?

Wikipedia says : <<The Fischer–Tropsch process is a collection of chemical reactions that converts a mixture of carbon monoxide and hydrogen into liquid hydrocarbons.>> However, you can obtain alcohols in small quantities, but that won't be satisfying for a synthesis. Do you ask about ethanol or ethane? Give more details about your research, and how did you think about this question.

The question is not about how you can produce ethanol, but why it cannot be done selectively. You can produce ethanol using a Rh catalyst I think but it can't be done selectively and that is what i want to know the reason for. I am not enquiring about ethane.

The Fischer Tropsch reaction between carbon monoxide and hydrogen has the potential to give a wide range of products. However by judicious choice of catalytic metal it is possible to produce methane, methanol and alkanes selectively. Fe and Co are most commonly used catalysts where Fe produces less methane, more oxygenated products and more alkenes and Co produces more methane, less oxygenated products and more alkanes. There are a number of other catalysts that can be used, for example Rh which can produce C2 and higher oxygenates and hydrocarbons. This includes the production of ethanol.

However, it is not selectively produced but is produced in a mixture of other products and in small amounts. My question is why can ethanol not be produced selectively?

in Fischer-Tropsch synthesis-like reaction there are (roughly) following processes:

$\ce{CO \rightarrow  Surf-CO}$
$\ce{H2 \rightarrow  2 Surf-H}$
$\ce{CO + Surf-(CH2)_{n}-H \rightarrow  Surf-CO-(CH2)_{n}-H}$
3'. $\ce{CO + Surf-H \rightarrow  Surf-CO-H}$
$\ce{Surf-CO-(CH2)_{n}-H + 4 Surf-H \rightarrow
    Surf-(CH2)_{n\,+1}-H + H2O }$
$\ce{Surf-CO-(CH2)_{n}-H + 3 Surf-H \rightarrow
    HO-(CH2)_{n\,+1}-H }$
$\ce{Surf-CH2-CH2-(CH2)_{n}-H \rightarrow Surf-H  +
    CH2=CH-(CH2)_{n}-H}$
$\ce{Surf-(CH2)_{n}-H + Surf-H \rightarrow H-(CH2)_{n}-H}$

All processes happen simultaneously, and tuning catalyst and reaction conditions it is possible to hinder some processes. However, as you can see, as long as you have chain growth at all (that is necessary to have $\ce{C2}$ products, like ethanol), you cannot stop $\ce{C2}$ surface particles to participate in further chain growth, producing $\ce{C_{3+}}$ products. So, if chain growth happens at all, a mixture of products with different chain length is always obtained, and there is no way around it. On the other hand, on proper catalyst in proper conditions it is possible to prevent chain growth, but still allow reduction of carbon monoxide, gaining either methane or methanol depending on catalyst. So, in practice it is possible to obtain one of the following

Methane
Mixture of terminal alkenes and alkanes
Methanol
Mixture of linear terminal alcohols (OK, this case is a bit more complicated, as other oxygenates may be produced as well, giving rather complex mixture, but that's beyond the point)

It is still possible to obtain pure ethanol indirectly (using hydration of isolated ethylene), or using other processes. It is also possible to obtain product mixture where some products, say C2-C4 alcohols, will be dominating.

The mechanisms you provide do offer a good explanation why ethanol isn't a common product, but your answer doesn't really draw this out.

authors: አቡበክር መሃመድ አቡበክር


text:
Terminology of redox reactions: which species are the oxidizing and reducing agents?
I've got a question regarding 'notation' for redox reactions. Suppose you are asked to identify the oxidizing and reducing agents for a simple reaction like this one:
$$\ce{Cl2(aq)}+\ce{2KI(aq)} \longrightarrow \ce{2KCl(aq)}+\ce{I2(aq)}$$
Now are you supposed to say $\ce{Cl}$ is the oxidizing agent or $\ce{Cl2}$? Also, is $\ce{KI}$ the reducing agent or $\ce{I}$ or $\ce{I^-}$?
By the way, if you were instead asked to identify which species got oxidized, would you again say that $\ce{KI}$ is oxidized or $\ce{I}$ or $\ce{I^-}$?

Related 2714, 9329, 1315

Ignoring spectator ions,
$\ce{Cl2 +2 I- ->2Cl- +I2}$
So, $\ce{Cl2}$ is the oxidant and $\ce{I-}$ is the reductant.

So is it incorrect to, for instance, say that $\ce{I}$ is the reducing agent?

Yes - you must say I- is the reducing agent.


text:
Effect of temperature on equilibrium
Why is exothermic reactions are favoured by low temperature while endothermic reactions are favoured by high temperature?

http://chemistry.stackexchange.com/questions/15758/determining-if-reaction-is-endothermic-or-exothermic-using-equilbrium-constant/15762#15762

I am not able to understand the answer you have written. Could you please explain it a little more simply.

I'll try and simplify the answer.

Looking at the image above, imagine the products sitting above the letter D and the reactants sitting below the letter B.  Now, imagine the particles involved in the reaction moving since they have kinetic energy.  It is much easier for the reactants to climb up the top of the hill labeled C then it is for the products to climb up the hill.  Once a reactant climbs the hill and slides down to products it will not be able to go back to reactants by climbing up the hill unless we give them more energy by raising temperature.  Raising temperature allows the products to climb up the hill and go to reactants.
The forward reaction in the diagram is exothermic and the reverse is endothermic.  So, the exo is favored at low T and the endo at high T.
You can also look at the equation delta G = delta H - Tdelta S.
Make up some numbers for delta H and delta S.  You will see that endo reactions becomes more spontaneous as you raise T and exo reactions become more spontaneous as you lower T regardless of delta S.

You have to know that when you have a chemical equilibrium that is subjected to a change in concentration, temperature or pressure,then  this equilibrium will readjust itself to counteract the effect of the applied change and a new equilibrium is established.
So, if the reaction is exothermic. This means it produces heat. So, if we decrease the temperature (low temperatures), the equilibrium will tend to counteract this effect and favor the reaction that produces heat to increase the temperaure of the system.
On the other hand, if the reaction is endothermic. This means it consumes heat. So, if we increase the temperature (high temperatures), the equilibrium will tend to counteract this effect and favor the reaction that consumes heat to decrease the temperaure of the system.
In sum, exothermic reactions are favoured by low temperature while endothermic reactions are favoured by high temperature


text:
What would be the effect of the addition of an inert gas to a reaction at equilibrium?
Why does a dissociation reaction shift to the right with the addition of an inert gas?

http://chemistry.stackexchange.com/questions/15553/applying-the-le-chateliers-principle/15555#15555

Please specify the nature of this dissociation reaction. Does it contain gas phase?

@YomenAtassi I am asking in general. I still don't know much about the topic.

Suppose there are two groups of people, your friends and enemies, that are forced to occupy the same room.
Most probably there would be a fight which would spread.
Now, if I add lots of pillows, so that there is no real mixing of your friends and foes, would there be a fight? Yes, but as the only thing to fight is using pillows, there would be no harm done.
So, basically you see adding something neutral, prevents association, in other words, it increase dissociation. It needs not to be inert gases, any chemical gaseous enough and neutral enough should do the job.

Dissociation obviously increases the number of moles.
The addition of an inert gas can affect the equilbrium, but only if the volume is allowed to change.
There are two cases on which equilibrium depends. These are:

Addition of an inert gas at constant volume:
When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change.

Hence, when an inert gas is added to the system in equilibrium at constant volume there will be no effect on the equilibrium.

Addition of an inert gas at constant pressure:
When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase. Hence, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases.

Consider the following reaction in equilibrium:
$$\ce{2 NH_3(g) ⇌ N_2 (g) + 3 H2 (g)}$$
The addition of an inert gas at constant pressure to the above reaction will shift the equilibrium towards the forward direction(shift to the right) because the number of moles of products is more than the number of moles of the reactants.
Please read about the Effect of adding an Inert Gas

I met a question about adding an inert gas at constant temperature. What will happen in that case?

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

@AaronJohnSabu Can you tell the complete question??

@AnanyaPateriya This was a long long long time back. I'm sorry I can't recall it

But will the equilibrium constant change? If not, why?

@Aditya $Q < K$ at constant pressure: https://youtu.be/kBGNRloiZq0?t=886
See Ben's answer below.
The $K$ value will change based on temperature and may change based on pressure. So, I assume constant temperature and $K$ does not change.
Please read: https://bit.ly/3baepEv
As I've not visited this topic in quite some time, I have doubts about $K$ in the case of addition at constant volume.
Also, to clarify other comments, Ideal Gas Law states $PV = nRT$, so note changes from that and "number" of moles is valid vernacular as is number of kilograms albeit they're clunky expression.

Nick's answer is good. Let's add a little maths.
Let's take an example dissociation reaction $\ce{A<=>B + C}$ for which $K_p=1$. Since
$$K_p = \frac{P_B P_C}{P_A}$$
one equilibrium scenario is $P_A =P_B = P_C = 1 \text{ atm}$. The total pressure $P_T = 3 \text{ atm}$.
If we add an inert gas at constant pressure, the total pressure cannot change. Thus, the partial pressures of the three components $\ce{A}$, $\ce{B}$, and $\ce{C}$ must decrease. If we add some inert gas $\ce{D}$ so that $P_A=P_B=P_C=P_D=\frac{3}{4} \text{ atm}$, the total pressure is still $P_T=3\text{ atm}$. The reaction quotient $Q_p$ is now:
$$Q_p = \frac{P_B P_C}{P_A}=\frac{(\frac{3}{4}\text{ atm})(\frac{3}{4}\text{ atm})}{\frac{3}{4}\text{ atm}}=\frac{3}{4}\text{ atm}<K_p$$
Since $Q_p<K_p$, the equilibrium will shift toward products.
At constant volume, the partial pressures of the three components $\ce{A}$, $\ce{B}$, and $\ce{C}$ can remain constant and equilibrium will be unchanged. Adding inert gas $\ce{D}$ will increase the pressure of the system.

excellent answer sir, is there any other reason for the change in equilibrium?


text:
Are the rules proposed by Slater for the calculation of effective Z right?
Today I studied Slater's rules for calculating the effective nuclear charge. One particular line, "The shielding power of the d and f orbital is lesser in comparison to the p and s orbitals", seems to appear in every single book and website I have studied from.
Slater's rules does not seem to take into account the orbitals for calculation of $Z_\mathrm{effective}$. Nevertheless, I don't disagree with the logic behind the statement.
Thus my question is: Do Slater's rules actually account for this logic? If they don't, is it faulty? And, how do Slater's rules account for the shielding effect of the individual orbitals?

Slater's rules are pretty good, but they're not perfect. Consider that he published his work in 1930: Phys Rev 1930 v. 36, 57
There are a number of people who have modified Slater's scheme to better reflect shielding effects and correct for experimental ionization energies and electron affinities (and similar properties) in later rows of the periodic table.
If I remember correctly, one such modification is that d and f orbitals shouldn't have as much shielding as suggested by Slater, but the correction is small. There's another scheme that uses an exponential correction instead.
For example, Wikipedia gives effective nuclear charges from Clementi and Raimondi, who did full self-consistent quantum calculations and you can compare these to Slater's.
The main take-home from Slater's rules is that they work pretty well for a wide range of elements, including 1st, 2nd, and 3rd row. I think the corrections generally tweak things when we get into the 4d and 5d transition metals and some of the lanthanides and actinides. Not bad for an 84-year old paper!


text:
Using solubility guidelines to predict a reaction of two ionic, water-soluble salts

Will a reaction occur in the following case? If so, write a net ionic equation for it:
$\ce{(NH4)2SO4 (aq) + ZnCl2 (aq) -> ?}$

The book says that all possible combinations of positive and negative ions lead to water soluble compounds, all of the ions remain in solution. No reactions occur.
Now I am a little confused. Why didn't $\ce{(NH4)2}$ join with $\ce{Cl2}$ and why didn't $\ce{SO4}$ join with $\ce{Zn}$? Or did they?

Of course $\ce{NH4+}$ can join $\ce{Cl-}$ and $\ce{Zn^2+}$ can join $\ce{SO4^2-}$, but the formed salts are completely soluble in water, and they completely dissociate in water to gives the corresponding ions.
On the other hand, the starting salts:  $\ce{(NH4)2SO4}$ and  $\ce{ZnCl2}$ are completely soluble in water, and they completely dissociate in water to gives the corresponding ions.
So, all of the ions remain in solution, as if no reaction occurs.


text:
Which solution has the highest total molarity of ions and why?
(a) 0.008 M $\ce{Ba(OH)2}$
(b) 0.010 M $\ce{KI}$
The answer is "a" but I do not understand why. The book says that "a" is a strong electrolyte with a total ion concentration of 0.024. How did they calculate the concentration from the information that they gave? And why is it a strong electrolyte based on that calculation. I thought strong electrolytes were (solutes) ions that dissolved completely in a solvent/solution?

Barium hydroxide is a slightly soluble salt. It dissociates in water and gives 3 ions (one ion of barium and two ions of hydroxide) with the concentration of $0.008 \ce{ M}$ for each one. This makes the total concentration of ions to be $0.024 \ce{ M}$. While the total concentration of $\ce{KI}$ is $0.02 \ce{ M}$. So the solution of barium hydroxide is a stronger electrolyte as it has a higher concentration of dissociated ions.
In your book, they treat barium hydroxide as a strong electrolyte, i.e. it dissolved completely in water and dissociates to its constituent ions. This is of course an approximation. While $\ce{KI}$ is a true strong electrolyte; it dissolved completely in water and dissociates to its constituent ions.
So, I agree with you with the definition of strong electrolyte, and I hope I clarify the ambiguity.

You are right in thinking that a strong electrolyte is something that dissociates completely in solution.$\ce{Ba(OH)2}$ and $\ce{KI}$ then, being a strong electrolytes, will effectively dissociate into $\ce{Ba^2+}$+ $\ce{2OH-}$ and $\ce{K+}$+ $\ce{I-}$, respectively.
Going into your example, it's a just a matter of multiplication. We can assume that the volume of the solution is 1L for simplicity. Since you begin with 0.008 M $\ce{Ba(OH)2}$, we can say that you begin with 0.008 mol of $\ce{Ba(OH)2}$. In solution this becomes 0.008 mol of $\ce{Ba+}$ and 2(0.008) mol of $\ce{OH-}$, or 0.024 mol of ions in total. Likewise for KI, you are left with 0.010 mol of $\ce{K+}$ and 0.010 mol of $\ce{I-}$, or 0.020 mol of ions in total.
Converting these values back to molarity (by dividing by 1), we get that the concetration of ions in the $\ce{Ba(OH)2}$ solution is 0.024 M, whereas it is only 0.020 M in the $\ce{KI}$ solution.

authors: Александър Марковски


text:
How do I write a valid CIF, PDB or XYZ file from the coordinates listed below?
I want to create a 3D molecular file using coordinates taken from:

Hrynchuk, RJ; Barton, RJ; Robertson, BE (1983). "The crystal structure of free base cocaine, C17H21NO4" (PDF). Canadian Journal of Chemistry 61 (3): 481–487. doi:10.1139/v83-085.

Here are the coordinates:
              X             Y            Z
C(1)          0.5419        0.1631       0.7095
C(2)          0.4579        0.2047       0.5345
C(3)          0.3080        0.1713       0.5168
C(4)          0.2903        0.0286       0.5727
C(5)          0.3950        0.0004       0.7409
C(6)          0.3870        0.1080       0.8678
C(7)          0.4858        0.2192       0.8450
C(8)          0.1028        0.2512       0.3219
C(9)          0.0323        0.2713       0.1411
C(10)         0.0857        0.2213       0.0238
C(11)         0.0189        0.2435       -0.1419
C(12)        -0.0995        0.3166        -0.1875
C(13)        -0.1528        0.3671      -0.0719
C(14)        -0.0880        0.3462      0.0951
C(15)         0.5172        0.1331      0.4127
C(16)         0.7261        0.1009      0.3465
C(17)         0.6418        -0.0439     0.8543

N(1)          0.5324        0.0158      0.7182

O(1)          0.2257        0.1933      0.3470
O(2)          0.0560        0.2836      0.4306
O(3)          0.4616        0.0531      0.3097
O(4)          0.6481        0.1713      0.4393

H(1)          0.6210        0.1840      0.7230
H(2)          0.4640        0.2980      0.5170
H(3)          0.2820        0.2340      0.5830
H(4)          0.1930        0.0110      0.5730
H(5)          0.3050        -0.0240     0.4930
H(6)          0.3850        -0.0960     0.7770
H(7)          0.4090        0.0810      0.9760
H(8)          0.2870        0.1410      0.8610
H(9)          0.4440        0.3140      0.8150
H(10)         0.5560        0.2330      0.9400
H(11)         0.1580        0.1680      0.0530
H(12)         0.0640        0.2050      -0.2240
H(13)         -0.1470        0.3330      -0.3060
H(14)         -0.2320        0.4140      -0.103
H(15)         -0.1280        0.3810      0.186
H(16)         0.7330        0.0190      0.3810
H(17)         0.8130        0.1470      0.3710
H(18)         0.6990        0.1060      0.2370
H(19)         0.7250        -0.0340      0.836
H(20)         0.6470        -0.0070      0.961
H(21)         0.6230        -0.1410      0.861

Converting this to XYZ coordinates, namely to:
43
Cocaine 1983
C          0.5419        0.1631       0.7095
C          0.4579        0.2047       0.5345
C          0.3080        0.1713       0.5168
C          0.2903        0.0286       0.5727
C          0.3950        0.0004       0.7409
C          0.3870        0.1080       0.8678
C          0.4858        0.2192       0.8450
C          0.1028        0.2512       0.3219
C          0.0323        0.2713       0.1411
C          0.0857        0.2213       0.0238
C          0.0189        0.2435       -0.1419
C         -0.0995        0.3166        -0.1875
C         -0.1528        0.3671      -0.0719
C         -0.0880        0.3462      0.0951
C          0.5172        0.1331      0.4127
C          0.7261        0.1009      0.3465
C          0.6418        -0.0439     0.8543
N          0.5324        0.0158      0.7182
O          0.2257        0.1933      0.3470
O          0.0560        0.2836      0.4306
O          0.4616        0.0531      0.3097
O          0.6481        0.1713      0.4393
H          0.6210        0.1840      0.7230
H          0.4640        0.2980      0.5170
H          0.2820        0.2340      0.5830
H          0.1930        0.0110      0.5730
H          0.3050        -0.0240     0.4930
H          0.3850        -0.0960     0.7770
H          0.4090        0.0810      0.9760
H          0.2870        0.1410      0.8610
H          0.4440        0.3140      0.8150
H          0.5560        0.2330      0.9400
H          0.1580        0.1680      0.0530
H          0.0640        0.2050      -0.2240
H          -0.1470        0.3330      -0.3060
H          -0.2320        0.4140      -0.103
H          -0.1280        0.3810      0.186
H          0.7330        0.0190      0.3810
H          0.8130        0.1470      0.3710
H          0.6990        0.1060      0.2370
H          0.7250        -0.0340      0.836
H          0.6470        -0.0070      0.961
H          0.6230        -0.1410      0.861

Opening this with Accelrys gives this structure. https://i.stack.imgur.com/ijb19.png

@AbelFriedman It is not cartesian coordinates. While xyz file is quite easy to construct, I can't guess an easy method to convert them, so it seems more convenient to find structure of compound of interest in more suitable form.

@permeakra: Wikipedia has the transformation matrix for fractional coordinates to cartesian coordinates.  The crystal system is monoclinic, which makes the transformation rather painless, nothing that a few lines of awk won't do.

Although its now old rasmol and rasmac is very relaxed about exact format as long as the atom type appear after the coordinates.  Its also free, and very fast to use if you need to rotate large protein models. It will do most simple angle and distance calculations etc. Basic, but very useful for quick look stuff.

Generally speaking, babel from the Openbabel suite will convert chemical file formats.  The openbabel wiki has a page with examples on input in free-form fractional coordinate format, and the abstract of the paper has the unit cell dimensions.
Single CIF files can be requested from the Cambridge Crystallographic Data Centre.

+1 Indeed, this would be my exact answer. It's the reason I added "free form fractional" to babel, since you run across this problem in older papers.

I have a first year chem knowledge, so at the moment I haven't the foggiest as to what "free-form fractional coordinate format" means. I made that request but they don't have such a CIF file at this data centre.

@BrentonHorne: It's an oblique coordinate system formed by the unit cell of the crystal.  The angles between the axes are the angles of the unit cell, and the units are dimensionless and expressed in terms of the edges of the unit cell.

OK, how do I do that for my example? I have no idea what to set the unit cell as.

@BrentonHorne: Look no further than the abstract for unit cell parameters: a, b, c, \beta. \alpha == \gamma == 90 deg, since the crystal system is monoclinic.


text:
Applying the Solubility Guidelines is AgClO4 soluble or insoluble?
The book I am using is "General Chemistry: Principles and Modern Applications
by Ralph H. Petrucci. On page 158 the Solubility Guidelines for Common Ionic Solids is given. Rule 3 states that, "Salts of silver, lead, and mercury(I) are insoluble, however, my book states that $\ce{AgClO4}$ is soluble. Can someone explain?

As you say, these are only guidelines, and nothing is absolute.

Silver perchlorate is voluble.

Different guidelines will say slightly different things to convey the same idea. Many I've seen don't explicitly mention silver, but still produce the same result for most compounds. I don't have that book but if it's anything like these, there's probably another line that says salts of anions like nitrate and chlorate are soluble.
This table is worded a bit better to make the possibility of exceptions more explicit:

Salts containing Group I elements are soluble ($\ce{Li+, Na+, K+, Cs+, Rb+}$). Exceptions to this rule are rare. Salts containing the ammonium
  ion ($\ce{NH4+}$) are also soluble.
Salts containing nitrate ion ($\ce{NO3-}$) are generally soluble.
Salts containing $\ce{Cl^-, Br^-, I^-}$ are generally soluble. Important exceptions to this rule are halide salts of $\ce{Ag^+, Pb^{2+}}$, and $\ce{(Hg2)^{2+}}$.
  Thus, $\ce{AgCl, PbBr2}$, and $\ce{Hg2Cl2}$ are all insoluble.
Most silver salts are insoluble. $\ce{AgNO3}$ and $\ce{Ag(C2H3O2)}$ are common soluble salts of silver; virtually anything else is insoluble.
Most sulfate salts are soluble. Important exceptions to this rule include $\ce{BaSO4, PbSO4, Ag2SO4}$ and $\ce{SrSO4}$ .
Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements
  ($\ce{Ca, Sr,Ba}$) are slightly soluble. Hydroxide salts of transition
  metals and $\ce{Al^{3+}}$ are insoluble. Thus, $\ce{Fe(OH)3, Al(OH)3, Co(OH)2}$ are not
  soluble.
Most sulfides of transition metals are highly insoluble. Thus, $\ce{CdS, FeS, ZnS, Ag2S}$ are all insoluble. Arsenic, antimony, bismuth, and lead
  sulfides are also insoluble.
Carbonates are frequently insoluble. Group II carbonates ($\ce{Ca, Sr, Ba}$) are insoluble. Some other insoluble carbonates include $\ce{FeCO3}$
  and $\ce{PbCO3}$.
Chromates are frequently insoluble. Examples: $\ce{PbCrO4, BaCrO4}$
Phosphates are frequently insoluble. Examples: $\ce{Ca3(PO4)2, Ag3PO4}$
Fluorides are frequently insoluble. Examples: $\ce{BaF2, MgF2 PbF2}$.

Since the author has made the table freely available, I incorporated it into the answer.


text:
Given volume of reagents find the chemical formula of nitrogen oxide
$$\ce{NH3 + O2 -> N_xO_y + H2O}$$
I know that there is $\pu{1 m^3}$ of both the nitrogen oxide and the ammonia and the volume of oxygen on the left side is $\pu{1.25 m^3}$.
I think that I should use the ideal gas equation. It is also obvious that there is only one atom of nitrogen in the oxide - I guess that there is also only one oxygen.
Could anyone give me any hints?

As a hint, you could write balanced equations for oxidation to N2O, NO, NO2 and N2O5.  (Did I forget any nitrogen oxide?)  Then you can compare moles (or volumes, since we are talking gases) of starting materials and products.


text:
Why does sugar make dough turn to liquid
While making dango with a friend we decided to try to make the dough a bit sweeter.  We already had a large firm lump of dough (rice flour, glutinous rice flour and water) so we sprinkled a bit of sugar on top.  While kneading the dough in my hand I started to feel it run through my fingers.  The entire lump had turned into a goopy drippy mess.
Our best guess was that the sugar soaked up all the water, turning it into a bunch of flour particles in free flowing sugar water, but I cannot find any sources on the internet to explain what exactly happened.  Can anyone explain the reaction that causes this?

Possibly a question worth asking on Seasoned Advice SE


text:
Hydrolysis of cyclic acetal

What happens to 1,5-dimethyl-6,8-dioxabicyclo[3.2.1]octane when it is hydrolyzed under acidic conditions?
I know that under acidic conditions the first thing to likely happen is protonation. Does it matter which oxygen is protonated first?
Once that oxygen has been protonated it is going to want to return to neutrality.
Now we have our carbonyl restored (sort of). However, we still want to return to neutrality. Water can attack.
Then we deprotonated the attached water and regenerate the acid-catalyst and we have our original starting compound.
Is this the correct mechanism and is the product correct?

In your middle structure, you show water attacking the methylated carbon on the left side of the drawing, yet it is the methylated carbon on the right side of the structure that carries some positive charge due to resonance.  Examine nucleophilic attack by water at this positive carbon and see what happens

Your product is correct, but the mechanism isn't quite right. In the first step, one of the oxygens will be protonated (either is possible). The other oxygen "kicks out" the one activated in the first step. So far so good. However, in the next step, water will attack the "carbonyl" carbon to give a hemi-acetal and a hydroxyl group. Repeating a similar set of steps reveals the ketone and second hydroxyl.

The key difference here is that by adding to the pi system, water reacts with the C-O pi-antibonding orbital, which is the LUMO of the system

Thanks! I had water initially attack on the side you suggested, but I just couldn't see what would happen next! This helps a ton!


text:
How do I figure out the relative polarity of organic compounds?
It's a question on my quiz which I guessed right and couldn't do the reasoning:

Based on the intermolecular forces, predict the boiling points of the following substances and list them in the order of increasing boiling point. State your reasons accordingly.

Dimethyl ether $(\ce{CH3OCH3})$
Ethanol $(\ce{CH3CH2OH})$
Propane $(\ce{CH3CH2CH3})$

Right away, I see the hydrogen bond in ethanol, so it must have the highest boiling point.
I couldn't decide on the other two, so I drew lewis structures (kind of) for both

I ignored the hydrogens because they don't affect the polarity (correct me if my lewis strctures are wrong).
This is what I guessed, and I was right. When my teacher takes this question up, he said that propane is non polar, which means it only interacts via london forces, which means it has the lowest boiling points out of these three.
But how is propane non polar though, when I asked, he said "I don't have an answer for this, life is hard."
I mean, if I draw the structure in the bad way, it does look symmetrical and balanced out:

But if I draw it in this way, it doesn't help me when I'm trying to figure out the polarity, because everything will look non polar. I can even say water is in a linear shape because the H's and the O line up perfectly in a line, which is not the correct picture of reality.
My biggest question is, how do I figure out if something is polar or not.

General steps to figure out the polarity (or at least relative polarity) and subsequently the relative boiling point of compounds:

Acquire the correct geometry
Estimate dipole moments
Sum up estimated dipole moments
Check for other intermolecular interactions

Acquire Correct Geometry
This step requires some knowledge about the geometry of hydrocarbon compounds. But it seems that you already know about the tetrahedral bonding centers of saturated hydrocarbons. This is then important later on for the dipole forces. Never try to linearize the carbon backbone just to try to make sense of this, it will do more damage than good.
Estimate Dipole Moments
The dipole moments that you have to estimate will be those along bonds. There is a larger dipole moment in the ether than the hydrocarbon without heteroatoms, because the oxygen atom is strongly electronegative. As such, there will be a dipole along the $\ce{C-O}$ bond.
Sum Up Estimated Dipole Moments
These dipoles can be understood as vectors along bonds, pointing from positive to negative partial charges, and can be summed up for the total dipole moment of a molecule. The larger the absolute value of this dipole vector is, the larger the dipole-dipole interactions between molecules: Subsequently, the higher the forces between the molecules of this substance.
Check for Other Intermolecular Interactions
One other main intermolecular action is, as you already pointed out in your question, hydrogen bonding. This one is quite strong, but isn't found for every compound out there. Basically, it is when a highly polarized $\ce{H-X}$ bond can be formed, where $\ce{X}$ can be oxygen, nitrogen, chlorine or some other electronegative atom (with respect to carbon). The hydrogen that is sandwiched in the hydrogen bond has to be (formally) bonded to such an electronegative atom as well, otherwise it wouldn't be as free to create those bonds (as you also have realized, because the hydrocarbon hydrogens are really quite boring from an intermolecular interaction perspective).
Result
Based on the steps above, we can generate a table:

The more "Yes" a substance has, the higher the intermolecular forces. Based on these data, assumed by us, we can generate a list (from highest boiling point to lowest):

Ethanol
Dimethyl ether
Propane

Look for atoms connected to other atoms with different electronegativities.  The larger the difference in electronegativity, the greater the polarity of a bond.  In propane, we just have carbon (EN=2.55) connected to carbon and hydrogen (EN=2.20).  Propane is weakly polar, it has a very small dipole moment, around 0.08 D. Dimethyl ether, on the other hand has carbon connected to oxygen (EN=3.44) and hydrogen.  The electronegativity difference between carbon and oxygen is 0.89 D, pretty large.  Dimethyl ether has a dipole moment of 1.3 D.


text:
Write a Net Ionic Equation to Represent the Reaction of Aqueous Ammonia with Propionic acid
I would like to know how do I begin to solve a problem like this?
That is finding the formula equation?
I would assume that it would be
$$ \ce{NH3 + CH3CH2COOH -> NH3 + CH3CH2COOH + H2O} $$
I am really confused as to how I can balance this equation. Do I give the ammonia a hydrogen? Do I break up the acid into smaller units?
From there I would like to know the best way for me to switch the equation into an ionic equation and then finally a net ionic equation.
By the way the net ionic equation is
$$ \ce{CaCO3 (s) + 2HC2H3O2+ (aq) -> CO2 (g) + H2O (l) + Ca^2+ (aq) + 2C2H3O2^- (aq) } $$
I am just clueless as to how they got there.

everyday-chemistry is more for questions that are things that one might run into outside of a classroom setting, like (for lack of a handy example) "why does my toothpaste have tin fluoride in it?"  I think this is more of an acid-base question, so I will retag it as such.

We know that ammonia is $\ce{NH3}$ and propanoic acid is $\ce{CH3CH2COOH}$.
Ammonia is a basic substance, and propanoic acid is... an acidic substance. More specifically, ammonia fits the definition of a Bronsted-Lowry base, so it will act as an $\ce{H+}$ accepter in solution.
Arrhenius acids are expected to increase the concentration of $\ce{H+}$ in solution, and propanoic acid is no exception. In water as propanoic acid dissociates...
$\ce{CH3CH2COOH ->[H2O] CH3CH2COO- + H+}$
This is where the $\ce{H+}$ comes (and is later bonded to the ammonia via)...
$\ce{NH3 + H+ -> NH4+}$
After those two steps occur, the products $\ce{NH4+ + CH3CH2COO-}$ are formed. Your ionic equation is as follows...
$\ce{NH3 + CH3CH2COOH -> NH4+ + CH3CH2COO-}$
There appear to be no spectator ions (ions that appear on both sides of the equation), so there is nothing to cancel out. As such, this is ALSO the net ionic equation. They both appear to be balanced.
So, yes, you had to do both. You were supposed to add an $\ce{H+}$ to the ammonia, but that happens if you dissociate the acid.
As for the equation with the calcium, carbonate ions sometimes create carbon dioxide and water (via carbonic acid readily decomposing, like in sodas).

Write a net ionic equation to represent the reaction of aqueous strontium hydroxide with nitric acid: So I know that nitric acid is $\ce{HNO3}$ and strontium hydroxide is $\ce{Sr(OH)2} $, thus $\ce{Sr(OH)2 + HNO3}$ What should be my second step after this? Place water and these two elements on the product side of this equation and start balancing?

You're right that there will be water in the products, but it's more about the mechanism than just saying "water exists".
$\ce{Sr(OH)2}$ is going to dissociate into $\ce{Sr^{2+} + 2OH-}$, and the nitric acid will dissociate into $\ce{H+ + NO3-}$. The $\ce{H+ + OH-}$ will make water, and the other two are going to make $\ce{Sr(NO3)2}$.
But that's soluble in water, so you can't add those two. Your ionic equation is $\ce{Sr^{2+} + 2OH- + H+ + NO3- -> Sr^{2+} + 2NO3- + H2O}$. Cancel out the spectator ions and that's your net ionic equation. And of course, balance.

So my last step should be to balance the equation? I thought that was supposed to be the first step? Also, is there any rhyme or reason to dissociation. Do they always split into free elements or polyatomic ions?

Dissociation comes down to energy and bonding. When the reaction $\ce{Na+ + NO3- -> NaNO3}$ occurs (or even $\ce{Cl-}$), the ionic bond that holds the reaction in place comes from the sodium's sole valence electron which "fills the gap" and forms an octet with the 7 valence electrons of chlorine. Ionic bonds are really strong, but they break apart in water because it takes less energy (the energy to break the bond and form one with water, overall, loses energy and increases entropy). The $\ce{NO3-}$ wouldn't "dissociate" because it's bonded so well. There's also solubility charts to memorize.

I am just clueless as to how they got there.

Well, you're not the only one. I really don't see how you get to equation 2 from equation 1.
To find out how to get the acid-base reaction equation, you first need to know which compounds react as bases (i.e. take up a proton) and acids (i.e. deliver a proton).
In your case, ammonia is the base (because the nitrogen has a lone pair that can be accessed by a proton) and propionic acid is the acid (nomen est omen). Then you know that the final equation has to be
$$ \ce{NH3 + CH3CH2COOH <=>[][\ce{H2O}] NH4+ + CH3CH2COO-} $$

Also, how often does one need to balance these types of reactions or do they usually end up balanced or working themselves out as balanced?

Acids (and bases) can be multiprotic, which means that they can give off (or accept) more than just one proton. An example for a diprotic acid would be carbonic acid $\ce{H2CO3}$.
As a result of this, it is possible to write several (equilibrium) reactions. For example:
$$ \ce{2NH3 + H2CO3 <=> NH3 + NH4+ + HCO3- <=> 2NH4+ + CO3^{2-}} $$
Where this equilibrium then lies depends on the equilibrium constants for each of these reactions.

The second equation can be expanded some more to show "intermediate" steps:
$$ \ce{CaCO3 + 2 CH3COOH -> Ca^{2+} + H2CO3 + 2CH3COO- \\ <=> Ca^{2+} + 2CH3COO- + H2O + CO2 ^} $$
Here carbonic acid decomposes to give carbon dioxide which bubbles out of the solution, thereby making the reaction stay out of the equilibrium zone.
I hope this clarified it somewhat for you.

The second equation is the "net ionic equation". I just fixed it. Nonetheless, when it comes to "Acid Base Reactions" should the primary action I need to take to write (emphasis) one be to represent the reaction so that it shows that a hydrogen atom is being donated (positive proton) from the ionic acid? Also, how often does one need to balance these types of reactions or do they usually end up balanced or working themselves out as balanced?

@Cetshwayo I've edited my answer.


text:
How do Gilman reagents open epoxides?

From what it looks like, Gilman reagents are good nucleophiles, and will attack at the less-substituted side of an asymmetric alkane because Gilman reagents don't necessarily need a substantial amount of partial positive character at the site of attack.
However, don't strong nucleophiles open epoxides in an SN2 like manner? Where's the inversion?

In the first reaction, the reaction doesn't take place at a stereocenter, so there's no way to observe the inversion.
In the second reaction, the stereochemistry in the product is R, R. The Gilman reagent approaches opposite the C-O bond. After bond rotation, we see the product in the indicated conformation.

First one or second one? @jerepierre

@Dissenter I've clarified my answer to address both examples.

but how do Gilmans open epoxides? From the less substituted side ... but does opening proceed with inversion or retention of stereochemistry?

@Dissenter The reaction goes with inversion.

you were wrong; the second product is RR.

@Dissenter You're right, it's RR. However it's still inversion.


text:
Stoichiometry of Gaseous Reactions
The Question:
How many liters of $H_2S$ would be required to react with excess oxygen to produce $12.0$ $L$ of $SO_2$? The reaction yield is $88.5$%. Assume constant temperature and pressure throughout the reaction.
My Approach:
I used dimensional analysis to convert from liters of $SO_2$ to $moles$ of $H_2S$ and then used Avogadro`s Law ($\frac{V_1}{n_1} = \frac{V_2}{n_2}$) to find the volume of $H_2S$ in liters.
Dimensional Analysis:
$\frac{1}{.885} * \frac{12.0L}{1} * \frac{1 g}{1 mL} * \frac{1 mol SO_2}{64g} * \frac{2 mol SO_2}{2 mol H_2S} * \frac{1 mol H_2S}{34 g} = 6.23$ $mol$ $of$ $H_2S$.
Avogadro`s Law:
$\frac{12.0L}{166 mol} * \frac{ L }{6.23 mol}$
End Result = $.028 L$
^ The end result does not seem right and any help in remedying my approach or shedding light on another approach would be much appreciated. Thanks in advance!

Is the reaction $\ce{H_2S}+ \frac{3}{2}O_2 ->H_2O+SO_2$?

Is it given that the density of $\ce{SO_2}$ is 1 gram per mL? As it's a gas I think it would be much lower.

I never really learned the method of solving this stuff with Avogadro's Law, but here's an almost identical way.
Your balanced equation is as follows...
$\ce{2H2S + 3O2 -> 2SO2 + 2H2O}$, and I think you agree on that by looking at your calculations.
At a given temperature and pressure, $1 mol = x L$. For example, at STP (standard temperature and pressure) this is $1 mol = 22.4 L$. It wasn't exactly defined what temperature and pressure are, but they are constant, so we can use a variable $x$ to denote some conversion (which cancels out anyway, I assume giving rise to Avogadro's Law).
$12.0L_{\ce{SO2}} * \frac{1mol_{\ce{SO2}}}{xL_{\ce{SO2}}} * \frac{2mol_{\ce{H2S}}}{2mol_{\ce{SO2}}} * \frac{xL_{\ce{H2S}}}{1mol_{\ce{H2S}}} = 12.0L_{\ce{H2S}}$
But this is assuming $100\%$ yield. I see you divided by $0.885$ above, which is correct, but I've gotten a metric ton of these wrong before, so I prefer the proportions way.
Regardless, $\frac{12L_{\ce{H2S}}}{0.885} = 13.6L_{\ce{H2S}}$
I think you definitely overcomplicated things by assuming a density of $1g/mL$ among other things. Your dimensional analysis also seems strange near the end of the left hand side where you somehow cancel $g$ with $mol_{\ce{SO2}}$.
If I were to do it with Avogadro's Law (taking a guess here), I'd still get $12.0L$ because the mole ratio is the same. And then I'd divide like you did.


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Mass Spectroscopy peak assignments in but-3-en-1-ol
i have the mass spectrum for the molecule but-3-en-1-ol. The Mr of this is 72
There are four peaks that are labelled which I have to name the species responsible for forming that peak.
The peaks are:
m/z = 66
m/z = 67
m/z = 71
m/z = 73 - this peak is labelled as (M + H)^+
My answer so far...
The peak at m/z=73 is the but-3-en-1-ol with one added hydrogen, somewhere to the double bond?
For the other peaks am I just removing hydrogens as the mass only decreases by 8 in total so i couldn't remove a carbon or the oxygen.
Does this mean that the peaks are:
m/z=71 is the removal of 1H
m/z=67 is the removal of another 4H (5H in total)
m/z=66 is the removal of another 1H (6H in total)
is this is correct? Which hydrogens are removed first and what does each species look like for each peak, and are there any charges on the species produced. Also is there a special name for this sort of fragmentation, e.g alpha cleavage?

73 peak, if it has low intensity (~4-5% of molecular peak) may be a result of presence of C13. Same for other peaks: most of them should have low-intensity pairs one atomic mass unit heavier, coming from replacement of one atom with C13. For some heavier molecules those peaks may dominate.

Here is the mass-spectrum http://webbook.nist.gov/cgi/cbook.cgi?ID=C627270&Mask=200 and I don't see 66/67 m/z ions. Could you recheck and clarify?

As you have pointed out, the peak at M = 73 is M+H, protonation is however more likely to occur at the oxygen atom.

In my opinion, peaks at M = 66 and 67, i.e. with a rather small mass difference to $\mathrm{M^+}$, are negligible.
Common sense and the link in permeakra's comment suggest that the major fragmentation of but-3-en-1-ol occurs along the red dotted line, yielding an allylic and a hydroxymethyl fragment.


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How can I deduce the linearity of XeF2 from the IR spectrum?
The vibrational modes of $\ce{XeF2}$ are at $555$, $515$, and $213~\mathrm{cm^{-1}}$ but only two are IR active.
Explain why this is consistent with $\ce{XeF2}$ having a linear structure.
From what I know, since it has $3$ atoms it should have $3n - 5$ vibrational modes, thus $3(3) - 5 = 4$ modes. Yet the problem tells me it only has $3$, which would suggest a non-linear shape. How to explain this?

Let's start with the rule of mutual exclusion, it will come in handy.  It states that,  for centrosymmetric molecules (molecules with a center of symmetry), vibrations that are IR active are Raman inactive, and vice versa.
A linear $\ce{AB2}$ molecule will have $(3n-5) = 4$ vibrational modes, but because it is linear 2 of the modes are degenerate.  $\ce{CO2}$ is an example of a linear $\ce{AB2}$ molecule, look at the following diagram showing the 4 vibrations for $\ce{CO2}$ and notice that the bending vibration ($v_2$) is degenerate.  That is, there is no difference between those 2 bending modes, they are identical.

Notice too that $v_1$ is the only vibrational mode that does not change the dipole moment of the molecule, therefore it is the Raman active mode. Since the linear molecule is centrosymmetric, the Raman band does not appear in the IR, nor do any of the IR bands appear in the Raman spectrum. Hence, for a linear $\ce{AB2}$ molecule we would expect 2 IR active bands and 1 Raman band. This is consistent with the spectral data you reported.
If the $\ce{AB2}$ molecule is bent rather than linear (water for example), then there are $(3n-6) = 3$ vibrational modes as pictured below.  All of these vibrations result in a change in the dipole moment.  Therefore, for a non-linear $\ce{AB2}$ molecule, all 3 of the bands will be IR active. Since the molecule is not centrosymmetric all 3 bands may also appear in the Raman spectrum (note: while a change in dipole moment is required for an IR-active band to appear, a change in polarizability is required for a Raman band to appear; looking at the water vibrations pictured, they should alter both the dipole moment and the polarizability of the molecule). Hence, for a non-linear $\ce{AB2}$ molecule we would expect 3 bands in the IR and 3 in the Raman spectrum. This is not consistent with the spectral data you reported above.

In summary, your spectral data is consistent with $\ce{XeF2}$ having a linear geometry.
Note: Thanks to @porphyrin for pointing out that bent $\ce{AB2}$ molecules are indeed not centrosymmetric.

your comment 'Therefore, for a non-linear AB2 molecule, all of the bands will be IR active and since the molecule is centrosymmetric, none will be Raman active' is wrong. In fact in $C_{2v}$ all normal modes are Raman active.

The most reliable way to do this is to use point group tables.
A non-linear molecule such as type $\ce{AB2}$ e.g. water has $C_\mathrm{2v}$ symmetry. (The normal modes are given by symmetry species $\mathrm{a}_1$ ($v_1$ & $v_2$) and $v_3$, $\mathrm{b}_2$.) From the point group, IR transitions only belonging to $\mathrm{A}_1$, $\mathrm{B}_1$ and $\mathrm{B}_2$ symmetry species occur. (These are symmetry species in same rows as $x, y, z$ in the third main column of the point group table)
Both $\mathrm{A}_1$ and $\mathrm{B}_2$  are also Raman active as is $\mathrm{B}_1$. (The Raman active species are the squared terms in column 4). The symmetry species $\mathrm{A}_2$ is also Raman active but is not present as a normal vibrational mode. Thus all symmetry species are Raman active in $C_\mathrm{2v}$. Thus as only one species is Raman active in your example, the molecule is not bent so must be linear.
As a check any molecule with a centre of inversion (such as linear $\ce{FXeF}$) has no common transition in the IR and Raman spectrum. All the frequencies are unique so there is no overlap, and the molecule is linear. In fact this is all you need to know to determine that it is linear.

So … IR active vibrations are those which transform just like $x,y$ or $z$. But which irreducible representations are possible for Raman active vibrations? Only $x^2, y^2, z^2$?

Raman depends on a molecules change in polarisability. Polarisabilty is proportional to volume, but as far as the photon is concerned it only 'sees' an area so any product xy, etc as well as xx etc. gives Raman allowed species.

Ah, so it’s not only the squared, but also the product terms.

authors: فارس الطريقي