How can spike recovery be over 100%? I am confident enough about my knowledge about spiking and the process of standard addition but I came across this process which claimed to have a percent recovery of 115% and it was considered as a good recovery process. How it is considered a good recovery? Is it because you add too much of the spike and recovered most of them?

A spike recovery value of >100% does not mean that you added too much or the recovery process is too good. All it means is that the analytical method which is being followed "in letter and spirit" gives a positive error. There may be a bias in the method or the precision of the method is poor. We cannot say just by looking at one number. At this stage it is assumed that there was no operator error; only the analytical method is to be blamed. A single number like 115% in analytical chemistry has no meaning. Whoever called it a good recovery process needs to revisit that statement.

Whenever we develop a new method, recovery is done is several matrices and even different types of samples. As a result one gets a range of recovery values in different matrices or for different samples. It is better to say that a method shows gives a recovery between 95% to 115%. This statement is far more meaningful than just stating that a recovery of 115% is good. Certainly, that method must be pretty good.


Determining the reaction time of water synthesis by use of differential equations How do I determine the reaction speed?

I know you need a value, k, and order, which is only gained through experimentation, but how do the differential equations play in? How do I plot the result?

Assuming it's the first order ...

How do I proceed? I now have this function but I am not sure what to do with it. I do not have the means to observe the reaction and figure out its k-value or order. Is this reaction a known one, though? Is the order and k-value something you can find on the Internet?:

$$\ce{2H2 + O2 -> 2H2O}$$

How do I determine the reaction speed?

The rate of reaction can be determined experimentally or theoretically, the latter case requiring that a reaction mechanism be known. Proposing a reaction mechanism requires knowledge of the reaction at multiple regimes and some guesswork. Evidence for a certain reaction mechanism is usually indirect.

Theoretically speaking, once the reaction mechanism and rate constants are known, we can write down a coupled system of kinetic rate equations and solve them simultaneously to find the rate of reaction.

  I know you need a value, k, and order, which is only gained through experimentation, but how do the differential equations play in? How do I plot the result?

The differential equations govern the reacting system, and their solution completely specifies the system, enabling you to find the rate of reaction. You would then simply plot, say, the rate of reaction vs. time.

  How do I proceed? I now have this function but I am not sure what to do with it. I do not have the means to observe the reaction and figure out its k-value or order.

I don't believe you can proceed, because your initial assumption was wrong. The combustion of hydrogen is far more complicated than a first-order reaction, which you have assumed; you should not, therefore, expect that your first-order analysis should describe your system well.

  Is this reaction a known one, though? Is the order and k-value something you can find on the Internet?

It has certainly been studied. I will refer you to Reaction mechanism of combustion of H2 and the links therein.


How much yeast is necessary to produce the same amount of carbon dioxide that I exhale? I think I might have bed bugs. I read that bed bugs are attracted to $\ce{CO2}$ and heat, I also know that yeast produces $\ce{CO2}$ as a waste product. Knowing that, I did some research and found this article on intructables.com which creates a "trap".

My gut instinct is that a 1/2 teaspoon of yeast will not produce anything close to the amount that a person exhales, but I don't know My question is, how much yeast would I need to create the same amount of $\ce{CO2}$ that I exhale on average in a day, per day?

What I know so far:

We exhale $\pu{0.02 m^3/h}$ while resting or low activity (Oscar W. Richards and Florence W. Haynes, Plant Physiol. 1932, 7 (1), 139–144.)
This graph (Figure 3 and 4 of above citation) shows the average $\ce{CO2}$ production per minute by yeast, but I don't know enough about chemistry to know what the units are or how to convert them to something useful (in my case, "useful" would be I need to buy x grams of yeast).

The trap definitely works as I've caught 2 in 24 hours. However, catching them isn't my interest (I wouldn't be able to get rid of them all with a trap).

However, please don't be distracted by the bugs. The bugs themselves aren't the problem of this question; I'm trying to determine the quantities. I'm fairly naive when it comes to chemistry, but in my mind this shouldn't be a hard problem to solve. Unfortunately, I don't know how do the conversions or read the paper's to arrive at the answer but I'm still actively trying to figure this out.

I think this is a chemistry question, so I will answer it here.

First, I would question whether the amount of $\ce{CO2}$ is relevant or the concentration (or if more convenient, partial pressure). There is no way for the bed bugs to integrate over the entire trap vs. the bed. If you have a concentration gradient from bed to trap, it should work.

Second, yeast is a catalyst. The amount of $\ce{CO2}$ made depends on the amount of available food (the yeast will multiply to keep up). In the simplest terms, if you want the yeast to produce as much $\ce{CO2}$ as you do, you have to feed it the same amount of food that turns into $\ce{CO2}$.

Because yeast and humans have a slightly different diet, you have to interpret "same amount of food" a bit. In human metabolism, all carbon from foods with calories (i.e. not fiber) is used in biosynthesis or ends up as exhaled $\ce{CO2}$. In an adult who doesn't gain or lose a lot of weight, you can estimate the amount of $\ce{CO2}$ produced from the amount of food taken in. In yeast metabolism, depending on the conditions, either most carbon turns into $\ce{CO2}$ (when there is plenty of oxygen) or into ethanol (when there is a lack of oxygen). For bed bug traps or baking, you want the former, while for brewing, you would want the latter.

For practical purposes, though, I think the total amount of $\ce{CO2}$ is moot (see my first point). As long as you have a source of $\ce{CO2}$ in the trap and remove the source of $\ce{CO2}$ in the bed (i.e. sleep elsewhere) and keep the room warm, there should be a nice $\ce{CO2}$ concentration gradient for the bugs to follow.


Where can I find a downloadable spreadsheet of element properties? Sorry for the FAQ, but I really don't know where to find a full spreadsheet for element properties. I mean a table that has one row for each element, and the columns are its properties, such as name, atomic number, density, melting point, etc. I know that there are many sites or references for that, but I need it in tabular format, because I want to analyze data, make charts, etc.

Well, here's a CSV file that I parsed out of the JSON data provided by Paul Nathan's website, which in turn was produced from gPeriodic data in response to this question. gPeriodic is FOSS, so I can only assume that the element data contained within is good to share, though I have no idea of its provenance.

The parser uses the python json module to read the data, which I then wrote into comma separated values, which should be readable by any halfway competent spreadsheet program.

Here it is: Pastebin Link

save it as elementdata.csv and you're good to go.

Some gotchas:

This reproduces the gPeriodic data, warts and all. I make no guarantees regarding its correctness, recency, etc. I just munged the data.
Some of the data is augmented with tildes, notes about temperature/polymorph/state etc. I've left these as they are but you may need to trim them if you want to plot them as numeric values.
Argon had an atomic radius of '2-', whatever that means. I cut it.
Excel is joke software and habitually interprets numbers wrapped in parentheses as negatives, because apparently some accountants decided that surrounding numbers in parens is a more sensible option than using a minus sign. Needless to say, this is wrong. I've addressed the problem by wrapping the affected numbers in angle brackets.
There are no ionic radii in the source data. At all. Not my fault.
The columns inherit the units of the source data. I've left the units out on purpose because I wanted each column to have no spaces in the name for ease of processing in R or whatever. Furthermore, the units in the source file have some inconvenient characters from an encoding perspective.

Tried plotting Z versus covalent radius in R from this data - looks alright:

Some of the data points are missing, some are not read by R because they are wrapped in <> - pre-process to your heart's content.


What are radicals and how do they act in radical chain reactions? What really is a radical? For example, when $\ce{Cl2}$ receives enough energy to dissociate the bond and turn into $\ce{ 2Cl.}$, the two chlorine atoms each carry one of the electrons that form the bond. What is the difference between the $\ce{Cl}$ radical and the original $\ce{Cl}$? They have exactly the same number of electrons. And how do they act as a chain carrier?

free radical is a molecule with one or more unpaired electrons. Despite this definition, d-element compounds with unpaired electrons on d-element, like $\ce{CrCl3.6H2O}$ are not usually considered as free radicals because their unpaired electrons are relatively unreactive. Free radicals are especially reactive, because their reactions usually go through activated complex with relatively small energy. The reasons behind it are not that easy do describe.

In general most reaction with free radicals in organic chemistry falls into two categories: one, where free radical reacts with exposed hydrogen/heteroatom and one where it reacts with exposed multiple bond. In both cases, orbitals of interests are three orbitals of the radical in consideration and two atoms of the organic molecules. MO theory says, that such interaction results in three MOs: bonding, nonbonding and antibonding. With three electrons occupying this orbitals, the energy of the system is mostly same as for the radical and the molecule separated, so the activation energy of the reaction is near zero.

In case of nucleophilic/electrophilic reactions in , the interaction in activated complex are similar, with two electrons on bonding and zero or two electrons on nonbonding orbitals. However, in case of aliphatic derivatives, the reaction usually goes throw sterically hindered state with 5-coordinated carbon  and/or requires generation of hi-energy particles that are not preserved after the molecule reacts. In cases when hi-energy intermediate may be preserved (for example, ionic polymerization), and/or reaction goes through sterically unhindered intermediates (for example, nucleophilic substitution in acyl halides), and/or one of the reagents easily produces active intermediates (for example, nitrating mixture in aromatic substitution) reactions without free radicals as intermediates may proceed swiftly.

Example of chain reaction for aliphatic radical substitution:

$\ce{Cl2 + CH4 -> CH3Cl + HCl}$

in case if $\ce{CH4}$ is presented in huge excess, resulting in reduced rate of multichlorination, the reaction proceeds as following:

Chain initiations: $\ce{Cl2 -> 2.Cl}$
Chain growth 1: $\ce{.Cl + CH4 -> HCl + .CH3}$
Chain growth g 2: $\ce{.CH3 + Cl2 -> CH3Cl + .Cl}$

In case of ionic polymerization

Chain initiation: $\ce{H2SO4 + CH2=CH-Ph -> HSO4- + H-CH_2-\stackrel{+}{C}H-Ph}$
Chain growth: $\ce{H-(CH2-CHPh-)_{n}-CH2-\stackrel{+}{C}H-Ph + CH2=CH-Ph ->} $  $\ce{H-(CH2-CHPh-)_{n + 1}-CH2-\overset{+}{C}H-Ph}$

Both reactions proceed smoothly at room temperature if source of active particles is available (UV-light for breaking $\ce{Cl-Cl}$ bond for the former and concentrated sulfuric acid for the latter)

Compare nucleophilic substitution $\ce{KOH + CH3I -> CH3OH + KI}$

$\ce{KOH -> K+ + OH-}$
$\ce{OH- + CH3-I -> (HO-CH3-I)- -> HO-CH3 + I-}$
$\ce{I- + K+ -> KI}$

As you see, the reaction proceeds through sterically hindered state with 5-coordinated carbon. Compare to acyl halide case $\ce{KOH + CH3COCl -> CH3COOH + KCl}$

$\ce{KOH -> K+ + OH-}$
$\ce{OH- + CH3-(C=O)-Cl -> CH3-C(O^{-})(OH)Cl -> CH3-COOH + Cl-}$
$\ce{Cl- + K+ -> KCl}$

in this case the reaction proceeds through sterically unhindered state with 4-coordinated carbon, so acetylchloride reacts with water at room temperature.

TL;DR : free radicals are particles with unpaired electron, readily available for covalent interactions, they reacts with molecules with easy because they are ones of big family of 'active particles' with low energy of activation of many reactions, and, unlike many other active species, they usually regenerate themselves in reaction, allowing chain reactions to proceed. Other active particles or plainly active substrates also exists, so free radicals are not special. The fact of presence of unpaired electron does not necessary make the particle an active free radical, many relatively inactive free radicals exists, like $\ce{NO}$.

Further reading: MO theory (case of 3-atomic interactions) and kinetics (case of chain reactions) parts of your favorite advance chemistry textbook, parts on reaction mechanisms and active particles of your favorite advanced organic chemistry textbook.

Despite above answer based exclusively on organic chemistry examples, everything stated above is true for inorganic chemistry. However the set of possible kinds of elementary acts, active particles and interactions is much wider than for basic organic chemistry, leading to many special and unique cases and consequently to much less predictable reactions in complicated cases.


Refractive Index (RI) variation with Temperature & disagreement between two standards I am comparing two different international standards for the same organic compound (clear and colorless liquid) and for the refractive index one standard says

  1.531 to 1.534 (@ 20 °C)

The other one says

  1.526 to 1.527 (@ 27 °C)

Now the ranges don't even overlap.

Couple of  questions:

Would the change in testing temperature (20 °C vs 27 °C) be enough to explain this difference in standards? Would an increase in Temperature typically reduce the RI or increase it? By what percent per degree change in Temperature?
Any other testing intricacies for RI that I ought to be aware of? Is there a methodological / protocol difference that may explain this?
Is this a difference large enough to worry about? The difference in ranges is only about 0.3 %. What is the typical least count / accuracy of RI measurements in the lab?

The refractive index of a pure liquid varies with the wavelength of the light (this dispersion causes the splitting of white light into rainbow colours) and with the density and thus temperature of the liquid (this causes the schlieren patterns when heating a liquid on a hot plate).

The section “Index of Refraction of Liquids for Calibration Purposes”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL. lists values for several organic liquids at various wavelengths and temperatures.

Reference values close to the values given in the question can be found for toluene:

$$\textbf{Index of refraction of toluene}\\
\begin{array}{lll}
\hline
&\lambda/\mathrm{nm} &T=20\ \mathrm{^\circ C} &T=25\ \mathrm{^\circ C} &T=30\ \mathrm{^\circ C} \\
\hline
&667.81 &1.49180 &1.48903 &1.48619 \\
&656.28 &1.49243 &1.48966 &1.48682 \\
&589.26 &1.49693 &1.49413 &1.49126 \\
&546.07 &1.50086 &1.49803 &1.49514 \\
&501.57 &1.50620 &1.50334 &1.50041 \\
&486.13 &1.50847 &1.50559 &1.50265 \\
&435.83 &1.51800 &1.51506 &1.51206 \\
\hline
\end{array}$$

The uncertainty of the refractive index is estimated as $\pm0.00003$.

For example, the refractive index measured for toluene using the characteristic wavelength of a sodium-vapour lamp $(\lambda\approx589\ \mathrm{nm})$ at $T=20\ \mathrm{^\circ C}$ is $1.49693$. When increasing the temperature, the refractive index is decreased to $1.49413$ $(-0.19\ \%)$ at $T=25\ \mathrm{^\circ C}$ and to $1.49126$ $(-0.38\ \%)$ at $T=30\ \mathrm{^\circ C}$.

The range of values given in the question corresponds to a decrease between $-0.26\ \%$ and $-0.52\ \%$ when the temperature is increased from $T=20\ \mathrm{^\circ C}$ to $T=27\ \mathrm{^\circ C}$.


Which preposition to use for the reducing agent? Which expression is correct when we use the prepositions with or by for indicating that the reduction is carried out using $\ce{NaBH4}$ as a reductor:

  It is reduced to sodium phenylselenolate with $\ce{NaBH4}$ in alkaline ethanol.

or:

  It is reduced to sodium phenylselenolate by $\ce{NaBH4}$ in alkaline ethanol.

Either is fine.  In the most literal interpretation of the statement I would say by (as it does the actual action of reduction), however I personally would likely use with, to a scientist there is little difference though.  If you were filing a patent, the term might might matter a bit more, which a quick survey of the patent literature suggests with.  Again though this is minutia and nit-picky.

I would suggest phrasing your sentence more as:

  ...is reduced to sodium phenylselenolate with a solution of NaBH4 in alkaline ethanol .


Electrolytic rust removal I'm currently removing rust on some bike parts so decided to use electrolysis. It's working fine, but the solution has turned a deep/dark blue whilst in most online videos it turns orange with the rust. My sacrificial anodes aren't stainless or chromed, and I dont believe they're galvanized either.

Is the blue colour any cause for concern (am I making anything toxic accidentally)?

If your anode/cathode are from copper, then the blue chemical in the solution is copper chloride (CuCl2).  Copper chloride can be toxic. So you should have safety equipment.

As for rust removal in general, I'd recommend you to use Citric acid (Squeezing lemons for the lemon juice isn't efficient since it will contain 5-6% Citric acid - so its better to buy powdered Citric acid from a store).


Does an increase of area lead to more collisons in a confined jar of gas? I'm reading a lecture by Richard Feynman in which he is talking about the behaviour of gas molecules/atoms within a confined vessel.

The confined vessel is a cylinder with a piston to stop the water molecules from flying out.

However this follow sentence has me stumped

"Clearly, the force is proportional to the area, for if we increase the area but keep the number of molecules per cubic centimeter the same, we increase the number of collisions with the piston in the same proportion as the area was increased."

If the area of the piston is increased, but number molecules remain the same, would there not be less force as the collisons are spread out?

I'm thinking of this in the same vain as shoes exerting less pressure than high-heels due to the increased area.

Pressure is force over an area, or as an equation you might state that as:

$$P=\frac{F}{A}$$

Which can be rearranged as

$$F=P \cdot A$$

You are correct in saying that if the number of molecules are held constant, that the force would be over a larger area and therefore decreased. As you increase the area but keep the number of molecules constant, the pressure decreases proportionally, and no change in force is seen, but what Feynman was talking about was keeping the amount of molecules per cubic centimeter constant, or in other words keeping pressure constant. This way as area is increased, you see a proportional increase in force.


Number of geometrical isomers? How many geometrical isomers will exist for compound of this type which has cycloalkene having carbon atoms greater or equal to 8 in the middle of chain?

I know that due to ring in middle of chain geometrical isomerism exist and also due to cycloalkene having carbon atoms greater or equal to 8. But I am confused in a case when a  cycloalkene exist in middle of the chain. Whether we will consider geometrical isomers due to cycloalkene or due to both, i.e. the ring and also due to cycloalkene case?

Geometrical isomers refers to isomerism in which the connectivity of atoms is preserved, and changes refer only to the spatial arrangement of the atoms. In other words, it only allows transformations around certain features that can be arranged in different ways in three dimensions (and which aren't just conformers that will easily transform into each other due to molecular motion).

I mention this because your reference to a "cycloalkene [...] having [8 or more] carbon atoms" suggest that you are looking for constitutional isomers (i.e. compounds with the same atoms but different connectivity) and not just geometrical isomers (i.e. compounds with the same connectivity but different spatial arrangement).

If we consider geometrical isomers of the compound you gave, 1) the cycloalkene ring, 2) the side chains and 3) the substitution sites in the ring are fixed (as changing them would produce constitutional isomers), so we are left with isomerism due to cis/trans double bonds.

You have three double bonds in the side chains that will produce $2^3=8$ isomers. Cyclooctenes are in principle large enough rings that their trans form can be isolated, so this leads to $16$ isomers; you can figure them out from your posted example (which is cis in the cycloalkene and trans in the side chains) by changing double bonds.

If you want to figure out constitutional isomers, that's a far more complex task; you'll have to account for ring size, number of side chains, ring substitution pattern, branching, position of insaturations and so on; with a hydrocarbon of this size, the number of possibilities will be huge.


Why is the current draw dropping in water electrolysis We've done an experiment regarding water electrolysis. Basically, we just separated the oxygen and hydrogen gas in a beaker that contains a magnesium sulfate solution. We set the power supply voltage to 12.46V with an initial .09A and final .01A over 45 mins. I was wondering as to what causes the drop in current draw. I have my suspicion that the concentration of the solution is decreased since we're using copper wires, and there was a displacement reaction between copper and magnesium sulfate.

If you look at the copper wires after the electrolysis, the cathode (connected to the negative pole of the current source) will be white. This electrode is coated with Mg(OH)2.

The low activity of copper on the electrochemical scale makes it unlikely (impossible) for copper to displace magnesium from an aqueous salt.

However, the initial voltage is high enough to theoretically evolve metallic magnesium, which would instantly react with water to produce H2 and Mg(OH)2, which will precipitate because of its low solubility. The Mg(OH)2 adheres, at least to some degree, on the copper electrode and interferes with current flow. The resistance to the electrolysis at the same voltage leads to a lower resulting current.

Even at electrode potentials below that needed to evolve metallic Mg, current carried by the aquo Mg++ ion will result in an increased pH and deposition of Mg(OH)2 around the cathode. An interesting question is whether the deposit from a low voltage electrolysis has the same adhesion as from a high voltage experiment. Maybe low voltage electrolysis would deposit Mg(OH)2 farther out in the solution and not be so blocked off by adhesion to the electrode.


Why do only some metals make protective oxide layers? Almost all Alkali metals and Alkaline Earth metals react with air to form their respective oxides, but why does only Magnesium forms a protective oxide layer? What properties of magnesium oxide allow it to form a layer?

The simple answer is that metals which form sufficiently insoluble oxide layers stifle the very process which caused the oxide layer in the first place.
It's not just a matter of energy of formation of the oxide, but whether the first reaction product formed will be soluble or insoluble, then whether it will be adherent or non-adherent. In the case of the alkali metals, the oxide may be adherent (reaction with oxygen produces a layer that sticks), but is so soluble that it dissolves in water and refreshes the metallic surface. Likewise in the case of the alkaline earth metals - even $\ce{CaO}$, which is almost insoluble, is soluble enough to give a pH of 12.5, and soluble enough to be non-protective.
Now, magnesium oxide is quite insoluble, but probably is on the surface as $\ce{Mg(OH)2}$, a secondary product. You might consider that $\ce{Mg(OH)2}$ is $\ce{MgO}$'s failed attempt to dissolve and go away from the surface. Instead, it precipitates, and not only that, it adheres quite well. So further reaction with water is inhibited. However, if you heat the water, the $\ce{Mg(OH)2}$ becomes more soluble, and corrosion proceeds, although not as fast as sodium reacts. In flowing hot water, where the $\ce{Mg(OH)2}$ could be carried away, corrosion would be faster; in a flask, stirring would exert a similar rate increase.
The effectiveness of $\ce{Mg(OH)2}$ as a stifling passivator may be related to its layer structure:
 (File from Commons, by User:Smokefoot)

Like a clay layer, the hydroxide ions fuzz out of each layer of close-packed magnesium ions, and the metal surface could be like a half-layer of $\ce{Mg(OH)2}$, with $\ce{OH-}$ ions sticking out and bonding to separate layers of $\ce{Mg(OH)2}$, or perhaps the metal itself just hold onto a layer of $\ce{Mg(OH)2}$ - or maybe a little of both. This adherence is necessary to stifle corrosion. The clay layer-like characteristics of $\ce{Mg(OH)2}$ are also evident in milk of magnesia, which is a stable suspension - if the suspension were not stabilized by clay-like interactions, it would settle fairly rapidly and need good shaking before use. (I shake it anyway.)
Corrosion on aluminium is stifled by an adherent, quite insoluble oxide, but not completely: the oxide layer continues to thicken, and corrosion continues, diminishing, but never stopping completely. Nature never sleeps.
Iron rusts with $\ce{H2O}$. Iron oxide could be adherent (it is on stainless steels, when mixed with some chromium), but unfortunately, pure iron oxide tends to form a hydrate which has a greater volume than the oxide, and as it forms, it peels away from the metal surface, leaving bare metal or a very thin layer of oxide, which can still corrode. Chloride ion is especially effective in promoting the continuation of corrosion on iron and steel (even stainless) and aluminium.


Labware to handle hot aqueous NaOH This question regarding appropriate handling of hot concentrated aqueous $\ce{NaOH}$ - with particular concern for avoiding dissolution of borosilicate glassware and contamination of the $\ce{NaOH}$ solution and other reagents - led me to think of two possible solutions:

Coat the glassware with a silane, providing say a polyfluorinated or other inert protective coat
Avoid using glass, choose instead metal or plastic, if possible

Is either of these choices reasonable or common, or is there a better alternative? If so, which is best (considering chemical inertness, safety, complexity (including potentially cost), in that order of priority)?

In the case of choice 2, what commercial sources are there for such labware (for handling hot caustic solution for extended period). A cursory search - eg google and Merck website - did not turn up such products.

Note this question is about relatively small-scale labware (benchtop), not for reactors and the like. Volume max $\pu{5 L}$, $\pu{1 M}$ $\ce{NaOH}$, $\pu{>100^oC}$, exposure time >1 h.

Instead of glassware, you might have to think about metallic containers. A 1 M NaOH solution should not be that harsh for stainless steel for a short period of time. Imagine, in the chlor-alkali industry, nothing is made of glass. The electrodes are made of titanium as far as I remember. If we were rich, a platinum lined crucible would have worked! Titanium might work too, very resistant to bases.

Here is a useful link: Selecting Stainless Steel for Caustic Soda

Regarding silanes, the Achilles heel is not the perfluoro functional group on the silane but the siloxane bond itself- i.e., the Si-O-Si ---- (functional group). Base will chew this bond apart into pieces even at pH 8-9. This is the biggest drawback of HPLC silica columns. We have not been able to synthesize any base stable silica bonding.


What is the group reagent for the zero group (ammonium cation) in inorganic salt analysis? Some texts suggest sodium hydroxide is the group reagent for the zero group.

However, some say there's no group reagent for zero group

So which one's right?

Ammonium salts are identified by the release of ammonia upon warming with base.
$$\ce{NH4+ + OH- ->[\Delta] NH3 ^ + H2O}$$
Vogel's Qualitative Inorganic Analysis 5th ed., p 400 writes

6. Test for ammonium ions Boil a little (c. 0.1 g) of the substance with sodium hydroxide solution. The evolution of ammonia, detected by its odour and its action upon red litmus paper and upon filter paper soaked in mercury(I) nitrate solution,* indicates the presence of an ammonium salt.
* Drop-reaction paper, treated with tannic acid and silver nitrate solution, may also be used.


Types of bonds in a molecule For example in dinitrogen pentoxide, $\ce{N2O5}$, covalent as well as coordinate bonds (type of covalent bonds) are present, but it appears that it contains only covalent bond.

What is a proper method to find out which type of bonds are present in a molecule?

Electrovalent bonds are easiest to identify. If a compound is made up of a metal and non-metal/non-metallic radical (like carbonate), then, 99.99% times, it contains electovalent bond. If a compound is made up of 2 or more non-metals/non-metallic radicals, then it contains covalent bond. Coordinate covalent bonds appear mostly with compounds containing Hydrogen element. To identify the coordinate covalent bonds, you can draw the branched structural formula of the compound and see if the shared pair of electrons are coming from the same molecule.


How to search substructures with explicit hydrogen in RDKit? For a substructure search I would like to search for structures containing unfused benzyl.
The idea was to explicitly add hydrogen. But apparently this does not give the expected results.

So, I assume either my understanding or expectations are wrong or I'm using RDKit not properly.
I just learnt (Substructure search with RDKit) that I better use SMARTS expressions with Chem.MolToSmarts() where I get the expected results.

However, if I have much more complex search structures, how can I get from the SMILES to the SMARTS? Do I have always to figure out the SMARTS code myself? Is there maybe a way just drawing a molecule and add explicit hydrogen and search with this somehow? I guess it should be pretty clear (at least to a human) what I want to find (or not find) with structure 2. Are there maybe any tools or code which can assist here?

Code:

from rdkit import Chem

smiles_strings = '''CC1=CC=CC=C1
CC1=C([H])C([H])=C([H])C([H])=C1[H]
C1(C2=CC=CC=C2)=CC=CC=C1
C12=C(CC3=CC=CC=C3C2)C=CC=C1
C12=CC=CC=C1C=C3C(C=CC=C3)=C2
C12=C(C(C(C=C3)=CC=C4)=C4C=C2)C3=CC=C1
'''
smiles_list = smiles_strings.splitlines()

def search_structure(pattern):
    found = []
    for idx,smiles in enumerate(smiles_list):
        m = Chem.MolFromSmiles(smiles)
        if m.HasSubstructMatch(pattern):
            found.append(idx+1)
    print("Structures found: {}".format(found))

smiles_1  = smiles_list[0]
pattern_1 = Chem.MolFromSmiles(smiles_1)
smiles_2a  = smiles_list[1]
pattern_2a = Chem.MolFromSmiles(smiles_2a)
smiles_2b  = Chem.MolToSmiles(pattern_2a)
pattern_2b = Chem.MolFromSmiles(smiles_2b)
smiles_2c = smiles_2b
pattern_2c = Chem.MolFromSmarts(smiles_2c)
smarts_3 = '[cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]'
pattern_3 = Chem.MolFromSmarts(smarts_3)

print("\nSMILES 1 : {}\n# my expectation: [1, 2, 3, 4, 5, 6]".format(smiles_1))
search_structure(pattern_1)
print("\nSMILES 2a: {}\n# my expectation: [1, 2, 3]".format(smiles_2a))
search_structure(pattern_2a)
print("\nSMILES 2b: {} (Mol from SMILES)\n# my expectation: [1, 2, 3]".format(smiles_2b))
search_structure(pattern_2b)
print("\nSMILES 2c: {} (Mol from SMARTS)\n# my expectation: [1, 2, 3]".format(smiles_2c))
search_structure(pattern_2c)
print("\nSMARTS 3 : {}\n# my expectation: [1, 2, 3]".format(smarts_3))
search_structure(pattern_3)

Result:

SMILES 1 : CC1=CC=CC=C1
# my expectation: [1, 2, 3, 4, 5, 6]
Structures found: [1, 2, 3, 4]

SMILES 2a: CC1=C([H])C([H])=C([H])C([H])=C1[H]
# my expectation: [1, 2, 3]
Structures found: [1, 2, 3, 4]

SMILES 2b: Cc1ccccc1 (Mol from SMILES)
# my expectation: [1, 2, 3]
Structures found: [1, 2, 3, 4]

SMILES 2c: Cc1ccccc1 (Mol from SMARTS)
# my expectation: [1, 2, 3]
Structures found: [1, 2, 4]

SMARTS 3 : [cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]
# my expectation: [1, 2, 3]
Structures found: [1, 2, 3]

If you work in a Jupyter Notebook you can visualize your substructure patterns and the search results, so you can see how the patterns work.

from rdkit import Chem
from rdkit.Chem import Draw

smiles = ['CC1=CC=CC=C1','CC1=C([H])C([H])=C([H])C([H])=C1[H]',
          'C1(C2=CC=CC=C2)=CC=CC=C1','C12=C(CC3=CC=CC=C3C2)C=CC=C1',
          'C12=CC=CC=C1C=C3C(C=CC=C3)=C2','C12=C(C(C(C=C3)=CC=C4)=C4C=C2)C3=CC=C1']

params = Chem.SmilesParserParams()
params.removeHs=False # draw and work with explicit Hs
mols = [Chem.MolFromSmiles(s, params) for s in smiles]
Draw.MolsToGridImage(mols, molsPerRow=3)

This will give you the compounds and especially the one with the explicit Hs.

Now we get the patterns and especially the one with the explicit Hs.

patt1 = Chem.MolFromSmiles('CC1=CC=CC=C1')
patt2 = Chem.MolFromSmiles('CC1=C([H])C([H])=C([H])C([H])=C1[H]', params) # search with explicit Hs
patt3 = Chem.MolFromSmiles('Cc1ccccc1')
patt4 = Chem.MolFromSmarts('Cc1ccccc1')
patt5 = Chem.MolFromSmarts('[cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]')

patts = [patt1,patt2,patt3,patt4,patt5]
leg = ['CC1=CC=CC=C1','CC1=C([H])C([H])=C([H])C([H])=C1[H]','Cc1ccccc1 smiles',
       'Cc1ccccc1 smarts','[cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]']
Draw.MolsToGridImage(patts, molsPerRow=3, legends=leg)

Search and highlight the result.

ms = []
patt = []
allsubs = []
for m in range(len(mols)):
    for p in range(len(patts)):
        sub = mols[m].GetSubstructMatches(patts[p])
        if len(sub) > 0:
            ms.append(mols[m])
            patt.append(leg[p])
            allsubs.append(sub[0]) # substructures could be find multiple time - just take the first

Draw.MolsToGridImage(ms, molsPerRow=3, legends=patt, highlightAtomLists=allsubs)

Compounds 5 and 6 have no nonaromatic bonds, so there is nothing to find.


How many grams of solid sodium bicarbonate will be consumed? Excess $\ce{NaHCO3 (s)}$ is added to 510 mL of $\ce{Cu(NO3)2 (aq)}$ $\pu{0.240M}$ for the reaction
$\ce{Cu(NO3)2 (aq) + 2NaHCO3 (s) -> CuCO3 (s) + 2NaNO3 (aq) + H2O (l) + CO2 (g)}$

$$\ce{Cu(NO3)2_{(aq)} + 2NaHCO3_{(s)} -> CuCO3_{(s)} + 2NaNO3_{(aq)} + H2O_{(l)} + CO2_{(g)}}$$
The equation is telling us that for every $1$ mole of $\ce{Cu(NO3)2}$ consumed, there are $2$ moles of $\ce{NaHCO3}$ correspondingly consumed in the reaction.

So, let us first find how many moles of $\ce{Cu(NO3)2}$ is consumed. Here,
$$\text{Molarity} = \frac{\text{#Moles of } \ce{Cu(NO3)2}}{\text{Volume of Solution}}\\ \implies \text{#Moles of } \ce{Cu(NO3)2} = 0.240 \text{ mol.L}^{-1}\times 510  \text{ mL} \times \frac{1\text{ L}}{1000 \text{ mL}} = 0.1224 \text{ mol}$$

Now, think.

$$1 \text{ mole } \ce{Cu(NO3)2}  \iff 2 \text{ moles } \ce{NaHCO3}\\
0.1224 \text{ moles } \ce{Cu(NO3)2} \iff 0.2448 \text{ moles } \ce{NaHCO3}$$

$\text{Weight of } \ce{NaHCO3}= 0.2448 \text{ mol} \times 84.007 \text{ g mol}^{-1} = 20.5649136 \text{ g}$


Why do alkenes react with oxidising agents? My book says :

  The high electron density associated with the double bond means that
  alkenes are attacked by oxidising agents.

And I dont understand how the high electron density in the double bond causes that.

As per my knowledge, oxidation is loss of electron(s) by any species and an oxidising agent is an acceptor of electron(s). So, high electron density on alkenes make them likely to be attacked by oxidising agents.


Trouble understanding interplanar spacing using miller indices We know that two parallel planes have the same Miller indices, this implies that we can have an infinite number of parallel planes close to one another all of which have the same Miller indices.
But it's claimed that

the interplanar distance between two adjacent planes is given as $$d_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}}$$.

I can't understand what adjacent means here?
to me the next parallel plane to any plane is infinitely close to it, hence the $$d_{h k l }=0$$
Can someone please explain what exactly is interplanar spacing?
Thank you.

The lattice planes are a property of the lattice based on dividing up the unit cell. The cell can be divided up indefinitely to give all the Miller indices for example $(\bar{10}\;20\;0);(10\;23\;20)$ etc. Atoms do not need to be on a lattice plane to add to the diffracted signal. The structural information is in the intensity of the diffracted spots not their position. (2 2 4) is parallel to (1 1 2) etc. as shown in the figure. In the general case it is usual to use vectors to find the perpendicular distance using formulas for finding a point at a given distance from a plane.
(Edit. I added (2 2 4) to indicate that it is a different plane but parallel to (1 1 2), in crystallography it is conventional to make all parallel planes have the same Miller indices, (h k l) (as they all scatter in the same way) by removing any common factor. )


Using Beer-Lambert law to calculate light intensities Supposing the mean absorption coefficient of seawater in the visible region is $\pu{0.8 m-1}$, calculate the depth at which a diver will experience half the surface light intensity.
I know that you have to use $A = -\log \left(\dfrac{I}{I_o}\right) = \epsilon cl$ but I'm not sure how to proceed

In my opinion, Beer–Lambert law is no use. I would rather write
$$\frac{\mathrm dI}{I\,\mathrm dl} = \pu{0.8 m^-1},\tag{1}$$
which leads to the integrated formula
$$\ln\left(\frac{I_0}{I}\right) = 0.8 l.\tag{2}$$
And, if $\displaystyle\frac{I}{I_0} = 0.5,$ the corresponding depth is
$$l = \frac{\ln 2}{0.8} = \pu{0.846 m}.\tag{3}$$


What property of zinc chromate makes it a good anticorrosion paint What properties of zinc chromate make it a good anticorrosion paint?  I know that stainless steel is protected by a layer of chromium oxide.  Is it because the zinc chromate doesn't flake off the metal it is protecting?

Long story short, if it's scratched in the presence of water(humidity) the chromium ion inside the chromate will get reduced to Cr 3+ from Cr 6+. This 3+ ion will then form a Chromium hydroxide compound,which is insoluble in water and corrosion resistant, inside the scratch. This reduction of the chromium ion first is what inhibits the oxidation of the underlying metal.

Here's the full article http://www.electrochem.org/dl/interface/wtr/wtr01/IF12-01-Pages34-38.pdf

The answer: Being water resistant in the first place and being capable of producing ions that can be reduced forming another corrosion resistant compound( chromium(III)oxy-hydroxide.)


Question regarding the colour of Benzoic acid and Benzyl alchohol This question is primarily based on the Cannizzaro reaction, where a base induces a disproportionation reaction of non-enolizable aldehydes (benzaldehyde, formaldehyde, etc) to form an alcohol and a carboxylic acid.
In the video by Nile Red exploring the Cannizzaro reaction (https://www.youtube.com/watch?v=qEBC204WTKs), the product mixture contained both Benzoic acid and Benzyl alcohol and had an orange color. I am aware that the benzoic acid was in the form of sodium benzoate due to sodium hydroxide also reacting with the acid produced.
My question is, where does the colour come from?
I am guessing it is either from the benzyl alcohol or sodium benzoate, but I have learned that most organic compounds are colorless.
Moreover, if the aldehyde used was something like formaldehyde, will the orange color also form?
Thanks in advance for the clarifications.

none of the components you mentioned (Benozic acid, Benzyl alcohol, sodium benzoate) has a yellow colour.
The colour in organic compouns is closely tied to ther delocalized sytems of pi-electrons and depends on the gap between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO).
When the energy of this gap corresponds to a wavelength that is visible, the compound absorbs this wavelength and the colour of the compound is the complementary colour.
A good example are condensated aromats. For benzene or naphthalene the gap between HOMO and LUMO is too big therefore, the corresponding wavelength is in the area of UV light.
For anthracene, the absorbed wavelenght is violett and therefore the colour of the compound is yellow.
Back to your question. Because the reaction happends under strong basic conditions, high temperature and with many aromatic compounds, I think sidereactions will occure where multiple aromats condensate together and therefore the solution turns yellow to orange.


What drives vibrational cooling in an excited stated? As we excite a molecule from its ground state, $S_{0,v=0}$, to some excited state in a higher vibrational state, i.e. $S_{1,v'=3}$, what drives vibrational cooling within that excited state manifold (such as $S_{1,v'=0} \leftarrow S_{1,v'=3} $?) This is a fast process, typically on the order of hundreds of femtoseconds, and it occurs prior to any photorelaxation (as per Kasha's rule). Due to the vibrational density of states being quite high near the first excited state, is it the anharmonic coupling to these states that drives vibrational cooling? Furthermore, is there a quantum mechanistic/mathematical way of depicting the probability/rate of this process (as in, what would the relevant perturbation term $\hat{H}{'}$ for dealing with this problem in the context of Fermi's golden rule)?

Normal modes are technically orthogonal so no energy could flow from one to another, but of course this is just a mathematical construct to make life simple for us, and in reality  'anharmonicity' (physical and electronic) will allow the energy to move about. This is to say that these modes are able to couple to one another.  Fermi golden rule is the usual starting point although the density of states may not be that high at low vibrational levels. The calculation involves calculating the Franck-Condon weighted density of states from the initial to final level. The process is called IVR or intramolecular vibrational redistribution.

In many types  molecules, when isolated in the vapour phase, the initially excited level can still remain excited for many nanoseconds and the fluorescence lifetime and spectrum  from these levels can easily be measured, e.g. naphthalene, pyrene, benzanthracene etc. This is particularly the case when the FC excited mode has a distinctly different frequency and displacement to other modes so that the FC factors to other vibrational levels is small. An interesting case is that of many types planar aromatic dyes which show persistent wavepacket behaviour of the FC excited mode even in solution at room temperature.

The theory of radiationless transitions has been thoroughly worked out mainly for intersystem crossing but the theory is essentially the same with different operators. Look for papers from the 1970's by S. H. Lin et al, W. Siebrand et al,  J. Jortner et al. and K. Freed et al.


Storing hydrofluoric acid before the invention of plastics The first person to synthesize hydrofluoric acid in large quantities was Carl Wilhelm Scheele in 1771. This acid is known for its ability to corrode glass and metals.

What materials were the containers in which it was stored before the invention of plastics?

In the original 1771 experiment, Scheele used a very simple setup consisting of a glass retort with a glass receiver (round-bottom flask).

Yes, the glass was etched to some degree by the fumes, but it was not drastic enough to destroy the apparatus.

From Anders Lennartson's The Chemical Works of Carl Wilhelm Scheele [1, p. 22]:

  3.1 Publication 1. Investigation of Fluorite and Hydroﬂuoric Acid

  Original publication: Kongl. Vetenskaps Academiens Handlingar. 1771, 32,
  120–138.
  Original title translated to English: Investigation of ﬂuorspar and its acid

  …

  The second paragraph of the paper deals with the action of acids on ﬂuorite, and
  this is where the important discovery is made. Scheele mixed powdered ﬂuorite
  with sulphuric acid (oil of vitriol) and heated it in a retort (Fig. 3.3) whereupon corrosive fumes were emitted. He repeated the experiment with some water in the receiver. He found that a white crust was formed on the water surface, and that the water contained a dissolved acid. The inner surface of the retort was etched by the fumes. Scheele identiﬁed the white crust as silicon dioxide (siliceous earth) and the residue in the retort as calcium sulphate (selenite or gypsum). This lead Scheele to the conclusion that “Fluor [ﬂuorite], therefore, consists principally of calcareous earth [calcium oxide], saturated with a peculiar acid; the nature of which we shall investigate hereafter”.

  Fig. 3.3 Schematic view of the apparatus Scheele used for preparing “acid of ﬂuorspar”

In response to critics, to prove that the glassware itself was the source of the white silicon dioxide powder, Scheele improved the setup, protecting the inner walls with tin and wax.

Continuing from [1, p. 98]:

  3.60 Publication 60. Experiments with Hydroﬂuoric Acid

  Original publication: Chemische Annalen für die Freunde der Naturlehre.
  1786, part 1, 3–17.
  Original title translated to English: New evidence for the characteristic nature of the acid of ﬂuor spar

  …

  Scheele powdered ﬂuorite in a brass mortar, mixed the powder with sulphuric
  acid (oil of vitriol) and transferred it to a retort lined with tin. The receiver was lined with wax, and partly ﬁlled with water. This time Scheele obtained a distillate free from precipitated silicon dioxide (siliceous earth), and addition of potassium hydroxide gave no precipitate (i.e. no $\ce{K2SiF6}$). Thus, this was pure hydroﬂuoric acid, $\ce{ HF_{(aq)} }$; similar results had been reported by Scheele’s friend, J.C.F. Meyer in Stettin.

References

Lennartson, A. The Chemical Works of Carl Wilhelm Scheele; Springer Berlin Heidelberg: New York, NY, 2017. ISBN 978-3-319-58180-4.


Relationship between dipole moment of a molecule and electronegativities of constituent atoms I want to know if I can calculate the dipole moment of a molecule given the electronegativity values for the atoms comprising it.

I think that dipole moments determine the polarity of molecules, so if one molecule has a larger dipole moment than another, the one with the larger dipole moment is more polar than the other. Is this correct?

Can the magnitude of how polar a molecule is be calculated from the difference in electronegativities among the elements comprising the molecule?

Yes you can. By looking at the different electronegativities of each element in a molecule, you are able to determine the dipole moment for each bond. Then by adding the dipole moments for each bond, you are able to determine whether a molecule is non-polar or polar and if it is polar, how polar it is.

For example, in $\ce{CH4}$, the carbon atom is bonded to four hydrogen atoms. The electronegativity of carbon is 2.55, while the electronegativity of hydrogen is 2.2 This means that for each $\ce{C-H}$ bond, there will be a dipole moment from the hydrogen to carbon atom. However, since methane has a perfect tetrahedral structure, the four dipole moments cancel each other out so that there is no net dipole moment. Hence, methane despite having polar bonds, is actually not a polar molecule, shown below.

Another example that we can consider is ammonia, $$\ce{NH3}$$. Ammonia consists of 3 $\ce{N-H}$ bonds. The electronegativity of nitrogen is 3.04. That means that ammonia consists of 3 polar $\ce{N-H}$ bonds. The molecular structure of ammonia is trigonal pyrimadal and hence unlike methane, when you add up the 3 dipole moments, they don't cancel each other out. Instead, they result in a net upward dipole moment. Hence, the molecule itself is also polar, shown below


Ka relating to Dissociation of Weak acid and Water I am studying $K_a$ currently and there is a passage in the book that doesn't appear to make sense.

It states that from the equation

$$\ce{HA <=> H+ + A-}$$

that $$K_a = \frac{[H^+][A^-]}{[HA]}$$

The following passage says When dealing with weak acids, you can assume that dissociation of the acid is much greater than the dissociation of water.

It continues to say that you can assume all the H+ ions come from the acid.

hence $$K_a = \frac{[H^+]^2}{[HA]}$$

What I don't understand is the dissociation of water.

The book gives the equation of dissociation of water as the following -

$$\ce{H_2O(l) + H_2O(l) <=> H3O+ + OH-}$$

but where does the H+ come from from this dissociation - it's made a Hydroxonium ion?

Is it because the $H_3O^+$ can act as the conjugate acid and release the proton?

Please help!

Yes, water autodissociates according to the equation $\ce{2H2O <=> H3O+ + OH-}$.

But I should point out that the equilibrium constant is written as though the concentration of water itself is a constant, so the reactions becomes:

$\ce{H2O <=> H+ + OH-}$.

and the equilibrium is expressed as:

$K_\rm{w} = 1\times10^{-14} = \ce{[H+][OH-]}$

The pH is $\text{-log}\ce{([H+])}$ thus the pH of pure water is 7.

This also has to do with significant figures. Few pKa's are known to better than two significant figures. So:

For weak acid with pKa < 5 you can ignore water contribution.
For acid with pKa > 9 you can assume all the  $\ce{H+}$ is from water and that all of the acid is essentially present as the HA form.
For 9 > pKa > 5 you have to consider both ionization of the acid and water.


Will sulphuric acid react with water produced in an esterification process? Sulphuric acid acts as a catalyst to catalyze the reactions of alcohol and carboxylic acid to form ester. Alcohol reacts with carboxylic acid to produce ester and water.

However, as sulphuric acid have strong afinity towards water, will sulphuric acid react with the water produced in an esterification process?

If so, will it mean that the equilibrium is shifted to the right as water is reacted with sulphuric acid? And will this mean that there would be less sulphuric acid present as a catalyst to catalyse the reaction?

Sulfuric acid acts as catalyst meaning it is regenerated at the end of the reaction. In the presence of water or other bases, sulfuric acid will primarily be deprotonated. $$\ce{H2O + H2SO4->H3O+ + HSO4-}$$

It is occasionally more correct to consider the water acting as a catalyst. Any time that you have a strong acid acting as a catalyst, because its deprotonation is much more likely than the organic reaction, it is nearly completely deprotonated at a given time. Thus the protonated $\ce{H3O+}$ will act as your catalyst. protonating the carboxylic acid and deprotonating the ester.


Why can't LiAlH4 react with alcohols? In the reaction of $\ce{LiAlH4}$ with carboxylic acids, deprotonation is followed by a step in which $\ce{O-AlH2-}$ acts as a leaving group. The mechanism is given here in this answer.
$\ce{LiAlH4}$ does not, however, reduce alcohols. Why doesn't a similar reaction occur with alcohols - deprotonation, followed by substitution by a hydride ion and removal of $\ce{O-AlH2-}$?

The main issue is that the Al needs to remove its hydride. With a carboxylic acid and/or an aldehyde, it can stick its hydride onto the carbonyl carbon without issue. But the carbon bonded to the alcohol cannot take on a hydride.


Why is Acetic acid (pKa = 4.76) stronger than carbonic acid (pKa = 6.36)? In order to try and understand this, I evaluated it via stability of conjugate bases. The acetate vs carbonate ion. The only difference is what's attached to the carboxyl group, the $\ce{-CH3}$ and the $\ce{-OH}$. Both anions seem to have equal resonance structures as I think resonance mainly only happens on the carboxyl group.

I think that because the methyl group is just larger than the hydroxyl group, it is able to better stabilize the negative charge on the molecule overall. But I also remember hearing that carbon can have a positive inductive effect, meaning it pushes electron density toward the carboxyl group, which would make it less stable.

I originally thought that carbonate would be more stable because the oxygen would pull the electron density toward itself, making the charge more evenly distributed.

If somebody could explain with diagrams I would appreciate it. I lost access to my software that did electron modeling functions for me.

Well actually the $\mathrm{pK}_\mathrm{a1}$ for carbonic acid at 25 °C is 3.6. The pka = 6.36 is also including the equilibrium with dissolved $\ce{CO2}$ in aqueous solution.


Equation for reaction of tin and iodine to produce tin(IV) iodide Tin(IV) iodide is prepared by direct combination of the elements.
Add 2.00 g of granulated tin to a solution of 6.35 g of iodine. Write an equation for this reaction.

Attempt:
1. Determine the amount of tin using moles = mass/molar mass. This yields a result of 0.017 mol $\ce{Sn}$.
2. Determine the amount of iodine using moles = mass/molar mass. This yields a result of 0.5 mol $\ce{I}$.
3. Determine the ratio of tin to iodine. Using relative amounts, for every 1 mol $\ce{Sn}$ there are 3 mol $\ce{I}$, so the ratio is 1:3. Because iodine is a diatomic molecule, I will adjust accordingly, so the ratio becomes 2 mol $\ce{Sn}$ to 3 mol $\ce{I2}$ or 2:6.
4. Construct the formula:
$$\ce{2Sn + 3I2 -> SnI4}$$
with excess tin and iodine.
Is my procedure correct?

The balanced chemical equation is independent of the amounts of each reactant added. In this case we can easily deduce the equation to be: $$\ce{Sn + 2I2 -> SnI4}$$

What the given masses tell us is that there is excess tin being added. We can see this because the $\ce{Sn:I2}$ ratio added is $1:1.5$ whereas the stoichiometric ratio is $1:2$ and so there is less iodine than is required to completely react with the added tin - or in other words there is an excess of tin.


Solution for Electroplating Selenium? I need to electroplate selenium onto a conductive metal (for an experiment). The end goal is a thin film of pure, elemental selenium. Unfortunately I can't find anything on the internet: Ideally, I'd like a simple, inexpensive, and easy to make and/or find solution. Does anyone have any suggestions. (Also tips on suitable anodes would be helpful too!)

I'm not sure of any conventional ways of electroplating selenium, but you could use a technique such as magnetron sputtering to spray a thin film of selenium onto another metal. Heck, you could even use glass or wood if you wanted to do it that way. And if you're going for thin, this technique can get you some of the thinnest, smoothest layers of metal out there. A major downside is that your source metal sample has to be pretty pure selenium. But still, I think It's completely worth it.

However, If you don't have access to one of these devices, They can get expensive. Most major colleges with significant science majors will have one of these or something similar to coat samples for electron microscopy, or other reasons. Be sure to check there before going off to buy your own. Also, there are legitimate YouTube videos out there that show how to make your one device for magnetron sputtering. Check out this one if you're interested.


Reaction between ruthenium and bleach I have recently seen a video of ruthenium on youtube. It says that ruthenium is invulnerable to many acids, especially aqua regia, but it gets destroyed by a simple household chemical, bleach. The video did not give a proper explanation about why this happens, it only gave the reaction mechanism. What is the reason for this Achilles' heel of ruthenium?

This paper actually explain the reason for which ruthenium is vulnerable to bleach. Bleach actually contains $\ce{OCl^-}$ ions which acts as an oxidising  agent and thus oxidises ruthenium. But the research on this phenomenon was done by using sodium hypochlorite, $\ce{NaOCl}$ in presence of a caustic alkali.

The experiment was done in two-fold process. First only using $\ce{NaOH}$ and the second using both $\ce{NaOCl}$ and $\ce{NaOH}$. In the first attempt, there was nothing occuring. The solutions with  $\ce{NaOH}$ were stable for a considerable time, but after several weeks began to break down, giving a deposit of the black oxide. In the second process, when a little of hypochlorite was given, it gave a orange-red solution as you can see in the above video.

Some more attempts were made by repeating the experiments and found that concentrated alkali hypochlorite more or less dissolved any ruthenium present. From these experiments, the conclusions were made that ruthenium, is rather soluble in concentrated alkaline solutions of hypochlorites and this might help to distinguish ruthenium from other metals. The overall reaction of this process is(from the video):-

$$\ce{Ru + 3NaOCl + 2NaOH -> Na2RuO4 + 3NaCl + H2O}$$

$$\ce{Ru + 7NaOCl + 2NaOH -> 2NaRuO4 + 7NaCl + H2O}$$

$\ce{NaOCl -> 2NaCl + H2O}$ (ruthenate catalyzed)

$$\ce{Na2RuO4 + NaOCl + H2O-> RuO4 + NaCl + 2NaOH}$$


IUPAC of heterocyclic compounds terminology During my studies of heterocyclic compounds I learned that monocyclic compounds has prefix+ stem+ suffix.
the prefix is coming from the heteromolecule name (e.g oxgen --> oxa- )

however the stem and suffix is ambiguous to me. I can link some of the stem (e.g et --> from tETra)  but what about all other stems and suffixes. Where do they came from?

It will be very helpful if the source of terminology is documented here.

This way of naming heterocycles is known as the Extended Hantzsch-Widman system of nomenclature (the above link is a free reproduction of a publication that revised this system : W.H. Powell, Pure Appl. Chem., 1983,55, 409-416).

The stems do actually come mainly from numeric prefixes. " 'ir' from tri, 'et' from tetra, 'ep' from hepta, 'oc' from octa, 'on' from nona, and 'ec' from deca."Source. They are mainly taken from the middle letters of the numeric prefixes. As for "ol" and "in" for 5 and 6, these are holdovers from the intial system from 1887 that was developed for nitrogen containing rings, so my guess is they were kept to maintain the connection to the endings of pyrrole and pyridine.


Where does the charge on an electrode, responsible for the formation of a double layer once submerged in an electrolyte, come from? I've started exploring electrical double layers and electrode-tissue interfaces whilst dealing with deep brain stimulation and stimulus artefact. I've heard many times that what causes the double layer to form is a net charge at the surface of the submerged electrode that then quickly attracts ions (cations or anions depending on the surface charge sign).

I always thought that in general, objects are neutral and don't carry charge around, where then, does the surface charge on electrodes come from?

This can be answered on more than one level. Where did the word "electrode" come from, to begin with? Supposedly it is due to the fact that you apply some electric potential to that metal needle (stuck into the brain of some creature or other electrolytic medium), and that's why the electrode gets charged.

Now what if we apply no potential? Well, for example, if two pieces of different metals are in contact, they would form a galvanic couple and develop some potential on their own, resulting in galvanic corrosion.

But wait, what if there is just one metal in the electrolyte? It turns out that some contact potential would develop even then. As the link says:

  Electrode potential appears at the interface between an electrode and electrolyte due to the transfer of charged species across the interface, specific adsorption of ions at the interface, and specific adsorption/orientation of polar molecules, including those of the solvent.

Think of a metal in a distilled water (which, BTW, is a poor electrolyte). Inevitably, a tiny portion of metal would dissolve, for such is the nature of chemical equilibrium: it is almost never shifted all the way to one side. Now, that metal would be in the form of cations, and the extra electrons would stay in the bulk, giving it a negative potential (not so tiny, because the electric forces are quite powerful). And if we consider other possible effects, especially if the water is not distilled but contains some ionic species, then the resulting potential on the metal can be pretty much anything (and not necessarily negative).

All in all, a random piece of metal in a random electrolyte will most likely have some potential difference with the electrolyte and hence form a double layer.


How can I build the wyoming like montmorillonite unit cell in avogadro for a molecular dynamics simulation? I am trying to build a unit cell for Wyoming-like montmorillonite, using Avogadro software. I am not sure how can I get access to the atom coordination and how to generate the .xyz file at the end? Is there any file that I can use as guidance during this process? Actually, I would appreciate any suggestions in this regard? please do not hesitate to share your knowledge with me.

I would suggest you try and find the unit cell (.cif file) of the compound in question from the CCDC database, and then use Mercury (a free version is available at the CCDC website) to create the system.
You can of course use Avogadro if that is what you like, but in my experience the coordinate files it produces for super cells are quite messy.


Why does lower concentration cause the metal atoms to dissipate more? In a concentration cell, one can measure a voltage because in one of the half cells (the one with lower concentration electrolyte), the atoms tend to dissipate more into the electrolyte as ions, leaving more electrons behind (compared to the other half cell), thus creating a current.
I am not sure if I understand, why a lower concentration causes the atoms to dissipate more. Is it because the atoms of the metal have more contact to water?

It is not primarily that the process $\ce{M(s) -> M^n+(aq) + n e-}$ is faster for more diluted solutions.
The main reason is that the opposite process $\ce{M^n+(aq) + n e- -> M(s)}$ is slower for more diluted solutions.
As the net effect, there is dissolution of $\ce{M^n+(aq)}$ ions and the electrode gaining more negative potential, until the rate of both processes gets equal and the electrode reaches its equilibrium potential.
Because the rate of ion dissolution grows with increased potential,
while the rate of ion depositions grows with decreased potential.
The final effect is, the electrode in then more diluted ion solution has the more negative potential.


Is the condenser for a Soxhlet extraction left open on top My question is about the condenser used in a Soxhlet extractions. It's unusual looking compared to the rest of condensers, used in distillation for example, I've noticed it has a thin and short tube on top and my question is, do one leave it open when performing a Soxhlet extraction or should it be stopped with a stopper or something?

First thing, you should NEVER perform any heated reaction with a stoppered condenser. You shouldn't even be allowed to work independently in the lab if this rule is not coming automatic to you. I am not sure what you mean by a special condenser being used in Soxhlet extractions. You can use any condenser that fits. Since you don't do a distillation you can use a shorter condenser so your whole Soxhlet set-up has a reasonable height.


Why does buffer capacity of tartrate buffer generally decrease as the buffer pH increases? From what I see, the buffer capacity of tartrate buffer is generally decreasing as buffer pH increases. Why is this? Also, does this mean that tartrate buffer doesn't have a maximum buffer capacity at pH=pKa?

Or is there a more reliable source where I can get the buffer capacity vs. buffer pH graph for buffers of tartaric acid? This image is from a blog, and I can't find the real source of this image, even with Google image search.

Generally, the maximum buffer capacity is at $\ce{pK_a}$ . The tartaric acid is somewhat special for 2 reasons:

It is a diprotic acid with both $\ce{pK_a}$ very close, with the $\ce{pK_{a1}}$ rather low, being affected by the reason 2. :

$$\ce{pK_{a1}}=2.89,\ce{pK_{a2}}= 4.40 (L+)$$
The solution buffer capacity (not limited to presence of specific buffer substances) generally increases toward $\ce{pH}=0$. It means, the buffer capacity is not given by concentration of  of conjugated acid and base, but also by concentration of $\ce{H+}$ itself.
That in large extent masks the $\ce{pK_a}$ maximum. It means, at a slope, a local peak must be big enough to be a peak.


Why are tertiary carbon atoms of polymers more prone to photo-oxidation and -degradation? In Polymer Chemistry by Seymour-Carraher:

  Saturated linear polymers such as HDPE are resistant to degradation, but slow degradation will occur in presence of $\ce{O2}$, UV, heat. Since tertiary carbon atoms are more readily attacked, PP is less resistant to degradation than HDPE.

It means a tertiary carbon atom makes polymer less resistant to photo-oxidation and photo-degradation. What makes tertiary carbon atom more likely to be attacked?

This is admittedly a very brief and somewhat tempting answer and it might not be complete.

Photo-oxidation and -degradation occurs mainly via a radical mechanism. This is outlined on Wikipedia in more detail. In principle the initial step produces bond cleavage via UV radiation and with it radicals. Let $\ce{[P]_{$n$}}$ be a polymer, then $\ce{.[P]_{$n'$}}$ and $\ce{.[P]_{$n''$}}$ are decomposition products of this polymer, both of radical nature.
$$\ce{P ->[h\nu] .[P]_{$n'$} + .[P]_{$n''$}}$$

In a saturated non chained polymer a homolytical bond cleavage would lead two primary radicals, i.e. $\ce{R-CH2.}$, which are very unstable. In a chained polymer, like PP, if the bond break occurs at the tertiary carbon, there would be a secondary radical $\ce{R-\overset{\bullet}{C}H-CH3}$ and another primary radical. Homolytical bond cleavage of a hydrogen carbon bond can even lead to a tertiary radical, $\ce{R-\overset{\bullet}{C}(CH3)2}$.  All in all this can lead to a higher concentration of radicals, and therefore more access points for oxygen to attack, hence propagating the chain. One of the driving forces is the gain in entropy of this process.

$$\begin{align}
\ce{.[P]_{$n''$} + O2 &-> [P]_{$n''$}OO.}\\
\ce{[P]_{$n$} + [P]_{$n''$}OO. &-> [P]_{$n''$}OOH + .[P]_{$n$}}\\
\ce{[P]_{$n''$}OOH &->[h\nu] [P]_{$n''$}O. + .OH}\\
\ce{[P]_{$n$} + [P]_{$n''$}O. &-> [P]_{$n''$}OH + .[P]_{$n$}}\\
\ce{[P]_{$n$} + .OH &-> H2O + .[P]_{$n$}}\\
\text{etc.}&\phantom{\ce{->}~~}\text{pp.}
\end{align}$$


Why mass spectra of 2-methylbutane do not show big peak at 71 coming from tertiary carbocation? Mass spectra of 2-methylbutane show strong peak at 57 by loss of methyl radical to give secondary carbocation while a loss of hydrogen radical from 2-position would form tertiary carbocation with m/z = 71 peak.

Since rules suggest that it is the stability of carbocation that determines the fragmentation pattern.So this process must be more favorable process still we do not get base peak at 71 instead we get a small peak at 71.

Same pattern is observed in case of 2-methylpentane where M-1 peak is almost absent. Other compounds also do not favor loss of hydrogen radicals to give M-1 peak.

I had read about this somewhere that it had something to do with energy of overall process but not able to recall. Kindly help me to explain the reason behind this phenomenon.

How do you know that losing a CH3 radical gives you a secondary carbocation?
Maybe it will perform a hydride shift and you land at the tertiary carbocation.
Always take this into account, because you can shoot yourself in the foot if you assume carbocations to be too stable. In EI, they rearrange all the time (e.g. benzylium <-> tropylium).
However, seeing that the ion with $m/z~57$ subsequently looses $\mathrm{\text{:}CH_2}$, it seems reasonable to assume that the secondary carbocation is formed here.

The important point is the following: You propose that instead of loosing the methylradical, yielding a secondary carbocation (or a tertiary, considering the argument from above), there should be a loss of a hydrogen radical.
However, the hydrogen radical is far less stable than the methyl radical, and I assume this is the reason for the regioselectivity, here.

Also keep in mind that you supply a lot of energy (70 eV). Hence, the induced fragmentation processes are highly endothermic, and should be under kinetic control.

If you really want to find out what is going on, you should probably have a look at McLafferty's/Turecek's classic textbook "Interpretation of Mass Spectra".


Dissolving Tylenol Science Experiment - why is it fastest in distilled water? My son has a science experiment for class to dissolve Tylenol (325mg acetaminophen) in 3 different solutions (125 mL volume each):

Distilled water
Vinegar (1 part white vinegar to 4 parts water)
Salt water (1 tablespoon sea salt)

All solutions were at room temperature.  As far as I could tell, the pill had no fillers and there was no coating on them.  On the bottle, the pills are described as 325mg acetaminophen, and there is no mention of non-medical / filler ingredients.

We used pH test strips, and it seemed the vinegar solutions was a pH of about 6, the water 7 as expected, and the salt water 8 or 9. Edit: it may be the pH was 7 for the salt water, see comment below.

His hypothesis was the Vinegar solution, being acidic would dissolve the fastest.  However, the Tylenol dissolved fastest in the distilled water.  The salt water was slowest, and the vinegar solutions was second.

We also tried pure vinegar, and this dissolved even more slowly than the diluted vinegar used at first.

I should add - it didn't really dissolve in any of the solutions, the pill just came apart into small particles.  Even stirring the solutions afterwards didn't seem to dissolve completely.

I am trying to help my son to determine why distilled water broke up the pill the fastest, but I'm afraid I don't know enough to help.

Any pointers, or websites, where I can help him to research greatly appreciated.

Water can only dissolve a certain amount of solids at a given temperature before it reaches saturation. Distilled water has been boiled and captured in gas phase, almost completely eliminating the presence of dissolved solids.  This lack increases the number of ' water to solid' molecular interactions that will take place and so speeds up the disolving.  The same would be true of any solute.


Separating plant extract from ethanol solution I have to do an experiment which involves getting the extract of certain leaves. I had decided to macerate it in ethanol (70%) solution and then to evaporate the solvent using water bath/rotary evaporator.
Due to the pandemic, I couldn't get access to labs to conduct this experiment. Is there any other way to evaporate the solvent at home without the need for water bath/rotary evaporator?
This extract will be used for testing antibacterial effects. So, I'm kinda concerned about simply heating up the solution because it may destroy the necessary compounds. Would it work if I try to heat the solution in lower temperatures?
For what it's worth, the compounds I'm trying to extract is Quercetin and ß-sitosterol.
Thank you,

Free evaporation or using a water bath ?
Both ethanol and water evaporate at any temperature, just with the exponential-like rate. ( Washed clothes will dry up even at freezing temperatures ).
If living in area with currently cold weather, consider using evaporation on the central heating radiator, parhaps partially isolated.
Free evaporation could be a safe bet, but slow evaporation can support oxidation of dissolved substances, so I would try both free and heated ways.
I remember that in my young-chemist days, I had bought a water stream vacuum pump, based on the Bernoulli equation.


What is ideal water hardness for brewing coffee? While procrastinatingworking, I am wondering whether it is optimal to use softened (or deionized) water for brewing coffee, rather than tap water which is in my area rather hard.

My gut feeling is that extraction of the coffee beans' aromatic molecules will be made more efficient if there are no ions on the water, but is this backed up by science?

According to a discussion I had on this matter with a friend in the coffee processing industry, you should definitely consider yourself lucky for your naturally hard water.  We live in a region with very soft water and he pre-treats all of his coffee-making water with a bed of ground granite (as a physical substrate) and powdered gypsum $\ce{(CaSO4 *2H2O)}$.  For individual use he suggests mineral water, defined in Wikipedia as:

  "The U.S. Food and Drug Administration classifies mineral water as
  water containing at least 250 parts per million total dissolved solids
  (TDS), originating from a geologically and physically protected
  underground water source."

The "taste testing" conferences attended by my friend have rigorously determined that DI water gives the "blandest" taste and water high in calcium and sulfur containing compounds to be the best.

Unfortunately, the chemical basis for this is poorly understood.  It could certainly be pH related, and that my friend's treatment bed just raises the pH which in turn better extracts the desirable components of coffee relative to the undesirable.  The same could be true of the mineral water and hard water in general should have a higher pH than soft water.


Conversion of carboxylic acid to alcohol Is it feasible to convert carboxylic acids to alcohols? I know the reverse is done during the oxidation of alcohols (e.g. in the production of vinegar), but is it possible to do it the other way round?

For example: the conversion of methanoic/formic acid ($\ce{HCOOH}$) to methanol ($\ce{CH3OH}$)?

Directly using $\ce{LiAlH4}$ works, but borane is probably a better choice:

  Borane ($\ce{BH3}$) is particularly good for carboxyl groups and permits selective reduction of them in the presence of many other groups (although the reaction with double bonds takes place at about the same rate in ether solvents). For many years, borane was the reagent of choice for this reduction. Borane also reduces carboxylic acid salts. (March 7ed, p 1515)

Clayden 2ed, pp 531-2 also discusses the borane reduction. Here's a small portion of it:


Reaction of Halogens with water The reaction of fluorine with water is very different from that of chlorine with water.
$$\ce{2F2 + 2(H2O)->O2 + 2HF}\tag1$$

$$\ce{Cl2 + H2O ->HCl + HOCl}\tag2$$

I want to understand two main things:

Why hypochlorous acid is formed in second reaction why doesn't it proceed like the first one?
Has it got to anything with reduction potentials? chlorine and oxygen are very close and is that reason?

I found  this where a related question was discussed here I expect the answer to describe what happens in reaction of any halogen with water and how does its reduction potential play a role in it also are $\ce{HOI}$ and $\ce{HOBr}$ also formed?

Why hypochlorous acid is formed in second reaction why doesn't it proceed like the first one?
  Has it got to anything with reduction potentials? Cl and O are very close and is that reason?

Your intuition in 2. is a good guess and could answer the question mathematically but would be an indirect explanation. If you look closer at reaction 1 you will see that oxygen is oxidized by fluoride where as in reaction 2 oxygen is not oxidized. From a re-dox perspective in the first reaction your equation 1 looks something like:

$$\ce{H2O -> 2H+ + 1/2O2 + 2e-}\tag A\\
\text{and}$$
$$\ce{F2 + 2e- -> 2F-}\tag B$$

Here the fluorine molecule takes 2 electrons from the oxygen atom in an irreversible process. But for equation 2 the redox reaction is more like:
$$\ce{Cl2 <=> Cl- + Cl+}\tag C$$
Here the chlorine is oxidize by the other chloride molecule and reacts with the water to form $\ce{HCl}$ and $\ce{HOCl}$

$$\ce{Cl+ + OH- <=> HClO} \qquad\text{and}\qquad \ce{Cl- + H+ <=> HCl}\tag D$$

  side note: $\ce{HOCl}$ would more properly be presented as $\ce{HClO}$ since chlorine has a $+1$ formal charge but not $\ce{ClHO}$ since $\ce{ClO-}$ is the conjugate base.

Note that the reactions for clorine are revresable now whereas fluorine was not. Combining equations C and D give your original equation 2

$$\ce{Cl+ + Cl- + OH- +H+ \implies Cl2 + H2O <=> HCl + HOCl}\tag2$$

Now for the why. The answer comes from the electronegativities of the elements. $(X_\ce{F} = 4.0 > X_\ce{O} = 3.5 > X_\ce{Cl} = 3.2)$ as you can see fluorine's electronegativity is greater than oxygen's which allows flourine to oxidize oxygen and oxygen's electronegativity is greater than chlorine's which mean chlorine cannot oxidize it. Further Hydrogen is already oxidized and so the only thin for chlorine to oxidize is itself.

Since the other halogens also cannot oxidize oxygen, this equation can be generalized for all of the halogens that are not fluorine as well:

$$\ce{X2 + H2O -> HX^{(I-)} + HX^{(I+)}O}\tag{X = Cl, Br, I}$$


Is there a chemical compound can transform heat into light at 70 °C? Is there is compound that can transform heat into light as phosphorescence / thermally generated photons at around 70 °C, in a reversible reaction?

For example, is there a phosphor which can be visible at 70'C and not at 20 degrees?

I'd very much appreciate some information. Thanks!

It might be possible in a few ways:

In the process of thermoluminescence, a previously-activated material gives off light when warmed. This is used to date materials such as ceramics, which accumulate energy from terrestrial radiation and cosmic rays. Though ceramics have to be heated quite a bit to release this stored energy, some substances exhibit thermoluminescence at room temperature or lower. $\ce{BeO:Zn}$ has peak emission at 307K, ~34oC.

N.B thermoluminescence is the release of energy previously stored in a material, not the direct transformation of heat energy into electromagnetic.

In an anti-Stokes process, a substance absorbs two or more photons of lower energy to release one of higher energy. This is used in IR beam visualization cards. However, it is unlikely that thermal energy would be sufficient to produce visible light.
Non-blackbody radiators can emit more energy at short wavelengths than would be expected according to Planck's law. "Limelight" owes its intense white light to candoluminescence, or non-blackbody radiation. Using a "heat mirror", the efficiency of incandescent lamps can be greatly improved. That said, though theoretically a material might exist that could prevent emission of almost all wavelengths longer than visible light, it is doubtful that a detectable amount of visible light would be emitted at 70oC.


Why does the ionization enthalpy of elements across a period not follow a regular pattern while the atomic size always decreases? First of all, I would like to mention that I am only talking about elements that aren't from the d or f blocks.
In order to further elaborate on my question, I would like to take the third period as an example.

As we can see, the atomic radii decrease across a period constantly, whereas the ionization energies increase and decrease depending on the element. I just fail to understand how this can be. As far as my understanding goes, isn't work done against the electric field to remove the electron called ionization energy? And if this is true, then it must mean that the closer an electron is to the nucleus, the stronger the force of attraction experienced by it would be, which would in turn translate into a higher ionization energy. However, we can see that this does not seem to be the case. I would really appreciate it if someone could tell me where I'm going wrong and why this is the case.

Put very roughly, we can boil this down to why aluminum and sulfur are depressed with respect to the average of the surrounding elements.
Aluminum is $\ce{[Ne]{3}s^2{3}p^1}$.  Its first ionization removes the $3p$ electron while leaving the more stable $3s$ orbital intact, whereas magnesium has to break its more stable $3s$ orbital.  So by comparison, the first ionization energy of aluminum drops off.
This does not mean aluminum will form ionic compounds more readily than magnesium.  Despite aluminum's subshell advantage, both elements have first ionization energies that are too high to be offset by the electron affinity of a nonmetallic element and electrostatic attraction between the resulting ions.  There is the possibility of retaining metal-metal bonding between +1 ions, which has become fairly well-known with magnesium, but ordinarily a reasonably low second or third ionization energy is needed to get a favorable energy balance with these elements.  So to form ionic compounds aluminum ultimately has to give up its $3s$ electrons, which are more tightly bound than magnesium's, and thereby the subshell difference is obscured.  Magnesium ultimately wins when it comes to ionic compound formation.
Turning now to sulfur, that element has the ground state configuration $\ce{[Ne]{3}s^2{3}p^4}$.  In this case loss of an electron gives a half-filled $3p$ subshell with all the electrons unpaired and having identical spins.  This stabilizes the half-subshell configuration through a quantum-mechanical exchange interaction between the electrons with like spin and total angular momentum.  Sulfur forms this favored configuration when it ionized once, while phosphorus has to break it, and so the first ionization energy of sulfur (but again, not second or higher ionization energies) is depressed as we saw with aluminum.  Note that the effect here is less than that involving full or empty subshells that we saw with aluminum versus magnesium.


Does silicon dioxide react with alkalis? In my science class, I have been taught that silica (silicon dioxide/sand) has three properties, namely

It does not react with water.
It does not decompose with heat.
It does not react with acid.

However I think that if it does not react with acid, will it react with bases? I am just curious because if it does react, why does my textbook not say anything about that?

It really does react:
\begin{align}
\ce{4HF + SiO2 &-> SiF4 + 2 H2O}\\
\ce{SiF4 + 2 HF &-> H2[SiF6]}\\
\ce{SiO2 + 2 NaOH &-> Na2SiO3 + H2O}\\
\ce{SiO2 + 2 F2 &-> SiF4 + O2}\\
\ce{SiO2 + 2 C &-> Si + 2 CO}\\
\ce{(2,1)Na2O + (1,1)SiO2 &-> Na4SiO4/Na2SiO3}\\
\ce{SiO2 + Si &-> 2 SiO}
\end{align}


Crystallization of zinc sulfate monohydrate I'm currently running an electrolysis experiment and I have a few questions regarding crystallization of zinc sulfate monohydrate $\ce{ZnSO4 * H2O}$.

I have prepared a saturated solution of zinc sulfate, will it crystallize as monohydrate, or only heptahydrate $\ce{ZnSO4 * 7 H2O}$ will?

And if so, what is the reason behind it? Does it has to do with the monohydrate and heptahydrate part?

TL; DR: obtaining a reasonably big single crystal of $\ce{ZnSO4 * H2O}$ seems to be a tricky task, whereas one can easily obtain $\ce{ZnSO4 * H2O}$ as a result of thermal dehydration of heptahydtare above $\pu{70 ^\circ C}$ in a form of melt, and above $\pu{120 ^\circ C}$ – as a polycrystalline solid material.

I finally got across to suitable references to back up my comments. In nature zinc sulfate monohydrate occurs as mineral gunningite, always in a form of efflorescence deposition on the surface of sphalerite and other minerals and doesn't form single crystals [1]:

  Surface and ground waters carrying dissolved oxygen attack
  the sphalerite, yielding soluble sulphates of zinc [...]:

  $$\ce{ZnS + 2 O2 -> ZnSO4}$$

  [...] In the presence of abundant water all of the soluble salts are removed. If, however, the conditions are such that the supply of water is restricted or evaporation takes place, the metal-bearing solutions become supersaturated with the consequent precipitation of a number of hydrates of zinc, manganese, and iron of which gunningite, $\ce{ZnSO4 * H2O}$, is one.

Thermal dehydration among zinc sulfate hydrates has been studied with DTA in several publications. The most relevant and thorough investigation [2, p. 471] suggests the following scheme:

$$\ce{ZnSO4 * 7 H2O (s) ->[\pu{37 ^\circ C}] ZnSO4 * 6 H2O (s) ->[\pu{70 ^\circ C}] ZnSO4 * H2O (l) ->[\pu{120 ^\circ C}] ZnSO4 * H2O (s) ->[\pu{285 ^\circ C}] ZnSO4 (s)}$$

Original text in German:

  Festgestellt werden vier endotherme Effekte: bei etwa $\pu{37 ^\circ C}$ verwandelt sich das Heptahydrat in Hexahydrat; bei etwa $\pu{70 ^\circ C}$ geht das Hexahydrat in Monohydrat über, wobei teilweise Schmelzen im Kristallwasser auftritt. Bei etwa $\pu{120 ^\circ C}$ siedet die Schmelze, Wasser scheidet
  sich ab, bis das feste Monohydrat zurückbleibt; bei etwa $\pu{285 ^\circ C}$ erfolgt der Übergang des Monohydrates in wasserfreies Salz.

The tricky part is that monohydrate can be isolated in crystal form, but it initially isolates as a melt, and only above $\pu{120 ^\circ C}$ makes a polycrystalline solid phase (occurs as white powder or granules).

Aqueous solubility $\omega_2$ for zinc sulfate $\ce{ZnSO4}$ in mass% of a solute ($\omega_2 = m_2/(m_1 + m_2)$, where $m_2$ is the mass of solute and $m_1$ the mass of water) as a function of temperature ($T$) from CRC Handbook [3, p. 5-172]:

\begin{array}{r|rrrrrrrr}
T,\,\pu{^\circ C} & 0 & 10 & 20 & 25 & 30 & 40 & 50 & 60 & 70 & 80 & 90 & 100 \\
\hline
\omega_2,~\% & 29.1 & 32.0 & 35.0 & 36.6 & 38.2 & 41.3 & 43.0 & 42.1 & 41.0 & 39.9 & 38.8 & 37.6
\end{array}

With this in mind, what can be suggested is to prepare a saturated at $\pu{70 ^\circ C}$ solution, and then slowly increase the temperature (e.g. using a thermostat), hoping that zinc sulfate monohydrate starts to crystallize (using the decrease in solubility starting from $\pu{60 ^\circ C}$).

According to [4, entry 3575, p. 470], solubility of $\ce{ZnSO4 * H2O}$ also decreases in this interval: $\pu{101 g}/\pu{100 g}~\ce{H2O}$ at $\pu{70 ^\circ C}$, and $\pu{87 g}/\pu{100 g}~\ce{H2O}$ at $\pu{105 ^\circ C}$. The difference is rather small, so I wouldn't expect rapidly-grown large crystals.

References

Jambor, J. L.; Boyle, R. W. The Canadian Mineralogist 1962, 7 (2), 209–218.
Balarew, C.; Trendafelov, D.; Gerganova, M. Monatshefte für Chemie 1971, 102 (2), 465–473. DOI 10.1007/BF00909340 (in German).
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC handbook of chemistry and physics: a ready-reference book of chemical and physical data.; 2017; Vol. 97.
Perry, D. L. Handbook of inorganic compounds, 2nd ed.; Taylor & Francis: Boca Raton, 2011. ISBN 978-1-4398-1461-1.


Is there a free, super simple app or website for IUPAC names from structures? I am looking for a free, very simple tool for IUPAC names of organic molecules.

I teach high school chemistry, and it would be fantastic to have an app or web tool that students could use to draw molecules and verify the systematic names of branched alkanes and simple alcohols, aldehydes, ketones, possibly acids; simple ethers, esters, cycloalkanes would be nice but aren't strictly necessary.

I do have MarvinSketch, but I find it too complex for absolute beginners, both to install and to use.
Chemdraw JS does not work well on mobile.

Suggestions?

A bit more intuitive to interact with may be Chemdoodle Webcomponents which is a JavaScript library set-up to integrate well within webpages and mobile devices like phone or tablet, too.  It comprises 2D/3D sketcher, a NMR predictor, and is freely available (open source GPL license).  Derived from Chemdoodle, there is a structure -> name utility speking multiple dialects (e.g., IUPAC / CA, Hantzsch nomenclature), too.

Example:

The interaction with Chemdoodle is nicely explained by Pritt Balagopal here on ChemSE.


What is the Purpose of an Electrolyte in a Galvanic Cell? I understand how a galvanic cell works and the purpose of erverything such as the ssalt bridge and the electrodes, however I don't understand why an electrolyte is required. For example, consider this scenario.

I have a copper electrode in one beaker, connected by a wire with a voltmeter to a graphite electrode in another beaker. The copper electrode is immersed in water while the graphite electrode is immersed in lead ions. There is also a salt bridge connecting the two beakers. Now my question is, will there be a reading in the voltmeter?

I believe that there should be one since copper is more reactive than lead so copper ions will form while solid lead will form on the graphite electrode. However my school teacher said it won't work as there is no electrolyte in the first beaker. Why does it make a difference if there is an electrolyte or not?

Two issues are at play here.

1) You will measure a potential without any supporting electrolyte. As soon as you have everything connected, a potential will develop. Because a self-respecting voltmeter doesn't require much current (ideally zero), the potential will be measured.

2) As soon as you try to pass any current, as comments have mentioned, that current has to be carried from the electrode through the solutions and the salt bridge. Without any electrolyte in the solution, the solution resistance will be huge(as Geoff mentioned) and the amount of current passing will be cell voltage/cell resistance. This will essentially limit the current passing through the solution to very low values.


Half life of Nitrofurantoin Both wikipedia and Toxnet list the half-life of Nitrofurantoin as 20-60 minutes in the blood plasma. Can anyone explain how this is so short, especially for an antibiotic, and how it still has time to be effective.

Answer to your question is within the same Wikipedia page, which states following:

  Mechanism of action: Nitrofurantoin is concentrated in the urine, leading to higher and more effective levels in the urinary tract than in other tissues or compartments. With a $\pu{100 mg}$ oral dose, plasma levels are typically less than $\pu{1 \mu g/ml}$ while in the urine it reaches $\pu{200 \mu g/ml}$. The mechanism of action is unique and complex. The drug works by damaging bacterial DNA, since its reduced form is highly reactive. This is made possible by the rapid reduction of nitrofurantoin inside the bacterial cell by flavoproteins (nitrofuran reductase) to multiple reactive intermediates that attack ribosomal proteins, DNA,[38] respiration, pyruvate metabolism and other macromolecules within the cell. Nitrofurantoin exerts greater effects on bacterial cells than mammalian cells because bacterial cells activate the drug more rapidly. It is not known which of the actions of nitrofurantoin is primarily responsible for its bactericidal activity. The broad mechanism of action for this drug likely is responsible for the low development of resistance to its effects, as the drug affects many different processes important to the bacterial cell.

High availability of nitrofurantoin in the urine ($\pu{200 \mu g/ml}$) compared to that in plasma ($\pu{1 \mu g/ml}$) is due to its high solubility in the urine. At this concentration, it exceeds the requirement for minimum inhibitory concentration for bacteria ($\pu{32 \mu g/ml}$), and as a result, nitrofurantoin becomes bactericidal in urine at therapeutical doses ($\pu{100 mg/capsule}$). Keep in mind that reduced nitrofurantoin (its metabolite by nitrofuran reductase) is much more reactive than nitrofurantoin, thus shorter half-life make it highly lethal to bacteria.


Which lone pairs/bonding pairs go into which pi molecular orbitals? I was just doing some practice midterms and I came across this question. The answers were posted, but there are no explanations for anything. I wasn't able to find anything about this online, either.

Here are pictures of the question.

Here's the first part:

Here's the second part:

Part e) is what is throwing me off. I was flitting back and forth between O and N, and O, N, and C, but in the end got it wrong. It says to base the answer off of part d), but there are two filled orbitals! How do I know which pair of electrons occupies which orbital?

Does the sp3 lone pair of the nitrogen occupy the lowest occupied MO or the highest occupied MO? Because it appears from the correct answer that it occupies the higher one - but why is that? The other delocalized pair of electrons is the pi electrons in the C=O double bond, and I get that maybe they have different energies, but I can't think of any reason one pair should occupy one particular orbital over the other.

Furthermore, according to the movement of electron pairs in the resonance structures of f) (which were all correct according to the answer key), the third structure suggests the lone N pair is shared between the carbon and the nitrogen, not the oxygen and the nitrogen. What gives?

Also, as a side question - the reason why the solvent is polar aprotic, despite having the capacity to shed a hydrogen from the carbon to form an induction-stabilized carbanion, is because protic solvents have the labile hydrogen attached exclusively to an oxygen or a nitrogen, correct?

This is a common attempt to visualise what π systems actually mean for the overall Lewis structure. But let me note a couple of things first:

Obviously, there is no $\mathrm{sp^3}$ lone pair on DMF’s nitrogen. Instead, the atom is $\mathrm{sp^2}$ hybridised and the lone pair (if it is there) occupies a p-type orbital.
Remember that the π analysis only deals with the π orbitals (obviously); there is an underlying σ system which is implied but hardly mentioned. Thus, if ‘one bond’ is created, it is actually a double bond, while ‘no bond’ means ‘only the underlying single bond’.

If you want, in simplistic terms, you can add the π orbitals one after the other. The first is fully bonding and has contributions by all three atoms. Since it is fully bonding across two bonds but only contains two electrons, we can say that it contributes $\frac22$ bonding electrons to each bond or a bond order of ½.

The second orbital is not really bonding as it has a perpendicular nodal plane through the central atom (carbon). Thus, we cannot really analyse it in terms of the $\ce{C-O}$ or $\ce{C-N}$ bond order directly. However we can say that the electron density will be stronger at the ends, i.e. on nitrogen and oxygen. Each of these atoms practically experiences $\frac22$ lone pair electrons (do not participate in bonding) or ½ a lone pair.

Consider what would happen if you had filled the third orbital. It would have been antibonding with respect to both the $\ce{C-O}$ and the $\ce{C-N}$ bond. It would have contributed the same bond order of ½ to both — but this time in negative. That would contribute to an overall bond order of zero or the bonding electrons being on either side of the bond as lone pairs.

For your question, this means:

the fully bonding orbital cannot contribute to any lone pairs since it is not compensated for by an antibonding one
the nonbonding orbital contributes the lone pair. Since it only experiences contributions from oxygen and nitrogen, this is where the lone pair will be located.
it would be wrong to look at the Lewis structure, see a lone pair on nitrogen and immediately associate that with the ‘lone pair’ of the question. If anything, consider all relevant resonance forms and all their lone pairs. Also, as soon as it has resonanted to form a bond, it is no longer a lone pair.
finally, as per your side question, note that the $\ce{C-H}$ bond in DMF has a very, very low tendency to deprotonate. That proton is not even mentioned on both the Evans table and the Bordwell data. While it might lead to a carbene, the carbene is not exactly stable. A much more likely deprotonation would be that of a methyl group which would extend the π system. Note that the sequential treatment with butyllithium and DMF is a method to convert electron-rich aromatic rings (e.g. thiophene) into their corresponding carbaldehydes — in this reaction, the proton of DMF is less acidic than that of a thiophene or proton exchange would occur in place of nucleophilic attacks.