What are dipole moments in a molecule supposed to act upon? In a $\ce{CO_2}$ molecule, a total of four electron pairs are shared between the carbon and oxygen atoms, such that 2 pairs are shared between the carbon atom and each oxygen atom. Oxygen has a greater electronegativity hence the atoms of oxygen will spend greater time with shared pairs and are partially negative. Since both oxygen atoms will exhibit this same behavior, the carbon atom will become partially positive.

I was reading about the dipole moment of $\ce{CO2}$ (e.g., as here), in which it said that the net polarity of the $\ce{CO_2}$ molecule is zero because if we add the individual dipole moment vectors directed from the positive pole (carbon) to the negative poles (the oxygen atoms), the resultant will be zero which implies the polarity of the molecule is zero.

My problem is this:

If these individual dipole moments (vectors) are showing the opposite and equal pull of shared pairs of electrons toward more electronegative atoms (oxygen atoms), then the carbon atom should stay partially positive with the oxygens remaining partially negative.

Given this, though, it seems that the molecule should remain polar on whole.  This is because these dipole moment vectors are acting on different shared pairs: the first vector acts on the two shared electron pairs between the carbon atom and one of the oxygen atoms, and the other acts on the two shared electron pairs between the carbon atom and the other oxygen atom. According to the following law, then, these vectors cannot cancel each other and thus $\vec u \neq\vec 0$:

  Two equal and opposite vectors do not cancel each other out when acting on different bodies.

My best attempt to resolve this contradiction:

Since in reality it is the case that the net polarity of $\ce{CO_2}$ is null, I think that both vectors (viz., the 'pull' from both oxygen atoms) must be acting individually on all electrons of carbon atom and therefore they cancel each other out resulting in molecule to be non-polar on whole. This explanation isn't all that satisfying to me, however.

Your confusion appears to arise from trying to consider dipole vectors as representing forces, which is incorrect.

Dipole vectors don't "act on" anything; they describe the displacement of the "center-of-charge"‡ of a given region of space (including both the electron cloud and the nuclei) from a particular reference point. In particular, I believe you are confusing dipole moment vectors with electric field vectors:

The electric field vector at a point describes the force experienced there by a point charge
The dipole moment vector at a point describes the displacement of the "center of charge" of the system from that point

Thus, whereas the electric field contributes to the forces present in a system, the dipole is simply a post facto descriptor of a system.

Further, in this case the law you quoted is not applicable:

  Two equal and opposite vectors do not cancel each other out when acting on different bodies.

It only makes sense to calculate an overall dipole moment if you use the same reference point for all of the 'subsystem' dipole vectors. Thus, in this example of $\ce{CO2}$, the two $\ce{C-O}$ 'subsystem' vectors point in exact opposite directions from the same reference point, since $\ce{CO2}$ is linear.  Thus, since the outside atoms are both oxygens, the magnitudes of both 'subsystem' vectors are equal and their sum is the zero vector.

‡
For a non-electroneutral region of space, by center of charge I mean the charge-weighted centroid of the system: (1) Choose a reference point $\vec r_0$. (2) At each point $\vec r$ in the region of interest, multiply the charge at that point, $\rho\!\left(\vec r\right)$, by the distance from the reference point, $\vec r - \vec r_0$. (3) Integrate the resulting product, $\rho\!\left(\vec r\right)\cdot\left(\vec r - \vec r_0\right)$, over all space. (4) Divide by the net charge of the region of interest. Dipole moment calculations are more complicated for electroneutral systems, but the principle is similar.


Have the shapes of synthesized anthropomorphic molecules (nanoputians) been verified experimentally? Have the shapes of synthesized anthropomorphic molecules (nanoputians) been verified experimentally?

X-ray diffraction, atomic force microscopy imaging of single molecules, anything?

I was tempted to use the title Has anyone ever seen a nanoputian?

Edit: The IBM video World's Smallest Movie is not of a molecule, but it does contain an anthropomorphic shape built from individual atoms, visualized at atomic resolution using scanning atomic force microscopy.

below x2: Simple illustration of AFM imaging of organic molecule shape. From here. For more recent experiments and videos, see 30 years of atomic force microscopy: IBM scientists trigger and observe Reactions in an individual molecule and also IBM & Warwick Image Highly Reactive Triangular Molecule for the First Time.

below: Lewis structure of a nanoputian; "Chemical structure of 2-(2,5-bis(3,3-dimethylbut-1-ynyl)-4-(2-(3,5-di(pent-1-ynyl)phenyl)ethynyl)phenyl)-1,3-dioxolane." From here.

below: Lewis structure of a nanoputian; "Ball-and-stick model of the nanoputian molecule, also known as nanokid, an synthetic compound whose molecules are human-shaped." From here.

I think to solve the question "seeing one of the nanoscholars, nanotexan, etc." in this particular conformation shown requests (at least) two steps:

The deposit of a single molecule on a clean, flat, and cold surface.  From the outsider's perception of work done @IBM Zurich and other groups, preparing such a pad accommodating an organic molecule is something understood well; perhaps some adjustments in how the molecules are then deposited are needed.  Are nanoputains as substance well sublimable, would be OMBE feasible?
The cold surface tightens the contact between molecule and substrate, lowers thermal vibrations of the molecule deposited; contributes that the then collected AFM is crisp.  Provided the earlier mentioned conformational flexibility of the material (not necessary flat molecule, either), how do you -- if necessary -- decoil the members of a nanoputain?  To move small molecules entirely, optical tweezers come to my mind, but here it were moving just a portion of a molecule representing a leg, redressing bow tie and hat that I assume as even more delicate.

So my speculation is the second part is the more challenging one.  Maybe it was already addressed as the original work was published in 2003, a citation analysis may shed some light on this aspect.


Why does the Born equation give the Gibbs free energy of solvation rather than enthalpy of solvation? The Born equation gives the difference in energy required to charge a particle in a vacuum and in solution which results in the work required to transfer an ion from a vacuum into solution. It is derived thus:

$$w=\int_0^Q\frac{Q}{4\pi\epsilon r}dQ-\int_0^Q\frac{Q}{4\pi\epsilon_0 r}dQ$$
Where $\epsilon=\epsilon_r\epsilon_0$ ($\epsilon_r$ is the relative permittivity of the medium/solvent - a dimensionless quantity). $w$ is work done.

$$w=\frac{1}{4\pi\epsilon r}\int_0^QQdQ-\frac{1}{4\pi\epsilon_0 r}\int_0^QQdQ$$
$$w=\frac{Q^2}{8\pi\epsilon r}-\frac{Q^2}{8\pi\epsilon_0 r}$$
$$w=\frac{z^2e^2}{8\pi\epsilon r}-\frac{z^2e^2}{8\pi\epsilon_0 r}$$
So for a mole of ions the work done is equal to:
$$\frac{N_Az^2e^2}{8\pi\epsilon r}-\frac{N_Az^2e^2}{8\pi\epsilon_0 r}=\triangle_{solv}G^{\theta}$$
My textbook seems to suggest that the last step makes it Gibbs free energy of solvation rather than just work done. I have two issues with this. Firstly, why is it Gibbs free energy and not, say, enthalpy of solvation or internal energy of solvation? Secondly, why would scaling the number of ions to one mole make it into Gibbs free energy? Isn't Gibbs free energy an extensive property with units of Joules? I know that the Gibbs free energy (and enthalpy) are often given with units of $Jmol^{-1}$ but isn't this strictly molar gibbs free energy/chemical potential?

1) Because $-\Delta G$ is the maximum possible nonexpansion work done by a system in a constant temperature and pressure process. You can write $$\Delta G \le w_{non PV}\quad (\text{constant T and P, closed system})$$
where the equality sign holds for reversible changes.

Here's proof:
$$\begin{align*}
H &= U + P V&\text{(definition of enthalpy)}\\
dH &= dq + dw + d(PV) &\text{(because dU = dq + dw)}\\
dG &= dH - T dS - S dT &\text{(because G = H - TS)}\\
&=dq + dw + d(PV) - T dS -S dT &\\
&= dq + dw + d(PV) - T dS & \text{(when the change is isothermal)}\\
&= dw_{rev} + d(PV) &\text{(if reversible,}\ dq_{rev} = T dS\text{)}\\
&= dw_{electrical,rev} + dw_{expansion,rev} + P dV + V dP & \text{(partitioning the work)}\\
&= dw_{electrical,rev} + V dP & (dw_{expansion,rev} = - P dV)\\
&= dw_{electrical,rev} & \text{(if pressure is constant)}\\
\end{align*}$$

2) It isn't the scaling to one mole that makes the result a Gibbs free energy (see 1); that just makes it a molar Gibbs free energy. You can use either extensive or intensive Gibbs free energies in calculations; whatever is most convenient. Just make it clear (either by context or notation) which one you're using.


How can I perform structure alignment on an inorganic species? I'm attempting to perform atomistic optimal structure alignment on an inorganic species. Can anyone point me in the right direction to software or a technique for performing this?

There exists a wide range of structural alignment software, but this really appears to be geared towards aligning protein backbones, whereas I'm interested in aligning unlabelled Cartesian coordinates of small molecules.

Inorganic structures – that could be proper solid or molecular metalloorgnic. For the later one you can use xp (program written G.M.Shedrick - comes with Bruker diffractometers). If you have running license of Cambridge Crystal Data Base there is program called IsoStar, and I think that it should do the job. I haven't used it my self but apparently Olex2 can make aligning of fragments (this program is free so that's the place were you can start). I'm not quite sure but I think that old GROMOS code could also make a fit of orthogonal coordinates.


How to calculate the vapour density of a mixture? I have a gaseous mixture of $\ce{H2}$ and $\ce{CO2}$ which contains $66\%$ by mass of $\ce{CO2}$. I have to figure out the vapour density of the mixture (defined as mass of a certain volume of a substance divided by mass of same volume of hydrogen).

My approach: let's assume $\pu{100 g}$ of mixture contains $\pu{66 g}$ of $\ce{CO2}$ and $\pu{34 g}$ of $\ce{H2}$. Hence, the amount of substance of $\ce{CO2} = \pu{1.5 mol}$ and the amount of substance of $\ce{H2} = \pu{17 mol}$. The total amount of substance $= \pu{18.5 mol}$.

The mass of $\pu{1 mol}$ of mixture is 5.4 g (mass of 18.5 mol is 100 g), and the vapour density $\text{VD} = 2.7$.

I got my answer right, but now I am doubtful of my approach. Was it safe to assume that the amount of substance of mixture equals that of the amount of substance of both the gases?

Was it safe to assume that the amount of substance of mixture equals that of the amount of substance of both the gases?

Amount of substance is additive, just like mass. You have to be careful with volume, though. So yes, it is safe to make that assumption.

  I have to figure out the vapour density of the mixture

I'm not sure you are starting with a safe question, i.e. if the vapour density quantity was intended to describe mixtures. Wikipedia has "vapour density = molar mass of gas / molar mass of H2", and I'm not sure whether a mixture has a defined molar mass. The same source states that vapour density is sometimes based on the density of air rather than dihydrogen gas (to assess whether certain gases collect on the floor or ceiling of a room when released).


Why does eutectic point have three phases, but consolute point have only two phases? At eutectic point, there are three phases: solid A, solid B and homogenous liquid phase containing both A and B.

But at consolute point, we are told it has 2 phases: liquid A and liquid B.

Why isn't there a homogenous phase containing both liquid A and B at consolute point just like an eutectic point?

I just realised, the 2 phases present in a liquid-liquid solution system with temperature below its solution temperature is a) conjugate solution enriched in A
b) conjugate solution enriched in B

No pure liquid A or pure liquid B phase exists.


Comparing energies of isoelectronic species, but one protonated Let's say we have the following reaction:

$$\ce{RH -> R- + H+}$$

By modelling it using quantum chemical calculations in the gas phase I would make calculations for both $\ce{RH}$ and $\ce{R-}$, but $\ce{H+}$ has no electrons.

On the other hand, $\ce{RH}$ and $\ce{R-}$ are isoelectronic and zero energy is assigned for infinitely separated particles.
I have a gut feeling that the energetics of $\ce{RH}$ and $\ce{R-}$ can't be compared directly in this case, but I can't think of a good argument for this.
So, can I compare them? Why or why not?

I think they can and should be compared. Consider the $\ce{H_2^+}$ molecular ion. When dissociating this correctly (that is, including symmetry breaking), one should eventually end up with $\ce{H}+\ce{H^+}$. One can then extract the dissociation energy of the bond from the calculations. This is equivalent to the proton affinity of atomic $\ce{H}$. By applying this logic to larger species, one reaches the conclusion, that yes, $\ce{H+}$ by definition has zero energy and should enter calculations of relative energies as such.

As an aside, the calculation of negatively charged species in the gas-phase can lead to erroneous results, but any argument made for $\ce{A-/AH}$ will certainly hold for $\ce{B/BH^+}$.


When we say "the energy in a system" (in chemistry) do we mean the energy stored in the chemical bonds in the compounds present in that system? Is it the energy stored in the chemical bonds of the compounds present in a system, that we refer to as the energy in/of a system ?

The answer to this question depends on the nature of the system we have defined.

Firstly, to define the term system according to this Wikipedia article:

  “...a physical system is a portion of the physical universe chosen for
  analysis. Everything outside the system is known as the environment.
  The environment is ignored except for its effects on the system. The
  split between system and environment is the analyst's choice,
  generally made to simplify the analysis. For example, the water in a
  lake, the water in half of a lake, or an individual molecule of water
  in the lake can each be considered a physical system.”

Every system has some relationship to it’s surroundings.  As another way of stating the above, we can describe the surroundings, or environment, as all parts of the universe that is not part of the system. We can then generally categorize the system as being isolated, closed or open:

Isolated system: Neither matter nor energy may pass into or out of the system.
Closed system: The exchange of energy is allowed but matter may not pass into or out of the system.
Open system: Both energy and matter can pass into and out of the system.

In the context of your question (as I think I understand it), we are talking about isolated systems.  It wouldn't make much sense to discuss the energy of chemical bonds without constraining the system, even if hypothetically, in this way. To use the lake analogy, we talk about the chemical bonds within a single water molecule in a manner that excludes the matter or energy in the rest of the lake (and the rest of the universe).  If we are talking about hydrogen bonding between the water molecules for example, then we are considering there to be some hypothetical container within the lake that excludes all matter and energy from outside the container.  The contents, matter and energy, of the container, comprise the system.


Stability of canonical forms Which of the following ions is more stable? Use resonance to explain your answer.

When we have to compare stability of resonance structures we generally check for the following features in the resonance forms (in sequence):

Neutral molecule;
More covalent bonds;
Complete octet of all the atoms;
Less separation of opposite charges;
Positive charge (if any) on more electro positive atom and negative charge (if any) on more electronegative atom.

In the above problem:

Both the molecules are charged;
Both have same nuber of covalent bonds;
Both have the octet of one carbon atom incomplete (the positively charged one);
Both have positive charges on carbon atoms.

From all of these considerations the molecules should be equally stable. But the answer given in my book is (A).
What is that I can't figure out?

It has been noted in the comments and I shall repeat that the premise of the question is already bogus. The question does not consider two different ions but two mesomeric depiction of the same ion. This is also evident by the mesomeric arrow between the two forms. Since mesomery strictly means that no geometrical change takes place, both depictions are identical in all aspects and have the same (resonance-influenced) energy. Thus, any book claiming that either of the two have a lower energy is wrong.

However, what is commonly done is to determine a more likely resonance structure. This is, for example, commonly used to explain the strongly stabilised nature of certain aromatic systems, e.g. when DMAP (4-N,N′-dimethylaminopyridine) is used as a carboxylic activating agent. We can attempt to do that for your structure, too. If we compare the two we see — of course — that a lot of features are identical. The principal differences are:

A has a trisubstituted double bond while B only as a disubstituted (and terminal) double bond
B has a secondary carbocation while A has a primary one (both, however, are allylic).

Since allylic stabilisation for cations is much stronger than hyperconjugation, we can discard the second bullet point as being essentially equal for both resonance structures. Thus, we should argue with the location of the double bond. Both double bonds can benefit from hyperconjugation twice (the primary cation of A cannot hyperconjugate due to geometry), but in A the two hyperconjugations stabilise different carbons. Along with the general double bond stability given as:

$$\text{monosubstituted} < \text{terminal disubstituted} < \textit{cis} < \textit{trans} < \text{trisubstituted} < \text{tetrasubstituted}$$

we can identify A as being slightly better than B and therefore contributing slightly more towards the overall structure.

You mentioned in the comments that your book states:

  structure B is not planar and hence conjugation is not possible in it

Your book probably wants you to assume that the following depiction in figure 1 is adequate for the two resonance forms:

Figure 1: structures B and A under the assumption of a chair-configured cyclohexyl structure for B.

Under the underlying assumption of a different geometry for both structures, this may be valid. However, there is no way that cation B will adopt a chair cyclohexane configuration. Instead, as I mentioned above, the following structures in figure 2 are much more likely and even show that we are dealing with resonance:

Figure 2: depiction of the mesomeric forms without geometry change.

You can easily see that the argument of non-conjugation is bogus, since there very well is a way to optimise geometry so that the orbitals may form an allyl system.


What makes carbon special? We are home schooling my son, and right now we are covering chemistry- ionic bonds, covalent bonds, etc.  We've talked about how carbon is "special" in that it can form very complex molecules (organic chemistry), and I have done some hand waving by saying that it is its ability to make four bonds that causes that specialness.

He, quite reasonably, asked why nitrogen, with its ability to make three covalent bonds, couldn't be almost as special as carbon.  I told him that it probably can make pretty complicated molecules, but that only being able to make three bonds is a limiting factor.  That didn't seem to satisfy him, and to be honest, it doesn't totally satisfy me.

So why is carbon so special?  Is it just the four bonds vs. three, or is it something else?

I think that the link that Ron mentioned answered the question adequately, although let me add my own take.  (What makes carbon special and versatile?)

I heard that the reason Carbon is able to make such stable multiple bonds is that its orbitals are close enough to its nucleus that they form extremely effective overlap when bonding.  It is therefore the best atom because it is the smallest atom that can form four bonds.  When you have 5 valence electrons already and can only make 3 bonds with 1 lone pair, that usually leads to weird bent shapes (although Nitrogen is also pretty common in organic compounds).

It's kind of like, which hotel is more versatile in booking large families—a hotel with four open rooms right next to a breakfast bar (Carbon), a hotel with three open rooms and a full room right next to a breakfast bar (Nitrogen), or a hotel with any amount of open rooms, except you need to walk a mile for breakfast (any other element)?


Can aluminium hydroxide be split into Al 3+ and OH-? When writing a net ionic equation containing aluminum hydroxide, can the aluminum hydroxide be split into ions of $\ce{Al^3+}$ and $\ce{OH-}$, or does it stay as $\ce{Al(OH)3}$?

Aluminum hydroxide has poor solubility in water (solubility product $3 \times 10^{-34}$).

However, $\ce{Al(OH)3}$ is amphoteric and soluble in acids and bases. An excess of $\ce{OH-}$ will form tetrahydroxidoaluminate:

$\ce{Al(OH)3 + OH- <=> [Al(OH)4]-}$


The test of metal hydroxides and metal oxides Is testing metal oxide different from testing metal hydroxide using a litmus paper?

Litmus paper does not test for either metal oxides or metal hydroxides; indeed, it will not change color at all. Try it: put dry $\ce{CaO}$ on dry litmus paper. Litmus does change color with pH+, the negative log10 ratio of $\ce{H+}$ ions to $\ce{OH-}$ ions (the ions may actually be more complex, such as $\ce{H3O+}$, but the idea is the same). This means the substance is in aqueous solution, i.e. in water.

Litmus also changes color with non-metals in solution, such as $\ce{NH3}$, which forms $\ce{NH4OH}$ in water, or with acetic acid, having an excess of $\ce{H+}$ ions.


Why is sodium sulfate neutral while its conjugate acid is weak? It is commonly taught in chemistry class that the conjugate base of a weak acid (well, at least if it's not weaker than water) is a weak base, as shown by the relation

$$\mathrm{p}K_\mathrm{a} = 14 - \mathrm{p}K_\mathrm{b}  \iff K_\mathrm{a}=\frac{K_\mathrm{w}}{K_\mathrm{b}}$$

where the $K_\mathrm{a}$ is of the conjugate acid, and $K_\mathrm{b}$ of the base. For weak monoprotic acids this means that their conjugate bases are weak, e.g. various carboxylic acids, phenol and hydrofluoric acid. For polyprotic acids, if the conjugate base's next $K_\mathrm{a}$ is smaller than its $K_\mathrm{b}$, then it would also form an alkaline solution but not fully dissociate, e.g. bicarbonate ion (carbonic acid), bisulfide ion (hydrogen sulfide), ammonia (ammonium ion).

However, according to Wikipedia as well as my teacher, sodium sulfate $\ce{Na2SO4}$ is neutral, but Wikipedia also says the acidic hydrogen on the bisulfate ion has $\mathrm{p}K_\mathrm{a} \approx 2$, so sulfate would have $\mathrm{p}K_\mathrm{a} \approx 12$ and a $1\ \mathrm{M}$ solution of sulfate with a perfectly non-acidic counterion would have $\mathrm{pH} \approx 8$. Is sodium sulfate in fact neutral? If so, what would be the reason? Could it be that $\ce{Na+}$ is slightly acidic? Or is the formula for conjugate basicity inapplicable here?

Given $\text{pK}_{a_1} = -3$ and $\text{pK}_{a_2} = 1.9$ for sulphuric acid, the pH for a 1M solution is indeed expected to be $7.9$, $7.4$ for a 0.1M, and $8.0$ for a saturated solution. So it is slightly basic.


Length of copper wire in 1 lb spool I was asked the following question in my introductory chemistry class:

  The diameter of metal wire is often referred to by its American wire gauge number. A 16-gauge wire has a diameter of $\pu{0.05082 in}$. What length of wire, in meters, is there in a $\pu{1.00 lb}$ spool of 16-gauge copper wire? The density of copper is $\pu{8.92 g/cm^3}$.

First I converted the imperial measurements to metric. $\pu{0.05082 in} = \pu{0.12908 cm}$, $\pu{1 lb} = \pu{0.45359 kg}$, according to Google. So I can use $\pi r^2 h$ to calculate the volume (and thus, using density, calculate the mass) of one meter of wire, I figure the radius to be $\pu{0.06454 cm}$. The equation becomes $V = \pi \times (\pu{0.06454 cm})^2 \cdot \pu{100 cm}$, which equals $\pu{0.41654 cm^3}$. So that's the volume of one meter of wire.

Using the density of copper $(\pu{8.92 g/cm^3})$, I figure the mass of one meter of wire to be $\pu{3.72 g}$. If $\pu{1  lb} = \pu{453.59 g}$, then converting $\pu{3.72 g}$ to lbs gives me $\pu{3.72 g} \cdot \frac{\pu{1  lb}}{\pu{453.59 g}} = \pu{0.01 lbs}$. So assuming I did everything correct, one meter of wire is a hundredth of a pound, so in one pound of wire there is $100$ meters.

Find a function of length depending on mass

Conversion factors:

$2.54 ~\mathrm{\frac{cm}{in}}$
$454 ~\mathrm{\frac{g}{lb}}$

Constants:

Wire diameter $d = 0.129~\mathrm{cm}$
Density of copper $\rho = 8.92~\mathrm{g\, cm^{-3}}$

Thus, our function $l(m)$ for the wire length from the wire mass in pounds reads:
$$ l(m) = m \cdot 454~\mathrm{\frac{g}{lb}} \cdot \frac{1}{\pi d^2 \rho}$$

Evaluate the function

Since I'm too lazy to evaluate something like this by hand (and since I somehow had to record this thought process and did not want to waste the processed remnants of a dead tree) there is a small python script available to do this for you, which you can find here and which, if you run it using the command python3 wires.py, will give you the following output:

  Wire length of 1.00 lb of 16-gauge wire: 9.71 m

(Bonus: You can run the command as python3 -i wires.py and will then be dropped into an interactive python shell, where you can call the wirelength function as many times as you wish using different masses for the wire.)


Chemical reactions in SFE? What are the chemical reactions and/or chemical processes that take place during SFE(supercritical fluid extraction)? Namely supercritical $\ce{CO2}$.

Supercritical fluid technology is used primarily to enhance the chemical extraction process.  The solubility of a compound in a supercritical solvent increases as the system pressure increases at constant temperature. This is due to the highly pressurized supercritical solvent disrupting the intermolecular attractive forces between the molecules of the compound to be extracted, and replacing those forces with increased attractive forces between the compound and the supercritical solvent.  Carbon dioxide is a very common material used for supercritical extraction.  It is cheap, not flammable, has low toxicity and, after the extraction is complete you simply return the system to STP, allow the carbon dioxide to evaporate and you are left with only the extracted material.  Some examples of commercial uses for supercritical carbon dioxide include caffeine extraction from coffee beans, sterilization of surgical equipment, fracking and dry cleaning.

Chemical reactions involving supercritical carbon dioxide are now being intensely studied.  One reaction that has been commercialized involves reaction of supercritical carbon dioxide with hardened, alkaline cements. A harder cement containing carbonate groups throughout the entire volume of cement is formed along with water (environmentally friendly process).


Determination of pKb of a mono acidic base $20$ mL of a weak monoacidic base($\text{BOH}$) requires $12$ mL of $0.3$ M $\text{HCl}$ solution for the equivalence point. During titration, the pH of the base solution was $10$ upon the addition of $4$ mL of $0.3$ M $\text{HCl}$ solution. What is the $\text{pKb}$ of the base($\text{BOH}$)?

My attempt:

The concentration of the base is given by,
$c * 20 = 12 * 0.3$ -- (equating the moles of the acid and base)
$c = 0.18$ M

So, initial moles of the base in the container $= 20 * 0.18 = 3.6$ mmol
Moles of acid that is added $ = 4 * 0.3 = 1.2$ mmol

Since the base is monoacidic, they will react in a 1:1-mole ratio. The acid is the limiting reagent, so it will be fully consumed. Therefore the moles of base left is,

$3.6 - 1.2 = 2.4$ mmol

The concentration of the base is,
$\frac{2.4}{24} = 0.1$ M

Since the pH of the solution is $10$, therefore the concentration of $\text{OH}^- = 10^{-4}$

Applying the approximated formula for calculating the $\text{k}_b$ of a weak base,

$$\text{K}_b = \frac{x^2}{c}$$

Where,
x = concentration of $\text{OH}^-$ ions
c = concetration of the base

In our case,
x = $10^{-4}$
c = $0.1$

Plugging in the values I got $pK_b = 7$. But the answer is given $4.3$.

Any help would be appreciated.

I appreciate your effort you showed in your question, but you missed one important thing that when a weak base is reacted with a strong acid a buffer solution is formed and the $\mathrm{pH}$ is not only contributed by base but also by the acidic salt ($\ce{BCl}$ in this case).

So, simply applying the Henderson–Hasselbach equation (derivation can be found here)

$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b} + \log{\frac{[\ce{B-}]}{\ce{[BOH]}}}$$

we get

$$4 = \mathrm{p}K_\mathrm{b} + \log{\frac{1.2}{2.4}},$$

which gives us

$$\mathrm{p}K_\mathrm{b} = 4.3010$$


Near-IR spectroscopy measures the broad overtone and combination bands of some of the fundamental vibrations (only the higher frequency modes) I am currently studying the textbook Infrared and Raman Spectroscopy by Larkin. In the first chapter of the textbook, the author opens (probably prematurely) with the following statement:

  Near-IR spectroscopy measures the broad overtone and combination bands of some of the fundamental vibrations (only the higher frequency modes) and is an excellent technique for rapid, accurate quantitation.

This will likely be covered in later chapters, but the statement piqued my curiosity: For someone who has not yet encountered concepts such as "broad overtone", "combination bands", and "higher frequency modes", what does this all mean? Essentially, I am seeking a brief, but clearer and more elaborate, explanation of what the author has described here.

I would greatly appreciate it if someone who has knowledge of this area would please take the time to clarify.

Indeed that sentence should be taken as merely anticipating what the book will treat in following chapters.

Nevertheless, to satisfy your impelling curiosity the statement trace a combination of facts that leads to the actual spectrum of a given molecule.

Overtones IR absorption bands are those resulting from breaking of selection rules. Instead of having just transitions between adjacent vibrational levels, basically the 1<-- 0 one, transitions with a delta nu of 2 are observed as well. In other words the vibrations aren't those of an harmonic oscillator but those described by a Morse's potential.

Combination modes arise from the combination of normal modes, ie they happen when different vibrations are simultaneously excited. Normal modes which are unseen can contribute to these combination modes, as the coupling alter the symmetry.

High frequency: we are speaking about NIR, which is the high frequency range for molecular motions, and it is where vibrational overtones typically occur. The lower frequency part of the spectrum is dominated by the fundamental modes, and except for its fingerprints part and fine details, it has a strong bond related character rather than a molecular one.

If this does not help but instead brings out new terms or doubts, I would suggest that you treat the introductory part of the book for what it is, namely an introduction.

Alternatively you can refer to this link which provides a condensed explanation and includes schemes, further links, and definitions:

http://bit.ly/31Mc6lT


Electrolysis used to cause a Chemical Reaction? Electrolysis is used to separate compounds like Water into Hydrogen and Oxygen.

Can it be used to induce a synthesis reaction rather than splitting reactants apart? For example, could you use electrolysis to recombine the Hydrogen and Oxygen?

You wouldn't need to supply electric energy to get oxygen and hydrogen to recombine into water, because the free energy change for that reaction is negative. In other words, the products have a lower potential energy than the reactants, and so the reaction will proceed "downhill" on its own. You would need to overcome the very small activation energy, which you could do with electricity or with a flame or spark. But once it got started, you would not need to supply additional energy, and could in fact generate electricity from the reaction.

This is the basis of how hydrogen fuel cells work - hydrogen and oxygen react in a device that is set up to extract the energy that is produced in the form of an electric current.

An easy way to think about it is that if you need to supply energy to make a reaction happen, then the reverse reaction will produce the same amount of energy (if we ignore losses due to inefficiency) or less (if we don't). The reverse is true as well.

On a side note - electrolysis specifically refers to splitting a compound. The "-lysis" suffix means "break apart," and "electro-" prefix means "with electricty." In your case, you are thinking of "electrochemistry" - which just means "using electricity with chemistry." Electrolysis is one thing you can do with electrochemistry.


Variable oxidation state of transition metal ions Transition metals can form stable ions with different oxidation states. But I am confused why doesn't only the most stable state exist. Let me clarify my question more:

$$\ce{Ti^{2+} -> Ti^{3+} + e^-} $$
With $E^0 = +0.5~\mathrm{V}$.

This just shows that $\ce{Ti^{3+}}$ is more stable. So my question is that why doesn't $\ce{Ti^{3+}}$ only exist in its most stable form?

Thermodynamics proposes, kinetics disposes.  The most stable form can have a high activation energy barrier surrounding it.  The medium matters.  Hard Lewis bases (water) stabilize high oxidation states (Cu(II), tungstate).  Soft Lewis bases (acetonitrile) stabilize low oxidation states (Cu(II); $\ce{W(CO)6}$ and http://www.google.com/patents/EP1995347A1?cl=en).

$\ce{Ti^{+4}}$ is the basement, especially with oxygen.  Now rationlalize titanium suboxide Magnéli phases.  They are spectacular as inert electrodes under extreme conditions, re Ebonex.

DOI:10.1016/j.electacta.2010.05.011
http://eprints.soton.ac.uk/cgi/request_doc?eprintid=152857&_action_null=Request+a+copy


What is the difference between enthalpy of reaction and standard enthalpy? What is the difference between enthalpy of reaction ($\Delta H$) and standard enthalpy ($\Delta H^\circ$)? I was told that the standard enthalpy of reaction is the change in heat when one mole of matter is transformed. I don't quite get what this one mole refers to in a chemical equation.

For example, I came across this equation in my textbook: $$\ce{N2 + O2 -> 2 NO \tag{ΔH = +181kJ/mol}}$$

Does the $\Delta H$ here mean the amount of energy absorbed during the formation of 1 mole of $\ce{NO}$? But since 2 moles of $\ce{NO}$ are produced, shouldn't it be $181\times 2 = 362$?

Yes, the standard enthalpy of reaction ($\Delta_\mathrm{r}H^\circ$) is the enthalpy change that occurs in a system when matter is transformed by a given chemical reaction, when all reactants and products are in their standard states. The only condition is that the participants have to be in their standard states, ie. usually at 1 bar pressure.

On the other hand, enthalpy of reaction ($\Delta_\mathrm{r}H$) is amount of energy absorbed or released at any condition. Considering your example,
$$ \ce{N2 + O2 -> 2NO} \qquad \tag{$H = \pu{+181kJ}$}$$
Here, $\Delta$H represents the amount of heat absorbed in the formation of 2 moles of $\ce{NO}$. For the formation of one mole of a substance from constituent elements, it is called enthalpy of formation($\Delta_\mathrm{f}H$)and if in standard states, standard enthalpy of formation.
$$ \ce{ 1/2 N2 + 1/2 O2 -> NO} \tag{$\Delta_\mathrm{f}H$}$$
As we can see, $\Delta_\mathrm{f}H$ = $\frac{1}{2}\Delta H$. Hence it should be $\Delta_\mathrm{f}H$= +90.5 kJ/mol.

Note: Unit of enthalpy is that of energy but unit of enthalpy for a reaction is of energy per unit mass or amount.


How can I purify Ferric or Ferrous acetate with a very basic lab? I’ve been trying to make Ferric acetate by mixing vinegar with steel wool and allowing it to react. The result is successful in dying leather or wood black and acting as a mordant. Theoretically there are ways of crystallizing, precipitating, filtering, or separating it with solvents.

There is only one problem: I can’t remove it from the vinegar solution without it oxidizing. Adding Hydrogen peroxide causes it to create iron oxide, heating it causes it to hydrolyze and create iron oxide. Leaving it in solution turns some of it into iron oxide.

One idea I had was using glacial acetic acid instead of vinegar but I’m hoping I don’t have to because that’s fairly expensive. I could also do the entire thing under vacuum but it still wouldn’t fix the hydrolysis reaction. Could I maybe crystallize it? I’m open to suggestions with an emphasis on stuff that doesn’t require expensive reagents or obscenely overpriced labware.

Both the Handbook of Chemistry and the Merck Index state that $\ce{Ferric Acetate}$ exists only in solution. If this solution is evaporated, a basic ferric acetate is obtained that is insoluble in water, with a formula $\ce{Fe(OH)(CH_3COO)_2}$. This basic acetate is soluble into acetic acid.

Nevertheless there has been recent reports stating that solid Trinuclear Ferric Acetate may be obtained in rather concentrated solutions with the following structure : $\ce{[Fe_3O(OAc)_6(H_2O)_3]OAc}$. Its synthesis has been described in A. Laurikenes et al., Lithuan. J. Phys. 56, 1. p.35 (2016).
They mix $11.2$ g Iron powder ($0.200$ mol) and $12$ mL $\ce{CH_3COOH}$  $50$% ($0.100$ mol) at $75°C$. Why this choice ? The contrary would have been a much better choice (interchange $0.1$ and $0.2$)... They obtain a concentrated green solution, which they filtrate, and they add $6$ mL $\ce{H_2O_2}$ $30$%, and filtrate the obtained red mixture. The obtained filtrate is very concentrated. It is then evaporated under vacuum. And they obtain $2.95$ g of the trinuclear compound. The yield is miserable.  But they say that they have been able to recrystallize this compound in ethanol.

It is interesting to know that this substance has three equivalent $\ce{Fe}$ atoms in the apexes of a triangle, with an Oxygen atom in the center of the triangle, and that each $\ce{Fe}$ atom is included in the center of an octahedron, where all apexes are occupied by Oxygen atoms.

See also : K. I. Turte et al. Synthesis and Structure of Trinuclear Iron Acetate, J. Struct. Chem. 43, 108-117 (2002)


Can I break starch down into glucose units? I was wondering if starch can be broken up into individual glucose units. Because on a site I saw, it said that through acid hydrolysis starch can be broken up into amylose and amylopectin. However, as I researched amylopectin I found out it is branched and had a different bond than amylose so just to simplify everything:

Can starch be broken up into individual glucose units and through what processes?

Starch is what plants primarily use as a glucose storage. As such, it is essential that they can break it back down into glucose otherwise it would defeat its own purpose and be removed by evolution very quickly. Think about it: Requiring a lot of energy to build up something that serves no further purpose — not exactly giving a plant an advantage in survival.

There are different methods how to break it down. Plants, of course, use enzymes. Humans can also break down starch with the help of enzymes (which is why potatoes give us energy), much like most other animals, fungi and many single-cell organisms.

The first amylose-breaking enzyme is present in saliva, so it’s that early that the breakdown begins. However, distinct glucose usually isn’t liberated until the gut.

Chemically, the easiest method would be acidic hydrolysis, i.e. use hydrochloric acid and heat. The downside is, that this method might destroy the individual glucose units, too; so using enzymes is preferred if you want glucose.


Is there a known relation between the diffusion of a molecule and the diffusion of the reactants? I have a plasma consisting of elements A and B (which have a similar mass). These elements can react to form molecules AB.

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

Now if my plasma or gas is expanding based on pressure gradients, I will have some sort of diffusion of these two element concentrations. In fact, this diffusion is the dominant process affecting the concentrations A and B. Based on my spectroscopic analysis, the rate of decay of both elements is similar/the same, so I model them as
$$
[\ce{A}](t) = A_0 \exp(-k_0t) \,, \\
[\ce{B}](t) = B_0 \exp(-k_0t) \,.
$$
Then the concentration of AB changes with
$$
\frac{d[\ce{AB}]}{dt} = k_1 A_0 B_0 e^{-2k_0t}\ - k_2 [\ce{AB}] \,,
$$
where $k_2$ is the rate constant for diffusion. The obtained formula for [AB] is then
$$
[\ce{AB}](t) = \frac{k_1A_0B_0}{2k_0-k_2} \left(e^{-k_2t} - e^{-2k_0t} \right) \,.
$$
This actually fits my data quite nicely, but one thing confuses me: When fitting my data for $[\ce{AB}]$ with this model, the parameter $k_2$ is exactly $2k_0$.

Or rather, they are very close, since the formula has a singularity at $k_2 = 2k_0$.  But if the formula is derived with this assumption directly,  one obtains
$$
[\ce{AB}](t) = (k_1A_0B_0)\, t\, e^{-2k_0t} \,,
$$
which also fits my data very nicely.

I don't understand why $k_2 = 2k_0$ and I am worried that I might say something very stupid in my paper when I try to explain this observation. But this is what I am thinking: $k_2$ is the rate of diffusion for AB, which is twice as heavy as A and B. It will have twice the momentum, which could lead to a doubling of the diffusion rate as well. Does this seem like a probable explanation? If not, which other effects will play a role?

Note that the plasma I am looking at is short-lived and changes constantly, which makes it difficult to apply textbook chemistry to it. Basically all signals I am looking at can be modeled as exponential functions or sums of exponential functions, and sometimes multiple processes might be involved. It's a lot of guesswork, and I am worried that I am actually on the wrong track here. But at least intuitively it makes sense to me that the diffusion of a heavier molecule should happen faster.

Since I was asked why I don't use a more complex diffusion/reaction model: Occam's razor, basically. I don't want to assume too much about my plasma. And I fear that a complex model might contain too many free parameters which would lead to overfitting of the data. (If I'm wrong about this approach, feel free to correct me and explain to me how to go about this. What I do not like are downvotes without a comment telling me what I'm doing wrong. None of the downvotes I have received bothered to do this, which is not helpful.)

More a set of queries and comments than an answer.

I don't understand why you make A and B vary with time irrespective of your reaction scheme.

You should really analyse using Fick's diffusion equations with the reaction scheme added and do so for each species. This will have to be a numerical calculation in space and time. (There are well established methods to do this. e.g. Crank-Nicholson). By the symmetry of your experiment you may be able to reduce this to one dimension and time.

You should be able to 'guestimate' the diffusion coefficient from kinetic theory at least their ratio which should be quite accurate.

Your second equation ('new formula') is that for a reaction $\ce{A + B ->[k_1] AB ->[k_1] }$ where both rate constants are the same. In your case there seems to be no reason why the two rate constants are the same, only you know enough about your experiment to explain this :)


Reasons for existence of eutectic point for alloy systems Why does a eutectic point exist for a system of alloy ? Why is it that there is such a definite ratio of composition at which this occurs ?

First, there are several other answers that you might be interested in looking at first (full disclosure - they include some by me).

A question on Physics SE on eutectics

Ideal vs regular solutions and eutectics on Chemistry SE

These together go in to some detail on why the eutectic forms (when the liquid free energy drops down and touches the tie-line between the two solid phases - the Physics one), and the interactions one needs to get the liquid phase to drop below the tie line between the solids (the Chemistry one). To get a eutectic you need an interaction term in the liquid Gibbs free energy.

As for why it happens at such a definite composition, the glib answer is the Gibbs phase rule. In a system at fixed pressure, three phases can coexist at only one point.

Given that this question is, in general, a duplicate of the ones listed above, particularly of the Chemistry SE one, I have no particular need of the bounty. Go read those and see if they help.


Jahn Teller Effect for d4 low spin configurations Why does Z-in take place in $d^{4}$ low spin configuration whereas in Z-out one can get all the 4 electrons in $d_{xz}, d_{yz}$ which would lead to lower energy

First off, I don't think all 4 electrons will be occupying the $d_{xz}$ and $d_{yz}$ orbitals in z-out distortion. If you do that, you'll be changing spin multiplicity on distortion. That doesn't happen. Rather, all the (originally) $t_{2g}$ levels would fill up, leaving you with a paired up electronic configuration in either $d_{xz}$ or $d_{yz}$ for z-out and a spin paired $d_{xy}$ for z-in.

Now, calculate the relative stabilization such that:

a. z-out:  $d_{xz}$ and $d_{yz}$ are stabilized by 1/3$\delta$ and $d_{xy}$ is destabilized by 2/3$\delta$.

b. z-in:  $d_{xz}$ and $d_{yz}$ are destabilized by 1/3$\delta$ and $d_{xy}$ is stabilized by 2/3$\delta$.

Your electronic configurations would then read ($e_g)^3$($b_{2g})^1$ for z-out, ($b_{2g})^2$($e_g)^2$ for z-in. Now calculate the relative energies. You will find that z-in is favoured by a factor of 1/3$\delta$.


Electrolysis Anode equation In the electrolysis of $\ce{Na2SO4}$, is it valid to write the anode half-reaction (oxidation) as
$$
\ce{OH-} \rightarrow \ce{H+} + \frac{1}{2}\ce{O2} + 2\ce{e-}
$$?

My book says it's
$$
\ce{H2O} \rightarrow 2\ce{H+} + \frac{1}{2}\ce{O2} + 2\ce{e-}.
$$

Other than the additional $\ce{H+}$, there's no difference between the two.

This depends on the medium in which you do the electrolysis. In the case of a basic medium your equation would be more correct, but in the case of an acidic medium your answer would be wrong, since the concentration of $\ce{OH-}$ ions would be far too small.

Generally: In acidic medium you balance your redox equations using $\ce{H+}$ and $\ce{H2O}$, in basic medium you balance with $\ce{OH-}$ and $\ce{H2O}$.

Note: Since you generate $\ce{H+}$ ions it is possible to argue that starting with a solution at $\text{pH}=7$ gives you an acidic medium, resulting in the equation in the book.


What is the appropriate way of writing the Iron/thiocyanate equilibrium? I understand the theory, but am struggling to answer the following discussion questions:

(i)
Why isn't this represented by: $\ce{Fe^{3+} (aq) + 3 SCN- (aq) <=>  Fe(SCN)3 (aq)}$
Why do we dilute the solution with water?

(ii)
I don't understand what the question is asking, can someone please explain

(iii)
I understand this.

(iv)
I am unsure, can someone suggest a reason.

(v)
I understand this.

(vi)
Constant volume of container?

ADDITIONAL QUESTIONS:

I thought ALL equilibrium reactions needed to be in a closed vessel - so if this experiment was conducted in an open test tube, wouldn't this be an error? How would this affect the accuracy of the results - would it be under/over estimated and why?

The second representation is more accurate because in water there can be no ``naked'' $\ce{Fe^3+}$ ion; as it goes into solution, iron combines with water to form an ion consisting of
a $\ce{Fe}$ atom surrounded by six molecules of water. It is the oxygens that form bonds with the iron atom, in a roughly octahedral arrangement: two oxygen
atoms in each coordinate direction, one on either side. The reaction
consists of the replacement of one of these coordinated water
molecules by the $\ce{SCN-}$ ion; now the iron has bonds to 5 oxygens
and 1 nitrogen atom.

The addition of $\ce{H2O}$ to the formula doesn't mean we are
diluting the solution with water.  It just means we're letting the
water that's  present anyway play a role in the equation, to reflect reality.

It's not clear to me what they are asking in (ii).  The colors change
dramatically when you mix $\ce{[Fe(H2O)6]^3+}$ (pale yellow) with $\ce{SCN-}$
(colorless); the result is deep red. Perhaps some context will help.  Read the discussion of the reaction in the text to see what they say about color intensity.

(iv) Again, what is the context?  If the text is about the spectrospic
measuring of the equilibrium constant, low
concentrations ensure that the reation obeys the ideal law.  In
concentrated solutions the law doesn't work quite right unless you
replace concentrations by activities, something that is only
indirectly measurable (it's whatever makes the formula work).

(vi) Again, we need to know the context.

No, not all equilibrium reactions need to be in a closed vessel. An
aqueous reaction that only takes a short time (so that for example
water won't evaporate appreciably, messing up the concentration) is
just fine in a test tube.


Experimentally finding order of a reaction For the reaction of bromocresol green and bleach, I was supposed to conduct an experiment to find the order of the reaction with respect to bleach. To do this, I used a fixed amount of bromocresol green and varied the amount of bleach in each trial. I recorded the amount of time it took for each reaction to finish.

To find the order, I was told that if the order = x, $\frac{time_1}{time_2} = (\frac{[bleach]_2}{[bleach]_1})^x$. However, I'm having trouble understanding why this is true. For example, say we begin with $[bleach]_1=y$ and $[bleach]_2=2y$. If $\frac{time_1}{time_2} = 2^x$, this means that the rate of reaction 2 is always $2^x$ times faster than the rate of reaction 1.

In the beginning, this is definitely true, since rate = $k[bleach]^x$. Initial $rate_1 = ky^x$ and initial $rate_2=k(2y)^x$. But at a time $dt$ later, $[bleach]_1=y-ky^x\cdot dt$ and $[bleach]_2=2y-k(2y)^x\cdot dt=2y-2^xky^x\cdot dt$. At this time, $\frac{rate_2}{rate_1} = (\frac{2y-2^xky^x\cdot dt}{y-ky^x\cdot dt})^x$, which does not equal $2^x$ unless $x=1$.

Thus, I'm not understanding why $\frac{time_1}{time_2} = (\frac{[bleach]_2}{[bleach]_1})^x$ is true for x other than 1. Any clarifications would be much appreciated.

There are a couple of odd things here.

the time for a reaction to be complete is usually infinite in theoretical kinetics, so I don't see how it could be useful and/or relevant. One usually looks at the initial rate of a reaction.
if bleach is in large excess, you may have a degenerate order. That means that the concentration of bleach barely varies during the reaction and can be considered a constant.
you use an expression with an infinitesimal amount of time ($\mathrm{d}t$), when $\mathrm{d}t$ is everything but infinitely small...

Let me note bleach $\mathrm{ClO^-}$, and bromocresol green "BCG".

I assume that the reaction is monitored through the changes in absorbance, that is, the concentration of BCG. Its rate $r$ is the rate at which BCG is disappearing.

$$r = - \frac{\mathrm{d}[\mathrm{BCG}]}{\mathrm{d} t} = k \times [\mathrm{ClO^-}]^\alpha \times  [\mathrm{BCG}]^\beta $$ where $\alpha$ and $\beta$ are the partial orders.
If you use two different concentrations of bleach ($C_1$ and $C_2$), the ratio of rates will be :
$$\frac{r_1}{r_2} = \frac{k \times (C_1)^\alpha \times  [\mathrm{BCG}]^\beta}{k \times (C_2)^\alpha \times  [\mathrm{BCG}]^\beta} = \left(\frac{C_1}{C_2}\right)^\alpha$$

A simple way to get the initial rate is to measure the slope of the tangent to the graph of  ${[\mathrm{BCG}]}$ along time (dotted line).

This line crosses the time axis at $t=\Delta t$.

In this case,
$$ r_i(t=0) = \frac{\mathrm{[BCG] (t=0)}}{\Delta t_i}$$
Therefore, one can write  :
$$\frac{r_1}{r_2} = \left(\frac{C_1}{C_2}\right)^\alpha =  \frac{\Delta t_2}{\Delta t_1}$$
which kind of looks like the formula you were given.


Chemical structure of SiO2 vs CO2 why does SiO2 form a crystal structure while CO2 forms a linear one? Would it be possible to make a CO2 crystal or SiO2 gas in a lab or is there something preventing that from happening.

Thanks

$\ce{SiO2}$ forms a tetrahedral covalently bonded network , while $\ce{CO2}$ is a strictly molecular compound that forms double bonds between carbon and oxygen. This is because Si is in the next period, so it has 3p orbitals instead of 2p orbitals. Orbitals of a higher shell (greater principal quantum number) are physically bigger, but oxygen still has smaller 2p orbitals, therefore there is less overlap between silicon and oxygen’s p orbitals than there is between oxygen and carbon’s p orbitals. This makes a Si-O π bond much higher in energy, so it does not form. Instead, only a lower energy tetrahedral network of σ bond forms, giving us silica as we know it. This explains why $\ce{SiO2}$ cannot form double-bonded molecules like $\ce{CO2}$. Also, single Si-O bonds are especially stable (466 kJ mol-1) and short. This is attributed to the overlap between oxygen’s p orbitals and silicon’s empty d orbitals. Additionally, silicon is a larger atom than carbon, so the bonding electron pair repulsion between the oxygens bonded to silicon in SiO2 is much lower than what would occur if such a tetrahedrally coordinated CO2 network could exist. Silica’s behaviour can, thus, be explained by enthalpy, MO theory and VSPER.

Crystal $\ce{CO2}$ can definitely be made in a lab, and it is known as dry ice. $\ce{CO2}$ freezes at -56 °C. It boils at -78 °C and you will often see this temperature written under arrows in synthesis because dry ice is the most common method of carrying out reactions at such a low temperature. At atmospheric pressure, the dry ice crystal structure does not, however, resemble network covalent silica, but some high-pressure solid $\ce{CO2}$ phases have network covalent structures. Here is a phase diagram:

Phase I $\ce{CO2}$, shown in the bottom-left corner, is the most common dry ice crystal structure, and it is molecular, exhibiting a face-centered cubic arrangement with respect to the carbon. The high-pressure Phase V $\ce{CO2}$ (top-right), on the other hand, has a "partially collapsed cristobalite" structure. Isostructural with one of silica's polymorphs, it is non-molecular.

As for gas-phase $\ce{SiO2}$, it would certainly be hard to make, as it boils at close to 3000 °C, presumably for the bond-stability reasons listed above. I haven't read amy literature about its gas-phase structure, but I assume it would be similar to that of $\ce{CO2}$ but extremely unstable.


How to derive composite acidity constant for H2CO3*? I see the equation (19) and (20) on Stumm and Morgan's Aquatic Chemistry book (Fig. 1 here). They said "a combination of equations 19 and 20" gives Equation 21. But I don't know how the Equation (21) is generated? Because if I make summation of (19) and (20), it will be:

$$\ce{2H2CO3 <=> H2O + CO2(aq) + H+ + HCO3-}$$

[OP] Because if I make summation of (19) and (20), it will be: [...]

They meant they combine the mathematical expressions, not the chemical equations.

To express

$$K_\ce{H2CO3^\Huge{*}} = \frac{K_\ce{H2CO3}}{1 + K}$$

in terms of concentrations, you substitute equations (1) and (2) into the expression:

$$K = \dfrac{\ce{[CO2(aq)]}}{\ce{[H2CO3]}}\tag{1}$$
$$K_\ce{H2CO3} = \dfrac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}}\tag{2}$$

So

$$\frac{K_\ce{H2CO3}}{1 + K} = \frac{\dfrac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}}}{1 + \dfrac{\ce{[CO2(aq)]}}{\ce{[H2CO3]}}}$$

Multiplying top and bottom of the fraction by $\ce{[H2CO3]}$ gives:

$$\frac{\ce{[H2CO3]}}{1 + K} = \frac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]} + \ce{[CO2(aq)]}}$$

If you want to make the intention clearer, you define a total concentration of carbon dioxide as

$$[\ce{H2CO3^\huge{*}}] = \ce{[H2CO3]} + \ce{[CO2(aq)]}$$

and then write

$$K_\ce{H2CO3^\Huge{*}} = \frac{K_\ce{H2CO3}}{1 + K} = \frac{\ce{[H+][HCO3-]}}{[\ce{H2CO3^\huge{*}}]}$$


What factors other than Gibbs free energy determine formation of a compound? The $\Delta G_{f}^{\circ}$ value for $\ce{Fe2O3}$ is less negative than that of $\ce{Fe3O4}$ in JANAF table at room temperature but still, $\ce{Fe2O3}$ is what's formed. How does one explain this given that it's not the thermodynamically favored one?

Thermodynamics only dictates whether or not the process is favorable. It gives no indication as to whether the process will actually take place or how quickly if it does.

You're asking about chemical kinetics which measures how quickly reactions proceed. In this case, the formation of the more stable product under ambient conditions is slow because the barrier to reaction is high.


Why do electrons follow from Zn to Cu but not Cu to Zn in the lemon battery experiment? In the lemon battery experiment, as I understand, both $\ce{Zn}$ and $\ce{Cu}$ meet acid in the lemon, cause $\ce{Zn}$ and $\ce{Cu}$ release electrons to form $\ce{Zn^{2+}}$ and $\ce{Cu^{2+}}$.

So why do the electrons follow from $\ce{Zn}$ to $\ce{Cu}$ but not from $\ce{Cu}$ to $\ce{Zn}$?

In fact $\ce{Cu}$ doesn't release electrons, only $\ce{Zn}$ does. For the usual explanation see this Wikipedia article, I will try to explain it in other words to make it much easier.

One important thing that you should know is that energy comes from metal not from the acid.
Some species have the tendency to donate electrons other species the tendency to acquire them. Chemically speaking some species have more internal energy when their are in the reduced form (when they have the maximum number of electrons) others species reach their minimum internal energy when they are in the oxidized form (when they have the least number of electron). In this world every system tend to go to a state of lower energy spontaneously so if two species are not in their lowest energy form if they can exchange their electrons freely they acquire (are reduced) or give (are oxidized) electrons until they reach the lowest energy.  There are also species that do not have any particular tendency or are only little more prone to acquire electrons or to donate them. In all these cases the exchange can occur only if the electrons can move easily from one specie to another, for this reason a good electrolyte is many times needed.
An easy comparison
We could imagine chemical species as men and electrons as money.
Imagine that there are different kind of men at the antipodes there are avaricious men that needs more money and charitable men that feel the urge to give money. I think that you have no doubt which of them do what when one meet the other. The problem is that many times the transition occurs only if there is the possibility that the flow of money goes from who want to give them to who want them so you need a third agents to create the right environment and conditions for the transition. Furthermore the loss of money must be counterbalanced by something similar.
Then there are others type of men that simply don't want to give or take money we can call them noble men and then there are other men that can take money without actually feeling the need, they simply take them, I will call them hydron men.
Your case
Lemon batteries are in fact one of the more complex and debated battery you could examine. When you plug the zinc and the copper in your lemon we can say that we have three main active species: copper, zinc and acid (citric acid). Zinc is one of that metals we compared to a charitable man it tends to donate electrons:
$$\ce{Zn <=> Zn^{2+} + 2e^{-} }$$
Copper however is a noble metal, physics explain this because it has the d-bands of the electronic structure filled, so it doesn't accept zinc's electrons. Instead hydrogen from acid, can accept the electrons in this way:
$$\ce{2H+ + 2e^{-} -> H2 ^}$$
And this is in fact what actual experimental results show, see J. Chem. Educ. 2001, 78 (4), 516 for reference.


Why does Boron only need 6 valence electrons unlike the standard 8? So I was a bit confused. Hydrogen and Boron seem to be the only outliers to the octet rule. Hydrogen makes sense because it has only one shell and 2 electrons complete its shell. Boron's configuration, however, is $1s^2 2s^2 2p^1$. Having 6 valence electrons would mean having a configuration of $1s^2 2s^2 2p^4$ which doesn't seem as stable as $1s^2 2s^2 2p^6$ or even as stable as $1s^2 2s^2 2p^3$ since one would think that there is repulsion going on between electrons in the first orbital of p in $1s^2 2s^2 2p^4$. Thanks in advance!

The principle of full octet is that most reactivity patterns are described well if there is a driving force / energetic gain for full octets in main-group atoms.

However, the only facet of an atom that dictates anything about the number of valence electrons is its nuclear charge.  Boron has a charge of 5.  This is balanced by 5 electrons.  Two of them are core electrons and the remaining 3 are valence electrons.  The valence electrons may participate in bonding through sharing with other atoms, to make three bonds.  Three bonds = six electrons.

That is the simple answer. The way this question is posed indicates you are very new to the study of chemistry, and this next bit goes well beyond an introductory chemistry course.

For a fourth bond to form, another atom would have to donate two electrons to the bond.  This happens in many cases, such as when an amine reacts.  In other cases, an atom already bonded may donate electrons, such as in boron trichloride.  Finally, borane gas itself can engage in 3-center two electron bonding which is really weird, but provides a full octet when two boron atoms share the bonding electrons in a boron-hydrogen bond.

Don't distress, boron has many ways to get is full octet.


Does H3CO3+ ion exist and what is its structure? Why doesn't $\ce{H3CO3+}$ ion exist and why doesn't the carbonyl group in carbonic acid $(\ce{H2CO3})$ seem reactive?

A protonated carbonic acid (trihydroxymethyl cation) $\ce{H3CO3+}$ does exist, and is in fact incredibly stable in the solutions of superacids [1] up to $\pu{0 °C}.$ One way of obtaining $\ce{H3CO3+}$ is dissolving inorganic carbonates and hydrogen carbonates in magic acid at $\pu{-80 °C}$ [2]:

  $$\ce{CO3^2- or HCO3- ->[FSO3H-SbF5/SO2][\pu{-80 °C}] H3CO3+ ->[\pu{-10 - 0 °C}] CO2 + H3O+}$$

As for the structure, the $C_\mathrm{3h}$-symmetrical molecule (triskelion shape, structure 7 below) was found to be the global minimum [2] (reference numbers updated):

  We also carried out ab initio and IGLO calculations […] on bicarbonate as well as free and mono-, and diprotonated carbonic acids to rationalize the observed experimental results. Initially the possible geometries were optimized at the HF/6-31G* level. The most stable isomers were subsequently determined from
  relative energies at the MP2/6-31G*//HF/6-31G* level. Finally, the lowest energy structures were further optimized at the MP2/6-31G* level. MP2/6-31G* optimized structures were used to calculate $\ce{^{13}C}$ NMR chemical shifts.

  […]

  Protonated Carbonic Acid. The symmetrical $C_\mathrm{3h}$ structure 7
  was found to be the global minimum. This structure was suggested for the observed species in superacid solution by Olah et al. [1] on the basis of $\ce{^1H}$ and $\ce{^{13}C}$ NMR spectroscopy. The ion at $\pu{-80 °C}$ shows single sharp peaks at $δ(\ce{^1H})~12.05$ and $δ(\ce{^{13}C})~165.4$ in the proton and $\ce{^{13}C}$ NMR spectra, respectively.

  Protonation of carbonic acid on the hydroxyl oxygen atom (8) was shown to be unfavorable by $\pu{23.4 kcal/mol}$ over protonation on the carbonyl oxygen (7). Frequency calculation at the HF/6-31G*//HF/6-31G* level showed that the structure 8 is not a minimum, as it contains two imaginary frequencies.

$\ce{H3CO3+}$ shares structural similarities with its triaza-analog, the guanidinium ion, as both are possessing resonance stabilization via their onium forms [3, p. 60].

References

Olah, G. A.; White, A. M. Stable Carbonium Ions. LXIV. Protonated Carbonic Acid (Trihydroxycarbonium Ion) and Protonated Alkyl (Aryl) Carbonates and Hydrogen Carbonates, and Their Cleavage to Protonated Carbonic Acid and Carbonium Ions. The Possible Role of Protonated Carbonic Acid in Biological Carboxylation Processes. J. Am. Chem. Soc. 1968, 90 (7), 1884–1889. https://doi.org/10.1021/ja01009a036.
Rasul, G.; Reddy, V. P.; Zdunek, L. Z.; Prakash, G. K. S.; Olah, G. A. Chemistry in Superacids. 12. Carbonic Acid and Its Mono- and Diprotonation: NMR, Ab Initio, and IGLO Investigation. J. Am. Chem. Soc. 1993, 115 (6), 2236–2238. https://doi.org/10.1021/ja00059a020.
Dewar, M. J. S., Hafner, K., Heilbronner, E., Itô, S., Lehn, J.-M., Niedenzu, K., Rees, C. W., Schäfer, K., Wittig, G., Boschke, F. L., Series Eds.; Topics in Current Chemistry; Springer Berlin Heidelberg: Berlin, Heidelberg, 1979; Vol. 80.


How does salt dough work? Sorry for coming in here with little chemistry knowledge and a pretty inane question, but my googling skills have completely failed to cut through the enormous combined output of the baking brigade and all the lovely people making lovely things with this stuff.  Unsurprisingly not one of whom has paused to mention what's actually happening on a chemical level.

For the uninitiated, salt dough is made by mixing half a part table salt with one part wheat flour, and one half to one part water, though recipes vary greatly. This is then moulded into the desired shape and baked in an oven at anywhere between 60 and 180 °C, resulting in a hard and surprisingly tough (to me anyway) finished article which is often used as a decoration on Christmas trees. But what is actually happening?

I can guess at something involving the re crystallization of the sodium chloride, but wheat flour, whilst being so incredibly common, is to me a completely incomprehensible mix of proteins, starches and all manner of other stuff. All of which involves chemistry way over my head.

I would greatly appreciate an explanation of what's going on in this hardening process.

What you get is a polymeric (starch) composite, with salt as inorganic filler.

The salt particles give toughness to the resulting composite, while the starch phase, which is just the same as in your bread, minus the unnecessary blowing agent (baking powder, yeast, whatever), connects the particles in an elastic, flexible way. You get a material that is nearly as hard as a piece of rock salt, but much less brittle.

(Elastic here means more elastic than salt. Dry bread is quite flexible, compared to rock salt ;-) .)

See also https://en.wikipedia.org/wiki/Composite_material


Does 1-ethylidene-2-(propan-2-ylidene)cyclobutane show geometrical isomerism? I think it should be yes, but my book says no.
I think the two geometrical isomers are

Z-isomer

E-isomer

The two isomers are distinct substances, differing only in the orientation of a double bond, so they must be geometrical isomers. However, as a practical matter, they may interconvert under mild conditions and/or one may be much less stable than the other.


How to get a precise answer to a textbook precipitation problem? This question concerns a textbook problem* on precipitation. Briefly, there are $2\cdot10^{-4}$ of Mn$^{2+}$ and Cu$^{2+}$ ions in a liter of solution with $0.003$ M HClO$_4$. This solution is then saturated with H$_2$S. The solubility of H$_2$S is given as 0.1 mol/L and assumed independent of the presence of other substances in solution. K$_{sp}$ for MnS is $3\cdot 10^{-14}$, K$_{sp}$ for CuS is $8\cdot 10^{-37}.$  K$_1$ and K$_2$ for H$_2$S are given as $1.0\cdot 10^{-7}$ and $1.2\cdot 10^{-13}$ respectively.

In part one of the question it is found that Mn remains in solution and  Cu$^{2+}$ will precipitate because the product [Cu][S] = $2.6\cdot 10^{-20}$ far exceeds K$_{sp}$ of CuS. In part two the problem is to determine how much Cu remains in solution.

The solution begins by stating that "most of the Cu$^{2+}$ will precipitate" and explains that a corresponding increment of [H$^+$] will be added to the solution from the H$_2$S which has lost S to precipitate, requiring a correction which leads to $8\cdot 10^{-21}$ moles Cu remaining in solution vs. $6\cdot 10^{-21}$ moles without the correction.

To make the correction the text assumes all the Cu precipitates, so here is my question.

The question being asked is: How much Cu remains in solution? To answer the question, it is assumed that none remains in solution...(!) This leads to a slight correction (from $3\cdot 10^{-21}$ % to $4\cdot 10^{-21}$ %) Cu remaining. The text does ignore the small change in [S] that accompanies the change in [H$^+$].

  If there is a situation in which this very small difference matters, what is the right way to find it?

My guess is that there would be a simple ODE governing this problem--but I'm not sure.

The details above are included for completeness but the answer I think does not depend on the numbers here.

*The source of this problem is Shaum's College Chem. (10 ed.), 18.13-14.

Ok, let's do the "complete" and "exact" problem.

First, I'm just going to ignore the $\ce{Mn^{2+}}$. Using a first pass with the simplifying assumptions would be enough to show that MnS will not ppt.

Second the problem statement notes that:

  This solution is then saturated with H2S. The solubility of $\ce{H2S}$ is given as 0.1 mol/L and assumed independent of the presence of other substances in solution.

Chemically $\ce{H2S}$ would be bubbled through the solution to saturate it. While the $\ce{H2S}$ was bubbling $\ce{CuS}$ would be ppt'ing. So when the final concentration of $\ce{H2S}$ is 0.1 molar the ppt of $\ce{CuS}$ is complete.

Let:

The final $\ce{[H2S]_f = 0.1\bar0}$ molar
This means that the final concentration of all the sulfur species ($\ce{[H2S]_f, [HS-]_f,}$ and $\ce{[S^{2-}]_f}$) will be a bit more than $0.1\bar0$ molar.
The initial and final $\ce{[HClO4]} = 3.\bar 0\times10^{-3}$ molar
From $K_{sp}$ for CuS we know that:
$$K_{sp} = [Cu^{2+}]_f[S^{2-}]_f$$ or $$\ce{[S^{2-}]_f} = \dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}}$$
For $\ce{H2S}$ we can multiply $K_{a1}\times K_{a2}$ to get an overall ionization product $$K_t = K_{a1}\times K_{a2} = 1.2\bar0\times10^{-20} = \dfrac{\ce{[H+]_f[HS-]_f}}{\ce{[H2S]_f}}\times\dfrac{\ce{[H+]_f[S^{2-}]_f}}{\ce{[HS-]_f}} = \dfrac{\ce{[H+]_f^2[S^{2-}]_f}}{\ce{[H2S]_f}}$$
also that
$$\ce{[H+]_f} = \sqrt{\dfrac{K_t\ce{[H2S]_f}}{\ce{[S^{2-}]_f}}} = \sqrt{\dfrac{K_t\ce{[H2S]_f[Cu^{2+}]_f}}{K_{sp}}}$$
and
$$\ce{[HS-]_f} = \dfrac{K_{a1}\ce{[H2S]_f}}{\ce{[H+]_f}} = K_{a1}\ce{[H2S]_f}\sqrt{\dfrac{\ce{[S^{2-}]_f}}{K_t\ce{[H2S]_f}}}=K_{a1}\sqrt{\dfrac{\ce{[H2S]_f[S^{2-}]_f}}{K_t}}= K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}}$$
$\ce{[Cu^{2+}]_f}$ be the final concentration of copper.
The initial $\ce{[Cu^{2+}]_i}=2\times10^{-4}$ molar which we will assume to be $2.\bar0\times10^{-4}$ molar. Thus the amount of copper that is removed from solution is $2.\bar0\times10^{-4} - \ce{[Cu^{2+}]_f}$
For the final $\ce{[H+]}$ we know that
$$\ce{[H+]_f = [HClO4] + [HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - \ce{[Cu^{2+}]_f}) - (10^{-7} -[OH-])}$$ or $$ 0 = \ce{[HClO4] + [HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - [Cu^{2+}]_f) - (10^{-7} -\dfrac{K_w}{\ce{[H+]}}) - [H+]}$$
substituting for $\ce{[HClO4]}$
$$0 = 3.0000\times10^{-3} + \ce{[HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - [Cu^{2+}]_f) - (10^{-7} -\dfrac{K_w}{\ce{[H+]}}) - [H+]}$$
and collecting terms
$$0 = 3.3999\times10^{-3} + \ce{[HS^-]_f + 2[S^{2-}]_f - 2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$
substituting for $\ce{[S^{2-}]_f}$
$$0 = 3.3999\times10^{-3} + \ce{[HS^-]_f} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$
substituting for $\ce{[HS^-]_f}$
$$0 = 3.3999\times10^{-3} + K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$
and finally substituting for $\ce{[H+]_f}$
$$0 = 3.3999\times10^{-3} + K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +K_w\sqrt{\dfrac{K_{sp}}{K_t\ce{[H2S]_f[Cu^{2+}]_f}}} - \sqrt{\dfrac{K_t\ce{[H2S]_f[Cu^{2+}]_f}}{K_{sp}}}}$$

Since $\ce{[H2S]_f = 0.10}$ per problem statement, the only variable left is $\ce{[Cu^{2+}]_f}$. I'll leave it to the reader to solve this mess for $\ce{[Cu^{2+}]_f}$, but to 2 significant figures the answer should be $7.7\cdot10^{-21}$.

The whole point in using significant figures to simply the problem is not to get the "wrong" answer, but to avoid a computational mess that doesn't yield any better answer.


Spontaneous explosion of a bottle containing alcohol and milk A friend of mine gave me a bottle of home-made liquor which is a concoction of pure alcohol and sweetened milk. I accepted his gift happily and then I put the bottle in my cocktail cabinet. After an hour, we happened to hear a terrible noise of a some sort of explosion. We were indeed right - the bottle had exploded with actually no reason! Could someone please explain this strange phenomenon?

I am not sure whether this is important but he claims he had been keeping it in his fridge, although the difference of temperatures (in the fridge and in the cocktail cabinet) does not seem terribly significant to me. Also, the bottle was almost full meaning there was little room for the air in the bottle.

Please find attached the full list of ingredients:

0.5 litre of condensed, unsweetened milk,
0.5 litre of condensed, sweetened milk,
0.25 litre of pure alcohol,
few sweets.

The bottle can't have exploded for "no actual reason"; there may have been no apparent reason, but everything does indeed happen for a reason, at least at this level. But this isn't English.SE...

My bet is on a combination of things. As with many things in everyday chemistry, the substances involved are never pure enough to pinpoint one exact cause. Case in point, milk is an emulsion containing a significant percentage of fats and proteins. So, possible causes include:

Hygroscopy and acidity of the alcohol causing the milk to curdle, by taking the water from the milk to form an alcohol solution that the complex milk proteins and fats precipitate out of due to lower solubility. This may produce a "greater than the sum" volume of product by leaving these proteins and fats in a porous gel (basically cottage cheese, though usually a stronger acid than ethanol is used in cheesemaking).
Esterification. Milk, as stated, contains fats including fatty acids. Add alcohol, and the fatty acids react with the alcohol to produce esters (neutral fats). The byproduct is water, formed from the acid stripping the alcohol's hydroxyl group. Again, the products of this reaction can take up more volume than the reactants.
Biological reduction. I actually do not think the concentration of alcohol in this solution would be antiseptic; if the alcohol is pure then we're only talking about a 20% solution by volume, and that would have to be a lab-distilled product like Everclear; "natural" distillation in open air only gets you to about 150 proof, and using even 100-proof vodka gives this mixture only a 10% concentration by volume of ethanol. According to at least one source, ethanol needs to be concentrated to at least a 65% solution by volume to be effective as an antiseptic. So, to a bacteria's point of view, the alcohol is an overworked police officer; it'll slow down bacterial growth, but there's way too much fat and sugar here for the alcohol to be much of a deterrent. Consumption of organic fuel by cellular metabolism, as a rule, produces CO2 gas, and in a closed container at room temperature, well, you get the picture.


How can I make a better whiteboard cleaning solution? I just started work as a lab assistant in a chem lab. The whiteboard the instructor uses for classes gets very dirty very quickly. The woman who started at the same time as I did spent over 3 hours scrubbing it with a solution of a powdered soap-like product in a milk-carton style container to get it anything like clean.

I'd rather not watch her do that again (or, worse, do it myself). Can I make a better cleaning agent without using too much of an expensive chemical?

So far  I've seen large quantities of Ethanol, Acetone, Alcohol, and milliQ water. What else we may have, you folks would probably know better than I.

Andselisk's recommendations are great, but you'll save yourself a lot of trouble if you just throw all the erasers out. They get crud on them quickly and after a while, they put more back on the board than they wipe off.

A microfiber cloth is much more effective, and works for a shockingly long time. I would bet on a filthy cloth over a brand new eraser, and unlike erasers, they can be thrown in the wash. You can get them specifically for whiteboards or just stop by an auto parts store.


Why do different substances have different boiling points? For example, why does for example oxygen turn into gas at a much lower temperature than water?

Does it have anything to do with the molecular structure? A water molecule does have a more complex structure than oxygen, though the R-410A (a mixture of two gases commonly used in heating pumps) is much more complex than water, and it boils at -48.5 degrees Celsius.

The boiling point of a liquid depends on the intermolecular forces present between the atoms or molecules in the liquid since you must disrupt those forces to change from a liquid to a gas. The stronger the intermolecular forces, the higher the boiling point.

Two oxygen molecules are attracted to each other through London dispersion forces (induced temporary dipoles between the molecules) while water molecules are attracted to each other by hydrogen bonding (attraction of the + dipole on H in one molecule to the – dipole on an oxygen in an adjacent molecule) that is relatively strong. (Hydrogen bonding is an important intermolecular force for molecules where H is directly covalently bonded to F, O or N, which are quite electronegative and thus form bond with H with a relatively strong dipole.) London dispersion forces become more important for atoms and molecules with more electrons. Dipole–dipole attractions are also important in some molecules.


On the notion of a component in thermodynamics Wikipedia article on a component (in thermodynamics) first mentions that:

  In thermodynamics, a component is a chemically-independent constituent
  of a system. The number of components represents the minimum number of
  independent species necessary to define the composition of all phases
  of the system.

A bit later it is said in the same article that:

  The number of components is equal to the number of distinct chemical
  species (constituents), minus the number of chemical reactions between
  them, minus the number of any constraints (like charge neutrality or
  balance of molar quantities).

How does the second statement follows from the first one?

Well, a reaction comes with an equilibrium constant, i.e., mathematically speaking, some equation on constituents. Now we have N unknowns (constituents) and M equations (constraints and equilibrium constants). Supposedly N>M. This means we can't determine all unknowns. To do so, we must fix N-M free variables. These are our components.


Difference between actual position of electron and Radial Distribution Probability Its known that the radius of maximum probability of 2s orbitals is more than that of 2p orbitals. It means that the maximum probability of finding an electron in an 2s is further away from electron than that of 2p. This means 2p electron are mostly found nearer to atom than the 2s atom. Well! thats totally wrong.

So I want to know the factors due to which 2p orbital lie further to nucleus than the 2s one.

Ok, let's start out by looking at the hydrogenic atom. The wave function $\Psi(r, \theta, \varphi)$ can be split into a radial part $R_{n,\ell}(r)$ and an angular part $Y_{\ell,m} (\theta, \varphi )$, so that $\Psi(r, \theta, \varphi) = R_{n,\ell}(r) Y_{\ell,m} (\theta, \varphi )$, where $n$, $\ell$ and $m$ are the Principal, Azimuthal and Magnetic quantum number, respectively, and the functions $R_{n,\ell}(r)$ are the solutions of the radial Schroedinger equation

\begin{equation}
\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} }  - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R_{n,\ell}(r) = 0
\end{equation}

with the nuclear charge $Z$ and the mass of an electron $m_{\mathrm{e}}$.

The probability of finding the electron in the hydrogen-like atom, with the
distance $r$ from the nucleus between $r$ and $r + \mathrm{d}r$, with angle $\theta$ between $\theta$ and $\theta + \mathrm{d}\theta$, and with the angle $\varphi$ between $\varphi$ and $\varphi + \mathrm{d}\varphi$ is

\begin{align}
  | \Psi(r, \theta, \varphi) | \, \mathrm{d} \tau = \left[R_{n,\ell}(r) \right]^{2} \left|Y_{\ell,m} (\theta, \varphi ) \right|^{2} r^{2} \sin \theta \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\varphi \ .
\end{align}

To find the probability $D_{n, \ell}(r) \, \mathrm{d}r$ that the electron is between $r$ and $r + \mathrm{d}r$ regardless of the direction, you integrate over the angles $\theta$ and $\varphi$ to obtain

\begin{align}
  D_{n, \ell}(r) \, \mathrm{d}r &= r^{2} \left[R_{n,\ell}(r) \right]^{2} \, \mathrm{d}r \underbrace{\int_{0}^{\pi} \! \! \int_{0}^{2\pi} \left|Y_{\ell,m} (\theta, \varphi ) \right|^{2} \sin \theta  \, \mathrm{d}\theta \, \mathrm{d}\varphi}_{= \, 1} \\
  &= r^{2} \left[R_{n,\ell}(r) \right]^{2} \, \mathrm{d}r \ .
\end{align}

Since the spherical harmonics are normalized, the value of the double integral
is unity.

The radial distribution function $D_{n, \ell}(r)$ is the probability density for the
electron being in a spherical shell with inner radius r and outer radius $r + \mathrm{d}r$.
For the $2\ce{s}$ and $2\ce{p}$ states, these functions are

\begin{align}
  D_{2, 0}(r)  &= \frac{1}{8} \left( \frac{Z}{a_{0}} \right)^{3} r^{2} \left( 2 - \frac{Z r}{a_{0}} \right)^{2} \mathrm{e}^{-Zr/a_{0}} \\
  D_{2, 1}(r)  &= \frac{1}{24} \left( \frac{Z}{a_{0}} \right)^{5} r^{4}  \mathrm{e}^{-Zr/a_{0}} \ ,
\end{align}

where $a_{0}$ is the Bohr radius.
The graph of those two function can be seen here:

The most probable value $r_{\mathrm{mp}}$  of $r$ is found by setting the derivative of $D_{n, \ell}(r)$ with respect to $r$ equal to zero. From the graph above it can be seen that $r_{\mathrm{mp}}(2\ce{p})$ is smaller than $r_{\mathrm{mp}}(2\ce{s})$ (and if you calculate it you can check that for yourself).

But $r_{\mathrm{mp}}$ is not the quantity you are actually interested in. Because although $r_{\mathrm{mp}}$ gives you the most probable distance of the electron from the nucleus you want to know the average distance $\langle r \rangle$ between the electron and the nucleus as that is the distance you will get when you measure $r$ repeatedly and average over your results.
The radial distribution functions may be used to calculate the expectation values
of functions of the radial variable r. You can get them via

\begin{align}
  \langle r \rangle_{n, \ell}  &= \int_{0}^{\infty} r D_{n, \ell}(r) \, \mathrm{d} r \ .
\end{align}

If you evaluate this integral for the $2\ce{s}$ and $2\ce{p}$ orbitals you get

\begin{align}
  \langle r \rangle_{2\ce{s}} = \langle r \rangle_{2, 0}  &= \frac{6 a_{0}}{ Z } \\
  \langle r \rangle_{2\ce{p}} = \langle r \rangle_{2, 1}  &= \frac{5 a_{0}}{ Z } \ .
\end{align}

You find that the $2\ce{s}$ electrons are on average further away from the nucleus than the $2\ce{p}$ electrons.

So, for the hydrogenic atom your claim that $2\ce{p}$ electrons are further away from the nucleus than the $2\ce{s}$ electrons is wrong. But in real atoms you have more than one electron and those electrons will influence each other. One important effect here is, that the inner electrons (or electron density) screens nuclear charge from the outer electrons. Thus the outer electrons feel a lower effective nuclear charge $Z^{\mathrm{eff}}$ than the inner electrons. The extend of this screening is distance dependent, so $Z^{\mathrm{eff}}
= Z^{\mathrm{eff}} (r)$.
Now, if you look back at the formula for $\langle r \rangle_{n, \ell}$ things will certainly change since $D_{n, \ell}(r)$ will have a different form and the additional $r$-dependence of $Z^{\mathrm{eff}}$ will make things more difficult. Assuming that the general form of $D_{n, \ell}(r)$ will be similar in the real atom, it can be expected that the parts of $D_{n, \ell}(r)$ for small $r$ now have a higher weight in the integral of $\langle r \rangle_{n, \ell}$ than before since these parts of the electron density suffer less from the screening effect. Since $2\ce{s}$ electrons have a small portion of the electron density that lies very close to the nucleus ("penetrates deep into the core region") and $2\ce{p}$ electrons don't this will lead to the observation that $2\ce{p}$ electrons are on average further away from the nucleus than the $2\ce{s}$ electrons.

A more involved explanation of the effects at work in multi-electron atoms can be found at this answer of mine. The important bit starts at the paragraph "Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?"


Why is the melting point of KBr higher than that of CsCl? I thought that when compounds have similar forces and charges, the one with the higher molecular weight has the higher melting point. If that is true, then why is the melting point of $\ce{KBr}$ higher than that of $\ce{CsCl}$?

It may be helpful considering molecular weight for say $\ce{KBr}$ vs $\ce{KCl}$ or $\ce{CsCl}$ vs $\ce{CsBr}$, and actually melting point would go down with increasing molecular weight, but it actually has nothing to do with molecular weight despite the trend. It would be more helpful to look at lattice energy, electronegativity, electron affinity, and atomic radii, but its a little more complicated to compare a $\ce{Cs}$ salt of a halide with a $\ce{K}$ salt of a different halide as you've changed more factors.  Although chlorine has a higher electronegativity and smaller atomic radius than bromine, caesium has an even larger atomic radius than potassium (relative to the size difference between chlorine and bromine) as well as a lower electronegativity than potassium.  All of these factors will affect the lattice energy and therefore the melting points.  But, like I said, when comparing two similar salts, make sure one of the elements stays constant.  Its much more tricky to compare $\ce{KBr}$ with $\ce{CsCl}$ than it is to compare ($\ce{KBr}$ with $\ce{KCl}$) or ($\ce{CsBr}$ with $\ce{CsCl}$), or even ($\ce{KBr}$ with $\ce{CsBr}$) or ($\ce{KCl}$ with $\ce{CsCl}$).  When comparing $\ce{KBr}$ with $\ce{CsCl}$, you must determine what will affect melting point more: the ($\ce{Cs}$ vs $\ce{K}$) difference or the ($\ce{Cl}$ vs $\ce{Br}$) difference.


Why does my filtered water turn thick and almost slimy when refrigerated? I just started using a water filter.

I have 2 containers to store the filtered water; one goes in the fridge and the other stays out on the filter.

The water that stays out is fine and stays good for days.  The water that is stored in the fridge gets thick and almost tastes gluey.

Why is this happening?

Your setup does not effectively filter nutrients. Nutrients can be problematic in climates like Charleston, SC which are humid.  Humid climates plus a nutrient content in the water create the potential for fungal growth.

Part of living in humid climates is dealing with fungal growth. Fungal growth starts from spores (capable of growing in water) which are probably living in the refrigerator.  Disinfecting the refrigerator should be done by the instructions in the manual which came with the refrigerator.

This should go without saying, the water with the starchy taste should probably be thrown out. Any drinking water left to sit has the potential to grow micro-organisms, so storing water in a refrigerator is a bad idea. Try making ice cubes in the freezer and adding it to water if you want cold water.

In humid climates, anything stored in the refrigerator should be sealed so spores cannot grow.  The refrigerator must be cleaned regularly.

Filtering water on an as-need basis may be best with your setup because a carbon filter is effective at removing anything biological.  Personally, I use a pitcher because I can refill it as needed, let the whole setup dry periodically, and easily throw out saturated cartridges.


Why is there no standard temperature? The standard state for a gas is defined at 1 bar of ideal gas.  Why isn't there a corresponding standard state temperature?

The short answer

The standard pressure is not an experimental condition, while the temperature is.

The long answer

The main reason the standard state is defined is because it leads to this very useful equation
$$K_p = e^{-\Delta G^\circ/RT}$$
Say you have this reaction: $A \rightleftharpoons B + C$ One way to use this equation is to compute the free energy of 1 mol of $A$, $B$, and $C$ at 1 bar using equations derived for an ideal gas, compute $\Delta G^\circ = G^\circ (B) + G^\circ (C) - G^\circ (A)$, and use that value to predict $K_p$.

If the gasses behave like ideal gasses "in real life" then the measured  $K_p$ will match the $K_p$ computed from $\Delta G^\circ$.  You can do the measurement at any pressure you want, not just at 1 bar.* The standard state refers to the pressure you use when computing $\Delta G^\circ$.  The only thing it has to do with the experimental measurement is that it defines the units you should use for your partial pressures when computing $K_p$

$\Delta G^\circ$ does also depend on temperature, but the temperature you chose should be the same as the experimental conditions.  So the temperature is not part of the standard state definition.

But what about "Standard temperature and pressure (STP)?"

Standard temperature and pressure (STP) refers to STP conditions under which $K_p$ is measured, not the pressure used to compute $\Delta G^\circ$. I know, they couldn't have made it more confusing if they tried when they named these things.

*Of course if you do the measurements at very high pressures or low temperatures, then the assumption that the gasses behave ideally will be less valid and the measured $K_p$ will differ more from the $K_p$ computed from Equation 1.  However, that is a separate issue unrelated to the standard state because the $K_p$ in Equation 1 refers to the $K_p$ you would measure if the gasses behaved ideally at the pressure and temperature used in the experiment.


Why do some reactions need to be maintained at a particular temperature for prolonged periods of time? Some reactions require that temperatures of 70–100 °C to be maintained for periods of time as long as 24 hours. Why is this so?

Quite simply, reactions happen faster at higher temperatures (because who likes waiting), and some may not proceed at all until heated. From a statistical viewpoint, you are giving more reactant molecules more energy, which thereby increases the number of collisions, which in turn increases the number of successful collisions (those which produce product). A typical way to attain such conditions is to place a reaction under reflux.


Naming of quantum chemistry basis sets For some DFT calculations I performed with localized (Gaussian) basis sets, I wonder how the basis sets I used are properly named. The names I could gather are:

H, C and O: 6-311G(d,p)
Al: 8-511G*
Ga: 86-4111d41G

(They all come from this website.)

However, I am not sure that the above names are consistent. Specifically:

Shouldn't I write 6-311G** instead of 6-311G(d,p), for consistency with 8-511G*?
86-4111d41G looks weird to me, but I don't know how it could otherwise be named. My knowledge of basis sets naming conventions didn't really cover anything beyond the third period of the periodic table!

Regarding 6-311G**:

The * notation (one for polarization functions on the heavy atoms and ** for polarization functions on the heavy and hydrogen atoms) is an older notation belonging to the split-valence basis sets of John Pople. Many people still use this notation today but if you want to be more precise and more descriptive, the (d,p) notation is what you would want to use.  In fact, we only ever use the explicit (d,p) type notation and never use the *'s in my lab.

X-YZg

Split-valence basis sets have the aforementioned structure for us to quickly and easily determine how that basis set was constructed. The X simply is the number of primitive Gaussians that describe the core electron shells of an atom. Most of the time you will be freezing core in your computations so you split the core and valence regions of your atoms and assign various basis functions to each one.  The Y and Z components mean you have two sets of primitive Gaussian functions which describe your valence region. Because you have two sets here, you can denote this basis set as a double-zeta. In your case, a 6-311G** is in fact a triple-zeta quality basis set, which is better than a double-zeta.

For the sake of interest, a 6-31+G* includes a + symbol. This simply means that diffuse functions would be added to the heavy atoms.  A ++ symbol would extend these diffuse functions to the hydrogen atoms. Of course we would never want to use the '*' notation so I should have written this as 6-31+G(d,p).  Of course polarization functions can come in sets as well such as 6-31+G(2df,2pd), and other combinations exist.

Lastly, when using split-valence basis sets such as 6-311G*, you must be very careful to check the manual and see what function space the program is going to use by default.  I will use Gaussian 09 as an example.  By default, Gaussian 09 uses spherical harmonics as its default... EXCEPT FOR A FEW DIFFERENT TYPES OF BASIS SETS including the family of 6-31G* type basis sets which uses the CARTESIAN SPACE BY DEFAULT.  This means that rather than the 5D, 7F stuff that you may be used to using on a regular basis, invoking the 6-31G* type basis set will automatically use the 6D, 10F function set which inevitably changes the size of your basis set which means that you may have data that is not readily comparable to other things.  I always opt out for the 5D, 7F type functions.