text:
Why does CO2 lowers the pH of water below 7?
Ok so I understand that having a pH below 7 is considered to be acidic and I also understand that part of the definition of an acid is that it gives off H+ when dissolved in water(even though not all acids necessarily need to give off H+ when dissolved in water)
But I was the rather confused when I came upon this ( when they refer to the 2 gases they are talking about CO and CO2)

What I dont understand is how the equation came to be ? ( the product side of the equation)
And how would It have looked if CO was dissolved in H2O, would it have still lowered the pH??

Carbon Dioxide (CO2) readily dissolve in water and form Carbonic Acid (i.e H2CO3 (aq) )

This is the formation of bonds.

Then Carbonic Acid (i.e H2CO3  (aq)  ) dissociate in water as follows.

So water gets H+ ions, so that cause water acidic.
The following shows dissociation of Carbonic Acid (i.e H2CO3 (aq) ) more clearly.

Carbon Monoxide (CO) do not readily react with water in room temperature in standard conditions. So it do not dissolve in water.

I once did a detailed calculation of this problem. Let me show it here. The partial pressure of $\ce{CO2}$ in the atomsphere is $3.8\times 10^{-4}\,\mathrm{atm}$. The solubility of an ideal gas in water is proportional to its partial pressure over the water surface. The proportionality constant is $0.034\,\mathrm{mol/(L\cdot atm)}$ at $25\,^\circ\mathrm{C}$. Among the dissolved $\ce{CO2}$ molecules in water, about $K_h=1.7\times 10^{-3}$ (at $25\,^\circ\mathrm{C}$) of them combine with $\ce{H2O}$ molecules to form $\ce{H2CO3}$, which then dissociates and releases $\ce{H+}$ in two steps with $\,pK_{a1}=3.58\,$ and $\,pK_{a2}=10.32$, i.e.,
\begin{align}
&\ce{CO_{2(aq)} + H2O <=> H2CO3},\quad K_h=\frac{[\ce{H2CO3}]}{[\ce{CO_{2(aq)}}]},\\
&\ce{H2CO3 <=> H+ + HCO3-},\quad K_{a1}=\frac{[\ce{H+}][\ce{HCO3-}]}{[\ce{H2CO3}]},\\
&\ce{HCO3- <=> H+ + CO3^{2-}},\quad K_{a2}=\frac{[\ce{H+}][\ce{CO3^{2-}}]}{[\ce{HCO3-}]}.
\end{align}
Therefore, we have
$$\ce{[CO_{2(aq)}] + [H2CO3] + [HCO3-] + [CO3^{2-}]} = 0.034\times 3.8\times 10^{-4}\,\mathrm{mol/L},$$
and from charge neutrality
$$\ce{[H+] = [HCO3-] + 2[CO3^{2-}] + [OH-]}.$$
We then obtain the equation for $[\ce{H+}]$ given by
$$[\ce{H+}]-\frac{K_w}{[\ce{H+}]}=\frac{\frac{K_hK_{a1}}{[\ce{H+}]}+\frac{2K_hK_{a1}K_{a2}}{[\ce{H+}]^2}}{1+K_h+\frac{K_hK_{a1}}{[\ce{H+}]}+\frac{K_hK_{a1}K_{a2}}{[\ce{H+}]^2}}\times 1.29\times 10^{-5}\,\mathrm{mol/L},$$
where $K_w=[\ce{H+}][\ce{OH-}]$ is the dissociation constant of water. The exact equation can be difficult to solve. Discarding insignificant terms, we obtain
$$[\ce{H+}]\approx\frac{K_hK_{a1}}{[\ce{H+}]}\times 1.29\times 10^{-5}\,\mathrm{mol/L},$$
which reduces to the familiar formula for the acidity of weak acids
$$[\ce{H+}]=\sqrt{K_hK_{a1}\times 1.29\times 10^{-5}\,\mathrm{mol/L}}=2.4\times 10^{-6}\,\mathrm{mol/L}.$$
Therefore the $\mathrm{pH}$ value of saturated carbonic acid is $\,\mathrm{pH}=5.6$, which is why we define rain water with $\,\mathrm{pH}<5.6\,$ as acid rain.

authors: PJ Food 優質食材網購批發


text:
When is it permissible to use the stoichiometry of a reaction to determine rate laws?
I have read in a general chemistry textbook that rate laws can only be determined by experiment. This seems intuitive because if a mechanism is not elucidated and tabulated reaction data would be one way of bypassing that info to reach the rate law. However, I have seen instances where the rate law is simply determined by using the reactants and raising them to their respective stoichiometric power...which I thought contradicted the idea of needing an experiment? I highly speculate that this is only acceptable if it a singular reaction that the reaction order can be determined experimentally? For example:

$$\ce{2A -> B_2}$$
Can be surmised as:
$$\text{rate} \  = k[A]^2$$
Is this true? My speculation is based on the notion that if one reactant is splitting to a product and that stoichiometric equivalent is needed for the conversion of reactant to product, then the rate law is determined by stoichiometric equivalent number of molecules hence the justification for the "raise to the power" method. Or is there another underlying factor when the exponents have been raised and no detail is given to rate laws determine by experiment?

Your reasoning is correct. When the reaction you are investigating is an elementary reaction step, you can write down the rate law using the stoichiometry of the reaction. Mathematically, this is done as follows.
Consider a reaction as
$$
\sum_{i} \nu_{i} R_{i} \rightarrow \sum_{j} \nu_{j} P_{j}
$$
Where $R$ and $P$ are reactants and products respectively and $\nu_{i}$ and $\nu_{j}$ are the stoichiometric coefficients.
If this reaction is an elementary reaction step, than the rate law in the forward direction is given by:
$$
r = k \prod_{i} [R_{i}]^{\nu_{i}}
$$
where $[R_{i}]$ is the concentration of reactant $R_{i}$ and $k$ is the rate constant.
For example, if you consider the reaction
$$
A + 2B \rightarrow 3C,
$$
then the rate in the forward direction is
$$
r = k[A][B]^{2}.
$$

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authors: Румба Румбик

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text:
How to disperse salicylic acid in sodium lauryl sulfate solution?
Anti-dandruff shampoo contains salicylic acid as an active ingredient, but when salicylic acid powder comes in contact with a base solution such as sodium laurylsulfate and water (1:3 wt/wt) caking of the particle occurs, forming a non-dispersible aggregate.
I know from the study of pharmaceutics that decreasing the contact angle between particles and solution leads to a decrease in surface tension and thereby dispersion of the particles.
How can one disperse the salicylic acid particles specifically in this base solution?

Sodium laurylsulfate isn't basic.

I don't mean by "basic" the PH of the solution, but i mean it is the main component

I don't have a lot of experience with salicylic acid as I'm just starting to work with it myself at the moment. However, I find that it disperses better if it is added to the solution at a high temperature (before surfactants) and then homogenised (once thickening agents, if any, have been added).
The other thing which may help is by dissolving in denatured alcohol, although I haven't personally tried this myself. Although it's not a shampoo, I have recently formulated a facial cleanser and as described above I firstly added the thickening agents to the water and then salicylic acid, heated to approximately $70~\mathrm{^\circ C}$ and then homogenised.

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text:
Corey -House synthesis vs Wurtz reaction
In Corey House synthesis the first step is formation of Alkyl Lithium. why can't we react alkyl lithium with alkyl halide to produce alkanes. Rather we choose to prepare Gilman's reagent? Does it have something to do with the reactivity of alkyl lithium?

As $\ce{R-Li}$ is very reactive ( even more than $\ce{RMgX}$ ). To decrease the reactivity of $\ce{R-Li}$ , it is reacted with $\ce{CuX}$ to form Gilman Reagent which is less reactive than $\ce{R-Li}$ to produce alkanes.
If  $\ce{R-Li}$ reacts directly with $\ce{R'-X}$, there will no formation of alkanes, but interchange reaction occurs.
$$\ce{RLi + R'X --> RX + R'Li}$$

thanks for the answer... can you also tell me how to determine the reactivity of organometallic compounds as u said R-Li is more reactive than RMgX .. and Happy New year !!!!

As Li is more electropositive than Mg. There is easy breaking of R-Li bond more instant than in case of RMgX to give a strong base R(-)ve. if u satisfy. please, approve and upvote the answer.

there will no formation of alkanes || O'Rly ? Wurtz synthesis with lithium is a thing.


text:
Measuring the effect of increasing salinty on the concentration of Dissolved oxygen
I want to investigate the effect of increasing salinity on the concentration of dissolved oxygen, in a water sample being distilled water.
Would that work?
My concern is, that since its not a true fresh water sample, I am worried that changing the salinity of distilled water, to create different saline solutions, would impact the oxygen content significantly?
I want to model the direct relationship, so I can apply it to a real life example. For instance the effect of road salting on oxygen concentration in fresh water habitats.

Yes, salinity impacts dissolved oxygen content, as a little Googling will readily reveal (and deeper reading will show why). Your biggest problem would be - do you have the equipment needed to measure it properly?


text:
Would hydrogen bond form in these scenarios?
My notes say that hydrogen bonding occurs between a hydrogen atom covalently bonded to a highly electronegative atom ($\ce{N, O, or F}$) and the lone pair of a highly electronegative atom of another molecule. I am wondering if hydrogen bond would form between a molecule with hydrogen atom covalently bonded to $\ce{N/O/F}$ and a molecule with $\ce{N/O/F}$ that has lone pair but is not bonded to hydrogen. For example, it seems like oxygen, ozone, carbon dioxide are soluble in water so they can form hydrogen bond with water though the $\ce{O}$ atom is not bonded to $\ce{H}$. How about $\ce{CH3CH2COCH2CH3}$, the $\ce{O}$ atom has a lone pair, but it seems like it is not soluble in water and cannot form hydrogen bonding with water.

Why, it can. It's just that these hydrogen bonds are not enough to overcome the influence of the hydrophobic part.

The molecule your example, 3-pentanone, actually has a water solubility of $\pu{50 g/L}$ at $\pu{20^oC}$. The water solubility of it's pure hydrocarbon analog, n-pentane, is $\pu{40 mg/L}$ at $\pu{20^oC}$. A simplistic interpretation of this comparison is that the addition of a single hydrogen-bond acceptor functionality to n-pentane increases it's water solubility by greater than three orders of magnitude! So the answer to your question is pretty clearly that yes, hydrogen bonding is happening in this scenario.


text:
Why is the boiling point of CH3COOH higher than that of C2H5OH?
Why is the boiling point of $\ce{CH3COOH}$ higher than that of $\ce{C2H5OH}$ ? Both are polar molecules held by hydrogen bond.

CH3COOH is more polar.

Related: How to rationalise the difference in the melting points of acids and alcohols with inter molecular forces?

Yes you are correct. The molecules are polar in nature and are bound by intermolecular hydrogen bonding. But you must pay attention to the extent of polarization in both the molecules. Consider the alcohol. Here the carbon bearing the $\ce{-OH}$ group is the only polarizing group present. But for $\ce{CH3COOH}$ , the carbonyl carbon is polarized by an $\ce{-OH}$ group as well an $\ce{=O}$ group attached to it, thus increasing its effective polarization more than the alcohol. Therefore $\ce{CH3COOH}$ has greater boiling point.


text:
Is Hydrolysis same as dissolution?
I have book that contains the following paragraph:

Due to high dielectric constant, it (water) has a very strong hydrating tendency. It dissolves many ionic compounds. However, certain covalent and some ionic compounds are hydrolysed in water.

Is this statement self-contradictory? Doesn't 'hydolysed in water' mean dissolved in water?

Absolutely not. $\ce{NaCl}$ dissolves, but never hydrolyzes. $\ce{Al2S3}$ hydrolyzes, but never dissolves.

What's the difference ?

I think that under the IUPAC definition of dissolution, hydrolysis is (or at least can be) a type of dissolution.

Hydrolysis is the chemical breakdown of a compound due to reaction with water.
Dissolution is a process by which the original states of matter turn into a liquid or other solvent and become solutes - dissolved components in solution.

Thanks for the response , but I don't think it answered my question effectively. It will be a great help for me if you could explain the statement given in the book(written in the blockquotes).

OK wait for a while..

Water has high dielectric constant because due its polar nature, it has more tendency to decrease the attraction between the ions in an ionic compound. It means that water is dissociating or ionising the ionic compound into its constituent particles i.e. it is known as hydrolysis which is done on some ionic and covalent compounds (not for all ionic compounds).

Dissolution simply means mixing. As all ionic compounds can mixed in water to from a solution which my be homogenous or heterogenous.  IF you like my answer, please up vote me.

On hydrolysis , the ionic compound is split into its constituent ions but on dissolution , the ions are mixed with water so what forms is not the original ions...Am I right ?

I unable to understand your question

I have edited the question...

Yeah.. Correct. You got it..

Hydrolysis is a broad chemical term that means reacting with the solvent water in some way. So for examples esters can be hydrolysed according to the following reaction:
$$\ce{H3CCOO-C2H5 + H2O -> H3CCOOH + HO-C2H5}$$
(Ethyl acetate and water react to form acetic acid and ethanol.)
Dissolution is the process of something being dissolved in water, intermixing with the water molecules so that each atom/ion/molecule of the substance is more or less completely surrounded by water molecules. This process is usually understood to happen without any reaction, thus the term hydrolysis is not used.
However, the above is often a simplification. For example, some water-free salts such as $\ce{AlCl3}$ or $\ce{CuSO4}$ will react with the water to form the corresponding hexaaquacomplexes $\ce{[Al(H2O)6]^3+ + 3 Cl-}$ or $\ce{[Cu(H2O)6]^2+ + SO4^2-}$. These are technically reactions, best exemplified in the aluminium example where the metal was originally coordinated by chlorido ligands and after dissolution is coordinated by aqua ligands.
The only thing that changes is the immediate vicinity of the metal ion in these cases, so speaking of dissolution rather than hydrolysis (essentially the metal cation remains the same as do the corresponding anions) is often tolerated.

So, putting it in a simple language, dissolution involves no reaction, just the constituent atom/molecule/ion of the substance is surrounded bu water molecules, whereas, hydrolysis involves the reaction. This is what I got from your answer. Am I correct ?

@user3503781 yeah.


text:
The mechanism of demethylation of a quaternary ammonium
The problem is such:

I know that the t-Bu locks the ring in equatorial position as I have drawn out:

Could someone please point me in the right direction as how I can get rid of a Me in the NMe3+ group?
Is it possible that tBuO- is acting as a nucleophile in this case?

t-butoxide can certainly act as a nucleophile, but it would much prefer to do an elimination. Why do you think the elimination in this case to form a cyclohexene is not favoured? What do you need for an elimination reaction (specifically an E2 reaction) to occur?

@orthocresol we need the NMe3+ to be axial and anti to the H removed which it is not

Due to steric interactions, the bulky base – triethylamine cannot readily abstract the proton. In these situations, a less sterically hindered proton is preferentially abstracted instead. As a result, the Hofmann product is typically favored when using bulky bases.

@SujithSizon there is no lone pair on N in the quaternary ammonium salt, there is no elimination going on in this reaction.

As you point out, the tert-butyl group wants to remain in an equatorial position, else both the tert-butyl group and the trimethylammonium group end up being axial which is disfavoured for two bulky groups.
The obvious role of tert-butoxide:
The tert-butoxide would obviously rather act as a base than a nucleophile (in-fact under ordinary circumstances we would say that tert-butoxide is a non-nucleophilic base in the same way we refer to LDA).
In order for the tert-butoxide to act as a base, and do the elimination, the conformation of the ring would first have to put both groups axial, which we have already said isn't favourable.
There is an additional reason why this is unable to happen however, since the protons that you need to abstract to get elimination are 1,3-diaxial to the tert-butyl group. This would be highly sterically disfavoured and so is unlikely to happen.
The less obvious role of tert-butoxide:
Since the tert-butoxide can't act as a base, there is the possibility of it attacking one of the methyl groups on the trimethylammonium group to give the product as drawn in an SN2 type reaction. Whilst this seems a little odd, it does have the advantage of neutralising the charges on both the tert-butoxide and the trimethylammonium group to give two neutral molecules.


text:
Why is NO2 formed when N valency is 3
Sorry if the question is silly but this is troubling me more than enough. Here's what I know about chemical formula: Reverse the valencies of the combining elements. But in this case, $\ce{NO2}$ is the formula for nitrogen dioxide while N valency is 3. Please explain it to me if I am wrong somewhere.

What do you think the formula should be?

N2O3 - this should be the formula

Your rule of swapping the valencies only works for a very specific set of compounds. In general there are a huge number of compounds where this is not the case and indeed the rule is merely a coincidence and has no real theoretical basis.

Then why is this rule present in every chemistry textbook of middleschool

Because it is useful for quickly getting the formulae of many compounds that you are likely to meet at that level.

So How is the formula decided at all. Is there any other theoretical rule which may be used ?

Let us continue this discussion in chat.

@Shashwat  i think NO2 exists in the form of N2O4(solid), whereas your suggestion N2O3 is a mixture of NO and NO2

@SujithSizon $\ce{N2O3}$ is a separate compound. It is formed from the reaction of $\ce{NO}$ and $\ce{NO2}$ but it is not a mixture of them.

@bon do they have to react? I mean NO and NO2 are unstable odd electron species right? Also NO2  exists as a N2O4 dimer (kinda in equilibrium) right?

@SujithSizon  $\ce{NO2}$ exists in equilbrium with $\ce{N2O4}$. The position of equilibrium varies with temperature, pressure etc.

The rule of swapping the valencies of the atoms to get their ratios only works for a very specific set of compounds and is more of a convenient trick than a rule based on any sort of hard theory.
In particular, it works for binary 'ionic' compounds such as $\ce{NaCl}$, $\ce{CuCl2}$, $\ce{Al2O3}$ etc. because the valencies of the atoms correspond to the charges on the ions and the net charge has to be zero.
However, it obviously doesn't work for a huge range of other compounds where there is more complicated bonding. For example there are a whole range of covalent nitrogen oxides of varying degrees of stability which do not fit the rule ($\ce{N2O}$, $\ce{NO}$, $\ce{NO2}$, $\ce{N2O3}$, $\ce{N2O4}$ etc).
Ultimately the only way to determine the formula of any arbitrary substance is through experiment, typically with a mass spectrometer.
If you are interested in finding out more about the different types of substances that are possible I suggest that you start reading about different types of bonding, in particular ionic versus covalent, to begin to understand the different types of compounds that exist.

Well. Here is an explanation which is valid at the secondary level. At the university level, it is more correct to speak of bonding and anti-bonding molecular orbitals.
Let's go ! $\ce{NO2}$ is a unusual molecule that has one lonely electron on the $\ce{N}$. This should not happen. But look ! It also contains $4$ $\ce{N-O}$ bonds, with a  negative partial charge on the oxygen side. As a consequence, there are $4$ partial positive charges on the central $\ce{N }$ atom. This is much ! Now, experience shows that  $\ce{NO2}$ exists only at high temperatures. When cooling it at low temperatures ($T < 0°C$), it gets dimerized and becomes $\ce{N2O4}$. Unfortunately this molecule $\ce{N2O4}$ has a drawback. It contains a bond $\ce{N-N}$ between two atoms holding both $4$ partial positive charges. Similar electric charges are repelling one another. So the bond $\ce{N-N}$ is rather weak in $\ce{N2O4}$, and does not resist at absolute temperatures higher than $\pu{270 K}$. At room temperature for example, $\ce{N2O4}$ is dissociated into $\ce{2 NO2}$, at least partially (as it is an equilibrium).
As a consequence, $\ce{NO2}$ can be considered as a half-molecule ! Exactly like $\ce{H2}$ or $\ce{Cl2}$ which produce $\ce{H}$ or $\ce{Cl}$ atoms when heated to rather high temperatures.

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text:
How can off-diagonal Hamiltonian matrix elements be nonzero when the overlap matrix is diagonal?
While studying about Hückel theory, I got accustomed to the approximation of making the overlap matrix an identity matrix; that is making the off-diagonal elements zero as $S_{AB}= S_{BA}= 0\;;$ this implies the use of orthogonal base states of AOs. The off-diagonal elements of the Hamiltonian matrix are still taken as constants that may be non-zero: $H_{AB}= H_{BA}= \beta_{AB}$, where $\beta_{AB}$ is a negative quantity.
Then I wondered why it isn't the case that $H_{AB}= H_{BA}= 0$ strictly, as it represents the expectation value - the average energy contribution of the overlapping region of AOs $A$ and $B$.  But, it would seem that overlapping is not actually possible, as is evident from the fact that $S_{AB}$ is zero.
In this question, when I asked about this, ifilot replied:

Indeed, this seems rather counterintuitive, but it is not. Another way of looking at $S_{ij}=δ_{ij}$ is saying that all atomic orbitals are orthonormal to each other. So if you would evaluate the overlap integral of two different orbitals, it would result in zero. This does not necessarily mean that evaluating the Hamiltonian integral $\langle ϕ_i|\hat H|ϕ_j\rangle$ results in zero, because first applying the Hamiltonian operator on the wave function and then integrating might result in a non-zero outcome.

I know he is right in this point; but I'm still having trouble seeing how this is possible.
As Peter Atkins in his book Elements of Physical Chemistry wrote:

[...] The integral $H_{AB}$ depends on both $\psi_A$ and $\psi_B$, and we can interpret it as the contribution to the energy due to the accumulation of electron density where the two atomic orbitals overlap, including, for instance, the Coulombic attraction between the extra accumulation of electron density and both nuclei.

Evidently, this phrase makes clear that $H_{AB}$ is non-zero iff there is overlap between the AOs.
So, how can $H_{AB}\ne 0$ and $S_{AB}= 0$ both be true at the same time? And, what are the physical implications? While the former means there is overlap, the latter means the opposite; it seems really contradictory.

Please don't use Math Jax in the title. It does not render in the mobile app and it comes with multiple more problems. I can't provide the link at the moment, but there are recent discussions on Meta. (Cc @brian)

@Martin-マーチン Meant to change the title when I edited; forgot.  Will tweak here shortly.

I believe the confusion lies in some ambiguity in the use of the term overlap.
For the $S_{AB}=0$ approximation of Hückel theory, overlap is used in the specific mathematical sense of the "overlap integral":
$$
S_{AB} = \left<\varphi_A|\varphi_B\right> =
\int_{\vec r}{\varphi_A^*\!\left(\vec r\right)\varphi_B\!\left(\vec r\right)\mathrm{d}\vec r}
$$
However, this does not mean that at each point in $\mathbb{R}^3$ only one of $\varphi_A$ or $\varphi_B$ is allowed to bear a non-zero value.  In fact, at most points in $\mathbb R^3$, $\varphi_A$ and $\varphi_B$ will both bear non-zero values. Since $\varphi_A$ and $\varphi_B$ will in general have "co-located regions of non-zero values," the two functions can be described qualitatively as overlapping, in the same sense as in your second quote. It is the overlap of the functions in this more general, qualitative sense, that allows $H_{AB}=\left<\varphi_A^*\middle|\hat H\middle|\varphi_B\right>\neq0$.
(To note,  in order for Hückel's approximation to be exact, the orbitals would have to be selected such that the specific integral $S_{AB}$ over all $\mathbb R^3$ equals zero$^\ddagger$. I believe Hückel did not actually do this, but instead just coerced to zero any nonzero $S_{AB}$.)
To help explain more explicitly why this is, I'll first write out the integral implied by the Dirac notation for $H_{AB}$:
$$
H_{AB}=\left<\varphi_A^*\middle|\hat H\middle|\varphi_B\right>=
\int_{\vec r}{\varphi_A^*\!\left(\vec r\right)\hat H\!\left[\varphi_B\!\left(\vec r\right)\right]\mathrm{d}\vec r}
$$
Then, I'll introduce two (real-valued) one-dimensional functions defined on the closed domain $x=[0,1]$ to use for demonstration purposes:
$$
\varphi_A\!\left(x\right) = \sin\!\left(2\pi x\right) \\
\varphi_B\!\left(x\right) = \sin\!\left(3\pi x\right)
$$
$\hspace{10mm}$
It is straightforward to show that $S_{AB}=0$ for these two functions.  One way is to plug the overlap integral into Wolfram Alpha. Another option is to do it by hand.  In the below figure, the green curve is a plot of the product $\varphi_A\!\left(x\right)\cdot\varphi_B\!\left(x\right)$, and the orange curve is a plot of the cumulative integral of this product:
$$
\int_0^x{\varphi_A\!\left(x'\right)\cdot\varphi_B\!\left(x'\right)\mathrm{d}x'}
$$
$\hspace{10mm}$
As can be seen, the integral over the entire domain is zero, satisfying $S_{AB}=0$.
Now, let's introduce a simple "Hamiltonian"$^\dagger$:
$$
\hat H = x
$$
The next figure is a replicate of the first one above, with $\hat H\!\left[\varphi_B\right]$ also plotted as the dashed red curve:
$\hspace{10mm}$
As in the second figure above, one last plot shows the local value of $\varphi_A\!\left(x\right)\hat H\!\left[\varphi_B\!\left(x\right)\right]$ along with the cumulative integral:
$$
\int_0^x{\varphi_A\!\left(x'\right)\cdot\left[x'\cdot\varphi_B\!\left(x'\right)\right]\mathrm{d}x'}
$$
$\hspace{10mm}$
The value of $H_{AB}$ here is clearly not equal to zero.  (Wolfram Alpha concurs.) Thus, it is absolutely possible to have $H_{AB} \neq S_{AB} = 0$.
To note, the principle exhibited here on a closed domain in $\mathbb R^1$ applies equally well to orbitals on an infinite domain in $\mathbb R^3$. Note also, if I were actually performing a quantum mechanical calculation here, I would have been a lot more careful about normalizing my wavefunctions (viz., ensuring $S_{ii}=1$).
$^\ddagger$Numerous methods are known for generating orthogonalized orbitals with $S_{AB}=0$ strictly, from a general set of orbitals where $S_{AB}\neq0$ occurs for some non-null subset of the $\left(\!A,\!B\right)$. To my knowledge, the most commonly used orthogonalization for atomic orbitals is Löwdin's method (see this PDF). The Gram-Schmidt orthogonalization (usually in its "modified" form) is used in a variety of other areas of quantum chemistry. Wikipedia also has a pretty solid article on orthogonalization methods in general.
$^\dagger$You'd never use this as a Hamiltonian operator. Further, I'm not sure if it technically even is a Hamiltonian operator. (I think it'd work as a dipole moment operator?) But, whatever, leave me alone, it's for purposes of illustration.

Okay, this means at most points $\varphi_A$ and $\varphi_B$ are non-zero; but still the sum of them over all the volume is $\approx 0,$ isn't it? Then how it makes the difference; overlap region is still on the verge of null; how could there be $H_{AB}$ as it is also the integral over all the volume? If the overlap is zero, then why should $H_{AB}\ne 0$,irrespective of how the overlap integral became zero-either by one of $\varphi$s zero at a point or the net sum being zero?

Cool! I am looking forward to seeing your example. In my previous answer, I already referred to my website (http://www.ivofilot.nl/posts/view/29/Introduction+to+Electronic+Structure+Calculations%3A+The+variational+principle) where you can see in Example 2 a scenario where an orthonormal basis gives off-diagonal Hamiltonian integrals. But perhaps this example is too complicated and you can come up with something simple and really to the point. :-)

I agree with the general principle illustrated by your example, i.e., that for two arbitrary orbitals $A$ and $B$, $S_{AB}=0$ doesn't necessarily imply $H_{AB}=0$. However, it should be emphasized that for two atomic orbitals $A$ and $B$ belonging to different atoms, it is almost invariably (if not always) true that $S_{AB}=0$ iff $H_{AB}=0$.

We still ignore $S_{AB}$ because most qualitative features of the solution don't change by this approximation (a counterexample: http://chemistry.stackexchange.com/a/42636/7841). But it would be a poor approximation if we really wanted to get accurate numbers (e.g., bonding energy) from Hückel theory.

@higgsss Mmm, so actually $S{AB} \neq 0$ holds generally, it's just that the Hückel approximation treats them as equal to zero?  Hadn't appreciated that.  Good points.

Right. It is hard to make $H_{AB}\neq 0$ with $S_{AB}=0$ with two atomic orbitals localized about different atoms. On the other hand, Wannier functions (https://en.wikipedia.org/wiki/Wannier_function) have the nice property that they are localized about different atoms as well as being orthogonal to one another. But finding such functions is another complicated problem.

@higgsss . I was picturing Loewdin-orthogonalized orbitals as a good example of an $H_{AB} \neq S_{AB}  = 0$ function space, but you're right, they're not single-center atomic orbitals.

@Brian:+1;  Really thanks for those plottings; I got the point that though $S_{AB}= 0,;;H_{AB}\ne 0$; I could extrapolate these graphs to 3D domain at ease. So, does that mean even if there is no overlapping, there can be accumulation of electron density in the internuclear-region?

@user36790 I think you should ask that as a new question, linked to this one, with a more full explanation of what you're asking. It'll be too much to try to answer in this comment thread. Also, make sure you read higgsss's comments above, in particular the one emphasizing that in actuality $S_{AB} \neq 0$ in Hückel theory,  but the values are taken as equal to zero as a key part of Hückel's approximation.

Maybe a little example from linear algebra (just imagine it to be spin space and we're back in the quantum world) can also add something to the discussion.
Suppose you have two wavefunctions
\begin{equation}
\psi_A = \frac{1}{\sqrt{2}} \begin{pmatrix}
1 \\
1 \end{pmatrix} \qquad \psi_B = \frac{1}{\sqrt{2}} \begin{pmatrix}
1 \\
-1 \end{pmatrix}
\end{equation}
Then obviously the overlap of the two is
\begin{equation}
S_{AB} = \langle \psi_A | \psi_B \rangle = \frac{1}{2}(1 \cdot 1 + 1 \cdot (-1)) = 0
\end{equation}
Nevertheless, if you consider the Hamiltonian
\begin{equation}
H = \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
\end{equation}
then you get $H_{AB} = \langle \psi_A | H | \psi_B \rangle = 1 \neq 0$.
My suggestion is to forget about the interpretation of these matrix elements given in Atkins, since it is very difficult to imagine how quantum-mechanical operators act (he also seems to forget that there is not only Coulomb interaction but also kinetic energy present in the Hamiltonian, the latter acting as a derivative operator on orbitals).


text:
Why is partial pressure proportional to mole fraction in Henry's law?
According to Henry's law of partial pressure,  partial pressure of a gas above a solution is directly proportional to the mole fraction of the gas in the solution.
But my question is: If the mole fraction of a gas in solution is greater, then that should mean there are less gas molecules in the air and so the partial pressure should be less. Why is it the other way round?

First, Henry's Law of Partial Pressure states:

"At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid."

Quoting from Wikipedia:

Henry's law is a limiting law that only applies for 'sufficiently dilute' solutions. The range of concentrations in which it applies becomes narrower the more the system diverges from ideal behavior. Roughly speaking, that is the more chemically 'different' the solute is from the solvent.
For a dilute solution, the concentration of the solute is approximately proportional to its mole fraction $x$, and Henry's law can be written as:

$$p = K_{\rm H}\,x$$
where $p$ is the partial pressure of the gas above and in equilibrium with the solution, and $K_{\rm H}\,$ is the proportionality constant called the Henry's Law constant.
This is a conceptual conclusion as steeper the concentration gradient, the rate of diffusion (and hence the amount of diffused substance) will be greater. Here, the concentrations of the "solute" is represented by partial pressure of the gas in the gaseous state and by it's corresponding mole fraction in the solution state.
What you have supposed stands only for a closed system. As in the second quote from Wikipedia, you understand that the law is applicable only for conditions in which the behavior/condition which you supposed is either not relevant or is negligible.
If you go through the linked article, you'll find that Raoult's Law for Vapor Pressure of Solutions appears to be a special case of Henry's Law for more concentrated solutions, and can be used as such.
Note that both Raoult's Law and Henry's Law are limiting laws and they are applicable each as when:
Henry's Law: $x$ → $0$
Raoult's Law: $x$ → $1$

Henry's law is not specifically formulated for a fixed amount of the dissolving substance.  If there is a fixed amount of the substance, it will partition between the gas phase and the liquid phase according Henry's law.  So, once you specify the volumes of gas phase and liquid phase, you can determine quantitatively how the substance partitions between them using Henry's law.


text:
What is the hybridisation of titanium in the hexaaqua- and hexaamminetitanium(III) complex?
I want to check if their hybridizations are as follows:
$\mathrm{sp^3d^2}$ in  $\ce{[Ti(NH3)6]^3+}$ and $\mathrm{d^2sp^3}$ in  $\ce{[Ti(H2O)6]^3+}$.

What exactly is the difference between $d^2sp^3$ and $sp^3d^2$?

In d2sp3 3d is involved while in sp3d2 4d is involved in case of Ti.

Do you REALLY need to use hybridization here?

Immediately stop using the hybridisation concept for transition metals. Don’t even take it a step further. It is not helpful here at all.
Both complexes have essentially identical electronic structures except that the relative orbital energies differ slightly. The following image — originally taken from Professor Klüfers’ internet scriptum for his coordination chemistry 1 course (LMU Munich; in German) — shows a typical octahedral coordination compound and only includes the ligand–metal σ interactions.

You can see the metal’s orbitals on the left-hand side; from bottom to top they are 3d, 4s, 4p. 4d would be even higher in energy, possibly even higher up than 5s. These virtual energies can be really hard to calculate or measure accurately. On the right-hand side, you can see the six donor orbitals which transform as $\mathrm{a_{1g} + e_g + t_{1u}}$. (For some reason, the u is missing.) The ligands’ p-orbitals that can take part in π bonding would transform as $\mathrm{t_{1g} + t_{1u} + t_{2g} + t_{2u}}$, and would therefore form another bonding-antibonding interaction with the metal’s $\mathrm{t_{2g}}$ orbitals. The ligand orbitals would again be bonding and the metal’s antibonding. If you check out the link and look for the image you’re seeing, you can also see pictures of the metal orbitals, and further down the scheme that includes the π interactions (only the relevant part highlighted).
The metal orbitals that interact with ligand orbitals in a σ fashion are two 3d, 4s and three 4p. The remaining three 3d orbitals interact in a π fashion. So if you really, really, really must use hybridisation in this case (did I say it is not recommended?) then you should report titantium’s hybridisation as $\mathrm{d^2sp^3}$ and not the other way round: 4d orbitals do not play any role.
The only difference between the hexaaquacomplex and the hexaamminecomplex is that the former has lower-lying ligand orbitals than the latter so the energy difference between $\mathrm{t_{2g}^*}$ and $\mathrm{e_g^*}$ will be slightly different. That’s all.

It should theoretically be the other way round. $\ce{NH3}$ is a strong field ligand as it is higher in the spectrochemical series than $\ce{H2O}$. Hence, $\ce{NH3}$ can cause force-pairing of electrons in inner shell ($\rm 3d$ orbitals) while $\ce{H2O}$ cannot, as it is a weak field ligand. This utilizes concepts from Crystal Field Theory.
Both $\ce{[Ti(H2O)6]^3+}$ and $\ce{[Ti(NH3)6]^3+}$ have $\rm d^2sp^3$ hybridization.
Thus, purely theoretically and naively, $\ce{[Ti(H2O)6]^3+}$ should involve the $\rm 4d$ subshell for accepting electrons donated from ligand molecules and have $\rm d^2sp^3$ hybridization while $\ce{[Ti(NH3)6]^3+}$ should involve the $\rm 3d$ subshell for accepting electrons donated from ligand molecules and have $\rm sp^3d^2$ hybridization.
The above is merely a theoretical and naive discussion. In fact, the $\ce{Ti^3+}$ has only 1 electron in its $\rm 3d$ subshell, and force-pairing of electrons does not occur in it. We find that in fact it has $\rm d^2sp^3$ hybridization in both cases, as it has vacant $d$ orbitals in the inner $\rm 3d$ subshell which can accept electronic pairs donated from the ligands. Here, the higher (in octahedral geometry) $\rm e_g$ subshell takes part in forming the $\rm d^2sp^3$ hybrid orbital, leaving the lower $\rm t_{2g}$ subshell as unhybridized. This happens for both $\ce{[Ti(H2O)6]^3+}$ and $\ce{[Ti(NH3)6]^3+}$
The take-home lesson here is that force-pairing of electrons only affects the overall hybridization if the penultimate $\rm d$ subshell has more than 3 or less than 8 electrons.
Source: Basic Concepts of Inorganic Chemistry.

I think your answer is exactly opposite to what i wrote, and the logic also seems correct. But according to you  [Ti(H2O)6]3+ must have hybridisation sp3d2 (4d).                                                                                click already registered .

Really? 4d subshell for titanium, a 3d metal? $-1$. Would minus more if I could.

@Jan, did you bother reading the 3rd paragraph? Also, there are 3d metals from iron onwards for sure which involve the 4d orbitals in complex formation, especially zinc.

The hybridisation concept is more than wrong in this case. I support @Jan's comment.

@TamoghnaChowdhury The 3rd paragraph did not contain any hint at the time of my comment that you wanted it to be understood as 3d, not 4d. Another thing I strongly disagree with, as a side note, is calling $\ce{NH3}$ a strong-field ligand. It is at best a very mildly stronger ligand than water.

@Jan that is true, definitely, but for most high school chemistry purposes NH3 behaves as a strong field ligand. Also, comments don't support LATEX formatting, and would you consider at least deleting your comment, if not retracting your downvote?

Of course comments support $\rm \LaTeX$ formatting; don’t lie.$$\ce{NH3}$$See, it works. $\ce{NH3}$ doesn’t behave as a strong field ligand in any purposes, since it is not strong enough to convert a high-spin complex into a low-spin complex; the typical definition of strong/weak field. Even a naive mention of 4d is wrong. I would also love to see your explanation of 4d orbitals for zinc complexes. All things considered, there is still enough in this answer that makes me keep my downvote and not retract it.

@Jan well at least they don't in the SE app on Android. Well, that's your decision to keep your voting, I won't argue with you. You're welcome for your feedback, which does help expand my concept of coordination compounds (I'm no expert, just a high school student). I won't be deleting this answer, but I'll restrict myself to questions more in my field in the future.


text:
What chemical measurements are related to the degree of flavor extracted by alcohol?
A recent project of mine includes making infusions of various ingredients to make liqueurs and bitters. Infusions, I've learned, vary widely in the amount of flavor extracted from an ingredient. Citrus zest, for example, can overbear most other ingredients in ways it wouldn't in similar amounts in other recipes.
To better understand the infusing process, I taste a series of infusions every few days, rating the perceived strength of their aroma and flavor. This is clearly subjective, and I'd like an objective, empirical means to check the amount of flavor compounds extracted from an ingredient over time.
The interest isn't in comparing different infusions, but in comparing a single infusion's strength at different times, and against human ratings. Here are a few questions I'd like to be able to answer:

Is this infusion done? Has it extracted as much as we can expect it to?
Has this really changed since yesterday?
Given this measurement, how strongly would my tasters rate this?

Here are some things I've considered:

A refractometer. (Presumably the flavor extracted is the form of dissolved solids, no?)
A proof hydrometer, though I'm unsure if we can reasonably expect proof to change much over the process.
Arduino biometric sensors for things like e.g. pH, though I'm again unsure of the expected relationship.

Are these, or any others, reasonable measures of an infusion's progress?

It’s really really hard because flavour, as perceived by humans, can originate from all kinds of molecules that generally have little in common (except for at least one carbon atom). Each of these methods suffer from you not knowing what the actual substance is, that you are isolating, and that creates a taste. NMR or IR (wait, did I actually just suggest IR?) may be the most promising attempts that help in identifying in some mediocre way, but they too work best on pure compounds, not really on mixtures.

Many of the compounds are volatile, so I'd suggest looking at what's in the headspace gas (GC comes to mind, but I'm not sure what the DIY/maker equivalent looks like, if there is one) of whatever vessel you're storing your solutions in. You'll have less water in the gas phase (and you'll know your solvent signature), so maybe @Jan's dreaded IR could be used without water dominating the spectrum where you want to look.

Thanks @Jan, that confirms my intuition about the difficulty. I'll read up on IR.

@ToddMinehardt Thanks, I hadn't considered headspace gas. A quick search suggests the DIY options for GC are pretty sparse, but I'll keep digging.

I don't think you understand the enormity of the question. Look at all the hoopla about coffee. How to choose the right beans, how to roast the beans,  how to grind the beans, how fresh the grind/roast must be, and finally how to brew the grind - all to get a cup of coffee. What you have to realize is that "coffee" is not A chemical, but a myriad of chemicals that are affected by all the choices made in producing the cup of coffee.

@MaxW Appreciate the feedback, though I think that's rather a more enormous question than I intended. To take up your coffee analogy, I'm not interested in anything so detailed as particular, individual chemicals in a blend: My immediate, practical interest ends with what quantity of compounds (whatever they may be) have been extracted in a particular cup. Incidentally, thanks to your comment I stumbled onto several discussions of comparing coffee using total dissolved solids. Cheers!

I obviously didn't make my point well. A "good" cu of coffee and a "bad" cup of coffee contain the same chemicals. It is the proportion of the chemicals that is different. So an alcohol extraction of vanilla beans isn't extracting one chemical but a multitude of them. So you need to measure what you do want,and what you don't want. Since there is a huge mixture of chemicals in such extractions I don't think instrumental testing is going to be terribly useful. I think you need to consider something more like the  Scoville taste test for capsaicin concentration.


text:
Alpha hydrogen and hyperconjugation
Why do we say that an alpha hydrogen is necessary for hyperconjugation? What exactly is an alpha hydrogen? From what I read online, it is the hydrogen attached to a functional group. Why is that required?
For example, in $\ce{CH3-CH=CH2}$, hyperconjugation takes place, but I don't see clearly any functional group that would facilitate it.
Alternatively, in $\ce{CH3-COOH}$ it does not take place.
Why is this?

To remember:
Hyperconjugation is the stabilising interaction that results from the interaction of the electrons in a σ-bond (usually C-H or C-C) with an adjacent empty or partially filled p-orbital or a π-orbital to give an extended molecular orbital that increases the stability of the system. Further information.
The proposed systems to take into account.
Looking at the resonance structures (which is a synonym of mesomeric effect) of the propene

(source: tutorvista.com)
and acetic acid

(source: nonsibihighschool.org)
Therefore, in hyperconjugation analysis you need to look at the position of the electron density. Not the vicinal hydrogen.
Moreover and with the aim to give a complete answer, the hyperconjugation of the acetic acid can be very well understood at here.

Your pic is for mesomeric stabilisation of acetate not hyperconiugation of acetic acid.

But  why does hyperconjugation not take place in acetic acid? Besides, why can only alpha hydrogen participate? Can you explain more about the electron density part?

@Shodai, I invite you to take a look at: http://chemistry.stackexchange.com/questions/27799/hyper-conjugation-in-acetic-acid

@Shodai, also, the relevance of the alpha hydrogen to a functional group with a carbonil is when you are doing a synthesis through an electrophilic substitution at this hydrogen. If this is what you want to know. I recommend you this site: https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/crbacid3.htm


text:
Use of aqueous KOH and alcoholic KOH in dehydrohalogenation reactions
What is the difference between aqueous KOH and alcoholic KOH? How do they react differently in dehydrohalogenation?

Aqueous $\ce{KOH}$ is alkaline in nature i.e. it dissociates to produce a hydroxide ion. These hydroxide ions act as a strong nucleophile and replace the halogen atom in an alkyl halide.
$$\ce{RCl + KOH (aq) -> ROH + KCl}$$
This results in the formation of alcohol molecules and the reaction is known as nucleophilic substitution reaction.
Alcoholic, $\ce{KOH}$, specially in ethanol, produces $\ce{C2H5O-}$ ions. The $\ce{C2H5O-}$ ion is a stronger base than the $\ce{OH-}$ ion. Thus,the former abstracts the ß-hydrogen of an alkyl halide to produce alkenes. This reaction is known as elimination reaction.
$$\ce{CH3CH2Br + KOH (alc) -> H2C=CH2 + KBr + H2O}$$

A good answer would also mention the order of each reaction (SN1, SN2, E1, or E2) for completeness.


text:
If a gas absorbs heat, then how can its temperature remain constant?
My book has this line:

In an isothermal expansion, the gas absorbs heat and does work

But, if a gas absorbs heat, then how can its temperature remain constant? (as the process is isothermal)

From the first law, $\Delta U=Q-W$.  If the process is isothermal, then $\Delta U=0$ (assuming an ideal gas), so Q = W.  The heat energy is consumed by your gas doing work, so there is no change in internal energy (temperature).

Hopefully I can provide you with some physical intuition on this phenomenon through a simple example.
Imagine a gas heated by a flame in a closed cylinder with a piston on the top balanced by the same pressure as the gas inside. As you heat the gas, the particles move faster (gain kinetic energy) and collide with the walls of the container harder, increasing the pressure of your system. This pressure increase is then used to drive the piston out. As the particles of your hot gas collide with the piston wall to move it, they necessarily dump some of their kinetic energy into it, slowing down (i.e. cooling) in the process. On the other hand, kinetic energy is not lost when the gas particles bump into the other rigid walls of the container, since those collisions are basically elastic.
So overall you have a heat source dumping kinetic energy into the gas particles from one side of the cylinder, and the gas particles dumping their kinetic energy into the surroundings while pushing a piston on the other side. Now run this process very carefully and slowly, and the temperature of the gas can be held more-or-less constant!

This answer is basically the same as the shorter answer by Chet Miller on Dec. 30 '15.

I was attempting to provide some physical insight into why the other answer makes sense, but I guess that isn't appreciated here.


text:
What is the most acidic proton of 2-methylcyclopent-2-en-1-one?
In an acid-base reaction of $\ce{HO-}$ with 2-methylcyclopent-2-en-1-one, which is the most acidic proton?

I had assumed it was one of them coming off C-5 according to IUPAC numbering (the sp3 carbon α to the carbonyl group). However, the answer book claims it to be C-4 (γ when going in the direction of the double bond).
I assumed C-5, since it seemed it would distribute the charge due to resonance more with the $\ce{O}$ atom. What am I doing wrong?

There are two possible enolates that can be formed from your compound: the α deprotonation (left equilibrium arrow in the scheme below) and the γ deprotonation. Both protons have roughly the same $\mathrm{p}K_\mathrm{a}$ values, so deciding which one is more acidic is non-trivial.

However, experimental procedures give us a hint. Usually, the α deprotonation is achieved with strong bases at low temperatures — kinetic conditions. Typically conditions such as LDA in THF at $-78~\mathrm{^\circ C}$. This leads to the conclusion, that the α deprotonation is kinetically favoured. This is more evident in open chain compounds and especially if the α carbon is a methyl group.
The γ deprotonation is generally achieved under more thermodynamic conditions: Weaker bases, warmer temperatures. I didn’t do it myself, so I cannot state conditions off the top of my head, but I would guess conditions such as $\ce{$t$BuOK}$ and $0~\mathrm{^\circ C}$. Therefore, this system must be the thermodynamically favoured one and the γ proton must be slightly more acidic. (But only a tad, or the kinetic product would not be so easily accessable.)
I cannot provide a proper explanation, but it would seem to me like the non-branched π system achieved by γ deprotonation is more stable than the branched one following α deprotonation.


text:
Units of rate constant (cycles per second vs radians per second) in Eyring equation
In the Eyring equation (EE),
$$k = \frac{k_\mathrm B T}{h} \exp\left(\frac{-\Delta G_{\mathrm f}}{RT}\right),$$
the units of $k$ are $\mathrm{s^{-1}}$. However, in general rate constants are usually expressed in $\frac{\mathrm{rad}}{\mathrm{s}}$. For instance, in expressions for damped oscillations of the form $\exp[(\mathrm{i} \omega - k)t]$, where $\omega$ is by definition in $\frac{\mathrm{rad}}{\mathrm{s}}$ and $k$ has to bear units providing consistent dimensions.
By expressing in the EE the energy $k_\mathrm B T$ as $\nu h$, where frequency $\nu$ is in cycles per second $\left(\mathrm{Hz}=\frac{\mathrm{cyc}}{\mathrm{s}} = \mathrm{cps}\right)$, it appears that $k$ is also given in $\mathrm{cps}$ in this equation. The desired units of $\frac{\mathrm{rad}}{\mathrm{s}}$ would be obtained if $h$ were replaced by $\hbar$. However, I have never seen the EE formulated that way. Has anybody an idea in this matter?

Related: Rate Constant Units and Eyring Equation

I don't believe this question is a duplicate. This question relates to the unit of the rate constant. The other SO question pertains to the order of the reaction and how the dimension of $k$ is accomodated for that.

Concur, this is not a duplicate. The problem of $\mathrm{Hz}$ units carrying an implicit "cycles" in the numerator $\left(\mathrm{Hz}=\frac{\mathrm{cyc}}{\mathrm{s}}\neq\mathrm{s}^{-1} \equiv \frac{ \mathrm{rad} }{\mathrm{s}}\right)$ is rife throughout chemistry, spectroscopy in particular. It explains why we use both $h$ and $\hbar = \frac{h}{2 \pi}$, for example.

slaw, radians are dimensionless units: $\theta$ in radians is the arclength in meters traversed by a point per meter of distance away from the center of rotation about which $\theta$ is applied.  So, $\mathrm{rad} = \frac{ \mathrm{m}}{\mathrm{m}} = 1$, and thus $\mathrm{rad \over s} = {1 \over \mathrm{s}}$.  That said, the units collision in your question stands unaffected.

See my answer to http://chemistry.stackexchange.com/questions/10115/rate-constant-units-and-eyring-equation which explains the units among other things.

I have looked at the original article of Henry Eyring (J. Chem. Phys. 1935, 3, 107). In that article, Eyring explains the derivation of his famous equation. It basically boils down to the following:
You assume an equilibrium between the initial and transition state and consider that once a species crosses the barrier, it goes to the product state:
$$R \leftrightarrows R^{\dagger} \rightarrow P$$
The rate expression is then simply:
$$k = \nu K$$
where $k$ is the reaction rate constant, $\nu$ the crossing frequency and $K$ the equilibrium constant between the IS and TS.
I will not go into detail about how to calculate $K$, but just remember that $K$ can be obtained by calculating the quotient of the partition functions of the system in the transition and initial state, respectively. How $\nu$ is calculated, will hopefully answer your question.
Let us assume that in the transition state, the bonds to be formed or broken are weak. In that case, we can model the reaction coordinate (the direction of the reaction) as a loose vibration. The partition function of a vibration is
$$
f = \frac{1}{1 - \exp \left(\frac{-h \nu}{k_{b}T} \right)}
$$
Herein, $\nu$ is the frequency of the loose vibration. We assumed that this vibration was very weak (loose). In that case we can apply a Taylor expansion to the exponent and cut off the series after the first term. This gives us:
$$
f = \frac{1}{1 - 1 + \frac{h \nu}{k_{b}T}} = \frac{k_{b}T}{h \nu}
$$
Plugging this result back into our original rate expression gives
$$
k = \nu \frac{k_{b}T}{h \nu} K^{\dagger} = \frac{k_{b}T}{h} K^{\dagger}
$$
Here, I have put a $\dagger$ after the equilibrium constant to make clear that I have extracted one partition function (that of the loose vibration) out of the equilibrium constant.
From the dimensional analysis of $\frac{k_{b}T}{h}$, you can see that $k$ should have units of $s^{-1}$ and not something else.

Except your last sentence is not the case: the units of $h$ are $\mathrm{\frac{J}{Hz}} \equiv \mathrm{\frac{J~s}{cyc}}$, which should provide units of $\mathrm{\frac{cyc}{s}}$ to $k$.

In the original paper of Planck (Ann. d. Phys.,  1901, 4, 553), the units for $h$ are given as $erg \cdot s$, which in S.I. would be $J \cdot s$. Since then, I believe we agree upon the dimensionality of this constant. For instance, IUPAC also defines it as $J \cdot s$. Where do these cycles originate from?

It originates in the difference between angular and cyclic frequency. A wave with cyclic frequency $f = 1~\mathrm{\frac{cyc}{s}} = 1~\mathrm{Hz}$ has an angular frequency of $\nu = 2\pi~\mathrm{\frac{rad}{s}}$. 'Cycles' are generally taken as dimensionless, and radians are strictly dimensionless; thus we end up with two sets of units that, propagated incautiously, are both written as "$\mathrm{s}^{-1}$" but that differ numerically by a factor of $2\pi$.

Sure. That I understand. But neither Eyring nor Planck actually mention these in their publications. To be honest, I also never have heard of this during my own education and this (the cycles part) is also not mentioned in the course books that I used. I also looked in the books of Pauling (Introduction to Quantum Mechanics) and the one of Glasstone & Eyring (The Theory of Rate Processes); it is also not mentioned in these. What am I missing?

I don't know. :-) It could be me that is missing something.  However, one specific example I know of where such tracking of 'cycles units' is crucial is in the derivation of the formula for molecular rotational constants: it explains precisely the factor of $8\pi^2=2\left(2\pi\right)^2$ in the denominator, and the presence of $h$ instead of $\hbar$ in the numerator.

To note, I think I reversed $f$ and $\nu$ two comments above; I think $f$ (or $\omega$) should be angular frequency, and $\nu$ cyclic frequency.

This whole discussion mystifies me.  What is the source for the claim that $h$ has "units" of Joule-seconds per cycle as opposed to just Joule-seconds?  The Planck time has units of seconds, not seconds per cycle, right?

@CurtF. Yes, but the Planck time incorporates $\hbar = {h \over 2\pi} \left[=\right] \mathrm{rad\over s}$ in its definition -- the factor of $2\pi$ is actually a conversion factor: ${\mathrm{cyc} \over 2\pi~\mathrm{rad}} = 1$.

@CurtF. The source is the expression for photon energies: $E = h\nu$.  $\nu$ in this expression bears $\mathrm{Hz}$ units, which are $\mathrm{cyc\over s}$.  In order to yield energy units, $h$ must be $\mathrm{J~s \over cyc}$.

In an attempt to resolve this debate: let us simply focus on the numerical value of the pre-exponential factor in the Eyring Equation assuming that all other contributions from the partition functions are unity. In other words, the pre-exponential factor is then $\frac{k_{b}T}{h}$. The numerical value at room temp. is according to Wolfram Alpha then $6.2 \cdot 10^{12} s^{-1}$. This is fully in line with several research articles also mentioning this numerical value.There is no factor of $2\pi$ difference..

Ivo, I'm not arguing that $h$ isn't what Eyring used in his development; or that $h$ hasn't been used continuously since then; or, even that $h$ isn't the correct value to use. I'm merely concurring with slaw that the resulting equation appears to have a subtle units problem when $h$ is present rather than $\hbar$.

For going to the effort of tracking down Eyring's original paper and working through a bunch of the math: BOUNTY!

The Eyring equation is numerically correct, despite the apparent units problem.
To understand the origin of the problem, one must go all the way back to the underlying statistical and quantum mechanics, since Eyring treated the motion across the transition state as being effectively a translation (J Chem Phys 3: 107, 1935, p109, emphasis added):

The activated state is because of its definition always a saddle point with positive curvature in all degrees of freedom except the one which corresponds to crossing the barrier for which it is of course negative. ... A configuration of atoms corresponding to the activated state thus has all the properties of a stable compound except in the normal mode corresponding to decomposition and this mode because of the small curvature can be treated statistically as a translational degree of freedom.

The starting point is the Hamiltonian for a particle in a box:
$$
H = - {\hbar^2 \over 2m}{d^2 \over dx^2} \tag{1}
$$
Additional to Eq. $\left(1\right)$, of course, is that the particle is confined by an infinite potential to the domain $x=\left(0,L\right)$.  After a standard undergraduate physical chemistry derivation, the wavefunctions $\Psi_n\!\left(x\right)$ of the time-independent Schrödinger equation $H\Psi_n=E_n\Psi_n$ are:
$$
\Psi_n\!\left(x\right) = \sqrt{2\over L}\sin{\left(n\pi {x\over L}\right)}\tag{2}
$$
Differentiating Eq. $\left(2\right)$ twice, substituting into the Schrödinger equation, and comparing the result to $E_n\Psi_n$ yields the following energy levels for the particle:
$$
E_n = {\hbar^2 n^2 \pi^2 \over 2mL^2} \tag{3}
$$
Eq. $\left(3\right)$ can be converted to the more commonly used form, that of Eq. $\left(53\right)$ of Salzman, by a simple substitution of $\hbar = {h \over 2\pi}$:
$$
E_n = {\hbar^2 n^2 \pi^2 \over 2mL^2} = {h^2 \over 4\pi^2}{n^2\pi^2 \over 2mL^2}
 = {h^2n^2 \over 8mL^2}\tag{4}
$$
This is where the units problem in the Eyring equation originates. The factor of $\pi^2$ in the numerator derives from the form of $\Psi_n$ and $H$, where the Hamiltonian requires twice-differentiating a sine function with the factor $n\pi$ in the argument.  This $n\pi$ is a fully unitless scaling factor for the non-dimensional position $x/L$ that is needed for $\Psi_n\!\left(0\right) = \Psi_n\!\left(L\right) = 0$ to hold, as required by the particle-in-a-box problem definition and the mathematical properties of the sine and cosine functions. I assume a key motivation for performing the transformation of Eq. $\left(4\right)$ is cosmetic, as it removes an apparently superfluous factor of $\pi^2$. But, it admixes into the overall expression the $4\pi^2 \rightarrow \left({2\pi\ \mathrm{rad} / \mathrm{cyc}}\right)^2$ factor in the denominator that is required to maintain the correct units downstream, obfuscating the dimensionality.
The next step is to obtain the translational partition function $q_\mathrm{t}$ which, per Eqs. $\left(54\right)$ and $\left(55\right)$ of Salzman, is:
$$
q_\mathrm{t} = \sum_{n\,=\,1}^\infty{\exp\!\left[{-{1\over k_\mathrm{B}T}{\hbar^2n^2\pi^2 \over 2mL^2}}\right]} \left\{= \sum_{n\,=\,1}^\infty{\exp\!\left[{-{1\over k_\mathrm{B}T}{h^2n^2\over 8mL^2}}\right]}\right\} \tag{5}
$$
$$
q_\mathrm{t} \approx \int_0^\infty{\exp\!\left[-{1\over k_\mathrm{B}T}{\hbar^2n^2\pi^2 \over 2mL^2}\right] dn} \left\{= \int_0^\infty{\exp\!\left[-{1\over k_\mathrm{B}T}{h^2n^2\over 8mL^2}\right] dn}\right\} \tag{6}
$$
In the above equations, I have provided the results for $E_n$ of Eq. $\left(3\right)$ first, with the final results using the $E_n$ of Eq. $\left(4\right)$ following it in curly brackets.  I will continue with this convention below as needed.
Per, e.g., Wolfram Alpha, the general form of the Gaussian integrals of Eq. $\left(6\right)$ is:
$$
\int_0^\infty{e^{-ax^2}dx} = \frac{1}{2}\sqrt{\pi \over a}
$$
Thus:
$$
q_\mathrm{t} \approx {L\over \hbar} \sqrt{m k_\mathrm{B}T \over 2\pi}
\left\{= {L\over h}\sqrt{2\pi m k_\mathrm{B}T}\right\} \tag{7}
$$
The bracketed expression in Eq. $\left(7\right)$ matches exactly the expression given by Eyring on p108.  Using manipulations I have not taken the time to retrace in detail, Eyring on p110 asserts the following expression for the prefactor of his now-eponymous equation:
$$
\left({\sqrt{2\pi m^* k_\mathrm{B}T} \over h}\right)\cdot {\overline{p}\over m^*} = {k_\mathrm{B}T \over h} \tag{8}
$$
This is made possible by derivation of the following expression (p110 and preceding):
$$
{\overline{p} \over m^*} = {k_\mathrm{B}T \over \sqrt{2\pi m^* k_\mathrm{B}T}} \tag{9}
$$
The expression in parentheses on the LHS of Eq. $\left(8\right)$ was apparently obtained by substituting $m^*$ for $m$ and setting $L=1$ (p108) in the bracketed expression of Eq. $\left(7\right)$:

If we set the length $l_i\!=\!1$ we have the number of unit cells per cm of length, a quantity frequently used in what follows.

Using instead the unbracketed expression of Eq. $\left(7\right)$, transformed as in Eyring, with Eqs. $\left(8\right)$ and $\left(9\right)$ gives:
$$
{1\over\hbar}\sqrt{m^*k_\mathrm{B}T\over 2\pi}\cdot{\overline{p}\over m^*} =
{1\over\hbar}\sqrt{m^*k_\mathrm{B}T\over 2\pi}\cdot{k_\mathrm{B}T \over \sqrt{2\pi m^* k_\mathrm{B}T}} = {k_\mathrm{B}T \over 2\pi\hbar} \tag{10}
$$
Thus, the result of Eq. $\left(10\right)$ is numerically equal to the prefactor reported by Eyring. So, happily (and unsurprisingly!), none of the work performed with it in the last eight decades needs to be revisited.  However, the cycles units discrepancy in Eyring's version arises because, as noted above, the factor of $2\pi$ appearing in the denominator of Eq. $\left(10\right)$ is a consequence of differentiation of the trigonometric translational wavefunction $\Psi_n$ underpinning the prefactor derivation, and not present as a conversion factor of $2\pi\ \mathrm{rad}\over\mathrm{cyc}$.

In the derivation of the Eyring equation (EE) quoted above by Ivo Filot, substitute consistently $\nu$ by $\omega = 2\pi\nu$ and $h$ by $\hbar$ wherever these two quantities appear. This will do no harm to the entire reasoning, but in the final equation the prefactor would be $$\frac{k_{b}T}{\hbar}$$ and $k$ would be expressed in $rad/s^{-1}$, as it should be. I dare to claim that this would  be the correct answer to the problem. I do not have enough courage to submit such a note to a peer-reviewed chemical journal indexed by ISI.

This unfortunately doesn't work numerically: $h\nu = \hbar\omega$; but $h\nu \neq \hbar\left(2\pi \omega\right)$. If one were to substitute $h=2\pi \hbar$, the final prefactor would be $k_\mathrm{B}T \over 2\pi \hbar$. My hunch is that there is a factor of $2\pi$ in the denominator that arises somewhere in the manipulation of the translational partition function, that was numerically (but not dimensionally) sensible to combine with $\hbar$ to yield the $h$ of the EE. My stat mech isn't strong enough to retrace the derivation, though.

Thanks for joining Chem.SE to post an answer, though! :-)

It's OK, @Brian. If in the standard EE $h$ is substituted by $2\pi\hbar$ then the rate constant in the l.h.s. will keep its units, i.e., Hz or cps, unchanged. Then, multiplying both sides of EE by $2/pi$ will remove the offending term $2\pi$ in the denominator of the prefactor while $k$ will be converted to $k' = 2\pi k$, where $k'$ will be in rad/s.

I have converted your other answer to a comment here, please have a look at the [help] for more information.

Thanks for giving it a shot, user24239, but this answer just notes what one might do to alter the units of Eyring's expression. It doesn't reach into Eyring's derivation, such as calculation of the partition function &c., to explain the presence of $h$ and how one might resolve the apparent units conflict.

@Brian - your explanations appear to resolve the unit problem. However, in the original Eyring paper of 1935 quoted by you the units of $k_{1}$ in equation (2) apearre cc molecules${-1} s$-1$ (2\pi)^{-3/2}$

I'm unclear on the point you're trying to make -- if one were to substitute $h=2\pi\hbar$ into that equation, the term with $\pi$ would end up as $\left(2\pi\right)^{1/2}$, in which case it would presumably be a fully non-dimensional term likely arising from some manner of mathematical origin.  The units are then $\mathrm{m^3\over molec~s}$, just as one would expect.


text:
Why is the electrode with the lower reduction potential the cathode?
Wouldn't it make sense that the electrode with the higher reduction potential be the cathode, since reduction occurs at the cathode?

No you misunderstand the meaning of reduction potential. Reduction potential refers to the voltage required to reduce a material under standard conditions. If a material has a higher reduction potential it takes more energy to reduce it than a lower reduction potential material. Therefore the higher reduction potential material is actually oxidized to reduce the lower reduction potential material.

A solution with a higher (more positive) reduction potential than some other molecule will have a tendency to gain electrons from this molecule (i.e. to be reduced by oxidizing this other molecule) and a solution with a lower (more negative) reduction potential will have a tendency to lose electrons to other substances.

This response does not add substantially to the previous ones.

Don't you notice that the previous answer is wrong? The answer is YES.

The reduction potential of copper is 0.52V and the reduction potential of Zinc is -0.76V. Copper is the Cathode and Zinc is Anode in a Cu-Zn Sulfuric Acid battery.


text:
Product of lead and tin alloying
It is common knowledge that tin melts at $449.5~\rm^{\circ}F$ and Lead at $621.5\rm~^\circ F$, and that when these are melted together you simply have an alloy.
What happens, or what product occurs, if you take pure lead to red heat ($1,400$ – $1,500~\rm^\circ F$) in a muffle furnace and cast in pure tin?
The mixture "Blooms" with somewhat small cubic crystals that, when scooped off the top into a receptacle, appears as a sulfur colored scoria that I was not able to re-melt into metal at even $2300\rm~^\circ F$.
I have a photo if I can figure out how to post it.

Must be some oxide.

You're probably creating a toxic miasma: though that is below the boiling point of lead, some would evaporate.

What is the material of the crucible you're using?

I think you have insufficient reputation to insert images yourself.  Post your photo to the image sharing service of your choice, and edit your question to include a link to it. One of the more established members of the community can then edit your post to include the image directly.

You're all very kind. I'm using a ceramic dish to molten the metals and a clay crucible as a receptacle for the material I gently scrape from the top of the molten metal. I'm trying to understand you're instructions for posting a photo but it's over my head experience-wise. I can say that the "Bloom" occurs quickly and the surface of the molten metal is completely covered with the forming "Crystals" which are much brighter than the molten metal, so much so that they appear incandescent.

I concur with Ivo's answer, that @IvanNeretin has it right. It seems unlikely that the melt could be extracting something from the ceramic dish to form the bloom at a high rate.

I think Ivan Neretin has given the right answer.
Any alloy between Sn and Pb melts at relatively low temperatures. You stated that the compound you fabricated did not even melt at 2300 °F. I have been looking at the stable lead oxides and tin oxides. The only compound of this set that has a melting temperature above 2300 °F is tin dioxide at 2970 °F.
Please note that I cannot be certain of this, as I only checked a small set of compounds I found to be the best candidates. It could be the case that you have created something more complex. (beautifully phrased by DrMoishe Pippik as a 'toxic miasma' :-) )


text:
Can we break the covalent bonds of water molecules using resonance frequencies?
Can we break the covalent bonds of water molecules $\ce{H2O}$ using resonance frequencies? What is the wavelegth of the water's molecule? Can we create a vibration like this and use it to separate hydrogen from oxygen?

Yes this is possible. Your question does not state the source of the frequency, but in this internet article they have used laser light.
I am not sure at what frequency you need to irradiate water to break the O-H bond, but I would speculate that a good frequency would be close to the IR stretching frequencies of the O-H bond in water.

I think a better frequency would be one that excites electrons …

You'll need to get above the ground electronic state, as @Jan notes. And I think the OP needs to chime in with respect to the part about separating hydrogen and oxygen: do you mean as charged species, radicals, diatomic gases, or what?

@Jan. That was also my initial guess, but as soon as you start exciting the electronic levels, the geometry of the molecule gets distorted and you are no longer 'pumping' at the right frequency to break the bond. I don't think it is possible then to break a specific bond. That's why my suggestion was to slowly 'pump' the vibrational levels until the bond breaks. In other words, you would like to stay as close to ground state chemistry. I haven't found the article though where the procedure is given.

In the article, they excite surface Si-H bonds. I suspect that being on a surface really helps prevent transfer of energy to other motions by collision. In liquid H2O, I suspect you would just heat the solution rather than break bonds very often, though maybe there's a laser that could deliver energy fast enough to overcome this.

The reference is a press-release without details provided, so I highly suspect there is a lot of gotchas that might considerably change the meaning. I particular, I want to point that molecular vibrations are unharmonic and at higher energy levels the vibration levels (and associated frequencies) deviate from expressions for harmonic oscillator.

Yes, but not through direct excitation into a dissociated state. This is explained in Ref. 1:

Unfortunately,  measuring D0 by  direct photoexcitation  of  H2O  from  its  vibrational  ground  state  to the  dissociation  threshold  is  prohibited  by  the  negligibly small intensity of such an overtone transition.

Instead the authors accomplished the measurement  by exciting the molecule through a sequence of rovibrational states:

We have performed a direct measurement of one of the most fundamental thermochemical values: the O-H bond energy in water. Using a triple-resonance laser excitation scheme, we excite the molecule through a series of vibrational overtone transitions to access directly the onset of the dissociative continuum. The dissociation energy obtained from our experiments, $\pu{41145.94 ± 0.15 cm−1}$, is ∼30 times more accurate than the currently accepted value and has important implications for other thermochemical quantities linked to the bond energy of water.

Please see the cited reference (behind paywall) for more information.
References

Pavlo Maksyutenko, Thomas R. Rizzo, and Oleg V. Boyarkina). A direct measurement of the dissociation energy of water. J. Chem. Phys. 125, 181101 (2006)


text:
Self aldol condensation of butanal under basic conditions
The reaction is such:

I've counted the number of carbons on each side: LHS=4, RHS=8
My working is such:

The reaction is a self aldol condensation. The enolate of the aldehyde attacks a molecule of un-enolised aldehyde. The resulting intermediate eventually loses water to give your product, as drawn.   Give the curly arrows a try and post what you end up with :)

You do see that your last intermediate would not arrive at your product, don’t you?

Conceptually, we can think of this reaction as occurring in three stages. I've provided diagrams for the self-condensation of acetaldehyde, however you should be able to draw an analogous mechanism for the self-condensation of butanal (your aldehyde).
Your mechanism is essentially correct, except for the final elimination step, which as you've drawn it gives the wrong product.
1. Enolisation of the aldehyde

Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

As you guessed, the hydroxide is basic enough to cause enolisation of the aldehyde. Using KOH/MeOH, we form the enolate under thermodynamic control, essentially meaning the enolate is in equilibrium with the aldehyde.
2. Addition of the enolate into another molecule of aldehyde

Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

The newly formed enolate may then attack an un-enolised molecule of the aldehyde (this is okay, since we've established that the aldehyde and the enolate are in equilibrium under these conditions). This nucleophilic attack leads to a beta-alkoxy aldehyde (the structure on the right), which will quickly pick up a proton to form a beta-hydroxy aldehyde.
Under these basic conditions (KOH, MeOH), we may be able to isolate the hydroxyaldehyde, however, they are prone to elimination (see below), which is what must happen to form the product you require.
3. Elimination of hydroxide

Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

The beta-hydroxyaldehyde formed (see above) is itself able to undergo enolisation, to form a new enolate. Rather than attacking another molecule of aldehyde intermolecularly, it may instead do an intramolecular E1cb type elimination, kicking out hydroxide (which is okay under these basic conditions), and giving you the product as drawn.
Your elimination step isn't the most favourable pathway to get rid of that alcohol, and infact the elimination only occurs under these conditions due to assistance form the carbonyl. The quote below puts this rather well.

You cannot normally eliminate water from an alcohol in basic solution as hydroxide is a bad leaving group. It is the carbonyl group that allows elimination here: these are E1cB reactions.
Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

In your last elimination step the OH- should attack the hydrogen on the carbon between the alcohol and the ketone.
This then forms an enol, which then reverts to the ketone. This pushes the pi-bond across one carbon. The alcohol then leaves.
You should end up with almost the same molecule you've put down, but the alkene should be one to the left.


text:
How can I make a fairly saturated copper (II) sulfate solution by heating?
I have an investigation to do for my chemistry class. I'm looking at copper complexes and using ammonia to create a precipitate in the copper (II) sulfate solution, then redissolving the precipitate to complete the ligand exchange. When I tried the experiment the first time, I was unable to make a good solution. I dissolved about $3.5$ grams of $\ce{CuSO4·5H2O}$ into $200$ mL of water, and couldn't get any more to dissolve. I titrated in $1$ mL of ammonia and the solution turned cloudy (I assumed this was the precipitate). After $4$ mL I stopped, as the solution hadn't changed at all. This was all carried out at room temperature.
I know the two best ways to make a more concentrated solution are time and heat. I don't have that much time, as I'm on winter break and I'll need to complete the experiment as soon as possible when I go back next week. It would probably take a week to get the solution concentrated enough, so I've decided to use heat to help the dissolution along. After looking around the internet, I found very little that was helpful. One page outlined how to make a supersaturated solution, which was not what I wanted. It advised heating to $100^{\circ}$ Celsius, so I suspect that will be too hot for my experiment. A video done by students showed them making a solution at $50^{\circ}$ Celsius, but they used copper (II) oxide. I'm not sure if that makes the procedure significantly different.
I'm not sure what temperature is necessary for this to work. If all else fails, perhaps I'll change my topic to an investigation of the relationship between heat and solubility rates. I'll finally know how to make a saturated solution in a reasonable amount of time.
Other notes: I'll be using a hot plate (Fisher Scientific, 11-500-4H) for the heating, not a Bunsen burner. The ammonia is stock/commercial, and the concentration isn't on the label.
Thanks in advance.

You're much below what the solubility of the pentahydrate should be. // Copper (II) sulfate will form a hexammine complex. Initially, the light blue hydroxide precipitates, but with the addition of more ammonia the precipitate will dissolve.

copper sulfate dissolves slowly. I suggest to move to chloride if possible and work with smaller volumes (25-50 ml at most for test experiments). Also, please post concentration of ammonia solution you used.

@MaxW I know the precipitate will form. That's what I'm looking for. It's useless, though, if the precipitate appears as soon as I add a few drops of ammonia. I need a solution with a higher concentration.

@permeakra Do you mean copper chloride? And I said at the end that the concentration of the ammonia isn't on the label. I'll be able to determine the concentration through the titration, because the reaction for the formation of the precipitate and its dissolution is known.

@kierlani Yes, I mean copper chloride. And with no knowledge of ammonia concentration it is possible that you didn't use enough of it. To my knowledge, a rather concentrated solution of ammonia is required.

I suggest to use a combination of heating, dissolving, shock cooling and filtrating off the precipitate.
As was already stated, copper(II) sulphate dissolves slowly and is better soluble in the heat. The exact temperature doesn’t matter, but why not just go up to $80$ or $100~\mathrm{^\circ C}$? This will also make the dissolution faster. You can look up the solubility of copper(II) sulphate to calculate how much you would need and use some more to be on the safe side.
Then, cool the solution down rather quickly (but not too quickly or the glass will crack). This will initially create a supersaturated solution but the supersaturated bit will crystallise out rather quickly. You can filter off this crystalline precipitate to get your saturated copper(II) sulphate solution. Do not filtrate too quickly, or the solution will still be too warm and slightly supersaturated.
Then finally remember that if you add too much ammonia, rather than a precipitate of $\ce{Cu(OH)2}$ you might get the redissolved $\ce{[Cu(NH3)4(H2O)2]^2+}$ tetraammindiaquacopper(II) complex. You can tell this by the intensive dark blue colour.

Thank you so much. This is just what I needed. I knew the supersaturated solution would naturally precipitate for equilibrium, but it didn't occur to me to filter that out. And actually I'm titrating to both the precipitation point and the ligand exchange point. (I'm ridiculously happy about this answer, thank you.)


text:
1-5 dicarboxylic acid to lactone with SOCl2?
I came across a reaction whereby a 1-5 dicarboxylic acid is converted to lactone with $\ce{SOCl2}$. I was wondering what the mechanism of this was.
I know that $\ce{SOCl2}$ converts carboxylic acids to acyl chlorides. Is it the case that one carboxylic acid undergoes this transformation, then the oxygen from the $\ce{OH}$ of the other carboxylic acid attacks the acyl chloride, making the $\ce{Cl}$ a leaving group?
Is this the case or am I missing something?

If you reacted a carboxylic acid with an acyl chloride you would get an acid anhydride, not a lactone (which is an ester).

Are you sure it was a lactone and not a cyclic anhydride? I could see the latter as more likely. Could you provide a source or a link?

Thank you so much, turns out it was a cyclic anhydride. I was happy with the mechanism but getting the naming of the functional group wrong made it very hard to look up if I was correct. Thanks again!

The simplest 1,5-dicarboxylic acid is glutaric acid (pentanedioic acid in a more systematical naming). If you react it with $\ce{SOCl2}$, you will indeed intermediately arrive at a compound that has one carboxylic group and one acid chloride group. Then, as you suggest, the other carboxylic acid can attack the acid chloride and $\ce{HCl}$ is eleminated.
However, the resulting product is not a lactone but rather glutaric anhydride; a cyclic acid anhydride. To convert it to a lactone you would need to remove exactly one carbonyl group. Whether this is done before or after the reaction with $\ce{SOCl2}$ does not matter, but before sounds easier to me.


text:
Which statement is correct about electron orbitals and energy levels?

Which statement is correct about electron orbitals and energy levels?
A. Yttrium, $\ce{Y}$ ($Z = 39$) is the first element in the periodic table with an electron in a f sub-level.
B. The maximum number of electrons in one d orbital is 10.
C. The maximum number of electrons in the 4th main energy level is 18.
D. In a main energy level, the sub-level with the highest energy is labelled f.

I thought B and D are all possible solutions to this answer.
B--the maximum number of electrons in 1 d orbital is 10.
D--in main energy level, the sub-levels in order of increasing energy is s $<$ p $<$d $<$f
So why is the answer only D?
Please advise and sorry in advance for the wrong tags or title name, as I am new to the group and am trying to improve on it.

You mixed up a d orbital with the d subshell. One d subshell contains five d orbitals.

@orthocresol, so for B, the maximum number of electrons in one d orbital should be 2?

Meowth, that's right!

If you really want to blow your mind (and, incidentally, prove D as technically sort-of incorrect), look into g-orbitals, and h-orbitals, and i-orbitals, and ....

Although @orthocresol got this first, and it should be credited to him, I'm posting it as an answer.
In option (B), the statement is incorrect as it refers to $d$ orbitals, which can (according to Pauli's Exclusion Principle) hold 2 electrons each, rather than the $d$ subshell, which can hold 10 electrons, as it has a total of 5 $d$ orbitals which can hold 2 electrons each, and $5*2=10$.
Thus, only option (D) is correct.


text:
Enthalpy change and equilibrium
I don't understand why enthalpy change won't change at an equilibrium?
The reason I ask this is because of this question:

At first I thought the answer is C, but I realised that the amount depends on where the position of equilibrium lies. The correct answer is D, but doesn't enthalpy change (delta H) also depend on where the position of equilibrium lies?

(C) can be incorrect as amount might refer to mass formed, which won't be the same at all.

Trick questions. (D) is correct since the equilibrium position is reversed as the reaction itself is reversed, so the enthalpy change is also reversed (by your own logic).

@TamoghnaChowdhury, in chemistry the word amount refers to number of moles, not mass. C would be correct only if the system were to begin with reactants and no products, so it is not always true, as stipulated by the question.

@Rafael while your first point is likely to be correct (depending on textbook localizations and whatnot), the second is unnecessarily contrived - equilibrium implies no net progress of the equation in either direction, microscopic transient fluctuations in the amounts of the reactants and products can always occur.

The answer should be D. Because of the principle of microscopic reversibility, the forward enthalpy is opposite to the reverse enthalpy. This also follows from the conservation of energy if you will.
In this StackExchange question, there is a nice picture given of a potential energy surface. There you can readily see that the forward reaction enthalpy is opposite to the backward reaction enthalpy.

The answer to this question is related to the definition of $\Delta H^0$.  It is precisely defined as the enthalpy change between the following two thermodynamic equilibrium states:
State 1:  Pure reactants (say, in separate containers) in stochiometric proportions at 1 bar and 25C
State 2:  Pure products (say, in separate containers) in corresponding stochiometric proportions at 1 bar and 25 C
From this, you can see that the enthalpy change from State 2 to State 1 is minus the enthalpy change from State 1 to State 2


text:
What is the typical range for the molar absorptivity of paint?
For example, $\ce{TiO2}$ is used a lot as pigments in paint, what is the range of molar absorptivity for that? Would I be able to determine this from this graph:

Ideally I could estimate a range for acrylic paints. I really just need a range, and it only needs to be as accurate as a few orders of magnitude.

Well, I cannot say much from the above. Estimation of molar absorptivity experimentally would involve some measurements with different concentrations. I think single measurement may not give that information.
$TiO_{2}$ if present in paint then it is only one component ( ~10-40% or more maybe). Other components may affect the molar absorptivity too.
 If you want to find molar absorptivity of paint (with several components combined) you can just measure paint highly diluted in a solvent  and measure UV-spectra at different known concentrations and use the Beer-lambert law.
For molar absorptivity of only $TiO_{2}$ I have different proposal.
If you have some
$TiO_{2}$ then you can weigh 5 different amounts of it. Then dissolve it in acid (sulphuric acid or mixture with hydrochloric acid or you can check yourself which is better ) to make 5 solutions (should be dilute solution ). You know the concentration for these solutions.
Then, measure absorption with UV spectrometer.
Finally, use the Beer-Lambert law to get the molar-absorptivity based on the absorption peak (~350 nm).
Anyway this is a weird idea. (I am not sure if absorption of acid species, say sulphate ions will affect this measurement.)
And if you really try, let us know.


text:
How to calculate denticity of ligands?
When looking at an ligand, how do I know its denticity? For instance, I know that though there are 2 lone pairs in $\ce{H2O}$, its denticity is only 1 due to geometrical reasons as the 2 of them can't be facing the metal ion at the same time. However, when I look at other ligands, especially unfamiliar ones, I cannot determine their denticity.

You can never tell for sure. A ligand may have different denticity in different complexes.

I also want to point out that even if water used both lone pairs, it would still be monodentate because only one atom in the ligand is bonded to the metal. Denticity describes how many different atoms in the ligand are bonded. If one atom uses two lone pairs, then we have denticity 1 and bond order 2. See http://chemistry.stackexchange.com/questions/7366/why-cl%E2%81%BB-cant-act-as-bidentate-ligand/7368#7368


text:
what contributes to the temperature of a vacuum
what is the temperature of Vacuum since temperature of a system is related to the average of the molecular kinetic energy and there is no molecule in a vacuum? i know there could be radiations but i want to know how much the contribution is?

There are no perfect vacuums, either man-made or in nature. So, there are particles (and as pointed out below, photons) in the volume.

This is not about chemistry...

I'm voting to close this question as off-topic because it's about physics, not chemistry.

It really depends on how the vacuum is defined. If you, for instance, pump all the air out of a steel container, you will have a vacuum in there. Yet, there will be photons constantly radiated off of the walls. This interplay of photons will be in thermal equilibrium with the walls and therefore will have a contribution to the temperature. For instance, if you would place a thermometer inside this container, the material of the thermometer would be hit by the photons of the container wall and therefore the thermometer would give a temperature equal to the temperature of the wall. As a matter of fact, even in the deepest of deep space, there is some radiation to be found which gives a temperature of about 3K. (This radiation is left over from the Big Bang)
Nevertheless, if you make a true vacuum inside some container of which the walls are at 0 K that cannot be penetrated by any kind of radiation or matter, then the temperature therein is 0 K.

There's no perfect vacuum and nothing has 0 Kelvin

Well... yes and no. Indeed, the Casimir effect would prevent the creation of a true vacuum, yet see this Science article for an example of a sub-zero Kelvin temperature. :-)

Heard about it earlier and well it's not "true" temperature.


text:
Dissolving in DMSO semantics
What does it mean when a substance (powder) is 'dissolved in DMSO and used at $\mathrm{5~\%}$'
Does is mean that the substance is first dissolved in DMSO and then added to another solution where it makes up $\mathrm{5~\%}$ of that total solution?
Or does it mean something else?

When adding a solute to a solvent the only way that a "5% solution" makes sense is by weight percent. When mixing two liquids it could mean "volume percent" (vol. %) or "weight percent" (wt. %) so the writer should specify which is intended.

See my comment on the answer below

Typically, the chemical substance in question is first dissolved by adding 100% DMSO - to some extent for practical reasons, i.e. all the advantages of working with stock solutions. The main reason (to my mind) is, however, that compounds often do not dissolve easily in (aqueous) solutions of DMSO for kinetic reasons, but are perfectly happy when diluting with water afterwards.
For your downstream application, e.g. some biological or biochemical assay,  the 100% solution is diluted with the assay buffer to e.g. 5% DMSO. The latter is almost exclusively volume per volume%.

A good author should express whether percents are weight $(wt./wt.)$, volume ($vl./vl.$) or weight/volume ($wt./vl.$).  This unfortunately does not happen often. Typically for solids dissolved in liquids the percent is weight/volume given as :
$$\frac{weight}{volume}$$
and expressed as:
$$5\% (wt./vl.)$$

Thanks but I am confused as to whether 'dissolved in DMSO and used at 5 %' means that the powder is first dissolved in DMSO and then that liquid is used in an aquas solution at 5% (vl/vl) OR if it means the powder makes up 5% (wt/vl) of a DMSO based solution.


text:
Is it possible to have accumulation of electron density in the inter-nuclear region even if the overlap integral is zero?
Recap:
Lately, I've been studying Hückel theory where I learned about the approximation of neglecting all overlaps and making overlap matrix an identity matrix that is $S_{AB}= 0\;_; \;S_{AA}= 1$ or $S_{ij}= \delta _{ij}\;.$ He then replaced the off-diagonal Hamiltonian elements with parameter viz. $H_{AB}= \beta_{AB}\;_,$ where $\beta_{AB}$ is a negative quantity.
Question then arose if the overlap is zero as is evident from $S_{AB}$ being zero, how can there be a non-zero value of $H_{AB}$. It was known to me that $H_{AB}$ is the

contribution to the energy due to the accumulation of electron density where the two atomic orbitals overlap, including, for instance, the Coulombic attraction between the extra accumulation of electron density and both nuclei.

So, it is contradictory to simultaneously have the condition $H_{AB}\ne 0$ with $S_{AB}= 0.$
I asked this querying for the same; there I got a nice answer from Brian where he showed that even if $S_{AB}=0$; it need not be necessary to have $H_{AB}= 0$; the graphs he provided in his answer prove this otherwise a counterintuitive fact.

Present thinking & associated query:
Brian's answer confirms the fact that $S_{AB}= 0$ and $H_{AB}\ne 0$ can go with each other. But doesn't that mean there is no need to have a net overlapping of AOs between the internuclear region in order to have accumulation of electron density in the internuclear region?
For example, while using Simple Hückel Theory for ethene, Atkins put all $S= 0$ as usual per the approximation and then wrote:

[...] Finally, as explained in the derivation, he also supposed that the terms $H_{AB}$ and $H_{BA} ,$ which represent the energy of the interaction of the two nuclei with the accumulation of electron density in the internuclear region .......

Really, if he writes this after considering $S_{AB}= 0\;_,$ doesn't that mean there can be accumulation of electron in the internuclear region even if the the participating AOs are orthogonal and there is no net overlapping? He is considering the case of ethene $\pi$-network which means there is electron density that leads to $\pi$-bonding even if $S=0$.
So, is it right to conclude that there can be accumulation of electron density in the internuclear region even if there is no net overlap in the atomic orbitals?

Yep, definitely better as a new question. :-)

@Brian: Also, I did ponder a while & wondered how Hückel actually made this approximation of $S= 0$; making this would make the basis sets orthogonal. But orthogonality of AOs depends on the geometry of the molecule, isn't it? I googled a bit & didn't found anything regarding this. Nevertheless I do have a book Computational Chemistry by Errol Lewars which briefly mentioned it citing  'a twisted allyl species'. So, definitely, Hückel Theory is really eerie: he didn't consider those overlaps & didn't pay heed to the geometry of the molecule:/

Oh, certainly Hückel's bases had $S_{AB}\neq 0$ in actuality -- this was higgsss's point in the comments on my answer at the other question, I believe.  In order to make for a tractable calculation, though, I think Hückel just made his approximation by editing the matrices to coerce $S_{AB}=0$. Absolutely, this makes the mathematical problem diverge from the physical system, but in many cases it still gives qualitatively reasonable results.

Using approximations to simplify calculations and having the given parameters physically zero are two completely independent problems. If neglecting S makes no big difference in the results, or simple parametrisation can mostly compensate this simplification, it is justified irrespective if the actual S is zero or not.

If you define overlap as given by the overlap integral,
$
S_{ij} = \langle \psi_{i} | \psi_{j} \rangle
$
than you can have an accumulation of electron density in a specific part in space even though $S_{ij} = 0$.
Of course, if the complete wave functions of the two AOs are completely non-overlapping in all space, for instance because the centers of the two AOs are seperated by a long distance (say more than 5 A), then there can be no accumulation of electron density anywhere. This is of course logical. There is hardly any accumulation of electron density between two atoms that are quite far apart.
An interesting scenario exists where two wave functions have opposite sign and an overlap of these wave functions actually gives a decrease of electron density. This is termed deconstructive interference opposite to constructive interference when there is an accumulation of electron density.

Thanks for answering the question; so that means even if the net overlap is zero, if there is, at some place, non-zero $\psi_i\psi_j$, it can have a net  Hamiltonian $H_{ij},$ is it so?  Also, in ethene, the $p$ orbitals do overlap to give $\pi$ bonding; although Hückel put $S=0$.

Another thing I want to ask you is: if the the sum(integral) of $\psi_i\psi_j$ is zero, even though there may be places where $\psi_i\psi_j$ is non-zero; does that mean even $\psi_{i} \hat H\psi_j$ is non-zero at some places, but the net sum is zero since the net overlap is zero? Brian in his answer has shown that this doesn't necessarily happen. So, can I conclude that $H_{ij}$ gives the average energy of those $\psi_{i}\psi_{j}$ which are non-zero?

There are two things to keep in mind here; you have the product of two wave functions $\psi_{i} \psi_{j} (\vec{r})$ and you have $S_{ij} = \langle \psi_{i} | \psi_{j} \rangle$ which is the integral over all space of this product. Even if this $S_{ij} = 0$, the former can have a non-zero value at a specific $\vec{r}$. Obviously, if the value of $\psi_{i} \psi_{j} (\vec{r}) = 0$ at all $\vec{r}$, then the integral will be zero as a consequence of that.

Furthermore, if $\psi_{i} \psi_{j} (\vec{r}) = 0$ at some $\vec{r}$, then  $\psi_{i} \hat{H} \psi_{j} (\vec{r}) = 0$ as well. Because if there is no overlap density at some point in space, then there can be no energy density at that point. This is a consequence of the following: If $a \cdot b = 0$, then either $a=0$, $b=0$, or $a = 0$ and $b=0$.