How to determine the composition for a sucrose/ zinc nitrate solution from its freezing point? A $4~\mathrm{g}$ mixture of sucrose ($\ce{C12H22O11}$) and zinc nitrate ($\ce{Zn(NO3)2}$) is dissolved in $150~\mathrm{g}$ of water. If the resulting solution freezes at $-0.768~^\circ\mathrm{C}$, what is the mass percent of sucrose in the mixture? Assume that zinc nitrate dissociates completely in water, $K_\mathrm f = 1.86$. (Answer: $2.8\%$)

First find the molality of solute particles.  Assume i = 1; we're just counting how many things are dissolved and it doesn't matter whether they are ions or molecules at this point.  I'm using m for mass and b for molality throughout to avoid confusion.

$\Delta T = ibk_f\implies\frac{\Delta T}{k}=b$

$b=\frac{0.768}{1.86}=\pu{0.413 molal}$

Next find how many moles of particles were dissolved.

$b = \frac{n_{\text{solute}}}{m_{\text{solvent}}}$

$n_{\text{solute}}=b\times m_{\text{solvent}}=0.413\times 0.150 = 0.0619~\text{moles of particles}$

Now determine the mass of each species based on this total number of moles.
$n_{\text{particles}}=3n_{\ce{Zn(NO3)2}}+n_{\ce{C12H22O11}}$

$n_{\text{particles}}=3\frac{m_{\ce{Zn(NO3)2}}}{M_{\ce{Zn(NO3)2}}}+\frac{m_{\ce{C12H22O11}}}{M_{\ce{C12H22O11}}}$

You know the total number of moles of particles, 0.0619, and you know the molar masses of both zinc nitrate and sucrose, but you don't know either the actual mass of zinc nitrate or of sucrose.  However, you do have one other piece of information that allows you to set up a system of equations.

$m_{\text{solute}}=m_{\ce{Zn(NO3)2}}+m_{\ce{C12H22O11}}=\pu{4g}$

Solving this system of equations should give you the masses of both solutes.


What decides the nature of emulsifying agents? I read

  The emulsifying
  agent forms an interfacial film between suspended particles and the
  medium. The principal emulsifying agents for O/W emulsions are
  proteins, gums, natural and synthetic soaps, etc., and for W/O, heavy
  metal salts of fatty acids, long chain alcohols, lampblack, etc.

What specific property of each type of emulsifying agent makes it suitable for the particular application for O/W or W/O emulsion?

An emulsifier is a surfactant. A surfactant is characterized by the fact that its molecule has one or more hydrophilic parts and one or more hydrophobic parts. The ratio between the molar mass of the hydrophilic parts and the molar mass of the hydrophobic parts in the molecule is expressed by its HLB value (Hydrophilic-lipophilic balance).

Higher HLB means higher water solubility, and lower HLB means lower water solubility.

The emulsifier should have a certain solubility in the outer phase of the emulsion to be distributed in the emulsion.

An emulsifier for forming o/w-emulsions should have higher water solubility, that means higher HLB, and an emulsifier for forming w/o-emulsions should have lower water solubility (= higher oil solubility), that means lower HLB. The corresponding range of HLB is given in the Wikipedia artivle above.


How is it possible to obtain a value of 8/3 as an oxidation state? Iron's oxidation state in $\ce{Fe3O4}$. I really can't get the logic behind this.

The formula of $\ce{Fe3O4}$ is a shorthand for $\ce{Fe^{II}Fe^{III}2O4}$, so there are two sorts of atoms of iron here in +2 and +3 oxidation state.

However, it is still possible to gain truly fractional oxidation states. The simplest example probably would be potassium superoxide, $\ce{KO2}$. Here, the anion is symmetrical $\ce{O2^{-}}$ anion, so obviously both oxygen atoms are in $\ce{-1/2}$ oxidation state. There are more complicated examples as well.

Often fractional oxidation state comes from symmetrical structures with charge, that would have unpaired electrons otherwise. However, systems with truly fractional oxidation states usually have bonding, hard to describe in terms of common Lewis structures (and with little relevance of idea of oxidation state), and mostly are considered in semi-advanced organic chemistry and advanced inorganic chemistry.


How to keep a flask submerged in water? I am trying to keep a 1000 ml Erlenmeyer flask submerged in a deep cooking pot filled with water. I tried tying the flask to a metal plate, but it was too light unfortunately.

Is there a purpose built piece of equipment for this or a commonly accepted method?

The vinyl coated lead rings suggested in the comment are a great solution. They are, however, expensive and take time to order. There is also the cost of shipping.

A quick and less expensive route is to use a PVC flange and attach what ever heavy metal you have (steel, cast iron, lead fishing weights). Epoxy to flange or drill holes and tie. You may want to paint the metal to avoid rust or other reactions with the metal.

Home Depot has inexpensive 3" flange or 4" flange.


Hand Sanitizer Chemical Changes I keep a bottle of hand sanitizer on my desk at work, as I touch a lot of things throughout the day. When I used it yesterday, a bit must've been left in the pump track or opening. So I was about to use it and I see that in the opening it looks opaque, kind of like lotion. I press the pump and it releases the opaque portion followed by the normal, gooey sanitizer.

What caused this chemical change? The air conditioning in our building is still on, but at about 6 p.m., the HVAC is turned off. Our lowest temps at night are between the 40s and the 50s.

Please clarify what is going on, particularly if something is wrong with this sanitizer and I need to throw it away.

Given the list of ingredients in Pat's comment, I would strongly doubt that a chemical reaction has occurred here, and suggest that most likely the volatile solvents ethanol and isopropanol have evaporated, leaving a cloudy residue of the remaining ingredients. Your concern of a possible hazard is appreciated, but I would say no worries.


Does lead(II) chloride dissociate into Pb²⁺ and Cl⁻ ions in HCl? While solving JEE Advanced 2020 (Paper 2) questions, I came across this particular question:

Q.10 Choose the correct statement(s) among the following.
(A) $\ce{SnCl2 · 2 H2O}$ is a reducing agent.
(B) $\ce{SnO2}$ reacts with $\ce{KOH}$ to form $\ce{K2[Sn(OH)6]}.$
(C) A solution of $\ce{PbCl2}$ in $\ce{HCl}$ contains $\ce{Pb^2+}$ and $\ce{Cl-}$ ions.
(D) The reaction of $\ce{Pb3O4}$ with hot dilute nitric acid to give $\ce{PbO2}$ is a redox reaction.

According to the final answer key, its answer is options (A), (B).
Coming to the option (C), after doing some research, I have been able to conclude the following:

$\ce{PbCl2}$ is sparingly soluble in water.
$\ce{PbCl2}$ + little amount of dil. $\ce{HCl}$ decreases its solubility due to common ion effect.
Upon further addition of dil. $\ce{HCl}$ or $\ce{PbCl2}$ + conc. $\ce{HCl}$ leads to formation of soluble lead complexes: $\ce{[PbCl3]-}$ and $\ce{[PbCl4]^2-}.$
Hence, option (C) is incorrect as in one case, $\ce{PbCl2}$ doesn't dissociate at all and even in the other, $\ce{Pb^2+}$ and $\ce{Cl-}$ ions are not formed (so the dilemma of the $\ce{HCl}$ doesn't exist anymore as this option is incorrect regardless of the $\ce{HCl}$ being dilute or concentrated).

Are the above four points accurate? It'd be appreciated if someone could add more details and correct the above points if needed.

Your analysis about $\ce{PbCl2}$ in four points is correct.
Then ($\ce{A}$ and ($\ce{B}$) are correct, if they are in aqueous solution. The following equation shows that $\ce{SnCl2}$ is a reducing agent that is easily oxidized by air : $$\ce{6 SnCl2 + O2 + H2O  -> SnCl4 + 4 Sn(OH)Cl}$$ and, in ($\ce{B}$),  $\ce{SnO2}$ is soluble in $\ce{KOH}$ solutions according to $$\ce{SnO2 + 2 KOH + 2 H2O -> K2[Sn(OH)6]}$$ (D) is wrong, because the reaction is the following $$\ce{Pb3O4 + 4 HNO3 -> PbO2 + 2 Pb(NO3)2 + 2 H2O}$$ This reaction can be deduced by dealing with $\ce{Pb3O4}$ as if it was a mixture of $\ce{PbO2 + 2 PbO}$ so that only the two $\ce{PbO}$ react with $\ce{HNO3}$, without changing its oxidation number $$\ce{2 PbO + 4 HNO3 -> 2 Pb(NO3)2 + 2 H2O}$$


Why are anti aromatic compounds unstable? Just like aromatic compounds, anti aromatic compounds have several resonance structures and some of them are also conjugated ( 1,3-cyclobutene).

Why is it that just having 4n electrons makes them unstable?

In conjugated-pi ring systems of N atoms, you typically have a single low energy orbital and then paired (degenerate) energy levels until you have N orbitals total. This has to do with the number of nodes in the conjugated pi system, and the number of ways you can lay them out with N atoms.

Now, when you add in electrons you add two electrons per orbital, so your 4n+2 electrons fill 2n+1 orbitals -- the single low energy orbital and then n complete sets of degenerate pairs.

If you only have 4n electrons, you can fill the single low energy orbital, then 2(n-1) complete sets of degenerate pairs ... but with two "extra" electrons which aren't paired up. This allows those two electrons to "spread" out between the pair of equal-energy orbitals. Those unfilled/partially filled orbitals mean that the anti-aromatic compound is quite reactive.

The other issue is that "stability" is not an absolute property. Stability is always in reference to some other state. The typical anti-aromatic systems you'll encounter tend to have those frontier orbitals as non-bonding or anti-bonding orbitals, meaning the electrons in the orbitals aren't really contributing to the molecular stability. Generally, if you build the energy level diagrams for alternate systems (e.g. a bent ring where you don't get localization) you'll find that - while the lowest energy orbital isn't as low energy as in the ring-conjugated system - the sum of the energies across all orbitals is indeed lower. And it's the total energy of the system which is optimized. (For example, in cyclobutadiene the ring-conjugated form would have two electrons in a low-energy bonding orbital and two in non-bonding orbitals. In contrast, linear butadiene bonding orbitals are not as low energy as for the ring-conjugated cyclo form, but there's two of them. And the sum of the energies from two linear bonding orbitals is lower than what you get from a single cyclo bonding orbital.)

(Final note: this is all high-level generalization. There's more sophisticated levels of quantum mechanical theory where things get more complex.)


Rotational Barriers - Amide and Hydroxamic Acid Why the barrier to rotation about the $\ce{C-N}$ bond higher in an amide than it is in a hydroxamic acid? (MO-based explanation preferred)

You can check out molecular orbitals of N-methylacetamide here. Unticking the spin-box will stop the molecule spinning. The important orbitals are $\ce{HOMO} -1$ and $\ce{HOMO} - 2$. (The actual HOMO is not part of the $\unicode[Times]{x3C0}$-system.)

The same site also features a low-level Simple Hückel Molecular Orbit Calculator. I used that to create the four images below of the $\unicode[Times]{x3C0}$-system of N-hydroxyformamide. From left to right the atoms are $\ce{O-N-C=O}$

The lowest $\unicode[Times]{x3C0}$-orbital is always bonding across the entire $\unicode[Times]{x3C0}$-system. In N-methylacetamide, the second $\unicode[Times]{x3C0}$-orbital has a nodal plain almost directly on the carbon, only slightly moved towards the carbonyl oxygen (theory tells me that, I can’t actually see that good in the model). The third $\unicode[Times]{x3C0}$-orbital is empty and thus does not participate in bonding. This means that each bond ($\ce{N-C}$ and $\ce{C=O}$ can be regarded as something between a single and a double bond — in accordance with the Lewis resonance structures. There is considerable overlap to be lost on rotating the system around $\ce{N-C}$.

In the hydroxamic acid, the second and third $\unicode[Times]{x3C0}$-orbitals (the occupied ones) only have very little electron density in the $\ce{N-C}$ bond. In fact, the lowest MO is largest for the $\ce{O-N}$ bond, the second is (weakly) antibonding concerning $\ce{O-N}$ and has the largest contributions on $\ce{C=O}$. The third is again antibonding along $\ce{O-N}$. Altogether this means that most of the electron density is between $\ce{O-N}$ and between $\ce{C=O}$ — the $\ce{N-C}$ bond only carries very little density and is thus much better described by a single bond. Much less resonance is lost when rotating, so the rotational barrier is lower.


Enantiomer of Amphetamine I saw this graphic in Organic Chemistry by Klein (2nd edition too!) and this has to be wrong, correct?

The mirror images should have the double bond on the far side of the ring in both images, right? I am sure, unless my visualization skills have totally failed me.

Here's the result when I flip the entire image using an online mirror effect creator:

The graphic from the book is correct, those two molecules are mirror images.  Same for your drawing, those are the same two mirror images as in the book.  Remember that the electrons in the aromatic ring are delocalized.  The double bonds in the aromatic ring are not fixed.  Where you choose to draw the double bonds in aromatic rings is your choice and does not play any role in determining if molecules are enantiomers.  Bottom line, the following three structures all represent the same thing.


What molecules are produced in carbon arc lamp and how do they produce light? I am curious about the chemistry of the arc lamp (an indirectly for the incandescent bulbs) thus I would like to ask the following.

It is a common understanding that the electrons pushed inside an arc lamp -- let's assume a carbon one, that is with electrodes in graphite -- will collide on the atoms of the carbon crystal and increase their kinetic energy rising those in the cathode to about 3600 degrees and those in the anode to 4200 degrees. My questions are:

Why is the cathode, where the electrons arrive first and at full energy, cooler than the anode?
What kind of molecules are produced at the electrodes? It is said that carbon evaporates at the electrodes when the current is allowed to flow between them. What is the vapour of carbon? From wikipedia, dicarbon is listed, but what would be the reaction from graphite (pure carbon) to dicarbon?
If dicarbon is produced by evaporation on the arc lamp, would that be produced also in the old incandescent lamps that used carbon filaments, such as that made by Edison?
Does dicarbon react with oxygen? Does it oxidise to carbon dioxide or monoxide?
And finally, how does dicarbon produce light? I imagine that it must be at a high energy state and photons are released with the decay of the electrons to their ground levels, is that correct? Is there some kind of quantitative measure of why arc lamps are more luminous than incandescent or halogen lamps?

Thank you

Answers:

Consider the mass and kinetic energy of an electron or of an ion of nitrogen or oxygen, and the reactivity with carbon.
Good question! It would be interesting to see the relative abundance of allotropes of carbon deposited from a filament in vacuo. There is a metallic glisten to the deposit, so there might be graphite, carbon nanotubes or buckminsterfullerene. $\ce{C2}$ would not be there for long, though.
There are a great many chemical species produced by an arc lamp operating in air. Depending on the gases in which it operates, an arc can produce complex organic molecules. Even something as ornate as a buckyball can be made!
I've not seen the relative abundance of products of an arc lamp, but I assume carbon oxidizes rapidly to the monoxide and dioxide, depending on abundance of oxygen. Charcoal briquets oxidize at a far lower temperature.
Most of the light is black-body radiation, though there would also be some spectral lines produced by recombination, such as from nitrogen. With a spectroscope, you'd see a continuum with a few bright lines added (or dark lines from absorption, as the gases cool).


Looking for a compound TL;DR;
I need a compound that is harmeless to the human body, odorless, tasteless, causes some kind of visible reaction (urine coloration perhaps) AND can be bought without medical recomendation.
LONG VERSION;
Food stealing in my company fridge has become a serious problem. It's not only a recurring issue but a targeted one... our thief has been targeting "high" value, particularly coconut water. HR seems to ignore our complains... so I decided to take matters in my own hands... since I have access to timesheets, I thought that if I can scare out thief to leave work or skip it, I can narrow-it-down/point-him-out because we have about 30 people in our office...

Urine typically is yellow, but in combination with blue dyes, a green colour can be obtained.

Noteworthy compounds are methylene blue, thymol, Cimetidine (antacid), or Mitoxanthrone (used in cancer therapy).

I've mentioned these just for the sake of completeness - I think it is irresponsible and stupid to spike food with anything that doesn't belong there.

Hiding a webcam in a shelf and running zoneminder probably isn't legal either and will get you in trouble with the privacy officer, but at least that doesn't mess around with the health of anybody else.


Can somebody explain the low temperature Schottky anomaly for two level systems? Statistical thermodynamics question. Can somebody explain the low temperature Schottky anomaly for two level systems? Why is this anomaly not seen in all materials?

The Schottky anomaly is a general feature of two levels systems. The heat capacity is defined as $\displaystyle C= \left( \frac{\partial U}{\partial T} \right)_\Delta$ where $U$ is the internal energy and the energy gap between the two levels is $k_B\Delta$, $k_B$ being the Boltzmann constant.

The gradient of the internal energy is thus important. At constant energy gap and at very low temperatures  only the lowest level is populated and the internal energy is small, and heat capacity small as the internal energy increase is small, i.e gradient $U$ vs $T$ is small. However, as the temperature is increased the upper level begins to be populated and so the internal energy rapidly increases, then the gradient $U$ vs $T$ is large and so is heat capacity.

When both levels are almost populated equally, which occurs at very high temperature, no more energy can be absorbed and the gradient $dU/dT$ effectively becomes constant and so the heat capacity becomes small and tends to zero.

Similar behaviour is observed when there are three or four etc i.e. a small  number of levels. In the 'normal' case of a molecule there is effectively no upper limit on the number of levels that can be populated as temperature increases, and the slope of $U$ vs $T$ has a constant finite value which leads to $3R$ as the heat capacity per mole at high temperatures.

The figure shows the internal energy (in units of $k_B\Delta $)  and heat capacity (in units of $k_B$) vs $T/\Delta$. To make the plots more general $T/\Delta$ is plotted as this ratio occurs in the final equation.

The equations are:

partition function $ Z=1+e^{-E/k_BT} = 1+e^{-\Delta/T} $ when energy gap $E=k_B\Delta $.

The Internal energy is $\displaystyle U=k_BT^2\left( \frac{\partial \ln(Z)}{\partial T} \right)_\Delta$.

The heat capacity is $C\displaystyle=\left( \frac{\partial U}{\partial T} \right)_\Delta= k_B\frac{ \Delta^2}{T^2}\frac{e^{\Delta/T}  }{(1+e^{\Delta/T})^2}$


With molecular formula, how do I know if it linear or one element surrounded by other elements? If I was given a molecular formula (i.e, PoOBr2), how do I know if the Lewis structure is linear (4 in-a-row, basically a straight line) OR it is an element surrounded by other 3 elements?

I know the answer for this is Po in the central and surrounded by other elements.

The formula is insufficient. Consider pentane, $\ce{C5H12}$, which occurs in straight and branched isomers, the extreme case being neopentane, in which the central carbon is linked only to four methyl groups.


Reaction of Fe2+ and S2- to form FeS Consider this reaction:
$$\ce{Fe^{2+} (aq) + S^{2-} (aq) <=> FeS (s)}$$
The equilibrium constant for the reaction is $1.6 \times 10^{17}$ at $\pu{298K}$. If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^2+}$ and $\pu{0.2 M}$ of $\ce{S^2-}$ are mixed, the concentration of $\ce{Fe^2+}$ at equilibrium is found to be $A \times 10^{-17}$. Find $A$.

Let $x$ amount of $\ce{Fe^2+}$ is converted to $\ce{FeS}$. Then
$$K = \displaystyle {\frac{1}{(0.06 - x)(0.2 - x)}}$$ Which then leads to the quadratic equation
$$ x^2 - 0.26x + 0.012 - 6.25 \times 10^{-18} $$
Solving this the $A$ I got is not the answer.
Answer is $8.93$

This is very poorly worded question, and really not just an equlibrium question. This is actually a solubility product question. Consider the given reaction:

$$\ce{Fe^{2+} (aq) + S^{2-} (aq) <=> FeS (s)} \tag{1}$$

If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^2+}$ and $\pu{0.2 M}$ of $\ce{S^2-}$ are mixed, since the equilibrium constant for the reaction is $1.6 \times 10^{17}$ at $\pu{298K}$, we can assume that all $\ce{Fe^2+}$ are reacted because it is the limiting reagent and would give $\pu{0.06 M}\times V$ amount of solid $\ce{FeS}$ (where $V$ is the volume of $\ce{Fe^2+}$).

However, the author of the question ignore that $\ce{FeS}$ would gives solubility product of $\ce{Fe^2+}$ and $\ce{S^2-}$.

Since most sulfides of transition metals are insoluble in water, following equilibrium would establish after the original reaction:

$$\ce{FeS (s)  <=> F^2+ (aq) + S^2- (aq)}\text{;} \,\,\, K_\mathrm{sp}= [\ce{F^2+}][\ce{S^2-}]=4.9 \times 10^{-18} \tag{2}$$

Here, $K_\mathrm{sp}$ has not be provided, so I found it in here at $\pu{298K}$ ($\pu{25 ^\circ C}$). The concentration of $\ce{Fe^2+}$ at this equilibrium is given as $A \times 10^{-17}$. We can also calculate the concentration of $\ce{S^2-}$ from equation $(1)$ that $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ (since equal volumes has made final volume dobuled and hence concentration of remaining ions are halves). Since this value is much larger than the $[\ce{S^2-}]$ generated by dissociation of $\ce{FeS}$ ($\approx \sqrt{4.9 \times 10^{-18}}$), we can assume $\pu{0.07 M$ is the same concentration of $\ce{S^2-}$ at this equilibrium. Accordingly:

$$K_\mathrm{sp}= [\ce{F^2+}][\ce{S^2-}]= A \times 10^{-17} \times 0.07 = 4.9 \times 10^{-18}$$

$$\therefore \; A = \frac{4.9 \times 10^{-18}}{1.0 \times 10^{-17} \times 0.07}  = 7.0 $$

Note that this final answer is highly depends on the $K_\mathrm{sp}$ value. Thus, the deviation of this answer from your given answer is based on what value should have been given at the first place.

Suppose we ignore the solubility product as author may be suggesting: Then, once the original reaction is completed, significant amount of $\ce{S^2-}$ would remain unreacted since it is in excess (compared to $[\ce{Fe^2+}]$). That amount would be $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ as indicated above. Since at this point, $[\ce{Fe^2+}]$ is $A \times 10^{-17}$, the equilibrium constant is set up as follows as OP has done:

$$K = \frac{1}{[\ce{Fe^2+}][\ce{S^2-}]}=\frac{1}{[A \times 10^{-17}][0.07]}= 1.6 \times 10^{17}$$

$$\therefore \; A = \frac{1}{1.0 \times 0.07 \times 10^{-17}\times 1.6 \times 10^{17}}= 8.93$$

Although, this setup gives the expected answer, it is not the correct way to solve this (at least to my knowledge).


Heat of combustion of stereoisomers of 1-tert-butyl-2,4-dimethylcyclohexane Which of the compounds below has the smallest heat of combustion?

I know that the heat of combustion is indirectly proportional to the stability of a molecule. And here both $\ce{CH3}$ and $\ce{C(CH3)3}$ are +I groups (inductive effect) but apart from that, I am stuck on how to compare these compounds.

Which of the compounds below have the smallest heat of combustion?

The question is asking, which of these 4 compounds is the most stable.  The most stable compound will have the lowest heat of combustion.

In the cyclohexane ring system, bulky substituents prefer the equatorial position rather than the axial position.  This is due to steric (this can also be described in electronic terms since it is due to the repulsion between interpenetrating electron clouds) destabilization. For example, in methylcyclohexane, the methyl group can be in an axial or equatorial position.  In the axial position, it is destabilized by the steric (electronic) interactions it has with the axial hydrogens on carbons 3 and 5.

These interactions are nicely described by cyclohexane A-Values.  The A-Value measures the energy difference (in kcal/mol) between having the substituent in the axial and equatorial positions. The larger the A-Value, the larger the preference the substituent has for the equatorial position. The A-Value for a methyl group is 1.7 kcal/mol. The A-Value for a t-butyl group is so large (>4 kcal/mol) that the t-butyl group is effectively locked into the equatorial position. t-Butyl substituted cyclohexanes do not undergo chair-chair interconversion - this would place the t-butyl group in an axial position, hence they are conformationally locked into the chair with the t-butyl in an equatorial position.

It is this last piece of information that helps simplify our problem.  Here are the structures of the 4 cyclohexanes redrawn in the chair conformation.

Each structure has the t-butyl group in the equatorial position.  Compound 1 has one methyl group in an axial position; this adds 1.7 kcal/mol to the energy of the molecule. Molecule 2 has two axial methyl groups, therefore it is 3.4 kcal/mol higher in energy.

I'll leave molecules 3 and 4 to you to figure out.  Once you determine their relative energies, which one of the 4 has the lowest energy relative energy?

Note: There are exceptions to the additivity of A-Values. See near the bottom of this page for a few examples.


Negative numbers for ionic radii? I was looking for a good table of ionic radii for an experiment on the van't-hoff factor when I stumbled upon R. D. Shannon, Acta Cryst. 1976, A32, 751.

In table 1, it lists 1 of the ions of carbon with as so:

  C +4 1S 2 III .06 -.08

The bold numbers are of concern here. The first number, 0.06 Å is what the paper refers to as the crystal radius, while the second number, -0.08 Å, is referred to as the effective ionic radius. I was also unable to find any good distinction between the two in the paper, except that the radius of $\ce{O^2+}$ for the crystal radius is 1.40 Å, while the effective ionic radius for $\ce{O^2+}$ is 1.26 Å.

I also doubt that this is a typo since it agrees with this statement made later in the paper:

  crystal radii differ from traditional radii only by a constant factor of 0.14 [ Å].

Ionic radii are not measured on individual ions by some sort of mini-micrometer, but bond distances can be measured and then radii are assigned to the individual atoms/ions. The fuzziness of the assignments can be seen by the variation in ionic radii as the coordination number changes: if you give the ion more room (greater coordination number), it will fill the space - if you assume that the ligands have the same size as usual (whatever that is).
(There are materials (ferroelectrics) in which the coordinated ion does not completely fill the space allotted. In $\ce{BaTiO3}$, the $\ce{Ti^{+4}}$ ion is so small that it can shift its position in the octahedral cage of oxygen anions in response to an external electric field.)
The data in the Shannon paper are extensive, certainly a credit to the Central Research Department of DuPont in 1976, but the trends reported seem to be much more valuable than any specific datum.
Nicolau mentioned covalency as one factor in determining radii. Purepani brought up negative ions not having an effect of negative radii. Assigning a radius to a proton other than zero suggests that you are allowing it to have some electron density surrounding it, able to repel an anionic ligand. Electrostatically, a positive ion will attract some electron body (anion or atom) and come closer; depending on your initial assignment of radii, you can adjust to keep your idea of proton radius constant, or adjust, even into the negative region (because we assume you have a good imagination). Negative ions have a larger size than protons, and are less hard, so they can dent a little. One reason to use negative radii for the cations is that the anion may be dented only at one site, but has a "normal" radius everywhere else around it.
My approach to handling a negative radius is to conclude that the proton (or other mini- cation) is so electrostatically demanding that it can begin to bury itself in the outer shell of the anionic ligand, if the coordination number allows such a deviation in distance to be measured. And since we don't want to redefine the anionic radius, a negative radius is assigned to the proton, and the mathematics works out OK.
I guess you could say atoms and ions are somewhat mushy, and mathematics can deal with it.


How can you calculate ring strain in beta lactams for different penicillin drugs? I am currently working on a lab, where I am trying to calculate whether the R group in penicillin type drugs has an affect on the ring strain of the beta lactam. I know how to find ring strain in cycloalkanes through the molar heat of combustion, however I can't find any information about the combustion of beta lactams as I'm sure it doesn't happen often, so is there another way I can calculate the ring strain of the beta lactams in different penicillin type drugs?
Thanks,

For the calculation of ring strain for ring-opening metathesis polymerization, I used DFT calculation.
In generell you open your molecule at the position you want to and saturate the bonds. Then you compare the enthalpy of products and educts.
As an example you can look at cyclopentene.If it is opened during Metathesis and you saturate the doublebonds at the end of the chain with two $\ce{CH2}$ or one ethene molecule.

This has also been reported in literature for example:
https://onlinelibrary.wiley.com/doi/full/10.1002/pola.28695
I have also done a quick search and in this paper they have used the same technique (they just used ethane instead of ethene because they saturated single bonds instead of double bonds). They used post Hatree Fock B3LYP to calculate the enthalpy. I think you are fine to use either one.
https://chemistry-europe.onlinelibrary.wiley.com/doi/10.1002/cphc.201900173


Proving 1 = 2 by POAC( Principle of Atom Conservation ) I have been told that applying POAC in any equation means conserving the number of moles of an element in both reactant and product .
For example :
$$\ce{H2 + O2 -> H2O}$$
Here when we apply POAC on oxygen we write :

Equation 1 :

No. Of Atoms(or moles) of "O" in reactant =  No. Of Atoms(or moles ) of "O" in product

Equation 2: (obtained from equation 1)

2(moles of $\ce{O2}$) = 1(moles of $\ce{H2O}$)

Where equation 2 gives us the necessary data required in solving questions of stoichiometric calculations .
I understand how equation 2 comes from equation 1 but I don't understand why equation 1 is correct even when the chemical equation is unbalanced ?

As I think equation 1 can be applied only when the chemical equation is balanced .
As here :

No. Of Atoms of "O" in reactant =2

No of Atoms of "O" in product = 1

Thus by equation 1 we can say that  $1=2$ .

And my book says POAC can be applied even when equation is not balanced . So here equation is not balanced and when I apply POAC I get $1=2$ . So what's is wrong with my reasoning?

As said by many others ,  The equation 1 given above is simply put very wrong  . As it can be applied only when a balanced chemical equation is given to us ,  otherwise we will reach at all sorts of contradictions as in the question above .

However many non- rigorous teachers , textbooks often write equation 1 in POAC (which is always true in a balanced chemical equation ) and then say now equation 2 is obtained from equation 1(which is partially incorrect) . It is true equation 2 comes naturally from equation 1 , but only for a balanced chemical equation.

So then how come we apply equation 2 in POAC in all sorts of question ? And still come up with the right answer ?

The answer is that equation 2 is true in itself always whether the chemical equation given ( in POAC ) is balanced or not . So it is not related to equation  1 for its validity .

How is equation 2 always valid ?

As pointed out by @iammax in the comments actually we (sort of )  essentially balanced the chemical equation given to us in the equation 2 itself .

For example in the $KClO_3$ question linked above , the moles of $O_2$ , moles of $KClO_3$ given to us are actually according to the stoichiometric coefficients , which enable us to use POAC without balancing the reaction (even though we have done that indirectly by using No. of moles of $O_2$ , $KClO_3$).

To conclude we can say after using No. Of moles of reactant , product we have indirectly balanced the chemical equation as the Ratio of their number of moles will be
Equal  to Ratio of their respective stoichiometric coefficients . After which we just use Conservation of Atoms( as the name says it) and multiply the number of moles of the molecule with the number of Atoms present  on which POAC is being applied ( For example we multiply no. Of moles of $O_2$ by 2 when we apply POAC on oxygen atom ).


How to measure a distance along the surface of a protein in a PDB I would like to measure the distance between two active sites in a protein complex (pdb), but I do not find a straightforward way to do it.

To be more precise I need the distance along the surface of the protein and not the linear distance that would go through the protein, and which the distance wizard will give.

I have access to Pymol and would prefer a solution using that software although I would consider other as well. Thanks.

I would use linear distances between a suitable amount of fixed points as an approximation. You cannot measure exact distances along a rough surface, as the distance will vary depending on minimum measuring size. This is known as the coastline length paradox in math. https://en.wikipedia.org/wiki/Coastline_paradox

In theory the length would approach infinity as the measuring size approaches 0.

In practice you can get around the issue if you use a step size compatible with the smallest stable feature.

A suitable measuring size would probaly be in the order of ~10^-11 meters. (1/10 the size of an carbon atom).


Does a catalyst increase atom economy? Somebody I know insists that the use of a catalyst increases the atom economy. They did chemistry at school and were told that a catalyst increases the atom economy. He pointed me to several past exams where this was written as a valid answer.

However, I am convinced that this is false. After all, atom economy is:

$$\text{atom economy} = \frac{\text{molecular mass of desired product}}{\text{molecular mass of all reactants}} \times 100\%$$

I feel this has nothing to do with a catalyst.

Am I missing something?

There are a couple of ways to think about this. First, considering your atom economy equation. Typically a catalyst is used in place of a stoichiometric reagent. A catalyst is (strictly speaking) not a reagent, so its effect on the atom economy is neutral. The alternative stoichiometric reagent is a reactant, so it will reduce the atom economy. This leads to the conclusion that a catalytic method offers a better atom economy than an alternative, non-catalytic method.

From a slightly different point of view, atom economy is one aspect of measuring the amount of waste that comes out of a reaction. A more relevant quantity is really the ratio of the mass of the desired product(s) to the total mass of all of the reaction inputs (including reactants, catalysts, solvents, purification materials, etc). Since catalysts can be used in smaller amounts than stoichiometric reagents, there will physically be less mass put into the process, which increases that ratio.


In the first example, the 4(2+2), if you were ABSOLUTELY going to adhere to significant figure rules, then 20 would be the correct answer, though I don't believe many people would fault you for saying 16. In your second case, 2.0(0.5+0.5), the appropriate answer here would indeed be 2.0.

In any case, significant figures are not really valuable when referring to purely mathematical equations like you did. We use them in order to avoid making errors when it comes to the accuracy of a result, in other words, significant figures are there so that your answer is not more specific than your measurement was. In mathematics, this is not an issue, as you are not making any measurements. So, while you COULD use significant figures in a situation like this, it really is not quite the appropriate situation in which to do so.


Current and reversible heat in battery reaction In the book Battery Technology Handbook by Kiehne [1] (Google Books) on page 21, equation (32):

The reversible heat effect per time unit can be related to current flow, because each multiple of the cell reaction requires the current amount $n\cdot F$:"
$$\frac{\mathrm dQ_\mathrm{rev}}{\mathrm dt} = \frac{Q_\mathrm{rev}}{nF}\cdot i \label{eqn:32}\tag{32}$$
with $n$: number of exchanged electrons; $F$: Faraday constant $(\pu{96485 A s/equivalent})$; $i$: current in $\pu{A}$.

Two questions:

How is the current related to $nF$? current and $nF$ have different dimensions?

How can one derive equation \eqref{eqn:32}?

References

Battery Technology Handbook, 2nd ed.; Kiehne, H. A., Ed.; Electrical engineering and electronics; Marcel Dekker: New York, 2003. ISBN 978-0-8247-4249-2.

Update in order to respond to comments
I know that $\Delta G$ is connected with a reversible process and in principle, your derivation makes sense. My problem lies in the formula:
$\Delta S = \frac{Q_{rev}}{T}$ is a basic thermodynamic relation and the Gibbs Energy is defined as $\Delta G = \Delta H - T\Delta S$. So, following your derivation $\Delta H$ would be zero. Since this problem is connected with my other question, which is linked in the comments, I would expect two different heat terms, a reversible one and a irreversible one. And from the formulas there, I expect $Q_{rev}$ to be connected only with entropy.
Therefore, $Q_{rev} = -T\frac{\text{d}G}{\text{d}T}$ rather than $Q_{rev} = -\Delta G$ as the combination of your equations (2) and (4) implies.
I is quite hard to explain my problem...
There is another way to get to the equation in question, but this seems to be too simple:
Since, $\frac{\Delta G}{nF} = \text{Voltage}$, $\frac{T\Delta S}{nF} = \text{formal Voltage}$. And $\textit{Voltage}\cdot I = P = \dot{Q} = \frac{\text{d}Q}{\text{d}t}$, equation (32) seems perfectly valid, but this is a kind of a weak derivation.

How is the current related to nF? current and nF have different dimensions?

$nF$ is the charge $C$ transferred during the reaction, while current $i$ is the rate of charge transfer ($dC/dt$).

The following is a derivation, now edited to be more rigorous. The equation you provide is an expression for electrical Joule heating, which follows in the steady-state from the expression for the power (rate of work) generated by the battery. From Ohm's law

$P=\left(\frac{dw_{rev}}{dt}\right)=-i^2R=-i E_0 \tag{1}$

where $E_0$ is the (open-circuit) electric potential, $i$ the current, $w_{rev}$ is the reversible electrical work. Note that $E_0$ is the potential when reversible work is done.

Now assume a steady state, with constant P and T constraints, with work done and heat generated cancelling (so that the internal energy $U = constant$), that is

$Q_{rev} = -w_{rev} \tag{2}$

and therefore also power and rate of heat dissipation equal:

$\left(\frac{dQ_{rev}}{dt}\right)=-\left(\frac{dw_{rev}}{dt}\right)=i E_0 \tag{3}$

Now according to the Nernst equation,

$\Delta G=w_{rev} =-nFE_0 \tag{4}$

In the steady-state the dissipated heat is equal to the electrical work, so that, combining (2) and (4), we have that

$\frac{Q_{rev}}{nF}=-\frac{w_{rev}}{nF}=E_0 \tag{5}$

which leads, combined with (3), to the equation in the book:

$\left(\frac{dQ_{rev}}{dt}\right)=i \frac{Q_{rev}}{nF} \tag{6}$

Aside

Note I apply the opposite sign convention for work from that in the OP (apologies) - in the convention I use work is positive when performed on the system (charging a battery is positive work, discharging negative work). This does not affect the result of the derivation, since I apply the same sign convention for heat (for an exothermal process heat is negative).

On $\Delta H$ and an alternative (longer) derivation

For a process at constant p, it is common to encounter the expression

$$\Delta H = Q_p$$

However if there is non-PV work, the more general form of this equation is

$$\Delta H = Q_p + w_{non-pV}$$

This expression is general and applies when the process is carried out either reversibly or irreversibly. In differential form,

$$d\Delta H = dQ_p + dw_{non-pV} \tag{a1}$$

Note also that in the preceding derivation non-pV work is electrical.

When the process is carried out reversibly a maximum amount of work is done and

$$w_{non-pV,rev} =\Delta G = \Delta H - T \Delta S = \Delta H - Q_{rev}$$

This leads to the following expression at constant T, equal to that in the linked problem (but note the different work sign convention here):

$$ d\Delta H = dQ_{rev} + dw_{rev} = d\Delta G + Td\Delta S \tag{a2}$$

where I dropped the "non-pV" subscript since the work is assumed to be electrical.

Since H is a state function it must be equal for reversible and irreversible processes, and we can equate (a1) and (a2), leading to the following general expression:

$$ d\Delta H = dQ_p + dw_{ele}=  d\Delta G + Td\Delta S $$

which leads finally to

$$ dQ_p = - dw_{ele} +  d\Delta G + Td\Delta S  \tag{a3}$$

Taking the time derivative of this equation results in the final expression in the linked problem (making sure to account for differences in the work sign convention):

$$\dot{Q} = \dot{Q}_\text{rev} + \dot{Q}_\text{irrev} = IT\,\frac{\mathrm dE_0}{\mathrm dT} + I(E-E_0) \tag{a4}$$

To obtain the equation in this problem it is only necessary to apply the reversibility condition starting from either (a3) or (a4). Starting from (a3),

$$ dQ_{rev} = - dw_{ele,rev} +  d\Delta G + Td\Delta S  $$

But $ dw_{ele,rev} =  d\Delta G  $ which leads to (somewhat trivially)

$$ dQ_{rev} = Td\Delta S  $$

Taking the time derivative and inserting the Nernst expression for $\Delta S$ (twice!)  gives

$$\dot{Q}_\text{rev} = \left(\frac{dQ_{rev}}{dt}\right) = IT\,\frac{\mathrm dE_0}{\mathrm dT} = I\frac{Q_{rev}}{nF}$$

which is the desired expression. The same result can be arrived at by applying the reversibility condition to equation (a4).


What happens to a molecule when it is unstable? I'm just starting OChem and I see that for a SN1 reaction, the rate of the reaction changes with the number of alkyl groups attached to the carbocation that is formed. The textbook (Klein) explains that this is due to the electron-donating nature of alkyl groups which helps to stabilize the carbocation. I was wondering, why doesn't instability help to increase the rate of reaction? Wouldn't the instability of a primary carbocation give it more energy to attract a negative substituent given that it is now more positively charged than a tertiary carbocation?

Thanks,

Reference: Klein, David. Organic Chemistry as a Second Language, First Semester.

Think of a reaction-energy diagram.  Specifically, think of a two-step reaction energy diagram linked like this one:

.

The carbocation is the intermediate $\mathbf{I}$.  The relative energy for alkyl-substituted carbocations will be lower.  This also practically means that $\mathrm{TS1}$ will have a lower relative energy as well.  The fact that $\mathrm{TS1}$ and $\mathbf{I}$ are of lower energy for a substituted carbocation (vs. a primary or methyl carbocations) means that it will be formed more often under the quasi-equilibrium assumption of transition state theory.  Basically, if the barrier is lower, then then the intermediate can form more often.  And thus the reaction can happen more often.

  I was wondering, why doesn't instability help to increase the rate of reaction? Wouldn't the instability of a primary carbocation give it more energy to attract a negative substituent given that it is now more positively charged than a tertiary carbocation?

I think you are correct as far as your thought goes.  If you could somehow instantaneously create a beaker full of primary carbocations, they might well react more quickly than other carbocations.  (And that's not only via the SN1 reaction we are considering but also via other pathways such as elimination as well.)  But instantaneously creating a beaker full of primary carbocations is way more difficult than creating a corresponding beaker full of secondary or tertiary carbocations.  So (referring to the diagram again) yes, iff you have lots of $\mathbf{I}$, for $\mathbf{I}=$ primary carbocations, reactions would probably be quicker than the case where $\mathbf{I}=$ higher carbocations.  But in reality in the primary carbocation case essentially no $\mathbf{I}$ is ever generated.  So the overall reaction rate is slower.


Particle suspension and re-suspension of nano-particles I am doing a senior project that synthesizes PbS quantum dots and our particles keep crashing out of solution. I was wondering if anyone can give me some advice or information on stabilizing the colloidal solution. The solvent is DI water. I am interested in the chemistry behind stabilization. like what needs to be added to lower the surface energies of particles? why it is added? and how it stabilizes the colloids?

Try mercaptoacetate surface treatment to put a negative charge on their surfaces for repulsion around pH 8 or 9 basic solution.  Uncoordinated mercaptoacetate is oxidatively dimerized to the disulfide by air (oxygen).  The drug S-mercaptocysteine as its salt (basic pH) should be more aggressive toward binding.  3-Mercapto-1-propanesulfonic acid will bind at lower pH.

The dose must be low enough that you do not disaggregate your particles, especially over time.  You can permeable membrane desalt an excess.


Why does the gauche conformation of ethane-1,2-diol show hydrogen bonding? Why does the gauche conformation of ethane-1,2-diol show hydrogen bonding even though there is no 6-membered ring system? If we don't consider one of the $\ce{H}$ (which is what is usually done), then there is only a 5-membered ring formed. How then does hydrogen bonding take place?

(Also, I feel that there must be some reason apart from 6-member ring for hydrogen bonding which I'm not aware of)

Cyclopentane has a relatively low strain energy, something on the order of 6.5 kcal/mol more than cyclohexane (reference).  In the case of ethane-1,2-diol, the strain energy in the 5-membered ring formed by hydrogen bonding will be even less because some of the $\ce{H-C-C-H}$ and $\ce{C-C-C-C}$ interactions that produce the strain in cyclopentane are no longer present; the oxygens don't have the hydrogens attached like a methylene group does and the hydrogen used in the hydrogen bond is also much smaller than a methylene group.
A typical $\ce{O-H..H}$ hydrogen bond can stabilize a system by ~7 kcal/mol (reference).  So to form a 5-membered ring in ethane-1,2-diol and create a hydrogen bond it will cost something less than 6.5 kcal/mol to overcome the strain energy in the ring, but we get 7 kcal/mol back due to hydrogen bond formation - a net stabilization.
$%edit$


Why does electronegativity increase as effective nuclear charge increases? I know that electronegativity is the ability to attract shared electrons and that effective nuclear charge is the pull of the nucleus on outer electrons based on my notes.

But I'm not really sure how to explain the connection. Please help.

The effective nuclear charge can be thought of the charge of the nucleus minus the charge of the core electrons. For an element such as fluorine, the nuclear charge is $+9$ and the core electrons have a charge of $-2$ so the effective nuclear charge is $+7$. Similarly for carbon it would be $+6 - 2 = + 4$.

Now let’s assume a $\ce{C-F}$ bond. There are two electrons in this bond, each charged $-1$. On the left, from the carbon atom, they ‘see’ a charge of $+4$; on their right, from the fluorine atom, they ‘see’ s charge of $+7$.

Electrostatic interactions are greater at shorter distance and at greater charge separation, so the interaction (i.e. force) the bonding electrons experience from fluorine’s side is almost double that from carbon’s side. (The carbon and fluorine atoms can be considered to be roughly of the same size.) It should now be clear why the electrons are drawn to the fluorine.

Electronegativity is just a name humans gave the observed effect of ‘this atom can pull electrons towards it better than others’. It takes a look at the result of the above process and rationalises it. The effective nuclear charge approach is the underlying physical force.

Note: This explanation is extremely simplified, does not take any quantum effects or the like into consideration and should thus only be used at introductory levels.


Are bond angles constant? I'd imagine the answer to the question posed in the title is no - bond angles are not constant. Covalent bonds are not infinitely rigid; hence IR as a spectroscopic technique.

If so, how can we accurately determine bond angles? What margin of error might we expect from an experimentally measured bond angle?

No, bond angles are not constant. For example, the low-energy IR absorption of water (at $1595 \:\mathrm{cm^{-1}}$) is from the H-O-H bend

There are a few techniques to measure bond angles. For polar compounds, microwave rotational spectroscopy can often be used on small molecules in the gas phase.

I'd guess though, that most bond angles are measured from crystal structures using X-ray diffraction.

Keep in mind that spectroscopic techniques always have a timescale associated with them. So any experimental measurement is time-averaged. In the case of bond angles, while the angle will change, the average value is the equilibrium - at some points, the vibration will create a larger bond angle, and at some points in time, the bond angle will be smaller.


Reactions between enantiomers in a racemic mixture As a follow up to this question I started thinking about another question. During racemization, enantiomers are mixed in a equal ratios to form an optically inactive mixture.

What type of reactions can take place between the enantiomers and how would this affect the mixture, if at all?

If a chiral molecule (an enantiomer) can react with itself, dimerize for example, to form a new compound, then formation of this new compound would contaminate the racemic mixture.

(image source)
Perhaps more importantly is interaction of the racemic mixture with itself in the solid state.  Just as the interaction between a right hand and left hand is different than the interaction between two right hands, the interaction between the R form and S form of the enantiomer will be different than the interaction between two R (or two S) forms.
In liquids and gases, molecular encounters are fleeting and cannot be preserved or "locked in" for significant periods of time as they can be in a solid. In other words, any favorable, lower energy molecular encounters are quickly broken apart by the energetically larger effects of thermal motion.  However, when the racemic mixture is in a solid phase, if the interaction between R and R (or S and S) is lower in energy then the interaction between R and S, then the R and S enantiomers will tend to form separate crystals and a mixture of pure R and pure S crystals will result.  Such mixtures are often termed "conglomerates".  On the other hand if the interaction between R and S is more favorable (lower in energy), then a racemic compound or racemate is formed where each crystal contains R-S pairs.
A conglomerate, being a simple mixture, will show a melting point depression compared to the pure enantiomer.  A racemic compound may have a melting point higher or lower than the pure enantiomer.
Further reading on racemic conglomerates and racemic compounds can be found here and here.


Why are there only less than 120 elements out there ? Why stop at 118 , Why aren't any more stable elements possible ?

They haven't stopped at 118. The researchers at CERN and Dubna are looking into heavier elements. Such as element 120 and 122 etc. It's all about the "island of stability". https://en.wikipedia.org/wiki/Island_of_stability
Where the right amount of protons and neutrons increase the stability of the nucleus. It's going to be very difficult to get there due to the increased mass of the target nuclei and increased mass of colliding nuclei. So really the only limit to the elements scientists can create, is the tools they use


acid alpha glucosidase vs glycogen phosphorylase What’s the difference between acid alpha glucosidase and glycogen phosphorylase? Both break down the same bonds in glycogen so I thought for a long time that they are exact same enzyme until I struggled through glycogen storage disease types and found out that each one is different than the other and that deficiency of either one causes a different type of GSD.

Acid alpha glucosidase  (sometimes called α-1,4-glucosidase or acid maltase) is not among those enzymes involved the pathways of glycogen synthesis and breakdown. It however occurs in lysosomes,where it functions to hydrolyze the disaccharide maltose  and linear oligosaccharides, as well as the outer branches of glycogen,thereby yielding free glucose.

This "second pathway" of glycogen metabolism is not quantitatively important as it concerns a minor fraction of glycogen. The reason that lysosomes normally take up and degrade glycogen granules is unknown perhaps for the disposal of structurally aberrant glycogen molecules.

Glycogen phosphorylase (a dimeric enzyme)
catalyzes glycogen phosphorolysis (bond cleavage by the
substitution of a phosphate group) to yield glucose-1-
phosphate (G1P).

$\ce{Glycogen + Pi <=>  glycogen + G1P}$

This enzyme will only release a glucose unit that is at least
five units from a branch point. (Note: I am not going to describe the reaction mechanism of glycogen phosphorylase here is much more complex than it sounds)

Having briefly described the enzymes, the most important thing to note is that:

structually these two enzymes are different,
active in the different locations
the mechanism of glycogen breakdown are different (i.e phosphorylation, hydrolysis)
different regulation mechanisms


Is tetraammine dichloro cobalt (III) an optically active compound? Is
$$ \text{Co}[{(\text{N}{\text{H}}_{3})}_{4}{\text{Cl}}_{2}]$$
optically active? If so, write its stereo-isomers.

I know that for a compound to be optically active, it should rotate the plane of Plane Polarized Light, but is there a way to do this without experiment?
Thanks!

$\ce{Co[(NH3)_4Cl2]}$ can exist as cis and trans isomers as shown in the following figure.

image source

If a molecule contains a plane of symmetry it cannot be optically active or chiral.

In the cis isomer there is a plane of symmetry containing the $\ce{Cl-Co-Cl}$ atoms, therefore this isomer cannot be optically active.  With the trans isomer there are several planes of symmetry (one contains the $\ce{Co}$ atom and the four $\ce{NH3}$ groups, and there are four more planes of symmetry containing the $\ce{Cl-Co-Cl}$ atoms and either containing or bisecting the pairs of amino groups), therefore the trans isomer is also optically inactive.


How to separate sodium iodide dissolved in water? Am fairly new to chemistry and apologize if the question and the description are too vague.If you think I lack basic background knowledge to deal with something like this, please point me to a resource, a book or anything that can help me in this regard. So in case this question is off-topic, please suggest where I should begin- at least starting point for my research will be helpful. However, in that case, if possible, please be specific, cause its too much of an effort to read entire chemistry textbooks to find out the answer to this one question.

Would like to know how to seperate NaI dissolved in water. Please suggest a way to do so without heating and evaporation. Like, what compound should I add to the solution so that the NaI can react with that resulting in solid products, maybe in the form of lumps/signicantly sized suspended particles that can be easily filtered out.

Additionally I would like to know how to find out, given a solvent (or any liquid compound), whether NaI dissolves in it or not. For eg: say that I have methanol, will NaI dissolve in it? For methanol i know the answer and its a yes. But how do i test solubility for any other compound. Of course chemists should not have to carry out practical experiments to find out this(it must probably be part of a known set of facts(about a particular compound) to chemists or they will be able to deduce it from some other known facts. How will a chemist answer this question?)

And finally, once I have any solution with NaI dissolved, how to separate it without evaporation(but by some chemical reaction). For eg: how to separate NaI from methanol? how to figure out the reactants I'll need in case the solvent is known?

I summarize some answers given in comments.

Evaporation is really one of the standard methods in any combination of solid (cristalline, amorph, polar, unpolar) and solvent (polar, unpolar, flammable...).
But you must not think of it as the pot heated on a flame.
In modern synthesis you often already work with a nitrogen or noble gas atmosphere in a closed apparature and use varying pressure for controlled destillation.
So it is no problem to just reduce the pressure and evaporate your (organic) solvent.
In your special case you even can think of a pot heated on a flame, there is no risk and disadvantage. It is simply an effective method.

Do you know the concept of polarity? A good rule of thumb is that polar substances mix with polar ones but not with unpolar ones.
So water dissolves salt but not paraffin wax.
Benzene on the other side dissolves parrafin wax but no salt and water and benzene do not mix with themselves.

=================Edit (Addition of Information, because of question in comment)

So let's see how to regain the solvent.
First of all there is not only one boiling temperature. The boiling temperature is always a function of pressure. Approximately it can be described by the Clausius-Clapeyron-Equation.
$$
T_2(p_2)=(\ln(\frac{p_1}{p_2})k+\frac{1}{T_1})^{-1}
$$
Here $k$ is some constant depending on the specific solvent and $T_1$ and $p_1$ is a corresponding pair of known boiling temperature and pressure. For example $100^\circ$C and 1bar for water.
In general the boiling temperature goes down when pressure goes down. (Water boils faster in the Himalaya)

So if you have a typical organic solvent like aceton with a boiling temperature of $56^\circ$C under 1bar it evaporates already at $20^\circ$C when applying a pressure of e.g. 200mbar. (it is not the exact number)
But it also condenses again when cooling down. Using this you can make it liquid again and regain it. Compare this image of a typical distill.
http://www.ssc.education.ed.ac.uk/BSL/pictures/distill.jpg

Summarizing: you use a pump to get into your desired range of boiling temperature and cool the gas in the condenser again to a temperature below the boiling temperature. Depending on your system you have boiling temperatures below $0^\circ$C and cool with liquid nitrogen in the condenser or normal temperature range and cool with water or...


How to estimate HCN content in a volatile mixture? How can you accurately qualitatively and quantitatively determine the HCN in a volatile product from water?

From a quick search of the literature, it seems that as you correctly thought, gas chromatography mass spectrography is a method that can be used to quantitate small amounts of cyanide in vapour.

While I am certainly no expert on this field or the precise method that you would use, this paper, which describes quantitation of cyanide and other chemicals in tobacco smoke may be of interest: A simple GC–MS technique for the analysis of vapor phase mainstream cigarette smoke


Finding sulfate contamination in crystallized silver nitrate What is the best way to check for a small amount of sulfate contamination (if any) in silver nitrate? Usually, I would use dilute barium chloride to precipitate out the resulting insoluble white barium sulfate in an unknown aqueous solution that contained sulfates. I tried this and got a white precipitate, but now I believe the precipitate I'm getting is insoluble silver chloride.

Is there a better/different method to check for sulfates in silver nitrate? I was researching for a better solvent to use, but haven't been able to find any conclusive answers so far.

Barium nitrate has a water solubility of $\pu{10.5g/100mL}$ at $\pu{25^oC}$. It isn't specified in the question what concentration of sulfate you suspect might be present, but given that you are trying to check for sulfate by precipitating with barium, barium nitrate should be the way to go.


Red glow in the dark pigments I want to do some spectrometry experiments with glow in the dark paints. The red paints I've read product details for all seem to require a uv light to charge. Are there any red glow in the dark paints that can be charged with visible light?

I was also wondering why most red phosphorescent pigments don't have the ability to glow as long as green pigments.

The wikipedia page on Phosphoresence has a red pigment. Calcium Sulfide. I can imagine that the lower energy (higher wavelength, lower energy) is more stable because the triplet is not as high in state energy as lower wavelength emitters. Because of the stability the time to emit is longer and the intensity lower.


How does bombarding a molecule with electrons cause ionization? I am a chemistry student, and my current are of study is mass spectrometry.

In the ionization process, atoms are bombarded with electrons, but how does this work? I assume the flow of electrons attract the electrons of the atom, thereby removing them, but where do the atom's electrons go?

This type of ionization is known as electron ionization (there are many types of ionization methods used for MS). One possible reaction for EI is:
$$\ce{M +e- ->M^{•+} +2e-}$$ source

If the incoming electron has the correct energy to interact and passes close enough to the molecule, it's electric field can impart enough energy to one of the molecule's electrons for it to be ejected. What happens to the electrons after that isn't particularly important as far as ionization is concerned, but in most EI sources, electrons are emitted perpendicular to the axis of the sample source and are collected on the other side by an electrode, preventing charging of the source chamber and focusing the emitted electrons into a beam.


Why can't this be another structural isomer of ethanol? Are the two structures depicting the same molecule? Why? Why not?

Try to imagine it in a 3D fashion. You can look it at molcalc.org. The molecule rotates around the $\ce{C-C}$ bond converting one structure into the other. These rotations are very fast.

Another point is, that the representations you chose are 2D, so the could still represent the same molecule:


Would gas come out of solution in an infinite volume? Say I have an infinite volume of water with some level of dissolved air in it. I lower the pressure to below the gas' saturation pressure. Does the air come out of solution if there are no interfaces (nuclei) for it to begin? If so, what is the driving force behind the formation of the bubbles? Are there any estimations of the time that it would take to form a bubble of a given size, or an estimation of the equilibrium size of said bubble? Presumably it would be a balance between the surface tension, radius and the water pressure (like $P + 2S/R$)...

To give some motivation, I have a cavitation tunnel where I can apply a pressure field to a large volume of water. In the literature, people speak of homogeneous nucleation where fluids below saturation pressure start to release their gas as bubbles, and I'm struggling to understand how that occurs.

According to this abstract from a paper titled "Why Is It Much Easier To Nucleate Gas Bubbles than Theory Predicts?":

  Bubble nucleation from supersaturated gas solutions generally takes
  place at much lower supersaturations than are expected from the
  theory. Furthermore, the same theory predicts that the threshold
  concentration of gas needed to cause nucleation should be essentially
  independent of the gas species used, a finding contradicted by
  experiment. There are two general explanations:  first, that the
  theory is wrong, or second, that there is a previously unidentified
  factor which is influencing the results of the experiments. Given the
  success of the fundamental theory in other areas, the second
  explanation is preferred. The previously unrecognized factor is
  identified here as being the surface activity of the gases which form
  the bubbles.

Although this is discussing conditions of low supersaturation and not subsaturation, I assume the same concepts apply, though it would cost me $40 to get the full article text and find out for sure.

I'm pretty sure what's happening here is that through the random motion of dissolved gas molecules, occasionally enough of them adhere to one another to form the nucleus of a bubble.  Particularly at very small radii, the high surface tension of water dominates otherwise destructive forces, and a micro-bubble begins to form.  But by the definition of saturation with respect to bubble formation, it would seem that you cannot have "macro-scale" bubble formation and growth persisting under conditions of subsaturation as equilibrium is reached.

If I can paraphrase your question as:

  "If I lower the gas pressure to below its saturation pressure, does
  the gas come out of solution if there are no interfaces (nuclei) for
  bubbles to form?"

then the answer is yes, regardless of the presence or absence of bubbles.  The dissolved gas molecules will partition at the gas-liquid interface until equilibrium concentrations are achieved in each phase.  If some gas bubble began to form, it would quickly redissolve under conditions of subsaturation with respect to the gas solubility.

Unless I'm missing something about the importance of bubbles in your question, this would pretty clearly be the bottom line answer.  Please comment if I've missed the crux of your question.


Open Database of Phase Diagrams? Is there some kind of database that has information about phase diagrams of different substances? I'm mostly looking for phase diagrams of elements (silicon, iron, hydrogen, etc.) because I'm trying to write a planetary generator, but I cannot find any information about the mathematics of a phase diagram.

Better yet, is there some way to simply approximate the phase diagrams of different compounds? I'm looking for simulation complexity and variation more than I want realism.

I believe you can compare many of the models with a search engine.
http://www1.asminternational.org/asmenterprise/apd/AdvancedSearchAPD.aspx


What is meant by forward reaction? While reading my chemistry textbook I came across the phrase:

  forward reaction is favoured by decrease in temperature in exothermic reactions.

I was tripped off by the term 'forward reaction', can someone explain in layman terms what is meant by this?

This phrase comes from Le Châtelier's Principle. Given a reversible reversible reaction, you may (arbitrarily) choose which chemicals to consider the reactants/products, since the reaction goes from both the reactants to the products and from the products to the reactants. Simply put, the forward reaction starts with what you consider to be the reactants. The reverse reaction starts with the products and ends with the reactants. If you flip the equation these names change.


Why is CO practically nonpolar? This question was in my book. According to me CO should be polar as it should have a dipole moment.

But I found that the $\sigma$-electron drift from C to O is almost nullified by the $\pi$-electron drift from O to C. What is about the $\pi$-electron drift towards the Carbon atom?

In Google it is given CO is polar, but my book says it is practically nonpolar ... now I am confused.

Preliminaries: I am using the wrong (but still common) notation of the dipole moment. Please see the question about the direction of the dipole moment.

The reason why carbon monoxide is often referred to a being practically nonpolar is because of its very small dipole moment.
\begin{align}
\ce{{}^{\ominus}\!:C#O:^{\oplus}} &&
\text{Dipole:}~|\mathbf{q}|=0.11~\mathrm{D} &&
\text{Direction:}~\longleftarrow\hspace{-1ex}+
\end{align}
Unfortunately the explanation of the bonding in carbon monoxide is far from simple, and it is perfectly understandable that you are confused. The dipole moment can be very well explained with molecular orbital theory, which I have already written about.
I will include a brief explanation at this point as I saw that the explanation on Wikipedia is from my point of view incomplete (not to say wrong).
Let's first state the obvious: Oxygen is more electro-negative than carbon. Therefore one can expect that all bonding orbitals are polarised towards the oxygen. That is of course only true when we are staying in the Lewis bonding picture. Based on this one would expect that the lone pairs cancel each other and since all orbitals of the triple bond are polarised to the oxygen, the dipole moment should also point in that direction.
This is obviously wrong, since the experiment tells us something different. There are ten valence electron in this molecule. They all occupy orbitals with bonding character, i.e. there is no nodal plane perpendicular to the bonding axis between the elements. The low lying orbitals are all polarised towards oxygen, because of electronegativity. The HOMO on the other hand is strongly polarised towards the carbon and is very large. In addition the nuclei contribute a dipole moment because of their asymmetry of charge.  With less positive charge on the carbon nucleus than on the oxygen nucleus, the dipole component of the nuclear charge contribution also has its negative end on carbon.  The combination of HOMO and nuclear dipoles in the reverse direction is larger than the combination of all the other occupied orbitals.

It is important to understand that it is often not possible to know about the properties of a compound simply by looking at a structure on paper. Very often the simple models we love and use break down easily for small molecules. A bonding situation has to be explained as is and not in terms of what would have been when we look at individual atoms and how does it change.


Is the 5d orbital involved in triiodide ion? Is the $5\mathrm{d}$ orbital involved in the triiodide ion, $\ce{I3-}$?
There are $5$ electron pairs around the central iodine.

(Almost) all of the hypervalent compounds involve the most electronegative elements: $\ce{F}$, $\ce{O}$, $\ce{Cl}$.

Also, the bonds involving the electronegative elements are often almost ionic:

  For sulfuric acid, computational analysis (with natural bond orbitals) confirms a clear positive charge on sulfur (theoretically $+2.45$) and a low $3\mathrm{d}$ occupancy.

So, we can say that $3\mathrm{d}$ is (almost) not involved in the central sulfur atom of sulfuric acid.

Is there any experimental data to show if $5\mathrm{d}$ is involved in the triiodide ion?

Professor Klüfers presents a structure of the $\ce{I3-}$ ion in his general and inorganic chemistry introductory course at the LMU Munich. The ion is linear and the $\ce{I\bond{...}I}$ distances are identical, measured experimentally to be $293~\mathrm{pm}$.

In neutral iodine, the $\ce{I-I}$ distance is measured to be $272~\mathrm{pm}$.

We see a notable increase in bond length which indicates a decrease in bond order. A reduction of bond order — say from a bond order of $1$ (single bond) to one of $0.5$ (four-electron three-centre bond) — means that we cannot add an additional bonding orbital to the mixture. Otherwise we would have something along the lines of two single bonds (bond order $1$), which is how the ‘octet expansion’ traditionally was explained.

The reduction in bond order is best explained — as mentioned — by a four-electron three-centre bond and thus without d-orbital participation.


In High Performance Liquid Chromatography, why are ratios of solvents used? How do they allow for better separation? For example, why isn't 100% hexane used?

Not sure if I can fulfill Ed's "real answer". I like the word you used for tuning solvent polarity- a knob. Modern students may understand this better.

There is no theoretical restriction in chromatography to use multiple solvents or a single solvent. For example in gas chromatography, you always use a pure gas. The reason for using a mixture of solvents in liquid chromatography originates from the concept of "general elution problem of chromatography" which basically says that there is no single column or a single mobile phase which can separate everything. Not only separation scientists use a single mixture composition of mobile phase (isocratic), they can also continuously change the mobile phase composition as a function of time. This is called gradient elution.

When you do real separations, you have no idea beforehand about the solubility of analytes injected into the column. If these injected solutes are not soluble in a pure solvent they will permanently absorb on the column. A real sample may contain 30-40 compounds. Say a petroleum sample, literally have thousands and thousands of different molecules. Now think yourself: will they all dissolve in 100% hexane?


Getting a cleaner GC Peak Resolution I have a GC with a peak pair I'm interested in separating but are not getting resolved cleanly at the moment. Any tips on what temperature programming tweaks might help me?

The problem pair elutes at about 3.3 mins. The temp program was 120 C for 2 min followed by a 8 C/min ramp to 180 C & a final hold of 5 min.

Changing the column, detector or carrier gas is not an option. I want to just get the best I can by temp. programming optimization.

I've tried isothermal runs at 120 & 100 C but not much help. The peaks still come unseparated.

Since it is just this two peaks that interest me mostly, theoretically a ramp is superfluous isn't it? Just figuring out the right isothermal run ought to give me the separation of interest?

Edit:

The column is an OV-351 packed col. 2 m long and 1/8" in diameter. N2 carrier gas with FID. The molecules are:

I just thought that the rearrangement might be occurring during the analysis I.e. The thermal ramp, implying a lower temperature would be better for you. The bp of each compound is too close to separate purely in that way.


Charge on Nitrogen in NCl3 What is the charge on $\ce{N}$ in $\ce{NCl3}$? I want to know if it is positive or negative. I feel it should be positive but I am not sure.

First of all charge of atom is not well defined (maybe except AiM). Formal charge of both is zero, but that does not help us to understand chemistry. Above mentioned reaction shows that nitrogen is more negative, and if you do simple QM computation both Mulliken and NBO gives ~-.44 for N and ~+.147 for Cl. If you want to have a bit feeling about it, think that way: both have similar electronegativity but Cl is bigger so it's easier to polarize.


Could a safe water ocean exist on a world with predominantly chlorine atmosphere? I am researching for a Si-Fi book.  Would a water ocean on a world with a predominantly chlorine ($\ce{Cl2}$) atmosphere be safe to touch?  Would the chlorine over time saturate the water creating a "laundry bleach' solution, or would it create a ocean of hydrocloric acid ($\ce{HCl}$) and hypoclorous acid ($\ce{HOCl}$), or would the chlorine just not react with the water that much?

Chlorine gas is dissolved in drinking water at low concentrations for disinfection purposes.  Chlorine gas has also be dissolved in somewhat higher concentrations in swimming pools.

The amount of chlorine dissolved will strongly depend upon pressure.  If the atmospheric pressure on the planet is low the concentration of chlorine gas in the water will be low.

Yes, most of the dissolved chlorine will reversibly react to form HCl and HOCl.

For quantitative information about the equilibrium constant and solubility as a function of pressure and temperature see Solubility of Chlorine in Water Industrial and Engineering Chemistry vol. 33, pages 741–744.


Why does precipitate always form so naturally? Whether be it from a textbook or teacher, precipitation reactions always occur by rearranging the elements:

For example

But then, aren't there other factors you have to bare in mind when predicting such reactions? Like electronegativity: obviously the potassium ion is much more electronegative than Silver ion, so it wouldn't be that easy for potassium to split up with iodine.

The ions combine while in solution, and since the resulting compound is not soluble, it precipitates. This happens so quickly because as the precipitate forms, it takes those individual ions out of solution, thus pulling the reaction toward the products side, according to Le Chatelier's principle.


Why does water spills on our shirt change its colour? While drinking water, some of it spilled on my shirt and as usual, as has happened to many of us, the colour darkened. So why does this happen? Is it due to some chemical effect or physical one?

This is not so much a change (physical and especially not chemical), it is more about perception of the diffuse reflection and refraction of light. According to Australia's ABC Science page Darker when wet:

  When your clothes are wet, all those air gaps (between the fibres) get filled with water. So when light hits a wet patch it's moving in and out of water and fibres, not air and fibres. The refractive index of fabric is a lot closer to that of water than it is to air, so the light doesn't change speed quite as dramatically going between water and fibres. And that means it bends at a smaller angle when it goes from one to the other.

This is summarised in the diagram below:

The information on the linked page is from Professor David Jamieson from the School of Physics at The University of Melbourne, Australia.

Another explanation according to Dr. Christopher Baird of the website Science Questions with Surprising Answers is that we perceive a darkening due to the fabric becoming more transparent - an explanation from the website:

  Spill some water on your white shirt and the situation changes. In the
  spot that is wet, the water fills all the little pockets in the fabric
  that used to be filled with air. In terms of its interaction with
  light, water behaves very similar to fiber material. Adding water to
  fabric effectively removes the multiple reflecting surfaces (at least
  as far as the light is concerned) and turns it back optically to a
  solid chunk of material, which is transparent.

Essentially, what we are perceiving when the fabric appears to darken, is due to the colour of what is underneath - view the cloth with a light source behind it, it appears 'lighter', due to being more transparent.


What compounds are produced when you heat CuO and sugar? I'm wondering on what compounds are produced when you heat CuO and sugar?

If there is enough copper oxide, all you obtain is metallic copper, $\ce{CO2}$ and water. This mixture ($\ce{CuO}$ + organic substance) was the standard way of analyzing any new organic substance in the $19$th century. The formula of the compound was derived from the weight of $\ce{CO2}$ and $\ce{H2O}$ collected. This is is still a standard question to be solved in high school chemistry courses all over the world.


My sample of gallium does not melt I held gallium under a heat lamp. It didn't melt. I didn't even find a thin film of liquid gallium. Can anyone explain why? Does gallium rust?

Having recently had an opportunity to play around with some gallium of my own, I have a few thoughts.

It's possible you have a different metal than what you thought you had, but frankly, I don't think that's very likely.

It takes a surprising amount of absorbed heat to melt gallium. Putting a sizable chunk of gallium in $35~^\circ\mathrm C$ water doesn't cause it to melt immediately. Even boiling water doesn't melt it instantly. Because gallium is a metal, it conducts heat fairly well, so when heated gently, it won't melt until the temperature of the whole piece gets close to the melting point.

As mentioned by user5134, a shiny metal like gallium isn't going to pick up heat very quickly from a radiative source like a heat lamp. If you wait long enough, it'll definitely melt. If you want to melt the gallium relatively quickly, then you'll want to put it into physical contact with something warm. The standard method for melting solid gallium in a container is to just immerse the closed container in hot water (not boiling water, especially if the container is glass).


Is 3,5-diethyltoluene an accepted IUPAC name? My book says that

  Monocyclic substituted hydrocarbons are named as derivatives of benzene or compounds like cumene, toluene, xylene, etc. However, if the substituent introduced into such a compound is identical with the substituent already present in the compound, then the substituted compound is named as tge derivatives of benzene.

But I guess that names like toluene have stayed due to historic reasons and are not accepted in IUPAC nomenclature. Is my book correct?

Names like toluene are "retained names".

In chemistry, a retained name is a name for a chemical compound that is recommended for use by a system of chemical nomenclature (for example, IUPAC nomenclature), but that is not fully systematic. - Wikipedia
‘Retained names’ are traditional or common well-established names that may be used either as preferred IUPAC name or as an allowed alternatives in general nomenclature, for example, naphthalene, benzoic acid and pyridine. IUPAC's Blue book,  P-12.3

Later in the same draft, we read

Toluene, xylene and mesitylene are specific parent hydrides that are composed of two components, one cyclic and the other acyclic and saturated. These names are retained due to a long and well established tradition. Toluene and xylene are preferred IUPAC names, but are not substitutable; toluene is substitutable under certain conditions for general nomenclature. P-22.1.3
Furthermore, anisole and toluene cannot be substituted by other
methoxy or methyl groups, respectively. P-46.2.3.2

Conclusion:

Names like toluene, anisole and xylene are retained names. They are preserved for historical reasons; however, their unsubstituted forms are PINs (Preferred IUPAC names).
The name 3,5-diethyl toluene isn't acceptable according to the systematic IUPAC nomenclature.
If you want to name the compound below, it's called 1,3-diethyl-5-methylbenzene.

Source


Is there any reason to fear personal exposure to rain with a pH of 3.1? Where I live now there can be significantly acidic rain. The government posts measurements for many cities and there can be stretches for days or longer with reported pH of 3.2 or 3.1. I even saw 2.9 reported once.

While this can have quite serious environmental effects depending on a number of factors including total rainfall, geology, geography, soil composition just to name a few, my question is about the direct effect of rainfall this acidic on people and their possessions.

Just how dangerous or deleterious to people and their immediate possessions (e.g. clothing, other objects) would getting a bit of pH 3.1 rain on oneself actually be? Would it sting at all if it got in the eye? Discolor hair? Is there any basis at all to being collectively phobic of being touched by this rain specifically because of its pH?

I've tried searching the internet but there is a jumble of information there. As far as I can tell, pH 3.1 seems to be somewhere between a soft drink, and tomato juice, but this is from unreliable sources.

So I thought I would ask in Chemistry SE for a solid, scientific answer. I'm not looking for an opinion, or a reasoned answer (e.g. "if it was that dangerous, then..."). I'd prefer factual data, or references related to safety and effects of topical exposure to (mildly) acidic water.

According to a United States Environmental Protection Agency report on the effects of acid rain on human health:

Walking in acid rain, or even swimming in a lake affected by acid rain, is no more dangerous to humans than walking in normal rain or swimming in non-acidic lakes. However, when the pollutants that cause acid rain $\ce{—SO2}$ and $\ce{NOx}$, as well assulfate and nitrate particles— are in the air, they can be harmful to humans.
$\ce{SO2}$ and $\ce{NOx}$ react in the atmosphere to form fine sulfate and nitrate particles that people can inhale into their lungs.  Many scientific studies have shown a relationship between these particles and effects on heart and lung function, such as asthma and bronchitis.


Is there a chemical test for gluten? I am looking for a chemical test for the presence of gluten: however, having looked online for such a thing, all the results seem to be about 'home testing kits' (presumably for people with a gluten intolerance). Is there a simple chemical test which can be used for a school science project (meaning that the chemicals involved should be easily available), something similar to how iodine can be used to detect starch?

Gluten describes a subset of proteins (glutenins and gliadins) found in wheat grain and making up a large fraction of the protein found in common wheat flour (>70%). Since the protein fraction of wheat flour can represent >10%/wt, the gluten fraction commonly represents >5%/wt of the flour.
Given the requirement for an easily implemented chemical test this eliminates more advanced and regularly used methods such as immunoassays (eg ELISA), and maybe even dye-binding assays such as described by the AOAC Official Method 2011.04.
One frequent chemical method to measure the amount of protein in a food sample is known as the Kjeldahl method, which detects nitrogen (which is absent in the carbohydrate). Other assays for nitrogen would also do the trick.
Büchi manufatures equipment for performing the required assay (Ref. 1), and provides an application note based on standard method AOAC 920.87. Although it is unlikely that you will have a Büchi SpeedDigester K-436, K-439, or Kjeldahl Sampler System K-370/K-371, available in your school lab, you'll still require the chemicals and equipment for performing a high temperature digestion and distillation, which should be similar to that listed in the application note. Note this is not a simple experiment.
Unfortunately the Kjeldahl method is likely to be difficult to implement in some educational laboratories given the required equipment, chemicals and experience. An alternative family of less challenging methods to determine protein are biuret methods (BCA and Lowry assays, which exploit protein-copper chelation and secondary detection of reduced copper), as described for instance on the ThermoFisher Scientific website. These however employ absorbance measurements to determine the amount of chelated or reduced copper.
If you are looking to distinguish gluten against a background of other proteins, you may have to isolate the gluten fraction (see eg Ref 2.). In general your approach would depend on the expected concentration and amount of gluten in the sample. Since this is a question about an educational application, it is assumed that you are not concerned with a non-gluten protein background or with detection of low concentrations.
References

Büchi Application Note 027/2010. Nitrogen and Protein Determination in Flour according to the Kjeldahl Method.

Ravinder Kaushik, Naveen Kumar,Manvesh Kumar Sihag, and Aradhita Ray. Isolation, characterization of wheat gluten and its regeneration properties. J Food Sci Technol. 2015 Sep; 52(9): 5930–5937.
doi: 10.1007/s13197-014-1690-2


Basic behaviour of bromide ion with sulfuric and phosphonic acid I  recently came across these two reactions:

  $$\begin{align}\ce{2KBr + 3H2SO4 &-> 2KHSO4 + Br2 + SO2 + 2H2O}\tag{1}\\[0.6em]
\ce{KBr + H3PO4 &-> KH2PO4 + HBr}\tag{2}\end{align}$$

I understand that the bromide ion being the conjugate base of a strong acid $\ce{HBr}$ is a very weak base and therefore it cannot take $\ce{H+}$ from a very strong acid $\ce{H2SO4}$ in the first reaction. But the second reaction is an acid-base reaction.

I checked the $\mathrm{p}K_\mathrm{a}$ values of $\ce{H2SO4}$ and $\ce{H3PO4}$ to find that $\ce{H2SO4}$ is a way better acid than $\ce{H3PO4}$, then why is that bromide ion could take $\ce{H+}$ from $\ce{H3PO4}$ but not from $\ce{H2SO4}$?

The first reaction is a redox reaction and I understand that redox reactions take place when acid-base reactions are not possible because acid-base reactions are quick. Then why is the first reaction not an acid base reaction?

A common misconception is that the more acidic component will always protonate the less acidic componend and displace it from its salts. That misconception would be exemplified by equation $(1)$

$$\ce{HBr + KH2PO4 -> KBr + H3PO4}\tag{1}$$

This is incorrect. There is a constant acid-base equilibrium and even bromide ions will be protonated to a non-neglegible extent. The principal difference between $\ce{HBr}$ and $\ce{H3PO4}$ is that the former is a gas at room temperature and standard pressure. Thus, in an open vessel, any $\ce{HBr}$ in solution is again automatically in equilibrium with $\ce{HBr}$ in the gas phase. Thus, the equilibrium should better be written in form $(2)$:

$$\begin{multline}\ce{KBr (aq) + H3PO4 (aq) <<=>\\ <<=> HBr (aq) + KH2PO4 (aq) <=> \\ <=> HBr (g) ^ + KH2PO4}\tag{2}\end{multline}$$

As you can see, the gaseous $\ce{HBr}$ can diffuse away. This reduces the partial pressure of $\ce{HBr}$ above the solution, which affects the right-hand side equilibrium by Le Chatelier’s principle, reducing the concentration of $\ce{HBr}$ in solution, which in turn affects the left-hand side equilibrium, reducing the concentration of $\ce{KBr}$. Thus, if left to equilibrate in a well-vented area, large portions of $\ce{HBr}$ will diffuse away leaving a solution of potassium phosphate or potassium hydrogen phosphate.

The principle alluded to in the first sentence is thus correct, if rather than using acid strength it is quoted as:

  The less volatile acid will displace the more volatile acid from its salts.

It is worth noting that an acid-base equilibrium $(3)$ also is observeable in your first reaction. However, the redox reaction is faster than the removal of hydrogen bromide. It is also irreversible, hence those are the final products obtained.

$$\ce{KBr + H2SO4 <=> HBr + KHSO4}\tag{3}$$


In the presence of an electron-rich species, why is Br2 the stronger electrophile than water? In the electrophilic addition of aqueous bromine to ethene, the first slow step involves the electron rich (pi bond) of ethene attacking the bromine molecule to form a carbocation and Br-:

And, then water will react with the carbocation to form 2-bromoethanol. But, why does bromine react with ethene first and not water(which is of higher concentration)?

I think it's because Br2 is the stronger electrophile than water. But why is that? Oxygen is much more electronegative which makes H more electron deficient than Br and thus a stronger electrophile?

Short answer: you are correct that oxygen is more electronegative than bromine, but the oxygen's thirst for electrons is fairly well satisfied by 2 hydrogens, whereas the bromine -- aack! -- is attached to another electron-hungry bromine.

Bromine will snap up electrons from moderately active metals (even from water!), but water doesn't (or else does it much more slowly). Consider iron, zinc, magnesium - all react more rapidly with bromine than with water.


Why do atoms "want" to have a full outer shell? Okay, so I know that this is about filling the orbitals of the atom, and I understand that. What I don't understand is why? For example, an Oxygen atom has 8 protons and 8 electrons spinning around it. Oxygen wants to have 8 valence electrons so it takes 2 electrons from other elements when bonding. But, since Oxygen has 8 protons and now has 10 electrons around it, it has a charge of -2. I don't understand how having a charge is more stable than having no charge at all, shouldn't the most stable version of an element be its neutral state? Why do the atoms want to look like the next noble gas?

You are attaching too much importance to Lewis structures. The 8-electron rule and Lewis structures which are derived from it are only rough guidelines for working out the electronic structure of a compound in very broad strokes. Often these broad strokes are accurate enough to make some meaningful statements about molecular properties but it does not accurately describe the true electron or charge distribution in a compound. Take water for example. As you say, the 8-electron rule would predict that the hydrogen atoms each transfer one electron to the oxygen molecule which would then carry a charge of -2. But experiments show a different result: here you can find a paper that determined

  that a charge of approximately $0.5 e$ is transferred from each
  hydrogen atom to the oxygen

which would amount to a partial charge of $-1 e$ on oxygen and here you can find a similar claim:

  The charge distribution depends significantly on the atomic geometry and the method for its calculation but is likely to be about
  $-0.7e$ on the O-atom (with the equal but opposite positive charge equally divided between the H-atoms) for the isolated molecule.

And there not being a $-2e$ charge on oxygen is quite understandable if you look at the electron affinities of oxygen (e.g. here): You see that the first electron affinity is $-142 \, \mathrm{kJ}/\mathrm{mol}^{-1}$ and thus energy is gained by the process of transferring a free electron to a neutral oxygen atom. But the second electron affinity is $+844 \, \mathrm{kJ}/\mathrm{mol}^{-1}$, so another electron transfer is very costly. The quite extreme situation of a -2-charged oxygen described by the 8-electron rule is very rarely found in real compounds. Even in ionic compounds like $\ce{NaCl}$ the electron is not 100% completely transfered from the electropositive to the electronegative element.


What is "dancing resonance"? While searching the Internet for information about the tricyclopropylmethyl carbocation, I found a site (askiitians.com) where it said that its stability was due to "dancing (superfast) resonance".

What is "dancing resonance"? How does it take place, and what are the prerequisites for it?

Resonance is not ‘dancing’. Resonance is often confused by undergraduates (and sometimes even professors) as some kind of oscillation back and forth between multiple structures. That is not the case.
When explained properly, resonance is a consequence of our unability to draw adequate depictions of what the bonding situation in the molecule truly looks like. For example, take a benzene ring. We can only draw single and double bonds. The structures should be identical whether the double bonds are in 1,3,5 or 2,4,6 positions. Our depictions predict different structures because double bonds are shorter than single ones. The actual benzene structure, however, has equal bond lengths between all carbons which could be described as 1.5-bonds. It is incorrect to think of the double bonds "resonating" between 1,3,5 and 2,4,6; rather, both depictions should be considered as two extreme pictures that by themselves cannot capture the truth.
This may be relatively easily understood in simple cases such as benzene or the cycloheptatrienyl cation, where the resonance depictions are all very similar. It is a lot harder to understand it in the case of a cation such as the tricyclopropylmethyl cation. There are a number of answers here explaining exactly which resonance forms stabilise it; it can and should also be seen as a nonclassical cation of the pentavalent carbonium type. You can, for example, check this question.
However, that does not mean that the resonance in these non-classical cations has a different nature. It is not correct to say that the resonance here is "faster" than in benzene, because resonance is simply not a oscillation or vibration between different structures: there is no "speed" or "frequency" to speak of.


Why do strong acids have weak conjugate bases? This question is based on the Brønsted-Lowry model. I saw many answers to similar questions online. However none of them have given a detailed answer that is based on molecular structure (the answers all told the OP to rely on a reference table for a way to identify strong and weak acids).

All chemical processes are reversible at least to some extent.

Keeping this in mind, a strong acid must have a weak conjugate base (I think you mean strong acids have weak conjugate bases).

Why? Let's try to prove by contradiction. If a strong acid had a strong conjugate base, then the base would quickly re-associate itself with a hydrogen proton. This reverses what the acid just did - dissociate itself from a hydrogen proton!

To illustrate this concept, consider hydrochloric acid - the oft-cited strong acid. $\ce{HCl}$ is a strong acid because in water solution it dissociates (nearly) completely into chloride ion and hydrogen protons (which are then solvated by water).

$\ce{HCl + H_2O ->H_3O^+ + Cl^-}$

That is the reaction of hydrochloric acid with water. Note the one-way arrow; the reaction is a one way street. $\ce{HCl}$ is the acid; $\ce{Cl^-}$ is the conjugate base; water is the base and hydronium ion is the conjugate acid. In other words, this is the generalized reaction:

$\ce{(strong)~acid + base -> conjugate ~acid + conjugate ~base}$

In bite-size pieces:

$\ce{(strong)~acid -> conjugate ~base}$

and

$\ce{base -> conjugate ~acid }$

Add the above two "half-reactions" together and you get a complete Brønsted acid/base reaction.

So back to the dissociation of $\ce{HCl}$.

If this reaction (the reverse reaction) happened to a significant extent:

$\ce{HCl + H_2O <- H_3O^+ + Cl^-}$

Then that would be antagonistic to the dissociation of $\ce{HCl}$! $\ce{HCl}$ would no longer be nearly completely deprotonated in water solution since the reverse reaction is significant!

Therefore, we conclude that if one has a strong acid, then its conjugate base must be weak. The converse is also true; if an acid's conjugate base is weak, then the acid must be strong.