CO2 removal from mushroom growing tent I'm trying to grow oyster mushrooms that like high temperatures. They look great in an indoor terrarium, and can eventually yield several tasty meals :)

However oyster mushrooms are hard to grow indoors, because they need low $\ce{CO2}$ levels (<1000 ppm according to Paul Stamets), AND high humidity (>95%).

Warm climate oysters tend to grow fast, and can generate quite a lot of $\ce{CO2}$ fast.

If I vent the air in tent about 10 times a day, humidity will suffer (and I can't do that all the time).

If I close the tent, mushrooms will be stunted and club-like, and yields will be next to nothing.

I'm thinking about putting a tray with $\ce{NaOH}$ crytals on it, probably with a fan to slightly move the humid air around a bit.

An other method would be to use an aquarium pump to bubble air through NaOH solution.

My questions are:

Does this look like something feasible? Can some $\ce{NaOH}$ remove enough $\ce{CO2}$ from the air to keep it under 1000ppm?
Which of the above methods seem to be the better one given the aquarium pump can bubble about 100-200 litres of air in every hour and the tent is about 500 L ?

Please stay away from sodium hydroxide, it's way too nasty to use in non-lab setup. You can handle it safely in dry conditions, but in contact with water, it becomes sticky and causes bad burns.

As a substitute, I suggest $\ce{Ca(OH)_2}$, readily available as slaked lime. You can structure it into relatively high-surface sculptures and it solidifies into common plaster upon binding enough $\ce{CO_2}$. It is often used for this purpose due to its low price, low toxicity and ease of handling.

In your situation, I would just put some vertically fixed mesh soaked in very viscous suspension of calcium hydroxide. In the high humidity conditions you'd have no problems with normal drying, so once it solidifies it has done its job and you can just hit the mesh against the ground to crumble the plaster away and put on new layer. It is advisable to use gloves and goggles, but it is not as necessary as it would be in the case of sodium hydroxide (look at all the masons around).


Does (2-fluoroethyl)benzene undergo elimination via E1cb or E2? I have seen in two different sources claiming that $\ce{PhCH2CH2F}$ gives styrene $\ce{PhCH=CH2}$ when treated with alcoholic $\ce{KOH}$, but one source suggests an E1cb mechanism, while the other suggests an E2 mechanism.

I think that the phenyl group should not be able to promote the formation of a carbanion in an E1cb mechanism. All the examples of E1cb mechanisms I have seen involve deprotonation of compounds which have $\mathrm{p}K_\mathrm{a}$ values of 20 or less, and toluene does not fit this.

I think that it is easier for the molecule $\ce{Ph-CH2-CH2-F}$ to undergo E1cb mechanism than E2 mechanism. It can be also possible that the molecule undergoes elimination through a mixture of E1cb and E2, but it can be said  that majority portion of the molecules will prefer the E1cb rather than any other pathway.

First of all, an E2 mechanism goes via simultaneous cleavage of the $\ce{C_{\beta}-H}$ and $\ce{C_{\alpha}-LG}$ bonds (LG being the leaving group). So, the rate of the reaction depends on the ease of breaking these two bonds. In this case the leaving group is fluoride ($\ce{F-}$), which has a very strong bond with the $\alpha$-carbon, making it a bad leaving group. So, due to this reason, the rate of E2 should lower for this molecule.

Now, if we check the plausibility of E1cb reaction pathway, we will find that it is most suitable in this case.

Firstly, the proton abstraction by the base gives a highly resonance stabilised benzylic carbanion intermediate, and this carbanion formation is the slow step. As this intermediate's energy will decrease, the activation energy will decrease resulting in an increase in the rate constant as, $k = A\exp(-E_\mathrm a/RT)$, helping to increase the rate for E1cb.

Secondly, the breaking of $\ce{C-F}$ bond doesn't happen in the rate determining step, which also doesn't alter the rate of the reaction  in this case.

So, observing the above mentioned cases, it can be definitely concluded that  E1cb is the more preferable pathway to undergo $\beta$-elimination to form Styrene.


Can an atom bond with more than 8 other atoms? Is it possible for an atom to bond with 8 other elements (same or other type)? If yes, then please give some examples. If no, then what could be the possible reason for it?

My question is not about valence shell electrons. My question is about whether an element could combine with more than 8 other same or different elements? It could be rather covalent bonds or any bond.

Yes, there are coordination complexes of large elements which have coordination numbers greater than eight. Some examples are:

$\ce{[ReH9]^2-}$ with a tricapped trigonal prismatic structure. The nine hydride ligands are small enough to fit around the relatively large rhenium atom fairly easily. This ion can be isolated as a potassium salt $\ce{K2ReH9}$.

$\ce{[U(NO3)6]^2-}$ with an impressive coordination number of 12, consisting of six nitrate ligands each forming two coordinate bonds to the central uranium ion.


What am I assuming when I make this approximation? Suppose you dissolve $\ce{AgBr}$ in a $1$ litre solution of M molar $\ce{Na2S2O3}$. What is the solubility of $\ce{AgBr}$?

I am given that the $\beta_2$ of $\ce{Ag(S2O3)2^3-}$ is $2.8*10^{13}$, and the $\ce{Ksp}$ of $\ce{AgBr}$ is $5.4*10^{-13}$.

So, I have the equations:

$\ce{\frac{[Ag(S2O3)2^3-]}{[S2O3^2-]^2*[Ag+]}} = 2.8*10^{13}$

$\ce{[Ag+][Br-]} = 5.4*10^{-13}$

$\ce{[Br-] = [Ag+] + [Ag(S2O3)2^3-]}$ (mass balance for $\ce{AgBr}$)

$\ce{M = [S2O3^2-] + 2[Ag(S2O3)2^3-]}$ (mass balance for $\ce{S2O3^2-}$)

How would I solve this system of equations for any $M$? What approximations could I make?

The approximations you can do are the following :

1 - Practically all silver atoms are included in the complex ion. So that the concentration of that ion is equal to [Br-]

2 - The amount of AgBr dissolved is negligible with respect to the amount of Na2S2O3 available. So that, at the end, $\ce{[S2O3^{2-}]}$ is equal to the concentration M of the thiosulfate.

With these approximations, you may rewrite $\beta{_2}$, so that [Ag+] is :
$$\ce{[Ag+]= {[Ag{_2}(S_2O_3^{3-}]}/{(M·{\beta_2})}} = [Br^-]/(M·{\beta_2})$$
With the solubility product of AgBr , you have a system of two equations with two unknowns. The first solution [Ag+] will be very small. But the second solution, namely [Br-] gives you the solubility of AgBr you are looking for.


Reaction of metals with water What does metals form when they react with water,for instance sometimes they form metal oxide but sometimes they form metal hydroxide.so how does this vary

In general, the reaction of metals with water could produce either the oxide or the hydroxide.  Roughly speaking, most metals are capable thermodynamically of displacing hydrogen from water to form the hydroxide, but in many cases you need such a high temperature to run the reaction that the hydroxide is not stable and thus we go over to the oxide.

To show the variety of possible reaction routes let's look at a few cases:

Iron

Iron does not react with water alone under ambient conditions; a simple reaction between iron and water requires steam at a high enough temperature to destabilize any hydroxides.  You therefore get only oxides, generally $\ce{FeO}$ or $\ce{Fe3O4}$.

But if you add the supplemental oxidizing power of air, then iron can react with both the water and the air to generate rust -- a mixture of hydroxide and oxide compounds, or even  oxide-hydroxides like $\ce{FeO(OH)}$ with iron(III).  The hydroxides become possible precisely because the combination of air and water lowers the reaction temperature to make it so.  See Wikipedia for a fuller explanation of the many reactions involved in the rusting of iron.

Magnesium and Calcium

Magnesium is something of an intermediate case between metals that clearly react with water alone at ambient temperatures and those that do not.  Magnesium reacts slowly at ambient temperature, possibly generating enough hydrogen bubbles to temporarily float pieces of the light metal in the water.  Reaction is faster in boiling water.  In both of these cases the hydroxide $\ce{Mg(OH)2}$ is a stable product; but this decomposes at $350 \, \pu{^\circ C}$   From that temperature upwards, including burning the magnesium in steam, you will get $\ce{MgO}$.

Calcium more clearly reacts with water under ambient conditions to give $\ce{Ca(OH)2}$, but this also transitions to forming $\ce{CaO}$ as the hydroxide decomposes at $580 \, \pu{^\circ C}$.

Alkali metals

In general, alkali metals react so rapidly with water and the hydroxides are so stable that at least for all practical purposes, we expect exclusively the hydroxides.


How much Carbon Dioxide would be required to displace enough oxygen to prevent ignition? In air, how much $\ce{CO2}$ would be required to prevent a match to light or a butane lighter to work?

I have a basement office that is unusable because of an apparent lack of oxygen, currently it is my guess that the oxygen is being displaced by $\ce{CO2}$.

A bit research suggests that it would take a reduction to around 14% for a safety match to fizzle out without any flame. But my naive estimate suggests 7% $\ce{CO2}$, and at that level isn't $\ce{CO2}$ deadly?
And this isn't even accounting for the displacement of nitrogen, surely that would mean the levels of $\ce{CO2}$ would be even higher.

Would a 6% increase in $\ce{CO2}$ displace oxygen by only small percentage say 2% as it's only a smaller ratio of the atmosphere? Or would the presence of $\ce{CO2}$ contribute to the lack of ignition compounding the lack of oxygen?

See my related question on the DIY stack exchange

UPDATE

An insepector from the local council has installed a data-logger, whilst he was in the basement he took a reading that I've attached below. When this reading was taken the air felt poor, but by no means the worst it's been, a candle wouldn't stay alight. An breathing during conversation seemed laboured. It looks like the assumption of displacement doesn't look correct, there's more to it.
Here's the readings:

Gas - Reading (Normal)

$\ce{CO2}$ 3.3% (0.04%)
$\ce{O2}$ 17.5% (20.8%)
Other 79.2% (unchanged?)

The parameter you are asking about is called the Minimum Oxygen Concentration (MOC), and is a measurement of the minimum amount of oxygen present in air to propagate a flame. The amount depends on the fuel being burned, and varies from about 4% for hydrogen to about 18% for some solid combustibles.

Let us consider your example of butane, and round some numbers to make our calculations simple. Butane has a MOC of 12%. Normal oxygen content in air is 20%, with nitrogen and sundries making up the remaining 80%. Assume uniform mixing and displacement of gases is equal. In order to reduce the oxygen content to 12%, you need to introduce enough CO2 to make up 40% of the air content:

$O2 + N2 = 100$  =>20 + 80 = 100%

$O2 + N2 + CO2 = 100$    => 12 + (12*80/20) + CO2 =100%

However, let us consider a couple of other very significant points.

An oxygen level of 12% is unsafe. Anything below 19.5% is unsafe. At levels of 12%, a person will have impaired judgement and coordination, and experience increased pulse rates. Prolonged exposure can have fatal consequences.
A CO2 level of 40% is fatal. Full stop. Levels above 7% can cause loss of consciousness and just 10% can be fatal.
CO2 is actually denser than air, so mixing of the main components is unlikely to be uniform. This means that higher CO2 concentrations will be found lower down in the room.

I hope you stay well clear of your basement until you have resolved this problem and have appropriate active ventilation installed, as well as possibly an oxygen monitor.

In the very least, get yourself a canary.


Why does a carbonyl group form instead of alkene in alcohol oxidation? I am looking over mechanism of the oxidation of alcohol by chromic acid and I am confused on why $\ce{H2O}$ abstracts hydrogen from alpha carbon and form double bond between $\ce{C}$ and $\ce{O}$ atom rather than beta hydrogen and form alkene like in usual elimination reaction we learned. Like to start from beginning i am confused why in chromate ester, which is an intermediate, entire group of $\ce{O-CrO3H}$ does not leave as learning group. Why does only $\ce{CrO3H}$ leave, leaving $\ce{O}$ atom behind which thus makes $\ce{H2O}$ abstact hydrogen from alpha carbon instead of abstracting beta hydrogen and creating alkene.

Technically, it would be possible for an entire chromate group $\ce{CrO4^2-}$ to be eliminated along with a β-hydrogen to give a double bond. However, that elimination would have to proceed via either an $\mathrm{E2}$ or an $\mathrm{E1_{cb}}$ mechanism as the chromium(VI) centre by itself is much more happy being reduced to chromate(IV) (the intermediate chromium species) in the electrocyclic breakdown of the complex you have been taught.

The entire reaction proceeds in acidic conditions, so any conjugate base mechanism is basically impossible. And likewise, there is no sufficiently strong base present that would allow an $\mathrm{E2}$ mechanism to occur, either.


O-H infrared band dependence of alcohol concentration in water With regard to IR spectroscopy, how would the O-H band change as an alcohol solution was diluted with water? I am referring to any alcohol, but for the sake of example we can use ethanol. I think first that the $\ce{O-H}$ band shown would increase in intensity. I believe this is the case because the water is becoming a greater proportion of the mixture and water is explicitly made from $\ce{O-H}$ bonds whereas the $\ce{O-H}$ bond in ethanol only represents a fraction of the bonds in the molecule. Is this correct? Is there also a wave number shift that would occur too that I am missing?

The easiest thing to do is look at some example spectra. All of them are taken from the Spectral Database for Organic Compounds.

This is 1-propanol, first as a liquid film and then in $\ce{CCl4}$.

In the liquid, the broad peak at $\pu{3333 cm^{-1}}$ is due to intermolecular hydrogen bonding, just as you would see in an IR spectrum of bulk pure water. The $\ce{CCl4}$ spectrum shows that as you reduce the probability of 1-propanol (or any alcohol) molecules contacting each other, the free stretch appears at $\pu{3637 cm^{-1}}$. This is similar to water in a gas-phase spectrum.

  water is explicitly made from $\ce{O-H}$ bonds whereas the $\ce{O-H}$ bond in ethanol only represents a fraction of the bonds in the molecule

Close. It isn't about the fraction of total bonds, in this case it's simply that water has two hydroxyl stretches and ethanol only has one. As water is added, the intensity of the hydrogen bonding $\ce{O-H}$ band should increase over the pure alcohol hydrogen bonding band, but we cannot know quantitatively what the increase will be without performing an experiment.

  Is there also a wave number shift that would occur too that I am missing?

Yes. In the gas phase, the water symmetric and asymmetric $\ce{O-H}$ stretches are at $\pu{3657 cm^{-1}}$ and $\pu{3756 cm^{-1}}$, which then red-shifts to $\pu{3404 cm^{-1}}$ in the bulk due to hydrogen bonding. Compare this to 1-propanol, where the "free" and hydrogen bonding values are $\pu{3637 cm^{-1}}$ and $\pu{3340 cm^{-1}}$, respectively. The water stretches are blue-shifted (higher in energy) compared to the alcohol.

Water vibrational frequency values taken from http://www1.lsbu.ac.uk/water/water_vibrational_spectrum.html.


What is the highest known bond order? In science we learn about single, double, and sometimes triple bonds. From a quick search I have found up to sextuple bonds.

Is there a maximum bond order? If yes/no, what causes this?

In the paper that you've mentioned$^{\ast1}$, there is something described, which is called effective bond order ($\mathrm{EBO}$).
$$\mathrm{EBO} = \frac{1}{2}\left(\sum_{i=1}^N \eta_{b,i}-\eta_{a,i}\right)$$
... with $\eta_b = 2 − x$ as the occupation number of an occupied orbital and $\eta_a = x$ as the occupation number of the corresponding antibonding orbital. (In multireference quantum chemistry methods, the resulting averaged orbitals can have occupation values of $0 \leq x \leq 2$)
In 2013 Ruipérez, et al. calculated the effective bond order for several homo- and heteroatomic dimers.$^{\ast2}$ Based on their very high quality calculations, the

[...] results show that the effective bond order ($\mathrm{EBO}$) of the $\ce{MoU}$ dimer (5.5) is higher than that for the tungsten dimer (5.2), known to date as the molecule with the highest $\mathrm{EBO}$.

So somewhat arbitrarily, $\ce{MoU}$ seems to be one of the best candidates for the highest bond order.

References:
$^{\ast1}$. Björn O. Roos, Antonio C. Borin, Laura Gagliardi, "Reaching the Maximum Multiplicity of the Covalent Chemical Bond," Angew. Chem. Int. Ed. 2007, 46(9), 1469–1472 (https://doi.org/10.1002/anie.200603600).
$^{\ast2}$. Fernando Ruipérez, Gabriel Merino, Jesus M. Ugalde, Ivan Infante, "Molecules with High Bond Orders and Ultrashort Bond Lengths: $\ce{CrU, MoU,}$ and $\ce{WU}$," Inorg. Chem. 2013, 52(6), 2838–2843 (https://doi.org/10.1021/ic301657c).


If copper has 2 valance electrons and sulfur 6, why don't they bind in pairs? Simple question, the average neutral copper atom has 2 valance electrons, and sulfur 6. Wouldn't the two electrons jump from copper and bind to sulfide?

The empirical formula is: $\ce{Cu + S -> Cu2S}$.

So apparently two copper atoms bind with every one sulfur atom. Does that mean that the copper atoms each gives one electron away and leaves them with 1 valance electron each?

the average neutral copper atom has 2 valance electrons

Nope. Only one:

In the transition metals, the previous shell is being filled. Generally, the valence shell can't have more than 8 electrons, but lower shells can. So, you have the two outer electrons filled in in Group 1 and 2, and the rest are filled in in the inner shell. With copper, we have 17 electrons in the inner shell, and 2 on the outer shell by this rule. But--it is sometimes more favorable for the inner shell to be "complete" with 18 electrons, leaving the outer shell with only one.

This is much better explained if you understand what atomic orbitals are. Think of these as as "subshells". In copper, we have two possible shell configurations: 10 electrons in $3d$ and one in $4s$, or 9 in $3d$, and 2 in $4s$. The numbers signify the shell, and the letters signify the orbital type. There are $5$ d orbitals per shell (not all shells have $d$ orbitals, though), and 1 $s$ orbital. Each orbital holds two electrons. Here, we have the trade-off between having a fully filled $3d$ orbital set and having a fully filled $4s$ orbital. Fully filled orbital sets lead to stability. Anyway, since both choices lead to a fully filled orbital set, a lot of other external factors decide which electronic configuration copper uses. So you have both $\ce{Cu^+}$ and $\ce{Cu^{2+}}$.

As a side note, $\ce{CuS}$ exists as well.


Do dark lines on porcelain dishes indicate usage of aluminium for the color? "I heard" that gray lines forming on porcelain indicate usage of aluminium in the coating. I would primarily like to verify this information.

Bowl #1 and bowl #2 have been used both for several months and were bought in different shops. The same observation goes for other dishes. Some don't feature lines, other dishes feature those after days already. The dishes who tend to display those patterns have been bought all at the same place (a large chain of stores specializing on furniture). The lines are most likely caused by cutlery as in cups the same lines are found forming ellipses at the bottom from the stirring. Again some cups don't feature those patterns others do.

Bowl #1 with no lines despite usage for a while:

Bowl #2 with lines:

Bowl #2 with lines - picture taken HDR mode:

If the two bowls look different although treated similar, the material must differ.

Your bowls are almost fully covered with a glaze - except for the ring on the bottom on which the ceramic was standing while fired in an oven.

Here, water can enter the ceramic core. If the ceramic material of one bowl is more porous, water can be taken up easier. Upon heating, as in a dishwasher, the water absorbed expands. The glaze can't withstand the tension build up and breaks.

This results in the cracks that you observe.


Why do subshells and orbitals exist? I'm just a curious high school student. Sorry if this sounds dumb.
How exactly did the concept of atomic subshells and orbitals come about? And why exactly are there n-1 subshells and 2l+1 orbitals?
Based on what I've heard, these things were a step forward after the Sommerfeld atomic model. But orbitals???
Again, sorry if this sounds dumb.
I just want to know how these things came about.

Very nice question. I also had the same question. These orbitals are nothing more than functions (wavefunctions). They are solutions to Schrodinger's equation, a second order differential equation. So you solve the equation and the solutions you get are determined by $4$ numbers also called quantum numbers $n, l, m_l, m_s$. A shell consists of all orbitals with same value of $n$ whereas a subshell consists of all orbitals with same values of $n$ and $l$. As others have pointed out the solutions are exactly only for the hydrogen and hydrogen like atoms.


Limiting molar conductivity of multiple solutes Suppose there is a water solution at $\pu{25 °C}$ with $\ce{Na+},$ $\ce{Mg^2+}$ cations and $\ce{Cl-},$ $\ce{HCO3-}$ anions. Concentrations $[\ce{Na+}],$ $[\ce{Mg^2+}],$ $[\ce{Cl^-}]$ and $[\ce{HCO3-}]$ are known.

I cannot understand how to apply the Kohlrausch's law of the independent migration of ions here:

$$\Lambda_\mathrm{m}^0 = \nu_+\lambda_+ + \nu_-\lambda_-$$

How this formula will transform for the solution with many kinds of ions?

Edit:

Electroneutrality condition holds, so:
$$[\ce{Na+}]+2 \cdot [\ce{Mg^2+}]=[\ce{Cl^-}]+[\ce{HCO3-}]$$
I've found the table with the limiting ionic conductivities $\lambda$ of each individual ion in water at $\pu{25 °C}$:

$\lambda_{\ce{Na+}} = \pu{50 \cdot 10^{-4} S m2 mol-1}$

$\lambda_{\ce{Mg^2+}} = \pu{106 \cdot 10^{-4} S m2 mol-1}$

$\lambda_{\ce{Cl-}} = \pu{76 \cdot 10^{-4} S m2 mol-1}$

$\lambda_{\ce{HCO3-}} = \pu{45 \cdot 10^{-4} S m2 mol-1}$

But I don't know what values for $\nu$ I should take.

I've found a answer for my question from this question.

The way to compute the conductivity of electrolyte with multiple ion types is given from the work Pawlowicz, Rich, ( 2008), Calculating the conductivity of natural waters, Limnol. Oceanogr. Methods, 6, doi:10.4319/lom.2008.6.489.

For general case consider the system, which consist of $N_+$ number of cation types, $N_-$ number of anion types and $N_{types} = N_+ + N_-$ total number of types, fully dissolved in solvent. This system can be viewed as a weighted sum of all possible pairwise combinations between cation and anion types. Then the conductivity $\sigma,\pu{[S m-1]}$ of the solution will be determined by:
$$\sigma = \sum_{i=1}^{N_+}\sum_{j=1}^{N_-}{\frac{c_i^+ z_i c_j^- z_j}{C_{eq}} \Lambda_{eqm, \, ij}}$$
where $c_i^{\pm}$ - molar concentration of corresponding ion type, $\pu{[mol m-3]}=\pu{[mM]}$;

$z_i$ - valency of corresponding atom;

$c_i^{\pm} \cdot z_i$ - equivalent ionic concentration of corresponding ion type $\pu{[mol m-3]}=\pu{[mM]}$;

$\Lambda_{eqm, \, ij}$ - equivalent molar conductivity of binary subsystem of cation type $i$ and anion type $j$, $\pu{[S m^2 mol^{-1}]}$.

Equivalent ionic concentration $C_{eq}$ is defined as:
$$C_{eq} = \sum_i^{N_+}c_i^{+} \cdot z_i = \sum_j^{N_-}c_j^{-} \cdot z_j = \frac{1}{2}\sum_k^{N_{types}}c_k^{\pm} \cdot z_k$$

In general a binary solute dissociates according to formula:
$$\ce{A_{\nu^+}B_{\nu^-} -> {\nu^+}A^{z^+ +} + {\nu^-}B^{{z^-}-}}$$
where $\nu^+$ and $\nu^-$ are coprime numbers and represent the moles of ions  for 1 mole of solute. It follows:
$$\nu^+ = z^-$$
$$\nu^- = z^+$$

Equivalent molar conductivity $\Lambda_{eqm}$ of single binary electrolyte is defined as:
$$\Lambda_{eqm} = \frac{\sigma}{c\nu_+z^+} = \frac{\sigma}{c\nu_-z^-}$$

For a binary solute, dissolved in water at normal pressure, $\pu{25 °C}$ and infinitely diluted, the law of the independent migration of ions can be written to find a limiting equivalent molar conductivity $\Lambda_{eqm}^{0}$:
$$\Lambda_{eqm}^{0} = (\frac{\nu_+}{z^+ } \lambda_{+}^{0} + \frac{\nu_-}{z^- }\lambda_{-}^{0}) = (\frac{z^-}{z^+ } \lambda_{+}^{0} + \frac{z^+}{z^- }\lambda_{-}^{0})$$
where $\lambda^{0}$ - limiting ionic conductivity of the ion, $\pu{[S m^2 mol^{-1}]}$.

Here interaction between ions is ignored.

Change of equivalent molar conductivity from changing of the solute equivalent concentration is determined by Debye-Hückel-Onsager equation of the form:
$$\Lambda_{eqm} = \Lambda_{eqm}^0 (1-A \sqrt{I}) - B \sqrt{I} =  \Lambda_{eqm}^0 K_A - K_B$$
where $A, \, B$ - Debye–Hückel–Onsager coefficients;

$I$ - stoichiometric ionic strength;
$$I = \frac{1}{2} \sum_{k=1}^{N_{types}}c_i z_i^2$$
$$ A = \frac{z^2eF^2}{3 \pi \eta}\left(\frac{2}{\varepsilon RT}\right)^{1/2}$$
$$ B = \frac{qz^3eF}{24 \pi \varepsilon RT}\left(\frac{2}{\varepsilon RT}\right)^{1/2}$$
where $\eta$ - viscosity of solvent, $\pu{[Pa s]}$;

$\varepsilon$ - dielectric permittivity of solvent;

$q$ - coefficient depends on $z^+/z^-$ of binary solute.

The expression for the conductivity of electrolyte (solvent - water at $\pu{25 °C}$ and normal pressure, ion pairng reduction factor is ignored ($\alpha_{ij}=1$ for all $i,j$)) will be:
$$\sigma = \sum_{i=1}^{N_+}\sum_{j=1}^{N_-}{\frac{c_i^+ z_i c_j^- z_j}{C_{eq}} (\frac{z_j^-}{z_i^+} \lambda_i^0 \, K_{A, \, ij} + \frac{z_i^+}{z_j^- } \lambda_j^0 \, K_{A, \, ij} - K_{B, \,ij})}$$

In my case I have 2 cations and 2 anions, their pairwise combination give 4 different solutes, which are assumed to be completely dissociate in water. Relation between solute molar concentration and its ions molar concentrations can be established.

$\ce{NaCl -> Na+ + Cl-}$
$\ce{NaHCO3 -> Na+ + HCO3-}$
$\ce{MgCl2 -> Mg^2+ + 2Cl-}$
$\ce{Mg(HCO3)2 -> Mg^2+ + 2HCO3-}$

Conductivity of my electrolyte will be:
\begin{align*}
\sigma &= \frac{1}{[\ce{Na+}]+2 \cdot [\ce{Mg^2+}]} \cdot \\
           &\cdot \Bigl(\ce{[Na+]}  \ce{[Cl-]}       K_{A(\ce{NaCl})}        ( \lambda_{\ce{Na+}}^0 + \lambda_{\ce{Cl-}}^0      - \frac{K_{B(\ce{NaCl})}}{K_{A(\ce{NaCl})}}      )+\\
           &+             \ce{[Na+]}  \ce{[HCO3-]}  K_{A(\ce{NaHCO3})}  ( \lambda_{\ce{Na+}}^0 + \lambda_{\ce{HCO3-}}^0 - \frac{K_{B(\ce{NaHCO3})}}{K_{A(\ce{NaHCO3})}})+\\
           &+             2\ce{[Mg^2+]}  \ce{[Cl-]}       K_{A(\ce{MgCl2})}        ( \frac{1}{2}\lambda_{\ce{Mg^2+}}^0 + 2\lambda_{\ce{Cl-}}^0      - \frac{K_{B(\ce{MgCl2})}}{K_{A(\ce{MgCl2})} }      )+\\
           &+             2\ce{[Mg^2+]}  \ce{[HCO3-]}  K_{A(\ce{Mg(HCO3)2})}  ( \frac{1}{2}\lambda_{\ce{Mg^2+}}^0 + 2\lambda_{\ce{HCO3-}}^0 - \frac{ K_{B(\ce{Mg(HCO3)2})}} {K_{A(\ce{Mg(HCO3)2})}})
           \Bigr)
\end{align*}

All the values here depend from given concentration of ions or from values found in tables. Hope it is correct, please, correct me if I'm wrong.


How to find the percentage contribution of a compound via gravimetric analysis? You have a sample of silver alloy and would like to know how much silver is in it. Dissolve the silver sample in nitric acid. Filter to remove any insoluble material. Add an excess of $\ce{KCl}$ solution, which causes the formation of silver chloride precipitate. Filter, dry, and weigh the precipitate. What do you estimate the percentage of silver in the original sample is if its mass was $5.2~\mathrm{g}$, and the mass of silver chloride was $6.4~\mathrm{g}$?

The atomic mass of silver is $M(\ce{Ag})= 107.87~\mathrm{\frac{g}{mol}}$. That of chlorine is $M(\ce{Cl}) = 35.45~\mathrm{\frac{g}{mol}}$.
My answer is $81\%$, because $5.2/6.4 \cdot 100 = 81$. However, I am quite unsure.


Water flow in condenser How should the water flow in your condenser? From the flask to the top of the condenser, or from the top of the condenser back to the flask? I could imagine both: flask to top will fill the entire condenser with water, because of the resistance of gravity, enlarging cooling area, or top to flask, because this will generate counter current exchange increasing heat flow, but then it is almost impossible to get a decent cooling surface without extreme water pressure.

This picture suggests the first.

In general the cooling water is attached according to the countercurrent exchange mechanism to maximize the efficiency. Of course this doesn't work well for the Liebig condenser you pictured, but you generally wouldn't use this kind of condenser for reflux anyway. The most common condenser for reflux is the Dimroth condenser which can be used for countercurrent exchange without any problems as the entry and exit of the water are both at the top of the condenser.

Here is an image from Arbeitsmethoden in der organischen Chemie (page 43) that shows where to attach the cooling water for different condenser types.


Can a zeroth order reaction be reversible? After recently studying about chemical equilibrium, I was convinced that the forward and backward reaction rates meet each other at equilibrium.

However thinking about zeroth order reaction annoyed me a little, as I could not figure out how the rates would become equal, if at all, in cases where the rate remains static throughout. Here's what I guessed:

The backward reaction of a zeroth order reaction must not be a zeroth order reaction in itself, so that it may reach the reaction rate equivalence point at equilibrium.But then this generalization is difficult to digest (for me).

or

A zeroth order reaction can not attain equilibrium (highly unlikely).

A zeroth order reaction does not achieve equilibrium as a zeroth order reaction.  Instead the kinetics will go away from zeroth order when we get close to the equilibrium condition.

Zeroth order kinetics is an approximation we get to when some aspect of the reaction approaches a saturation condition, for example the reaction mechanism may involve adsorption on a surface and there is enough reactant to saturate all the adsorption sites.  Such conditions are perforce relatively far from equilibrium; "saturation" and "reversibility" do not come about simultaneously.


Why does cesium have a positive electron affinity? When an electron is added to cesium, why is energy released?

All elements have a positive (first) electron affinity, except noble gases, for which it has not been measured conclusively, but might be either slightly positive or slightly negative.

You can more clearly see the reason why electron affinity is positive if you think of the opposite process: detaching an electron from the negative ion X– to form neutral atom X. Because the anion X– is stable, it takes energy to take away the electron.


How to use molality and molarity to determine the molar mass of a compound? An aqueous solution of a certain organic compound has a
  density of $1.063~\mathrm{g/mL}$, an osmotic pressure of $12.16~\mathrm{atm}$ at
  $25.0~\mathrm{^\circ{}C}$, and a freezing point of $-1.03~\mathrm{^\circ{}C}$. The compound is known not to dissociate in water. What is the molar mass of
  the compound?

The molarity and molality can be determined with the given information.

$$m=\dfrac{\Delta T_{f}}{k_{f}}=0.554m$$ and
$$ M=\dfrac{II}{RT}=0.497M$$

$$\text{mass of }\ce{H2O}=\dfrac{1000~\mathrm{g}~\ce{H2O}}{0.554~\mathrm{mol}~\text{of solute}}\cdot 0.497~\mathrm{mol}$$

I don't understand how the last equation is used to determine the mass of $\ce{H2O}$. How are we justified in using the molarity and molality to determine the mass of water?

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

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

Same result gives my (preferred) method in the comments


Thermodynamic properties of enantiomers Pairs of enantiomers have the same chemical and physical properties (except rotation of plane polarized light) within an achiral environment. However, I am curious about the thermodynamic properties of pairs of enantiomers: say a molecule has a certain $\Delta H_f$, $\Delta S$, and $\Delta G_f$; will its enantiomer have the same exact values?

All the normal thermodynamic properties of a pair of enantiomers are the same.
Unless there is some interaction with a chiral environment. The only way any properties will differ is if the environment in which they are measured is chiral.
For example, the heat of combustion of D glucose is the same as for L glucose (as combustion doesn't care about the stereochemistry). But only D glucose is usable by higher organisms in metabolism as all the enzymes involved are chiral (despite the thermodynamics of the overall reaction being exactly the same).


What are the products of the reaction between methanamine and nitrous acid? What are the products of the reaction between methanamine and nitrous acid?

I'm finding this particular reaction very problematic. Here's why:

My teacher tells me that methanamine, when treated with nitrous acid (created by reacting sodium nitrite and hydrochloric acid) converts to an the unstable (at room temperature) intermediate: methane diazonium chloride, which decomposes in the (aqueous) solution to yield methanol, hydrogen chloride and dinitrogen.

A friend of mine (who happens to be a fairly trustworthy senior), tells me that the actual products are methoxymethane and dinitrogen, since methanamine is an exception and doesn't give the same kind of products the other primary aliphatic amines do (he read it up a long time ago, but doesn't remember where).

In hope of clearing up this mess, I resorted to my trusty copy of Organic Chemistry (Morrison and Boyd, 7th edition), according to which, the products of a reaction between a primary aliphatic amine and nitrous acid should be: dinitrogen and a mixture of alcohols and alkenes.

It doesn't mention anything about methanamine being an exception, and "a mixture of alcohols and alkenes" is a little ambiguous. Also I didn't discard this idea because maybe - through some method I don't even know - the methyl bits from the amines might couple and somehow result in an alkene. I wouldn't know since I'm relatively new to amine + nitrous acid reactions, so I won't take the risk.

I couldn't find any trust-worthy reference on this issue for this reaction. I need help resolving this ambiguity.

No chance to get alkenes from methanamine because there's no $\beta$-hydrogen for E2 elimination. And E1 elimination isn't going to work because you're not going to generate a methyl cation from the diazonium. The mixture of alcohols is probably because the alkyl cation could rearrange, but you're not creating the cation intermediate in this reaction.

The diazonium is a very good leaving group, so this complex is certainly going to be susceptible to SN2 reactions. The best and most available nucleophile you have in your mixture is water (the solvent), so you're going to get methanol.

Now, methanol could be a nucleophile that reacts with diazonium, but you're in aqueous solution. That means that the amount of water is going to dwarf any methanol that is produced. A negligible amount of the dimethyl ether will be produced.


Why is it easier to determine the radius of inert/noble gases? My textbook says it is possible to determine the atomic radius of the inert gases like He, Ne, Ar etc because they roam freely and are incapable of forming bonds. However it is very difficult to find out the atomic radius of elements that are chemically active because the position of electron in an atom is uncertain (according to Heisenberg's principle I guess). I am just curious that the principle is true for the electrons of inert gases too, then why did they say it was possible for those gases but very difficult for the chemically active atoms?

There are several definitions of atomic radius. Guessing from context of your answer, I assume you are talking about  van der Waals atomic radius. By definition, it is a radius of atom in respect of equilibrium position with other atoms when only van der Waals forces are acting between the atoms. Since atoms of inert gases do not tend to make covalent bonds, the equilibrium distance between two in solid equals double atomic radius - plain and easy. Not so easy for other elements.

Since most metals and some non-metals form closely packed lattice in elemental form, there is no simple way to obtain their vdW radius. It is required to find a compound with the atom sterically unhindered and forming a crystal lattice containing close contact with said atom involved. Since most metals are electron-deficient and tend to interact with almost anything, no such experimental data are available for metals in a strict sense (though I can come with few examples when it may make sense). As such, vdW radii can be measured (and, actually, are meaningful) only for atoms of top-right corner of periodic table plus hydrogen. Furthermore, even for them vdW radius may fluctuate from compound to compound depending on the oxidation state and atomic neighborhood. Furthermore, the radius is pressure-dependent and, for chemically bound atom, may be unequal in different directions.

For covalent radii the problem is even larger, as the radius is dependent on near neighborhood and valence state in significant way (N-N bond legnth may vary from 1.10 Angstrom in dinitrogen to 1.45 in hydrazine)

As such, even if people talk about larger and smaller atoms, larger and smaller substituents, there is no strong definition for such relations.

As for atomic radius no related to vdW and/or covalent interaction, it depends on the definition of atomic border, which does not exist. A free atoms does not have a clear border, its electronic density fades exponentially with distance, but is non-zero for all finite distances.


Reaction of orthocresol with butyrolactone in presence of AlCl3 and heat What will be the major product of the following reaction?

Please also help me with the mechanism. I'm not sure whether C- acylation or O- acylation will occur here. And if O- acylation does occur, I think the heat will cause a Fries' rearrangement. What will be the major product of this rearrangement?

Also, if simple Friedel Crafts acylation occurs, will AlCl3 affect the reaction due to formation of -O-AlCl3 (Lewis acid-base interaction), something like what happens in the case of Friedel Crafts reactions with aniline? (I know aniline cannot undergo FC reactions due to deactivating effect of -NH-AlCl3 interactions)

I'm unable to upload a pic of the question due to some technical problem, but what I mean is reaction of ortho cresol and butyrolactone in AlCl3, along with heat.

I can take a guess since no one else has provided an accepted answer.
First, I don't think lactones will take part in FC-acylation on an arene, but rather prefer alkylation as mentioned in this paper. In the case of o-cresol and heat, this could be a possible explanation for your question why the O-acylation takes place instead of the C-acylation.
Second, when the Fries rearrangement occurs, and the acylium ion is released in the intermediate, the phenolic oxygen is complexed with the aluminum chloride (second to last step here). If you imagine an ortho-methyl group, as well as the complexed aluminum chloride, it's plausible substitution to the remaining ortho-site is not sterically favorable. This would force it to substitute elsewhere on the ring.
As for why "elsewhere" prefers para to the methyl group and not the oxygen—going by what you wrote in your comment—I'm not sure. The para alumino-oxy substitution should produce the more stabilizing resonance structure for the arenium intermediate. At least, in one other reaction of electrophilic aromatic substitution with ortho-cresol, namely its nitration, the 3/5-nitro products are not found. Maybe it's a solvent matter. Wikipedia states a non-polar solvent prefers ortho in the Fries rearrangement; perhaps if the ortho site is blocked, it will settle for para to a methyl substituent.
Edit
In a summary of the comments, the m-hydroxy-ketone was called into question as the correct product. I mentioned none of the papers I found of similar reactions furnished anything but the para or ortho product. Here are the papers I looked through:
paper 1 - first paragraph: o-tolyl acetates yield predominantly p-hydroxy-ketones.
paper 2 - tables 2 and 3
paper 3 - table 1 (page 378)
paper 4 - 5th paragraph
A few others: book p 223, paper, paper


What is the difference between O & O2 What is the difference between $\ce{O}$ and $\ce{O2}$.

If C is carbon and then why $\ce{O2}$ is oxygen.

The common confusion here is that two very different things have the same name.  $\ce O$ is a free oxygen atom and $\ce{O2}$ is two oxygen atoms chemically bound to form an oxygen molecule.

There is no common analogy for $\ce C$, but $\ce{N2}$ is called nitrogen, $\ce{H2}$ is hydrogen and $\ce{Cl2}$ is chlorine, each having the same name as that of their constituent elements.

So, to reiterate, the difference between oxygen $\left(\ce O\right)$ and oxygen $\left(\ce{O2}\right)$ is that the former is an oxygen atom while the latter consists of two $\ce O$ atoms bound together, forming a molecule also called oxygen.


Are Na and Cl ions from NaCl in water separate charged particles? Suppose I pour some sodium chlorine into water. So, what happens is that

$$
\ce{NaCl (s) -> Na+ (aq) + Cl- (aq)}
$$

meaning that the ionic bond between Na and Cl breaks up (correct this far?)

Now, does this mean that the water actually contains separate charged Na and Cl particles? So... since chlorine boils at −34.04 °C according to Wikipedia, why is there then no chlorine gas evaporating? Because it is chlorine ions there, not chlorine atoms?

If I feed electrons some how into the solution, will chlorine gas start forming?

Also, could I use this so that I pour NaCl into water and get Na and Cl ions, and then (since they are separate) add something more to create Na[something] or Cl[something]? Something that would not form if I just mixed NaCl + [something else]. I hope I am not completely lost here..

Most of what you wrote is correct. It is difficult to feed many electrons into the salt water because the Coulomb forces are extremely strong, see 1 and 2, but you can feed electrons in if you drain them out elsewhere so there is no net gain or loss of charge. that is called an electric current.

You can use that to electrolyze salt water, and get chlorine as you surmise; NASA has a nice lab. BTW, you would expect to get Cl gas at the positive electrode, and do get it, but you do not get Na metal at the negative, because hydrogen bubbles out before Na. You can electrolyze molten salt, though, to get a bit of Na (or at least you see it form and then catch fire with a yellow flame).


Solar still and how it works I use a solar still that holds 182 cubic feet of water. Here is how it works:

There is an insulator on both sides of the metal bowl. On the outside it covers the whole thing. on the inside it covers the sides but not the bottom. This collects the rain and also heats up the rainwater during sunny days so that the evaporation rate is faster. I have the pipe nearest the metal bowl the largest it can be while the metal bowl still collects rain and gets heated up by the sun. The water and any gases pass through there and into the condenser. I have this cold enough to condense water into a liquid but not so cold that the water turns solid and the $\ce{CO2}$, $\ce{O2}$, $\ce{N2}$ etc. turn liquid. Last but not least there is a spigot at the very end and a bucket below it. This separates atmospheric gases from liquid water with $\ce{CO2}$ being the main concern.

There is also dust and pathogens from the air in rainwater.

The dust most likely will stay behind and just make the bowl dusty. The pathogens get killed by the heat that there still is when all the water has been evaporated.

I would say the maximum loss with this system is from 1 pint or so to 1 gallon or maybe more depending on the water level in the bowl.

However I think there is a way to calculate maximum and minimum loss(neither of these is 0).

How could I calculate maximum and minimum loss based on the depth of the water(because obviously at different depths there is a different amount of loss) and the size of everything(which by the way the pipe slowly tapers off to a certain size and so you have to take pressure into account as well as volume)?

Also how could I calculate the pressure in the pipe based on the fact that it tapers off and starts at the maximum size possible with the bowl still collecting rain and being heated by the sun?

There is no way to answer your question accurately for this setup. It will depend on the temperature of the water in the bowl, the exposed surface area, the convective flow rate of air near the surface, and the relative humidity of the air - none of which are constant or even very predictable.

Your best bet is to set it up and measure the amount before and after distillation. The difference would be your loss.


How did Thomson calculate the momentum of electron when he didn't know the mass of electron? This was proven in 1903 by J. J. Thomson who calculated that the momentum of the electrons hitting the paddle wheel would only be sufficient to turn the wheel one revolution per minute. source

J. J. Thomson measured the charge to mass ratio of electron in year 1897, and R. K. Millikan measured charge of electron in year nearly about 1910, i.e., Thomson knew only the charge to mass ratio of electron in 1903.
Then how did he was able to determine the momentum of electron when he didn't knew the mass of electron?

Thomson described the mechanical effect produced by cathodic rays on the first edition of its book Conduction of Electricity Through Gases (1903, pp. 501–502) and again later on the second edition (1906, pp. 629-630).
As you mentioned it refers to the Crookes experiment with the paddle wheel in which the electron beam collides with the mill coausing it to move (fig.below).

The electron beam carries a current ($I$) that can be measured and is equal to the number of electrons hitting the paddle every second (N) multiplied by the charge of the electron (e).
$$ I =N\cdot  e$$
Considering that the electrons are much smaller than the paddle you can consider that they bounce back with the same velocity (like when you hit a wall with a ball) and therfore the momentum transferred to the paddle per unit time ($p_{tot}$) is twice the momentum of the electron beam.
The momentum of the electron beam per unit time can be calculated with the well known formula $m\cdot v\cdot N$ with N to account for the number of collisions per unit time.
Therefore the total momentum per unit time transferred to the mill is:
$$ p_{tot} = 2m\cdot v\cdot N$$
This is equivalent to (multiplied and divided by the same quantity $e$):
$$ p_{tot} = 2\cdot \frac{m}{e}\cdot v\cdot (N\cdot e)= 2\cdot (\frac{m}{e})\cdot v\cdot I$$
Thompson considered an hypothetical best case scenario with a very high current of $\pu{10^{-5}A}$ carried by the electrons and a very high velocity of $\pu{10^8 m/s}$.
These are the highest reasonable values he could think could be applied in the Crookes' experiments (he even mentioned that the actual current measured was only $\pu{10^{-7}A}$.
The current can be easily measured while the velocity can be estimated from various methods using magnetic and electric fields (i.e. see section Charge-to-mass ratio of the Wikipedia where v is derived). And $\frac{m}{e}$ was already determined (around $\pu{6e-12Kg/C}$.
With those data, you can estimate that in the best case scenario (very high velocity and many particles hitting the paddle per unit time) the momentum transferred to the paddle per unit time is:
$$p_{tot} = 2\cdot (\frac{m}{e})\cdot v\cdot I = 2\cdot 6\cdot 10^{-12}\cdot 10^8\cdot 10^{-5}= \pu{1.2e-8 N} $$
This corresponds to $\pu{1.2e-3 Dyne}$ in the cgs units (units used by Thompson).
If you consider a paddle of $\pu{1cm2}$ then the pressure on the paddle is $\pu{1.2e-4 Pa}$, that is extremely small (an atmosphere is around $\pu{101 325 Pa}$!). Because of this it is reasonable to  attribute the movement to heating transfer.
As you see here, Thompson couldn't calculate the momentum defined as $m\cdot v$ but instead he calculated the  momentum per unit time  (that is a force)! He also considered the momentum carried by many electrons so that he could use the current in the formula of $p_{tot}$.


Is lithium bicarbonate an aqueous solution of lithium carbonate? This question is related to:

Are Lithium bicarbonate and Magnesium bicarbonate both soluble in water?
What type of chemical reaction is this?
How can bromine water be called a solution?

In the first question, the page from JD Lee (Concise Inorganic Chemistry 5 ed) states:

  Lithium is exceptional that it does not form that it does not form a solid bicarbonate, though $\ce{LiHCO3}$ exists in solution. All the carbonates and bicarbonate are soluble in water.

and then in the sideline(the most important line):

  If it exists in water, it must definitely be soluble in water.

Now, Lithium carbonate is prepared by bubbling $\ce{CO2}$ gas to a solution of lithium hydroxide, $\ce{LiOH}$.(source)

$$\ce{2LiOH + CO2 → (Li2CO3 + H2O)}$$

But in another source (endmemo), it shows that lithium hydroxide and carbon dioxide directly combines to form lithium bicarbonate.

$$\ce{LiOH + CO2 -> LiHCO3}$$

So, basically lithium bicarbonate is aqueous solution of lithium carbonate. Isn't it?

Now, what I gather from the fact that lithium carbonate is soluble in water and form a solution where the components do not react with each other. However, when the two reaction are compared, we observe that lithium carbonate reacts with water to form lithium bicarbonate. It seems contradictory with the above statement. How is this possible?

In the 3rd question, I said that if bromine reacts with water to form a mixture of acids, how can it be called a solution? Components of solution do not react with each other. Same case goes on with this reaction. If lithium carbonate is in aqueous solution, it should dissociate into component ions i.e $\ce{Li+}$ and $\ce{CO3^2-}$ but instead it goes on form $\ce{LiHCO3}$ and thus the solution has $\ce{Li+}$ and $\ce{HCO3-}$ ions. Why?

Lithium bicarbonate solution is something like a lithium carbonate solution in the sense that precipitation would produce the carbonate instead of the bicarbonate.  But what is in solution is mostly lithium and bicarbonate ions.  Precipitation is accompanied by a reaction in which the highly polarizing lithium ions react with bicarbonate:

$2\ce {Li+}+2\ce {HCO_3-}\rightarrow \ce {Li2CO_3}(s)+\ce {H2O}+\ce {CO2}$

Other alkali metals give less polarizing ions and do not react in the same way; they just precipitate the bicarbonate salt.


Optical rotation of mixture The specific rotation of optically pure adrenaline is $-53$. A mixture of (R)- and (S)-adrenaline was found to have a specific rotation of $-45$. Calculate the % ee of the mixture (enter numeric value only, without % sign, rounded to the nearest whole percent).

This should be $\%~\text{ee}= \text{specific rotation sample}/\text{specific rotation enantiomer}$

But $100\times (-53/{-45})=117.78~\%$

The fraction the other way is $84.9~\%$ but doesn't stick to the formula.

Enantiomeric excess (ee) is defined as:

$$\text{ee} = \frac{\alpha(\text{sample})}{\alpha(\text{pure enantiomer})}\tag{1}$$

Here, we are given the specific rotation of our sample ($-45^\circ$) and that of a pure enantiomer ($-53^\circ$). To calculate the ee, all you need to do is to plug these two values into equation $(1)$ and work out how to turn a fractional number into a value in $\%$.

   The answer is $85~\%~\text{ee}$.


How can I dissolve a precipitate when stuck to the side of a boiling flask I’ve been experimenting with producing Iron Acetates and I’ve run into a snag. Here’s my process so far:

  I started by adding white vinegar and ~50% Hydrogen Peroxide to a reaction vessel with steel wool sponges and allowing it to react for a month. After that I ran the solution through a filter multiple times until it was fairly clear to remove most of the unreacted solids. After that I tested an aliquot of the fluid by placing some leather in it to see if the leather turned black. It did quite quickly. After that I placed it in a boiling flask and heated it with a flame directly (because I don’t have a big enough beaker or other apparatus to indirectly heat it) and it turned orange. I’m guessing this happened because the elevated temperature allowed it to react with atmospheric oxygen. I continued heating till the solvent was gone leaving a red-brown-black solid on the side of the boiling flask.

I want to dissolve the solid so I added ~100mL of ~50% Ethanol (Smirnoff blue label vodka) to the flask to dissolve it and left it overnight, but it’s barely dissolved at all. I thought about heating it to create pure ethanol vapor which would do the trick but I don’t have a condenser to condense the ethanol again. Is there a better way of dissolving it?

This is all rust which has precipitated on the walls and it is very hard to remove it. You might have to use oxalic acid to remove rust stains on glass. Oxalic acid plays a dual role, it dissolves the rust like an acid, but it also complexes the iron, which prevents further hydrolysis and re-settling.


Is it possible to normalize the concentration equilibrium constant between different reactions for comparison? A follow up question from this answer:

  Compare:
  $\ce{A(aq) +B(aq)->AB(aq)}$
  $\ce{2A(aq) +B(aq)->A2B(aq)}$

  We have two similar reactions but the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants.  Similarly, with a constant solid/liquid concentration (activity = 1) the equilibrium constant effectively ignores that product or reactant.

Is there a way to "normalize" the $K_c$ of reactions like that shown above, such that a fair comparison can be made, regarding whether one reaction is more product favoured than the other?

Since the equilibrium constant depends on the concentration of the species involved, there is no way to normalize it in the way I think you imagine it. A fair comparison of equilibrium constants is already given. The equilibrium constant is, as its name gives away already, constant. As such there is no normalization to be made.

Also, the quote is slightly off the point where it says that

  the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants

because the equilibrium constant might actually be the same (except for the unit, of course). The way you calculate the constant changes, but inferring from that that the value changes drastically is simply overreaching.

If $K > 1$ then the equilibrium lies on the side of the products, if $K<1$ then the equilibrium lies on the side of the reactants.


Is it possible to predict the color of the flame test result based on the element's structure? The flame test is quite simple to do and very useful on the chemistry world, but to understand what your results mean you need to know previously a range variety of element-color relations. So, since the wavelength of the photon emitted reliess on the quantum jump done, I was wondering if, somehow, is it possible to make a fairly certain prediction of the results, based on the compound's structure. If someone knows a relation, it would be very useful.

The question is why would you want to predict that? All elements have been screened for allowed and forbidden transitions ranging over the complete visible and non-visible spectrum. You can just look it up there. There are also multiple possible transitions which lead to the often complex spectra.

And then there is the problem with the actual jump. In a simple shell-model where you only add electrons in shells this jump is just to the next level. And you can introduce Rydberg-atoms and get element specific Rydberg-constants. But I remember to have tried that for a Lithium atom and if I didn't do a mistake back then it would have required quite a high energy level to get the red emission while in reality the principal quantum number (n) doesn't even change for lithium. You can see this pretty easily if you look at the energy diagram for the famous yellow sodium emission line. It's a jump from the $3p$ to the $3s$ orbital.

But in the end, as I said, they have all been calculated and confirmed. You can just look at lists on the net where every possible transition is given with the orbitals, energy, wavelenght and something like an intensity.

And engines like WolframAlpha even use that. You can just search there for let's say sodium emission spectrum and it will give you the spectrum. And there are also websites where all spectra are given.

I'm a bit confused with your question. It sounds like you don't have such a list. I guess, as you are talking about flame colors and not spectra you are only looking for the flame color itself. This should be mentioned in most analytical chemistry books and I think there was even a Wikipedia article with pictures. But at some point this will become quite hard. Lithium vs Calcium vs Strontium is such a case where I wouldn't be too sure if I could distinguish them by their flame color without looking at spectra. In the end the observed emission will be mixture of many weak and strong emissions.


Electrochemical concentration cells -- how anion levels change It seems clear that in the anode, the anion concentration will increase as it enters from the salt bridge to balance the cations leaving from the oxidized electrode. But what is happening to the anion concentration in the cathode?  Cations are leaving the salt bridge to replace those being reduced on the electrode.  So is the anion concentration just constant?

Considering you have chosen a $\ce{Zn}$ concentration cell:
The cell representation is:
$$\ce{Zn(s)/ZnSO4(M1)||ZnSO4(M2)/Zn(s)}$$

Anodic reaction puts more $\ce{Zn^2+}$ ions in the solution:
$$\ce{Zn(s) -> Zn^2+(aq) + 2e-}$$
To balance these extra $\ce{Zn^2+}$ ions salt bridge pulls some anions into anodic compartment.

Cathodic reaction removes $\ce{Zn^2+}$ ions from cathodic compartment:
$$\ce{Zn^2+(aq) + 2e- -> Zn(s)}$$
To make up this equivalent amounts of cations are pulled off from the salt bridge.

So activity of $\ce{SO4^2-}$ is not changed in either of the solutions. It is only those of $\ce{Zn^2+}$ ions in both the compartments, which undergoes a change leading to redox reaction.


Perturbation Theory in Orbital Interactions in Chemistry In the entire book, perturbation theory is used as a qualitative tool to rationalise some chemical phenomena. The authors write that
$$\psi_i = \sum_\mu{T_{ji}\psi_j^\circ}\tag{1}$$
The proof of the derivation for $T_{ji}$ involves the following statement:
$$T_{ji} = \left(\mathbf{C}_j^\circ\right)^\mathrm{T}\mathbf{S}^\circ\mathbf{C}_i \tag{2}$$
where
$$\left(\mathbf{C}_j^\circ\right)^\mathrm{T} = \left(C_{1j}^\circ \quad C_{2j}^\circ \quad \cdots \quad C_{mj}^\circ\right)\tag{3}$$
(indexing the coefficients for the jth unperturbed MO), and
$$
\mathbf{C}_i =
\begin{pmatrix}
C_{1i} \\
C_{2i} \\
\vdots \\
C_{mi}
\end{pmatrix}
\tag{4}
$$
(indexing the coefficients for the $i$th perturbed MO); $\mathbf{S},$ the overlap matrix, is defined as usual.
I'm not quite sure how the aforementioned statement is obtained. Could someone please walk me through how
$$T_{ji} = \sum\sum C_{nj}^\circ S_{nm}C_{mi}?$$
Source: Orbital Interactions in Chemistry, 2nd edition, page 794-802

Rather than focusing on the individual matrix entries, it may be more intuitive to look at the matrix $\mathbf{T}$ formed by putting together all entries $T_{ij}$. In this case, $\mathbf{T}=\mathbf{C}^T\mathbf{S}^0\mathbf{C}$, where $\mathbf{C}$ is the coefficient matrix with column $k$ denoting how atomic orbitals mix together into molecular orbital $k$.

Now the point is that molecular orbitals, or the columns of $\mathbf{C}$, are orthonormal and are eigenfunctions of $H_{eff}$ (see page 34). This orthonormality makes sense if you think of the Hamiltonian matrix, which is certainly self-adjoint, and solving the eigenproblem is equivalent to diagonalizing the matrix with an orthonormal basis of eigenvectors.

Hence, $\mathbf{C}$ is unitary, and $\mathbf{T}=\mathbf{C}^T\mathbf{S}^0\mathbf{C}$ can be simply seen as a change-of-basis transformation done on the overlap matrix from atomic basis to the molecular orbital basis.

I hope this rather mathematical perspective may be of help to you. Thanks!


Variation of Arrhenius equation with concentration In the Arrhenius equation do Arrhenius constant and activation energy depend on concentration ? (Their units do)

The Arrhenius equation gives you a rate constant $k$.

For a second order reaction, $k$ has units of $1/Ms$.

For a first order reaction, $k$ has units of $1/s$.

The Arrhenius prefactor $A$ has the same units as $k$.

The rate, as opposed to the rate constant will depend upon concentration. For the first order reaction, the rate constant is multiplied by one concentration to get the rate, and for a second order reaction the rate constant is multiplied by two concentrations to get the rate.

If $k$ and $A$ do depend upon concentration, that would imply that the reaction is not accurately modeled by the rate constant (for example not really first order when you are modeling as first order or not really second order when you are modeling as second order).

As far as activation energy $E_a$, the equation can be written in terms of energy per mole divided by $RT$ or energy per molecule divided by $k_BT$ (where $k_B$ is Boltzmann constant).  $E_a$ does not depend upon concentration.


Mechanism of enol ether formation I assumed the the $\ce{TsOH}$ would dissociated to produce hydronium ions that would protonate the carbonyl. This would make the carbon more delta + and so susceptible to attack by $\ce{MeOH}$. After proton exchange I thought the $\ce{H2O}$ group would leave and we would have a carbocation left over. I can't work out how to resolve this carbocation with the changes in double bonds.

Based on what you have in the question, you are good up until formation of the carbocation by loss of a water leaving group. It looks like you are missing one of two things:

The carbocation is resonance stabilized.
One of the carbon-carbon double bonds in the product is new.

The scheme below has two of the major resonance contributors to that carbocation. Perhaps one of give you some insight on how to show the formation of the second alkene.


Naming a given organic compound according to IUPAC naming rules Would the given organic compound be called "3,4-diethyl-3,7,7-trimethylnonane" with IUPAC naming rules?

I don't really know if this should be called 3,4-diethyl-3,7,7-trimethylnonane, because it could also be called 6,7-diethyl-3,3,7-trimethyl something.
I am trying to do this by IUPAC rules.

Yes, your naming would be correct. For saturated hydrocarbons, the first point of difference rule takes precedence after the longest carbon chain. To apply this rule, the locants of each set are ordered numerically in a list from lowest to highest, and the series that has the lower number at the first occasion of difference is determined as the lower locant set, regardless of what comes after this first point of difference. In this case, "3,3,4,7,7" would be the lower locant set than "3,3,6,7,7", because 4 is lower than 6.


Birch reduction of anisole Hepworth, Waring and Waring (2002) mentions that:

  In general, electron-withdrawing substituents make the arene more susceptible to reduction, while the opposite applies for electron-donating substituents.

They then go on to say:

  Methoxybenzene (anisole) is reduced more rapidly than benzene under Birch conditions. This is an exception to the rule, the reasons for which are not clear.

Why is this the case? Does anyone know of any possible explanations?

Source: Wikipedia Commons

Also, I was thinking, since the methoxy group donates electron density mesomerically to the ortho and para positions, wouldn't it be unfavourable for the carbanion to form that the ortho positions? This seems to happen in the mechanism shown above.

Lastly, I understand that the dispute between whether it is "Mechanism O" or "Mechanism M" was settled a while ago but I've read that the evidence pointing to "Mechanism O" seems to be computational studies and experiments. However, there has yet to be any qualitative explanations, based on chemistry principles and concepts, as to why "Mechanism O" is the one which occurs to a greater extent. I was wondering if anyone knows of any such explanations?

Reference

Hepworth, J. D., Waring, D. R., & Waring, M. J. (2002). Aromatic Chemistry. United Kingdom: The Royal Society of Chemistry.

Chemistry is an experimental science supported by computational methods. Qualitative explanations for reaction mechanisms follow as a result. More on that issue later.The mechanism of the Birch reduction has been investigated by Zimmerman and Wang.1 What are the experimental results? The Birch reduction is first order in substrate, electrons and alcohol. Therefore, the rate-limiting step is the protonation of the radical anion 2 of anisole 1. Based on the observation that the dianion of anthracene is more basic than its radical anion, 2 the authors designed an experiment utilizing 2% d1-t-BuOH as the alcohol source. The argument is that the radical anion, which is less basic than the anion, will be more discriminating,3 favoring protonation rather than deuteration (isotope effect). The more basic anion, 4 or 7, which is less discriminating, will give rise to a higher D/H ratio. Employing deuterium NMR, the meta position in 8 was found to have 7-times the amount of deuterium than the ortho-position. Thus, the ortho-position (2a --> 3) is protonated first; mechanism O (ortho) predominates over mechanism M (meta). --continued--

Calculations showed that the methoxy ortho-substituted cyclopentadienyl radical is more stable than the meta- and para-isomers (red box). But your request was for a qualitative rationale for preferring radical 3 over 6. Radical 3 has an advantage over radical 6 because radical 3 may be written with the radical on the ipso-carbon, the same radical that would be expected to be generated by auto-oxidation of 5-methoxycyclohexa-1,3-diene. Cross-conjugated radical 6 does not have this option.

1) H. E. Zimmerman and P. A. Wang, J. Am. Chem. Soc., 1993, 115, 2205.; H. E. Zimmerman, Acc. Chem. Res., 2012, 45, 164.
2) V. D. Parker, M. Tilset and O. Hammerich, J. Am. Chem. Soc., 1987, 109, 7905.
3) Consider by analogy that chlorine atoms are less selective than bromine atoms in the radical halogenation of alkanes.


Conflict Between Organic and Physical Chemists--explaining the intensities of IR absorptions I am currently a Chemistry undergrad and I love to fact check everything my O Chem professor says against my research advisor, a physical/computational chemist. Well, my O Chem professor, when discussing IR absorptions, said something which was a simplification for teaching purposes but which seemed rather unusual to me. He said that high intensity peaks are due to a large change in dipole (clealy true) which often corresponds to more electronegative bonds and then gave the example of carbonyl absorptions which are intense, or nitrile absorptions, or $\ce{CO}$ single bond absorptions, etc.

What he then said, and says he has no good explanation for, is the following:

The absorption of an $sp^3$ hybridized $\ce{CH}$ stretch is high-medium intensity.

The absorption of an $sp^2$ hybridized $\ce{CH}$ stretch is medium intensity.

The absorption of an $sp$ hybridized $\ce{CH}$ stretch is medium-low intensity.

Of course the hybridizations given refer to the hybridization state of the carbon to which the hydrogen is attached.

The reason he has no good explanation for this phenomenon is that in most other situations, an $sp$ hybridized carbon will behave as if it is more electronegative than an $sp^2$ or $sp^3$ hybridized carbon.

An example of this would be that $pK_a$ of an $sp < sp^2 < sp^3$. That is a characteristic of how electronegative they are in an acid-base sense.

Here, however, we see the opposite trend. That is, when the more electronegative stretch should have a higher intensity peak, it is weaker, and the less electronegative stretch has a high intensity peak.

For the record, the arbitrary assignment of more and less electronegative based on behavior isn't super satisfactory to me, and is probably why I'm digging deeper on this.

So, my advisor essentially said all that matters is this equation: $$f=\left\langle\psi_0\right| \mu\left| \psi_v\right\rangle^2$$
Basically, he said I can ignore the wavefunction details of the equation because the wavefunctions of the three stretches are not going to differ by much because they all have identical reduced masses and some other stuff I can't remember.

So, all that matters is the dipole moment change as the bond is displaced from its equilibrium position.

His physical explanation of the intensities described above is that there is a smaller dipole moment change for $sp^2$ and $sp$ hybrid carbons because the electron density is trapped in the double and triple bonds more so than in the single bond. That seems reasonable to me.

The Organic Chemist fires back, however. He says that he would expect intensity to increase with the double and triple bonds because the electron in pi bonds are in a higher energy state than the sigma bond electrons and are thus more easily displaced which ought to result in a larger dipole moment change. That also makes a lot of sense I think.

So, is my advisor correct or is there a more complicated explanation?

Sorry for being so long-winded, I just wanted to get past the simple explanations so that we can have a deeper, better answer.

Edit: A very good answer would also explain why $sp$ hybridized C-H stretches are sometimes very high intensity. Even as high as $sp^3$ stretches at times.

In considering the C-H stretch, let's consider that the C atom is attached to something else via an "immobile" bond(s).

For $sp^3$ hybridized the C would have 3 other bonds besides the immobile one.

For $sp^2$ hybridized the C would have 2 other bonds besides the immobile one.

For $sp^1$ hybridized the C would have 1 other bonds besides the immobile one.

For a non linear molecule there should be 3N-6 modes of vibration, but a linear on has 3N-5.

   Hybridization  # Atoms  Linear   Vibration Modes
       sp3            4       No         6
       sp2            3       No         3
       sp1            2       Yes        1

The more vibration nodes, the more microstates for the molecule in both its current state and its excited state. Thus having more microstates increases the probability that there are more transitions from the current state to some excited state.

For an analog on the microstate overlap, think about atomic spectra vs UV-VIS. Atomic spectra are line spectra, UV-VIS is band spectra.


Is there a relationship between London dispersion forces and hydrophobic interactions? Looking at examples of these two molecular interactions makes me wonder what the differences or similarities are exactly.

To summarize the excellent but rather lengthy and detailed Stack Exchange answer referred to by Andy above, hydrophobic interactions do not require London forces between the hydrocarbon chains. They are a result of the very strong attraction of water for itself.

I would further quote the following comment to that entry (by https://chemistry.stackexchange.com/users/5633/thomij), which I think sums it up nicely:

"This is a good answer, but the last part about van-der waals forces being required to pull the hydrophobic solvent together is incorrect. If you had particles with no attractive forces between them in a solvent that has attractive forces, as long as the enthalpy drop for phase separation is larger than the entropic penalty, they will separate. Van der waals forces help to make the enthalpy drop larger, but they aren't necessary. –  thomij Jun 23 '14 at 19:02"


Enthalpy and Bond Dissociation Energies I'm not sure if I understand the reasoning behind this particular question.

The sigma and pi bonds of $\ce{C=C}$ have a combined bond dissociation energy of 632 kJ/mol. Using this information, predict whether the following reaction is exothermic or endothermic.

$\ce{C2H4 + H2O -> CH3CH2OH}$

My attempt:

The bonds being broken are the $\ce{C=C}$ pi bond and the $\ce{H-OH}$ bond. I realize there are two bonds being formed, but I am unable to correctly identify the second one. One is $\ce{CH_3CH_2-OH}$. I think the other should be $\ce{CH_2-H}$. The textbook says this bond is $\ce{CH_2H-R}$. I think R is referring to the side group, which varies (and in this case, is $\ce{CH_2-OH}$?), but the enthalpy value for breaking the $\ce{CH_2H-R}$ is listed as 410 kJ in the solutions. According to the table in my book, the value for a $\ce{CH_3CH_2-H}$ is 410 kJ/mol. I do not understand how these two bonds are equivalent. Can someone please explain this?

Tabulation table such as the ones you reference, describe 'typical' bonds when using generic groups such as R.  You are right in your discomfort that the numbers should be somewhat different, but that would require a thermochemical measurement for every known molecule, which is difficult to achieve in practice.

Calculations such as the one you are performing are 'estimates', and while good ones, they are never as credible as an actual experiment.


How is the unit of measure "equivalent" abbreviated? I usually use "eq" but I have been told to use "equiv." in order to avoid confusion with "equation". Is there a reference I can check?

The first place to check would be standards that are used frequently globally, such as CAS. According to the American Chemical Society page CAS Standard Abbreviations & Acronyms, the abbreviation for the unit of measurement 'equivalent' is equiv, while that of the adjective 'equivalent' is equiv. (with a dot).

Also, the ACS page Standard Abbreviations and Acronyms, shows equation abbreviated to eq and equivalent as equiv.

However, some journals, such as the Journal of Biological Chemistry page Table I: Abbreviations of units of measurement and of physical and chemical quantities state that the abbreviation of equivalent is as you used, eq, so if you are intending to publishing, just be aware of what the journal expects in terms of abbreviations.


Is teflon a poison if ingested? Nowadays, a lot of frying pans are made with a non-stick coating. That's usually teflon, or chemically speaking polytetrafluorethylene - $\ce{(C2F4)}_n$. Can you be poisoned if some of that teflon gets in your meal by scraping the pan or in any other way? And if so, what happens (chemically speaking) in your body after ingestion?

I hope I am not breaking any rules of this forum by directly quoting, but here you go:

"The cold polymer is harmless on ingestion, in contact
with the skin, or implanted in living tissues. When
heated above 250° C. it begins slowly to give rise to
small amounts of decomposition products, some of
which are toxic.
The accidental inhalation of fumes from the overheated
polymer by workmen produces an influenza-like
illness which has been described elsewhere. The
illness follows a latent interval of a few hours and
resolves within a day or two with no subsequent illeffects.
This may happen, for example, when the
polymer is heated to comparatively high temperatures
(350-400 C.) in an oven, an extruder, or some other
equipment used to fabricate it, but it is easily prevented
by the intelligent use of exhaust ventilation."

In: Harris, Kenwin. “Toxicity of Polytetrafluoroethylene.” British Medical Journal 1.5285 (1962): 1146.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1958415/pdf/brmedj02864-0082a.pdf


Acidity of thiamine - how comes hydroxide anion with pK about 15 ionizes thiamine with pK about 18? Textbooks say removing proton from thiamine's C atom will produce a resonance stabilized carbanion which is hence bioactive. That's OK, but how comes that 1) we abstract the acidic proton with the OH base whose pK is lower that for thiamine, and 2) the remaining anion acts - in bio context - in aqua medium (?), hence will immediately deprotonate water and lose its negative charge.

In the ionized structure shown, which looks like an ylid, the carbanion is formed on an $\mathrm{sp^2}$ orbital and no resonance stabilization of this charge is possible.
Regarding your question about the ionization, remember that biochemical reactions don't always happen in water, they may happen in the active site of enzymes, which have a completely different environment: the dielectric constant is different, the $\mathrm{p}K_\mathrm{a}$ of substances change and the proximity of certain groups stabilize reactive intermediates that would not be formed in water.


What's the mechanism for the oxidation of primary alcohol to carboxylic acid with HNO3? Can someone help me find the mechanism for this reaction. I've been looking for it, but I
haven't found it. Thank you.

http://connection.ebscohost.com/c/articles/25511593/oxidation-nitric-acid-aliphatic-alcohols-diols-carboxylic-acids

http://pages.towson.edu/ladon/orgrxs/reagent/oxidizer.htm

It is a dangerous thing to do for the easy detonation of nitrate esters.  Activated and stabilized alcohols, e.g., benzoin to benzil, are reasonable.


Ground state electron configuration of chromium What is the ground state electron configuration of chromium?

Is it $\ce{[Ar]}4s^23d^4$ or

Is it $\ce{[Ar]}4s^13d^5$

WebElements page about chromium (and a number of resources) agree with the comment by @Philipp:

  The ground state electron configuration of ground state gaseous neutral chromium is  $\ce{[Ar]}3d^54s^1$

Which in some resources is written as $\ce{[Ar]}4s^13d^5$

Based on the Royal Society of Chemistry article The trouble with the aufbau principle:

  it appears that the most stable configuration for atoms of chromium, copper, niobium, molybdenum, ruthenium, rhodium, silver, platinum and gold involves only moving one electron into an $s$ orbital.

Chromium is one of a handful of Transition elements that share this electron configuration.


Why can't a reaction go to completion? What is the reason for why an exergonic reaction would still have some remaining reactants (or an endergonic reaction to have any products)? The explanation given on this page (see Chemical reactions and mixing) is that there is a Gibbs free energy change caused by the entropy change by mixing; therefore, it is always energetically favorable to have at least some of each.

At another page (see Dissociation of weak acids) on the site, the explanation as to why weak acids (with positive $\Delta G^{\circ}$ for reaction with water) dissociate at all seems to be that some molecules do have the energy required for the reaction, analogous to molecules overcoming activation energy in the context of kinetics.

Are these two explanations equivalent? The first one is a macroscopic view that involves entropy of the whole system; the second one involves the distribution of speeds of the molecules themselves.

You alluded to the answer when you mention activation energy. Kinetically the equilibrium constant is $K_e = k_f/k_b$ where $k_f$ and $k_b$ are the forward are reverse reaction rate constants in the reaction $\ce{A <=> B}$. The reason that there is a finite and not zero back reaction rate constant, is that the activation barrier going B to A is not infinitely high. Quite the opposite in fact and  while larger than the barrier A to B, (if the reaction is exothermic), may nevertheless be of such a magnitude that the products can return to reactants and thus an equilibrium is established. In this case molecules are continually transforming between one another. The ratio of concentrations at equilibrium is the equilibrium constant.
The reason that the activation barrier can be surmounted is that there is a Boltzmann distribution of energies in all molecules and this means that the number of molecules in either A or B with energy $E_i$ is given by $n_i=n_0\exp(-E_i/(k_BT))$. This is the Boltzmann distribution for molecules with energy $E_i$ and $k_B$ is the Boltzmann constant.
The total number of molecules of type A is given by the sum of those in each level $E_i$ from level 0 to the maximum energy level, which is
$$\frac{n_A}{n_0}=\sum_i \exp(-E^A_i/(k_BT))$$
The equation for molecules of type B with energy $E^B_i$ is
$$\frac{n_B}{n_0}=\sum_i \exp(-(E^B_i+\Delta E_0)/(k_BT))=\sum_i \exp(-E^B_i/(k_BT))\exp(-\Delta E_0/(k_BT))$$
where $\Delta E_0$ is the difference in energy between the lowest levels of A and B.
At equilibrium the constant K is the ratio of the two concentrations which is therefore the same as the ratio of the two populations and is
$$K=\frac{Z_A}{Z_B}\exp(-\Delta E_0/(k_BT))$$
where  Z represents the summations and are called the partition functions. This equation represents the equilibrium constant in terms of the molecules energy levels.
The reason that there is a Boltzmann distribution, but a fixed energy at a give temperature, is that by far the the most likely way of populating a range of energy levels is according to the Boltzmann distribution. In fact the chance of having a Boltzmann distribution is greater than all other possible ways added together.
In classical thermodynamic terms, a plot of free energy G vs extent of reaction $\zeta$ (zeta) which ranges from 0 (entirely A) to 1 (entirely B), has a minimum at some intermediate value of $\zeta$ and this value naturally depends on the particular reaction.  The minimum value is where the reaction is at equilibrium and at the minimum the slope of G vs. $\zeta$ is zero which is also when $(\partial G/\partial \zeta)_{T,P} = 0$.
The minimum in G occurs because the entropy term $-TS$ has a minimum as the mole fraction of B increases (at constant T). Initially it is the entropy of A alone in solution, but at equilibrium is a mixture of A and B. If the reaction goes entirely to B then the entropy is that of B alone. The entropy of a mixture is naturally greater than that of one species thus $-TS$ has a minimum vs. $\zeta$.
If $dn_A$ and $dn_B$ of A and B molecules react the the infinitesimal change in zeta is $d\zeta = dn_A-dn_B$. The total change in free energy is $dG=\mu_Adn_A+\mu_Bdn_B$ where $\mu$ is the chemical potential or the free energy / mole. From these two equations the change in free energy of the reaction is
$$dG=(\mu_b-\mu_A)d\zeta$$
The reaction proceeds until $(\partial G/\partial \zeta)_{T,P} = 0$ and thus to proceed further expressions for the chemical potential are needed. For a perfect gas $\mu=\mu^0+RT\ln(p) $  thus
$$(\partial G/\partial \zeta)_{T,P} = \mu^0_B-\mu^0_A+RT\ln\frac{p_B}{p_A}$$
which can be recast as
$$(\partial G/\partial \zeta)_{T,P} = \Delta G^0+RT\ln\frac{p_B}{p_A}$$
and at equilibrium as the derivative is zero
$$\Delta G^0=-RT\ln\frac{p_B}{p_A}=-RT\ln K_p$$
which connect the equilibrium constant ($K_P$ for the gas phase reaction), with the free energy.

Edit:
Notes: The shape of the plot of G vs. extent of reaction can be shown by considering a general reaction of the form $\ce{A + B <=> 2C }$ . The Gibbs function is then
$$G = n_a\mu_a + n_b\mu_b + n_c\mu_c $$
for $n_i$ moles of species i with chemical potential $\mu_i$.  If we assume that the reaction is one of perfect gases then at total pressure P and with mole fraction x, the chemical potential of species i is
$$\mu _i = \mu_i^0+RT\ln(P)+RT\ln(x_i)  $$
where $\mu_i^0$ is a function of temperature only.  G can now be written as
$$G= n_a\mu_a^0 + n_b\mu_b^0 + n_c\mu_c^0+RT\ln(P)+RT(n_a\ln(x_a)     +n_b\ln(x_b)+n_c\ln(x_c) )$$
If we assume that the total pressure is 1 atm. then the $RT\ln(P)=0$ and can be removed from the equation. Using the relationship between number of moles $n_a=n_b$ and $n_c=2(1-n_a)$ gives
$$G= n_a(\mu_a^0 + \mu_b^0-2\mu_c^0) + 2\mu_c^0 + 2RT(n_a\ln(n_a/2)     +(1-n_a)\ln(1-n_a) )$$
The $\mu^0$ are the properties of the pure components so do not change during the course of the reaction which means that G is a function of $n_a$ only.
A plot of $G(n_a)-G(0)$ vs. $n_a$ is shown below.

Plotting this way gives a value of zero when $n=0$, which is the condition should all reactants be converted into product, point S. The equilibrium value is at point E. The energy T-R is that due to mixing reactants and R-E that extra energy of mixing when C is formed. The dashed diagonal line is $n_a(\mu_a^0 + \mu_b^0- 2\mu_c^0)$ which is how the reaction would behave if no entropy of mixing is included.
The important point is that there is a minimum in the free energy curve because the creation of a mixture is an irreversible process and this mixing produces a decrease in free energy.
The equilibrium constant is calculated using $-RT\ln(K_p)=2\mu_c^0-\mu_a^0-\mu_b^0$


Spiralling of electrons into the nucleus A drawback (and a major one) of Rutherford's model was that the electrons, being accelerated charged particles, would gradually spiral into the nucleus and collapse into it. My question is how did Bohr rectify this drawback in his own model.

In Bohr's model too, electrons move in a circular path, which means they are accelerated. Thus, going by Electromagnetic theory of Maxwell, those accelerated electrons should radiate out their energy, and collapse into the nucleus, just like they did in Rutherford's model. However every book mentions that this does not happen in Bohr's model. Why not?

Please tell me where I am wrong, because this problem has been disturbing me for quite long now.

As a matter of fact, Bohr literally went past around this issue. In 1913, he proposed his model postulating electrons could only have certain classical motions:

Electrons in atoms orbit the nucleus.
The electrons can only orbit stably, without radiating, in certain orbits (called by Bohr the "stationary orbits") at a certain discrete set of distances from the nucleus. These orbits are associated with definite energies and are also called energy shells or energy levels. In these orbits, the electron's acceleration does not result in radiation and energy loss as required by classical electromagnetic theory (This was based upon Planck's quantum theory of radiation).
Electrons can only gain and lose energy by jumping from one allowed orbit to another, absorbing or emitting electromagnetic radiation with a frequency ν determined by the energy difference of the levels according to the Planck relation:
$$ \Delta {E}=E_{2}-E_{1}=h\nu \ $$

where h is Planck's constant.

As remarked above, these are postulates (statements assumed to be true from which further reasoning follows), and (even though that part was left unexplained), it's acceptance was due to the fact it explained correctly many experimental results. It was further displaced by the modern quantum theory.

Hope it helps, it also puzzled me a few years ago! :P

Reference: https://en.wikipedia.org/wiki/Bohr_model


Can I make a mirror on PET plastic using Tollen's reaction? I want to make a very shiny surface, like a mirror. I tried to use chrome paint but it's barely shiny. It's more like gray even though I bought one of the most expensive chrome looking paint.

I know mirrors can be produced using Tollen's reagent on glass (seen some demonstrations on YouTube).

But glass is quite expensive for what I want - I would like to make the mirror on plastic - more specifically on PET - the same material that a normal Coca Cola bottle is made from. I have big sheets of this PET material and I intend to transform them into mirrors.

Will silver deposit on the PET as it does on glass?

I need to order the ingredients online, then wait for it a couple of days - also because they are quite large quantities, it will end up being quite expensive. So, before making a mistake, I thought to ask here.

Could I use a Tollen's reaction to make a mirror on plastic?

I found a video on YouTube making a mirror on PET bottles.

Just that it consumes quite a lot of substance, and is not using Tollens reagent. My question is specifically about Tollens reagent.

What she uses is some "secret mixture" called RA-205.

It just proves that Ag can precipitate on PET bottles, which further supports the fact that will work with Tollens reagent. I'm will edit this answer once I execute my experiment, so this finding may only be somewhat helpful.


Real gas equation at high molar volume How would the van der Waals equation of state look like when molar volume of gas is high? Does high molar volume correspond to low pressure?

My textbook says that

  A real gas behaves similar to an ideal gas in the limit of large molar volumes.

What is the reason behind that?

The ideal gas equation of state assumes all of the particles are not interacting. This is a bad assumption when particles are close together, such as in a liquid. However as particles get further apart this becomes a more realistic model since particles collide less and less often.

In general low pressure and high temperature leads to conditions where the ideal gas equation will work. We see this in combustion reactions actually. You can model a gasifier fairly well with the ideal gas equation of state because of how high the temperatures are.

At high temperatures, molecules move with higher energy, so they bounce of walls harder. Thus to attain a given pressure, you need fewer molecules if they are hitting the walls harder. Thus, at a higher temperature, there are fewer molecules in your vessel, making them further apart, which makes them more ideal.

In terms of molar volumes (I prefer pressure but won't begrudge your textbook), yes as the molar volume gets bigger that means the same number of particles take up more space, meaning they are farther apart from each other. In the extreme limit they will become infinitely far apart and be exactly like an ideal gas. In practice the molar volume does not have to be extreme for the ideal gas to start modeling the system well.


Why is entropy favorable? I cannot seem to grasp the logic behind it. We say that more entropy (or more disordered system) is favorable over less entropy. But why? Why is randomness preferred over proper arrangement of atoms/molecules? The only thing I can think of is more degrees of freedom. Molecules in a more random system would have more degrees of freedom, and would thus be favorable. But, again, why? Why would more degrees of freedom be favorable? I've already asked my teachers and looked up internet, but haven't found any satisfactory answer. Also, is there any way you could also explain diffusion by the same concept (diffusion is entropically favorable, right?)?

PS: if this question is better suited for physics stack exchange, I'd be more than happy to have it migrated.

Thermodynamics. The second law of thermodynamics states that entropy always increases in an isolated system. This is taken as a fundamental postulate---we simply accept this statement as a fact regarding how the world works, and our justification is that no experiment has ever shown the second law incorrect. In the framework of macroscopic thermodynamics, there is no deeper principle from which we can hope to derive the second law.

Statistical mechanics. Statistical mechanics brings a microscopic perspective into thermodynamics. We talk of macrostates---macroscopic states of our system, characterized by (macroscopic) observables like pressure, temperature, volume,  etc.---which may be realized by a number of microstates---a complete microscopic description of our system, consisting of the positions and momenta of each particle. Many different microstates can correspond to the same macrostate, since a macrostate only deals with macroscopic properties.

In statistical mechanics we can go one level deeper, taking as a fundamental postulate the principle of equal a priori probabilities, the idea that the system has an equal probability of being in any given microstate. This implies that the most likely macrostate for the system is that with the greatest number of corresponding microstates. At this point, knowing also from statistical mechanics that the entropy is a monotonically increasing function of the number of microstates (specifically the logarithm of the number of microstates in an isolated system, but the functional form is unnecessary here), we can motivate the second law: the entropy of a macrostate of an isolated system is a measure of the number of its corresponding microstates; the greater this number, the more likely this macrostate will be observed. The equilibrium state of a system is simply its most probable macrostate.

Example. (Microscopic diffusion.) Let us consider a microscopic model of diffusion: a $10$-by-$10$ grid of particles, $50$ red and $50$ blue. The particles are indistinguishable except by color. In addition, we will consider two sets of macroscopic states, 'ordered' and 'disordered'. Take as initial state the configuration with the two groups of particles separated by color into two equal halves by a horizontal boundary, the blue particles on the top and the red on the bottom. The diffusion process is akin to removing this artificial boundary.

There is one initial ordered microstate and $100!/(50!)^2$ final microstates. (Note that indistinguishable microstates should be considered the same microstate.) The entropy of our system has increased, because we have relaxed a constraint and allowed more microstates into our system. Most of these final states look disordered. Diffusion is therefore an entropically favorable process that brings an ordered system into a disordered one.

Additional remark. I find statements like "entropy drives the system towards a particular state" to be somewhat misleading, because they imply that entropy behaves like a force. But entropy, as we've seen, is really just statistics.


Equivalent weight in case of disproportionation reaction We all know that

$\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}}$

However consider a reaction like:

$\ce{Br2 + OH- -> Br- + BrO3- + H2O}$
where oxidation numbers of $\ce{Br}$ are $0, -1, +5$ respectively.

  What should here be the equivalent weight of $\ce{Br2}$?

PS: I understand that I should also show what I've attempted. But, this is just a question that came up into my head
and I've made it as descriptive as possible. I don't really see anything else that I must add to make this question complete.

Thank you!

UPDATE: I just came across a formula for calculating the n-factor for a dispropportionation reaction $n_f=\frac{n_1\cdot n_2}{n_1+n_2}$ (sources, 1 and 2), for $n_1$ the magnitude of n-factor in oxidation reaction and $n_2$ the magnitude of n-factor in reduction reaction. Can anyone prove how to arrive at this formula?

The $n$-factor of a compound in a reaction gives the number of moles of electron lost or gained by it (during oxidation and reduction respectively) per mole of the given compound. In a disproportionation reaction, the compound undergoing both the oxidation and reduction is the same.

It is important to understand that the $n$-factor depends not just on the change in oxidation state of the element but also the number of moles of the element undergoing that change.
This is the reason why we cannot simply take the average of the two $n$-factors as the $n$-factor of the disproportionate. It is necessary to also consider how much of the reactant is getting oxidised and how much of it is getting reduced, something along the lines of a weighted mean.

It is quite simple to derive the expression you have found for the $n$-factor of a disproportionated compound. Consider the following general reaction:

$$\ce{nA -> xB + $(n - x)$C}$$

Where $x$ moles of $\ce{A}$ (for which $n$-factor is to be calculated) oxidises to $\ce{B}$ and the rest of it reduces to give $\ce{C}$. Let the $n$-factors for the oxidation and reduction respectively be $n_1$ and $n_2$.  Which means $n_1 x$ moles of electrons go into oxidation, and $(n - x)n_2$ moles into reduction.

We have 2 equations: one by conservation of electrons,
$$\begin{align*}
n_1 x &= n_2(n - x) \\[7pt]
\implies x &= \frac{n_2\cdot n}{n_1 + n_2}
\end{align*}$$
and the other from the definition of $n$-factor:
$$\begin{align*}
n_f &= \frac{\text{moles of electrons transferred}}{\text{moles of reactant}} \\
&= \frac{n_1 x}{n} \\
&= \frac{n_1 n_2}{n_1 + n_2}
\end{align*}$$

An interesting observation here is that upon simplifying the obtained equation, you get the formula for the equivalent resistance of resistors connected parallelly in an electric circuit. The correspondence between them is intuitive.
$$ \frac{1}{n_f} = \frac{1}{n_1} + \frac{1}{n_2} $$


Substances that pass visible light but absorb microwaves Specifically, the substances should be water-miscible or water-soluble, but not water itself. Additionally, they should be absorptive but not opaque.

Sure - ethanol and methanol (among others) are listed here as being "high absorbers" of microwave energy.