Saturated vapor pressure From the table of the saturated vapor pressure of the water, at 20 degrees C, the saturated vapor pressure is 2.3388 kPa.

My question is: If I put a water in a closed container, and I pressurize the container up 2.3388 kPa (above the atmosphere pressure) by nitrogen or any other inert gas, will the water evaporate, or it will never evaporate because we are already above its saturated pressure?

In other words, is the saturated vapor pressure of a substance affected only by the partial pressure of the substance itself, or by the amount of pressure above its surface regardless of the substances cause that pressure?

This is purely an equilibrium phenomenon. The vapor pressure of a fluid is generally just a function of the temperature. This pressure is just the partial pressure of the fluid, not anything else. To rationalize why this is, consider what the reaction looks like:

$$\ce{A(l) <=> A(g)}$$

$$\mathrm{K}_{eq} = \frac{P_{\ce{A(g)}}}{\ce{[A(l)]}}$$

The partial pressure is only dependent on factors that affect the activity of the pure substance. Since the liquid is essentially incompressible, the total external pressure does not have an effect.


Why does pouring soda slowly and smoothly seem to keep more of the carbonation? So, I caught myself gently pouring a can of soda into a cup of ice (making sure that there was room for air to move through so the stream of soda from the can to the cup was consistent and even) instead of just haphazardly tilting and dumping the entire thing in there as quickly as possible. I was doing this in part to prevent spills at my desk, but also to keep the carbonation present in the soda. This got me wondering: why does pouring soda quickly cause it to fizz more and lose carbonation?

Because your soda is a supersaturated solution of $\ce{CO}_2$, and such solutions are non-equilibrium and hence unstable. Any physical disturbance (shaking, touching a rough surface, etc.) causes it to emit the gas and return to equilibrium state. Throw in some salt, and you will see a reaction even more vigorous.

Supersaturated solutions of solid compounds exhibit similar behavior, but they are not that easy to come by (I mean, starting from the typical household items). Sugar solution is too viscous, and table salt is no good because its solubility does not depends on temperature all that much, so you can't really supersaturate it.

There is, however, another pretty common example of non-equlibrium state, and that is superheated water. If you pour some bottled water into a clean cup, heat it in a microwave for a while, and then touch it not too gently, it may almost literally explode in your face, so beware.


Why does it yield iron(II) chloride instead of iron when copper meets iron(III) chloride? Complete the equation:
$$\ce{Cu + FeCl3 -> ? + ?}$$

I wrote $\ce{CuCl2}$ and $\ce{Fe}$, but it turned out to be $\ce{Cu}$ and $\ce{FeCl2}$. I really wonder why.

Take a look at standard reduction potentials at $\pu{25 ^\circ C}$:

$$
\begin{array}{clr}
\hline
\text{Reaction} & E^\circ~(\pu{V}) \\
\hline
  \begin{align}
    \ce{Fe^3+ + e- &-> Fe^2+} \\
    \ce{Cu^2+ + 2e- &-> Cu} \\
    \ce{Fe^3+ + 3e- &-> Fe} \\
    \ce{Fe^2+ + 2e- &-> Fe}
  \end{align}
&
  \begin{array}{r}
    +0.77 \\
    +0.34 \\
    -0.04 \\
    -0.41
  \end{array} \\
\hline
\end{array}
$$

A more positive potential means a greater tendency to be reduced, so $\ce{Cu(II)}$ is easier to reduce to copper metal than either $\ce{Fe(II)}$ or $\ce{Fe(III)}$ being reduced to iron metal. So, copper metal can't force the formation of iron metal.

But $\ce{Fe(III)}$ can be reduced just to $\ce{Fe(II)}$ at a relatively high potential, high enough to allow the copper to be in its oxidized form as $\ce{Cu(II)}$. So, we can have $\ce{Cu(II)}$ and $\ce{Fe(II)}$ at equilibrium and thus we can get the corresponding chlorides $\ce{CuCl2}$, $\ce{FeCl2}$.


Why does HCl ionize in water? Why does the Hydrogen proton "prefer" to join the $\ce{H2O}$ molecule instead of remaining bonded to the chlorine atom?

Chlorine is a very large atom compared to hydrogen, which means the overlap between the hydrogen $1s$-orbital and the chlorine $sp^3$ orbital is poor, which translates into a low bond energy.

Additionally, chlorine is very electronegative, which helps pull the H-Cl bonding pair onto the chlorine atom. The chloride anion solvation in water also plays a big part.

It's simply because it's energetically favourable. Maybe someone knows in greater detail than I do.


What makes a species "fluorescence quencher"? I just came across an article where they said that sodium fluoride is not quencher (in the context of fluorescence). Why is NaF not a quencher but NaBr is one?

A fluorescence quencher is any species that causes your fluorophore to stop fluorescing. There are several ways this can happen, but there are two broad categories: chemical reactions and energy transfer. I will give simple examples of each category below. However, which method is operating is going to depend on the fluorophore. Bromide and fluoide will not behave the same way with every fluorophore. Some might be quenched by fluoride and not bromide. Some might be quenched by both. Some might be quenched by neither.

However, in the case of the specific fluorophore in the article, it is quenched by bromide and not be fluoride. The article likely provides an explanation or hypothesis about this behavior, or it cites something that does. Likely this article then is about a selective sensing application. Since bromide turns the sensor off (quenches fluorescence) and fluoride does not, the sensor is selective for bromide over fluoride.

  Quenching by chemical reaction

Any chemical reaction will change the relative energies and energy spacing of the frontier molecular orbitals, which could lead to quenching. This chemical change could be something irreversible like oxidizing or reducing the fluorophore (probably not what is happening here). Other chemical reactions are possible, and it could be that bromide reacts and fluoride does not.

This change could also be a reversible one, like protonation/deprotonation, ligand binding, etc. If the fluorophore in your article is a transition metal complex, than it could be quenched by complexation with bromie anion (which displaces some ligand that has charge-transfer absorption/fluorescence with the metal). Fluoride might not do this.

This behavior could be more complex. Perhaps the fluorophore only fluorescences in the presence of some analyte. Maybe bromide reacts with the analyte (and fluoride does not) or bromide competes with the analyte but does not promote fluorescence.

  Quenching by energy transfer

Fluorescence works as follows. The fluorophore absorbs a photon and is excited from its ground state to its excited state. When the excited state relaxes back to the ground state it emits another photon.

Fluorescence can be quenched if the excited state can be intercepted. An example is as follows. You have a system with your fluorophore and another molecule. Your fluorophore has a smaller HOMO-LUMO gap than this other molecule, so the fluorophore selectively absorbs photons of a certain frequency range and is excited. However, if the other molecule has a lower energy excited state than your fluorophore, then energy transfer can occur between them if they are close. Now your fluorophore is back in its ground state and the other molecule is in its excited state. If this molecule does not fluoresce, then fluorescence is quenched. This is probably not happening in your example. Note this phenomenon can be given various names like "resonance energy transfer" or "Forster energy transfer."

This type of quenching can get more complex as well. The energy acceptor could go and do chemistry or it could be so eager to be reduced that it not only steals the energy from the fluorophore, but it grabs an electron as well. This is not likely what is going on with bromide and fluoride.

The other type of behavior that can go on here is analyte-induced aggregation. Aggregation of the fluorophore causes the electronic properties to smear out over the aggregate, which likely means no more fluorescence (and perhaps no more absorbance) at the frequencies you care about. This phenomena could be happening in your example. Bromide could induce aggregation while fluoride does not.


How to determine the pressure to output a gas to achieve a particular gas flow rate? I have a 420-liter (14.9 cubic ft) helium cylinder and a pressure regulator with a gauge supporting up to 160 psi. My requirement is to ensure a flow of 10 to 15 liters / min. How do I determine the pressure setting on my pressure regulator to fulfill my flow rate requirement?

Well, since the pressure in the tank will decrease with usage, you can't ultimately do what you seem to want to do.

At a given backing pressure, there are equations to determine the orafice size to get a flow rate - google 'choked flow'. Downstream of the regulator, your orafice should be good for quite some time.

For an experimental setup, one would use either a mass flow controller (MFC) to actively control the flow rate, or (much cheaper) one of those little flow indicators with the ball floating in the He (air) stream. Note the second option here is basically an adjustable orafice with integrated flow indicator.


What is the correct definition of hydration enthalpy and why is it always negative? The first definition of hydration enthalpy I found on the net is:

  The hydration enthalpy is the enthalpy change when $1$ mole of gaseous
  ions dissolve in sufficient water to give an infinitely dilute
  solution. Hydration enthalpies are always negative.

Another definition I found in a book is:

  Hydration enthalpy is the energy released when $1$ mole of a compound is
  dissolved in sufficient water to give an infinitely dilute solution.

Now I have three questions:

Which of the two definitions is the correct one? Why?
Why exactly are hydration enthalpies always negative? Suppose I have hydrated copper(II) sulphate which is mostly stable as $\ce{CuSO4.5H2O}$. I convert it to $\ce{CuSO4.6H2O}$ by adding more water forcefully. Shouldn't the hydration enthalpy be positive in this case as it will make the final product less stable than the initial one?
Which definition of hydration enthalpy applies for $\ce{CuSO4}$? I do not think $\ce{CuSO4.5H2O}$ is "infinitely" dilute by any extension of logic. Please explain this if possible.

P.S: I am a novice in thermochemistry. It would be helpful if you could keep the explanations as simple as possible.

Hydration is defined as the following process (with any salt but using copper(II) sulphate as an example):

$$\ce{CuSO4 + 5 H2O -> CuSO4 . 5 H2O (s)}\tag{1}$$

This reaction will only happen if the associated enthalpy $\Delta_\mathrm{hyd} H^0$ — or more precisely, the associated Gibbs free energy $\Delta_\mathrm{hyd} G^0$ is negative. For some compounds, that is not the case; for example sodium chloride does not form hydrates: reaction (2) does not occur spontaneously.

$$\ce{NaCl + n H2O} \nrightarrow \ce{NaCl . n H2O}\tag{2}$$

Thus, we cannot measure a hydration enthalpy. If we can measure it, the process must be spontaneous and thus the enthalpy negative. Thus, all measurable hydration enthalpies are negative.

Sometimes, salts can form multiple hydrates. However, not every hydrate is always possible. Hydration enthalpies only exist for those hydrates which are possible.

Your definitions are basically identical only that they do not measure hydration enthalpies but solvation enthalpies. That is the process as shown in equation (3).

$$\ce{CuSO4 (+ H2O) -> Cu^2+ (aq) + SO4^2- (aq) (+ H2O)}\tag{3}$$

For that process, the enthalpy does not have to be negative. In fact, quite a few ionic compounds dissolve endothermicly, e.g. $\ce{NH4Cl}$.


Has a non-carbon-based form of life been discovered since 2010? In 2010 NASA announced the discovery of microorganisms that, after being treated in laboratory, could keep growing only using arsenic, instead of the (though very small amount of) phosphate they also had in nature.
However, this was confuted by further studies in the following two years.
Has there been success in finding "alien" life on Earth, somewhere else, since then?

Edit: the NASA article from back then says what gilleain sais below. I naively hoped a couple of articles from different (not scientific, but not bad) journals were not modifying the facts. Thus the question actually is if non-carbon-based life has been found.

Just because an organism might need a nutrient, whether arsenic or cobalt (which homo sapiens needs for making RBC's) does not mean that the organism is not carbon-based. What had been announced was that an organism used arsenic to replace phosphorus, not carbon, which has since been amended.

There is also the question whether the GFAJ-1 microbe is a facultative or obligate user of arsenic, or simply excludes arsenic. Though the organism can survive high concentrations of arsenic, it appears to do so by discriminating between As and P and using the phosphorus in its metabolism.

There are organisms such as Sulfurospirillum arsenophilum and S. barnesii that are at least facultative As users, deriving energy from oxidation or reduction of arsenic compounds, but the arsenic is not replacing carbon, much less phosphorus, in their biochemistry.


Conversion of units for concentration: mol/l to ppm The concentration of an aqueous solution of $\ce{Mg(SO4)2}$ is  $0.001\, \textrm{mol} \cdot \textrm{dm}^{-3}$.

Find out the concentration of solution and the concentation of $\ce{Mg^{2+}}$ ion in ppm.

How to do this? How to find moles in ppm?

If anyone could clearly describe, it would be great.

Convert mol-dm^2 to grams/volume.  A decimeter is 10 cm.  A dm^3 is then a liter.  Water has density ~1 g/cm^3.  Multiply the g(Mg)/g(water) by a million.


Why do you use acetic acid instead of water to create your standard riboflavin solutions? I must create my own lab and I was going to use water, however, a similar experiment online says that acetic acid should be used to create my standard riboflavin (vitamin B2) solutions.
Why is that?

From Sigma product information:

  The flavins are light sensitive and unstable in alkaline solutions.  The dry solid is stable to diffuse light, but is highly photolabile in solution, especially in alkaline solutions. Neutral and acidic solutions are stable in the dark (3% decomposition per month at 27 °C at  pH 6.0).  Acidic solutions may be sterilized by autoclaving.

So making a slightly acidic solution probably helps to prevent decomposition. If you dissolve it in pure water without a buffer, the pH is not well defined.


Does a C=C double bond donate electron or withdraw it? I have learned that a $\ce{C=C}$ double bond is electron-rich because of the p-p overlap and hence can easily donate an electron pair: I agree with this statement.

However, in a $\ce{C=C}$ double bond, the carbon is $\ce{sp^2}$ hybridized and I've also learned that more '$\ce{s}$-type' character increases electronegativity. Does this mean that the carbon atom - being more electronegative - tries to withdraw electrons?

For example consider the double bond connected to a positively charged Carbon, then will it donate an electron and stabilize it? or will it (being more electronegative) withdraw an electron?

I guess it depends on the way you quantify "electron donation/withdrawal" (because it is a useful concept, but not really a measurable observable). I'm sure someone will find a counterexample, but if you consider methane and ethylene as Brønsted acids, you find that methane is a much weaker acid. Thus, ethylene is better at stabilizing the negative charge. (This is in line with your sp2 argument.)


In Diazonium Ion why does N contain 4 bonds? Nitrogen's valenc number is 3. Sometimes it can form a co-ordinate bond by donating it's lone pair electrons. In diazonium ion N has 4 bonds, then one of those 4 bonds must be a co-ordinate bond. Actually I can't define the co-ordinate bond in diazonium ion.Again In diazonium structure, it is shown the first N with 4 bonds is positive but why??

One way to think about this structure, the benzenediazonium cation, is to consider the structure of benzonitrile - C6H5CN - just replace the positive nitrogen with a neutral carbon atom. This has the same basic electronic structure as the diazonium ion, but differs in that the carbon nucleus has one less positive charge than the nitrogen nucleus.

I don't think the concept of "coordinate bonds" is very helpful here. There are many different species with positively charged nitrogen, such as tetramethylammonium salts, that have four full covalent bonds to nitrogen.


is this the proper identification of this cyclic molecule? Attempt:

(1S,3S)-1,3-Dichlorocyclohexane

I'm just not sure about the nomenclature for R and S configurations (our teacher was a little unclear). I gave it my best shot after looking through some textbooks but would like just a quick confirmation or someone to tell me if it's wrong.

In addition, if I exchanged the positions of the Cl and H groups on each side with each other would I get an enantiomer? ((1R,3R)-1,3-Dichlorocyclohexane)

It's correct, as already mentioned in a comment.

Some might find more or less helpful to redraw the structure to:

Your additional question can be tricky, understood a different way. It's possible to exchange H and Cl positions, or rather axial to equatorial bonds and the other way around – by switching to the other boat conformation, i.e. without changing the configuration:

(The two structures are identical, one of them can be rotated to superimpose the other.)


Is there anything that doesn't mix with water OR oil? It's well known that water and oil don't mix, and if you put them together, the oil will float on top of the water in a distinct layer. My understanding is that this is because water is polar due to oxygens high electronegativity and oil is not. Most liquids and many solids are capable or mixing or dissolving into water, and I don't know of any other nonpolar liquids, but I imagine they would mix with oil.

In electromagnetism, there are magnetic materials that will be affected by magnetism, and nonmagnetic materials that aren't, but there is also diamagnetism which opposes influence of external magnetic fields.

So is there any kind of liquid substance that is neither polar nor nonpolar, and therefore doesn't mix with either water or oil? Could you have a beaker full of 3 liquids that has 3 distinct layers?

And if that was the case, are things that dissolve into one layer ONLY able to dissolve into that later, or is there the ability to diffuse between layers?

In short, yes, many highly fluorinated liquids are miscible with neither water nor organic solvents.
If you look at my answer to this question, you can see that I made a mixture of water, hexanes, and a fluorinated solvent called HT-110.
Likewise, the HT-110 will not dissolve most compounds, apart from fluorinated ones like Teflon AF.
Many compounds will have appreciable solubility in more than one layer and if they are in contact, there can be transfer between them.

Polarity plays a large role in whether two liquids are miscible.
Basically, for two liquids to mix, it must be energetically favourable to disrupt the intermolecular forces of the individual liquids in favour of the new intermolecular forces between the components of the mixture.
For your example of water and oil, water on its own has significant hydrogen bonding, but since the oil can't participate in hydrogen bonding, there would be a lot less hydrogen bonding in a homogeneous mixture of oil and water—the oil molecules essentially block water molecules from accessing each other.
Because the mixture would greatly reduce the intermolecular forces holding the liquids together, it is more energetically favourable to remain separate.
In contrast, a mixture of water and ethanol is energetically favourable because ethanol can hydrogen bond and is similarly polar, so the disrupted water-water intermolecular interactions are replaced by similar water-ethanol interactions.

In a little more detail, a mixture is favourable when its Gibbs energy of mixing is negative: $$\Delta G_{mix} = \Delta H_{mix}-T\Delta S_{mix}$$

In an ideal mixture of two components A and B, A-B interactions are the same strength as A-A and B-B, so the enthalpy of mixing, ∆H, is 0.
If A and B have similar properties, A-B interactions tend to be similar and ∆H will be close to 0.
If A-B interactions are weaker (e.g. water and oil), ∆H will be positive.
∆S is almost always positive for common mixtures because each component has a larger volume to spread into than if the components were separate. (the exception is when A-B interactions are much stronger than A-A and B-B interactions which prevents A and B from moving randomly through the volume)
This means that mixtures can still form if the enthalpy of mixing is positive, if the entropic contribution can overcome it.
This explains why some mixtures are possible at elevated temperatures but not room temperature.


Solid aluminum powder I am curious if there is a way to dissolve or break down solidified aluminum powder. I have tried to break it apart but it is a solid chunk and I need to figure out how I can break it down to 150 pound pieces.

Can a block of hardened aluminum powder be physically or chemically broken down back to a soft form.

If you had a sealed container of aluminum powder that became exposed to air, your prospects for turning your current material back into aluminum powder are grim.

Your block probably appears like something that became clumpy when exposed to humidity (table salt, sugar, etc.), and that you aught to be able to just break it apart with a little force to get the powdery stuff again.

Unfortunately, aluminum reacts with the oxygen in air to form a very hard oxide layer, which is likely what is now binding the formerly pure aluminum powder together. In other words, you don't just have aluminum "sticking" to aluminum, but you've formed strong chemical bonds, actually creating a new compound altogether (aluminum oxide) There is likely no simple solution for returning your 150 pound blocks of mixed aluminum / aluminum oxide back into pure aluminum powder.


In crystal field theory, why do all the metal electrons enter the d orbitals and not the s orbital? Suppose we have the complex $\ce{[Fe(H2O)6]}$. Iron has 26 electrons, and its electronic configuration is therefore $[\ce{Ar}]\,\mathrm{4s^2 3d^6}$, but in crystal-field theory, we say that the $\mathrm{3d}$ orbitals have 8 electrons instead of 6. Why?

Your example is a little questionable. I would think you are talking about $\ce{[Fe(H2O)6]^2+}$, in which case you are actually supposed to have an $\ce{Fe^2+}$ ion with a $[\ce{Ar}]\mathrm{(3d)^6}$ configuration. There are no $\mathrm{4s}$ electrons in the $\ce{Fe^2+}$ ion.

Nevertheless, it is true that in crystal field theory, electrons are assumed to be entirely in the d-orbitals, even for low-oxidation state complexes. For example, nickel(0) in $\ce{[Ni(CO)4]}$ is assumed to have a "$\mathrm{d^{10}}$" configuration. There is a reason for this, the explanation of which requires a molecular orbital treatment. Consider a generic octahedral complex, $\ce{[MX6]^{n-6}}$, which is formed from a $\ce{M^n+}$ ion and six $\ce{X-}$ ligands. We could construct an MO diagram:

(Source: Miessler & Tarr, Inorganic Chemistry 3rd ed., p 347)

Each $\ce{X-}$ ligand contributes one pair of electrons, which gives rise to a total of six pairs; these six pairs of electrons enter the low-energy bonding MOs $\mathrm{a_{1g} + e_g + t_{1u}}$. These MOs are similar in energy to the ligand AOs, and therefore are likely to be mostly centred on the ligand, which means we can essentially ignore these MOs when we are describing the properties of the complex. (Of course, these MOs are important in the sense that they are responsible for the bonding between metal and ligand. If these bonding MOs did not exist, the complex would not form. However, we are usually much more interested in the higher-energy orbitals.)

The electrons in the $\ce{M^n+}$ ion can be in any of the AOs ($\mathrm{3d}$, $\mathrm{4s}$, or even $\mathrm{4p}$) - it does not matter. The only important thing is that, once the complex is formed, these electrons will enter the lowest-energy MOs according to the aufbau principle. Since those low-energy bonding MOs are already occupied by ligand electrons, the metal electrons will have to go into the $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$ MOs, which in the simplistic treatment of crystal-field theory are called the "d-orbitals".

The truth is that those two MOs have quite a lot of metal d-orbital character (as you can see from the dotted lines), so crystal-field theory is not entirely wrong. Nevertheless, you can see from the MO diagram why it would be wrong to assume that the metal $\mathrm{4s}$ orbital is filled: what CFT calls the $\mathrm{4s}$ orbital is actually the $\mathrm{a_{1g}^*}$ orbital in the MO scheme. This orbital can only be filled if the $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$ MOs are already filled, which is not the case unless we are talking about some kind of zinc(0) complex (very unlikely!).

Similar diagrams can be constructed for other geometries (tetrahedral, square planar etc.) and the same point will hold. Therefore, even though $\ce{Ni}$ itself has a $[\ce{Ar}]\mathrm{(3d)^8(4s)^2}$ configuration, the tetrahedral $\ce{[Ni(CO)4]}$ complex will have a $\mathrm{(e)^4(t_2)^6}$ configuration (as you are probably familiar with, in tetrahedral complexes, the d-orbitals are split into $\mathrm{e + t_2}$). Again, in the language of crystal-field theory, both the $\mathrm{e}$ and $\mathrm{t_2}$ orbitals are simply called "d-orbitals", so this is equivalent to having a "$\mathrm{d^{10}}$" configuration.


How to speed up flash powder flaming reaction Do you have an idea how can be flaming of flash powder (probably $\ce{KClO_4} $ (60%) + $\ce{Al}$ (40%) ) can be speeded up? Of course we consider the pure oxygen-in-air conditions. So how? With some kind of catalyst?
Thanks.

First, we postulate safety is not an issue. If you try any of this you will probably die.  If you need a short duration event-triggered flash, explode a wire.  A Pockels cell plus a polarizer is a nanoseconds shutter.

Ball mill potassium perchlorate all by itself.  Mix well with micronized aluminum.  Compact.  (Impact, friction, or a static discharge may kill you.)

Aluminum has a high boiling point, 2519 C.  Magnesium is 1091 C and burns under nitrogen.  As above, with micronized Mg, adjusting stoichiometry for the new equivalent weight.  Same warning.

Use a more potent oxophile, like titanium, adjusting stoichiometry for the new equivalent weight.  Micronized titanium can flash in air, nitrogen, or carbon dioxide all by itself.  Same warning.

DO NOT TRY CHLORATE, as opposed to perchlorate.  There is almost no activation energy barrier to reaction.


How to make permanent ink that appears after some time? I'd like to ask,  how to make permanent ink, that will be invisible when used, but after like 30 minutes it will appear (preferably dark color)? Thanks.

Depending on what you want to do (what do you want to do?) you might be looking for gall ink.  Briefly, the chemistry works like this: the recipe consists of soluble polyphenols (typically gallic acid) and a Fe(II) salt.  The preparation must be stored in an air-tight container and even then decomposes over time.  As soon as it is applied to paper and exposed to air, the Fe(II) is oxidized to Fe(III), which forms insoluble deeply coloured complexes with your polyphenol.  The permanent colour takes about a day to develop.  That's why often a non-permanent dye is added so people can see what they are writing.


Bromination of 1,3,5-cyclooctatriene followed by nucleophilic substitution by excess dimethylamine Write the mechanism for the following reaction (bromination of 1,3,5-cyclooctatriene followed by nucleophilic substitution by excess dimethylamine):

My answer so far:

I have only shown one of the reactions, I presume it is the same for adding the other $\ce{Me2NH}$ on the other side.
How do I get rid of the second bromine, or does just one of the Br's add? If it is only one, why, because you usually use $\ce{HBr}$ for that?

This is a quote from Solomons and Frylhe Organic Chemistry 10th Ed, Chapter 13, p.614

Reactions of this type are quite general with other conjugated dienes. Conjugated trienes often show 1,6-addition. An example is the 1,6-addition of bromine to cycloocta-1,3,5-triene.

A plausible mechanism for the bromination step would be the initial formation of a bromonium ion which is attacked distally by bromide:

Sequential nucleophilic substitution with dimethylamine then leads to the desired product.
Echoing Martin's comment on Siddharth Porwal's answer, there is a distillation step between the bromination and the substitution with dimethylamine. It is not possible for a putative bromonium ion to survive. Also, the isolation of the 1,6-dibromide also disproves the possibility of 1,2-addition mentioned in RE60K's answer.
The reaction above was originally reported by Cope et al.:[1]

Bromine (3 g) in 20 ml of chloroform was added to a stirred solution of 2 g of 1,3,5-cyclooctatriene in 20 ml of chloroform at −15 to −25° in a nitrogen atmosphere in a period of one hour. The product was distilled rapidly under reduced pressure; b.p. 95° (0.3 mm), n25ᴅ 1.5954. [...] Previous work had shown that dibromocyclooctadiene prepared by the two routes gave the same bis-dimethylaminocyclooctadiene by reaction with dimethylamine.

Reference

Cope, A. C.; Stevens, C. L.; Hochstein, F. A. Cyclic polyolefins. v. preparation of bromocyclooctadienes and 1,3,5-cyclooctatriene from 1,5-cyclooctadiene. J. Am. Chem. Soc. 1950, 72 (6), 2510–2514. DOI: 10.1021/ja01162a047.


How to decide at which end to start the numbering of the parent chain when the options for locant sets are the same? I would like to know whether the blue or purple way of numbering is preferable? Following the two possible ways, I get the following names:

4-(1,1-dichloro ethyl)-5-(1-chloro-2-bromo ethyl)-octane
4-(1-chloro-2-bromo ethyl)-5-(1,1-dichloro ethyl)-octane

In each case, the locant set is (4, 5), so from which side should I start numbering from?

If numbering from each end of the parent chain gives the same set of locants, the next criterion stipulates that lowest locants are assigned to the substituent that is cited first as a prefix in the name.

On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Note that Rule f takes precedence over Rule g. If no decision can be attained by application of P-14.4(f), the lowest set of locants is selected according to P-14.4(g) in their order of their citation in the name.

This means in effect that alphanumerical order is used to establish the numbering when a choice is possible since alphanumerical order is used to establish the order of citation of substituent prefixes in the name.

  P-14.5 ALPHANUMERICAL ORDER

  Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved

  Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.

  (…)

  P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

For example ‘dibromo’ comes before ‘chloro’ since ‘bromo’ begins with ‘b’, ‘chloro’ begins with ‘c’, and the multiplicative prefix ‘di’ does not change the alphabetical order.

However, the example given in the question does not include simple prefixes but the compound substituents ‘1,1-dichloroethyl’ and ‘2-bromo-1-chloroethyl’. Such compound names are considered to begin with the first letter of their complete name.

  P-14.5.2 The name of a prefix for a substituent is considered to begin with the first letter of its complete name.

Therefore, the correct name for the compound given in the question is 4-(2-bromo-1-chloroethyl)-5-(1,1-dichloroethyl)octane since ‘2-bromo-1-chloroethyl’ starts with ‘b’ and ‘1,1-dichloroethyl’ starts with ‘d’.


What really happens atomically in an explosion? Let's say a room is filled with butane, I then throw a cigarette into the room. What happens to the atoms/molecules of the butane when they are in contact with the heat from the cigarette?

The combustion of alkanes like butane is fearsomely complicated involving dozens of transient compounds and hundreds of different reaction. If you have a few spare hours there is a dissertation that presents a nice summary of the process here (this is a 1MB PDF).

A butane molecule is pretty stable and doesn't react with oxygen on contact so you need some way to get the reaction going. Typically the reaction is started by the generation of free radicals. These will react with stable molecules to split them up and generate more free radicals so in effect we have a chain reaction.

So the ignition process is basically the generation of free radicals, and that's what the lit cigarette does. The heat of the burning tobacco generates free radicals that then start the butane oxygen reaction.  You can also generate free radicals in lots of other ways e.g. using shock, from an electric spark, or by using a suitable catalyst such as the catalytic convertor on your car exhaust.


How accurate would be the following conclusion on 'Critical Temperature'? "Boiling does not occur when liquid is heated in a closed vessel. On heating continuously vapour pressure increases. At first a boundary is visible between the liquid and vapour phases because liquid is more denser than vapour. As temperature increases more and more molecules go to vapour phase and density of vapours rises. At the same time liquid becomes less dense. It expands because molecules move apart. When the density of liquid and vapours becomes the same; the clear boundary between the two phases disappears. This temperature is called critical temperature"

That is EXACTLY what my textbook says. And that is where I am having doubts. It's because - a gas might still be found in its gaseous phase below the critical temperature if the pressure is considerably below its critical pressure ($P_c$) and it has a volume sufficiently larger than its critical volume ($V_c$).

Just consider any point in the Pressure-Volume isotherms of $\ce{CO2}$ given above, which is in the region of gas-liquid equilibrium (within the shaded dome shaped curve) but has a volume greater than the critical one. For e.g a point that lies

Within the shaded region.  and
On a line passing through B and parallel to the pressure axis.

Let that point represent the state of liquid inside the vessel.
From there if we increase the temperature, we would arrive at a point on the boundary of the dome shaped curve, (viz. B) but that point would be on a lower isotherm than the critical isotherm. And it looks like B is also a point where the two densities becomes equal.(isn't it?)
Based on that premise, we can't say that the temperature arrived at in the above (heating liquid in a clossed vessel) situation would necessarily be the Critical temperature.

What I want to ask is if my reasoning is correct or am I missing anything?

My question can be rephrased in the following manner also
Is it true that critical temperature is the only temperature at which the density of the gaseous phase of a gas becomes equal to its liquid density?

Ok, I added a point Z to the diagram which I've added below, and I'm going to quote you text in italics below the image...

Just consider any point in the Pressure-Volume isotherms of $\ce{CO2}$ given above, which is in the region of gas-liquid equilibrium (within the shaded dome shaped curve) but has a volume greater than the critical one. For e.g a point that lies:

Within the shaded region. and
On a line passing through B and parallel to the pressure axis.

Let that point represent the state of liquid inside the vessel.

Question 1 - My point Z meets your two conditions doesn't it?

... From there if we increase the temperature, we would arrive at a point on the boundary of the dome shaped curve, (viz. B) but that point would be on a lower isotherm than the critical isotherm.

Assuming you agree with point Z, I agree with the idea above. The next two sentences is where you're wrong.

And it looks like B is also a point where the two densities becomes equal.(isn't it?) Based on that premise, we can't say that the temperature arrived at in the above (heating liquid in a closed vessel) situation would necessarily be the Critical temperature.

Let's start at point A, and follow the purple isotherm. At point A all the carbon dioxide is gas. At point B all of the carbon dioxide is still gas. But as you start to goto point C from point B the carbon dioxide vapor becomes supersaturated and liquid carbon dioxide forms. Now when you get to C there is so little volume that all the carbon dioxide is liquid. Thus only in the blue shaded area under the curve do you have two phases gas and liquid.

Liquid carbon dioxide is denser than gaseous carbon dioxide. So the gas will be on top and the liquid below.


Calculating Avogadro's Number For a school project, I am in a group tasked with calculating Avogadro's Number in multiple manners.  Other than the Millikan Oil Drop Experiment, we know of no ways to compute it.  A Google search yielded no methods that could be of use (either complex beyond the high school level, or through using a mole of materials, which defeats the point of deriving the number).  What methods can we (intelligent high schoolers on a limited budget) use to compute Avogadro's Number?

Placing a drop of oleic acid on the surface of water contained in a pan is a relatively straightforward experiment that has been widely used to determine Avogadro's number.  Here are two links that provide good background on a) how to run the experiment, b) how to analyze the data, and 3) the science behind the experiment.  The two links are just variations on the same experiment, but if you read them both it will really enhance your understanding of the experiment.

link 1

link 2


Why is the entropy of container A the greatest? Question

  9.35 Container A is filled with 1.0 mol of the atoms of an ideal monatomic gas. Container B has 1.0 mol of atoms bound together as diatomic molecules that are not vibrationally active. Container C has 1.0 mol of atoms bound together as diatomic molecules that are vibrationally active. The containers
  all start at $T_\mathrm{i}$ and the temperature is increased to $T_\mathrm{f}.$ Rank the containers in order of increasing change in entropy. Explain
  your reasoning.

Answer

   B < C < A

My Approach

I thought that you were supposed to assume that container C had the most number of atoms and therefore a greater entropy automatically.

An accurate summary of the relative molar heat capacities of the different species is given in another answer and recapitulated in the following table together with the total heat capacity for each gas:

$$\begin{array}{|case|Cv|} \hline \textrm{molecule} & C_{vm}/R & NC_{vm}/R \\ \hline  \mathrm{A} & 3/2 & 1.5 \\ \mathrm{B} & 5/2 & 1.25  \\\mathrm{C} & 7/2 & 1.75 \\ \hline \end{array}$$

where $N$ is the total amount of gas, which based on the problem described in the OP is either $\pu{1 mol}$ of atoms ($\mathrm{A}$) or $\pu{\frac{1}{2} mol}$ of homonuclear diatomic molecules ($\mathrm{B}$ and $\mathrm{C}$).

Since the entropy change for the isochoric temperature change is $\Delta S = C_v \log\left( \frac{T_2}{T_1} \right)$, it follows that $\Delta S_C > \Delta S_A > \Delta S_B$.


How do I figure out the amount of water (mL) and amount of Aluminum (g)? We have a project where we have to extract copper from copper sulfate. But my group is given a powder version ($5~\mathrm{g}$), so we have to first dilute it with water and then combine it with aluminum to extract the copper. If this is the equation:

$$\ce{3Cu(SO4) + 2Al + H2O + salt -> Al2 (SO4)3 + 3Cu}$$

Salt is acting as a catalyst for the experiment.

The first step you need to take is to determine the moles of your copper sulphate.

This is simple arithmetic. $$moles = \frac{mass}{molar.weight}$$
The mass of the substance is your measured 5 grams. The molar mass of your compound can be obtained from Wikipedia. In this case, anhydrous copper sulphate has a molar mass of $159.61\frac{g}{mol}$.

Plugging into the equation: 0.0031 $moles = \frac{5}{159.61}$

Now that you know the moles of your copper sulphate, you should be able to work out the the needed amounts of your other compounds using stoichometry.


What is the mechanism for this intramolecular SN1 reaction? What is the mechanism for the following intramolecular $\mathrm{S_N1}$ reaction?

For this question, my professor hinted at an intramolecular reaction, and he asked us to recall the bromonium ion as well. As such, I believe the $\ce{SMe}$ group attacks the open carbocation when $\ce{OTs}$ leaves. I think I understand how the left-side enantiomer is formed: $\ce{SMe}$ attacks the open carbocation, then water attacks from the back because the front-side is occupied by the hydrogen (not shown). However, what is the mechanism for the right-side enantiomer? How does the $\ce{SMe}$ end up in front (on a wedge) along with water after hydrolysis?

You are correct on your initial assumption. This hydrolysis is assisted by the neighboring thiomethoxy group ($\ce{SCH3}$) as depicted in the following scheme:

The cyclic methylsulfonium intermediate, which is achiral in this case, is a mimic of bromonium ion in bromination of alkenes (also see Chemistry of Sulfur Mustard on releasing chlorine through cyclic alkylsulfonium ion). Realistically, this intermediate is mimic of bromonium ion of of cis-2-butene (also achiral), and hence would give a racemic mixture. Similar to bromonium ion intermediate, the upcoming nucleophile ($\ce{H2O}$) can attack at two places: through path $a$ or through path $b$. Keep track on the (R,S)-configurations of two stereo centers. The original compound has (S,S)-configuration. The (S)-configuration on carbon attached to $\ce{SCH3}$ is unchanged but carbon attached to $\ce{OTs}$ has been changed to (S)-configuration in cyclic intermediate, because it is necessary to have an anti-periplanar conformation for attacking group ($\ce{SCH3}$) and leaving group ($\ce{OTs}$), similar to $\mathrm{E2}$ mechanism.
During the path $a$ mechanism, $\ce{H2O}$ attack the carbon where $\ce{OTs}$ originally was. Therefore, the carbon bearing $\ce{SCH3}$ group does not change its (S)-configuration. However, the $\ce{H2O}$ attack happens in the opposite face of $\ce{SCH3}$ group, the (S)-configuration retain in the carbon bearing $\ce{OH}$ (as in the original compound).
However, both stereochemistry would be reversed during the path $b$ mechanism (see the scheme). Accordingly, $\ce{H2O}$ attack the carbon where $\ce{SCH3}$ originally was, which has (S)-configuration. Since the $\ce{H2O}$ attack happens in the opposite face of $\ce{SCH3}$ group, the (S)-configuration would be changed to the (R)-configuration in this carbon, now bearing $\ce{OH}$. (as in the original compound). The second carbon, now bearing $\ce{SCH3}$ group, would not change its (R)-configuration (as in the cyclic intermediate).
As a result, you get racemic mixture.


Rates of hydrolysis in bicyclic and monocyclic compounds Which of the following compounds, 1-bromo-1-methylcyclohexane 1 or 1-bromobicyclo[2.2.2]octane 2, hydrolyzes faster (under comparable conditions)? It may be acid catalyzed or base catalyzed hydrolysis, but the conditions must remain the same for each pair.

In my view because of the steric hindrance in first compound, hydrolysis should be easier for second, but I am not sure about it.
See part one of my question here.

I’m going to shamelessly steal electronpusher’s drawing for clarification:

Two general hydrolysis mechanisms are possible: $\mathrm{S_N2}$ and $\mathrm{S_N1}$. First, let us consider $\mathrm{S_N2}$ since the analysis is more simple.

Both of the two species are tertiary bromides meaning that $\mathrm{S_N2}$ reactions are already severely inhibited. However, the second compound is even worse: if the nucleophile tries to attack from behind to interact with the $\sigma^*_{\ce{C-Br}}$ orbital, it would need to come from inside the bicyclo[2.2.2]octane system. This is completely impossible, not only due to sterics. Thus, while compound 1 is very slow in $\mathrm{S_N2}$ reactions, it is not possible for 2 to react in an $\mathrm{S_N2}$ manner.

Conclusion: 1 should react faster by $\mathrm{S_N2}$ — albeit still unmeasurably slow.

To undergo an $\mathrm{S_N1}$ reaction, first the halide must dissociate to give a carbocation. Both species are, as noted above, tertiary, meaning that carbocations are well stabilised by triple hyperconjugation. Initially, we expect both $\mathrm{S_N1}$ mechanisms to be fast.

A carbocation when formed must practically always adopt a planar, $\mathrm{sp^2}$ hybridised configuration. Thereby, the unoccupied orbital has the highest energy and all bonding electrons are as low as possible in energy. Any distortion away from the perfect $\mathrm{sp^2}$ geometry requires the empty orbital to be stabilised at the cost of populated orbitals — very unfavourable.

For bridgehead carbons as in 2, especially those that bridge small rings (which is the case here), the general rule is that they do not undergo reactions that require planar transition states for exactly that reason: they are too strained to adopt an $\mathrm{sp^2}$ configuration. (This configuration would not only require the bridgehead to ‘move inwards’, but would also strain the bonds of the three neighbouring carbons, whose angles would have to be decreased.)

Conclusion: while 1 reacts rapidly via $\mathrm{S_N1}$, compound 2 does not react via $\mathrm{S_N1}$.

Complete conclusion: 2 is unreactive towards hydrolysis. 1 therefore reacts much faster.


How to choose alcohol solvent for organic extraction? Assuming I have an organic substance that is known to be "soluble in alcohol" (e.g. capsaicin) I want to extract, which alcohol should/can I use? I know that both ethanol and methanol are commonly used, but what about:

isopropanol
glycerol
n-Butanol
1,4-Butanediol
tert-Butyl alcohol
1-Pentanol
tert-Amyl alcohol

In general, which alcohols are suitable/unsuitable for organic compound extraction (e.g. which should be the best solvent in terms of amount of solute dissolved in given volume of solvent), and why? Is there any a priori way to determine e.g. if isopropanol extraction will be OK for a given "soluble in alcohol" organic substance?

Note: I am aware of this discussion, but it doesn't cover the specific nature of alcohol extraction.

The optimal solvent will extract the greatest amount of desired product and the least amount of difficult-to-remove substances.

The extraction is only the first step; separation from solvent and other extractives follows. Try one, then another solvent to see if there is a difference. Maybe number of carbons or hydroxyls is important; you can also mix alcohols, and/or add water.

Capsaicin has a phenolic group; perhaps it would be extracted from its source with alkali (NaHCO3 or Na2CO3).


Why does microwaving napalm make it nonflammable? In this video, microwaving napalm made it nonflammable. What caused this to happen?

I assume it was some sort of chemical reaction. Could someone explain? I understand some chemistry, but a not-too-technical answer would be appreciated.

In a very similar video I found, some gentlemen microwave gasoline, the flammable component of napalm, along with aluminium foil. The aluminium sparks when exposed to the microwaves, igniting the gasoline vapors in the microwave. Once all the vapor in the microwave is combusted, the microwave must then once again fill with gasoline vapor. When it once again comes into contact with the ignition source, the vapor combusts again.

Since the gentleman in the video you used did not use an ignition source in the microwave, the gasoline in the napalm vaporized, but did not ignite. The vapor filled the microwave, and what was left was the much less flammable gelatin.


Stability of Chromic Acid I am told that chromic acid is exceedingly unstable and that it cannot be purchased as a pure reagent.

Is that so?

I don't find any mentions of stability or lack thereof on Wikipedia, and hey, if I wanted to, I could purchase it here:

http://www.grainger.com/product/LABCHEM-Chromic-Acid-8FV43

So what's the deal with having to make chromic acid in situ? Does this have to do with chromic acid's stability? Or something else?

EDIT: oops I realized that the linked reagent for purchase is just chromic trioxide, which can be used to make chromic acid.

"chromic acid" is also a nickname for chromic trioxide.  It is stable enough to be sold commercially.  Chromic acid in the strict sense ($\ce{H2CrO4}$) is not sold commercially, I think.  For most purposes what you're looking for is Cr(VI) plus acidity, so the usual practice of adding concentrated sulfuric acid to sodium or potassium dichromate (which have multiple uses and can be found in every lab) is preferable to having to buy another reagent.


Confusion with definitions of mass What is the difference between:
atomic mass
relative atomic weight
relative isotopic weight
standard atomic weight
atomic weight
relative atomic mass
I am told by Wikipedia that:
relative atomic weight = atomic weight
and a the start of the article it says:
For a single given sample, the relative atomic mass of a given element is the weighted arithmetic mean of the masses of the individual atoms (including their isotopes)
so standard atomic weight = relative atomic weight
But at https://en.wikipedia.org/wiki/Relative_atomic_mass#Standard_atomic_weight#Definition it says:
It is a synonym for atomic weight, though it is not to be confused with relative isotopic mass. Relative atomic mass is also frequently used as a synonym for standard atomic weight and these quantities may have overlapping values if the relative atomic mass used is that for an element from Earth under defined conditions. However, relative atomic mass (atomic weight) is still technically distinct from standard atomic weight because of its application only to the atoms obtained from a single sample; \textbf{it is also not restricted to terrestrial samples, whereas standard atomic weight averages multiple samples but only from terrestrial sources}
I am told here that:
The standard atomic weight (Ar, standard) of a chemical element is the weighted arithmetic mean of the relative atomic masses (Ar) of all isotopes of that element weighted by each isotope's abundance on Earth.
But in this video I am told this is the definition of relative atomic mass.
Here I am told:
Atomic mass (ma) is the mass of an atom. A single atom has a set number of protons and neutrons, so the mass is unequivocal (won't change) and is the sum of the number of protons and neutrons in the atom.
Atomic weight is a weighted average of the mass of all the atoms of an element, based on the abundance of isotopes. The atomic weight can change because it depends on our understanding of how much of each isotope of an element exists.
In other words:
Atomic weight = standard atomic weight
Also, which of these definitions is used to convert grams of a substance into moles

Relative Atomic Mass vs Atomic Weight
Relative Atomic mass ($m_a$) is the measurement of the mass of an atom relative to carbon-12. It's dimension is M.[1] It can be measured in Da/u, kg or g.* [3]
$$m_a \approx \mathrm{Mass\ Number} \times {1\ \mathrm{u}}$$
except for carbon-12 where they are the same.
Atomic weight is:
Relative atomic mass vs Standard atomic mass
Relative atomic mass or Relative atomic weight is the weighted average of the isotopes of an element according to their abundance in a given sample compared to the atomic mass of carbon-12. It's dimension is M/M or 1. [2] [4]
Standard atomic mass is the same except that the samples are restricted to terrestrial samples normally considering/averaging results from many different samples or results given by the CIAAW.
Relative Isotopic Mass vs Relative Atomic Mass
Relative Isotopic mass considers the mass of an isotope compared to the atomic mass of carbon-12. This is also dimensionless.
*Universal mass constant vs kg , g ,etc
The universal mass constant ($m_u$) is defined as $\frac{1}{12}$ of the mass of carbon-12 which defines 1 Da or 1 u. All measurements in Da are comparisons to the mass of a carbon-12 atom. Since the discovery of Avogadro's constant, we know: [3]
$$ 1\ {\rm{Da}} = 1.660\ 539\ 066\ 60(50)\times 10^{-27}\ \mathrm{kg}$$
Molar Mass vs Molecular Mass (or Formula Mass)
Molar Mass is the grams of a substance in 1 mole of it measured in gram per mole. It's dimension is $\mathbf{M N^{-1}}$.
Molecular mass or molecular weight is the sum of the atomic masses of all the component atoms of each element measured in u. It's dimension is M. Formula mass or formula weight is the sum of the atomic masses of component ions (not sure of the difference between this and molecular mass - in terms of quantity).[2]
[1] see the answer to this Q by M Farooq
[2] What is the difference between "molecular mass", "average atomic mass" and "molar mass"?
[3] Mass number, (relative) atomic mass, average mass
[4] Quick and simple explanation of molar mass, molecular mass and atomic mass


Calculating the energy used during electrolysis of NaCl Suppose we are getting the necessary $\ce{Na}$ for $\ce{HCl}$ out of $\ce{NaCl}$. We need to end up with $V = 100~\mathrm{m^3}$ of $\ce{HCl}$ at standard conditions. Assuming that $\ce{HCl}$ is an ideal gas, and using the ideal gas law, I found there are $$ n = 4272~\mathrm{mol}$$ of matter in those $100~\mathrm{m^3}$. The applied voltage to the cell is $$U = 3.6~\mathrm{V}$$ and the current utilization factor is $\mu=88\%$.
Then we also have Faraday's law of electrolysis:
$$\frac{m}{M}=\frac{q}{F}$$(the oxidation state of chlorine is $z=1$).
Since there is a constant voltage being applied to the cell, the energy used in this reaction is
$$E = \frac{UI \Delta t}{0.88} = \frac{Uq}{0.88} = 470~\mathrm{kWh}$$
However the correct answer is $$E=48.9~\mathrm{kWh}.$$ Where did I go wrong? Perhaps it's something blatantly obvious.

Let's reformulate your equations a little bit.
The ideal gas law was just fine:
$$n = \frac{p V}{RT}$$
A problem lies in your formulation of Faradays law.
We are not interested in masses, but in moles. Of course it is interchangeable by using the molar mass, but why make things complicated. As a rule of thumb you can remember for such tasks: Convert from mass in the beginning and convert to mass in the end (if needed). Nearly all equations in "the middle" are based on the concept of fixed relationsships between the number of involved particles and all equations are much easier to understand if you use $n$.

Faradays law means that the overall charge is just the product of charge per particle times the number of particles:
$$
Q = nzF
$$

The energy is just the product of $Q$ and $U$ divided by $\mu$. You don't need time and amperage.
This gives:
$$
E = \frac{QU}{\mu} = \frac{zFpVU}{RT\mu}
$$

Unfortunately when putting values into these equations you get another result. The following python code will give 448.4 kWh as result. I think that your result is correct and our difference results from differently used constants...:

constants = {}
constants['F'] = 96485.33289
constants['R'] = 8.314
conversion = {}
conversion['to kWh'] = 2.7777e-07

n = lambda p, V, T : p * V / (constants['R'] * T)
Q = lambda n, z : n * z * constants['F']
E = lambda Q, U, mu : Q * U / mu
# Using p=1.01325*10^5 Pa, V=100m^3, T=298K, z=1, U=3.6V, mu=0.88
print(E(Q(n(1.01325e5, 100, 298), 1), 3.6, 0.88) * conversion['to kWh'], 'kWh')


What is the formula for finding the average atomic mass of an element based on the abundance of its isotopes? In my class my teacher showed us how to find the average atomic mass of an element with a method, but he didn't really state a formula one could use. I came  up with a formula of my own, and from what I observed it works:
$$
\frac{i_1x + i_2y}{100} = A,
$$
where $A$ is the atomic mass,
$i_1$ is the first isotope's atomic weight,
$i_2$ is the second isotope's atomic weight,
$x$ and $y$ are the percentages of the isotopes, respectively,
and they add up to 100, i.e. $x+y = 100$.

Therefore: $$
\frac{i_1x + i_2(100-x)}{100} = A
$$

I'm just wondering, is there an official formula, or is this method it?

The average relative atomic mass of an element comprised of $n$ isotopes with relative atomic masses $A_i$ and relative fractional abundances $p_i$ is given by:

$$
A = p_1 A_1 + p_2 A_2 + \dots + p_n A_n = \sum\limits_{i=1}^n p_i A_i
$$

For example carbon:

\begin{array}{lrrr}
\text{Isotope} &
\text{Isotopic Mass $A$} &
\text{Abundance $p$} &
A\times p\\\hline
\ce{^{12}C}:   & \pu{12.000000 u} & 0.98892 &= \pu{11.867 u}\\
\ce{^{13}C}:   & \pu{13.003354 u} & 0.01108 &= \pu{00.144 u}
\end{array}

Since we have $p_1A_1$ and $p_2A_2$, we add those together to find $A$, therefore
 the chemical relative atomic mass of carbon is
$$A = \pu{00.144 u} + \pu{11.867 u} = \pu{12.0011 u}.$$


Why an asymmetric geometry with sp3d and sp3d3 hybridization? A molecule in which the central atom is $sp^3d^2$ hybridized ($\ce{SF_6}$), for example, acquires an octahedral shape, which can be explained by assuming that the hybrid orbitals arrange themselves to minimize repulsion amongst themselves. An octahedron makes sense, it the only way to symmetrically arrange six hybrid orbitals on an atom. The case for an $sp^3$ hybridized atom is similar which from a similar logic can be explained to have a tetrahedral geometry.

But consider the case of $sp^3d$ hybrid orbitals which arrange themselves in a trigonal bypyramidal shape ($\ce{PCl5}$). This is a gross breach of the usual adherence to symmetry. Two bonds are at the axial location and three at the equatorial location. In fact, the axial bonds are slightly longer than the equatorial ones. The case for $sp^3d^3$ hybridization is similar ($\ce{IF7}$) with five orbitals crammed into the equatorial plane and two sticking out above and below it. Why are these arrangements preferred in favor of one with a simple symmetric arrangement of orbitals (which, I think, would minimize the repulsion)?

As already stated in the comments, there is a considerable debate as to whether d-orbitals can be considered to take part in main group elements’ hybridisation at all.

d-Orbital hybridisation was invented by Linus Pauling to explain the bonds in sulfate and sulfur trioxide because he didn’t like the idea of separating charges. However, d-orbitals are far removed energetically compared to s- and p-orbitals — also the reason why 4s is filled before 3d. Technically, before a d-orbital hybridisation can be considered, the next shell s-orbital should be hybridised in first, following the path of lowest energy.

Conveniently, all the cases you presented in your question can be adequately explained using only s- and p-orbitals. See the links for $\ce{SF6}$, $\ce{PCl5}$ and $\ce{IF7}$ respectively. Many thanks to ron for providing them in his comments!


What chemical is responsible for the distinictive odor of Scotch tape? Scotch Tape has a distinctive odor which is noticeable if you hold a piece close enough to smell. What chemical is responsible for this odor? The wikipedia page doesn't seem to mention composition.

Is the odor due to the the kind of plastic used in the strip of tape itself? Or is it due to the adhesive?

This is a long answer, but according to this:

  While some pressure sensitive tapes are still prepared with natural
  rubber, the majority are now made using mostly synthetic materials.
  The backing for cellophane tape usually consists of cellulose acetate,
  a synthetic derivative of cellulose, which comes from wood pulp or
  cotton seeds. The cellulose is chemically treated with acetic acid and
  anhydride, and the side that won't receive an adhesive coating is
  treated with a release agent that enables the tape to be wound and
  unwound without sticking together. Although this compound varies among
  manufacturers, some commonly used substances include stearato chromic
  chloride and polystearic carbonate. Prior to the application of the
  adhesive, the adherend side of the backing may be primed with a
  solvent or aqueous dispersions such as nitrile rubber or chlorinated
  rubber. To produce the final adhesive substance, some manufacturers
  use as many as 29 raw materials that go through various stages of
  production. However, the generic adherend is made up of acrylic
  resins, petroleum byproducts that are broken down into alcohols and
  acids before being fused into a polymer compound. This compound is
  then mixed with mineral spirits or a hydrocarbon solvent, creating an
  aqueous emulsion (a solution in which the microscopic resin particles
  are held suspended) that is applied to the backing.

In other word, there are many candidates.  I consider the top possibilities of those listed above to be: compounds composing the cellulose, the nitrile-based rubber compounds, and the semi-volatile organics composing both the adhereant and the base material.  With so many compounds going into a single product, it is likely the mixture that gives the distinctive smell rather than any individual compound.


Why doesn't this β-hydroxycarboxylic acid undergo dehydration upon heating? This is a β-hydroxyacid, but why doesn't it get dehydrated to give an α,β-unsaturated acid on heating?

The mechanism which I think should occur is:

In order to obtain an α-β-unsaturated carbon acid from a β-hydroxy acid, you need a secondary or tertiary carbon in α position, because otherwise you would have to break a carbon-carbon bond, which requires a lot more energy then simply heating.

The mechanism is basically self catalysed, which is why this may happen easily with a little bit of heat. The acid may protonate intramolecular (or intermolecular) the hydroxy group. You have water as a nice leaving group, creating a carbocation, and abstraction of a proton in α-position leads to the olefin.

If you do not have that α-hydrogen, you are stuck with the carbocation. You might be able to obtain a lactone,like Waylander suggests, but I doubt this would easily happen with a little bit of heating, as spiro-compounds with four-membered rings tend to be quite unstable.

In your attempt you choose the less acidic hydrogen to transfer and then break a carbon-carbon bond resulting in a very unstable molecule which would rearrange to cyclohexanecarboxylic acid. I'd suspect you would need insane temperatures to overcome the activation barrier.


In an SN1 reaction, 1-bromohexane reacts with NaOH, but would the Na or H be the nucleophile that substitutes the Bromine? The answer says that the Bromine would be replaced with OH and NaBr would be formed, along with CH3(CH2)4CH2OH, but why? Does it have to do with stability? And if so, how does the OH replacing Br make the product more stable? Thanks.

SN1 is very less likely to be followed in 1-bromohexane, rather a SN2 pathway will be followed. OH- replaces Br as Br- as Br is a good leaving group. This has nothing to do with product stability. The haloalkane substrate is reactive so it reacts, as simple as that.


Does the priority go for the longer substituent, the closer one, or one which contains a pi-bond in naming Alkenes and Alkynes? According to the new IUPAC nomenclature, the longest chain is given the priority over that which contains the most number of double bonds.

In naming this alkene:

From which end would we start numbering? Because of what substituent?

You have already correctly identified the parent structure to be the longest chain, i.e. dodecene. As usual, the most important simplified criteria for the numbering of locants for substituent groups in such compounds are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds (first to double bonds if there is a choice)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
| (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The compound that is given in the question doesn’t have a principal characteristic group that is expressed as a suffix. Therefore, Rule (c) is not relevant in this case.
A low locant is assigned to the double bond according to Rule (e). In this case, this rule always results in the same locant for the double bound (the ‘ene’ ending); it doesn’t matter at which end of the chain you start counting; i.e. the name of the parent structure is dodec-6-ene.
Finally, low locants are assigned to all the substituents as a set according to Rule (f). Therefore, the systematic name is 8-ethenyl-3-ethyl-5-propyldodec-6-ene rather than ‘5-ethenyl-10-ethyl-8-propyldodec-6-ene’ since the locant set ‘3,5,8’ is lower than ‘5,8,10’.


How is a side-group that contains a double bond named? In my chemistry lecture, we are working on the naming and nomenclature of organic molecules. When we got to the naming of alkenes and alkynes, we were told that the functional group had to be in the main, name giving chain of carbons. This brought up the question: What if there are multiple functional groups arranged in such a fashion as for it to be impossible for them to all be on the main chain (see example below)? How would one go about naming such a molecule? I would much appreciate any clarification that can be provided. I have already checked the IUPAC rules, and unless I totally missed it, it isn't addressed.

Here is the Picture of the molecule in question:

$\hspace{45 mm}$

Thanks in advance for any light you can shed on this question.

When there is a choice among acyclic parent structures, a principal chain must be chosen. The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) stipulates various criteria for the selection of the principal chain.

You are right in so far as functional suffixes that express the characteristic group of a compound (e.g. the suffix ‘ol’ for alcohols) are always attached to the name of the parent structure (i.e. the principal chain). However, since the given structure does not include any characteristic group that is expressed as suffix, the order of seniority of suffixes is not relevant in this case.

The first relevant criterion in this case provides that the principal chain has the greater number of skeletal atoms; i.e., the principal chain is the longest chain (see Subsections P-44.3 and P-44.3.2 in the Blue Book).
Only if this rule does not effect a choice of a senior parent structure, the principal chain has the greater number of multiple bonds (see Subsections P-44.4.1 and P-44.4.1.1).
(Note that the order of seniority between length of chain and unsaturation given in earlier IUPAC recommendations was reversed. Thus, the first criterion to be considered in choosing the preferred chain was unsaturation; the length of the chain was the second criterion.)

In this case, the longest chain has eight carbon atoms; i.e. the unmodified parent hydride is octane. (Note that two equivalent chains with eight carbon atoms exist in the given structure.)

Then

  P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.
  (…)

Therefore, the correct name of the parent hydride is oct-3-ene.

The preferred names of the substituent groups are derived from the respective parent hydrides. The presence of free valences formally derived from the loss of one, two, or three hydrogen atoms from a parent hydride is denoted by replacing the ending ‘ane’ of the parent hydride name by ‘yl’, ‘ylidene’, and ‘ylidyne’, respectively (see Subsection P-29.2).

Therefore, the preferred names for the substituent groups of the given structure are
methylidene $(\ce{=CH2})$,
ethylidene $(\ce{=CH-CH3})$, and
ethyl $(\ce{-CH2-CH3})$.

With regard to numbering of locants within the same compound class, multiple bonds have seniority over simple substituent groups (e.g. methyl, ethyl, etc.) that are used for naming of branched alkanes.

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (e) saturation/unsaturation:

  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (…)

Note that Rule e takes precedence over Rule f.

Therefore, the structure given in the question is named as 3-ethyl-5-ethylidene-6-methylideneoct-3-ene rather than 6-ethyl-4-ethylidene-3-methylideneoct-5-ene since low locants are given first to multiple bonds (i.e. ‘ene’) and the locant ‘3’ is lower than ‘5’.

Furthermore, note that the double bond in the ethylidene group gives rise to cis-trans isomerism. As drawn, the configuration is ‘E’.


Why do metallic elements have a very small band gap while nonmetallic elements have a large band gap? I recently was going over solid state bonding and conductivity. I was wondering why metallic elements (such as sodium, iron, etc.) have a small band gap / no band gap (as conductors that have the fermi level in the overlap between the valence and conduction band) while elements such as carbon have such large band gaps. What distinguishes carbon and sodium, for example, such that carbon is an insulator while sodium/iron is a conductor of electricity? If you could explain in terms high school to undergraduate level chemistry, that would be great.

The energy of the orbitals may be the first point to explain it at this level. You can use the example of a simple 1-D LCAO to form the bands(check page 723 of Atkins physical chemistry 8th edition). For instance, the energy difference between the bands formed by s orbitals and p orbitals will be small or large depending on the atom.


Why is the Debye Hückel Limiting Law called the "limiting law"? I could not find any source which explains this point well.

I looked up Atkin's physical chemistry and Castellan's physical chemistry book but could not quite understand the logic bhind it being called the limiting law.

The Huckel limiting law allows us to determine the activity coefficient of an ion in a dilute solution of a known ionic strength.

Here's an appealing answer which makes perfect sense to me that I found on EveryScience

  The name 'limiting law' is applied because in the limit of arbitrarily low molalities (ie as the concentration of the solute gets
  closer and closer to zero), all solutions are expected to behave in a
  manner consistent with the expression.


Does the position of the functional group on the terminal carbon change the nomenclature? Consider two alcohols:

$$\rm{CH_3 - CH_2 - OH }$$

and

$$\rm{CH_3 - CHOH - H }$$

In the second alcohol, the hydroxyl radical instead of sticking out from the molecule, is bonded below the terminal carbon, like this
$$\rm{H}$$

$$|$$
$$\rm{R- C- H}$$
$$|$$
$$\rm{OH}$$

Now, this doesn't affect the nomenclature (as far as I know), both will be called $\rm{1-ethanol}$. Clearly, these are isomers. But IUPAC nomenclature is supposed to be completely unambiguous. Or did I go wrong somewhere and there actually is a difference in their naming?

PS: Is there any way to produce structural formulas more efficiently in latex on chemistry stackexchange?

They are not isomers, they are the same molecule, ethanol (the "1" is superfluous, there is no 2-ethanol), CH3CH2OH


Is phosphorine (C₅H₅P) aromatic? Phophorine seems aromatic as it has 6 conjugated electrons. But the answer given is that it is not. This seems odd since pyridine has a similar structure and is also aromatic. Thus I ask is phospohrine aromatic or not?

Phosphorine (IUPAC: phosphinine) actually has aromatic character nearly as great (88%) as that of benzene.  According to the reference, phosphorine is sufficiently stable to be handled without air-free techniques; and it undergoes electrophilic substitution reactions similar to those of benzene.


How can I know if there is a phospholipid (phosphatidylcholine) in a solution? I have a solution which probably contains a phospholipid (phosphatydylcholine), I would like to know if there is an easy method to show the presence of the phospholipid.

I came across TLC, but I couldn't perform it since it's too difficult and expensive to buy chloroform.
So, If you guys know an easy method to show the presence of phospholipids please reply.

Your question kind of asks two different things:

1) Whether there is an easy method to show the presence of hosphatydylcholine
and
2) Whether there is an easy method to show the presence of phospholipids in general.

My first comment is that if TLC with chloroform is too expensive, then you are probably out of luck. This is not a simple thing to do, particulary the specific analysis of a single particular phospholipid in the presence of others. This Wikipedia article gives some insight on this:

  There are no simple methods available for analysis of phospholipids
  since the close range of polarity between different phospholipid
  species makes detection difficult. Oil chemists often use
  spectroscopy to determine total Phosphorus abundance and then
  calculate approximate mass of phospholipids based on molecular weight
  of expected fatty acid species. Modern lipid profiling employs more
  absolute methods of analysis, with nuclear magnetic resonance
  spectroscopy (NMR), particularly $\ce{{^{31}P}}$-NMR, while HPLC-ELSD
  provides relative values.

Unfortunately, these methods are going to be orders of magnitude more expensive to do yourself, as compared to TLC, though outsourcing may or may not be a viable alternative in your case.

The same article discusses a phospholipid characterization technique based on refractive index measurements that, again, may or may not be a viable type of analysis for your specific use case.


Redox Combustion Reactions Consider a combustion reaction like the complete combustion of ethane. Here is what I have for the complete combustion as well as the oxidation and reduction:
$$\ce{\overset{-3}{C}_2\overset{+1}{H}_6 + \frac{7}{2} \overset{0}{O}_2 -> 2 \overset{+4}{C}\overset{-2}{O}_2 + 3 \overset{+1}{H}_2\overset{-2}{O}}$$
$$
\begin{align}
\ce{2 H2O + C2H6 &-> 2 CO2 + 10 H+ + 10 e-}\tag{ox}\\
\ce{\frac{1}{2} O2 + 10 H+ + 10 e- &-> 5 H2O}\tag{red}
\end{align}
$$

Why is it that when writing the reduction half equation for the reduction of oxygen that we don't consider oxygen being reduced into carbon dioxide? i.e. why isn't the reduction equation something of the form $\ce{O2 -> H2O + CO2}~?$

Also the oxidation reaction suggests that water is required initially for the reaction to proceed, is this true?

Edit
I realise my reduction half equation is not balanced and so I now have this equation:
$$\ce{\overset{-3}{C}_2\overset{+1}{H}_6 + \frac{7}{2} \overset{0}{O}_2 -> 2 \overset{+4}{C}\overset{-2}{O}_2 + 3 \overset{+1}{H}_2\overset{-2}{O}}$$
$$
\begin{align}
\ce{2 H2O + C2H6 &-> 2 CO2 + 10 H+ + 10 e-}\tag{ox}\\
\ce{\frac{7}{2} O2 + 14 H+ + 14 e- &-> 7 H2O}\tag{red}
\end{align}
$$
which is clearly not correct as the reduction and oxidation half equations do not add together to give the combustion reaction and the reduction equation does not have the $\ce{CO2}$ product as well. I guess my question now stands as how might I write (or begin to write) the half equations for a combustion reaction?
Edit 2
I realise I wrote my equations wrong above
$$
\begin{align}
\ce{4 H2O + C2H6 &-> 2 CO2 + 14 H+ + 14 e-}\tag{ox}\\
\ce{\frac{7}{2} O2 + 14 H+ + 14 e- &-> 7 H2O}\tag{red}
\end{align}
$$
But my question still remains. Why do we not consider the reduction of $\ce{O2}$ into $\ce{CO2}$ as well?

Your redox equations look awkward (and wrong, too) because you are trying apply the half-reactions method as if it  were happening in aquatic media to the combustion reaction, which effectively occurs in the gas phase.
Of course, there should be no requirements for water to be present.
As it's been noted in the comments, ionic half-reactions method is not the most elegant way of find coefficients in this case.

If you really want to use it, I'd suggest to write the elements undergoing redox reactions separately from the corresponding compounds (and yes, we do consider oxygen being reduced into carbon dioxide too):

$$
\begin{align}
\ce{2 \overset{-3}{C} &→ 2 \overset{+4}{C} + 14 e-} &|\cdot 2 \tag{ox} \\
\ce{\overset{0}{O}_2 + 4 e- &→ 2 \overset{-2}{O}} &|\cdot 7 \tag{red} \\
\hline
\ce{4 \overset{-3}{C} + 7 \overset{0}{O}_2 &→ 4 \overset{+4}{C} + 14 \overset{-2}{O}}
\end{align}
$$

resulting in a balanced equation:

$$\ce{2 C2H6 + 7 O2 → 4 CO2 + 6 H2O}$$

Note that in general there is an average oxidation number of carbon involved, and it might as well be fractional.


Finding average Mr of a gaseous mixture A 2 g mixture of cyclohexanol, acetone and pentanal is completely burnt in oxygen. When the gaseous product is passed through $\ce{P2O5}$, the mass of $\ce{P2O5}$ increased by 1.998 g. Find the average molecular mass of the mixture.

$\ce{P2O5}$ absorbs water, so the amount of water absorbed is 0.111 mol. I attempted constructing an equation for the combustion reaction but realized that the proportions and thus the mole ratios of each gas are unknown.

I suspect one might utilize the fact that all three compounds – cyclohexanol $\ce{C6H12O}$, acetone $\ce{C3H6O}$ and pentanal $\ce{C5H10O}$ – share the same empirical formula $\ce{(CH2)_xO}$, which is also applicable to the mixture of reactants, and so instead of writing a set reaction equations

\begin{align}
\ce{C6H12O + 8.5 O2 & -> 6 H2O + 6 CO2} \tag{R1}\\
\ce{C3H6O + 4 O2 & -> 3 H2O + 3 CO2} \tag{R2}\\
\ce{C5H10O + 7 O2 & -> 5 H2O + 5 CO2} \tag{R3}
\end{align}

one can avoid finding volume fraction $\phi_i$ of each individual component and use a single reaction scheme such as

$$\ce{(CH2)_xO + $\frac{3x-1}{2}$ O2 -> x H2O + x CO2} \tag{R4}$$

and find relation between average molecular weight $\bar{M}$ and $x$ solely from (R4). Lets denote the mass and the total amount of gas mixture components as $m$ and $n$, respectively. Then according to (R4) total amount of gas mixture is $n = \frac{n(\ce{H2O})}{x}$, and the average molecular weight is established as

$$\bar{M} = \frac{m}{n} = \frac{x \cdot m}{n(\ce{H2O})} = \frac{x \cdot m \cdot M(\ce{H2O})}{m(\ce{H2O})} = \frac{x \cdot \pu{2 g} \cdot \pu{18.015 g mol-1}}{\pu{1.998 g}} = (18.033 x) \,\pu{g mol-1} \tag{1}$$

At the same time using molecular weights of the elements:

$$\bar{M} = x \cdot (M(\ce{C}) + 2M(\ce{H})) + M(\ce{O}) = (14.027 x + 15.999)\,\pu{g mol-1} \tag{2}$$

Equating (1) and (2):

$$18.033 x = 14.027x + 15.999 \tag{3}$$

and solving the resulting equation (3) for $x$, one finds that $x = 3.994$. Solving either (1) or (2) with obtained $x$ yields in average molecular weight $\bar{M} = \pu{72.023 g mol-1}$.


Is the given statement regarding Radial Probability Distribution Function correct or not? Radial Distribution Function $(4πr^2R^2(r))$ gives the probability of the electron being present at a distance $r$ from the nucleus.

Answer: The given statement is correct.
My Query:
According to me, isn't the Radial Probability Distribution Function ($RPDF$) used for finding the probability of an electron being present at a distance $r$ from the nucleus within a region of thickness  $dr$?
Therefore, shouldn't the above statement be wrong?

Conceptually you are right as the commenters have mentioned, but since we are on a thread about nitpicking, we might as well go the extra distance.
Technically, $r^2R^2$ itself is not a probability but a probability density. In order to get the actual probability, you need to integrate it over a region. The probability of finding an electron between $r = r_1$ and $r = r_2$ would therefore be
$$P = \int_{r_1}^{r_2}r^2R^2 \, \mathrm dr \tag{1}$$
For an infinitesimal region, it suffices to multiply by $\mathrm dr$, such that the infinitesimal probability of finding an electron in the region $(r, r + \mathrm dr)$ is
$$\mathrm dP = r^2R^2 \,\mathrm dr \tag{2}$$
Notice how if you integrate $(2)$ over a finite region you get $(1)$.
Also, some books give this extra factor of $4\pi$, which I never really understood. You have two choices: either the probability density function is $r^2R^2$ (which works for all orbitals), or $4\pi r^2\psi^2$ (which works only for spherically symmetric orbitals, i.e. s-orbitals). I've written about this in more detail at  What is the exact definition of the radial distribution function?, so won't repeat myself here.


Positive deviation from ideal behaviour I am confused about Positive behaviour from ideal behaviour as to why it occurs.

1) Molecular attractions plays a dominant role in positive deviation

OR
2) Molecular volume plays dominant role in positive deviation

I am confused these two facts.

If in a binary solution of two components A and B, the interaction between unlike components A-B is weaker than the interaction between the like components A-A and B-B, then it means that in such solutions molecules of A (or B) will find it easier to escape in the state A-B, than when they are in their pure states (A-A or B-B). It means that molecules of A (or B) will find it easier to escape from the A-B form.

This will result in having greater vapour pressure of each component of the solution than expected on the Basis of Raoult's law, and hence the total vapor pressure will also be higher than in the case of ideal solution. Thus the deviation will be positive.

In such solution $\Delta_{mix}H$ is positive because energy is required to break A-A and B-B bonds. For such solutions, the dissolution process is endothermic, i.e, the solubility will increase with temperature (Le Chatelier's principle) .

$\Delta_{mix}V$ is also positive because there is decrease in the magnitude of intermolecular forces in the solution, the molecules are loosely held, and therefore the volume of mixing increases.


Can we synthesize alcohols by the photolysis of hydrogen peroxide? As I was going through my revision of organic chemistry, I came across free radical substitution of alkanes with halogens. In a similar process where we use hydrogen peroxide, instead of chlorine, so that photolysis would yield a hydroxide radical $\ce{OH.}$ instead of chlorine radical $\ce{Cl.}$. The rest of the steps would be the regular mechanism of This way we can get an alcohol instead of alkyl chloride. Here is what I mean:
\begin{align}
  \ce{H-O-O-H &->[h\nu]H-O. + .O-H}\\
  \ce{CH3-CH3 + .O-H &-> CH3-CH2. + H2O}\\
  \ce{CH3-CH. + .O-H &-> CH3-CH2-OH}\\
\end{align}

Does my hypothesis work in reality?

I am aware of the fact that dialkyl peroxides cannot be used since they are highly explosive. However, hydrogen peroxide isn't that explosive, so I think that its safe to use it. Since $\ce{H2O2}$ is a liquid, while alkanes and $\ce{Cl2}$ are gases, the reaction with $\ce{H2O2}$ could be done at low pressure. We could photolyse the $\ce{HO-OH}$ bond using light of a sufficient frequency, which probably should initiate a free radical reaction.

I did a few calculations and got the maximum wavelength of light needed to photolyse the $\ce{O-O}$ bond is $\pu{0.0084m}$, which is pretty low. So I think it's possible to photolyse $\ce{H2O2}$ and carry on the reaction.

I do not believe so.

The mechanism of decomposition of $\ce{H2O2}$ is:

$\ce{H2O2(g) -> 2HO.(g)}$

$\ce{H2O2(g) + .OH(g) -> H2O(g) + HOO.(g)}$

$\ce{HOO.(g) + .OH(g) -> H2O(g) + O2(g)}$

$\ce{.OH(g), HOO.(g)}$ are both radical compounds.

If you take a look at the first step in the mechanism, it is the rate limiting step and it is the same as your first step in the mechanism, the photolysis step.  Therefore if $\ce{H2O2}$ undergoes partial photolysis all that will happen is that the sample will undergo decomposition quicker.

Additionally, your mechanism of oxidation has been proposed before.  You can take a look at this article, in the section Mechanism of Oxidation by Peroxides: Radical or not Radical, That is the Question, in the end of subsection 4.1 and 4.2 they refer to the mechanism you proposed.  They end up proving that it proceeds through a concerted mechanism that requires the presence of a suitable metal catalyst.  It involves oxidative insertion or some type of $\ce{C-H}$ insertion by the metal followed by oxidation of the product of that reaction.

But say the generation of hydroxyl radicals does work out, the corresponding alcohol may not be the only product that forms.  Hydroperoxides have a tendency of forming as well.  This is because hydrogen peroxide tends to decompose into oxygen and water, the oxygen can interact with the alkyl radical to form hydroperoxides.  Along with the fact that the alcohol formed will tend to undergo oxidation in the presence of hydrogen peroxide and hydroxyl radicals.

Also if you are still going to try the reaction, the bond dissociation energy of the $\ce{O-O}$ single bond is around 146 kJ per mole. So the energy per molecule is $\ce{2.433\times10^{-19} J}$.  So the maximum wavelength of light to ionize this is $\ce{8.21\times10^{-7} m}$.  Your calculation is a little off, it is in the near infrared range (you may have forgot to convert from kilojoules to joules or something like that).  This is why hydrogen peroxide is stored in dark bottles and told to not be put in the presence of light.


Pre heat coal before burning So I was wondering if pre-heating coal will improve the efficiency of a steam boiler.
I did find that intake air pre-heater and water pre-heater do increase efficiency, but I could not find anything related to pre-heating coal.

If we were to take the hot exhaust, make a hot iron platform and keep the coal on top of it, it would heat it and maybe reduce the moisture content also?!

Chemistry troll idea or actually feasible?

Note: it is a traditional hand fired boiler

EDIT:

Coal type: Lignite

Application: Industrial/Factory

The update of your question changes everything!

In Germany, around 25% of the base load is generated in lignite-fired power plants.
Lignite (brown coal) is awfully wet (50% water) and it is definitely a good idea to apply drying.

In the power plants, powdered lignite is used. Traditionally, the milled lignite is dried using the hot exhaust gas at temperatures up to 1000 °C, but modern plant like the one in Niederaußem operated the fluid bed drying at temperatures as low as 110 °C.

In summary, pre-drying of lignite is a good idea, if not essential, but it open to question how efficient the drying process of stacked lignite is.


Empirical and molecular formula I was doing a homework assignment and I got the question wrong because I was supposed to multiply the empirical formula of $\ce{C6H10S2O}$ by 3 to get the molecular formula.

I put $\ce{(C6H10S2O3)3}$ and the teacher said it should be $\ce{C18H30S6O3}$.

My question is: isn't what I put the same thing?

$\ce{(C6H10S2O3)3}$ implies that there are $3 ~ \ce{C6H10S2O3}$ molecules bonded together instead of $1~ \ce{C18H30S6O3}$ molecule. Although they are the same thing, the correct notation is your teachers way because it shows the isometric formula for the molecule. Also if you were using that notation you should have put $\ce{(C6H10S2O)3}$ because $\ce{(C6H10S2O3)3}$ has 6 too many oxygens.


What comes after f-orbitals, and in what order are they filled? I know about s, p, d, and f-orbitals. However, my physical chemistry textbook does not mention what comes after f-orbitals. What are the next letters, and in what order are these higher orbitals filled?

After s, p, d, and f, the next letters follow in alphabetical sequence: g, h, i, k, l, and so on. Note that j is omitted from the sequence: see Why is the letter J omitted in the spdf... sequence? for more information. Since g-subshells would have $l = 4$, the lowest g-subshell would be the $\mathrm{5g}$ subshell, bearing in mind $l$ must be strictly less than $n$.

According to the aufbau principle, orbitals are filled in increasing order of $(n+l)$. Amongst orbitals with similar $(n+l)$, those with smaller $n$ are filled first:

$$\begin{array}{ccc}
\text{Orbital} & \hline (n+l) & n \\ \hline
\mathrm{1s} & 1 & 1 \\ \hline
\mathrm{2s} & 2 & 2 \\ \hline
\mathrm{2p} & 3 & 2 \\
\mathrm{3s} & 3 & 3 \\ \hline
\mathrm{3p} & 4 & 3 \\
\mathrm{4s} & 4 & 4 \\ \hline
\mathrm{3d} & 5 & 3 \\ \hline
\end{array}$$

Following this principle you would theoretically have the order

$$\mathrm{\underbrace{1s}_{\text{Period 1}} \to \underbrace{2s \to 2p}_{\text{Period 2}} \to \underbrace{3s \to 3p}_{\text{Period 3}} \to \underbrace{4s \to 3d \to 4p}_{\text{Period 4}} \to \underbrace{5s \to 4d \to 5p}_{\text{Period 5}} \to \underbrace{6s \to 4f \to 5d \to 6p}_{\text{Period 6}}} \\
\mathrm{\to \underbrace{7s \to 5f \to 6d \to 7p}_{\text{Period 7}} \to \underbrace{8s \to \mathbf{5g} \to 6f \to 7d \to 8p}_{\text{Period 8}} \to 9s \to \cdots}
$$

However, at such large atomic numbers, the aufbau principle starts to fail. The subshells are much more closely spaced in energy and the exact electronic configurations are not so clear-cut: this is already evident within the d- and f-blocks where some anomalies are present.

On top of this, relativistic effects (which become significant at large atomic numbers) cause subshells to split into two distinct energy levels (except for the s-subshell). See also: Shape of the P1/2 Orbital.


What value should I use for P in the ideal gas equation for an unknown molar quantity? I did a chemistry lab, in which baking soda and vinegar were reacted in a ziplock bag, and I was to calculate the proper quantity of reactants to fill the bag with CO2. The molar quantity of gas was found using the ideal gas law, where volume, temperature, and pressure were all known. Pressure, however, was set to the rooms air pressure, and I was wondering why that would be the case. Wouldn't a plastic bag full of gas have a fairly high pressure? If its pressure were in equilibrium with the air, wouldn't none of the gas flow out when the bag is opened?

Strictly speaking, the pressure inside must be certainly higher inside due to the restoring forces generated by the stretching of the plastic bag. It is this force that causes the gases to gush out when punctured or opened.

Bottom line: the assumption of the pressure inside being equal to atmospheric pressure can be considered correct with a considerably large error margin, provided we are talking about a ziploc bag and not the tyre of a tractor.


Why neutrons are neutral? If electrons have a negative charge and protons have a positive charge
then how come neutrons have zero charge without consisting of protons and electrons?

Protons and neutrons are both baryons, subatomic particles composed of three quarks. Quarks are smaller fermionic particles that make up protons, neutrons, and other hadrons.

Electrons are leptons, fermions that are not composed of quarks.

Quarks have mass, spin, charge, and (confusingly) color charge. Color charge is a quantum property that quarks have, and all hadrons need to have a net zero color charge.

The proton is composed of three quarks: two up quarks and one down quark. Each up quark has a spin of $+\frac{1}{2}$, a mass of $2.3~\text{MeV/c}^2$, and a charge of $+\frac{2}{3}$. Each down quark has a spin of $-\frac{1}{2}$, a mass of $4.8~\text{MeV/c}^2$, and a charge of $-\frac{1}{3}$. The total charge of the proton is $2(+\frac{2}{3})+(-\frac{1}{3})=+1$. The total spin of the proton is $2(+\frac{1}{2})+(-\frac{1}{2})=+\frac{1}{2}$. The masses of the quarks do not add up because most of the proton's mass is in its confinement energy.

We can do the same analysis for a neutron, which is composed of one up quark and two down quarks. The charge of the neutron is $(+\frac{2}{3})+2(-\frac{1}{3})=0$. The total spin of a neutron is $(\frac{1}{2})+2(-\frac{1}{2})=-\frac{1}{2}$. Again the masses do not add up.

Protons and neutrons are different because the are composed of different quarks. However... enough physics. Now for the nuclear chemistry.

  Interconversion of protons, neutrons, and electrons

The proton is a stable particle with a lifetime at least $2.1\times 10^{29}$ years.

However, protons can be converted into neutrons (and electron neutrinos) by electron capture:

$$\ce{p+ + e- -> n +\nu}_e$$

This requires an investment of energy because the rest mass of the neutron is higher than the rest mass of a proton by $1.29~\text{MeV/c}^2$.

Free neutrons are not stable, undergoing beta decay to a proton, electron, and electron antineutrino with a halflife of about 10 minutes.

$$\ce{n -> p+ + e- +\bar{\nu}}_e$$

Most importantly, and fortunately, most neutrons bound in nuclei are stable. The protons and neutrons in a nucleus can be treated quantum mechanically analogously to electrons from several atoms bound in a molecule. There are not a collection of independent particles that do not interact. The nucleons occupy quantum mechanical energy levels. Neutrons can only decay if the proton it would become can occupy a lower energy level, which is usually not possible except for those radioisotopes which tend to undergo beta decay, thanks to the Pauli Exclusion Principle! Thus, the bound neutron is stabilized.


Reduction of nitrobenzene with LiAlH4 and NaBH4 After searching a lot on the internet I've found that nitrobenzene is reduced by LiAlH4 to the following compound whose naming I couldn't do.

I couldn't find the mechanism for this reaction anywhere.
Can someone tell what is the mechanism for it?

The structure you provided is known as azobenzene . It is a parent compound in a well-known family of aromatic azo compounds. The compounds usually are not very thermally stable, but often have intense color and are used as light duty dies (or as pH or other indicators) Aliphatic azo compounds are not stable at all and very prone to fragmentation with formation of molecular nitrogen and free radicals and some of them are used exactly for that.

  I couldn't find the mechanism for this reaction anywhere

The reduction of the nitro-group might procede through attack of hydride onto the nitrogen atom or electron transfer onto the group with subsequent reaction of the anion with proton (the exact mechanism depends on conditions and reagents). It can produce a wide amount of monomeric products, in particular: nitrosocompounds, hydroxilamines and amines. It is known, that amines can react with nitrosocompounds forming the $\ce{N=N}$ bond. It looks like the first step in the reaction is a nucleoplilic attack of the amine nitrogen onto nitroso nitrogene with subsequent elimination of water.


What is Q-carbon? Does it exist? Glowing press releases and news articles in 2015 proclaimed a new allotrope of carbon. However, even the journal article is light on chemical detail (e.g. no structural formula).

What is its bond structure?
Why is it ferromagnetic? fluorescent? harder than diamond? etc?
Or is it an erroneous result?

Update: as of late 2020, Bhaumik & Narayan have published further papers about Q-carbon, but I still have not seen any reports of replication by others.

Ref: Jagdish Narayan and Anagh Bhaumik J. Appl. Phys. 2015, 118, 215303.

There's no structural formula because it's what we call an amorphous solid. That is, it's like glass. It has no clearly defined crystal structure, and is likely to be extremely hard, but also extremely brittle.

As to why it's ferromagnetic or fluorescent, I'm unsure. It's likely they don't quite know, either. I suspect it has something to do with the curious bond structure of Q-carbon, which is a mishmash of both 3-way and 4-way bonds. In comparison, graphene is uniformly 3-way bonds, while diamond is a fairly uniform composition of 4-way bonds.

I speculate (don't quote me on this, I can't find a reliable source) that the way the electrons interact across the carbon atoms has something to do with why it illuminates (as photon emission is not an uncommon effect of electrons changing energy levels) and with why it's magnetic (as ferromagnetism explicitly refers to the quantum mechanical properties of electrons).

Either way. Please take this explanation with a grain of salt, but, if you're able to find more reliable sources, feel free to tack it on.