text:
Please explain the shapes of the orbitals
For a lot of years, I had been believing that sphere was the most stable 3-dimensional shape. But after coming across the p,d and f-orbitals, I am unable to comprehend the fact that these orbitals have such crude shapes. Can we prove that these shapes(dumb bell and crossed dumb bell) are the regions in which the electrons would be most stable (No wave functions please. I am only in high school)?
P.S.: An intuitive answer would be much appreciated than a mathematical one.
P.S.S: This is not a duplicate question. I don't want to know how the shapes are like that. I know it is due to the probability functions, I want to know if it is just the rule of nature, or can we show it is because of stability.

I know you said no wavefunctions, so I really hate to say this, but it has to be said. You absolutely cannot rationalise the shapes of the orbitals without quantum mechanics. In fact, you cannot even rationalise the existence of orbitals without QM, because when you say that electrons can only go into the 1s, 2s, 2p, 3s.... orbitals, you are effectively saying they are only allowed to adopt a specific set of energies. And only QM can predict, and explain, this quantisation of energy. Any classical model of the atom (i.e. using the laws of electromagnetism) is doomed to failure.

@ShankRam Stop thinking about the orbitals, or you will invent some wild misconception and instill it in your head forever. I am serious. You are talking of orbitals as if they were real, which is not quite so. You are talking of their shapes as if they had shapes, which is not quite so. You are talking of them being stable, but in fact, the very word "stable" is absolutely inapplicable here. You may have called them "green" or "smiling" as well; that would make just as much sense.

Well, the shapes of s, p, d, etc. orbital come about because atoms are spherically symmetric. The spherical symmetry let's you split the wavefunction/orbital into 2 parts, a radial one (which basically just determines how many nodes your orbital has) and an angular one (which determines the shape), which are then multiplied with each other. The angular functions (called spherical harmonics), loosely speaking, correspond to different angular momenta and indeed the spherical shape is the one with the lowest angular momentum and in that sense "the most stable one" (don't take that too literal)...

... But, as has already been mentioned by others, quantum mechanics dictates that each energy level/orbital can only be occupied by 2 electrons of different spin (Pauli Exclusion Principle). Thus the electrons of each shell can't all go into the "most stable" spherical s orbital. Only 2 are allowed and the rest has to occupy the other differently-shaped orbitals. The same principle applies to the different shells (those are linked to the above-mentioned radial part of the wavefunction): not all electrons can go into the first and "most stable" shell.

@orthocresol, I do understand that the orbitals are just plots of the square of probability functions in 3--dimensional  space, but I don't get how the electrons are in such an arrangement. I am a little bit familiar with QM (like Arthur-Beiser modern physics book level), but the mathematical analysis is too much for a high school student. If you could interpret the same in an elegant way, I will surely get it. Btw, this is not a duplicate question. I don't want to know how the shapes are like that. I want to know if it is just the rule of nature, or can we show it is because of stability.

I'm guess that you have read about orbital shapes in the wikipedia article, or done a google search on the term.
In general, such "orbitals" shown are typically calculated for a lone electron, not any "real" multi-electron configuration.
Think of an orbital like a single loop of cotton fiber wound into a cotton ball. The planetary model of the electron-nucleus pair indicates that the electron is a solid ball following the fiber. The quantum model of the electron-nucleus pair indicates that the electron has no fixed position, but that it is essentially at all places on the loop at the same time.
So how big is the cotton ball in diameter? It has no limiting diameter, it depends on how hard you squeeze! So if I catch a Hydrogen atom with my magic tweezers, there is a finite probability according to the Schrödinger wave equation that its electron can be in orbit even further away than the dwarf planet Pluto, or inside the nucleus! So we we talk of an the "size" of an ion it is somewhat of an artificial abstraction, and the Schrödinger wave equation can't be "the gospel truth." (I don't mean to badly disparage the Schrödinger wave equation for it is very useful.)
Not to leaving you totally confused, the size of an ions does have some real physical significance. For instance consider table salt, $\ce{NaCl}$. This is really $\ce{NaCl}$. Using x-ray diffraction we can measure how close the $\ce{NaCl}$ are, so we know their "size."
Starting with the Aufbau principle for the atoms, chemists can predict the electron structure of an atom, say carbon. Given the structure of bonds in the atoms, say carbon and hydrogen, then chemists can predict how carbon and hydrogen will bond to form molecules. The fact that chemists can make predictions about molecular structure is the "proof" that the models work.
Howver weird things do have with real electron orbitals. For instance if there was a single "normal electron configuration" for chromium, then there would only be one chromium chloride. However chemists have synthesized three.

Chromium(II) chloride, also known as chromous chloride.
Chromium(III) chloride, also known as chromic chloride or chromium
trichloride
Chromium(IV) chloride

So there must be weird configurations that are stabilized by some sort of "hybridization." As another example all the C-H bonds in methane, $\ce{CH4}$, are the same because the four carbon orbitals (one 2s, and three 2p) orbitals of the carbon atom hybridize into four $sp^3$ orbitals that are equivalent.

To put such orbital shapes into perspective, a chemist thinks a unicorn is simply a hybrid of a horse and a rhinoceros. So a model helps making predictions, but you can't have a fixated belief that the model is the whole
truth.

Another example would be showing you a picture of a blueberry pie. Would just such a picture "explain" a blueberry pie? In order for the picture to have meaning you must have some underlying knowledge.

You know in general what a pie is.
You have know that pies contain fruit pieces, so this kind is made
from blueberries.
You know that pies are sweet.


text:
Does enthalpy equal heat when PV work is done?
The derivation that enthalpy equals heat at constant pressure goes like:
$$\begin{align}
H &= U + P_{\mathrm{int}}V \\
\Delta H &= \Delta U + \Delta (P_{\mathrm{int}}V)
\end{align}$$
If $P_{\mathrm{int}}$ is constant, then
$$\Delta H = \Delta U + P_{\mathrm{int}}\Delta V$$
We also have
$$q = \Delta U - w$$
If external pressure is constant, $w = -P_{\mathrm{ext}}\Delta V$. Then
$$q = \Delta U + P_{\mathrm{ext}}\Delta V$$
So in order for $\Delta H$ to equal $q$, we need $P_{\mathrm{int}} = P_{\mathrm{ext}} = \mathrm{constant}$. This is true when the system is open to a constant external pressure and no PV work is done. But when there's PV work done (and the temperature is still constant), $P_{\mathrm{int}}$ clearly changes, so does $\Delta H$ still equal $q$?

The external pressure applied by the surroundings on the system at the interface between the system and the surroundings is always equal to the force per unit area exerted by the system on the surroundings at the interface, even if the process is irreversible.  It's just that, in the irreversible case, the pressure within the system is not uniform (so there is no single unique value for the pressure) and the force per unit area exerted by the system on the surroundings at the interface also includes viscous stresses. So, if the external pressure applied by the surroundings is constant, the work done by the system on the surroundings is equal to $p_{ext}\Delta V$.  If the constant external pressure applied by the surroundings is also equal to the initial pressure p exhibited by the system in its initial thermodynamic equilibrium state, then the enthalpy change is $\Delta H=\Delta U + p_{ext}\Delta V=\Delta U + p\Delta V$.  But, by the first law, this is just equal to the heat added to the system.  So, irrespective of what the thermal conditions are (say, isothermal or adiabatic) or whether chemical reaction is occurring, if the external pressure is held constant at the initial pressure of the system in its initial thermodynamic equilibrium state, the heat transferred is equal to $\Delta H$. If the process is such that the initial and final temperatures of the system are the same, we call the corresponding $Q = \Delta H$ the heat of reaction (at least for ideal gases for which the heat of mixing is zero).

So since $\Delta H=\Delta U+\Delta(p_{int}V)$, don't we need $p_{int}$ to be held constant as well (in addition to the external pressure being held constant at the initial pressure of the system in its initial thermodynamic equilibrium state)?

No.  It isn't correct to even talk about $p_{int}$ for a system experiencing an irreversible deformation since the pressure isn't even constant spatially within the system, plus, viscous stresses in the system contribute to the force of the system on the boundary with the surroundings.  However, the force per unit area that the system exerts on the surroundings will be constant at the boundary and equal to $p_{ext}$ throughout the deformation.  Only at the initial and final states will $p_{int}$ be relevant and equal to $p_{ext}$ throughout the system.

I shall start off by mentioning that your question is not so much about p-V work being done, but rather about the pressure not being constant. It is perfectly possible for p-V work to be done when the pressure is constant, and indeed under these circumstances, the equation $\Delta H = q$ is still obeyed. However, this would necessitate a change in temperature.

Well, you can always try to come up with a counterexample.
Your hypothesis would be that if pressure is not constant and p-V work is done, then $\Delta H = q$. So, let's try calculating those two quantities for a reversible isothermal expansion of $1\mathrm{~mol}$ of an ideal gas, from an initial state of $(p_1, V_1)$ to a final state of $(p_2, V_2)$. Let's just arbitrarily set $V_2 = 2V_1$. Of course, since this is an ideal gas we are talking about, $p_2$ will be equal to $p_1/2$. And since this is a reversible process, $p_\mathrm{int} = p_\mathrm{ext}$ throughout, so let's drop the int and ext subscripts.
Let's calculate the work done (on the system) first.
$$\begin{align}
w &= -\int \! p\,\mathrm{d}V \\
&= -\int \! \frac{nRT}{V}\,\mathrm{d}V \\
&= nRT \ln{\left(\frac{V_1}{V_2}\right)} \\
&= -RT \ln 2
\end{align}$$
Now the internal energy change of the system has to be zero, since this is a closed system with an ideal gas and therefore $U = U(T)$. $\Delta T = 0$ therefore implies $\Delta U = 0$. So:
$$\begin{align}
q &= \Delta U - w \\
&= RT \ln 2
\end{align}$$
However, the enthalpy $H$ is also only a function of temperature: $H = H(T)$. Therefore,
$$\Delta H = 0 \neq q$$
Hypothesis: disproved!

So if internal pressure changes but external pressure is constant, then enthalpy change can't equal heat right?

@swenger In general, it will not be equal.

Then I don't know why in my textbook, it says that when pv work is done at constant external pressure and constant temperature, enthalpy change equals heat?

That makes no sense, especially if you consider an ideal gas system. Regardless of whether the process is reversible or irreversible, constant temperature of the system means $\Delta H = 0$ and $\Delta U = 0$. And, if $w \neq 0$ then $q = -w \neq 0$ (regardless of what the expression for the work is).

Actually, in the example in my textbook, there's a reaction occurring which causes the system to expand under a constant temperature and external pressure. So in this case, $\Delta H \neq 0$. But can we say that $\Delta H=q$?

I am not 100% sure, but I don't think so. As far as I know, there is no reason for $\Delta H$ to be equal to $q$ if the conditions you laid out in your question aren't met.

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text:
Why does capping after each coupling in Solid phase peptide synthesis improve the results?
Why does capping after coupling improve results during solid phase peptide synthesis (SPPS)?
The Case: I have conducted a lab experiment where we coupled three Fmoc-protected amino acids, and did a final capping (acetylate) at the end of the synthesis. However, I was told that one can theoretically cap after every single amino acid coupling to ensure that the final product is correct. That is, one can "eliminate" or terminate the synthesis of peptides where the amino acids are innocently coupled (?).
I don't really understand why, because if you cap then you "block" all of the peptides that you are synthesizing (?). Therefore, how can you add further onto the chain if you cap after each amino acid? This seems counterintuitive, or I might be a bit confused. If you are able to de-protect the cap to further couple on more amino acids, then what is the point of capping?
Is there someway that capping only “caps” onto amino acids that are not coupled or that did not couple the previous amino acid somehow?

I'd guess that it means that at each step if you capped the amino acid by acetylation that you could purify each intermediate product. Thus in the end you'd have a higher purity product, but probably less of it.

if you cap then you “block” all of the peptides that you are synthesizing.(?)

No, you don’t, the order of steps is important. First you try to couple your new amino acid to the solid phase. You will end up with uncoupled ends with free amino groups and coupled ends with a protected amino group. Then you cap, e.g. with phenoxyacetic acid anhydride. This will only cap the uncoupled ends with free amino groups. Then the Fmoc or whatever protective group is removed rendering free amino groups for new couplings.
Say you wanted to create the sequence $\ce{CHEMISTRY}$. And say at one specific spot on the resin, you failed to couple the serine residue; you have $\ce{H2N-TRY-CO-resin}$ at that one spot. If you now do not cap, you might generate the sequence $\ce{CHEMITRY}$ because isoleucine will couple to threonine’s free amino group etc. By capping, you essentially generate $\ce{Ac-NH-TRY-CO-resin}$ which won’t disturb you anymore.
And the benefits don’t stop there. Without capping, at the end you would end up with the sequences $\ce{CHEMISTRY}$ (the one you want), $\ce{HEMISTRY}$, $\ce{CEMISTRY}$, $\ce{CHMISTRY}$, $\ce{CHEISTRY}$, $\ce{CHEMSTRY}$, $\ce{CHEMITRY}$, $\ce{CHEMISRY}$, $\ce{CHEMISTY}$ and maybe even $\ce{CHEMISTR}$. These are all of similar length to your desired product and hard to purify away via HPLC. But if you cap, you can only get $\ce{HEMISTRY}$, $\ce{EMISTRY}$, $\ce{MISTRY}$, $\ce{ISTRY}$, $\ce{STRY}$, $\ce{TRY}$, $\ce{RY}$, and $\ce{Y}$ — most of these are significantly shorter and easier to purify away.
Also, assume a constant amount of amino acids always being added to the resin to couple. If only the so far correctly built sequences are available, these will ‘see’ a higher ‘concentration per site’ of their coupling partners maybe resulting in slightly increased yield, too. (Note that this paragraph used layman’s terminology.)

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text:
Demonstrating orthonormality for wave functions for ground state and excited state of He

The electron configuration of the ground state $\ce{He}$ atom is $\ce{1s^2}$, and the excited state configuration under study is $\ce{1s^2 2s^1}$. Show that these states are orthonormal, i.e.,
$$\langle\Psi_0 \vert\Psi_0\rangle = 1$$
$$\langle\Psi_1 \vert\Psi_1\rangle = 1$$
$$\langle\Psi_0 \vert\Psi_1\rangle = 0$$

I start off by defining the wave functions as Slater determinants,
$$
\Psi_0 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\phi_1(1)\alpha(1) & \phi_1(1)\beta(1) \\
\phi_1(2)\alpha(2) & \phi_1(2)\beta(2)
\end{vmatrix}
$$
$$
\Psi_1 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\phi_1(1)\alpha(1) & \phi_2(1)\beta(1) \\
\phi_1(2)\alpha(2) & \phi_2(2)\beta(2)
\end{vmatrix}
$$
I am able to show that the wave functions are normalized, but I have trouble showing that they are also orthogonal. I get that $\langle\Psi_0 \vert\Psi_1\rangle = 1$. I set up the integral over all coordinates, carry out the multiplications, and add together the terms that are equivalent due to indistinguishable electrons (identical when all coordinates are interchanged). I end up with the expression
$$
\int d\tau \left[  \phi_1^*(1)\alpha(1)\phi_2^*(2)\beta(2)\phi_1(1)\alpha(1)\phi_2(2)\beta(2) - \phi_1^*(1)\alpha(1)\phi_2^*(2)\beta(2)\phi_1(2)\alpha(2)\phi_2(1)\beta(1)   \right]
$$
where $d\tau$ represents all coordinates. Separating the spin coordinates from the spatial coordinates, I get that
$$
\langle\Psi_0 \vert\Psi_1\rangle = \\
\sum_{m_{s1}=-1/2}^{1/2} \sum_{m_{s2}=-1/2}^{1/2} \alpha(1)\alpha(1)\beta(2)\beta(2) \cdot \langle\phi_1(1) \vert\phi_1(1)\rangle \cdot \langle\phi_2(2) \vert\phi_2(2)\rangle \\
- \sum_{m_{s1}=-1/2}^{1/2} \sum_{m_{s2}=-1/2}^{1/2} \alpha(1)\alpha(2)\beta(2)\beta(1) \cdot \langle\phi_1(1) \vert\phi_2(1)\rangle \cdot \langle\phi_2(2) \vert\phi_1(2)\rangle \\
= 1 - 0 = 1
$$
The first term reduces to $1$, I find, while the second term reduces to $0$. To get $\langle\Psi_0 \vert\Psi_1\rangle = 0$, either both terms need to be $1$, or both terms need to be $0$. Something must be incorrect. Is there some obvious flaw in my derivations?

page 8 of the pdf at this link discusses this problem, but it has been so long since I did this that I don't really remember how to do the math. https://www.chem.tamu.edu/rgroup/hughbanks/courses/673/handouts/antisymmetry.pdf Obviously the gist is that $1s^12s^1$ to $1s^2$ is a forbidden transition.

First prove a well-known fact that any two spin orbitals $\psi_i$ and $\psi_j$ are orthogonal if the corresponding spatial orbitals from which they are constructed are themselves orthogonal.

For two spin orbitals constructed from two different spatial orbitals orthogonality follows from orthogonality of spatial orbitals regardless of spin components.
For two spin orbitals constructed from the same spatial orbital orthogonality follows from orthogonality of spin functions.

Then you can work with spin orbitals to (re)establish a well-known result that for two Slater determinants differ in just one spin orbital, the overlap integral $\langle \Psi_0 \vert \Psi_1 \rangle$ is zero.
$$
\Psi_0 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\psi_1(1) & \psi_2(1) \\
\psi_1(2) & \psi_2(2)
\end{vmatrix} \, ,
\quad
\Psi_1 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\psi_1(1) & \psi_3(1) \\
\psi_1(2) & \psi_3(2)
\end{vmatrix} \, ,
$$
where
$$
\psi_1(1) = \phi_1(1) \alpha(1) \, ,
\quad
\psi_2(1) = \phi_1(1) \beta(1) \, ,
\quad
\psi_3(1) = \phi_2(1) \beta(1) \, .
$$
Algebraically, it is very easy to show that,
$$
  \langle \Psi_0 \vert \Psi_1 \rangle =
  \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \langle \psi_1(1) \psi_2(2) - \psi_2(1) \psi_1(2) \vert \psi_1(1) \psi_3(2) - \psi_3(1) \psi_1(2) \rangle \\
  = \frac{1}{2} \Big(
    \langle \psi_1(1) \psi_2(2) \vert \psi_1(1) \psi_3(2) \rangle
  - \langle \psi_1(1) \psi_2(2) \vert \psi_3(1) \psi_1(2) \rangle \\
  - \langle \psi_2(1) \psi_1(2) \vert \psi_1(1) \psi_3(2) \rangle
  + \langle \psi_2(1) \psi_1(2) \vert \psi_3(1) \psi_1(2) \rangle \Big) \\
  = \frac{1}{2} \Big(
    \langle \psi_1(1) \vert \psi_1(1) \rangle
    \langle \psi_2(2) \vert \psi_3(2) \rangle
  - \langle \psi_1(1) \vert \psi_3(1) \rangle
    \langle \psi_2(2) \vert \psi_1(2) \rangle \\
  - \langle \psi_2(1) \vert \psi_1(1) \rangle
    \langle \psi_1(2) \vert \psi_3(2) \rangle
  + \langle \psi_2(1) \vert \psi_3(1) \rangle
    \langle \psi_1(2) \vert \psi_1(2) \rangle \Big) \\
  = \frac{1}{2} \Big(c_1 \cdot 0 - 0 \cdot 0 - 0 \cdot 0 + 0 \cdot c_1 \Big) = 0 \, .
$$
In the case when spin orbitals are normalized (to one) $c_1 = 1$, but it is not required here: the overlap integral will vanish for spin orbitals that are just orthogonal and not necessarily normalized (to one) as well.

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text:
How can I change the pH level?
I am conducting a experiment that needs a pH 5 sulfuric acid but I only have pH 3 sulfuric acid. I have some distilled water and a graduated cylinder but I do not know how I can change it.

Can you convert pH 3 into a concentration of sulfuric acid?

yes @orthocresol

Do so, then. And do the same for the pH 5 sulfuric acid..

If you wish, you can answer your own question using whatever method you have found works for you. Answering your own question is recommended here.

authors: Йоана Генова


text:
Families of some columns
From previous lessons, I know that column 1 are alkali metals, column 2 are alkaline earth metals, column 3 till column 12 are transition metals, column 17 are halogens and column 18 are inert gases. But till now I don't know the properties of elements in columns 13, 14, 15 and 16... Are they metals or non-metals? I have not faced such a question before.

Are you asking for the common names of the groups e.g. Group 1 is commonly called "alkali metals", or are you asking for their physical properties i.e. whether they are metals, metalloids, or non-metals? To be honest, you can just look at the individual Wikipedia pages of the groups, or the elements to easily find your answer. The trend is that there is a general increase in metallic character going down the groups. So carbon is a non-metal, silicon and germanium are metalloids, tin and lead are metals.


text:
Are there magnetic rocks, and not hematite?
I dug up some sand and small rocks in a stream bed near me and moved a powerful magnet through the wash picking up numerous small rocks and black sand.
Now, the black sand is definitely small crystalls of magnetite, but what about the small rocks which tend to be gray in color, some being brownish?
I thought pretty much anything you would be likely to encounter commonly (like in a stream bed) that is magnetic is magnetite. In other words, my understanding is that the only COMMON mineral that is magnetic is magnetite. But these small rocks do not appear to be magnetite crystals, and the brown ones definitely are not.
Does this mean there are magnetite crystals INSIDE the rocks? Like if I break them up will I find little tiny magnetite crystals within?

Magnetite is indeed very magnetic. But other minerals are magnetic too, and given a strong enough magnet you will attract more than magnetite. Chromite is one that comes to mind that's also common in some black sands.
You might not always see the magnetite. Magnetite is mixed valence of iron 2 and 3 and as such it isn't stable in our oxygen rich atmosphere. Given enough time it will alter to a variety minerals such as hematite and iron hydroxides. So you might have magnetite in the core of a grain mantles by non magnetic hematite. It will look brown but it will still be attracted to a magnet.
Same thing with your grey rocks: they most likely have magnetite inside, hidden from view.

There are many minerals that are magnetic. You used a relatively "strong" magnetic to extract them.

Does this mean there are magnetite crystals INSIDE the rocks? Like if I break them up will I find little tiny magnetite crystals within?

Perhaps. "Rocks" aren't homogeneous.
Hematite is relatively uncommon in my area, but if you're in northern Minnesota, then it is a common mineral. So what's common/uncommon depends where you're looking.

None of the minerals on that page (except magnetite) are magnetic. My understanding is that magnetite is almost universally found everywhere, although in some areas only smaller particles are found. Why are you referencing hematite? Hematite is not magnetic.

I just used hematite as an example for a common vs uncommon mineral.

Since you used a strong magnet to collect the rocks you have to consider "weakly" magnetic rocks. // So a "lodestone" will pick up a paper clip which is not magnetic itself. A paper clip doesn't have a magnetic field itself. The magnetic field in the paper clip is induced by the lodestone. So if paper clips were a mineral then they would be "weakly magnetic."

Depending on particle size even paramagnetic particles may show visible attraction to magnets. Also, various iron oxides/hydroxides may form inhomogenous lumps of magnetic microcrystalls. Iron is everywhere.


text:
Suitable acid/base combo
I've been trying to get any chemical reaction that with the least volume of reagents (including possible water) would generate the highest amount of a safe gas (at a reasonable cost and with a fast reaction).
The final implementation would be having small, portable and stable reagents (preferably solid) that could be mixed at any moment to fill up a balloon. To be used by the general public without much reagent handling (e.g. irritants would be ok).
Currently by best guess has been citric acid with any of the 1st group (bi)carbonates, adding water before to each, and then mixing them.
Does anyone have another suitable reaction?
Any other safe acid?
Any more soluble option to reduce the water volume?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site and visit the [help] if questions about it remain. I’m not sure about your question, as it stands it is rather broad. It may get closed for being too broad. If you can narrow it down more, you should. E.g. are you restricted in which safe gas is generated?

dry ice would be about as good as you could do relative volume wise. "safe" is relative. I wouldn't hold the dry ice in my hand.

thanks for the tips already and the welcome. edited further for clarity


text:
What happens to the attraction of glycerol to a positively charged acetate rod if water is added
I know that glycerol is more polar than water because it has stronger intermolecular forces and its also more viscous. If I were to add water to a glycerol solution, would the solution get less polar with weaker intermolecular forces? Also if I were to add water, would the solution be more attracted to a positively ...
I mean if I place a positively charged acetate rod next to the mixed solution of water and glycerol, would the solution get more attracted to the rod or less attracted or would there be no change?

This is a non sequitur: "I know that glycerol is more polar than water because it has stronger intermolecular forces and its also more viscous." Polarity is determined by distribution of charges on a molecule. Mineral oil is viscous but not polar.

A good rule of thumb (as long as you don't work with $\ce{HF}$) is that nothing is more polar than water.

Ok, but what happens if I mix water and glycerol? Does the solution get more attracted to the acetate rod or less attracted and why?


text:
Orthogonal Wavefunctions
My current understanding of orthogonal wavefunctions is: two wavefunctions that  are perpendicular to each other and must satisfy the following equation: $$\int\psi_1\, \psi_2\, \mathrm{d}\tau =0$$
From this, it implies that orthogonality is a relationship between 2 wavefunctions and a single wavefunction itself can not be labelled as 'orthogonal'. They must be orthogonal with respect to some other wavefunction. However I have seen some textbooks refer to single wavefunctions as being orthogonal.
Am I missing some crucial thing in my definition of orthogonal wavefunctions? Could anyone provide me with a more complete definition of orthogonal wavefunctions from a chemistry point of view?

Could you provide a little bit more context i.e. examples of said claims in said textbooks?

@orthocresol Well, it said something along the lines, 'all wavefunctions are orthogonal', and in this question, http://chemistry.stackexchange.com/questions/7029/why-are-wave-functions-orthogonal, they talk of single wavefunctions as being orthogonal

It is not really talking of a single wavefunction as being orthogonal, but rather a set of wavefunctions. By having a set of orthogonal wavefunctions $\psi_1 , \psi_2 \ldots$ it means that any pair of wavefunctions that you pick will be orthogonal to each other. This will always happen if the wavefunctions are eigenfunctions of a hermitian operator (it is one of the standard "elementary" proofs in QM). I think the question that you linked does use the word a little loosely, although I still understood it to refer to a set of wavefunctions.

@orthocresol Right , that makes a lot more sense. So is my understanding of what orthogonal wavefuntions correct? Is there more of a chemistry way at explaining orthogonal wavefunction, rather than using mathematical expressions?

Somewhat. Have you done MO theory? Say in HF, the H 1s orbital doesn't overlap with the F 2px or 2py orbitals. That's because those two wavefunctions are orthogonal to each other.

Note that the term "perpendicular" has no meaning for (wave) functions. Two are orthogonal if, and only if, said integral is zero.

@orthocresol Yes, I am familiar with MO theory. So any set of  wavefunctions that don't overlap each other are orthogonal to each other?

Yeah, pretty much. In MOT, there is an expression called the "overlap integral" and denoted $S$: $S = \int \psi_1^* \psi_2 \mathrm{d}\tau$. Of course, that's just the same expression as above (apart from the complex conjugate which you missed out). The difference in energy between the MOs and the constituent AOs can be said to be proportional to $S^2$. You will notice that, if two wavefunctions are orthogonal (such as the H 1s and F 2px), $S = 0$ and therefore the constituent MOs are no different in energy from the AOs - exactly what you would expect since their symmetries do not match.

The notion of orthogonality in the context of the question referrers to the very well-known general concept of linear algebra, the branch of mathematics that studies vector spaces. Instead of going deep into the mathematics (that requires at least 50 textbook pages) let's just clear some OP's doubts.
First, a small (but important) correction: two wave functions $\psi_1$ and $\psi_2$ are called orthogonal to each other if
$$
  \int \overline{\psi}_1 \psi_2 \, \mathrm{d} \tau = 0 \, ,
$$
where the first function is complex-conjugated as indicated by a bar on top of it. Wave functions are complex-valued functions and complex-conjugation of the first argument is important.1
So, yes, orthogonality is a not a property of a single wave function. It either refers to a pair of them being orthogonal to each other as described above, or, in general, to a set of them, being all mutually orthogonal to each other, i.e. to a set $\{ \psi_i \}_{i=1}^{n}$ such that for any $i \neq j$
$$
  \int \overline{\psi}_i \psi_j \, \mathrm{d} \tau = 0 \, .
$$
In the last case it is said that the whole set $\{ \psi_i \}_{i=1}^{n}$ is orthogonal.

1) More precisely, one of the two functions has to be complex-conjugated in this expression, where which one is the matter of convention: in physical literature it is often the first function, while in mathematically oriented literature it is usually the second one.

I wanted to add one other important thing in my answer but you were faster; feel free to include if you think its important ;). It often occurs that textbooks write orthogonal set but mean orthonormal set and normalisation is indeed a property of a wavefunction on it's own.

@Wildcat Does  $\int \overline{\psi}_1 \psi_2 , \mathrm{d} \tau = 0 \Rightarrow \int \overline{\psi}_2 \psi_1 , \mathrm{d} \tau = 0 $?

@adosar yes you can conjugate the expression.

@adosar, yes. In fact, $\int \overline{\psi}_i \psi_j , \mathrm{d} \tau = \int \overline{\psi}_j \psi_i , \mathrm{d} \tau$.

As described in Wildcat's answer, a single wave function cannot be orthogonal, but a set of wave functions can all be mutually orthogonal.
To address the second part of the OP's question, the physical meaning of orthogonality is that a pair of mutually orthogonal wave functions are mutually exclusive; observing one precludes the possibility of observing the other unless the system is changed.

The same wave functions can not be said to be orthogonal. the comparison must be between two different function.

This is not true. You can measure the overlap of any wavefunction with itself. If it's normalized, it should equal 1.


text:
Which is the Electrolyse Reaction?
$$\ce{NaOH +H2O->H2 +O2 +NaOH}$$
We know that $\ce{H+}$ and $\ce{OH-}$ go  to the cathode and anode, but how can I write the total reaction?
Is it: $$\ce{2NaOH +2H2O->2H2 +O2 +2NaOH}$$
I've put NaOH both sides because if the mass of NaOH is given, how can I find the mass of H2 for example?

Why would you need to write $\ce{NaOH}$ on both sides?

Because if I'm given the mass of NaOH, how should I find the mass of other substances?

Electrolyse of water is a redox reaction, meaning that there is a transfer of electrons between atoms. Now, every redox reaction can be written as a combination of 2 half-equations. One half-equation shows one species getting oxidised (losing electrons) and the other half-equation shows the other species getting reduced (gaining electrons). When you combine these two half-equations you get the overall reaction, which is what you are asking for. Here is guide that should help you find the overall equation of any redox equation, not just this one example.
Step 1: Determine what species gets oxidised and species gets reduced
Now, what I think the reaction you are trying to describe is the electrolyse of water. Correct me if I am wrong. If that is the case, there is no need for $\ce{NaOH}$ in the equation as it appears on both side, meaning that they cancel each other out (also $\ce{NaOH}$ has nothing to do with electrolyse of water). Secondly, it is $\ce{H2O}$ that gets oxidised and reduced, not $\ce{H+}$ and $\ce{OH-}$.
Step 2: Determine  half-equations
These can be easily found by using your table of standard reduction potentials. Oxidation half-equation: $$\ce{2H2O -> O2 + 4H+ + 4e-}$$
Reduction half-equation: $$\ce{2H2O + 2e- -> H2 +2OH-}$$
Step 3: Balance each equation so that there is the same number of electrons
We can't immediately add the half-equations as the electrons won't cancel each other out. This means that we need to multiply the reduction half-equation by 2:$$\ce{4H2O + 4e- -> 2H2 +4OH-}$$
Step 4: Add the half-equations
When you add each half-equation, you get: $$\ce{6H2O   -> 2H2 + O2 + 4H+ + 4OH-}$$ The $\ce{H+}$ and $\ce{OH-}$ combine to form water which then cancel out with the water the other side. Therefore the overall reaction is: $$\ce{2H2O  -> 2H2 + O2}$$

What if I am given the mass of NaOH, how am I supposed to find the mass of other elements?

I do not understand what you mean. How is NaOH related to the electrolysis of water? Can you explain your problem in more detail?


text:
Oxidation of alkenes by selenium dioxide to diketones
I know that $\ce{SeO2}$ readily affects the allylic position. The reaction mechanism of this reaction goes like this:

(Source: Wikipedia; License: Public Domain)
But, in my book, there's another reaction given as an exception:

Here, an $\alpha$-diketone is formed.
Why and how is this alternative product formed? What is the mechanism for this product?

$\ce{\alpha}$-diketones are often formed when $\ce{SeO2}$ reacts with alkynes that don't have any hydrogens on the carbons adjacent to the acetylenic carbons.  Are you sure your book showed an alkene?

Yes, for sure, I have checked it again and its an alkene not alkyne. I know that alkyne reaction but this is not that.

I found formation of an $\alpha$-diketone from a ketone, but not from an alkene.


text:
Calculating the charges of anions and cations
I am new to chemistry. Could someone explain me how to calculate the charges on anions and cations. I've heard about some crossing method (the left subscript becomes the right superscript and the right subscript becomes the left super script.)
Example: $\ce{Rb2S -> 2Rb+ + S^2-}$
But what about:
$\ce{K2MnO4}$ and $\ce{Zn(ClO)2}$
This trick doesn't seem to apply here, so how can I find out the charges?

@Close flagger, I don't think this is a homework question. It looks more like questioning a [stupid] rule resulting from oversimplification to me.

Some basic knowledge of the atoms in the periodic table is necessary to extract the ionic charges. In the case of $\ce{K2MnO4}$, for example, we know that the potassium ion is positive, and that the manganate ion is negative. We see that the compound is neutral: has zero net charge. So how can we place charges to get the molecule to be neutral, given the subscripts?
Potassium is a Group 1 element, meaning it only has 1 valence electron, placed in the  4s orbital. Potassium is eager to lose this electron to fulfil the octet rule. Therefore potassium ions have one less electron than protons, and the charge is +1. Since we have two $\ce{K^+}$ ions, the manganate ion must have a charge of -2 to make sure the molecule has zero net charge.
$$\ce{K2MnO4 <=>[\ce{H2O}] 2K+ + MnO4^2-}$$
Zinc is one of two transition metals that only have one (natural) oxidation state: +2. The electron configuration is [Ar]3d$^{10}$4s$^2$. When zinc becomes a cation, it loses the two 4s electrons. Knowing this, we see that the hypochlorite must have a charge of -1: the two hypochlorite -1 charges counter the 2+ charge of zinc.
$$\ce{Zn(ClO)2 <=>[\ce{H2O}] Zn^2+ + 2ClO^-}$$
I recommend not relying on simple rules, such as the one you mentioned. Rather, try to understand why the rule came about, and understand when the rule fails, and why. It really helps to be able to "read" the periodic table, as a lot of basic information is incorporated into it.

Potassium problem: Is there a way to know that Mn is 2- except by deduction like you did?

@privetDruzia Each potassium ion has to have a 1+ charge, so $x$ in $\ce{[MnO4]^{$x$-}}$ has to be 2.

Thx. This semi-deductive method seems to work indeed. But not for more "complex" molecules like: KnaCO3 or Ba(BrO2)NO2, ... Or am I wrong?

This may not be the best reply, but you should just memorize the charge of common ions. There are ways of estimating the charge in molecular ions, such as "Formal charge". The formal charge assumes, however, that all bond electrons are shared equally, regardless of electronegativity differences, so such an approach may not always be appropriate. In quantum mechanics, the electrons don't necessarily belong to a single atom, but rather the whole molecule. Thus, fractional charges are  more common than integer charges. See this pdf: http://www.chem.ucla.edu/harding/tutorials/formalcharge.pdf

Elements like K, Li, Cs and Rb can only have +1 charge on them (Due to very high second ionization energy). Also, they can't have negative charge due to their low electron gain enthalpy (they are metals and are electropositive in nature). Also, Oxygen can only have -2 (in oxides, eg: H2O), -1 (in peroxides, eg: H2O2) and -0.5 (in superoxides, eg: KO2). Also, H can only have +1 or -1(if H is more electronegative of the two elements, as in the case of hydrides, eg: LiH) and Fluorine can have only -1 (since it is the most electronegative element). Similarly, Ca, Mg.. have +2 charge. You need to remember these basic rules and a few others (refer to any inorganic chemistry text book and you can find them), to calculate the charges.
Take the case of K2MnO4 : K has +1. O has -2 (in is an oxide). Now, let the charge on Mn be x. So, x + 2*(+1) + 4*(-2) = 0 (Since the net charge on the species is zero). Hence, x = +6. I hope this answer helped.

Note that alkali metals have been reported to have a negative oxidation state though.

Also, oxygen has an oxidation state of +2 in oxygen difluoride.

I clearly mentioned these are the few basic rules and the person who asked this question has to refer a book to know all the exceptions and other such rules.


text:
Obtaining deuterium directly from water
I know that there is 2:6400 deuterium and hydrogen molecules respectively in water.
When electrolysis is done, hydrogen and oxygen will be obtained. The hydrogen obtained should contain small amounts of d2 and t2.
Is there a way to separate it by diffusion or effusion (or any other way)?

Deuterium is already obtained by electrolysis of water. Deuterium is a little less reactive than Hydrogen, So when we repeatedly electrolyze water, we get heavy water. Then we can obtain deuterium from that.


text:
heat of fusion in an equation

You add $100.0\ \mathrm g$ of water at $55.0\ \mathrm{^\circ C}$ to $100.0\ \mathrm g$ of ice at $0.00\ \mathrm{^\circ C}$.  Some of the ice melts and cools the water to $0.00\ \mathrm{^\circ C}$.  When thermal equilibrium is established at $0.00\ \mathrm{^\circ C}$, what mass of ice has melted?
  $\Delta H$ of fusion for water is $333\ \mathrm{J/g}$.

My question is I got a negative for $q$.  Would this change to a positive because you need heat to melt the ice?  I no you can't have a negative. So I am not sure if it should be a neg or a positive
$q = m \times \text{specific heat} \times (T_\text{final} - T_\text{initial})$
specific heat of water is $4.186$ (should memorize or on a chart)
$q= 100\ \mathrm g \times 4.186\ \mathrm J \times (0 -55) = -23\,023\ \mathrm J$
mass ice × heat of fusion
$q= m \times H_\text{fusion}$
$-23\,023\ \mathrm J = m \times 333\ \mathrm{J/g} = 69.12\ \mathrm g$

The negative q is the amount of heat you need to add to the water to get its temperature from 55 to 0.  The negative value means you are removing it from the water and adding the same amount of heat (with a plus sign) to the ice.

If you go by thermodynamic conventions,
$$\begin{align}q > 0 &\implies \text{ heat is gained by system}\\
q < 0 &\implies \text{ heat is lost by system} \end{align}$$

You initially considered water as your system, and it lost heat, therefore  $q$ was negative.
Now when you do calculation for ice, it becomes your system, and the heat is gained by ice, $\implies$ positive $q$


text:
How to define a flame in numerical combustion?
I am running a CFD simulation (computational fluid dynamics) opposed jet case, where a stream of fresh reactants is opposed to a stream of burnt products. I am looking at the effects of stratification (different equivalent ratio on each jet) on the flammability limits of methane.
In some cases, in the reaction zone, the heat release is about $10^3$ to $10^4$ lower than compared to my reference case, which is far from the lean or rich flammability limits. Since I am sitting on or close to those limits, it is important for me to define one (or multiple) parameter for which I can say that the fuel is indeed burning.
Here is my reasoning right now, but it might be incomplete or even erroneous. To check if the fuel is indeed burning, I have been looking at the distribution of key radicals. If the mass fraction of say $\ce{H2}$, does not have peak in the reaction zone and only diffuses from one stream to another, then I can definitely say there is no flame. But a contrario, if there is a peak, can I definitely say there is a flame?
If there are some key parameters I can look at, please share. Note that since this is a numerical simulation, I virtually have access to all of the thermodynamic or chemical properties (chemical composition, reaction rates of each reaction etc...) at every point of the flow.


text:
How do I calculate the molecular mass of a complex in a complexometric titration with EDTA?
I had to titrate 0.35 mmol (3.2 g) of $\ce{Co2F2(Hdmpz)6(BF4)2}$ in a solution with 10 drops of 37 % $\ce{HCl}$, 10 mL of 4 M $\ce{NaOAc}$, 25 mL water and xylenol orange as indicator with  a 0.05 M EDTA solution.
Now I'm supposed to calculate the percentage $\ce{Co^2+}$ and then calculate the molecular weight of the whole complex.
Let's just say I saw color change at 16 mL EDTA. How do I calculate the percentage and the molecular weight? I'm also not sure about the molar ration $\ce{Co^2+}$:$\ce{EDTA}$.
If I just count the atoms it should be about 906 g/mol.

The stoichiometric ratio is 1:1. It forms a $\ce{[Co(edta)]^2-}$ complex.

I also thought the stoichiometric ratio was 1:1, but my teacher said it was wrong. But she didn't explain me why.

You titrated $0.7\mathrm{~mmol}$ of $\ce{Co^2+}$ with EDTA and you reached the endpoint at $(0.05\mathrm{~M})(16\mathrm{~ml}) = 0.8\mathrm{~mmol}$ of EDTA, so I don't see why it's not 1:1. I am guessing you might have added just a bit too much EDTA, especially if it is your first time doing an EDTA titration (I know I did so).

Yes, I have most likely used too much EDTA, and my 3,2 g wasn't measured perfectly.  So it is 1:1, I don't understand my teacher.


text:
Why does the COO- ion is pulled at during critical Micelle concentration?
During the Micelle formation of a soap (made of sodium), when soap comes in contact with water, it dissociate into Na+ ions and RCOO-  ions(where R is an hydrocarbon chain). Then in the RCOO- ions, R is  hydrophobic and COO is hydrophylic. So at critical Micelle concentration, what happens to this RCOO- ion?

"R is hydrophilic and COO is hydrophobic" -- You sure?

I am sorry. It is the opposite. R is hydrophobic and COO is hydrophylic.

Nothing happens to the single ion. The ion still remains pretty much the same, with hydrophylic head and hydrophobic tail, and the forces acting upon it are also the same. Consider a simpler analogy: you put $\ce{NaCl}$ in water, it dissociates into certain ions, you add more and more of it, everything goes well, until at about 360g/L... something happens.


text:
What is a possible side reaction during the oxidation of fluorene?
What is a possible side reaction during the oxidation of fluorene? 9-fluorenone is synthesized, but what else?

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.


text:
Are Isopropyl Alcohol (2-Propanol) and 1-Propanol tautomers?
They both are structural isomers, but are they tautomers too?

First, how do you define "tautomer"?

Tautomers need to be able to interconvert readily and quickly. This is done via proton transfer. For example in keto-enol tautomerism (https://en.wikipedia.org/wiki/Keto%E2%80%93enol_tautomerism#/media/File:Keto-Enol-Tautomerie.png) the proton on the alpha carbon donates its electrons to form a pi bond between the 2 carbons. As a result the pi bond between the carbonyl carbon and oxygen breaks and the now negatively charged oxygen picks up the proton which was originally lost.
In propan-1-ol and propan-2-ol, the oxygens are on different atoms and there is no way they can interconvert, so they are not tautomers.


text:
C6H5CO2H dimerization in Benzene
Why does $\ce{C6H5COOH}$ in benzene form dimers? Also why only in benzene, why not in other solvents such as water?
In general, what are the properties of compounds which form dimers or polymers in benzene?

Hint: somehow it didn't occur to you that your asking about hydrogen bonds.

Yes, but why in benzene, and why not in water?

https://en.wikipedia.org/wiki/Carboxylic_acid

I see. So they will form dimers in non-polar solvents.
Can you explain this please? "whereas higher carboxylic acids are less soluble due to the increasing hydrophobic nature of the alkyl chain."
Thanks!

That should be an entirely separate question. However, if you bear in mind that water is a pretty polar solvent, and alkyl groups are non-polar (hydrophobic basically means non-polar), I suppose you could deduce the answer.

Is it because since the salt is non-polar, water is unable to break it. But then how does Benzene dissolve the salt? What exactly is meant by solubility in Benzene?

Note that the acetic acid forms the dimer even in gaseous phase, it is not benzene specific. The acid does not form dimers, if the solvent is able to form strong enough hydrogen bonds so the molecules of the acid preferably  bonds with the solvent.

In polar solvents such as water, the water molecules will solvate the carboxylic acid. Thus, hydrogen bonds will form between the 2 molecules and the carboxylic acid will dissolve.
The water is in large excess as it is the solvent so it will solvate most of the carboxylic acid molecules.
In a non-polar solvent such as benzene, no such hydrogen bonds can take place with the carboxylic acid and benzene molecules. As a result, the carboxylic acids simply interact with themselves and form dimers.


text:
Why does fluorine atom have a higher ionization energy than oxygen atom?
The book I'm working from says $\ce{F}$ has a higher ionization energy than $\ce{O}$ because the nucleus has more protons so there is a higher attraction on the electrons. Isn't that balanced out by the higher number of electrons or is it because the atomic size is smaller?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site and visit the [help] to learn more about how it works.

Of course it is balanced to some extent, otherwise the ionization energy would depend monotonically on the atomic number, which in fact is not quite so.

At first approximation, we need to consider the orbital we are extracting an electron from. Both oxygen and fluorine (and nitrogen, carbon, boron and neon) have a $\mathrm{2p}$ orbital as their highest occupied, so we expect to ionise from that. If you draw a trend of the ionisation energies, you observe that the energy rises from boron to nitrogen, is lower for oxygen and rises from oxygen to neon. Disregarding the slight decrease at oxygen at first approximation we can say that it becomes increasingly hard to ionise an electron from a $\mathrm{2p}$ orbital with higher element number.
This is because essentially the shape and size of the $\mathrm{2p}$ orbitals are (close enough to, at first approximation) equal. So the wavefunction of the electrons we are removing does not change significantly. However, in their vicinity we have a nucleus which changes charge. For boron, we have an effective nuclear charge of $+1$ or $+3$ (depending on whether you count the $\mathrm{2s}$ electrons as shielding or not). For every step we take and every move we make to the right, the effective nuclear charge ($n(\ce{protons}) - n(\ce{core~electrons})$) will increase by one. We need to counter that nuclear charge to remove an electron which, naturally, because increasingly harder the stronger the charge holding it back.
The slight drop in ionisation energy at oxygen warrants a separate explanation. If you observe closer, you realise that the first three p-electrons (boron to nitrogen) are added as α spins while oxygen is the first element to obtain a β spin electron. A single β electron at first approximation has three degenerate orbitals it can populate. This is an unfavourable situation — compare with the Jahn-Teller effect in coordination chemistry that exists to counteract this situation. Thus, the energy required to remove this seemingly extraneous electron is lower than it would be just looking at effective nuclear charge. On the other hand, three α spins are more stable since the system is no longer degenerate. Thus it is slightly harder to remove an electron from nitrogen than it should be.


text:
CO2 dry cleaning what is the process
From what I understand CO2 drycleaning is essentially heating and pressurizing Co2 until it is at a supercritical state and then using it as a solvent to wash clothing. I just don't understand why clothing that comes out of the resulting process would be cold. Doesn't supercritical carbon dioxide require temperatures of at least ~300 Kelvin to form? Anyway, I would greatly appreciate it if someone with more knowledge on the subject could provide a basic outline as to how CO2 dry-cleaning works. There are very few resources that go into detail on the process.
Thanks!

Like aliphatic cleaning solvents such as hexane, both liquid and supercritical $\ce{CO2}$ are effective solvents for fatty stains. According to Green America, though, "[L]iquid CO2 is pumped into the mix. Clothes are rotated in a cycle that lasts five to 15 minutes at room temperature."
The phase diagram for $\ce{CO2}$ shows that it is a supercritical fluid above ~300 K, just above room temperature, at ~80 kg/cm2, so that it may be either liquid or supercritical during the cleaning. However, at the end of the cycle, "[T]he liquid CO2 is pumped back into the storage tank, to be reused again, if possible. The remaining CO2 is released in the air." That implies a bit of a pressure drop at the end, with concomitant adiabatic cooling.


text:
Entropy change of surroundings
The entropy change of a system is given by
$$\mathrm dS=\frac{\mathrm dq_\text{rev}}{T}$$
which at constant temperature is
$$\Delta S=\frac{q_\text{rev}}{T}$$
I also learned that the entropy change of the surroundings is
$$\Delta S_\text{surr}=-\frac{q_\text{actual}}{T}$$
Why do we use $q_\text{rev}$ for the system but $q_\text{actual}$ for the surroundings? How do we derive
$$\Delta S_\text{surr}=-\frac{q_\text{actual}}{T}$$

The equation $\Delta S_{surr}=-q_{actual}/T$ is based on the assumption that  the surroundings consists of an infinite reservoir at constant temperature T (i.e., a reservoir with an infinite capacity to absorb heat without its temperature changing).  So any heat transfer to the reservoir will always take place at the temperature T.  In this equation, T is temperature of the reservoir, and not the temperature of the system.  The  only place that the temperature of the reservoir matches the temperature of the system is at the boundary between them.


text:
Phosphate buffer system
Can anyone show me how to find the masses of $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$ needed to prepare $1.0~\mathrm{L}$ of a $0.1~\mathrm{M}$ phosphate buffer solution with $\mathrm{pH} = 7.4$ using the simultaneous equations method?
This is what I have so far:
$$\ce{NaH2PO4 + H2O <=> H3PO4 + NaOH} \\
\ce{Na2HPO4 + 2H2O <=> H3PO4 + NaOH + H3O+} \\
\ce{NaOH -> Na+ + OH-} \\
\ce{H3O+ + OH- <=> 2H2O} \\
\ce{H3PO4 + H2O <=> H3O+ + H2PO4-} \\
\ce{H2PO4^- + H2O <=> H3O+ + HPO4^2-} \\
\ce{HPO4^2- + H2O <=> H3O+ + PO4^3-}$$
The species present are: $\ce{HPO4^2-}$, $\ce{PO4^3-}$, $\ce{H2PO4-}$, $\ce{H3PO4}$, $\ce{Na+}$, $\ce{OH-}$, and $\ce{H3O+}$.
I have two equations:
$$\begin{align}
K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}] &= 1.0 \times 10^{-14} \\
[\ce{OH-}] + 2[\ce{HPO4^2-}] + 3[\ce{PO4^3-}] + [\ce{H2PO4-}] &=  [\ce{H3O+}] + [\ce{Na+}]
\end{align}$$
I need 5 more equations for this system. How do I get them?

Are you not allowed to just simply use the Henderson-Hasselbalch equation anymore?

No lol my prof is against it

Okay. There are three more equations that would come from the successive dissociation of $\ce{H3PO4}$. One more equation comes from the fact that the total concentration of phosphate-containing species must be 0.1 M. And I believe the last equation involves the total concentration of $\ce{Na+}$ ions. I must admit, this is a horribly disgusting system of equations to solve.

@orthocresol what then?

Make logical approximations to simplify the system. Also, these systems of equation normally arise when you are trying to find the pH of a given system. Now you are trying to do the reverse process, which is to find the system that corresponds to a given pH. I recommend you let $x$ be the initial concentration of $\ce{NaH2PO4}$ and $(0.1-x)$ be the initial concentration of $\ce{Na2HPO4}$. You then solve the system to find out how the pH varies as $x$ changes. That allows you to work out the value of $x$ that gives you the pH that you want.

By the way, some of the chemical equations you wrote out are either wrong, or pointless. In the first one, there is no point including the spectator ion $\ce{Na+}$. If you exclude the spectator ion, then it becomes exactly the same as [- equation (4) - equation (5)]. The second one is just plain wrong. The third one is pointless, because there is no NaOH added to the system. I would also recommend to just use $\ce{H+}$ instead of $\ce{H3O+}$.

Things to consider:

Sodium is merely a spectator ion. It will not take part in any reactions, so the initial amount of sodium you weigh in will remain constant. We can thus ignore it.
The concentrations of $\ce{H3PO4}$ and $\ce{PO4^3-}$ will be sufficiently small to entirely ignore them.

This reduces your system to the following relevant chemical equation:
$$\ce{H2PO4- + H2O <=> HPO4^2- + H3O+}$$
With the unknown concentrations of $\ce{H2PO4-}$, $\ce{HPO4^2-}$ and $\ce{H3O+}$. For these three we only need three equations and luckily we have them. Remember that we know:

the final phosphate concentration we want;
the final $\mathrm{pH}$ we want; and
the $\mathrm{p}K_\mathrm{a}$ value of our acidic species.

This gives us three equations that I am going to hide behind a spoiler tag:

 $[\ce{H2PO4-}] + [\ce{HPO4^2-}] = 0.1~\mathrm{M}$
 $10^{-7.198} = K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]}$
 $10^{-7.4} = [\ce{H3O+}]$

We can now put the third equation into the second resulting in exactly the Henderson-Hasselbalch equation, because this exactly is how that equation was derived. The maths of one possible way to approach this problem is shown in the following spoiler tag.

 $10^{-7.198} = \frac{10^{-7.4} \times [\ce{HPO4^2-}]}{[\ce{H2PO4-}]}$
 $10^{-7.198 + 7.4} = \frac{[{HPO4^2-}]}{[\ce{H2PO4-}]}$
 $[\ce{H2PO4-}] \times 10^{0.202} = [\ce{HPO4^2-}]$ ← Here! You have the Henderson-Hasselbalch right here! (In a different form but equivalent.)

 $[\ce{HPO4^2-}] + [\ce{H2PO4-}] = 0.1~\mathrm{M}$
 $[\ce{HPO4^2-}] = 0.1~\mathrm{M} - [\ce{H2PO4-}]$

 $[\ce{H2PO4-}] \times 10^{0.202} = 0.1~\mathrm{M} - [\ce{H2PO4-}]$
 $[\ce{H2PO4-}] \left (10^{0.202} + 1 \right ) = 0.1~\mathrm{M}$
 $[\ce{H2PO4-}] = \frac{0.1~\mathrm{M}}{10^{0.202} + 1} = 0.0385~\mathrm{M}$

The last steps to calculate the required concentration of $\ce{Na2HPO4}$ including calculating the respective masses of the water-free salts should be trivial and can be found in another castle answer.
I can only speculate why you are not allowed to directly use the Henderson-Hasselbalch equation since mathematically, every optimal approach will simplify itself to exactly that equation. Maybe the point of the excersize is to derive that equation; only your prof will know.


text:
pH level of fabric with regard to inhibiting fungal growth
Someone on another forum made the point that doing laundry with sodium hypochlorite (bleach) will kill any fungus such as athlete's foot fungus, but that after the fabric dries, the pH level of the fabric is conducive to more fungal growth, since bleach is a base.  The person made the point to substitute bleach with an acid like lemon juice, or to first do the laundry with bleach and then a second time with lemon juice.  Exact quote below:

"Bleach removes fungus and mold but also promotes it’s later development. I had jock itch on and off for years until I stopped using bleach in my wash. Once I stopped using bleach it never returned. Fungus thrives in high PH environments caused by Bases like bleach. My jock itch returned for a short time when my wife used bleach again. If you must use bleach you should neutralize it with a post acid rinse like vinegar or lemon juice."

Is there any scientific rationale to backup this statement?  Is lowering the pH of fabric with an acidic substance a workable means of preventing fungal growth?  Or does it not matter- in other words, washing your clothes with bleach should neither promote nor inhibit future fungal growth (emphasis: athlete's foot fungus)

I doubt that anyone can point to a specific scientific study about this. As a chemist it doesn't make sense. I know whites smell like chlorine after a bleach wash. I suspect that the 2nd lemon juice wash just gives the clothes a nice smell. Rather than spending money on lemon juice I'd suggest some foot powder would be vastly more effective against athlete's foot. I also have used dipped my feet in a dilute bleach solution to kill it.

Hypochlorite bleach is a potent biocide, so throughly washing clothes with it should kill all fungi. The pH part is true, bleach is alkaline, but so are all common detergents. But this should not even matter, if the clothes are properly rinsed with clean water, that should remove almost all of the bleach and restore the pH to close to neutral.
What I think happened, is that his clothes were not rinsed well enough, so traces of the bleach and the detergents stayed in the clothes, causing or exacerbating skin problems. Some people have skin especially sensitive to bleach and allergic reactions to detergents can happen to.
So I would recommend washing with bleach, but making sure the clothes are well rinsed. You may need to set your washing machine on an extra rinse cycle to get the desired effect. Using lemon juice should not be necessary.


text:
Photochemistry of beta, gamma-unsaturated ketones
Is there a Norrish reaction for beta, gamma-unsaturated ketones?
I know that Norrish reaction is used for saturated ketones, but in my homework I found this question. I have looked a lot in books and websites but I haven't found anything

Beta, gamma-unsaturated ketones can undergo Norrish photoreactions from the excited singlet state.  For example, in the case (1) below $\ce{R'}$ would be an allyl radical if we had started with a beta, gamma-unsaturated ketone.

From this solvent caged biradical several things can happen.  The acyl radical can decarbonylate (lose carbon monoxide) and then recombine with either end of the allyl radical to produce hydrocarbons.  Also the radical pair could simply recombine.  This would regenerate the starting material, or if the allyl radical has different substituents on each end, then it could recombine at the other end of the allyl radical to produce an isomeric beta, gamma unsaturated ketone.
However, the really interesting photochemistry observed with beta, gamma-unsaturated ketones occurs from the triplet state and is known as the oxa-di-pi-methane rearrangement.  The following figure (2) provides an example showing the "characteristic" 3-membered ring in the final product.

It is just the "oxo" analogue of the all-carbon photo di-pi-methane rearrangement.


text:
Calculate the mole fraction of component B in this solution? (partial pressures)
The vapour pressure of a pure liquid solvent A is $\pu{0.80atm}$. When a non volatile substance B is dissolved in the solvent, its vapour pressure drops to $\pu{0.60atm}$. Calculate the mole fraction of component B in this solution.
All I can think of is that $P_A=x_A * P_A^*$, which is Raoult's Law. But I don't know how to apply it.

Raoult's Law is correct. Think of it as a math problem Y=a*X.

Raoult's law is the correct approach to solving this problem. The definition of Raoult's law is explained well in this Wikipedia page as:

a law of thermodynamics established by French chemist François-Marie Raoult in 1887. It states that the partial vapor pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.  Mathematically, Raoult's law for a single component in an ideal solution is stated as:
  $$P_i = P_i^*x_i$$
  where $P_i$ is the partial vapor pressure of the component $i$ in the gaseous mixture (above the solution), $P_i^*$ is the vapor pressure of the pure component $i$, and $x_i$ is the mole fraction of the component $i$ in the mixture (in the solution).

Because you will be solving for mole fraction, you can rearrange the equation to:
$$x_i = P_i/P_i^*$$
$$x_i=\pu{0.60atm}/\pu{0.80atm} = 0.75$$
Of course the value calculated above is for the volatile solvent A. Since the total mole fraction of both components combined is equal to unity, the mole fraction of the non-volatile solute B must be $1 - 0.75 = 0.25$.


text:
Different ion effect
I need to have the highest concentration of (bi)carbonate ions in a water solution.
Thus far, my only idea was to use Potassium Bicarbonate (safe and cheap) as it has the highest solubility but it is yet not enough.
I was thinking whether adding both Sodium AND Potassium Bicarbonate, to saturation, would increase the total amount of dissolved Carbonates.
Another solution, which might be more promising is using a complementary salt to make use of the 'Different ion effect' and increase the solubility.
Would any of these options be useful for my purpose? Thanks alot for the help.

In theory, but at such high concentrations theory goes out the window. I'll point out that the pK value of bicarbonate will change in such a concentrated salt solution too.  // By adding $\ce{NaHCO3}$ to a saturated solution of $\ce{KHCO3}$ I'd expect a little to dissolve but not much. Easy enough to try.  // Details on what you're investigating might  allow us to suggestions about other approaches.

Dissolving $\ce{KHCO3}$ in hot water might give you an unstable supersaturated solution. In other words it might be stable just long enough to do some experiment.

I have no idea what would happen with mixtures of $\ce{K2CO3}$, $\ce{CO2}$ and $\ce{H2O}$ under  very high $\ce{CO2}$ pressures.

The final use is to have the solution react with another solution quickly to generate gas. This means temperature is out of the question as it is a volume of solution you need to carry around. Any idea on terms/nomenclature on saturated experiments and Ks changes I can look up (chemistry is not my field)??   Thanks

You can buy liquefied $\ce{CO2}$ in canisters. Pop that open and it all goes to gas.


text:
What is the exact definition of state functions? How to determine state functions?
How do you know if something is a state function? I know Internal Energy (U) is a state function since it is the sum of energies, which is constant at a given time.
However I don't know for sure if $\Delta$U is a state function. I know that $\Delta$U$= U_{final}-U_{initial}$, and so any path from final to initial will yield the same $\Delta$U. So by this definition any $\Delta$ function should be a state function.
As a result I would expect Heat ($q$) to also be a state function since $q=\Delta$TE, but this is not the case.
Where am i going wrong in my understanding? I understand the "altitude" analogy, however i need a more technical definition with chemistry terms.

(source: cnx.org)
]

Where am i going wrong in my understanding?

In short: everywhere. I mean all possible ways. Trying to cheat by randomly throwing symbols and words out there without deep thinking is a way to nowhere. Thermodynamics is a tough subject because it requires a strong mental discipline, so to start from, you should know and understand the definitions. Funky pictures won't help unless you understand the actual meaning of the analogies used there.
State function
A state function is a property of a system that depends only on the current equilibrium state of the system, not on the way in which the system acquired that state.
Using this definition we could answer the question

How do you know if something is a state function?

trivially: if a quantity is uniquely determined by the equilibrium state then it is a state function, otherwise it is not. $\Delta U$ is clearly not a state function, since by its very definition ($\Delta U = U_\mathrm{final} - U_\mathrm{initial}$) it is not determined uniquely by the state of the system: neither by the final, nor by the initial one. $\Delta U$ is not even a characteristic of the state (since it clearly refers to two of them, not just one), rather it is the characteristic of a thermodynamic process of transition from one state into another.
Internal energy

I know Internal Energy ($U$) is a state function since it is the sum of
  energies, which is constant at a given time.

Sorry, but this sounds like a nonsense to me. What do you mean by constancy of internal energy at a given time due to it being sum of some energies? From a microscopic point of view, internal energy of a system can indeed be subdivided into kinetic energy of motion of all the system's particles and potential energy due to all kinds of interaction between particles in the system,
\begin{equation}
 U = E_{\mathrm{micro,k}} + E_{\mathrm{micro,p}} \, .
\end{equation}
But what kind of constancy follows from this subdivision? Do you want to say that $\Delta U$ is always zero or what?
Note also that such definition of the total energy belongs already to the realm of statistical thermodynamics, not the classical one. In classical thermodynamics the total energy is postulated to be a state function. Essentially, the first law of thermodynamics tells us that there exist a state function called internal energy, $U$, the increase in which for a closed system during a thermodynamic process is equal to the heat supplied to the system plus the work done on the system,
\begin{equation}
 \Delta U = q + w \, ,
\end{equation}
or in infinitesimals for a quasistatic process,
\begin{equation}
    \newcommand{\dif}{\mathrm{d}}
    \newcommand{\difbar}{\text{đ}}
 \dif U = \difbar q + \difbar w \, .
\end{equation}
Note that infinitesimal heat and work transfers in a quasistatic process are denoted by $\difbar$ rather than $\dif$, used to denote infinitesimal change in internal energy, to emphasize the fact that heat and work in contrast to internal energy are not state functions. Mathematically speaking, $\dif U$ is an exact differential, while $\difbar q$ and $\difbar w$ are inexact differentials. For more on that story, check this answer of mine. What is really important for our current purposes is the very final sentence there: essentially it was experimentally found that for a closed system $\oint \difbar q \neq 0$ and $\oint \difbar w \neq 0$, but $\oint (\difbar q + \difbar w) = 0$. This finding implies that there exist a state function $U$ such that $\dif U = \difbar q + \difbar w$. So, the law that postulates that $U$ defined by $\Delta U = q + w$ is a state function is based on experimental evidences.

Thanks. Does $q=\Delta$TE. Also what does this mean about Hess's law and $\Delta$H. Since $\Delta$H is the same for any chemical process from A to B.

Cavalier use of equilibrium. A state need not be at equilibrium.


text:
Finding moles of different compounds using balancing
$$\ce{K2Cr2O7 + $x$\,H2SO4 + $y$\,SO2 -> Cr2(SO4)3 + K2SO4 + $z$\,H2O}$$
For $\ce{Cr}$ change in oxidation number is: $+12$ becomes $+6$
For $\ce{S}$ change in oxidation number is: $+4y$ becomes $+6$
For $\ce{H}$ change in oxidation number is: $+2$ becomes $+2z$
since $\ce{Cr}$ decreases by 6, $\ce{S}$ and $\ce{H}$ have to increase by 6, so
$$(4y+2)-(6+2z)=6$$
$$2y-z=5$$
Also using the principle of atom conservation (POAC) for $\ce{H}$ atoms from $\ce{H2SO4}$ to $\ce{H2O}$, we get:-
$$2x=2z$$
$$x=z$$
I can't find any more equations to solve this.

Your equations for the changes in oxidation number are all wrong (apart from that of chromium). Hydrogen is not undergoing any redox anywhere. Why do you multiply sulfur in reactants by y, but in the products you just ignore the stoichiometric coefficient? You could just use the conservation of atoms for H, S and O, which gets you 3 equations in 3 unknowns which you can then solve.

POAC = Principle of Atom Conservation

Change in oxidation number? Cr goes from +VI to +III, sulphur from +IV to +VI. I have an idea where you got your numbers from, but they are highly confusing at least to anybody who studied chemistry in my part of the world …

While calculating the change in oxidation number for $S$ why are we not considering the source from Sulphuric acid?According to me ideally why is it not $6x+2y$?

Never having heard of the Principle of Atom Conservation (POAC) before I'm curious.
Looking up the concept, I understand it simply to mean that atoms can't be created or destroyed in a chemical reaction. (So we're ignoring radioactive decay for example.) I start blindly with a variable for each reactant and product, and will then look for linear relationships in the variables.
$$\ce{$a$\,K2Cr2O7 + $b$\,H2SO4 + $c$\,SO2 -> $d$\,Cr2(SO4)3 + $e$\,K2SO4 + $f$\,H2O}$$
So with $a$ $\ce{K2Cr2O7}$ molecules, I must get $a$ $\ce{Cr2(SO4)3}$ molecules to balance $\ce{Cr}$, and $a$ $\ce{K2SO4}$ molecules to balance $\ce{K}$. So:
$$\ce{$a$\,K2Cr2O7 + $b$\,H2SO4 + $c$\,SO2 -> $a$\,Cr2(SO4)3 + $a$\,K2SO4 + $f$\,H2O}$$
Now $\ce{H}$ is only in $\ce{H2SO4}$ on the left and $\ce{H2O}$ on the right so $f=b$.
$$\ce{$a$\,K2Cr2O7 + $b$\,H2SO4 + $c$\,SO2 -> $a$\,Cr2(SO4)3 + $a$\,K2SO4 + $b$\,H2O}$$
To balance $\ce{S}$:

$$b + c = 4a$$

To balance $\ce{O}$:

$$7a + 4b + 2c = 16a + b$$ which reduces to $$3b +2c = 9a$$

At this point just balancing atoms leaves you with 2 linear equations and 3 unknowns. So there are not enough equations to explicitly solve the problem using linear equations. However relationships can be established.
\begin{array}{cccc}\\
3b& +2c & = & 9a \\
-2(b& +c & = & 4a)\\ \hline
 & b & = &a \\
\end{array}
$\therefore c = 3a = 3b$
The final step in this problem is to realize that an integer solution (a Diophantine problem in mathematics) is desired. So $a=b=1$ and thus $c=3$. So the final chemical equation is:
$$\ce{K2Cr2O7 + H2SO4 + 3SO2 ->Cr2(SO4)3 + K2SO4 + H2O}$$

PS - Many thanks to user Mathew Mahindaratne for suggesting a cleaner solution in the comments.

When I took chemistry I don't remember a high falutin name for this. I remember mostly doing this in my head to figure out where the sticking point was in balancing the equation.

Indeed there are two linear equations in three unknowns. The solution to that "problem" is that you are free to multiply the entire chemical equation by any number you like and it would still be correct. For example, we could either write $$\ce{H2 + 1/2 O2 -> H2O}$$ or $$\ce{2H2 + O2 -> 2H2O}$$ and both would still be correct. Therefore, you have the freedom of arbitrarily choosing the value of one of the remaining three unknowns, say $a$, and finding the corresponding values of $b$ and $c$ using the last two equations. It's analogous to normalising a wavefunction in QM.

@orthocresol - Thanks for the comment. I had left the solution hanging. So I added a bit about needing an integer solution and finished the solution. To me the overall idea was that even using POAC you still got to a point where you more or less "guess" then check the guess. So $$\frac{1}{2}\ce{H2} + \frac{1}{4}\ce{O2 ->} \frac{1}{2}\ce{H2O}$$ might work in a math class, but not in a chemistry class.

@MaxW: Assuming OP is not familiar with mathematics, it'd be nicer to show $a = b$ and $c=3a=3b$ by simple mathematics using your two linear equations.

@MathewMahindaratne - Thanks for the suggestion! I changed the solution and it is a cleaner result.


text:
Stability of carbocations: CF3+ vs CH3+
Is $\ce{CF_3+}$ more stable than $\ce{CH_3+}$?
In $\ce{CF_3+}$, there is $\ce{C-F}$ back bonding that increases  stability of the carbocations. But -I effect of $\ce{F}$ dominates +R effect and this decreases the carbocation stability. How can I compare these two competing effects to find the more stable carbocation?
Also, is $\ce{CF_3+}$ more stable than $\ce{CH3-CH+-CH_3}$ and tert-butyl carbocation?

So what is the question?

Is CF3+ more stable that CH3+

Yeah, but your first sentence answers that question... You said that the -I effect outweighs the +R effect and therefore the $\ce{CF3+}$ cation is less stable. So your question is probably how to judge which of the inductive or resonance effect is larger. I mean, I think this could be a good question about competing effects if you are clear about what you are asking.

@orthocresol: Is CF3+ more stable than propyl carbocation (+ charge at secod carbon of propane) and tert-butyl carbocation?

Alright, I don't know the answer to that. You'll have to edit it into the post and hopefully someone else will answer.

Is there evidence that CF3(+) is more stable than CH3(+)? My intuition would be that methyl cation is more stable.

@jerepierre : at DMSO, pka of methane is about 56 while pka of CF3H is 25-30

But deprotonation forms the anion, not the cation. The $\ce{CF3-}$ anion is trigonal pyramidal, so there's no backdonation from fluorine, and it's stabilised by the -I effect of fluorine, which is what leads to the smaller $\mathrm{p}K_\mathrm{a}$.

So the pka values only explain the carbanion stability....

In the gas phase, the trifluoromethyl carbocation is more stable than the methyl carbocation, See p. 170 here

and p.54 here

$\ce{CF3+}$ is more stable than $\ce{CH3+}$. The comparison between -I and +R effect is made while talking about activation or deactivation of halogens when substituted on benzene. Fluorine always stabilises a carbocation to a large extent because of a very good overlap of $\ce{2p -~2p}$ orbitals.

So should I consider that net  electron donating tendency of halogens to carbocations is more than the net electron withdrawing tendency?

@user1825567: I guess the extent of effect follows, Mesomeric > Resonance > Inductive. No?

@Rahul Verma  Sorry, I am no longer fluent in this topic. BTW...All the best for JEE !

@RahulVerma: Aren't 'mesomeric' and 'resonance' effects the same thing? According to the Gold Book 'mesomeric effect' is an alternate term for 'resonance effect'. If so, can you tell what did you mean by 'Mesomeric > Resonance'?

@GuruVishnu It seems like a mistake, mesomeric effect and resonance are essentially the same thing, and this effect is stronger than inductive

@GuruVishnu, the mesomeric effect is included / subsumed in resonance effect category. It is a special case of resonance, where $\ce{\pi}$ orbitals are in conjugation. Other cases of resonance can include $\ce{C^+}$ or $\ce{C^-}$

@RahulVerma  Is there any explanation for the trend? I would like to know why resonance > Inductive in determining organic cation stability.  Thanks

@234ff, have a look at this answer. It has a basic explanation of the trend. If you find it useful, don't forget to vote it

@234ff, or if you need a very detailed insight, you can look up this chapter.

Despite the strong $\sigma$-electron withdrawing effect of fluorine, $\ce{CF3+}$ has been stabilized relative to that of $\ce{CH3+}$. For example, the reaction:
$$\ce{CH3+ + CHF3 -> CF3+ + CH4}$$ is exothermic by $\pu{13.2 kcal mol-1}$ (Ref.1). It was also found that hydrogen abstraction in following reaction is even more exothermic than previous one:
$$\ce{CH3+ + CH2F2 -> CHF2+ + CH4} \qquad \Delta H = \pu{-28.3 kcal mol-1}$$
According to authors, the only order which can be constructed that is consistent with all observations gives the relative hydride affinities as $\ce{CH3+ \gt CF3+ \gt CH2F+ \gt CHF2+}$. It is also noteworthy that within the references in Ref.1 has given that $\Delta H_f (\ce{CH3+}) = \pu{276.1 kcal mol-1}$ while that of $\Delta H_f (\ce{CF3+}) = \pu{99.3 kcal mol-1}$.
References:

R. J. Blint, T. B. McMahon, and J. L. Beauchamp, "Gas-phase ion chemistry of fluoromethanes by ion cyclotron resonance spectroscopy. New techniques for the determination of carbonium ion stabilities," J. Am. Chem. Soc. 1974, 96(5), 1269–1278 (DOI: https://doi.org/10.1021/ja00812a001).


text:
Are there any free pharmacophore modelling software?
I would like to know if there is any free and relatively updated software to getting started in pharmacophore modelling.

Have you checked the ones listed here: https://en.wikipedia.org/wiki/Pharmacophore ?

I think they are not free though


text:
Determining Overall Equation and Oxidation Potential

Cell 1 is composed of a reference chloranil half-cell (standardised against SHE as $+0.680 \mathrm{V}$) connected to an $\ce{Ag/AgCl}$ half-cell containing saturated ($0.125\ \mathrm{mol\ L^{−1}}$) $\ce{NaCl}$ in glacial acetic acid.
$\ce{Ag_{(s)}\ |\ AgCl_{(s)}\ |\ Cl^{-}_{(CH3COOH)}\ ($0.125$~M)\ ||\ H+(aq)\ ($0.5$~M),\ C6Cl4O2_{(aq)(sat)},C6Cl4(OH)2_{(aq)(sat)}|\ Pt_{(s)}}$
Cell 2 is composed of the same reference chloranil half-cell connected to an Ag/AgCl half-cell containing $\mathrm{1.0 ~mol~ L^{-1}}$ $\ce{HCl}$ in glacial acetic acid.
$\ce{Ag_{(s)}\ |\ AgCl_{(s)}\ |\ HCl_{(CH3COOH)}\ ($1.0$~M)\ ||\ H+(aq)\ ($0.5$~M),\ C6Cl4O2_{(aq)(sat)},C6Cl4(OH)2_{(aq)(sat)}|\ Pt_{(s)}}$
a) Write the equation for the overall reaction occurring in both cells.
Cell 1 is used to determine the value of $\mathrm{E^o}$ for the reduction of $\ce{AgCl_{(s)}}$ to $\ce{Ag_{(s)}}$ and $\ce{Cl^{–}}$ in glacial acetic acid –
  $\mathrm{E^\circ}\ce{(AgCl/Ag)}$. The cell potential of Cell is $0.422\ \mathrm{V}$
b) Calculate $\mathrm{E_{OX}}$ – the reduction potential of the oxidation half-cell – of Cell 1.

My Attempt
a) Electro chemistry is not my strong point and I am not too sure on the notation for the cell. I have never seen a cell that has 3 species in the anode and in the cathode half-cell. Could someone explain to me what it means?
I am guessing the oxidation half-equation is: $$\ce{Ag_{(s)} + Cl^{-}_{(aq)} -> AgCl_{(s)} + e^{-}}$$
My overall equation is then:
$$\ce{2Ag_{(s)} + 2Cl^{-}_{(aq)} + C6Cl4O2_{(aq)} + 2H^{+} -> 2AgCl_{(s)} + C6Cl4(OH)2_{(aq)}}$$
Is this correct? Also will the overall reaction for cell 2 be the same as the overall reaction for cell 1 as the only difference is that $\ce{NaCl}$ is replaced with $\ce{HCl}$.
b) Is the question a typo and really means: "Calculate $\mathrm{E_{ red}}$ – the reduction potential of the oxidation half-cell – of Cell 1." or am I missing something?
If the question is a typo, I get that the answer is $-0.258\ \mathrm{V}$, is this correct?


text:
How is the Felkin-Anh model of stereoinduction correctly explained with MO theory?
One method to predict stereoinduction in asymmetric reactions is the Felkin-Anh model (derived from the more basic Cram model) that applies to nucleophile attacks on α-chiral aldehydes. In a simple case such as (S)-2,3,3-trimethylbutanal, assignment of large, medium and small residues is trivial and we predict nucleophile attack as indicated in scheme 1 to generate (2R,3S)-2-hydroxy-3,4,4-trimethylpentanenitrile.

Scheme 1: Felkin model for all-carbon systems.
In some cases — I was taught ‘in cases of a group with high electronegativity on the α-carbon’ — we may no longer consider simply the size of the substituents but instead need to consider the electron-withdrawing substituent to be orthogonal to the carbonyl bond axis. Examples would be α-haloaldehydes, α-silyloxyaldehydes or α-aminoaldehydes as shown in scheme 2 with (S)-2-substituded butanal.

Scheme 2: Felkin-Anh model for systems containing electronegative $\ce{X} = \ce{Br, Cl, OTBS, NHR, etc}$.
How is the Felkin-Anh model explained on a molecular orbital basis? Please include diagrams showing the orbitals involved and what exactly is being stabilised.
Also, if you can confirm (or unconfirm) my list of possible electron-withdrawing substituents, please do so in comments to the question.

From the organic chemistry textbook by Clayden, Warren, Wothers and Greeves on pp. 890-891:

  The $\pi^{*}$(C=O) and $\sigma^{*}$(C–X) orbitals add together to form a new, lower-energy molecular orbital, more susceptible to nucleophilic attack. But,
  if X is not a leaving group, attack on this orbital will result not in nucleophilic substitution but in addition to the carbonyl group. Again, this effect will operate only when the C–X and C=O bonds are perpendicular so that the orbitals align correctly.
What does this mean for stereoselectivity? Conformations of the chiral carbonyl compound that place an electronegative atom perpendicular to the C=O bond will be more reactive-size doesn't matter.

Update
In response to the requests specified in the bounty description:

1) What about the mixing of sigma(C-X) and pi*(C=O)? 2) What about pi(C=O) and sigma*(C-X)? Shouldn't that raise the sigma*'s energy?

As you probably know, the above treatment of the effect of one fragment orbital interacting with another one is basically just a pictorial way of applying perturbation theory to the problem. From this theory you get the result that the interaction strength of two fragment orbitals gets stronger the better the overlap and the closer the orbitals are in energy - a very(!) detailed derivation of the fundamental equations involved has been done by @Martin here (an answer he truly deserves much more upvotes for :)). Overlap will not be an issue, but the energy differences will be. So, when thinking about which orbital interactions are most important for the problem at hand we have to look at their supposed energy levels. Thus, I've done a rough sketch, taking into account that:

the $\ce{C}$ atomic orbitals are lower in energy than the $\ce{O}$ or $\ce{X}$ atomic orbitals but that the $\ce{O}$ and $\ce{X}$ orbitals will be rather close in energy (similar electronegativities etc.) and
$\pi$-interactions are weaker than $\sigma$-interactions due to less overlap thus resulting in a lower level-splitting between the bonding and antibonding orbitals.

On the right side of the picture I added arrows indicating the energy differences between the most important orbitals.

As you can see the energy differences between $\sigma(\ce{C-X})$ and $\pi^{*}(\ce{C=O})$ and between $\pi(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$ are very large, much larger than the one between $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$. Thus the two interactions you ask about will only have a very minor influence on the LUMO and the single most important interaction is the one between $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$. Certainly, some mixing will occur (see below for some more on this) but when employing this method you are mostly interested in the big effects at work and can ignore the small ones.

3) Shouldn't we better consider the entire thing as an allyl cation system?

The system at hand has much lower symmetry than the allyl cation because it consists of 3 different atoms/fragments. That makes drawing analogies between this and the simple pure carbon case difficult. It's certainly possible but the treatment shown in the book by Clayden is much easier. But even if you do a more rigorous perturbational treatment not only considering the first order interactions between $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$ but also the second order mixing of $\pi(\ce{C=O})$ (which in effect would be like doing something similar to the allyl cation) you get qualitatively the same picture as drawn above in the book by Clayden. Basically the only effect the mixing of $\pi(\ce{C=O})$ has is a slight down-shift in energy of the low-lying $\pi(\ce{C=O})$ orbital and a slight up-shift in energy of the new LUMO you get from Clayden's method, i.e. the bonding combination of $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$.

4) If you can perform calculations on a simple system or know where to find one that would be greatly appreciated

You could try to do calculations for the following molecule

whereby you do one calculation with the $\ce{F}$ of the $\ce{CH2F}$ group perpendicular to the $\ce{C=O}$ plane and one with one of the $\ce{H}$ atoms of said group in that position. Then you can compare the LUMOs of both conformers and see if the position of $\ce{F}$ atom actually has an energy-lowering effect.

The answer that doesn't invoke mixing of empty orbitals is given in Carey, Sundberg Advanced Organic Chemistry (X is a group capable of accepting a significant amount of electron density):

There is a stereoelectronic (hyperconjugation) effect, which involves the interaction
  between the approaching nucleophile and the LUMO of the carbonyl group. This
  orbital, which accepts the electrons of the incoming nucleophile, is stabilized when
  the $R_L$ group is perpendicular to the plane of the carbonyl group. This conformation
  permits a stabilizing interaction between the developing bond to the nucleophile and
  the antibonding sigma antibonding orbital associated with the C-$R_L$ bond. Because this is a sigma bonding-sigma antibonding
  interaction, it should increase in importance with the electron-acceptor capacity of X.


text:
How do these solutions "react"?
If I were to combine solutions of $\ce{(NH4)2CO3}$ and $\ce{HNO3}$ in water, the only ionic compounds that could form are $\ce{H2CO3}$ and $\ce{NH4NO3}$ right? But since both are soluble no precipitate would form correct? So does that mean that no reaction occurred? Does it also mean that an ion like $\ce{NO3-}$ will never react in an aqueous solution because it is always soluble in water?

You picked a difficult question because this question involves acid-base reactions. It is no longer a simple "double displacement" reaction where the cations and anions "swap partners".

If I were to combine solutions of $\ce{(NH4)2CO3}$ and $\ce{HNO3}$ in water, the only ionic compounds that could form are $\ce{H2CO3}$ and $\ce{NH4NO3}$ right?

No, not really. You have to realise that the protonation of $\ce{CO3^2-}$ occurs in two stages: firstly, to $\ce{HCO3-}$, and then only to $\ce{H2CO3}$. Also, the $\ce{NH4+}$ ion is acidic, and can react with water to form $\ce{NH3}$.
On top of that, $\ce{H2CO3}$ will spontaneously decompose into $\ce{H2O}$ and $\ce{CO2}$, the latter of which is released as a gas.
Since these species are all soluble in water, and soluble ionic compounds exist as ions in water, a more appropriate way to think about this question is "what species are present", not "what ionic compounds are present". Species is just a general term that includes ions as well as molecules (since, for example, $\ce{CO2}$ and $\ce{NH3}$ are not "ions").

But since both are soluble no precipitate would form correct?

That is correct.

So does that mean that no reaction occurred?

No, there is a reaction happening because of the $\ce{CO2}$ gas formed. You will observe a lot of effervescence if you were to mix these two together.
However, if this question did not involve $\ce{CO2}$... Let's say you are reacting $\ce{NaCl}$ with $\ce{KBr}$. Since all the possible products ($\ce{NaBr}$ and $\ce{KCl}$) are soluble, you could say that in this case, no reaction would occur.

Does it also mean that an ion like $\ce{NO3-}$ will never react in an aqueous solution because it is always soluble in water?

No. You could say it will never be precipitated out. However, $\ce{NO3-}$ can also react in other ways apart from precipitation. For example, it can be used as an oxidising agent. One famous example is the dissolution of a copper coin in concentrated nitric acid.

Great answer, I'd just add one fine point. $\ce{NH4NO3}$ does have a finite solubility in water. The reason that no ppt is formed is because $\ce{NH4NO3}$ is much more soluble than $\ce{NH4CO3}$.

Would the formation of $\ce{CO2}$ form bubbles? Or would it happen very slowly because the $\ce{H2CO3}$ has to decay first, and the bubbles wouldn't be visible?

It is very fast. Watch this: https://www.youtube.com/watch?v=sHhT9tmePbk


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Acidic and alkaline salt solutions: why do some salt form acid/bases while others don't
I have a question regarding salts and their solutions. How come some salts, like sodium carbonate and ammonium chloride, produce acids/bases when dissolved in water? Moreover, how can you predict whether these solutions will act as an acid or as a base just by looking at the salt?

I believe you have some idea as to what is hydrolysis...

Without trying to go into too much detail this has a two part answer.
First a salt by definition has an anion and a cation. So ordinary table salt, $\ce{NaCl}$ is a chemical salt, but table sugar is not a salt. So lets think of the cation as $\ce{C^+}$ and the anion as $\ce{A^-}$.
So we start out with solid salt which in chemistry is written as $\ce{CA_{(s)}}$. When the salt dissolves the two ions are hydrated which means that water molecules surround each ion and separate the pair. the chemical reaction is written like this: $$\ce{CA_{(s)} -> C^+_{(aq)} + A^-_{(aq)}}$$
Typically the (aq) subscripts are left off, but implied.
The second part of the answer involves the nature of the cation or anion. To understand if the pH of the solution will change depends on whether the cation is able to "donate" a hydrogen ion, or the anion is able to accept one.
$\ce{NH_4^+}$ is a cation and is able to donate a hydrogen ion to the solution. So the pH will drop and the solution becomes more acidic. $$\ce{NH_4^+ + H_2O -> NH_3 + H_3O^+}$$
$\ce{CO_3^{2-}}$ is a anion and is able to accept a hydrogen ion from the solution. So the pH will increase and the solution becomes more basic. $$\ce{CO_3^{2-} +H_2O-> HCO_3^- + OH^-}$$
When water, $\ce{H_2O}$, makes or breaks bonds the process is called hydrolysis.


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Does the acid dissociation constant fluctuate or stay constant during titration?
I'm doing titration experiment, where I record the volume of base added and the pH value, then use these two variable to make a titration graph. I'm suppose to extrapolate any data needed from the chart, and find the acid dissociation constant at 0% to equivalence point, 50% to equivalence point, and 75% to equivalence point. The ka value I determined respectively are 0.000165, 0.000457 and 0.000239. The first, and possibly the last value (by a stretch) are the only acceptable values. So my question is, does ka stay constant or can it fluctuate during titration, before settling down to the literature ka value?
I'm leaning towards it stays constant regardless and that the discrepancy is due to errors I made, but then I also think that during titration, not all of the molecules neutralizing each other are going to meet and react instantly, and that could also affect something somehow.

As the name implies, it is constant. When the only thing happening is proton exchange, the equilibrium is reached pretty rapidly.

There is a pretty good explanation of titration and how you can use it to determine Ka on the "Titration curve" Wikipedia page. The pKa can be determined as the pH value measured at the "50% to equivalence" point. To be accurate, you might want to repeat the experiment a few times. I am not seeing how you would use the 0% and 75% values.