How determine the number of significant figures of a feet/inch system?

Significant figures is a rule of thumb, to avoid giving overly precise results when the input doesn't warrant it.

In actual scientific calculations - where keeping track of the precision matters - significant figures aren't used. Instead, explicit error ranges (confidence intervals) are used. Notation of these varies, but often takes the form of a plus-minus format (14.5±0.3 m). The concept of significant figures uses implicit error ranges, conventionally set to one half of the last significant digit.

When error ranges are explicitly given, the error of a calculation result is found with formal propagation of uncertainty, rather than the rule-of-thumb significant figure rules. Sometimes this is based on a model of the distribution of error around the central value (for example, assuming that the error is normally distributed around the central value with a given standard deviation), but most commonly the error is taken as the minimum and maximum of an acceptable range of values around the main measurement. Under this scheme, the propagation of uncertainty often takes the form of the "crank three times" method, whereby you run the calculation once for the central value, once for the minimum value, and once for the maximum value.

So if you were doing a calculation like $\pu{14.5\pm0.3 m} - \pu{1.5\pm0.2 m}$, you would calculate once for the central value ($\pu{14.5 m - 1.5 m = 13 m}$), once to give you the smallest value you could get with the range ($\pu{14.2 m - 1.7 m = 12.5 m}$), and once to give the largest value your could get with the range ($\pu{14.8 m - 1.3 m = 14.5 m}$). You'd then convert this into a range, normally rounding both values such that the error value only has 1-2 significant figures: $\pu{13\pm0.5 m}$

It turns out that the rules for significant figure calculations are just a quick-and-dirty rule of thumb of doing more-or-less the same thing. I'd encourage you to try it out with a number of examples. If you assume that the error in each quantity is one half of the last significant digit, and you do the crank three times error propagation, you should come up with a result that has an error that is about one half the last significant digit of the answer by the significant figures approach - give or take a digit.

So really what you want to do is to convert the measurement of $\pu{5 ft 10 in}$. into one which has an explicit error associated with it. For this measurement, I'm guessing you'd have an error of $\pu{\pm0.5 in}$. - I say this because rounding to the nearest 10 inches seems rather silly to me. (Feel free to work the calculations below with different assumption, say with ±5 in., if you disagree.) So you have a measurement of $\pu{5 ft 10 in. \pm 0.5 in}$ in other words $\pu{5 ft} (\text{exact}) + 10\pm\pu{0.5 in}. = \pu{5 ft} (\text{exact}) \times \pu{12 in./ft }(\text{exact}) + \pu{10\pm0.5 in.} = \pu{70\pm0.5 in.}$ Note that is the same as $70$. in with two significant digits.

Since we have explicit error bars, we can use the crank three times method to convert it to feet: $70~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.8\bar{3}~\mathrm{ft}$ and $69.5~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.791\bar{6}~\mathrm{ft}$ and $70.5~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.875~\mathrm{ft}$. This is $5.8\bar{3}±0.041\bar{6}~\mathrm{ft}$. Convention holds that we round the error to one or two significant figures, and also round the main value to the same decimal place: $5.83±0.04~\mathrm{ft}$. -- This is basically what you'd have from $\pu{5.8 ft}$ with two significant digits.

The upshot is that you'd never use $\pu{5 ft 10 in.}$ in a calculation. You'll want to convert it to a single consistent unit first. When you do that, keep in mind the actual error associated with the measurement, and use that error as a gauge at how precise the converted quantity is.

(Note that if this is a question on a test or on a homework, ask your teacher how they think you should get the "correct" answer. When dealing with rule-of-thumb things like significant figures, intelligent people can vary in their interpretations. This means an answer valid in one context can be marked as wrong on the answer key because it doesn't match the formalism the teacher is using.)


With constant volume heat capacity, it is defined as, for an ideal gas, that has the internal energy equation
$$dU=TdS-\rho dV= \frac{\nu}{2} Nk_{\beta} T\ (1)$$ which is the Ideal Gas Law, obtained from the first law of thermodynamics. Using moles, (1) is equivalent to
$$ \frac{\nu}{2}nRT $$
Now, onto heat capacities.
$C_V$ is defined as,
$$C_V=(\frac{\partial U}{\partial T})_V= \frac{\partial}{\partial T}\left(\frac{\nu nRT}{2}\right)=\frac{\nu}{2} nR $$

Constant pressure heat capacity is defined as
$$C_{\rho}=C_V + nR=\frac{\nu+2}{2} nR$$, where $n$ is the amount of substance and $R$ is the universal gas constant. $\nu$ is the degrees of freedom of the molecule, which is easily obtained by thinking about a monoatomic ideal gas and the ways it can move.

Below is some fun theory surrounding the heat capacity for constant volume.
Note that constant-pressure specific heat is also defined as
$$C_{\rho}=(\frac{\partial U}{\partial T})_{\rho}=dU+\rho dV+Vd\rho=
(\frac{\partial H}{\partial T})  $$
Where $H$ is the enthalpy of heating the gas, which is
$$\mathrm{d}H=\mathrm{d}U+p\mathrm{d}V+V\mathrm{d}\rho=T\mathrm{d}S-\rho\mathrm{d}V+\rho\mathrm{d}V+V\mathrm{d}\rho=T\mathrm{d}S+V\mathrm{d}\rho$$
Enthalpy is commonly defined with a constant pressure process, thus $\mathrm{d}\rho=0$.
So, for a constant pressure, we have $$\mathrm{d}H=T\mathrm{d}S$$,
exploting the definition of entropy for a reversible change,
$$\mathrm{d}S=\frac{\mathrm{d}Q}{T} \rightarrow \mathrm{d}Q=T\mathrm{d}S $$
so, we finally get $$\mathrm{d}H=\mathrm{d}Q$$
which is a standard thermodynamic identity, and can also be used to get to the answer at the beginning of the post (though it is admittedly longer and more complicated to do so.)

Hope these hints help you!


Which units do I use in calculating heat capacity?

Yes, you would have to change it to grams.

Units must be consistent. Specific heat is defined as the energy required to raise one gram of water by one kelvin. At atmospheric pressure in a closed container, the specific heat of water is 4.184 joules.

One mole of water is 18.016 grams. So:

$E =mc \Delta T $

Where $m$ is the mass of the water, $c$ is the specific heat capacity of water, $\Delta T $ is the change in temperature, and $E$ is the energy required. Substitute in your numbers, and you shall get the right answer:

$E = (18.016\ \mathrm{g})(\frac{4.184\ \mathrm{J}}{\mathrm{g\ K}})(363\ \mathrm{K} - 293\ \mathrm{K})$

$E = 5280\ \mathrm{J}$

If you wanted to use moles directly instead of converting to grams, you would have had to use the molar heat capacity. Generally, people prefer to use specific heat. Engineers sometimes use the volumetric heat capacity as well.


An electric potential is generated when a liquid moves with respect to a solid (streaming potential). Is the reverse true?

Your assumption is correct. Electrokinetic pumps use a potential difference across an osmotic membrane, i.e. through narrow channels, the inverse of flow causing a streaming potential difference. There are other aspects of the phenomena such as electrophoresis, capillary osmosis etc.; see:

Electrokinetic Phenomena a

On the Efficiency of Electrokinetic Pumping...

and

short video demo


How to prove it mathematically that either rhombic or the monoclinic sulfur is denser?

$$\frac{\mathrm dP}{\mathrm dT} = \frac{\Delta_\mathrm{trs}S}{\Delta_\mathrm{trs}V}$$

$\Delta_\mathrm{trs}S$ is positive. $\displaystyle\frac{\mathrm dP}{\mathrm dT}$ is said to be positive as well. Thus, $\Delta_\mathrm{trs}V$ has to be positive, i.e. monoclinic sulfur has a greater molar volume, and thus rhombic sulfur is denser.


Influence of vapor pressure on surface tension

High surface tension is caused by strong intermolecular forces, the same are responsible for a low vapour pressure.

It is however quite a different matter to put a molecule on a surface and to completely strip it from a condensed phase and put it into the gas phase. There sure is no (inverse) proportionality or similar between vapour pressure and surface tension, and they don't "affect" each other.

A really stupid test question, imo.


Full equation when using vinegar, hydrogen peroxide, and salt to etch copper

The acetic Acid is used to dissolve the copper, however it cannot do so directly as the reaction of copper with protons is not spontaneous.

$$\require{cancel} \ce{Cu_{(s)} + 2H+_{(aq)} \cancel{->} Cu^2+_{(aq)} + H2_{(g)}}$$
The hydrogen peroxide is used to oxidize the copper since copper does not spontaneously react with acids directly.
$$\ce{\frac12 H2\color{red}{O2} +e- -> OH-}\\ \ce{Cu_{(s)} -> Cu^2+_{(aq)} + 2e-}$$
However, hydrogen peroxide is unstable in acid so the acetic acid is used because it reacts with hydrogen peroxide to form a peroxyacid which is more stable.

$$\ce{CH3C(O)OH + H2\color{red}{O2} -> CH3C(O)\color{red}{OO}H + H2O}$$

This peroxyacid can react directly with copper to dissolve it.

$$\ce{CH3C(O)\color{red}{OO}H_{(aq)} + Cu_{(s)} -> Cu[CH3C(O)O]OH_{(aq)}} \\ \ce{Cu[CH3C(O)O](OH)_{(aq)} + CH3C(O)OH_{(aq)} -> Cu[CH3C(O)O]2_{(aq)} +H2O}$$

As far as the sodium chloride goes, I'm not sure what that is for. My best guess is that it is catalytic in dissolving the copper similar to how salt corrodes steel

$$\ce{CH3C(O)\color{red}{OO}H_{(aq)} + Cu_{(s)} ->[\text{cat.} Cl-_{(aq)}] Cu[CH3C(O)O]OH_{(aq)}}$$

As far as the final species goes, you probably have a complex from the excess chloride

$$\ce{Cu[AcO]2 (aq) + 2NaCl (aq) -> [Cu(AcO)2Cl2]^2- (aq) + 2Na+ (aq)}$$
which is in dynamic equilibrium
$$\ce{[Cu(AcO)2Cl2]^{2-} (aq) <=>[Cl-][AcO-] [Cu(AcO)Cl3]^{2-} (aq) <=>[Cl-][AcO-] [CuCl4]^{2-} (aq)}$$

The color comes from the excess chloride that acts as a ligand to the copper From Wikipedia:

  Aqueous solutions of copper(II) chloride. Greenish when high in [$\ce{Cl−}$], more blue when lower in [$\ce{Cl−}$].


Electrophilic aromatic substitution reaction of dihydrofuran-2,5-dione with anisole

Nice! You've brought everything you need.

If you can't identify your electrophile immediately, do it the other way around: Assign function to all components. The one that is left must be your electrophile - or its precursor.

Do you remember the role of $\ce{AlCl3}$ in electrophilic aromatic substitutions? Where could it possible attack here?

Here is another hint: What is the trivial name for dihydrofuran-2,5-dione?


I conducted some experiment and I want to know the resulting redish black thing

Solvated copper(II) ions for the blue color, copper(II) oxide for the black (as hinted by IvanNeretin). To prove this hypothesis, one could dissolve the copper oxide in sulphuric acid, neutralize the solution and add ammonia. The deep blue color indicates the $\ce{[Cu(NH3)4(H2O)2]^{2+}}$ complex. The latter can also be done for the blue solution currently present.


Temperature & time necessary to carbonize all organics

There's a lot of exciting, and not well defined, chemistry that happens with seasoning cast-iron cookware.  Essentially, you are creating a cross-linked polymer film that is hydrophobic, providing a "stick-free" surface.  There are many many websites claiming the "best way" to season (or re-season) cast-iron cookware.  Few of these sites (and I'll let the OP and visitors search for those on their own) are based on science.  (Although I am not going to question tried and true family secrets, passed down from generation to generation.)  One site in particular makes a fairly convincing argument for seasoning that is science-based.  I suggest this approach can help restore your pot.

I suspect what has happened is that the scrubbing did remove the seasoning and exposed base metal to water and oxidation - your pot rusted.  The "soapy contaminants" is probably iron oxide rather than something from the dishwater, although there are such products as metal soaps which are metals chelated to long-chain fatty acids.

As for heating the pot to burn off all the residue - I doubt this is the most effective use of your time.  Scrubbing with acidified water to remove the iron oxide and a long seasoning session with an easily cross-linkable oil should get your pot back to (or better than) original condition.


Bomb calorimeter enthalpies of straight chained alcohols

I searched 'combustion enthalpies of straight chain alcohols' in Google Scholar and found a couple of links to papers that look relevant, see below. Unfortunately one looks like you have to pay for it, but if you're associated with an institution they may give you free access. The other involves cyclic alcohols, obviously not straight chain but maybe helpful nonetheless.

https://link.springer.com/article/10.1007%2FBF02885455?LI=true

http://actachemscand.dk/pdf/acta_vol_16_p0046-0052.pdf

I would imagine the molar enthalpies of combustion would increase fairly linearly as chain length increases. Each extra carbon in the chain means an extra molecule of CO2 is produced, and the strong C=O bonds formed are the major contribution to the exothermicity of the reaction. Steric or torsional effects in the larger molecules may lead to a slightly smaller than expected increase per carbon.

Hope this helps.


Does exothermic solvation mean solute is more soluble at low temp?

Think of this in terms of equilibrium constant.

$$K=e^{-{\Delta G\over RT}}=e^{-{\Delta H-T\Delta S\over RT}}=e^{-{\Delta H\over RT}}\cdot e^{{\Delta S\over R}}$$

Now it is pretty clear that exothermic reaction means negative $\Delta H$, which in turn means positive coefficient in $-{\Delta H\over RT}$, which consequently is a decreasing function, as is the whole exponential expression. The entropy change (or indeed the sign of it) does not matter.

The example of $\ce{CaCl2}$ is irrelevant, for anhydrous $\ce{CaCl2}$ is never the phase in equilibrium with the solution. Roughly speaking, one simply cannot just up and dissolve $\ce{CaCl2}$. In effect, there would be two reactions: first the hydration, which produces $\ce{CaCl2.6H2O}$ (that's the exothermic part), and then the dissolution of that.

The implicit approximation we relied upon is that our $\Delta H$ and $\Delta S$ won't change much throughout the dissolution process. This fails when it comes to insanely well soluble things (think of $\ce{NaOH},\;\ce{KOH}$) where by the time we reach equilibrium, we might even have more solute than water. Naturally, the properties of such solution would be quite different from those of pure water.


How do I list these elements according to their radius?

We have these elements:(just for ease, I'm naming them, you can just avoid it)

  $$
\begin{array}\\
Z=33&\text{Arsenic}&\ce{As}\\
Z=37&\text{Rubidium}&\ce{Rb}\\
Z=9&\text{Fluorine}&\ce{F}\\
Z=31&\text{Galium}&\ce{Ga}
\end{array}
$$

Now, let's see the periodic table (the useful region, thanks to Martin) :
$$
\begin{array}{ll|cccccccccccccccccc}
&\text{Gr.} &1 &2 &13  &15  &17 \\
\text{Per.}&\\\hline
2&&. & .& .& .& \color{red}{\ce{\tiny^9F}}\\
4&& .& .&  \color{red}{\large\ce{^{31}Ga}}& \color{red}{\small\ce{^{33}As}}& .\\
\hline
5&& \color{red}{\huge\ce{^{37}Rb}}\\
\end{array}\\
$$

Now consider this:

While going down or towards left, size increases. Equivalent, is to say the size is larger towards bottom left corner

It wouldn't be now difficult to conclude that size is as follows:
$$37>31>33>9$$


Propyl is a simple hydrocarbon with three carbons in it. Propane is $\ce{CH3CH2CH3}$.

Propyl is the unit formed by attaching that chain to something else (which implies replacing one of the carbons with another bond). There are two ways to do this: one attaches the propyl unit using the end carbon on the chain (giving $\ce{X-CH2CH2CH3}$); the other involves attaching the middle carbon to something else (giving $\ce{CH3CHXCH3}$). The first is called (at least in older terminology) normal-propyl, n-propyl or $\ce{n-Pr}$; the second is called iso-propyl, i-propyl or $\ce{i-Pr}$. These reduce the amount of space required to write the structure.

In your compound helicene I'd guess that the propyls hanging off the rings are there to make the compounds far more soluble in organic solvents as flat aromatic structures tend to be fairly insoluble due to the way the pack in crystals.


Buffer Power of the amphiprotic substance

As the curve that represents the buffer capacity does not simply rely on one fraction of the whole system but on all species of the acid-base system it is not enough to only consider the derivative of $c(\ce{HA-})$.

You also have to take care of the other deprotonated species in your system. For the case of a diprotic acid this means that $c(\ce{A^2-})$ is also important albeit not much as you can see in the following figure.

Figure 1: Buffer capacities for a diprotic acid with different amounts of assumptions taken into account – denoted by their plot labels. The lower graphs are zoomed versions of the upper graphs.

As you have come to the mass balance for $c(\ce{HA-})$ I don't have to show you how to come to the mass balance for $c(\ce{A^2-})$:
$$c(\ce{A^2-}) = \frac{c_0(\ce{H2A})~k_{a1}~k_{a2}}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+k_{a1}~k_{a2}}$$

This means that you have to derive the following equation:
$$C_B=\frac{c_0(\ce{H2A})~\left(c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}\right)}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}}$$
or if you want to include he self-ionization of water it has to be
$$C_B=\frac{k_W}{c(\ce{H3O+})}-c(\ce{H3O+})+\frac{c_0(\ce{H2A})~\left(c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}\right)}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}}$$

Let's derive (with the help of D. D. van Slyke, J. Biol. Chem. 1922, (52), 525-570 and Mathematica (or by hand if you've got time for that))
$$\beta=-\frac{\mathrm{dC_B}}{\mathrm{dpH}}=-\ln(10)~c(\ce{H3O+})~\left(\frac{\mathrm{d C_B}}{\mathrm{d}c(\ce{H3O+})}\right)$$
$$\frac{\mathrm{d C_B}}{\mathrm{d}c(\ce{H3O+})}=-\frac{k_W}{x}-1-\frac{c_0(\ce{H2A})~k_{a1}~c(\ce{H3O+})~(2~c(\ce{H3O+})+k_{a1})}{(c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+k_{a1}~k_{a2})^2}+\frac{c_0(\ce{H2A})~k_{a1}}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+k_{a1}~k_{a2}}$$

Using this final equation you should come up with plots like mine and see that the buffer capacity (for $c_0(\ce{H2A})=0.1~\ce{mol/L}$) at $\ce{pH}=\frac{1}{2}\left(\ce{pK_{a1}}+\ce{pK_{a2}}\right)$ is not zero albeit not being very high.

complete: $\beta=0.0014~\ce{mol/L}$
complete w/o water: $\beta=0.0013~\ce{mol/L}$
w/o $c(\ce{A^2-})$: $\beta=0.0001~\ce{mol/L}$
w/o $c(\ce{A^2-})$ and water: $\beta=1.1 \cdot 10^{-10}~\ce{mol/L}$

So ... is it correct to say that at maximum concentration of the amphiprotic substance the buffer capacity is zero? No but for your to acidic constants it's nearly zero.

But is it always zero at $c_\text{max}(\ce{HA^-})$?

Let's look at those systems a little bit closer. Imagine a few systems and their corresponding buffer capacity curves with $pK_{a1}=4$ and $pK_{a2}$ values between 4 and 10 like in the following figure.

If you create a plot with the buffer capacity over $\frac{1}{2}\left(pK_{a1}+pK_{a2}\right)$ you would end up with something like this

What you can see now is that the buffer capacity at $c_\text{max}(\ce{HA-})$ depends clearly on the distance between both $pK_a$ values. If they are close together the buffer capacity at $c_\text{max}(\ce{HA-})$ is high and if $\Delta pK_a \approx 3$ the buffer capacity at $c_\text{max}(\ce{HA-})$ is nearly zero but never exactly zero.


Attempting to balance a redox reaction using half reactions leaves an unbalanced component

It seems likely that you are starting with the wrong half reactions.

Since this reaction happens in water, separate the ionic species into their ions:

$$\ce{Mg(NO3)2 -> Mg^{2+} + 2NO3-}$$

$$\ce{HNO3 -> H+ + NO3-}$$

Now, one half reaction should involve magnesium (and have no nitrogen anywhere). The other half reaction should involve nitrogen only.

$$\ce{Mg -> Mg^{2+}}$$
$$\ce{NO3- -> N2O}$$


Why do copper (II) complexes contain so many valence electrons?

I'll address $\ce{[Cu(NH3)4(H2O)2]++}$, but the reason is similar for $\ce{[Cu(NH3)4]++}$

The 9 d-electrons in the complex are in non-bonding and antibonding orbitals.

See pages 6-9 of Ligand Field and Molecular Orbital Theory.

  the ligand σ orbitals are derived from AOs with energies much lower than those of the metal d-orbitals. Thus, the six bonding MOs of the complex are mostly ligand-orbital in character. These six bonding orbitals can accommodate the 12 electrons provided by the six ligand lone pairs. Therefore, the electrons provided by the ligands are largely confined to the ligands in the complex.

So, to a large degree, the 12 electrons from the ligands are in bonding orbitals, but the 9 d-electrons are in nonbonding or antibonding orbitals.

  The d electrons of the metal centre...occupy the lower energy, non-bonding metal $t_{2g}$ set and the higher energy, antibonding $e_g$ set. Therefore, the metal based d electrons tend to reside largely on the metal atom. In summary, the frontier orbitals of the complex are the non-bonding, entirely metal based $t_{2g}$ orbitals and the antibonding $e_g$ orbitals, which are mainly metal in character anyway.


Why can't this be another structural isomer of ethanol?

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

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


The diference between covalent molecule lattice and covalent network lattice

Covalent molecular compounds are actual molecules. Examples include $\ce{H_2O, NH_3, CO_2, N_2, I_2, SO_3, N_2O_4}$ etc. They are made of molecules which are made of covalently bonded atoms. Covalent network compounds form a network of bonds, and the whole object can almost be considered a "molecule," such as diamond (C network), silicon (Si network), graphene (C network), and black phosphorous (P network).

It's not a coincidence that all but one of the covalent molecules that I listed are gasses at room tenperature and pressure. The intermolcular bonds between molecules are relatively weak. On the other hand, every molecule in a network solid is surrounded in a network of strong covalent bonds.


Why do they use graphite electrodes in the Hall–Héroult process?

According to this source, graphite is used as anode material because carbon is easier to oxidize than $\ce{O^2-}$. The anodic oxidation of carbon

$$\ce{C + 2O^2- \rightarrow CO2 + 4e-}$$

proceeds more efficiently than the oxidation of $\ce{O^2-}$ at an inert anode

$$\ce{2O^2- \rightarrow O2 + 4e-}$$

$\ce{C}$ is a much better reducing agent than $\ce{O^2-}$ whose oxidation state is not changed in the reaction with graphite. As a consequence, anodic current density and the deposition rate of aluminium at the cathode will increase.


Synthesis of potassium trioxalatocobaltate(III) trihydrate

It appears that Sørensen was the first to publish the synthesis of trioxalatocobaltate(III) trihydrate [1].
A summary written in English (and that should answer both of your questions) including chemical reactions can be found in the very first volume of Inorganic Syntheses1 [2, p. 37]:

C. POTASSIUM TRIOXALATOCOBALTIATE
$$\ce{K3[Co(C2O4)3]·3H2O}$$
$$\ce{H2C2O4 + CoCO3 → CoC2O4 + H2O + CO2↑}$$
$$\ce{2 CoC2O4 + 4 K2C2O4 + PbO2 + 4 HC2H3O2 → 2 K3[Co(C2O4)3] + 2 KC2H3O2 + Pb(C2H3O2)2 + 2 H2O}$$
Twenty-three and eight-tenths grams (0.2 mol) of cobalt carbonate is dissolved in a solution of 25.2 g. of oxalic acid $(\ce{H2C2O4·2H2O})$ and 73.7 g. of potassium oxalate $(\ce{K2CrO4·H2O})$ in 500 cc. of hot water.
When the solution has cooled to 40°C., while it is vigorously stirred, 23.9 g. of lead dioxide (see synthesis 16) is added slowly, followed by 25 ml. of glacial acetic acid added a drop at a time.
The stirring is continued for an hour, during which time the color changes from red to deep green.
After the unused lead dioxide is filtered out, the trioxalatocobaltiate is precipitated by the addition of 500 ml. of alcohol.
The material appears as emerald-green needles, which are sensitive to both light and heat; The yield is 70 g. (71 per cent).

Notes

Please disregard obsolete typography and notations used in this textbook such as full stops after unit symbols, as well as the lack of white space between the numerical value and a degree symbol (as seen in “40°C.”) and use of notations such as “cc.” (cubic centimeter).

References

Sörensen, S. P. L. Kritische Präparatenstudien. III. Darstellung von Einigen Kobaltidverbindungen. Z. Anorg. Chem. 1896, 11 (1), 1–5. https://doi.org/10/fh47nq. (in German)
Inorganic Syntheses; Booth, H. S., Ed.; John Wiley & Sons, Inc.: Hoboken, NJ, USA, 1939; Vol. 1. ISBN:9780470131602, https://doi.org/10.1002/9780470132326.


Animate molecular vibrations as multiple xyz coordinate files

If I remember correctly, frequency data are in the stdout clusterfuck usually named log file, as well as in the (formatted) checkpoint files. There shouldn't be any need to prepare a cube file for this.

If GaussView is not available, GM's comment has the answer: Avogadro can visualize vibrations and I'm sure I've done this. Unfortunately, I don't have an output file at hand here.


Determining the charge on an ion through mole calculation in electrolyis

The suppose total and half reaction are:

$$\begin{align}
\ce{2 MCl_n &-> 2 M(s) + $n$ Cl2} \\
\ce{M^n+ + $n$ e- &-> M(s)} \\
\ce{2 Cl- &-> Cl2 + 2 e-}\\
\end{align}$$

The question is rather tricky.

Note that I am addressing the supposed question in hand, not the electrochemical evaluation.

In some sense, if we consider large scale motion, no electrons passed through the electrolyte, they just got on one side in and on the other side out.

In usual sense, there is 3:1 electron:halogenide molar ratio for $\ce{MCl3},$ therefore result would we $\pu{0.06 mol}.$

But in yet other sense, if we consider nanoscale motion, the $\pu{0.06 mol}$ of electrons passed through the electrolyte at the anode, and the same amount through the electrolyte at the cathode, therefore in total $\pu{0.12 mol}.$

By other words, $\pu{0.12 mol}$ is total molar amount of electrons involved in net electrochemical reaction, passing through the electrolyte between electroactive substances and electrodes.

I would say, the question is either intentionally tricky either badly formulated.


Phenol slowly oxidizes on exposure to air and turns pink in color. At first, quinone (p-benzoquinone) is formed which again reacts with phenol to form a polymerized product phenoquinone which is red in color.

There has been research on the oxidation of phenol like wet-air oxidation and catalytic oxidation but those are out of scope.
Oxidation of phenol using light has been studied using UV/$\ce{H2O2}$/ozone system. From this paper:

The photolysis with 253.7 nm radiation practically does not degrade
phenol. But the combination UV/ozone gave better results than the
using only ozone. In the degradation of phenol using ozone there are
two main reactions: the direct reaction of ozone with phenol and its
products and the reaction of the hydroxyl radical generated. Some
intermediate products as benzoquinone, catechol and hydroquinone were
detected during the degradation of phenol.

So, yes. Necessary precaution are taken such as phenol are kept at brown bottles.

I think solid phenol is less prone to oxidation due to air/light than liquid/azeotropic phenol because of $\ce{\pi}$ stacking, efficient packing of crystals and hydrogen bonding leading to stability.


Reaction of aluminum hydroxide and nitric acid

Aluminum oxide could be the product of fusing Al(NO3)3 at high enough temperature. Aluminum metal would not be a product, in water or in fused salt.

2.Aluminum metal does not readily dissolve in nitric acid because of formation of passive oxide/hydroxide film. The nuclear power industry had need of large amounts of aluminum nitrate, and it was made by dissolving aluminum metal in hydrochloric acid (very fast) and treating the solution with nitric acid, boiling off the HCl.

Aluminum metal is often prepared for welding by removing the oxide film with dilute hydrofluoric acid. The aluminum surface is bare metal for, oh, maybe a microsecond, before oxidizing back to the oxide. But the aluminum oxide film will be very thin for a few hours and easy to weld. The oxide film grows thicker in air as it ages.


An alternative approach might be something like this. The tertiary alcohol product with two identical groups suggests that a Grignard reaction with an ester might be a useful step.


Consequence of interactions between filled orbitals

On p. 261, Atkins & Friedman (2005) mention two criteria when considering the significance of orbital interactions. On the first criterion, they write:

  First, to participate significantly in bond formation, atomic orbitals must
  be neither too diffuse nor too compact. In either case, there would be only
  weak constructive or destructive overlap between neighbouring atoms, and
  only feeble bonds would result. It follows that in Period 2, ($\ce {1s,1s}$)-overlap can be largely neglected in comparison with ($\ce {2s,2s}$)-overlap, for $\ce {1s}$ orbitals are too compact to have significant overlap with each other. Indeed, it is generally safe, for qualitative discussions at least, to consider only overlap between orbitals of the valence shell, for only these orbitals are neither too compact nor too diffuse to have significant overlap.

This means that it is likely that for the $\ce {I2}$ molecule I mentioned in the question, the destabilisation from the interaction of filled inner shell orbitals (e.g. $\ce {1s, 2s, 2p}$ etc.) is probably very insignificant. As orthocresol writes in the comments, this can also be related to the overlap integral $\ce {S}$.

In orthocresol's answer here, we note that the reason behind antibonding MOs being more antibonding than bonding MOs are bonding is that the value of the overlap integral is not negligible (i.e. $\ce {S}$ is not much less than $\ce {1}$ such that $\ce {S}$ $\approx$ $\ce {0}$). However, in the case of the $\ce {1s-1s}$ overlap in the $\ce {I2}$ molecule, the orbitals may be too compact to interact such that $\ce {S}$ may actually become negligible. Thus, the antibonding and bonding MOS are destabilised and stabilised to the same extent respectively. Thus, these filled orbital interactions of the inner shell orbitals do not constitute much destabilisation.

Reference

Atkins, P.; Friedman, R. Molecular Quantum Mechanics (4th ed.). Oxford University Press, 2005.


Would a carbamate be attacked twice by an alkyl lithium cuprate reagent?

I don't think so, carbonyls don't typically react with Gilman's reagent (dimethyllithium cuprate) unless the carbonyl happens to be $\ce{{\alpha}-{\beta} }$ unsaturated. So I would expect no reaction under normal reaction conditions.


What is the difference in Da of ESI analytes with multiple charges?

More realistically if you have a molecule with Br, then the distance between isotopes:

Will be 2 $m/z$ units in case of 1 charge
And 1 $m/z$ units in case of 2 charges

You can use this information to determine if there are molecules ionized twice — their isotopes appear closer than expected.

Eventually since ESI/APCI adds $\ce{H+}$ for each charge, given original EMW was 100 Da, then:

$m/z$ in case of 1 charge: $(100+1)/1=101$ and $(102+1)/1=103$
$m/z$ in case of 2 charges: $(100+2)/2=51$ and $(102+2)/2=52$

Or in case of negative ionization you'll need to subtract 1 and 2 Da, respectively.


How to calculate the pH of a 0.325 M solution of pyridine hydrofloride?

I think part of the issue is that you are looking at the wrong reaction. I believe the reaction should be $$\ce{C5H5NHF_{(s)} ->C5H5NH+_{(aq)} + F-_{(aq)}->C5H5N_{(aq)} + HF_{(aq)}}$$ When looking at this, we can start by noting that $\ce{HF}$ is a stronger acid than $\ce{C5H5NH+}$, so the reaction will favor the reactants. Then, we can compare the reactants and note that $\ce{C5H5NH+}$ has a greater acid strength than the base strength of $\ce{F-}$  (ie $\mathrm{K_a} $ of $\ce{C5H5NH+} >\mathrm{K_b} \text{ of } \ce{F-} $) so the solution will be acidic. Its always useful to have this sort of sanity to check to make sure your answer makes sense. (For more details on this portion, Cengage has a chapter discussing acid/base mixtures)

A good approximation for the pH can be found in this paper, as shown here: $$[\ce{H3O+}]=\mathrm{K_a}*\sqrt{\frac{\mathrm{K_w}}{\mathrm{K_a * K_b}}}$$ where $\mathrm{K_a}$ is for $\ce{C5H5NH+}$ and $\mathrm{K_b}$ is for
$\ce{F-}$. This leads to $[\ce{H3O+}]=4.85*10^{-5}\to\mathrm{pK_a}=4.314$

As a comparison, the $\text{p}\ce{H}$ of $.325$M pyridinium is $2.86$, which it makes sense that the $\text{p}\ce{H}$ is higher when it is mixed with a base.


Storing hydrofluoric acid before the invention of plastics

In the original 1771 experiment, Scheele used a very simple setup consisting of a glass retort with a glass receiver (round-bottom flask).

Yes, the glass was etched to some degree by the fumes, but it was not drastic enough to destroy the apparatus.

From Anders Lennartson's The Chemical Works of Carl Wilhelm Scheele [1, p. 22]:

  3.1 Publication 1. Investigation of Fluorite and Hydroﬂuoric Acid

  Original publication: Kongl. Vetenskaps Academiens Handlingar. 1771, 32,
  120–138.
  Original title translated to English: Investigation of ﬂuorspar and its acid

  …

  The second paragraph of the paper deals with the action of acids on ﬂuorite, and
  this is where the important discovery is made. Scheele mixed powdered ﬂuorite
  with sulphuric acid (oil of vitriol) and heated it in a retort (Fig. 3.3) whereupon corrosive fumes were emitted. He repeated the experiment with some water in the receiver. He found that a white crust was formed on the water surface, and that the water contained a dissolved acid. The inner surface of the retort was etched by the fumes. Scheele identiﬁed the white crust as silicon dioxide (siliceous earth) and the residue in the retort as calcium sulphate (selenite or gypsum). This lead Scheele to the conclusion that “Fluor [ﬂuorite], therefore, consists principally of calcareous earth [calcium oxide], saturated with a peculiar acid; the nature of which we shall investigate hereafter”.

  Fig. 3.3 Schematic view of the apparatus Scheele used for preparing “acid of ﬂuorspar”

In response to critics, to prove that the glassware itself was the source of the white silicon dioxide powder, Scheele improved the setup, protecting the inner walls with tin and wax.

Continuing from [1, p. 98]:

  3.60 Publication 60. Experiments with Hydroﬂuoric Acid

  Original publication: Chemische Annalen für die Freunde der Naturlehre.
  1786, part 1, 3–17.
  Original title translated to English: New evidence for the characteristic nature of the acid of ﬂuor spar

  …

  Scheele powdered ﬂuorite in a brass mortar, mixed the powder with sulphuric
  acid (oil of vitriol) and transferred it to a retort lined with tin. The receiver was lined with wax, and partly ﬁlled with water. This time Scheele obtained a distillate free from precipitated silicon dioxide (siliceous earth), and addition of potassium hydroxide gave no precipitate (i.e. no $\ce{K2SiF6}$). Thus, this was pure hydroﬂuoric acid, $\ce{ HF_{(aq)} }$; similar results had been reported by Scheele’s friend, J.C.F. Meyer in Stettin.

References

Lennartson, A. The Chemical Works of Carl Wilhelm Scheele; Springer Berlin Heidelberg: New York, NY, 2017. ISBN 978-3-319-58180-4.


Problem with the solubility and Ksp of zinc hydroxide

There are no simple relation between the solubility and the solubility product when a doubly charged ion is involved. The measured solubility is always much bigger than the value obtained from the solubility product. This is due to the doubly charged ion.

Doubly charged ions like $\ce{Zn^{2+}}$ are usually hydrolyzed in water, and are partly transformed into basic ions like $\ce{[Zn(OH)]^+}$ by the following reaction : $$\ce{Zn^{2+} + H2O -> [Zn(OH)]^+ + H+}$$ If $\ce{Zn(OH)_2}$ is dissolved in water, it will not only produce $\ce{Zn^{2+}}$ ions but also basic ions by the following equation :
$$\ce{Zn(OH)_2 -> [Zn(OH)]^+ + OH^-}$$
As a consequence, the concentration of non complexed $\ce{Zn^{2+}}$ may be calculated from the solubility product of $\ce{Zn(OH)_2}$ but this $\ce{[Zn^{2+}]}$ is much smaller than the measured solubility of $\ce{Zn(OH)_2}$, because the measured solubility of $\ce{Zn(OH)_2}$ is the sum of $\ce{[Zn^{2+}]}$ and $\ce{[Zn(OH)]^+}$. It is even possible that ion pairs like  $\ce{Zn^{2+}(OH^-)_2}$ are dissolved in water without being dissociated.

In the case of $\ce{Zn(OH)_2}$, the measured solubility is $4.2 ·10^{-6}$ M. And the concentration of $\ce{Zn^{2+}}$ taken from $\ce{K_{sp} = 3·10^{-16}}$ is $\ce{[Zn^{2+}] = 1.95·10^{-6}}$ M. This is much less than the measured solubility. It means that the total concentration of the dissolved ion pairs $\ce{Zn^{2+}(OH^-)_2}$ and of the basic ions $\ce{Zn(OH)^+}$ is equal to the difference $(4.2 - 1.95)·10^{-6} = 2.25·10^{-6}$ M. Only $46.4$% of all the dissolved $\ce{Zn(OH)_2}$ produce  simple $\ce{Zn^{2+}}$ ions (which may be $\ce{[Zn(H_2O)_6]^{2+}}$ by the way).


I have run this exact reaction with pTSA in CH2Cl2 many times. As @NotEvans says above, wash with saturated  sodium hydrogen carbonate solution, dry over anhydrous sodium sulfate, filter and concentrate under reduced pressure


This is just an unproven guess on my part.  Whether compounds are soluble or not depends on 1) how strong the bonding (including lattice) forces are that hold the compound together (e.g. how much energy do we have to put in to liberate the ions, and 2) how much energy is recovered through solvation of the resulting ions.  I don't see anything odd about the ionic radius of the silver ion so I'm guessing that the energy of solvation of the resultant ions would be fairly normal; so it is probably not factor "2" that makes most silver salts insoluble.  Here is a link to an article that suggests that bonding in silver compounds may have a covalent component, it says, for silver salts "the structure has some contribution from covalent bonding. This appears to be bonding the structure together more tightly, as in a giant covalent structure."  The words "more tightly" catch my eye.  This seems to suggest that factor "1" may be in play in that silver salts may have increased bond\lattice energies.  If true, this would help explain the insolubility of many silver salts.  Finally I'll add that some of the soluble silver salts ($\ce{AgF, AgNO3, Ag(CH3CO2)}$) seem to involve anions that would be expected to be strongly hydrated.  The extra energy generated through solvation would further assist in solubilizing these silver salts.


Reaction between anisole and concentrated HI in excess

The correct answer here should be C. $\ce{I-}$ is an excellent nucleophile, it would attack the $\ce{C-O}$ bond in the methly group because phenoxide is a good leaving group as it is resonance stabilized and quite polarizable. So, the products are phenol and methyliodide. As for the excess of conc. $\ce{HI}$, it is just there so that some other nucleophile like $\ce{OH-}$ doesn't grab the limelight. I don't think it is energetically favourable to remove hydroxide from phenol to make it iodobenzene.


The amount of $\ce{Mg2P2O7}$ in 1.054 gram of a pure sample is:
$$ n = \frac{\pu{1.054g}}{\pu{222.57g mol-1}} = \pu{0.004736 mol}$$

There are two atoms of $\ce{P}$ in $\ce{Mg2P2O7}$, but only one in $\ce{NaH2PO4}$. Without considering the two reactions done in detail, we can infer a 2:1 ratio:

$$\ce{2 NaH2PO4 + ... -> Mg2P2O7 + ...}$$

Thus, the mass of $\ce{NaH2PO4}$ is
$$  m_\ce{NaH2PO4} = 2 \cdot \pu{0.004736 mol} \cdot \pu{120.0 g mol-1}= \pu{1.137g} $$


Your working equation is correct.
$$\ce{H2SO4 + 2NaCl ->Na2SO4 + 2HCl}$$

Find the amount of substance of protons of each solution via $n=c\cdot V$.

   $n(\ce{H+})=2~\mathrm{mol}$, $n(\ce{Cl-})=0.25~\mathrm{mol}$

What is the limiting agent?

   Chlorine

How many moles of hydrogen chloride can only be formed?

   $n(\ce{HCl})=0.25~\mathrm{mol}$

Calculate the mass of hydrogen chloride via $m = n\cdot M$

   $M(\ce{HCl})=36.5~\mathrm{g/mol}$, $m(\ce{HCl})=9.1~\mathrm{g}$


Condenser shooting out hexane in a soxhlet extraction

Heating without boiling chips or stirring will cause vapor to burst. Your condenser is not the efficient type either. Suggestion: add boiling chips (after it is cooled down); change a more efficient condenser; heat up slowly.


I don’t know where you heard about the water molecule interacting with d orbitals before displacement takes place, but the idea is wrong. Phosphorus’ d orbitals are energetically too far removed to take part in bonding. Remember that according to the aufbau principle, 3d orbitals are higher in energy than 4s! They exist and you can excite electrons into them but that’s it (and you need high energies for that excitation).

Rather, the oxygen of water interacts with the $\sigma^*_{\ce{P-Cl}}$ orbital. Since phosphorus uses an orbital with a very high p-content, it has a notable lobe on the far side allowing for interaction. An intermediate tetracoordinated phosphorus species is formed which displaces the better leaving group (chloride). Water can lose a proton at any point in this mechanism.

The mechanism is basically identical to that of a carbon-centred $\mathrm{S_N2}$ reaction except that the intermediate breaks down much faster in carbon-based reactions.


Is it possible predict the oxidizing/reducing nature of a compound?

You may or may not be familiar with the concepts of electronegativity and oxidation states, but ultimately here is the rule of thumb you seek. If your compound contains highly electronegative elements not in their most negative oxidation state, expect it to be an oxidizer. If it contains elements with low electronegativity which are not in their most positive oxidation state, expect it to manifest at least some reducing properties.

On top of that, just like you anticipated, there is a whole lot of exceptions and subtleties which together constitute the magnificent field of elementary inorganic chemistry.


What is the explanation for this stereochemical product?

No, you’re overthinking this. It’s just sterics.

The iodine will form a iodonium ion (three-membered ring with positive charge on iodine) where the double bond was. Since iodine is huge, it will come from the less sterically hindered side which is from behind the monitor plain.

The carboxylic acid then attacks the iodonium rom above closing a five-membered lactone ring (kinetically favoured against a six-membered one). The side chain carbon retains iodine pointing backwards (you can twist along the $\ce{C-C}$ bond to push the carbon into the monitor plain and iodine backwards) while the lactone oxygen coming from above means that the methyl substituent on the ring must be pointing behind the plain, again.

Also, stereochemistry on the hydroxide should be retained since it does not take part in the reaction.

Scheme 1: A not-well drawn mechanism for the reaction.


Czakó/Kürti$^{[1]}$ suggest a [1,2] shift, which in your case would mean a shift of the 5-6-bond to the 1-carbon; in concert with a free electron pair of the anionic oxygen neutralizing the 6-carbon. The carbonyl oxygen becomes anionic and accepts a proton in the next step. So instead of the 1-6-bond breaking, the 5-6-bond breaks.

Responding to a comment regarding yield: Czakó/Kürti give several references and cite one example yielding 63% over the two steps (addition of base and acid).$^{[2]}$ Other examples are given, but without yield.

1) L Kürti, B Czakó: Strategic Applications of Named Reactions in Organic Synthesis, Elsevier (2005).

2) P A Grieco, J L Collins, J C Huffman, J. Org. Chem., 1998, 63 (25), pp 9576–9579, DOI: 10.1021/jo981419z (as given by Czakó/Kürti).


Why is DBU the most popular base with an amidino group? Does the basic strength have something to do with the ring size being 7 for one of the rings?

DBU is a popular amidine base because it can be deemed to be of lower nucleophilicity than cheaper amidines e.g. formamidine.  It has been sometimes erroneously labelled as "non-nucleophilic" but seeing as it reacts with carbon dioxide and sulfonyl halides and anhydrides,  this cannot be supported. As has been pointed out, the price of bulk chemicals such as caprolactam (made from cyclohexanone which in turn comes from phenol) and acrylonitrile makes DBU accessible.
The 7-member ring structure isn't special per se but assists with steric bulk to reduce nucleophilicity of the free base and increase the nucleophilicity of the counter anion when acting as a (large) amidinium ion. This occurs through a reduced association of the ion pair.
For cost weighed up against basicity,  tetramethylguanidine is better value as an organic base if the extra strength is required.


Finding new pH after adding potassium hydroxide to citrate buffer

$\mathrm{p}K_\mathrm{a} = 3.06.$

Citric acid is a triprotic acid:

$$\ce{H3C6H5O7 <=>[K_1] H2C6H5O7- <=>[K_2] HC6H5O7^2- <=>[K_3] C6H5O7^3-}$$

The pKa given is for the second step. Citric acid is the fully protonated species, and citrate the fully deprotonated one. When you add the two, they neutralize to give the intermediate species.

  I'm trying to calculate it like initial $\mathrm{pH}$

The easiest way is to calculate the amount of protons available (3 x 10.26 mmol = 30.78 mmol) and the total amount of organic species (17.24 mmol). There would be no citrate (17.24 mmol of protons to protonate all with one proton, 13.54 mmol protons left), and no citric acid. 17.24 mmol - 13.54 mmol = 3.70 mmol would be monoprotonated and 13.54 mmol would be diprotonated. You can plug those numbers in (assuming that your volume is not crazy high or low) to get the initial pH.

  then if we add $\pu{4.72 mmol}$ $\ce{KOH},$ it's $1:3$ to citric acid.

No, because citrate is not a major species before or after adding the KOH. Instead, you would deprotonate 4.72 mmol of the diprotonated species to get 4.72 mmol more of the monoprotonated species.

$$\mathrm{pH_1} = 3.06 - \lg\frac{\pu{13.54 mmol} - \pu{4.72 mmol}}{\pu{3.70 mmol} + \pu{4.72 mmol}} = 3.04$$

  but this answer is incorrect. Where is the mistake?

I think both answers are wrong, for the reasons given above.


What are the equations giving the molar flow of different molecules linked by a reaction?

I can't tell you precisely where your reasoning starts to be wrong, but here is another way to think about the problem.

Notice that for a specific reaction as you gave here, the term $rV$ is always the same independently of the molecule you choose to determine it. It does vary for sure over time though (don't read I said it was a constant).

As you wrote:

$$\begin{cases} F_{A,in} + 0 - arV = F_{A,out} \\ F_{B,in} + 0 - brV = F_{B,out} \\ F_{C,in} + crV - 0 = F_{C,out} \\ \end{cases}$$

Thus, $$rV = \frac{F_{A,in} - F_{A,out}}{a} = \frac{F_{B,in} - F_{B,out}}{b} = \frac{F_{C,in} - F_{C,out}}{-c}$$

You can write what happens for $C$ in terms of $A$: $$\frac{F_{A,in} - F_{A,out}}{a} = \frac{F_{C,in} - F_{C,out}}{-c} \Leftrightarrow F_{C,out} = F_{C,in} + \frac{c}{a} \left(F_{A,in} - F_{A,out}\right)$$

Which boils down to, $$F_{C,out} = F_{C,in} + \frac{c}{a} X F_{A,in}$$

And if you write equations in function of a particular molecule, $A$, as you did here, make sure that the conversion you wrote $X$, is to be more precise, the conversion of $A$ at the exit of the reactor. Otherwise, it doesn't make sense.

EDIT:

Note as well that your equation for $B$ is wrong as well, using the same reasoning as before you should find: $$F_{B,out} = F_{B,in} - \frac{b}{a} X F_{A,in}$$ which make sense because the flow of $B$ must be less at the end of the reactor if this reaction really occurs.


How do dibromopyrazolones undergo base-promoted decomposition?

Although a mechanism was not reported in Carpino's paper of 1958, there is a more concise mechanism than the one offered by @Soumik Das. There is no need to dissemble structure 1 to a tricarbonyl compound. Vinylogous elimination of the elements of HBr from dibromide 1 leads to species 2. [Alternatively, attack of hydroxide at the carbonyl of 1 would also suffice.] Attack of hydroxide at the carbonyl group of 2 produces 3, which via a Grob-like fragmentation affords the propargylic carboxylate 4 of the acid 5. ADDENDUM: The ring opening of 3 should not be construed as concerted. Just saving space.

Carpino, L. A., J. Am. Chem. Soc., 1958, 80, 599.


Why is Q/nF equivalent to the amount of substance in electrolysis?

For a general electrolytic reaction,

$$\ce{M^{n+} + $n$ e^- -> M}$$

where $n$ moles of electrons are consumed in depositing 1 mole of substance ($n$ is also known as $n$-factor). So, for $x$ moles of substance, $nx$ moles of electrons would be consumed. Therefore,

$$Q = nxF,$$

dividing by $nF$ on both sides, we have

$$\frac{Q}{nF} = x,$$

which is the amount of substance.


PEG is the term usually reserved for shorter chains, while PEO is  used to refer to longer chain molecules. When polyethylene glycols are dissolved in water, they increase the viscosity of the solution. While it might be easy to model short-chain polymers in dilute solution as some kind of folded chain approaching a spherical volume, as the chains get longer and the solution becomes more concentrated (>1%), there is significant entanglement.

The picture above is for a long-chain polyacrylate, which can be adjusted to be very hydrophilic by adding alkali to generate anions all along the chain. The images show various shapes in water: some fairly tight, some looser (spread out), some spread out but very much entangled. At some point in increasing chain length or increasing concentration, it becomes difficult to project the image of a single molecule of PEO into useful predictions of solution properties. Measurements of dilute or shorter chain molecules would include viscosity; as the solution becomes more concentrated or the chain longer, the property includes some elasticity, so we have viscoelasticity, and at very high chain lengths, we have pituosity (more commonly known as stringiness or ropiness). A video shows this effect dramatically:
https://www.neatorama.com/2017/10/09/Fun-with-Polyethylene-Glycol/.
Since PEGs and PEOs are so soluble and compatible with water, the first molecules to be added just fill all the volume of water; you could calculate how long a PEO molecule would be if completely stretched out. Stirring can extend the molecule, showing effects due to viscoelasticity and pituosity.


Is water "hardness" dependent more on calcium or magnesium?

The one with the magnesium would be harder (assuming no other hardness-inducing ions) because magnesium has a lower molecular mass, so the molarity of magnesium would be higher. Assuming a 1:1 molar ratio of magnesium:calcium would give the same hardness (i.e. the same $ppm\ \ce{CaCO3}$ rating).


Does isopropyl chloride undergo SN1 or SN2 with sodium acetate?

Actually in these type of reactions, where the reactant can both possibly undergo $\mathrm{S_N1} $ and $\mathrm{S_N2} $, the reactions actually proceeds through a mixture of both type of reaction mechanism. In these type of reactions, rate is related as $$\text{Rate} = k_1 [\ce{R}] + k_2 [\ce{R}][\ce{X-}]$$ So, there is also the stereochemically inverted product and also racemised product.
Fraction of inverted product is $k_2/(k_1+k_2)$ and the fraction of racemised product is $k_1/(k_1 +k_2)$. But the actual values of $k_1$ and $k_2$ is experimentally determined and probably very difficult to determine by logical approach only.


Equilibrium concentration, where does the 5.00 for iron thiocyanate complex come from?

You are just making a small sign error:
\begin{align}
ax^2 +bx +c &= 0 & x_{(1;2)} &= \frac{\color{\red}{\pmb{-}}b \pm\sqrt{b^2 - 4ac}}{2a}
\end{align}

Taking $K=\frac13$:
\begin{align}
\frac13x^2 - \frac{21}{3}x +\frac{80}{3} &= 0 &
 x_{(1;2)} &= \frac{\color{\red}{\pmb{-}}(-\frac{21}{3})\pm\sqrt{\left(\frac{21}{3}\right)^2 - 4\cdot\frac13\frac{80}{3}}}{\frac{2}{3}}\\
&& x_{(1;2)} &= \frac{7\pm\sqrt{49-\frac{320}{9}}}{\frac{2}{3}}\\
&&           &= \left(7\pm\frac{11}{3}\right)\cdot\frac{3}{2}\\
&& x_{(1;2)} &= 16; 5
\end{align}

The first value obviously does not make any sense, hence $c(\ce{[Fe(SCN)]^2+})= 5~\mathrm{mol\, L^{-1}}$.


What is the effect of dilution on the degree of dissociation of a weak acid or base?

Your argument is valid, though I believe the scenario you describe is not within the range of applicability of Ostwald's law. Allow me to recast your example in a more obvious form.

You have a solution of weak base. You add a stronger base. The increase in hydroxide concentration leads, by Le Chatelier's principle, to a lower extent of dissociation.*

This is beyond the scope of Ostwald's law, which tacitly assumes the lack of what amounts to a common-ion effect. Ostwald's law holds if the solvent which you're using to dilute your solution does not interfere (or interferes only weakly) with your electrolyte's solubility; the added solvent must serve to decrease the concentration of your electrolyte. Evidently, then, adding a stronger base will interfere with the solubility of your weak base, and is not governed by Ostwald's law.

*There is perhaps some subtlety here. As an example, consider the salt $\ce{MX}$, which dissociates weakly into $\ce{M+}$ and $\ce{X-}$. The intuition behind the greater degree of dissociation is that (1) upon dilution, the equilibrium constant for dissociation doesn't change, whereas (2) the concentration of each species decreases. Because dissociation produces more ions than was originally present, this always leads to $Q < K$, and more dissociation will occur.

If we  add a salt $\ce{MY}$, then, we are introducing two effects: (1) decreased concentration of $\ce{X-}$ and (2) increased concentration of $\ce{M+}$. Because these effects pull the equilibrium in opposite directions, we can't conclusively determine what exactly will happen. Going back to your example, if we add a slightly stronger base, then we get exactly these two contrasting effects, and the change in the degree of dissociation is indeterminate. In contrast, if we add a much stronger base, then certainly (2) dominates and the degree of dissociation is reduced.


Example of an achiral molecule without a plane of symmetry or inversion center?

Part 1

Are there any molecules lacking plane of symmetry/center of inversion but that are achiral due to presence of an axis of improper rotation?

Yes, although such examples are very rare. In general, the point groups $S_{2n}$ $(n \geq 2)$ are where you should look. These molecules have a $S_{2n}$ rotation axis, but no plane of symmetry, and no inversion centre.
Molecules in the $S_6$ and $S_8$ point group are are exceedingly rare, and generally need to be constructed specifically to satisfy these criteria. The best examples, therefore, come from the $S_4$ point group. You can find several examples on these websites:

http://gernot-katzers-spice-pages.com/character_tables/S4.html
http://csi.chemie.tu-darmstadt.de/ak/immel/tutorials/symmetry/index7.html#Sn

In particular, one of the nicest examples is 2,3,7,8-tetramethyl-spiro[4.4]nonane from the second link. In the graphic below, I've taken the liberty of using four different coloured atoms in place of the methyl groups. The reader should therefore bear in mind that the third image is the same thing as the first, even though the colours are different.

The XYZ coordinates that I used for the pictures above are as follows. You can download these and view them in software such as Avogadro or ChemCraft, although I don't think you can perform a "reflection" per se (I manually swapped the colours between the second and third pictures to achieve this effect).
C         -2.32545        1.12231        0.91714
C         -2.19292        2.47894        0.24035
C         -0.83350        2.34003       -0.44571
C         -1.75438        0.10430       -0.11277
H         -3.31782        0.88874        1.35769
H         -1.62382        1.17900        1.77968
C         -0.98281        0.99078       -1.13371
H         -1.65567        1.19844       -1.99603
H         -0.05915        0.56089       -1.57585
C         -2.87426       -0.73718       -0.79187
C         -0.83536       -0.95866        0.55714
C         -2.33811       -2.16138       -0.80738
H         -3.27791       -0.35456       -1.75318
H         -3.74616       -0.78244       -0.10108
C         -1.65652       -2.23999        0.55910
H         -0.36215       -0.67792        1.52187
H          0.00882       -1.17785       -0.13479
B         -1.40058       -2.29939       -1.82101
H         -3.13035       -2.92594       -0.95624
O         -2.60331       -2.18962        1.57227
H         -1.04117       -3.15429        0.69961
H         -2.24954        3.33056        0.95160
F         -3.19594        2.63169       -0.70627
N          0.17857        2.27493        0.50135
H         -0.60168        3.16745       -1.14998

Part 2

Are there are also examples of vice-versa, where a molecule has a plane of symmetry/center of inversion, but lacks an axis of improper rotation and is thus chiral?

An improper rotation $S_n$ is defined by a rotation about $360/n$ degrees, followed by reflection in a plane that is perpendicular to that rotation axis.
A plane of symmetry ($S_1$) and an inversion centre ($S_2$) are special cases of an improper rotation ($S_n$). It is easier to convince yourself of the $S_1$ case: according to the definition above, $S_1$ means rotation through $360^\circ$ followed by reflection in a plane. Since rotation through $360^\circ$ obviously does nothing, this is the same as a reflection in a plane.
So, the answer to this question is no. if a compound has either a plane of symmetry or an inversion centre, that automatically means it has an improper rotation axis.


What is the correct structure of sodium thiosulfate?

Both of your Lewis structures are wrong.

In both of the shown structures, the solid line from sodium to oxygen/ sulfur implies that there is a discrete bond between these atoms. This is not the case.

The sodium ions and the thiosulfate ions form ion pairs (or ion triples) and are attracted to each other via diffuse electrostatic interactions. Usually that is what we consider ionic bonding.

In aqueous solution the ions will be very mobile. Sodium ions may coordinate to one or more thiosulfate ions, and therein to more than one site of the thiosulfate ion. At the same time thiosulfate ions may coordinate to more than just two sodium ion, or fewer. In such solutions, the solutes also coordinate to the solvent.
Ignoring any external field, the oxygen within the thiosulfate ion are equal.

The Lewis structure best describing the thiosulfate molecular ion is having a +2 formal charge at the central sulfur, and a -1 charge on each of the ligand atoms oxygen/ sulfur. (See below for details.)
One possible (and more accurate) Lewis structure is therefore:

(Note that the dashed lines do not represent bonds, but weakly coordinating interactions, but not necessarily all of them.)

More common are probably the following Lewis structure representations:

In solid state coordination of the ions depend on various factors, especially how much water is incorporated into the crystal structure. In any of these cases, it would be wrong to draw any bonds between the (molecular) ions. I'll include a few links to publications at the end of the post.

Additional thoughts of your input structures

  Which structure of sodium thiosulfate is correct? [...]

You noted correctly that in the first structure the central sulfur needs to "expand the octet". This would only be possible if d orbitals of said sulfur are involved. It has been proven many times that the contribution of d-orbitals in these hyper-coordinated species is minimal (usually below 1%).

The second structure is technically, or strictly speaking, not a Lewis structure, because dative bonds are not part of this description. However, if we allowed them as an extension, which has been proposed multiple times, then the arrow would need to point from the oxygen towards the central sulfur, as in those occurrences the more electronegative element would assume the higher electron density.

Disclaimer: The following section is a bit more advanced, and might not be appropriate to the casual reader, as another user pointed out that it might obfuscate the most important points about the structure of the anion.

Gas phase approach to determining the most likely Lewis structure

If we assume this molecule to be an ion-pair and calculate it in the gas phase, then we can draw a couple of conclusions, how a more accurate representation in the Lewis framework would look like.
I have performed a quick calculation on the DF-BP86/def2-SVP level of theory with Gaussian 09 and performed a natural bond order analysis (NBO 6.0) on it. The following picture shows the optimised geometry, alongside some inter-atomic distances.

As you can see, the sodium ions are coordinated to two oxygen, or one oxygen and a sulfur, respectively. The natural atomic charges reproduce well what I have initially stated, and reproduce well the Lewis structure I have suggested.

\begin{array}{rrr}
\text{Atom}& \text{No}& \text{Natural Charge}\\\hline
  \ce{S}  & 1 &   1.98 \\
  \ce{S}  & 2 &  -0.73 \\
  \ce{O}  & 3 &  -1.01 \\
  \ce{O}  & 4 &  -1.01 \\
  \ce{O}  & 5 &  -1.01 \\
  \ce{Na} & 6 &   0.91 \\
  \ce{Na} & 7 &   0.86 \\\hline
\end{array}

In terms of the NBO analysis, we can back this up, as it indicates no lone pair at the central sulfur (S1), three lone pairs at the oxygen (O3, O4, O5) and the ligand sulfur (S2), and four single bonds. There are vacant acceptor orbitals at the sodium ions (Na7, Na8). Here is an abridged version of the analysis:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
(skipping core orbitals)
  24. (1.98855) LP ( 1) S  2            s( 90.65%)p 0.10(  9.34%)d 0.00(  0.01%)
  25. (1.88728) LP ( 2) S  2            s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)
  26. (1.84883) LP ( 3) S  2            s(  1.80%)p54.53( 98.09%)d 0.06(  0.11%)
  27. (1.97170) LP ( 1) O  3            s( 77.50%)p 0.29( 22.49%)d 0.00(  0.01%)
  28. (1.83558) LP ( 2) O  3            s(  0.67%)p99.99( 99.25%)d 0.11(  0.07%)
  29. (1.82232) LP ( 3) O  3            s(  0.03%)p99.99( 99.89%)d 2.42(  0.08%)
  30. (1.97170) LP ( 1) O  4            s( 77.50%)p 0.29( 22.49%)d 0.00(  0.01%)
  31. (1.83558) LP ( 2) O  4            s(  0.67%)p99.99( 99.25%)d 0.11(  0.07%)
  32. (1.82232) LP ( 3) O  4            s(  0.03%)p99.99( 99.89%)d 2.42(  0.08%)
  33. (1.96648) LP ( 1) O  5            s( 77.67%)p 0.29( 22.33%)d 0.00(  0.01%)
  34. (1.83753) LP ( 2) O  5            s(  0.02%)p99.99( 99.91%)d 4.78(  0.07%)
  35. (1.82147) LP ( 3) O  5            s(  0.00%)p 1.00( 99.93%)d 0.00(  0.07%)
  36. (1.95903) BD ( 1) S  1- S  2
               ( 56.47%)   0.7515* S  1 s( 25.58%)p 2.86( 73.06%)d 0.05(  1.36%)
               ( 43.53%)   0.6598* S  2 s(  7.70%)p11.86( 91.35%)d 0.12(  0.95%)
  37. (1.97694) BD ( 1) S  1- O  3
               ( 34.82%)   0.5901* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 65.18%)   0.8073* O  3 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  38. (1.97694) BD ( 1) S  1- O  4
               ( 34.82%)   0.5901* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 65.18%)   0.8073* O  4 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  39. (1.97758) BD ( 1) S  1- O  5
               ( 34.64%)   0.5886* S  1 s( 25.28%)p 2.91( 73.66%)d 0.04(  1.06%)
               ( 65.36%)   0.8085* O  5 s( 22.34%)p 3.47( 77.48%)d 0.01(  0.18%)
 ---------------- non-Lewis ----------------------------------------------------
  40. (0.05958) LV ( 1)Na  6            s( 95.91%)p 0.04(  3.91%)d 0.00(  0.18%)
  41. (0.10019) LV ( 1)Na  7            s( 96.97%)p 0.03(  2.93%)d 0.00(  0.10%)
  42. (0.25853) BD*( 1) S  1- S  2
               ( 43.53%)   0.6598* S  1 s( 25.58%)p 2.86( 73.06%)d 0.05(  1.36%)
               ( 56.47%)  -0.7515* S  2 s(  7.70%)p11.86( 91.35%)d 0.12(  0.95%)
  43. (0.24584) BD*( 1) S  1- O  3
               ( 65.18%)   0.8073* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 34.82%)  -0.5901* O  3 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  44. (0.24584) BD*( 1) S  1- O  4
               ( 65.18%)   0.8073* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 34.82%)  -0.5901* O  4 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  45. (0.24610) BD*( 1) S  1- O  5
               ( 65.36%)   0.8085* S  1 s( 25.28%)p 2.91( 73.66%)d 0.04(  1.06%)
               ( 34.64%)  -0.5886* O  5 s( 22.34%)p 3.47( 77.48%)d 0.01(  0.18%)

Key: LP = lone pair, BD = bond, LV = lone valency, BD* = "antibond"

Alternative coordinations

There are at least two other structures of the stoichiometry $\ce{Na2S2O3}$,  essentially with the same bonding motif, but different coordinations of the sodium atoms. The first is about $\pu{2.4 kJ mol-1}$ more stable than the previously shown.

The second is about $\pu{3.2 kJ mol-1}$ higher in energy.

These are electronic energies only, but they show that the sodium atoms are not discretely bound and may interchange positions quite freely.
Given that disodium thiosulfate probably only exists in a crystal lattice which cannot be described by Lewis structures, or in solution where we have to worry about other effects and equilibria, it is obvious, that there cannot be a single Lewis structure for this compound. It simply does not exist as a discrete molecule.

Literature (selection) of crystal structures

The crystal structure of anhydrous sodium thiosulphate. Sandor, E.; Csordas, L.; Acta Cryst. 1961, 14, 237-243. DOI: 10.1107/S0365110X61000802 Crystallography online: Crystal Structure of Na2 (S2 O3)
On Polymorphism of $\ce{Na2S2O3}$. von Benda, H.; von Benda, K.; Z. Naturforsch. B 1979, 34, 957-968. DOI:
10.1515/znb-1979-0716 Alternative direct link (pdf) Crystallography online: Crystal Structure of Na2 (S2 O3), Crystal Structure of Na2 (S2 O3), Crystal Structure of Na2 (S2 O3)
Rerefinement of sodium thiosulfate pentahydrate. Prasad, S. M.; Rani, A.; Acta Cryst. E, 2001, 57, i67-i69. DOI: 10.1107/S1600536801011175 Crystallography online: Crystal Structure of Sodium thiosulfate pentahydrate
Structure of a new sodium thiosulfate hydrate. Hesse, W.; Leutner, B.; Böhn, K.-H.; Walker, N. P. C.; Acta Cryst. C 1993, 49, 363-365. DOI: 10.1107/S0108270192009399 Crystallography online: Crystal Structure of sodium thiosulphate 2/3-hydrate
The crystal structure of sodium thiosulphate pentahydrate. Taylor, P. G.; Beevers, C. A.; Acta Cryst. 1952, 5, 341-344. DOI: 10.1107/S0365110X52001027


Distinguishing Brown Ring Test for Nitrates and Nitrites

Yes, you are quite right. It has something to do with prevention of oxidation of nitrites. By controlling the pH, the nitrite present is reduced to nitric oxide ($\ce{NO}$) which reacts with ferrous sulfate to form a brown-colored complex ion, $\ce{FeSO4.NO}$. Nitrates do not react in such way and thus not registered in the test. It uses a special agent to prevent color formation or precipitation of common
interfering ions. Here is the relevant information[1]:

  The high range nitrite test is a modification of the classical brown
  ring test for nitrate using ferrous sulfate. By controlling the sample
  pH, the nitrite present is reduced to nitrous oxide, which reacts with
  the indicator to form a greenish-brown color. Nitrates are not
  registered in the test. [...]

  In an acidic medium ferrous sulfate reduces nitrogen in nitrite
  ($\ce{NO2–}$) to form nitrous oxide ($\ce{NO}$). Ferrous ions combine
  with the nitrous oxide to form a brown-colored complex ion, the color
  intensity of which is in direct proportion to the nitrite present in
  the water sample. Color development follows Beer’s Law.

  $$\ce{2Fe^2+ + 4H+ + 2NO2‾ → 2Fe^3+ + 2NO + 2H2O}$$

  $$\ce{NO + FeSO4 → FeSO4·NO}$$

Reference

https://www.hach.com/asset-get.download.jsa?id=7639984196


Does 30% hydrogen peroxide eat away at platinum?

Yes, as @Ivan said, hydrogen peroxide will not corrode platinum, instead hydrogen peroxide will get decomposed.

Catalytic decomposition of hydrogen peroxide by metals and alloys of the platinum group:

  The catalytic decomposition of dilute aqueous solutions of hydrogen
  peroxide has been studied on a number of Group VIII metals and binary
  alloys, including palladium-gold. The kinetics were measured at 27 °
  as a function of catalyst composition and pH of the solution. In
  neutral solution, the activity of the individual metals of Group VIII
  was generally greater than those of Group VIII2, the order of specific
  activity (per unit surface area) decreasing in the sequence $\ce{Pt > Os >
 Ir > Pd > Ru > Rh}$. (...)

Other references:

http://ac.els-cdn.com/S0926337306001561/1-s2.0-S0926337306001561-main.pdf?_tid=a114888e-1c29-11e7-9622-00000aacb360&acdnat=1491635258_a3e20984dda9eaaf07e78f4814383b96
http://pubs.acs.org/doi/abs/10.1021/ed080p788
http://jcsp.org.pk/ArticleUpload/2541-11513-1-CE.pdf


Is hybridization used in ab initio valence bond calculation?

(i) do people still use hybridization to construct the wavefuntion?

Yes and no. No: Only atomic orbitals are included in the base set. Yes: if the wavefunction is optimized, linear combinations of atomic orbitals are allowed, and we all know that one S-orbital + some P-orbitals give hybridized orbitals.

For $\ce{H2O}$, the lowest 3 molecular orbitals are generally these (see below):

At first, this result seems unlike the SP$^3$ bonds between hydrogen and carbon. There is a trick, however. It is called "localization of the orbitals", meaning that we generate linear combinations of molecular orbitals to generate new ones. Below, I'll give you an example in which we linear combinations of molecular orbitals $\varphi_2$ and $\varphi_3$ generate orbitals that look more like the SP$^3$ orbitals from text books.

(ii) if the answer to question (i) is yes, does the result differ from the textbook description? For instance, the lone pair electron in the $\ce{H2O}$ molecule? (is there any unitary equivalent result like orbital localization in MO-based method?)

As you may have noticed above, the results differ slightly, but I can elaborate.

Lone pairs are given by $\varphi_4$ and $\varphi_5$ below.

For $\varphi_4$, note that p$_z$ is along the z-axis, which is in the plane of the oxygen and hydrogen atoms: it is the 2-fold rotation axis.

For $\varphi_5$, note that p$_y$ is along the y-axis, which is perpendicular to the plane of the oxygen and hydrogen atoms: it is out of the plane.

For $\varphi_6$ and $\varphi_7$, note that they are unoccupied orbitals (i.e. lumo and lumo+1)

Now, orbital localization can be applied, linear combinations of $\varphi_4$ and $\varphi_5$ will give 2 lone pair orbitals, that stick slightly out of plane. I haven't tried drawing those yet, as it is not so easy to do so in this representation in 2D, so I'l try to switch to another representation...

As you see, those localized molecular orbitals closely resemble the SP$^3$ lone pair orbitals on oxygen, as predicted by valence bond theory. The only difference is the shape & probability of the small counter-part of the lobe; I included phase, not probability density here.

Does this answer your question?


What is the scaling for % completion time in a Michaelis–Menten scenario as a function of substrate concentration?

In short, yes.

To obtain the relation between τ and v, we need the actual time evolution of the concentration of the product by solving the differential equations (DEs) and then some way to get steady-state concentration. The DEs are non-linear. So we have to find approximate solutions. This paper develops a solution. It's not nice..

I will probably come back to add more. For the time being, I highly recommend those interested to browse through the paper.

And what do you mean by τ≈1 leading to failure of steady-state approximation please?

Edit: The tittle of the paper is "Analytical solution of coupled nonlinear rate equations : I. Michaelis-menten kinetics" by Phillipson


Understanding the rate of change in titration curves

This answer from a member of Chemical Forums who seems to have been quite useful to other posters in the past given his or her "karma" would appear to confirm that the shape of titration curves owes in part to their logarithmic nature:

http://www.chemicalforums.com/index.php?topic=6140.0


In the simplest terms, what is exchange current density?

Roughly speaking, it is a measure of how "dynamic" the "dynamic equilibrium" is when ancelectrode is at its equilibrium potential.

Let's say you are dealing with a copper electrode in a slightly acidic copper sulfate solution, a common situation in copper electroplating.  If you apply a high positive potential to this electrode, then of course the copper dissolves.  Upon measuring the rate of this dissolution, as the current density you are producing, you find that this current density increases exponentially as you crank up the potential, as long as you don't get any interference from mass transfer limitations or other electrode reactions.

Now switch to a highly negative potential.  Now copper deposits, and again (in the absence of mass transfer limits or other reactions) the current density varies exponentially with how much you are driving the electrode potential downwards.

What happens at intermediate potentials? The copper dissolution goes along with its exponential rate law and the deposition follows its rate law, but now these reactions are competing against each other so you macroscopically see only the difference between the plating and fissolution current densities.  You might have something like $10 \mu A/cm^2$ for dissolution and $1 \mu A/cm^2$ (these are just numbers I grabbed for this argument, not necessarily true kinetics), in which case you see a difference of $9 \mu A/cm^2$ dissolution.

At the equilibrium potential, the two current density numbers generated by the opposing rate laws are the same so you would not see any net reaction, even though in fact both are occurring together at the atomic level.  That common amount of current density being "exchanged" between the opposing reactions when you macroscopically see equilibrium, is the "exchange current density".


Given:

HClO4= 100.5, d HClO4=1.8
(ACO)2O=102, d (ACO)2O=1.1
H2O=18, d H2O=1

A) Preparation of 100 ml perchloric acid 0.1 Molar aqueous solution.

  I had to do some poking around to understand the density number. Without a temperature the density of liquids is nebulous. Wikipedia lists a density of 1.768 g/ml which seems to be for the 72.5% azeotrope.

So just blundering with the information given assume density is for 70% by weight perchloric acid.

Molarity $\ce{HClO4}$

$$\dfrac{0.70\times\pu{1.8 g/ml} \times \pu{1000 ml/L}}{\pu{100.5 g/mol}} = 12.5 M$$

Volume of conc $\ce{HClO4}$ for 100 mls of 0.1 molar

$$\dfrac{\pu{0.1 L}\times \pu{0.1 M}}{\pu{12.5 M}}= \pu{0.0008 L} = \pu{0.8 ml}$$

B) Preparation of 100 ml perchloric acid 0.1 Molar non-aqueous solution in acetic acid.

This is a bit gnarly.

Volumes are not typically additive. So typically an excess of the acetic acid would be made and this would be used to dilute the 0.8 ml of perchloric acid to 100 ml. However in this case the 0.8 ml of perchloric acid contains water too. Also no density is provided for glacial acid acid, nor for the 0.1 Molar perchloric acid in glacial acetic acid. So the problem isn't really solvable as is.

The other wrinkle here is that mixing exact stoichiometric proportions is impossible. If an anhydrous solution was wanted, then an excess of acetic anhydrous would be used or some other means to dry the solution would be necessary.

Never the less, let's blunder through some sort of solution, as poor as the problem is...

Assume that the volumes are additive.

  So we'll get the right proportions of the three components but only somewhere near 100 ml of solution, not 100 ml exactly.

$$V_{solution} = V_{anhydride} + V_{water} + V_{\pu{conc acid}}$$

$$V_{\pu{conc acid}} = \pu{0.8 ml}$$

$$\therefore V_{anhydride} + V_{water} = \pu{99.2 ml}$$

However as mentioned before, the concentrated perchloric acid contains 30% water by weight.

$$\pu{g water in acid} = 0.3 \times \pu{0.8 ml}\times \pu{1.8 g/ml} =\pu{0.4 g}$$

Since water has a density of 1 g/ml that is also 0.4 ml.

We were given that (ACO)2O=102, d (ACO)2O=1.1 so for the anhydride the moles per ml is:

$$\dfrac{\pu{1.1 g/ml}}{\pu{102 g/mol}} = \pu{0.0108 mol/ml}$$

We were given that H2O=18, d H2O=1 so for the water the moles per ml is:

$$\dfrac{\pu{1 g/ml}}{\pu{18 g/mol}} = \pu{0.0556 mol/ml}$$

Now since the perchloric acid contained 0.4 ml of water we need to figure out how many ml of acetic anhydride we need to neutralize that water.

$$\dfrac{\pu{0.0556 mol/ml}\times\pu{0.4 ml}}{\pu{0.0108 mol/ml}} = \pu{2.1 ml}$$

So we now need 99.2 - 2.1 = 97.1 ml of water/acetic anhydride mixture in a 1:1 molar ratio. For 1 ml of acetic anhydride the ml of water needed is

$$\dfrac{\pu{0.0108 mol/ml}}{\pu{0.0556 mol/ml}}= \pu{0.194 ml water}$$

Let $x$ be the multiplicative factor

$$x = \dfrac{97.1}{1.194}= 81.3$$

So:
0.8 ml conc perchloric acid
83.4 ml acetic anhydride
15.8 ml water

I loathe that the problem is inferring that volumes are additive.

Looking up some data...

\begin{array}{|c|c|c|c|} \hline
 & Mol. Wt. & Density\ (g/ml) & mol/ml \\ \hline
\pu{Water} & 18.015 & \pu{0.9950 @ 25 °C} & 0.05523\\ \hline
\pu{Acetic anhydride} & 102.09 & \pu{1.082 @ 20 °C} & 0.01052 \\ \hline
\pu{Acetic acid} & 60.05 & \pu{1.0446 @ 25 °C} & 0.01740 \\ \hline
\end{array}

  The density data is out of Pubchem. I looked for a while and couldn't find density values at same temperature, but the values are close enough to make the point...

assume 1.0000 ml of acetic anhydride and appropriate water to give 1:1 molar ratio

\begin{array}{|c|c|c|} \hline
 &  ml & g \\ \hline
\pu{Water} & 0.1904  & 0.1894\\ \hline
\pu{Acetic anhydride} & 1.0000 & \pu{1.082 g} \\ \hline
\pu{Totals} & 1.1904 & 1.271 \\ \hline
\end{array}

So density of acetic acid by combining volumes is
$$\dfrac{1.271}{1.1904} = 1.067$$
which gives an error of
$$\dfrac{1.067-1.0446}{1.0446}\times 100\% = +2.1\% $$

Thus the total solution as calculated above would be closer to 98 ml than 100.


How does this method of calculating the water equivalent of a calorimeter work?

So you're mixing hot and cold water in a calorimeter, and from the temperature change read from a time versus temperature curve, you're finding the heat capacity of the calorimeter. You're not understanding why the heat capacity is computed the way it is.

First, identify all of your heat flows. You've got heat coming from the hot water, heat absorbed by the cold water, and heat absorbed by the calorimeter. If energy is conserved, $$q_{hot} + q_{cold} + q_{calorimeter} = 0$$

One thing that's confusing you is that last term, $q_{calorimeter}$. It isn't zero. It's very small, since your calorimeter is Styrofoam, but it will be measureable. Measuring it is a central goal of this experiment!

Now, you have a time versus temperature curve for the contents of the calorimeter. You're extrapolating the temperature plateaus before and after mixing to the time of mixing to figure out the initial and final temperatures. Since you had the cold water sitting in the calorimeter initially, and you dumped the hot water into it, the final temperature minus the initial temperature gives you the temperature change for the cold water. It also gives the approximate temperature change for the calorimeter if you can assume that the temperature of the calorimeter is the same as the temperature of its contents.

The heat capacity of the calorimeter is $$C = \frac{q_{calorimeter}}{\Delta T_{calorimeter}}$$
You know $\Delta T_{calorimeter}$ from your time vs temperature graph. You know $q_{calorimeter}$ from the energy conservation equation above; $q_{calorimeter} = -q_{hot}-q_{cold}$. Therefore you know $C$. You should know how to get $q_{hot}$ and $q_{cold}$ from the masses and initial and final temperatures of the hot and cold water (and the specific heat of water). Can you take it from there?


In which pH range are silica-based TLC plates stable?

Assuming "normal" TLC with silica as stationary phase, and a mix of ethyl acetate / hexane as eluent, charged compounds (i.e., salts) do not elute well; their adsorption is too tight to move forward.

Hence, if your reaction is performed under acidic conditions (sulfonation reaction, for example), drop a small sample of your reaction into a test tube or small vial, gently quench it against an aqeuous solution of $\ce{Na2CO3}$, extract it with some ether / MTBE / methylene chloride (well...) and spot the extract on the TLC plate.  With some training, this works easily around o.5 mL of total volume.

Conversely, if your reaction proceeds under basic conditions (like a Grignard reaction, for example), bring the sample back to neutral, for example with dilute HCl (or, with more fragile products like tertiary alcohols, with an aqueous solution of $\ce{NH4Cl}$), and then spot the organic extract on the plate.

I referred to "normal" TLC as there are other stationary phases than silica, too.  Alumina (basic, neutral, acidic) works with the same principle as silica.  These two contrast to applications of RP-TLC, were the stationary phase is coated with alkyl chains, and much more polar solvents are deployed as eluents, hence with much lesser issues of free salts / free acids. But the price tag for these plates is higher than for normal phase TLC.


How to apply Faraday's law to calculate the charge of metal ions?

When you're confused by a problem like this it is definitely time to use dimensional analysis.

For the amount of electrons $\text{mol}_e$

\begin{align}
\text{The ampere} &= \text{Amp} = 1.00 \frac{\text{coulomb}}{\text{second}} \approx 6.242\times10^{18}~\frac{\text{electrons}}{\text{second}}\\
\text{Farady} &= 26.801 \frac{\text{Amp}\times\text{hours}}{\text{mol}_e}\\
30~\text{minutes} &= \frac{30 \text{minutes}}{60~\text{minutes/hour}} = 0.500~\text{hours}\\
\text{mol}_e &= \frac{0.50~\text{Amp} \times 0.500~\text{hours}}{26.801~\text{Amp}\times\text{hours}\text{/mol}_e} = 9.328\times10^{-3}~\text{mol}_e
\end{align}

For the amount of substance of metal $\text{m}_m$

$$\text{mol}_m = \frac{0.700~\text{g}}{150~\text{g/mol}_m} = 4.667\times10^{-3}~\text{mol}_m$$

Charge per atom = $z$

$$z = \frac{\text{mol}_e}{\text{mol}_m}= \frac{9.328\times10^{-3}}{4.667\times10^{-3}} = 1.998 = 2.0 \frac~{\text{electrons}}{\text{atom}}$$


Why does Co2+ have 7 electrons in the 3d orbital, and not 5 like Mn?

Well, you can find some kind of explanation why the electronic configuration of $\ce{Co^2+}$ is $[\ce{Ar}] \, 3d^7$ here and there, but at the end of the day, you just need to * memorize it*.

  When d-block elements form ions, the $4s$ electrons are lost first. Source

The $n+l$ rule tells you the order in which atomic orbitals are filled, and according to the rule the $4s$ orbital is occupied before the $3d$ orbital because it has lower energy.
Thus, the electron configuration of $\ce{Mn}$ is $[\ce{Ar}] \, 3d^5 4s^2$ while that of $\ce{Co}$ is $[\ce{Ar}] \, 3d^7 4s^2$.
But, the $n+l$ rule, as many other rules of old quantum theory, is not 100% working, and thus, sometimes gives wrong electron configuration.
For instance, in general, it is applicable only for neutral atoms in their ground state, and thus, if you apply it to cation $\ce{Co}^{2+}$ you will get the wrong electron configuration [\ce{Ar}] $3d^5 4s^2$.

As it is sometimes explained, the statement that $4s$ orbital is lower in energy than $3d$ orbital is true only when the orbitals are unoccupied. But while you fill $3d$ orbital with electrons it becomes lower and lower in energy and eventually ends up lower in energy than the $4s$ orbital. Thus, when electrons are lost from $\ce{Co}$ atom, they are lost from the 4$s$ orbital first because it is actually higher in energy when both $3d$ and $4s$ are filled with electrons.

P.S. And if I remember correctly this has nothing to do with stabilization in ligand fields.


Calculate the cathode electrode potential in this redox reaction

The equation that corresponds to the standard reduction potential is:

$$\ce{Mg^{2+} + 2e- -> Mg}$$

Notice that this is not the actual half reaction in your total equation, where magnesium is oxidized. But the book is determining the reduction potential, so the magnesium ion is a reactant not a product. In the reaction quotient, the concentration/activity of magnesium ion should be in the denominator.

I'm not sure I agree that we should choose the standard reduction potential as the electrode potential for the electrode where oxidation occurs, but at least the book is consistent.


What is the purpose of a boiling stick when heating a reaction mixture?

Boiling chips and boiling sticks are practical solutions to problems chemists face in the lab.

When boiling liquids they often observe the the boiling does not happen evenly. Instead of a gentle warming giving gentle bubbling we find that nothing happens for a while until, suddenly, a great deal of bubbling happens and this causes the liquid to splash overwhelming the capacity of the vessel and throwing the liquid into places we don't want it to be (during distillations, for example, the splash may throw liquid to the top of the distillation column, contaminating the product with raw liquid; during reactions the product can be expelled from the vessel entirely which is rarely what you want).

Boiling chips or sticks ensure the boiling happens evenly,  keeping things under control.

The problem they solve is that, in a smooth vessel, there are often not many rough patches to initiate bubbles. Bubbles are much easier to initiate if there is a small rough patch on the vessel wall. Chips or sticks provide such a rough patch. In the absence of a rough patch some liquids will superheat and may spontaneously vaporise but suddenly and violently. Roughness ensures bubbles start to form evenly and in many places avoiding superheating and making the vaporisation much more even.

The idea that things won't happen this way as "If the liquid is above the boiling point, why hasn't it turned into a gas then?" is a mistake from applying the ideas of thermodynamic equilibrium to a situation that is clearly not in equilibrium. If you are heating something it is clearly not in equilibrium and you have to worry about the practical effects not the ideal thermodynamic view. In practice, vaporisation is encouraged by nucleation sites (rough patches on the glass, for example). If there are not enough of those, it may not happen evenly.


Understanding gas stoichiometry for the reaction of xenon and fluorine

Yes, this is a rounding error.
Where you calculate:
$$n = \frac{pV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.18 mol}$$
The actual unrounded answer is $\pu{0.181072885 mol}$ - a little bit higher than the rounded value.
So, in your final equation:
$$\pu{0.181072885 mol} \times \pu{207.1 g//mol} = \pu{37.5 g}$$
I find using the full value of an intermediate step usually eliminates these types of errors in the final calculation.


Interpreting equilibrium equations correctly

However, each reaction at a different temperature has a different equilibrium constant. With that said, the amount of each molecule will change which also changes the concentration so the amount of NH3 would not necessarily be 2.

This is true. I have not watched the video, but if he is suggesting that the stoichiometric coefficients define the ratios of the equilibrium concentrations of the compounds, then he is wrong. Stoichiometric coefficients do play an important role in equilibrium expressions, but not that one.

  He raised each concentration to the power of the starting coefficient for that molecule. If you start out with $\ce{3H2}$ and you need a wildly different concentration to reach equilibrium, why are we raising the concentration to the power of the starting amount of $\ce{H2}$?

This is how the equilibrium constant (and more broadly the reaction quotient) is defined.

Take your generic reaction:

$$\ce{aA + bB <=> cC + dD}$$

The reaction quotient $Q$ is defined as the product of the concentrations of the products divided by the product of the concentrations of the reactants. Since $\ce{a}$ equivalents of $\ce{A}$ are in the reactants, $[\ce{A}]$ is multiplied $\ce{a}$ times for $[\ce{A}]^{\ce{a}}$.

$$Q_c = \frac{\prod\lbrace[\text{Products}]_i\rbrace}{\prod\lbrace[\text{Reactants}]_i\rbrace} = \frac{[\ce{C}]^{\ce{c}}[\ce{D}]^{\ce{d}}}{[\ce{A}]^{\ce{a}}[\ce{B}]^{\ce{b}}} $$

The pressure version of $Q$ replaces concentrations with partial pressures $P_i$.
$$Q_p = \frac{\prod\lbrace (P_{\text{products}})_i\rbrace}{\prod\lbrace( P_{\text{reactants}})_i\rbrace} = \frac{P_{\ce{C}}^{\ce{c}} P_{\ce{D}}^{\ce{d}}}{P_{\ce{A}}^{\ce{a}} P_{\ce{B}}^{\ce{b}}}$$

The equilibrium constant ($K_c$ for concentrations and $K_p$ for pressures) is defined as the reaction quotient at equilibrium concentrations/pressures. At equilibrium:

$$Q_c = K_c = \frac{[\ce{C}]^{\ce{c}}[\ce{D}]^{\ce{d}}}{[\ce{A}]^{\ce{a}}[\ce{B}]^{\ce{b}}}$$

$$Q_p = K_p =\frac{P_{\ce{C}}^{\ce{c}} P_{\ce{D}}^{\ce{d}}}{P_{\ce{A}}^{\ce{a}} P_{\ce{B}}^{\ce{b}}}$$

The values of $K_c$ and $K_p$ are determined by experimentally measuring the concentrations or pressures of $\ce{A}$, $\ce{B}$, $\ce{C}$, and $\ce{D}$ and plugging them into the expressions above.

For the specific reaction of nitrogen and hydrogen to form ammonia, the equilibrium constant (at a given temperature) is determined by measuring the equilibrium concentrations/pressures and plugging them into the expressions below.

$$\ce{3H2(g) + 2N2(g) <=> 2NH3 (g)}$$

$$K_c = \frac{[\ce{NH3}]^3}{[\ce{H2}]^3 [\ce{N2}]^2}$$

$$K_p = \frac{P_{\ce{NH3}}^2}{P_{\ce{H2}}^3 P_{\ce{N2}}^2}$$

Once we know the value of $K_c$ or $K_p$, we can then use it and some partial concentration/pressure data to predict the equilibrium concentrations/pressures.

At 300 K, the $K_c$ of this process (the Haber Reaction) is $4.34\times10^{-3} \text{ M}^{-3}$. If we started with $[\ce{H2}]=3.0 \text{ M}$, $[\ce{N2}]=2.0 \text{ M}$, and $[\ce{NH3}]=2.0 \text{ M}$, we can calculate the equilibrium concentrations of all three species when the dust has settled. First we calculate $Q_c$ and compare to $K_c$. If $Q_c$ is greater than $K_c$, there is too much product, and the reaction will shift toward reactants. If $Q_c$ is less than $K_c$, there is too much reactant, and the reaction will shift toward products.

$$Q_c = \frac{[\ce{NH3}]^3}{[\ce{H2}]^3 [\ce{N2}]^2} = \frac{(2.0 \text{ M})^2}{(3.0 \text{ M})^3 (2.0 \text{ M})^2}= \frac{1}{27 \text{ M}^3}=3.70\times10^{-2} \text{ M}^{-3}$$

$$ 3.70\times10^{-2} > 4.34\times10^{-3} \space \therefore \space Q_c > K_c  $$

At this combination, the reaction shifts toward reactants. $[\ce{H2}]$ and $[\ce{N2}]$ increase, and $[\ce{NH3}]$ decreases, all by a factor of $x$ multiplied by the stoichiometric coefficients of each species. The new concentrations are $[\ce{H2}]=3.0+3x \text{ M}$, $[\ce{N2}]=2.0+2x \text{ M}$, and $[\ce{NH3}]=2.0-2x \text{ M}$. The equilibrium expression is:

$$K_c 4.34\times 10^{-3} = \frac{(2-2x)^2}{(3+3x)^3 (2+2x)^2}$$

Since I chose the starting concentrations to be those values implied by the stoichiometry, I ended up (coincidentally) with a reaction near equilibrium. Thus, I cannot make assumptions about the value of $x$ relative to the concentrations, so no simplification is possible. To avoid mistakes, I will let WolframAlpha handle the algebra and solve for $x=3.18\times 10^{-1}$. The equilibrium concentrations are thus:

$$[\ce{H2}]_{eq} = 3+3(3.18\times 10^{-1})=3.95 \text{ M}$$
$$[\ce{N2}]_{eq}= 2+2(3.18\times 10^{-1})=2.64 \text{ M}$$
$$[\ce{NH3}]_{eq} = 2-2(3.18\times 10^{-1})=1.36 \text{ M}$$

Note that if we had started at different initial concentrations, we would end with different equilibrium concentrations. The maths are easier if you start with zero of one or two species.

I could also go on about the other way to determine the equilibrium constant (i.e. from the relationship $\Delta G^o=-RT\ln{K}$), but I have gone on long enough.


When finding the liquid density which gives the most accurate results, a 25 ml Pycnometer or a graduated 10 ml cylinder?

Cylinder is always a "secondary standard" in volumetric measurements. It is never used in quantitative transfers. A pycnometer is designed to hold known volumes and it should always be used for density measurements at known temperature. Good quality pycnometers come with a certificate of quality assurance.

If you are interested in a deeper investigation, analytical glasswares come in two grades, Class A and Class B defined by so-called tolerances by various organizations. If you look closely you will see these labels on quality glassware. Similarly, look for TD and TC label on cylinders and other volume measuring glasswares. TD means to deliver and TC means to contain. If you don't see any A or B label, chances are that this glassware just good for educational purposes (or economical version). European union and other countries have set limits on volumetric accuracy.


Nomenclature priorities with ethers, acyl groups and ketones

In principle, your assumptions about the order of priority for classes is correct. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

  (…)
  7. acids
  (…)
  9. esters
  (…)
  15. aldehydes
  16. ketones
  (…)
  42. ethers
  (…)

Thus, it is correct that the names for all compounds given in the question have the suffix ‘oic acid’.

However, the names proposed by you still contain a few errors.

Your name “(2-oxopropionyl)oxy butanoic acid” is ambiguous since the locant for the complex substituent group ‘(2-oxopropionyl)oxy’ is missing.
The complete name is “4-[(2-oxopropionyl)oxy]butanoic acid”.
However, the preferred name is 4-[(2-oxopropanoyl)oxy]butanoic acid.

Your other names include the correct suffixes and locants; however, the suffixes are not cited in alphabetical order. The correct names are

4-methoxy-3,4-dioxobutanoic acid (not “3,4-dioxo 4-methoxy butanoic acid”)
6-oxo-3-(2-oxopropyl)heptanoic acid (not “3-(2-oxopropyl),6-oxo heptanoic acid”)
6-oxo-3-propanoylheptanoic acid (not “3-propionyl,6-oxo heptanoic acid”)
6-oxo-3-(3-oxopropyl)heptanoic acid (not “3-(3-oxobutyl),6-oxo hexanoic acid”)


Something about the idea of Molarity is confusing me

Correct me if I'm wrong, but don't you need at least 2 solutes to make a solution?

You are, unfortunately, incorrect. In wet chemistry, you need just a solvent$^\dagger$ and at least one solute to make a solution.  While this does mean that you need at least two constituents to make up a solution, you do not need at least two solutes.

  ... is it saying that the ONLY chemical in the bottle is this $\ce{H2SO4}$?

It depends on your definition of a chemical:

If you consider water to be a chemical, then no: there are two chemicals in the bottle$^\ddagger$, $\ce{H2O}$ and $\ce{H2SO4}$.
If you are not considering water to be a chemical, then yes: the only chemical in the bottle$^\ddagger$ is $\ce{H2SO4}$.

$^\dagger$ If not specified, the solvent is almost always water.

$^\ddagger$ Except for small amounts of inevitable impurities.


Why is methoxy group an electron donating group?

The lone pair on the oxygen delocalizes into the conjugated aromatic ring.

Edit (based on Jan's suggestion):

Notice that the answer is actually a lot more complicated. A methoxy substituent actually impacts the ring electronics via two competing effects. The oxygen's lone pair is well-placed to delocalize and increase electron density within the ring's conjugated system. This allows delocalization to better stabilize positive charges. So the methoxy is electron-donating from a resonance perspective.

On the other hand, because oxygen is quite electronegative, the methoxy group is electron-withdrawing in an inductive sense via the $\sigma$ bonds.

Take a look at a linear free energy relationship (LFER) plot. For acidity of benzoic acids, the methoxy group has a positive $\sigma_{\mathrm{meta}}$ value but a negative $\sigma_{\mathrm{para}}$ value.

The meta $\sigma$ value measures the effect when there is no resonance structure that delocalizes effectively to the carboxylic acid, and its positive value tells us that the methoxy is electron-withdrawing in that position. The negative para $\sigma$ value tells us that when there is effective conjugation into the carboxylic acid, the methoxy is electron-donating. Notice that it is still electron-withdrawing inductively, but the resonance donation is much stronger.


Do the names of some elements have anything to do with planets?

Neptunium (93) and plutonium (94) were so named because they followed uranium (92) on the periodic table.

Uranium was named after Uranus. German chemist Martin Heinrich Klaproth discovered uranium in 1789, eight years after William Herschel discovered Uranus. At the time, uranium was the densest element known and Uranus was the farthest planet from the sun known.


Carbon monoxide bonds to transition metals using "synergistic π* back-bonding." The bonding has three components, giving rise to a partial triple bond. A sigma bond arises from overlap of the nonbonding (or weakly anti-bonding) sp-hybridized electron pair on carbon with a blend of d-, s-, and p-orbitals on the metal. A pair of π bonds arises from overlap of filled d-orbitals on the metal with a pair of π-antibonding orbitals projecting from the carbon atom of the $\ce{CO}$. The latter kind of binding requires that the metal have d-electrons, and that the metal is in a relatively low oxidation state ($<+2$) which makes the back donation process favorable.

As electrons from the metal fill the π-antibonding orbital of $\ce{CO}$, they weaken the carbon-oxygen bond compared with free carbon monoxide, while the metal-carbon bond is strengthened. Because of the multiple bond character of the $\ce{M-CO}$ linkage, the distance between the metal and carbon atom is relatively short, often $< 1.8~\mathrm{Å}$, about $0.2\ \mathrm{Å}$ shorter than a metal-alkyl bond.