Why are multiple oxidation states useful for a catalyst?

The answer has a lot to do with "what is a catalyst in the first place?"
A catalyst is a compound that can affect the rate of a chemical reaction
by providing an alternative and lower energy profile or pathway. That is,
it is a substance that makes it so that A -> B at a lower energy cost. But
it only has an effect on the rate of the reaction. That is, it only changes
the cost of the activation energy. It is not related to the thermodynamics
of the process and hence, the final product distribution. That is, if the
above process gives 90% B via a non-catalyzed pathway in 4 days, then a
catalyst will help you get 90% B in say 4 hours.

Okay, so a catalyst affects the transition state and activation path. How
does it do this? Typically, by complexing one of the reagents. Complexation
by transition metals affords access to a wide variety of oxidation states
for the metal. This has the property of providing electrons or withdrawing
electrons from the transition state of the reaction. That is, if the
transition state is electron rich, then the transition metal might hold
some of that electron density and those prevent too much from building up
on the reagent. This would then facilitate the reaction. Or the transition
metal might undergo formal oxidation/reduction to achieve electron transfer
to a substrate, thereby allowing a reaction to occur. This is "complexation
and electron storage" taken to the extreme but is a common mechanism in
organometallic chemistry. Indeed, a variety of catalytic pathways rely on
a two electron transfer between the metal and the substrate (e.g.
hydroformylation). It is the ability of the transition metal to be in a
variety of oxidation states, to undergo facile transitions between these
oxidation states, to coordinate to a substrate, and to be a good
source/sink for electrons that makes transition metals such good catalysts.

Of course, this is for transition metal complexes. Most industrial used
catalysts are the transition metal in a bed, as a metal or bound structure.
The above considerations are important but also the physical properties of
absorption/adsorption and the electron band structure of the material.


In a crystal lattice, there are two contributions to the entropy: the vibrational entropy and the configurational entropy. The first is related to the kinetic energy; atoms with higher kinetic energies have tend to have larger displacements from equilibrium, and have a larger "footprint" in phase space, resulting in a higher entropy. The second contribution, configurational entropy, arises from disorder in the crystal lattice, namely the number of ways of arranging the atoms. In a perfect crystal, there is only one way, so the configurational entropy ($k\ln W$) will be 1.

An counterexample to the suggestion you make is any disordered solid, whether it's amorphous, a random alloy or some kind of crystal with geometric frustration. The most obvious example of this is water ice. Each water molecule occupies a lattice position and is tetrahedrally coordinated with its neighbours. In a defect-free ice crystal, the ice rules state that each molecule must accept exactly two hydrogen bonds and donate exactly two hydrogen bonds. This is quite a strong constraint, but allows orientational disorder --- the molecules can be oriented in one of six possible ways at each lattice position, resulting in an orientationally disordered structure. This will have a finite entropy even at absolute zero.


Where does Free Energy go in a Redox Reaction?

Free energy is a state function. No matter how you run the reaction, if you start with a certain state (set of concentration) and end with a certain state (set of concentrations at equilibrium), the change will be the same.

The free energy of reaction is the maximal amount of non-pV work (electrical work in this case) the reaction can do. If it does not do any work, that energy will be released as heat instead. For example, if you short the battery, the wire and the electrochemical cell will heat up while the reaction is going on.

If you want an argument from first principles, you should look at the first law of thermodynamics. The change in energy of a system is equal to the sum of the heat and the work exchanged with the surrounding. When making optimal use of the free energy, you maximize the work (even sometimes transferring heat from the surrounding to do even more work).


Does unsubstituted benzoic acid not show resonance due to steric effects?

I have never heard of this effect occurring with an ordinary, unsubstituted benzoic acid. There is, however, a phenomenon known as the ortho effect that occurs when bulky groups occupy the ortho positions on the ring. In that case, the steric hindrance between the bulky groups and the carboxylic acid moiety forces the carboxylic acid group to rotate out of the plane of the aromatic ring, thereby breaking the conjugation.

One of the consequences of this, which may not be readily apparent, is that this often produces a substantial increase in the acidity of the carboxylic acid, even when the substituents are not electron-withdrawing (viz. 2,6-dimethylbenzoic acid, which is disubstituted at the two ortho positions, having a $\mathrm{p}K_\mathrm{a}$ roughly one unit lower than ordinary benzoic acid, while similar para- and meta-substituted benzoic acids have higher $\mathrm{p}K_\mathrm{a}$ values — that is, are less acidic).

While I've seen references to this effect in multiple sources, I've never actually found a thorough explanation for it. I think it could be argued that the benzene ring is electron-donating within the $\pi$-system (i.e., by resonance), which would only serve to destabilize the carboxylate conjugate base already bearing a negative charge. By rotating out of the plane, the electron-donating effect is removed. Rather than consider the stability of the resulting carboxylate, I suppose one could also consider the stability of the carboxylic acid directly. Here, I'd expect that the reactivity of the carboxylic acid flanked by bulky substituents should be greater, as the loss of $\pi$-conjugation with the carboxylic acid moiety is inherently destabilizing. My reasoning may be flawed, though, so if anybody has some corrections or a better explanation to offer, that would be entirely welcome.

Whether this effect occurs with ordinary benzoic acid or not, I can't be certain. At least one good reason I can think of to doubt that it does is the observed regioselectivity of electrophilic aromatic substitution reactions on benzene rings conjugated to carbonyl groups. They are known to be deactivating (i.e., they withdraw electron density) via their conjugation with the $\pi$ system of the aromatic ring. If they were not coplanar with the ring, the conjugation would be impossible, and the deactivating effect would not be observed.


Why are atoms with eight electrons in the outer shell extremely stable?

First, this isn't quite true.  It is true for the first row of the periodic chart (from lithium to neon). It is almost true for the second row (from sodium to argon.  But there are exceptions there.  Beyond that it really isn't true at all for the elements beyond the first two columns.

The reason for the increased stability for the first two rows lies in quantum mechanics.  Classically we can note that there are no $d$ electrons there.  Another ways of looking at it from a classical point of view is that the early elements are too small to allow too many other atoms or groups of atoms around them.  That tends to go away as you go down the periodic chart and the atoms get "fatter".   A typical example is chloroplatinic acid which has six chlorines around it.

Most transition metals also can have more than four groups around them as well.

I suspect that this isn't an exceptionally useful explanation.  As I said, the answer really lies in quantum mechanics.  In looking up "molecular orbital theory",  one reference can be found here.


Why don't particles of a colloid settle?

Particles in colloid absorb either positive or negative ion preferentially from solution depending on there surface. This develops a potential on colloid surface called Zeta potential. Due to this potential colloid particles repel each other and can't coagulate.

Particle size also play a major role. With decreasing size surface to volume and hence surface to weigh ratio increases. So, same volume  of colloid  particles  have higher surface area than suspension. So, colloid surface have enough charge to keep them apart whereas suspension don't have.

Conclusion Both the point you've mentioned is responsible for prevention of colloid coagulation.


Why does the alkyl group anti to the hydroxyl migrate in Beckmann rearrangement?

Oximes can undergo the Beckmann rearrangement.  Oximes also exist as stable syn and anti isomers.  In the figure below, $\ce{R_1}$ is anti to the hydroxyl group; another isomer exists where it is syn the the hydroxyl group.

In the first step of the Beckmann rearrangement the protonated hydroxyl group makes for a good leaving group.  As the hydroxyl group begins to depart, the $\ce{R}$ group which is anti to the leaving group begins to migrate and form a bond with nitrogen.

(Wikipedia)

This step is said to be concerted (rather than stepwise) with the $\ce{N-O^{+}H2}$ bond being broken and the new $\ce{R-N}$ bond being formed at (more or less) the same time.  Whether you think of the $\ce{R}$ group as participating in something similar to an SN2 reaction or whether you think of the $\ce{R}$ group as starting to bond to the available $\sigma^{*}$ orbital of the elongating $\ce{N-O}$ bond (equivalent descriptions), you see that having the $\ce{R}$ group approach the nitrogen from a 180° angle from the breaking $\ce{N-O}$ bond would be preferred. Hence, the $\ce{R}$ group anti to the $\ce{N-O}$ bond migrates preferentially.

There are many reactions showing this geometric preference, the Baeyer-Villiger oxidation being an example.


Is the carbanion formed in benzyne mechanism non-planar?

I have given one of these molecules a little spin on the computer. With whatever I tried, I was unable to find a non-planar solution. I really tried my best. Below is an animation (sorry for the poor quality) of an attempt which uses a chair like starting geometry. If there was any non-planar geometry lower in energy, this would have found it.

The level of theory is DF-BP86/def2-TZVPP and the program used is Gaussian 09 rev. D.01.
There is really not much more I can add to the discussion.

  I was told by my chemistry teacher that the carbanion formed in the picture above is non-planar. I don't understand why. It looks like benzene with an extra electron pair at one of the carbons so shouldn't it be planar?

To be absolutely, brutally honest here: I have no clue why your teacher had this idea. If this wasn't a "naked" carbon, we could talk about it, but it is essentially a deprotonated phenyl ring, and as such, you are absolutely right in assuming the lone pair being in phase with the ring. The π system is hardly different from benzene. To fill this post a little bit more, have those fancy π-orbitals.

(I could not find the completely anti-bonding pi-orbital.)

And for good measure, here is the HOMO, which is the lone pair at the carbon:


Creation and Combustion of Nitrocellulose

Flash paper is commonly used in classroom demonstrations, I think, and it is made of nitrocellulose.  A journal article from 1995 describes rapid creation of flash paper through nitration with nitric acid.

I have never done this procedure myself, but looking through it the biggest safety risk to my eyes is using and handling concentrated nitric and sulfuric acids.  $\ce{H2SO4}$ and $\ce{HNO3}$ are very dangerous especially when concentrated.  Full PPE when handling these is an absolute must.  I'm glad to hear in your question that you are already thinking about safety requirements at this step.

Another important risk is the inherent risk of lighting anything on fire.  Nitrocellulose is an energetic material.  Since it is heavily nitrated, it needs less oxygen gas to completely burn than things like wood, paper, or even gasoline: a lot of oxidizer is already built into the molecule.  This risk is best managed by only ever igniting small quantities of nitrocellulose, and of course by storing any larger quantities of the nitrocellulose far away from sources of ignition.  Full PPE is a good idea too.  Perform the combustion in a well-ventilated area.

I'm less worried about the particular risk incomplete combustion forming dangerous levels of $\ce{NO_{x}}$, especially if you limit the combustion to small sheets of nitrocellulose at a time.  High levels of $\ce{NO2}$ are unlikely to form via incomplete combustion; if large amounts did form, since $\ce{NO2}$ is a strong oxidizer, it would react with remaining nitrocellulose to less toxic, more reduced nitrogen oxides or more likely to nitrogen gas.

Small amounts of $\ce{NO2}$ do form from high-temperature combustion of anything in air, but if you use small amounts of flash paper, exposure of your students to $\ce{NO2}$  is likely to be background sources in the atmosphere will vastly exceed anything produced by your demonstration.

All that said, don't take too much safety advice from the internet.  Talking to experienced chemists in person who have done this demonstration is probably the single best thing you can do to stay safe.

Update: I should explicitly say that although I have never made flash paper, I have burned small pieces of it on numerous occasions, never with any problems or even noticeable amounts of smoke or fumes.


How to calculate crystal field stabilisation energy?

Step 1: Look up Nickel Carbonyl and find out what geometry it has.

We need the geometry to know how the $d$ orbitals will split in the ligand field. The geometry can also be predicted: late transition metals with strong field ligands tend to be tetrahedral.

  Step 2: Find the appropriate crystal field splitting diagram for this geometry.

  Step 3: Figure out how many $d$ electrons there are.

Nickel is in Group 10, with a configuration of $\ce{[Ar]}4s^2 3d^8$ or $\ce{[Ar]}4s^1 3d^9$. In coordination compounds we will consider this configuration to become $\ce{[Ar]}4s^2 3d^8 \implies \ce{[Ar]}3d^{10}$. The $s$ electrons are considered to move to the $d$ sub-shell in bonding - bonding need not represent ground state atomic electron configurations just group state molecular electron configurations. For more information on $d$ electron count, go to this Wikipedia page.

  Step 4: Fill in the $d$ electrons.

You can do this part yourself. How do you put 10 $d$ electrons in the orbital diagram?

  Step 5: Determine $\Delta E$.

$\Delta$ is the energy different between the $e_g$ and $t_{2g}$ sets of orbitals. This Wikipedia page walks through an octahedral complex. In tetrahedral complexes, the energy of the $e_g$ orbitals is lower and the energy of the $t_{2g}$ orbitals is higher. The energies are:

$e_g \ (d_{x^2-y^2},\ d_{z^2}) \ E=-\frac{3}{5}\Delta_\textrm{tet}$ (stabilized)
$t_{2g} \ (d_{xy},\ d_{xz},\ d_{yz}) \ E=\frac{2}{5}\Delta_\textrm{tet}$ (destabilized)

Then multiply the numbers of electrons in the orbitals by the stabilization/destabilization values, and sum.

$$\Delta E = 4\times\left(-\frac{3}{5}\Delta_\textrm{tet}\right)+6\times\left(\frac{2}{5}\Delta_\textrm{tet}\right)=?$$


The solvent used is a polar protic solvent which heavily solvates the OH- radical thereby decreasing it's nucleophilicity as the solvent forms a cage around the radical anion .Even though the basic strength of the anion is also decreased by solvation still the decrease in nucleophilicity is dominant as it is more difficult for a large solvated anion to attack at a carbon atom than to sneakily abstract an acidic hydrogen and hence the enolate is formed which then , proceeding through the SN NGP (neighbouring group participation) mechanism eliminates the Cl ion forming a cyclopropane.
Hope that this helps.


How to decide whether to pick 4-bromo-2-methoxyphenol or 4-bromo-6-methoxyphenol as the name for the compound?

When choosing which name to use when the only difference is the numbering, we choose the combination that gives the lowest possible numbers, 2 and 4 in this case.
So the name will be 4-bromo-2-methoxyphenol. The substituents are given in alphabetical order.


Is Al2O3 (aluminum oxide) diamagnetic or paramagnetic?

Actually the Lenz effect doesn't rely on paramagnetism to sort metallic objects from nonmetallic. See for example this video, where the moving magnets actually impel the nonferrous metals, permitting separation from nonmetallic components. If you take a strong magnet, like the NdFeB magnets you can get at the hardware store or even the Dollar Store, and move it rapidly over the bottom surface of an aluminum frying pan without it touching the surface you can actually feel the force between the two. This dynamic force (dynamic in that you can't feel the force between pan and magnet if they aren't moving relative to each other unless the pan is actually iron) is sufficient to impart more momentum to the metal waste in the video above and so achieve separation.
Looking at this table we can see that $\text{Al}_2\text{O}_3$ is in fact diamagnetic with $\chi_{\text{m}}=-37\times10^{-6}\text{cm}^{3}/\text{mol}$ and $\text{Pd}$ is paramagnetic with $\chi_{\text{m}}=+540\times10^{-6}\text{cm}^{3}/\text{mol}$. I have a strong pyramid magnet that can pick up some of the granules from crushed Cheerios and has a visible pull on some of the lettering on U.S. currency. I suspect it could separate out Pd-coated from noncoated beads statically.


What is the characterstics of additives that it must posses to make any mixture hypergolic?

The ideal formulation of rocket fuel is a balanced mix of  oxidizer and fuel. A strong oxidizer and a highly flammable compound or strong reducer, which is termed the "fuel", Is the Ideal combination for a fuel that will combust hypergolically when mixed. The compounds in hypergolic fuels are ideally liquids, because fluidity allows for easier mixing.

As oxidizers, Nitrates work well because the large quantity of oxygen atoms attached to the single nitrogen atom makes for a strong oxdizer. Other compounds, such as Sulfates, Phosphates, Chromates & Dichromates, Chlorates & Perchlorates, Peroxides, and Permanganates also serve as fairly good oxidizers for this reason. However, strong oxidizers do not always require high oxygen content. For example, Chlorine Trifluoride is a very strong oxidizer and is commonly used in many different hypergolic rocket fuel combinations

As far as the rest of the mixture, the fuel, most flammable liquids work well. Aniline, Liquid H2, Hydrazine and it's analogues, Kerosene, Gasoline, and most short-chain alkenes work well with the right oxidizer. An ideal choice when it comes to Hypergolic fuels is fairly small alcohols and ethers, like ethanol, diethyl ether, furfuryl alcohol, and especially Glycerol. It seems that the occasional oxygen atom in the fuel makes it easier and faster to oxidize when the time comes.

In conclusion, Adding a small amount of a stronger oxidizer in solution with your main oxidizer should help spark the reaction if it has trouble getting started. Also, while adding a fuel booster doesn't always work, sometimes adding a little ethanol or glycerol can help get the reaction in motion as well as help it run more smoothly. This is similar to how ethanol is often added to gasoline to increase the efficiency of the burn.

I kind of have an interest in the chemistry of rocket fuels, so if you ever find yourself looking for another answer, feel free to mention me directly.


Rate of reaction and dependence on concentrations

If we carefully see them ozone formation is a complex ( not elementary) reaction.
Yes the slowest step would be rate determining .
so for slowest step ;

Rate =$   k [O] [O _3] $

where $k$ would be rate constant.

but  final reaction is :
$$\ce { 2 O_3 -> 3O_2} $$
and so ,
equillibrium constant K(for 1st reaction)

$K= \frac{[O][O_2]}{[O_3]}$

thus substituting $[O] $from above equation to the rate equation .

We get

Rate =$\frac {  k × K [O _3]^2 }{[O_2]}$

And hence if $[O_2]$ is increased it causes rate to decrease.


How to rationalise with MO theory that CO is a two-electron donor through carbon?

The electrons in the frontier orbital(s) play a special role for the chemical reactivity. In $\ce{CO}$, the HOMO is the $ 5 \sigma $ orbital (ref: your diagram), and has mainly $\ce{C}$ character. The two electrons in it act like a lone pair on the carbon.

The predominant $\ce{C}$ character of the HOMO accounts for the reactivity via carbon ($\sigma$ donor

Side note: $\ce{CO}$ is also a $\pi ^*$ acceptor due to the LUMO (which is also predominantly $\ce{C}$ in character.

To answer your comment:

  So overall, how could I tell exclusively from the MO diagram that HOMO electrons are centered on C and not on O?

Don't think you can. The contribution to MOs is determind by the coefficients in the linear combination. You should see Martin's calculated values in the answer link he provided.


You are likely looking for trace levels, so I do not suggest any at home kit. At best, is it colorimetric and sensitive to interferences.
If you really want to know, find an environmental lab and ask them to run ICP-MS to determine the lead content. Atomic absorption spectroscopy would be the second best method.


Is there a way to use free software to convert SMILES strings to structures?

According to the website, Open Babel should do the trick: Documentation - SMILES, Sourceforge.

For example, the following code will give you a neat SVG file of the molecule benzene:

obabel -:"c1ccccc1" -O benzen.svg

If you experience problems using it, you are welcome to ask more specifically.

Alternatively, you can use a web-query from the national cancer institute. It is easily accessible by the following code

http://cactus.nci.nih.gov/chemical/structure/"structure identifier"/"representation"

For example: benzene, "structure identifier"=c1ccccc1, "representation"=image.

Another open source solution, where you can directly export the structure into a molecular editor is Avogadro. (It uses Open Babel though.)

Depending on the actual problem, however, there might already be more advanced routines.


Do sugar alcohols have the potential to get a person drunk?

Sugar alcohols are only partially absorbed. It is very unlikely anyone will become intoxicated with sugar alcohols. Also, their effects are less likely to cause intoxications.


What is the flow of electrons, cations, and anions in a voltaic cell (please see image attached)?

According to my notes and many sources on the internet, electrons and cations both travel from the anode (A in the image) to the cathode (B in the image).

The idea of the salt bridge is to prevent electrolytes mixing while providing ion flow. When you have a high concentration of inert ions in the salt bridge, cations in the salt bridge will flow into B, and anions in the salt bridge will flow into A. The solution in the salt bridge is not mixed, so it would take a very long time for the electrolyte in A to diffuse all the way to B.

To conclude, ions flow out of the salt bridge into the solutions A and B, but virtually no ion flows all the way through the salt bridge. This tells you that the salt bridge will be depleted after a certain amount of current flows through the wire. Accordingly, electrochemical cells used in batteries usually don't have a salt bridge but a semipermeable membrane instead, or a clever choice of half reactions where mixing of the electrolytes is not a problem (lead acid battery for example, where the only soluble species, sulfuric acid, is common to both half reactions).


How does dimethyltryptamine cross the blood-brain barrier?

Although serotonin and dimethyltryptamine (DMT) have similar chemical structures (see Jori's answer), serotonin is not transported across the blood brain barrier (BBB) ("serotonin does not cross the blood–brain barrier").  On the other hand, not only does DMT cross the BBB, it is one of the few compounds that is actively transported across the BBB.  While most compounds that cross the BBB do so by being lipid soluble (relatively small non-polar compounds) some molecules are assisted across the barrier through a variety of mechanisms.  Some of the known assistive paths include

glucose carrier: transports glucose and several other sugars across the BBB
phenylalanine carrier: transports a number of neutral amino acids across the BBB
arginine carrier: transports several cationic amino acids across the BBB
lactate carrier: transports several monocarboxylic acids across the BBB
adenosine transporter: transports purine nucleosides as well as uridine
organic cation transporter: transports a variety of cationic compounds, including monoamine neurotransmitters

There are more classes and sub-classes within each.  The following two references provide nice, readable overviews

Small Molecular Drug Transfer across the Blood-Brain Barrier via Carrier-Mediated Transport Systems
Drug transport across the blood–brain barrier

I don't know which of the above transporters is active in the case of DMT, but I strongly suspect it is the "organic cation transporter".  Here are some further references that relate specifically to DMT active transport across the BBB

Takahashi T, Takahashi K, Ido T, Yanai K, Iwata R, Ishiwata K,
Nozoe S (1985) 11C-labeling of indolealkylamine alkaloids and
the comparative study of their tissue distributions. Int J Appl
Radiat Isot 36:965–969
Yanai K, Ido T, Ishiwata K, Hatazawa J, Takahashi T, Iwata R,
Matsuzawa T (1986) In vivo kinetics and displacement study of
a carbon-11-labeled hallucinogen, N, N-[11C]dimethyltryptamine.
Eur J Nucl Med 12:141–146

They were found in this article, "Bufotenine: Toward an Understanding
of Possible Psychoactive Mechanisms"

P.S.  We've all read fairy tales where kissing a frog turns the frog into a prince.  When the skin of the Bufo toad (Colorado river toad, sonoran desert toad and other members) is licked, a psychedelic experience results.  It is not illegal in the US to own the toad, although the Bufotenine is a controlled substance!  Anyhow, just wondering if this, or something similar, is what the fairy tales are based on.


Why is it important to remove the used up electrons in oxidative phosphorylation?

I get that oxygen is the terminal electron acceptor in oxidative phosphorylation. But why is it essential to accept these electrons and remove them as H2O?

A redox reaction is made of two half reactions. Neither one of them is more or less important; the combination of both determines the energetics of the reaction.

  Why is having an efficient garbage collector (oxygen picking up the used electrons) so crucial to cellular respiration, and to ATP production?

The electrons are not "used" as in an old car or worn-out clothes. There as shiny and new as before. Without oxygen, the reaction would not happen in the first place (electrons are transferred from less tightly bound to more tightly bound). The Gibbs energy of proton transport matches what is necessary for ATP synthesis, which in turn matches the Gibbs energy of the redox reaction. With a different electron acceptor (specificities of the enzymes involved aside), the proton gradient might be too weak to make ATP.


If it is permanganate, the brown $\ce{MnO2}$ stain will decolorize and solubilize in acidic reductant - lemon juice, perhaps boosted with a little crushed and dissolved Vitamin C.   I'd go for crystal/gentian violet or Disperse Red 9 as a marker, or a reactive Cibacron or Procion dye.

Cthulhu for Fearless Leader!  Why vote for the lesser evil?


How to decide whether +M effect or -I effect will operate in this case?

The extent of stabilizing effect follows the order: $\ce{\text{Mesomeric} > \text{Hyperconjugation} > \text{Inductive}}$.

In general, this order is based on extent of $\ce{e-}$ transfer. In mesomeric effect, $\pi$-bonds are in conjugation which completely transfers $\ce{e-}$ density to carbocation. On the other hand, in inductive effect, induction of charge takes place, which partially transfers $\ce{e-}$ density through $\sigma$-bond. Hence, it is weaker.

  why +M effect of 'N' is more than that of 'O'?

As $\ce{N}$ has lower electronegativity than $\ce{O}$, hence it has greater $\ce{e-}$-donating tendency, therefore greater $\ce{+M-\text{effect}}$


My Lab First Aid book$^{*1}$ tells the following:

  The local therapy mainly consists of the instant, intense rinsing (at least 10 minutes) with tap water (ideally 15 to 20 °C). Don't try to neutralize the chemical burn, because the heat generation could lead to further tissue damages.

This means, that your lab instructor is absolutely right.

The rinsing water should wash the cold water away faster than the heat from the acid base reaction could harm you. It's not for nothing that there are emergency showers and eye douches in every lab.

I have a friend who is a chemical lab assistant and once had to carry concentrated sulfuric acid in a big glass beaker. Suddenly the bottom of the beaker broke off and all the acid poured over her etc. She not even thought of anything else than getting out of her clothes and under the next emergency shower.

$^{*1}$ The book is: R. Rossi, Erste hilfe bei akuten Notfällen - Begleitheft zu Jander - Blasius, S. Hirzel Verlag, 2005, p. 11f.. The text is in german and I translated it.


Are children's sparklers based on a magnesium reaction?

To answer the question, let us understand sparkler chemistry. A sparkler consists of:

An oxidizer: potassium nitrate or potassium chlorate. They burn off a mixture and are part of oxidation-reduction reaction:
$$
\begin{align}
\ce{\underset{potassium nitrate}{2 KNO3 (s)} &-> \underset{potassium nitrite}{2 KNO2 (s)} + O2 (g)}\\
\ce{\underset{potassium chlorate}{2 KClO3 (s)} &-> \underset{potassium chloride}{2 KCl (s)} + 3 O2 (g)}
\end{align}
$$
Fuel: firework fuel is charcoal and sulfur. Both sulfur and carbon (charcoal) are reducing agents forming sulfur dioxide and carbon dioxide.
Metal powder: aluminum, iron, steel, zinc, magnesium, titanium or ferrotitanium dust or flakes. These metal flakes become incandescent and shine brightly in the presence of oxidizers. If temperatures are high enough, these metal flakes burn. The temperatures of these sparks range from 1800 °F to 3000 °F (1000 °C - 1600 °C)(Ref. I). Many elements heat up and emit at a characteristic wavelength (Ref. II).

A combustible binder: binders hold oxidizers, fuel and metal powders together. Common binders are dextrin (a sugar) dampened by water.
Colorants: aluminium, magnesium and titanium all give near brilliant white sparks. Iron produces orange sparks.

Putting all together: these chemicals are mixed together to form a slurry. The slurry is coated on to wire and allowed to dry. When ignited the metal flakes  aluminum, iron, steel, zinc or magnesium dust or flakes create the bright, shimmering sparks. As to your question, children sparklers are based on metal oxidizing reactions.

References

Ref. I https://www.wired.com/2014/07/the-awesome-physics-in-a-simple-sparkler/
Ref. II https://www.forbes.com/sites/startswithabang/2016/07/01/the-physics-of-fireworks/#3895e0711eee
http://www.compoundchem.com/2014/11/04/sparklers/
http://www.compoundchem.com/2013/12/30/the-chemistry-of-fireworks/
http://www.compoundchem.com/2014/07/02/the-chemistry-of-gunpowder/
http://www.skylighter.com/fireworks/how-to-make/sizzler-sparklers.asp


Geometry optimization of a small molecule inside of a larger one

I don't think you've provided nearly enough detail about your problems, particular "a".

But let me comment about "exporting from ChemDraw." Keep in mind that ChemDraw is a 2D tool. Consequently, the lengths between atoms are designed for depiction on paper, not necessarily for accurate representation of the 3D geometry necessary for calculations.

Avogadro can import 2D geometries from ChemDraw and attempt to assign reasonable 3D geometries. In your case, this seems especially troublesome since you want to have one molecule inside another.

I'd either:

Sketch the two molecules separately in ChemDraw and then use Avogadro to "insert" the smaller one into the larger after geometry optimization.
Use Avogadro to create the full 3D geometries of each molecule and then, select the smaller molecule and use the manipulate tool to move it into position.

As several people have commented, if you want to preserve the appearance of bonds, XYZ is not the right format. It only stores coordinates, not bonding information. Avogadro and other tools will attempt to re-perceive bonds and bond orders, but this is not a perfect task, particularly in the presence of multiple nested molecules.

Instead, I'd keep a file in CML format with the bond orders intact. This is more-or-less the "native" file format for Avogadro.


to convert ppm gas (N2O) concentrations in microgram nitrogen per gram dry soil basis?

I assume the ppm value you are reporting is ppm by volume (or equivalently for ideal gases, ppm by mole).  If that is true, then:

The total moles of all gas in the container can be calculated from the ideal gas law:

$$ N_{tot} = \frac{PV}{RT}=\frac{(101325 \rm{~Pa})(\frac{80}{1000^2} \rm{~m^3})}{{(8.314 \rm{~\frac{J}{mol\cdot K}})}{(298 \rm{~K}})}$$

Here I have assumed that the temperature was 298 K, or about 25 °C.  Note that the units work out because pascals ($\rm{Pa}$) are equivalent to joules per cubic meter, $\rm{\frac{J}{m^3}}$.  Also, I assume your experiment was done at approximately atmospheric pressure, which is 101325 Pa.

The moles of $\ce{NO2}$ in this 80 cc of gas can be calculated from the ppm measurement:

$$N_{\ce{NO2}}=\frac{\rm{ppm~value}}{1,000,000}\times N_{tot}$$

The mass of the nitrogen, in grams, in the $\ce{NO2}$ gas can be calculated from the formula weight of a nitrogen atom, $\rm{FW}_{\ce{N}}$:

$$m_{\ce{NO2}}={\rm{FW}_{\ce{N}}}\times N_{\ce{NO2}}=14\frac{\rm{g}}{\rm mol} \times N_{\ce{NO2}}$$

The mass of soil used was 5 grams, so the mass ratio of $\ce{NO2}$ to soil, $x_{\ce{NO2}}$  can be calculated by simply dividing the masses:

$$x_{\ce{NO2}} = \frac{m_{\ce{NO2}}}{5 \rm{~grams}}=\frac{m_{\ce{NO2}}\times \frac{1,000,000 \rm{~\mu g}}{\rm{g}}}{5 \rm{~grams}}$$

The last part of the equation above means that if you want the mass ratio in units of micrograms (of $\ce{NO2}$) per gram of soil, you have to multiply by 1,000,000.

Also, the question says you want a "per gram dry soil basis", but you didn't mention if the 5 g of soil you used is "dry".  If it isn't, then you will first have to figure out how many grams of dry soil you added to your reaction, if it wasn't 5.


What is the relation between absolute temperature, and rates of diffusion and effusion of a gas?

The self diffusion coefficient in a gas is given by $\displaystyle D=\frac{\lambda}{2}\bar c$ and the rate of effusion by $\displaystyle \frac{\rho}{4}\bar c$ where $\bar c$ is the average velocity of the gas molecules of mass $m$, and $\displaystyle \bar c= \sqrt{\frac{8k_BT}{\pi m}}  $, $\lambda$ the mean free path and $\rho$ the gas density. The diffusion constant is $\displaystyle D=\frac{1}{\pi \sigma^2 n}\sqrt{\frac{k_BT}{\pi m}}$ where $\sigma$ is the molecule's diameter, and the rate of effusion $\displaystyle \rho \sqrt{\frac{k_BT}{2\pi m}}$. The diffusion constant can also be written as $\displaystyle D=\frac{m}{\pi \sigma^2 \rho }\sqrt{\frac{k_BT}{\pi m}}$ thus the dependence on temperature is the same for diffusion and effusion if the gas density is held constant.

[ The rate of effusion is calculated by  considering the gas bouncing around in a container and then suddenly opening a tiny trap door of area $dA$ in the wall so that some minute proportion of the gas streams out in time $dt$. The trap door diameter should be no more than 1/10 of the mean free path.   If each molecule travels with average speed $\bar c$ the number leaving would be $n\bar c dA dt$ if there are $n$ molecules /unit volume. The efflux in terms of the mass/unit area/unit time is then $mn\bar c =\rho \bar c$.

However, half the molecules are moving away from the trap door at this instant so the rate is divided by 2. Additionally molecules come out at different angles so it is necessary to integrate over the volume element $\sin(\theta)d\theta d\phi$ and this leads to a further factor of 1/2.]


How can you estimate how much of a substance will dissolve in a given lipid?

The major issue with the blood-brain barrier is that aside from small hydrophobic/lipophilic compounds (including $\ce{O_{2}}$ and $\ce{CO_{2}}$), almost everything else is prevented from passing due to cellular tight junctions and will only (potentially) pass through via active transport.

However, for small lipophilic molecules passing through a phospholipid bilayer (such as on a cell surface in the blood-brain barrier), this post provides information on equations: http://www.ncbi.nlm.nih.gov/books/NBK21626/

Where:

$\frac{dn}{dt}$ is diffusion rate (mol/s)

$K$ is the partition coefficient (the equilibrium constant for its partition between oil and water) - e.g. for Urea, K = 0.0002 while for diethylurea which is more hydrophobic, K = 0.01

$C^{m}$ is the concentration in the hydrophobic part of the phosopholipid bilayer (mol/L)

$C^{aq}$ is the concentration in aqueous solution (mol/L)

$P$ is permeability coefficient

$D$ is diffusion coefficient of the substance within the phospholipid membrane

$A$ is the area of the membrane (in $m^{2}$ I think?)

$x$ is the membrane thickness (in m I think?)

$C_{1}^{aq}$ and $C_{2}^{aq}$ are the aqueous concentrations on either side of the layer (one will be the serum or CSF concentration, the other will be the intracellular concentration)

$K = \frac{C^{m}}{C^{aq}}$
$\frac{dn}{dt}=PA(C_{1}^{aq}-C_{2}^{aq})$ and $P = \frac{KD}{x}$
$\frac{dn}{dt}=A\frac{KD}{x}(C_{1}^{aq}-C_{2}^{aq})$

You would have a similar equation to 2. on the other side of the cells lining the blood-brain barrier.

It seems from my reading that $P$ tends to be better known than $D$, which is actually usually calculated using equation 2. from $P$, $K$ and $x$.

Hope that helps!


I think your teacher is correct, numbering of carbons is done acording to the lowest locant rule.

The red numbering gives you : 2, 3, 3
The green numbering gives you : 2, 2, 3
We see that the first point of difference between the two numberings occurs at second number, so the green numbering is preferred.
Consequently, the correct name should be : 3-Bromo-2,2-bis(4-chlorophenyl)butane


Why do two solutes permitted to diffuse simultaneously across a semi-permeable membrane both end up at equal concentrations on both sides?

If we assume that the solvent is the same on both sides of the membrane (or that the solvent itself can penetrate the membrane) then the final equilibrium will have the same concentration across the membrane.  That only translates to "50% of solute b to occur on side 1 and 50% on side 2" if the volumes are equal.

The rate at which solute A moves from left to right is proportional to the rate at which molecules of solute A on the left side randomly collide with the membrane.  This is proportional to the concentration of A in the left side solution.  The presence of molecules for solute B does not affect this process.

The reverse process is also happening with molecules diffusing from right to left at a rate proportional to their concentration in the right side solution. As the concentration on the right side increases to be equal to the concentration on the left, so the diffusion rates become equal and there is zero nett diffusion and the system approaches equilibrium.

Note that this assumes a "perfect" system where there is no chemical reaction occurring between the solutes or between the solutes and the membrane.  In practice this means that either the interaction between solutes A and B is the same as the interaction between the solutes and the solvent or that the solute molecules are so greatly outnumbered by the solvent molecules that the solute-solute interactions are not significant.

The rate of diffusion of solute A may be different from B (i.e. the proportionality constant between rate and concentration may be different).   This means that before reaching equilibrium the relative concentrations of A and B may change but at equilibrium, the relative concentration will be the same as initially.

If we define "reaching equilibrium" as having some fraction (say 99.99%) of the final concentration then increasing the initial global concentration will increase the lag for both solutes equally and will not change their relative concentrations.


Calculating temperature after adding heat to ice

It is an enthalpy balance that requires knowledge of the enthalpy change at each step: (The steps you have listed are not correct.)

Solid heat capacity of ice of the form e.g. $C_p^\mathrm{solid}(T)$. This allows you to calculate the amount of heat (or energy) required to warm the sub-cooled ice from $-5~^{\circ}\mathrm{C}$ to $0~^{\circ}\mathrm{C}$.
(Latent) heat of fusion, $\Delta H_\mathrm{fus}$ then must be added to transform the solid ice to liquid water at $0~^{\circ}\mathrm{C}$.
Next we must warm the liquid water from $0~^{\circ}\mathrm{C}$ to its boiling point at $100~^{\circ}\mathrm{C}$, which requires knowledge of the liquid heat capacity of water e.g. $C_p^\mathrm{liquid}(T)$.
Now at the boiling point, we must account for the required amount of energy to transform the liquid water into its vapor state at $100~^{\circ}\mathrm{C}$ using the (latent) heat of vaporization of water $\Delta H_\mathrm{vap}$.
The final step then involves super-heating the vapor from $100~^{\circ}\text{C}$ to $T_\mathrm{final} = ?~^{\circ} \mathrm{C}$, which requires knowledge of the vapor heat capacity of steam e.g. $C_p^\mathrm{vapor}(T)$.

You know the overall enthalpy change ($53.2~\mathrm{ kJ}$) and you know the starting temperature ($-5~^{\circ}\mathrm{C}$), so all you have to do is add the results from each of the steps and solve for $T_\mathrm{final}$ e.g. $303~^{\circ}\mathrm{C}$.


Protoporphirinogen IX formation in Heme synthesis

So, will a water molecule form in this step. 2H+ will come from the two carboxylic groups and oxygen will combine with it to form water. Will water molecule form in this way?

No.

Firstly, note that there are 4 more hydrogen atoms in the reactant than in the product (one from each carboxylic acid group and one where each double bond forms).

However, at physiological pH, the two protons on the carboxyl groups will not really be there.  The equation in the question also ignores the 4 electrons transferred.

As for the other two protons, a 5’-deoxyadenosyl radical abstracts a hydrogen atom (proton + electron) from each propionate side chain.

See Coproporphyrinogen III Oxidase Enzyme and references cited therein for more information.


There are two isomers of chromene (1-benzopyran): 2H-chromene and 4H-chromene.

According to Subsection P-14.7 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the name can be made specific by indicating the position of the extra hydrogen atom in the structure, i.e. the position where no multiple bond is attached. This is accomplished by adding to the front of the name an italic capital ‘H’ preceded by the appropriate numerical locant for this atom.

In 2H-chromene, the ‘indicated hydrogen’ locates one extra hydrogen atom in position ‘2’ of the chromene structure; and in 4H-chromene, the ‘indicated hydrogen’ indicates an extra hydrogen atom at position ‘4’, i.e. one hydrogen atom more than the number present if there were a double bond in the ring at that position.


Diluting acetic acid to obtain a solution of pH 5

This is a dilution problem. So in this instance, you want to add deionized water to your acetic acid. However, the question is, how much water do you add?

It is necessary first to find the molarity of the acetic acid. On PubChem, we see that the molecular weight is $60.05196 \ \frac{\text{g}}{\text{mol}}$ and density is $1.0446 \ \frac{\text{g}}{\text{cm}^3}$ @ $25\ ^\circ \text{C}$.

If we have 1 L of acetic acid, then

$$ 1.0446 \ \frac{\text{g}}{\text{cm}^3} \cdot \frac{1000\ \text{cm}^3}{1\ \text{L}} \cdot \frac{1\ \text{mol}}{60.05196\ \text{g}} = 17.394\ \frac{ \text{mol}}{\text{L}}$$

We know that,

$$ \text{pH} = -\log{\ce{[H3O+]}} $$

which implies that the concentration of hydronium ions to make the solution with pH equal to 5 is, $\ce{[H3O+]} = 1 \cdot 10^{-5} \ \text{M}$. We make our RICE table,

$$\ce{HC2H3O2 + H2O -> C2H3O2- + H3O+}$$

\begin{array} {|c|c|c|c|c|}
\hline
\text{Initial conc.} & x \ \text{mol} & - & \text{0 mol} & \text{0 mol}\\
\hline
\text{Change conc.} & -1 \cdot 10^{-5}\ \text{M} & - & 1 \cdot 10^{-5}\ \text{M} & 1 \cdot 10^{-5}\ \text{M}\\
\hline
\text{End conc.} & x - 1 \cdot 10^{-5}\ \text{M} & - &  1 \cdot 10^{-5}\ \text{M} & 1 \cdot 10^{-5}\ \text{M}\\
\hline
\end{array}

$$K_\text{a} = \frac{[\ce{H3O+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{(0.00001)(0.00001)}{(x - 0.00001)} = 1.75 \cdot 10^{-5} $$

$$\frac{(1 \cdot 10^{-10})}{(x - 0.00001)} = 0.0000175$$
$$1 \cdot 10^{-10} = 0.0000175x - 1.75 \cdot 10^{-10} $$
$$2.75 \cdot 10^{-10} = 0.0000175x$$
$$x = 1.6 \cdot 10^{-5}\ \text{M} = [\ce{CH3COOH}] $$

Thus, you go from a $17.394\ \text{M}$ solution of acetic acid, to a $0.000016\ \text{M}$, which is a factor of $1.09\cdot10^{6}$. You can then dilute the solution by the appropriate amount by placing the amount of acid you want in a volumetric flask, and then successively dilute it to the required pH.

Two assumptions,

Deprotonation is small enough that the equilibrium concentration of the acid is approximately equal to the same as its initial concentration.
Autoprotolysis of water does not significantly contribute to pH.


Is the bond dissociation enthalpy of Cl-F greater than that of Cl-Cl?

From this source, it can be seen that the $\ce {Cl-F}$ bond (BDE $\approx\pu{251 kJ/mol}$) is indeed stronger than the $\ce {Cl-Cl}$ bond (BDE $\approx\pu{243 kJ/mol}$). This can be explained by the partial ionic character of the $\ce {Cl-F}$ bond. The Pauling electronegativity value of fluorine ($\chi = 3.98$) is appreciably higher than that of chlorine ($\chi = 3.16$). Thus, there are significant partial positive and negative charges on the fluorine and chlorine atoms in the $\ce {ClF}$ molecule. Although the difference in orbital energies and size would mean that the covalent interaction between the chlorine $\ce {3p}$ and fluorine $\ce {2p}$ orbital would be less strong compared to that between 2 chlorine $\ce {3p}$ orbitals, the ionic character of the bond overcompensates for this and in fact, results in the $\ce {Cl-F}$ bond being stronger than the $\ce {Cl-Cl}$ bond.


Does mercury have an NFPA 704 classification?

According to Sigma-Aldrich

Some symbols:

GHS06: Acute toxicity (oral, dermal, inhalation), categories 1,2,3
GHS08: Respiratory sensitization, category 1; Germ cell mutagenicity, categories 1A,1B,2; Carcinogenicity, categories 1A,1B,2; Reproductive toxicity, categories 1A,1B,2; Specific Target Organ Toxicity – Single exposure, categories 1,2; Specific Target Organ Toxicity – Repeated exposure, categories 1,2; Aspiration Hazard, category 1
GHS09: Hazardous to the aquatic environment; Acute hazard, category1; Chronic hazard, categories 1,2

And according to ScienceLab:

Health: 3
Fire: 0
Reactivity: 0

I believe the ScienceLab information is the 'NFPA 704 classification'.

EDIT****

An email exchange between NIST and myself asking for clarifiation:

My Email (Pre-filled form on the website, I don't have the original):

Something along the lines of "Hello, I see NIST and a few other manufactures
of have different NFPA classifications (I provided links to the SDS's),
could you explain the difference between your ratings and theirs?"

Their response:

Good afternoon,
Our NFPA rating Health = 2 Fire = 0 Reactivity = 0 matches other manufacturers.
You can check Aldrich website for 99.99 Mercury
http://www.sigmaaldrich.com/catalog/product/sigald/215457?lang=en&region=US

Thank you,
Office of Reference Materials

Other manufacturers (Note, plural) apparently now equals 1 example


The naming of organic compounds are governed by IUPAC guidelines, specifically the way the names are formed are by some general principles, which are given on the IUPAC website at their page Guide to Name Construction: General Principles, with naming examples here.
Another rather nice summary is found on the  webpage Nomenclature (Reusch, 1999), states that a organic compound name will have the following features, based on several common chemical characteristics of organic compounds:

•  A root or base indicating a major chain or ring of carbon atoms found in the molecular structure.
•  A suffix or other element(s) designating functional groups that may be present in the compound.
•  Names of substituent groups, other than hydrogen, that complete the molecular structure.

(Specific IUPAC guidelines for different hydrocarbons are included in the link).
Another summary is found on the handout Short Summary of IUPAC Nomenclature of Organic Compounds (Dr. Jan Simek, California Polytechnic State University at San Luis Obispo), which state that the general nomenclature principle is:

IUPAC nomenclature is based on naming a molecule’s longest chain of carbons connected by
single bonds, whether in a continuous chain or in a ring. All deviations, either multiple bonds or
atoms other than carbon and hydrogen, are indicated by prefixes or suffixes according to a
specific set of priorities.


Whether a substance is a strong or weak acid/base depends on the solvent it's in. For example, $\ce{HNO_3}$ is a strong acid in water. This is partly because $\ce{HNO_3}$ has a relatively high tendency to donate a proton (the conjugate base $\ce{NO^{-}_3}$ is quite stable and therefore has a low tendency to accept a proton back), and partly because water has a certain basic character, accepting protons to create $\ce{H3O+}$. However, the same $\ce{HNO_3}$ in pure liquid $\ce{H3CCOOH}$ (also known as glacial acetic acid) is actually a weak acid. This change in character is largely due to the fact that $\ce{H3CCOOH}$ is a much, much weaker base than water, forming the $\ce{H3CCOOH^+_2}$ ion with significantly more reluctance. Even though $\ce{HNO_3}$ can easily lose its proton, $\ce{H3CCOOH}$ is even moreso unwilling to accept it, and thus $\ce{HNO_3}$ becomes a weak acid. An amusing situation arises in pure liquid $\ce{H_2SO_4}$. Adding $\ce{HNO_3}$ there, we find that $\ce{HNO_3}$ actually behaves as a weak base. $\ce{H_2SO_4}$ itself also has a high tendency to donate protons, and of the two possible reaction products from protonation, $\ce{H_2NO^+_3}$ is somewhat more stable than $\ce{H_3SO^+_4}$.

Hence, acidity is dependent on the solvent, not only on the solute. A strong acid such as $\ce{HNO_3}$ in aqueous solution at high dilution is completely dissociated, as we expect of a strong acid. However, if more and more water is removed, then the solution starts to behave differently. Even though $\ce{HNO_3}$ has a strong tendency to protonate water, you can have a solution that contains more moles of $\ce{HNO_3}$ than water. Then it becomes obvious that not all $\ce{HNO_3}$ molecules are deprotonated. Practically all of the water left in the solution will be in the form of $\ce{H_3O^+}$, but there will be left over undissociated $\ce{HNO_3}$, so most of the excess stays undissociated, except for a small amount that undergoes self-protonation, generating a bit of $\ce{H_2NO^+_3}$.

All in all, in extremely concentrated solutions, even strong acids/bases will not show 100% dissociation, and saying that "strong acids completely ionize" is just a rule of thumb which shouldn't be taken too far.


Does carbon monoxide form or carbon dioxide forms when carbon reduces metal oxide?

If both carbon monoxide and carbon dioxide are thermodynamically favored, then carbon monoxide comes "first" in the sense that the metal oxide first oxidizes carbon to $\ce{CO}$.  This is because the reaction is carried out at relatively high temperature where any potential carbon dioxide would itself react with carbon to form the monoxide.  If more oxide is available after all the carbon is converted to $\ce{CO}$ and further oxidation to $\ce{CO2}$ is favorable, the latter reaction then occurs.


I think you need to make at least one more assumption along the way to determining the relative densities. But let’s take it stepwise:

We are given a primitive cubic cell for nickel and a face-centred cube for copper.
We therefore conclude that per unit cell one atom of nickel and three atoms of copper are present.
We know from the periodic table that copper is a bit heavier than nickel ($63.5$ versus $58.7$)
Therefore, it is clear that copper’s unit cell is more than four times as heavy as nickel’s. But we don’t know their sizes.
From mathematical considerations we can also calculate how much of each unit cell is filled with atoms ($74~\%$ for fcc; somewhat less which I am too lazy to calculate now for primitive cubic) — this again means that there should be more weight per volume in the copper unit cell.
What seems to be lacking in my eyes is a way to compare the unit cell sizes, whether by atom size or unit cell dimensions. However, since nickel and copper are right next to each other in the periodic table, it may be safe to assume their sizes to be almost identical.
If that assumption is safe, we can use either route to clearly show how copper’s density is higher. But note that if copper’s atoms were significantly larger than nickel’s, this reasoning would break down extremely quickly.


Your calculation of $[\ce{OH-}] = 10^{-5.51}\ \mathrm{mol\ l^{-1}}$ is correct.

The pH is well above 7. Therefore, the present $[\ce{OH-}]$ is mainly a result of the dissociation of $\ce{Cr(OH)2}$:

$$\ce{Cr(OH)2 <=> Cr^2+ + 2 OH- }$$

Thus, you know
$[\ce{Cr^2+}] \approx \tfrac{1}{2} [\ce{OH-}] = \tfrac{1}{2}\cdot10^{-5.51}\ \mathrm{mol\ l^{-1}} $.

$K_\text{sp} = [\ce{Cr^2+}][\ce{OH-}]^2 = \tfrac{1}{2}\cdot10^{-5.51} \cdot \left(10^{-5.51}\right)^2 = 1.48\cdot 10^{-17}$


Does changing the volume of a reactant affect the rate of reaction? Does changing the volume of HCl affect the rate of a Mg & HCl reaction?

More volume of HCL does not increase the concentration and therefore does not increase the likelihood and so rate of reaction.
Mg and HCl forms a metal salt MgCl and produces Hydrogen gas.

There may be a very slight increase in rate of reaction but this is permissible/ very negligible due to the fact that metals and acids form adducts and these can further react/ delay further reactions between the acid and metal molecules.

The rate would definitely be affected by factors such as temperature of the (aq) HCL. This is until the reactant Mg is used up then the reaction will cease leaving you with a flat curve on a rate graph.

Now, for Higher concentrations- this implies that more reacting molecules are at higher proximity to each other therefore the intermolecular collisions between the Mg and HCl molecules in your case are more frequent therefore forming more product per unit time.

Control variable: mass of Mg to at least 1.d.p, concentration of acid moldm-3, volume of HCl cm3 - even though this doesn't affect the rate of reaction enough - they would want you to keep this constant for valid test etc.

Hope this helps!


Why did my iron nails not dissolve in hydrochloric acid?

It seems like there is not enough acid on the nails.

Perhaps a depleted layer is forming next to the nails. To avoid that issue, use a magnetic stirrer or provide a source of vibration.
There might not be enough $\ce{HCl}$ for the large amount of iron in the nails. Check the total amount of $\ce{HCl}$, given the volume and that ~2/3 is "other ingredients" (likely water). Weigh the nails. Calculate the number of moles of  $\ce{HCl}$ needed to react with the number of moles of  $\ce{Fe}$. Is the product  $\ce{FeCl2}$ or  $\ce{FeCl3}$?
Have you removed all the grease (from the manufacturing process) from the nails? A little soap or detergent and rubbing may be needed.
Could there be some difference in the nails you used? If some are galvanized (zinc coated, to prevent rust), then that would affect the reaction.

Let us know how you do: we learn more from our mistakes than from successes.


Identifying the D/L form of any aldose or ketose in cyclic form

Your structure is a ketose because it is a hemiketal of a ketone. It is a furanose because it is a 5-membered (furan-oid) ring. Fructose, which your structure is not, is usually drawn as a furanose as in sucrose but fructose itself exists principally in the pyranose (6-membered ring) form.

How do you recognize the D-configuration? Your furanose 1a is in the D-series. Reverse your furanose to the keto form 1b by following the red bonds. Rotate about the $\ce{C4-C5}$ bond by $120^\circ$ as illustrated by the blue arrow in 1b. This rotation produces the Fischer projection 1c having the $\ce{C5}$-hydroxyl on the right. Thus, furanose 1a is of the D-series.

There is another method that can be used with qualification. At any carbon that determines D vs. L in a carbohydrate, as long as the substituents are $\ce{O, CH2OH, CH(OH)C}$ and $\ce{H}$, the D-series will be of the R-configuration. Conversely, the L-series will be of the S-configuration. In furanose 1a, the priorities are $\ce{O>C4>C6>H}$. This determination is easy using this method.

Structures 2a, 2b, and 2c are different representations of $\alpha$-D-glucose. All of the $\ce{C5}$-carbons are, necessarily, of the R-configuration and therefore in the D-series.

Structures 3a and 3b are Haworth renderings of sucrose. Both the glucose (pyranose) and fructose (furanose) moieties are of the R-configuration and in the D-series.


The mass of a carbon-12 atom is 12 amu. What percent of the mass is located in the nucleus?

I can only give a simple answer, so if others have better ones, great, but here's my shot:

An electron's mass is a lot less than a proton/neutron.
"When talking about relative masses, we use the term atomic mass unit (amu). Using this unit, a proton has a mass of 1.007 amu, a neutron a mass of 1.008 amu, and an electron a mass of 5.45 X 10-4 amu."

Basically, the proton and neutron is close to 1, and the electron is close to 0.

Carbon-12 means it's got a total of 12 protons and neutrons, and since carbon has 6 protons, that makes 6 neutrons. Since it's a normally charged carbon atom too, that makes 6 electrons.

Can you do the calculations yourself with that? Because now you know the masses.

Also this value should be really close to 100% because of how electrons are so much lighter than protons and neutrons; protons and neutrons are located in the nucleus and the rest of the atom is mostly empty space + some electrons, so that's why the bulk of the mass is there.


How is the process of converting calcium carbonate (CaCO3) into calcium oxide (CaO)?

From wikipedia article of calcium oxide

  Calcium oxide is usually made by the thermal decomposition of
  materials, such as limestone or seashells, that contain calcium
  carbonate ($\ce{CaCO3}$; mineral calcite) in a lime kiln. This is
  accomplished by heating the material to above 825 °C (1,517 °F), a
  process called calcination or lime-burning, to liberate a molecule
  of carbon dioxide ($\ce{CO2}$), leaving quicklime.

  $$\ce{CaCO3(s) → CaO(s) + CO2(g)}$$

  The quicklime is not stable and, when cooled, will spontaneously react
  with $\ce{CO2}$ from the air until, after enough time, it will be
  completely converted back to calcium carbonate unless slaked with
  water to set as lime plaster or lime mortar.


Why does tea dissolve better when put in water rather than milk?

Milk is an emulsion containing small bubbles of oil/fat. Whenever you boil milk or do something drastic, this fat tends to separate. You can do this nicely, and get cream or butter, preserving the bubbles, only part of the water content is separated, or can do more drastically, when these bubbles collapse to a big oily blob.

When you put your filter in the milk, most probably some of the oil/ fat bubbles hit the surface and cover it with an oily film. This film prevents the easy dissolution of caffeine, etc from the tea leaves in water.


Paramagnetic relaxation in $\mathrm {~^{13}C}$ NMR:

Quantitative $\mathrm {~^{13}C}$ NMR requires full relaxation of all carbons at each scan.  Adding paramagnetic relaxation reagents could speed up the relaxation.  The most common reagent you can find online is chromium acetylacetonate, $\ce{Cr(acac)3}$. However, the amount you need to add to your solution is very critical. It is tricky that if your added amount to your solution is too little, $\mathrm T_1$ of the $\mathrm {~^{13}C}$ will not be short enough. I you added too much of $\ce{Cr(acac)3}$, the signal will be too much broadened due to too much shortening of $\mathrm T_2$, thus you will lose resolution. Typically used recommended amount by experts is Between $\mathrm{0.1wt\%}$ and $\mathrm{0.5wt\%}$ (https://blogs.umass.edu/weiguoh/?p=71). With that amount, the solution will have a light purple color [1] (courtesy of https://blogs.umass.edu/weiguoh/?p=71).

Above picture is a picture of a standard sample with $\mathrm{0.5wt\%}$ $\ce{Cr(acac)3}$, which is an excellent reference for you. It'd provide you the easiest way to avoid the mess of weighing $\ce{Cr(acac)3}$. What you have to do is to look at the color of your solution.  Following the color of the picture, you can add $\ce{Cr(acac)3}$ a little by little so that your sample has a similar or slightly lighter color than this.

To read about othe reagents download this thesis (available online): https://www.diva-portal.org/smash/get/diva2:793434/FULLTEXT01.pdf


Will freezing water still expand width-wise in a pipe (possibly breaking it) if it's free to expand lengthwise?

It might do.

Imagine you have a U-shaped pipe. The two ends are above ground. The pipe runs into the ground for a short length in the middle, then back out again. It's free at both ends but filled with water.

You have a problem there because in cold temperatures, the two above-ground parts are going to freeze first. Then, if it's really cold, the underground part will freeze also. When that happens, the already-frozen above-ground ice won't move and your underground pipe may well go pop. What was initially "free at both ends" became blocked.


The technical definition of 'freezing' implies a phase transition from liquid to solid, and therefore one assumes your interpretation requires that a table can at some point be a liquid.

Under normal circumstances, being a solid does not necessarily imply something is frozen, because not all materials can exist as liquid. Heating some materials will simply lead to (i) phase transition through sublimation or (ii) combustion or decomposition (a chemical reaction) leading to a both gaseous and solid products.

Presumably, the table you are pondering is made of wood. Wood is a mixture, composed of mostly cellulose, hemicellulose and lignin. The ratios depend on the wood type, and there are a few other bits in there also. Importantly, wood, even tables, still contain water. Under standard conditions one might expect to find a table (normal pressures), simply heating will not cause your table to melt. Above about 100oC, decomposition products are slowly released from the bulk wood; noncombustible carbon dioxide, water vapour and trace organics. Cellulose decomposes above about 250oC, and hemicellulose somewhat below that. Lignin begins to decompose above about 150oC, but has a large decomposition range.

Now, if you could heat a table under immense pressure, you may be able to influence the decomposition rates to a point where you could possibly get these materials to melt rather than simply decompose. Heating wood under moist high pressure conditions is known to impart an element of plasticity to some woods, so that it can be bent. I'm not confident that you could melt it though.

So I would call your table simply solid, not frozen.


How metal-plating on plastics is performed?

I couldn't comment on the name.
I am familiar with vacuum deposition, but i have only come across deposition on silicon, not plastic. the substrate ad metal are put in a vacuum chamber, the vacuum means that now the metal can be evaporated at a much lower temperature, and is then deposited onto the substrate.(see link)

I did an experiment of electroless plating of plastic with nickel. I can't say what plastic it was, probably PS or PP. usually electro plating requires electrodes and current to flow between them. electroless plating uses a electrochemical reaction in which a metal precipitates. Due to the treatment of the plastic the metal prefers to precipitate onto the plastic (this is a very simplistic information see liks for more information if you're interested)

the method I used in my experiment:
the plastic is cleaned by diping it in a base bath (sodium carbonate, trisodium phosphate, sodium hydroxide), this is mostly to remove oily debris .

Etching: ...etches the plastic to create microscopic pores that improve
adhesion of metal and following steps
Sensibilization: plastic is
dipped in ti solution, keeps the palladium at its reduced state (next
step)
Activation: plastic is dipped in a palladium solution, Palladium
catalyses the formation of nickel particles.
Plating: since there is no current flowing another particle has to be oxidized for the nickel to be reduced. in this case the reaction used is oxidation of  hypophosphite to phosphite

The result was a very nice metallic finish.

more information:

Electroless nickel plating on Wikipedia
plating kit
plenty of information on electroless deposition
evaporation-deposition


There is little reality to such an extrapolation. Attaching much significance will just lead you astray. There are a number of real gas equations which purport to correct for this or that "problem" with the ideal gas equation. The greater truth is that the equations just add extra fitting coefficients which make the equation more accurate. Different equations will work for different gases in different ranges.


What type of bonding is there among d-block metals?

Unpaired electrons are not the only thing that are used for bonding in d-block elements. The outermost ns electrons are also present for bonding.
     and when you ask what type of bonding is present in d-block elements, it means you mean to ask the type of bonding present only amongst these metals, and not with some other element(like p-block elements, say F, Cl). If that is the case, then only metallic bonds are present in d-block elements. For example, copper atoms are held by metallic bonds in the metal crystal.

The factors affecting the strength of a metallic bond are-

No. Of valence electrons (more number means higher attractive forces, and hence stronger bond)
Distance of ionic core from the valence electrons (lesser distance means a smaller size of the metal ion, and hence higher value of the force)
Presence of vacant d-orbitals (the free valence electrons in metal travel through the metal crystal using these vacant d-orbitals. So, if there are more number of vacant d-orbitals, the free electrons are more easily "accessible" for attraction by the ionic cores)


Perchlorate salts: (explosive) reactivity with methane?

In general, explosion of alkali metal perchlorates (at least in small quatities) is deflagration, with the perchlorate providing oxygen for combustion of another component of a mixture, rather than detonation (due to decomposition of the perchlorate). However, perchloric acid itself and other inorganic and organic perchlorates such as cis-bis-(5-nitrotetrazolato)tetraminecobalt(III) perchlorate can transition from deflagration to brissant detonation.

So, to answer your specific question, gaseous methane would burn more readily in the presence of Group I perchlorates, but the flame-front would be limited by the small contact area and low concentration of methane at the solid/gaseous interface. On the other hand, if one were to synthesize methyl perchlorate or  ethyl perchlorate, one might not be able to perform subsequent syntheses.

So, to avoid maiming or perhaps a terminal experiment, eschew perchlorates.


Why starch (amylose) and cotton (cellulose) are so different?

Your structures don't clearly show the different configurations between α- and β- linked glycopyranosides. Cellulose is a non-branching (poly) β-glycopyranoside. Amylose (a component of starch) is a non-branching (poly) α-glycopyranoside. The enzymes in your mouth and digestive system are good at breaking apart the α-glycopyranoside linkage, but not the β-glycopyranoside (it doesn't fit into the enzyme).
Amylopectin (the other component of starch) is a branched (poly) α-glycopyranoside. It branches with a 1-6 linkage. Glycogen is another branched (poly) α-glycopyranoside.
Cellulose is not the only (poly) β-glycopyranoside; there is also β-glucans.


Balanced equation for neutralisation reaction betweeen potassium carbonate and hydrochloric acid

They are all correct and they will all stay in equilibrium with one another, whereas the products of the second equation will be the most favoured, since carbonic acid is an unstable weak acid and will likely dissociate into $\ce{CO_2}$ and $\ce{H_2O}$. At the same time potassium hydrogen carbonate will get protonated easily: $$\ce{HCl + KHCO_3 \rightarrow KCl + H_2CO_3}$$ again resulting in more carbon dioxide and water.


You get sulfinic acids $\ce{RSO2H}.$

This paper by Dowson et al. [1] states that Grignard plus $\ce{SO2}$ gives sulfinic acids and as a reference quotes an older article by Allen Jr. et al. [2] which states

  The reaction between sulfur dioxide and Grignard reagent has been widely used for the preparation of aromatic and aliphatic sulfinic acids.

I would suggest you read [2] and the references it quotes.

References

Dowson, G. R. M.; Dimitriou, I.; Owen, R. E.; Reed, D. G.; Allen, R. W. K.; Styring, P. Kinetic and Economic Analysis of Reactive Capture of Dilute Carbon Dioxide with Grignard Reagents. Faraday Discuss. 2015, 183, 47–65. https://doi.org/10/gf9hvn. (Research Gate — PDF)
Allen, Jr., Paul.; Rehl, W. Richard.; Fuchs, P. E. Sulfone Formation during Sulfination of the Alkyl Grignard Reagent. J. Org. Chem. 1955, 20 (9), 1237–1239. https://doi.org/10/d42z3x.


What exactly makes a carbon atom "α" in a protein residue?

This nomenclature is due to the fact that amino acids are carboxylic acids.

Near the carboxylic acid moiety, the carbon chain is unbranched and simple, so the positions are named like an unbranched, simple aliphatic carboxylic acid.

The carboxylic acid ($\ce{-CO2H}$) is not indicated with a position. But the carbon immediately next to it is $\alpha$. The one next to that is $\beta$. That should be enough to explain the $\alpha$ in your context because that carbon is the one immediately next to the carboxylic acid. In a protein or polypeptide, the carboxylic acid is most frequently converted to an amide in a peptide bond, but the Greek letter nomenclature is unchanged.

This nomenclature is used in other cases for example, with an $\alpha$,$\beta$-unsaturated carbonyl.

The most publicly visible use of this nomenclature is when we use $\omega$ to denote the end of the carboxylic acid chain. This is the origin of the term $\omega$-3 [unsaturated] fatty acid.


Precipitation of CaCl2 and FeCl3 in Na2CO3

This must have been a very good question for a geochemist. They'd give you a complete answer with details of why $\ce{Fe2(CO3)3}$ does not but $\ce{Fe2O3}$ exist in our surrounding. Yet, I would like to give only a simple answer for the question: No, $\ce{Fe2(CO3)3}$ does not form (in sort) when you added $\ce{Na2(CO3)3}$ to an aqueous solution of $\ce{FeCl3}$. The reason?

An aqueous solution of $\ce{FeCl3}$ is acidic because $\ce{FeCl3}$ is a salt of a strong acid $(\ce{HCl})$ and a weak base $(\ce{Fe(OH)3})$. Thus an aqueous solution of $\ce{FeCl3}$ is in following equilibrium:
$$\ce {Fe^{3+} + 2H2O <=> [FeOH]^{2+} + H3O+ }$$
Thus, $\ce{H3O+}$ ions in solution can react with added $\ce{CO3^{2-}}$ ions to form carbonic acid $(\ce{H2(CO3)3})$,which can decomposed to form $\ce{CO2}$  and $\ce{H2O}$. Remaining $\ce{Fe^{3+}}$ will combine with left behind $\ce{OH-}$ ions to make $\ce{Fe(OH)3}$, which would decomposed to $\ce{Fe2O3}$. Thus, when $\ce{Na2CO3}$ is added to an aqueous solution of $\ce{FeCl3}$, one can expect the following sequence of reactions to be taken place:
$$\ce {FeCl3 (aq) + Na2CO3 (aq) -> [Fe2(CO3)]^{3+} (aq) -> Fe2O3 (s) + CO2 (g) + H2O (l)}$$


What is the formula for heat of vaporization under reduced pressure?

You are looking for the Clausius-Clapeyron equation.  It comes in various forms:

The exact version is

$$\frac{\mathrm dp}{\mathrm dT}=\frac{\Delta H}{\Delta V T}$$

or, since the volume of the gas phase is so much larger than the volume of the condensed phase

$$\frac{\mathrm dp}{\mathrm dT}=\frac{p \Delta H}{R T^2}$$

or even more succinctly, assuming $\Delta H$ remains constant

$$\frac{\mathrm d \ln p}{\mathrm d \frac{1}{T}}=-\frac{\Delta H}{R}$$


How to use molality and molarity to determine the molar mass of a compound?

You can use 3 different approaches to solve that. One I've given in the comments, another one that I will use here that hopefully sheds some light to the equation which by itself out of context here doesn't make much sense.

In say 1000g of water you have 0.554 mol of compound (definition of molality)
In 1000 ml of solution you have 0.497 mol of compound (definition of molarity)
Thus in 1115 ml of solution you have 0.554 mol (linearity)
So you have 1000g of water and 0.554 mol of compound in 1115 ml of solution or 1115*1.063 = 1185 g of solution (using the density)
Obvioulsy enough you have 1185-1000 = 185g of compound. But we know already that there are 0.554 mol of the compound. So its molecular weight is 185/0.554 = 333.8 (so it's not sugar?)

Same result gives my (preferred) method in the comments


The first step is a conversion of the nitrile into a carboxylic acid, in the literature Martin Olbrich, Peter Mayer, and Dirk Trauner (J. Org. Chem., 2015, 80, 2042-2055) used KOH in ethylene glycol at 155 oC for a couple of hours. The next reaction is attack of iodine on the alkene. The cyclic cationic intermidate will be attacked by the carboxylic acid.

I am not sure what SBH is, but it is some reducing agent which functions in the same way as hydrogen with platinium and a base. This combination of reagents removes the iodine by converting the alkyl halide into a sp3 CH. This gives us P.

Another option I would consider is to use nickel chemistry, I would put some hydrated nickel(II) chloride in a flask with ethanol, I would then add a solution of sodium borohydride in water. Do this with care as it emitts a lot of hydrogen gas. It will change the nickel solution from a nice green clear solution to an ugly black slurry. This is not a problem. Add the organic material to this. This black slurry acts like raney nickel. I have used it a few times to remove things like thioether groups from aromatics. For example if you react 2-acetyl pyridine with a base and carbon disulfide and methyl iodide you can make a new compound. If this is then reacted with 2-acetyl pyridine and base, before being boiled up with ammonium acetate in acetic acid you can form terpy with a SMe group on the 4 position of the central pyridine ring. The way to convert this SMe version of terpy to terpy is to boil it with the black slurry made from nickel and sodium borohydride. The problem is that it is hard to do oxidative addition of alkyl halides, many reagents do not work well.

If we consider the Monsanto process for acetic acid then the rate determining step is the reaction of the methyl iodide with the rhodium, if we move to the more modern BP version which uses iridium then the oxidative addition is much more easy. I think that the platinum catylst will work a lot lot better than either the nickel method I mentioned or palladium on carbon.

I worry that sodium borohydride with no modification will be unable to deal with the alkyl iodide in the moelcule.

Another option would be to use tributyl tin hydride and AIBN, this is a selective radical based reducing agent which will convert alkyl iodides into alkyl hydrides (alkanes) without touching alcohols, esters or other carboxylic acid based groups. This could be a method of putting a deuterium label into the product if we very much wanted it.

Keep in mind that if you are ever asked for an example of a free radical chain reaction other than halogenation of an alkane that you can always offer reduction of an alkyl halide by tributyl tin hydride or a free radical polymerisation reaction.

Now a lot of people hate tributyl tin chemistry, there are several disadvantages of it. Firstly it is often hard to separate the tributyl tin side products from the product. Secondly organo tin chemistry is rather toxic, the worst effects are on marine life such as oysters. Tributyl tin compounds can cause female marine organisms to turn into an intersex creature which then dies. But in our case it is not so bad. What we can do is to use a tributyl tin reducing agent and the use the acid base properties of the carboxylic acid to separate it from the neutral tributyl tin iodide.

If shaking a solution of the product in benzene with sodium hydroxide was to fail, I would suggest making a solution of the carboxylic acid in methanol and then passing this through a pad of a weak base ion exchange resin (amino groups on a polymer such as polystyrene). I would then elute the column with more methanol and then with formic acid dissolved in methanol.

The next thing is that we use aqueous caustic potash to break open the lactone ring. The lactone is then oxidised with permanganate. I would have used a chromium oxidant as they tend to be a little more gentle.

Next we react the ketone with ethylene glycol using acid and some dehydration method. One good method is to use toluene and a dean-stark head. Another one which I like is to use triethyl ortho formate with a little acid. This triethyl orthoformate method works under more gentle conditions to remove the water. We might get some ester formation but in the ideal world I would want to do the next step with an ester not a carboxylic acid.

We reduce the carboxylic acid using the cyclic acetal to protect the ketone from the lithium aluminium hydride. Then using acid in water we deprotect the ketone. We then convert the alcohol into a leaving group with tosyl chloride. We then form an enolate with the LDA which then will attack the tosylate to form the ketone.

It is important to keep in mind that under some conditons that enolates can be attacked on the oxygen by hard electrophiles such as tosylates. But I suspect in this case that the tosylate will be unable to reach the oxygen. It would be best to run this reaction at a high dilution. One method would be to slowly add the ketone-tosylate to a large flask containing a large amount of solvent and some LDA. Another method would be to convert the tosylate to the alkyl iodide by reacting the tosylate in acetone with sodium iodide first.

It may be a good idea to read the paper by H. W. Whitlock in Journal of the American Chemical Society 1962, 84 (17), 3412-3413. This is the original synthesis of the compound.


Why are antibonding molecular orbitals formed?

Why should they make also an antibonding molecular orbital?

Orbital symmetry must be conserved. Thus, when you mix orbitals, the number of orbitals must also be conserved. As a consequence, when two orbitals are operated on to produce new orbitals, two new orbitals must be created.

Think about it conceptually this way: When two orbitals that have the same symmetry and appropriately matched energy come together, a bonding orbital is formed. This bonding orbital contains all of the "good orbital stuff" from that pair of orbitals. But, because there were two orbitals originally, there has to be "orbital stuff" left over. This leftover "orbital stuff" forms the antibonding orbital.

Now, the screwy part is that orbitals are not real. They are mathematical constructs to describe the wave-like probabilistic properties of electrons. If there are no electrons in the orbital, then the orbital is not really "there". So, while an antibonding orbital is formed, it is not real unless it has electrons in it. In the case of water above, the antibonding orbitals are not populated, so they don't exist.


Does s stand for sigma bond and p for pi bond?

You got it backwards.

The $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, $\mathrm{f}$ orbitals stand for sharp, principal, diffuse, and fundamental. Wikipedia: Electron Configuration § Notation:

  The choice of letters originates from a now-obsolete system of categorizing spectral lines as "sharp", "principal", "diffuse" and "fundamental" (or "fine"), based on their observed fine structure: their modern usage indicates orbitals with an azimuthal quantum number, l, of $0$, $1$, $2$ or $3$ respectively.

Then, $\sigma$, $\pi$, $\delta$, $\phi$, being the Greek equivalents of $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, $\mathrm{f}$ respectively, came to designate bonds or molecular orbital symmetry, because $\mathrm{s}$ is the first orbital to form $\sigma$ bonds, $\mathrm{p}$ the first orbital to form $\pi$ bonds, $\mathrm{d}$ the first orbital to form $\delta$ bonds, and $\mathrm{f}$ the first orbital to form $\phi$ bonds.

The $\mathrm{s}$ and $\mathrm{p}$ in $\mathrm{sp^3}$ are precisely the $\mathrm{s}$ orbital and the $\mathrm{p}$ orbital. $\mathrm{sp^3}$ means that $1$ $\mathrm{s}$ orbital mixes with $3$ $\mathrm{p}$ orbitals to create $4$ hybrid orbitals known as $\mathrm{sp^3}$ orbitals.


How to calculate the oxidation number of sulfur in H2SO5?

$\ce {H_2SO_5}$ has an oxygen-oxygen bond. This means that two of the five oxygen atoms have an oxidation number of $-1$. Same case as in $\ce{H2O2}$.

$$2\cdot (+1) + 1\cdot(x) + 3\cdot(-2) + 2\cdot (-1) = 0\\2 + x - 6 - 2 = 0\\
x = +6$$

You have 3 oxygens with oxidation number $-2$ and 2 oxygen atoms with $-1$.


Why there is no temperature scale tied to normal human body temperature?

Why nobody used human body normal temperature as 0?

That has been done.
See Medical Thermometry and Human Temperature at page 12:

37° Centigrade scale
29.6° Reaumur
98.6° Fahrenheit
77° Walferdin's tetracentigrade
0° of the physiological scale

Then the physiological scale is further discussed starting on page 265, which includes a drawing of a physiological scale thermometer.

physiological 0 is placed in the centre of the drama, whose acts are health, sickness, and death


This is a fundamentally very difficult question. Your example of of $\ce{HClO + NaOH}$ hides several assumptions, e.g., time frame and aqueous solution, which really means some number of $\ce{HClO}$, some number of $\ce{NaOH}$, and presumably a huge number of $\ce{H2O}$. When mixing all three components a wide variety of reactions take place, e.g., disproportionation of water ($\ce{H2O + H2O -> H3O^{+} + OH^{-}}$), solvation of $\ce{NaOH}$ etc.

In general, a full exploration of all possible reactions and their barriers have to be assessed in order to make a prediction. For instance, $\ce{HClO + H2O -> HCl + O2H2}$ is a possible reaction, or $\ce{NaOH + 2 HClO -> NaCl + H2O2}$ even in the absence of extra water. The number of possible recombinations of the input materials generally increases exponentially with the number of atoms (e.g., the number of possible isomers of an n-alkane).


Does evaporation reverse salt hydrolysis?

Simple cyanide salts

In aqueous solution, simple cyanide salts more or less completely dissociate into their constituent ions.  E.g. for potassium cyanide:

$\ce{{KCN}(s) + H2O(l) -> K+(aq) + CN- (aq)}$

Those solutions are relatively stable.  There is an acid-base equilibrium between $\ce{CN-}$ and $\ce{HCN}$, but in unbuffered or in alkaline aqueous solution this equilibrium will be almost completely on the left:

$\ce{CN- + H2O <<=> HCN + OH-}$

Over very long times, if the solution is left in an open container, some of the tiny, minute amount of $\ce{HCN}$ formed in this way could evaporate, leading to a very slow loss of cyanide from the solution.  If stored closed this will not be a problem.

(Mixing simple cyanide salts with an acid will result in an immediate and dangerous release of gaseous $\ce{HCN}$, so in this question I assume that no acids are involved.)

Thus, for simple cyanide salt solutions stored appropriately, yes you should indeed be able to recover all of the initially added the solid salt after boiling off or otherwise removing any water.

Cyanide complexes of transition metals

Salts such as potassium ferricyanide $\ce{K3Fe(CN)6}$ consist of potassium cations and complexed ferricyanide anions, $\ce{Fe(CN)6^{~3-}}$.  They are soluble in water as well, but unlike their simple cyanide anions, they can be hydrolyzed.

The hydrolysis is catalyzed by UV light and results in the formation of free $\ce{CN-}$ anions as well as iron hydr(oxide) solid precipitates.

See this book for more information.


In any case, you should calculate the mean $\overline X$, the standard deviation $\sigma$, and the standard error of the mean $\sigma_{\overline X}$ for the six trials.

However, the estimated uncertainty $u(y)$ of the final result $y$ depends on what is meant by “all with the uncertainty of $\pm0.1\ \mathrm{ml}$”, i.e. $u(x)=0.1\ \mathrm{ml}$.

For example, if this uncertainty $u(x)=0.1\ \mathrm{ml}$ is caused by the unknown error of a pipette and the same pipette was used for all six trials, each of the six individual results includes the same systematic error. You can neither eliminate this error nor reduce the caused uncertainty by making additional trials. In this case, you should combine (by using the rules for error propagation) the calculated standard error $\sigma_{\overline X}$ and the assumed uncertainty $u(x)=0.1\ \mathrm{ml}$. Therefore, the estimated uncertainty $u(y)$ of the final result cannot be smaller than $u(x)=0.1\ \mathrm{ml}$.

If the uncertainty $u(x)=0.1\ \mathrm{ml}$ is caused by a random error of each individual trial, it is already included in the observed standard deviation $\sigma$  and thus already taken into account for the calculated standard error $\sigma_{\overline X}$. Therefore, the observed standard deviation $\sigma$ should be greater than (or at least equal to) the assumed uncertainty $u(x)=0.1\ \mathrm{ml}$. In this case, you may assume that the estimated uncertainty of the final result $u(y)$ is approximately equal to the calculated standard error $\sigma_{\overline X}$.


Why are acetylene and air commonly used for AAS?

Because it gives a very high temperature: 2200-2400 $^\circ$C.
This is due to its high heat of combustion, around 1300 kJ/ mol.

If still higher temperatures are required, as for Al, Si, Ti, etc., a combination of acetylene and $\ce{N2O}$ is used, which reaches 2600-2800 $^\circ$C.


How are the hybrid orbitals of sulfur hexafluoride shaped?

If you limit consideration to hybridization of atomic orbitals, a good reference to see is On the role of d orbitals in sulfur hexafluoride J. Am. Chem. Soc., 1986, 108 (13), pp 3586–3593.

First, it is found that of the 6 valence electrons that atomic sulfur has (two 3s and four 3p), in SF6 a total of 3.1 electrons worth of electron probability density remain in sulfur atomic orbitals.

Of the 3.1 electrons:

0.97 are in the 3s orbital

1.82 are in 3p orbitals

0.24 are in 3d orbitals

and small fractions are in 4s, 4p and 4 d orbitals.

For a more recent study, see Theory of Hypervalency: Recoupled Pair Bonding in SFn (n = 1−6) J. Phys. Chem. A, 2009, 113 (27), pp 7915–7926


Great question! I don't believe it'll wear out in the short term. Light is produced when the electron is excited to a higher orbital and drops back down; this is completely reversible. Neon tubes work on the same principle and last for years!
Eventually (probably after many years) a whole bunch of effects will probably kick in; if there's some impurities in the krypton, they might react with the plasma; the walls of the container will degrade and become opaque, etc.
Whether a certain voltage will cause the krypton to glow seems to be a pretty complicated problem! I believe that would depend on the frequency of the tesla coil (V = dB/dT), how it's coupled to the gas (is there a wire in the tube?), and how large the volume of gas that you want to light.


Predicting the temperature of an object using the heat of its surroundings after a certain amount of time?

You're basically looking at radiation, conduction, and convention as methods to dissipate heat.

Heat loss by radiation can be computed from Wein's law or something similar.

Conduction requires knowing the thermal properties and dimensions of the materials in contact.

Convection is the trickiest one since it requires some model for how the air or other fluid around the object is moving and coupling this to conduction.

All three are subject to significant errors in computation if you're not careful though if you're trying to estimate rough order of magnitude, you might not care.

All in all, this is not an easy problem to compute from first principles.

Note also that your example is not nearly well-defined enough to solve this problem. What is the surface area of the bottle and the size of the refrigerator. How well does the refrigerator maintain its internal temperature? Is there a fan? Is the water inside a thermos bottle?


Are β-sheets formed from one or more polypeptides?

A β-sheet is formed as the result of hydrogen bonding between the β-strands in the secondary structure of a protein. The hydrogen bonding occurs between the carbonyl oxygen and the amide hydrogen of adjacent strands, and can take on either a parallel or an antiparallel arrangement. As can be seen in the diagrams, antiparallel arrangement of the polypeptides results in more stable interactions between the strands, but both arrangements are possible and can be found in nature. Regardless of the type of arrangement, β-sheets can be formed by either interactions between different polypeptides or intramolecular hydrogen bonding within a single polypeptide. In fact, intramolecular hydrogen bonding creating β-sheets is really common. One example of it occurring in β hairpin motifs.


Is chirality a property of only molecules with exactly 1 chiral centre?

I don't want to try to interpret the intent behind the wording of the phrase you quoted, so I'll just state the situation and you can read into it as you wish.

There is no truly general relationship between the number of chiral centres in a compound and whether it displays chirality overall.

However, if a compound has exactly one chiral centre, then it must be chiral. Compounds with more than one chiral centre may or may not be chiral, depending on what those chiral centres are and their relative stereochemical relationship. And as has already been rightly pointed out (4 seconds before me), compounds with zero chiral centres may still be chiral.


Shouldn't enthalpy be constant measured at constant volume?

Enthalpy is heat flow associated with a constant pressure process.  Here is how:

Start with the familiar first law of thermodynamics for energy change:

$\Delta E=q+w$

$E$=energy

$q$=heat flow into the system

$w$=work done on the system

The work done is rendered thusly:

$w=-\int_\text{start}^\text{end}P\,\mathrm dV$, $P$=pressure, $V$=volume

Pressure is deemed constant so:

$w=-\int_\text{start}^\text{end}P\,\mathrm dV=-P\Delta V=-\Delta(PV)$

And then our first law balance equation becomes:

$\Delta E=q-\Delta(PV)$

$q=\Delta(E+PV)$

So heat flow is the change in a combination of state variables involving energy, pressure and volume.  This combination is what we define as enthalpy and usually label $H$.  Thereby, this enthalpy variable is designed to automatically include (mechanical) work inherent to a constant pressure process.


You can't remove the acids, because they're dissolved in the liquid. You can, however, neutralise it.

Acid is basically $\ce{H+}$ ions dissolved in the water. By reacting it with a base, you do the following reaction:

$\ce{H+ + OH- -> H2O}$

So the acid reacts to form harmless water.

The problem is that you can't just add $\ce{OH-}$, you have to add it as some salt. Also, the citric and phosphoric acid you have in the cola were added are also dissociated from a salt or molecular form:

$\ce{H3PO4 -> H+ + H2PO4−}$ for phosphoric acid, for instance.

So whatever you put in there that reacts with the $\ce{H+}$ should also react with the $\ce{H2PO4−}$ ion to make something that's not soluble.

Usually, neutralising acids is done with carbonates such as sodium carbonate or calcium carbonate (also known as soda and lime, respectively). Both of which are safe materials to handle. The problem is that by doing that, you're introducing carbonate ($\ce{CO_3^2−}$) which might alter the taste.

A better solution is to add in hydroxides, such as sodium or calcium hydroxide ($\ce{NaOH}$ or $\ce{Ca(OH)2}$). The problem is that those are much more hazardous things which can easily hurt you if you don't use them properly. I would recommend using calcium hydroxide because then the calcium will react with the phosphate anion in the cola to make insoluble calcium phosphates and with citrate to make (somewhat) insoluble calcium citrate. Then you can just filter the stuff out and have your acid-free cola.

Basically, by adding in $\ce{Ca(OH)2}$ to your cola, you react the "acid" bit to make water, and you react the "phosphoric" and "citric" bit to make a white solid precipitate in the bottom of your glass.

DISCLAIMER

This is all in theory. I highly recommend against adding chemicals to your food and drink. If you don't like the acidic taste of cola, don't drink it. Find another brand or another drink that you do like. Tea? Anyway, do not mess with your food unless you know exactly what you're doing and you know that your chemicals are food-grade and safe. Don't go and do stuff with your cola just because someone on the internet (me) told you it will work.


Electron configuration of iron(I) cation

Isolated in a vacuum, in the absence of external fields, the first configuration is correct - $[\ce{Ar}]\mathrm{(3d)^6(4s)^1}$ is the ground electronic state of the iron(I) cation $\ce{Fe^{+}}$. More specifically, the ground state also has the term symbol $\mathrm{^6D_{9/2}}$.
NIST's Atomic Spectra Database has compiled the ground state electronic configurations of almost all possible ions of almost every known element (including such exotic things as $\ce{U^{58+}}$), obtained from a combination of experimental data and theoretical calculations. For example, search for iron, and you will be presented a table with all ground state electronic configurations, from the neutral atom $\ce{Fe^{0}}$ to the hydrogen-like $\ce{Fe^{25+}}$. You can find a description of the output data here.
I'm sure you'd like a reason for why $\ce{Fe^{+}}$ has the first configuration you suggest and not the second, but there is almost certainly no explanation which is both simple and correct. Unfortunately the predictive power of the aufbau principle is somewhat overstated, even when taking into consideration qualitative corrections. Depending on your notions, there are arguably several dozen "unexpected" electronic configurations, across all possible ions of all elements. The worst offenders are the heavier elements (say, the sixth period and onwards), which are rarely (if ever) covered in early chemistry classes.


Difference between Side Chains and Functional Groups

So the example you present is of a poor question. It's a classic case of a "best answer" MCQ that requires you pick the answer that is most accurate, rather than an answer that is unambiguously right. (I hate these.) Alkanes are listed in most textbooks and guides to organic chemistry as a "functional group." The reality is that it's hard to do chemistry on an alkane in most circumstances, so it's probably the least "functional" of the functional groups.

That said a side chain is an imprecise term. It's most consistently used in amino acid chemistry to refer to the differentiating moiety. But it's really used to indicate any moiety that hangs off of a common structure. What constitutes the side chain versus the primary structure is a function of context. A large polymeric structure can have side chains that consist of aromatic rings, while with a smaller structure consisting of a single benzene ring with some alkanes hanging off of it, the alkanes will be considered the "side chain." There's no substantive difference between a side chain and a functional group except for emphasis in a discussion.

So in brief: Functional groups are the things you learned about like amines, amides, carboxylic acids, etc. Side chains are parts of a molecule that are considered to be "hanging off" the "main" part of the molecule, and are highly dependent on the context of the discussion. Sometimes these side chains are single functional groups, and sometimes they're much bigger structures. It really depends. But functional groups are always functional groups. Though in general, chemists tend to "forget" alkanes as functional. But as I indicated earlier, the question isn't all that great, so don't feel too bad if it confused you.


at constant volume why the change of enthalpy doesn't equal the change of heat

At constant volume, the pressure is not necessarily constant. The pressure will usually increase when the system is heated. This makes the $PV$ term not constant. For this reason, you get:

$$\Delta H = \Delta (E + PV) $$
$$\Delta H = \Delta E + \Delta (PV) $$
$$\Delta H = \Delta E + V\Delta P $$

Now heat is just $\Delta E$, because there is no work. So,

$$\Delta H = q+ V \Delta P $$


Why does ionisation energy to remove electrons from the same subshell increases?

The ionisation energies would be same if there was no shielding effect but it exists. The electrons due to mutual repulsion  prevent the outer electrons from the nuclear pull. Thus the force with which the nucleus attracts the electrons decreases as we move away from the nucleus both because of the increasing distance and also due to the increasing shielding effect of the inner electrons . Now in the same shell when one electron is removed the shielding effect decreases . The shielding effect due to the electrons of the same shell is the greatest. Thus the successive ionisation energies of any atom increases due to decreasing shielding effect.