How to arrive at the preferred IUPAC name for 1,2-diphenylethane? I am reading a textbook that mentions the molecule 1,2-diphenylethane, $\ce{C14H14}$. A Google search leads to PubChem, where it is mentioned that the IUPAC name for the molecule is 2-phenylethylbenzene. I'm trying to understand how one names the molecule, using IUPAC nomenclature. However, I'm struggling to find any material that describes how apply IUPAC nomenclature to these "double benzene" (polycyclic aromatic hydrocarbon?) molecules. I would greatly appreciate it if someone could please redirect me to a page that describes this and describe the naming process for this particular molecule.

The essential problem in this case is finding the parent structure. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical units that are linked by di- or polyvalent groups can be named in two ways:

by ordinary substitutive nomenclature, in which one of the parent structures is chosen as the senior parent structure and the remainder of the structure is expressed by substituent prefixes
by multiplicative nomenclature, in which two or more parent structures are connected by symmetrical or unsymmetrical single or concatenated substituent groups

Both methods yield correct names, which may be used in general nomenclature.

1. Substitutive nomenclature

In many related simple cases, the compound consists of just a ring and a chain so that the parent structure could be the ring or the chain.

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

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

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

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

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

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

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

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

These rules apply to compounds that consist of one ring and one chain. The compound that is given in the question, however, consists of a symmetrical chain with two rings. Extending the use of the above-mentioned rules to this case, we could pick one of the rings as the parent structure in accordance with Rule P-52.2.8. This approach would result in the correct name (2-phenylethyl)benzene. Since this method usually yields the preferred IUPAC in many simple cases, it should not be a surprise if some computer-generated names follow this pattern also in this case. Note, however, that the name “2-phenylethylbenzene” suggested in Pubchem is not correct and ambiguous (“2‑phenylethylbenzene” could be interpreted as “(2‑phenylethyl)benzene” or “2‑phenyl‑1‑ethylbenzene”).

In accordance with P-44.1.2.2, the context may favour the chain as parent structure. For the compound that is given in the question, the context may favour the chain in order to highlight the symmetrical shape of the structure, in order to treat the equivalent rings alike, or because a following reaction happens at the chain. This approach would result in the correct name 1,2-diphenylethane.

2. Multiplicative nomenclature

The IUPAC recommendations explain when multiplicative nomenclature is used to generate the preferred IUPAC name (PIN) as follows.

  P-45.1 MULTIPLICATION OF IDENTICAL SENIOR PARENT STRUCTURES

  P-45.1.1 Multiplicative nomenclature is senior to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical senior parent structures, other than alkanes, in the name of the parent structure (see P-51.3.1). In most cases multiplicative names are shorter than regular substitutive names. A preferred IUPAC name is generated by multiplicative nomenclature when the following criteria for its use are met (see P-51.2.3).
  (1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical; and
  (2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
  (3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.

  When these conditions are not met, substitutive nomenclature generates preferred IUPAC names.

The compound that is given in the question fulfils these requirements; therefore, the preferred IUPAC name is generated by multiplicative nomenclature as follows.

  P-15.3.1.3 Multiplicative name formation
  Multiplicative names are formed in accordance with the number of occurrences of identical structural units as defined in P-15.3.1.1 and the relationship of the linking multiplicative substituent group to the identical structural units.
  When a compound contains identical structural units as defined in P-15.3.1.1 linked by a symmetrical simple, compound, complex, or concatenated multiplicative group (a di- or polyvalent substituent group), it is named by stating successively:

  (a) the locants for the positions of substitution of the linking multiplicative substituent atom or group to the identical parent structural unit (the locant 1 is omitted when alone in the name of a mononuclear parent hydride);

Here: 1,1'

  (b) the name of the linking multiplicative substituent atom or group;

Here: ethane-1,2-diyl

  (c) the numerical prefix ‘di’, ‘tri’, etc.; and/or ‘bis-‘, ‘tris-‘, etc., with no elision of the final vowel before the name of the identical parent structural unit;

Here: di

  (d) the name of one of the identical structural units including the principal characteristic group and substituents, if any, enclosed in appropriate enclosing marks (see P-16.5).

Here: benzene

  The numbering of the identical parent structural unit is retained and, when there is a choice, the locants of the point of substitution by the linking multiplicative substituent groups on the identical parent structure are as low as possible.

Therefore, the preferred IUPAC name is 1,1'-(ethane-1,2-diyl)dibenzene.


Is the absorbent substance in nappies biodegradable? I am looking to use the absorbent material found in nappies to improve the water retention of the soil in my garden.

To this end, I have two questions that I have not found definitive answers too.

Are disposable nappies made with Sodium or Potassium Polyacrylate?
Is Sodium/Potassium Polyacrylate biodegradable?

To keep it simple: there are as many different polymers as suppliers of the superabsorbent polymer used in diapers.https://en.wikipedia.org/wiki/Superabsorbent_polymer#Current_synthesis.

So, answer to #1. Some are, some aren't. As far as #2, define what you mean by "biodegradable". Take a diaper, put it into a composting bin, and turn it over every week for a million years, and I'd expect little to remain. It is likely based on your question that you are unaware that some diapers are marketed to be "biodegradable" and some aren't. The reality is that the term has no standard meaning - it's marketing hype. However, it's also true that some polymers will be inherently much more prone to decomposition than others and that some of these are incorporated into some diapers.

Your best bet is to call (or visit) the customer service of the specific diaper manufacturer you are interested about. Note that the nih link provided in a different answer doesn't mention any use of the sodium salt for any major brand.  Pampers (P&G), I think, dominates the market and sells 9 different disposable diaper brands. I didn't see any mention on their site about biodegradability, which was an important controversy some generations back. Since even "highly biodegradable" materials can take centuries to decompose in landfills, the question is a bit meaningless. (If you want biodegradable, use toilet paper (not seriously, since TP is designed to fall apart rapidly, too rapidly for practical use in a wearable product).)


Binding forces in crystalline solids I am referring to rocks in this instance. Individual crystals grow together in an igneous rock as it solidifies. I know that the $\ce{SiO4}$ tetrahedron is the basic building unit of silicate minerals, and metal cations attach to ensure electroneutrality.

What I am wondering, however, is what force keeps the crystals together (as in bound to one another) in the solid? For comparison, sediment grains are held in place by a silica or calcite cement in fragmental rocks, but what stops individual crystals from breaking free in crystalline solids?

Converting to answer: While the energy to introduce a grain boundary into a single crystal depends on the relative orientation of the two resulting grains, there will be some atomic rearrangement (strains or reconstruction), and you will end up with a number of bonds across the grain boundary. The boundaries are usually weaker than the bulk crystal, which is why they preferentially fracture along the grain boundary.


Physical gelation result in a single massive polymer chain? Physically crosslinking PVA polymers using the freeze-thaw method (Peppas 1975) results in regions of interconnected PVA polymers through crystalline regions formed when frozen and which persist after thawing. These physical crosslinks are sufficient to produce a homogenous hydrogel of predictable bulk characteristics. The formation of Jello gelatin uses a similar phenomenon.

My question is:
Are these crystalline regions only available to form at the free ends of the polymer chains, or can these lamellae form along any portion of the polymer chain?

If crystallization were only able to occur at free ends of the chain, this would result in a massively long polymer chain (assuming tip to tail occurred for many polymer chains in the solution) and as such the bulk polymer might be defined as a tangled network.

The physical alignments can occur anywhere. Keep in mind that these regions are both inter- and intra- molecular. (CH4 can't align with itself, but CH3(CH2)1000CH3 can and does.)


Electronegativity of noble gases SO, I stumbled upon a periodic table of electronegativity-

http://www.thecatalyst.org/electabl.html

Now I don't understand why the first 3 noble gases have blank spaces instead of the electronegativities.

If it is because their octet is complete then why doesn't the noble gases Kr, Xe and Rn have blank spaces under them?

They don't have values there because they aren't on the pauling scale of electronegativity, as they don't form any compounds with other elements. However, argon and neon can technically form compounds with other elements; it is just extremely unlikely. For example there is a compound called Argon Fluorohydride, HArF, that can exist, but is only stable at around 27K (-246C). I don't think Helium can form any compounds other than with itself.
The other noble gases also can react with fluorine (As well as other elements), and make slightly more stable compounds. Hope that helps :)


Catalyst that combines oxygen and hydrogen Are there any known catalysts that can combine hydrogen and oxygen into water?

Does this type of reaction have a name? Trying to search for information, it seems to be all about splitting hydrogen and oxygen, is there any work on combining these and some catalysts that could be used?

Yes

Not only is this possible it can be done while extracting electricity from the reaction. This is the key reaction in a hydrogen/oxygen fuel cell.

Fuel cells have been known for well over a century and Hydrogen oxygen fuel cells have been used commercially since the 1930s. Platinum is often used as a catalyst for the key reaction in the cells.


Coupling constant in 1H NMR with phosphorus I observed a doublet with peaks at 5.6336 and 5.5886 ppm with the NMR instrument running at 300 MHz, and got a $J_\ce{PH}= \pu{13.5 Hz}$. Is this correct for this molecule (ethoxycarbonylmethyl(triphenyl)phosphonium bromide)? This value is larger than the hydrogen coupling constants for the C-C bonds. Is this larger coupling constant a consequence of the phosphorous?

Yes, this is the 2JP-H coupling. You could verify this (if you had access to a NMR spectrometer) by running a 1H{31P} experiment; that is, a proton with 31P decoupling. This doublet would collapse back to a singlet. You would also notice some changes to the phenyl protons, as they all have some coupling to the 31P also. A 2J coupling of 13.5Hz is very typical for this type of 1H-31P coupling.

The magnitude of 1H-31P couplings varies enormously, depending on bond order, oxidation state of 31P and bonding arrangement.

1JPH of PH3 = ~175Hz.
1JPH of HP(O)(OMe)2 = ~1200Hz
2JPH P(CHx)y are typically 1-20Hz

It is sometimes observed in 31P-1H (and 31P-13C) that high bond order couplings are larger than lower bond order couplings (ie 3JPH > 2JPH)


Which of the two structures is more stable? Which of the two structures is more stable?

As far as I know, alkenes with a greater number of hyperconjugating structures are more stable. Hyperconjugation has a greater stabilizing effect than inductive effect. So, (d) should be more stable as it has more α-hydrogens.
But the answer given in the book is (c).
Is the answer wrong? Or am I missing something?

The explanation can[1] be given by analysing their respective resonance[2] structures which is as follows:

(Credits: ChemWriter)

Now, inductive effects would work.

In case of first (c's resonance), positive charge would be stabilised by $\ce{+I}$ effect of methyl group; whereas in the second (d's resonance) one, $\ce{+I}$ effect would become dead upto the positive charge.

Hence, less stable would be (d).

Notes:

[1]: Firstly, I'm not sure if hyperconjugation will stabilize carbocation in resonance structures or not, because it will separate the charges even more. Secondly, in the given structures, it is very hard to differentiate the stability, as contribution of inductive and hyperconjugation are resulting in comparable stabilities. Hence, I tried it explaining by other canonical form.

[2]: resonance will happen due to conjugation of $\ce{\pi}$ bonds in both of the molecules


Hybridization of Na in [Na(H2O)6]+ In the complex ion $\ce{[Na(H2O)6]+}$, the sodium cation forms 6 coordinate bonds with water ligands. Typically this octahedral form is associated with $\mathrm{sp^3d^2}$ hybridization as far as I know, but in the case of sodium the $d$ orbitals aren't readily available, so how can this be described as $\mathrm{sp^3d^2}$ hybridization?

I would guess that its hybridization is $\mathrm{s^2p^4}$?

On further thought I think the d orbitals probably are accessible and that it is indeed $\mathrm{sp^3d^2}$ but this has a high hybridisation energy and so we dont see $\ce{[Na(H2O)6]+}$ in large amounts instead we see the ion solvated by partial charges and electrostatic rather than dative covalent interactions

Solvation is the process by which a species is dissolved in a solvent. The most classic example of this is when a metal ion is dissolved in water. The electronegative oxygens in the water molecules are attracted electrostatically to the positive charge on the metal ion. A solvation shell of water molecules result.

According to this article, which written about the hydration of alkali metal ions:

  In spite of many conducted studies the knowledge of the structures and bonding properties of the hydrated alkali metal ions in aqueous solution is scarce and deviating.

Also according to another letter, which reports research done on the stability of the different possible forms of the ion, there is not a consensus on only one form. This letter discusses three possible forms of the ion all in the form of $n_1 +n_2$, where $n_1$ is the number of molecules in the first shell of hydration and $n_2$ is the number of molecules in the second shell of hydration. The forms discussed are:

demonstrating that the water molecules do not necessarily have to be coordinately bonded to the metal itself, but in fact that the second shell of hydration could include water molecules interacting with the water molecules in the first shell of hydration.

In short the hydration of sodium (and the other alakali metals) is not well understood yet, though it is under study.


How do I find the ratio in which the residue of a reaction combines with HCl? A mixture weighing $4.08\ \mathrm g$ of $\ce{BaO}$ ($M(\ce{Ba})$ = $138\ \mathrm{g\ mol^{-1}}$) and an unknown carbonate $\ce{XCO3}$ was heated strongly. The residue weighed $3.64\ \mathrm g$. This was dissolved in $100\ \mathrm{ml}$ of $1\ \mathrm M$ $\ce{HCl}$. The excess acid required $16\ \mathrm{ml}$ of $2.5\ \mathrm M$ $\ce{NaOH}$ solution for complete neutralization. What is the molar mass of $\ce{X}?$

My approach:

$16\ \mathrm{ml}$ of $2.5\ \mathrm M$ $\ce{NaOH}$ means $0.04\ \mathrm{mol}$ of $\ce{NaOH}$. Thus, the excess acid was $0.04\ \mathrm{mol}$ of $\ce{HCl}$. $100\ \mathrm{ml}$ of $1\ \mathrm M$ $\ce{HCl}$ means $0.1\ \mathrm{mol}$ of $\ce{HCl}$. Thus, $0.06\ \mathrm{mol}$ of $\ce{HCl}$ reacted with the residue. Now, how can I find the ratio in which the residue reacted with $\ce{HCl}?$ What is the composition of the residue?

In the comments I mentioned one way to solve the problem, but it turns out that was the roundabout way! There is a simpler solution, as follows.

From the neutralization reaction, it is possible to calculate that the mixture reacted with $0.06\ \mathrm{mol}$ of $\ce{H+}$. The carbonate and the oxide react in a proportion of 1:2 with the hydrogen ions, so we know that $n_{\ce{BaO}}+n_{\ce{XO}}=0.03\ \mathrm{mol}$. The mass balance from the thermal decomposition then suggests that $m_{\ce{XCO3}}-m_{\ce{XO}}=0.44\ \mathrm g$. These two equations are already enough.

Notice that the for the molar masses of the unknown oxide and carbonate, we have $M_{\ce{XCO3}}-M_{\ce{XO}}=44\ \mathrm{g/mol}$. If one converts the masses in the second equation into expressions in $M$ and $n$, the result is $n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XO}}M_{\ce{XO}}=0.44\ \mathrm g$, which can be turned into $n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XO}}\left(M_{\ce{XCO3}}-44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$. It is easy to see that the thermal decomposition does not alter the number of moles of compound containing $\ce{X}$, so that $n_{\ce{XCO3}}=n_{\ce{XO}}$. Thus:

$$n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XCO3}}\left(M_{\ce{XCO3}}-44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$$

$$n_{\ce{XCO3}}\left(M_{\ce{XCO3}}-M_{\ce{XCO3}}+44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$$

$$n_{\ce{XCO3}}\times 44\ \mathrm{g/mol}=0.44\ \mathrm g\quad\Rightarrow\quad n_{\ce{XCO3}}=0.01\ \mathrm{mol}$$

Now the problem unravels easily. We get that  $n_{\ce{BaO}}=0.02\ \mathrm{mol}$, which implies $m_{\ce{BaO}}=3.08\ \mathrm g$ and $m_{\ce{XCO3}}=1.00\ \mathrm g$. Hence, $M_{\ce{XCO3}}=100\ \mathrm{g/mol}$, and from there $M_{\ce{X}}=40\ \mathrm{g/mol}$. A quick look on a periodic table yields calcium as the most likely candidate, and indeed its carbonate has the formula $\ce{CaCO3}$ and a molar mass of $100\ \mathrm{g/mol}$.


Asking about definitions of word "sugar" in biochemistry In a lecture about anabolic pathways of sugar, the lecturer was not clear when stating the name of a multiple sugar carrier & it sounded like "dolichol"so is it correct?

Absolutely correct!

  N-Linked oligosaccharides are initially synthesized as lipid-linked
  precursors. The lipid component in this process is dolichol, a
  long-chain polyisoprenol of 14 to 24 isoprene units (17–21 units in
  animals and 14–24 units in fungi and plants; isoprene units are C5
  units with the carbon skeleton of isoprene , which is linked to the
  oligosaccharide precursor via a pyrophosphate bridge.

Dolichols apparently anchor the growing oligosaccharide to the endoplasmic reticulum membrane. These compounds have strong hydrophobic interactions with membrane lipids, anchoring the attached sugars to the membrane, where they participate in sugar-transfer reactions.

The pathway of dolichol-PP-oligosaccharide synthesis involves stepwise addition of monosaccharide units to the growing glycolipid by specific glycosyltransferases to form a common “core” structure:

Dolichol pyrophosphate glycoside. The carbohydrate precursors of N-linked glycosides are synthesized as dolichol pyrophosphate glycosides. Dolichols are longchain polyisoprenols (n = 14–24) in which the α-isoprene unit is saturated.

References

Voet and Voet Biochemistry
Lehninger Principles of Biochemistry


Understanding strong acid titrated with weak base graph I don’t understand why the conductivity doesn’t change after the equivalence point. To my understanding, at the equivalence point there is $\ce{NH3}$, $\ce{NH4+}$ and $\ce{Cl-}$ in the solution. As more $\ce{NH3}$ is added it only partially dissociates, thus conductivity stays the same. But doesn’t the concentration of $\ce{Cl-}$ ions also contribute to the conductivity, and its conductivity decreases are more aqueous $\ce{NH3}$ is added as it becomes more dilute? Where is the issue in my understanding. Thanks for the help.

You are missing an important point. Conductometric titration is not just plotting the conductance after adding the titrant. One cannot plot conductance as it is, because you are significantly diluting the solution.  You have to apply a correction to take dilution into account. Once you apply a correction, the curve will remain flat because further addition of ammonia will not enhance conductivity.
See for example,

Conductometric titrations
Throughout a titration the volume of the solution is always increasing, unless the
conductance is corrected for this effect, non linear titration curves result. The
correction can be accomplished by multiplying the observed conductance either
by total volume (V+V') or by the factor (V+ V')/V, where V is the initial volume
of solution and V' is the total volume of the reagent added. The correction
presupposes that the conductivity is a linear function of dilution, this is true only
to a first approximation.


Peroxide and 2 extra electrons? Trying to figure out where peroxide gains 2 electrons and why it has a 2- charge? On the Lewis dot structure it shows each O has 6 electrons and that it has a single bond between the two Oxygen atoms.

To me, each O atom would have 4 electrons and have a double bond between them?

Well, I might take a different approach at an answer, and try to get to a reasonable Lewis structure with data we already should have had from drawing structural diagrams! It's kinda upside down, since none of what's written below is happening in real-life chemistry.

The structure you have in mind is the diatomic oxygen molecule, with 12 electrons in its system overall.

$\hspace{33ex}$
$\hspace{45ex}$Source

While the peroxide ion overall has 14 electrons in its system, since each oxygen has 6 valence electrons and the system also has two extra electrons indicated by the $2-$ charge of the ion. Now, let's oversimplify this a bit: Imagine that in peroxide also, the oxygen atoms need to fulfill their octet configuration, hence each of them would have 8 electrons in its outer shell.

Let $N_b$ be the number of bonding electrons and $N_{nb}$ the number of non-bonding electrons of each oxygen atom. We know that

A single oxygen atom would have non-bonding electrons and bonding electrons that fill the octet.
All of the electrons in the system, which are the non-bonding electrons for each of the two oxygens, and the bonding electrons, make 14 electrons.

Hence we solve:

$$\Bigg\lbrace\begin{array}\\ N_b + N_{nb} = 8 \\ N_b + 2N_{nb} = 14 \end{array}\Bigg\rbrace \Rightarrow \boxed{N_b = 2},~\boxed{N_{nb} = 6} $$

Hence we could figure out how many bonding and non-bonding electrons there are in compounds we can draw Lewis structures for, by finding out the overall number of electrons (taking into account the number of electrons in the valence of each atom and the number of electrons lost or gained due to the charge of the ion) with a rudimentary approach.

Try to find out how many bonding and non-bonding electrons there are for the azide, permanganate, manganate, sulfate, oxalate, salicylate, and superoxide anions and thionyl, sulfuryl, nitrosyl and methanium cations.


Temperature measurement using thermometer I have two samples of water – 'A' and 'B'. My thermometer measured them to be 90 °C. They both have different volumes. So, the kinetic energy of the particles of sample A and B is same or different? Through this question, I want to ask whether temperature measurement results depend upon how many particles hit the bulb of my thermometer or it is independent of this? If it is independent of this, then how thermometer works?

Please provide me the simplest answer as I am in class 9 only.

As another example, let me consider a gas which has the density of 2 mol/22.4 L. Its temperature recorded by my thermometer is 60 °C. Then its density is reduced to about 10 atoms/22.4 litre. Now, its temperature is the same?

I have two samples of water – 'A' and 'B'. My thermometer measured them to be 90 °C. They both have different volumes. So, the kinetic energy of the particles of sample A and B is same or different?

The total kinetic energy is different. The average kinetic energy per particle is the same.

  Through this question, I want to ask whether temperature measurement results depend upon how many particles hit the bulb of my thermometer or it is independent of this?

In your specific example, the same number of particles hit the bulb because it just depends on the area of the bulb, as long as it is covered in water. Not only is water hitting the bulb, the atoms in the bulb are also hitting the water. After they have done this for a while, water and bulb are at the same temperature.

  If it is independent of this, then how thermometer works? As another example, let me consider a gas which has the density of 2 mol/22.4 L. Its temperature recorded by my thermometer is 60 °C. Then its density is reduced to about 10 atoms/22.4 litre. Now, its temperature is the same?

For the example with the two gases at the same temperature but at different pressure (i.e. particle density), the situation is a bit different, but the thermometer still works. There are a different number of "hits" in this case, but if you wait long enough, the gas and the bulb will be at the same temperature.

  Please provide me the simplest answer as I am in class 9 only.

I tried, but if the question addresses a complicated phenomenon, the answer can't be super-simple.

  [From the comments:] So, if i have 10,000000 particles in 1 L and 10 particles in 1L, temperature is going to be same or different?? (I mean the readings)

Ideally, there should be more sample than thermometer. This means that no matter what the temperature of the thermometer was to start, it will have the temperature of the sample afterwards. With only 10 particles in the sample, the 10 particles will take on the temperature of the thermometer (and the enclosure), so you will be unable to measure the initial temperature of the sample.

  [My own question] If the sample keeps hitting the thermometer, why doesn't it get hotter and hotter?

As I said above, the thermometer also keeps hitting the sample. There is some net transfer of energy while the temperatures are different (from hot to cold, warming up the colder and cooling down the hotter). Once the temperatures are the same, sample and thermometer keep "hitting each other", but the temperature of both stays constant (if the entire setup is insulated against the surrounding).


How can I increase the current on a voltaic cell for high school lab? I'm creating a voltaic cell project for my high school chemistry students. This is a design challenge for them to use electronegativity differences to produce enough voltage and current to power a small DIY motor and fan. So far, I have been able to generate about 1.7V but only 0.4mA using lots of spirals of Mg ribbon along with a Cu electrode using porous cups (and not a salt bridge). I've increased the concentration of Copper nitrate. What methods might increase the current? We're have enough voltage to run 1.5V DIY motors but they won't run due to lack of current.

We're doing this: http://static.nsta.org/files/tst1409_30.pdf

Increase the area of the electrodes (e.g more ribbon).
Decrease the resistance of the electrolyte (e.g. reduce the spacing between electrodes, increase concentration of salts).
Reduce polarization of electrodes (bubbles on surface, etc.). Scrape or lightly sandpaper the electrodes and test immediately after immersion, before bubbles form.


How can a ketone be enantioselectively reduced, in the presence of an ester? I'd like to reduce enantioselectively a ketone in presence of an ester (which should NOT be reduced to an aldehyde and further to an alcohol).

Adding $\ce{LiAlH4}$ would also reduce the ester and gives no enantioselective reduction of the ketone.
In the following reaction (Frater-Seebach Alkylation) they used Bakers yeast:

Probably S-BINAP also give 2. I think there is a $\ce{RuCl2BINAP}$ transition metal complex in combination with $\ce{H2}$ used.
I would first use KOH, then adding Evans auxiliary, adding LiCl for chelation of acid carbonyl to Evans auxiliar carbonyl, then adding $\ce{NaBH4}$ and to a standard Felkin Anh reduction. And as a last step I would use $\ce{LiOEt}$. Does this work or will the chelating ketone also be reduced?

May be my own suggestion is easier to reproduce in lab than the use of asymmetric reduction using $\ce{H2}$ under high pressure.

I'm looking for further ways to get the (S)-hydroxy compound 2. Are there easier ways?

I have had great results for enantioselective ketone reduction using the Noyori Ru-TsDPEN system.
This has advantages through being catalytic with low stoichiometry and doesn't require high pressure, or molecular hydrogen for that matter. The downside might be the initial cost of the catalyst / chiral ligand but you don't need very much. The typical reduction reagents are triethylamine and formic acid, making this a very mild system to work with in terms of conditions.
It might be cheaper to purchase the RuCl2(p-cymene) and chiral TsDPEN separately. That way you can also play with the ligand-metal ratio if you need to.
This saves a lot of hassle rather than adding on and removing chiral auxiliaries


What is the difference between hydroxyl, hydroxide and hydroxy? Is there any difference between hydroxide, hydroxyl, and hydroxy used in chemistry or are they all the same?

Hydroxide - Hydroxide is a diatomic anion with chemical formula $\ce{OH-}$. It consists of a hydrogen and an oxygen atom held together by a covalent bond, and carries a negative electric charge.

Hydroxyl radical- {hydro- + oxy(gen) + -yl + radical} - The hydroxyl radical, •OH, is the neutral form of the hydroxide ion (OH-1). It is the monovalent group -OH which contains hydrogen and oxygen. The suffix -yl indicates it's a radical just like carbonyl.

Hydroxy or hydroxy group: A hydroxy or hydroxy group is the entity with the formula $\ce{-OH}$. It contains oxygen bonded to hydrogen. Hydroxy is also prefix which means containing hydroxyl group.


Hydration number: Carbonic acid in water Carbonic acid ($\ce{H2CO3}$) is known to be soluble in water, and that this particular compound is unstable, so much so that according to the 2014 article from Berkeley Lab:
New Insights on Carbonic Acid in Water, it only exists for a fraction of a second (300 nanoseconds according the article).

Newly reported experiments and observations from the Berkeley Lab found that:

  “Our results support an average hydration number of 3.17 with the acid’s two protons each donating a strong hydrogen bond to solvating waters, the carbonyl oxygen accepting a strong hydrogen bond from solvating water, and the hydroxyl oxygen molecules accepting weak hydrogen bonds from the water”

What is not clear is how is the hydration number calculated and how does this relate to the short existence of carbonic acid in water?

You may wish to refer to the publication of Saykally et al., The hydration structure of aqueous carbonic acid from X-ray
absorption spectroscopy

Basically, to determine hydration number, X-ray spectroscopy was used. This relates to the short existence of carbonic acid as, since it is so short lived, it is necessary to analyze it in such a time frame. As such, this was "the first measurement of X-ray absorption spectra for aqueous carbonic acid, enabled by the use of rapid-flow liquid microjet technology" which allowed it to be captured.

Then, by analyzing the spectra, it is possible to calculate the approximate hydration number, as "the increasing H-bond strength is proportional to the hydration
number (measured for distances up to $2.5\mathrm{\overset{\circ}A}$): $\ce{Na2CO3}$ (5.55), $\ce{HCO3}$
(4.26), $\ce{H2CO3}$ (3.17)".


Reaction of aluminum sulfate in water I dissolved aluminum sulfate in DI water and expected to see just a clear water solution with aluminum sulfate dissolved as ions. However, a precipitate was visible. I addd only 0.6g of Aluminum sulfate so there is no way that the precipitated is saturated aluminum sulfate. Furthermore, it does not seen to be aluminum hydroxide, so then what is the precipitate? Aluminum sulfate has the property of filtering out impurities in water, however, I am using DI water so then what is the precipitate?

Also, the precipitate is not aluminum hydroxide because aluminum hydroxide is dissolvable in sulfuric acid; however, when I added sulfuric acid to the precipitate solution, the precipitate did not disappear.

According to Wikipedia, "When dissolved in a large amount of neutral or slightly alkaline water, aluminium sulfate produces a gelatinous precipitate of aluminium hydroxide, Al(OH)3."


How is hair tissue mineral analysis performed? This might be a mixed chemistry / physics question, but I've always wondered what methods were used to precisely determine the levels down to trace amounts of all minerals in human tissues. I know there's a tissue preparation phase, in which I am interested though less than in the details of the actual final testing methods and processes to determine those trace amounts of all minerals.

The analytical technique is called inductively coupled plasma optical emission spectroscopy. It was invented by a chemist V. A. Fassel, although it is pure physics. The specimen is digested in an acid, and a fine spray is created. That "spray" is heated to a temperature, which is higher than the surface of the Sun. Most of the elements of the periodic table emit specific wavelengths in that hot plasma- the characteristic emission of each element is just like a fingerprint. Not all elements can be detected easily but it is very good for metals. This is the mineral analysis. Organic analysis is even more complicated and it is a separate story.


Why does NaOH catalyze H2O2 decomposition? I had a lab practice in which I had to test the effectiveness of different catalysts for the decomposition of $\ce{H2O2}$. Among them, there was: $\ce{KI(aq)}$, $\ce{FeCl3(aq)}$, $\ce{MnO(s)}$, $\ce{NaOH(aq)}$, $\ce{HCl(aq)}$ and $\ce{I2(s)}$.

In different test tubes, I added some $\ce{H2O2}$ and some catalyst. After a while, I added $\ce{NaOH}$ as well. I observed an increase in the effervescence for $\ce{KI}$, $\ce{FeCl3}$, $\ce{I2}$ and $\ce{MnO}$. However, when the mixture was only $\ce{H2O2}$ + $\ce{NaOH}$, nothing happened. I've made some research, but can't find an explanation for this...

I believe $\ce{NaOH}$ increases the rate of the reaction of the other catalysts, but why?

According to THE HOMOGENEOUS BASE-CATALYZED DECOMPOSITION OF HYDROGEN PEROXIDE J. Phys. Chem., 1961, vol. 65 , pages 304–306:

$\ce{H2O2 + OH- <=> HO2- + H2O}$

$\ce{H2O2 + HO2- ->}$ [6-membered ring with 2 "H"s bridging 2 "O-O"s, with another H bonded to one of the "O"s] $\ce{-> H2O + O2 + OH-}$

  It is clear that the rate will go through a maximum
  when the peroxide is 50% ionized, that is
  $[\ce{H2O2}] = [\ce{HO2-}]$.

So OH- catalyzes decomposition of peroxide, but a maximum is reacted when the amount of hydroxide added is half the amount of peroxide.  If more hydroxide than peroxide is added, the catalytic effect is lost.


How does the number of electrons increase in beta decay? So, I am having trouble understanding beta decay. Particularly the number of electrons.

For example; Carbon-14 decaying into Nitrogen-14.

Carbon-14 has 12 neutrons, 6 protons and 6 electrons.
When it decays 1 neutron is converted into: 1 proton, 1 electron and 1 anti-neutrino.

So after the neutron breaks down, and before emission, there should now be:
11 neutrons, 7 protons, 7 electrons and 1 anti-neutrino.

But now 1 electron is emitted from the atom (as a beta ray), alongside the anti-neutrino. Which means the numbers should now be:
11 neutrons, 7 protons and...6 electrons.

Since 1 electron is created in the process and 1 is emitted out of the atom, then surely the number of electrons shouldn't change from the original 6 of carbon-14.
But Nitrogen-14 has 7 electrons?? Can someone explain?

Thanks.

The two commenters have pretty much answered your question: after beta emission, which is a sufficiently energetic process to tear apart a molecule, the resulting fragments/ions would likely undergo a series of additional reactions with any nearby molecules to finally produce stable species. The emitted electron also would be captured by nearby molecules, leading to a cascade of reactions, also eventually producing stable species. In any case you do indeed end up with a nitrogen atom and an extra electron.

Example: if decay occurred in a bottle of liquid methane, you would get ammonium cation (isoelectronic with methane) and an extra electron loosely attached to methane - a methane radical anion. These could recombine to give neutral ammonia and a hydrogen atom, and then further reactions are possible.

I would also point out that carbon 14 has 6 protons and 8 (not 12) neutrons, but in any case your question remains valid.


An ionic compound dissociates into 3 ions of 3 different elements? When an ionic compound dissolves in water and becomes ions.

1) Typically we see:

$\ce{AB ->}$ Ions $\ce{A+ + B-}$

$\ce{ABC ->}$ Ions $\ce{AB+ + C-}$

2) But how about the one below, is there any case that will fit the situation

$\ce{ABC2 -> A+ + B+ + 2C-}$

or $\ce{AB2C-> A+ + 2B+ + C^3-}$

Yup, they are definitely possible. Such salts are called as mixed salts. They furnish more than two ions when dissolved in water. A popular example is Mohr's salt, popularly used as a standard titrant to measure concentrations of oxidizing agents. It's (anhydrous) formula is:

$$\ce{Fe^{II}(NH4)2(SO4)2}$$

This salt on dissolving in water would split as:

$$\ce{Fe^{II}(NH4)2(SO4)2 <=>> Fe^2+ + 2NH4+ + 2SO4^2-}$$

This is one such example. There are many others, notably alums. Alums are a general type of mixed salts, of the form:

$$\ce{A^{I}M^{III}(SO4)2}$$

where $\ce{A^{I}}$ is a monovalent metal ion, and $\ce{M^{III}}$ is a trivalent metal ion. Clearly, these are also mixed salts.

These are just a few practical examples. There's literally quite nothing that can stop imagination of course.


How do I explain the existence of these 2 peaks in this mass spectrometry? Question about HRMS:Phosphatidylcholines and phosphatidylethanolamines are common types of glycerophospholipids found in biological membranes. Because of the high amount of heteroatom content, electron impact mass spectrometry is a bad idea because the ions fragment too fast. For biological molecules, electrospray ionization is the most common method. A compound in the glycerophospholipid class was analyzed by low-res electrospray ionization. It produced peaks: 773 Th(100 percent abundance), and 795 Th(about 20% abundance). The question says these are not isotope peaks. The same compound was analyzed by high-rest EI mass-spec to have peaks at 772.5856 Th and 794.5680 Th. The question says to explain the 2 peaks’ presences in the high-resolution mass spectrometry.

For question #6, I came up with the molecular formula for C40H80NO8P for both classes of lipids but I was unsure of how to use that to come up with the actual molecular formula. The first lipid I found was dipalmitoylphosphatidylechloline with a molar weight of 734. But it's smaller than either of the 2 fragments. Then I added 6 methylene groups to come up with a molecular formula of C46H92NO8P. But I'm not sure how to get the molecular formula for both fragments(or if we actually need them). Can you help with this question or let me know of what I'm missing?

In electrospray ionization, a molecule can end up with an extra proton or an extra cation. Either one of these +1 charged species is accelerated and detected in the mass spectrometer (negative charged particles or neutral species are not; if there are multiple cations, they would supposedly repel each other once the solvent is gone, and not fly together).

So a molecular formula that fits the data is $\ce{C43H82NO8P}$. The lighter species detected in the spectrometer would be $\ce{C43H82NO8PH+}$, and the heavier species would be $\ce{C43H82NO8PNa+}$. The masses given by https://www.sisweb.com/referenc/tools/exactmass.htm for these species match the given values pretty well. The difference in mass is not due to fragmentation, but to a cation piggy-backing with the molecule of interest (instead of the usual proton).

The data does not distinguish between phosphatidylcholines and phosphatidylethanolamines. You can, however, infer the number of C=C double bonds in the fatty acids attached to the glycerol.

I had to look up the unit "Th" on wikipedia:

  The thomson (symbol: Th) is a unit that has appeared infrequently in scientific literature relating to the field of mass spectrometry as a unit of mass-to-charge ratio. The unit was proposed by Cooks and Rockwood naming it in honour of J. J. Thomson who measured the mass-to-charge ratio of electrons and ions.

HRMS stands for high resolution mass spectroscopy, I think.


Hydrogen bonding in sugar acids I recently learnt that the reason for low solubility of mucic acid as compared to glucaric acid was more hydrogen bonding in the crystal state of mucic acid as compared to glucaric acid.

I would like to know why mucic acid has more hydrogen bonding? Does symmetry of the molecule play a role? If so how?

From the abstract of an article titled "The crystal structure of galactaric acid (mucic acid) at −147°: An unusually dense, hydrogen-bonded structure"

[...] D$_m$ = [1.790] $\pu{g cm−3}$ [...] The crystal structure has a system of strong, intermolecular hydrogen-bonds, which accounts for the high crystal density and low solubility in water.

The crystal structure was solved at very low temperature, but the density D$_m$ is quoted for room temperature. For comparison, the density of sucrose or cellulose is around 1.5 $\pu{g cm−3}$.

For glucaric acid, the volume of the asymmetric unit is 421.9 Å$^3$ / 2, while for mucic acid it is 194.0 Å$^3$. As the molar mass of the two stereoisomers is the same, there is a difference in density of about 9%.

A lower density implies more attractive interactions or less repulsive interactions in the crystal state. For a complete analysis, you would also have to compare how well these compounds are solvated in aqueous solution. Typically, it is impossible to predict solubility by just looking at structures because both solvation and crystal structure are hard to predict.


Which material would work best as "sacrificial material" for melting aluminum powder in a kiln? I need to melt aluminum powder in a kiln, but before it melts it gets oxidized almost completely. My kiln is almost airtight, but it doesn't have inert gas inlet.
Which material would work best as sacrificial material, which would react with oxygen (and water vapor) before aluminum?

You need some oxygen scavengers before you start heating up the system. However, please note that you are going to need a lot of scavengers. To get rid of all the oxygen, you will have to put in a large excess of these scavengers into the chamber.

There is some information on oxygen scavengers on Wikipedia


Moles of chlorine gas needed to make up an equivalent amount of saturated salt solution I’m trying to solve a question about a saturated salt solution but am unsure of how to answer one of the parts or what it’s asking for.
The Problem
The maximum solubility of table salt (NaCl) in water at 25 degrees Celsius is approximately 359 g/L. Assume the volume of the solution does not change with the addition of salt.
a) What is the molar concentration of the mixture in mol/L?
b) What is the molal concentration of the mixture in mol/kg? Assume that the density of water is 980 kg/m3.
c) How many moles of Cl2 gas would need to be present in order to make up an equivalent amount of saturated salt solution?
d) What is the density of the mixture?
My Attempt
a) molar concentration
359 g NaCl * 1 mol NaCl / 58.44 g NaCl = 6.14 mol => 6.14 mol/L
b) molal concentration
(1 L H2O * 1 m3 H2O * 980 kg) / (1000 L H2O * 1 m3) = 6.14 mol / 0.980 kg = 6.26 mol/kg
c) moles of Cl2 gas
I thought it was asking how many moles of chlorine gas is equivalent to the amount of salt in the 1 L solution (359 g NaCl). There are 70.9 g in 1 mole of Cl2, so that gives me an answer of 5.06 mol Cl2 gas (359 g NaCl / 70.9 g Cl2).
But the answer key states the correct solution as 3.07 mol Cl2 gas. How should I go about this part?
d) density of the mixture
density of water = 980 kg/m3
density of salt = 359 g/L => 359 kg/m3
density of mixture = (980 + 359) kg/m3 = 1339 kg/m3

a) molar concentration
There are indeed 6.14 moles of $\ce{NaCl}$ in 359 grams of $\ce{NaCl}$.
I'd prefer to write the molar mass as "$\pu{58.44 g mol^-1 }\ \ce{NaCl}$" or "$\pu{58.44 g/mol}\ \ce{NaCl}$" to keep it all together.
b) molal concentration
You got the right value, but you did it the wrong way. For the solvent:
$$\dfrac{\pu{980 kg/m^3}}{\pu{1000 L/m^3}} = \pu{0.980 kg/L}$$
c) moles of $\ce{Cl2}$ gas
The book is correct. There are 6.14 moles of $\ce{NaCl}$. Each molecule of chlorine gas contains two atoms of chlorine, so you need half as many moles to get the same number of atoms of Cl. So 6.14/2 = 3.07
This part of the question has poor phrasing. You can't dissolve 3.07 moles of $\ce{Cl2}$ in water and get a solution of $\ce{NaCl}$.
d) density of the mixture
Your answer is correct.


Why does Brady's reagent include sulfuric acid and methanol? I'm supposed to explain why sulfuric acid and methanol are used in combination with 2,4-dinitrophenylhydrazine to form Brady's reagent. Are they just used as a solvent system for the hydrazine? Also, wouldn't sulfuric acid react with methanol?

2,4-DNPH is usually dissolved in an alcoholic solvent, together with sulfuric acid, when used as a reagent to detect aldehydes and ketones (Brady's reagent). The sulfuric acid is a source of $\ce{H+}$, which catalyses the reaction between the hydrazine group and the carbonyl compound, as depicted in the mechanism below:

The final hydrazone product is the precipitate which signals a positive test.

The solvent used is typically methanol or ethanol. You are correct that these could theoretically react with sulfuric acid: dimethyl ether is the most likely product. However, the concentration of sulfuric acid is typically not sufficiently high, and these side reactions do not take place.


Why is it aqueous and ethanolic conditions for nucleophillc substitution of cyanide ion with benzyl chloride? I was doing an exam question where benzyl chloride was undergoing a nucleophilic substitution with cyanide ion to form benzyl cyanide.

Please can someone enlighten me as I have never seen both conditions before just one or the other. I have Googled it but no sites talk about the conditions of nucleophilic substitution with a cyanide ion in a benzene context. Does the benzene alter the conditions?

Since $\mathrm{p}K_\mathrm{a}$ of $\ce{HCN}$ is 9.21 (Wikipedia), the cyanide ion in water is pretty alkaline and contains significant amounts of hydroxide ions, according to the following equilibrium:
$$\ce{NC- + H2O <=> OH- + HCN}$$
That's why you smell strong $\ce{HCN}$ in aqueous  $\ce{KCN}$ or $\ce{NaCN}$ solution.  Thus, when you perform the reaction of $\ce{KCN}$ or $\ce{NaCN}$ with the haloalkane (including allyl and benzyl) in water you tend to get the substitution by $\ce{^-OH}$  instead of $\ce{^-CN}$. To get major product as cyanoalkane, you may need to do the reaction in absolute alcohol.

Not relevant, but note worthy that $\ce{^-CN}$ is an ambident nucleophile, meaning a nucleophile which can attack acitve site of a molecule through more than one sites (here either by $\ce{N}$ or $\ce{C}$). Thus, a reaction of alkyl halide ($\ce{R-X}$) with $\ce{KCN}$ gives alkyl nitrile ($\ce{R-CN}$) while a reaction of alkyl halide with $\ce{AgCN}$ gives an alkyl isonitrile ($\ce{R-NC}$).


Can antimony pentafluoride be made by adding hydrogen fluoride to antimony trifluoride? I was looking at how the strongest acid was made and I found something I did not understand. On the Wikipedia page for antimony trifluoride it shows the production of $\ce{SbF3}$ using $\ce{HF}$:
$$\ce{Sb2O3 + 6 HF -> 2 SbF3 + 3 H2O}$$
The page also shows production of $\ce{SbF5}$ using $\ce{F2}$:
$$\ce{SbF3 + F2 -> SbF5}$$
However on the page for fluoroantimonic acid it shows production of:

$$\ce{SbF5 + 2 HF -> SbF6- + H2F+}$$

So my question is why the production of $\ce{SbF5}$ seems to require fluorine insted of hydrogen fluoride? And is it possible to use hydrogen fluoride to make $\ce{SbF5}$?

It can be prepared using hydrogen fluoride, in a double displacement reaction:

  $\ce{SbCl5 + 5 HF -> SbF5 + 5 HCl}$

However, hydrogen holds onto fluorine more tightly than does antimony in $\ce{SbF5}$.

Consider the reverse of your proposed reaction, which is preferred energetically:

  $\ce{SbF5 + H2 -> SbF3 + 2HF}$


Cheap and not messy solvent for motor oil I have an old oil cooler of  an old Volkswagen engine (the type where the oil itself is cooled, and uses no water) stained with decades of usage. If I pass water through it, it comes out clear, but if I pass a clear oil through it, it comes out black.

What is the best method to clean it that uses cheap and easy to find substances? I want to use it to cool clean mineral oil, I want my oil to remain clean.

My father suggested to pass mineral oil through it until what comes out is clean, then install it in the system, but that would be horribly expensive and messy, and I would need to discard all that dirty oil properly afterwards. I myself thought of using gasoline, filling halfways, shakink and discarding, until clean enough, but although that would be less expensive, it would be as messy as the first option, and probably dangerous. My sister suggested using water and kitchen wash-up liquid. That would be much less messier, because it is water soluble, but I don know the proportion of water and kitchen detergent tI must use for it to be effective. My last idea was to use some grese remover like Ajax, that would not be so.much messy, either.

So, what do you suggest?

In terms of inexpensive, safe, and a little polar to move oxidation products, I'd try soaking in and flushing with hot soybean oil (replace elastomer hoses and seals).  Then flush with kerosene (odorless white spirits, hardware store  paint section), then blow out with compressed air.

Carburetor cleaner was mixed cresols - dangerousy on skin (heavy gloves), but great against deposits.  If it is really nasty in there, I'd go for gun bore cleaner soaking, specifically Ed's Red without the lanolin and substituting MEK (methyl ethyl ketone) for the acetone to lower flammability. At the end, blow it out, flush with kerosene, blow it out.

Contain all wastes.  Beware of skin contact (gloves) .  Do not breath the fumes (ventilation, and maybe a charcoal latex paint mask, then discard).  Eye protection (goggles).  VAPORS ARE FLAMMABLE AND HEAVIER THAN AIR.

http://www.youtube.com/watch?v=2OfaCCJjBEU
Ed's Red Gun Bore Cleaner  (wear gloves!)

1 part Dexron II, IIe or III ATF, GM Spec. D-20265 or later.
1 part Kerosene - deodorized, K1 (can be Kroil or WD-40)
1 part Aliphatic Mineral Spirits, Fed. Spec. TT-T-2981F, CAS #64741-49-9, or may substitute "Stoddard Solvent", CAS #8052-41-3, or equivalent, (aka "Varsol")
1 part Acetone, CAS #67-64-1.
1 lb anhydrous lanolin or  Lanolin, Modified, Topical Lubricant.

Note that Ed's Red is rapid death to plastics, polymers, varnishes, paints, and coatings.


How much KMnO4 is required for the complete oxidation of ferric/ferrous oxalate in acidic medium? Find out the amount of $\ce{KMnO4}$ required for the complete oxidation of $\pu{5 mol}$ of an equimolar mixture of ferric oxalate and ferrous oxalate.

I understand that $\ce{MnO4-}$ will be oxidised to $\ce{Mn^{2+}}$ and ferrous ions to ferric ions. I wrote the following reaction:
$$\ce{FeC2O4 + Fe2(C2O4)3 + MnO4^- + 8H+ + e- -> 2Fe^3+ + 4CO2 + Mn^2+ + 4H2O}$$

Then, I applied equivalence concept but could not get a correct answer. Any ideas?

I'd like the question to be tackled in two parts. In each part, we're going to deal with one of ferric oxalate and ferrous oxalate.

Since the mixture ($\pu{5 mol}$) given to us is equimolar, we have $\pu{2.5 mol}$ of ferric oxalate and $\pu{2.5 mol}$ of ferrous oxalate. I second your understanding that $\ce{MnO4-}$ will be oxidised to $\ce{Mn^{2+}}$, because the medium is acidic, and ferrous ions to ferric ions.

Ferric oxalate doesn't contain any $\ce{Fe^3+}$ ions, so our main focus here is the conversion of oxalate ($\ce{C2O4^2-}$) into carbon dioxide. The oxidation state of carbon in oxalate is $+3$, while it's $+4$ in carbon dioxide. So there's a net change of $+1$ per carbon atom. Oxalate contains two carbon atoms, so that gives us a valency factor of $2$ per mole of oxalate ion. However, for $1$ mole of $\ce{Fe2(C2O4)3}$, you'll have $3$ moles of oxalate ions that are undergoing the aforementioned change. This gives us an overall valency factor for $\ce{Fe2(C2O4)3}$ as  $2 \times 3 = 6$. The valency factor for the $\ce{MnO4- -> Mn^2+}$ change is $5$. Now, we're going to equate the equivalents as per the Law of Equivalence.

$$ \begin{align*}
n_{\ce{KMnO_4}} \times V_{f_1} &= n_{\ce{Fe2(C2O4)3}} \times V_{f_1'}\\
x \times 5 &= 2.5 \times 6 \\
x &= 3 \end{align*}
$$

For the part of the mixture that contains $\ce{FeC2O4}$, we need to change our valency factor. This is because here we're taking into account the conversion of ferrous ions into ferric ions; this conversion has a valency factor of $1$. So the overall valency factor, in this case, would be $2+1=3$. Now, since this has been done, we proceed to apply the same equation as in the previous case.

$$ \begin{align*}
n_{\ce{KMnO_4}} \times V_{f_2} &= n_{\ce{FeC2O4}} \times V_{f_2'}\\
x \times 5 &= 2.5 \times 3 \\
x &= 1.5 \end{align*}
$$

Thus, overall, it can be concluded that you'll need $\pu{3+1.5=4.5 mol}$ of $\ce{KMnO4}$ for the complete oxidation. The beauty of this? I didn't use any reaction-equilibrium equations.

This is the reason we use equivalent concept, to save ourselves from the hassle of writing (and perhaps balancing) chemical reactions.


Requirements for electroplating I understand that electroplating involves using an anode (copper in this case) and an electrolytic solution containing Cu(II) ions with some sulfuric acid. Why exactly does the anode need to be copper if you want to plate something with copper; couldn't something like a graphite electrode be used instead, as the $\ce{Cu^2+}$ ions can be provided by the solution?

You are right in saying that the electroplating could equally well be carried out with any electrode. Just offhand though, I could think of a couple of reasons why you would want the anode to be made of copper.

Choosing copper as the anode makes the overall cell potential equal to zero. The reaction at the anode is $\ce{Cu -> Cu^2+ + 2e-}$ and the reaction at the cathode is $\ce{Cu^2+ + 2e- -> Cu}$ - they are identical but one is flipped, so $E^\circ_{\mathrm{cell}} = 0~\mathrm{V}$. Compare this to if you used graphite - the reaction at the anode would then be $\ce{2H2O -> O2 + 4H+ + 4e-}$ and you would have $E^\circ_{\mathrm{cell}} = -0.892~\mathrm{V}$, meaning you would need more energy to run the reaction.
More importantly, it also ensures that the concentration of $\ce{Cu^2+}$ ions in solution remains constant. If you use a graphite electrode, the solution can initially provide the $\ce{Cu^2+}$ ions required for electroplating, but as the concentration of $\ce{Cu^2+}$ decreases, $E_{\ce{Cu^2+}/\ce{Cu}}$ becomes smaller (by the Nernst equation) and it becomes harder and harder to reduce $\ce{Cu^2+}$ at the cathode.


What (non toxic) light sensitive materials do not burn away at high heat? Specifically related to photoceramics, I am wondering what photosensitive materials exist that can be bonded to glass at high heat in a kiln (a microwave kiln to be more exact).
By non-toxic I mean a material that does no give off hazardous fumes, so silver, cyanotype and dichromates would be out of the question as far as I am aware.
The process I had in mind is, coat on glass, expose image via contact print using UV, then kiln fire to make the image permanent with the glass.
If there exists a photosensitive glaze out there, I would love to know!

There are a number of types of photosensitive glass which change in appearance and chemical reactivity upon exposure, and of photochromic glass, which undergoes reversible change in color with radiation.
If you are using this as a surface applique on ceramic, one key factor would be having therml expansion coefficient compatible with the ceramic, or the object will crack on cooling. Consult the literature from a manufacturer, such as Schott or AZO Materials.
There are also techniques for making a photosensitive layer on ceramics, such as Pyrofoto, that do not involve applying glass. A small Pyrofoto kit is ~US$20.
BTW, you might find that microwave oven kiln adapters do not reach high enough temperature to be useful in this work -- literature states a limit of ~900°C, cone 010.


Validity of the leveling effect of water in nitration reactions Wikipedia gives this as the levelling effect of water

  Any acid that is stronger than $\ce{H3O+}$ reacts with $\ce{H2O}$ to form $\ce{H3O+}$. Therefore, no acid stronger than $\ce{H3O+}$ exists in $\ce{H2O}$. For example, aqueous perchloric acid ($\ce{HClO4}$), aqueous hydrochloric acid ($\ce{HCl}$) and aqueous nitric acid ($\ce{HNO3}$) are all completely ionized and are all equally strong acids.

So from what I can understand, if I take an aqueous medium, then $\ce{H2SO4}$ and $\ce{HNO3}$ both should be equally strong.

But then comes the nitration reaction, where the nitronium cation acts as the real attacking electrophile, which means that in this reaction, $\ce{HNO3}$ acted as a base and $\ce{H2SO4}$ acted as an acid.

I am also aware concentrated acids are taken in the nitration reactions, but even the most concentrated acids have some amount of water.

How are these two concepts contradicting, where am I going wrong?
What are the limitations of the levelling effect and where all is it applicable? Is it not applicable to concentrated solutions, and if so, how concentrated?

What you have gone wrong in this question is you completely disregard the fact that mentioned stronger acids than water are all in aqueous medium, in other words, dissolved in water. Thus, fast acid-base reaction ($\ce{H2O}$ acts as a base here) happens to give $\ce{H3O+}$ as the only acid in these mixtures (leveling effect of water).

On the other hand, concentration of sulfuric acid is $\approx 98\% (w/w)$. That means, approximately only $\pu{2 g}$ of $\ce{H2O}$ in $\pu{100 g}$ of concentrated solution. Therefore, mole ratio of $\ce{H2O : H2SO4}$ would be about $0.1 : 1$. Thus, ($\ce{H2SO4}$ is not leveling with water. That extra remaining $\ce{H2SO4}$ can be acted on $\ce{HNO3}$ in nitrating mixture as shown in your scheme (vide supra).

Note: As a rule of thumb, you may simply consider $1:1$ mole ratio of water and any stronger acid than water would be the limiting level for the leveling effect.


Radioactive Isotopes form Compounds? The Hershey-Chase Experiment Our chemistry teacher taught us last year, that radioactive isotopes of elements don't form compounds with other elements "because they're unstable" and since we're high-schoolers (I'm assuming that's why) she didn't go into details.

But today in our Biology class, we were taught about the Hershey-Chase experiment (https://en.wikipedia.org/wiki/Hershey%E2%80%93Chase_experiment) , where we come across radioactive P-32 being incorporated into the phosphate group of nucleic acids and radioactive S-35 being incorporated into amino acids in proteins.

When we reached that bit, what our Chem. teacher said came back to me and it conflicted with the idea being propounded in class. Naturally I raised that issue with our Bio teacher, but she said she wasn't sure about' radioactive isotopes not forming chemical bonds', because that was precisely what was happening in the Hershey-Chase experiment.

So can anyone tell me whether:

a) My Chem. teacher was correct (perhaps within limits), and that the incorporation of radioactive isotopes in nucleic acids and proteins is a phenomenon that occurs outside of the limiting case she mentioned.

b) My Bio. teacher was right. Radioactive isotopes do form compounds.

c) None of the above (could you specify why?)

Radioactive isotopes have similar chemical properties to that of nonradioactive element of same atom. They actually do from chemical bond with compound. You even heard the name of radio-pharmaceutical that involved the complex of radioactive metal with ligand or the radioactive isotope is attached to the pharmaceutical through covalent bond.

The simplest example is FDG is 18F-labled glucose analogue.

Carbon-11 is another radioactive isotope that form covalent bond similar to Carbon-12 and used in Nuclear medicine as PET imaging agent. C11-Methionine is an example used in the field of Nuclear Medicine for the diagnosis of brain tumour

Iodine-131 is another sample of radioactive isotope that is used for the detection of hypothyroidism and also form bond with compound. Like in metaiodobenzylguanidine is used in neuroblastoma

.

I can gave you lots of others examples but i think these are enough for your understanding. I hope, this will be helpful.


How will these hydrogen atoms interact? Assume that we have a space with just two $\ce{H}$ atoms and their distance to each other is $d$. Let's say they don't have initial velocity. What is the force with which they will act on each other?

This question might have some mistakes in itself, but the main thing I'm trying to learn is how an atom of element $X$ interact with an atom of element $Y$.

How one atom interact with another atom is an difficult question and requires quite some calculations, even for a system with only two hydrogen atoms.  Basically, this is done by solving the Schrödinger equation for the system.  The answer will also depend on the electronics state of the system, especially the spin state.

If the system is in a triplet electronic state, the two atom will experience a repulsive force that approach infinity as you pull the two atoms closer, and that force approaches zero as the distance approaches infinity.

If the system is in a singlet state, the force is zero at the equilibrium bond length of $H_2$ atom(~0.74Å).  As the distance gets smaller the atoms experience an increasing repulsive force which approaches infinity at 0.  As the two atoms are pulled apart from the equilibrium there is a decreasing attractive force that approaches 0 at infinity.


Why are alpha-carbonyl relatively less stable? So if we have an alpha halogen ketone, why is an Sn1 type mechanism so unfavorable for that species?

I don't understand. It seems like the C+ would be resonance stabilized by the carbonyl's double bond.

The carbon atom of the carbonyl group has a partial positive charge due to electron withdrawal by the electronegative oxygen atom. An $S_{N}1$ mechanism would result in an intermediate with a positive charge next to a partial positive charge, which would be high in energy.


solution concentrations I think I'm making this more complicated than it needs to be, but here it is:

I have 7 drug solutions, all 1 mg/mL which need to have the following concentration in urine:

A: 25,000 ng/ml

B: 25,000 ng/ml

C: 100,000 ng/ml

D: 100,000 ng/ml

E: 10,000 ng/ml

F: 1000 ng/ml

G: 1000 ng/ml

How much of each 1 mg/ml solution do I need to add to urine to get a stock with the above concentrations? My ultimate goal is to take several 75 ul aliquots of urine for testing. Any suggestions?

Since you did not tell us how much of this final solution you would like to make, let's imagine that you want to make a generic $V_f\ \ce{mL}$ of it.

Take $\text{A}$ as an example:

$$
n_A \ \text{in final solution} =  25000\ \ce{ng/mL} \times V_f\ \ce{mL}
$$

$$
V_{A}\ \text{stock to take} = \frac{n_A}{c_{A\ stock}} = \frac{25000\ \ce{ng/mL} \times V_f\ \ce{mL}}{1\ \ce{mg/mL} \times 10^6\ \ce{ng/mg}} \frac{1000\ce{\ \mu L}}{\ce{mL}}= \frac{25000}{10^3}\ V_f\ \ce{\mu L}
$$

So let's suppose that you want to make $10\ \text{mL}$ of such a solution containing $\text{A}$ at $25,000\ \text{ng/mL}$, then you need to measure out $25000 \times 10\ /\ 1000\ \ce{\mu L} = 250\ \ce{\mu L}$ of the $\text{A}$ stock and bring it up to the final volume with urine.

You will want to carry out these calculations for all your compounds, then add all the corresponding quantities to a calibrated vessel, e.g. a volumetric flask, then bring this mixture up to the final volume with your solvent, i.e. urine.


Experimental measurment of thiosulphuric acid decomposition kinetics Problem

I am trying to do a kinetics analysis on the following reaction:

$$\ce{Na2S2O3 {(aq)} + 2 HCl {(aq)} -> S {(s)} + 2 NaCl {(aq)} + SO2 {(g)} + H2O {(l)}}$$

Suggested Method

However, most practicals online suggest I use an 'X' Mark and let the sulphur precipitate cloud the mark (Here's an Example) . To measure the rate of the reaction, I would begin timing as soon as I mixed the two solutions and end as soon as the 'X' was completely obscured.

Alternatives

However this form of time measurement is somewhat qualitative.

I considered looking at:

pH or temperature:       Those readings would not provide substantial data.
Change in volume or mass:   For the gas released, but $\ce{SO2_{(g)}}$ is dissolvable in water.
Conductance:         Although this is viable, I was told there is an alternative.
Colour (Spectrometry):    This probably won't work because the spectra observed and                  measured will suddenly reach its maximum as solids are                becoming present that have complete absorption.

I was told that the last option (colour or spectrometry) was "on the right direction".

Question

So, with all this in mind...

Are there any quantitative means of measuring the rate of the above reaction?

Measuring conductivity over time might do the trick (7 ions (8 charge equivalents) in, 4 ions/charge equivalents out).

Regarding spectrometry: you could deliberately make use of the light scattering caused by the precipitating sulphur. Should be somewhere in the 600+ nm range. It's a standard method in biological sciences to follow the growth of bacteria.


Confusion in IUPAC Nomenclature I was trying to name the following compound but I got stuck in naming the aldehydic part of the compound.

I looked for the answer and found it as 6-bromo-2-oxocyclohexanecarbeldehyde. I had come to know earlier that carbeldehyde suffix is used when there are more than two aldehydic group in the compound but why carbeldehyde is used here? I am not able to understand this. If someone can explain which rule is used here. I will be very grateful for this.

Why …carbaldehyde?
It's similar to $\ce{C6H11-COOH}$ cyclohexanecarboxylic acid (not cyclohexanoic acid). It's the case where the functional group's carbon is not included in the parent structure. (Such case are also nitriles/carbonitriles, $\ce{R-CN}$)
Why not …hexyl… or rather …cyclohexyl… ?
This (wrong name) would be constructed similarly to a different, non-preferred type of nomenclature, a radical name, which names e.g. ethanol as ethyl alcohol.
There are more than two aldehydic group in the compound.
They are not. Alhedyde group is ${\ce{-CHO}}$. The other one, $\ce{>C=O}$ is an oxo group (ketone). The group for suffix is chosen by the pre-defined priority, ${\ce{-CHO}}$ > $\ce{>C=O}$.
Why 2-bromo-6-oxo-… and not 6-bromo-2-oxo-… ?
The groups named by prefixes (after highest priority - suffix group is assigned lowest locant possible) are not numbered by their priority, but such that the assigned locant set is lowest possible. If there are more equivalent possibilities, then they are numbered in the alphabetically sorted order (which is used in the name).


Degradation mechanism of hydroxymethylfurfural to 2,5-dioxo-3-hexenal and 4-oxopentanoic acid I want to understand the degradation mechanism of hydroxymethylfurfural (HMF). According to the Wikipedia article, HMF is degraded in honey. Someone said that the degradation proceeds via 2 possible routes and that the products are 2,5-dioxo-3-hexenal and 4-oxopentanoic acid.

What is the mechanism by which these compounds form? If you add a picture of the mechanism, it will be better.

Here is a link to a detailed reference on HMF.  In section 3.3 (page 32) they show a mechanism for the acid catalyzed degradation of HMF.  As you noted, they do propose two pathways for decomposition, one leads to polymeric materials, the other leads to 2,5-dioxo-3-hexenal, which then can react further to produce 4-oxopentanoic acid. The work they describe was done by Horvat (Horvat, J.; Klaič, B.; Metelko, B.; Sunjic, V. Tetrahedron Lett. 1985, 26, 2111).

I don't want to reproduce their drawing, but follow the link and look for Scheme 17.  Most of the steps involve protonation of either

a double bond to form a resonance stabilized carbocation followed by addition of water and elimination of a proton to yield an alcohol
a hydroxyl group followed by elimination of water to yield a resonance stabilized carbocation followed by elimination of a proton to yield a double bond
or protonation of the furan oxygen followed by ring opening to generate a protonated, resonance stabilized carbonyl

If you have any specific questions on the mechanism, just post back here.


Stereospecificity of the bimolecular electrophilic substitution reaction We know that $\mathrm{S_N2}$ reactions always go with inversion at the saturated centre (they're stereospecific) and $\mathrm{E2}$ reactions are always anti (also stereospecific).

I recently observed in a paper an example of what was called an $\mathrm{S_E2}$ reaction, but can't understand how the stereochemistry could be predicted in this case? Is it also stereospecific, or is there some other factor involved?

The $\mathrm{S_E2}$ mechanism is most commonly found using organometallic reagents ($\ce{R-Li, R-MgBr}$ etc.) in which the metal is the electropositive element react with electrophiles.
Unlike with $\mathrm{S_N2}$, there are multiple stereochemical outcomes, which can make it difficult to determine whether the reaction is indeed stereospecific. This is described in Modern Physical Organic Chemistry (emphasis mine):

One of the most interesting aspects of these reactions is stereochemistry. Varying results are obtained depending upon the metal, the solvent, and the R-group.
Source: Modern Physical Organic Chemistry, Ansyln and Dougherty [1]

Criterea for a stereospecific reaction
Before going any further, it may be instructive to define what we mean by stereospecific:

stereospecific is used for those reactions where the configuration of the starting material and the configuration of the product are related in a mechanistically constrained way
Source: Molecular Orbitals and Organic Chemical Reactions (Reference Edition), Fleming [2]

What this means in practice is that if one enantiomer of a starting material gives one enantiomer of product, then the opposite enantiomer of starting material must give the opposite enantiomer of product. What stereospecific does not mean is high selectivity, many reactions are highly stereoselective but in no way mechanistically stereospecific (and equally stereospecific reactions are not always $100~\%$ stereoselective).
The empirical evidence
Unlike the $\mathrm{S_E2}$ reaction, where many examples have been studied, it has been considerably more challenging to examine the stereochemical cause of the $\mathrm{S_E2}$ reaction. In general, chiral alkyl-metal species are prone to racemisation (chiral Grignards, as an example, are barely known), and those chiral species that are stable and usually such for steric reasons (such as a constrained ring system).
Many of the experiments that have been conducted are described in March[3], and the overwhelming majority of them go with retention (note that this doesn't necessarily imply a stereospecific process in itself and it has often been hard to study both enantiomers, for various reasons).
The orbital explanation
If we consider the orbitals involved during an $\mathrm{S_E2}$  reaction, it becomes clearer why there are multiple stereochemical outcomes, consider first $\mathrm{S_E2}$  at a saturated centre, such as $\ce{MeLi}$ attacking an electrophile :

Source: Molecular Orbitals and Organic Chemical Reactions (Reference Edition), Fleming [2]
In this reaction, the electrophile is quite able to interact in such a way as to give rise to overall inversion or overall retention, in order to distinguish, these two electrophilic substitutions are labelled $\mathrm{S_E2}_\text{back}$ and $\mathrm{S_E2}_\text{front}$ respectively, though both are formally designated DEAE if the IUPAC system is followed.[4]
This differs from the $\mathrm{S_N2}$ reaction in which inversion is almost always observed due to an anti-bonding interaction in the transition state if the nucleophile approaches in such a way as to give retention:

Source: Molecular Orbitals and Organic Chemical Reactions (Reference Edition), Fleming [2]
Conclusion
The short answer is that it doesn't appear that the reaction is stereospecific. Orbitally speaking, there are two pathways leading to two different products (front or back attack), which may be observed in poor selectivity.
It appears, with the reactions studied so far, that the outcome is far more likely to be down to sterics/electronics as to whether the electrophile approaches from one side or the other

References and Notes:
[1]: Ansyln, E.V.; Dougherty, D.A.  Modern Physical Organic Chemistry; University Science Books:California, 2004
[2]: I. Fleming  Molecular Orbitals and Organic Chemical Reactions (Reference Edition); Wiley:Chichester, 2012
[3]: March, J.; Smith, M.B.  Advanced Organic Chemistry; Wiley:New Jersey, 2013
[4]: IUPAC Recommendations for the Representation of Reaction Mechanisms: Acc. Chem. Res. 1989, 22, 343


What are scattering losses? In UV spectroscopy, when a beam of light is shone into solution, what is scattering loss during this process?

Scattering is the redirection of the 'beam' of light, as is illustrated in the diagram below:

Source: University of Wisconsin

The description from the document Fundamentals of modern
UV-visible spectroscopy (Figure 46) states that

  Scattering causes an apparent absorbance because less light
  reaches the detector

They also detail two main types of scattering in UV spectroscopy (Figure 47):

  • Rayleigh scattering: Particles small relative to wavelength
  • Tyndall scattering: Particles large relative to wavelength


What are the resonance structures for chlorite anion? What are the possible resonance structures for $\ce{ClO_2^-}$? Assigning one double bond to the structure makes for formal charges of $\ce{O(-1)-Cl(0)=O(0) <-> O(-1)=Cl(0)-O(-1)}$. It appears that the same set of formal charges can be achieved with two double bonds (expanding the octet on $\ce{Cl}$ to 12 electron capacity, using the d orbital). That is, $\ce{O(0)=Cl(-1)=O(0)}$.

Is this feasible, and if not, why not?

Disclaimer
The here used methods are also only approximations and I do not claim that the provided numbers are accurate. However, they are used here to make the point, that bonding situations involving resonance are not easily explained.

Oxygen is slightly more electronegative than chlorine, therefore the bonds will always be slightly polarised towards the oxygen.

The following resonance scheme is based on an NBO analysis on the DF-BP86/def2-SVP level of theory. The percentage value gives an estimate how close the given conformation is to an idealised Lewis structure. From this you could also derive how much such a structure would contribute to a resonance situation (in terms of Valence Bond theory).
What we do see here is, that the electronic structure is better described with more formal charges. The contribution of a π bond is best described by an ionic donor-acceptor structure, rather than breaking the octet rule at chlorine.
Noteworthy is also that the doubly ionic donor-acceptor configuration is as equally good a lewis structure than any that breaks the octet rule.
The worst representation is the one with a four electron three centre bond (which is also not part of the Lewis bonding picture anyway).

Videos like the one mentioned in the Wikipedia article are easy to follow, but often they are too simplified to be correct.


Nomenclature of an Alkane Is the correct name for the following structure 3-tert-butylhexane or 3-ethyl-2,2-dimethylhexane?

I think it is 3-tert-butylhexane, but the textbook says it is 3-ethyl-2,2-dimethylhexane. Can someone please explain?

According to IUPAC (source):

  2.6 - If chains of equal length are competing for selection as main chain in a saturated branched acyclic hydrocarbon, then the choice goes in series to:
  (a) The chain which has the greatest number of side chains.

Therefore the correct name is 3-ethyl-2,2-dimethylhexane as opposed to 3-tert-butylhexane because the former has more side chains.


Flame Ionization Detector versus Mass Spectrometer Why is chromatographic separation more important with a flame ionization detector than with a mass spectrometer?

FID requires separation as it is not specific and cannot distinguish 2 non separated compounds.

MS is specific and can determine the particular compound even qualitatively.

For complex matrices, it is always good to have saparation even for MS, as MS spectrum of mixture would be confusing.


Is there a difference between energy levels and electron shells? Is there any difference between energy levels and electron shells, or do they mean the same thing? Does the number of energy levels remain constant when progressing across a period?

Introduction

The absolute energy levels of all electrons in an atom are different, and will shift subtly when electrons are added or taken away. However, it is true that, generally speaking, sets of electrons in an atom will tend to have energy levels very closely clustered together, with big differences in energy to other sets of electrons in the atoms. We say that these electrons are all in the same energy level. On the other hand, election shells have to do with the mathematics of the electrons 'orbitals', and are closely related, but are not the same. If you want to avoid 'hair-splitting' detail at this point, the difference is this -

Energy levels

Each electron in an atom belongs to an energy level in which all the electrons in that level have nearly exactly the same energy, but there are big energy differences between the levels. If you were to strip away all the electrons from an atom and then return them one by one, you would find that you could only put 2 in the lowest level and then 8 in the next. The capacity for the first five levels are 2, 8, 8, 18, 18 in that order. This corresponds to the number of atoms in each of the first five periods of the periodic table and helps explain the properties of elements in the periodic table. This makes 'energy levels' a very useful idea. So energy levels go from 1 to 5 for the first 5 periods of the periodic table, and in simple terms, full energy levels tend to be 'desired' by atoms.

Electron shells

Electron shells are slightly different, but closely related. They are numbered, like the energy levels. The first shell, 'is' the first energy level. Likewise, the second shell, is the second energy level. But after this, it becomes more complicated, and this is where the confusion can come in.
With the exception of the first shell, each shell is made up of sub-shells. These sub-shells make up each energy level, but sub-shells can belong to different numbered energy level from shell 3 onwards.
To add some detail. There are 3 kinds of shell, 3 ways of satisfying the quantum maths for bound electrons, in the first 5 periods. There are labelled, 's' shells, 'p' shells and 'd' shells. In the mathematics for these, there are 'shell numbers'. For shell number '1', the mathematics only allows for 's'-shells. For shell number 2, 's'-shells and 'p'-shells are allowed, and for shell number 3, s, p and d-shells are allowed. S-shells can take 2 electrons, p-shells 6 electrons and d-shells 10. So shell level 1 has 2 electrons, level 2, 8 (2+6=8) electrons, and level 3, 18 (2+6+10=18) electrons, giving the pattern 2, 8, 18. This is clearly different from the energy levels, and to know why, we have to know more about the sub-shells.
The sub-shells are written down with the shell number first, the shell type second, and the number of electrons actually in that shell, third, (usually as a superscript). So Hydrogen's electrons would be 1s1, and Helium's 1s2. For the second energy level, Li would be 1s2 2s1, and further across the period, Carbon, 1s2 2s2 2p2, having six electrons. At the end of the period, Neon would have two full energy levels, written as 1s2 2s2 2p6. (Often lower full shells are ignored, so Neon would be written 2s2 2p6.) So far, electron 'shell numbers' coincide with energy levels. However, the third energy level has only 3s and 3p shells, not 3d. This is because the 3d shell has an energy level between 4s and 4p. So the 3d sub-shell belongs to energy level 4. Similarly, the 4d sub-shell has an energy level between the 5s and 5p sub-shells. These extra 10 electrons in the energy levels 4 and 5 produce the transition metal block in the periodic table.

Conclusion

Shells and energy levels come from the properties of atomic electrons, but one comes from the mathematics of electon orbitals, and the other from the energy that the electrons take in these sub-shells. This means that referring to energy levels as ‘shells’ will produce confusion, when an understanding of ‘sub-shells’ is needed for a better understanding of the periodic table chemistry.


How does a chemical "know" which other species are present and where equilibrium is? Please forgive me if this is a vague question. I have always wondered how a chemical reaction "knows" where its equilibrium should be. For example, using a basic example of Le Chatelier's principle, for some theoretical reaction
$$\ce{A + B <-> C + D}$$
 If I were to add extra A and B to the tube, the reaction would proceed to the right until a new equilibrium was established. How do the chemicals "know" that I have added extra A and B?

Am I thinking about this incorrectly? Is it not actually a matter of the chemicals "sensing/knowing" which other species are present, but rather their own individual properties and how likely they are to exist in one form or another?

Related to this quandary, I learned in my metabolism course that NADH and NADPH are used to regulate different reactions because the NAD+/NADH and NADP+/NADPH ratios need to be different, and that merely adding the phosphate allows these separate species to have different equilibria. Can NAD+ really distinguish NADH from NADPH?

What we got here is a dynamic equilibrium, not a static one. For a static one the molecules would need to know when to stop reacting or when to react in a certain way, in a dynamic equilibrium you don't need to know.

Here's a ELI5 explanation of what's happening:

  Let's say you've got a garden with an appletree and hundred of apples
  are on the ground. You don't want them there so you throw them to your
  neighbor. he's an old man and doesn't want them either, so he throws
  them back. You are much faster than him and there are hundreds of
  apples around you, so you don't even need to move to throw them and
  you can throw them very fast. He on the other hand has only a few
  apples laying around and has to move quite far, so he's throwing them
  back much slower. But what happens after time?

  Well, there will be a point where there are more apples on the old
  mans side than on yours, so you have to run a lot to get to those
  apples while the old man has to move only some steps to reach a new
  apple. Eventually it will level out and both of you will throw apples
  at the same rate. That's a dynamic equilibrium. Because there are more
  apples on the old mans side he can throw them back as fast as you can
  throw them to him, even if you are faster.

Now here the chemical explanation:

Reaction rates are dependent on the concentration. High concentration equals fast reaction, low concentration equals slow reaction. At the beginning you got only starting material, it will react fast to the product. The product will react back but there is extremly low concentration so it's slow. But after some time you will reach a point where the concentration of the starting material is so low that the reaction is so slow, that it matches the speed of the back reaction which got faster because you got more and more product. At that time there will equal amounts react in both directions, making it look like the reaction is standing still.

For example: If we got the reaction $\ce{A + B \rightleftharpoons C + D}$ and the reaction rate of the forward reaction is $r_1=k_1[A][B]$ and the backwards rate is $r_2=k_2[C][D]$ then we will reach an equilibrium $r_1=r_2$ and that's the case if $k_1[A][B]=k_2[C][D]$.


Are phase transitions reversible when performed at supercooled temperatures? For the state transition of supercooled water to ice, is the process reversible? I wonder this because I want to know if I can use entropy change of the universe is 0. If it isn't reversible, how do you find the entropy change of the system? I know you can find the entropy change of the universe by calculating the heat of the process and divide by the temperature. Also how does the heat capacity factor into this problem? You can calculate the new enthalpy change at the new temperature but how does that help you with the entropy?

The entropy change for a system is the same whether or not the process that takes the system from its initial to final states is performed reversibly. That's because the entropy is a state function. The difference between a reversible and irreversible process, all else being equal with the terminal states of the system, lies in what happens to the surroundings.

Therefore, to find the change in entropy of the system you start by describing a reversible path for which you can compute the entropy change. Such a path might consist of the following steps:

(1) Warm supercooled water from the initial temperature ($\pu{-20^\circ C}$) to the normal freezing point ($\pu{0^\circ C}$)

(2) Freeze the water to ice at the normal freezing point ($\pu{0^\circ C}$)

(3) Cool the ice from the normal freezing point to the final temperature ($\pu{-20^\circ C}$)

The path involves reversible heating and cooling of supercooled water  and ice, respectively, and the intermediate step of freezing water to ice at the normal freezing point of water ($\pu{0^\circ C}$).

You can use the heat capacities to compute the entropy changes during heating and cooling of the system, respectively, from the initial and to the final states. For one mole of water:

$$\Delta S_{m,T_i\rightarrow T_f}=C_{p,m}\ln\left(\frac{T_f}{T_i}\right)$$

making sure to use the appropriate heat capacity (assumed constant over the temperature range) for water or ice.
For the first step, you warm up the supercooled water, and compute the entropy change by assuming that the supercooled state is stable. For the final cooling step, you compute the entropy change of the cold ice.

For the intermediate freezing step

$$\Delta_fS_m=\frac{\Delta_f H_m}{T_f}$$

The molar change in entropy for the system undergoing the irreversible freezing process is then just equal to the sum of the above changes for the three steps.

As for the computation of the entropy change of the surroundings: you compute the enthalpy change for the alternate path; then, since the enthalpy change for the actual irreversible process is the same as that for the reversible path, and the process is isothermal and isobaric (the intial and end states have same T and p), you can equate the enthalpy change with the heat, and can compute the change in entropy of the surroundings by dividing the heat by the temperature, making sure to use the correct value ($\pu{-20^\circ C}$).


Relativistic effects in element 137 (Feinmanium) and above Here I saw that if an element above atomic no 137 has to exist, it must have electron speed greater than speed of light. My question is , has this calculation  been done keeping in mind Einstein's relativity? {I am just asking, I have not done this calculation.}

Yes, the value 137 occurs even considering Einstein's theory of special relativity.

137 comes from considering Sommerfeld or Dirac relativistic theories, when the nucleus is modeled as a point.

See equation 1 of A new method for solving the Z > 137 problem and for determination of energy levels of hydrogen-like atoms


What will be the colour of HCl with methyl orange? I am new in chemistry and this question can be silly, but I am troubled with this question.

According to the pH range of methyl orange, if I add methyl orange into water, it gives a yellow color (pH 7). Then, if I add some HCl drops (a very little amount) into the water, what will be the color?

Usually, the colour of HCl is red with methyl orange (because of acid), but when HCl is added to water, will it be a colour within yellow and red, or pure red in colour?

Methyl orange (MO), as you know, is an acid-base indicator.

The problem with those is that they do not distinguish between acidic and basic medium, but rather between different pHs. MO has a transition interval of 3.1-4.4, which means it's red below 3.1 and yellow over 4.4. Within this interval it's all about transition colors (red through orange to yellow).

Answering your question: a few drops of HCl would affect water differently. Let's say, you had 1 L of water and you added 3 drops (0.15 mL) of concentrated saturated HCl (~40% w/w). That leads us to  ~0.002 mol of HCl added and to pH lower than 3, so the MO would turn red. If you had diluted solutions, like 0.01M or so, the calculations would tell you the pH would still be higher than 4.4, so the MO would still be yellow.


Why don't free radicals undergo rearrangement like carbocation/carbonium ions do? Why don't free radicals and carbanions undergo rearrangement like carbocation/carbonium ions undergo in a reaction?

First off, your observation is correct, rearrangements involving free radicals and carbanions are noticeably less common than rearrangements involving carbocations.

Many common carbocation rearrangements involve 1,2 shifts.  If we look at the MO diagram for a 1,2 shift and compare shifting a substituent to an empty p orbital (carbocation, case A in diagram) or to a filled p orbital (carbanion, case B in diagram), we see that the carbocation process involves a 2-electron system whereas the carbanion process is a 4-electron system.  The 2-electron process is allowed to proceed themally according to orbital symmetry, but the 4-electron process would need to proceed with inversion of configuration (bonding to the other lobe of the p orbital as we transform from reactant to product) in order to be thermally allowed - and this would be geometrically difficult.

(Source - V8rik at the English Wikipedia via Wikimedia Commons)

If we consider hydrogen shifts and replace the p orbital in our digram with an s orbital, we see that a 1,2 hydrogen shift in a carbocation is thermally allowed, but the corresponding shift in a carbanion is thermally forbidden (with an s orbital, there is no "other" lobe and proceeding with "inversion" is not possible)

Since thermal 1,2 shifts with retention at the migrating center are not allowed in carbanions by orbital symmetry, and a 1,2 shift with inversion at the migrating center is geometrically difficult, such processes would involve higher energy transition states.  As a consequence of the higher energy, 1,2 shifts in carbanion systems are less common than orbital symmetry allowed, thermal 1,2  shifts involving carbocations. One might reasonably expect that a radical process would be somewhere in between these two cases; and again the transition state would be of higher energy (since it is not orbital symmetry allowed) making such rearrangements less common.

While radical and carbanion shifts are not as common as carbocation shifts, they do occur.  In the case of radicals it requires high temperatures (hydrocarbon cracking) or weak bonds (peroxides for example). Carbanionic rearangements are more common when the carbanion is stabilized.  This can be achieved by having an electronegative heteroatom attached to the carbanionic center (as in the Wittig rearrangement) or by placing the anion on an $\ce{sp^2}$ or $\ce{sp}$ hybridized carbon, rather than an $\ce{sp^3}$ hybidized carbon (the Fritsch rearrangement provides an example here).


Heat of combustion per mole and per kilogram I need to calculate the heat of combustion per mole and per kilogram of complete combustion of, say, X.

I write reaction : $\ce{X + O2 -> ... + Q}$
Find Q using Hess law

Now, what do I do after that?

Q is the heat of combustion per mole. Next, you would use the formula weight to determine the heat of combustion per g (or kg).


Why is vitamin C called an antioxidant? According to Wikipedia, vitamin C (ascorbic acid) "may also act as an antioxidant against oxidative stress," and is "widely used as a food additive, to prevent oxidation."  But acids are, by definition, oxidizing agents, so how is it that ascorbic acid is used to prevent oxidation?

Ascorbic acid is considered an antioxidant because it can supply an electron to a species with an unpaired electron (a "radical", labeled "X" in the below diagram) to stabilize the radical. Of course, since ascorbic acid has now lost an electron, it is itself a radical - but it is a very stable radical due to resonance. Below I have included a diagram of this process below, from UC Davis.

Note how the we can "electron-push" to get the unpaired electron in the middle structure to move to the oxygen on the other side of the ring. This resonance delocalization of the unpaired electron in the ascorbyl radical makes it much more favorable to have the unpaired electron in the ascorbyl radical rather than the X.

Because radicals normally poach electrons from surrounding molecules to pair with their lone electron, they normally cause a chain reaction where their "victim" goes on to pull electrons from another neighboring molecule (oxidation), and so on and so forth. However, because the ascorbyl radical is stable, it is less likely to poach an electron from another molecule in its environment, preventing the chain reaction from occurring. Thus, it is an antioxidant.


Calculating the cell potential of a half cell from two half cell equations I understand the premise of the question and how to solve it but I don't quite understand why I'm a factor of 2 out:

Construct a potential diagram (at $\mathrm{pH} = 0$) for the reduction of aqueous $\ce{HNO2}$ to $\ce{NO}$ and then to $\ce{N2O}$ given that $E^0$ for the $\ce{HNO2/NO}$ and $\ce{NO/N2O}$ couples are $\pu{+0.98 V}$ and $\pu{+1.59 V}$ respectively. Calculate $E^0$ for the following half-reaction:
$$\ce{2HNO2 + 4H+ +4e- <=> N2O + 3H2O}$$

So by constructing the two half-equations:
$$\ce{HNO2 + H+ +e- <=> NO + H2O~~~~~~~~~~~~(E^\circ= +0.98V)}$$
$$\ce{2NO + 2H+ +2e- <=>N2O +H2O~~~~~~~(E^\circ= +1.59V)}$$
And so by adding 2 times the first equation to the second, you obtain the desired overall equation. And so you would add the individual reduction potentials to obtain the overall potential:
$$\pu{+0.98 V + 1.59 V} = \pu{+2.57 V}$$
The answer stated however is $\pu{+1.29 V}$, ie, half of my answer.
Why do we need to divide by $2$ at the end here? The standard reduction potential doesn't depend of stoichiometry, and when doing similar questions combining two half cells into a full equation (with no electrons in the equation), simply subtraction or adding the equations is sufficient. Why the difference here?

Josh, you and others have missed the key point. It is not a simple question! They are asking you to draw a potential diagram not Ecell, basically a Latimer diagram.
So take the approach of using a Latimer diagram. The book answer is correct. Nitrous acid when goes to nitrous oxide, the potential must be +1.29 with respect to SHE. Check the web for solving Latimer diagrams.
You would work on the portion starting from nitrous acid to nitrous oxide.