text:
Why is ethanol distilled off before adding water in the first step of a malonic synthesis?
I am doing a malonic synthesis shown below, and I wonder why I should distill ethanol off before adding water in the first step of the reaction (First I add absolute ethanol and solid sodium).

I see no reason to distill off EtOH before adding water for the base hydrolysis step. The EtOH facilitates the reaction as the butyl malonate is readily soluble in it but not in water.

I think it is because of this http://www.passmyexams.co.uk/GCSE/chemistry/carboxylic-acids-reaction-alcohol.html i.e alcohol will react with your product (carboxylic acid) to form an ester.

Could you clear up the confusion and detail steps in more detail. Your reactions don't indicate when you boil off the ethanol...

I think the $\ce{KOH}$ hydrolysis is not necessary. Acid hydrolysis would do it. It is necessary to remove ethanol to prevent resterification (Le Chatelier's principle).

related: https://chemistry.stackexchange.com/a/37195/102629

I assume this is a real procedure for actual synthesis.
Distilling ethanol off does have an effect on the workup of this reaction. Usually one needs to separate the reaction product and educts or side products afterwards by separation between an aqueous and an organic phase. Having ethanol in the mixture prevents the two phases from separating. In order to isolate your product you will need to distill ethanol off before adding water, because water would also be distilled off, if you added it first.

I disagree. You remove the ethanol after the base hydrolysis.

Where does it say the ethanol is distilled off after base hydrolysis?

It doesn't. My point is that it makes little sense to remove the EtOH before the base hydrolysis, all you need do is add some water and an excess of KOH to conduct the hydrolysis, then concentrate as the first part of the workup.

If you distill it off before, then do the hydrolysis, you can isolate your acid by acidifying your solution and extraction with apolar solvent. I think you just have a different procedure that you like to stick to, but OP describes the procedure like i explained.

The synthesis adds two carbon units from malonic acid to butyl bromide, followed by a de-carboxylation step, to form caproic acid.
There are 2 reasons:
 1. Water and alcohol form an azeotrope, why not recover some absolute ethanol before the second step?
 2. Reflux temperature for de-esterification will be higher with a greater percentage of water as ethanol is removed.
Butyl bromide boils around 101 C, so it can be distilled off during KOH reflux.  A possible by-product, n-butanol, which boils at 117 C but forms an azeotrope with H2O at 92 C, can also be removed with higher temperatures.
So, alcohols and butyl bromide could be fractionally distilled off during the KOH de-esterification (this technique would be practiced on a lab scale to perfect it), leaving the water soluble malonate adduct for subsequent decarboxylation.
Interestingly, after decarboxylation step, caproic acid could then be extracted back into non-polar solvent for further purification.


text:
What force attracts electrons from another atom to a non-closed valence shell?
As I understand the balance in an atom is reached when the number of protons is equal the number of electrons. And the force responsible for this is based on each proton attracting one electron.
But why non-full valence shell attracts electrons from other atoms? Looks like it exerts a force independent from proton-electron bondage force. Does the shell itself have some force?
For example, Na and Cl form an ionic bond resulting in NaCl. But why a balanced atom of Cl (17 protons VS 17 electrons) attracts an excessive 18th electron from Na? If it's not the proton of Cl attracts it then what? Looks like the valence shell has its own force.

P.S: I haven't found the answer in the referred possible duplicate topic.

The shell itself is not a thing at all. Also, the picture of each proton attracting one electron is a wrong one. Instead, each proton attracts all electrons, and vice versa.

I'll go a bit further. It is possible for a neutral atom, or even an anion, to still "attract" electrons. In other words the extra electrons which make the atom negative will still have a binding energy. This is because the valence electrons don't shield each other perfectly.

Then don't ask about "force". If you don't get ionic bonding read https://chemistry.stackexchange.com/questions/32533/why-is-an-ionic-bond-a-chemical-and-not-a-physical-bond

My guess is this attraction happens because when two atoms "touch" each other, their valent orbitals slightly overlap and electrons on the valence orbitals fall under influence of opposite atoms. So that now one of the protons attracts only half of an electron. And to find the balance again it attracts half of the electron of another atom.
For example, two H atoms share electrons so that each proton attracts half of its electron and half of neighbor's electron thus maintaining the balance (+1 VS -1).

source: Stack Exchange

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

sort: votes


text:
Why should you round to proper significant figures after you complete a calculation rather than while doing the calculation?
In class, my teacher mentioned that you should complete a calculation without considering significant figures and then at the very end, round to the proper number of significant figures. Why couldn't we just round the numbers after each step of the calculation? Is there a difference?

The notion is that "excessive" rounding in the intermediate calculations results in a rounding error in the final result. Consider $$1.01^{5}\times 6.7 = 7.0$$ but if you round  1.01 to 1 at each multiplication step then $$1.0\times1.0\times1.0\times1.0\times1.0\times6.7 = 6.7$$

No rounding during calculation is a bit impractical, but you should always keep one additional digit in the numbers until properly rounding the final result.

@Karl - I'd carry two extra digits. That doesn't seem to be much of a burden for calculations done with a modern calculator. I couldn't do that though when I was working out problems with a Post Versalog slide rule which I used from high school through first year of graduate school.

@MaxW Or do the obligatory error estimation. ;-)

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


text:
HOCl + ethyne reaction-hypochlorous acid with acetylene
What is the product of acetylene ($\ce{C2H2}$) and hypochlorous acid ($\ce{HOCl}$) reaction?
I gathered that $\ce{Cl}$ acts as electrophile and so $\ce{HOCH=CHCl}$ is formed. Now two things can happen: it could tautomerize to keto form since enol form is unstable, or another molecule of $\ce{HOCl}$ can add itself, if it adds again there are two ways of adding them, both chlorine atoms could end up on the same carbon or on adjacent carbons. How do we decide what will be the final product?

Tautomerization is usually rapid in which case chloroacetaldehyde is formed. If the enol were reacting with more HOCl, the enol would react at the Cl bearing carbon with the electropositive chlorine to afford dichloroacetaldehyde. Depending on the stoichiometry of the reaction, chloroacetaldehyde may re-enolize to give dichloroacetaldehyde.

Hypochlorous acid is strong oxidant, so resulting products would be various chloroacetic acids, depending on conc. of reagents, temp. etc.


text:
Why Benzene Cannot be chlorinated using HOCl?

Why HOCl doesnt do the job please , can anyone show me the mechanism ?

One possibility:  When you chlorinate with, let us say, $\ce{Cl2 + FeCl3}$, you generate a "leaving group", $\ce{FeCl4^-}$, that comes off easily because it is highly polarizable and only weakly basic.  The $\ce{HOCl +H^+}$ combination would have to displace a more strongly basic and less polarizable water molecule.
Chlorination can be effected with hypochlorous acid by adding silver ion as another catalyst/co-reagent.  See here.

Oscar Lanzi Phenol is brominated with bromine water .How do we explain this ?

A more activated ring, and bromide ion is the displaced species.


text:
What chemicals are often confused for different chemicals?
Recently I asked a similar question on Chinese SE: Why would Antimony (Sb) and Aluminum (Al) be confused?

It seems, though, that this is by no means a Chinese phenomenon.
The Wiki page for Tin foil mentions that:

aluminium foil, [...] is still referred to as "tinfoil" in many regions.

What other chemicals are often confused for different chemicals?
I'm especially interested in things that have retained their incorrect names.

Molybdenum is named after a Greek word which means "lead". There are other elements with names chosen like that.

Questions asking for list of examples aren't a good fit for the StackExchange Q&A model. That's particularly true if the inclusion criteria has a somewhat subjective component to it.

One especially serious example is confusing methanol with ethanol. Methanol is highly toxic and sometimes is deliberately added to or inadvertently present in bootleg alcoholic beverages, causing many deaths and blindness in survivors every year around the world. It would be a life saver for someone to develop a cheap dipstick test for methanol.


text:
E227 Food Additive effectiveness
I have been researching on E227 (Calcium Bisulfite), and could find minimal information about how it acts as a preservative and how effective it is. I could only find that it preserves by converting itself to sulfurous acid in an acidic liquid, but could not find out how effective and how exactly it preserves. I would appreciate any help.

It is a reducing agent and preserves by reacting preferentially with any oxidising agents present

Sodium bisulfite is one of several sulfur-based chemicals commonly known as ‘sulfites’. Others include potassium bisulfite, potassium metabisulfite, sodium metabisulfite, sodium sulfite, or sulfur dioxide, and also your calcium bisulfite. They have fungicide properties and stop the natural browning process of produce. Dried apricots stay orange, and golden raisins stay yellow. Sulfites stop the browning and further ripening of fruit by blocking the natural enzyme, polyphenol oxidase. Sulfites are also added to some wines.
Sulfites destroy vitamin B1 (thiamin) and so the FDA does not allow sulfites on food containing vitamin B1 (thiamin). They estimate that 1 in 100 people are sulfite sensitive, so it is illegal to use sulfites on foods sold raw, except potatoes. They get an exception because there is no alternative to keeping french fries from looking brown.
The FDA lists sulfites as generally recognized as safe, but sulfites are linked to neurological, gastrointestinal, and respiratory problems. Trace amounts of sulfites have been known to cause hives, vomiting, wheezing, and anaphylaxis. I know from experience that too many dried apricots can cause intestinal cramping.
Modified, but essentially reporting from  https://traditionalcookingschool.com/preserving/dehydrating-preserving/sulfites-and-food-preservation/


text:
How can the solid state structure of cyclopentadienyllithium be explained?
In the solid state, LiCp adopts a polymeric multidecker structure where Li atoms are sandwiched between two Cp rings (shown in this diagram from Organometallics 1997, 16 (17), 3855–3858):

If we consider the η5-binding mode of the Cp rings, then it seems that the number of electrons around Li is more than eight. However, Li can't extend its octet. How can the structure and bonding be explained?

It is ionic bonding not covalent bonding. A lithium cation is between two cyclopentadienyl anions in the 3D structure of the solid, but the chemical formula is 1:1.

I think the question may well be more interesting than "it's ionic bonding". MeLi and other alkyllithiums, for example, can hardly be described as purely ionic - the structure and bonding is quite complicated. I wouldn't be surprised if there was significant covalency in CpLi, too, although I might of course be wrong, and a cursory flip through my usual bunch of references doesn't turn up any more useful info.

Well, as you can easily count it's six electrons per 1 Li as all rings are bound to two Li atoms. This type of boding is present in multidecker sandwich complexes.

@Mithoron...I understand your point but in that case, its octet is not fulfilled, then how this structure is so stable? why the molecule does not adapt any more complicated structures?

It didn't ping me (because of dots presumably). Li makes electron deficient molecules almost all the time. It's middle ground between, say, Rb with close to zero valence electrons and carbon with octet most of time.

Or very simply speaking, it doesn't need the full octet, because the bonding still has a large ionic contribution.

Whats more interesting is that the rings are not rotated off-set from each other.

This summary identifies several calculations indicating that covalent interactions between lithium and cyclopentadienyl moieties do occur; the interaction is mostly but not purely ionic.
Draw the occupied $\pi$ orbitals of the cyclopentadienyl groups and, for comparison, draw the $s, p_x, p_y, p_z$ orbitals on lithium with the $z$ axis along the length of the chain.  Observe:
• The lithium $s$ and $p_z$ orbitals overlap the lowest energy cyclopentadienyl $\pi$ orbitals.
• The lithium $p_x$ and $p_y$ orbitals overlap the bonding degenerate pair on the cyclopentadienyl ring.
With these overlaps, we can construct three lithium-cyclopentadienyl bonding combinations per cyclopendienyl group, one corresponding to each cyclopentadienyl $\pi$ bonding orbital.  We then occupy these orbitals with six electrons per ring.  In effect we have extended the aromatic coupling of the cyclopentadienyl ion into the third dimension with the lithium orbitals.
Heavier alkali metals do not work as well because their orbitals are too diffuse, but they could overlap larger rings.


text:
carbon dioxide in liquid and gaseous phase with liquid water
What are the differences between carbon dioxide solubility (miscibility, reaction) in liquid water when in gaseous ($\pu{<38 bar}$) and liquid ($>\pu{38 bar}$) phase? The temperature in my experiments is usually between $0$ and $\pu{5\!^\circ C}$. Is there more phase - interface for the reaction between water and $\ce{CO2}$ in the gaseous phase?
According to Henry's law for real gases (including a Poynting correction), the solubility should rise with the fugacity of the vapor. How is it however, when the carbon dioxide is liquid? When observing a pressure drop in a closed stirred vessel with liquid water and $\ce{CO2}$ on top of it, the pressure drop is much larger for gaseous $\ce{CO2}$ than for the liquid $\ce{CO2}$, this does not however imply a larger solubility? In my understanding it just means, that the liquid $\ce{CO2}$ is not that compressible and will mix in a different manner than gas and liquid water, resulting in a lower pressure drop.

The water doesn't care wether there is also some liquid CO2 in the same vessel. I'll doubt you find any significant change of CO2 solubility at 38 bars.

related, though I do not have access: https://link.springer.com/article/10.1023/A:1015294905132

At a given temperature a mixture of gaseous $\ce{CO2}$ and liquid $\ce{CO2}$ will have a particular pressure. Thus more liquid $\ce{CO2}$ doesn't mean that more $\ce{CO2}$ will dissolve in the water.
Obviously liquid $\ce{CO2}$ has a lot more matter per unit volume than gaseous $\ce{CO2}$. So a little liquid $\ce{CO2}$ makes a lot of $\ce{CO2}$ gas.


text:
Does a molecule need to be placed symmetrically in the unit cell?
I was watching an OCW by MIT (3.091X) on solid state. My previous notion of lattice point and unit cell got shattered after this example.
Where are the unit cell and lattice points in this picture?

Unit Cell is the smallest group of atoms which has the overall symmetry of a crystal, and from which the entire lattice can be built up by repetition in three dimensions.
Lattice is the arrangement of points in space such that any point us identical to any other.

So using these definitions, the answer is

Green is the unit cell, and red ones are the lattice points.
Previously I used to think that a molecule should be symmetrically placed in a unit cell such that its center of mass lies on the lattice point. But in this example, the center of mass (shown by the yellow-green circle) is way off the lattice point.
My question is, Are there any other similar examples where the center of mass of molecule doesn't coincide with the lattice point or at least doesn't placed symmetrically in a unit cell (like on the face center, body center)?

Don't think of it so much as putting the center mass of an atom on a vertex but as matching the unit cell symmetry to the crystal symmetry. Take the definition from google below, I've bolded key parts:

The smallest group of atoms of a substance that has the overall symmetry of a crystal of that substance, and from which the entire lattice can be built up by repetition in three dimensions.

While the red boxes satisfy the first and last bolded criteria, they fail the second. The red boxes have 2-fold rotational symmetry and inversion symmetry, but the green boxes have 2 mirror planes that the red boxes do not. This lack of mirror symmetry with the red boxes does not have the same overall symmetry of the lattice and therefore the green box is the unit cell.
Think of Sodium Chloride; its unit cell face is represented as:
$\ce{\color{red}{Na}\  Cl\ \  \color{red}{Na}}$
$\ce{Cl\ \  \color{red}{Na}\ \ Cl}$
$\ce{\color{red}{Na}\  Cl\ \  \color{red}{Na}}$
instead of
$\ce{\color{red}{Na}\  Cl}$
$\ce{Cl\ \  \color{red}{Na}}$
Because the first example has 4-fold rotational symmetry where as the second has 2-fold and the first one has 2 mirror planes that the second example does not. Both have inversion symmetry. That said the atoms at edges/corners are centered because that is what allows for the symmetry otherwise the unit cell would be unsymmetrical which does not reflect the symmetry of the lattice.
As an example of a sort of off centered lattice there is perovskite. You can't see it in the image but the black atom is $\ce{Ti^4+}$ and is too large to actually center in the lattice so it will reside in an off-centered configuration (the titanium ion is shifted off-center towards a face). This property of off-centering makes perovskites excellent as dielectric materials for capacitors as the ion can change is off-centering by an electric field.

But in the definition, there is no mention of the mirror symmetry !

@AdityaShrivastav Yes but a lack of mirror symmetry does not reflect the symmetry of the crystal.

Of course to replicate in 3 dimensions the ball and stick model shown is wrong. There is only 1/2 of an atom for the red spheres and 1/8 of an atom for the blue spheres as shown.

@A.K. You may have your explanation out of order in your answer. You say it satisfies "the first two bolded criteria, but fails the third", but "has the overall symmetry" is your 2nd bolded reason. This might be what is causing the misunderstanding

@AdityaShrivastav note Tyberius's comment. apologies for the confusion.


text:
Why is the Kelvin Temperature Scale Based on the Triple Point of Water?
Why would the Kelvin temperature scale not just be based on the freezing point of water, which is only .01 degrees Celsius off from it? Is this just a more accurate way of measuring and quantifying the value?

It's independent of pressure, otherwise someone could ask if melting in, say, one bar or one atmosphere.

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

The triple point is unique, the melting temperature isn’t.

I don't see how the selection of a standard is not about chemistry.

In reality all the "primary" scientific values have changed over time as experimental techniques have improved. Science has evolved to the point where a set of seven "SI base units" (ampere, kelvin, second, meter, kilogram, candela, and mole) have been selected and defined so that better measurements won't cause disruptive changes in scientific data.
So the gist is that the "temperature of ice water" depends on the atmospheric pressure. So it is necessary to define that 0 °C is the temperature of ice water at "standard atmospheric pressure." That means that if the "standard atmospheric pressure" changes, then so does the "temperature of ice water at standard pressure."
However the triple point of water doesn't change and has a unique value. So by defining the triple point of water as exactly 0.01 °C in 1954, then the Celsius scale is anchored as much as the Kelvin scale. By also defining absolute zero as 0 K and −273.15 °C then the temperature scale itself won't change if better ways to measure temperature are devised.
Thus by the time the whole notion of the primary base units was sorted out there was a massive amount of scientific data that had been accumulated using the old definition of the Celsius scale that was based on ice water at a standard atmosphere. Letting the triple point be exactly 0.01 °C kept all of that "old" temperature data "reasonably accurate" since 0.01° is only a 0.01/273.15=+0.0037% change.

It's less than 0.0037% change.  The old melting point standard was 0°C=273.15 K at one atmosphere, whereas the triple point is a little higher than the one atmosphere melting point.  You guessed it -- that difference is almost exactly 0.01°C or K.


text:
When an acid is added to water, why does the hydroxide ion concentration decrease?
At equilibrium in pure water, we have
$$\ce{[H_3O+][OH-]} = 10^{-14}$$
Since $\ce{H3O+}$ and $\ce{OH-}$ ions are produced in pairs, we may conclude
$$\ce{[H_3O+]}=\ce{[OH-]} = 10^{-7}$$
So far so good. But shouldn't things change when we introduce a new substance into water ? I mean why does the first equation above hold no matter what ?
Also when I introduce $\ce{H2SO4}$ into the water, it doesn't just give a $\ce{H+}$ ion, it also gives $\ce{HSO4-}$ ion. Shouldn't these new negative ions change the behavior of water? Why does my textbook never talk about these new negative ions? Help appreciated. Thanks!

The ionic product of water ($K_w$) has constant value of $10^ {-14} $at temperature of $25^o $ C. Unless you change the temperature, nothing is going to happen to $K_w$.

Your question title is a bit misleading, but i try to answer all the small questions in you question text.
The equation $[\ce{H3O^+}][\ce{OH^-}]$ holds true, if other parameters (like T) are constant. Keep in mind, the power of hydroxide decreases, whereas the power oxonium increases. Being equal in the equation and considering how logs are being computed,bthey will add up to 14 every time.
Regarding the introduced $\ce{HSO4^-}$, they don't contribute to pH by definition. On the other hand, they alter the behaviour of the water, by increasing its conductivity.
Your textbooks don't talk about the other negative ions in acidic or alkaline solutions, because they don't directly contribute to the values of pH or pOH by definition. In cases of polyacids like sulfuric acid, ions like $\ce{HSO4^-}$ are accounted for by using a different formula to calculate the actual pH value, but the "not-hydrogen" part is largely irrelevant in the behaviour of the solution itself.

Ohk since the $pH$ and $pOH$ must add up to $14$, I see why $pOH$ decreases with addition of acid. Thank you :) But I still don't get why we care so much about $pH$, and never bother about other ions that may present in the water like  $pHSO_4$. Why change in $H^+$ is a big deal ?

It is the "biggest deal" in many cases, because the concentration of $\ce{H^+}$ has a high impact on the reactivity (behaviour) on everything else in the solution. It is a bit like asking, why T-shirt sizes are only expressed as being "l" or "m", even though they do not account for the type of material used for the shirt.

Oh ty for the quick response! I think I get $H^+$ is a big deal, but I don't see how the concentration of other solutes like $HSO_4^-$ don't influence the reactivity ?

Hydrogensulfate can donate another proton, but then it is a rather unreactive anion. It is fully oxidized sulfur with a very stable (empty) electron shell. Becoming a stable and unreactive compound is actually an important part of the driving force behind the deprotonation reaction itself.

If I understand correctly, you're saying $HSO_4^-$ is stable and unreactive; whereas $H_3O^+$ and $OH^-$ are not so stable. So they affect the reactivity and this is the reason we bother about $pH$ a lot ?

Not quite. $\ce{HSO4^-}$ is still active, because it can deprotonate. Also, sulfate is not unreactive in an absolute sense, but in contrast to the rest of the solution, if you only add sulfuric acid to water. There are many reactions that include sulfate. For instance, you can reduce sulfate to sulfite. But i think we are losing focus of what you were originally asking.

I feel I understand this better now. Thank you so much :) If I may ask one last question - suppose I somehow  extract only $HSO_4^-$ ions and add them to pure water. Clearly this wouldn't change the $pH$ of the water. So is it safe to drink this water ?

It would change the pH, because $\ce{HSO4^-} is still an acid with a pka of 1.99, which will make your pure water acidic. It will also not be safe to drink!

As long as you stay in the dilute regime, that is if water molecules have most of the time to interact just with each other, $K_w$ won't change. Add 20% of anything soluble (salt, acid, alcohol, acetone, ...) and things look different.

Oh, so if I add NaCl or other solutes to the water, the $H_3O^+$ and $OH^-$ ions will not notice the newly added solutes ? Is this because the solute concentration is low ? I think I get it... Thank you :)

Yes. Added OH- or H3O+ ions take part in the autoprotolysis equillibrium, the general properties of the solvent are however mostly uninfluenced.


text:
Why are some ions mutagenic and others are not?
It is my understanding that when ionic species (such as $\ce{He^2+}$, or alpha particles) enter the human body, they tend to react with biomolecules, causing damage by (for example) mutating DNA.
If such ions can cause severe damage to the human body, why are other ions like $\ce{Na+}$ or $\ce{Cl-}$ safe to consume and not prone to causing similar mutagenic effects?

$\ce{He^2+}$ whats his electrons back very badly. $\ce{Na+}$, not so much. There are plenty of ionic species in the body. To say they are harmful is a really unjustified generalization, much like saying that all *.exe files are viruses.

Hmm.. That's a good analogy :) But can you please elaborate why these other ions are not causing any harm to dna molecules or any other effects for that matter?

Because chemistry. Sorry, but there is no short answer.

Because living systems have developed based on ionic chemistry.

Plus, an alpha particle typically has a lot of energy and loses it by bashing electrons and other nuclei around pretty violently. If you sent an MeV Na ion into you, it would do similar things.

It's the difference between holding a bullet in your hand and getting shot by one.

So maybe, correct me if I am wrong it has something to deal with octet/duplet configuration? Na+ or Cl- ion has achieved its octet configuration in its outermost shell so it has very less energy to form any sort of bond and hence behaves as noble.

@VivekKumarPathak - No you have entirely the wrong notion. An alpha particle isn't just a $\ce{He^{2+}}$ ion, it is a $\ce{He^{2+}}$ ion with tremendous kinetic energy. In other words it is zipping along like a bullet. The alpha particle sheds its kinetic energy by knocking into electrons which creates all sorts of ions, and rips apart molecules in ways that won't happen via any normal chemical process in our bodies.

You are confused about the cause of mutagenesis in alpha particles.
The reason why an alpha particle is mutagenic is not that it is an ion, it is that it is an ion carrying a great deal of kinetic energy. Alpha particles are produced in energetic nuclear reactions and carry a great deal of energy away from the nucleus that emits them (a typical alpha particle has about 5MeV of energy, enough to break a typical chemical bond). It is the kinetic energy that matters not what the ion consists of. Heavier ions with that sort of energy would be just as bad for biological systems as are other forms of energetic radiation (beta particles are just electrons but will still do damage, gamma rays are just powerful photons but will dump enough energy into molecules to break DNA).
Non-energetic helium nuclei would pick up electrons extremely rapidly from the first molecules they encountered (probably from air even before reaching your skin) and would probably be as harmless to you as the helium filling a balloon).

I am certainly no expert in this area but this is my understanding.
Let's consider an analogy of electrons. There are low energy electrons that move reasonably freely through a lump of iron for example. The electrons in it are not harmful, we can even eat this if it is small, in fact it is nutritious. We wouldn't normally call these electrons beta particles. Then there are high energy electrons released by a nuclear process. We call these beta particles, and they can smash through numerous atoms, causing many random chemical changes. This is a cause of radiation sickness and possible death. As you can see these have completely different characteristics because of the different amount of energy they have.
The situation is the same for ions. Common, low energy ions, trapped in a chemical substance, for example sodium ions in water, are very different from a high energy He2+, liberated by a nuclear processes, not bound up in a chemical substance.
Ultimately the important difference is physical rather than chemical.
There is one other aspect: ions like Cl- are only safe when they are in a substance that is neutral overall (for example Cl- with Na+). If you were to be exposed to pure Na+ it would be extremely harmful.


text:
Balancing the reaction of nitric acid with zinc using oxidation number method
In a previous question, there was confusion relating to what this "oxidation number method" is. So I will illustrate that before my question.

Write the skeleton equation representing chemical change.
Assign oxidation numbers to the atoms in the equation and then write separate equations for atoms undergoing oxidation and reduction.
Find the change in oxidation number in each equation. Make the change in both equations by multiplying with suitable integers. Add both the equations.
Complete the balancing by inspection. First balance those that have undergone change in ON and then other atoms except H and O. Finally, balance the Hs, the Os will be automatically balanced.

Example:
  $$\ce{Cu  + HNO3 -> Cu(NO3)2 + NO2 }$$
  $$\ce{Cu^0 -> Cu^{+II}(NO3)2} \tag{1}$$
  $$\ce{HN^{+V}O3 -> N^{+IV}O2} \tag{2} $$
To make increase and decrease in ON equal, equation 2 is multiplied by $2$. Adding we get,
$$\ce{Cu + 2 HNO3 -> Cu(NO3)2 + 2NO2 + H2O} \implies \ce{Cu + 4 HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O}$$

Question:
Balance $\ce{Zn + HNO3 -> Zn(NO3)2 + H2O + N2O}$
I am trying to use oxidation number method but unable to reach the right answer.
The half equations are:
$\ce{Zn -> Zn^2+(NO3)2 \tag{1}}$
$\ce{2HN^{+V}O3 -> N^{+I}2O \tag{2}}$
Thus we should multiply the first equation by 2 and then add.
$\ce{2Zn + 2HNO3 -> N2O + 2Zn(NO3)2}$
Then I tried balancing the nitrogens by adding one 2HNO3s to LHS but after that the oxygens ust won't balance!
How do I go about balancing this then?

First, I'd write all half-equations with ionic compounds in ionic form, even if asked otherwise. Second, if an atom costs 10 bucks, then two atoms cost 20; now replace bucks with electrons. Third, how would you go about finding the formula of zinc nitrate, were it not given to you in the problem statement?

Here is an alternate-
Let the final balanced equation be $\ce{yZn + 2xHNO3-> yZn(NO3)2 +xH2O +zN2O}$
Now For balancing nitrogen we must have $2x=2y+2z$ and for balancing oxygen atoms we should have $6x=6y+x+z$ Now we get an equation $x=5z$. Put $z=1$ you get $x=5$ and $y=4$.
Hence the balanced equation is:
$$\ce{4Zn + 10HNO3 ->  4Zn(NO3)2 + 5H2O + N2O}$$


text:
Why is diazobenzene named so and not phenyl diazobenzene?
I have searched for this answer for a long while but haven't got it. Why is diazobenzene named so and not phenyl diazobenzene or something? The group with two bonded nitrogens is the diazo group. Why does the term diazobenzene refer to a molecule with two benzene rings?

Does it?$;!!$

Diazo group is $\ce{=N2}$.
Example: diazomethane $\ce{CH2=N2}$
(detailed structure: $\ce{H2C=N+=N- <-> H2C^{-}-N+\bond{#}N}$)
$\ce{-N=N-}$ is azo group.
Example: azomethane $\ce{CH3-N=N-CH3}$
However, preferred names are based on diazene $\ce{HN=NH}$, so better name for azomethane is dimethyldiazene.
So, note that “diazobenzene” does not exist. You obviously mean “azobenzene”, $\ce{C6H5-N=N-C6H5}$, whose preferred name is diphenyldiazene.

Actually my book had written diazobenzene, I think it was a misprint.

@Hema I wonder what the structure was.. there are similar compounds, diazonium salts, like $\ce{C6H5-N_{2}^{+} X^{-}}$

the structure in my book was of azobenzene


text:
The reaction of iron with sulfur to produce Iron(II) sulfide is inconvenient (for me)
I am taking a lot of iron properties this semester, but I have found some contradicting piece of information about the reaction of iron with nonmetals illustrated with two cases
\begin{align}\ce{ **First Case** }\\\end{align}
If we react the iron with chlorine, only FeCl3 will result as following:
$$\begin{align}\ce{2Fe +3Cl2 &->[\space \Delta ]2FeCl3}\tag{1}\\\end{align}$$
No FeCl2 has produced because iron prefers to have oxidation state at +3 as sublevel 3d will be half-filled.
One case that FeCl2 will result when H2 exists as
$$\begin{align}\ce{Fe +2HCl &->FeCl2 +H2}\tag{2}\\\end{align}$$
In this situatuon if any FeCl3 formed, H2 will reduce it to FeCl2
\begin{align}\ce{ **Second Case** }\\\end{align}
Although, Iron favor +3 oxidation state (as illustrated in 1), If we react iron with sulfur; it will actually favor +2 oxidation state as follows:
$$\begin{align}\ce{Fe +S &->[\space \Delta ]FeS}\tag{3}\\\end{align}$$
I tried to find a convenient way that explains how this happen (as the way illustrated at 2), but
I failed to do so. I am really confused:
Does the iron actually favor +3 state or not?
If yes, Why 3 happens? Why there isn't any Fe2S3 in the products?
If no, Why 1 happens?
Is this has to do anything with some hidden conditions or unobvious reaction mechanism?

Iron is not the only party to decide how many electrons will be transferred.

Better stop thinking about "preferences" for ox. states or "stability" - while such thinking might have some use, it's leading you astray. Fe can "be made" to have ox. state as high as +6, or lower then 0. It's just a matter of right reagents.

Not an answer per se (I lack sufficient reputation to comment): are these reactions in aqueous solution, or dry gases? Just a suggestion, but by omitting the charges on these species and not writing them as half-reactions you are obscuring the electron exchange in each. Thus in your first example (iron reacting with chlorine gas), you take metallic Fe, whose oxidation half-reaction can be written as
$$\ce{Fe^0 -> Fe^3+ + 3e-}$$
The equivalent for reduction of chlorine would be
$$\ce{2e- + Cl2_{(g)} -> 2Cl-}$$
Normalizing half reactions to electrons transferred ($6$), adding and simplifying stoichiometry gives
$$\ce{2Fe^0 + 3Cl2 -> 2FeCl3}$$
Thus chlorine is the electron acceptor (strong oxidizer) and iron the donor. In your reaction (2), iron is again a donor (this time of only $2$ electrons), hydrogen is the acceptor, with chloride ion just along for the ride:
$$\ce{Fe^0 -> Fe^2+ + 2e-}$$
$$\ce{e- + H+ -> \tfrac{1}{2}H2} $$
In contrast to these simple reactions, those involving sulfur are exceeding complex because of the diversity of species that can form. In your reaction to form iron monosulfide, one possible way for Fe to be oxidized but native sulfur (zero-valent S actually present as stable ring, $\mathrm{S}_8$) to be both reduced and oxidized is via a disproportionation reaction, with water reduction to hydrogen consuming additional electrons in an anoxic environment, thus yielding both FeS and iron sulfate:
$$\ce{Fe^0 + \tfrac{1}{8} S8 + 2H2O -> \tfrac{1}{2}FeSO4 + \tfrac{1}{2}FeS + 2H2}$$
Amended answer:
If you're heating metallic iron and S to significant temperatures, the elemental sulfur will form a gas, $\mathrm{S}_{n}(g)$ ($n$ indicates which T-dependent allotrope is present). To synthesize FeS however, I think you would need an oxygen-free system. If $\mathrm{O_2}$ is initially present, consumption could be accomplished by sacrificing some of the reactants via in situ oxidation of the metal, or production of $\mathrm{SO_2}$. Then this can proceed directly, using the remaining $\mathrm{S}_{n}(g)$, e.g.:
$$ \tfrac{1}{2}\mathrm{S_2}(g) + \mathrm{Fe}(s) = \mathrm{FeS}(s) $$
I believe the ratio of S/Fe of the product will increase with temperature, and at sufficiently high temperature reach 2 ($\mathrm{FeS_2}$, pyrite).

Thanks for clarification, but my textbook just mentions that If we heated sulfur and iron (without any water involved), only Fes will result. Also, I saw the full procedure on a video and there is no water in the process.

Amer, sorry for the confusion. I thought for some reason you were referring to room temperature reactions in solution (my bias). I've amended the answer, above. Hope this clarifies. The reaction you're describing is quite commonly used as a showy lab demo of an exothermic reaction (remembering from many years ago).

@Amer: Also, I forgot to answer your question re why Fe2S3 is not present in the products: probably due its poor stability at temperature. In their solution syntheses of ferric sulfide by addition of ferric chloride with sodium sulfide nonahydrate, Stansberry et al. (1993) stated that Fe2S3 disproportionated to mixtures of pyrite (FeS2), pyrrhotite, and elemental S above temperatures of 10C.  Apparently there was significant interest in Fe2S3 as a catalyst for coal liquefaction.


text:
Why does ozone have higher entropy than oxygen?

$$\ce{3O2(g) -> 2O3(g)} \qquad \Delta H^\circ (\pu{298 K}) = \pu{+142 kJ mol^-1}$$

Since the RHS of the equation contains less number of gaseous molecules, naturally I expected the $\Delta S$ to be negative. In fact, while solving many thermodynamics questions relating to sign of $\Delta S$, I used this concept to get the correct answers (we were even taught this, that's why).
However, this appears to be an exception as this reaction does happen (thermodynamically possible due to negative $\Delta G$). It's given in my textbook (page 186, NCERT Chemistry) and the textbook accounts for it stating that the reaction has $\Delta S > 0$ without mentioning the reason for it.
Why is the entropy of ozone more that that of oxygen? Is there any simple explanation to it?

Not clear what you mean by "why?" If you're asking why, intuitively, why the relationship in nature exists, I suspect it's because you are placing too much emphasis on the change in number of molecules: you're getting new rotational states in ozone that you don't have in oxygen.

And vibrational states as well...

To put it more plainly - it's molar entropy and mole of O3 is more then mole of O2.

You are mixing up two different quantities (entropy of reaction, and molar entropies of the two gases involved in the reaction) and either your text book is in error, or you have misread it somehow. For that reaction, as written, the formation of ozone from oxygen,
$\Delta G$ is positive and $\Delta S$ is negative (the opposite of what you wrote in your question). It is unfavourable on both enthalpic and entropic grounds. Ozone is thermodynamically unstable with respect to oxygen under standard conditions.
Your experience, regarding the sign of $\Delta S$ and its relation to the change in the number of molecules in a gas-phase reaction, is by-and-large a good guide, because the translational contribution to the entropy typically dominates over the rotational and vibrational contributions in a gas. This reaction is consistent with this rule of thumb.
The relevant thermodynamic quantities are the standard Gibbs free energy, enthalpy, and entropy, of formation of ozone, since the oxygen is in its standard state. Taking the figures from this copy of CRC we have
$$
\Delta_{\text{f}} G^\circ (\text{O}_3) = 163.2 \, \text{kJ}\,\text{mol}^{-1}
\quad
\Delta_{\text{f}} H^\circ (\text{O}_3) = 142.7 \, \text{kJ}\,\text{mol}^{-1}
$$
and for the standard entropies
$$
S^\circ (\text{O}_3) = 238.9 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}
\quad
S^\circ (\text{O}_2) = 205.2 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}
$$
So indeed the standard molar entropy of ozone is a little higher than that of oxygen (as the comments say, if you want to explain this precisely you need to take into account internal degrees of freedom, as well as properly calculating the translational contributions), but this does not lead to a positive entropy of reaction. To calculate this, i.e. the entropy of formation of ozone, we need
$$
\Delta_{\text{f}} S^\circ (\text{O}_3)
= S^\circ (\text{O}_3) - \tfrac{3}{2} S^\circ (\text{O}_2)
= -68.9 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}
$$
remembering that all these standard quantities of formation are per mole of ozone. Equivalently you can calculate this number from $[\Delta_{\text{f}} H^\circ (\text{O}_3)-\Delta_{\text{f}} G^\circ (\text{O}_3)]/T$ giving the same answer.
So the reaction has a negative $\Delta S$ and a positive $\Delta G$.

[EDIT following OP comment]
This means that, in the oxygen-ozone equilibrium, both forward and backward reactions are happening, at the same rates,
but the rate constants are such that the equilibrium concentration (or partial pressure) of O$_3$ is much lower than that of O$_2$.
This is not how things happen in the stratosphere, nor in most methods for preparing ozone. Generally, these involve a multi-step process, in which
the first step is the dissociation of O$_2$ into two oxygen atoms,
by absorbing a photon of UV light, or through an electrical discharge (for example). Wikipedia provides a reasonable summary of the Ozone-oxygen cycle, or Chapman cycle, with links to further resources, while various production methods are described on the Ozone page. The above equilibrium thermodynamics relations cannot be applied to processes like this, because of the injection of energy.
Instead, in the case of the Chapman cycle, a steady-state kinetic scheme can be set up for creation and destruction of O$_3$, and O atoms, where some of the rate constants depend on the intensity of UV light. The net result is a steady (small) population of O$_3$, and O atoms, accompanied by the conversion of energy from UV light into heat which is dissipated in the atmosphere. In real life, the situation is more complicated still, because of the roles of other species and reactions: books such as RP Wayne's Chemistry of Atmospheres give a lot more detail.

[2nd EDIT following further OP comments]
For the equilibrium between oxygen and ozone, the standard Gibbs free energy change for the reaction $\Delta_r G^\circ$ may be related to the equilibrium constant $K$ (see e.g. here)
$$
\Delta_r G^\circ = -RT \ln K
$$
where $K$ is expressed (for ideal gases) in terms of the partial pressures
relative to the standard pressure $p^\circ$ (1 bar)
$$
K = \frac{(p_{\text{O}_3}/p^\circ)^2}{(p_{\text{O}_2}/p^\circ)^3}
$$
For large positive $\Delta_r G^\circ$, $K$ is extremely small,
so the equilibrium partial pressure of ozone is extremely small,
if we are given that the partial pressure of oxygen has a reasonable value.
Indeed, if we take (for the reaction as written, producing two moles of ozone)
$$
\Delta_r G^\circ=2\times\Delta_{\text{f}} G^\circ (\text{O}_3) = 326.4 \, \text{kJ}\,\text{mol}^{-1}
$$
then $K=\exp(-\Delta_r G^\circ/RT)\approx 6\times10^{-58}$;
if the partial pressure of oxygen is $\approx 0.21$ bar
(as in the atmosphere around us),
then the equilibrium partial pressure of ozone is $p_{\text{O}_3}\approx 2.4\times 10^{-30}$ bar.
It is wrong to conclude that the reaction producing ozone "cannot happen"
because the free energy change is positive.
$\Delta_r G^\circ$ is simply the free energy change associated with
the complete conversion of reactants into products.
In reality, an equilibrium is always established somewhere in between.
In this case, it is very far over to one side, but not 100%.
If $\Delta_r G^\circ$ were not so large (but still positive) the
equilibrium would not be so far in favour of reactants.
Thermodynamically, you can think of the position of equilibrium
as being determined by both $\Delta_r G^\circ$ and the entropy of mixing of the reactants with the products:
this will always result in an equilibrium position that is neither 100% reactants
nor 100% products.
It is also possible to give a dynamical interpretation of chemical equilibria
in terms of forward and reverse reaction rates,
but I emphasize that this argument only applies, at least in this simplified form, to one-step chemical reactions. The oxygen-ozone equilibrium certainly involves multiple steps
(nobody is pretending that three molecules of oxygen miraculously collide
together at once). To see why this more complicated mechanism
does not affect the argument above about the equilibrium constant, see
How is it that the equilibrium constant does not depend on the mechanism? and you might also find Transition state and free energy useful. So consider the dynamical equilibrium
$$
\text{A} + \text{B} \leftrightharpoons \text{C} + \text{D}
$$
and assume that the forward and reverse reactions are simple one-step processes
obeying the rate equations
$$
\text{Rate}(\rightarrow) = k_\rightarrow [\text{A}][\text{B}],
\qquad\text{and}\qquad
\text{Rate}(\leftarrow) = k_\leftarrow [\text{C}][\text{D}].
$$
At equilibrium both forward and reverse reactions are happening at the same rate because otherwise the concentrations would be changing.
We can write
$$
\text{Rate}(\rightarrow) = \text{Rate}(\leftarrow)
\qquad\Rightarrow\qquad
K = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}
= \frac{k_\rightarrow}{k_\leftarrow}
$$
So, if $\Delta_r G^\circ$ is large and positive, $K\ll 1$,
and $k_\leftarrow \gg k_\rightarrow$.
The rate constant for the reverse reaction is much larger than that for the
forward reaction.
However, both reactions are still happening, and the rates of these reactions
are equal.
As I said, you can't apply this argument directly to a multi-step reaction scheme, but the two links I gave above show how the same ideas apply in such a case.
Also, most of the above material is covered in standard physical chemistry texts.

Yes, I realised that I had badly misread my book. Could you please tell me how the reaction happens despite having positive gibbs free energy?

I've added a few lines to my answer, to try and address this. Please let me know if it is not clear.

I m not sure your edit answered my query. My question is how can a reaction with positive value of Gibbs free energy happen.

I think negative value of Gibbs free energy is essential for anything to happen

I've added some more material which I hope covers this. There's quite a lot of detail. I suggest that you take some time to digest this, have a look at the links I've given, possibly consult a physical chemistry textbook to confirm the details, and then let me know if there are still any points that are not clear.


text:
Deperoxidation of ethers via alumina: what happens?
I was reading about the procedures for the removal of peroxides from ethereal solvents, and a procedure which seems to have survived to our days is filtration through a column of "activated alumina" (which, if my understanding is correct, is alumina in its "gamma" form).
What I was wondering is: what happens to alumina? My guess is that it could form a sort of stable complex with (hydro?)peroxides, just like urea or carbonates do.
Yet, I couldn't find an explanation.
If a complex is formed, how stable is it? For instance, if the same alumina employed was dried, either via evaporation at room temperature, or at high temperature in an effort to obtain, again, its gamma form, would it decompose violently? Or maybe would it release oxygen, or gaseous peroxides?
And, if alumina has to be handled with care after use, is there a way to dispose it safely, or (better) to "reduce" it, maybe making it usable again for the same process?


text:
How to control the Friedel–Crafts alkylation reaction to methylbenzene as the major product?
My question is due to a discrepancy I found in reactivity and organic conversions.
In reactivity of organic compounds it has been established that an alkyl group activates the benzene ring, and, hence for example, methylbenzene is more reactive than benzene.
Then consider the following situation. Suppose that there are two benzene molecules in a sample, and we convert one of them to toluene (methylbenzene) using Friedel–Crafts alkylation. Then we would see the following change in the mixture:

Now, consider we carry out alkylation once more. As toluene is more reactive than benzene, it will react first forming dimethylbenzene. By continuing this logic we would get hexamethylbenzene as the major product regardless of the amounts we add.
So, how is that we carry out conversions with Friedel–Crafts alkylation regarding toluene as the major product?
Feel free to correct me if my logic is incorrect and if so please specify what should be corrected.

Indeed the reaction would be hard to control. A "dirty" way for obtaining a somewhat good yield of toluene would be that of using a large excess of benzene (both as a reactant and as a solvent), and a smaller amount of alkyl halide: this way, chances are that the reactants "meet" a benzene molecule, instead of an "active" toluene molecule

I agree. @The-Vinz you should make this an answer

@The_Vinz the solution is indeed using an excess of benzene. It is not a dirty way but the usual way of dealing with this kind of problem.

Can deactivating benzene by sulphonation work here?

@The_Vinz Can you please explain how one will get hexamethylbenzene. Isn't methyl group ortho para directing? Surely friedel craft wont add methyl group to meta positions? Isn't it?

The solution to this problem of having chance of poly-alkylation of Benzene rings by Friedel-Crafts alkylation has a general way, and the way is to first perform Friedel-Crafts Acylation and then reduction of the introduced Carbonyl carbon to saturated carbon atom.

Now, proceeding via this way helps to avoid all the problems that arise in a Friedel-Crafts alkylation. Firstly, after the first step, in the main product, we are having one Benzene ring attached to a carbonyl carbon , which deactivates the ring for participating in further electrophillic aromatic substitution and reduces any chances of poly-acylation which upon reduction also doesn't give any poly-alkylated product. Secondly, Friedel-Crafts acylation helps to reduce the possibility of rearrangement which occurs very often in the alkylation step. That's because the positive charge is not on any primary,secondary or tertiary carbon atom but on the acyl carbon which will not undergo any changes in its carbon skeleton structure. So, we can add the unrearranged carbon chain to the Benzene ring and reduce to get less-substituted carbon chains attached to the ring.

To solve your problem, I would rather prefer that you first react Benzene with a mixture of $\ce{CO}$ and $\ce{HCl}$ in presence of anhydrous $\ce{AlCl3}$ (Gatterman-Koch Aldehyde Synthesis) to get Benzaldehyde. Then you can perform a Clemmensen reduction ($\ce{Zn -Hg / conc. HCl}$) or a Wolff-Kishner reduction ($\ce{H2N-NH2/KOH}$) to get Toluene as the exclusive product.

Please note that, you can always control the reaction conditions to get Toluene by Friedel-Crafts alkylation only, still there remains a good amount of chance of getting di-substituted product as those kinds of controls need to be optimised and difficult to carry out and you have to perform purification techniques anyways. However, if you proceed via acylation and reduction, your side products will  be in lesser amount and product can be easily obtained.

It is much better to use an excess of benzene as @The_Vinz has said and to perform only one separation through distillation. Note also that benzaldehyde is made from toluene and not the other way round

We can definitely do that. But this is a kind of standard and nicer way to do it so I preferred to say this.

I would not call $\ce{CO}$, $\ce{Hg}$ and $\ce{NH2-NH2}$ nice; quite the opposite

Meh, it's not like anyone's gonna make toluene from benzene, practical details aren't that important for getting good grade on test, and that's very test-like not practical example.

true - no-one does this in the real world


text:
Why is the probability of a didentate ligand to form a chelate with a central atom, high?
When one ligating group of didentate ligands (like ethylenediamine) is coordinated with a central atom, then the probability of the other ligating group being coordinated, and hence forming a chelate, is high. This is seems to be because one of the ligating groups is bound to be closer to the central atom when the other is coordinated. Does this always hold true? If so, why?

Speaking in general, when a molecule is in a "energetic" environment (in this case, for instance, the energy comes from the temperature of the room, about 298 kelvins away from the "zero energy" point), the molecule moves A LOT. It bends, rotates, and changes its conformation.
When EDTA has no complexed metal, its "two arms" just keep moving randomly, until they catch a cation. When one of the arms "catches" a cation, the arms keep moving randomly. But this time, it's much more likely that the other arm will find the cation, being it so close (the cation won't move too far: the other "arm" keeps it close!).
When both arms find the cation, the rigid complex is formed.
To make a stupid example, think about a person with two magnets on his hands, moving randomly in a room in the hope of capturing with both his hands a metal ball: the "first arm" has to search around all the room, but the second one only has to find the first one

Great! So that's why there's a huge change in entropy accompanying chelation! But why is there a very small change in the heat energy produced?

Indeed, in "normal" conditions complexation by EDTA is exothermic (see: Energetics of Ca(2+)-EDTA interactions: calorimetric study. https://www.ncbi.nlm.nih.gov/pubmed/17030316) what makes the developement of heat negligible is, I guess, the low concentration which is usually employed


text:
Help with naming molecule
I am trying to name the molecule pictured below. I thought it would be 5-bromo-4-isopropyl-2-methylhex-2-ene, but googling that name does not show any results, which makes me think it is not right. Can anyone suggest what I might be doing wrong?

This is how I would be happy to name it, but there are IUPAC rules that you need to look up about systematic naming. You could look at ChemSpider and try the molecular formula; you will find very many compounds, however.

The name is really not the correct one, from two major reasons - substituents given by prefix, like methyl and bromo, or bromoethyl, have the same priority (there's no priority), and because of rules P-45.2.1, P-45.2.2

The suggested name 5-bromo-4-isopropyl-2-methylhex-2-ene is not wrong; however, it is not the preferred name according to current IUPAC recommendations.
Since the compound has a double bond but no principal characteristic group that would be expressed as a suffix, it is clear that the parent structure is a 2-methylhex-2-ene. However, there is a choice between two different principal chains, which lead to different names:

5-bromo-2-methyl-4-(propan-2-yl)hex-2-ene and
4-(1-bromoethyl)-2,5-dimethylhex-2-ene

Both possible chains have two additional (but different) substituent groups in the same positions; therefore, the usual nomenclature criteria related to number and location of substituent groups do not help. Finally alphabetical order is used to identify the preferred name as 4-(1-bromoethyl)-2,5-dimethylhex-2-ene.


text:
Why is the first affinity energy negative?
So, I've found a number of somewhat obscure responses to this question but I'm looking for something concrete:
Why is the first affinity energy negative for most elements?
I appreciate that this is because energy is lost but why is energy being released rather than gained?
Am I also correct in thinking that the second affinity energy is positive because it's 'harder' to push an electron into an electron shell (due to electron shielding)?
Thank you in advance for your help :)

Could you state your level of education? If you are say a 5th grade student, one has to answer this completely different than if you are say a chemistry undergrad in their 1st year. The first step of solving your problem would be clarifying several of the words you used. If you are for example a Ph.D. student in organic chemistry, using concepts like electron shells and shielding, neglecting most of the advances in quantum mechanics in the last 100 years, is highly inappropriate. If you are middle school student however, one can only draw from what you should know

Thank you for your reply, Raditz_35. I'm in my first year of medical school and this is for a sort of 'intro to the hard sciences' module.

The planetary model of orbitals would have the 2s orbital perfect shielding the three 2p orbitals of neon since the three 2p orbitals are degenerate. However the quantum model of the atomic orbitals would show that the 1s, 2s, and three 2p orbitals are all distributed over space rather than having specific lanes like the planets orbiting the sun. So the 1s orbitals doesn't perfect shield the 2s orbital, and the 2s orbital doesn't perfect shield the three 2p orbitals, and the three 2p orbitals don't perfectly shield each other. Slater's Rules give a rough approximation to the shielding of orbitals.
So if you look at the electron affinity data page for free atoms in space, you can see that most elements have a positive electron affinity. That means that the capture of an electron by the neutral atom releases energy.

An atom is composed of a positive nucleus and some number of electrons that are located around the nucleus. Regardless of the model, this is true.
The nucleus has a charge $+Z$ which there are $Z$ electrons, each with charge $-1$.
If you introduce a new electron, the electron is attracted by the positive charge in the nucleus and repelled by the negative charge of the electrons.
However, these are not perfectly balanced out. Generally speaking, the electrons are distributed such that roughly half of the electrons are closer than the nucleus and half are farther away. It works out that in many cases, the electrons farther away can't cancel out enough of the nuclear charge, so there is still a net attraction between the nucleus and the new electron.
You can arrive at this conclusion even without any fancy quantum mechanics.

Thanks a lot, Zhe! This makes a great deal of sense. A follow-up question, if I may: how does adding an extra energy lead to a release in energy (e.g. a photon of light)?

Sorry, I don't understand the follow-up...


text:
Can Molisch test be used to separate monosaccharides from polysaccharides?
Do the results of Molisch test on monosaccharides have any differences with when it's done to disaccharides or polysaccharides?

No. Mono and poly saccharides both give a positive test.


text:
Finding mass percentage of components of a solid mixture
A workbook question asks:

$\pu{2.184g}$ of a solid mixture containing only $\ce{K2CO3}$ ($\mathrm{FW = 138.2058}\pu{ g/mol}$) and $\ce{KHCO3}$ ($\mathrm{FW = 100.1154} \pu{g/mol}$) is dissolved in distilled water. $\pu{31.29mL}$ of a $\pu{0.742\!M}\ \ce{HCl}$ standard solution is required to titrate the mixture to a bromocresol green end point. Calculate the weight percent of $\ce{K2CO3}$ and $\ce{KHCO3}$ in the mixture.

I started this problem by finding the amount of moles of $\ce{HCl}$ added to the titration, assuming that the solution was completely neutralized at the endpoint.
In $\pu{0.03129L}$ of a solution of $\pu{0.742\!M}\ \ce{HCl}$, there are
$$\mathrm{0.03129\; \pu L \times \frac{0.742\; mol\; \ce{HCl}}{\pu L} = 0.0232\; mol\; \ce{HCl}}$$
Then I set up the balanced equations:
$$\ce{2HCl_{(aq)} + K2CO3_{(aq)} -> 2KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$
$$\ce{HCl_{(aq)} + KHCO_{3(aq)} -> KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$
This should indicate that the $\ce{HCl}$ will combine with the carbonate and bicarbonate in a 2:1 ratio, favoring the carbonate, since two moles of HCl are required to neutralize one mole of carbonate.
Then:
$$\ce{\pu{0.0232mol}\ HCl \times \frac{1\; mol\; CO_3}{2\; mol\; HCl} = 0.0116\; mol\; CO3}$$
$$\ce{0.0232\; mol\; HCl \times \frac{1\; mol\; HCO_3}{1 \pu{mol} HCl} = 0.0232\; mol\; HCO3}
$$
And:
$$\ce{0.0116\; mol\; CO3 \times \frac{1\; mol\; K2CO3}{1\; mol\; CO3} = 0.0116\; mol\; K2CO3}$$
$$\ce{0.0232\; mol\; HCO3 \times \frac{1\; mol\; KHCO3}{1\; mol\; HCO3} = \pu{0.0232\; mol}\; KHCO3}$$
Converting moles to grams:
$$
\ce{0.0116\; mol\; K2CO3 \times \frac{138.2058\; g}{mol\; K2CO3} = 1.60\; g}
\\
\ce{0.0232\; mol\; KHCO3 \times \frac{100.1154\; g}{mol\; KHCO3} = 2.32\; g}
$$
When I add these masses together, I get $\pu{3.92 g}$, which is greater than $\pu{2.184 g}$.
Where did I go wrong?

HINT - This is linear algebra with two equations and two unknowns. So there are two equations relating moles(K2CO3) and moles(KHCO3).

Welcome to Chemistry.SE Please note some of the edits made. It will make your life easier notably $\ce{}$ for chemical equations and formulas, $\mathrm{}$ for mathmatical equations and $\pu{}$ for units.

We first start by writing the system of equations in this reaction, assuming that we don't know the ratios that the elements combine in.
Because we know the mass of the unknown mixture, we'll use $x$ to represent the mass of $\ce{K2CO3}$ and $y$ to represent the mass of $\ce{KHCO3}$, in grams. This is our first equation:
$$x + y = 2.184$$
Next, we know the final amount of moles of $\ce{HCl}$ that reacted. It was
$$\mathrm{0.03129\; \pu L \times \frac{0.742\; mol\; \ce{HCl}}{\pu L} = 0.0232\; mol\; \ce{HCl}}$$
Now, let's reconsider the balanced equations for the reaction. We have
$$\ce{2HCl_{(aq)} + K2CO3_{(aq)} -> 2KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$
$$\ce{HCl_{(aq)} + KHCO_{3(aq)} -> KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$
We can see that it takes $\ce{2 HCl}$ to neutralize one $\ce{CO3}$, but it only takes $\ce{1 HCl}$ to neutralize one $\ce{HCO3}$.
We also know the formula weights of the individual components. Now, we can write the second equation as an expression of the sum of the moles of the reactants. Our coefficients for $x$ and $y$ become:
$$
\ce{\frac{2\; mol\; HCl}{1\; mol\; K2CO3} \times \frac{1\; mol\; K2CO3}{138.2058\; g}} \times x
$$
$$
\ce{\frac{1\; mol\; HCl}{1\; mol KHCO3} \times \frac{1\; mol\; KHCO3}{100.1154\; g}} \times y
$$
Now we can cancel out units and write out the second equation:
$$\ce{\frac{2\; mol\; HCl}{138.2058\; g}}x + \ce{\frac{1\; mol\; HCl}{100.1154\; g}}y = \ce{0.0232\; mol\; HCl}$$
After that, this is straightforward linear algebra. This system of equations is simple enough to be solved by basic substitution and elimination. However, you can choose to use a matrix. In that case, we have
$$
\mathrm{
\begin{bmatrix}
    1 & 1 \\
    \frac{2}{138.2058} & \frac{1}{100.1154}
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
}
=
\begin{bmatrix}
2.184 \\
0.0232
\end{bmatrix}
$$
Rearranging gives
$$
\mathrm{
\begin{bmatrix}
    1 & 1 \\
    \frac{2}{138.2058} & \frac{1}{100.1154}
\end{bmatrix}^{-1}
\begin{bmatrix}
2.184 \\
0.0232
\end{bmatrix}
=
\begin{bmatrix}
x \\
y
\end{bmatrix}
}
$$
Using a calculator we find that
$$\mathrm{
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
\begin{bmatrix}
0.309\\
1.875
\end{bmatrix}
}
$$
$$x = \ce{0.309\; g\; K2CO3}$$
$$y = \ce{1.875\; g\; KHCO3}$$
Finally, the mass percentages are determined to be:
$$\ce{\frac{0.309\; g\; K2CO3}{2.184\; g\; total} \times 100{\%} = 14.1{\%} K2CO3}$$
$$\ce{\frac{1.875\; g\; KHCO3}{2.184\; g\; total} \times 100{\%} = 85.9{\%} KHCO3}
$$


text:
Automated Identification of Isomeric Relationship from 3D Structure?
Is there a software package that when given two sets of 3D structures can return the isomeric relationship between them?  Even if it was only able to state that they are conformational isomers that would be enough.
It appears that Jmol has the ability, but I am seeking a non-graphical solution that could be easily included in scripts that would compare many such pairs of structures - in addition, I'm not a java programmer.

You might become interested in OpenBabel.  Written in C++, it offers bindings to Python (ref).  It looks like stereochemistry is covered more in detail (here, here) than the obchiral command (here).
OpenEye's Chemistry Toolkit, on the other hand, equally capable to read numerous chemical file types (ref), offers several approaches to detect stereochemistry (examples).  Written in Python, it may offer an alternative starting point.  Perhaps you qualify for one of their academic licenses (ref).


text:
Number of configurations/states for electrons in orbitals
I'm performing complete active space calculations for the prediction of X-ray absorption L edge spectra. This involves an excitation from 2p orbitals to the 3d orbitals of the metal. For the electronic configuration of 2p5 3d5 I can calculate the total number of microstates from "n choose k" to be 1512 however the total number of pure spin states is 600. Could someone please explain how to obtain this number?

What do you mean by "microstates"? From $2K$ spin-orbitals with $N$ electrons, you get $\begin{pmatrix}2K\N\end{pmatrix}$ Slater determinants but not all of them are eigenfunctions of the total electron spin operator. You can take appropriate linear combinations of determinants that form spin-adapted configurations or configuration state functions (CSFs). For configurations with a large number of unpaired electrons and a low total spin, the number of CSFs is significantly smaller than the number of determinants.


text:
How to find the current transferred to the anode and cathode of a lithium ion battery at different voltages?
I am trying to find the current transferred to the anode and cathode of a lithium ion battery at different voltages. I know the temperature (approximately) of the battery, the battery's starting voltage, the voltage at 30sec intervals of a constant power discharge (from which current can be deduced), the dimensions of the battery and the initial amount of charge in the battery. What could I do to find this current transfer?

Use ammeter instead?

Do you mean "current" (e.g. mA) or "charge transfer" (e.g. mAH)?

Current so it would be mA and can't really use an ammeter because that would require me opening up the battery which is kind of hard at this stage.

Exactly what is the question? Isn't this a series circuit and the current flow the same thruout whether electrons in the wire or ions in the battery? Are you interested in the cell internal resistance or the electrode efficiencies?

What is a "current transfer" ? Current do exists. It is not transferred.

Power is the energy transferred per unit time ($\frac{J}{s}$) and is determined by the potential drop (V) and the current (I) at a given point in time as $P = IV$ so to get the current, solve for $I=\frac{P}{V}$. This gives you the current at a given point in time.
Your question has a few points that might be worth addressing. As Dr Moishe Pippik commented, "current transferred" is a little confusing, either instantaneous current ($I$), or total charge transferred ($Q$), which is the current integrated over time: $\sum_{t} I(t)\Delta t$ in your case. So if what you really want is the total charge transferred, you would have $Q = \sum_{t}\frac{P\Delta t}{V(t)} $. Strictly speaking you should exclude t=0 point to avoid double counting.
You also asked about both the anode and the cathode. The current for both will be the same number but with opposite sign.


text:
Entropy change during flow of heat

If heat $Q$ flows from the surrounding nature temperature at $T_\text{surr}$ to the system at temperature $T_\text{system}$, then $$\Delta S_\text{total}=\frac{+Q}{T_\text{system}}+\frac{-Q}{T_\text{surr}}$$

This definition is from my book in which heat is exchanged between system and surrounding. This formula for me would only be valid when both the system and the surrounding is undergoing reversible processes but nothing about it is said over here. Does this mean this formula assumes both the heat gained and heat lost are rebersible processes? Isn't entropy defined for reversible processes?

I agree with your assessment; the statement is imprecise and even "hinky."  It might be considered correct if the amount of heat transferred was small enough  that it changed neither the temperature of  the system nor surroundings.

Entropy is a state function, which means its value does not depend on how it gets from state A to state B, rather it only depends on the final state B itself.  In this case, we can imagine a process which takes the system from state A to state B via a reversible process, and your expression for the change in entropy would be correct:
$$\Delta S = \frac{q_\text{rev}}{T_\text{sys}} - \frac{q_\text{rev}}{T_\text{surr}}$$
where I have explicitly labelled the heats as reversible heat flows (this quantity does depend on the path taken).
Even if in reality such a reversible path would not be taken, the final change in entropy would still have the same value, so it is still valid to use that expression.

Note that this expression for the entropy assumes an isothermal process.


text:
Why are 5-ringed saturated chelates more stable than unsaturated ones while 6-ringed unsaturated chelates are more stable than saturated ones?
5-ringed saturated chelates seem to be more stable than unsaturated ones. On the other hand, 6-ringed unsaturated chelates are more stable than saturated ones. Why do these hold true?


text:
Why is 5-(aminomethyl)-3-ethylheptan-2-ol named so?
Why is 5-(aminomethyl)-3-ethylheptan-2-ol named so and not 6-amino-3,5-diethyl-heptan-2-ol? I.e why is the amine part not taken in the longest chain since it is a functional group with a definite priority?

As a substituent prefix, 'amino' does not have any inherent priority over 'methyl', etc.

Note that 5-(aminomethyl)-3-ethylheptan-2-ol and 6-amino-3,5-diethylheptan-2-ol are different compounds (10 vs 11 carbons).

@orthocresol but shouldn't carbon attached to amino group be included in main chain?

Yes, but that is only a requirement if you don't already have a more senior functional group in the molecule. Here you already have an alcohol, so the amine doesn't matter.

The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for substituents corresponding to the suffix
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)

Thus, the principal chain shall include the $\ce{-OH}$ group corresponding to the suffix “-ol” according to Rule P-44.1.1.
Furthermore, the principal chain shall be the longest possible chain according to Rule P-44.3.2.
These two rules are already sufficient to identify the principal chain of the given compound as heptan-2-ol. The further rules are not relevant in this case.
Thus, the position of the amine group is not relevant for the principal chain. It would be relevant (as a prefix "amino") if the Rules P-45.2.1 or P-45.2.2 are be needed for a decision, or (as a suffix "-amine") if the amine is the principal characteristic group.


text:
What are the magnetic quantum numbers for specific d orbitals?
I mean I know they’re $\ce{d_{-2}}$, $\ce{d_{-1}}$, $\ce{d_{0}}$, $\ce{d_{1}}$, $\ce{d_{2}}$ but how do these numbers relate to each one of $\ce{d}$ orbitals $\ce{d_{z_2}}$, $\ce{d_{xz}}$, $\ce{d_{yz}}$, $\ce{d_{xy}}$, $\ce{d_{x_2 - y_2}}$)?
I couldn't find a definitive answer anywhere.

They don't. $,$

Katie, $\ce{d_{z_2}}$, $\ce{d_{xz}}$, $\ce{d_{yz}}$, $\ce{d_{xy}}$, $\ce{d_{x_2 - y_2}}$ are linear combinations of $\ce{d_{-2}}$, $\ce{d_{-1}}$, $\ce{d_{0}}$, $\ce{d_{1}}$, $\ce{d_{2}}$. See around page 338 of Quantum Chemistry (MCQUARRIE, D., 2nd ed.).

On p. 238 of my copy of McQuarrie (1983; his Table 6-6), he lists the following for the hydrogenlike atomic wave functions (expressed as real functions), describing these as "commonly used":
\begin{align*}
n=3, l = 2, m=0:     & \ \psi_{3d_{z^2}} \ ... \\
n=3, l = 2, m=\pm 1: & \ \psi_{3d_{xz }} \ ... \\
       : & \ \psi_{3d_{yz }} \ ... \\
n=3, l = 2, m=\pm 2: & \ \psi_{3d_{x^2-y^2 }} \ ... \\
       : & \ \psi_{3d_{xy }} \ ... \\
\end{align*}


text:
Is sodium ion an acid?
As sodium ion has a 1+ charge and it can form the ionic compound NaOH with OH1-, hence it is a lone pair accepter (is this correct?). Therefore sodium ion should be an acid?

Yes, it's a Lewis acid, but a weak one.

But $\ce{Na+}$ isn't an acid in the Brønsted–Lowry theory, so it depends...

I thought sodium was in part of the alkali metals,but ions clearly don't work the way you think they're supposed to,but yes it's technically and "acid".

Sodium ion in aqueous chemistry is essentially nonacdic (contrast this with lithium ion, which displaces carbon dioxide when the bicarbonate is  "precipitated" and the precipitate is actually lithium carbonate). But in organic chemistry sodium ion shows some Lewis acid character, as demonstrated in the incompletely ionic structure and the solvation of sodium cyclopentadienide:

The nature of $\ce{NaCp}$ depends strongly on its medium and for the purposes of planning syntheses; the reagent is often represented as a salt $\ce{Na^+C5H5^-}$. Crystalline solvent-free $\ce{NaCp}$, which is rarely encountered, is a "polydecker" sandwich complex, consisting of an infinite chain of alternating $\ce{Na^+}$ centers sandwiched between $\ce{μ-η5:η5-C5H5^-}$ ligands.[1] As a solution in donor solvents, $\ce{NaCp}$ is highly solvated, especially at the alkali metal as suggested by the isolability of the adduct $\ce{Na(tmeda)Cp}$.[2]

Cited References

Robert E. Dinnebier; Ulrich Behrens & Falk Olbrich (1997). "Solid State Structures of Cyclopentadienyllithium, -sodium, and -potassium. Determination by High-Resolution Powder Diffraction". Organometallics. 16 (17): 3855–3858. doi:10.1021/om9700122.

Elschenbroich, C. (2006). Organometallics. Wiley-VCH: Weinheim. ISBN 978-3-527-29390-2.

According to the bronsted lowry definition of an acid, the sodium ion is not an acid. This is because the bronsted lowry definition of an acid states that an acid is a proton donor. The sodium ion is not a proton donor and so according to this definition it is not classed as an acid.
I hope this answers your question :)

But the question doesn't specify Bronsted-Lowry. It actually has descriptions related to Lewis and Arrhenius acids, so the Bronsted-Lowry definition is the least related to the question.

But in water the ion will be hydrated, and the hydrated ion can most definitely act as a (weak) Bronsted-Lowry acid

@IanBush exactly


text:
Nuclear Fusion and the worlds water supply
First off, I'm probably missing an important part of the argument regarding nuclear fusion.
What I think I know about it;
Nuclear fusion (using heat instead of gravity) is taking Deuterium and Tritium(sp?) and heating them until they fuse. This causes a release of energy that is used to generate heat for the mundane use of steam power generation.
My issues;
I know that deuterium is a very rare isotope of hydrogen. However class one stars aren't made from deuterium. That means it's possible; however unlikely, that should fusion power become feasible that hydrogen-0 will eventually be fusible.
Water is made from hydrogen and oxygen. Fusion turns hydrogen into helium. Using helium for fusion is infinitely more difficult than using hydrogen. If we start start harvesting hydrogen from water then more oxygen would be release into atmosphere which would drive world temperatures down, but not as fast as water removal would lead to super heating in the thinner crust of the ocean; see Venus. Losing the oceans are a consequence all their own and I don't feel the need to go into detail of how bad a 1% oceanic loss would be on ecological systems as well as future habitation of the Human home world.
I haven't seen any argument against nuclear fusion outside of conspiracy theory regarding aliens and some god awful eldritch abominations. Read as, no one seems concerned about this.
I'm assuming that my thought process on this subject must be flawed since I would assume the end of the world would be higher on the priority list than clean energy. Yes there is a lot of water on Earth, but at one point crude oil was though inexhaustible. Better refinement processes and higher demand proved that statement wrong.
Why are my thoughts regarding this wrong, or am I right and why isn't anyone concerned.
P.S. I considered putting this on World Building but it feels more legitimate than it should be.

The gist is that it is possible to fuse deuterium. However the reactors presently don't generate as much power as they consume to create the fusion reaction. If/when scientists do get a fusion reactor working that creates a net surplus of energy then it will be a remarkable source of clean energy.

Solar and wind energy with a peak battery storage facility can replace power plants and they are completely renewable. Fusion power isn't renewable as it destroys the hydrogen atom. However this is off question. I'm wanting to know what I'm missing that makes destroying hydrogen+1 (deuterium)  and eventually hydrogen a good idea. It seems like a bad plan in the long run. Maybe this is the wrong stack for this question?

@Kayot Yes, this is definitely a physics question. The answer is that we could power the world from deuterium and tritium (which is made from lithium, actually) for thousands of years without making a noticeable dent in the supplies available on earth ($5x10^{13}$ tons of deuterium in the worlds oceans $2x10^{11}$ tons of Li).

Nothing "wrong" with using solar and wind. There is a humongous amount of water on the planet. Using a tiny fraction of it would not be a problem for a long time. Whether we have enough water to provide power till the sun novas and destroys the planet in 5 billion years is another matter.

So it's not that this isn't an issue, more that it's like, "It won't be a problem in my life time so why care?" scenario?

The energy released by a the fusion reaction is exorbitantly higher than that of chemical reactions, according to
https://www.iter.org/sci/FusionFuels
four million times more energetic (per mass) than burning conventional chemical fuel. If the plan would be to get the deuterium by elektrolysis from ocean water the normal hydrogen would not be used.
Arguments against nuclear fusion that I have heard of are:
-can create radioactive products (if I remember correctely it has something to do with the generated neutrons)
-big initial starting energy requiered (how do you get them back running in case of an blackout)
-expensive costruction which needs to be able to withstand high temperatures
and this list is by no means conclusive...

You forgot the one about a massive nuclear explosion because of reasons. I had a thirty minute discussion with a friend explaining that it would only slag the reactor.

@Kayot A fusion reactor just cannot explode. Fission reactors explode if they are built and run by idiots who refuse to understand that a molten reactor core produces hydrogen gas from the coolant water.

Or if your reactor gets hit by a tsunami and all of your emergency systems fail. Whereas the fusion reactor's plasma would expand if the magnets holding it in place would fail and cooles which stops the fusion. There are other types of fission reactors with heavy water as moderator (to slow down the neutrons) which are said to be saver because the fission stops if the heavy water would be replaced with water or removed...

That's what I said. I had to explain that in a worse case scenario it would slag the reactor. That means it would simply melt the inside and then cool off.

@BenyBenz Which happens if you first deny the danger of tsunamis at the japanese coast, and then when the inevitable has happened don't fly in emergency pumps and power generators. And still then the reactors wouldn't have exploded if they had had the catalytic converters for hydrogen which were installed in e.g. german reactors (enforced by law) thirty years ago. (Heavy water reactor cores also melt if they are inadequately cooled.)


text:
How to without calculations understand the liquid-liquid extraction efficiency?
I'm asking myself this question for a very long time and I haven't found a simple an good explanation yet for the fact that it's better (if you want to clean them) to wash your hands two times with one millilitre of soap rather than one time with two millilitres.
If you are an organic chemist you will probably more ask yourself,

Why is it better to extract a solute in a solvent $\ce{S1}$
  (non-miscible with a solvent $\ce{S2}$) doing two extractions using
  $\ce{10 mL}$ of $\ce{S2}$ rather than one extraction using $\ce{20 mL}$
  of $\ce{S2}$?

This is what I learnt at school and what I always used in organic chemistry lab sessions.
As mention above I am looking for a simple and good explanation, that is, without using any calculation. I am looking for a reasoning a bit like Feynman. I tried to resonate using dots but every time I fail and it leads me to the same result, that is, there is no better way to extract a solute...
EDIT : I put the below parts in very small because I think some people misunderstood my question. The question is just above, even though there is not interogation point.

For those who wonder if that is true or not, and as I wasn't convinced at all, I made some calculations before to be "sure" about this fact.
So let's start and imagine you want to extract iodine from water using benzene. In the following, the subscripts $aq$ will stand for the aqueous phase and $org$ will stand for the organic phase. $C_T$ will refer to the total concentration. $A$ will refer to the molecule we want to extract.
Let's call $D$ the partition coefficient between the organic and the aqueous phase, $$D = \frac{C_{T, org}(A)}{C_{T, aq}(A)}  = \frac{C_{T, org}}{C_{T, aq}}\tag 1$$
The quantity $Q$ of the solute $A$ in both phases, is determined by, $$\begin{cases} Q_{T, org} = C_{T, org} \times V_{org} \\ Q_{T, aq} = C_{T, aq} \times V_{aq} \end{cases} \tag 2$$
At the equilibrum, for the first extraction, a mass balance gives, $$C_0 \times V_{aq} = C_{1, aq} \times V_{aq} + C_{1, org} \times V_{org} \tag3$$
leading to, $$C_{1, aq} = C_{0, aq} \times \frac{1}{1+Du} \tag 4$$ where $u = V_{org}/V_{aq}$.
Then after $n$ extractions, $$C_{n, aq} = C_{0, aq} \times \left( \frac{1}{1+Du}\right)^n \tag 5$$

For the case of the extraction of $\ce{I2}$ from water to benzene, I used the solubility values found on the French Wikipedia page to calculate the partition coefficient (I am not taking in consideration the activity coefficients as I don't know them). It leads to $D = 373$ (assuming my calculations are correct)
The values I used are, $C_0 = 1.300 \times 10^{-3} \text{ mol/L}$, two extractions with $u = 0.2$ compared to one with $u = 0.4$.
For the two extractions, $C_{2, aq} = 2.27 \times 10^{-7} \text{ mol/L}$
For one extraction, $C_{1, aq} = 8.66 \times 10^{-6} \text{ mol/L}$
We clearly see the difference which is huge, even though these calculations are approximative.

Basically, it all boils down to $\left(1+{x\over2}\right)^2>1+x$.

@IvanNeretin could you elaborate, please? I don't understand what your $x$ refers to. And if it refers to the concentration of the solute I don't understand your reasoning that boils it down to this inequality.

My $x$ stands for your $D\cdot u$. If you split the solution in halves, then you make your $x$ twice smaller, but your $n$ twice bigger.

@IvanNeretin Oh ok., but my question is about finding a logic reasoning leading to this fact. My calculations are just for non-convinced people. Because even though my calculations are correct this fact it absolutely not logic for me. Imagine you want to explain this to a non-scientific person.

That's why I made it a comment and not an answer.

@IvanNeretin I thought I didn't ask my question correctly which wasn't a problem. I wasn't sure so I preferred to edit it. No offence.

I tried a solution from thermodynamics point of view: https://physics.stackexchange.com/q/430495/153036


text:
Expansion work by a gas when pressure is not explicitly constant?
How is it possible to calculate the work done by a gas when pressure is not explicitly stated to be constant? The question states:

Calculate $\Delta H$, $Q$, and $W$ when $1\ \mathrm{mol}$ of He expands from $V = 5\ \mathrm L$ at $T = 298.15\ \mathrm K$ to $V = 10\ \mathrm L$ at $T = 373.15\ K$.

I am not told that the process is isobaric, adiabatic, or even to treat He as an ideal gas (although I know it roughly approximates one).
If I treat it as an ideal gas, I can find $\Delta H$ by using the equations $$H = U + pV$$ $$p = nRT/V$$ and $$U = (3/2)RT$$ but I have no ideas about finding $Q$ or $W$ since I cannot make any other assumptions. Is there an implicit assumption I should make about He gas (or ideal gases in general) expanding, or is there simply not enough information to solve this problem?

As you’ve said, the problem statement, which does not specify a path, is insufficient to calculate the path functions $q$ and $w$.

To add to what @a-cyclohexane-molecule said, it is possible to determine Q-W for any unspecified path between the two end states, but not Q and W individually (which are both non-unique).

You can calculate the work done and Heat exchange if you assume that the process is quassi-statically performed and it is reversible and most importantly done in a single step.

Assume that the process through which the $\ce{He}$ is expanded obeys the equation, $PV^x = K$(constant). This equation can also be alternatively stated as, $TV^{x-1} = K$. Now, suppose the gas is transformed to the state $(T_2,V_2)$ from the state $(T_1,V_1)$. So, we can thus write, $$T_2V_2^{x-1} = T_1V_1^{x-1}$$ so, putting the given conditions, we can find $x$ as, $$x= 1+ \frac{\log\left(\frac{T_2}{T_1}\right)}{\log\left(\frac{V_1}{V_2}\right)} \approx 0.676$$ Thus, after knowing $x$ ,you can calculate the work done by the gas as,  $$W= \int_{V_1}^{V_2} P\mathrm{d}V = \int_{V_1}^{V_2} \frac{K}{V^x}\mathrm{d}V = \frac{P_2V_2 - P_1V_1}{1-x} = \frac{R(T_2-T_1)}{1-x}$$ We know all the value of $T_1, T_2 $ & $x$, and work done can be easily calculated. and we already know the change in Internal energy  and thus heat exchange can be easily calculated and also $\Delta H $ can be calculated   ($Q = (C_V + \frac{R}{1-x}) \Delta T$ and $\Delta H = C_P \Delta T $) and thus all the quantitites can be known.

While $PV^x = C$ is perhaps the simplest general assumption you can make, it is certainly not implied by the question statement. I know you mention that this is an assumption in your answer, but I’d like to emphasize it again.

Another path would be to assume that temperature is varied linearly with volume along the process path.  This would lead to a different amount of work and heat.

@SoumikDas. So T = 223.15+15V doesn't satisfy the two end conditions?

@Chester Miller, Oh Sorry, I had some misunderstandings. Yes it is possible. Sorry for that...


text:
Molecule identification
I'd really like to know what compound this formula is and don't know if there's any software for that matter, if anyone could help me :)
(With a sw suggestion or the compound identification)

Many software, both online and downloadable executables, are available for identifying a molecular structure - either by giving the official IUPAC name, or matching up the structure to a large database and returning IUPAC/trade names/properties/etc...
Online: NIST, Pubchem, ChemSpyder, Reaxys
Download: ChemBioOffice Software
A variety of options are available to search for structures such as by molecular formula, structure, identification values, etc... Easiest for most approaches is using any of the software's "Draw" features, to draw the exact structure and then search the database for it. Online options may have difficulty with the exact structure searches on molecules as large as yours above, so as much detail as possible if useful (stereochemistry, for instance, using dashes and wedges).
ChemBioOffice (specifically ChemDraw) is able to give you the IUPAC name, which would then let you look it up in any of the online databases.
If the drawing of the structures is causing you difficulty, you can also search with molecular formulas, but it may take some trial and error to identify the actual structure you want as you look for larger or more complex molecules this way. NIST is a great choice for this approach.
Outside these options, if you have any interest or experience coding, you may be interested in using RDkit (Python module), which has a large array of functions for molecular data interaction. For instance, if you use those above structure "Draw" features online, most will also show a SMILES code which is a way of shorthanding a molecular structure's features; place a SMILES code into a script that accesses RDkit and you can request it to draw the structure, look up information related to it (if it can find the SMILES code in its databases), etc... Warning - it is mostly just development manuals, no tutorials last I checked, so perusing the module can be rather arduous.
As an aside, your molecule in particular gave those online database structure searches a bit of difficulty when I attempted it, so you may need to find a way of giving it more structural information, or perhaps searching for sub structures first (such as just the fused and linked cyclohexanes without the Chlorines or amine group).
However, I found one very similar to your structure, but with aromatics instead for two of the rings: 1,1-dichloroheptadecane

Maybe you have drawn the structure incorrectly (some unsaturation/aromaticity missing), and meant rather

Which belongs to sertraline.
(No software used, just the experience/expertise/memory :) )

The double ring could also be naphthalene which to me seems more likely, then the compound would be a  methyl amino  dichloro phenyl naphthalene.

@porphyrin right, if the original structure was drawn incorrectly, there are more possibilities what is could be, but I guess your compound, properly named 4-(3,4-dichlorophenyl)-N-methylnaphthalene-1-amine, is much less notorious.

mykhal and @porphyrin: I agree, and I'm glad I wasn't the only one thinking this. On top of instinct and for the purposes of the question, I also found that the only "matching" structures were the ones you've both mentioned here.


text:
How do significant figures work on this computation?
This isn't a homework question although it may seem like one. We've never really discussed significant figures and it was just introduced to us when we were learning how to solve word problems regarding gas laws.

Find $V_2$ when $V_1=\pu{400 mL}$ and $T_1=\pu{32^\circ C}$ and $T_2=\pu{-12^\circ C}$.

My computation goes like this:
$$V_2=\frac{T_2V_1}{T_1}=\frac{(\pu{261 K})(\pu{400 mL})}{\pu{305 K}}=\pu{342.295082 mL}$$
As per the given, the least number of significant figures is one (from $\pu{400 mL}$). How do I approximate $\pu{342.295082 mL}$ to one significant figure?
I didn't know how to create a search query about this so I went here. Thanks.

400 is three significant figures. 0 is as good a figure as any other.

342 the number with least figures determines result figures in this operation. Usually 400 should be written with a bar above the zeros. If we took 400 with just one figure, the result would be 300.

@IvanNeretin No it isn't. The zeros might be significant. And to solve this question (i.e. report a sensible answer), you'll have to decide for one or the other.

Technically $400$ has 1 significant figure. $400.$ has 3. Though for this problem I think assuming 3 significant figures in $\pu{400mL}$ is reasonable

Assuming the precision of the given volume, $V_1 = \pu{400 mL}$, is only to 1 significant figure (i.e. only the hundreds place digit of "4" is precise), then your final answer would be rounded to $V_2 = \pu{300 mL}$.*
If we were to say that $V_1$ had 3 significant figures, then the final answer would round to the ones place, resulting in $V_2 = \pu{342 mL}$. However, the value would technically need to be written as "$\pu{400\!. mL}$," including the decimal place to denote the precision going out to the ones place, in order to imply 3 significant figures instead of 1.
In general, zeros (to the left of the decimal place) not surrounded by non-zero values are not denoted as "significant." For further rules and guidelines for identifying what numbers are significant, how many significant figures there are, and how addition/subtraction or multiplication/division impact the final answer's significant figures, check out this web page from Northern Kentucky University (recommended search terms for future information on this subject would be "understanding significant figures" or "significant figures rules").
*If difficulty was also found in the process of where to begin rounding the answer from, the key is finding the "rounding digit," which, if you have only 1 significant figure to work with, will be the furthest numerical value to the left. See this page for further tips.