How can I check the sodium content of food? What is an easy (i.e., home project) way to measure sodium in food?

For example, the nutritional label of a tahini brand that I often consume shows very little sodium (around 10mg/100g). I suspect the sodium content is much higher.

What is the chemical way to measure the salt content?

Sodium is a difficult ion to estimate at home.

Since most sodium salts are water-soluble, colorimetric or spectroscopic methods are often employed, both of which I assume are not within arms' reach.

Some alternatives:

Speaking of sodium estimation in foodstuff, it is usually determined by Mohr's
method which involves the precipitation titration of $\ce{Cl-}$ ions. This is done under the assumption that sodium is mainly present as sodium chloride. Additionally, it will also register $\ce{Cl-}$ from other chlorides, and hence may give an elevated reading. You might be able to do this at home. Here is a page
to help you with the process.

It is done in the presence of chromate ion ($\ce{CrO4^2-}$), and $\ce{Ag+}$ ions are added gradually. This precipitation happens initially:

$$\ce{Ag+ + Cl- <=> AgCl(s)}$$

When the solution runs out of $\ce{Cl-}$ ions, precipitation of the red-brown $\ce{Ag2CrO4}$ begins:

$$\ce{Ag+ + CrO4^2- <=> Ag2CrO4(s)}$$

This signals the end point.
There are sodium ion meters, you can get them for ~300$.
There is a volumetric estimation[1] for sodium directly, but it is not expedient to do it. It is based on the precipitation of sodium zinc uranyl acetate.

Conclusions

IMO it is difficult to do this as a home project, as it will require stuff that is not usually found at home to do it precisely.

References

[1] J. Am. Chem. Soc. 1931,  53, 9, 3288-3291 DOI: 10.1021/ja01360a011


Most probable point for finding an electron in the 1s orbital of a Hydrogen atom My professor says that the most probable point for finding an electron
in a 1s orbital of a hydrogen atom is at its origin.

He explains this by citing the fact that the square of the wave function which gives the probability density is maximum at the origin.

At the same time, we all agree that the Bohr radius is the distance at which probability of finding the electron is maximum for 1s orbital.

I can understand the reasoning he gave about the origin being the highest probable point for finding an electron, but then why are the graphs of the radial distribution function so?

It makes a difference if you look in a point or in a volume element.

For the radial distribution we sum up the probabilities within (very thin) shells at different distances from the core.
If the radius get's bigger the volume in this shell gets bigger.
Remember that the surface of a sphere goes by $4 \cdot \pi \cdot r^2$ so the volume of the shell is essentially for all practical purposes $4 \cdot \pi \cdot r^2 \cdot \Delta r$.
At the core the radius is zero which means the volume is zero and thus the probability is zero too.

At the same time the probability of finding the electron in a point at a certain distance goes down, but the volume goes up.
Because of the scaling of the volume and the "decay" of the probability to find it in a point further away it turns out that there's a maximum of finding the electron at a certain distance away from the core (but not a single point at that distance).


Calculation of pH in acid base reaction This problem(from ionic equiibrium) states:

We are given $\pu{40 mL}$ of $\pu{0.1 M}$ $\ce{HCl}$ and $\pu{10 mL}$ of $\pu{0.45 M}$ $\ce{NaOH}$ in two separate beakers. These two solutions are mixed. Find the $\mathrm {pH}$ of the resultant solution formed.
Consider the concentration of the ions formed by self dissociation of water to be negligible.

My method
Number of moles of $\ce{H+}$ ion in the first solution is $\pu{0.004 mol}$ and the number of moles of $\ce{OH-}$ in the second solution is $\pu{0.0045 mol}$.
Now, after mixing the two solutions, the resulting concentration of $\ce{H+}$ ion becomes $\pu{0.08 M}$ and that of the $\ce{OH-}$ will become $\pu{0.09 M}$. Now these $\ce{H+}$ and $\ce{OH-}$ ions will react with each other such that the equilibrium constant of their reaction becomes $10^{-14}$.
So I formed the equation
$$(0.08-x)(0.09-x) = 10^{-14}$$
From here we can calculate $x$ and subsequently we can get the concentration of $\ce{H+}$ ion in the final solution.
I want to know is this the correct way to solve the problem?
The answer to this problem is given as $12$, which I doubt can be obtained from my way of doing the problem.

No, you should not do the problem your way.
Given
$$(0.08−x)(0.09−x)=10^{−14}$$
You can solve for $x$ here, but its value will be quite close to $0.08$. This is the only way to make the product a small number, by making the one of the multiplicands very small.
However, $x$ will be so close to 0.08 that will have trouble figuring out the value of $0.08 - x$, which is what you want. Therefore, the right answer is obscured by your method of solving the problem.
In this problem, you can work around that by calculating the pH value from the hydroxide ion concentration, which, by the argument above, will be roughly $0.01\ \mathrm{M}$, consistent with a pH of 12. However, in the most general case, you may not be able to do that to determine what the final equilibrium concentrations are.


Can 4-pyridones be synthesised analogously to 2-pyridones? I understand that 2-pyridone derivatives can be synthesized by oxidizing an N-methyl pyridinium salt with $\ce{[Fe(CN)6]^3-}$ under basic conditions. Can a similar method be used to produce 4-pyridones ?

A reliable and old reference on the synthesis of 1-methyl-2-pyridone (Org. Synth. 1935, 15, 41, DOI) does not mention the formation of 1-methyl-4-pyridone as a side product and I doubt that it and can be formed that way.

The synthesis of 1-methyl-2-pyridone starts by reacting equimolar amounts of dry pyridine and dimethyl sulfate to form the N-methyl pyridinium salt. Then, the mixture is diluted with water and solutions of $\ce{NaOH}$ and $\ce{K3[Fe(CN)6]}$ are simultaneously added.

This seems to do the trick: the $\beta$-amino alcohol formed by adding $\ce{OH-}$ ortho to the nitrogen atom is immediately oxidized, which prevents the opening of the pyridinium salt.

For the synthesis of 1-alkyl-4-pyridones, a somewhat classical approach would be to react the corresponding 4-pyranone with a primary amine under more or less drastic conditions, as shown in Can. J. Chem., 1976, 54, 3377.

Altenatively, Grignard reaction of a 2-substituted 4-benzyloxypyridine-N-oxide, followed by debenzylation/methylation with methyl triflate and subsequent treatment with $\ce{NaOH}$ will yield the desired a 1,2,5-trisubstituted 4-pyridone (DOI).


Won't the net effect of a catalyst be zero if it creates a new path with lower activation energy? A catalyst will provide a new path with a lower activation energy (Figure 1). Won't this mean the forward and backward reactions will both speed up (as they both have a lower activation energy path to take)?

So then what is the point of using a catalyst? Or alternatively, how does the catalyst provide a net benefit for the forward reaction if both the forward and backward reaction are sped up?

I am assuming it might have something to do with the reactants having a higher energy than the products, but I can't think up a proper formal reason that a catalyst has a net benefit for the forward reaction.

Figure 1 (source: Chemguide.co.uk)

Your realisation is correct and something chemistry teachers try to hammer into their students’ heads time and time again (and yet, the point is still often lost):

Catalysts will never change the thermodynamics of a reaction. They only ease the path of the reaction. Forward and backward reactions will be accelerated equivalently.

So what is the benefit of a catalyst? There are multiple ones.

Speed

Take for example the Haber-Bosch process to synthesise ammonia from nitrogen and hydrogen.

$$\ce{N2 + 3 H2 <=> 2 NH3}\tag{1}$$
$$\Delta_\mathrm{r} H^0_\mathrm{298~K} = -45.8~\mathrm{\frac{kJ}{mol}}$$

This reaction is exothermic and thus should, theoretically or thermodynamicly, proceed spontaneously, e.g. if you mixed nitrogen and hydrogen in the appropriate ratio and added a spark. It does not, however. Significant activation energy is required to cleave the $\ce{N#N}$ triple bond.

Typical methods to add activation energy include heating. In the Haber-Bosch process, the mixture is heated to $400$ to $500~\mathrm{^\circ C}$ to supply the required activation energy. However, since the reaction is exothermic, heating will favour the reactant side. Increasing the pressure improves the entropic term of the Gibbs free energy equation, hence why pressures of $15$ to $25~\mathrm{MPa}$ are used.

Catalysts, based on iron with different promotors, are used to accelerate the reaction. By using catalysts, one can lower the temperature required in a trade-off between speed of reaction and favouring the product side of the equilibrium. With the conditions and catalysts used, one achieves a yield of $\approx 15~\%$ of ammonia within a reasonable timeframe. Not employing a catalyst would give much lower yields at much longer timeframes — economically much less feasible.

Direct reaction path not accessable

This is mainly true for many transition-metal catalysed carbon-carbon bond formation reactions, but is also true for some inorganic processes like the disproportionation of hydrogen peroxide as in equation $(2)$.

$$\ce{2 H2O2 -> 2 H2O + O2}\tag{2}$$

Hydrogen peroxide is a reactive chemical that cannot be stored forever, but the direct disproportionation path is not typically what degrades it. However, you can add $\ce{MnO2}$ to it. Upon addition, oxygen gas vigourously bubbles out of the solution. In this case, there was a kinetic barrier impeding the direct transformation due to reactants and products having different multiplicities (oxygen gas’ ground state is a triplet, all others are singlets). The $\mathrm{d^3}$ ion manganese(IV) is a radical itself that can partake in different radical reactions, allowing the diradical oxygen to be liberated.

Selectivity

This is exceptionally true for transition-metal catalysed organic carbon-carbon bond formation reactions. Note first that the action of a catalyst is frequently depicted as a catalytic cycle: A reactant reacts with the catalyst to some intermediate species, this rearranges or reacts with other reactants/additives/solvents in a set of specific steps until finally the products are liberated and the catalytic species is regenerated.

Many such reactions require organic halides as one of the reacting species. And the first step is typically an oxidative addition as shown in equation $(3)$, where $\ce{X}$ is a halide ($\ce{Cl, Br, I}$).

$$\ce{R-X + Pd^0 -> R-Pd^{+II}-X}\tag{3}$$

Palladium typically prefers oxidatively adding to bromides or iodides and tends to leave chlorides alone. I myself have performed a reaction with near-quatitative yield in which a reactant contained both a $\ce{C-Br}$ and a $\ce{C-Cl}$ bond — selectively, only the $\ce{C-Br}$ bond took part in the palladium(0) catalysed Sonogashira reaction.

Although I did not try it myself, I am pretty sure that switching to a nickel(0) catalyst species would shift the reaction in favour of reacting with the carbon-chlorine bond rather than the carbon-bromine one.

Mildness

This is basically a reiteration of the first point albeit with different intentions. Many a time in organic synthesis, one has a rather sensitive reactant that would degrade or undergo side-reactions if subjected to standard reaction conditions, such as high pH-value or elevated temperatures. As an example, consider a transesterification as shown in equation $(4)$.

$$\ce{R-COO-Et + Me-OH <=> R-COO-Me + EtOH}\tag{4}$$

This reaction is, of course, an equilibrium and by using methanol as the solvent we can shift it to the product side. For the reaction to happen, one would need a base strong enough to deprotonate methanol, giving the methanolate anion, which can then attack the ester functionality. However, methanolate being a strong (and nucleophilic) base itself can introduce undesired side-reactions, including epimerisation of the α-carbon.

One can catalyse this reaction by using $\ce{Bu2SnO}$, which will activate the carbonyl group, making it more susceptible to a nucleophilic attack. The reaction speed is the same but the conditions are milder (no additional base required) and the number of side-reactions is those strongly limited. In particular, I noticed no epimerisation of the α-carbon in the tin(IV) catalysed method.


LiAlH4 / H2O reaction energy I'm dealing with a free-energy scam on another forum. It is well-known that $\ce{LiAlH4}$ reacts explosively with water. My chemistry days are long behind me, so I thought I'd come here to ask:

$1~\text{mol}$ of $\ce{LiAlH4}$ reacted with $4~\text{mol}$ of water gives how much energy?

I'm not about to do this, mind you. I was just curious about the specific numbers.

So you have the reaction :

$$\ce{LiAlH4 + 4H2O <=> LiOH + Al(OH)3 + 4H2}$$

You want to get the "energy" given by this reaction, which is nothing but the reaction enthalpy.

$\Delta_rH^0(T) = \sum_{i}\nu_i.\Delta H_{i,f}^0(T)$

Now you get this, just look for the theorical values of each $\Delta H_f^0$ you need for the 5 compounds you have (they are tabulated) and sum them up pondered with the stoechiometric coefficients.


What does a radical mean here? I'm really confused as none of the definitions of radical on the internet compare to this.

This is from my college textbook of Applied Chemistry-

  Radicals

  Every inorganic compound is made up of two electrically charged parts
  called radicals. One part has positive charge and is called basic
  radical, while the other part has negative charge and is called acid
  radical. Hence,

  An atom or a group of atoms which forms a part of an inorganic compound is called a radical.

  For example :

  Sodium Chloride is made up of two parts:

  (i) Sodium radical (Na+)  and   (ii) Chloride radical (Cl-)

  The two radicals combine to form sodium chloride molecule.

  Copper sulphate is made up of two parts:

  (i) Cupric radical (Cu2+)  and   (ii) Sulphate radical (SO42-)

  The two radicals combine to form copper sulphate molecule.

  It should be noted that a radical cannot exist independently.

This definitely doesn't mean the same as free radicals. I don't think it means "functional group" because how can sodium be a functional group?!. Functional groups are in organic chemistry, right?

Can someone explain what the term "radical" means here?

A radical, according to the IUPAC, has an unpaired electron. Radicals may or may not be ions.

  A molecular entity such as $\ce{.CH3}$, $\ce{.SnH3}$, $\ce{Cl.}$ possessing an unpaired electron.

http://goldbook.iupac.org/R05066.html

Contrary to your book's claim, the sodium ion is not considered a radical. It actually has all its electrons paired. It is isoelectronic with helium.

Regarding your book's usage of "radical:" your book's terminology is outdated. Here is the IUPAC statement on your book's usage of the term:

  In the past, the term 'radical' was used to designate a substituent
  group bound to a molecular entity, as opposed to 'free radical', which
  nowadays is simply called radical. The bound entities may be called
  groups or substituents, but should no longer be called radicals.


Suggest the product from two-substituted benzene ring Which are the possible products from this reaction?

The answer in the mark scheme shows that it should be as demonstrated as pic A.
However, I think it could be either A. or B. since both substituted groups of the benzene ring, NH2 and Cl, direct the incoming electrophiles to attack the 2,4 and or 6 positions. Do I understand correctly?

I have never come across two substituted groups before. Should I consider both NH2 and Cl as substituted groups or only one of them, but which one? Also, how to count the position of carbon on the benzene ring in this case?

Please help! Any help would be much appreciated.:)

On substituting an already substituted benzene ring, you need to know which kind of substitution takes place; It is electrophillic substitution that is taking place.

Then can you now think about the correct choice?
$\ce{-NH2}$ has a greater positive mesomeric effect (+M effect)
in comparison to the Halogen groups.

Since the para position with respect to $\ce{-NH2}$ is already occupied by chlorine the ortho position would be attacked.

What we come to know is that since $\ce{-NH2}$ creates more electron availability in comparison to halogen it leads to "electrophilic substitution" on ortho position with respect to $\ce{-NH2}$.

And moreover $\ce{-NH2}$ group since a very very strong electron donor so once again another substitution of next bromine takes place at the other ortho position with respect to $\ce{-NH2}$.


Changing solubility of metal complex by adding ethanol After creating a solution of copper(II)-sulfate-pentahydrate, water and concentrated ammonia, the newly formed complex (tetraamminecopper(II)-sulfate-monohydrate) is completely dissolved in water. By adding ethanol, one is able to decrease the solubility of that complex in water and have crystals percipitates out of solution.

Since the complex is not soluble in ethanol, I am having trouble understanding why adding ethanol changes anything about the solubility of the complex. Would adding some organic phase like petrol ether or a non-soluble solid such as ceramic pellets do the same?

Would adding some organic phase like petrol ether or a non-soluble solid such as ceramic pellets do the same?

No. But adding other organic solvents which are miscible with water would.

The idea behind the "trick" is to reduce the polarity of the aqueous phase. With other words, you're making water more ethanol-like.


Does the temperature of liquid ammonia stored at high pressure, room temperature suddenly drop to −33.35 °C when poured into a bowl? Boiling point of ammonia is $-33.35\ \mathrm{^{\circ}C}$ and its critical temperature $132.4\ \mathrm{^{\circ}C}$ is well above the room temperature. So we can liquify the ammonia gas by compressing it into a bottle, and store it at room temperature for ever.
(Even though the bottle is kept at room temperature, the ammonia will stay in liquid state due to high pressure. )

Hope I'm correct with the above thinking. Next, I open the bottle and pour some liquid ammonia into a bowl. Since this bowl is at 1 atm, I expect something to happen. Does the temperature suddenly drop to $-33.35\ \mathrm{^{\circ}C}$? If so, why?

I'm trying to relate this to adiabatic expansion of gas – the gas does work on the surrounding as it expands and loses its kinetic energy.
But here ammonia is in liquid state as its temperature drops to $-33.35\ \mathrm{^{\circ}C}$. I don't see any work getting done by the ammonia, so I don't see how the temperature drops. I feel I'm missing some important concept.

The key to understanding what happens is that evaporation costs energy.

Changing the state of a liquid to a gas requires the input of energy: this is basic thermodynamics. This is in addition to any effects related to a change in temperature which also costs energy or releases energy.

When liquid ammonia evaporates to gaseous ammonia, energy is required from somewhere. If liquid ammonia at room temperature is poured into a bowl at room temperature it will want to evaporate as it is above its boiling point. But that evaporation requires an input of energy. That energy will come from the environment which consists of the bowl and the rest of the liquid. The only way to provide that energy is to cool the rest of the liquid and the bowl. This happens quickly and the temperature will often fall locally to below the boiling point of ammonia as there isn't enough time to equilibrate the temperature of the liquid with the temperature of the environment surrounding the bowl.

The net effect is that you will end up with some liquid ammonia at some temperature below its boiling point and a lot of gaseous ammonia (don't do this outside a good fume hood).

Forced evaporation of volatile liquids used to be used to freeze water via a similar process. For example, place a small bowl of water in a larger bowl of ether. Blow air onto the ether, which will evaporate, drawing energy from its environment (including the remaining ether and the bowl of water). The water will freeze if you used enough ether. I don't recommend doing this as ether is very flammable, but it was once a common demonstration of this process in chemistry classes.


How can I go about obtaining the activation enthalpy of a reaction using the Arrhenius Equation experimentally? I plan to do a string of experiments on the kinetics of the decomposition of hydrogen peroxide. I’m going to see the effect of catalyst concentration on rate and the effect of varying hydrogen peroxide concentration on rate, which will then allow me to formulate a rate equation for the reaction.

I would also like to calculate the activation enthalpy of the reaction specifically using the Arrhenius Equation $k=A\operatorname{e}^\tfrac{-E_\mathrm{a}}{RT}$. How would I do this? As far as I’m aware I should use varying temperatures in the reaction to get different values of the rate, but I’m not sure.

So what kind of experiment should I do (varying temperature to see the change in rate?)?

What should I do with my results?

Is the value of $E_\mathrm{a}$ obtained from analysing a specific graph that will be drawn up, and if so what graph would this be?

Experiments at various temperatures will allow you to calculate $E_a$. This is clear from a putting the Arrhenius equation in a different form:

$$\ln{k}=\ln{A} + \frac{-E_a}{RT}$$

A plot (or linear regression) of $\ln{k}$ vs $\frac{1}{T}$ yields $\frac{-E_a}{R}$ as the slope. The same experiment with a catalyst will show the activation energy is decreased.


Mechanism for oxidation of primary alcohols to carboxylic acids My teacher told me that the following is the mechanism for oxidation of primary alcohols to carboxylic acids:

I've searched in books and online and didn't find a similar mechanism. For example, in Wikipedia:

Which is correct?

Clayden et al., Organic Chemistry (2ed), p. 545 gives the reaction pathway as:

Here's some evidence that supports this pathway. One of the most well-known methods to selectively oxidise primary alcohols to aldehydes, without further oxidation to the carboxylic acid, is by using pyridinium chlorochromate in dichloromethane as solvent. This presumably works because water is excluded, which prevents the hydrate from being formed. Clayden writes:

  Aqueous methods like the Jones oxidation [n.b.: the Jones oxidation is $\ce{CrO3}/\text{aq. }\ce{H2SO4}$] are no good for this, since the aldehyde that forms is further oxidized to acid via its hydrate. The oxidizing agent treats the hydrate as an alcohol, and oxidizes it to the acid. The key thing is to avoid water, so PCC in dichloromethane works quite well. The related reagent PDC (pyridinium dichromate) is particularly suitable for oxidation to aldehydes.

If the reaction pathway was as your teacher taught you, then there would be no point in excluding water, since in that pathway water is not needed for the over-oxidation to the carboxylic acid.

The actual mechanism for the oxidation step is as follows (Clayden, p. 195):

If water is present, then the aldehyde product simply forms the hydrate and the mechanism for oxidation to the carboxylic acid is exactly the same, except that one of the hydrogens is replaced with an $\ce{-OH}$. Note that you need an $\ce{-OH}$ group on the starting material to form a chromate ester - that means that aldehydes will not undergo oxidation, but their hydrates (which are geminal diols) will.


UV radiation on IBr (Radical Substitution) I came across a radical substitution which Methane reacts with $\ce{IBr}$ to form iodomethane and hydrogen bromide and many more products. I understand the whole process but I don't quite see how the $\ce{IBr}$ molecule will break via homolytic fission instead of heterolytic. Since Bromine is more electronegative than Iodine, wouldn't $\ce{Br}$ takes both electrons and form $\ce{Br-}$ ions?

The presumed stability of $\ce{Br-}$ is overcompensated by the instability of $\ce{I+}$. While some polyatomic iodine cations like $\ce{I2+}$ exist, $\ce{I+}$ is found only to be stable in the gas phase and in the absence of a counterion. Due to the low difference of electronegativities (0,5) of Br and I, the I-Br bond is predominantly covalent, which also makes heterolysis unlikely. Homolytic cleavage of the I-Br bond is therefore energetically more favorable.


Synthesis of ethyl methyl ketone from 2-(4-hydroxyphenyl)ethyl methyl ketone Is it possible to reduce raspberry ketone (4-(4-Hydroxyphenyl)butan-2-one) to ethyl methyl ketone? How would I do this in practice?

There are a number of ways to accomplish your goal, but they all involve several synthetic steps. As mentioned in the comments, it's always easier to buy a compound then run a reaction, but if you had a small amount of some exotic derivative, you could convert it to methyl ethyl ketone.

One possible route would be to convert the raspberry ketone to an ester using the Baeyer-Villager reaction.  In this reaction the secondary (ethyl) branch migrates in preference to the primary (methyl) side.  The resulting ester can then be converted to the desired ketone by any number of methods such as the Weinreb route.


What is the meaning of '(en)'? What is the meaning of $\ce{(en)}$ on this two coordination compounds:

$\ce{[Co(en)_2(NO_2)_2]Cl}$

$\ce{[Cd(en)_2(CN)_2]}$

Is it a type of compound? If yes, what is the meaning?

It's an ethylenediamine ligand.

Edit - elaborating as per your request: Unfortunately, you don't provide any information as to how much you already know about complex chemistry, so I'm going back to the roots here. If you already know what a complex is, skip this paragraph. As you probably know, atoms and molecules may bind to each other, forming new compounds. We all know covalent bonds in which two atoms move close together and mutually share their electrons, and ionic bonds wherein one of the partaking partners is more egocentric in terms of electrons, "stealing" it from the other but staying in the vicinity nontheless. What these types have in common is that each atom or molecule provides an electron for the formation of the bond. However, it may also happen that both electrons come fron one partner alone; an electron-rich molecule or ion approaches another atom or ion, "donating" his electrons without really giving them up. This is the situation in coordination chemistry: A "central atom" or ion -- usually, but not necessarily a transition metal -- is surrounded by a varying number of "ligands" which donate electron pairs to the central atom, all while these electron pairs remain "located" at their respective ligands (very dangerous expression, I know ...). The compound that is formed is called a complex.

So ligands are atoms or molecules which can provide lone pairs for coordination bonds, i.e. electrons in non-bonding, doubly filled orbitals. In the above example of ethylenediamine, the binding centers of the molecule are the nitrogen atoms. The en ligand is an example of a multidentate (or chelate) ligand which may form more than one coordinating bond per molecule, while the $\ce{NO2^-}$ and $\ce{CN-}$ are monodentate ligands. The chloride anions in your first compound does not bind to the copper ion, but rather stay outside its "coordination sphere" to balance the positive charge of the cobalt complex (which you might then interpret as an ionic bonding between the complex cation and the chlorine anion).

Note that there are, of course, many quirks to this. Complexes with multiple central atoms, different theories about bonding, bridging or radical ligands, charge-transfer complexes, etc. This is a very wide and important field of chemistry, after all; nothing you could exhaustively explain in one or two paragraphs.


Hydrolysis of starch by means of amylase to unclog bathtub drain After my son decided to pour a substantial amount of uncooked rice down the bathtub drain, I find myself in the situation of removing it.

After trying some common methods - to no avail - I thought I could resort to chemistry.
Specifically, I thought about dissolving (hydrolysis) the main component of rice (starch) by means of amylase (possibly bought in a shop, not salivating over the drain).

Would this be a good solution to unclog the drain?
Under what conditions should I operate to achieve the result hoped?
Would there be a better solution?

As you might guess, I'm not a chemist.

The problem you'd face using any chemical to dissolve the rice is that rice forms a thick gelatinous surface coating as it dissolves, so that it would take an extremely long time for the amylase to penetrate... perhaps months or years. You'd be better off using some of the standard mechanical drain-cleaning techniques:

Disassembly of the sink trap (and drain, if needed) and cleaning or replacing it
Drain auger
Drain snake
Water-pressure drain cleaner

Note that augers and snakes can scratch sinks or toilet bowls, and that water-pressure cleaners can cause pipes to come apart.

N.B. If you've already poured a potentially dangerous cleaner into the drain, remove it safely before trying anything else!


Find the valences of Z=26 I've tried this:

But its hows that the valence of Fe in normal state is 4, when in the periodic table it isn't. Why does this happen? Can you show me the answer with configuration formula in the states with energy?


Finding out the number of atoms in the sun. This is the question from the text book:

This is what I did, since it worked with all the previous questions (similar to this).

$$
\frac{\frac{4}{3}\pi(7*10^8)^3*1.4}{1.008} * 6.022*10^{23}
$$

This equals $1.2 *10^{51}$, but the answer is $1.2*10^{57}$

What have I done wrong? Thank you in advance.

The problem is that you did not include units with your numbers. If you had, you would have easily seen that you are multiplying $m^3$ and $cm^{−3}$ and coming up with 1 instead of 1000000.


Variation in atomic radii of elements in different blocks? If we look at the values for the atomic radii (look at the table here), we can see that they rapidly decrease across the period initially. Looking at the second period,

The graph is pretty steep early on. But further down the period, as we enter the p-block elements, the graph levels out. This trend is repeated across the third and higher periods as well. Why is that?

The variation of atomic radii across the d-block and f-block elements is even more gradual. Why is that again?

Valence electrons experience an electrostatic force from the nucleus. The nucleus has its character positive charge, but due to shielding by core electrons the total positive charge is not completely felt by the electrons, so the actual net positive charge felt by the valence electron has its own name and is called the effective nuclear charge $Z_{eff}$. This effective nuclear charge, largely effects the size of the atom. The general trend of the periodic table is that as you go down a group, and go from right to left, the $Z_{eff}$ decreases, and you see an increase in the size of the atom. The easy way to think of it, is that as you go from left to right, the number of valence electrons, and  protons increases but the number core electrons stays the same; so, you see an increase in $Z_{eff}$.

$Z_{eff}$ is calculated using Slater's Rules. Where
$$Z_{eff}=Z-s$$
Where $Z$ is the atomic number and $s$ is the shielding constant.

The rules for calculating are as follows (from wikipedia):

Firstly,(1)(4) the electrons are arranged into a sequence of groups in order of increasing principal quantum number n, and for equal n in order of increasing azimuthal quantum number l, except that s- and p- orbitals are kept together.

$\mathrm{(1s) (2s,2p) (3s,3p) (3d) (4s,4p) (4d) (4f) (5s, 5p) (5d)}$ etc.
Each group is given a different shielding constant which depends upon the number and types of electrons in those groups preceding it.

The shielding constant for each group is formed as the sum of the following contributions:

An amount of 0.35 from each other electron within the same group except for the (1s) group, where the other electron contributes only 0.30.
If the group is of the (s p) type, an amount of 0.85 from each electron with principal quantum number n one less than that of the group, and an amount of 1.00 for each electron with principal quantum number two or more less.

If the group is of the (d) or (f), type, an amount of 1.00 for each electron "closer" to the atom than the group. This includes
i) electrons with a smaller principal quantum number n and
ii) electrons with an equal principal quantum number and a smaller azimuthal quantum number l.


How to calculate Madelung's constant for NaCl? Madelung's constant for $\ce{NaCl}$ is $1.748$.
What I tried to so is sum up all the individual interactions into $1$ component where the $X$s are the nucleus and the electron if it was a $2$ atom system. Is this the correct way to go about it? Thanks!

The series $\displaystyle \frac{6}{1}-\frac{12}{\sqrt{2}}+\frac{8}{\sqrt{3}}-\frac{6}{2}+\frac{24}{\sqrt{5}}\pm\ldots$ diverges as has been proven in 1951.

The correct value can be obtained by summation over lattice cubes with edge length $2n+1$ and getting the limit for $n \to \infty$. This has been shown in 1985.

Wikipedia provides more information and further reading on this.


When using a "dilute hydrochloric acid" solution, how diluted should it be? Quite often when reading through some instructions or documentation or an outline that I've found online (EG: acid-base extractions), theres a step that involves using a "dilute hydrochloric acid solution", but oddly enough, it often doesn't detail exactly how diluted the HCl should be. Heres one example (little past half way into the 3rd paragraph)

  Dilute hydrochloric acid is often used in the extraction of basic substances from mixtures or in the removal of basic impurities. The dilute acid converts the base such as ammonia or an organic amine into water soluble chloride salt.

So my question is: How does one know exactly how diluted the HCl should be? Is there a commonly used concentrated amount of HCl that I just don't know about? Sometimes it'll suggest using Muriatic acid instead, which I believe is basically 31.45% concentrated HCl, is that a commonly used concentration?

The exact concentration doesn't matter that much. The stock solution that I used to use was 1 M aqueous HCl (~3%). If your compound is acid-sensitive then you might want to try using even more dilute HCl (e.g. 0.01–0.1 M), or skip this step altogether and just purify it using a different method. The important thing is to make sure that you have enough HCl to protonate all the basic stuff you want to remove – a relatively simple stoichiometry exercise.

I would suggest not using concentrated HCl (12 M, or ~37%) as for a typical acid-base extraction it's simply not necessary. You should save that for other circumstances, such as when you want to adjust the pH of a solution without diluting it too much with water.


Why do nanoparticles have a different color than their macro counterparts? It astonished me when I learned that at nanoscale, gold is no longer "gold",rather it's red. Other elements' nanoparticles also have a different color than their macro counterparts.

Why does this phenomenon occur?

There is a fairly simple explanation of why small-enough particles are different from bulk materials. Once the particle becomes similar in size to the wavelengths of light involved then quantum effects start to matter for how the particle behaves.

The actual mechanisms that give specific colour may vary, but the main point is that the size of the particle comes to dominate the bulk properties of the material. In a bulk semiconductor, for example, the electronic properties depend on the band gap between occupied electron levels and unoccupied electron levels. When the particle is small enough, however, the electrons are more constrained and become more like an electron confined to a small box. This can mean that the possible energy levels of the electrons are determined by the physical size of the box not the bulk properties of the material. Hence any emission or absorption of light will be different to the bulk material. The additional physical constraints impose a different set of energy levels on the electrons.

The actual details can be quite complicated but the basic idea is that when particles are small compared to the wavelengths of light, you have to take into account the extra constraints on the system when calculating possible energy levels involved in electron transitions which are what causes colour.


What do s, m, w, m-w, and w-s mean in the context of infrared spectroscopy? I am referring to Donald Pavia's Introduction to Spectroscopy, and I found this table that tells about the intensities of the peaks of vibrations of certain bonds.

I can't understand what s, m, w, w-s, etc. are supposed to mean. My guess is that s could mean small, m could mean medium and w could mean weak. But that's crazy, since small and weak mean the same thing? So what do they mean actually?

Given the general agreement surrounding my comment on the question and the similar abbreviations and descriptions in the chart on the page linked by Tyberius in his comment, the meanings appear to be:

strong
moderate (or medium)
weak

Notations such as m-w and w-s presumably indicate a peak intensity that is variable in the indicated range, depending on the properties of the particular molecule of interest.


Determine if the following change is an oxidation, a reduction, or neither The following reaction is not a complete reaction. Determine if the change is an oxidation, a reduction, or neither:
  $$\ce{CrO4^{-2} -> Cr2O7^{-2}}$$

I know that the answer to this problem is neither, but I don't have any idea as to how that answer is reached. Could anyone give me a hint? I am familiar with oxidation, reduction, and oxidation numbers.

As pointed out in the comments and the other answer, it is neither an oxidation, nor a reduction.

Q: So, what is it?

A: It is a condensation: two molecules of an ortho acid combine under loss of one molecule of water to yield a diacid (pyro acid).

This is frequently found in oxoacids, see

$\ce{H3PO4}$ (phosphoric acid) and $\ce{H4P2O7}$ (diphosphoric acid)
$\ce{H3AsO4}$ (arsenic acid) and $\ce{H4AsO7}$ (diarsenic acid)
$\ce{H2SO4}$ (sulfuric acid) and $\ce{H2S2O7}$ (disulfuric acid)


If a chemical formula for a mineral contains a comma, are there limits on the amounts of each element? I have a chemical formula for the mineral actinolite which is: $$\rm{Ca}_2({Mg,Fe})_5{Si}_8{O}_{22}({OH})_2$$

The formula contains a "," which I understand means that the magnesium and the iron can have different combinations (up to 5 total).

The question is, if magnesium is the dominant atom, is it possible for the amount of magnesium to be less than iron and still be actinolite, or must magnesium always have the greatest amount?

Actinolite is a mineral that is part of a so-called solid-solution series, which means that you get different but related minerals depending on the amounts of (in this case) iron and magnesium that are present. The magnesium-rich end member is tremolite; the iron-rich end member is ferro-actinolite; and what lies between is actinolite.

According to Wikipedia's article on actinolite, the formula is more precisely:
$$\rm Ca_{2}(Mg_{4.5-2.5}Fe^{2+}_{0.5-2.5})Si_{8}O_{22}(OH)_{2}$$

So, for your question:

  is it possible for the amount of magnesium to be less than iron and still be actinolite, or must magnesium always have the greatest amount?

the answer is no: at best, Mg and Fe can be present in equal amounts for the mineral to still be called actinolite, but note that in that particular case you are on the "border" between actinolite and ferro-actinolite, so perhaps what you choose to call it is up to you. And as mentioned above, if magnesium is dominant, the mineral is tremolite.

In the case where iron is the dominant atom, the mineral is called ferro-actinolite with the formula:
$$\rm Ca_{2}(Mg_{2.5-0.0}Fe^{2+}_{2.5-5.0})Si_{8}O_{22}(OH)_{2}$$


what is the difference between how taste and smell are detected? Taste and smell are both chemical processes

  ... parts of the food dissolved in saliva come into contact with the taste receptors.[1] These are located on top of the taste receptor cells that constitute the taste buds. The taste receptor cells send information detected by clusters of various receptors and ion channels to the gustatory areas of the brain via the seventh, ninth and tenth cranial nerves. (https://en.wikipedia.org/wiki/Taste_bud)

, is there any difference in the way taste and smell receptors work?

Taste receptors work with water (or saliva, or whatever other liquid that happens to occur in your mouth) and the things dissolved therein.

Smell receptors work with air and airborne chemicals.

Short of that, they are similar in that both are wide-range chemical sensors, and different in that they send different signals to our brain.


Question about the definition of an isotopes My answer was as follows : atoms of the same element with differrnt relative atomic masses.

However in the mark scheme the answer was simply : atoms of the same element with different masses.

Was there something wrong with my answer?

Why does it matter?

Usually the term "relative atomic mass" for an element is a weighted average of the atomic masses of the various isotopes of an element, with the weighting determined by their abundance. So, since the atomic masses of various isotopes enter into this determination, your first answer is not too good.


finding out fractions of ions(different valencies/charges) of an element in non-stoichiometric compounds Analysis shows that nickel oxide has the formula $Ni_{0.98}O_{1.00}$. What fractions of nickel exist as $Ni^{2+} $and $Ni^{3+}$?

What i did:

Let the number of $Ni^{2+}$ and $Ni^{3+}$ atoms be $x$ and $y$ respectively

one  atom/ion of $Ni^{2+}$ satisfies one  $O^{2-}$ atom

$2 \over 3$ atom/ion of $Ni^{3+}$ one  $O^{2-}$  atom

and then i wrote down the equations:
$x + $$3 \over 2 $$y =1$  and  $x+y=0.98$

Then i solved the equations to get the answers:

$x=0.94 $  $and$ $y=0.04$

Is what i did correct? If not then guide me how to solve the question

What you have done seems to be correct. But you might want to convert those numbers to percentiles. I can tell you another method based on charge conservation.
Let us take  100 oxygen atoms and 98 Nickel atoms into our consideration.
Total negative charge$= 200$
Total positive charge$= 2x+3y$ where $x$ and $y$ are what you have assumed to be in the question . Now Net charge =0.Equate the positive and negative charges.
Also $x+y=98$. Solve them and you will get the same number. This of course has an high degree of
similarity to what you have done


First and last unit of polyethene Is a hydrogen atom bonded to the first and the last unit of polyethene?
If yes, where does the hydrogen come from?

Yes, in The Characterization of Linear Polyethylene SRM 1475. II. Determination of Total Methyl Content by Infrared Spectrophotometry and references cited therein, it is confirmed that the polymer chains terminate with methyl groups (the repeating unit plus an additional hydrogen as you phrase it).


Na2HPO4 is amphoteric: write the two reactions $\ce{Na_2HPO_4}$ is amphoteric, which means it can act as a base or as a acid depending on which substance they react with.

This is the reaction in which disodium phosphate appears to be an acid:

$$\ce{Na_2HPO_4 + H_2O \rightarrow H3O+ + Na_2PO_4-}$$

This is when it acts as a base:

$$\ce{Na_2HPO_4 + H_2O \rightarrow OH- + Na_2H_2PO_4+}$$

Did I write those two reactions currently? If not, what's wrong with my reasoning?

$\ce{Na2HPO4}$ is dissociated in solution. The amphoteric species is actually the hydrogen phosphate dianion:

Reaction as an acid:
$$\ce{HPO4^{2-} + H2O <=> PO4^{3-} + H3O+}$$

Reaction as a base:
$$\ce{HPO4^{2-} + H2O <=> H2PO4- + OH-}$$


Molecularity of Saponification Reaction I had a doubt while reading about the saponification reaction of esters in basic medium. What is the molecularity of the reaction and why?
P.S - I think its bimolecular and heres why -
The RDS of the reaction is the formation of the tetrahedral intermediate after the attack of  the OH ion. In this step both the ester and the hydroxide ion are involved (The ester is shifting the carbonyl bond electrons to the oxygen and the hydroxide is attacking the electrophilic carbon). Therefore, the molecularity should be two.

You should know that the molecularity is not relevant in your case and has no significance for multi-step reactions, it is just applicable to elementary reactions. The relevant parameter for multi-step reactions is the, reaction order wich is originally an experimental parameter  indirectly related to the molecularity of all elementary reactions that makeup the reaction mechanism.
So for saponification the proposed mechanism includes 3 steps, the last one of them is irreversible and controls the reaction. The order of the reaction is considered 2, the molecularity for the step that you have mentioned is also 2 but it is just one step in the mechanism.
Also, there are two types of mechanisms through which this reaction takes place: a base catalyzed mechanism (with simple base or alkoxides) and an acid catalyzed mechanism.


Does exotic potassium exist? I was reading a book called 'Hundred and Seven Stories of Chemistry'.

There is a chapter about how the attributes of different elements can change in presence of enormous pressure and temperature. The book says in presence of enormous pressure and temperature potassium's 19th electron can enter the 3d orbital. Then, it would no longer be a metal.

Is it possible? If yes, is there any theoretical calculation possible for the pressure and temperature needed for this exotic behaviour?

First, a point that is more like a comment: I'm not sure that just because electrons get put in to d-orbitals (and thinking of crystal band structures in terms of atomic orbitals is of questionable value anyway) that it would not be a metal. Many elements (one could even say most elements) with d-electrons are quite happy as metals.

You might want to start with L.F. Lundegaard et al., Phys Rev B80 020101(R) (2009), where they do DFT calculations on potassium at different pressures. I will quote from their conclusions section:

  The observation of the oP8 structure in K reveals it to
  have a pressure-induced structural behavior like that of Na,
  as well as having behavior like Rb and Cs as indicated by the
  observation of the tI4 and oC16 phases. The similarity to Na
  is unexpected in relation to the recent electronic structure
  calculations of Ma et al.$^{19}$ As noted, they found K, Rb, and Cs to have the same high-pressure structural sequence, and
  they attributed this to these heavier alkalis’ being dominated
  by d electrons arising from s-d transfer, whereas the ultrahigh
  pressure behavior of Na arises, they suggest, from p-d
  hybridization. Thus our results question this picture of a
  simple dependence of structural stability on d-electron dominance
  or p-d hybridization.


How to make oil more heavy? I would like to increase the density of a given oil (or anything that repels water) so that it is slightly heavier than water; to be able to precisely control how slow the oil would fall in the water.

What can I add to the oil (or the water) to accurately set its density?

I am looking for a cheap solution, that I can easily do at home.

Edit

Another way to create slow and tiny drops would be to drop water into oil. Yet, the oil surface tension prevents water from falling down ; I asked another question for this problem.

Halogen-substituted aliphatic hydrocarbons are usually miscible with oils. Pick the compound by its density, price, safety, boiling point and other characteristics.

For example, trichloroethylene (trico) has a density of ~1.6 g/cm3, and a Material Safety Data Sheet shows a vapor pressure of ~60mm Hg at room temperature and a time weighted average (TWA) toxicity limit of 50 ppm (269 mg/m3) from ACGIH and OSHA. Though it's readily available, trico might be too toxic and volatile for your use. A similar chemical is tetrachloroethylene, sold as "perc", with the same toxicity issue.

In general, going down the halogen group increases density, so bromine or iodine substituted compounds would be denser than those with fluorine or chlorine.


Nomenclature of cycloalkane with substituent having the same number of carbons What is the IUPAC name of the following molecule? More importantly, what is the rule for cycloalkanes having a substituent with the same number of carbons.

The undergraduate organic chemistry texts that I use when teaching tend to hide from complex nomenclature issues that cannot be addressed by the more common rules. Most texts clearly state that the parent chain in cases like these is the one with the larger number of carbon atoms.

For example, the following is butylcyclopentane.

The following is 1-cyclopentylhexane.

The addition of another substituent also makes it clearer, since we choose parent chains to maximize the number of substituents. The following is 1-cyclopentyl-3-methylpentane.

Many compounds can have two or more systematic names in accordance with several methods recommended by IUPAC. Concerning systems composed of rings and chains, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

The corresponding names for the examples given in the question are:

Method (1): “pentylcyclopentane” (preferred IUPAC name; ring preferred to chain)
Method (2): “1-cyclopentylpentane”

Method (1): “butylcyclopentane” (preferred IUPAC name; ring preferred to chain)
Method (2): “1-cyclopentylbutane”

Method (1): “hexylcyclopentane” (preferred IUPAC name; ring preferred to chain)
Method (2): “1-cyclopentylhexane”

Method (1): “(3-methylpentyl)cyclopentane” (preferred IUPAC name; ring preferred to chain)
Method (2): “1-cyclopentyl-3-methylpentane”


Relationship between solubility and enthalpy of solution I'm reading my lab manual and there is a part where it goes through Van't Hoff's isochore. That's fine but then it states

  The variation of the solubility of a substance with temperature
  may thus be given by the relation

  $$\dfrac{d \ln Sol}{dT}=\dfrac{\Delta H}{RT^2}$$

where do they derive this from. Is it from chemical potentials and mole fractions. I can't really find a good derivation and explanation online.

For equilibria such as $\ce{[C6H5COOH]_{(s)} <=> [C6H5COOH]_{(aq)} }$.  The equilibrium constant $K_{eq}$ can be written in the very simple form:

$$K_{eq}\equiv \ce{[C6H5COOH]_{(aq)}} \equiv \text{solubility}\,  (sol)$$

That's more than likely where the equation comes from. Then you get

$$\dfrac{d \ln Sol}{dT}=\dfrac{\Delta H}{RT^2}$$

when you separate the variables and integrate with respect to both sides you get:

$$\int_{\ln sol_1}^{\ln sol_2}d(\ln sol) = \int_{T_1}^{T_{2}}\dfrac{\Delta H}{RT^2}\, dT$$

$$\ln sol_2 - \ln sol_1 = \dfrac{\Delta H}{R}\Bigg{(}\dfrac{1}{T_1}- \dfrac{1}{T_2}\Bigg{)}$$

Which gives the useful result

$$\ln \dfrac{sol_2}{sol_1}= \dfrac{\Delta H}{R}\Bigg{(}\dfrac{1}{T_1}- \dfrac{1}{T_2}\Bigg{)}$$


How can I specify pyridine as a substituent? You know how when you carry out a reaction with many compounds, when drawing the reactions you just draw a line at a right angle to the bond with a 'R' at the end (see image)?

Then you can specify what group R is in a table (eg: $\ce{OMe}$). However, if instead of adding a substituent to a benzene ring, you have pyridine? How can you specify it?

The issue, I think, is to specify how the pyridine is connected. This means you have to identify which carbon on the pyridine ring is connected. This picture shows the numbering:

So, using your example the structure and names of the various possibilities would be (using connections on the 3-position of the benzonitrile ring as an example):

I'm sure you can generalise from this example.


Removing Heavy Rust from Very Old Carbon Steel Carving Knife on the Cheap I am trying to remove heavy rust from a carbon steel 10-inch carving knife. Mothers Polish did not get rid of much of the rust. I turned to vinegar soaking and scrubbing: At first it took off a good bit of the rust leaving visible residue in the bottom of a glass carafe. With a lot rust left I turned to vinegar and salt (see below). I have been soaking the knife using this solution in the 1-quart glass carafe for three or four days. Every day I pull out the knife and scrub it vigorously with a wire brush. Lots of black stuff comes off on a paper towel but there is still a lot of stubborn rust. When I take the knife out of the carafe the knife is blacker then when I put it in. After a vigorous wiping an alligator-like pattern remains on both sides.
I am using the same solution daily without changing it save for topping off the carafe. Also there seems to more chemical activity going on at the top of carafe than at the bottom, for some reason: is there an explanation of the phenomena?
If the rust is indeed being dissolved slowly I am wondering if the solution is loosing potency? Is the stoichiometry changing as the rust is being removed? Would changing the solution daily/ frequently help speed this process? Would frequent stirring help uniform the gradient? Is there a better method using household products?
I do not want to resort to expensive OTC cleaners or high-end sandpapers. Oh! FYI: pun intended-I am many decades rusty in Organic and Physical Chemistry.
Thanks for any thoughts.
Source of the idea: Hunker....Fill a large container made of a non-reactive material, such as plastic or glass, with 1 gallon of distilled white vinegar. Add 1 cup of table salt and stir the mixture to combine. The salt will increase the potency of the acid in the vinegar, making it even more effective. Use 1/2 gallon of vinegar with 1/2 cup of salt for smaller objects

First, read my answer here where I cite the electrochemistry background. The latter include source links to galvanic corrosion and, in particular, in the current instance, a classic case of pit corrosion.
Stop long-term soaking, as this in the presence of salts and an H+ together with dissolved oxygen will just further promote the slowly evolving electrochemistry.
I, personally, ignored stake-knives left in a sink (water and oxygen exposure) that were used to cut a mango. The sticky mango juice, rich in Ascorbic acid (Vitamin C) is apparently particularly notorious for promoting pit corrosion.
Never have completely removed the stains, so do not have high expectation here,


Derivation and application of the Henderson-Hasselbalch equation Write down an expression for the dissociation constant, $K_{\mathrm{a}}$, of a very weak monobasic acid $\ce{HA}$ in an aqueous solution in terms of the concentrations of $\ce{H+ (aq), A- (aq) and HA (aq)}$ in the solution.
  Hence, show that

  $$
\mathrm{p}K_{\mathrm{a}} = \mathrm{p}\ce{H} - \mathrm{log}_{10} \frac{[\ce{A- (aq)}]}{[\ce{HA (aq)}]} \text{ where } \mathrm{p}K_{\mathrm{a}} = - \mathrm{log}_{10} K_{\mathrm{a}}
$$

  At a particular temperature, $\pu{2.00e-3 mol}$ of the acid $\ce{HA}$ was dissolved in water and the solution diluted until the volume was $\pu{75.00 cm^3}$. When $\pu{25.00 cm^3}$ of a $\pu{0.004 mol dm^{-3}}\,\ce{NaOH}$ solution was added to that acid solution, the $\mathrm{p}\ce{H}$ of the resulting solution was found to be 6.0. Calculate the dissociation constant, $K_{\mathrm{a}}$, of the acid $\ce{HA}$ at that temperature.

Below you can see what I have tried with my current answers.

For 1, I got $$K_{\mathrm{a}} = \frac{[\ce{H+ (aq)}][\ce{A- (aq)}]}{[\ce{HA (aq)}]}.$$
For 2, I got the same answer.

But I am stuck on 3. These are the things I already know:

Moles of NaOH added: $(25/10000) * 0.004 = \pu{0.001 mol}$
Moles of acid HA present initially: $\pu{0.002 mol}$

How can I proceed further?

The reaction taking place is the quantitative reaction:
$$\ce{AH + OH- -> A- + H2O}$$

The initial concentrations are:
 $$\ce{[OH- ]_0}=\frac{0.001}{0.1}=0.01  \mathrm{M}$$  $$\ce{[AH ]_0}=\frac{0.002}{0.1}=0.02 \mathrm{M}$$
The equilibrium concentrations:
 $$\ce{[AH ]}=\frac{0.002-0.001}{0.1}=0.01 \mathrm{M}$$
 $$\ce{[OH- ]}=10^{-8}  \mathrm{M}$$
 $$\ce{[A- ]}=0.02-0.01=0.01 \mathrm{M}$$
Using the equation of question (ii):

$\ce{pH = p}K_a=6$


Does granite become soft as modelling clay at 2000 °C? As Wikipedia tells us, it consists of $\ce{Al2O3}$ ($14.42\%$) (alumina) whose melting point is $\pu{2072^\circ C}$, and $\ce{SiO2}$ ($72.04\%$) (silica) whose melting point is $\pu{1713^\circ C}$. So does it soften to modelling clay (or at least partially soft, modelable via tools structure) at $\sim\pu{2000^\circ C}$, or does it stay hard and why?

Mixtures of different compounds do not usually work like that.  They generally melt at lower temperatures than the constituent pure compounds.  So I would expect your granite to be not like modeling clay, but molten even before getting to the melting point of silica.

Relevant:  https://en.wikipedia.org/wiki/Eutectic_system.

If you want modeling clay, try 1300 °C where the Earth's lithosphere gives way to the asthenosphere.  See the Characteristics section of https://en.wikipedia.org/wiki/Asthenosphere.


Molecular orbital explanation of why the protonation of the oxygen atom makes a carbonyl group more electrophilic I was working on nucleophilic addition reactions to the carbonyl group (namely, hemiacetal formation) when I came across the following explanation for the use of acid catalyst and how it increases reaction rate:

  In acid (dilute HCl, say), the mechanism is different in detail. The first step is now protonation of the carbonyl group’s lone pair: the positive charge makes it much more electrophilic so the addition reaction is faster.

The reaction at hand is the following:

Ignoring the presence of the positive formal charge on the oxygen, I am trying to understand on a molecular orbital theory level why this makes sense. I'm suspecting that this somehow lowers carbon's LUMO (${π}^{*}$ orbital), which would enhance the electrophilicity of carbon.

Would someone be able to give an explanation to the enhanced electrophilicity of the carbonyl group based on molecular orbital theory?

References

Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

Ignoring the presence of the positive formal charge on the oxygen

I think this is exactly what is responsible, even from the MO viewpoint. From the MO perspective, the electrophilicity of the carbonyl group can be viewed as how energetically-accessible is the $\ce {C=O \pi ^*}$ MO to the nucleophile's HOMO. The protonation of the oxygen atom effectively decreases the energies of all MOs of the carbonyl, including the $\ce {C=O \pi ^*}$ MO. Why? The proton is positively-charged. By attaching to the oxygen atom, there is a net decrease in the potential energy of the molecule due to there being more electron-nuclei attractive forces in the resultant entity. Hence, protonation makes the $\pi ^*$ LUMO more accessible to the attacking nucleophile, increasing electrophilicity.


Understanding the unit cell of Chalcopyrite vs its formula According to the Wikipedia article for Chalcopyrite it has a formula of: $\ce{CuFeS2}$ and a unit cell of:

But from the unit cell it has 13 Copper, 10 Iron and 8 Sulphur.  This doesn't match with the formula.

Some of the Copper and Iron lie on the faces of the unit cell.

Copper atoms

The 8 Copper atoms at the corners all represent the same Copper atom so that means only 13 - 7 = 6 Copper atoms.
Then there are an additional 2 repeats on the top and bottom faces and front and back faces bringing it to 4 Copper atoms.

Iron atoms

The Iron atoms similarly have a pair repeat on the left and right faces, the top and bottom faces and the front and back (10 - 3 = 7).
Additionally the 4 Iron atoms in the centre line (from left to right) represent the same atom so thats 7 - 3 = 4 Iron atoms.

This is how the formula and the unit cell are consistent with each other.


What is "dithreitol"? The article by Hu et al. [1] mentions a compound "dithreitol" in the Saliva Proteome Analysis section (emphasis mine):

  For the "shotgun" approach, saliva samples (1 mL) were pre-fractionated with the use of Millipore ultracentrifuge filters (Millipore Corp., Billerica, MA, USA).
  Individual fractions were treated with 10 mM dithreitol for 30 min and then 50 mM iodoacetamide for 30 min.

PubChem doesn't reveal any details as to what this compound exactly is and suggests alternative spelling, e.g. D-threitol.

Google Books search results in several mentions of this compound in the literature, but I didn't find any further details except that it's probably a sulfhydryl reagent.
This makes me think that "dithreitol" could be another name for dithiothreitol, but I didn't manage to find any solid references to support this proposition.

To sum it up: what is dithreitol, exactly, and what chemical and structural formulas does it posses?

References

Hu, S.; Li, Y.; Wang, J.; Xie, Y.; Tjon, K.; Wolinsky, L.; Loo, R. R. O.; Loo, J. A.; Wong, D. T. Human Saliva Proteome and Transcriptome. J Dent Res 2006, 85 (12), 1129–1133. DOI: 10.1177/154405910608501212.

For the "shotgun" approach, saliva samples ($\pu{1 mL}$) were pre-fractionated with the use of Millipore ultracentrifuge filters (Millipore Corp., Billerica, MA, USA). Individual fractions were treated with $\pu{10 mM}$ dithreitol for $\pu{30 min}$, and then $\pu{50 mM}$ iodoacetamide for $\pu{30 min}$.

This reaction sequence in Ref.1 has a clear intention: First, break disulfide bonds in saliva proteins using a thiol exchanging reagent and then trap the recently formed thiol groups by iodoacetamide (Ref.2). According to Ref.2:

  Alkylation of cysteine thiols with iodo, bromo or chloro substituted acetic acid or acetamide is a classic approach that has been exploited since the 1930′s.

Thus, it is safe to assume that this second reaction is intended to trap thiol group as quickly as possible (The relative reaction rates between glutathione and these halogenated acetates are $100:60:1$ for iodo-, bromo- and chloro-acetates respectively; Ref.3).

Also keep in mind that disulfide reduction to corresponding thiol is accomplished primarily by thiol exchange type reagents such as dithiothreitol (DTT) or 2-mercaptoethanol (ME). Thus, it is also safe to assume that dithreitol mentioned in Ref.1 is definitely a thiol exchanging reagent, possibly DTT. Therefore, I think it is a chain mistake as M. Farooq suggested in his comment.

References:

S. Hu, Y. Li, J. Wang, Y. Xie, K. Tjon, L. Wolinsky, R. R. O. Loo, J. A. Loo, D. T. Wong, "Human Saliva Proteome and Transcriptome," J. Dent. Res. 2006, 85(12), 1129–1133 (DOI: 10.1177/154405910608501212).
Jakob R. Winther, Colin Thorpe, "Quantification of Thiols and Disulfides," Biochimica et Biophysica Acta 2014, 1840(2), 838-846 (https://doi.org/10.1016/j.bbagen.2013.03.031).
Frank Dickens, "CLII. Interaction of halogenacetates and SH compounds. The reaction of halogenacetic acids with glutathione and cysteine. The mechanism of iodoacetate poisoning of glyoxalase," Biochem J. 1933, 27(4), 1141–1151 (doi: 10.1042/bj0271141).


How to find concentration of three species from given molar absorptivities If I am given the molar absorptivities of three species at a given wavelength, and I only have the concentration of one of those species, how can I determine the concentrations of each of the remaining two species? The distance light travels, the molar absorptivities and the amount of light absorbed are given.

The equation you need would seem to be $\ce{A=\epsilon [C] L}$ where A is the amount of absorption (optical density), L the cell path length, your 'distance light travels' and [C] concentration. As you have 3 species absorbing at the same wavelength you have $\ce{A=(\epsilon_1[C_1] + \epsilon_2[C_2] + \epsilon_3[C_3])L}$. You know L and the $\epsilon$'s but only one of the three concentrations and therefore you do not have enough information to determine the two remaining ones.


Are there any nonflammable thermal insulators which can be made at home? I am trying to build a homemade thermos. It needs to be able to conserve the heat in 125 ml of water for 15 minutes.

However, unlike a normal thermos, it also needs to be able to be heated. What homemade substances can I use that are not flammable, yet provide a relatively high degree of insulation?

Note that my thermos will be heated with a flame from a candle for 25 minutes.

Consider using a ceramic container.  You could adjust the density, thickness, and size to match your insulating requirements.

"home made' is loose, here, since you would need to acquire oven bake-able ceramics, or a similar product.  (unless you have a kiln lying around).


How to desalinate seawater at freezing temperature or high pressure? The deep ocean depths have very high pressure and freezing temperatures. Could one or both of these characteristics be used to desalinate seawater?

Reverse osmosis uses high pressure to separate salt from seawater to make
freshwater. Freezing seawater can also make freshwater. Less energy would
be utilized in both processes.

If there is a problem raising the freshwater to the top, why not make ice
cubes from the freezing ocean depth and have them float to the top via a vertical pipe? The high ocean depth pressure would make membrane technology
less energy intensive.


DFT vs. MP2 for stacked dimer Recently, I looked at these two papers analyzing the excited-state properties of  modified DNA bases (2-aminopurine and 8-vinyl-A) and how they are influenced by stacking with natural nucleobases:

Jean, J. M.; Hall, K. B. 2-Aminopurine fluorescence quenching and lifetimes: Role of base stacking. Proceedings of the National Academy of Sciences 2001, 98 (1), 37–41 DOI: 10.1073/pnas.98.1.37.
Kenfack, C. A.; Burger, A.; Mély, Y. Excited-State Properties and Transitions of Fluorescent 8-Vinyl Adenosine in DNA. J. Phys. Chem. B 2006, 110 (51), 26327–26336 DOI: 10.1021/jp064388h.

Briefly, the authors considered dimers comprising the respective modified base and each of the natural bases. The geometries of the individual monomers were optimized using MP2/6-31G(d,p) while maintaining their relative position within in the dimer. Afterwards, the authors examined the excited state properties with TDDFT (B3LYP/6-311+G(d)) to identify dark states that could indicate fluorescence quenching of the modified bases.

I'm not an expert in theoretical chemistry and so I was wondering how the choice of methods here was motivated. In particular, why was MP2 used for geometry optimization and not DFT? Does it have to do with a better description of dispersion interactions by MP2?

For their TDDFT calculations the authors used the 6-311+G(d) basis set. I want to try out some similar calculations using ORCA. According to the manual and online resources from the developers, Ahlrich's basis sets seem to be preferred over Pople style basis sets. Could I expect similar (or better/worse) accuracy with def2-TZVP? It's also triple-zeta and contains polarization functions on heavy atoms, but lacks the diffusion functions that were present in the studies mentioned above. Would these be important anyway in such a case were only valence transitions are investigated?

Both papers you refer to used methods which were basically state of the art at their time (Gaussian 98/03 compared to Gaussian 16 today). But that is almost twenty (or fifteen) years ago. Nowadays we thankfully have more developed methods available, and one should check whether the results found back then are still consistent. An (admittedly also a bit outdated) overview can be found here: DFT Functional Selection Criteria.

Long story short: You cannot reliably perform calculations on complexes where non-covalent interactions are dominant. You need to to use a method that treats dispersion at least at an empirically derived level. A popular example is Grimmes DFT-D3 program, which is also available in many ESS codes. (Starting point is The Journal of Chemical Physics 2010, 132 (15), 154104.) In the semi-empirical method xtb a newer version is already implemented.

For that reason I guess (since I only read the methods sections) they used MP2 for molecular structure optimisations, as it recovers dispersion interactions to some degree.

Whether or not DFT is suitable for your investigations, I cannot say. A thorough research of the recent literature will certainly offer far better insight.

The Pople basis sets, while an excellent example to understand the structure of them, are generally a terrible choice for almost everything practical. There are multiple basis sets available, which produce more consistent results, and are better optimised to perform in modern ESS codes. If implementations are missing from you ESS program of choice, you have a high chance to find it in the Basis Set Exchange Library.

For ORCA there are most (if not all) Ahlrichs' def2-variations are available, see the ORCA input library.
I would expect augmented basis sets (ma-def2-XVP or def2-XVPD) are necessary to investigate such complexes.


converting SMILES (or .mol2) to SMARTS The conversion between different chemical structure formats, e.g. from .mol2 to .xyz, by openbabel may tied to conditions to be met.  One example of such a filter is to allow such a conversion only for data sharing a certain scaffold, e.g.

babel -imol2 AHECAS.mol2 -oxyz AHECAS.xyz -s "c1ccccc1-[#16](=[#8])-[#6]-c2ccccc2"

So the program allows SMARTS as constraining input parameter.  In difference to SMILES, (apparently, based on babel -L formats) openbabel however does not offer a conversion of any format into SMARTS and my source to generate them is the on-line PubChem Sketcher.

Question:  Aiming to generate SMARTS locally (offline), I know Marvin by ChemAxon may export a structure into SMARTS.  However, is there a portable alternative (with lesser footprint than a complete new GUI sketcher), preferentially allowing a conversion in a pattern like SMILES2SMARTS (or mol2toSMARTS) from the terminal / by a script?

The thread indicated in the comment by @andselisk dates back to July 2014. It may be condensed to

  "A substructure querying with openbabel requires a good SMARTS string. In case you don't have one, often a good SMILES string may be used instead of a SMARTS string, too."

A check for a rather simple molecular structure of (benzylsulfinyl)benzene

there is more than one way to describe it in either form of a SMARTS, or a SMILES string.  Consequently, the strings per se sometimes differ from each other.  Consequently, recognizing the relevant entries in the collection of .mol2 with openbabel sometimes failed.

Below I summerize (status: 2019-May-15) a few experiences: the string offered by the software, its type, and source; eventually, if it worked (i.e., entries retrieved), or not (i.e., entries not retrieved) in a small test set of 419 model data and 7 relevant hits.

|-------------------------------------+--------+--------------------+--------|
| entry                               | type   | source             | status |
|-------------------------------------+--------+--------------------+--------|
| c1ccccc1-[#6]-[#16](-c2ccccc2)=[#8] | SMARTS | pubChemSketcher    | works  |
| C1=CC=CC=C1C[S](C2=CC=CC=C2)=O      | SMILES | pubChemSketcher    | fails  |
|-------------------------------------+--------+--------------------+--------|
| O=S(Cc1ccccc1)c2ccccc2              | SMILES | ACD ChemSketch     | works  |
|-------------------------------------+--------+--------------------+--------|
| C1=CC=C(C=C1)CS(C(C=CC1)=CC=1)=O    | SMILES | ChemDoodle         | fails  |
|-------------------------------------+--------+--------------------+--------|
| O=S(C1=CC=CC=C1)CC2=CC=CC=C2        | SMILES | ChemDraw Pro12     | fails  |
|-------------------------------------+--------+--------------------+--------|
| O=S(CC1=CC=CC=C1)C1=CC=CC=C1        | SMILES | Marvin JS          | fails  |
|-------------------------------------+--------+--------------------+--------|
| O=S(Cc1ccccc1)c2ccccc2              | SMILES | Peter Ertl JSME    | works  |
|-------------------------------------+--------+--------------------+--------|
| c1cccc(c1)CS(=O)c1ccccc1            | SMILES | openbabel, default | works  |
| c1ccc(cc1)CS(=O)c1ccccc1            | SMILES | dito, universal    | works  |
| c1ccc(cc1)CS(=O)c1ccccc1            | SMILES | dito, Inchified    | works  |
| O=S(c1ccccc1)Cc1ccccc1              | SMILES | dito, canonical    | works  |
|-------------------------------------+--------+--------------------+--------|

PubChem Sketcher: version 2.4, https://pubchem.ncbi.nlm.nih.gov/edit2/index.html
ACD ChemSketch:   Freeeware version 2015
ChemDoodle:       version 8.1.0, 2019 and webcomponents,
                  https://web.chemdoodle.com/demos/sketcher/
ChemDraw:         version 12 Pro, (then) CambridgeSoft, 2009
Marvin JS:        ChemAxon, v19.11.0, on-line test at
                  https://marvinjs-demo.chemaxon.com/latest/
                  Its SMARTS (very differently written) equally failed:
                  O=S([#6]-[#6]-1=[#6]-[#6]=[#6]-[#6]=[#6]-1)[#6]-1=[#6]-[#6]=[#6]-[#6]=[#6]-1
Peter Ertl JSME:  https://peter-ertl.com/jsme/JSME_2017-02-26/

Addition:
The approach mentioned in the question basically calls babel (or òpenbabel) as an iteration per file.  Especially if the data set is accessed more than once to identify different structures, and / or consists of multiple thousand models, this may be tediously slow.  Instead, it is worth to "bundle" the individual information

babel *.mol2 mymols.sdf --unique

creating a container like a multi-model .mol2 (this even works with .xyz files as input).  The .sdf may be indexed

babel mymols.sdf -ofs

by fingerprints (cf. chapter 5 of the manual).  Now, very quickly the names of the files about a strucure matching the query substructure are identifed, e.g.

babel mymols.fs -ifs -s"O=S(c1ccccc1)Cc1ccccc1" results.smi -xt

and stored in results.smi.  Consequently, any further work may focus on them.


How do ionic, polar, hydrogen-bonding and non-polar compounds compare in TLC? $R_\mathrm{f}$ decreases with increasing polarity. Does this mean a compound with an ionic bond would have the lowest $R_\mathrm{f}$? And nonpolar would have the highest $R_\mathrm{f}$?

Example: $\ce{RCOO- Na+}$ is ionic.  If the above is true it should have a lower $R_\mathrm{f}$ than cyclohexanol which has H-bonding?

No, your analysis is wrong.

The nice thing about TLC is that you can pick the "paper" and the solvent to suit your analysis. So if you use benzene as the solvent then it won't "drag" ionized compounds along behind the solvent front very well.  If you use water ionized compounds do get dragged along nicely.


Relative Energies between LGOs and 2s orbital of oxygen in H2O MO Diagram In this MO diagram, we find that the LGOs of the two premixed H1s orbitals are significantly higher in energy than oxygen's 2s orbital. Although oxygen's $Z_{eff}$ is considerably higher, is it really enough to make it so that the 2s orbital of oxygen is lower than the in-phase combination of two 1s orbitals?

You probably already understand that the oxygen 2s orbital is lower in energy than the hydrogen 1s to begin with. For the unconvinced reader, consider the fact that the ionisation energies of oxygen and hydrogen (1314 and 1312 kJ/mol respectively) are quite similar, which implies that O2p and H1s have (somewhat) similar energies.

The question is then about how much lower in energy the H1s + H1s bonding combination is, relative to the H1s orbital itself. However, the hydrogen atoms are quite far apart: using the H–O–H bond angle (104.45°) and H–O bond length (95.84 pm) quoted on Wikipedia, the H–H distance can be calculated as being 151.5 pm, over twice as long as in dihydrogen itself (74 pm). This means a very poor overlap between the 1s orbitals, so you should not have an in-phase combination that is significantly lower in energy than the original hydrogen 1s orbital.

For the same reason, the "antibonding" LGO should also have roughly the same energy as the original 1s orbitals. The only reason why we are looking at bonding & antibonding LGOs is because of symmetry; it doesn't have anything to do whether they are actually bonding or antibonding to an appreciable extent.


Predicting electronic geometry, observable geometry, and hybridization for any atom in a molecule Can someone please explain the intuition around the answers to this problem:

I'm finding it very confusing because it seems like you have to nitpick between two resonance structures to get the desired results. Do we just always look for the "lowest" possible VSEPR in any resonance structure? Why isn't the observable geometry for "X" linear instead of none?

Do we just always look for the "lowest" possible VSEPR in any resonance structure?

Yes. Consider the two resonance contributors below for a N,N-dimethylacetamide, which has a functional group in common with your molecule. The rule of thumb I teach is: **if an atom appears to be $sp^2$-hybridized in one resonance contributor and $sp^3$ in another, then that atom is $sp^2$-hybridized always.

In the figure above, I have used the curved arrow notation to represent how the lone pair on the nitrogen atom is delocalized into the $\pi$-system. In order for this delocalization to occur, that lone pair needs to be in a $p$ orbital. In order for that nitrogen atom to have a $p$ orbital, it must be $sp^2$-hybridized (or possibly $sp$-hybridized if applicable). All three of the atoms (N, C, and O) in the delocalized system need a p orbital, and thus must be $sp^2$-hybridized.

Once you have hybridization, then you have limited your choices of geometry:

$sp^3 \implies$ tetrahedral geometry group: tetrahedral, trigonal pyramidal, bent, "none"
$sp^2 \implies$ trigonal planar group: trigonal planar, bent, "none"
$sp^{\ \ \  }  \implies$ linear group: linear, "none"

  Why isn't the observable geometry for "X" linear instead of none?

You are confusing observable geometry at an atom with molecular geometry.

Observable geometries at an atom are a relationship describing the relative geometric positions of other atoms bonded to that central atom. The central atom is not counted.A linear arrangement means that the atom in question is bonded to two other atoms or groups and those two atoms or groups are $180^\circ$ apart. If there is only one atom or group bonded to the atom in question, we cannot define the geometric relationship between that atom or group and the other atoms or groups bonded to that atom because there are not other atoms or groups. Consider the following examples from the tetrahedral group. The last case is "none" or "undefined, because you need at least two things to have a relationship.

This is different than molecular geometry which deals with the shape of the entire molecule. Let's look at HF. The shape of the molecule is linear, because the two points define a line (segment). The geometry around F is "none" or undefined. We cannot describe the arrangement in space of the hydrogen atom relative to other atoms bonded to fluorine because there are none.


Which has greater dipole moment: cis-1,2-dichloroethylene or 1,1-dichloroethylene? Which has greater dipole moment: cis-1,2-dichloroethylene or 1,1-dichloroethylene?

I know that dipole moment depends upon the strength of dipole and angle between the dipoles.

1,1-Dichloroethylene has two chlorine atoms with 120° angle between them because of $\mathrm{sp^2}$ hybridization of carbon atom.

cis-1,2-Dichloroethylene has the same with the angle between chlorine atoms as 60° (or else the dipole moment of trans form would not have been zero).

So, the dipole moment of cis should be greater than that of 1,1-dichloroethylene, but the answer says the opposite (μ(1,1-dichloroethylene) > μ(cis-1,2-dichloroethylene)). Can anyone help me?

Dipole moment is a vector entity. Any applying dipole moment can be replaced by two perpendicular components:

For convenience to identify, I put $a$, $b$, $c$, and $d$ as the magnitudes of applying dipole moments of relevant $\ce{C-Cl}$ bonds. Yet they are theoretically identical: $a = b = c = d = \mu$.

Let's consider cis-1,2-dichloroethene first. The horizontal components of vectors $a$ and $b$, $a \sin 30^\circ$ and $-b \sin 30^\circ$, will cancel each other (negative sing is because of opposite direction of the vector). However, vertical components are in the same direction, thus they add together: $a \cos 30^\circ + b \cos 30^\circ = 2\mu \cos 30^\circ = 2\mu \times \frac{\sqrt{3}}{2} = \sqrt{3}\mu$. This is the net dipole moment of cis-1,2-dichloroethene.

Now, consider 1,1-dichloroethene. The vertical components of vectors $c$ and $d$, $-c \cos 30^\circ$ and $d \cos 30^\circ$, will cancel each other (again, negative sing is because of opposite direction of the vector). Here, horizontal components are in the same direction, thus they add together: $c \sin 30^\circ + d \sin 30^\circ = 2\mu \sin 30^\circ = 2\mu \times \frac{1}{2} = \mu$. This is the net dipole moment of 1,1-dichloroethene.

Therefore it is clearly, $\sqrt{3}\mu \gt \mu$. Realistically, $\mu_{1,1}$ is smaller than $\mu_{1,2}$ because of increasing electron density of two chlorine atom attached carbon in 1,1-dichloroethene compared to that of one chlorine atom attached carbon in 1,2-dichloroethene. This can be easily show using MaxW's values: $\mu_{1,1} = \mu = \pu{1.3 D}$ and hence, $\mu_{1,2} = \sqrt{3}\mu = \pu{1.3 D} \times \sqrt{3} = \pu{2.25 D}$, which is actually $\pu{1.9 D}$.


400–430 degrees Celsius heated bath I want to perform a reaction which needs the temperature to be maintained between 400–430 °C. Since it's the only practical solution, I opted for a heated bath.

I'm looking for a hydrocarbon fraction/blend with the boiling point in this range, however, I keep finding conflicting information. If anyone knows such a hydrocarbon product please, let me know.

If it doesn't specifically need to be a hydrocarbon, zinc metal melts at 419.5 °C. Could you do an "ice bath" of zinc chunks in molten zinc, maintaining the melt right at its melting point?


Why large basis sets give better approximations to the exact solution of the Schrödinger equation? The variational principle states that the energy of any approximate wave function will always be equal to or greater than the energy of the exact solution. Therefore the energy is minimized when performing approximations, in order to get as close as possible to the exact solution. Larger basis sets give the wave equation "more flexibility", but I am not sure I understand this fully. Here's a statistics analogy:

When performing multivariate regression analysis, we choose the regression equation of lowest order; "overfitting" the data is something to be avoided. In my head, choosing large basis sets for simple problems seems like "overfitting".

So why is it that a larger basis set leads to a better approximation? The statistics analogy may not be applicable, but it illustrates what goes on in my head.

The statistics analogy may not be applicable, but it illustrates what
  goes on in my head.

The overfitting analogy is indeed not applicable here. It is a totally different and unrelated phenomenon. Basically, when we are doing regression analysis we do not know the relationship between a dependent variable and independent variables. Then we can indeed have an overfitting problem when instead of fitting the actual relationship we start to describe some random noise. It generally happens when we do have too many parameters in a model, and we never know how many is too many.

But when we are doing LCAO-MO we know the relationship between a dependent variable (say, energy) and independent variables (LCAO-MO fficients). ANd we are certain that the bigger the number of AOs (and correspondingly of the coefficients), the better result could be expected with the the exact solution corresponding to using infinitely many AOs. So we should not worry about anything like overfitting. Yes, by increasing the number of AOs in a basis set we increase the number of variational parameters, but we know it can't do any harm: in the worst case1 we will get the same energy as before the just another expansion of the basis.

This is similar to a Taylor series expansion of a function. The exact result corresponds to an infinite series, and by using a finite number of terms a function can only be approximated. But the more terms we use, the better approximation is. There is even a formal proof, but I think it should be intuitively clear that if an infinite series is the exact solution, than the later you truncate it the better approximation you get.

1) The worst case would be, for instance, when you expand a basis by including new functions that can be expressed as linear combinations of basis functions from the previously used basis set. You won't improve energy by doing so, but only increase the computational time. Note, that most of the programs have a simple defense mechanism against such situations: they are smart enough to detect the linear dependence in the basis and get rid of it prior to doing the calculations.


Why does Birch reduction of naphthalene with isoamyl alcohol form tetralin instead of 1,4-dihydrodialin (as formed with C2H5OH)? I recently came across this question, which required me to know the difference between performing a Birch reduction on naphthalene using ethanol and using isopentanol. Why does using isopentanol lead to that extra hydrogenation?

As I noted in a Comment that there should be no significant difference between the use of ethanol or isoamyl alcohol in the reduction of naphthalene with sodium in that both alcohols are primary. The difference lies in the conditions of the reaction. The sequence is a classic case of kinetic vs. thermodynamic conditions.
The reaction conditions in the diagram are from this paper.
Naphthalene (1) reversibly adds an electron from sodium under ambient conditions to form the resonance stablized radical anion 2 that is shown in a canonical form. This species is readily protonated by an alcohol, be it primary or tertiary as in this instance, at the site of highest electron density to afford the resonance stabilized radical 3. The addition of a second electron to 3 in this 2-electron reduction gives rise to the resonance stabilized anion 4, which, in turn, is protonated to form the deconjugated 1,4-dihydronaphthalene 5, the kinetically formed product. Just like cyclohexene is unable to reduce under these reaction conditions, neither can alkene 5.  During the course of this reduction, something else is formed -- namely -- a base, sodium tert-butoxide.
To achieve reduction of the double bond in 5 it is not sufficient to conduct the reaction at reflux but rather to conjugate the double bond with the aromatic ring. This is the role of the base at elevated temperature. Deprotonation of 5 at the benzylic site provides resonance stabilized anion 4 again, which can reversibly protonate to form 1,2-dihydronaphthalene (dialin) 6. This styrene-like double bond has a low reduction potential. Addition of an electron produces the radical anion 7 that protonates to form stabilized benzyl radical 8. Subsequent addition of a second electron affords the benzylic anion 9. This anion is readily protonated by the alcohol to form 1,2,3,4-tetrahydronaphthalene 10 (tetralin).


Finding initial amount of ammonia using equilibrium constant and amount of nitrogen gas Question

When an amount ammonia is added at $\pu{600 K}$ in a $\pu{1 L}$ container the following reaction takes place:
$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$
The equilibrium constant $K_c = 4.20$ at $\pu{600 K}.$
At equilibrium, it is known that $\pu{0.200 mol}$ of $\ce{N2}$ gas exist in the container. What amount of ammonia was added at the start of the reaction? Choose from the answers below:
$\pu{0.826 mol};$ $\pu{0.482 mol};$ $\pu{1.226 mol};$ $\pu{0.400 mol};$ $\pu{0.800 mol}.$

My attempt
I created an ICE table, but I think it's wrong:
$$
\begin{array}{lccc}
\ce{&N2(g) &+ &3 H2(g) &<=> &2 NH3(g)} \\
\text{I} & 0 && 0 && y \\
\text{C} & +x && +3x && -4x \\
\text{E} & 0.2 && 0.6 && y-0.8 \\
\end{array}
$$
$$K_c = \frac{(y - 0.8)^2}{0.200\times 0.6^3}$$
Solving the equation with Maple gives me none of the answers above. I must have messed up on the table. Can anyone tell me where I messed up?

Since the volume is given and it's a constant, the initial amount of ammonia $n_0(\ce{NH3})$ can be found from its initial concentration $c_0(\ce{NH3}):$

$$n_0(\ce{NH3}) = c_0(\ce{NH3})\times V\tag{1}$$

To find $c_0(\ce{NH3}),$ an ICE table might indeed come in handy; however, yours needs corrections.
First, I suggest to rewrite it according to the process occurring in the system, namely dissociation of ammonia:

$$
\begin{array}{lccc}
\ce{&2 NH3(g) &<=> &N2(g) &+ &3 H2(g)} \\
\text{I} & c_0 && 0 && 0 \\
\text{C} & -2x && +x && +3x \\
\text{E} & c_0 - 2x && x && 3x \\
\end{array}
$$

Second, note that the provided equilibrium constant $K_c$ is given for the synthesis of ammonia, hence in our case a reciprocal value is relevant:

$$
\begin{align}
\frac{1}{K_c} &= \frac{[\ce{N2}][\ce{H2}]^3}{[\ce{NH3}]^2}\tag{2.1}\\
\frac{1}{K_c} &= \frac{x (3x)^3}{(c_o - 2x)^2}\tag{2.2}\\
\frac{1}{K_c} &= \frac{27x^4}{(c_o - 2x)^2}\tag{2.3}\\
\frac{1}{\sqrt{K_c}} &= \frac{3\sqrt{3}x^2}{c_o - 2x}\tag{2.4}
\end{align}
$$

$$c_0 = 3\sqrt{3K_c}x^2 + 2x\tag{3}$$

At the equilibrium there is $\pu{0.200 mol}$ of nitrogen in the $\pu{1 L}$ vessel, so $x = \pu{0.200 mol L-1},$ and the initial amount of ammonia can be found as follows:

$$
\begin{align}
n_0(\ce{NH3}) &= (3\sqrt{3K_c}x^2 + 2x) × V\\
  &= (3\sqrt{3\times 4.20}\times (\pu{0.200 mol L-1})^2 + 2\times \pu{0.200 mol L-1})\times \pu{1 L} \\
  &= \pu{0.826 mol}\tag{4}
\end{align}
$$


What are the parameters of protonation of a molecule? Adding an acid replaces the $\ce{-MgBr+}$ with an $\ce{H+}$. Why does this happen? Is $\ce{H+}$ a better electrophile than $\ce{MgBr+}$?

Or, to put it another way, why doesn't $\ce{MgBr}$ stays attached to $\ce{O}$?

We are taking about electrophilicity of ions.
It basically depends upon:
1. charge on the ion(+)
2. density of plus charge on it

So comparing $\ce{H+}$ and $\ce{CH3+}$, the former has a higher plus charge density so that's why it will be more attracted towards 'O'.


m-CPBA Selectivity between Baeyer Villiger Oxidation and Epoxidation? If both a ketone and an alkene are present in a molecule and you add one equivalent of m-CPBA, what are the factors that determine which group will be oxidized, the ketone through Baeyer-Villiger oxidation or the alkene through concerted epoxidation? Is one of the reactions particularly faster and better than the other?

EDIT: Since the literature seems to be conflicted and there is not a clear general answer, let us take the below as an example because I am interested in the competing factors here. For this example, I am leaning towards epoxidation just due to electron-rich alkene, and there is no base to make m-CPBA a better nucleophile for BV.

The answer appears to be "it depends" (perhaps this should not surprise an experienced organic chemist).

According to the abstract of the paper by Mendelovici and Glotter [1] in steroid chemistry:

  Treatment of 3β- and 3α-hydroxy-(acetoxy-)cholest-4-en-6-one and of 6β-and 6α-hydroxy-(acetoxy-)cholest-4-en-3-one with MCPBA gives two types of product, depending on the initial site of the peroxy acid attack. Attack at the carbonyl group gives a Baeyer–Villiger rearrangement leading first to enol lactones and then by epoxidation of the latter to epoxy lactones. Alternatively, attack at the double bond gives epoxy ketones which can subsequently undergo a Baeyer–Villiger rearrangement leading to epoxy lactones. With one exception (3α-hydroxycholest-4-en-6-one), the Baeyer–Villiger oxidation of the enone is the dominant process. Epoxidation of the double bond is suppressed in the presence of an axial 3α- or 6β-acetoxy group.

However, Kazmeier and Horn [2] working with unsaturated aldehydes find

  Purified m CPBA is a useful reagent for the oxidation of several classes of aldehyde. Although linear unbranched aliphatic aldehydes are oxidized to the corresponding carboxylic acids, α‐branched ones undergo Baeyer–Villiger oxidation to formates. α‐Branched α,β‐unsaturated aldehydes provide enolformates and/or epoxides, which can be saponified to α‐hydroxy ketones with shortening of the carbon chain by 1 carbon. Unbranched α,β‐unsaturated aldehydes undergo an interesting Baeyer–Villiger oxidation/epoxidation/formate migration/BV oxidation cascade, which results in formyl‐protected hydrates with an overall loss of two carbon atoms.

Overall it appears that both processes can occur so Murphy's Law applies. It would be a good idea to explore alternative epoxidation conditions for an unsaturated ketone.

References

Mendelovici, M.; Glotter, E. Epoxidation and Baeyer–Villiger Oxidation of γ-Hydroxy-Αβ-Unsaturated Ketones on Exposure to m-Chloroperbenzoic Acid. J. Chem. Soc., Perkin Trans. 1 1992, No. 13, 1735–1740. DOI: 10.1039/P19920001735.
Horn, A.; Kazmaier, U. Purified MCPBA, a Useful Reagent for the Oxidation of Aldehydes. European Journal of Organic Chemistry 2018, 2018 (20–21), 2531–2536. DOI: 10.1002/ejoc.201701645.


Does an acetal give a positive Tollens test? In Organic Chemistry (Wade) there is a question:

  Which of the following compounds would give a positive Tollens' test? (Remember that the Tollens' test involves mild basic aqueous conditions.)

The answers manual says this compound - $\ce{CH3-(CH2)3-CH(OCH3)2}$ - gives a negative result as it is an "acetal stable in base: no reaction"

Now, I would have thought this is positive. Earlier in the chapter it's taught that since the Tollens reagent removes any aldehyde present, even if at equilibrium the amount of aldehyde is very small, the Tollens reagent will nevertheless predominate due to Le Chatelier's principle (the small amount of aldehyde will be removed slowly but surely, until eventually all of the solution is converted to aldehyde and then to the tollens product).

In this example, whilst I respect that the acetal product may be by far the major product, surely a very, very small percentage of the acetal is nevertheless converted to the aldehyde, via protonation of a methoxy group, which leaves, and then attack by $\ce{H2O}$, followed by deprotonation of $\ce{H2O -> OH}$, and then protonation of the other methoxy group, which leaves, leaving a carbocation, which the $\ce{OH}$ attacks forming a double bond and a positive charge on the $\ce{OH}$, followed by deprotonation on the $\ce{OH}$  giving an aldehyde?

This may be unlikely, but according to Le Chatelier's principle, it just needs to occur at some small quantity in order to be continuously removed by the tollens reagent, and thus ends up predominating and removing all of the acetal product given enough time? Why is this not positive?

The description of the hydrolysis of an acetal is accurate. The trouble is that the initial protonation event happens to such a small amount that the reaction does not proceed. We can estimate where the equilibrium of that protonation lies by looking at the pKa's of the relevant acids [Evans pKa table]. In basic solution, water (pKa 16) is the strongest acid available to protonate the acetal. The protonated acetal will have a pKa close to protonated methanol (pKa -2). So the equilibrium for that protonation will lie to the water side by approximately $\ce{10^18}$. As Jori is alluding to in the comment, the rate moving forward is going to be exceedingly slow because the concentration of the protonated intermediate will be exceedingly low. This is where the rule of thumb that acetals are stable to base comes from. See also: Why are acetals stable to bases and nucleophiles?

If the Tollen's test was performed under acidic conditions, then you would be correct that the acetal would test positive for an aldehyde. A hemi-acetal would also test positive.


How long will it take for a measurable change in mass (corrosion) in an iron copper corrosion cell to happen? I am planning to do an experiment, in which I will test the affect of salt concentration on corrosion rate. Would there be a relationship? How long would it take for rust to form, that could be scrapped off and corrosion rate measured?

In order to get any measurements, you would probably need to leave the coupon in the solution for about a week or so.

To see the difference due to the salt concentration differences, I would recommend going even longer, probably at least 2-3 weeks.