How to calculate the potential via Hess's law? Is my textbook wrong about this? Given the two reduction half-reactions for the following fuel cell, determine the standard cell potential $E^\circ_\mathrm{cell}$,
  $$\ce{CH4 (g) + 2 O2 (g) + 2 OH- (aq) -> CO3^2- (aq) + 3 H2O}.$$
  Reduction half-reactions:
  \begin{align}
\ce{CO3^2- (aq) + 7 H2O (l) + 8 e- &-> CH4 (g) + 10 OH- (aq)}&
E^\circ &= \pu{-0.1 V}\\
\ce{O2 (g) + 2 H2O (l) + 4 e- &-> 4 OH- (aq)}&
E^\circ &= \pu{+0.55 V}
\end{align}
  Answers: (a) $\pu{+0.45 V}$ (b) $\pu{-0.45 V}$ (c) $\pu{-0.65 V}$ (d) $\pu{+0.65 V}$

It is my understanding that to apply Hess' Law to this equation the top equation sign would be reversed and the outcome would be $\pu{+0.65 V}$. Often the system produces incorrect answers. I was looking through my textbook to double check this, and there is an instance where it says
$$E_\mathrm{cell} = E_\mathrm{anode} - E_\mathrm{cathode},$$
but I believe that rule does not fit this instance. Could somebody validate that this answer provided is indeed correct or not?

I would like to clarify that my answer was (d) and the system provided that the correct answer was (a).

I sent an email to my teacher to contest the question using an example from the textbook. Given that Martin has provided me with an answer concurrent with the $\pu{+0.45 V}$, then there must be a mistake in the text or I am misinterpreting something. I would really like to understand this.

At first my teacher told me that after flipping the equation you still need to subtract the anode from the cathode. So I responded like this:

  I would like to reference the E Cell example from the text. In the text it explains that $$E_\mathrm{cell} = E_\mathrm{anode} - E_\mathrm{cathode}.$$

  The example given in the book is identical to the one provided in the exam:

  \begin{aligned}
 (1): &&\ce{3 Ag+(aq) + Fe (s)   &-> 3 Ag (s) + Fe^{3+} (aq)}\\
 (2): &&\ce{3 Ag+(aq) + 3 e-     &-> 3 Ag (s)} & E^\circ &= 0.80~\mathrm{V}\\
 (3): &&\ce{Fe^{3+} (aq) + 3 e- &-> 3 Fe (s)} & E^\circ &= -0.04~\mathrm{V}\\
\end{aligned}

  By using the same logic that I have applied to the exam question the $E_\mathrm{cell}$ was found to be $\pu{0.84 V}$. This is because the sign on the $E$ value of the $\ce{Fe}$ half reaction was reversed like I did to the top equation (which consequently is the cathode).

  As the given equations line up with the primary equation and signs are flipped whenever the equations are flipped, then it is not important to decipher which is the anode and which is the cathode. This is also in accordance with Hess' Law.

That being said, I am confused whether the textbook example was incorrect, the exam question was incorrect, or my understanding is incorrect.

You have to take into account, that the given reactions are only reductions, while you need to have an oxidation, too. So one of the equations has to be reversed.

\begin{aligned}
(A)E^\circ = -0.10~\mathrm{V}&&\ce{CO3^{2-} + 7H2O + 8e- &-> CH4 + 10 {}^{-}OH}  \\
(B)E^\circ = +0.55~\mathrm{V}&&\ce{O2 + 2H2O + 4e- &-> 4 {}^{-}OH}  \\\hline
\color{\red}{-1}\cdot(A) + 2\cdot(B)
   &&\ce{CH4 + 2O2 + 10{}^{-}OH + 4H2O&-> CO3^{2-} + 7H2O + 4 {}^{-}OH}\\
\implies
   &&\ce{CH4 + 2O2 + 2{}^{-}OH &-> CO3^{2-} + 3H2O}
\end{aligned}

Equation $(A)$ gives you the reduction of the carbonate ions, hence the oxidation of methane. Equation $(B)$ is the reduction of oxygen. You have a galvanic cell of the form $\ce{CH4|CO3^{2-}||O2|{}^{-}OH}$, hence the standard potential difference is
\begin{aligned}
\Delta E^\circ(\ce{CH4|CO3^{2-}||O2|{}^{-}OH})~
               &= E^\circ(\ce{O2|{}^{-}OH}) -E^\circ(\ce{CH4|CO3^{2-}})\\
               &= E^\circ(\ce{O2|{}^{-}OH})
                  -\left[\color{\red}{-1}\cdot E^\circ(\ce{CO3^{2-}|CH4})\right]\\
               &= +0.55~\mathrm{V}
                  -\left[\color{\red}{-1}\cdot (-0.10~\mathrm{V})\right]\\
\Delta E^\circ(\ce{CH4|CO3^{2-}||O2|{}^{-}OH})~
               &= +0.45~\mathrm{V}
\end{aligned}

Just to address the second edit of the OP and to clarify:
\begin{aligned}
 (1): &&\ce{3 Ag+(aq) + Fe (s)   &-> 3 Ag (s) + Fe^{3+} (aq)}\\
 (2): &&\ce{3 Ag+(aq) + 3 e-     &-> 3 Ag (s)} & E^\circ &= 0.80~\mathrm{V}\\
 (3): &&\ce{Fe^{3+} (aq) + 3 e- &-> 3 Fe (s)} & E^\circ &= -0.04~\mathrm{V}\\
\end{aligned}

The sign of the second equation has to be flipped and then subsequently subtracted from the first equation, hence the potential that should be found is $\Delta E^\circ = 0.74~\mathrm{V}$. Therefore, I believe your textbook is wrong.


Does adding mass to a liquid (not water) increase the volume The Question sounds stupid but i have a
legit problem

I mixed 1000 mL coconut water and 225 g of agar-agar to make a coconut water [CW] agar. It gave me more than 1000 ml. the medium went beyond the 1000 ml mark. the CW agar was also too soft to use

then i tried mixing 775 mL coconut water and 225 g agar-agar it gave me little below the 1000 mL mark. the CW agar was just right not too soft and not to hard :|

(I did heat after mixing during those two trials but the volume remained the same after cooling to room temp.)

Seems to break rules taught to us.  i though when you add mass to a liquid it won't increase the volume but the density will increase or does this only apply to water?

"The rule" is that an ideal mixture will have the volume of the sum of its components and the density will be the molar averaged density of the components. Real mixtures deviate from this rule to varying degrees.

The name of the property you're looking for that describes this deviation is "excess volume". Excess volume can be either positive (1L of A and 1L of B creates >2L of AB mixture) or negative (1L of C and 1L of D creates <2L of CD mixture). The deviation from the rule depends mostly on how the molecules of the components interact with themselves compared to with the each other. Without knowing much about this specific case, I would guess that the big agar polymer molecules interrupt the local organization of the water molecules. This causes the density to decrease.

For many fluids, such as ethanol and water, the densities have been well studied and you can look up tables of excess volume. On a quick google search, it doesn't appear that anyone has ever done this for coconut water and agar. Unfortunately, that means more work for you. Fortunately, it means you get to break new scientific ground!


Formating a chemical equation - Physical states I study chemistry in an University and I have two professors that says opposing things. One says that the physical state must be subscribed (1 at the picture) and other that it must be written to the left side (2). Since both of them are posdocs and have way more experience in chemistry than me, I really don't know to whom I should list. i've tried to look for a recommendation in the IUPAC site, but I couldn't find anything about that. Someone could help me with a link from some organization (preferably an international one) that recommends about this topic?

Overall science is not affected by whether you write subscript or a superscript or in line. Afterall all notation was invented by humans. When professor quibble over these things my suggestion for them is to read the problem of "How many angels can stand on the point of a pin?" Apparently when Constantinople was falling, allegedly scholars were discussing these problems. Anyway, the lesson is that science is more important, notation comes later.

The International Union of Pure and Applied Chemistry (IUPAC), makes recommendations to systematize notation. Show the book "Quantities, Units and Symbols in Physical Chemistry" Third
Edition, https://doi.org/10.1039/9781847557889. It is available for free.

There was a similar question What is the standard way to denote physical states in a chemical reaction?

but the reference is old. The above reference is recent but there is no difference. Just for your interest, to see how old that reference is, it looks like type-written text!

They suggest to write the state of aggregation as they call it, just next to your symbol like $\ce{Na(s)}$, not as a subscript.


Are there notable organic compunds which look very similar but have very different properties? I've seen people arguing for some questionable "miracle cure-all" alternative medicines not scientifically, but by their own logic. A common claim is, that if a compound looks very similar to an important compound found in our body, it must be very beneficial. By looks very similar I mean the drawings of their structures look similar, as in, similar as images.

I found this reasoning alone to be unscientific. (correct me if I'm wrong). I would like to explain it to people, but it's hard to do if they lack even basic understandings in chemistry, and have claims like "I think it might happen to be like this, so it's a proof that it's like this".

So, I'm looking for a counter-example. Are there famous and well-known (organic) compounds which look similar on a structure drawing, but have vastly different properties, for example one is either beneficial or a necessary component of our body, and the other is highly toxic?

I'm sorry for the possible unclear terminology, English is not my first language and I didn't find any better term to say "structure drawing" for something like this

One example of disastrously different effects of very similar molecules is thalidomide. There are two isomers, only differing in their stereochemistry. The (R)-isomer is responsible for the sedating effect it was initially used for. The (S)-isomer lead to horrifying birth defects in the children of woman that took that drug.

It is somewhat more complicated, as the isomers are racemized quickly in vivo, so even if you administer only the (S)-isomer, it will still be converted to the (R)-isomer in the body.


Why is a Grignard reagent not protonated by chloramine? When $\ce{RMgX}$ (Grignard reagent) reacts with chloramine ($\ce{NH2Cl}$) it forms $\ce{RNH2}$.

But why doesn't it form $\ce{RH}$ by taking a hydrogen atom from chloramine as the negative charge so developed on nitrogen will be more stable as nitrogen has high electronegativity and $\ce{Cl}$ being a electron withdrawing group further stabilise the anion.

So where am I wrong? Can anybody suggest a mechanism?

The $\ce{N-H}$ protons are relatively weakly acidic and this makes deprotonation kinetically less favored if an alternative nucleophilic attack mechanism is available.  With $\ce{NH3}$ no such alternative is there, due to lack of a decent leaving group, so deprotonation ultimately prevails; but $\ce{NH2Cl}$ allows displacement of a chloride ion and thus enables nucleophilic attack.


Why is solubility often listed as a chemical property? Isn't the action of dissolving just a physical change? I am making a list of basic physical vs chemical properties.
I have been taught that when salt dissolves in water it is simply a special kind of mixture called a solution.
I have been taught that mixtures do not create a chemical change or reaction.
But I found a lot of high quality chemistry websites that list solubility as a chemical property of compounds.

Why isn't solubility a physical property if dissolving is simply forming a mixture and not causing a chemical reaction?

The classification of physical and chemical changes has done more harm to students than good. This topic was taught in the 19th century but it still lingered in modern general chemistry textbooks which serve to repel good students from taking up chemistry as their field. For example, if we call the charge of electron red, and the charge of proton blue, will it change any fundamental thing?

You can try to understand physical vs. chemical change as follows: If you are changing the physical form (liquid, solid, gas) of one or more component, then you can safely label it as a physical change provided you can bring the original form back.

You can melt NaCl crystals into a liquid, this is a physical change because we can cool it and bring back the original material.

In the same way, solubility is just a property of matter. It is a very complex phenomenon. Whether you call it physical or chemical it does not matter in real life (most scientists will not worry about it and nobody will ever ask you these questions once you pass this course-so much so for the utility of this concept). For example, HCl is very soluble in water, but also reacts with water in such a way that it protonates water. Dry HCl gas is "neutral" but its solution in water is a powerful acid. There is no simple way to extract the gas back out of water. This solubility involved a chemical change.

On the other hand, salt NaCl when dissolved in water, it will not chemical react with water per se and you can recover the salt by evaporation. Should I call this a chemical change or a physical change? It does not matter.


Why does magnesium form covalent bonds with carbon? According to my professor, magnesium and lithium form covalent bonds with carbon but calcium does not. He did not elaborate.

Why is this so? The electronegativities are:

Mg — 1.31
Li — 0.98
Ca — 1
C — 2.55

So a $\ce{C-Mg}$ bond would have an electronegativity difference of 1.24 and a $\ce{C-Li}$ bond would have an electronegativity difference of 1.57. Both are considered covalent. So how come a $\ce{C-Ca}$ bond wouldn't be covalent? There exists an electronegativity difference of 1.55 — comparable to the difference in the $\ce{C-Li}$ bond.

I know that Mg likes to be tetra-coordinated (Grignards) but I don't think this has much to do with this discussion. So, what makes Mg and Li special compared to Ca?

Summary of possible explanations others have come up with

$\ce{Mg^2+}$ has an ionic radius 72 pm, $\ce{Ca^2+}$ has an ionic radius of 100 pm, and for $\ce{Li+}$, it's 76 pm. So perhaps the greater charge density of $\ce{Mg^2+}$ and $\ce{Li+}$ allow them to pull electrons toward themselves a bit more, thus making the bond more covalent in character? That being said, $\ce{Li+}$ has a single positive charge while $\ce{Ca^2+}$ has a double positive charge, and I haven't calculated the actual charge densities for any of these ions so I don't think we can come to a definitive conclusion here.

http://www.wiredchemist.com/chemistry/data/metallic-radii

The crystal structure of calcium carbide gives a subtle hint of covalent bonding.  It looks like a rock salt structure, but note that the carbon atoms in the anions have their lone pairs directed towards the calcium atoms and so are favorably oriented for covalent overlap. The structure might be considered either a rock salt structure or a series of parallel $\ce{-Ca-C#C -}$ polymeric chains.

Source


MO diagram of N2(2-) In the MO diagram of $\ce{N2^2-}$, does s-p mixing happen to a significant degree to change the ordering of orbitals - $\pi_\mathrm{2p}$ orbital below $\sigma_\mathrm{2p}$? As it is isoelectronic with oxygen, I think that the s-p mixing shouldn't change the ordering.

In order to test whether there may be $\ce{s}$-$\ce{p}$ mixing and also to check whether there may be actually a bound state, I optimised this molecule (at PBE0-D3/def2-TZVPD) using NWChem 6.6 and analysed it using natural bonding orbitals (NBO 5.9).
I found following minimum geometry:

Figure 1: Minimum geometry found for $\ce{N2^{-2}}$ (distance in angstrom)

So there seems to be a bound state.

Natural charge on each $\ce{N}$ is -1.0 as would be expected (natural electron configuration was found to be $\ce{1s^{2.00} 2s^{1.65} 2p^{4.17} 3s^{0.06} 3p^{0.05} 3d^{0.07} 4d^{0.01}}$).
Wiberg bond order in the NAO basis for the $\ce{N-N}$ bond was found to be 2.2511.

Two bonding NBOs were found:

    (Occupancy)   Bond orbital/ Coefficients/ Hybrids
-------------------------------------------------------------------------------
  1. (2.00000) BD ( 1) N  1- N  2
              ( 50.00%)   0.7071* N  1 s(  0.00%)p 1.00( 99.74%)d 0.00(  0.24%)
                                                 f 0.00(  0.02%)
              ( 50.00%)   0.7071* N  2 s(  0.00%)p 1.00( 99.74%)d 0.00(  0.24%)
                                                 f 0.00(  0.02%)
  2. (2.00000) BD ( 2) N  1- N  2
              ( 50.00%)   0.7071* N  1 s( 36.78%)p 1.70( 62.66%)d 0.01(  0.51%)
                                                 f 0.00(  0.06%)
              ( 50.00%)   0.7071* N  2 s( 36.78%)p 1.70( 62.66%)d 0.01(  0.51%)
                                                 f 0.00(  0.06%)

That is,

$$\pi_{\ce{N1-N2}} = 0.7071 \ce{p}_{\ce{N1}} + 0.7071 \ce{p}_{\ce{N2}}$$

Figure 2: $\pi_{\ce{N1-N2}}$ bond of $\ce{N2^{-2}}$

$$\sigma_{\ce{N1-N2}} = 0.7071 \ce{sp^{1.70}}_{\ce{N1}} + 0.7071 \ce{sp^{1.70}}_{\ce{N2}}$$

Figure 3: $\sigma_{\ce{N1-N2}}$ bond of $\ce{N2^{-2}}$

Furthermore,

NATURAL BOND ORBITALS (Summary):

                                                    Principal Delocalizations
          NBO                 Occupancy    Energy   (geminal,vicinal,remote)
===============================================================================
Molecular unit  1  (N2)
  1. BD ( 1) N  1- N  2        2.00000     0.18488
  2. BD ( 2) N  1- N  2        2.00000    -0.43475  10(g),42(g)

Thus, according to the simple Lewis picture provided by NBO,

$\ce{s}$-$\ce{p}$ mixing ($\ce{sp^{1.70}-sp^{1.70}}$) happens for the formation of $\sigma_{\ce{N1-N2}}$, as expected;
the formation of $\pi_{\ce{N1-N2}}$ happens through pure $p$ hybrids; and
$\pi_{\ce{N1-N2}}$ lies in energy above $\sigma_{\ce{N1-N2}}$ by 0.61963 a.u..


Difference between internal energy of combustion and enthalpy of combustion? I am having real trouble understanding the first law of thermodynamics and I think the question, below, highlights most of my issues:

1) (a) A bomb calorimeter was calibrated by igniting a $0.825\ \mathrm{g}$ sample of benzoic acid in the presence of excess oxygen. The heat given out made the temperature of the calorimeter rise by $1.940\ \mathrm{K}$ from $298\ \mathrm{K}$. The internal energy of combustion of benzoic acid is $-3251\ \mathrm{kJ\ mol}^{-1}$. Calculate the heat capacity of the calorimeter.

(b) In two separate experiments in the same apparatus, $0.498\ \mathrm{g}$ of fumaric and $0.509\ \mathrm{g}$ of maleic acid (both of RMM 116) were ignited and gave temperature rises of $0.507\ \mathrm{K}$ and $0.528\ \mathrm{K}$ respectively. Use the heat capacity of the calorimeter, calculated in (a), to calculate (i) the molar internal energy of combustion, (ii) the molar enthalpy of combustion, and (iii) the molar enthalpy of formation of fumaric and maleic acids. Comment on the difference. The standard enthalpy of formation of water is $285.8\ \mathrm{kJ\ mol}^{-1}$ and of $\ce{CO2}$: $-393.5\ \mathrm{kJ\ mol}^{-1}$.

Part (a) was fine - the heat capacity at a constant pressure = change in internal energy/temperature change. Then divide by number of moles.

Part (b) is where I struggle conceptually. I got the answer correct but only by following lecture notes. Sticking with just fumaric acid because there's no point talking about both. It's easy to find the molar change in internal energy using the constant volume heat capacity ($-1338\ \mathrm{kJ\ mol}^{-1}$).

However, to find the molar enthalpy change, the change in $H = U + pV$. but since $\ce{pV=nRT; H=U+nRT}$ where n depends on the change in the number of gaseous moles from products to reactants - in this case +1. So $H = U + RT$ giving $-1336\ \mathrm{kJ\ mol}^{-1}$ (less exothermic).

Firstly: Why are the values negative. If the system is the fumaric acid/products of reaction then surely there is an increase in internal energy because the particles will be moving faster after the bonds have broken. (Am I right in thinking that the system is the chemical)?

Secondly: Why is the enthalpy value less exothermic? (I have a very poor grasp for what enthalpy is especially in a situation like this).

Any help with answering my questions or helping me with this concept would be greatly appreciated. Also, do you know any good books/videos I could watch to help me undertand?

Combustion releases energy from the system usually as heat and/or work to the surroundings, that's why the standard energy of combustion has a negative sign. The bomb is the system in this case (alternatively you can see the chemical as the system, but you will need to include the gaseous products in your system too, changing the volume of the system, which I find a less preferable viewpoint).

The value for the enthalpy will be lower since energy is trapped in the form of increased pressure introduced by the reactions products (mainly by $\ce{CO_2}$) in the bomb. This pressure can be used to do work on the surroundings (in an open system this will be usually on the atmosphere). If you determine the final pressure, you can calculate your $\Delta U$ back ($\Delta U = \Delta H - (p_{end}-p_{atm})V_{bomb}$).

What book do you use for the lectures?


Finding molecules that you randomly create using a modeling kit I recently bought https://snatoms.com/ (which is just a real-life modeling kit for atoms/molecules) and was just keen on learning about chemistry in general. I often randomly join atoms to make up some molecule. Was wondering if there is an easy way to find the molecule i created?

Here's a list of all the stuff you can try:

For you, the common theme of all these steps would be to be able to correctly identify the skeleton of the molecule that you built yourself, and then being able to draw it on the online molecular editor.

Sign into Reaxys or SciFinder, draw your structure, and search for molecules or reactions. Note that these require institutional access (or a very hefty fee).
ChemSpider Structure search: It's a free tool, and my personal goto for getting access to several matching compounds. You can search for "exact matches", or partial matches like "tautomers", "same skeleton", or even "all isomers". There's search filters as well. But again, the first few options are sufficient for your snatoms kit.
eMolecules (in case you didn't notice, we also advertise it in our right sidebar) very similar to ChemSpider above.
PubChem Structure search - again, similar to the above two. Note that though this one has an odd (and old-looking UI) which you might not find intuitive.

The main difference between all four of these is primarily the level of detail and depth of their data. Functionally, they all do the same thing. But, they do so on different databases. That means for more complex molecules, their outputs may differ. However, this hopefully won't be a problem with your simple snatoms kit, so you can go ahead and try any of them. Good luck!

Addendum: generating a 3D structure from the molecule name

You may use this free online tool based on JSMol. Type the IUPAC name of the molecule (say (1-methylcyclobutyl)benzene) into the lower right box, and press Enter. Its 3D structure is then immediately transferred into the box above. You can zoom it in or out, or rotate it by 360deg. This would help when you're trying to create the 3D molecular structure given the IUPAC name.


What is oil smoke point? I understand that the smoke point is the temperature at which oil starts to produce smoke. But how exactly is that smoke produced? I've read that it is related to hydrolysis and oxidation, but I'd like to know a more precise link.

Typically smoke point has to do with the composition and amount of free fatty acids in the fat/oil. Most fats and oils that you see on the market are mostly composed of triacylglycerols, with only small amounts of free fatty acids. The more free fatty acids you have the lower the smoke point. Similarly unrefined oils have a low smoke point. If you have fats/oils that contain high levels of unsaturated fatty acids, then that product will also have a lower smoke point than say a fat that contains significant amounts of saturated fat. In regards to the chemical mechanism, upon heating the triacylglcyerol structure gets broken down into free fatty acids and glycerol via a hydrolysis reaction. The free fatty acids volatilize and cause the visible smoke. Unsaturated fatty acid are more sensitive to heat and will be hydrolyzed quicker than saturated fatty acids. For more info read this article: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4424769/


What accounts for the high spin state of the complex Tris(acetylacetonato)iron(III)? I understand that there's 5 d-electrons for $\ce{Fe^3+}$ ion, but why it doesn't fill up the lower energy orbitals first to form one unpaired electron, but rather, filling up all the orbitals to form 5 unpaired electrons?

I am unsure how the ligands in this complex affects the crystal field splitting. Do we consider [acac] to be a weak field ligand?
Could it be the geometry of this complex that give rise to high spin state?

I will go into the theoretical details first and come to the practical answer for your exact case later; separated from the theory by a horizontal rule.

Approaching the complex theoretically, our first observation is that we are going to have six atoms coordinating our $\ce{Fe^3+}$ centre. Six ligands means that we should consider an octahedric environment as initial approximation. For the next bit of the discussion, I am going to ignore the fact that two ligands each belong to the same molecule and treat them all as $\ce{O^{\frac{1}{2}-}}$ or maybe as $\ce{OH-}$ since the latter is close enough.

The crystal field theory, our first approximation, considers these ligands to be point-shaped negative charges that approach an $\ce{Fe^3+}$ ion. The lonely iron(III) in vacuum will have five d-orbitals of degenerate energy; the approaching negative charges will destabilise all those orbitals that point towards them — for simplicity, let us assume that the corners of the octahedron are the axes of a coordinate system, and thus the destabilised orbitals are $\ce{d}_{x^2 - y^2}$ and $\ce{d}_{z^2}$. (The other orbitals point between the ligands and are therefore unaffected.) This will give rise to the most basic of all transition metal complex energy diagrams as shown in figure 1.

Figure 1: splitting of d-orbitals in an octahedral complex according to crystal field theory. The energy difference always corresponds to $10~\mathrm{Dq}$

Crystal field theory, unfortunately, gives us no information about the energy difference between the two sets of d-orbitals — this is important, since we need a significant energy difference if we want to keep our electrons paired at a lower level. Figure 1 chose to keep a high-spin complex simply for spin retaining. Therefore, the again most simplistic approach to analyse different octahedral complexes by their actual energy differences (remember that $10~\mathrm{Dq}$ or $\Delta$ is only a variable) is to assume a certain contribution both from the metal’s side and from the ligands’ side:

$$10~\mathrm{Dq} = g_\mathrm{M} \times f_\mathrm{L}$$

The $g_\mathrm{M}$ contribution depends on the metal and needs to be experimentally measured, similarly for $f_\mathrm{L}$ on the ligands’ side. $f_\mathrm{L}$ is normalised so that $f_\mathrm{L}(\ce{H2O}) = 1$. $g_\mathrm{M}(\ce{Fe^{III}}) \approx 1.40 \times 10^4 ~\mathrm{cm^{-1}}$. By knowing that $f_\mathrm{L}(\ce{OH-}) \approx 0.9$ we could calculate the energy difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ to be $1.26 \times 10^4~\mathrm{cm^{-1}}$ and compare that to other iron complexes to find out that this would mean high-spin. (Using a value for $\ce{acac}$ should change the result slightly but not much.) Note that a typical low-spin iron(III) complex such as $\ce{[Fe(CN)6]^3-}$ has the following value: $10~\mathrm{Dq} = g_\mathrm{M}(\ce{Fe^{III}}) f_\mathrm{L}(\ce{CN-}) = 1.40 \times 10^4~\mathrm{cm^{-1}} \times 1.7 = 2.38 \times 10^4~\mathrm{cm^{-1}}$ — almost double the difference.

However, if you put more research into this phenomenon, it quickly becomes highly dissatisfying. The approximation for $10~\mathrm{Dq}$ up there is no real good, because the same ligand has different $f_\mathrm{L}$ values for different complexes — and even worse, the same metal-ligand combination can sometimes be found in both high- and low-spin states resulting in wildly different values for $10~\mathrm{Dq}$: For high-spin $\ce{[Fe(ptz)6]^2+}$, $10~\mathrm{Dq} = 1.23 \times 10^4~\mathrm{cm^{-1}}$ while for low-spin $\ce{[Fe(ptz)6]^2+}$, $10~\mathrm{Dq} = 2.00 \times 10^4~\mathrm{cm^{-1}}$. And whenever simple approximations don’t adequately explain the true picture, we need to resort to molecular orbital theory.

The simplest approximation in molecular orbital theory gives a similar picture as crystal field theory does: We have a set of six s-type orbitals approaching our metal ion, and these six orbitals will form σ-bonds with the metal. The main difference is that we can better explain why the metal orbitals will split: The ligand orbitals transform as $\mathrm{a_{1g} + t_{1u} + e_g}$ which means that they will interact with the metal $\mathrm{e_g}$ orbitals. Since the ligands are typically lower in energy, we realise that what we are in fact observing are antibonding $\mathrm{e_g^*}$ orbitals on the metal. Thus, electrons in these orbitals should weaken the metal-ligand bond (order); or complexes lacking electrons in $\mathrm{e_g^*}$ should have a shorter metal-ligand distance. However, this approach is also unsatisfactory: negatively charged ligands such as $\ce{OH-}$ or $\ce{acac-}$ should have donor orbitals of higher energy, meaning more similar to those of the metal, meaning a greater splitting – which we do not observe.

Therefore, in a second approximation we need to add the ligands’ p-orbitals to the picture. These transform as $\mathrm{t_{1u} + t_{2u} + t_{1g} + t_{2g}}$ — we have an additional set of orbitals that we can linear combine with the metal’s $\mathrm{t_{2g}}$ orbitals. We now have additional ligand to metal π-bonding and can assume the metal-centred $\mathrm{t_{2g}}$ orbitals again to be antibonding ($\mathrm{t_{2g}^*}$). This again raises their energy levels and thus lowers the difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g}$, therefore lowering $10~\mathrm{Dq}$. We finally have an adequate approximation of $10~\mathrm{Dq}$.

Now to get back to high-spin versus low-spin. Remember that it is not favourable per se to pair electrons — when applying the aufbau principle, you must never forget Hund’s rule which tells you to fill orbitals of the same energy with parallel spins first before adding the first antiparallel spin/pairing the first electron. In certain cases such as chromium, it is even favourable to have $\mathrm{3d^5 4s^1}$ — all valence spins unpaired. This is no different when going from elements to compounds: A certain energy gain is always required to make electron pairing favourable. Oftentimes, this energy gain is simply not supplied as is the case for most $\mathrm{3d}$ transition metal complexes that are not $\ce{d^8}$. The high-spin iron(III) complexes are the rule, not the exception, for the $\mathrm{3d}$ metals.

To raise the energy difference, we need strong-field ligands. The above paragraphs tell us what they need to: σ-bonding must be strong to raise the energy of $\mathrm{e_g^*}$ while π-bonding (if present at all) must be weak to not raise the $\mathrm{t_{2g}}$ level. Cyanide is good at doing this because its σ-donor orbital has roughly the same energy as the metal, so the bonding will be largely covalent with a large difference.

If, by the way, we accidentally manage to turn a high-spin into a low-spin centre, as was highlighted above with the ptz-ligand, a number of follow-up changes will also occur: Since the $\mathrm{e_g^*}$ orbitals are no longer occupied, the ligand to metal bond order increases and the ligands come closer to the metal. Therefore again the energy difference will rise because the covalent bond will be strengthend. This is the reason why the energy difference suddenly jumps from $\approx 1.3 \times 10^4~\mathrm{cm^{-1}}$ to $\approx 2.0 \times 10^4~\mathrm{cm^{-1}}$ when transitioning between high- and low-spin states.

In the case of the $\ce{acac-}$ ligand, for starters we need to realise that oxygen (the donating atom) is rather electronegative and thus we can assume its orbitals to be rather low-lying energetically. Low σ-donor orbitals means less destabilising of $\mathrm{e_g^*}$ and therefore a lower field split.

Secondly, we always have p-type orbitals on oxygen that do not participate in the π-system so that they can π-bond to the metal, too — or it could even be the π-system itself bonding to the metal. In any case, $\mathrm{t_{2g}}$ will be split and the metal component, $\mathrm{t_{2g}^*}$ will be raised in energy. This further reduces the field splitting, altogether putting $\ce{acac-}$ into the weak field department.

Finally, even though we have iron(III) and even though higher oxidation states are generally more prone to give low-spin complexes, iron(III) is not one of them. It takes significantly stronger ligands than water or even ammonia to generate low-spin iron(III) complexes, so generally any anionic oxygen-donor will not be enough to overcome the pairing energy.

Abbreviations:
acac: acetoacetone (or -ate if charged)
L: ligand
M: metal
ptz: 1-propyl-1H-tetrazol-κN4.

Acknowledgement:
Most of the knowledge is taken from Prof. Klüfers’ lectures on coordination chemistry and refreshed by checking his course’s internet scriptum.


Ketone/aldehyde synthesis from N-acylazetidines or aziridines A common principle in organic chemistry is to make sure the product is less reactive than the starting material, to prevent overreaction. So, for example, the treatment of an ester with a Grignard reagent will afford not the corresponding ketone, but the alcohol.

However there are some cases where the contrary is true. In particular, when the tetrahedral intermediate is stable, then the reaction can be worked up to 1) produce the desired ketone and 2) destroy any unreacted organometallic reagent, thus preventing overreaction. The classic example is a Weinreb amide, where the tetrahedral intermediate is stabilised by chelation, AFAIK.

Somewhat recently the synthesis of ketones from N-acylazetidines was described:1

and also the reduction of N-acylaziridines is known to produce aldehydes:2

Presumably the idea is the same: the tetrahedral intermediate is stable under the conditions used. Why is this so? Is it connected to the small size of the ring?

References

Liu, C.; Achtenhagen, M.; Szostak, M. Chemoselective Ketone Synthesis by the Addition of Organometallics to N-Acylazetidines. Org. Lett. 2016, 18 (10), 2375–2378. DOI: 10.1021/acs.orglett.6b00842.
Brown, H. C.; Tsukamoto, A. Selective Reductions. I. The Partial Reduction of Tertiary Amides with Lithium Aluminum Hydride. A New Aldehyde Synthesis via the 1-Acylaziridines. J. Am. Chem. Soc. 1961, 83 (22), 4549–4552. DOI: 10.1021/ja01483a013.

I think your suggestion about the stability of the tetrahedral intermediate is correct, but it's not that the tetrahedral intermediate itself has stability. I think it's because the collapse of the tetrahedral intermediate leads to a less stable intermediate.

My key insight for rationalizing this is that lithium hydride reduction of amides generally affords amines. We generally prefer to eliminate the lithium-oxygen adduct from the tetrahedral intermediate, forming an iminium ion.

With the two substrates you've provided, the iminium ion would be very highly strained because of the geometry of the small ring. I think collapse of the intermediate is thus disfavored. On work up, we actually kick out the amine to collapse the intermediate, providing an aldehyde or ketone.


Electron configuration of indium cation Just had a lecture there where the lecturer was doing revision on electronic configuration.

He stated that the electronic configuration of $$\ce{In^3+=[Kr] 5s^2 3p^1}$$

He didn't include the $\ce{4d^10}$ orbital. He said it doesn't matter and if we write the full E.C out it's incorrect. It basically goes against everything we studied in 1st year general chemistry. Also I believe he didn't take into account the fact that $\ce{In^3+}$ has $46 \, e^-$ not $49 \, e^-$.

Should the electronic configuration not be $$\ce{[Kr] 5s^0 4d^10 3p^0}= \ce{[Kr] 4d^10}$$

Can someone shed some light on whether he is right.

Yup, it's $[\ce{Kr}]4d^{10}$.  Post-transition ($p$-block) elements rarely if ever ionize or even covalently bond their $d$ subshell electrons.  Although we use the bracketed core designation only for noble gas cores, by the time we get to Group 12 or 13 the $d$ electrons are also effectively part of the core.


Turn lead into gold via radioactive decay I was looking at the isotope's table and noticed that one of lead's isotopes can actually turn into stable gold through the following mechanism:

$$\ce{_{82}^{197}Pb -> _{81}^{197}Tl -> _{80}^{197}Hg -> _{79}^{197}Au}$$

I know (or at least guess) that such a process must be awfully ineffective. Still, I was wondering:

How could we get a great quantity of $\ce{_{82}^{197}Pb}$?
How long would it take for it to turn into gold? Can that be accelerated?
Once most of it has been turned into gold, how can we extract the gold (there are still remaining $\ce{Pb,Tl,Hg}$ molecules)?
What would be the yield of the process? How much money would it approximately cost compared to how much we win in gold?

Interesting idea, but it has already been done, and not cheaply - read on.

  How could we get a great quantity of $\ce{_{82}^{197}Pb}$ ?

There would be two problems with getting a large amount of $\ce{_{82}^{197}Pb}$.  First, the parent nuclide of $\ce{_{82}^{197}Pb}$ is $\ce{_{83}^{197}Bi}$ which is unstable and has a half-life of only 9.33 minutes - so you can't get a large quantity of $\ce{_{82}^{197}Pb}$'s precursor to begin with.  Second, once $\ce{_{82}^{197}Pb}$ is formed, it has a half-life of 8.1 minutes, so it transmutes quickly to $\ce{_{81}^{197}Tl}$.

  How long would it take for it to turn into gold?

$\ce{_{82}^{197}Pb}$ half-life = 8.1 minutes

$\ce{_{81}^{197}Tl}$ half-life = 2.84 hours

$\ce{_{80}^{197}Hg}$ half-life = 64.14 hours

After 10 half-lives, ca. 0.1% of the starting material will be left
$$\ce{(1/2)^10 = $0.0009766$}$$
The last step is the slowest by far, so after about 641.4 hours (26.73 days), you should have something around 99.9% pure gold.

  Can that be accelerated?

Unlike chemical reactions that can heated, catalyzed, etc., this type of nuclear transformation keeps a set schedule.

  Once most of it has been turned into gold, how can we extract the gold
  ? (there are still remaining Pb,Tl,Hg  molecules)

As noted above, you can get whatever purity you desire, just wait.

  What would be the yield of the process?

It would be high for the 3 nuclear transformations you listed.  Each of the elements you listed decays directly and only to the daughter isotope you've shown.  However, as noted above, you can't start with $\ce{_{82}^{197}Pb}$, you generate it from $\ce{_{83}^{197}Bi}$, the decay of which adds some impurity along with lead. And then since the bismuth isotope is not long-lived you'd probably start with its precursor, and so on until you find something that has a long enough life that you could assemble a reasonable quantity.

Back around 1980 Glenn Seaborg actually transmuted bismuth to gold, but only a few thousand atoms (see this reference also).

  How much money would it approximately cost compared to how much we win
  in gold?

The Wikipedia article I referenced directly above notes, "the expense far exceeds any gain."  There are other ways (fission and fusion) to produce gold, but at least with the methods available today, the cost would be astronomical.


What happens at the inert anode in an electrolytic cell? Recently I did an experiment using graphite as the anode and iron as the cathode for an electrolytic cell. I tried plating different metals such as copper, nickel, and zinc with CuSO4, NiSO4, and ZnSO4 respectively. I know that normally when the anode is made from the same material as the metal being used for plating, the anode is decomposed. What I am wondering is what happens at the anode when it is made of an inert material like graphite? Is the SO4 ion oxidized there, what does it become?
During the experiment, I have also observed bubbles forming at the anode, and I hypothesized that it may be oxygen from the electrolysis of water, but I had no way to be sure.

There is oxidized oxygen in water
$$\ce{2 H2O -> 4 H+(aq)  + O2(g) + 4 e-}.$$
$\ce{HSO4-}$ in sulphuric acid solution can get oxidized as well at high current density and therefore high enough anodic potential:
$$\ce{HSO4-(aq) -> S2O8^2-(aq) + 2 H+(aq) + 2 e- }$$
This is the industrial way to produce potassium peroxodisulphate.


How do I find out the number of bonds an atom must link to? I've always been told that carbon bonds with four hydrogens and that nitrogen bonds with three hydrogens, like in benzene and hexazine, respectively.

What's the name of this type of bonding and how does one come up with the number?

It's named a covalent bond. The amount of hydrogen atoms that can be bond (or any other atom) can be  calculated most of the time using the octet rule, that states that any atom will try to get the electron configuration of a noble gas (i.e. 8 electrons in the outer shell). The electrons from the covalent bond count for both atoms that take part in the bonding.

As an example consider $\ce{N}$ that has five electrons in its outer shell, and $\ce{H}$ that has just one. $\ce{N}$ wants three more electrons and can get them from covalent bond with hydrogen. $\ce{H}$ want one more electron (this is an exception) in its outer shell to fill it up, so will kindly receive them from nitrogen.


What makes dimethyl compounds so extreme? I've noticed that dimethyl compounds tend to have extreme reactive, including bioreactive, characteristics.

Am I being misled by anecdotal selection bias, or is this generally true.  And if generally true, are there clear theoretical explanations as to why?

Examples that come to mind:

Dimethyl sulfoxide (DMSO) readily penetrates skin, and can carry lots of things across the skin barrier along with itself.
Miniscule amounts of dimethyl mercury will not only penetrate skin and many common PPE but are also irreversibly lethal.
Many dimethyl-metals are pyrophoric (i.e., they auto-ignite in a standard atmosphere).

This is because of two trivial facts, one from chemistry and another from math: (1) methyl is the smallest organic radical, and (2) it is not uncommon for a function to reach an extremum at the border of its domain.

Arguably, this is not even true in any significant way. Not sure about DMSO, but I firmly believe that organic compounds of mercury are all about as deadly as dimethylmercury. Well, those with bigger organic radicals might be slightly less dangerous because they are not quite as good at penetrating the cell barriers, or lower solubility, or because they contain less mercury per gram. The maximum is probably reached at methyl; but the maximum is not sharp.

All that applies to pyrophoric metalorganics as well.

So it goes.


What kind of metal alloys can resist corrosion from molten salt? Molten salt is pretty hot, 801 C. With such heat and the corrosive nature of salt, I am wondering what type of metal alloy or other material hybrid would be appropriate to handle and work with molten salt?

George Lai's High-Temperature Corrosion and Materials Applications (excerpts available on Google Books) is a good source. Very often for molten salts, Hastelloys are used; they're special-purpose alloys developed by Haynes International. I can't find a reference, but I believe Oak Ridge National Laboratories used Hastelloys for their molten salt reactors (which had fluoride salts, very reactive).

If you're working with sodium chloride (little potassium), according to Lai, Hastelloy-188 seems to be the best bet, although many others would work as well. One reality of chemical engineering is that there will always be corrosion, it's just a matter of keeping it at an acceptable rate for the application.


How does polydimethylsiloxane stick to glass? Polydimethylsiloxane (PDMS) is the main ingredient in Rain-X, a treatment that adheres to windshields.

How does PDMS adhere to class? It can't form covalent Si-O bonds with glass, according to one source. Makes sense, as the Si is already coordinated with four groups in PDMS and none of them seem to be easily removable.

A similar windshield treatment chemical, FAS17, was noted by the same source as being able to form covalent bonds with glass in the presence of an acid to cleave the ether linkages.

The source, however, didn't detail how exactly PDMS would adhere to glass. So what major forces would be involved in the adhesion of PDMS to glass?

Source:

http://hydrophobicsam.blogspot.com/2012/04/science-of-self-assembling-monolayers.html

If you ask a chemist about PDMS and its adhesion to glass, she will immediately think of the work done in constructing micro-reactors.  This is because one method of manufacturing micro-reactors is to stick a patterned piece of PDMS to a piece of glass; the PDMS is patterned such that it has reactor channels that are sealed when it is stuck to the glass.  The glass is usually a micro-scope slide or something similar.

With the above in mind, in order to answer your question I began to search through the literature which describes the preparation of micro-reactors from glass and PDMS.  What they were saying was that in order to form a bond between the molecules within the glass and the molecules within the polymer (PDMS) the glass had to be pre-treated,

  "most of the research papers mention the
  use of oxygen plasma for developing chemical (siloxane) bonds be-
  tween the participating surfaces"

from "Studies on Surface Wettability of Poly(Dimethyl)Siloxane (PDMS) and Glass Under Oxygen-Plasma Treatment and Correlation With Bond Strength", DOI: 10.1109/JMEMS.2005.844746.

Since the manner in which your material is applied does not have any sophisticated pre-treatment then the only alternative method of adhesion (as pointed out earlier in this thread) is Van der Waals forces.

If this isn't quite the answer you expected, at least it is a good start.


Why is one lobe of an sp3 hybridized orbital smaller than its other half? A hybrid sp3 orbital is drawn with one lobe smaller than its other half, the latter which is of equal size when drawing the p orbital. Why is it so?

When combined at a given atomic center, any two atomic orbitals add in a vectorial
manner. For example, consider the orbital $\phi$ defined by $\ce{p_{x}}$ and $\ce{p_{y}}$ atomic orbitals as

\begin{align}
\phi = c_1 \ce{p_{x}} + c_2 \ce{p_{y}}
\end{align}

The orbital addition can be pictured like this

for the two cases $c_1 = c_2 > 0$ and $c_l = -c_2 > 0$, respectively. The ratio $c_1 / c_2$ controls how much the orbital $\phi$ is tilted away from the $x$ (or $y$) axis.

The mixing of atomic orbitals with different angular momentum quantum number is
also controlled by a vectorial addition and leads to various types of hybrid orbitals
some examples of which are shown here:

The example a) would be an $\ce{sp}$ hybrid orbital. It gets its sense of direction from the $\ce{p}$ orbital used to construct it because the $\ce{s}$ orbital is isotropic. Its shape comes about because the parts of the orbitals that have an opposite sign (this is hinted at by the shading in the pictures) cancel each other out to some extent. So, if the left lobe of the $\ce{p}$ orbital is negativ and the right one is positive and the $\ce{s}$ orbital is also positive then the left half of the $\ce{s}$ orbital overlaps with the left lobe of the $\ce{p}$ orbital in a destructive manner and they cancel each other out, such that only a small portion of the initial $\ce{p}$ orbital lobe remains. But the right half of the  $\ce{s}$ orbital overlaps with the right lobe of the $\ce{p}$ orbital in a constructive manner and so the right lobe of the initial $\ce{p}$ orbital becomes larger.


Rate law equation - orders? Consider the following reaction:

  $$\ce{2 H2 (g) + 2 NO (g) -> N2 (g) + 2 H2O (g)}$$

  The elementary steps of this reaction are:

  $$
\begin{align}
&1^\text{st}~\text{step:} &\ce{H2 (g) + 2 NO (g) &-> N2O (g) + H2O (g)} &&\text{(slow)} \\
&2^\text{nd}~\text{step:} &\ce{N2O (g) + H2 (g) &-> N2 (g) + H2O (g)} &&\text{(fast)} \\
\end{align}
$$

  The reaction is exothermic overall.

  (a) What is the rate law for this reaction? Justify your answer.

  $$\text{rate} = k[\ce{NO}]^2[\ce{H2}]$$

  The 1st step of the reaction is the rate limiting step.

How did they know the order of the reaction with respect to $\ce{NO}$ was $2$, and $1$ for $\ce{H2}$? With other questions the orders are something they state in the question and are unrelated to the coefficients.

You are correct, the reaction order cannot be deduced from the brutto-equation and is usually determined experimentally.

However, here you are presented with the elementary reactions and their rates. Obviously, the slowest step is the bottleneck and controls the overall kinetics, hence the total rate matches the rate of the slowest step. And as this is an elementary reaction, you just apply the law of mass action and use the stoichiometric coefficients for the rate equation.


What is the product of the titration of potassium permanganate and iron(II) sulfate? I recently did an experiment at school, where I had to titrate KMnO4 with FeSO4. The solution was colourless, then turned yellowish until the end point was reached and it turned pink in one drop. After letting it sit for a while, the pink tinge disappeared. Can someone please explain what happened and show the products and this mysterious observation? Thanks

From your description I'd say you were titrating ferrous sulphate, $\ce{FeSO4}$ solution (the analyte), with potassium permanganate, $\ce{KMnO4}$ solution (the titrant), in acid conditions (dilute $\ce{H2SO4}$ present).

The permanganate ion $\ce{MnO4^{-}}$ is a strong oxidiser and oxidises the ferrous ion $\ce{Fe^{2+}}$ to the ferric ion $\ce{Fe^{3+}}$ very easily. In dilute solution both ferrous and ferric ions are more or less colourless.

In this process the very strongly coloured permanganate is reduced to the manganous $\ce{Mn^{2+}}$ ion, which is also colourless in dilute solution.

When all the ferrous ions have been oxidised, one more drop of the strongly coloured permanganate solution then turns the analyte solution pink and this is your end point.

  I recently did an experiment at school, where I had to titrate $\ce{KMnO4}$
  with $\ce{FeSO4}$.

In that case you'd have started from a purple solution (analyte) which would gradually, during the titration, have turned colourless. This method is also used.

As this is a redox reaction, I suggest you write out (or look up) the reduction (permanganate to manganous) half reaction and the oxidation (ferrous to ferric) half reaction, to complete your knowledge of this titration.


Does deuterium exhibit hydrogen bonding? I searched for this on Google but I only got research articles in the SERP. I know of the concept of hydrogen bonds, that they are a special class of van der Waals forces between highly electronegative atoms ($\ce{F/O/N/Cl})$ and the hydrogen atom. I am asking if deuterium also exhibits the same hydrogen bonding, and how does its "$\ce{D}$ bond" compare to the hydrogen's original "$\ce{H}$ bond"?

(Based on ron's answer here on the inductive effect accorded by deuterium - do give that a read, and consider upvoting it)

Since deuterium has a higher mass than protium, simple Bohr theory tells us that the deuterium 1s electron will have a smaller orbital radius than the 1s electron orbiting the protium nucleus.
The smaller orbital radius for the deuterium electron translates into a shorter (and stronger) $\ce{C-D}$ bond length.
 See the bottom half of ron's answer for the proof

In your case, that would mean a shorter $\ce{N/O/F-D}$ bond.

A shorter bond has less volume over which to spread the electron density (of the 1 electron contributed by $\ce{H}$ or $\ce{D}$) resulting in a higher electron density throughout the bond, and, consequently, more electron density at the carbon end of the bond. Therefore, the shorter $\ce{C-D}$ bond will have more electron density around the carbon end of the bond, than the longer $\ce{C-H}$ bond.
The net effect is that the shorter bond with deuterium increases the electron density at carbon, e.g. deuterium is inductively more electron donating than protium towards carbon.

So we can expect the $\ce{N/O/F-D}$ bond to be smaller and more polar than the corresponding bond with protium.
Karl very astutely pointed out that the boiling point of heavy water is higher than that of normal water. In light of the inferences drawn earlier, I took the liberty of interpreting this fact as being (at least somewhat) indicative of deuterium permitting a stronger hydrogen bond than protium.
As Linear Christmas pointed out, the variations in mass across different isotopes is known to affect boiling point (as in the case of helium). This is likely to contribute to some degree, to the increase in boiling point in heavy water. However, I'm still of the opinion that the higher dipole moment of $\ce{D2O}$ over $\ce{H2O}$ is the main cause for the higher boiling point.

Finis
Hydrogen bonding with deutrium does occur and should be, in theory (drawn from ron's answer), stronger than that with protium. The fact that heavy water $\ce{D-O-D}$ is more polar (and hence has a higher boiling point) than regular water $\ce{H-O-H}$, appears to be a consequence of this.


How are neutral amines effective bases in organic chemistry? I have found it to be rather strange that neutral amines, such as triethylamine  or DIPEA, find applications in organic chemistry in deprotonating alcohols or even α-carbonyl compounds. Amines are basic but they aren't really that basic, compared to negatively-charged hydroxides or alkoxides. According to Wikipedia, the $\mathrm{p}K_\mathrm{a}$ (in water) of the conjugate acid of triethylamine is $10.75.$ The same values for alcohols are in the range of $16 - 18.$ For water, it is $14.$ Thus, hydroxides and alkoxides are superior alternatives to these amines.

My motivation for asking about this is that I recently encountered a question in my school chemistry course on why the protection of alcohols with $\ce{MOMCl}$ uses large bulky amines to deprotonate the alcohols, rather than using hydroxides. The answer given was that hydroxides would not be able to deprotonate the alcohols to a significant extent as alcohols are not acidic enough. I find this answer very strange and inconsistent with what I know since as shown by the data above, amines are less basic than alkoxides and hydroxides.

There are several interesting aspects to your question.

  How are neutral amines effective bases in organic chemistry?

First of all, recall the definition of a base. Anything which accepts a proton (or donates a pair of electrons) is a base. Organic bases are very strong and corrosive! These bases should not be undermined, even though they may sound weak (and their stench is worse than inorganic bases). Your lye-free oven cleaner has an organic base, an neutral organic amine such as ethanolamine, which dissolves dirty grease stuck in the oven. This molecule has no charge yet it is a strong base, acting just like sodium hydroxide based oven cleaner in terms of performance.

  Thus, hydroxides and alkoxides are superior alternatives to these
  amines.

The second reason to avoid hydroxides and alkoxides is their solubility in organic solvents. Neutal organic amines easily dissolve in toluene and a variety of organic solvents. NaOH or NaH remain as a suspension.

  MOMCl  uses large bulky amines to deprotonate the alcohols, rather
  than using hydroxides

Again you have to think about the reaction solvent. Is an inorganic hydroxide soluble in your solvent? Then comes an important point. Why use bulky amines? Their sole purpose is to act as a proton sponge in the reaction mixture and prevent the formation of free acid. Bulky amines are preferred so that they don't react easily with the alkyl chlorides or acyl chlorides, thus prevent the formation of their own undesired side products.


What is a electronic configuration of scandium ion? The electronic configuration of Scandium is: $\ce{[Ar] 4s^2 3d^1}$

What is the electronic configuration of Scandium ion?

Also about the electronic configuration of Cobalt, which is $\ce{[Ar] 4s^2 3d^7}$. So What is the electronic configuration if Cobalt ion?

Is it $\ce{[Ar] 4s^2 3d^6}$ or $\ce{[Ar] 4s^1 3d^7}$ ? And why?

Thanks for your help.

The electronic configuration of the cobalt is $\ce{[Ar] 3d^1 4s^2}$ the electrons with the higher energy will be snatched and the electronic configuration of Argon is very stable then with scandium you will have easely $\ce{Sc^{3+}}$.

For the Cobalt it is a bit more difficult, like you write the configuration you cannot find a correct answer. For all element first you do first the configuration using the Klechkovsky rule and after you put all orbitals by principal quantum number growing.

So the cobalt configuration is $\ce{[Ar] 3d^7 4s^2}$ then if you snatch two electrons from the $\ce{4s}$ orbital you have a stable configuration for the $\ce{Co(II)}$ ion. You cannot snatch them from the $\ce{3d}$ orbital (even if $\ce{[Ar] 3d^5 4s^2}$ looks stable because the $\ce{3d}$ orbital is half full then the spin is maximal) because its energy is lesser than the energy of the $\ce{4s}$ orbital.

You can find the Cobalt at different oxydation states from $\ce{+I}$ to $\ce{+IV}$ but it depends on what you have in your solution or in your gas if you have a gas.

NB :
Remember that the configuration of the elements are given if gas phase, then for example the more stable configuration of the copper iron is for $\ce{Cu^+}$ and not for $\ce{Cu^2+}$, $\ce{Cu^2+}$ is stable in water, so the answer may depends on the problem you have.

Explaination for the copper :

Stability in aqueous conditions depends on the hydration energy of the ions when they bond to the water molecules (an exothermic process). The $\ce{Cu^{2+}}$ ion has a greater charge density than the $\ce{Cu^+}$ ion and so forms much stronger bonds releasing more energy.

The extra energy needed for the second ionisation of the copper is more than compensated for by the hydration, so much so that the $\ce{Cu^+}$ ion loses an electron to become $\ce{Cu^{2+}}$ which can then release this hydration energy.

I hope it can help you !


How does the crystal know where to stop stacking the unit cells? I tried to make different crystals in the cubic system by stacking cubes in a 3D software. However, I need to tell the software where to skip rows or stop stacking the unit cells in a specific direction else it will only make a cube.

How does this work in nature? How does the fluorite "know" where to skip rows in order to make the octahedron crystal? Does it have something to do with Bravais lattices and coordination numbers?

First I thought that the unit cell needs to look like the final shape, but I think I'm wrong there.

The ideal crystal is terminated by planes, which are described by
Miller Index. Some of these planes are more favorable than others, and as a result, the crystals do get macroscopic shape which is different from the shape of unit cell. For some discussion on reasons for different stability, I have found this article: iopscience.iop.org/0022-3719/12/22/036/pdf/jcv12i22p4977.pdf


How to determine which salt will precipitate from a solution containing multiple ions? I have a solution containing $0.2\:\mathrm{mol}$ of $\ce{Br^-}$ ions and $\ce{SO_4^2-}$ ions each.
Now if I add excess  $\ce{AgNO3}$ in this solution,
what will precipitate?
 $\ce{Ag2SO4}$ or $\ce{AgBr}$ or both?

And if I add just $0.2\:\mathrm{mol}$ of $\ce{AgNO3}$, what will precipitate?

(I don't want to calculate anything. Just wanted to know qualitatively that which compound precipitates.)

You need to know that $K_\mathrm{sp}$ is used to determine the solubilities of sparingly soluble salts usually. You can get the values of $\ce{AgSO4,AgBr}$ and as $K_\mathrm{sp}=[\ce{Ag+}]^2[\ce{SO4^2-}]$ or $[\ce{Ag+}][\ce{Br-}]$. Now concentration of silver ion is constant so try calculating the other two anions and try comparing them.


Basic strength of compunds Why is Ammonia stronger than Phosphine in terms of basic strength?
I am confused as Nitrogen is more electronegative than Phosphorous.

The strength of a Lewis base corresponds to the strength of its conjugate acid. As usual, a weak acid will give a stronger base, therefore parent acids with higher pKa's deprotonate to stronger Lewis bases. ammonium, $\ce{NH4+}$, has pKa=9.25, while phosphonium, $\ce{PH4+}$, is a much stronger acid with pKa=~-12. Basicity decreases down the group because proton affinity decreases as electronegativity decreases. The N-H bond is stronger than the P-H bond (hence $\ce{NH4+}$ is a weaker acid than $\ce{PH4+}$).

Lewis bases donate electrons from their HOMO, and for ammonia's MO, the HOMO, which corresponds to the lone pair (LP), is very similar in energy to the nitrogen AO, therefore the LP electron density can be considered to be more localized to the nitrogen. Presumably in a MO for an analogous species with a group 15 central atom lower on the periodic table (with lower electronegativity), the AO of the central atom would be higher in energy, so the HOMO electron density would be less concentrated on that atom.

Also, ammonia is more hybridized than phosphine. The s-p energy gap is greater for phosphorous, so the energy penalty of hybridization is higher, therefore the LP will have more pure s character than sp3 character, making it less directed.

Since the ammonia LP electron density is more concentrated on the central atom, and more directed, it makes it more nucleophilic.


What is the chemical composition of a Rubbermaid Roughneck container and is it feasible for me to make my own molds? I'm looking for more information on the composition of a Rubbermaid Roughneck container.  The description says "Roughneck containers are made with a durable polyethylene material for a lifetime of durability and strength." The Wikipedia entry indicates that polyethylene is the most common type of plastic with chemical composition $\ce{(C2H4)_{n}H2}$. But if this is so how are these containers able to flex instead of break like other plastic containers as well as tolerate greater temperature changes? I have a project that requires a custom shaped material of similar attributes. How feasible would it be to make my own mold at home? The shape I want to mold is very simple: a very thin rectangle about 35 width and 23 length. Are the chemicals involved dangerous to work with?

This is a pretty broad question and might be better split into several, however a general comments on polymers, such as poly(ethylene), is that the properties of polymers can be tweaked quite a bit by control over the mean molecular mass and dispersity (probability distribution of molecular mass) of the polymer chains, as well as the addition of all kinds of stabilisers and plasticisers. Plasticisers especially may dramatically alter the flexibility of plastics. The exact composition of the Rubbermaid Roughneck is unlikely to be forthcoming.

Molding of thermoplastics by injection molding or casting is carried out by melting down plastic pellets and squirting the melt into or onto a mold. A mold release agent is often used to help get the cooled plastic out of the mold. The safety of the plastic depends on its specific composition (read the material safety data sheet for the pellets) and how hot it is (don't get molten plastic on yourself, wear protective clothing etc.). I don't know anything about the safety of mold release agents except for the fact that miniature model kits typically still have the agent on the plastic when purchased. This suggests they are fairly safe, though their presence may interfere with paint adhesion which is why pros scrub kits down before painting.

The geometric simplicity of your custom part, however, suggests that you could cut it from a plastic sheet, obviating molding and the potentially expensive equipment that goes with it.


Why does concentrated HCl fume? It is a common observation that a concentrated solution of hydrochloric acid kept in the lab fumes. What is the chemistry behind this phenomenon? Does $\ce{HCl}$ somehow react with the moisture in the air due to its hygroscopic nature?

In the case of $\ce{HNO3}$, according the abstract of this paper$^\dagger$, nitric acid vapors act as nucleation sites for ambient water, causing the mists/fumes that are observed above concentrated nitric acid solutions:

  The gas phase reaction [forming nitric acid from other reagents] supplies $\ce{HNO3}$ nuclei on which the water vapor readily condenses to form strong nitric acid mist droplets.

According to this book$^\ddagger$:

  Hydrochloric acid has thermodynamic properties rather similar to those of nitric acid.

So, I would hypothesize that the physical chemistry phenomena of evaporation and hygroscopic nucleation are responsible for the fuming of $\ce{HCl}$.

$^\dagger$ Goyer, J Colloid Sci 18(7), 616 (1963)

$^\ddagger$ Kulmala and Wagner, "Nucleation and Atmospheric Aerosols," 1996, p591.


'Emulating' calcium hydroxide I posed this question at the Seasoned Advice SE and thought I might get some more information here.

Traditionally prepared corn is cooked in a calcium hydroxide solution to make it more digestible, but when my search for $\ce{Ca(OH)2}$ hit a dead end, I began thinking: could I make a substitute with stuff I already have (sodium hydroxide, which I use for pretzel/soap making; and calcium chloride, for brewing)?

My thought process was that, upon dissolving, sodium hydroxide plus calcium chloride would provide the same ions (though not necessarily in the same concentration) as calcium hydroxide and sodium chloride. Since I would be adding salt to this recipe at some point anyway, I didn't see any harm to this.

Is this feasible, or am I off-base in my chemical understanding about this? More importantly, is there some chemical consideration I've failed to take into account?

Your idea seems mostly sound, but one thing you should consider is the solubility of calcium hydroxide (a.k.a. slaked lime).  It is fairly low.

That low solubility provides you with an opportunity.  If you have calcium chloride and sodium hydroxide, you can mix saturated solutions of both and a white precipitate should form.  The precipitate is calcium hydroxide.  You can then remove the precipitated calcium hydroxide, wash it with ice cold water, and then you'll have $\ce{Ca(OH)2}$!  A YouTube demo of this idea provides some more details.

$$\ce{CaCl2(aq) +  2NaOH(aq) -> 2NaCl(aq) + Ca(OH)2(s)}$$

I think this method is preferred because the low solubility of calcium hydroxide compared to sodium hydroxide naturally limits its alkalinity and corrosiveness.  It would be too easy to use undesirable (and possibly even dangerous) levels of NaOH if you made a mistake in how much you used for nixtamalification.  Additionally, simply mixing NaOH and calcium chloride and using the resulting milky liquid as a substitute for calcium hydroxide will have far too much salt (NaCl) present to be a good substitute.  Or conversely, if you lower the amount of NaOH and calcium chloride to where the saltiness is tolerable, you will have too little alkalinity.


Can I deduct heteronuclear van der waals parameters from homonuclear parameters in a forcefield? I got a forcefield (OPLS-AA in this case) that lists lennard-jones parameters for all homonuclear van der waals interactions. Is there a way to deduct the heteronuclear vdw-interaction parameters from those?

Found the answer myself. In case someone else stumbles over this, the combining rules for Lennard-Jones parameters of van der Waals interactions are:

Distance parameter:
$r_{XY}=\frac{1}{2}(r_{XX}+r_{YY})$ or $\sigma_{XY}=\frac{1}{2}(\sigma_{XX}+\sigma_{YY})$

Depth parameter:
$\varepsilon_{XY}=\sqrt{\varepsilon_{XX}\varepsilon_{YY}}$


What's the mechanism for this isotopic substitution reaction? I'm curious to know what's the mechanism of:
$$\ce{HF + D^+ \rightarrow DF + H^+}$$

To answer this question you should understand the kinetic isotope effect.

Since you deliver no real data about reaction rate constants I assume that the dissociation of $\ce{HF}$ is faster (has a larger $k_\text{dis}$) than the dissociation of $\ce{DF}$. I postulate the following mechanism:

$\ce{HF + D+ <=>[k_\text{dis}^\ce{HF}] H+ + F- + D+ <=>[][k_{\text{dis}}^\ce{DF}] DF + H+}$

Essentially it is a pretty simple reaction mechanism, but it tends to forming $\ce{DF}$ over $\ce{HF}$ because the deuterium dissociates more slowly.


Why do electrolytic concentrations affect voltage in electrochemical cells? I know that the voltage of a cell is calculated via the Nernst Equation as $$E_\text{cell}=E^\circ-\frac{RT}{nF}\ln Q_\mathrm r$$ where $Q_\mathrm r$ represents $\frac{\text{concentration at the anode}}{\text{concentration at the cathode}}$.

However I don't understand how these concentrations actually affect the potential difference. Could anybody explain the physical processes that mean changing the concentrations affect the voltage of the cell?

Potential is work to move charge and this work is done by the chemical energy stored in the reactants, so if the concentration of reactants increases then there are more reactants which will do more work so this implies that potential increases with comcentration.


Calculate thr unit cell length of perovskite In a ideal unit cell of perovskite $\ce{KMnF3}$ (closed packed with purely ionic bondings),  calculate the unit cell length:

radius:  $\ce{K}=1.7~$  $\ce{F}=1.2~$   $\ce{Mn}=0.8$

The structure

Green $\ce{K}$, red $\ce{F}$, blue $\ce{Mn}$. (ignore the sizes in the picture)

Am I going right by using the $\ce{F-K-F}$ diagonal length to calculate the cell length  using pytagroas, as the Mn diameter is to small and therefore the $\ce{Mn-F-Mn}$ will have some spacing on it and are not fully adjacent.

I would get unit cell length =4.101.   can somebody confirm this?

Just a bit confused as I have the radius of $\ce{Mn}$ but don't use it for calculation

Many Thanks

Experimental X-ray crystallography data yields a value of 4.182 for the lattice constant for $\ce{KMnF3}$. See this reference for further details. Given that the paper was published in 1961, there are likely more recent refinements experimentally. That said, your number is within the error of observed values.

Reference: K. Knox. Perovskite-like Fluorides. I. Structures of $\ce{KMnF3}$, $\ce{KFeF3}$, $\ce{KCoF3}$, $\ce{KNiF3}$, and $\ce{KZnF3}$. Crystal Field Effects in the Series and in $\ce{KCrF3}$ and $\ce{KCuF3}$. $\it{Acta\,Cryst.}$ (1961) $\bf{14}$, 583.


Why did my iron nails not dissolve in hydrochloric acid? I was trying to dissolve iron metal in HCl to make iron chloride, but ran into some problems. I'll give a brief background of my experiment. I took iron nails and put them in a container filled with muriatic acid (chemicals for my pool). The acid contains an equivalent of concentrated HCl since it is composed of HCl by 37.5%. The other 62.5% is listed simply as "other ingredients".

The faint yellow acid began to turn a more solid green yellow. Gas immediately began to bubble on the nails' surface. I left it for two days to dissolve and when I checked back the iron nails were not dissolved. They look shiny like the rust was cleaned off. The solution is black/dark green and there is a white/green precipitate at the bottom. The nails have small green crystals growing on them

I did this experiment once previously (with the same type of nail and HCl) with just a single nail inside of a hollow plastic pen tube and it seemed to have worked very well; it formed a dark brown solution at the bottom and the rest of the nail looked very rusted to a point where I could not remove it because the rust had stuck it to the inside of the pen.

Is there any explanation for the two different results and the iron nails not dissolving in my other trial? Also is there any guess as to what my solution and precipitate contains? Let me know if any more information is needed.

It seems like there is not enough acid on the nails.

Perhaps a depleted layer is forming next to the nails. To avoid that issue, use a magnetic stirrer or provide a source of vibration.
There might not be enough $\ce{HCl}$ for the large amount of iron in the nails. Check the total amount of $\ce{HCl}$, given the volume and that ~2/3 is "other ingredients" (likely water). Weigh the nails. Calculate the number of moles of  $\ce{HCl}$ needed to react with the number of moles of  $\ce{Fe}$. Is the product  $\ce{FeCl2}$ or  $\ce{FeCl3}$?
Have you removed all the grease (from the manufacturing process) from the nails? A little soap or detergent and rubbing may be needed.
Could there be some difference in the nails you used? If some are galvanized (zinc coated, to prevent rust), then that would affect the reaction.

Let us know how you do: we learn more from our mistakes than from successes.


Peculiar behaviour of tea boiling as milk+Water While making tea last night I forgot all about it and left the kitchen. When I did remember about it, I ran back and I was surprised to see that it was still in the vessel (though it formed a peculiar foam-like mass on the top).

This pushed me to do the same with water and milk (as these were constituents this tea).
I was not so lucky in case of milk, When I boiled the milk it rose too much and overflowed.

Boiling water alone didn't help either. The level of water in the vessel rose a little but it was far from overflowing (I provided the highest temperature as I could provide to vessel).

So I observed three different phenomena for three different liquids:

Tea (having water and milk in proportion of 3 and 1) rose till the height of container.

Milk overflows.

Water doesn't even remotely show anything that would indicate it was responsible for this peculiar behaviour of tea.

I can't seem to figure out why this happens. Would anyone know why?Thanks for any help
EDIT
After listning to people saying that they need more experimental data I decided to go further.
I boiled the milk along with water in a proportion of 1 by 1 (without tea leaves and sugar to it) and did the same with different proportions of milk and water.
After doing so I got that As I increased the amount of water (in combo of milk and water) the level of solution started decreasing and ultimately at the ratio of 1 and 5 (for milk and water) the solution didn't overflow.
The results were as expected but I thought these can be helpful for you guys in answering the question.

Original Hypothesis

As mentioned in this post, milk will overflow thanks to the effect of surfactants, which allow for the formation of stable bubbles. Since tea contains both milk and water, it contains surfactants from the milk, but at a lower concentration thanks to the water. This would result in some bubble formation, but not as much as you would find in pure milk, explaining why the tea filled up the container, but did not overflow. Either it was a coincidence that the bubbling stopped right at the top of the container, or it relates to the fact that more surface tension is required to create bubbles without the support of the container walls. In addition, other components in the tea could be contributing to this effect, but I think it is a result of the fact that milk and water are mixed. If you want to experiment with this, I would try a different volume of your water/milk mixture and see if that also makes it to the top of the container without overflowing.

Results of Additional Experiments

The results of your additional experiments seem to indicate that something else is going on. It is interesting that you found a different ratio that filled the container when there was no tea involved. This seems to indicate that the tea is at least somewhat involved in this interaction. Also, the fact that past the ratio of 1:5 (milk:water) the container did not fill up with bubbles seems to indicate that my hypothesis about surface tension was wrong, and that there is nothing special about filling the container. In this case, you probably just got lucky that your original tea did not overflow.

These results are dependent on the fact that you used the same vessel for your new experiments and allowed adequate time for bubbles to form. More information on your experimental procedures will help me narrow down on a possible explanation.


What are the factors controlling the fate of post-adhesed volatile molecules of the olfactory epithelium? What happens to volatile molecules that reach the main olfactory epithelium in the nasal cavity after they bind and the neural stimulus fades?

To what extent do such factors as receptor kinetics and diffusion (or any others) direct where these molecules end up?

Simply, the odorant particles/molecules never actually enter the cell through endocytosis. Instead, these olfactory receptors, belonging to the G-protein-coupled receptor family, are transmembrane receptors whose structure passes through the cellular wall (several times actually) to functional complexes on the other side, and they use simple and reversible ligand binding as their primary activation mechanism.

Odorant particles are dissolved in the mucous lining the nasal cavity, and in this form come into contact with the receptors in the cell membrane of the epithelial cells. These bind as ligand complexes to receptor sites, triggering a change in the receptor's structure which in turn causes the coupled G-protein to release its bound guanosine diphosphate, and instead bind guanosine triphosphate from the interior cell environment. When that happens, the G-protein is "activated"; it decouples from the receptor, splits in half into Ga and Gbg subunits, and those two halves bind to effector enzymes which trigger the cell signalling. As a result of the protein decoupling, the charge balance of the receptor changes, releasing the complexed ligand from its binding to the receptor, back into the extracellular environment.

To reverse all this, an enzyme called RGS (Regulator of G-protein Signalling) binds to the Ga subunit, triggering hydrogenation of the GTP molecule it still has bound into GDP, causing the protein's subunits to recombine and recouple to the receptor.

The released ligand, meanwhile, is washed away again by the mucous flow from the Bowman's glands. It may complex with other receptors along the way, but eventually it makes its way past the nasal membrane to the pharynx at the back of the throat, where the mucous is swallowed and the odorant broken down in the stomach. This is the final fate of virtually all odorant agonists.

A select few, however, bind more strongly, and are not released by the structure changes of the receptor. When this happens, the receptor is stuck in the "on" position. One of two things then happens; either the receptor continues to activate, causing neural impulses which your brain eventually ignores (odor fatigue), or mechanisms inside the cell notice the faulty receptor, bring it into the cell in a process similar to endocytosis, and attempt to break down its components and fix the damage to reuse it. More often than not, the bound chemical ends up poisoning the cell by binding to something more vital than an ordinary G-protein receptor, eventually triggering programmed cell death. That's fine; the basal cells behind the surface of the olfactory epithelium divide fast enough to replenish every cell in the nasal membrane about once every two days.

Sources: Wikipedia - Olfactory epithelium, Olfactory receptor and G protein-coupled receptor.


Smoldering Combustion Does smoldering combustion only occur in an environment where oxygen levels are low? Will it always transition into flaming combustion (whenever there is ample supply of oxygen?

Or can it keep smoldering even when there is an ample supply of oxygen?

Some smoldering areas in forests continue to smolder for years where there is an ample supply of oxygen -> it's open air.

If you're smoldering, you're generally not producing enough heat to vaporize anything except in a small local environment. Only the vaporized fuel will burn.


Aluminium, Carbon, Aluminium Sulphate battery. What's the reaction? I made galvanic cells from the following items (while trying to do something else), and they give between 0.6 V and 1 V per cell.

Aluminium foil – negative terminal

Carbon (these work: charcoal sticks, graphite rods, carbon felt) – positive terminal.

Electrolyte – aluminium sulphate $\ce{Al2(SO4)3}$ in water.

With 9 cells – 3 series * 3 parallel – it lights a red LED.
Left on overnight it gets very dim, but given some recovery time, it comes back bright again.
So the current isn't high, but I wasn't expecting any.

I'm not observing any gas being evolved, or any change in the foil (yet).
The battery has been running about a week.

I know the Al sulphate reacts with water to make sulphuric acid + Al hydroxide.
Is that what's happening here?
What happens next in the reaction?
What's the final result?

Thanks for your help.
It's a long time since school chemistry lessons.

Very interesting: no gas, no change in the electrode, yet a significant electrical output. This could be the beginning of perpetual power (similar to perpetual motion!).

But no, probably the output power is barely significant, and something is happening somewhere, possibly in the solution. Aluminum sulfate doesn't usually produce a clear solution, so possibly you have missed an increased turbidity.

Possibly oxygen is reacting with the metal under the acidic conditions to produce Al(OH)3 which is removed by the acid solution (without reducing the acidity, because Al(OH)3 is not much of an alkali). The equation could be 4 Al + 3 O2 + 6 H2O --> 4 Al(OH)3. If you restrict oxygen access, the voltage and current should decrease. If you increase it, say with an aquarium pump, you might get a higher output.


Why do some covalent compounds form lattice while others do not? Why do some covalent compounds form lattice while others do not? I'm not only talking about amorphous solids but covalent compounds that form molecular lattice in general. For Example: Compounds like SiO2, SiC and AlN form covalent lattices while some covalent compounds like methane and SiF4 form molecular lattices. The question is - Why do only some covalent compounds do this and also, is there a way to identify which compound forms covalent lattice and which forms a molecular lattice? If so, then how?

NOTE: I've looked this up over Concise Inorganic Chemistry by JD Lee and also on the internet but have found nothing.

See, as far as I know, their is no fixed way to compare catenation properties of elements randomly without any context. Some general pointers however, to keep in mind are-:
Carbon has one of the highest catenation powers, as soap molecules like sodium stearate can contain up to 17 carbon atoms.
After carbon, some other elements having significant catenation tendencies are silicon, sulphur, boron etc. (roughly in that order). I will not be elaborating further here as I'm not very sure myself about this except perhaps a few rough pointers such as medium-ish size of atom and the presence of group 13 and 14(which have the maximum number of covalent sigma bonds formed without expansion of octet)
The thing you really need to understand is that the ability to form multiple bonds is more important here to predict the bonding nature of the crystal lattice. Oxygen and nitrogen can form multiple bonds, which would allow them to kind form cross links while satisfying the valency of the central atom. While on the other hand, fluorine will finish it's covalency in satisfying the needs of the central atom (Si) only, so the tetrahedral unit formed will majorly interact via Van der Waals forces (checking it with the facts, this is a pretty good prediction). The exact crystal structure of SiO2 and ALN are quartz and wurtzite respectively, which would be difficult to predict simply by making the covalent structures. However, a fair idea can be taken about the kind of interactions which will prevail in the crystal lattice, provided it is also covalent.


The "nose" of the periodic table My teacher said that on the periodic table there is a "nose" formed by Al, Zn, Ag, and Cd. She said that they are all fixed charged (+3, +2, +1, and +2 respectively), and said that if I write them in ionic equations, I just say Silver Nitrate instead of Silver (I) Nitrate. She also said to put all Al as +3 charge in all cases, etc. But I did some research and found out that you do say Silver (I) Nitrate (actually both are fine, but my teacher specifically said never put the (I) in such cases). So is my teacher wrong? And are there cases where the charges of the elements are not what my teacher says?

All of these elements can form compounds in other oxidation states. Aluminium forms some compounds in the +1 state (e.g. see the section in https://en.wikipedia.org/wiki/Aluminium_iodide), as does Zinc (see the section in https://en.wikipedia.org/wiki/Compounds_of_zinc) and Cadmium (e.g. https://en.wikipedia.org/wiki/Cadmium(I)_tetrachloroaluminate). But Silver is the element that shows this most often in the list given - for instance it forms fluorides in the +1, +2 and +3 oxidation states (https://en.wikipedia.org/wiki/Silver_fluoride). As such I wouldn't say including the oxidation states in the formula is wrong, but on the other hand if you omit it everybody who knows there is an ambiguity will understand what is implied.


Recrystallization of two unknown compounds? I managed to separate a binary mixture(in solid form) of two unknown substances into two individual solid compounds. For the separation, I used an acid-base extraction with dichloromethane, NaOH, and after separation HCl.  My next step is to purify the two separated solids using recrystallization(this is the technique mandated by my instructor for solid purification) but I am not sure which solvents to use. Can anyone suggest how I should proceed ? Thanks a lot !

Edit: I was thinking since I used NaOH and HCl on my aqueous layer, NaCl would have formed. Since compound doesn't dissolve in water, is it okay to just put the compound in water, let NaCl dissolve and vacuum filter the solid(which presumably should be pure solid) ?

Choosing a recrystallization solvent is as much an art as a science, even when you know what the substance is. Without knowing, it is a matter of making educated guesses. You want a solvent where it is much more soluble in the hot solvent than in the cold. Given that this is a lab exercise, we can assume that these are common substances. So, the one that dissolves in base is likely a carboxylic acid--perhaps benzoic acid or something close. That substance dissolves about 10 times as much in hot water as in cold - so I would try water first.  If it is too insoluble, then you could go to methanol, ethanol, etc., moving toward more nonpolar solvents until you get it to dissolve when heated. Then, hopefully it will crystallize out on cooling.  For the other substance, it is obviously soluble in dichloromethane and is not an acid. Perhaps it is non-polar. I would start with ethanol, and see how that works - if you need to go to a more non-polar solvent, toluene might be a good choice. Good luck!


Can the thermodynamic information of a compound be calculated or must it be found emperically? Do attributes such as heat of formation have to be found through experimentation or can they be calculated based on the attributes of the bonds and the atoms involved?

For gaseous compounds heat of formation can be calculated using quantum chemistry packages as a difference between energies, assigned by the package, though inaccuracies are possible. Entropy of formation is a bit trickier, it requires calculation of vibrational and rotational states, but still pretty straightforward. For condensed phase (liquids, solids) the situation is harder, as knowledge of structure of solid state must be known beforehand, or full exhaustive search of possible crystal structures must be performed.

In practice, however, both heat and entropy of formation are found experimentally.


Why is Electrostatic Potential Energy positive when the charges are like? We know that in an atom charge of an electron at infinity is zero. As it approaches the nucleus it become more and more negatively charged.
We also know that E.P.E is positive when charges are like and negative when charges are opposite.
Well does it mean that as an electron approaches nucleus there is repulsion which causes the approaching electron to go away. As the electron go away it is getting closer to to infinity point because of which electron become less and less negatively charged. So the E.P.E is positive .Nucleus of atom is like earth and the electron are objects.The farther the object move from earth it has more Potential energy.

There is no way to know the actual potential energy of a system, we can only determine it relative to some predefined zero value.  In the case of electrostatic potential, zero potential energy is defined as the energy between charges when their separation is infinite.

The charge of the electron doesn't change in this scenario; it is always $1.6\times 10^{-19}\ C$.

When oppositely charged particles near one another, they lose potential energy (their energy becomes more negative.)  When particles with the same charge near one another, they gain potential energy.  The attractive force between oppositely charged particles does work as their separation decreases, and the repulsive force between similarly charged particles requires work to be done on them in order to move closer to one another (so potential energy goes up.)


Nanoscopy Nobel Prize - why are all the tools in the visible light range? It seems like all the new advances in molecular imaging/microscopy are in the visible light range of the EM spectrum. An example would be the most recent Nobel Prize for Chemistry.

Is there some reason we are only looking for things in the visible part of the EM spectrum? I'd love to know why, and also what techniques look outside of this range?

As far as fluorophores go, if the excitation wavelength is too short (< 350 nm), the incident light will get stuck in the microscope.  Near-infrared fluorophores, on the other hand, suffer from several drawbacks, especially low quantum yields from competing non-radiative processes.


What are the side effects of adding dry ice to a swimming pool? My apartment complex pool was closed today due to a resident dumping a large quantity of dry ice into the pool. In the notification from my landlord, they stated:

  Dry ice is frozen carbon dioxide and not only imbalances the pool, but can cause burns due to the below freezing temperatures.

My understanding of dry ice being added to water is that it would sublimate and carbonate the water, causing a minor temperature decrease, and creating a low-hanging cloud of carbon dioxide gas on top of the surface (since it's heavier than air).

However, it would seem that if the dry ice indeed does cause "below freezing temperatures", that would just freeze the water; so, unless you already in the water, you shouldn't get a freeze-burn.

I've done a bit of searching, but most material talks about burns from solid dry ice, not from adding it to water.

One thread on Reddit suggests that freeze-burns are more possible from dry ice gas if the skin is wet, but there are no citations.
CCOHS states,

  Direct contact with the liquefied [carbon dioxide] gas can chill or freeze the skin (frostbite).

... but refers to liquefied gas, not gas in the gaseous state.
Wikipedia doesn't list any particular safety concerns aside from handling of liquid or solid carbon dioxide.

So, here's what I'm wondering: How does dry ice affect the pool's pH, and to what extent? How does it affect the temperature, and to what extent? And what other side effects (mostly harmful ones) would arise from the addition of dry ice to a swimming pool?

Solubility of $\ce{CO2}$ in water is $\pu{1.45 g/L}$ at SATP. Apparent dissociation constant for carbonic acid at the same conditions is $K_\text{a}(\ce{H2CO3}) = \pu{4.47e−7 mol L^{-1}}$.

Now lets speculate how much dry ice could be disposed and what the pool capacity might be, guessing that the water warmed back up. It's hard to tell which pool size your apartment complex (hopefully not this one) might have, so I'm going to propose some values (marked in $\mathrm{\color{red}{red}}$) which look plausible to me.

Assuming $m(\ce{CO2}) = \color{red}{\pu{10 kg}}$, and the pool volume of approx. $V = \color{red}{\pu{5e5 L}}$ (olympic pool is $\pu{2.5e6 L}$, I took $\frac{1}{50}$ of that), one notices that in order to completely dissolve $\pu{10 kg}$ of $\ce{CO2}$, $\pu{6.9e3 L}$ of water is required (based on solubility data), which means that we have more than enough water in the pool to completely dissolve all $\ce{CO2}$.

For the weak acid

$$[\ce{H+}] \approx \sqrt{K_\text{a} C} = \sqrt{\frac{K_\text{a} m}{M V}} = \sqrt{\frac{\pu{4.47e−7 mol L^{-1}} \cdot \pu{e4 g}}{\pu{44 g mol ^{-1} \cdot \pu{5e5 L} }}} = \pu{1.43e-5 mol L^{-1}}$$

so the pool water will indeed be acidified to the level of

$$\mathrm{pH} = -\log[\ce{H+}] = 4.85,$$

assuming the values I've willfully established, and that all carbon dioxide completely dissolved in water. You can put your values into the equation and argue with your landlord further.


Net ionic equations for an aqueous solution Given the reaction of $\ce{Ni^{2+}}$ with aqueous ammonia to give $\ce{Ni(OH)_2}$ Would this be the correct reaction?

$$\ce{Ni^{2+} (aq) + 2NH3 (aq) + 2H2O(l) -> Ni(OH)2 (s) + 2 NH4+ (aq)}$$

Your answer is probably correct. There are some extra details one might wish to consider. If a short yes/no answer was enough, you must not read any further.

First, let us remind ourselves that in aqueous solutions metal ions coordinate with water molecules. Octahedral complexes are most common.

$$\ce{Ni^2+(aq) + 6 H2O(l) <=> [Ni(OH2)6]^{2+} (aq)}$$

Then $\ce{NH4OH}$, better described as a hydrate $\ce{NH3.H2O}$, is added.
$$\ce{NH3 (aq) + H2O (l)<=>NH3.H2O (aq)<=>NH4+ (aq) +OH-(aq)}$$

Now either $\ce{NH3}$ or $\ce{OH-}$ steals a proton from $\ce{[Ni(OH2)6]^{2+}}$. Note that the resulting complexes are equivalent.

$$\ce{[Ni(OH2)6]^{2+}(aq)+NH3.H2O(aq) <=> [Ni(OH)(OH2)5]+(aq) + NH4+(aq) +H2O(l)}$$

$$\ce{[Ni(OH2)6]^{2+}(aq) + OH-(aq)<=> [Ni(OH)(OH2)5]+(aq) + H2O(l)}$$

If another proton is taken, we reach the required precipitate $\ce{Ni(OH)2}$. (Due to the equivalency mentioned, I will write only one option.)

$$\ce{[Ni(OH)(OH2)5]+ (aq) + OH-(aq)<=> [Ni(OH)2(OH2)4](s) + H2O (l) \tag 1}$$

Often, however, the $\ce{Ni(OH)2}$ precipitate is not visible. The whole point of this excercise was to give background to the explanation that follows. When $\ce{NH3.H2O}$ is in excess, the following processes occur. The precipitate could directly react, in which case we might end up with

$$\ce{[Ni(OH)2(OH2)4](s) + 6 NH3.H2O (aq)<=> [Ni(NH3)6](OH)2 (aq) + 10H2O(l)}$$

Generally, the preferred explanation is through Le Chatelier's principle. $\ce{[Ni(OH2)6]^{2+}}$ ions react via $(2)$

$$\begin{align}
\ce{[Ni(OH2)6]^{2+} + 4 NH3.H2O (aq)} & \ce{<=> [Ni(NH3)4(OH2)2]^{2+}(aq) + 8H2O(l)}\\
\ce{[Ni(OH2)6]^{2+} + 6 NH3.H2O(aq) } & \ce{<=> [Ni(NH3)6]^{2+}(aq) + 10 H2O(l)}\\
\ce{[Ni(NH3)6]^2+(aq) +2 OH-(aq) } & \ce{<=> [Ni(NH3)6](OH)2(aq)}
\end{align}$$

Since the $\ce{[Ni(OH2)6]^{2+}}$ ions are used up, the equilibrium of $(1)$ tilts to the left, thus the precipitate disappears.

Once more, this was simply to explain why showing $\ce{Ni(OH)2}$ as a precipitate might not correspond with reality when there is (excess) dissolved ammonia.


Ratio of copper(II) ions to EDTA? What is the ratio of copper(II) ions to EDTA in a $\ce{[Cu(EDTA)]^{2-}}$ complex?

I know that EDTA forms up to 6 coordinate bonds with the central atom, but is that what they are asking for with "ratio"?
According to this website, it seems that copper actually bonds with all 6 free electron pairs.

The ratio of Cu to EDTA is the number of copper ions per EDTA ion. In this case there is one EDTA ion for every copper ion so the ratio is 1:1.

Copper bonds with six electron pairs on EDTA to form the complex shown below where the central metal species is copper(II):


Can water oxidize zinc? According to this textbook, H₂O has lower reduction potential (-0.83 V) than Zn (-0.76 V). This mean zinc is unable to reduce water to hydroxide. If so, why this video demonstrate that by heating zinc in water, a reaction occurs producing hydrogen gas from the bubbles while also turning wet red litmus paper blue?

I've seen similar demonstrations with magnesium, and in both cases the explanation is the same.  The metal reacts to produce the hydroxide plus hydrogen; with zinc this would be
$\ce{Zn + 2 H2O -> Zn(OH)2 + H2}$
But, the zinc hydroxide is essentially insoluble in cold water and seals up the metal before the reaction gets very far.
Boiling the water speeds things up in two ways.  First, the metal hydroxide, with either zinc or magnesium, is inherently more soluble.  Second, the turbulent action created by the boiling (the bubbles you see are actually steam, not hydrogen) acts as a stirring mechanism, speeding up the mass transfer so the slight concentration of dissolved hydroxide gets to the litmus paper faster.
Magnesium differs from zinc because the former hydroxide is somewhat more soluble.  Therefore, the basic nature of the product with magnesium can be demonstrated with phenolphthalein, which requires a slightly higher pH to react than litmus.  Some demonstrations with magnesium show a rose coloring of the magnesium metal in cold water with the phenolphthalein showing that the magnesium hydroxide remains essentially on the surface.