Find the Structural Formula of Hydrocarbon Compound X There is a specific compound $\ce{C3H6O}$ that exists as hydrocarbon compound X. It does not react with metallic sodium, bromine does not lose it's color when applied to the compound, and does not give the silver mirror reaction when ammonium silver nitrate, in fact it doesn't react.

I've realized since it does not react with sodium, it cannot have a hydroxyl functional group, and that since bromine does not give the discoloration reaction, the Carbons of the compound cannot have double bonds. And obviously, if it does not give the silver mirror reaction, than it cannot be an aldehyde or a carboxyl acid. However, the answer that the book gave was $\ce{CH3COCH3}$. Am I doing something wrong here?

Your reasoning is correct. Are you confused because acetone is not a hydrocarbon? That has to be a mistake in the wording of the question. Another common mistake is to forget that there are some reactions that aldehydes can do that ketones are inert to.

There are seven constitutional isomers for $\ce{C3H6O}$, not counting the enols, which are the minor tautomers of acetone and propanal:

Acetone - a ketone
Propanal - an aldehyde
Cyclopropanol - an alcohol
Allyl alcohol - an alkene and an alcohol
Methoxyethene - an alkene and an ether
Oxacyclobutane - a cyclic ether
Propylene oxide - an epoxide

Your reasoning is correct based on the chemical tests:

Sodium reacts with alcohols - a negative sodium test means no alcohol functional group is present, eliminating allyl alcohol (and the enols if you consider them).
Bromine reacts with alkenes and alkynes (and a small number of other functional groups, like phenols) - a negative bromine test means no alkene functional group is present, eliminating allyl alcohol, methoxyethene, and the enols.
Ammoniacal silver nitrate (Tollens' reagent) reacts with aldehydes - a negative Tollens test means no aldehyde function group is present, eliminating propanal.

The results of the qualitative test leave you with three possibilities, acetone, oxacyclobutane, and propylene oxide. With the evidence you have, you might not be able to distinguish between the three.

An iodoform test would confirm or eliminate acetone.

A test using silver nitrate and periodic acid would confirm or eliminate propylene oxide.


pH of ammonium acetate solution I'm trying to calculate the pH of a 1M $\ce{NH_4CH_3COO}$. I know that I'll have these reactions:

$\ce{NH_4CH_3COO <=> NH_4^+ + CH_3COO^-}$
$\ce{NH_4^+ <=> NH_3 + H^+}$
$\ce{CH_3COOH <=> CH_3COO^- + H^+}$

I know the $K_a$s of the last two, so I'm able to calculate the $K$ of the first one (it's $\frac {K_{a1}}{K_{a2}}$), which gives me these equations:

$\frac {K_{a1}}{K_{a2}} = \frac {(x-y)(x-z)} {1 - x}$
${K_{a1}} = \frac {y(y+z)} {x-y}$
${K_{a2}} = \frac {z(y+z)} {x-z}$

But I have only 2 independent equations (the first one is just the ratio of the second and third) and three variables so I'm unable to solve for $[\ce{H^+}]$, which is $y+z$...

What do I do?

Ok, I will follow the assumption proposed by @Zhe above (possible incorrect as he states, but please do not be confused by that).
In order to solve this problem we need two acidity constants:  $\mathrm{p}K_{\mathrm{a}(\ce{NH4+})} = 9.25$ and
$\mathrm{p}K_{\mathrm{a}(\ce{CH3COOH})} = 4.76$.
First  we state the proton balance (the amount of protons taken up have to be equal to the amount of protons given off in the system): Initially we have $\ce{H2O}$ and $\ce{CH3COONH4}$.
Proton balance:
$[\ce{H3O+}] + [\ce{CH3COOH}] = [\ce{OH-}] + [\ce{NH3}]$
At $\mathrm{pH} = 7, [\ce{H3O+}] = [\ce{OH-}] = \pu{10^{-7} M}$:
At this $\mathrm{pH}$ the proton balance can be simplified as $[\ce{CH3COOH}] = [\ce{NH3}]$.
The simplified proton balance will be true only at a $\mathrm{pH}$ that is exactly in the middle of the two $\mathrm{p}K_{\mathrm{a}}$-values. We get $\mathrm{pH} = (4.76 + 9.25)/2 = 7.005 \approx 7$ (only one significant figure is given, since you have stated the concentration as $\pu{1 M}$).


Toxicity of Mercury (II) Iodide and dicopper (I) tetraiodomercurate (II) This is a continuation from my previous post.
Could you please give me some info on the toxicity of compounds $\ce{HgI2}$ and $\ce{Cu2[HgI4]}$? I found on sigmaaldrich.com that these are very dangerous but could not find any information on the dosages for skin contact and swallowing.

Following the @Nilay Ghosh recommendation, I found an SDS sheet for $\ce{Cu2[HgI4]}$ and there is SDS for $\ce{HgI2}$.
For $\ce{Cu2[HgI4]}$:

Acute toxicity estimate Oral (Expert judgment): 5,1 mg/kg
Acute toxicity estimate Inhalation (Expert judgment) 4 h: 0,051 mg/l
Acute toxicity estimate Dermal (Expert judgment): 5,1 mg/kg

For $\ce{HgI2}$:

LD50 oral rat: 18 mg/kg
Acute toxicity estimate Inhalation (Expert judgment) 4 h: 0,051 mg/l
Acute toxicity estimate Dermal (Expert judgment): 5,1 mg/kg


Is melting of gold a physical property? I read that melting of gold is a physical property, but I wonder if this is true.

If we want to melt gold we will need to add heat, and once we add heat to gold to melt it, there must be a chemical change, not a physical change, right?

It is a phase change - gold exists as a solid crystal (fcc I believe off the top of my head), as a liquid, and as a vapor (gas).  First order phase transitions exhibit a volume change, and an enthalpy of the phase change.  In the case of gold, both the crystal and the liquid are metallic, with similar numbers of nearest neighbors.

All elements exhibit various phases.  Since pure gold is, well, pure gold, I'm not sure how a phase change in gold could be considered a 'chemical change', particularly in the solid-to-liquid.  The type of bonding does not change, and there are no 'chemical reactions' going on. This is all fairly basic thermodynamics.


Is there a rule of thumb to predict when coupled cluster might dip below the variational limit? Coupled cluster is a non-variational method, meaning that it can give energies that are below the true FCI energy (the "variational limit" for variational methods).
Often coupled cluster still gives energies far above the true FCI energy, and sometimes it gives energies that dip below the variational limit.
Are there some guidelines on when this will happen and when we are safe to assume it will not? I'm aware that coupled cluster generally works best when there is one dominant determinant (single-reference cases), so I am asking about guidelines within the single-references case, when we might expect to get energies that violate the variational principle.

Let us begin with the following principle: the CC ansatz always returns an appropriate electronic wavefunction (meaning a combination of Slater determinants). Still, at least in a non-relativistic approach, the Hamiltonian does commute with the Spin operator
\begin{equation}
S^{2}=S_{+}S_{-}-S_{z}+S_{z}^{2}
\end{equation}
(I'm using atomic units here)
and we want the wavefunction to be an eigenfunction of total spin. The difficulty in doing so is due to the fact that the excitation operator to ensure spin purity is not easy to parametrize unless you are lucky enough to have a $S^{2}=0$ state and your system wavefunction is represented by a $S^{2}=0$.
So, when you're starting with a wavefunction which is not a spin eigenfunction (for example the ground state of the nitrogen atom with the unpaired electrons), CC might not be able to recover the spin purity and, by allowing superposition of functions with different spin, you might get an energy lower than the correct one because you're looking for the minimum in a space bigger than the one you are physically allowed to .
Hope I was clear


CO2 conversion to a harmless gas like oxygen I am doing some research about indoor purifiers and I was wondering if there is any approach to purify co2 or convert it to harmless gases like oxygen just like what plants do.

Submarines currently use NaOH/Ca(OH)2 to absorb CO2. They generate O2 separately either from gas tanks or by electrolysis of water. Peroxides seem attractive at first, but they are a fire hazard and you don't want fire on a submarine. British subs are using Zeolite. The role of zeolite isn't to deal with CO2, but to remove the smell of volitile organic compounds (of dirty socks, so to say). Air is first dried and purified.

What exactly are you trying to achieve?


What is the explanation of the changes in stability going down a group for carbonates, bicarbonates, fluorides, and chlorides? For carbonates and bicarbonates, I know that stability increases down the group, and for chlorides and fluorides, stability decreases down the group. Why does this happen?  Can someone explain this in detail?

(I am talking about S block alkali metals)

Carbonates

The quote from your text:

  Carbonates of alkaline earth
  metals are insoluble in water and can be
  precipitated by addition of a sodium or
  ammonium carbonate solution to a solution
  of a soluble salt of these metals. The solubility
  of carbonates in water decreases as the atomic
  number of the metal ion increases. All the
  carbonates decompose on heating to give
  carbon dioxide and the oxide. Beryllium
  carbonate is unstable and can be kept only in
  the atmosphere of CO2. The thermal stability
  increases with increasing cationic size.

So the stability that you are referring to is thermal stability. This is an important detail. So what is thermal stability? It's how resistant a molecule is to decomposition at higher temperatures.

What's happening to cause thermal instability?

So, lets look at the carbonate ion here:

This is just an illustration, and in reality the negative charge we see on the two $\ce{O}$ atoms is localized due to resonance.

Below the illustration shows where the negative charge is likely to be concentrated (colored in red).

So, when we create a carbonate complex like the example below, the negative charge will be attracted to the positive ion.

Because of this polarization, the carbon dioxide will become more stable and energetically favorable.

How does going down a group play into this?

Well as you go down the group, the charged ion becomes larger. The larger the ion, we see a lower charge density. Charge density is basically the amount of charge in a given volume. So, if a small ion has the same charge as a larger ion, the charge density will be greater for that small ion. Greater charge density, means a greater pull on that carbonate ion, and a greater pull causes the delocalized ions, and a more stable $\ce{CO2}$ molecule. So, the larger the ion, the lower the charge density, the less polarizing of an effect, and reduced stability of a $\ce{CO2}$ molecule, favoring the $\ce{CO3}$.

Chloride and fluoride stability

Stability of fluorides, chlorides, and other halogens, are likewise related to thier size. The halogens, specifically fluouride, is known for their electronegativity. Electronegativity, is the tendency to attract electrons to itself. As you move up the group, you see an increase in electronegtivity. This results in the creation of polar bonds.

Illustrated below, you see that as charge of the positive ions increase, polarizability increases (left), and as the halogen ion increases, polarizability and electronegativity decrease (right). When the ions electron cloud, is less polarized, the bond is less strong, leading to a less stable molecule.

Information and illustrations on carbonate ions were sourced from here.


Why is a strong base needed for an E2 reaction and not a an E1 reaction? For E2 reactions, why is a strong base like $\ce{NaOH}$ or $\ce{RONa}$ needed? Whereas for E1, even a weak base like $\ce{H2O}$ could be used.

Wikipedia states:

  E2 typically uses a strong base, it needs a chemical strong enough to pull off a weakly acidic hydrogen.

However it does not explain why. The lone pair of the base directly attacks the hydrogen, regardless of E1 or E2.

E2 is a concerted mechanism. The alpha proton in an E2 reaction substrate is also only weakly acidic because the bond between the alpha proton and the alpha carbon is relatively strong; there is only some inductive withdrawal of electron density from the C-H bond by the halogen.

Therefore, we need a strong base to rip the proton away from the alpha carbon.

E1 however involves two steps. In E1 the leaving group leaves, and then the base attacks.

When the leaving group leaves, a carbon lacking an octet is formed. This carbon is highly electronegative and will withdraw electron density. This makes any alpha proton more acidic as hydrogen is less electronegative than plain old carbon - much less a carbon deficient of an octet in a covalent framework.

In other words, electron withdrawal by the carbocation weakens the bond holding on to the alpha proton, and a weaker base suffices for taking away the proton.

Note that I don't particularly like the below picture because it doesn't show anything abstracting the proton. Still the general idea is correct; electron density from the alpha C-H bond flows toward the hypovalent carbon. Heck, we can draw valid, hyperconjugative resonance structures for this baby.


Coconut oil vs Olive oil: Spray dynamics Ps: If this would better fit Physics then please move it there or let me know. I put it here based on fact that viscosity and colloids were part of chemistry curriculum 10/15 yrs back in school.

Let's say we had to spray Edible oil.

How would spraying each differ?

How different are their Physical Properies when it comes to spraying and how would they affect the spray and compare with each other?

(You may point out Properies that matter in spraying that I may not be aware of, but these are few I can remember from school/ college such as viscosity, stickiness, cohesion, adhesion, density)

Also, I understand temperature would/ could play a role as well, so let's specify and put them at their ideal temperature for spraying.

Let's optimize this spraying experiment for both of them & compare.

I'd love to hear the science in play here.

Ps: I'm curious if any of the follow terms apply to these Virgin oils: homogenous, colloidal, plasma?

The fluid dynamics of spray calculations depend on many conditions including the nozzle used to create the spray.  The wikipedia article on spray nozzles has a good discussion on how different liquid properties tend to affect spray. Spray information tends to be empirical, so it is not really science at play, it is engineering.

The Institute of Liquid Atomization and Spray Systems is a professional society set up around the science and engineering of spraying.  Atomization and Sprays is a periodical journal set up around this entire topic.  A internet search of 'textbook on spray and atomization' turns up many textbooks on the topic.

Plenty of manuscripts written by scientists exist providing detailed results of experiments comparing sprays.  An example would be an experiment done comparing the spray of waste cooking oils in an engine when it is used as fuel.

Summarizing a discussion of all of these experiments is too broad and out of the scope of SE.  If you are interesting in learning about sprays make sure you have some understanding of fluid dynamics.  Conduct a literature review relevant to the spray experiment you have in mind.


Why was a Plimsoll symbol chosen to indicate standard state? Historically, the Plimsoll symbol (aka Plimsoll line) was created as hull mark that would serve as a ready indicator of whether a ship was overloaded and thus running too low in the water.  It was later adopted by chemists to indicate standard state (currently defined by IUPAC as 1 bar).

Why was the Plimsoll symbol chosen for this purpose?  According to Wikipedia,

  The plimsoll symbol (⦵ or o) ... is used as a superscript in the notation of thermodynamics to indicate an arbitrarily chosen non-zero reference point ("standard state").

The language used by Wikipedia suggests it was chosen because it looks like a zero with a line through it, thus communicating "not zero".  Applying this to enthalpy or free energy seems redundant and unnecessary, though, since by definition there is no true zero reference point for either (among state functions, this exists only for entropy).  It's like having a special symbol attached to $ΔU$ to indicate that $U$ is conserved, when, of course, $U$ is always conserved.

Indeed, the problem with attaching a symbol to $U$ saying it's conserved is that it suggests that sometimes $U$ is not conserved (otherwise why would you need the symbol?).  Likewise, attaching a symbol to $U$, $H$, or $G$ saying that the reference state is non-zero suggests sometimes it could be zero.  I.e., I understand they needed a symbol for reference state.  But I'm puzzled why they chose one that communicated "non-zero" instead of just choosing something that meant "reference", period.

Nevertheless, I'm sure these people knew what they were doing, and had good reasons for what they did.  I just don't understand what that those reasons would be.  I'd thus be interested to hear from anyone out there with knowledge of the history of this.

Perhaps the answer (here I speculate) is that they wanted to communicate the obvious (that thermodynamic reference states can never be true zero reference states) — and that the reason they wanted to do this was because they anticipated tables of thermodynamic values would be used by a broad audience, some of whom might not understand the underlying thermodynamic theory.

Prigogine's Chemical Thermodynamics gives a bit more context as to why a plimsoll symbol is somewhat more preferable and is indeed used to denote non-zero nature of standard values [1, pp. xxiv, 86]:

  The properties of a pure substance are denoted by a superscript $^0$. The use of this symbol to denote "standard" thermodynamic quantities is then permissible only when these refer to the pure substance as standard state. Some other convention is needed to denote the more general "standard quantities" defined in equation (7.51), where the standard state may for example refer to an infinitely dilute solution. After much consideration the symbol $^⦵$ has been introduced. This symbol is based on the circle, and is thus closely related to the more common (but occasionally confusing) notation for standard quantities. It seems useful, however, to associate it with the plimsoll mark which, appropriately enough, refers to a reference state of loading of a ship; this use of an ideogram has, we believe, some value in keeping the essential nature of "standard states" in prominence.

  Quantities denoted by the plimsoll symbol are still, in general,
  functions of temperature and pressure; the idea of "standardization"
  of temperature and pressure is not stressed. When necessary we denote
  quantities at the standard pressure (which for convenience is nearly
  always chosen as the unit of pressure) by the superscript $^†$, quantities at a standard volume (which for convenience is always chosen as the unit of volume) are denoted by $^‡$. Quantities referring to a (usually hypothetical) ideal system are denoted by $^\mathrm{id}$.

  [...]

  $$P = P^⦵(T,p) + P^M(T,p,x_1\ldots x_c) \tag{7.51}$$

It's also interesting that Lewis and Randall, known for defining the standard state in the beginning of XX century as we know it, used an upper circle [2, p.623]:

  $G^\circ$ A property of a substance in its standard state.

However, they referred to it as to a standard reference state (distinguished from a standard state) throughout the textbook [2, p. 88, 99]:

  While we cannot give the absolute magnitude of $\bar{H}$ for a substance in solution, we may ascertain how much greater or less this is than the heat content of the same substance in some chosen state. Thus, at any temperature, if we are dealing with the partial molal heat content of water in some solution in which water is the solvent, we may choose pure liquid water as the reference state, and denote its molal heat content by $H_1^\circ$.

  [...]

  It is highly convenient, especially for purposes of concise tabulation, to know the heats of reaction when various substances are formed from their elements. It is therefore desirable to choose some one standard reference state for each element. For this purpose, we shall, at all temperatures, take the element at a pressure of one atmosphere, and in that form which is most stable or most common at room temperature. Thus liquid mercury, gaseous oxygen, solid iodine, rhombic
  sulfur and graphitic carbon, all under a pressure of one atmosphere, will be considered to be in their standard reference state.

References

Prigogine, I.; Defay, R. Chemical Thermodynamics; Treatise On Thermodynamics; Longmans, Green & Co Ltd: Glasgow, 1954; Vol. 1.
Lewis, G. N.; Randall, M. Thermodynamics and the Free Energy of Chemical Substances, First Edition later Printing edition.; McGraw-Hill Book Company, 1923.


What can you use to remove hair gel? The hair gel is called "lets Jam Gel". It says it uses micro emulsion technology.
What chemicals can you use to break it down or undo the bonds?

I'm trying to get it out of my head. I washed my hair with T sal and cut my hair off but it still grows back patchy and easy to style like it has gel in it. I think it may be in the epidermis or even dermis by now.

ingredients: 42692 SS11 - Aqua / Water Isocteth-20 Paraffinum Liquidum / Mineral oil Glycerin Oleth-2 Propylene Glycol Parfum / fragrance Sorbic Acid Methylparaben DMDM Hydantoin Ethylparaben Limonene Hydroxycitronellal Geraniol Amyl cinnamal Linalool Simethicone CI 19140 / Yellow 5 CI 14700 / Red 4, CI 172200 / Red 3. F.I.L D45949/1

Yikes.

As far as I'm aware, a thorough application of ordinary soap or washing-up liquid should be able to wash this off using the normal surfactant processes (binding to organic components with lipophilic regions of the molecule). If you're experiencing long-lived after-effects (say, more than a week maybe), you might want to see a doctor -- it's not impossible you could have had some sort of allergic reaction. Oh, and if you're going to wash hair with soap or washing-up liquid, I'd suggest using conditioner afterwards, to reduce the long-term damage incurred by stripping the natural fatty acids and oils off.

I'm not a medical professional or a tricologist, though, so please don't take this as Official Medical Advice.


Can the green spots in these coins be copper chloride? My father collects coins as a hobby and lately has observed some green spots in some of his coins so he told me to ask here to see if anyone knows what it is.

The coins are made of silver (999 parts out of 1000 according to the description) so at first he thought that it might be some silver oxide but the silver oxide is dark brown and not green.

He looked it up in the Internet and he has read that it might be copper chloride but it doesn't seem right to him due to the purity of coin.

Is this feasible? Can the spots be copper chloride? If not, what can they be?

This phenomenon is relatively common among jewellers and silversmiths and they call it verdigris. This verdigris can be copper oxide, copper acetate, copper carbonate and even copper chloride. The former options are more likely, but copper chloride can actually be a possibility if you live close to the sea.

Perhaps relevant to your father: verdigris can be cleaned by applying a mildly acidic solution to it, such as lemon juice or vinegar. The formation of verdigris can be prevented by keeping the metal dry as this kind of corrosion is typically formed by contact of the metal with moisture.


Zero order half life Two substances A and B react with each other in such a way that A is 50% consumed in 33 minutes and 75% consumed in 66 minutes. Changing the concentration of B has no effect on the results. Which statement is true?

And the answer is:

  This reaction is 1st order in A and zero order in B.

I understand why A is first order but I don't get why B is zero order. The equation for the half life of a zero order reactant is $\displaystyle\frac{[\ce{B}_\text{inital}]}{2k}$.

Doesn't this mean that the concentration for a zero order reactant would affect the half life?

Changing the concentration of B has no effect on the results.

This statement could only mean one thing: B does not participate in the reaction. Obviously, if B does participate in the reaction, and you change the concentration of B to 0 the reaction won't happen.

The equation for the half life of a zero order reactant is
[Binital]2k.

Zero order reaction have no half-life. Only 1st order reaction has half-life.


What is the product of reaction between 3,3-dimethyl butanal and SeO2 followed by conc. OH- ions? So basically first step will be replacement of $\ce{-CH2 -}$ by a ($\ce{=O}$ bond) making a aldehyde and ketone side by side. Now what should I do after this? I tried attacking the aldehyde first as it is more electrophilic. I think both carbonyl groups will be attacked because of high concentration of hydroxyl ($\ce{OH-}$) ions. I even tried dehydration of  the formed product after attacking with $\ce{OH-}$ ions but I always end up with a ketone and a carboxylic acid by its side  after this.

In the first step, Riley Oxidation (selenium dioxide-mediated oxidation of methylene groups adjacent to carbonyls) is taking place. You will get A as:

A on reaction with conc. $\ce{OH-}$ (i.e. in strongly alkaline condition) undergoes redox reaction (famous Cannizzaro reaction). And terminal aldehyde will be oxidized to form the metal salt of acid, ketone will be reduced to form the alcohol, note that intramolecular reaction is taking place, to finally form the product B.


Why does NO2 react with HCl ? What is the mechanism of the reaction? I found the following reaction given in a certain book:
 $$\ce{2NO2 + 4HCl -> 2NOCl + Cl2 + H2O}$$

I cannot think of any reason why an acidic oxide like $\ce{NO2}$ will react with an acid like $\ce{HCl}$.

So, why will they react? And, what will be the mechanism of the reaction?

Yes, I believe it is a typical redox where nitrogen dioxide is an oxidising agent and hydrochloric is a mild reducing agent as Ivan noted.

The reaction proceed in two phases:

$$\ce{NO2 + 2HCl ->[\Delta] NO + Cl2 + H2O}$$

  Nitrogen dioxide react with hydrogen chloride to produce nitric oxide,
  chlorine and water. This reaction is proceeds by heating.(source)

Now some nitrogen monoxide react with water to form nitrosyl chloride.

$$\ce{Cl2 + 2NO → 2NOCl}$$

Finally, we get our main reaction

$$\ce{2NO2 + 4HCl -> 2NO + 2Cl2 + 2H2O ~~~~~~ (1)*2}$$

$$\ce{2NO2 + 4HCl -> (2NO + Cl2) + Cl2 + 2H2O}$$

$$\ce{2NO2 + 4HCl → 2NOCl + 2H2O + Cl2}$$

  Nitrogen dioxide react with hydrogen chloride to produce nitrosyl
  chloride, water and chlorine.(source)


Will squeezing lemon juice on top of foods with ferrocyanide additive release hydrogen cyanide gas? For example, will squeezing lemon juice on top of table salt with sodium ferrocyanide ($\ce{Na4[Fe(CN)6]}$) additive (E 535) release hydrogen cyanide gas?

Examples of ferrocyanide additives:

sodium ferrocyanide (E 535)
potassium ferrocyanide (E 536)
calcium ferrocyanide (E 538)

No. The hexacyanidoferrate(II) complex is a very inert complex that does not undergo ligand exchange easily. This is somewhat exemplified by the harsh conditions alchemists used to synthesise hexacyanidoferrate(II), which included high temperatures and strong bases.

Furthermore, whatever amount of lemon juice you add the resulting acidity will be much less than in your stomach. If you swallow a cyanide source such as potassium cyanide, this will get protonated by your stomach acid to liberate hydrogen cyanide and potentially kill you. Food additives — especially approved ones — undergo various testing procedures to make sure that does not happen.


Why doesn't fluoride ion have highest nucleophilic power (nucleophilicity)? As you can see from the image above,

'$\ce{R3C}$' anion has 1 lone pair
'$\ce{R2N}$' anion has 2 lone pairs
$\ce{RO}$ anion has 3 lone pairs
$\ce{F}$ anion has 4 lone pairs

According to VSEPR theory,

  lone pair-lone pair repulsion >  bond pair-lone pair repulsion >  bond pair-bond pair repulsion

As fluoride ion has the greatest number of lone pairs (= 4), it should suffer from highest number of lone pair-lone pair repulsions.

In order to reduce its  lone pair-lone pair repulsions and thus its unstability, shouldn't it donate its lone pair more readily than other anions? Thereby, shouldn't fluoride ion have more nucleophilic tendency than the other (mentioned) anions?

Nucleophilicity of an ion does not solely depend on the number of lone pairs it has. The solvent in which the ions are dissolved matters, too.

If you are using a polar protic solvent, then flouride ion would behave as a very bad nucleophile (it would be solvated in the case of water; water will form hydrogen bonds with it).

Among the nucleophiles you have stated, $\ce{R3C}$ anion would be the best nucleophile in a polar protic solvent. However, if you use a polar aprotic solvent, flouride ion would behave as a very good nucleophile (because aprotic solvents do not solvate anions).

Remember, nucleophilicity is a kinetic concept and basicity is a thermodynamic one. Your reasoning for comparing the nucleophilicities was based somewhat on basicity of the ions. (This would work well in an aprotic solvent.)

In very layman terms:

Nucleophilicity: How good is something at reacting (doesn't concern stability of the ion)
Basicity: How badly something would want to react (concerns stabilities)


Is it possible to estimate a substance's water solubility from data like the substance's partition coefficient, dissociation constants, etc.? If it is possible, how would you estimate a compound's water solubility from data like its partition coefficient, $ \log P $, its dissociation constant, $pK_a $, its molar mass, etc.?

It could certainly be estimated from those properties, the structure, etc.  A lot of companies have developed proprietary algorithms for things like that:  solubility, diffusion coefficient, polarity.... (For example of a company, see "ChemAxon")

The water solubility for many chemicals is already known, though.  Search for a chemical ChemSpider, and look under the "Properties" tag.

They've even got an API, in case you're interested in using water solubility in an app.


Why is the boiling point of a solution higher than that of a pure solvent? Why is the boiling point of a solution higher than that of a pure solvent?
I think when a non-volatile solid solute is added to a solvent, some of the liquid solvent molecules are replaced by solute molecules at the surface. The lesser the number of solvent molecules at the surface that fewer molecules evaporate, and thus the vapor pressure of the solution will be lower than the pure solvent.

Is this correct?

The change in vapour pressure with added solute is called Raoults law. The vapour pressure of a species i above a solution is proportional to mole fraction of this species (solute) $x_i$ and its pure vapour pressure $p_i^*$ or $p_i=x_i.p_i^*$.

As the solution now contains a fraction of a low vapour pressure solute then this component will not contribute much to the overall vapour pressure which will be less (in this case) than pure solvent.

It is not correct to say that solvent molecules are replaced at the surface as both solute and solvent are distributed uniformly in the solution. Thermal diffusion moves the solute and solvent around in a random manner. It is that there are now fewer solvent molecules as some have been replaced by low vapour pressure solute (in this example) and because they are of low volatility they need more energy to evaporate than the solvent does. As the temperature is constant, by the Boltzmann (energy) distribution fewer solute molecules populate higher energy levels compared to solvent and so have a lower vapour pressure.


Ionization energy comparison between K+ and Cl- So I have encountered a question which looks very suspicious to me.

If you have $\ce{Cl-}$ and $\ce{K+}$, Can you say that $\ce{K+}$ has more ionization energy than $\ce{Cl-}$?

We know for a fact that when you put more electrons it makes the the pulling force of the protons a bit scattered (because of the incremental of the electrons in $\ce{Cl-}$) so the ionization energy of chlorine should go down.

The opposite happens to $\ce{K+}$ when I remove electrons the pulling force of protons is stronger (Because it pulls less electrons) than it used to be so the ionization energy goes up!

So What I am proposing here that we can't just say that $\ce{K+}$ has more ionization energy than $\ce{Cl-}$ just because that happened. $\ce{Cl}$ had a really high ionization energy compared to $\ce{K}$ and the decrement of the ionization wouldn't be that big to make it less than the ionization energy of $\ce{K+}$.

(The answer was that $\ce{K+}$ has more ionization energy than $\ce{Cl-}$ in the exam paper but I questioning that)

So if there is a chart or something for this that would be awesome. I hope I can get some help here.

The general reaction we are looking at when discussion ionisation enthalpy is:

$$\ce{X -> X+ + e-}$$

With the electron removed into space (i.e. totally disappearing). An electron is a negatively charged particle, so removing it from something negative will be easier than removing it from something positive. Thus, you have a very stable $\ce{K+}$ ion and are trying to rip away another electron from this (bad) — and you need to rip it out of a core orbital — all valence orbitals are already empty, the last one when you removed that previous electron to generate $\ce{K+}$ (very bad). A combination of bad and very bad gives us a very high ionisation energy for $\ce{K+}$.

Discussing chloride, first of all we have an anion, so the removal of an electron is generally good for removing excess charge. Second, we again have something that has a full octet — but this time the octet is a valence octet, not a core octet, so removing the electron will be a lot easier. We have a combination of good and relatively good, so the ionisation energy of $\ce{Cl-}$ will be rather low.


How many molecules does it take to have a phase? A single molecule can't be solid, liquid or gas. It's just a molecule. A mole of something can be any of the three. So, how many molecules does it take for phases to be meaningful?

I realize that this may depend on the molecules in question. The minimum number that can form a solid might depend on the configuration that molecules adopt when solid. The smallest crystal of a lattice for example. How about a gas though? Can two molecules be called a gas if they exist under conditions where a larger quantity of the same moiety would be in a gaseous state?

In short, can a minimum number of molecules necessary to adopt a particular phase be defined?

This is actually an active area of research for water clusters. In principle, for $\ce{(H2O)_{n}}$ there should be a "melting" phase transition, much like for ice to liquid water.
So, in principle, if you had an accurate enough theoretical method, you could do molecular dynamics and see when the melting point of the cluster matches bulk water.
Defining a line between liquid and gas seems tricky to me, but I don't do statistical mechanics, and perhaps there are molecular dynamics techniques for defining a boiling point.
Defining the solid-liquid phase transition is much easier, because the radial distribution function of the system can be monitored to see the transition (i.e., the density changes).
At the moment, I don't think the best dynamics methods get the melting transitions accurately yet, but I think n~20 is about right. I'll revise this when I get a better answer.
EDIT: It turns out that 21 seems to be the number for water drops.
"How many water molecules are needed to solvate one?"
Chem. Sci. 2021

The minimal structure eventually responsible for proper solvation is made of a total of 21 water molecules and includes two complete solvation shells, of which the whole first one is tetrahedrally coordinated to the central molecule.


Are there any safety guidelines for mixing sulfate with chloride? Some ceramicists decorate pottery using soluble salts (usually chlorides and sulfates e.g. copper chloride, cobalt sulfate) which are used individually or mixed to produce different colors.

I have been exploring the use of these chemicals on porcelain.  From the MSDS of each chemical I've become aware of the safety concerns and the proper precautions to take when mixing, applying, and firing them in a kiln.

But I don't know much about chemistry and do not understand the reactions when mixing these chemicals.  Are there any chemicals which should absolutely not be mixed together?  For instance, the Wikipedia article on Potassium permanganate says "Concentrated sulfuric acid reacts with $\ce{KMnO4}$ to give $\ce{Mn2O7}$, which can be explosive."  Does this mean I should not mix Potassium permanganate with a sulfate like Copper Sulfate?  In general, what happens when a sulfate is mixed with a chloride?  Are there any general safety guidelines you could offer?  Is there a fairly simple way to understand these interactions without the requirement of years of Chemistry study?

Apologies for my ignorance!

Some chemicals that ceramicists have used before:

Color

gray

copper chloride (heavy application and heavy reduction can give pinks and reds)
palladium chloride
ruthenium chloride
selenium (selenous acid, selenium toner)
silver nitrate
tellurium chloride
vanadium (vanadyl sulfate, vanadium pentoxide)

blue

cobalt chloride
molybdenum (molybdic acid)

green

ammonium chromate
nickel chloride
potassium dichromate
sodium chromate

brown

iron chloride (iron chloride emits heat when mixed with water so the water    should be added gradually in small amounts)

pink/purple/maroon

gold chloride (1-5% solution, adding either cobalt, manganese or
tellurium will give different shades)

yellow

praseodymium chloride (very pale color)

black

cobalt chloride (50% solution) and iron chloride (100% solution)
cobalt chloride (50% solution) and nickel chloride (50% solution)

Overall, always wear disposable gloves, goggles, and a dust mask.  Add a lab coat, smock, or apron so as not to contaminate your clothing.  Avoid skin contact, don't ingest and don't inhale (this includes the dust).

Manganese heptoxide is a special case with concentrated sulfuric acid.  Avoid mixing strong oxidizers (permanganate, chromate/dichromate, perchlorate) with organics.  NEVER screw around with chlorate salts.  They are powerful oxidizers with very low kinetic barriers.  Finely divided metals (especially titanium) can go to white-hot burning in open air.  Titanium burns under air, carbon dioxide, nitrogen, water...

Palladium is a heavy metal, poison and tissue corrodant.  A microgram of tellurium by any contact gives "tellurium breath"

Local contact with chromates can cause ulceration.  Neodymium gives you purples, cerium yellow to orange.  I seem to remember that cobalt blue (cobalt aluminate after firing) in tin oxide applied to white ware gives bubblegum purple.

Tin Glazing

Gold via the Purple of Cassius is very nice.

Here's a pretty!  Suppose you mixed clay with hollow glass microbubbles, then fired.  Would the clay foam to an interesting surface or whole body?  This is the stuff, sold generally as lightweight resin filler, including hobby stores.  Most prices are outrageous when sold by volume.  Density of 3M Scotchlite K20 is 0.2 g/cm^3.  Goggles and dust mask.  Be careful or the stuff goes everywhere.

http://www.aircraftspruce.com/catalog/cmpages/bubbles.php
Good price.
http://solutions.3m.com/wps/portal/3M/en_US/Energy-Advanced/Materials/Product/?N=5002440+7570097&rt=r3


Clarifications on the calculation of the $\Delta\mathrm{G}$ of a reaction I'm trying to calculate the $\Delta\mathrm{G}$ of the following reaction. Are my calculations correct? I'm also a bit puzzled, 'cause this is a endoergonic reaction right? But the $\Delta\mathrm{G}$ is negative, so the reaction would be spontaneous if my calculations are correct and if the activation energy is reached right?

CH3OH + H2O --> HCOOH + 4H        Delta H = 99.49 kJ/mol

% Standard enthalpy of formation of every compound

% -238.7 kJ/mol CH3OH
% -285.830 kJ/mol H2O
% -425 kJ/mol HCOOH
%    0 kJ/mol H2
% -393.51 kJ/mol CO2

% Standard entropy of formation of every compound

% 126.8 J/mol*K CH3OH
%  69.91 J/mol*K H2O
% 129 J/mol*K HCOOH
% 130.6 J/mol*K H2
% 213.6 J/mol*K CO2

delta_H_1 = -425 - (-238.7 - 285.830);
delta_S_1 = 129 + 130.6 - (126.8 + 69.91);
T = 273.15 + 20;

delta_G_1 = delta_H_1 - T * delta_S_1;

Correct version

delta_H_1 = -425 - (-238.7 - 285.830);
delta_S_1 = 129 + 4*130.6 - (126.8 + 69.91);
T = 273.15 + 25;

delta_G_1 = delta_H_1 - T * delta_S_1;

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$

Gibbs Free energy does not only relate to the spontaneity of a reaction, the magnitude of Gibbs Free energy describes how much "free energy" is available, or how much work is possible.

$\Delta G^\circ$ can either be negative or positive even if $\Delta H^\circ$ is $+$. Therefore,
we have to look at all three terms, $\Delta H^\circ$, $T$, and $\Delta S^\circ$

The Gibbs Free Energy, enthalpy, and entropy are state functions. Although I won't derive it here, we can find $\Delta G$, or $\Delta H$, or $\Delta S$, by using Hess's Law.

The information provided above gives us the $\Delta H_{rxn}$ which is $99.49 \text{kJ/mol}$ and thus endothermic.

We calculate entropy from the a list of experimental tabulated values:

$$\Delta S^\circ_{rxn} = \Sigma \Delta S^\circ_{products} - \Sigma \Delta S^\circ_{reactants} $$

$$\Delta S^\circ_{rxn} = (129.0 \text{J}\cdot\text{mol}^{-1}\text{K} + 4*130.6 \text{J}\cdot\text{mol}^{-1}\text{K}) - (126.8 \text{J}\cdot \text{mol}^{-1}\text{K} + 69.91 \text{J}\cdot\text{mol}^{-1}\text{K})$$

$$\Delta S^\circ_{rxn} = 454.69 \text{J}\cdot\text{mol}^{-1}\text{K} = 0.45469 \text{kJ}\cdot\text{mol}^{-1}\text{K} $$

While solving for entropy, you forgot to multiply the entropy of hydrogen gas with the number of moles. Also, you should always include the states of your reactants and products, as thermodynamic data for compounds differs for different states.

Another thing that is incorrect is standard temperature for thermodynamic reactions is at $25^\circ C$ not $20^\circ C$ and thus your temperature should be $298.15 K$

  I'm also a bit puzzled, 'cause this is a endoergonic reaction right? But the ΔG is negative, so the reaction would be spontaneous if my calculations are correct and if the activation energy is reached right?

This is where a lot of people make mistakes. It is true, that in general, endothermic reactions with a positive enthalpy usually have a positive $\Delta G$ and are thus not spontaneous.

However, if we look at our equation:

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$

We have to consider our other terms and not only enthalpy, namely that of temperature and entropy.

A reaction is spontaneous if $ \Delta G < 0 $ such that:

If $ \Delta H < 0$ and (-) and $ \Delta S > 0$ (+), the reaction is spontaneous at any temperature.

If $ \Delta H > 0$ and (+) and $ \Delta S < 0$ (-), the reaction is NOT spontaneous at any temperature.

If $ \Delta H < 0$ and (-) and $ \Delta S < 0$ (-), the reaction is spontaneous at low enough temperatures.

If $ \Delta H > 0$ and (+) and $ \Delta S > 0$ (+), the reaction is spontaneous at high enough temperatures.


2p Orbital Occupancy What does the question mean by how many electron pairs are in 2p orbitals? How does one tell? Given that both the oxygen and the nitrogen are $\ce{sp^2}$ hybridized, wouldn't there be two 2p orbitals?

In the resonance structure drawn at the top of your image, the $\ce{p}$ orbital on carbon and the $\ce{p}$ orbital on oxygen share a bonding pair of electrons, a lone pair of electrons reside in a nitrogen $\ce{p}$ orbital, two lone pairs reside in two oxygen $\ce{sp^2}$ orbitals.  So for this case, the most commonly drawn representation, there is one lone pair in a $\ce{p}$ orbital.  In your middle resonance structure, there is a lone pair in an oxygen $\ce{p}$ orbital, and another lone pair in the nitrogen $\ce{p}$ orbital; hence, 2 lone pairs in two different $\ce{p}$ orbitals.  In your bottom resonance structure, there is a lone pair in an oxygen $\ce{p}$ orbital and a shared bonding electron pair in the $\ce{C-N}$ $\ce{pi}$ bond; therefor, one lone pair in a $\ce{p}$ orbital for this case.

This seems to be a poorly worded question.  Since the real molecule is represented by a combination of the 3 resonance structures, the correct answer would be "somewhere between one and two lone pairs in $\ce{p}$ orbitals."  Adding the detail contained in the above paragraph would probably help the grader understand your explanation.


What is the stochiometric equation of the sucrose and potassium chlorate reaction with sulfuric acid? I am researching this reaction for my chemistry assessment. Various sources I have dug up state different answers and opinions on what the stoichiometry is and what the products are, sometimes giving no answer at all.

In one source, sulfuric acid is a catalyst that speeds up the reaction while in another source, sulfuric acid is one of the reactants, and is therefore part of the stochiometry. The final products also vary from source to source, and include pure carbon or potassium chloride in addition to carbon dioxide and water.

Can anyone clarify the subject for me?

In this paper [Sugar Dehydration without Sulfuric Acid: No More Choking in the Classroom! T.P. Silverstein and Y. Zhang. J. Chem. Ed. 75 (6) 748-749 (1998)], the authors outline two reactions involving sucrose and sulfuric acid. The first one does not include $\ce{KClO3}$ and utilizes a significant amount of $\ce{H2SO4}$ in order to fuel the dehydration of sucrose: this is not the reaction you're interested in.

The second reaction detailed herein makes use of a very small amount of sulfuric acid to start the reaction (a drop is the commonly-cited amount): the sucrose is oxidized by $\ce{KClO3}$. This is the reaction you want to examine.

The authors give the reaction stoichiometry as
$$\ce{C12H22O11(s) + 8KClO3(s) \xrightarrow{H2SO4} 12CO2(g) + 11H2O(l) + 8KCl(s) + heat}$$

They note that the fumes from the reaction contain $\ce{Cl2, ClO}$, and $\ce{ClO2}$, but the amounts are not quantified. I suspect (but don't know for sure) that the amounts are finite but negligible. And even with the scant amount of sulfuric acid used here, I think you're safe going with the reaction as presented above.


Why can't the N- end of the nitrile anion attack the alkyl halide in an SN2 fashion during an alkylation reaction when OH- can? I'm reading my Clayden textbook and something keeps nagging me. It says that when we're carrying out alkylation reactions with an enol attacking an alkyl halide with strong base catalysis (with hydroxide), we should keep the base separate from the alkyl halide electrophile to prevent it from performing the SN2 attack instead of the nitrile anion formed from deprotonating the nitrile at the alpha position.
But, if OH- can attack the alkyl halide as a "competing reaction", why can't the N- end as well instead of the nitrle anion carbon-carbon double bond? They both (OH- and N-) seem like equally good nucleophiles. Is it because when we have the option to choose from the "soft" end (the double bond in the nitrile anion) and the "hard" end (the N-) nucleophile to attack the soft alkyl halide electrophile, we'd much rather have a soft nucleophile to soft electrophile attack?

It can.
According to this JACS paper here steric effects are a major factor in whether the ketenimine is formed or the C-alklyated product - the more hindered the nitrile, the more N-alkylation.


Henry's law Numerical I got very basic doubt. Henry law's Give relation between concentration vs Partial pressure.  So here is the Numerical Of my textbook

**Now My question is, why concentration is Considered only in Grams? shouldn't it be in Molarity, Mole fraction or Strength(gm/Vol) etc? **

In Other textbook, They simply used formula to solve it.
mass = Henry's Constant * pressure.

For this example, the volume of the solution doesn't matter. It's a thought experiment, doesn't say what sovent is used, and you could replace ethane by helium or miraculane.

As long as you talk about the same compound, mass is proportional to molar quantity. Plus the units cancel out in the calculation anyway.

Think of the hypothetic experimentator as a guy who only has one glass beaker, without graduation. ;)


Ozonolysis of nitronate What is the mechanism for the ozonolysis of a nitronate anion? What gas is produced in this reaction? Is there any chemoselectivity in the presence of a double C=C bond in the molecule?

What is the mechanism for the ozonolysis of a nitronate anion?

Probably very analogous to ozonolysis of a double bond.

A typical synthetic procedure is detailed in OrgSyn.  Note the addition of dimethylsulfide during work-up; this is common in ozonolysis reactions as it helps reduce the ozonide and also reduces any other dangerous, highly oxidized by-products.  The discussion section of the OrgSyn article says,

  "There are various methods available for transforming a nitro group
  into a carbonyl group, including the Nef reaction (strongly acidic),²
  permanganate oxidation of nitronate anions (basic, oxidative),³
  persulfate oxidation of nitronates (basic, oxidative),⁴ treatment with
  a mixture of organic and inorganic nitrite (neutral, oxidative),⁵ and
  treatment of either a free nitro compound or a nitronate anion with
  aqueous titanium(III) (neutral, reductive).⁶ Each method is limited,
  however, by poor yield, inconvenience, or lack of generality.

  With the proviso that the substrate not contain a reactive
  carbon-carbon double bond, the present ozonolysis procedure⁷ appears
  to provide a convenient and efficient method for carrying out the
  desired "

The Nef Reaction is mentioned above, it is worth taking a look at the linked article since the Nef reaction is complementary to the ozonolysis reaction of nitronates.  You may also find the "Reaction mechanism" section interesting.

  What gas is produced in this reaction?

According to the above mechanism, no gas would be generated, rather $\ce{NO3^{-}}$ is produced

  Is there any chemoselectivity in the presence of a double C=C bond in the molecule?

No, as mentioned in the OrgSyn reference, this approach can't be used when other double bonds are present.


Pressure Cooker at High Altitudes and Low Altitudes I just have a doubt if I'm right or wrong about this. Please explain more if possible.
Also, I have some questions regarding boiling point and vapour pressure.

 High Altitude Cooking
At high altitudes, the atmospheric pressure is lower than that at sea level, so the boiling point at high altitudes is quite low, which means water boils very fast and at low temperatures. The food inside it does not get enough heat to get cooked and thus food is difficult to cook at high altitudes.
Using a pressure cooker at such conditions helps increase the boiling time as the pressure inside the pressure cooker increases due to the vapour produced inside it. The boiling point increases and thus the time taken to reach the temperature increase and the cooking is better because the food is getting heat for a long period of time.

At Low Altitudes
The boiling point of liquid at low altitudes is more than that at high altitudes, so we do not have to worry much about less time being available for cooking the food. In fact, we try to decrease the cooking time in order to get food cooked faster. Even less amount of water can produce a good vapour pressure and get equal to the pressure produced inside the cooker. But less amount of water takes a lesser time to reach a particular temperature than more amount of water ( as $Q=mc{\delta}t$ ) and thus lesser the mass, lesser the time  to get the required heat, hence faster the cooking.

Questions
1. Have I understood this perfectly or is there something more to it?
2. Does the vapour pressure of the water inside the cooker have to become       equal to the vapour pressure produced due to itself in order to boil?
3. How does pressure cooker increase time for cooking at high altitudes and decrease time for cooking at low altitudes being basically the same thing?
4. The Critical temperature is the temperature at which gas just becomes liquid. The boiling point is the temperature at which all the bulk of the liquid starts turning into gas. How are Critical and Boiling Points different? How are they similar?Is there any mathematical relation between them?

Edit - As 1e9dB's answer points out I made a huge mistake in this answer. The inside pressure of the pressure cooker is ambient pressure plus the extra pressure of the valve. The pressure cooker does not cook at an absolute pressure independent of altitude.

You have it all backwards. The pressure cooker allows the pressure inside the cooker to be above the sea-level atmospheric pressure which also increases temperature inside the cooker and thus decreases cooking time. So the critical factor is the increase of the boiling temperature of water as a function of the pressure inside the cooker.

The pressure cooker raises the pressure inside the cooker to the same absolute pressure, regardless of the altitude at which the cooking is being done. A typical value for the internal pressure would be about 2 atmospheres absolute which makes the internal cooking temperature about 120 °C. (The steam is pushing against the pressure regulator (weight) to escape, not the outside atmosphere. It is like the safety valve on an old steam locomotive that keeps the pressure at a certain level so that the pot doesn't blow up.)

The inside temperature of a pressure cooker when steaming (120 °C) is way below the critical temperature of water (about 374 °C) so that factor doesn't enter into the cooking time at all.


Electron distribution in a C-S bond My lecturer said that in a $\ce{C-S}$ bond, sulfur is slightly $\delta^{-}$ and the carbon slightly $\delta^{+}$, although they have (almost) the same electronegativity. What is the cause of this phenomena if it is not electronegativity?

I think your lecturer would have been more correct had he said that the carbon-sulfur bond reacts as if the sulfur is slightly $\delta^{-}$ and the carbon slightly $\delta^{+}$.

Sulphur is a larger atom so it has more, loosely held electrons than carbon.  This means that a sulfur atom is more polarizable than a carbon atom.  Although carbon and sulfur have similar electronegativities, the effect of this increased polarizability for sulfur is that sulfur behaves or reacts as if the bond were polar.  Take the case of nucleophilic attack on a $\ce{C-S}$ bond.  As the nucleophile approaches, the $\ce{C-S}$ bond will tend to polarize because of sulfur's polarizability.  And in which direction will the bond polarize?  Electron density will prefer to shift such that it resides on the larger atom (charge to size effect), sulfur.

This is the same argument used in the case of a carbon-iodine bond.  Again the two atoms have similar electronegativities (carbon = 2.55, iodine = 2.66), so although the bond is not polar based on electronegativity considerations, the polarizability of the iodine allows the bond to react as if it were polar when a nucleophile approaches.


Why is there no non-bonding molecular orbital in the overlap between H1s and Be2s in BeH2 I've only recently started learning about this and I'm stating to get confused. (The orbital in the middle belongs to Be) So you have a + (+) + which gives a sigma bonding orbital and a + (-) + which gives a sigma anti-bonding orbital. Why isn't there a + (-) - which would result in a non-bonding orbital?

Well, you need two sigma bonds here, and two anti-bonding ones. And there are actually two non-bonding orbitals here.

First of all: how you write it down doesn't really work, since there are p-orbitals in Be, not only s-orbitals. You need to look at them too, not only the s-orbitals. And I'm not quite sure if you did, but don't confuse Valence Bond and MO theory.

So for Valence Bond Theory:

We know this molecule is linear. Therefore we need two orbitals on the Be which can form bonds with the hydrogens and are in an 180° angle. So we do sp-hybridization and with one sp-orbital from Be and the s-orbital from hydrogen we can form a $\sigma$ and $\sigma^*$ orbital on each side. Two sigma bonds and two empty p-orbitals on Be which are non-bonding.

For MO Theory:

We construct a $\ce{H2}$ first, in this case we could write + + and + -, so we got a bonding and an antibonding orbital. Now we put our Be in the middle. And now it get's quite complicate to write down, so here on page 10 you can find the MO of $\ce{BeH2}$. You see that the bonding "$\ce{H2}$" orbital was combined with the 2s orbital of Be in an bonding and antibonding way, and the antibonding + - orbital of "$\ce{H2}$" was combined with a p-orbital in a binding and antibinding way. The two other p-orbitals aren't used and are non-bonding, since they would have equal constructive and deconstructive interaction with both "$\ce{H2}$" orbitals. In the end same result as with the valence bond theory.


Oxidation State in Tetraaquadichlorochromium(III) chloride $\ce{[Cr(H2O)4Cl2]Cl}$, Tetraaquadichlorochromium(III) chloride

According to a book, the (III) indicates the oxidation state of chromium, but it doesn't explain why. Cr has 6 valence electrons. How does it get to +3?

I think the easiest way to see this is to count the formal charges on the ligands.

The four waters do not have net charge.  The remaining ligands--the three Cl's--are likely $\ce{Cl-}$, and so their total charge contribution is $-3$.

The only way the complex can be neutral is if the chromium has a $+3$ charge:  hence Chromium (III).


Why is this a Lewis structure of ethyl acetate? I read somewhere that the structure in the following picture is a Lewis structure of ethyl acetate.

However, I don't understand why:

The $\ce{C}$ has a positive charge, but only three bonds
The $\ce{C}$ does not satisfy the octet rule.

The two rules you cite are guidelines for the 'best' Lewis structure.  However, in the example provided, all electrons are accounted for and all atoms are connected making it a 'valid' Lewis structure.

In the context of resonance forms, this is likely a minor contributor.

However the electrophilic center in ethyl acetate is the carbon you marked as positive, this form has some descriptive  and predictive power in discussing reactions of ethyl acetate.  In general, for $\ce{R2C=O}$ molecules (esters, aldehydes, ketones, etc), The carbocation (sextet) carbon has some relevance.


Berry Pseudorotation I was at a chemistry competition, and one of the questions they asked was on Berry pseudorotation

Now the question gave 4 molecules, namely, $\ce{IF5, PF5, PCl5, and XeF5-}$, and asked which of these molecules is unable to undergo Berry pseudorotation. I immediately eliminated $\ce{PF5 and PCl5}$ as they have trigonal bipyramidal structure, so that left me with the other two molecules.

At this point, I am unable to determine which of the remaining molecules should be the answer, so I take a guess and select $\ce{IF5}$, as my reasoning is that it has lone pairs so it is unable to interconvert. (However this also could be used to explain for why $\ce{XeF5-}$ is a possible answer too)

After reading the Wikipedia article above, it turns out $\ce{IF5}$ is able to undergo such pseudorotation, which leaves me surprised, but in that case, could someone explain why $\ce{XeF5-}$, as opposed to $\ce{IF5}$, is unable to undergo such rotation? (I think it has something to do with the lone pairs on $\ce{XeF5-}$, but I'm not sure)

As per classic literature, PF5 shows identical bond lengths of all P-F bonds which are supported by data from NMR and IR spectroscopy which led to the development of berry-pseudo-rotation. But data's from electron diffraction shows that there are eventually two types of bond lengths (P-F (axial)> P-F(equatorial)) which supports the fact that a naive might predict the hybridisation to be sp²d and hence T.B.P. Even in JD lee, it has been clearly stated that this fluctuation may be called the "Paradox" of Chemistry


Electrolytic conductivity vs metallic conductivity in a galvanic cell I'm analyzing the current output of a galvanic cell with aluminum electrode in aluminum sulfate solution and platinum electrode in acetic acid solution.

What I am confused about is whether molar conductivity will account for current resistance in the cell even though the electrons are being transferred through the metallic electrodes. Or is it just the metallic conductivity that accounts for current resistance?

If molar conductivity is also one of the reasons for current resistance in the cell, how is it?

The resistance you'll find in your system basically corresponds to that of the solution (plus some charge-transfer resistance if your kinetics are somewhat sluggish).

Remember that for current to circulate, you need a closed loop; so the movement of electrons from anode to cathode is only part of the big picture. An equal ammount of current will be transported in the solution by ions approaching/leaving the electrode's surface. So, if you want to minimize your resistance effects, two intuitive approaches come to mind: i) make your ions move faster and/or ii) have a lot of ions moving at the same time. That means that the diffusion coefficient and the concentration of your ions (hence, the molar conductivity) will have direct effect on the solution resistance.

Since metals/alloys are, normally, excellent electronic conductors when compared to electrolytes (which, in turn, have ions as charge carriers), the overall cell resistance is pretty much due to the latter.


When heat is increased, does water move upward in a straw due to increased capillary action or increased pressure? I left a water bottle with an internal straw in my car throughout a long, hot day. When I opened the bottle at the end of the day, water immediately splashed out from the straw. This made me wonder if the increase in temperature caused the liquid to move with greater capillary action upwards in the straw due to higher movement of $\ce{H2O}$ atoms OR whether it was an increase in pressure that led the water molecules to automatically expel outwards from the straw once the bottle was opened.

I know this may seem like a silly question but what explanation would you give for the expulsion of the water from the straw? (I find this recurring event strange because the shape of the bottle never shrinks at all from the heat yet something forces the water outwards but I cannot discern what force causes this. By the way, this never happens to me with regular water bottles - only with ones containing straws/tubes on the inside.)

It almost certainly was due to thermal expansion of water and the air above it. Consider that the thermal expansion coefficient of water near room temperature is ~$0.000214$ K$^{-1}$. [Though it is not linear over a broad range, assume for the small temperature rise in your car that it is.]

If the water was cold (e.g. ~$277$ K, maximum density) and it warmed to ~$300$ K ($80$o F), a temperature change of ~$23$ K, it would have expanded by a factor $(300 \ce{K} - 277 \ce{K}) \cdot 0.000214 \ce{K^{-1}} = 1.005$.

Assume you have a one-liter bottle filled completely with water when you start; by the time you return it's $1.005$ L. In the whole bottle, the change would hardly be noticeable, but the narrow straw greatly magnifies the rise of liquid. If the straw is $6$ mm in diameter, the volume of liquid is given as: $$V=\pi\cdot r^2\cdot h$$

$$5 \mathrm{mL} = 5 \ce{cm}^3$$

$$r = 0.3 \ce{cm}$$

$$h = \frac{5 \ce{cm}^3}{\pi \cdot (0.3 \ce{cm})^2} = 18 \ce{cm}$$

So the water would have risen about $18$ cm in the straw!

This exercise only took into account the water's expansion... for any air above the water in the bottle, and for other (approximately) ideal gases, $PV=nRT$ and thermal expansion would be far greater.

To some extent, the bottle itself expands, decreasing the rise of the liquid, so these calculations are not very accurate. Most likely, air expansion contributed more to the "mass ejection" than did that of water... it depends on whether you view your container as half-full or half-empty.


What is the oxidation state of sulphur in sulphur dioxide? So for $\ce{SO2}$, here's the solution when finding for the oxidation number of sulfur.

$$x + 2(-2) = 0 \\
x - 4 = 0 \\
x = +4$$

$x$ here would be the oxidation number of sulfur, so the compound would now be
$\overset{+4}{\ce S} \overset{-2}{\ce{O2}}$.

Now if I tried using oxygen as the $x$ it would become:
$$1(-2) + 2x = 0 \\
-2 + 2x = 0 \\
2x = 2 \\
\frac{2x}{2} = \frac 22 \\
x=1$$

So, in this case the compound would actually be $\overset{-2}{\ce S} \overset{+1}{\ce{O2}}$.

I'm pretty confused about this since we just got into this today and haven't checked many examples. There is probably something I'm missing in the solution or a rule. Can someone explain to me the correct one? And also can't sulfur's and oxygen's oxidation number both be -2 as they're in group VI(A)?

Your teacher used an interesting example that doesn't obey the rules cleanly. Sulfur can have many oxidation states in this method. In general, larger elements can have oxidation states different than suggested by the group they are in.

If you have learned the electronegativity rules yet, here is a good place to use. If you haven't, higher electronegativity basically means that the element has stronger driving force to reaching an octet; you can look up an electronegativity table if you like. The more electronegative element takes the electrons first. So, in your example, $\ce{SO2}$ has $\ce{2O^2-}$, because of the group oxygen is in. Then, because the $\ce{SO2}$ is neutral, there must be a $\ce{S^4+}$.

But, in a compound such as $\ce{Na2S}$, you can see that S is closer to reaching an octet than Na. So, this becomes $\ce{S^2-}$ while each sodium become $\ce{Na+}$.

Try finding the oxidation states (numbers) for $\ce{PF5}, \ce{WO3}, \ce{SOCl2}$ for understanding. Solutions in the spoiler below.

   $\ce{PF5~is~P^5+,5F-}$
 $\ce{WO3~is~W^6+,3O^2-}$
 $\ce{SOCl2~is~ S^4+, O^2-, 2Cl^-}$


Boltzmann factor vs. graph of Maxwell–Boltzmann distribution In discussions of the Arrhenius equation describing how rates change with temperature, there is often a reference to collision theory.

In the Arrhenius equation,

$$ k = A \exp\left(-\frac{E_\mathrm A}{RT}\right)$$

there is a Boltzmann factor interpreted as the fraction of collisions of sufficient energy to overcome the activation barrier. If you plot the Boltzmann factor $\mathrm e^{-E_\mathrm{A}/RT}$ against temperature, it looks like a simple exponential decay (all quantities in the exponent are positive). However, most textbooks use a graph of the Maxwell–Boltzmann distribution (i.e. distribution of speeds of gas molecules), which is not a simple exponential decay but has a maximum near the average speed.

So how is a graph of the Maxwell–Boltzmann distribution relevant to the Arrhenius equation?

Rationale for posting an "answer" to my own question

[Comment from OP] I’m looking for an explanation and a graph that is better than what is currently in the first year college textbooks .

The comments to the question contain some excellent insight and references but there is no comprehensive answer to far, so I am collecting what was posted here. Maybe the depth of the comments and the lack of a highly rated answer reflects that it is impossible (at the first year undergraduate level at least) to make a meaningful quantitative connection between Maxwell-Boltzmann distribution of kinetic energy in a ideal gas on the one hand and the Arrhenius equation on the other hand.
Textbook treatment of the question
In first-year college textbooks, the Arrhenius equation is often rationalized for gas phase reactions via the collision theory. The quotes here are from OpenStax Chemistry as hosted by Libretext.org, but other textbooks have similar material:

Both postulates of the collision theory of reaction rates are accommodated in the Arrhenius equation. The frequency factor A is related to the rate at which collisions having the correct orientation occur. The exponential term, $e^{−E_a/RT}$, is related to the fraction of collisions providing adequate energy to overcome the activation barrier of the reaction.

This is great because it references collisions providing sufficient energy.
Then they go on to explain:

The Arrhenius equation (Equation 12.5.6) describes quantitatively much of what we have already discussed about reaction rates. For two reactions at the same temperature, the reaction with the higher activation energy has the lower rate constant and the slower rate. The larger value of $E_a$ results in a smaller value for $e^{−E_a/RT}$, reflecting the smaller fraction of molecules with enough energy to react. Alternatively, the reaction with the smaller $E_a$ has a larger fraction of molecules with enough energy to react. This will be reflected as a larger value of $e^{−E_a/RT}$, a larger rate constant, and a faster rate for the reaction.

This is pretty much correct if you don't over-interpret "fraction of molecules with enough energy to react". What is relevant is the energy made available by collisions, not the energy of the molecules, but if the molecules have higher kinetic energy, the collisions will be of higher energy. (There might be some energy associated with rotation and vibration, but since collision theory starts with a mono-atomic ideal gas and we are shooting for the simplest explanation, those will be ignored).
Then they show a figure (and this figure is fairly representative of the corresponding figures in other textbooks and those found online from various sources):

This figure suggests that the fraction of molecules with kinetic energy higher than the activation energy will react. It also suggests that if you integrate over the shaded area, you get the Arrhenius equation.
If you take a close look at the curve, you will see that the slope at the origin is too small. It looks like the x-axis is the velocity rather than the energy, and the graph is accordingly distorted.
As an aside, there are two aspects of this figure that are not representative of what first-year students will experience and observe in the lab.

The activation energy shown is on the order of a couple RT. This would be a reaction too fast to measure in an undergraduate lab, and it gives the impression that maybe 1 out of 10 collisions leads to a chemical reaction.
The two temperatures differ by almost a factor two. If $T_1$ is room temperature, the second experiment would have to be done at 586 K, i.e. in an oven where most solvents would evaporate.

Relevant distribution for collision energy
The comments were very insightful. Ivan Neretin addressed the relevant degrees of freedom:

When the particles are about to react, in fact you are only interested in the distribution of one component of the velocity vector (say, vx). This distribution is a simple exponential decay.

I made a diagram to illustrate what happens in an inelastic collision:

So you can't simply take the kinetic energy of one molecule (or add both kinetic energies together) but instead have to consider one degree of freedom each. The maximum energy provided by a collision is not the sum of the kinetic energies of the particle because the center of mass of the two particles combined continues on its path, so part of the kinetic energy is not available.
The wikipedia article on the Maxwell–Boltzmann distribution states:

Since the energy is proportional to the sum of the squares of the three normally distributed momentum components, this distribution is a gamma distribution; in particular, it is a chi-squared distribution with three degrees of freedom.
By the equipartition theorem, this energy is evenly distributed among all three degrees of freedom, so that the energy per degree of freedom is distributed as a chi-squared distribution with one degree of freedom.

So maybe the distribution of energies in a collision is a problem with two degrees of freedom ($v_x$ of either particle). The chi-square distribution with two degrees of freedom happens to be a pure exponential lacking a T-dependent pre-exponential factor, matching the Arrhenius equation.
Orthocresol gave a reference to the rigorous derivation of the Arrhenius equation from the Maxwell-Boltzmann distribution in one of his comments:

The TL;DR is that porphyrin's first comment is pretty much correct, the area under the graph does generate an exp(−E/kT) form, although the maths is slightly more nuanced than simply doing the integral from EA to +∞.

In this answer, orthocresol sketches the derivation, with two references that show the details.
Relevance to teaching at the first-year college level
Most reactions students will see and do will be in solution, where things are different than in the gas phase, but the Arrhenius equation still is a good description of rate vs. temperature. On the other hand, as porphyrin state in one of his comments, things are usually more complicated than a derivation of rates vs. temperature via the collision model suggests:

The Arrhenius eqn is only a v crude model, sure molecules have to collide in the correct way, even if this is inline the wrong part may still collide depending on the molecule's size. Also vibrational energy can aide or impede collisions even if every thing else is ok. Have a look at RRKM, Tolman's stat mech model or better still Kramer's model of reactions.

So maybe it is best to treat the Arrhenius equation as a definition of the activation energy in an empirical sense.

A graph to illustrate collision energy distribution at multiple temperatures might be helpful.

Following the logic of deriving the Arrhenius equation from the collision model, this would simply be a graph of $e^{−E_a/RT}$ at different temperatures. For values of temperature relevant to first year experiments (eg. room temperature vs water bath), the two traces would be very close to each other, and almost zero at energies equal to a typical activation energy obtained in a first year lab (e.g. 80 kJ/mol for iodination of acetone). To see anything, you have to add an inset zooming in near the activation energy of interest.

This graph is very different from the textbook graph. It shows the distribution of collision energies rather than the kinetic energy of individual particles, and the two temperatures chosen are explicitly given and more relevant to an undergraduate lab experience. It shows that the energy distribution is not so different between room temperature and body temperature (the average energy of a molecule is proportional to T, after all, and this is just a 6% difference in temperature). However, at typical activation energies (insets at 40 kJ/mol and 80 kJ/mol), this smaller difference results in a large difference in rate constants. It also shows that of the collisions of sufficient energy, most have an energy quite close to the activation energy (fast drop-off of probability with increasing energy).


Simultaneous equilibria Rephrasing:
The following reactions are a small example of my big system
$$(1)\  \ce{A + B <=> AB}$$
$$(2)\  \ce{AB -> CB}$$
$$(3)\ \ce{A -> C}$$
$$(4)\ \ce{C + B <=> CB}$$
with
$$\ce{\frac{[AB]}{[A][B]}} = 100 = K_e$$
$$\ce{\frac{[CB]}{[C][B]}} = 100 = K_e$$
I know the rate constants for reactions (2) and (3), and I know that
$$\ce{[A] + [AB] + [C] + [CB] = T}$$
with T =  some constant.
Now, although there is a fast equilibrium between $\ce{A + B}$ and $\ce{AB}$, and between $\ce{C + B}$ and $\ce{CB}$, I want to know how the concentration of each form changes with time.
But I dont know how to write down the differential equations of the system and include the equilibrium constant. Could someone please help me out? My only problem is how to include $K_e$ in the system of ordinary differential equations.

Ok.  You have five equations, two are differential equations and three are algebraic.  And you have the explicit assumption that the equilibria are fast enough so that the two equilibrium equations are always satisfied.

You also have five unknowns, A, B, C, AB, and BC.  One method of solution is to use the algebraic equations to eliminate three of the variables.  This should leave you with two differential equations with two unknowns.  If you are very lucky (I've not worked out the algebra) you will get two equations that can be integrated independently.  More likely you will have two simultaneous differential equations.

In either case the result is apt to be a mess.  So I'd cheat.  If this is a textbook problem or homework problem, there is probably a way to combine the equations to get simple enough expressions.  Otherwise, you may have serious troubles solving the set.


How to lose as little heat as possible to the surroundings from an endothermic reaction? For an experiment, I'm adding several substances (as powders) to hydrogen chloride, and measuring the temperature changes to calculate enthalpy. What is a container I could perform it in (other than a styrofoam cup) that would minimise the heat lost to the surroundings?

I looked up using a vacuum flask of some sort, but would that be appropriate for this sort of reaction? The temperature is being measured via a temperature probe.

What is the scale?
What is the budget?

Reactors are available from 250 mL up.

It might be cheaper to place a flask in a cut-off bottom of a suitable PE bottle and insulate it with PU foam from a DIY market.

UPDATE

A small Dewar cylinder might be an option too.
The model 00C has a volume of 100 mL and lids are available too. Usually, these Dewars are used to store liquid nitrogen and cooling mixtures (2-propanol + dry ice, etc.) so I don't know how resistant the lid is towards $\ce{HCl}$ over time. In addition, this setup isn't as tight as a flask with hollow stoppers and/or a silicone seal if that's an issue.


How many peaks of HCl are found in a mass spectrum? In a mass spectrometer some hydrogen chloride molecules will split into atoms. Chlorine has two isotopes. The hydrogen involved here is the isotope $\ce{^1_1H}$ only.
How many peaks does a mass spectrum will show?

My answer was two with $\ce{H^35Cl}$ and $\ce{H^37Cl}$ but I found the answer to be wrong. Where have I been deceived?
I also found the mass spectrum in CIE 9701/02 2003:

Why are there peaks for $m/z = 35$ and $37$? Even if I consider them to be chlorine's, why aren't there any for hydrogen?

Collision between an electron and the molecule during a MS-experiment usually results in two phenomena:

\begin{align}
\begin{cases}
\ce{AB + e &-> [AB]+ + 2e} &\text{electron ionization}\\
\ce{AB + e &-> [AB]-} &\text{resonance capture}
\end{cases}
\end{align}

Resonance capture is less frequent and usually doesn't add much to the spectra. For $\ce{HCl}$ considering electron ionization as the primary process the following process takes place:

$$\ce{HCl + e -> [HCl]+ + 2e}$$

Based on ionization method (e.g. applied energy) several fragmentation processes may accompany the formation of cation/cation-radical ($\ce{[HCl]+}$/$\ce{[HCl]^{+.}}$):

\begin{align}
\ce{[HCl]+ &-> H+ + Cl^. }\\
\ce{[HCl]+ &-> H^. + Cl+ }\\
\ce{[HCl]+ + e &-> H+ + Cl-}
\end{align}

Fragmentation is often followed by recombination acts of different degrees of probability, e.g.:
\begin{align}
\ce{H+ + Cl- &-> HCl}\\
\ce{H+ + HCl &-> [H2Cl]+}\\
\ce{H^. + [H2Cl]+ &-> [H3Cl]^.+}
\end{align}

One can possibly observe peaks of $\ce{[H]+}$ and $\ce{[H2]+}$, but those are usually have negligible intensities due to recombination processes.

Regarding the spectra OP provided, it looks somewhat strange as the base peak @ $m/z = 36$ is not $100\%$ (approx. $90\%$ instead). I would recommend to check for the experimental data in more reliable sources, e.g. in NIST database:

Considering the fragmentation processes and the fact that $\ce{^{35}Cl} : \ce{^{37}Cl} = 3 : 1$, one can perform the following peak assignment:

\begin{array}{rrr}
m/z & \text{r.i.}, \% & \text{fragment}\\
\hline
35 & 18  & \ce{[^{35}Cl]+}\\
36 & 100 & \ce{[^1H^{35}Cl]+}\\
37 & 6   & \ce{[^{37}Cl]+}\\
38 & 33  & \ce{[^1H^{37}Cl]+}\\
\hline
\end{array}

Alternatively, for $m/z = 37$ and $m/z = 38$ fragments $\ce{[^1H2^{35}Cl]+}$ and $\ce{[^1H3^{35}Cl]+}$, respectively, could be proposed, but it wouldn't be correct due to the following:

recombination acts leading to the formation of large molecules as detected end-products are less possible and usually result in much smaller peaks;
observed relative intensities are not in agreement with isotope distribution for chlorine ($\ce{^{35}Cl} : \ce{^{37}Cl} = 3 : 1$, which is in agreement with r.i. ratios for $\ce{[^{35}Cl]+} : \ce{[^{37}Cl]+} = 18 : 6$ and $\ce{[^1H^{35}Cl]+} : \ce{[^1H^{37}Cl]+} = 100 : 33$, but not with alternatively proposed fragments);
the problem itself states that

  hydrogen chloride molecules will split into atoms.


Why are bonded ions more stable than separate ions? I was reading on the Born-Haber cycle when I encountered a problem. The textbook seemed to suggest that since energy must be supplied to separate an ionic compound into it's composite ions, then the ions are more stable when they are bonded than when they are separate. My question is why?

What's meant by stability is whether the compound "wants" to be in current state. If it wants to change the state (separated ions want to bond) then the system is less stable.
In your example you can simply imagine 2 spherical magnets placed close to each other - they would rush towards one another. They had potential energy that turned into kinetic energy. And once they met - they don't have potential energy anymore, thus they are at lower energy state (more stable) and don't want to separate. Then you'd need to apply force to separate them.


Electrophilic aromatic substitution reaction of dihydrofuran-2,5-dione with anisole I am having trouble understanding the reaction of anisole with dihydrofuran-2,5-dione (succinic anhydride) in presence of $\ce{AlCl3}$ and acid works or how an electrophile is formed. I feel like the product should be something like $\ce{C11H12O4}$.

Nice! You've brought everything you need.

If you can't identify your electrophile immediately, do it the other way around: Assign function to all components. The one that is left must be your electrophile - or its precursor.

Do you remember the role of $\ce{AlCl3}$ in electrophilic aromatic substitutions? Where could it possible attack here?

Here is another hint: What is the trivial name for dihydrofuran-2,5-dione?


Dependency of Standard Gibbs free energy (using electrode potential) on stoichiometric moles of electron exchanged I have a question regarding the definition of the number of moles $n$ when using the equation-
$$\Delta G=-nF\Delta E$$
I understand that $n$ is the number of moles transferred in the overall redox reaction, however, it seems to depend on how you write the equation.
If I write an equation as-
$$\ce{Ag+ + e- -> Ag}$$ $$\ce{\frac{1}{2} Cl2 + e- -> Cl-}$$
Then, we have $n$=1.
But if I wanted one mole of chlorine and I write:
$$\ce{Cl2 + 2e- -> 2Cl-}$$
then I get $n$=2.
Which one is preferred and why?

It does not depend on how you write the reaction. You were right in saying that, $n$ is the mole number transferred in the reaction. So, it is not the number of electrons you write arbitrarily in the equation that represents the reaction, it is rather proportional to the number of other substances, which, even being all arbitrary always satisfy that proportional relation.
Think of $n$ as the factor needed to relate one intensive property with one extensive property.


2,4-dinitrophenyl hydrazine test I found this reaction over here:

which illustrates the carbonyl test but looks fuzzy.

Assuming that the R in the product is a typo for H, I couldn't conserve the number of Nitrogen atoms.

What is the correct reaction?
Is this correct?
Is there a possibility of a methyl substitution on the ring in a variant of this test?

Somebody messed up the scheme completely!

As you have already imagined, there is no magical $\ce{NO2}$ to $\ce{CH3}$ conversion in this classical reaction ;-)

In the times before modern spectroscopy, these reactions were valuable analytical tools: the melting points of the dinitrophenylhydrazones and/or other derivatives were used to identify the carbonyl compounds.


Is it safe to mix methylated spirits with isopropyl alcohol? I have about 200ml of methylated spirits and 500ml of 100% isopropyl alcohol that I want to mix together in a plastic drink bottle.  I will use it for fuel in an alcohol burning stove while camping.  I can't seem to find any other information about mixing them so...

Is it safe to do so?
Is a clear plastic drink bottle OK to keep them in? (worried about melting the bottle - adverse reaction)
If it is safe to mix them, will I end up with a more flammable mix?

Thanks for any help!

Yes, as far chemical safety is the concern. No dangerous reactions nor dangerous phenomena will occur. The mixture, a solution to be more precise, will retain at least qualitatively the properties of the two components, eg will be a disinfectant, a flammable liquid, a solvent, and so on.
Flammable materials are better stored in separated containers in separated places.
For the case in question, going camping with two small bottles or one double sized bottle in the backpack won't make any real difference.


Is There a Difference in the Definition of Complete and Stoichiometric Combustion? My question has to do with some combustion related definitions. When the term complete combustion is used, is it assumed that the combustion is stoichiometric (reactants combine perfectly), but dissociation does not occur, such that only carbon dioxide and water is the product? If so, does the term stoichiometric combustion mean the same thing except dissociation effects do occur?

Edit: Minor species of combustion as a result of dissociation

Stoichiometric, theoretical or complete combustion, is the idealized process where a hydrocarbon is completely reduced to carbon dioxide and water, consuming oxygen (or, if other elements are present, they get all also oxidated, like sulphur compounds yielding sulphur dioxide, but as you can see, not to its maximum oxidation state, sulphur trioxide is not produced in real combustions)

It is called stoichiometric for obvious reasons, stoichiometric amounts of the hydrocarbon and oxygen / water carbon dioxide are consumed / generated. It is supposed to be a complete reaction.

But is also called theoretical, because it does not take place like that in real life. In practice, long hydrocarbon chains do not break down completely as efficiently as it's expected, subproducts are obtained, not only carbon dioxide is formed, but also carbon monoxide, unburned elemental carbon and like in every chemical reaction, it does not proceed to completion.

Also, if there is not enough oxygen (not at least an stoichiometric amount), incomplete combustion takes place, yielding almost exclusively carbon monoxide (toxic stuff).
In some cases, such as the fuels combustion processes, another process called pyrolysis, occurs before combustion. In this incomplete combustion, products of pyrolysis remain unburnt and important contaminants are produced, like the product of partial oxidation of ethanol can produce acetaldehyde, which is really toxic (Considerably small LD50) and very very flammable (flash point under 0 Celsius degrees).


K°P vs KP in equilibrium thermodynamics I'm having a little trouble with definitions.

Here's a copy and paste from my hwk:

...K°P, and K°c have no units...Please enter the KP in 1/bar in  3 significant figures...

So these values even have different units, it's not a typo or anything.

Given: aA+bB $=>$ cC+dD

Is $$K°P=\frac {[C]^c[D]^d}{[A]^a[B]^b}$$
or is
$$KP=\frac {[C]^c[D]^d}{[A]^a[B]^b}$$

the question stated that K°P had no units, and the equation could be unitless if a+b=c+d, but that may not always be the case.

Can somebody tell me which "KP" matches the definition, and also provide, or link the formula for the other "KP"?

Sorry for asking such a trivial question but I'm having alot of trouble finding these definitions.

Thanks all.

The equilibrium constant should be dimensionless, it has to be to be used in equations such as the free energy for reaction at equilibrium
$\Delta G^0_R = -RT\ln(K_p)$.

If the reaction is
$\ce{\alpha A + \beta B \<=>\ \gamma C + \delta D} $
 if we write

$$K_p= \frac{P_A^{\alpha} P_B^{\beta}}{P_C^{\gamma}P_D^{\delta}} $$

where $P_a$ etc are partial pressures at equilibrium then $K_p$ is going to  have dimensions in general depending ion the values of $\alpha , \beta $ etc. as you state in your question.

In the past it was sort of implicitly assumed that each pressure was divided by 1 atm so then $K_p$ becomes dimensionless (or the units were simply ignored and numerical values used) but nowadays this is explicitly added so we obtain

$$K_p= \frac{\left(\frac{P_A}{P^0}\right)^{\alpha}  \left(\frac{P_B}{P^0}\right)^{\beta}}{\left(\frac{P_C}{P^0}\right)^{\gamma}\left(\frac{P_D}{P^0}\right)^{\delta}} $$

where $P^0 = 1 \pu{ atm}$  and so makes things dimensionless no matter what the values of $\alpha, \beta $ etc are but does not change any numerical value.

Although I have used partial pressures the same applies to concentrations $\ce{[A], [B]}$ etc where now each concentration is divided by  $\ce{[C^0]}$ where this is  $ \pu{1 mol/dm^3}$.

I'm not quite sure what your notation $K^0P$ etc. means but I suspect it refers to the two cases I have described. Hope this helps.


Comparision of ONO Bond angle in NO2 and N2O4 How can I compare the bond angle between $\ce{O-N-O}$ atoms in $\ce{NO2}$ and $\ce{N2O4}$?

In $\ce{NO2}$ there is lone pair-bond pair repulsion, and in $\ce{N2O4}$ there is bond pair-bond pair repulsion.

The bond angle in NO$_2$ is $\approx 134^\mathrm{o}$ and the NO bond length 120 pm, in $\ce{N2O4}$ the N-N bond is very long 175 pm and the angle $\approx 126^\mathrm{o}$.

Nitrogen dioxide is paramagnetic with the odd electron in the $\sigma$ bonding sp$^2$ orbital in the 'third' lobe as it were with the O atoms occupying the other two. Additionally there are 4 electrons in the $\pi$ system, two bonding and two antibonding. NO$_2$ also has several low lying excited states which greatly complicates matters.

Joining up two NO$_2$ naturally leads to $\ce{N2O4}$ which is planar and has a very long single $\sigma$ N-N bond. Why the molecule is planar is not clear as it might be expected to twist to remove oxygen-oxygen repulsion. Intuitively one would expect an MO to extend over the whole molecule and so prevent twisting, but the NN bond is very long and single.

Hopefully someone has more up to date data, otherwise you will have to search the literature for MO based (i.e. numerical quantum/chemistry) approaches to describe the bonding.


Entropy of Dissolution of Hydrocarbons Here is what I think I know: The entropy of dissolution reactions increases as methylene groups are added (i.e. butanol has higher entropy of dissolution than propanol). Also, acyclic saturated hydrocarbons become more hydrophobic as the number of methylene groups increases.
Here is where I get confused:
Larger hydrocarbons have a higher enthalpy of dissolution because they are composed of a greater number of bonds. BUT why does the CHANGE in entropy of dissolution become more negative as the number of carbons increases? Doesn't the dissolution of a larger hydrophobic compound yield a more positive entropy value/become less negative because the initial entropy of the solvent is very low (many large, ordered solvent cages) and then solvent entropy increases after the molecules aggregate by the hydrophobic effect?
Super confused...thank you.

I think that you should focus on what you wrote first, longer the hydrocarbon chain higher their entropy in solution. When dissolved, they can access a multitude of states both energetically and in spaces (think of all possible conformers).
This said, it does not mean that they must dissolve in, eg, water. Right because the H term.


Hydrogen venting: Why no flame arrestors? I was reading the guidelines for safe venting of Hydrogen. One item says "no flame arrestors".

Just wondering the scientific reason why flame arrestors may be frowned upon when for venting of most other flammable vapors / gases flame arrestors seem standard.

https://www.eiga.eu/publications/eiga-documents/doc-21117-hydrogen-vent-systems-for-customer-applications/

Hydrogen has a very broad range for combustion in air, with a LEL (lower explosive limit) of about 4% H2 in air and a UEL (upper explosive limit ) of about 96% H2 in air (or 4% air in H2 if you prefer) - well, actually a more recent reference from Matheson indicates 4% to 75% - I've learned something new today, thanks! It also has a very low ignition energy (energy input required to get it going). So, your exhaust system can fill up with lots of H2 and then get set off easily. This can quickly go from a small flame zone to deflagration in the exhaust system (sub-sonic) to detonation (supersonic reaction front), and the reaction will continue until you are outside of the LEL/UEL range.

Look over the Matheson reference, and you will see that there are very few other gases with such a broad LEL/UEL range. For example, methane (or natural gas) is only 5% to 15%, propane is 2.1% to 9.5%. So, if you fill up a space with methane or propane, only so much of it will burn, then it will quench as the mix goes out of the LEL-UEL limits.

The implication is that a hydrogen-air mixture can sustain detonation/deflagration across a much broader range, so it can (and by Murphy will) make a lot more flame and shock and awe. Bottom line, don't have anything in the way of all that excitement exiting the facility as quickly and completely as possible.

An article in Chemical Engineering discusses the issues briefly, although doesn't go into the engineering details (a bit odd for an engineering magazine?).


Decomposition reaction of tin(II) nitrate What are the products formed when $\ce{Sn(NO3)2}$ is subjected to high temperatures?

I searched over the internet but I didn't get any satisfactory answer.

No wonder you cannot find any literature on $\ce{Sn(NO3)2}$, because I don't think it exists in solid form. This is supported by the fact that I even can't find its CAS number online. This book (Ref.1) and relevant paper (Ref.2) support my suggestion, both of which state that:

  Attempts to prepare a covalently bound tin(II) nitrate, by the reaction of tin(IV) tetranitrate with anhydrous nitric oxide, have produced only a white solid solid of formula $\ce{SnN2O6}$, which gives a tin(IV) Mössbauer resonance ($\delta = \pu{0.29 mm\:s-1}$, $\Delta = \pu{0.96 mm\:s-1}$)...

Nonetheless, a study on thermal decomposition of metal nitrates (Ref.3) states that:

  Due to a back-donation of electronic cloud from the nitrate to an unfilled $\mathrm{d}$-orbital of transition and noble metals, their nitrates generally exhibited lower decomposition temperatures ($T_d \lt \pu{700 K}$) than those of the base metals ($\gt \pu{850 K}$).

They suggest those metal nitrates with lower decomposition temperatures decompose to corresponding oxides together with $\ce{NO2}$ and $\ce{O2}$. Thus, it is safe to assume that if $\ce{Sn(NO3)2}$ exists, it would decompose as following reaction suggest:

$$\ce{2 Sn(NO3)2 (s) + heat -> 2 SnO (s) + 4 NO2 (g) + O2 (g)}$$

However, keep in mind that this is pure speculation.

References:

R. Greatrex, “Chapter 8: Mössbauer Spectroscopy,” In Specialist Periodical Reports: Spectroscopic Properties of Inorganic and Organometallic Compounds, Volume 6; N. N. Greenwood, Ed.; The Chemical Society Burlington House: London, United Kingdom, 1973, pp. 494-622.
P. G. Harrison, M. I. Khalil, N.Logan, “Concerning anhydrous tin(II) nitrate,” Inorganic and Nuclear Chemistry Letters 1972, 8(6), 551-553 (https://doi.org/10.1016/0020-1650(72)80139-9).
Shanmugam Yuvaraj, Lin Fan-Yuan, Chang Tsong-Huei, Yeh Chuin-Tih, “Thermal Decomposition of Metal Nitrates in Air and Hydrogen Environments,” J. Phys. Chem. B 2003, 107(4), 1044-1047 (https://doi.org/10.1021/jp026961c).