Why is fluorine more reactive than iodine despite the weaker I-I bond? The atomic radius of halogens increases as we go down the group due to the addition of new shells. As a result, the bond length of halogen $\ce{X-X}$  increases down the group. So, less energy is required to break the bond (bond dissociation enthalpy decreases). So, as per this assumption, iodine should be the most reactive halogen. But this is not true. Rather,  fluorine is the most reactive halogen. It reacts violently with almost all chemicals. Bond cleavage of fluorine is much easier that that of other halogens, which may be due to the repulsive force of the lone pairs of electrons.

$$\ce{2F2(g) + 2H2O(g) → O2(g) + 4HF(g) + heat}$$

So, what about iodine? This source gives the values of bond dissociation enthalpy. It shows that the B.D.E value of iodine ($\pu{151 kJ/mol}$) is less than that of fluorine ($\pu{158 kJ/mol}$). So, iodine should be the most reactive halogen and not halogen. But this does not happen. In most of the reactions of iodine, equilibrium is maintained. To push the reaction forward, a catalyst is used.

$$\ce{I2(l) + H2O(l) <=> OI-(aq) + 2H+(aq) + I-(aq)}$$

Another similar phenomenon is that acidity of hydrohalic acid increases on going the group. $\ce{H-X}$ bond length increases on going down the group. Bond dissociation enthalpy decreases. Bond breaks into corresponding ions. So, $\ce{HI}$ is able to furnish $\ce{H+}$ ion faster than $\ce{HF}$. So, $\ce{HI}$ is more acidic than $\ce{HF}$. But, the reverse is observed. $\ce{HF}$ is so reactive that it reacts with glass and has to be stored in wax bottles.

$$\ce{SiO2 + 4 HF -> SiF4(g) + 2 H2O}$$

$$\ce{SiO2 + 6 HF -> H2SiF6 + 2 H2O}$$

In most of the reactions of hydroiodic acid, equilibrium is maintained. To push the reaction forward, a catalyst is used.

$$\ce{H2(g) + I2(g) <=> 2 HI(g)}$$

My question: Why is it so?

In comparing the reactivity of fluorine or iodine, there are a couple of factors to consider. Part of it is the strength of the $\ce{X–X}$ bond, which is a barrier that tends to discourage them from reacting. Obviously, the higher this barrier, the less reactive the halogen is (in general).

However, unless you're trying to consider the relative reactivity towards homolytic cleavage of the bond, it isn't sufficient to look at the bond strength. The other bit is the driving force for the formation of the products – in other words, something that encourages them to react.

In this case, the formation of fluorine-containing products is generally much more thermodynamically favourable than that of the corresponding iodine-containing products. Some factors influencing this are:

The electron affinity of fluorine is much larger, which makes reduction to fluoride much easier
$\ce{E–F}$ bonds are stronger than $\ce{E–I}$ bonds, in both an ionic and covalent sense.
Fluoride has a larger solvation enthalpy than iodide

For example, factors 1 and 3 lead to the large differences in the reduction potentials: $E^\circ(\ce{F2}/\ce{F-}) = \pu{+2.87 V}$ versus $E^\circ(\ce{I2}/\ce{I-}) = \pu{+0.54 V}$.

Silica is readily attacked by $\ce{HF}$ not because it is a strong acid, but rather because of the strength of $\ce{Si-F}$ bond thus formed.

That deals with the thermodynamic parts of "reactivity", in the sense that it explains why some reactions with iodine require elevated temperatures for any significant conversion, whereas reactions with fluorine proceed readily at room temperature.

The same factors, though, apply to the kinetic aspects of "reactivity". The activation energy is largely dictated by bonds that one has to break or form, for example, and the rate of electron transfer processes usually increases with $E^\circ$. Exceptions to this arise when – for example – the reactants have solid state structures which must be broken apart before reaction can occur. However, that is not the case here.


How can I separate these two alcohols from each other out of the aqueous solution? I have the following question for homework and am stumped. I am tasked with separating 3-(2-methoxyphenoxy)propane-1,2-diol and (1R,2S)-2-(methylamino)-1-phenylpropan-1-ol. They are both solvated in water.

I was thinking to work up with acid to cleave the ethers on the left compound, but that won't do much to reduce solubility for crystallization. Esterification with a large hydrophobic carboxylic acid wouldn't do much either because both compounds have hydroxyls. Anyone have suggestions?

I am tasked with separating these two compounds.

If you want to keep both compounds, degrading one of them is obviously not an option.

  They are both solvated in water.

This can be changed - reversibly ;-)

Your left compound is a diol. React the mixture with a ketone in the presence of an acidic catalyst.

Typically, these reactions are performed to protect ketones by heating them in with ethanediol in toluene in the presence of toluene sulfonic acid. There is no reason not to use it the other way around to protect the diol.

The reaction product is a cyclic acetal (1,3-dioxolane) which is much less  soluble in water that the parent diol.


A question about "post it note" I dipped a sticky note in water and there was a clear difference between the water absorption in the area with glue and the one without glue. The area with glue was more damp than the one without it. I would like to know why.

I like your "why?" question. So I dig down to find out why Post-It notes stick to any surfaces, yet, when removed, it's able to post again in another place. This is my finding and it makes sense Adhesives:

  Look at the back of a sticky note using an electron microscope and you'll see not a continuous film of adhesive but lots of microscopic glue bubbles, known as microcapsules, which are about 10–100 times bigger and much weaker than the glue particles you'd find lazing around on normal sticky tape. When you push a Post-it® onto a table, some of these relatively large sticky capsules cling to the surface, providing just enough adhesive force to hold the weight of the paper in the little note. Every time you attach and peel off the note, dust and dirt attach to the adhesive capsules, so they progressively lose their stickiness. But since there are so many capsules of all different sizes, a Post-it® note does go on sticking for quite a while.

I believe, when you dipped your sticky note in water, these microcapsules filled with water, which would be a thicker layer than the water in other areas. So, gum layer area contains more water, thus, that part is actually more damp than the rest of the paper.


How does the position of hydroxyl group in a nucleotide monomer affect the dehydration synthesis of nucleotides? In dehydration synthesis of nucleotides, the hydrogen atom from the 3' carbon on the deoxyribose sugar of one nucleotide reacts with the hydroxyl group on the phosphate group of another nucleotide to form water. As the water molecule forms, a new covalent bond comes into existence between the two nucleotides.
Hydroxyl group can be in different positions within the phosphate group, for example:

I'm wondering if different positions of the hydroxyl group have an impact on the dehydration synthesis between nucleotide monomers. For example, when the hydroxyl group is at the bottom, will the dehydration synthesis occur at all? Also, when the hydroxyl group is on the left, does the spatial structure of the sugar-phosphate backbone have a spatial shape different from when the OH group is on the top?

The phosphate ion is of vital importance in biological systems. Its molecular geometry is tetrahedral. Thus, based on the $\mathrm{pH}$ of the system, hydroxy groups are inter-changeable with $\ce{P=O}$ form and hence all three remaining $\ce{O}$s are equivalent (see Waylander's comment). Therefore no impact on the dehydration.
However, the stereochemistry of the hydroxy group on the 3'-carbon on the deoxyribose nucleus of the nucleotide has an impact on the enzymatic dehydration process. For instance, if you change the original $(S)$-orientation of 3'-hydroxy group to $(R)$-orientation, DNA-synthase would not pick that nucleotide to synthesize intended DNA sequence.


Why is filter paper fluorescent? Typical filter paper used in chemistry lab (e.g. Whatman #1) exhibits autofluorescence in near UV with excitation maximum at 365 nm and emission maximum at 440 nm. It was reported in literature1 that fluorophores from filter paper can be extracted by PBS buffer in wide range of pH, and thus they are at least partially soluble in water. However, cellulose, the major component of filter paper, is not fluorescent in this wavelengths range and is insoluble.

What is the source of filter paper fluorescence? I assume it comes from paper pretreatment and bleaching, but what is chemical identity of fluorescent compounds in it?

Danielson, T. L. et al.; Impact of filter paper on fluorescence measurements of buffered saline filtrates. Talanta 2003, 59, 601–604.

In the production of white paper, the bleached pulp is often treated with optical brightening additives (OBA), that absorb in the uv and emit in the blue range. If I remember correctly, a typical example is 4,4'-diaminostilbene-2,2'-disulfonic acid.

Here's an article from The Open Textile journal (OA) in which the compound mentioned above is used as a starting material in the synthesis of another optical brightener for textiles.


Why did my lemon and penny battery fail? I followed instructions in Popular Mechanics publication for making a battery using lemons and pennies but it didn't work. I have a few ideas on why it might not have worked and would like to know which of them are more or less reasonable.

Here is what I did:

I used sandpaper to remove copper from one side of pennies to reveal the zinc core. Supposedly, only pennies minted after 1982 have a zinc core so I made sure to use only such pennies. The instructions call for 5-10 pennies - I did 6.

Deviations from instructions:

I did not thoroughly remove all traces of copper from 2 of the 6 pennies. The amount of copper that was still visible covered less than 1% of the surface.
I used a lower grade sandpaper than the instructions called for.

I cut cardboard discs slightly smaller than the pennies in equal number then soaked them in lemon juice

Deviations from instructions:

the cardboard discs I cut separated into three layers, the outer two layers appeared to be very tightly pressed and minimally absorbent

I made a stack of alternating pennies and lemon-soaked cardboard discs with the copper-side of each penny facing down. On the top of the stack I placed an intact, non-sanded penny then wrapped electrical tape around the stack.

Deviations from instructions:

none that I am aware of

I attached an LED to the top and bottom on the penny stack. I tried both combinations of where to put the positive and negative ends but neither caused the LED to light up.

Why I think it Failed:

By not removing every last trace of copper from the sanded-side of the pennies, I created a kind of short-circuit that discouraged the charges from moving through the stack of pennies.
Since the cardboard was not very absorbent, it might have been particularly porous either and this might have prevented sufficiently free flow of particles between each penny in the stack.
By using a relatively low number of pennies, I guess I set it up to have a relatively low voltage. Maybe I could have overcome problems 1 or 2 if I used more pennies.

I want to know if I my assessment of what went wrong is reasonable and welcome any suggestions of things I might be overlooking.

By not removing every last trace of copper from the sanded-side of the pennies, I created a kind of short-circuit that discouraged the charges from moving through the stack of pennies.

This is the main problem. Until all of the copper is gone from the zinc side of each penny, there is little reason for the two halves of the electrochemical reaction between zinc and copper not to just take place locally.  Your battery likely had one or more ionic short-circuits, just as you say, on these incompletely sanded surfaces.

  Since the cardboard was not very absorbent, it might have been particularly porous either and this might have prevented sufficiently free flow of particles between each penny in the stack.

This is unlikely.  As long as the cardboard was soaked through with the lemon juice, it should function fine as an ion-transport medium.

  By using a relatively low number of pennies, I guess I set it up to have a relatively low voltage. Maybe I could have overcome problems 1 or 2 if I used more pennies.

This is also unlikely. With five junctions (one between each neighboring pair of pennies), at a theoretical $1.1~\mathrm V$ per junction your stack voltage should have been at least about $5~\mathrm V$. This should be plenty to light a simple LED.


pH of a buffer solution calculation Calculate the pH of a mixture of $75\:\mathrm{cm^3}$ of methanoic acid solution of concentration $0.10\:\mathrm{mol\:dm^{-3}}$ and $75\:\mathrm{cm^3}$ of sodium hydroxide solution of concentration $0.050\:\mathrm{mol\:dm^{-3}}$. The $\mathrm{p}K_\mathrm{a}$ of methanoic acid is $3.75$.

I got $0.00375\:\mathrm{mol}$ of $\ce{HCOONa}$, the salt, made. The concentration of that is $\dfrac{0.00375}{150/1000}=0.025\:\mathrm{mol\:dm^{-3}}$(?)

Using Henderson–Hasselbalch: $\mathrm{pH}=3.75+\log\dfrac{0.025}{0.10}=3.15$

The answer says $3.75$, is that a typo?

One will need the Henderson Hasselbach equation which is given as

$$\text{pH}=pK_a+\log\frac{\ce{A-}}{\ce{HA}}$$

where HA and A- are the concentrations of the conjugate acid and conjugate base.

First we find our moles of each.

$\mathrm{{mol_ {acid}}=75~cm^3\times(0.10~mol~dm^{-3})\times(10cm)^{-3}\ dm^{3} = 0.0075\ mol\ \times \dfrac{1000\ mmol}{1\ mol} = \ 750~mmol}$

same thing for base: $\mathrm {mol_{base} = 375} $ mmol

Set up an ICA (not ICE) to handle your limiting reagent: (Write down the chemical equation if this is hard to see. One should react acid with the best base available)

$\ce{HA}$: $\pu{750 mmol -375 mmol} = \pu{375 mmol}$
$\ce{OH-}$: $\pu{375 mmol - 375 mmol} = \pu{0}$
$\ce{H2O}$: --------------------
$\ce{A-}$: $\pu{0 + 375 mmol} = \pu{375 mmol}$

In the Henderson Hasselbach equation moles can be used instead of molarity because both conjugates are in the same volume so they cancel:

$$\mathrm{pH}=3.75+\log\dfrac{375\ \mathrm{mmol\ \ce{A-}}}{375\ \mathrm{mmol\ } \ce{HA}}$$

$\mathrm{pH}=3.75$


How can potassium hydroxide be prepared from available potassium chloride (KCl) and Quick Lime (CaO) I live in a small town where Potassium hydroxide is not readily available. (I need above for preparing liquid soap.) But Potassium chloride (KCl) and Quick Lime (CaO) are available as fertilizer and White-wash ingredients respectively. Can KOH be prepared from above ingredients? What will be reaction conditions?

Mixing $\ce{KCl + CaO}$, or $\ce{KCl + Ca(OH)2}$, will never produce pure $\ce{KOH}$ without $\ce{Ca(OH)2}$. And this Ca(OH)2 will prevent soap from being synthesized out of oil, as $\ce{(Ca(OH)2}$s destroy soap in case a little bit of soap has been synthesized. The only way of producing $\ce{KOH}$ out of $\ce{CaO}$ or $\ce{Ca(OH)2}$ is to mix it with potassium carbonate $\ce{K2CO3}$. Of course $\ce{Ca(OH)2}$ is not very soluble in water. But $\ce{CaCO3}$ is still more insoluble. So the following reaction is possible :$$\ce{Ca(OH)2 + K2CO3 -> 2 KOH + CaCO3(s)}$$ And the insoluble $\ce{CaCO3}$ can be filtrated to produce a KOH solution


What is a good resource for learning about the intermolecular bonding in water beyond high school? I'm writing a paper for school on boiling water at various temperatures and pressures, but only noting cases wherein high temperatures were achieved without boiling under low pressure (metastable states). I haven't been able to find a good explanation for this phenomenon other then the chaotic nature of boiling coming from heterogeneities. I believe it is because of the high surface tension, but I have not been able to find any in depth resources for intermolecular bonding (in water) other than at the high school level (hydrogen bonding). I'm just looking for some guidance, thank you.

Google Scholar is very helpful when you need to read advanced level of work. The key word you can search is Superheating of liquids (water is one of the common examples). Anyone who has been burnt by microwaved water knows the pain of superheated water. There are some excellent reviews on the phenomenon of superheating.

This is pretty good overview:
http://www.lassp.cornell.edu/sethna/Nucleation/

Starting from the original work in 17th Century to recent ones...

An experience to prove, that water, when agitated by fire, is infinitely more elastic than air in the same circumstances
Phil. Trans., 41 (1) (1739), pp. 162-166

Experimental superheating of water and aqueous solutions
https://www.sciencedirect.com/science/article/pii/S0016703709000994

Maximum Superheating of Water as a Measure of Negative Pressure
https://aip.scitation.org/doi/abs/10.1063/1.1722122


How does one tell (or conclude) if a substance is carcinogenic? It's common to read statements like:

  Tetrachloroethene ($\ce{CCl2=CCl2}$) was earlier used as a solvent for  dry cleaning, but since it contaminated ground water and is a suspected carcinogen, it has been replaced by ... .
  Various hydrocarbons harms plant tissues, shed leaves, flowers and twigs. Many of them are also carcinogenic.

How do we tell or suspect one compound to be carcinogenic? And are all carcinogens organic compounds?

How do we tell or suspect one compound to be carcinogenic?

As written in the comments to the question, this the result of large studies on the human population, correlating blood or urine levels of some substance to instances of cancer. Alternatively or concurrently, studies on animal models, such as mice, rats or even in-vitro studies may be carried out. As a result, regulatory bodies may include a substance in lists, which may preclude them from use in consumer products, manufacturing procedures etc.

Are all carcinogens organic compounds?

No, this does not have to be the case: plain arsenic and basically all inorganic cadmium compounds have been classified as suspected or definitively carcinogenic.


Why are the majority of cleaning solutions basic? When testing for pH for common household cleaning products, such as dishwasher fluid, I noticed that majority of them tend to be basic? Why is that? Do acid make good cleaning solutions as well?

Dishwashing agents are usually anionic detergents—i.e. they are alkylbenzenesulfonates.
They consist of two parts:

Sulfonate part which is hydrophilic(water loving i.e. dissolves in water)
Alkylbenzene part which is lipophilic(fat loving i.e dissolves in fats and oils)

Thus, when in water, they dissociate to give $\ce{RSO3-}$ and $\ce{Na+}$ .  $\ce{Na+}$ is from a strong base (making it a weak conjugate acid), while $\ce{RSO3-}$ is from a weak acid (making it a weak conjugate base), thus the resulting solution is basic.

The sulphonate ion bonds to water, while alkylbenzene bonds to oil, protein, or fat . Thus fats, oils, proteins, and grease dissolve in water

The following is the mechnism of a dishwashing agent:

These are the ingreadients mentioned on a dishwashing bottle in my house:

As you can see, sodium dodecylbenzenesulphonate (an alkylbenzenesulfonate) is one of the main ingredients.

  Do acid make good cleaning solutions as well?

Yes they can make a good cleaning solution, but not as a dishwashing agent.

According to Wikipedia ,

  Acidic washing agents are mainly used for removal of inorganic deposits like scaling. The active ingredients are normally strong mineral acids and chelants. Often, there are added surfactants and corrosion inhibitors.


Why trans-1-chloro-4-methylcyclohexane have a higher relative energy than trans-1-chloro-2-methylcyclohexane? Both can be A,a or e,e, in 1,4 the groups are further apart -> less steric strain, why is 1,4 less stable according to experimental values?
1,2 high energy conformer's relative energy (kj/mol) is 4.05 and
1,4 high energy conformer's relative energy (kj/mol) is 4.50

trans-1-Chloro-4-methylcyclohexane (1) has two chair conformations, 1-ee and 1-aa. Conformation 1-ee has two non-interacting functional groups, chlorine and methyl, both of which have anti-relationships with the $\ce{C2-C3 and C5-C6}$ bonds of the ring. Accordingly, the energy of this conformation is set at 0 kJ/mol. There is no need to try to set some absolute value to the energy because the only issue is how 1-ee compares with conformation 1-aa. Conformation 1-aa has both substituents axial in gauche-relationships with the aforementioned C-C bonds. This conformation behaves as a composite of axial methylcyclohexane and chlorocyclohexane. Its energy is merely the sum of their respective A-values, 9.7 kJ/mol.
Diaxial conformation 3-aa of trans-1-chloro-2-methylcyclohexane has the same energy value of 9.7 kJ/mol because both groups are axial and non-interacting with each other. However, the diequatorial conformation 3-ee has a gauche interaction between the chlorine and the methyl group. The magnitude of this interaction may be estimated.
The value of the $\ce{CH3/CH3}$ interaction in conformation 4-ee of trans-1,2-dimethylcyclohexane may be estimated by dividing the energy of 4-aa by 4 or by halving the A-value for a methyl group. The value of 3.8 kJ/mol is in accord with the difference in the heat of combustion of dimethylcyclohexanes 2 and 4. Indeed, the heat of combustion of 4 is greater than 2 by ~5 kJ/mol.  [https://nvlpubs.nist.gov/nistpubs/jres/39/jresv39n1p49_A1b.pdf
]. Similarly, division of the energy of 3-aa by 4 gives an estimated value for the $\ce{CH3/Cl}$ gauche interaction of 2.4 kJ/mol. The heat of combustion of 1,2-isomer 3 should be slightly greater than that of 1,4-isomer 1 by ~2.4 kJ/mol. In the event, trans-1-chloro-4-methylcyclohexane is more stable than trans-1-chloro-2-methylcyclohexane by ~2.4 kJ/mol. By more stable is meant, more negative heat of formation and smaller heat of combustion.


Does oxygenating tap water remove the tap water flavor? I know flavor is subjective and not exactly quantifiable, but I think there's merit to this question.

We've all experienced that tap water often tastes a lot worse than bottled water or filtered water. I've always felt, however, that cold water actually tastes much purer than warmer water.

For instance, I was recently in a house where I knew the water was so called "hard water" and I could taste that it had the typical mineral flavor straight from the tap.

So, I took a glass of this tap water and refrigerated it for about a half hour and then drank it and I couldn't taste the typical tap water flavor at all.

I'd encourage people to try this just to make sure I'm not crazy.

Additionally, I have family that lives in Ohio and if anyone has ever tasted Ohio's tap water after living in a place like Washington (where I live) which has extremely fresh water, you know that Ohio's water is disgusting. I have found, however, that it tastes much better after being refrigerated for a while.

So, there are a couple of questions here.

Is it even true that refrigerating water (and thus cooling it) would cause a significant amount of oxygen to be dissolved in the water?

If so, would the amount of time the water is being refrigerated (even at a constant temperature) continue to dissolve more oxygen into the water? In other words, how long might it take for oxygen to reach it's equilibrium level of dissolution?

Those last two are definitely answerable chemistry questions. This next one is a little hairier.

Is there any reason, related to chemistry, that oxygen rich water would taste better than water not saturated with oxygen? Meaning, does the oxygen actually have a well-defined effect on the flavor of water, or is this simply answered by the fact that most people like the feeling of cold water more than luke-warm water and hence I identify cold water as "tasting better"?

Many public water systems spray water into the air or trickle over rocks (aeration) to let some dissolved gases bubble out (e.g. odorous $\ce{H2S}$). This also may make the water less acid and thereby precipitate dissolved manganese and iron.

When tap water treated with chlorine, bromine, ozone or other disinfectant is allowed to stand, open, then those chemicals also can escape.

However, I doubt anyone actually tastes dissolved oxygen (the human body can sense $\ce{CO2}$, but not oxygen, which is why miners kept canaries. In addition, one can taste the acid of $\ce{CO2}$ in soda water, and feel the bubbles bursting.)

So the practice of letting water stand and refrigerating is useful in improving flavor, but $\ce{O2}$ has little to do with it. Chilling water does increase the solubility of gases, though.


Assumption behind Gibbs energy and maximum work I'm somewhat confused by this derivation (in Schroeder's Book of Thermal Physics) of the fact that in a closed system, the change in the Gibbs Energy is always less than the Non-Expansion work done on the system.

If we take the change in the entropy of the universe (system + ideal surroundings maintained at constant temperature and pressure also assuming that reactions are taking place in system only i.e. the composition of various species present in surroundings remain constant), then it turns out (in this derivation) that it equals to negative of change in Gibbs Energy divided by Temp of system (which is same as that of surroundings).

Now since entropy of Universe always increases it turns out that Gibbs Energy of system will always decrease. But this is only true in absence of Non-Expansion work.

However during this derivation nowhere it's mentioned that 'no non expansion work assumed' yet the result only holds true in absence of non expansion work.

I think that maybe it was assumed somewhere and I'm not able to spot where. Please help me to clear my doubt.

The whole point of Gibbs free energy is to quantify a bound on possible non-expansion work at constant p and T. That non-expansion work is implicit as part of the term "dU" which in the reversible case can be written as

$$dU = -pdV + TdS + dw_{\text{non-exp,rev}}$$

The $-pdV$ term is the expansion work if the process is carried out reversibly. You remove that contribution by adding a  $pdV$  term:

$$dU + pdV= -pdV + TdS + dw_{\text{non-exp,rev}} + pdV = TdS + dw_{\text{non-exp,rev}}$$

Finally if you subtract the entropy term you are left with the maximum possible non-expansion work (obtained when the process is performed in a reversible fashion):

$$dU + pdV - TdS= dw_{\text{non-exp,rev}}$$


Why is ozone O₃ diamagnetic? Please explain how to derive the magnetism  of polyatomic molecules? If we derive it using an addition of an oxygen atom to molecular oxygen ($\ce{O2}$) it will come to be paramagnetic like $\ce{O3}$?

Ozone's ground state is a singlet.  All electrons therein are paired,

However,

http://link.springer.com/article/10.1134%2FS0018143906040060#page-1
http://link.springer.com/article/10.1134%2F1.1870062#page-1


Why are hydrogen-bonded compounds (such as NH3) excellent solvents for most lab reagents? Hydrogen bonds are said to be one of the strongest molecular interactions in nature. Then why can most substances that interact through hydrogen bondig such as water and $\ce{NH3}$ able to dissolve a large variety of both ionic and covalent laboratory reagents? Does it have something to do with the structure of $\ce{NH3}$ or something else?

Recall that hydrogen bonding interactions result from the polarized nature of bonds between hydrogen and nitrogen, oxygen, and fluorine. This polarized bond results in a hydrogen atom with very low electron density—a property that allows the atom to interact very strongly with negatively charged particles.

Thus, anions and electronegative atoms in polarized molecules have very strong electrostatic interactions with the low-electron-density hydrogen atom in $\ce{NH3}$ or $\ce{H2O}$, for example. Cations and positively charged elements in polarized molecules can also be stabilized by the electronegative atoms in molecules exhibiting hydrogen bonding ($\ce{N}$ and $\ce{O}$ in this case).

These strong interparticular interactions ultimately result in polarized or ionic solutes generally being very soluble in solvents with hydrogen bonding such as $\ce{NH3}$ or $\ce{H2O}$.


What is the electrophile in Perkin's condensation reaction? In Perkin's reaction, an aromatic aldehyde reacts with an acid anhydride to give an alpha-beta unsaturated aromatic acid. Which reactant is the electrophile and which one is the nucleophile?

The reaction mechanism shown on the Wikipedia page for the Perkin Reaction is perfectly clear.

The first step is deprotonation of acetic anhydride by acetate anion.
The second step is nucleophilic attack by the anion of acetic anhydride on the aromatic aldehyde; thus the aldehyde is the electrophile.


What is the size of the rate constant dependent on? In the rate equation, consider k:

I know that $k$ is fixed unless temperature changes, but what else besides temperature is $k$ dependent on?

The rate constant $k$ includes all the factors other than the dependence on concentration.  A significant factor in the rate is collision frequency and energy, so temperature has a significant influence on the value of $k$.  Other factors like the range of angles of collision necessary for a successful reaction, the strength of dipoles, solvent effects, quantum effects, and whatever factor corrects the units of the equation are not determined.

However the rate constant can be expressed by the Arrhenius equation
$$k=Ae^{-{E_a/RT}}$$
Doing this decomposes the constant into an exponential relationship including the activation energy, $E_a$, the molar gas constant, $R$, and of course temperature, $T$, which represents the fraction of total molecules that have sufficient energy to overcome the activation energy.
This equation comes with it's own new constant, the pre-exponential factor, $A$, which now includes all the factors in the rate constant that aren't the exponential relationship with $E_a$, $R$, or $T$. There may still also have some temperature dependence hidden within $A$, but usually it can be assumed constant over small temperature ranges.  Factors like collision frequency and angle, or action of a catalyst, are still obscured within this constant.

Even if these obscured effects were determined there would still be some new constant, $C$, which would include subtle effects such as interactions with surfaces in a container or non-reacting particles, temperature and concentration gradients, ionic strength, entropy changes in the solution due to changing composition, and any other factor that influences the reaction between the particles that hasn't been explicitly considered.


Why does a reaction stop when turning into a solid? I recently was looking at a question about the volume of carbon dioxide produced when reacting calcium carbonate with dilute sulfuric acid versus using nitric acid. In the answer it says that calcium sulfate is produced, which is a solid, which stops the reaction, but I don't get why. Calcium nitrate produces an aqueous solution which causes the reaction to carry on reacting.
The reaction equations are:
$$\ce{CaCO3(s) + H2SO4(aq) -> CaSO4(s) + H2O(l) + CO2(g)}$$
$$\ce{CaCO3(s) + 2HNO3(aq) -> Ca(NO3)2(aq) + H2O(l) + CO2(g)}$$

The reason is due to the surface area of calcium carbonate available to react.

When reacting a block of solid calcium carbonate with dilute sulphuric acid, solid calcium sulphate is produced which is deposited on the surface of the calcium carbonate block (since that is where the reaction occurs). The calcium sulphate is insoluble in dilute sulphuric acid and so you end up with calcium carbonate coated in a layer of calcium sulphate, which stops the acid from getting at the carbonate and so stops the reaction.

In contrast, calcium nitrate, produced from the reaction with nitric acid, is highly soluble in dilute nitric acid and so no solid layer forms around the carbonate block and the reaction can continue until one of the reagents is used up.


What kind of unit is this [A*m^2/mol]? I found a term on this website that gives a unit of [A*m^2/mol]. The equation is:

$k_d^{eff}=\frac{2RT\kappa^{eff}}{F}(t^+-1)\left(1+\frac{\text{d}\ln(f)}{\text{d}\ln(c)}\right)$

where R is the gas constant [J/mol/K], K is ionic conductivity [S/m], T is the temperature [K], t+ is unitless, f is unitless and c is the concentration [mol/m^3], F is the Faraday constant [C/mol].

This equation gives the unit $\frac{A \cdot m^2}{mol}$. What kind of unit is this one? I can see that they define it as electrolyte diffusional conductivity, but I can't find this unit and better definition anywhere in the literature.

Can anyone help me with this?

First thing to say, I don't know much about electrochemistry. But concluding from thermodynamics, rewriting your equation a bit and making dimensional analysis there are some things to say:

It seems that you interchange the words dimension and unit. So an example: Length is a dimension and meter, inches... are units that measure length. You are probably asking for the physical interpretation of the electrolyte diffusional conductivity and not for the specific unit it is measured in.
The logarithm can be applied only to dimensionless numbers. In order to calculate it, you have to divide by some $c_0$ usually this is given in mol/l. In addition both terms in brackets have to be dimensionless, because there is a sum with a dimensionless number and in sums the dimensions have to match. So the only term relevant for the dimension, is the fraction.
The resulting dimension of the fraction is in SI units [A / m] and not [A m$^2$ / mol]


Which diagram shows the effect of catalysis on chemical equilibrium? My course book has the diagram in figure (1) to represent the effect of a catalyst on chemical equilibrium. I drew the second diagram in figure (2).

When reviewing the diagram in figure (1), I observe the following mistakes:

The rate of catalyzed and uncatalyzed forward reaction  at equilibrium will be the same, which seems to be wrong because the value of rate constant of catalyzed forward reaction ($k_f$ catalyzed) more than the value of rate constant of uncatalyzed forward reaction ($k_f$ uncatalyzed), so the rate of catalyzed forward reaction at equilibrium = ($k_f$ catalyzed) $[A][B] > $ ($k_f$ uncatalyzed) $[A][B]$, so  it isn't the same as the diagram.
The rate of catalyzed forward reaction before equilibrium less than the rate of uncatalyzed forward reaction before equilibrium, which seems to be wrong too.

I need feed back about the diagram or both are wrong and about my reasoning.

If (as suggested in the comments) the y-axis is supposed to represent concentration, such that the upper curve is "concentration of $\mathrm{A+B}$" and the lower curve is "concentration of $\mathrm{C+D}$", then Figure 1 is correct, albeit horribly mislabeled.

Your Figure 2 is correct if we want the y-axis to represent a rate-of-conversion instead, i.e., if the upper curve is "number of parcels of $\mathrm{A+B}$ being converted to $\mathrm{C+D}$ per second", and the lower curve is vice-versa. As you point out, a catalyst increases the conversion rate, and does so in both directions. So, exactly as you have drawn in Figure 2, those catalyzed rate curves should both be higher than their uncatalyzed siblings.

("Rate of reaction" is an awfully ambiguous term here – since at equilibrium, the rate of reaction must be zero!)


How to analyze a acid and salt mixture using titrimetry? Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, $0.1\:\mathrm{M}$ $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.

\begin{align}
\ce{ M(OH)3 &~<=> M(OH)^{+}_{2} + OH^{-} } &
\mathrm{p}K_{\mathrm{b}1} &= 0.5 \\
\ce{ M(OH)^{+}_{2} &~<=> M(OH)^{2+} + OH^{-} } &
 \mathrm{p}K_{\mathrm{b}2} &= 0.7 \\
\ce{ M(OH)^{2+} &~<=> M^{3+} + OH^{-} } &
 \mathrm{p}K_{\mathrm{b}3} &= 9.0
\end{align}

All of the species in above equilibria are water soluble.

Progress so far:

$\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

\begin{align}
\ce{ M^{3+} + {H2O} &~<=> M(OH)^{2+} + H+ } &
\mathrm{p}K_{\mathrm{a}1} &= 14.0 - 9.0 = 5.0 \tag{a} \\
\ce{ M(OH)^{2+} + {H2O} &~<=> M(OH)^{+}_{2} + H+ } &
\mathrm{p}K_{\mathrm{a}2} &= 14.0 - 0.7 = 13.3 \tag{b}\\
\ce{ M(OH)^{+}_{2} + {H2O} &~<=> M(OH)_3 + H+ } &
\mathrm{p}K_{\mathrm{a}3} &= 14.0 - 0.5 = 13.5 \tag{c}
\end{align}

Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
Looking at the $\mathrm{p}K_\mathrm{a}$ values of $\mathrm{(b)}$ and $\mathrm{(c)}$ it can be assumed that reaction $\mathrm{(a)}$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.

Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

Prologue

The following will show you not only how to sketch a titration curve but how to produce an analytical form of a titration curve. So this might not be an easy solution even if I try to keep it as easy as possible. There is a wonderful yet german book that has (to my knowledge) never been translated to english which describes everything of the following in detail:

Bliefert, Claus; Linek, Alfred; Morawietz, Gerd (1978): PH-Wert-Berechnungen. Weinheim: Verlag Chemie.

There might also be some imprecise terms because I don't know many of the english counterparts to the german terms I am familiar with.

Answer

Let's think a bit about your problem first to simplify it from scratch:

As you mentioned, reaction (a) will take place right after all protons from hydrochloric acid have reacted and before reaction (b) and (c) would start.
The pK$_a$ of (b) and (c) are so close to another that they are unable to be separeted by titration and
the pK$_a$ of (b) and (c) are so high that they are unable to be determined by titration in diluted aqueous solutions.

If we take a look at the titration curve of phosphoric acid, we can see what is meant by "too high to be determined". You can clearly identify the two inflection points for the first two protons but there is no third inflection point that could be determined. What is likewise to be seen, is that one can neglect this third deprotonation step in titration curves of diluted solutions (yellow vs. blue titration curve).

Right now we can reduce your problem to a mixture of hydrochloric acid and M$^{3+}$. To continue we need to establish a proton condition (Protonen-Bedingung):

$$c(H_3O^+) = c(OH^-) + c(A^-) + c(X^-) - c(HB^+) \tag{1}$$

It means that protons come from the autoprotolysis of water (c(OH$^-$)), from hydrochloric acid (c(A$^-$)), from your metal ion (c(X$^-$)) and that the protons react with the used base (c(HB$^+$)).

To define an equation for the acid residue anions a small equation system (where small c's denote actual concentrations in solution and the big C denotes the origin concentration)
$$\begin{align}c(HA) \cdot k_a - c(H_3O^+) \cdot c(A^-) =&~0\\c(HA)+c(A^-) =&~C(HA)\end{align} \tag{2.1}$$
needs to be solved to give
$$c(A^-) = \frac{C(HA) \cdot k_a}{c(H_3O^+)+k_a}. \tag{2.2}$$

This can be done analogously for the other acid and also for the base.

What remains is the concentration of the hydroxide ions from the autoprotolysis of water that is simply determined by rearranging the equation for the autoprotolysis itself:

$$c(H_3O^+) \cdot c(OH^-) = k_W \Leftrightarrow c(OH^-) = \frac{k_W}{c(H_3O^+)} \tag{3}$$

Combining everything so far we get the following proton condition:
$$x = \frac{k_W}{x} + \frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}} - \frac{x \cdot C(B) \cdot k_b}{x \cdot k_b + k_W} \tag{4}$$

where $c(H_3O^+)$ has been replaced by $x$ for the sake of simplicity.

The plot of titration curves usually shows the pH over the base volume (see eq. 8) or the degree of neutralisation ($\tau$ or $\gamma$) that is defined as $$\tau = \frac{c(B)}{C(HA)} \tag{5.1}$$

In the case of your mixture of two acids - and as always neglecting the increase in volume - it can be written as
$$\tau = \frac{c(B)}{C(HA)+C(HX)} \tag{5.2}$$

To sketch the titration curve let's say that we know the two concentrations of the acids. To calculate (5.2) we also need to know c(B) that can easily be obtained by rearranging (4) to get:
$$c(B) = \left(1+\frac{k_w}{x \cdot k_b}\right)\left(\frac{k_w}{x}-x+\frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}}\right) \tag{6}$$

If one decides to include the increasing volume the concentrations of the acids and the base have to be extended to give
$$\begin{align}C(HA) \rightarrow&~\frac{C(HA) \cdot V(HA)}{V(HA)+V(B)}\\C(B) \rightarrow&~\frac{C(B) \cdot V(B)}{V(HA)+V(B)}\end{align} \tag{7}$$

Inserting (7) in (4) and rearranging for V(B) we finally get a neat equation

$$V(B)=\frac{V(HA)~\left(\frac{k_w}{x}-x+\frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}}\right)}{\left(x-\frac{k_w}{x}+\frac{C(B) \cdot k_b}{\frac{k_w}{x}+k_b}\right)} \tag{8}$$

As the pH is defined as $pH = -log_{10}(c(H_3O^+))$ and $c(H_3O^+)=x$ you can replace every $x$ in equation 8 with $10^{-x}$.

Now you are able to calculate with e.g. Excel enough (V(B),pH)-pairs to draw a titration curve and get some information about your system of two acids during the titration with your desired base.

The following figure shows some examples of your mixture using equation 8.

Please let me know if someone is interested in the full equation including also the second and third deprotonation step with pK$_a$ values 13.3 and 13.5 and I will add it here.


Molality of the solution prepared from a hydrated salt A solution is made up by dissolving $\pu{13.8 g}$ $\ce{Na2CO3 · 10 H2O}$ in $\pu{103.5 g}$ of water. What is the molality of $\ce{Na2CO3}$ in this solution?

I can't solve it. I tried converting to moles with $\pu{13.8 g mol}/\pu{286 g} = \pu{0.04825 mol}$. Then I did $\pu{0.04825 mol}/\pu{0.1035 kg} = 0.466$ but it didn't work.

Let's denote a hydrated salt as $\ce{M_pA_q*xH2O}$ ($\ce{M}$ – metal, $\ce{A}$ – anion). Then molality $b$ of the aqueous solution of $\ce{M_pA_q}$ is defined as

$$b=\frac{n(\ce{M_pA_q})}{m_\mathrm{tot}(\ce{H2O})}\, ,\label{eq:1}\tag{1}$$

where $n(\ce{M_pA_q})$ – amount of water-free salt; $m_\mathrm{tot}(\ce{H2O})$ – total mass of water.

Note that the amounts of the hydrated and the water-free salt are equal:

$$n(\ce{M_pA_q})=n(\ce{M_pA_q*xH2O})=\frac{m(\ce{M_pA_q*xH2O})}{M(\ce{M_pA_q*xH2O})}\, ,\label{eq:2}\tag{2}$$

where $M$ – molar mass.

Total amount of water is found as the sum of water added to dissolve the salt and the water "stored" within the hydrated salt itself:

$$
\begin{align}
m_\mathrm{tot}(\ce{H2O}) &= m(\ce{H2O})+m_\mathrm{hydr}(\ce{H2O})\\
&= m(\ce{H2O})+n_\mathrm{hydr}(\ce{H2O})\times M(\ce{H2O})\\
&=m(\ce{H2O})+x\times n(\ce{M_pA_q*xH2O})\times M(\ce{H2O})\\
\eqref{eq:2}\quad\Rightarrow\quad&=m(\ce{H2O})+x\times m(\ce{M_pA_q*xH2O})\times \frac{M(\ce{H2O})}{M(\ce{M_pA_q*xH2O})}\label{eq:3}\tag{3}
\end{align}
$$

Now, putting \eqref{eq:2} and \eqref{eq:3} back in \eqref{eq:1}, one can obtain uniform expression for molality of a hydrated salt:

$$b=\frac{m(\ce{M_pA_q*xH2O})}{m(\ce{H2O})\times M(\ce{M_pA_q*xH2O})+x\times m(\ce{M_pA_q*xH2O})\times M(\ce{H2O})}\label{eq:4}\tag{4}$$

   Plugging in the numbers for sodium carbonate decahydrate:
 $$b=\frac{\pu{13.8 g}}{\pu{103.5 g}\times\pu{286.14 g mol-1}+10\times\pu{13.8 g}\times\pu{18.02 g mol-1}}=\pu{4.3E-4mol g-1}$$

A side note regarding the notations and units [1, p. 28]:

  The term molal and the symbol $\pu{m}$ should no longer be used because they are obsolete. One should use instead the term molality of solute $\ce{B}$ and the unit $\pu{mol/kg}$ or an appropriate decimal multiple or submultiple of this unit. (A solution having, for example, a molality of $\pu{1 mol/kg}$ was often called a $1$ molal solution, written $\pu{1 m}$ solution.)

Reference

Thompson, A.; Taylor, B. N. Guide for the Use of the International System of Units (SI). NIST Special Publication 811, 2008.


How does dimethylsulfoxide serve as a cryoprotectant? In cell culture, dimethylsulfoxide (DMSO) is added to prevent the formation of ice crystals which may lyse the cells. Exactly how does dimethylsulfoxide act as a cryoprotectant?

This question has a simple, but non-obvious answer. DMSO and water are miscible. Mixtures, and in particular homogeneous ones like that formed by water and DMSO, usually have lower melting (or more relevant in this case, freezing) points than either substance does alone. This is the reason salt melts ice: the salt solution has a lower freezing point than pure water. The physics behind this melting point depression are complex, but a simple enough approximation involves disruption of the intermolecular forces holding the solid crystal lattice together.

Water freezes at 0 °C, and DMSO freezes at 19 °C. Mixtures of the two have lower freezing points. More importantly, mixtures of the two likely form less disruptive crystals when frozen. Water is well know for having a solid state that is less dense than its liquid state. Ice crystals take up more room than the liquid water they formed from. Single water crystals can grow quite large, especially as smaller crystals merge. Additionally, water crystals tend to form this hexagonal fractal pattern (snowflakes) which have "pointy bits". DMSO, even at the 10% by weight customarily used, must prevent such large disruptive crystals from forming, especially when the everything is cooled to -80 °C.

DMSO is used for this purpose instead of other water-miscible liquids such as ethanol because it has very low toxicity.


Reaction of Iron(III) with hydroxide ion The observations:

1) Treatment of 1 mL of 0.05 M $\ce{Fe(NH4)SO4}$ solution with excess 1 M $\ce{NaOH}$ results in the formation of a rust-colored red-brown iron-based solution with many suspended particles in the solution. These particles match the description of $\ce{Fe(OH2)3(OH)3}$. This is the vial on the right in the picture.

2) Treatment of 1 mL of 0.05 M $\ce{Fe(NH4)SO4}$ solution with excess 6 M $\ce{NaOH}$ results in the formation of a clear but colored solution with no suspended particles in the solution or any sort of precipitate. The yellow-brown colored solution matches the description $\ce{Fe(OH2)4(OH)2+}$. This is the vial on the left.

Why is it that treatment with more concentrated base results only two apparent deprotonations rather than three?

More observations (after about 10 minutes):

1) The 1 mL of 0.05 M $\ce{Fe(NH4)SO4}$ solution treated with excess 6 M $\ce{NaOH}$ has developed suspended particles. This is the vial on the left.

2) The 1 mL of 0.05 M $\ce{Fe(NH4)SO4}$ solution treated with excess 1 M $\ce{NaOH}$ has started to become clear (the suspended particles are disappearing). It seems like much of the particles have settled on the bottom. This is the vial on the right.

It seems that the 6 M $\ce{NaOH}$ is eventually "getting the job done" by going through with the third deprotonation of $\ce{Fe(OH2)6^{3+}}$. I don't think that the 1 M $\ce{NaOH}$ is deprotonating $\ce{Fe(OH2)3(OH)3}$.

Still, why is it that treatment of iron(III) ions with concentrated base does not result in the formation of $\ce{Fe(OH2)3(OH)3}$ immediately but treatment with less concentrated base does?

Even more observations:

1) Treatment of 10 mL of 0.05 M $\ce{Fe(NH4)SO4}$ with excess 6 M NaOH results in immediate and obvious $\ce{Fe(OH2)3(OH)3}$ formation. This seems to support the nucleation explanation; the higher solution volume means lower concentration and lower supersaturation and therefore fewer and larger crystals of $\ce{Fe(OH2)3(OH)3}$ being formed.

Possible explanations I came up with:

1) I ran the results past a grad student and he said his best guess is that the excess $\ce{HO-}$ present stabilizes $\ce{Fe(OH2)4(OH)2+}$. He suggested a conductivity test to make sure that there are $\ce{Fe(OH2)4(OH)2+}$ ions in the clear solution. However, I did not have a conductivity meter handy.

2) A problem in the lab manual from an unrelated section notes that strongly electrolytic solutions, such as saturated $\ce{H4NCl}$ solution, can cause dispersion rather than dissolution of $\ce{Al(OH2)3(OH)3}$ in solution. A 6 M $\ce{NaOH}$ solution is strongly electrolytic, and as a result, the ions may be keeping the $\ce{Fe(OH2)3(OH)3}$ from clumping together.

What are your thoughts?

I wouldn't expect the reactions to yield different products but rather different particle size. The higher concentrated $\ce{NaOH}$ should hereby produce smaller particles†.

The "solution" in the left vial is probably rather a colloid of microscopically dispersed iron(III) hydroxide. It should show the Tyndall effect if you send a beam of light through it. However, the particles will aggregate over time to form larger particles that also precipitate, as seen in the second image.

† Higher concentration means higher supersaturation. Higher supersaturation leads to a higher nucleation rate, so more and thus smaller crystals are being formed.


Is there an online database of acid and basic dissociation constants? Is there an online database of acid and basic dissociation constants, all in one place, free to use?

There are a number of resources available that provide dissociation constants for acids and bases.  The following are examples:

UC Davis Chemwiki pages E1: Acid Dissociation Constants at 25°C and E2. Base Dissociation Constants at 25°C provide many examples.  These are based on information taken from CRC Handbook of Chemistry and Physics, 85th Edition.

The WikiBook Chemical Principles/Solution Equilibria: Acids and Bases provides a reasonable database of dissociation constants of acids and bases (Tables 5.3 and 5.4), as well as some further information about how the values are calculated.

The California State University webpage p-functions for common acids and bases, also at 25C, of a series of acids and bases.


How to identify whether a molecule is planar While solving for plane of symmetry of molecules I came across 1,3-dichlorocyclobutane (1) and I realized it does not have any plane of symmetry because we don't know whether hydrogen and chlorine are in same plane. How to judge which plane do the substituents belong to?
Similarly for compound 2, but in compound 3 it is known that both the planes of the rings are perpendicular. Is there some kind of rule for it?
<img src="https://i.stack.imgur.com/4iFNg.png" alt="1: 1,3-dichlorocyclobutane; 2: 2‐[(2E)‐1,3‐oxathietan‐2‐ylidene]‐1,3‐oxathietane; 3: 2‐[(P)‐1,3‐oxathietan‐2‐ylidenemethylidene]‐1,3‐
oxathietane" />

The concept of a plane of symmetry is easier to understand with molecules which are planar by itself.  Benzene, for example, has such a plane which may accommodate simultaneously all carbon and hydrogen atoms:

However, planarity of molecules is not a requirement to identify such planes.
Even if a construction of the cyclobutane molecule with model kit illustrates you the non-planarity of the molecule

(source)
you drawing depicts it like it were planar; because you defined drawing plane where carbon atoms alternate to be either closely above, or below this level in common.  For compound 1 illustrating 1,3-dichloro cyclobutane, however, the symmetry analysis is much affected by the configuration of the two special carbon atoms.  Either you find the two chlorine atoms on the same side of this plane (syn)

or you find them directing to different sides (anti):

The interactive training set by Symmetry@Otterbein may be a helpful resource for you.  Note that the representation of compound 2 is a simplification, and the symmetry attributed strongly depends on the underlying geometry.


Can H+ ion exist independently? Can H+ ion exist independently since it does need to form any bonds.Why do we say that it needs to be in H30+ ion form?

Be aware that $\ce{H+}$ is nothing else then a proton $^1_1\mathrm{p}$. Protons have an extreme charge density, typically $10^{15}$ higher than  any other ion ( but bare electrons or atomic nuclei).
Protons can exist free in vacuum or plasma, but even in gases they quickly react with many gaseous molecules, forming molecular ions, like dioxidanylium $\ce{O2 + p -> HO2+}$.
Protons act in aquaeous environment as extremely strong Lewis acid with an electron pair of $\ce{H2O}$(Lewis base), forming a ion called hydronium, a particular type of oxonium ions.:
$$\ce{p + H2O -> H3O+(aq)}$$
For convenience, we often use notation $\ce{H+}$ or $\ce{H+(aq)}$, supposing it is at least $\ce{H3O+(aq)}$.
Experiments mentioned in the above Wikipedia link confirm hydronium is usually coordinated with 6 water molecules as $\ce{[(H3O)(H2O)6]+}$, but multiple various hydration structures have been observed or assumed, like
Zundel ion $\ce{H5O2+}$,
$\ce{H9O4+}$($\ce{H3O+ . 3H2O}$)
or a cluster $\ce{H3O+ . 20 H2O}$
Note that the hydronium ion forms solid salts with acids with dissociation constant $K_\mathrm{a} \gt 10^{9}$.
E.g. a crystallic monohydrate of perchloric acid is hydronium perchlorate $\ce{[H3O+][ClO4-]}$


What is a sigma complex? According to Wikipedia, "sigma complex" refers to an arenium ion.

The suffix "-ium" denotes a positively charged ion.

So, it is correct to call the resonance structures in the below picture depictions of a sigma complex? It's distinctly anionic rather than cationic.

Yes, it's anionic sigma complex, more precisely Meisenheimer complex. They are much more stable then arenium ions (Wheland intermediates) and can be isolated much more easily.

There was actually synthesized a sigma complex which is zwitterionic combination of both Meisenheimer and Wheland complexes:

(picture created by Wikipedia User:V8rik)

The term "sigma complex" has also other meanings. Organometallic compounds with sigma bonds connecting with aromatic rings are called like that, too. There are also sigma complexes of cyclobutadiene with aluminium chloride.

Generally, molecules are called sigma complexes to stress the difference between them and pi complexes which can also be created from cyclic annulenes. First ones are bonded using sigma bonds, the other ones use pre-existing pi orbitals.


Is an SN1 mechanism feasible with allylic or benzylic halides as substrates? My teacher stated that allylic carbocations are comparable in stability to tertiary carbocations. $\def\SN#1{\mathrm{S_N}#1}$

With this in mind I am confused why question 13 of this MIT practice test states that these two molecules are only expected to undergo $\SN 2$ reactions:

An $\SN 1$ reaction on the first molecule with the phenyl substituent would yield a primary carbocation, yes, but this primary carbocation is also benzylic, and can be stabilized through resonance.

An $\SN 1$ on the second molecule similarly yields a carbocation, but this is an allylic carbocation, and can easily be stabilized through resonance and delocalization of pi-electrons.

So, would $\SN 1$ be valid pathways of reactions for both of these molecules, or is $\SN 2$ simply the only pathway for these molecules to react?

I planned to add this as a comment but I don't have enough reputation ... A definite answer can only come by referring to the literature and/or from the lab. In the context of an undergraduate exam it is reasonable to accept both possible answers ($\mathrm{S_N1}$ and $\mathrm{S_N2}$ ) for allylic and benzylic halides. $\ce{EtOH}$ solvolysis of benzyl bromide is a known reaction and the hypothesis that it goes via $\mathrm{S_N1}$ is the most reasonable one. Is it possible that the solvolysis acually goes via $\mathrm{S_N2}$ instead? The answer lies on performing the appropriate mechanistic studies.

However, in this case, having checked the actual exam question, it specifies acetone as the solvent. This polar aprotic solvent favours $\mathrm{S_N2}$ conditions, which is aprotic (see this question also). Incidentally, aside from the use of $\ce{KI}$ over $\ce{NaI}$, this halide exchange is basically the Finkelstein reaction, which occurs via $\mathrm{S_N2}$. (Credit to Greg E. who wrote this in a now deleted comment.)


Converting 1,4-benzenedimethanol to benzene and methanol? I have the molecule 1,4-benzenedimethanol:

How could I produce two methanol molecules and one benzene molecule? (Is there a catalytic reaction method available for this?)
I'm guessing that hydrogen ions could be used to split off the methanol molecules on both sides?

Oxidize to the dialdehyde (Jones reagent), decarbonylate with rhodium (do it at the bench before building the pilot plant), reduce the CO to methanol.  Why would you want to do that?


Ground state elevation for phase transition I have a question about the illustration from chem.libretexts.org:

the ground states of the liquid and gas are offset from that of the previous state by the heats of fusion and vaporization, respectively.

Why does the change in enthalpy increase the ground state energy?
I would be very grateful if someone can help me with this problem. I have been struggling with it for five days, but Google could not help me.
P.S. My question mark does not work.

In order to make the phase transition, you need to supply your sample molecules with an additional amount of kinetic energy (i.e., the heat of fusion or vaporization) to overcome their intermolecular bonding. This however means that after the transition, your system contains more internal energy than before. That additional energy is encoded in the vertical offset on the energy scale.

Note that if you look at each phase on its own, you could set the their ground states as the zero point on your energy scale. However, if you want to compare the system among different phases, then you need to take that offset into account. Otherwise you'd assume that your system can change from solid to liquid to gas without any effort.


The similarities and differences between electron transfer and charge transfer Please explain the similarities and differences among electron transfer,electron transport, charge transfer, and charge transport from the molecular, atomic and electronic levels.

Is electron transfer different from electron transport?

Is charge transfer different from charge transport?

I'm not sure what you are asking but electron 'transfer' is different to 'transport' in the sense that transport ultimately involves many electron transfers events one after the other. Charge transport is the same as electron transport.

There is is difference between electron transfer and charge transfer. Charge transfer is usually associated with charge transfer complexes, such as TTF-TCNQ (tetrathiofulvalene-tetracyanoquinodimethane) but there are very many others. In these complexes the charge transfer from one species to the other occurs as the photon is absorbed so that the lowest excited state is $\ce{D + A + photon -> D+A-}$. This often means that these complexes are highly coloured.

In electron transfer no photon is needed and the process follows Marcus theory whereby a barrier is crossed between $\ce{D + A}$ and $\ce{D+ + A-}$, i.e. this is a thermal process with an activation barrier. The unusual feature of Marcus transfer is the that activation energy has a parabolic dependence on energy thus the rate constant can be slow in both endothermic and very exothermic conditions.


Shape of a wavefunction I am trying to plotting the angular parts of a wavefunction ($Y_{(m,l)}(\theta, \phi)$) with Wolfram Mathematica. this is the table of spherical harmonics.

I think the shape of these function should looks round each part of p and d orbitals is a ellipsoid looks like an egg, like this figure:

I have 2 question with my result:

1. All my figure of p orbitals only have half part the orbital above(examples:
$p_0$ orbital and $p_y$ orbital). And all the d orbitals looks longer than that above (example: $d_{xz}$).

2. When I square the spherical haromics ($[Y_{(m,l)}(\theta, \phi)]^2$), The p orbitals looks the same as the top figure (example: $(p_y)^2$) ,but the p orbitals getting far longer and thinner than that before squared, also the figure in the top (example:$(d_{xz})^2$)

What's wrong with my plot or the shape is not looks like eggs? Do I miss something that make the figure only have half part?

You are plotting different functions. The figures have no reason to be more than vaguely reminiscent of each other.

Now, those nice puffy things above are the isosurfaces of $\psi$ function. This is a function of three variables, and it just so happens that it can be represented as a product:
$$\psi(r,\varphi,\vartheta)=R(r)\cdot Y_{\ell,m}(\varphi,\vartheta)$$
You fix a value, find all points in 3D where the function takes that value, and plot them. That's what an isosurface is.

(By now, you might have noticed that the radial part $R(r)$ never appears at all in your attempts, so they got to be producing something different. And so they do.)

OK, then what are your plots? These are the spherical graphs of the angular component alone. You stay at the origin, point at the direction $(\varphi,\vartheta)$ with a stick having length $Y_{\ell,m}(\varphi,\vartheta)$, and leave a dot there. You don't care at all about the value of $\psi$ at that location. Besides, when $Y_{\ell,m}$ is negative (which it is half of the times), you point in the opposite direction. This, BTW, is the reason for having only one lobe in your p-orbital-like plots. No wonder you've got both lobes after you've squared $Y_{\ell,m}$ and thus made it always positive.

It could have been worse, you know. Think of a cone with its tip at the origin. Is it an isosurface of any function? Of course it is. What function? For example, $\cos\vartheta$ would do. What is the spherical graph of $\cos\vartheta$? A sphere, as you already know. Does a sphere look like a cone? Er, well...


Gas solution - State of matter When a gas is dissolved in water does the state of gas changes ? We write it as aqueous but what is the state that exists ? does it has the properties of gases when it is dilluted ?

Good question! For a solid dissolved in liquid, obviously the solid behaves as a liquid rather than as a solid, since its particles are separated and moving freely.

A gas, though, is defined as free to fill its container, unlike a liquid with a definite volume. Since a dissolved gas (as opposed to one chemically bound to a liquid) is free to disperse, e.g. $\ce{CO2}$ bubbling out of seltzer, one might still consider it gaseous, rather than liquid, just loosely bound to the liquid.

That said, in a closed container in equilibrium, or over a short time, dissolved gases do behave as liquids. One can pour $\ce{NH4OH}$ ($\ce{NH3}$ aq.) into $\ce{H2SO4}$ ( $\ce{SO3}$ aq.), and the majority of the dissolved gases stay in solution -- though some $\ce{NH3}$ gets out, leaving one tearing and coughing, if not done under a hood.

It appears to be a matter of definition, then if dissolved gases are considered as gas or liquid.

BTW, gases dissolved in solids, such as methane clathrate, are more tightly locked in place and may behave as solids until warmed.


Why doesn't CaO decompose into Ca and O2? Many metal oxides decompose into the free metal and oxygen gas at high temperatures, but why doesn't $\ce{CaO}$ do that? What happens to $\ce{CaO}$ at high temperatures?

This is from problem 2 of the 2011 USNCO local exam:

  Oxygen gas can be produced by the decomposition of all
  of the following substances EXCEPT

  (A) calcium oxide. (B) hydrogen peroxide.
  (C) mercury(II) oxide. (D) ozone.

The correct answer to this question was (A).

Stability, the thermodynamic property, is defined as whether a decomposition reaction is exergonic or endergonic at a certain temperature. The reaction $\ce{CaO -> Ca + 1/2 O2}$ is endergonic across a very large temperature range. But that doesn’t necessarily help you.

In that list of four, we have to clearly molecular compounds and two mostly ionic ones. We need different discussions for both. Basically, the molecules are unstable towards decomposition into $\ce{O2}$ and something ($\ce{H2O}$/$\ce{O2}$) because they contain energetic bonds whose replacement by an $\ce{O=O}$ bond is favourable.

For the ionic ones, we have a number of different discussions we can do. For one, the metallic partner is calcium versus mercury; while the latter is known since ancient times and pretty stable in liquid form the former is not: It will readily react to form salts. For another, the $\ce{Ca^2+}$ ion is rather small and thus a hard Lewis acid according to Pearson’s concept. $\ce{Hg^2+}$ is much larger and thus a much softer acid. Therefore, the adduct of $\ce{Ca^2+}$ and $\ce{O^2-}$ is predicted to be more stable (hard–hard interaction) than that of $\ce{Hg^2+}$ with $\ce{O^2-}$. Finally, we can also consider heats of formation which should (I unfortunately don’t have values) be much more negative in the case of $\ce{CaO}$ than for $\ce{HgO}$ — therefore, the latter can be decomposed with enough heat due to $\Delta G = \Delta H - T \Delta S$ while the former cannot.


Does reduction/ oxidation take place separately at a half cell electrode? In the redox reaction chapter, we learn that oxidation and reduction go hand in hand.  The oxidising  agent oxidises by accepting the electron released by the reducing agent and gets reduced in the process.

However, in the half cell electrode (say, $\ce{Zn}$/$\ce{Zn^{2+}}$) only oxidation is  taking place and no simultaneous reduction is happening in the scene. Also, in a $\ce{Cu}$/$\ce{Cu^{2+}}$ half cell electrode, only reduction is taking place. Until  the half cells are joined by a  saltbridge, oxidation and reduction go on violating the basic  principle of  redox reactions.

In a (standard) Zn/Zn++ half cell, no oxidation (and no reduction) is taking place. A piece of zinc is sitting in a 1M solution of Zn++ ions (with their accompanying anions). It is not imaginary, and on a micro scale, an equilibrium is occurring (could be occurring!) where Zn + Zn++  --> Zn++ + Zn. No overall change occurs; this is a reference point. If it were not stable, some process would be taking place that degrades the half cell and it would be reacting itself away. In reality, half cells are only useful to the extent that they don't react themselves away until they are connected to the other half of the overall cell.

Same for Cu/Cu++. However, when the half cells are connected (thru a salt bridge so ions can flow, and a wire so electrons can flow), two processes occur: electrons leave the zinc (the zinc oxidizes) and electrons flow to the copper (reducing it). No "process" or reaction occurs in the half cells until they are connected.

The reality of oxidation and reduction require both to occur simultaneously - WHEN THEY OCCUR - but we can speak of each process independently of the other. When electrons leave an atom (oxidation), they have to go somewhere (reduction); if electrons are not flowing (no actual reaction) there is no oxidation or reduction. The half cell potentials then define places from which electrons will flow, or locations to which they will flow.


Ratio between heat tranfer coefficient and thermal conductivity Consider the Robin boundary condition for the diffusion/heat equation $u_t=a(t)u_{xx}+f(x,t)$:

$$-k(t)u_x(0,t)=h(t)u(0,t)$$

or

$$u_x(0,t)+\frac{h(t)}{k(t)}u(0,t)=0$$

where $k(t)$ thermal conductivity and $h(t)$ heat tranfer coefficient.

My Question: Is it possible that the ratio $h(t)/k(t)$ to be constant? Could anyone please help me? I have really no idea.

The heat transfer coefficient and the thermal conductivity can often be treated as constants in practical heat transfer problems.  However, the heat transfer coefficient can change if the exterior convective flow conditions change.


Why is sodium carbonate less soluble in water than sodium bicarbonate? Why is sodium carbonate less soluble in water than sodium bicarbonate? If you think about their structures, the only thing that is different is that sodium carbonate has two sodium atoms, while sodium bicarbonate has one sodium atom and one hydrogen atom instead of the second sodium atom. Sodium and hydrogen have different electronegativity values, giving sodium bicarbonate polar character.

Does the difference in solubility have to do with the dipole moment of the molecules? If so, how can it be explained in more detail?

Oops! The question contains an incorrect assumption. But chemists are amazing - they can still answer such a question:

Sodium carbonate is way more soluble than sodium bicarbonate at all temperatures above about freezing. $Na2CO3$ has several crystalline hydrates that are temperature sensitive and make the curve less smooth than most other salts. (I suppose the caffeine is in there to wake us up.)
So, I woke up to the fact that when we calculate the weight of "sodium carbonate" which dissolves at 0 C, it's the decahydrate, whereas sodium bicarbonate is anhydrous, so at 0 C, the solubility of sodium bicarbonate is 0.833 moles per liter, while the same weight of sodium carbonate decahydrate dissolved in a liter is only 0.245 moles.
Now I'm more puzzled by why sodium carbonate has such a complex hydrate system, and sodium bicarbonate doesn't.


Product operator analysis for CHn groups in HSQC and HMQC I'm taking and advanced NMR course this semester and we are learning about product operator formalism. I have a homework where I'm supposed to apply this formalism to an HSQC and HMQC experiment. The problem I have is that I should use a 3 and 4 spin systems $(\ce{CH2})$ and $(\ce{CH3})$ groups instead of just simply $(\ce{CH})$ and we only did two spin systems in class. I understand how to use the formalism and how to apply it to the experiment, but I need let's say a concept clarification. We use basic pulse sequences for HSQC and HMQC.

In the exercise, we assume that the hydrogens do not couple to each other, only to the carbon. I did the example a found out that the result (in the product operator formalism it was the same observable term which with I was left at the end) was the same for $\ce{CH3}$ group as for $\ce{CH}$ after optimization of delay time. After I thought about it, I feel it makes sense. Since the three hydrogens will do the same as one hydrogen, only I will have a different intensity afterward in the spectrum and of course different shifts. I even calculated that the optimal tau delay is the same for both $\ce{CH}$ and $\ce{CH3}$ groups.

And since HMQC basically gives the same info, I assume it will work the same, and results for $\ce{CH}$ and $\ce{CH2}$ group will also give me the same.

Is this correct? Is there any other difference in what happens during the experiment to the $\ce{CH}$ and $\ce{CH3}$ group (or $\ce{CH2})?$

Just to clarify, it is not a multiplicity editing experiment, just a simple HSQC/HMQC applied to $\ce{AX2}$ and $\ce{AX3}$ spin systems.

Your analysis is generally correct. In the case where all the protons are equivalent, you don't need to worry about proton–proton coupling. The main thing you need to worry about is that when you have a term such as $2I_{1z}S_x$ you will get evolution of different proton–carbon couplings, represented by (e.g.) the Hamiltonian $2\pi J_{IS} I_{2z}S_z$. This could, in theory, generate more complicated product operators such as $4I_{1z}I_{2z}S_y$. You probably already know that

$$S_x \xrightarrow{2\pi J_{IS} I_{2z}S_z \tau} (\cos \theta) \, S_x + (\sin \theta)\, 2I_{2z}S_y$$

where $\theta = \pi J_{IS}\tau$. A term on a different nucleus, such as $I_{1z}$, does not evolve under this coupling, so you can "multiply" by $2I_{1z}$ on both sides* to get

$$2I_{1z}S_x \xrightarrow{2\pi J_{IS} I_{2z}S_z \tau} (\cos \theta) \, 2I_{1z}S_y + (\sin \theta)\, 4I_{1z}I_{2z}S_y$$

Thankfully you do not need to worry about this, because the only times that these can evolve are during the $t_1$ periods of both experiments, when there is magnetisation present on carbon. Conveniently, both experiments have a proton 180° pulse in the middle of $t_1$ which refocuses all C–H couplings anyway. Outside of the $t_1$ periods the magnetisation of interest is single-quantum on proton, so the only coupling that evolves is that one proton–carbon coupling, which behaves in exactly the same way as for a CH system.

For a more detailed treatment I suggest consulting Chapter 12 of James Keeler's Understanding NMR Spectroscopy, 2nd ed. (2010).

However, if the protons are not equivalent, then you will start to get evolution of H–H homonuclear couplings, which can't simply be refocused by a 180° pulse. For the HSQC this is less important as it just leads to a reduction in intensity from magnetisation that goes down non-useful coherence pathways. Because the magnetisation during $t_1$ is single-quantum carbon, it does not evolve under homonuclear couplings. However, for the HMQC, the multiple-quantum coherences present during $t_1$ will evolve under homonuclear couplings, which means that you will get multiplet shapes in the indirect dimension after Fourier transformation. In practice it is difficult to resolve these splittings and so it is just manifested as line broadening.

There are also considerations to be made when using a sensitivity-enhanced HSQC sequence, but I assume those are not relevant for your current situation. The interested reader is directed to J. Biomol. NMR 1994, 4, 301–306.

* To be more precise, this is because $I_{1z}$ commutes with the Hamiltonian. So, it also commutes with the unitary propagator $U = \exp(-\mathrm i H\tau)$, and we can write

$$2I_{1z}S_x \xrightarrow{H\tau} U(2I_{1z}S_x)U^\dagger = 2I_{1z} \cdot US_xU^\dagger$$


Why did the Rio 2016 pool water turn green? One of the swimming pools at Rio 2016 Olympics has turned green:

While proliferation of algae is a likely culprit, there have been some alternative explanations offered by local organizers. CNN has the following quote from a Rio 2016 official:

  A change in alkalinity

  Mario Andrada, the communication director for the Rio 2016 local organizing committee, says a sudden change in alkalinity is the culprit.

  "We expect the color to be back to blue soon," Andrada said, adding there is "absolutely no risk to the athletes or anybody."

How could a change in alkalinity occur on such a rapid time scale (a few hours)? And why would it directly change the color of the water? I can imagine that a more alkaline water could lead to precipitation of green metal hydroxide compound, but you would need a massive concentration of e.g. nickel to start with — does not seem very likely.
This article (http://www.independent.co.uk/sport/olympics/rio-2016-green-pool-drained-as-olympic-organisers-reveal-hydrogen-peroxide-is-cause-of-colour-change-a7189616.html) claims, that the treatment of the water via hydrogen-peroxide is what has caused the rapid change in the chlorine levels (which in turn made it possible for the algae to pollulate coloring the pool). Is this chemically possible? If so, how (or if not, why not)?
Also, the organizers claimed, that if was safe to enter the pool even after it had turned green. Was it really safe, or were the organizers lying about the hazards to cover up what they had done?

Although not strictly alkalinity, this article quotes a changed pH level and an expert gives two possible explanations

  One: A low or high pH can bring out minerals. "Calcium would show itself as cloudy water, copper would show up as a blue-green," he says.

  Two: A too-high pH would render the chlorine ineffective, which could then allow for the proliferation of algae. (Hernandez says it's doubtful, though, that the green color is mainly from algae. It typically doesn't bloom that fast in such large pools.)


Resonance structures of 4‐aminobut‐3‐en‐2‐ylium Problem

How many resonance structures can be drawn for the following molecule?

(a) 1
(b) 4
(c) 3
(d) 2

Answer

 (d) 2

Question
I think that the molecule can be drawn in three ways, which contradicts the book's answer. Can anyone explain?

Just shift the pi-bond from the double bond to the right side. The carbocation then shifts to the left. That's the second resonance structure. The first one is already shown, making a total of two different structures (D).

You might be wondering about the allylic lone pair on the Nitrogen atom. Well, that could have resulted in another resonance pair, but if you shift the lone pair to the adjacent C-C bond on the right, there will be a total of 3 charges on the resonance structure. This is a highly unstable form, so it's not a significant resonance structure. Thus, it can be neglected.


How to identify if a chemical equation has a 'fractional order'? Consider the following reaction:
$$\ce{CHCl3 + Cl2 -> CCl4 + HCl}$$
The rate law expression is:
$$\text{Rate} = [\ce{CHCl3}]\cdot[\ce{Cl2}]^\frac{1}{2}$$
with an order of $1.5$. But 'how can we determine' this fact?
And why is it $1.5$ and not $2$?

It is not possible to determine the rate law of a reaction based on solely the balanced reaction. We must determine the rate law by experiment or by an established mechanism for the reaction.

What is meant by a mechanism is a series of elementary steps, a set of reactions whose linear combination yields the balanced reaction. Each elementary step represents a microscopic collision that happens during the process of the reaction, and these elementary steps obey the law of mass action, which allows us to write the rate law for each step in the reaction.

We then make a few approximations, perhaps guided by experimental observation, to obtain the rate law for the reaction. For example, we may conclude that one elementary step is rate-determining, so that the rate law for the reaction is governed by the rate law for that step; we may make a quasi-equilibrium approximation, which means that the species involved in an elementary step are equilibrated; or we may make a pseudo-steady-state assumption, which states that the net rate of production of an intermediate, presumed to be short-lived, is zero.

How do we know that a proposed mechanism is correct? Well, we don't. All we can do is prove that a mechanism is incorrect, by establishing that it posits the wrong rate law as found from experiment; we cannot distinguish between different mechanisms that result in the correct rate law. Nevertheless, we can often obtain much indirect evidence to support a given rate law, and we can frequently make good guesses based on patterns from experimental observations.

For example, while I wouldn't be able to accurately guess the rate law for the reaction you've provided, I can at least say that it is likely to be of half-integer order in chlorine, because such reactions often follow a radical mechanism, which often lead to non-integral powers in rate laws.


How can I tell if a certain carbon/hydrogen molecule can actually exist? My 11 year old daughter asked me if the following compound exists in real life, and if so, what is it called. Not only do I not know, I have no idea how to find the answer to such questions. Any ideas?

When drawing organic structures Carbon should have 4 "lines" coming from/to it while Hydrogen can have only 1 covalent bond. So if you could draw your molecule following these rules, it probably can exist. Here is an example:

H       H
 \     /
  C=C=C
 /     \
H       H

Alternatively there could be this structure with the same number of "lines" coming from/to Carbon:

  H   H
   \ /
    C     H
   / \   /
  H   C=C
     /   \
    H     H

As for the naming.. It's more complicated since it would depend on whether there is only one = bond or more than that. You could watch CrashCourse Organic Chemistry intro that talks about some naming conventions. But if you're stumbled, you could count the number of Carbons and Hydrogens and google for it: C3H4 or C3H6.


Could an extremely strong reducing/oxidising agent attack nitrogen molecules? Could a very strong reducing agent like lithium aluminium hydride or a very strong oxidizing agent like manganese heptoxide react with $\ce{N2}$? Usually when people talk about reactive substances they emphasize either strong electron acceptors or donors. I am wondering if using such compounds could be employed in a brute force method of dissociating $\ce{N2}$ molecules. I know there exist industrial methods for that already, but i'm poking around to see if maybe there's a way to do it under atmospheric pressure without tailored molecules like enzymes or organometallic complexes.

I know there exist industrial methods for that already, but im poking
around to see if maybe there's a way to do it under atmospheric
pressure without tailored molecules like enzymes or metalorgnic
complexes

Though Haber process which was used to industrially produce ammonia was well-known, some researchers have also attempted to split nitrogen molecules using metal complexes:

Chirik et.al (Nature, volume 427, pages 527–530(2004)) attempted to split nitrogen molecules by using a zirconium complex containing two bulky chemical groups called methylated cyclopentadienyl. The complex was the right size to snuggle up to both nitrogen atoms and give an electron to each atom. This double donation take place in an organic solvent at 100 °C and atmospheric pressure, disrupts the triple bond. Further reactions in the same solution then finish the job of splitting the nitrogen molecules and adding hydrogen atoms to make ammonia. (source)
Carbon monoxide and a molecule containing hafnium to break apart nitrogen and create a fertilizer called oxamide. (source)
John Mark P. Martirez and Emily A. Carter at Princeton University propose a gold-molybdenum catalyst that could split the nitrogen triple bond at room temperature using energy from visible light. That dissociation step is the primary limit on the reaction rate in the Haber-Bosch process. (source)


Should cyaphide create similar chemical compounds as cyanide? The cyaphide anion $\ce{(CP)-}$ should be chemically similar to its nitrogen analogue cyanide. $\ce{(CN-})$ But unlike it's counterpart it readily polymerizes  so it has only been found in outer space naturally. However, it should form compounds that should be the same  like cyanides.For example sodium cyanide ($\ce{NaCN}$),would be called sodium cyaphide ($\ce{NaCP}$) and hydrogen cyanide ($\ce{HCN}$) would be called hydrogen cyaphide or as it's called methylidynephosphane ($\ce{HCP}$) and the anions: phosphaethynolate (cyaphate) ($\ce{PCO-}$) and thiophosphaethyonlate (thiocyaphate), ($\ce{PCS-}$) and also potentially the cyaphogen ($\ce{P2C2}$) have been observed but as far as I know the only reliable info I found on cyaphides were on Wikipedia and they were very short as well. Which source on information do I go with?
More information: https://en.wikipedia.org/wiki/Methylidynephosphane

Methylidynephosphane or phosphaethyne is known and it has been achieved in 1961 by Thurman Gier. He passed phosphine gas at low pressure over an electric arc produced between two carbon electrodes. Condensation of the gaseous products in a –196 °C (–321 °F) trap revealed that the reaction had produced phosphaethyne along with acetylene and ethylene.
$$\ce{PH3 ->[electric arc][carbon electrodes] HCP + H2C=CH2 + HC#CH}$$
The ions $\ce{PCO-}$ and $\ce{PCS-}$ are also known. From the abstract of this paper1:

A range of λ3-phosphaalkynes $\ce{P⋮C−R}$ $\ce{(R = F, Cl, O-, OLi, OH,}$
$\ce{OSiMe3, S-}$, $\ce{SLi}$, $\ce{SH, Me,}$ $\ce{SiMe3, H)}$ has been geometry-optimized at the HF, MP2, and CCSD(T) level of theory. Vibrational frequencies as well as 31P and 13C NMR chemical shifts were calculated and compared to experimental values of the large variety of λ3-phosphaalkynes prepared over several years via differing synthetic routes. With a few obvious exceptions, a good agreement of calculated and experimental parameters was found. This also opens up the possibility to make predictions on
spectroscopic data of compounds not yet synthesized or fully
characterized, e.g., $\ce{PCOSiMe3}$ and $\ce{PCSH}$. The bonding situation
especially of the negatively charged species $\ce{P⋮C−O-}$ and $\ce{P⋮C−S-}$ was
studied via NBO analyses to determine the degree of phosphaalkyne and
heterocumulene contributions to the mesomeric anions.
$$\ce{^{-}X-C#P <-> X=C=P- (X=O,S)}$$

Cyaphogen ($\ce{(CP)2}$) has not been synthesized yet. It is considered a pseudohalogen. Structurally, it is formulated as $\ce{P#C-C#P}$.
Currently, no binary cyaphides are known but numerous organometallic cyaphide complexes are known. Some of them are listed in the Wikipedia article.
Reference

Theoretical Studies of NMR Chemical Shifts and Vibrational Frequencies in λ3-Phosphaalkynes P⋮C−R, Klaus Hübler and Peter Schwerdtfeger, Inorganic Chemistry 1999 38 (1), 157-164, DOI: 10.1021/ic9811291


Configuration, chirality I'm having trouble assigning priorities for this molecule:

I tried assigning priorities with the rules for doubly bonded atoms in mind. I know that the lowest priority group should be going into the page (not on a wedge) but I read that I can simply assign a priority with the fourth priority group on a wedge and then take the opposite of whatever I assigned as the true configuration of the molecule. Problem is, I get R. Answer key says S.

Your drawing is correct and yes, "I read that I can simply assign a priority with the fourth priority group on a wedge and then take the opposite of whatever I assigned" this is true too.  In your drawing 1->2->3 is a clockwise rotation (therefore "R" but), but the lowest priority item (4) is coming out towards you.  Therefor, the R/S assignment needs to be reversed (from "R") to "S".


What is the nature of chemical energy? Is chemical energy basically stored energy?

I'm not sure how to define "chemical energy" while I know many physical formulas for energy e.g. potential, movement and other.

Is chemical energy an energy that is "conserved" and if so, why is chemical energy not just potential energy?

Additional questions:

If chemical energy is an "own" form of energy, why does it not have a specific equation?
Did we have chemical energy before the big bang or when did it start?
Can you have chemical energy at or close enough to absolute zero?
How much of the total energy in the universe is chemical energy?
What's the point of many different definitions of energy?

Chemical energy is really nothing but electric potential energy.

The crucial difference is that chemical energy involves energy that can be extracted by breaking/making bonds.

For example, we say that the reaction $\ce{2H2 + O2->H2O}$ is accompanied by a release of energy. This energy is the difference of potential energies of the bonds in the reactants and products. Which potential energies? The bond potential energies. Bonds are, at the most basic level, formed due to attraction and repulsion of electrons and nuclei (coupled with some quantum mechanical effects). This carries some potential energy. The change in this potential energy achieved by breaking the bond is the "bond energy", and forms part of your "chemical energy". Some other parts are lattice energy, hydration energy etc, which are (in the end), due to electromagnetic forces/electromagnetic potential.

So yeah, chemical energy is just a "subset" of potential energy,

  If chemical energy is an "own" form of energy, why does it not have a specific equation?

Since chemical energy is an amalgamation of various types of electric potential energy in atoms, it cannot be given a formula (though you can calculate it for some compounds if you know some standard values).

  Is chemical energy an energy that is "conserved" and if so, why is chemical energy not just potential energy?

Chemical energy isn't really conserved anywhere unless there is no change to the chemical composition of the system. However, in a chemical reaction, most of the time any changes in energy are compensated for by absorption/release of heat. But it is just potential energy, as mentioned.

  Did we have chemical energy before the big bang or when did it start?

Meaningless question, there was no "before" the big bang. Also, energy really isn't something that you should classify into compartmentalized types.

But, since atoms formed a few minutes after the Big Bang, there wasn't anything to facilitate the existence of any chemical energy -- so the first occurrence of chemical energy was then.

  Can you have chemical energy at or close enough to absolute zero?

Yep. Bonds exist at absolute zero. Note that we can never reach absolute zero, but bonds exist both close to and hypothetically "at" absolute zero.

  How much of the total energy in the universe is chemical energy?

I can't answer that, and I doubt it can be answered to any reasonable approximation. I think the percentage would be rather low -- our universe is mainly stars (besides being empty space). The average star doesn't have many chemical bonds--it contains hydrogen/helium plasma, and plasma can't exactly form bonds.

  What's the point of many different definitions of energy?

In certain situations, you know that there is no method to facilitate conversion of one form of energy into another. In such a case, one can separately conserve that form of energy. So the forms of energy help in analyzing/solving a situation, but not much more.


Why do aromatic hydrogen atoms have a higher chemical shift value compared to aliphatic hydrogen atoms? In Nuclear magnetic resonance (H-NMR) spectroscopy, the chemical shift of aliphatic hydrogen atoms are very much closer to the 1.0 ppm than that of aromatic hydrogen atoms.

Aromatic hydrogen atoms have a chemical shift value of about 7.0-9.0 ppm whereas the chemical shift value of aliphatic hydrogen atoms ranges between 2.0-3.5 ppm.

Why is this so?

The chemical shift gives you information about how well shielded the nuclei are from the magnetic field. A proton at higher chemical shift values is deshielded, so the aromatic protons are obviously less shielded than aliphatic protons.

One effect that causes deshielding is the presense of electronegative atoms that draw electrons away from other atoms and thereby deshield them. But that is not the cause of the aromatic chemical shift.

The reason for that one is the ring current that is induced in a magnetic field. The induced current creates a local magnetic field that has the same direction as the external field outside the aromatic ring where the attached protons are, leading to the deshielding of the nuclei.


What causes passivity of specific metals in conc. nitric acid? I know  that specific metals like $\ce{Pb}, \ce{Sn}$ and $\ce{Al}$ become passive in conc. nitric acid. My questions are:

Why does this happen only with conc. nitric acid, and not other conc. acids?
Why is the passivity specific to these metals? Do they have something in common or are there individual reasons?

Aluminium

This link states that:

  Aluminium metal dissolves readily in dilute sulphuric acid to form solutions containing the hydrated Al(III) ion together with hydrogen gas, $\ce{H2}$. The corresponding reactions with dilute hydrochloric acid also give the hydrated Al(III) ion. Concentrated nitric acid passivates aluminium metal (due to formation of thin protective oxide layer, $\ce{Al2O3}$ on surface of metal which cuts off further reaction. )

$$\ce{2Al(s) + 3H2SO4(aq) → 2Al^3+(aq) + 2SO4^2-(aq) + 3H2(g)}$$
$$\ce{2Al(s) + 6HCl(aq) → 2Al^3+(aq) + 6Cl-(aq) + 3H2(g)}$$

Lead

This link states that:

  The surface of metallic lead is protected by a thin layer of lead oxide, $\ce{PbO}$. This renders the lead essentially insoluble in sulphuric acid. Lead reacts slowly with hydrochloric acid and nitric acid. In the latter case, nitrogen oxides are formed together with lead(II) nitrate, $\ce{Pb(NO3)2}$.

$$\ce{Pb + HCl -> H2[PbCl4] + H2}$$
$$\ce{Pb + 4 HNO3 → Pb(NO3)2 + 2 NO2 + 2 H2O}$$

Tin

This link states that:

  Tin is attacked only slowly by dilute acids such as hydrochloric acid and sulfuric acid. [...]Tin dissolves easily in concentrated acids.

$$\ce{Sn + 2H2SO4 → SnSO4 + SO2 + 2H2O}$$
$$\ce{Sn + 3HCl → H[SnCl3] + H2}$$
$$\ce{Sn + 4HNO3 → H2SnO3 + 4NO2 + H2O}$$

Summarizing as a table,

  Do they have something in common or are there individual reasons?

The three metals do not have common properties and reacts with different acids differently as we can see in the table.


Resonance Forms in Electrophilic Aromatic Substitution In most diagrams for electrophilic aromatic substitution, the following is represented:

What happens to the other two resonance forms of the sigma complex? It seems like most of the time the first form is simply chosen to illustrate the return of aromaticity.

Suppose I switched the resonance structure in the second step to the following:

Well I'm too lazy to open up chemdraw to make a nice drawing so my handwriting will have to do.

The way I see this kind of question is that, to interconvert resonance structures, electrons need to move. Of course, you'd represent that with curly arrows (top half of the picture).

Now for the deprotonation step, imagine you were using the first resonance structure. You'd draw the first two of the three arrows that I drew in the bottom half of the picture. (Or you might omit drawing the base, although for simple mechanisms like this I prefer having everything accounted for.)

What's the third arrow for (the one going from a double bond to the adjacent single bond)? It's the reverse of the arrow you used to interconvert the resonance structure. So you could think of the first two arrows being the "actual" reaction, and the third arrow being a "bookkeeping" arrow to make sure that everything ends up nicely aromatic.

Of course, that's kind of a flawed view because the reaction as shown has to have all three arrows or else you end up with a product that doesn't make sense. But using this idea of "the original reaction" + "reverse resonance arrow" works for every single instance of the question that you are asking, i.e. how to draw a mechanism involving an alternative resonance form. As long as you are able to draw arrows to show the interconversion between resonance forms, you should never run into a problem drawing a mechanism, regardless of what resonance form you use.

Now just a note on what you said - "It seems like most of the time the first form is simply chosen to illustrate the return of aromaticity." You're absolutely right about that. Using the first resonance form is just the simplest option since it involves the fewest arrows. Otherwise the writer of the book is just giving him/herself extra work... and nobody likes that.


What are Degenerate Rotational Energy Levels Physically? I'm in introductory chem, so I don't have much of the physical background to really understand what's going on yet, but we're learning about rotational energy levels, and how some rotational energy levels for molecules are degenerate.

As I understand it, degenerate energy levels correspond to different "ways in which the molecule could rotate" such that it gains the same amount of potential energy.

I have two questions...

Why does a molecule "gain potential energy" when it rotates? Does it want to stop rotating for some reson?
What do degenerate energy levels correspond to physically in terms of the molecule's rotation?

Thanks!

I have two questions...

  Why does a molecule "gain potential energy" when it rotates? Does it
  want to stop rotating for some reason?

  What do degenerate energy levels correspond to physically in terms of
  the molecule's rotation?

Do you remember the Newton's law of motion? A body in motion will always remain in motion until and unless there is a force acting on it. So you can imagine, that once a force caused the molecule to rotate, the molecule will continue to rotate forever. As a crude analogy can you recall the time when the Earth stopped spinning on its axis? It is spinning for billions of years. In the same way you can think that the isolated molecule, once excited, will rotate for a pretty long time (on a spectroscopic time scale).

You can think of degenerate rotation as a a way of physical rotation of a molecule in such a way that if you calculate the rotational energy in each case, it turns out to be the same number.

Third point, as pointed out by orthocresol, rotational energy is kinetic energy not potential energy. So no need to invoke potential energy.

Sometimes, we should not actually try to visualize what is happening physically at very small molecular levels. This is where mathematics enters. Many results originate out of pure mathematical calculations which can be hard to imagine mentally. For instance, Bohr's got correct H2 spectrum results by assuming the electron is rotating around the proton. The question is it actually rotating like a planet around a proton? A short story is worth sharing which will help you in future. Feynman (Physics Nobel) used to think a lot about of photons as a student and discussed them with his father. When he finished his PhD in physics, he realized, he still could not explain what a photon is. Sorry I cannot recall the reference. But mathematics has no problem. Yet on the web we can "see" photons in pictures as balls, which are obviously pure but meaningless imaginations.