text:
Pauli-forbidden term symbols for atomic carbon
Carbon has a $\mathrm{p^2}$ configuration, and within the Russell-Saunders coupling scheme, we have
$$\begin{align}
s_1 = s_2 &= \frac{1}{2} & S &= 1,0 \\
l_1 = l_2 &= 1 & L &= 2,1,0 \\
\end{align}$$
which gives rise to $^3D$, $^1D$, $^3P$, $^1P$, $^3S$, and $^1S$ terms. Out of these six, only $^1D$, $^3P$, and $^1S$ are allowed by the Pauli principle.
It's not hard to show that the $^3D$ term is forbidden, since this term has to contain one state with $m_{l1} = m_{l2} = 1$, which would correspond to a spatial wavefunction that is symmetric with respect to interchange of the two electrons. Pairing this with a triplet spin wavefunction, which is also symmetric, would lead to an overall symmetric wavefunction, which is not allowed.
But that's all the textbooks say - they don't describe why the $^1P$ term is forbidden, for example. I know how to construct a table of microstates, and I know I can use the direct product table of the full rotation group to find that $P \times P = S + [P] + D$, and I know that both of these methods show that the allowed term symbols are only $^1D$, $^3P$, and $^1S$.
Essentially, I am looking for a more intuitive explanation of why the $^1P$ and $^3S$ states are forbidden that preferably links back to the requirement that the wavefunction be antisymmetric - if that is possible.
Alternatively, an explanation of why $P \times P = S + [P] + D$ would also work. I know that from the Clebsch-Gordan series we get $S$, $P$, and $D$ terms, but not why the $P$ state should be antisymmetric. I do know the general idea of how to obtain the vector coupling coefficients (by repeated application of raising/lowering operators), and these do show that the $P$ state is antisymmetric, but I'd like a more interpretative answer than just "that's how the maths works out".

Okay, actually I managed to find somewhat of an explanation myself.
We could construct a state with $M_L = 1$ and $M_S = 0$, for example:
$$\begin{align}
\psi = \frac{1}{2}\left[\mathrm{2p_1}(1)\mathrm{2p_0}(2) + \mathrm{2p_0}(1)\mathrm{2p_1}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right]
\end{align}$$
where $\mathrm{2p}_m$ represents the 2p orbital with $m_l = m$.
However, such a combination of $(M_L,M_S)$ can only occur once, since the two electrons are indistinguishable. In other words, the following two other wavefunctions are not permissible because permuting the two electrons changes the wavefunction:
$$\begin{align}
\psi &= \frac{1}{\sqrt{2}}\left[\mathrm{2p_1}(1)\mathrm{2p_0}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right] \\
\psi &= \frac{1}{\sqrt{2}}\left[\mathrm{2p_0}(1)\mathrm{2p_1}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right]
\end{align}$$
and the $^1D$ state, which has $S = 0 \Rightarrow M_S = 0$ and $L = 2 \Rightarrow M_L = 2,1,0,-1,-2$, already has to encompass one state with such a combination of $(M_L,M_S)$. Therefore, a state with $(M_L,M_S) = (1,0)$ cannot contribute to a $^1P$ term.
It is logically not the strongest argument (sufficiency vs necessity), but it is kind of what I was looking for. If anybody can provide another viewpoint though I will be glad to accept the answer.


text:
Viscosity of branched vs. linear short alkanes
I was looking through the CRC Handbook of Chemistry and Physics section on liquid viscosities, trying to explain trends in viscosities. In higher alkanes, the viscosity increases with the degree of branching, which is logical. However, I noticed for lower alkanes, the opposite is true. For example, 1-methylbutane has a viscosity of 0.286 mPa-s at 25 C, whereas its linear counterpart pentane has a viscosity of 0.30 mPa-s at 25 C. The same trend holds for alkanes with 6 carbons. These trends go against my chemical intuition and I cannot think of a sound explanation, which brings me to my question: why is the viscosity lower for branched vs. linear alkanes in the lower alkanes?

Viscosity is related to Van der Waals/London forces between molecules.  This in turn is related to packing and surface area.
A branched 6-chain carbon, tumbling in solution cannot align as much of its surface area with another molecule as the linear hexane.
At some point the larger molecules are becoming able to be 'entangled' and form structures somewhat similar to knots.  When the molecules get this large, viscosity increasing branching becomes more likely.
Finally, when you get to true polymers, the chains are so long that branching again begins exclude other molecules from structure and again linear molecules are more likely to have higher viscosities (although for polymers, these are solution viscosities).

I don't think I quite follow your line of reasoning. I understand that viscosity is related to vdW forces. When you say "A branched 6-chain carbon, tumbling in solution cannot align as much of its surface area with another molecule as the linear hexane," it seems like molecular alignment in linear hexane would decrease the viscosity, since molecules can more easily slide past one another.
Also, linear polymers tend to have lower viscosities than branched ones, so I'm not sure I follow your last point, either.

Think is if more like strips of velcro.  The more they touch the stronger the bond is.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Why quaternary nitrogen but not tertiary oxygen?
Why do quaternary ammonium ions with a partially positive nitrogen form fairly readily and are often stable but tertiary oxygens, apparently called oxonium ions, are more rare/less stable? The trend seems to continue down the periodic table, too, with phosphonium and sulfonium.

Short answer could be that ethers are weaker nuclophiles and better leaving groups then amines.

This has to do with the electronegativity of the species and how good of a base (or how good an acid the conjugate acids) are.
$\hspace{9ex}$
$\hspace{7ex}$
[source]
To put this in perspective, nitric acid ($\ce{HNO3}$) has a $\mathrm{p}K_\mathrm{a}$ of $-1.3$. Yes, $\ce{HNO3}$, a strong acid, is more stable than $\ce{H3O+}$! The central oxygen in $\ce{H3O+}$ is too electronegative to effectively share these electrons and form a stable bond. In $\ce{NH4+}$, its lone pairs are far more available due to nitrogen's lower electronegativity.
How common each species is only has to do with the number of acids stronger than $\ce{H3O+}$ or $\ce{NH4+}$. As you can imagine, there are going to be a lot more acids stronger than $\ce{NH4+}$, which starts out around the middle of the $\mathrm{p}K_\mathrm{a}$ spectrum, than compared to $\ce{H3O+}$, which is near the end.

The electronegativity of Oxygen is higher than Nitrogen, so it is much less suited for holding a positive charge, which, in turn leads to Mithoron's comment about weaker nucleophilicity and better leaving group ability.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
RCSB PDB - how to choose between multiple copies of the same protein?
I want to map the location of missense mutations on a protein, SERPINB3. I intend to do this using PyMol and download the protein structure from RCSB PDB. However, searching for SERPINB3 reveals two molecular structures (as well as one homolog, SERPINB4).
One of the structures, 2zv6, presents the protein as a multimer of three identical chains. This structure is older (2009) and is used by computational tools on CbioPortal.
Another structure, 4zk3, presents the protein as monomer and has been more recently published (2015).
I am inclined to trust the more recent protein over the older one on the vague notions that:

Its newer so it was probably an improvement in terms of some aspect of the structural determination process
I've heard that its "unusual" for a multimer to form with three identical chains

...which are not particularly good rationalizations.
I have not found any literature comparing these competing molecular structures and honestly I am not entirely clear on how different they are from each other beyond the fact that one is presented as a multimer.
What justifications can be made based on chemistry and/or biology to support one of these structures over the other? I am interested in both choosing between these two structures and in general good practices for choosing between two competing structures.

Did you checked if monomer in first case differs from second deposit?

Yes, there are differences in the number of a-helices and B-sheets

2zv6 and 4zk3 are, upon inspection, completely different proteins. I'm afraid you made strange mistake thinking them describing the same thing.

What do you mean they are completely different proteins? Both turn up in a search for "SerpinB3", both are referenced to the same Uniprot ID (P29508), running a BLASTp search comparing one subunit of 2zv6 to 4zk3 returns an Score of 1640 and E-value of 0 indicating high degree of sequence similarity with an infinitesimally low chance of the similarity being due to random chance.

They are coded by the same gene but are different. Even length is different!


text:
Ballistic behavior of molecules on potential energy surfaces
I came across the following sentence in a publication$^1$: "At this point we have to remember that the computed MEP path [sic] represents the reaction in the absence of any kinetic energy, while in reality the true reaction path, to some degree, will be ballistic." Further, they write that, "This path is accessed because after the transition state the molecule is 'ballistic'."
A minimum energy path (MEP) is calculated "step-wise", taking the previous geometry as a starting point, and the program follows the gradient. In classical terms, this might be to place a ball on a tilted plane, releasing it, and stopping it after a given distance, and repeat.
When they say that, in a real reaction, the molecules will have kinetic energy and be ballistic, I think they mean that the potential energy surface (PES) may be explored in directions slightly different than the gradient. This can be compared to releasing a ball on the top of a changing surface and allowing it to roll freely, where the ball will not follow the gradient when its velocity is high.
I cannot find a decent explanation of this anywhere, and it would be nice to get some input.
$^1$ De Vico, Liu, Krogh, and Lindh, 2007. J. Phys. Chem. A 111:8013-8019. doi:10.1021/jp074063g.  Cited passages are on p. 8016, Section 3.3.

I think the description you lay out in your question has it basically right.
MEPs, IRCs, etc. all assume that the geometric rearrangements that occur over the course of a reaction strictly follow the gradient uphill from reactants to transition state, and then downhill from TS to products (or intermediates).  Energetically, this provides the "path of least resistance."
In the paper you cited, it looks like the MEP would require some atoms in the system to "take a right-angle turn." Quoting a broader excerpt around the first sentence you include (emphasis added):

As can be seen in Figure 3, after TS‘ S$_0$ the MEP takes a turn on the PES:  the O−O‘ distance increasing and the torsion around the O−C−C‘−O‘ dihedron nearly stopping, while the increase of the C−C‘ distance comes into action.
  ...
  At this point we have to remember that the computed MEP path represents the reaction in the absence of any kinetic energy, while in reality the true reaction path, to some degree, will be ballistic.
  ...
What would be the consequences if the reaction continued along the torsional mode? Section 3.2 already suggested that this is of interest—the torsional mode is the principal internal coordinate of the T$_1$ MEP.

Nuclei have mass (obviously), and thus have inertia. While Newton's relevance at the atomic scale can be fuzzy at times, classical treatment of atomic motion has sufficient validity that acceleration of nuclei can be considered to require a suitable accompanying force (viz., PES gradient).  If the system has enough "inertia" and the PES gradient provides too small an acceleration to overcome it, the progress of the reaction can absolutely move away from the MEP.

Alternatively, at least one additional mechanism can introduce  deviations from the MEP: molecules at finite temperature are always vibrating. Thus, even in a case where the key internal molecular coordinates (bond lengths, angles, dihedrals) for a given reaction strictly follow the IRC, the "uninvolved" vibrational motion is still occurring, and so the  system is actually always oscillating around the IRC in a high-dimensional sense.  If the energy of the system is high enough, these oscillations have the potential to be of sufficient magnitude to "bump" the reaction coordinate away from the gradient-following pathway.
Rotational motion may also contribute to this, but I would think in a more subtle way than vibrational motion.

For further reading: Steven Bachrach of the Computational Organic Chemistry blog has an entire tagged category, "Dynamics," dedicated to this phenomenon.  In no particular order, some posts of possible survey value include:

An approach towards identifying dynamic effect without trajectories
Insights into dynamic effects
Bifurcating organic reactions

The entirety of Chapter 8 of his book is also dedicated to this topic.

Addendum: FWIW, as a chemical engineer, the (imperfect) analogy that I like to use is to liken the MEP to the 'reversible process' of thermodynamics:

Start at the beginning state
Take an infinitesimal step
Let the system infinitesimally relax
Repeat 2 & 3 until you reach the end state

As I (very approximately) understand it, irreversibility comes from changing the state of a system "too much, too fast", leading to net generation of entropy as the system relaxes back to the reversible pathway, and it's "kinetic energy" in a qualitative, generalized sense that allows the system to depart from the reversible path.


text:
Can rate of reaction depend on both the reactants and the product?
In introductory chemical kinetics, an elementary reaction is taken for instance
$$\ce{A -> B}$$
Then, we go on to derive integrated rate kinetics equations for them by assuming
$$\text{rate} = k [\ce{A}]^n$$
Could it possible in a more generic case that :
$\text{rate}=k[\ce{A}]^n [\ce{B}]^m$ and the total order of the reaction be $m+n$? In other words, can the rate of reaction depend on both the molar concentrations of reactants and products independently?

For complicated mechanism, it is actually quite common to have a rate law which is dependent on the concentration of the products. However, their orders will be negative, meaning that they slow down the rate of the reaction which makes sense.

Not necessarily negative; autocatalytic reactions are known as well.

There exist autocatalytic reactions than can show such a behaviour. E.g. a reaction $\ce{A + B \rightarrow 2\ B}$ might show the rate $k[\ce{A}] [\ce{B}]$.
The reaction of permanganate with oxalic acid is a well known autocatalytic reaction.
$$
\mathrm{ 5\ (COOH)_2 + 2\ MnO_4^- + 6\ H^+ \rightarrow 10\ CO_2 + 2\ Mn^{2+} + 8\ H_2O}
$$
Here the generated manganese(II) ions catalyze the reaction. The reaction starts slowly if no initial manganese(II) ions are present, and then speeds up.

However, logically speaking if I have a fixed concentration of reactant A and I am varying concentration of B with reference to the general reaction A --> B. Then shouldn't the rate also depend on concentration of B?

No, because the reaction order is defined by how the rate of reaction changes when the concentrations of one reactant is changed (though rate of product formation can be considered as well), and can be thought of as a number the sum of the possible successful collisions that the atoms can make. For example, in the reaction $$\ce{A +B->products}$$
There order of the reaction can be:

[source]
If $\ce{B}$ is not a reactant, then it has no place in a rate equation with $\ce{A}$ because the two aren't colliding in order to react.

What about a reaction that is autocatalytic? (I just saw there is a new answer discussing this, not sure how I missed that, but the question still stands).

The rate equation would change if B is a reactant. In the question OP asked the equation was not $\ce{A +B ->2B}$

I was thinking more of a reaction that can be catalyzed by acid that makes acid as a product. For instance, one of the reactions I teach is the reaction of aniline and acetic anhydride to make acetanilide and acetic acid. I realize this reaction is quite fast without an acid catalyst and a wake acid like AcOH probably isn't a very strong catalyst, but it technically should catalyze the reaction to some extent.

You have given a reaction for an irreversible reaction: $$\ce{A -> B}$$ so the products would not really come into consideration. For a reversible reaction: $$\ce{A <=> B}$$ the relative rates of reaction of the forward and backward reactions would yield the equilibrium constant, $\ce{k_{eq}}$, which is given by: $$\ce{k_{eq} = \dfrac{[B]}{[A]}}$$. The forward reaction would be $$\ce{ r_{A} = k_{A} [A]^{n_{A}}}$$ and the reverse reaction or B would be $$\ce{ r_{B} = k_{B} [B]^{n_{B}}}$$ The quirk here is that $\ce{n_{A}}$ and $\ce{n_{B}}$ do not need to be integers for the rate equation, but in the equilibrium equation the exponents for the concentrations of A and B would be the same as the integers in the stoichiometry of the chemical equation.

Equilibrium ($\ce{ <=> }$), not resonance ($\ce{<->}$).

ok, fixed formatting to use double arrow...


text:
Does a wavefunction have to be an eigenfunction of an operator in order to obtain an expectation value?
I am reviewing an intro quantum chemistry book and I came across the line "If $\hat{A}$ corresponds to a physical property of the system, whether or not $\psi$ is an eigenfunction of $\hat{A}$, the average value of that property is given by the expectation value of the operator." Doesn't $\psi$ have to be an eigenfunction of the operator in order to obtain an expectation value?

No. The wave function will in general be a linear combination of some or all of the eigenfunctions. It does not have to be a specific eigenfunction. If it is a linear combination of eigenfunctions, than the expectation value will be a linear combination of the expectation values of the eigenfunctions.

If the wavefunction is a combination of only some of the eigenfunctions, than the rest of the functions that make up the wavefunction are not eigenfunctions? What are these other functions?

No, you have it backwards. The eigenfunctions are the orthogonal basis for the solution space. Any wave function in that solution space will be some linear combination of the eigenfunctions. That is why they are called 'eigenfunctions'.

Okay so if I understand correctly the wavefunction is a linear combination of basis functions (which are eigenfunctions) and this linear combination can be in a sense "incomplete" only consisting of a fraction of the basis functions, thus leading the wavefunction to not be an eigenfunction of an operator?

Well, life gets really complicated, because eigenfunctions are not unique. Given one set of functions that are a basis for a space, you can generate an infinite number of different, orthogonal basis sets through appropriate combinations of the first set. But, as an example, in chemistry one will often work with the eigenfunctions for atomic electron orbitals. Taken all together, they are a complete basis set. You can express any electron state around the hydrogen atom with some linear combination of them. But, that does not imply that some state has to have a contribution from the 5s orbital.

So if the state of an electron in a hydrogen atom is expressed as a linear combination of basis functions with the exception of the 5s orbital (in your example) this translates to the wavefunction not being an eigenfunction of the operator, but if it is represented by all of the basis functions in the set than it is an eigenfunction of the operator, is that correct. So in other words even for the Schrodinger equation the wavefunction does not have to be a "true" eigenfunction of the Hamiltonian

No, assuming that wavefunction is normalized, the expectation value $o$ of the operaror $\mathcal O $ is obtained as $\langle \Psi | \mathcal O | \Psi \rangle $.
EDIT
In response at your comment: I am not asking how to calculate the expectation value, I am asking why the wavefunction does not have to be an eigenfunction of the operator in order to calculate an expectation value
The fact that the expectation value is obtained with $\langle \Psi | \mathcal O | \Psi \rangle $ implies that there is no need that the wavefunction be an eigenfunction of the hermitian operator $\mathcal O$.
If it is an eigenfunction, then it is possible to get a simple expresion with:
$$ \langle \Psi | \mathcal O | \Psi \rangle = \langle \Psi | \mathcal O  \Psi \rangle =  \langle \Psi | \mathcal o  \Psi \rangle  = o  \langle \Psi | \mathcal   \Psi \rangle = o$$
If not the case, then you can not make the substiution  $\mathcal O  \Psi = o \Psi$. So, the application of the map $\mathcal O$ to $\Psi$ will be some function $\Omega$. Then, the expectation value will be  $\langle \Psi | \mathcal   \Omega \rangle$ that is an scalar that you can compute as
$$ \int d\mu \Psi^*\Omega$$

I am not asking how to calculate the expectation value, I am asking why the wavefunction does not have to be an eigenfunction of the operator in order to calculate an expectation value

@Stuff Your welcome


text:
How to calculate molar mass using the ideal gas law?

It is found that 250 ml of a diatomic gas at standard temperature and pressure (STP) has a mass of 1.78 g. What is the diatomic gas?

How can this question be solved? I know that I am supposed to use the ideal gas law
$$pV = nRT,$$
but am not sure how to apply it to this question.

You can first use the ideal gas law to calculate $n$:
\begin{align}
pV &= nRT\\
(\pu{101325 Pa})(\pu{0.250 l}) &= n(\pu{8.314 J K-1 mol-1})(\pu{273 K})\\
\end{align}
Solving for "$n$" gives $n = \pu{0.0111577 mol}$.
Now we use the connection between mass $m$, amount of substance $n$, and molar mass mass $M$:
\begin{align}
  M &=\frac{m}{n}\\
  n &=\frac{m}{M}\\
  \pu{0.0111577 mol} &= \frac{\pu{1.78 g}}{M},
\end{align}
which gives a molar mass of $M = \pu{159.53 g mol-1}$.
Looking at the Periodic Table, we can identify the diatomic gas as being bromine gas ($\ce{Br2}$), which has a molar mass of $\pu{159.8 g mol-1}$.

The question reads "at standard temperature and pressure (STP)"; therefore the pressure should be $p=100,000\ Pa$ and not $p=101,325\ Pa$; however, that's a common mistake in homework questions. Another mistake in this question is that bromine is a liquid at STP.

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


text:
Titration curve graph, finding exact point of the equivalence point

I plotted all my data points and made this graph in excel. Now I need to find the volume of titrant required to read the equivalence point, as well as the pH after 18.5 mL of titrant have been added. If there any... best/correct way to go about this or anything in excel I should use to point the exact points on the line I have? Or am I just taking my best guess looking at it here?

A common technique is to take the derivative where the place with the steepest slope is the equivalence point.

I'm not really sure what you mean by that...

Assuming the titration involves a strong acid and a strong base, the equivalence point is where the pH equals 7.  From inspection alone and the use of a ruler, you can approximate that to be at 25.88mL of NaOH.
To show the equivalence point on a the curve, just draw a line from where the pH is equal to 7 and line it up with the titration curve.  I show this in the attached image.
To calculate the pH at 18.5mL, you need to know the molarity of the acid, for which you first need the identity of the acid (for stoichiometric purposes) and the acid's concentration which you can from your titration data.  Then you calculate the number of excess moles of acid present after reacting with the 18.5mL of NaOH titrant.  The pH is the negative logarithm of the concentration of the protons.  Remember that the moles of protons is not necessarily equal to the moles of acid since acids can release more than one proton.


text:
dissociation of aluminum sulfate in water
I had a question as to what happens when aluminum sulfate is placed into water. After conducting research online, I got varied answers: some say that it dissociates into Al$^{3+}$ and SO$_4$$^{2-}$ ions, while others claim that aluminum sulfate will react with water to form aluminum hydroxide and sulfuric acid.
I even conducted a mini-lab by placing aluminum sulfate in water and sometimes there was precipitate (assumed to be aluminum hydroxide) and sometimes there was none.
What is really happening?

Aluminum sulfate has special properites and uses. It is true that aluminum sulfate is soluble in water, and it has specific solubility values for certain temperatures:

$31.2\ \mathrm g$ per $100\ \mathrm{mL}$ at $0\ \mathrm{^\circ C}$
$36.4\ \mathrm g$ per $100\ \mathrm{mL}$ at $20\ \mathrm{^\circ C}$
$89.0\ \mathrm g$ per $100\ \mathrm{mL}$ at $100\ \mathrm{^\circ C}$

You stated you used $12\ \mathrm g$ in $100\ \mathrm{mL}$, so it would seem to completely dissolve. However, it this situation, since it isn't being dissolved in very large amounts of water, it doesn't dissociate and form aluminum hydroxide. Aluminum sulfate can be used in water purification in small amounts, because it causes impurities to group up into larger particles and then settle to the bottom of the container (or be filtered out) more easily. The solid you see sometimes are the impurities of the water that have been grouped up and look like precipitate. If you use more pure water, you will see less of these impure solids, or possibly nothing if the water is pure enough (such as if it's distilled water).

i added 12g of aluminum sulfate to 100 g of water when there was a precipitate seen and the temperature was room temperature. SO the what is happening? Is the aluminum sulfate going to ions or forming aluminum hydroxide and sulfuric acid

If you adding that little amount, then it has to do with the properties of aluminum sulfate. Check my edited answer.

where is your edited answer?

ok so the aluminum sulfate is then just being dissociated into individual ions?

Yes, because the ions cause the impurities to group together in smaller amounts of water.

but aluminum ion is very attracted to the hydroxide group of the water so will it react with the hydroxide group?

Aluminum has a relatively low electronegativity, meaning it would be very hard to rip away 3 hydroxide ions from water.

ok so in my chemistry textbook, it is mentioned that aluminum salts dissolve in water, giving hydrated aluminum ion. Is this fact wrong?

The precipita is not impurities. The amount of water use is falling below solubility limits.

I am assuming that you are measuring 12 mL in a flask. This is very unusual because Sulfate is in a solid form therefore you are best to weight the sample in a balance on order to determine the solubility component to hold in the water volume.
The density of Alum assumed to be the unhydrate is 2.572 grams per mL therefore you have 32.064 g of the anhydrous form. The solubility i water if 31 to 36 g per 100 mL. You are adding in 100 Ml of water in laboratory analytical conditions. Therefore, you are closing too close to the solubility lists at room temperature.
Another effect to consider is the dramatic increase in specific gravity. The water volume in the solution will shrink to compensate the final specific gravity of 1.33. This will saturate the solution and will precipitate to accommodate specific gravity and normal solution solubility at the temperature.
This chart will help you to evaluate how the ratios of sulfate weight and water volume impacts the specific density the Aluminum Sulfate solution: Ind. Eng. Chem., 1945, 37 (10), 1016–1018

Aluminium sulfate does dissociate in water, forming ions. That's true. However, most cations with double and triple positive charge bind water very strongly. Strong positive charge make hydrogen atoms of the associated water molecule acidic, so it readily dissociates into hydrogen cation and hydroxide group, associated with aluminium ion. The resulting structure can associate with another aluminium ion, eventually forming large network of aluminium hydroxide.
However, the topic largely avoided in school chemistry for convenience, because the chemistry is large, counterintuitive and is not ruled by simple rules.
In general, it can be safely assumed that at least some hydrolysis occurs for aluminum sulfate unless a strongly acidic solution is involved. The hydrolysis may be complete for at least part of aluminum sulfate until enough free acid produce to stop further hydrolysis.


text:
Relative basicity of fluorine and oxygen in silyl ether deprotection

Why does the $\ce{O}$ take the $\ce{H+}$ instead of the $\ce{F}$?
What I took into consideration:

$\ce{O}$ pulls more $e^-$ from atoms near it: $\ce{O}$ is more basic
$\ce{F}$ has one more $e^-$ pair than $\ce{O}$: $\ce{F}$ is more basic
$\ce{F}$ is more electronegative than $\ce{O}$   : $\ce{O}$ is more basic(?)
$\ce{O}$ has more substituent group making it sterically hindered: doesn't matter(?)

Is it okay to think this way? What's the most important factor?

Because oxygen has a lower electronegativity than fluorine, oxygen is less stable with a negative charge (and more stable with a positive charge). This is why $\ce{OH^-}$ is more basic than $\ce{F^-}$ and why $\ce{HF}$ in water is an acid (hydrofluoric acid), while $\ce{H2O}$ in water is neutral.

The reaction mechanism is actually:

The $\ce{Si-F}$ bond is one of the most stable bonds in all of chemistry ($565\ \mathrm{kJ\ mol^{-1}}$). Therefore, $\ce{F-}$ easily displaces the oxygen, leaving it as an alkoxide ion.
Once in the tert-butyldimethylsilyl fluoride, the fluorine is a very poor base (due to its high electronegativity) and is very resistant to sharing any of its unbonded electrons with a hydrogen ion. The alkoxide, on the other hand, will readily accept the hydrogen ion to form the alcohol.

Thank you again^^ let me ask you a little bit more T_T. It's difficult for me to understand why higher bonding energy(kJ/mol) is associated with stability. because I always thought these high bonding E is usually from partial ionic bond b.w two atoms, thus making this bond susceptible to reaction. and as far as I know, OH bond is strong(467 kJ/mol) but seems to be broken pretty well. Si-F is higher but 'about 100kJ/mol' makes so much difference?

Are you sure that first reaction is an SN2 reaction? Silicon can (and often does) readily take a fifth bond.

Yes, there would be an intermediate that places 5 bonds on the silicon, but from what I've seen it does break off into the silyl fluoride and an alkoxide.

@NKYu When bonds form, the more energy that is released, the higher the bond energy of the bond is, and by default, the more stable it is. Being more stable by $100\ \mathrm{kJ\ mol^{-1}}$ is a lot. This would be an extremely exothermic reaction.

isn't it right that the reason Si-F bond is so strong is because of difference of electronegativity b.w SI and F? if so, wouldn't the electronegativity that made this bond strong also makes the bond vulnerable to reaction at the same time because F can hog e- pair shared b.w Si and F?


text:
Determine reaction rate constant from the reaction time?
I was doing some research and I came across an experiment similar to mine (potassium iodate with sodium bisulfite) from the University of Massachusetts (http://www.chem.umass.edu/genchem/chem122/122_Experiment_4.htm), but with the reaction of potassium permanganate with oxalic acid.
The rate equation is given as:
Rate = k[KMnO4][H2C2O4].
It was stated that the value of k could be determined by dividing the concentration of the potassium permanganate [KMnO4] by the reaction time to find the rate and then dividing the rate by [KMnO4] multiplied by the concentration of the oxalic acid [H2C2O4] to find the reaction rate constant.
They provide the following equations:
Rate = [KMnO4]/t
k = Rate/([KMnO4][H2C2O4])
Would this be applicable to my reaction (potassium iodate with sodium bisulfite), because this would allow me to find the activation energy without performing trials to find k experimentally?

Assuming the reaction is a one step reaction, and does not have any elementary reactions, this could potentially work. However, if there are elementary reactions, this would change the rate law, and it would not be Rate = k[KIO3][NaHSO3].
You would also need to know the reaction order for both the potassium iodate and the sodium bisulfate, as well as the coefficients for both in the original reactions, yes you would be able to do this.
Knowing the coefficients, you could figure out the rate using rate -dA/adT. You could plug this rate into the rate law, Rate = k[KIO3]^m[NaHSO3]^n. However, you would need to know the reaction orders for KIO3 and NaHSO3, to plug in for m and n, and these can only be determined experimentally. If you know these, however, this equation would work for you.

Thank you very much. I have finished collecting data for the reaction and will analyze it this weekend to try and determine the reaction orders for KIO3 and NAHSO3.


text:
Relationship between enthalpy change and endo/exothermic reactions

Something that confused me, is there a relationship between enthalpy change and endo/exothermic? Does region "x" represent the enthalpy change?
What represents enthalpy change, kinectic energy, activation energy and potential energy on the graph below?
But why must, for an endothermic reaction, the value of activation energy be higher than the enthalpy change? Then, how about the backwards in equillibrium, if the forward reaction is exothermic, is it the same?
And why is, for an exothermic reaction, the value of the activation energy equal to the enthalpy change? Is that a must? Can the activation be energy greater than the enthalpy change?

I will attempt to answer all your questions however I don't completely understand some of them. Could you perhaps reword them or explain them a bit more?
Question 1

is there a relationship between enthalpy change and endo/exothermic

Of course there is! What determines if a reaction is endothermic or exothermic is by looking at the sign of the $\Delta \mathrm{H}$ (enthalpy change). If $\Delta \mathrm{H}$ is negative, then the reaction is exothermic and when $\Delta \mathrm{H}$ is positive, the reaction is endothermic.

does x represent the enthalpy change

That is also correct as enthalpy change is a state function. In other words, it doesn't mater what happens in the middle of the graph, the enthalpy change is just the final value minus initial value.
Question 2

What represents enthalpy change, kinetic energy, activation energy and potential energy on the graph below?

Enthalpy change- as already discussed, x represents the enthalpy change
Activation Energy- can be seen as the difference between the maximum potential energy and initial potential energy. Hence it is w.
Potential energy- that is the y value at every point on the graph
I am not sure how you can find the kinetic energy. I suppose you can find the change in kinetic energy if assume that the potential energy completely converts to kinetic energy.
Question 3

why must for an endothermic reaction the value of activation energy be higher than the enthalpy change?

That is true for any reaction. That is because as I stated above, the activation energy is the difference between the maximum potential energy and initial potential energy. Hence there is no way the enthalpy change, the energy difference between the final and initial state, can be larger than the activation energy.

Then, how about the backwards in equilibrium, if the forward reaction is exothermic, is it the same?

I don't really understand what you mean here. Could you perhaps explain this more.
Question 4

And why is, for an exothermic reaction, the value of the activation energy equal to the enthalpy change? Is that a must? Can the activation be energy greater than the enthalpy change?

Like I have said above several times, the value of the activation energy  is always greater than the enthalpy change.

Thanks a lot  , for question 3 , i mean that if there is an equillibrium , such that A + B ⇌ C -49kJ mol-1, (backward reaction is endothermic ), is the value of activation energy be higher than the enthalpy change?But after looking your detail explaination , i get it , thx :) sorry for my poor english

I said the same thing, and I was the first one to post your answer

@YinTingNg No worries. I am glad I was able to help.

Yes the region x represents Enthalpy change.The relation between Enthalpy change and endothermic reaction  is that if a reaction is endothermic then Enthalpy change is +ve and if the reaction is exothermic then the Enthalpy change is -ve.
In an endothermic reaction the amount of energy you supply is lesser that the amount of energy released when product is formed so the amount of energy you supply to transform the reactants to activated complex will be more than the energy released when the activated complex transforms to product.
In the graph you have provided the reaction in forward direction is exothermic and in backward direction is endothermic. Whichever direction you take  activation energy will always be more than the Enthalpy change because Enthalpy change is the difference in the Enthalpy of the products and the reactants whereas the activation energy is the difference in the Enthalpy of the activated complex and the reactants. The Enthalpy of the activated complex is always more than the Enthalpy of the products and the reactants so both these quantities cannot be equal.
For both exothermic as well as endothermic reactions it not a necessary condition for the activation energy to be equal to the Enthalpy change as activation energy is always greater than the Enthalpy change.


text:
Buffer formation in weak acid/strong base titration?
I'm a little confused about what happens when you titrate a weak acid or base. For example, let's say that you are trying to titrate a weak acid. Based on this question, it seems like the weak acid isn't initially a buffer. But, as you begin to add strong base, conjugate base is created from the reaction between the weak acid and strong base. Does this mean that a buffer is created from the titration?

I think that the above diagram will explain a lot of your questions.

it seems like the weak acid isn't initially a buffer

That is correct as to form a buffer you need a solution containing a weak acid and its conjugate weak base. So in the example above, initially the solution only contains acetic acid ($\ce{CH3COOH}$, but its conjugate base, acetate $\ce{CH3COO^{-}}$, is not present.

But, as you begin to add strong base, conjugate base is created from the reaction between the weak acid and strong base

That is absolutely correct. For example, in the above diagram the reaction that is occurring when you titrate the acetic acid with sodium hydroxide is:
$$\ce{CH3COOH + NaOH -> CH3COO- + Na+ + H2O}$$
Now as you begin to titrate the acid, its conjugate base starts to form. This is supported by the fact that the pH doesn't really change when more $\ce{NaOH}$ is added to the solution.
To form an effective buffer solution you need roughly equal concentrations of the acid and conjugate base, hence that is why it says it is a buffer when half of the acid is titrated. However anywhere near that region can be considered as a buffer. Generally a solution is a buffer if the pH is within +/- 1 of its pKa. So for acetic acid, a buffer will have a pH between 3.76 and 5.76.

Does this mean that a buffer is created from the titration?

Yes, a buffer is created. However is doesn't stay as a buffer for long. As I mentioned above, when the ration of the concentration of acid and the conjugate base becomes too high or too low, it won't be a buffer. That is evident as when more $\ce{NaOH}$ is added to the buffer, the concentration of the conjugate base becomes much greater than the concentration of the acid, resulting in the pH to increase dramatically.


text:
Enthalpy change for exothermic and endothermic reactions
I understood the concept behind endothermic and exothermic but finding it hard to understand it numerically.Let me describe my confusion through example.
Suppose for a reaction,
reactants require energy=$\rm40~J$
which on forming product give energy=$\rm50~J$ (so yes it is exomthermic)
where change in enthaply=$\rm10~J$
But as per the equation:
"Enthalpy change=Enthalpy of products- Enthalpy of reactants."
since enthalpy at constant pressure is Total heat content.
So I should have
Change in enthalpy=[U(internal energy)-50]-[U+40]=-90J
Could You explain it where I am getting this wrong? Is my understanding of enthalpy correct?

According to the definition,
 Enthalpy = Enthalpy(products) - Enthalpy(reactants), this should be followed.
When you are saying that 40 J are used up it means that 40 J are used in transforming the reactants to an activated complex(reaction intermediate) and when you say that 50J are released then it means that 50J is liberated when this activated complex is transformed to the product.
Assume that the energy of the activated complex be 0, then the energy of the reactant must be -40J(because 40J are required to convert reactant to the activated complex) , and the energy of the products should be -50J( because energy is released in converting the complex to the product)
Enthalpy(products)=-50J
Enthalpy(reactants)=-40J
Therefore,
 Enthalpy = (-50)-(-40) J = -10J.


text:
How to apply Le Chatelier's principle?

The following reaction is at equilibrium $$\ce{4HCl(aq) + O2(g) <=>2H2O(l) + 2Cl2(g)}\quad \Delta H>0$$Which one of the following changes will result in an increase in the concentration of $\ce{Cl2(g)}$

adding some $\ce{H2O}$
decreasing the temperature
adding a catalyst
decreasing the pressure
removing some $\ce{HCl}$

Decreasing the pressure (4) is the answer but I don't understand why. I though decreasing pressure shifts the equilibrium to the side with the larger number of moles which is the left?

Welcom to Chemistry.SE! Your explanation would be correct if all the molecules would be gases. Are they? ;-)

Remember that $\ce{HCl}$ (aq) and $\ce{H2O}$(l) are liquids while the oxygen and chlorine are gases . As the number of moles of gases on the left  hand side (one mole of oxygen) is smaller than that of the right hand side (two moles of chlorine), decreasing the pressure makes the equilibrium position move to the right hand side, therefore the concentration of chlorine increases.

While checking for the effect of pressure on equilibrium, consider only the species which are in gaseous state. In above equilibrium only dioxygen and chlorine are in this state.
Ref:http://www.adichemistry.com/physical/equilibrium/le-chatelier/le-chatelier-principle.html


text:
Why is the boiling point of sulfuric acid much higher than that of phosphoric acid?
Why is boiling point of sulfuric acid much higher than that of phosphoric acid?
According to the data book,
• The boiling point of sulfuric acid is 337 °C (639 °F; 610
   K) When sulfuric acid is above 300 °C (572 °F), it will decompose
   slowly.
• The boiling point of phosphoric acid is 158 °C (316 °F; 431 K) When phosphoric acid is above 213 °C (415 °F; 486 K), it will decompose slowly.
However, sulfuric acid can only form 2 hydrogen bonds per molecule but phosphoric acid can form 3 hydrogen bonds per molecule. While both of them are strong acids which can undergoes complete ionization, what other factors that I've ignored may lead to the fact that boiling point of sulfuric acid is much higher than that of phosphoric acid?

Phosphoric acid is not a strong acid and it does not undergo complete deprotonization.
Both acids have a similar molar mass of around 98 grams per mole.
Another property that should be considered is the formation of a net dipole moment on the molecule.  With the phosphoric acid 3D structure attached, each of the three oxygen atoms in the $\ce{-OH}$ group pull with equal strength, cancelling each other out, leaving an net dipole moment going upwards coming from the double bonded oxygen.

With sulfuric acid however, both of the hydrogen bonding groups pull in relatively the same direction, while the double bonded oxygen atoms pull in the opposite direction, forming a stronger net dipole.

This is analogous to the net dipole properties of chlorinated methanes.  See the attached image and note the similarities in their structures and the structures of the acids we are discussing.

Another key reason for the difference in boiling point is the autoprotolysis of sulfuric acid.  Even without the presence of water, sulfuric acid can lose a proton as shown in this equation:
$$\ce{2H2SO4 <=> H3SO4+ + HSO4-}$$
Phosphoric acid is a weak acid and does not have such interactions and only loses protons (in equilibrium) in the presence of water.  The autoprotolysis is important because through autoprotolysis, sulfuric acid now forms ion-ion interactions, which are much stronger interactions than hydrogen bonding.
Therefore, as a combination of a slightly stronger dipole moment and more so due to autoprotolysis, sulfuric acid has stronger intermolecular forces, and thus a higher boiling point.

Pure H3PO4 also undergoes autoprotolysis and to even greater extent then H2SO4. Behavior of pure substances has little in common with their interactions with water, when they are highly diluted.

@Mithoron So, does this mean the pKa of $\ce{H3PO4}$ in $\ce{H3PO4}$ is higher than the pKa of $\ce{H2SO4}$ in $\ce{H2SO4}$? I've looked for a source for that but haven't been able to find any, do you have one, by chance? I trust that what you're saying is true, but if there is a consolidated source for this, I'd find it handy.

@DGS https://books.google.pl/books?id=corDCgAAQBAJ&pg=PA171&lpg=PA171&dq=h3po4+autoprotolysis&source=bl&ots=0hm6Qu3nxx&sig=bhR6K0Zg4jOSRetOlR47z3XnsFE&hl=pl&sa=X&ved=0ahUKEwio99DYt7fLAhVJCZoKHWxTCUQQ6AEIJDAB#v=onepage&q=h3po4%20autoprotolysis&f=false It's autoprotolysis is probably strongest of all known acids, but rather because of higher basicity not acidity (gets protonated easier).

what does the molar mass have to do with the BP?

@colnegn Molar mass is a rough indicator of how large molecules are and subsequently the strength of their London dispersion forces. However, this too is better understood through structure and electron clouds.

@Howsikan Mass turns out not to be the best indicator for dispersion forces because they are exclusively an electronic effect (think about isotopes--albeit molar mass takes care of that issue). Look up a table of atomic polarizabilities, and you will find that the trend is down-left on the PT, and is reflected in the atomic radius and ionization energy. Regarding molecules: effects from things like the molecule's surface area must be considered: two isomers of C16H32 boil at 513 and 554 K; 9 & 16 C main chains, respectively. You might imagine polarizability as how 'floppy' the species is

Agreed, but I initially viewed mass as a rough indicator of protons, subsequently electrons, provide a rough indicator of dispersion forces.

In addition to the factors Howsikan mentioned, my intuition leads me to believe there's a parallel with the boiling points of $\ce{NH3}$ and $\ce{H2O}$. I think the oxos ($\ce{O=P}$ and $\ce{O=S}$) are better acceptors than the hydroxyls, so the acids would prefer $\ce{OH...O=}$ over $\ce{OH...OH}$. $\ce{H3PO3}$ has three oxos to one hydroxyl whereas $\ce{H2SO4}$ has two of each, so $\ce{H2SO4}$  can create a favourable network more easily.


text:
Would adding acetic acid from household vinegar to washing machine make it too dilute to be effective?
Many people advise adding a cup of acetic acid (vinegar) to one's laundry during the clean cycle. From my calculations, considering one cup of acetic acid and a top gallon machine (at 40 gallons), it would seem this would completely dilute the concentration next to nothing.
Is this correct or am I missing something? Someone had mentioned something about acetic acid being a weak acid, but I'm not sure what that means in this context.

As a weak acid, acetic acid will act as a buffer in the washing machine solution.
Following similar calculations from the first answer of the question,What's the pH of Vinegar when it generally contains 5% Acetic acid?, we can calculate the concentration and pH of acetic acid in the washing machine solution. 1 Cup is ~ .237 L and 40 gal is ~ 152 L. Using the concentration of acetic acid in vinegar from the post above (0.847 mol/L):
(0.847 mol/L)(0.237 L)=(X mol/L)(152 L), X=0.00136 mol/L
pH of the washing machine solution = 0.5[4.76-log(0.00136)] = 3.80
So, although the acetic acid is very dilute, there is still a significant pH drop from the addition of 1 cup of white vinegar. Changes in pH can strongly affect the solubility of many compounds, which would seem to be helpful in the cleaning process. Furthermore, the denaturing of proteins is promoted at lower pH, so lowering the pH could be particularly useful in the removal of protein-based stains (grass, bodily fluid, egg, etc.).


text:
Arrangement of elements in the reactivity series
I have seen a variety of reactivity series of elements online and also have consulted some books regarding the arrangement of the elements. I understand that the arrangement of elements is based on degree of reactivity of elements with cold water, steam, oxygen, dilute acids etc.
Lithium is less reactive with respect to the criteria mentioned above yet it is placed on the top of more reactive elements in the reactivity series that I have observed.
What is the reason for the position of Lithium above let us say potassium and sodium which are more reactive.
Link to one of the references:
https://kaiserscience.wordpress.com/chemistry/metal-reactivity-series/

The electrode potentials in water, really aqueous acid, are free energy deltaG[0] changes of the reaction with H2 or H+ ions Everything in their standard states, to equilibrium. If it sounds complicated it is because it is complicated. It takes into account all the various heats and entropies. What is of limited importance is the apparent reactivity. Li reacts more sedately than Na or K or frightenly reactive Cs[The differences really are very small] but seems to go farther when done. The relative energies will change if products change by changes in solubility or by complexation

The reason that lithium is on the top of the activity series is due to its behavior in aqueous reactions with water and acid. Due to the lithium atom's very small size (only two energy levels), the lithium ion interacts well with water forming ion-dipole interactions.  The lithium atom's small size compared to sodium and potassium is more entropically favorable in the formation of ion-dipole interactions with water and consequently so are reactions involving lithium ion formation favored to to other alkali metals.
The reasoning behind this is that reactions favor a greater state of entropy (disorder) in the universe. With larger cations, more water molecules have to become ordered around the cation to form the appropriate ion-dipole interaction.  However, due to the lithium cation's small size, less water molecules are required to form the ion-dipole interaction, thus less order, being more favorable in terms of entropy.


text:
Identifying the dependence of various time-independent wavefunctions
I'm tasked with finding the variables that each of the following wavefunctions depends on: (1) a free particle in 2 space (2) a free particle in 3 space (3) a hydrogen atom (4) a lithium atom.
for situations (1) and (2) I get $\psi(x,y)$ and $\psi(x,y,z)$. However, regarding situations (3) and (4), I'm unsure how to proceed - advice?

Hydrogen atom has one nucleus (in 3D space) and one electron (in 3D space). Lithium atom one nucleus and 3 electrons. Help yourself!
Once done, try some coordinate transformations from laboratory frame to something more local.

@ssavec I would love to figure this out myself, but I don't know how to interpret one nucleus and one electron in terms of a wavefunction.

you do not need to write the wavefunction, just find the variables. So 3 coordinates for electron and 3 coordinates for nucleus.

@ssavec wait, but spin... ;)

@ssavec So I would have $\psi(r_e,\theta_e,\phi_e)$ for the electron and $\psi(r_n,\theta_n,\phi_n)$ for the neutron?

@Jay: almost, one wavefunction to rule them all.  $\psi(r_n,\theta_n,\phi_n,r_e,\theta_e,\phi_e )$

Okay I think I get it. So then lithium would have a wavefunction composed of  $r,\theta,\phi$ for 4 objects, specifically one nucleus and three electrons?

As @Wildcat mentioned there is an additional spin coordinate $\omega$ for the nucleus and each of the electrons, which complicates matters, but neglecting that you're right.


text:
Which substance(s) can solve gelatin (on film stock)?
I want to loosen the exposed layer on film stock which is made up of gelatin. Anything which would make gelatin dissolve that's not overly toxic? Thanks

Hot water, about 100 C, should do the trick. Also, this article has some information on the properties of gelatin layers in photographic films.
Properties and Stability of Gelatin Layers in Photographic Materials

Sorry about that  [Properties and Stability of Gelatin Layers in Photographic Materials] (http://albumen.conservation-us.org/library/c20/hendriks1.html)

thanks for that, interesting article. I tried boiling the film already but that hasn't caused any effect at all. Maybe I need more heat than boiling water in a cooking pot?

You might try heating the film in water for a longer period of time or , as you said, raising the temperature. This could be accomplished with a pressure cooker. Another idea is to try lowering the pH of the solution using a household acidic pH buffer, like vinegar (5% acetic acid).


text:
How much energy is stored when H+ is converted to H2?
I'm doing a write-up for this for my students:

Source: Amirav Research Group (Facebook)
The energy of formation of isopropyl alcohol is $-318.2~\mathrm{kJ/mol}$ (kilojoules per mole), that of acetone is $-249.4~\mathrm{kJ/mol}$, so $68.8~\mathrm{kJ/mol}$ are gained on this end.
But I'm getting confused about the hydrogen end. I'm seeing that The energy of formation of $\ce{H+}$ is $0~\mathrm{J/mol}$ (it can be treated as a free proton), and that of $\ce{H2}$ is also $0~\mathrm{J/mol}$. Of course you would say that oxygen has $0~\mathrm{J/mol}$ as well, so since  water has enthalpy of formation of $241.83~\mathrm{kJ/mol}$, I'd expect it to yield that much energy in the combustion.
But hydrogen gas has an “energy density” as a fuel of $71~\mathrm{kJ/mol}$. So the treatment that $\ce{H2}$ and $\ce{O2}$ have $0~\mathrm{J/mol}$ enthalpy of formation and water has enthalpy of formation of $242~\mathrm{kJ/mol}$ seems to not square with that. Where's the rest of the energy?
Particularly, in the diagram above you see the charge separation as the electron pair moves to the hydronium, producing hydrogen. But a charge separation (J = coulomb volt) is energy, so moving the electrons to the hydrogen has to be adding energy to them, which seems to indicate that the $\ce{H+} = 0~\mathrm{J}$ and $\ce{H2}$ also = $0~\mathrm{J}$ things is wrong. But I don't know where to go from here, does anyone feel confident helping with this?

@Nilay Please do not wrap every number you see in \ce{...}. Only use this environment for chemical formulae. I don't want to check every edit you make.

I'm seeing that The energy of formation of H+H+ is 0 J/mol0 J/mol (it can be treated as a free proton)

No. Nope. Noway.
Formation of proton from hydrogen molecule is extremely endotermic process. It requires to 1) dissociate H-H bond, which is, btw, 432 kJ/mol, and then ionization of hydrogen atom, which is around 1300 kJ/mol.
However. In solution hydrogen ion is bound to water molecules (or other solvent), which results in serious energy discount, around 1100 kJ/mol for the case of water.
However again :) this is of no significance here, since the dive into small details is not required here.
The gross process is $\ce{ Me2CHOH \space =\space  Me2CO + H2 }$, and since entalphy of formation of hydrogen is roughly zero, only change from isopropanol to acetone is relevant. Hence, the difference is 68.8 kJ/mol.

if the 2H+ + 2e- -> H2 transition is worth the association of a 432kJ bond, isn't the [H+] in solution 1. a limiting factor of this reaction and also 2. actually the energy source?
So this reaction works only in a maintained acidic environment is what I'm saying.

I guess you switched a minus sign there; it's endothermic as the product acetone is less negative in energy than the reactant i-propanol. So the energy is going into the acetone rather than the proton-hydrogen conversion.
I'm quite surprised that the energy density is not equal to the combustion energy, but these values are not really important here since there is no combustion in the first place.

yes but the combustion, and the [H+] consumption, is relevant because this is intended as the photocatalyst which supplies acetone for a fuel cell
So combustion is the other half of this apparatus

2H+ cannot be converted to $\ce{H2}$, because H+ has no electrons.

It depend on how deeply you want to divide the chemical reaction cycle.
The overall reaction is:
Isopropyl alcohol -> acetone + hydrogen gas
Typically the internal energy of hydrocarbon compounds is measured by burning them to carbon dioxide and water in an excess of oxygen to find the heat of combustion. Since this reaction produces hydrogen too, the energy of the reaction for one mole would be the difference between burning isopropyl alcohol minus that of burning acetone and hydrogen.
User permeakra gave the chemical reaction cycle broken a different way in terms of the heat of formation, but the net result is the same. That is the beauty of chemistry.
Now you can divide the chemical reaction cycle into all sort of intermediate species, but the net reaction has to be what was outlined above. In fact the problem is usually worked backwards. By knowing what the energy of intermediate species like a solvated $\ce{H^+}$ cation is and knowing what the energy of a solvated $\ce{OH^-}$ anion is we can work out what the energy of some unstable intermediate organic "molecule" must be.

DavePhD is absolutely right in his answer. You can't just convert two $\ce{H^+}$ cations into $\ce{H2}$, because a $\ce{H^+}$ cation has no electrons. You have to get two electrons from some other chemical species and that will cost some chemical energy to free the electrons, not to mention the energy contained in the electron itself (which isn't chemistry but nuclear physics).

There is a change in molar entropy as well as expansion work being done by the system in the transition from $\ce{H+}$(aq) to $\ce{H2}$(g). Assuming constant temperature and pressure, there will be a change in the internal energy of the system as well.


text:
How to store bulk gas products and reagents?
Suppose I'm doing a batch process where an intermediate (non-hazardous) reagent is a gas, and I need to produce and temporarily store something on the order of a cubic meter of it to later feed back into a reaction.  Assuming the gas can't be condensed at anywhere near STP, isn't there some sort of standard flexible lab gas storage apparatus?
I've been looking for something like a large mylar (or other relatively impermeable) bag with a tube insert I can put on a barb and valve.  Either I don't know what this would be called, or it doesn't exist and I should be using something else.
I realize I could try to pump out a large rigid container, but once we're talking about a few cubic meters, and trying to avoid contamination, that seems like a pain compared to a flexible gas bag.

I would condense nevertheless, via solid CO2 or liquid nitrogen. Still much simpler to keep cryocooled flask, than handling cubic meter of whatever gas wrapped in plastic. Or compressed into steel cylinder.

I haven't sprung for the cost of LN2 yet.  I wonder which would be cheaper: LN2 supply and condenser apparatus, or a pump and fittings capable of compressing significantly into a compressed gas cylinder....

Glassware condenser is commonly used for liquefying ammonia and should work for you. LN2 is about 10cents/L, so dirt cheap. Compressing specialty gas is nasty, even if not corrosive.

@ssavec - It's paying for a LN2 cylinder that's expensive.  My amateur lab isn't near any big institutions where I can just stop by with a small dewar I can stuff in a backpack, so I'd have to have it delivered.  Between cylinder rentals and resupplies that's a 4-figure-a-year proposition!

I don't know where you are based, but the LN2 suppliers are able to rent a Dewar flask for very acceptable price and at least in Europe, the liquid gas production sites are no more than 50km apart. Hospitals also use liquid gases, etc...

@ssavec - I can drive to suppliers, but I hadn't thought to see if I can rent the transportable dewars, (which last I checked started at $600).    Good idea.

It probably depends on local regulations and the properties of the intermediates!
Gas bags might indeed be a viable solution.
In the biogas community, see SolarCities for an example, PVC bags from PUXIN, a chinese supplier, are quite popular.


text:
Theoretical temperature change in lime hydration
Calcium oxide (quicklime) and water have an exothermic reaction
$\ce{CaO + H2O \rightarrow Ca(OH)_2} \quad \quad \Delta H^{\circ}_f=-65.3 \;  \mathrm{kJ \cdot mol^{-1}}$
Assume that I have 1 kg of lime which means that I would theoretically need 322 grams of water for complete hydration. However, I want to use different amounts of excess water and calculate the final temperature of the whole system after the reaction has finished.
The heats of formation are given at the standard temperature in the tables. Therefore, if I assume that the calcium oxide and water start the reaction at 25 °C I know that 65.3 kJ of heat per mole of reacted $\ce{CaO}$ has to be transferred to the $\ce{CaO}$/$\ce{H2O}$ mixture. With 1 kg of lime (molar mass of $\ce{CaO}$ is 56.08 g) this would mean 1164 kJ of energy released.
So, if for example 1 kg of water was added, the final mixture would consist of 1.322 kg of $\ce{Ca(OH)2}$ and 0.678 kg of $\ce{H2O}$. Using the specific heat capacities the heat capacity of the total system could be calculated and the final temperature rise obtained.
Now, assume that I want to take into account the effect of adding reactants at different temperatures. Say that my slaking water is 16 °C and the lime 35 °C. Should I first calculate the initial temperature by assuming that the reactants do not react until the temperature difference between the lime and the water has settled?
I obtained the heat capacities from here: http://www.update.uu.se/~jolkkonen/pdf/CRC_TD.pdf
$\ce{H2O}$ heat capacity 75.3 J mol$^{-1}$ K$^{-1}$ $\rightarrow$ 4.18 kJ kg$^{-1}$ K$^{-1}$
$\ce{CaO}$ heat capacity 42.0 J mol$^{-1}$ K$^{-1}$ $\rightarrow$ 0,749 kJ kg$^{-1}$ K$^{-1}$
$\ce{Ca(OH)2}$ heat capacity 87.5 J mol$^{-1}$ K$^{-1}$ $\rightarrow$ 1.18 kJ kg$^{-1}$ K$^{-1}$
The amount of heat is calculated from
$Q=cm\Delta t$
If I use the arbitrary initial temperatures of 16 °C for water ($t_1$) and 35 °C for the lime ($t_2$), and initial weights of 1 kg the starting temperature of the system without reaction would then be obtained from
$c_1 m_1 (t-t_1)+c_2m_2(t-t_2)=0$
and solving for final temperature $t$
$t = \frac{c_1m_1t_1+c_2m_2t_2}{c_1m_1+c_2m_2}=18.9 \;°C$
I would then calculate the heat of the reaction, account for the reduction of water and the formation of $\ce{Ca(OH)2}$ and finally calculate the heat distribution and the final temperature.
Is this the proper approach to this problem? How large error is made if the starting temperature of the reaction is not the normal temperature of 25 °C? The temperature range I am interested in would be between 10 °C and 100 °C.
The recommended final temperature for wet slaking is about 71-93 °C. Here I assumed that all of the quicklime reacts and none of the heat is lost to the environment. The reaction rate is dependent on the temperature and in too cold conditions (too much excess water) the hydration can be incomplete as the lime is "drowned". My ultimate goal is to create a model of a slaker and use it to improve the temperature control of a milk of lime plant.

Are you familiar with Hess's Law?  It says that the change in extensive properties (e.g., enthalpy) from the initial state of the system to the final state is independent of path.  Reading up on Hess's Law will help you overcome your confusion.  Incidentally, 65.3 kJ is not the amount of heat per mole of reacted CaOCaO has to be transferred to the CaOCaO/H2OH2O mixture.  It represents the amount of heat that would have to be removed from the reacting system to hold the temperature at 25 C.

@ChesterMiller I am interested in the final temperature of the system. 1:1 ratio of CaO and H2O might be a bad example as the temperature would rise over 100 degrees as the water evaporates. Nevertheless, if the heat of the reaction is not removed, it has to go to the system itself therefore increasing its temperature. By calculating the heat capacity of the system the theoretical temperature increase could be obtained. To my understanding the reactants have to start from the same temperature so I wanted to ask if my reasoning for the initial temperature was correct.

It doesn't matter whether you take that first mixing step or not.  The answer will be the same either way.  For the adiabatic situation you are interested in, all you need to do is multiply the enthalpy at the final temperature of each of the final reactor constituents by its mass, sum these, and subtract the sum of the enthalpies of each of the initial reactor constituents multiplied by their masses.  This should be zero.  This will give you an equation for the temperature.

Alternatively, you can calculate the amount of heat required to heat the initial reactor contents to 25 C, add the heat of reaction times the number of moles converted at 25C, and then add the heat to raise the temperature of the final reactor mixture to the final temperature.  This should all sum to zero.

Ok, now I think i got it. I didn't understand what you meant with multiplying the enthalpies with the masses but I understood the second explanation. So instead of calculating the initial temperature balance I can just calculate the amount of heat needed to be added or removed from the system to bring each reactant to 25C. Then calculate the amount of heat generated, account for the initial addition or removal of heat, and finally use the remaining heat to calculate the final temperature change from 25C. Thanks for your time!


text:
IR spectra and melting point range comparison
I did a Diels-Alder experiment in lab, and the desired product was 9,10-dihydroanthracene-9,10-α,β-succinic anhydride. According to my IR spectra, I believe I isolated the product. However, my observed melting point range is 269-272°C, while the literature melting point is 262-264°C.
How significant is my melting point discrepancy?
And am I overlooking peaks in my IR that would be indicative of an impurity in my sample?

Welcome to Chemistry.SE! For an impure sample, I'd rather expect a melting point depression. Did you heat up too fast and miss the start of the melting process? I'd check and calibrate the melting point apparatus with some reference samples.

Personally I don't think the IR spectrum is of much use in finding impurities.

Thank you @KlausWarzecha! Unfortunately, as a student I do not have access to the laboratory to check the apparatus. In general, how can I tell how significant my discrepancy in ranges is? In my case it is only a few degrees, but a few degrees seems significant to me.

Impurities only lower melting points, they never raise melting points. I agree that the most common causes of an elevated melting point are 1. increasing temperature too quickly then 2. an improperly calibrated thermometer.

You are right, a few degrees are significant. Too bad that you don't have further access to the melting point apparatus. Even measuring the melting points of your starting materials, anthracene and maleic anhydride, would give some hints on  how much the apparatus is off  - or whether you botched the measurement ;-)

Indeed! :)  Thank you so much @KlausWarzecha!

Calibrating the thermometer is certainly the right idea but reagent grade anthracene is only 97% pure. I'd want something purer to calibrate the thermometer. It would nice to bracket the desired MP of 260 C also. http://www.sigmaaldrich.com/catalog/product/aldrich/a89200?lang=en&region=US


text:
Perovskite reaction equation (KMnF3)
For the synthesis of $\ce{KMnF3}$ I said the reaction equation was
$\ce{3KF + MnCl2 -> KMnF3 + 2KCl}$
Is it this simple or are am I missing something with these particular types of reactions?

http://scripts.iucr.org/cgi-bin/paper?S0365110X61001868 Also what types?

First row transition metals with KF

Maybe I misinterpreted the reaction, so would it be the HF formed in solution with the KF and nitric acid that would be reacting instead? The ratio of reagents in my lab manual would suggest 3:1 like the equation above.

Hmm, I guess it's OK.

Quoting from Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed:-

$\ce{KMnF3}$ is prepared by the addition of $\ce{KF}$ to $\ce{Mn^2+}$ solution. [...] Fluorination of $\ce{MnF2}$ in presence of an alkali metal fluoride leads to the complex Mn(IV) fluorides of composition $\ce{M2MnF6 (M = K, Rb, Cs) }$.

So, the reaction is all right.


text:
Why are vegetable oils better than water in cooking?
In core science we have been learning about vegetable oils but we haven't covered why they are better than water.
Are vegetable oils better than water?
Why?
What can we use vegetable oils for other than in cooking?
Hope someone could answer this.
Thanks.

You might get better traction on Seasoned Advice than here, at least for the cooking-related part of your query. Your question is a little broad and opinion-based as it reads, so you might consider some editing to narrow the scope, too.

Define "better"

I'm voting to close this question as off-topic because this question belongs on Seasoned Advice

@Geoff meh, I think "better" is subjective in Seasoned Advice as well. :)

Our food is the source of many compounds that are needed to keep our body working. Some of these are directly available from the food (vitamins), others (provitamins) have to be converted first in our body to their final structure before we can actually use them.
Chemically, all these compounds belong to very different classes, but we can roughly divide them into two groups by one property, their solubility in water.
The vitamins that are insoluble in water are soluble in fat.
Did you ever step into a piece of tar on the beach and the tar wouldn't get off your foot no matter how often you jumped into the sea? But with a bit of tanning oil, which is insoluble in water, it was usually no problem to remove the tar from your foot.
The same principle applies for the vitamins.
Digestion of (vegetable) oils together with food, such as carrots, helps your intestines to take up the fat-soluble provitamins and vitamins more easily.

Found the answer. Vegetable oils have a higher boiling point than water so that means food can be cooked at higher temperatures and faster speeds.

Cooking time is only one of the many ways in which "better" can be measured.


text:
Hydrogen bonding in staggered vs eclipsed conformations
In my first semester of organic chemistry, we learned that staggered conformations have lower energy/higher stability.
However, my thought is that if you have interacting OH groups, the hydrogen bonding would make the eclipsed conformations more stable than the staggered conformation. Could this be the case? Why or why not?
For example, a small molecule containing two carbons, each with an alcohol group (or more attached).
--
Related question:
In a 1,3-diaxial interaction, would multiple OH groups (or anything that exhibits hydrogen bonding) lend to stability?

Eclipsing is still going to be very high energy. The transient 5-membered ring that would form from hydrogen bonding would probably be lower energy if it wasn't flat, so again, I would expect gauche would be preferred.

As DGS has noted, the gauche conformation of ethylene glycol is stabilized by the formation of an intramolecular hydrogen bond and remains the preferred conformation for this molecule  2) In cis-1,3-dihydroxycyclohexane, the diaxial conformation is stabilized because of the hydrogen bond formation; due to this stabilization the diaxial conformation is preferred over the diequatorial conformation.

"Last semester in organic chemistry 1, we learned that gauche conformations have lower energy/higher stability."  Note that in most cases the anti conformation is more stable than the gauche. In a few molecules such as ethylene glycol and difluoroethane the gauche is more stable then the anti.

@ron why difluoromethane? Won't there be repulsion due to the fluorine lone pairs and all?

@EashaanGodbole  They are outweighed by stabilizing electronic interactions. See here, for example.

Disclaimer: I am not a computational chemist, I am completely new to this stuff. Just a humble student of @pentavalentcarbon

However, my thought is that if you have interacting OH groups, the hydrogen bonding would make the eclipsed conformations more stable. Is this be the case? Why or why not? For example, a small molecule containing two carbons, each with an alcohol group (or more attached).

I looked at two model compounds, ethane, and ethane-1,2-diol. Also, I did preliminary molecular mechanics based geometry optimisations using the MMFF94s force field in Avogadro.
I performed a CCSD(T) calculation in psi4 using the aug-cc-pVDZ basis. For ethane, obviously two conformers were identified:
$$\begin{array}{cc}
\hline
\text{conformer} & \text{Energy (in hartrees)} \\ \hline
\text{eclipsed} & -79.5971\\
\text{staggered} & -79.6027\\
 \hline
\end{array}$$
The difference in single point energies between the staggered and eclipsed forms $\Delta E_{\text{staggered-eclipsed}} = -0.0056$ which means the staggered form is lower in energy by about $14.7\ \mathrm{kJ\ mol^{-1}}$ (or $3.5\ \mathrm{kcal\ mol^{-1}}) $
Now, for ethane-1,2-diol. A first preliminary calculation, and random rotor search in Avogadro gave me three distinct conformers (although, two were very close in energy), however, every trial after that just produced two. They are given below:
$$\begin{array}{cc}
\hline
\text{conformer} & \text{Energy}\ (\mathrm{kJ\ mol^{-1} }) & \ce{OCCO}\ \text{bond angle} \\ \hline
\text{conformer 1} & 72.0914 & 59.0266\\
\text{conformer 2} & 85.534 & -179.967\\
 \hline
\end{array}$$
I have attached some screen shots below:

Now, hydrogen bonding is not explicitly treated in this calculation, but based on my discussion with @pentavalentcarbon the Leonard-Jones term in the forcefield (and to a lesser extent the electrostatic term) does capture these interactions.
So clearly, some stabilisng influence is evident in the structure, because if I were to argue on sterics alone, conformer 2 should have had lower energy than conformer 1. As an additional test, I measured the distance between $\ce{H}5$ and $\ce{O}2$ and found it to be 2.62 Angstroms. This is indeed suggestive
Now, using the geometries of conformer 2 and conformer 1, and that of an additional conformer with completely elcipsed oxygens that I obtained from Avogadro, I performed another CCSD(T) calculation is psi4 (again in the aug-cc-pVDZ basis). The table given below summarises my results:
$$\begin{array}{cc}
\hline
\text{conformer} & \text{Energy (in hartrees)} \\ \hline
\text{conformer 1} & -229.744410904091524
\\
\text{conformer 2} & -229.740781143788695\\
\text{completely eclipsed} & -229.725483329882337
 \\
 \hline
\end{array}$$
Again, the difference in single point energies between conformer 1 and conformer 2, $\Delta E_{\text{1-2}} = -0.0036297603$, means that conformer 1 is more stable by about $9.5399\ \mathrm{kJ\ mol^{-1}}$ (or about $2.277\ \mathrm{kcal\ mol^{-1}})$
This is probably due to an intramolecular hydrogen bond, as suggested, which has an overall stabilising effect on the gauche form of ethylene glycol (conformer 1). This is thus one of the instances in which the anti conformation (conformer 2) is higher in energy relative to the gauche conformation.
Finally, the difference in single point energies between conformer 1 and the eclipsed conformer, $\Delta E_{\text{1-eclipsed}} = -0.0189275742$, this implies that conformer one is more stable by about $49.694\ \mathrm{kJ\ mol^{-1}}$ (or $11.877\ \mathrm{kcal\ mol^{-1}}$)
Thus, I would say the energy penalty of eclipsing is still much higher than any stabilisation that can be gained from hydrogen bonding in this conformation. Shifting the angle between the two $\ce{OH}$ groups by about 60 degrees relieves this eclipsing strain while still allowing for the formation of a hydrogen bond, and is by far the best compromise.
To be honest, I wanted to scan over several dihedrals and make a potential energy plot, and calculate an exact barrier to rotation but faced two hurdles: a) couldn't write down the z-matrix properly, and couldn't write the appropriate loop in the program. b) the calculations that I did perform and have quoted here, took a while to run and based on that, I am guessing what I had in mind would have been a little too ambitious for my poor,  little laptop.
Addendum @Jan pointed out in the comments that the gauche effect is responsible for the stability of the gauche conformer, which to be fair, I did not think about at all because I was so focused on the "hydrogen bond", however, it is worth looking into. Here's a post that discusses it (answered by @Jan himself ;) )

One thing to note: the stability of the gauche conformer is probably due to the (orbital-explained) gauche effect.

Since I've posted this question several months ago, I've come across some evidence in the form of published papers that explore cases where eclipsed conformations are favored over their staggered counterparts. One such paper is, "An eclipsed Csp3-CH3 bond? An ab initio investigation of an atypical rotation barrier."
This particular paper explores an earlier report of crystalline trihydrate tricyclic orthoamide in which an eclipsed Csp3-CH3 bond was observed using low-temperature X-ray analysis.
Calculations showed a 5.5 kcal-mol-1 energy difference between the eclipsed form and its staggered counterpart which is pretty cool.
This doesn't answer my question to the degree that multiple hydrogen bonds lend to a preferred eclipsed conformation (at least for my example), but it does show the potential for an eclipsed conformation with lower energy than its staggered counterpart.

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


text:
How to determine the hydroxide ion concentration in an acidic aqueous solution?
When doing a titration with $\ce{HCl}$ and $\ce{NaOH}$, both being strong and dissociating completely, if you add a volume of $\ce{NaOH}$ that is less than needed to reach the equivalence point, you say all the $\ce{OH-}$ reacted with an equal amount of $\ce{H+}$ from $\ce{HCl}$.
It would be amount of substance $\ce{HCl}$ - amount of substance of $\ce{OH-}$ = $[\ce{H3O+}]$, and all of the $\ce{OH-}$ is gone, so $[\ce{OH-}] = 0$. But if you take the pH found using the $[\ce{H3O+}]$ value, use that to find pOH and then use that to find $[\ce{OH-}]$ you get a very small value, but not zero. So if I need to report a $[\ce{OH-}]$ value, do I just put 0 or the one gotten from pOH?

The problem that you are facing comes from the fact that technically, not all of the $\ce {OH-}$ reacts with the $\ce {H+}$.  It is only because the amount of OH- that doesn't react is so negligible compared to the amount that does, that the $\ce {[H_3O+]}$ can be taken to be $\ce{[HCl]-[$\ce{OH-}$]}$.
So, yes, to find the remaining (negligible) amount of $\ce{OH-}$, you must use the "very small value" given by either subtracting $\ce {pH}$ from $14 =\ce{pOH}$ or doing $\ce K_w/[\ce{H+}]=[\ce{OH-}]$ (which amounts to the same thing).


text:
Chlorination of propene at 773 K
The first step of preparation of glycerol
with propene involves the chlorination of of propene at 773 K. It results to form allyl chloride. However the halogenation of alkenes  generally give vicinal dihalides
$$\ce{CH3-CH=CH2 + Cl2 -> CH2Cl-CH=CH2}$$

at that temperature it is a free radical rxn

Note that the value of 773 K suggest a precision that is not justified. The numerical value is just the result when a temperature of about 500 °C is expressed in kelvin.

What are you asking for? Are you asking why chlorination of propene leads to formation of allyl chloride and not vicinal dihalide?

One more thing- low halogen concentration leads to formation of allyl chloride but for high halogen concentration leads to formation of vicinal dihalide. http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/radikale/radi_subst_add_alkane_alkene.vlu/Page/vsc/en/ch/12/oc/radikale/radi_brom_allyl/radi_brom_allyl.vscml.html

@NilayGhosh Do you really think concentration is important in free radical rxn ? it's temp who determines ? Isn't?

@Nilay: Concentration is important. NBS in CCl4 delivers bromine in low concentration to facilitate a free radical reaction. Free radical halogenation is first order in halogen. Halogenation, at least for bromine, of a double bond is second order in halogen.

The reaction proceeds via a SFR (Substitution Free Radical) mechanism and yields an allylic or benzylic halogenated alkene. This result is specific for low concentration of halogen and high temperature.
The results of the same reaction of halogens with alkenes but at different reaction conditions, i.e. high halogen concentration and room temperature result into anti addition of the halogen atoms on the double bond. Thus the reaction takes place faster in presence of $\ce{CCl4}$, i.e. non polar solvents. The reaction proceeds via formation of cyclic intermediate. The result of the same reaction in water instead of $\ce{CCl4}$ will result in to anti addition of halide ion and $\ce{OH-}$ ion on the double bond.


text:
Why do tosylation and mesylation of alcohols follow different mechanisms?
Why do the tosylation and mesylation of an alcohol proceed via different mechanisms?
More specifically,

In tosylation, the lone pair of the alcohol attacks the sulfur in TsCl, but why does it not do so with MsCl? Surely, TsCl should be more sterically hindered due to the benzene ring.
Why is the sulfene intermediate even formed in mesylation? Can that position be deprotonated by $\ce{Et3N}$? It's clear that this mechanism cannot happen with TsCl which has no alpha protons, but I fail to see why it should happen with MsCl.

Tosylation with pyridine:

Mesylation with triethylamine:

I've never seen the methyl proton of MsCl be deprotonated like that by TEA. I'll look it up, but I'm skeptical.

My bad. Sulfene is formed using TEA, apparently, but I'm not sure that happens in in this reaction. The reason that reaction wouldn't happen with TsCl is because there are no alpha protons there. The methyl group protons are more acidic than they are in toluene, but less so than those on MsCl.

Nevermind, my whole world has flipped upside-down. Sulfene is indeed an intermediate. Never seen that before.

If you check out the text book by March,  it will verify sulfenes and give evidence that sulfonyl chlorides do not react like acyl chlorides, as you correctly depict.

The alcohol oxygen still attacks the sulfur atom. It's just that alpha protons on sulfonyl chlorides are quite strong. Apparently, similar reactions happen with acyl chlorides, affording the ketene, which a nucleophile can attack.

Note that the mechanism for substitution is still probably not concerted. The addition and fragmentation is likely two steps.

Just a comment about "acid-base reactions will always occur first since they are much faster than other reactions", this is obviously not true, if it was it would not be possible (for example) to add organomagnesium or organolithium to ketone or esters.

The only real difference between an alkanesulfonyl chlorides and arylsulfonyl chlorides is the $\alpha$-hydrogen. In a mechanistic study of the hydrolysis of methanesulfonyl chloride, the reaction was found to have a very small secondary kinetic isotope effect for $\mathrm{pH}<6.7$, indicating no bond breakage to the isotopically labelled hydrogen and thereby an $\mathrm{S_N2}$ reaction mechanism.

For $\mathrm{pH}>6.7$ there is a large primary kinetic isotope effect, indicating bond breakage to the isotopically labelled hydrogen, corresponding to the formation of the sulfene intermediate.$^{[1]}$ This shows that though methanesulfonyl chloride isn't incapable of direct attack by alcohols, in the presence of basic amines, which serve the additional purpose of driving the reaction to completion by accepting the alcohol proton, it is more quickly deprotonated to form the sulfene.
While pyridine isn't that great of a base ($\mathrm{p}K_\mathrm{b}=5.21$) and won't form the sulfene to any great extent, triethylamine ($\mathrm{p}K_\mathrm{b}=11.01$) and trimethylamine ($\mathrm{p}K_\mathrm{b}=9.81$) are much stronger bases and will deprotonate methanesulfonyl chloride to give the sulfene to a greater extent. The mechanism is just as dependent on the base as it is the sulfonating agent, so all these considerations must be taken into account.

$^{[1]}$ King, J. F.; Lam, J. Y. L.; Skonieczny, S. Journal of the American Chemical Society 1992, 114 (5), 1743–1749.

I finally got around to reading this paper that has already been cited multiple times. Thank you for not saying that "the pKa of MsCl is 6.7" - that is inaccurate and quite illogical, too (a typical sulfone has pKa ~25-30). I guess the bottom line is that a similar mechanism with TsCl would operate if there was an $\alpha$-hydrogen, but since there isn't, it is restricted to the direct displacement mechanism. I'll bounty later today.

1 & 2. An acid-base reaction can only occur on MsCl, since as you point out it is the only one with acidic alphaprotons. The protons are indeed quite acidic since the negative charge is stabilized by the two oxygens. (pKa ~6.7, thanks @Mrityunjay Gupta)
Et3N is quite basic (pKa ~11), and acid-base reactions will always occur first since they are much faster than other reactions. I'm quite sure the alcohol also attacks the sulfur on MsCl, but the a/b-reaction is just faster. The elimination of Cl- on MsCl follows an E1Cb-mechanism.
Source: Clayden Organic chem. 2nd ed pp. 403-404, J. Am. Chem. Soc., 1992, 114 (5), pp 1743–1749 (pKa of MsCl)

This answer is actually pretty reasonable, but it should be expanded on slightly.

The pKa of Mesyl Chloride can be looked at from the following work by J. F. King, J. Y. L. Lam, and S. Skonieczny in their paper - J. Am. Chem. Soc. 1992, 114, 1743 - 1749.
They have shown that in pH range 6.7-11.8, rate-determining attack by hydroxide anion forms sulfene which is quickly trapped by water. So the Pka of Mesyl Chloride can be said to be 6.7; this is less than the Pka of Et3N at 10.75. So Mesyl chloride will be forming the sulfene intermediate if treated with Et3N. And the SN2 attack by Oxygen happens in both cases, I don't know what you wanted to ask in your first question. Your mechanisms themselves show the attack in both cases by oxygen on the sulfur atom. I hope I am correct in my analogy. Apologies if I am wrong somewhere.


text:
What else other than zinc can reduce Cr(III) to Cr(II)?
I successfully completed the synthesis of bis(chromium(II)hydrate)tetraacetate (aka chromous acetate) using zinc as the reducing agent. For my lab report, I was asked to suggest at least two other reagents which may used in place of elemental zinc in the synthesis of Cr2(H2O)2(u-OAc)4.
Would magnesium metal and manganese metal work since both their standard reduction potentials are more negative than Zn(s)?
And why does zinc make the more practical choice?
I couldnt find ANY literature showing this synthesis using any other reducing agent other than zinc...

https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)

@permeakra Based on that, how does reduction stop at $\ce{Cr^{2+}}$ using zinc metal?

Wikipedia describes the reduction of $\ce{CrCl3}$ to $\ce{CrCl2}$ with zinc metal, $\ce{LiAlH4}$$^{*}$, and $\ce{H2}$. You could use any of these methods and stir with $\ce{NaOAc}$ to form $\ce{NaCl}$ and $\ce{Cr2(OAc)4}$:
$$\ce{2CrCl2 + 4NaOAc + 2H2O -> Cr2(OAc)4(H2O)2 + 4NaCl}$$
I'm not sure if magnesium or manganese would work well, but I would assume they would do the reduction. The problem is that you want something that will reduce $\ce{Cr^{3+}}$ to $\ce{Cr^{2+}}$ but not any further. Using something with a more negative reduction potential could cause overreduction. These are good ideas, and I would certainly give credit to my students for proposing them, they just aren't mentioned specifically anywhere I'm seeing.

$^*$You would need to isolate $\ce{CrCl2}$ in the case of $\ce{LiAlH4}$ reduction (or kill off any excess LAH, at least) because you don't want to reduce the acetate added after the reduction.


text:
Using absorbance to find amount of a substance in a sample
If you use absorbance to find the amount of one substance that a sample contains, and do three determinations, should you just average the three values and use that? Or just talk about them individually?

Usually, I would average the values. That is, unless one is much off. Then, I would repeat the measurements and ask myself what went wrong. Is the compound air-sensitive? Is one of the cuvettes contaminated inside? Did I clean the outside of the cuvette properly?

Alright. That's what I thought, but always best to be sure. Thanks!

My pleasure! Btw, welcome to Chemistry.SE! If you don't mind the hint, I suggest that you turn your unregistered, cookie-based account into a registered, login-based account. Less hassle, more fun. Have a look at Why register for details.


text:
Calcium sulfate soluble in water
In the volumetric estimation of calcium in a given solution as calcium oxalate. We convert the calcium oxalate to oxalic acid by dissolving the former in hot (~70 °C) 2 N sulfuric acid solution. This is titrated against $\ce{KMnO4}$.
The reaction would be
$\ce{H2SO4 + CaC2O4 -> H2C2O4 + CaSO4}$
The calcium sulfate formed is normally not soluble in water. But here we get a clear solution which we titrate against $\ce{KMnO4}$.
Why does the calcium sulfate not precipitate in the solution?

It is not "not soluble". It is poorly soluble.

The overall reaction is really better given as: $$\ce{2H2SO4 + CaC2O4 -> H2C2O4 + Ca^{2+} + 2HSO4^{-}}$$ since at that strong of a pH the predominate sulfuric acid species is $\ce{HSO4^{-}}$.
Better yet the reaction could be written as  $$\ce{2H^+ + CaC2O4 ->C[{0.2 N \ H2SO4, 70 °C}]\ H2C2O4 + Ca^{2+}}$$
As was already pointed out in the comments, calcium sulfate has a fair solubility 0.2 g per 100 ml of water at at 20 °C, and the $K_\text{sp} = 4\times10^{-5}$.
The solubility varies with temperature. The solubility of calcium sulfate dihydrate increases from 0.223 g/100 ml at 0 C to about 0.265 g/100 ml at 40 C, then decreases to 0.205 g/100 ml at 100 C.

I calculated the solubility product of CaSO4 at the required concentration and got as .031 which is higher than the Ksp . The question is would the Ksp increase so much with just a 50 degree temp rise? Added to it is the negative temperature coefficient of CaSO4 which should mean that the solubility decreases with increase in temperature..

you can look at calcium sulfate solubility as function of temperature here https://en.wikipedia.org/wiki/Solubility_table Did you account for the fact that most sulfate is $\ce{HSO4^{-}}$ ?

I would check on that "of course be more soluble in hot water" claim. I have read otherwise.

Calcium sulfate is somewhat soluble in acids (CRC Handbook: "s a"). The previous answer accurately states that the overall reaction is really better given as:
$\ce{2H2SO4 + CaC2O4 -> H2C2O4 + Ca^2+ + 2HSO4−}$.
If only one sulfuric acid per oxalate were used: $\ce{H2SO4 + CaC2O4 -> H2C2O4 + CaSO4}$. Then insoluble $\ce{CaSO4}$ would replace insoluble $\ce{CaC2O4}$, perhaps coating the oxalate and rendering the analysis inaccurate. Excess acid (and heat) is used to make the mixture go clear, so that the reaction can be seen to be complete.
Similarly, $\ce{CaCO3}$ is insoluble in water, but excess $\ce{CO2}$ gives $\ce{HCO3-}$, and $\ce{Ca(HCO3)2}$ is somewhat soluble - that's why we have stalagtites and stalagmites in some caves (when the $\ce{CO2}$ comes out of solution, the $\ce{CaCO3}$ precipitates).


text:
Find the amount of atoms for a given mass
The atomic mass of two elements A and B is 40 and 80 respectively. If x g of A contains y atoms, how many atoms are present in 2x g of B?
I have tried to solve by assuming that

x g of A has y atoms
so x g of B has 2y atoms, because atomic weight of B is 2 times higher than A
so 2x g of B contains 4 y atoms.

But the answer key says it was wrong, the correct answer is y. What am I doing wrong?

The atomic mass of two elements A and B is 40 and 80 respectively. If x g of A contains y atoms, how many atoms are present in 2x g of B? Since B has a mass double that of A, 2X g og B should have the same amount of atoms as 1 X g og A. The answer is y atoms.

For the elements A and B, the relation between the number of atoms n, the mass m, and the atomic mass M is given by

$$n_A = \frac{m_A}{M_A}\quad,\quad n_B = \frac{m_B}{M_B}$$

Your assignment states:
$$m_B = 2\cdot m_A $$
$$M_B = 2\cdot M_A $$
Solve for $n_B$ ;-)

I seriously suggest to practise these calculations a lot. Ask your tutor for more worksheets or find a workbook (with solutions). Solve it alone, or work in groups. No matter how, you have to practise.
If you are in doubt, where to divide and where to multiply, remember:
Units are your friends!


text:
What causes transition from smoldering combustion to flame combustion?
What causes the transition from smoldering combustion to flame combustion?
From what I have read about it is that, smoldering combustion gives off a lot of flammable gases, primarily, Carbon Monoxide, however, the lower flammability of carbon monoxide is 12.5% and it takes about 2-3 hours to reach a concentration of 1%, it is lethal to humans in concentration of 1%.
There have been reported experiment studies, where smoldering combustion transitioned into flame combustion, as quickly as 22 minutes. So, there must be some other cause. Carbon Monoxide is, I guess not the reason.

Other gaseous products of incomplete combustion may play a role (depends on the fuel, though). But essentially, I believe this is about the subtle balance between the heat generation and dissipation. It is mostly physics, not chemistry.

Did you delete and repost question?

Burning in open air isn't a steady state process, but a very dynamic one. Likely when the fuel is smouldering there isn't enough of a convection current in the air to bring in fresh oxygen. As the smouldering continues something about the geometry of the area smouldering changes and fresh oxygen is brought in by a strong convection current. At that point the fuel bursts into open flame.

Smouldering is a non-flaming, self-sustained combustion process, often described as a heterogeneous combustion, because the reaction occurs on the surface of the solid fuel, rather than in the gaseous phase of the released flammable vapours. It is a combination of partial combustion and pyrolysis. Most fuels that can undergo smouldering combustion are typically porous; cotton, tobacco, sawdust and peat are all good examples. Smouldering fires can self-propagate for many hours before transitioning to flame. Some wilderness smouldering fires have been burning for in excess of 6000 years, and given the correct conditions would still transition to flaming fire.
There has been a large amount of research into the transition from smouldering to flaming combustion.  A large number of fatalities in house fires can be attributed to smouldering fires that have transitioned to flaming fires after several hours (and everyone is asleep).
A couple of excellent references examining this transition mechanism include:

A.P. Aldushin, A. Bayliss, B.J. Matkowsky. Combustion and Flame 145 (2006) 579–606
C. Y. H. Chao and J. H. Wang. Combustion and Flame 127 (2001) 2252-2264

There are three requirements to transition from smouldering to flaming combustion:

the rate of release of combustible vapours must be sufficient to support flame
the ratio of air (oxidant containing medium):vapourised fuel is high enough to support flame.
heat of the system can support flame combustion.

You might note that these three factors are in fact the key ingredients in the fire triangle.
In smouldering fires, there is a complicated balance between the direction of smoulder travel, and direction and velocity of gas flow. Pathways of pyrolysis (which is endothermic) also serve to compete with heat production.
As the fuels are generally somewhat porous, there is an airflow created the burning solid fuel which influences heat dissipation and oxygen supply. Convections created by the smoulder travel introduce the gas flow. This may mean that the air flow pushes heat, volatile fuels and oxygen into a zone that will allow for flame combustion. Often the directions of travel are contradictory, and so can oppose each other. Air/gas flows can push volatiles away from the combustion zone, and also dissipate heat. Whilst the actual trigger mechanism appears to depend on the fuel type, composition and airflow of the system under study, it is suspected that a critical point is reached where the balance between oxygen content, fuel levels and heat is reached. It appears that oxygen concentration is the critical key to this process; not necessarily the existence of other flammable fuels.
Smouldering fires are usually low temperature fires, and sufficient heat must be present to support flaming combustion. Again, oxygen concentration is critical here, and plays a key role in the char oxidation (solid fuel combustion) which provides a rapid boost in heat to support transition to flame. Char oxidation is usually a second phase oxidation; it occurs at some point after initial partial combustion of the starting fuel.