Why does LiAlH(O-tert-Bu)3 not reduce acyl chlorides to alcohols? Why do $\ce{LiAlH(O-^{t}Bu)3}$ and similar bulky reducing agents not reduce acidic chlorine further?

$$\ce{CH3-COCl + LiAlH4 + H2O -> CH3-CH=O -> CH3-CH2-OH},$$
 but
$$\ce{CH3-COCl + LiAlH(O-^{t}Bu)3 + H2O -> CH3-CH=O -> no~reaction}.$$

Without being very specific it looks that $\ce{LiAlH(O-^{t}Bu)_3}$ is a lot less reactive than $\ce{LiAlH4}$, mostly because the $\ce{^{-}O-^{t}Bu}$ is a much weaker base than $\ce{H-}$. Also its much bulkier, further reducing reactivity.

The aldehyde being less reactive than the acyl chloride the reducing agent first eats up all the acyl chloride, and then starts reducing the aldehyde. If you were careful to add stoichiometric amounts, you can stop right when you run out of the acyl chloride.

See also this link.


How can I undo the tarnish on my aluminium headphones? I thought of cleaning my headphones, which has an outer aluminium casing, with Dettol liquid which I believe has strong oxidising agents. This was a bad idea. It left a tarnish on the aluminium and I'd like to know if I can undo this.

I would try any sort of weak acid that wouldn't cause damage to other parts of the headphones.

Citric Acid (found in lemon juice), Acetic Acid (found in vinegar), and Phosphoric Acid (found in cola) are probably all safe bets for sticking on a cloth and wiping off the casing. The primary difficulty you have here is the issue that comes from the fact submerging your headphones in anything would likely damage the electronics inside. You can rig up a shallow flat pan full of acid and resting the headphones on it's side in there, though I'd take caution.


Total valence electrons in a rhodium complex How many valence electrons does the following metal complex have?

I have read that benzene donates 6 electrons to a transition metals valence electrons. Rh has 9 from itself, and H and Cl add 1 each; hence in this assignment above I thought one would end up with: 9+1+1+2+2+6=21 valence electrons.
However, the given answer is 16. How is that so?

This is not a benzene ligand ($\ce{C6H6}$), but rather a phenyl ligand ($\ce{C6H5}$).

In the benzene complex, all the carbons have their hydrogens. The pi electrons of the ring act as donors to the metal. On the other hand, in the phenyl complex, the ring carbon bonded to the metal does not have its hydrogen. That carbon is formally negatively charged and the lone pair on carbon acts as a donor to the metal.

Aryl ligands, just like alkyl ligands, count as 1-electron donors within the "covalent" electron-counting system (which you seem to be using, since you counted H and Cl as 1e donors as well).

Therefore the total electron count is $9 + 1 + 1 + 2 +2 + 1 = 16$.


Why do we subtract the mass of one proton in mass spectrometry (negative mode)? This is probably very basic, but I have been given the understanding that we subtract the mass of only one proton when looking for our compound in a mass spectrometry (negative mode). I dont really understand why we are only subtracting the mass of one proton when our compound could potentially loose more than one proton at higher pH?

In example, the amino acid glutamate, would it not loose or gain more than one proton is a negative (higher pH) or positive (lower pH) mode mass spectrometry (MS)? Perhaps I have misunderstood the concept of negative and positive mode for MS?

Interesting question, which originates from a major misconception related to pH in LCMS. Nowadays, there are dozens of ionization techniques so in the question always mention the ionization method.

Anyway, LCMS is usually done with electrospray ionization (ESI) technique and so are amino acids. First of all nobody completely knows the exact process in  electrospray ionization as yet. This is not unusual in mass spectrometry where even a common ionization technique like ESI and electron impact, the picture is not 100% clear.

Second important point is the pH in ESI does not determine whether a negative ion or a positive analyte ion will be formed. I also used to have this misconception a decade ago about pH. The final ionization process in ESI takes place in a gas phase. However it is well known that certain ion can suppress ionization very badly. Don't try to connect mobile phase pH with the state of ionization of the molecule in the gas phase.

In negative ion, you are simply observing a net negatively charged species. Nothing

For details, I'd encourage you to read "pH Effects on Electrospray Ionization Efficiency" in the Journal of the American Society for Mass Spectrometry· Go to the journal website and locate the paper with this title. I am not providing a link so that students develop the habit of searching themselves.


Effect of anisotropy on electrical conductivity of crystalline solids I was reading through a chapter named 'Solid state' in the NCERT Chemistry book.
I learnt about anisotropy of crystalline solids.
And what I understood was, you measure a certain property of that solid in one direction and it could be X while you measure it in the other direction it could be Y.
I then learnt about how a solid could show the property of birefringence.

My question : So does that mean, even electrical conductivity isn't the same in all directions? Does that mean a copper wire is going to conduct differently in different directions?
How is that possible? It feels impossible?
Does that mean different copper wires conduct differently?

I understand the last part is like a chunk of questions altogether and our community doesn't like that, but then it is basically having the same origin with just a little counter questions kinda? so I just let that be. I believe the contributors can understand what confusion I'm dealing with if I ask it like that.

There are indeed compounds with very different conductivity in different directions
Depending on the structure of the crystal, it is possible that a compound has very different properties in different directions. Some crystal structures cannot show this as they are essentially the same in all spatial directions, but others can. Metals don't usually have anisotropic conductivity due to the nature of the conduction bands but other conductors can show it.
The archetypal compound that shows this property is graphite. Graphite consists of multiple flat layers of covalently bonded C-6 hexagons in an extended plane stacked on top of each other by much weaker forces. The delocalised electrons in the plane can conduct electricity well but there is little scope for electrons to move freely perpendicular to the plane. So the conductivity is about 1,000 times higher along the plane than it is perpendicular to it (see Electrical conductivity of graphite).
So it is certainly possible that electrical conductivity can be different in different crystal directions. But this requires particular structural features. Other forms of anisotropy are far more common (like birefringence).


Stereochemical aspect of heating of cyclohexane-1,2-diol Does stereochemistry play any role in the heating of Cyclohexane-1,2-diol?

If I remember correctly, different products are formed depending on whether the diol is cis or trans. (I think the products are cyclohexanone and cyclopentanecarbaldehyde). Please clarify this along with the reason why different products are formed depending on the stereochemistry.

The reaction you are thinking of is likely the thermal decomposition of the cyclic sulfites prepared from the cis- and trans-diols.  Here is a link to the abstract.

In both reactions $\ce{SO2}$ is extruded.  Whichever group on the vicinal carbon that is lined up in an anti-periplanar fashion to the departing oxygen is the group that migrates.  This is because the antiperiplanar geometry allows for good orbital overlap during the migration; if a choice is available then hydrogen migrates in preference to an alkyl group just as with carbocations. In the cis-isomer, the vicinal hydrogen is properly aligned (I darkened the affected bonds to highlight the geometric arrangement), its migration produces cyclohexanone.  In the trans-isomer, only a $\ce{C-C}$ bond in the cyclohexane ring is properly aligned, its migration produces cyclopentanecarbaldehyde.


Contradiction in variation of conductance and conductivity on dilution I do conceptually understand that on dilution conductivity decreases and conductance increases but if we look at the formula of conductivity, a contradiction seems to arise.

$$\kappa=GG^*$$
where $\kappa$ is conductivity, $G$ is conductance and $G^*$ is cell constant.

It appears that on dilution if conductivity decreases then conductance should also decrease which is obviously wrong. Where is the problem?

You get it wrong.

With the dilution, both conductance and conductivity ( specific conductance ) proportionally decreases, bound by the conductometer cell constant $G^* = \dfrac{\kappa }{ G}$.

What increases is molar conductivity $\Lambda_\mathrm{m}= \dfrac{\kappa }{ c}$ toward its infinite dilution value by the Kohlrausch equation $\Lambda_\text{m} =\Lambda_\text{m}^\circ - K\sqrt{c} = \alpha f_\lambda \Lambda_\text{m}^\circ$.

Note that conductance is NOT the ease with which ions move. You confuse it with electrical ion mobility, that is closely linked to the molar conductivity.

In fact conductance and conductivity have nothing to do with ions at all all.
Ionic solutions are just one special case how to manage these general current related quantities.

See conductance   and Conductivity on Wikipedia.


Is this the true expression for isochoric enthalpy of non-ideal gases? $\def\d{\mathrm{d}}$
$\def\D{\Delta}$

I recall seeing an equation in a thermodynamics lecture,

  $$\tag{non-ideal gas at constant $V$} \D H = n \int_{T_1}^{T_2}\! C_{P,\mathrm{m}}\ \d T$$

where $C_{P,\mathrm{m}}$ denotes molar heat capacity at constant pressure. There was not really a derivation of this, merely a statement whose mathematical equivalent is

  $$\underbrace{\D H = nC_{P,\mathrm{m}}\D T}_{\text{true for ideal gases}}\to\overbrace{\D H=n\int_{T_1}^{T_2}\! C_{P,\mathrm{m}}\ \d T}^{\text{the simplest generalisation}}.$$

I have to admit that the explanation made sense at the time. However, trying to justify the result from first principles$-$the $1^{\text{st}}$ law of thermodynamics$-$has proven unfruitful thus far.

$$\tag{$\D V=0, \text{def of}\ C_{V,\mathrm{m}}$} \d H=nC_{V,\mathrm{m}}\,\d T + V\,\d P$$
$$\tag{$C_{P}-C_{V}=\frac{\alpha^2TV}{\kappa_T}$}\d H=n\left(C_{P,\mathrm{m}}-\frac{\alpha^2TV}{n\kappa_T}\right)\,\d T+V\,\d P$$
Assuming the lecture formula holds, then

$$\tag{volume constant}\frac{1}{n}\int_{T_1}^{T_2}\!\frac{\alpha^2T}{\kappa_T}\,\d T=\D P.$$
Is this the case? Why?

The derivation for the first part of your question is shown below. I'm not sure what you are trying to do in the second part.

The heat capacity is the heat required to raise the temperature of  1 mole of substance by 1 K. Suppose  for the moment that heat capacity C is constant over a temperature range $T_1 $ to $T_2$, then the heat q required to raise 1 mole of substance  over this temperature range is
$$q=C(T_2-T_1) = C\Delta T$$
If the heat capacity is not constant, which is generally the case as real molecules have vibrational and rotational energy levels then,
$$ q=\int _{T_1} ^{T_2}C(T)dT$$

where in many cases the heat capacity can be approximated as $C(T)=a+bT+cT^{-2}$. For a gas the partition function can be calculated, if spectroscopic values are known, and $C(T)$ calculated that way.

The heat absorbed depends on the path taken to go between the two end points so that C is not well defined. To make it well defined either T or P must be constant.

At constant volume, $q_V=C_V\Delta T$ and the work done as expansion against external pressure is zero, thus from the first law with internal energy as U,
$$ \Delta U = C_V\Delta T $$

and a similar integral as described above is used when the temperature range is too large for $C_V$ to be considered a constant. (This equation shows also, in the limit of small changes,  that heat capacity is the slope of internal energy vs. temperature.)

In the case of constant pressure $q_P=C_P\Delta T$. The change in enthalpy is
$$\Delta H = C_P\Delta T$$
and so in general
$$ \Delta H=\int _{T_1} ^{T_2}C_P(T)dT$$

To demonstrate that $q_P=\Delta H$ we note that in this case the volume is not constant and so the heat transferred at constant pressure $q_P$  is related to internal energy change $\Delta U$ as
$$ \Delta U = q_P-P\Delta V$$
where $P\Delta V$ is the work done. Rearranging and expressing the $\Delta$'s directly gives
$$ q_P= (U_2+PV_2) - (U_1 +PV_1)$$
from which using the definition of enthalpy $H=U+PV$ gives $q_P=H_2-H_1$ and shows that $q_P=\Delta H$.


Physical implications of the simple harmonic oscillator quantum mechanical solutions Take a look at the wavefunctions for the different energy levels of a simple harmonic oscillator (a crude approximation for a diatomic).

The wavefunctions seem to make sense: they tend to zero as x tends to plus or minus infinity so it is normalisable. The number of nodes increases by one with increasing energy levels etc.

However, when you take a look at $|\psi(x)|^2$ which represents the probability density of the particle to be found at that point in space (or the diatomic to be in that particular state/bond length) I struggle to see how a node can exist. This is because a diatomic will stretch and contract through the full range of extension/contraction - the quantum representation does not seem to conflict this in the sense that there is a non zero probability when the potential is equal to the total energy. However, to reach this state from, for example, $x=0$ in the $n=2$ energy level it must pass through a node. What does this mean? surely a bond can't cease to exist as it stretches from equilibrium to some extended or contracted state? What's really going on here?

However, to reach this state from, for example, x=0 in the n=2 energy
  level it must pass through a node.

To start, ask yourself: "What should pass through a node?" The answer is: it is the probability that "passes through the nodes", not atoms (or, I should better even say nuclei). Probability is just a mathematical quantity ant the zero value is allowed for it, so there is nothing to worry about. Yes, quantum harmonic oscillator has some forbidden regions of displacement from the equilibrium. But it can also penetrate into classically forbidden regions. Quantum world is weird.

The root of this (and most of the other) apparent quantum paradoxes is the classical way of thinking: you think that you have some funky quantum balls on a funky quantum spring out there, but you don't. Quantum harmonic oscillator is a quantum system that essentially has the Hamiltonian similar in its form to the Hamiltonian of a classical harmonic oscillator. And this is it. No balls, no springs, just a similar Hamiltonian.

It should be mentioned as well that atoms (again, more precisely, nuclei) are not vibrating in molecules even in a classical sense: their motion can be only approximately separated into vibrational and rotational ones. Besides, the vibrational motion can only approximately be treated as being harmonic.


pKa and equivalence point I was doing a titration and using $\ce{NaOH}$ as the titrant. The analyte was a weak acid with a $\mathrm{p}K_\mathrm{a}$ of 4.90. Today I want to use another acid with a $\mathrm{p}K_\mathrm{a}$ of 4.50. Do I need to change the indicator from the last experiment? How close do the $K_\mathrm{a}$ or $\mathrm{p}K_\mathrm{a}$ values have to be in order for me to use the same indicator and still obtain accurate results.

$99.9\%$ of the acid $\ce{HA}$ is titrated at  $\mathrm{pH}=7.9$ resp.$ 7.5$.

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+\log \frac{[\ce{A-}]}{[\ce{HA}]}=\mathrm{p}K_\mathrm{a}+3$$

Considering the final concentration 0.05M,
titrating 0.1M acid by 0.1M hydroxide,
100% of the acid is titrated at:

$$\begin{align}
\mathrm{pH}&=7+\frac12 \cdot ( \mathrm{p}K_\mathrm{a} + \log  c_\mathrm{acid, total}) \\
&=9.45(9.25) + \frac12 \cdot (-1.3) \\
&=8.8(8.6)\\
\end{align}$$

Titration with $100.1\%$ of used hydroxide we can consider
as $0.05 \cdot 0.001=5\cdot 10^{-5}\ \mathrm{M} \ \ce{NaOH}$
with $\mathrm{pH}=-\log ( 5\cdot 10^{-5})=9.7$

The indicator  phenolphthalein with range $8.2-10.0$ works well in both cases.


How dangerous is cobalt(II) chloride used in silica gel balls? I was looking up silica gel beads on Wikipedia and noticed that some (and from experience most) are doped with a colouring agent to indicate that they have absorbed all that they can.

The Wikipedia page says that usually cobalt(II) chloride or methyl violet is used, both of which are toxic. But it mentions nothing about what quantity these become toxic? I assumed the quantity in a normal silica gel bag (that come in shoes for example) isn't enough to be harmful, otherwise why would they be using it?

Also, it mentions that

  Crystalline silica dust can cause silicosis but synthetic amorphous silica gel is indurated, and so does not cause silicosis.

Are the silica gel beads synthetic amorphous, i.e. do not cause silicosis?

Cobalt poisoning is apparently very rare. The only situation in which cobalt poisoning is a serious problem is when it is exposed directly to your bloodstream (Recently it has been an issue with people who have undergone certain hip implants -- see this page).
The amount of Cobalt chloride in silica gel balls would be absolutely harmless. According to most sources, you have to be exposed to significant amounts of cobalt (at least more than 20g for an average adult) for a long period of time for symptoms to appear.
Note: Cobalt is actually necessary for our body (in small amounts) in the form of Vitamin B12.


Why and when are bonds broken I am new to chemistry (I have only done a GCSE in it) so forgive me if I am missing something obvious. But I am trying to very roughly simulate some chemical reactions. My background is in Artificial Life and Adaptive Systems rather than computational chemistry or chemistry in general. My stumbling block is that I am unsure exactly why bonds break and when they break.

Here are the things I am confused about:

Is the breaking of bonds by using energy a prerequisite for new bonds to be made with the involved atoms? If so, is this because the more energy there is, the more the atoms will move/vibrate, eventually pulling the bond apart? Otherwise, can bonds be broken by a new bond sort of pulling it apart in some way?

Is the strength/length of the bond what determines the amount of energy needed to break it? I'm also unsure of what the strength or length of a bond is based on. Am I correct in thinking that bond order is about the number of shared electrons in a covalent bond? But what about ionic bonds? Is it the amount of electrons "taken" by the more electronegative atom?

I know that there are multiple questions here. But they all answer the bigger question. Why and when do bonds break exactly? I have had a hard time trying to figure this out by searching on the internet alone. Often I just get pages explaining the energy requirements of a reaction. Thanks in advance, any help would be appreciated.

Is the breaking of bonds by using energy a prerequisite for new bonds to be made with the involved atoms?

A: No, it is possible for a bond to form between two species without first breaking a bond. But it is more common for reactions to occur with a tradeoff between breaking old bonds and forming new ones.

If so, is this because the more energy there is, the more the atoms will move/vibrate, eventually pulling the bond apart?

A: Adding energy in the form of heat can lead to breaking of bonds, which in turn can lead to formation of new bonds.

Otherwise, can bonds be broken by a new bond sort of pulling it apart in some way?

A: You can have a "displacement" reaction where one atom replaces another in a structure: the bond to the new atom forms at the same time as the other bond is broken. That process has a lower energy barrier between reactants and products, relative to breaking an existing bond first and then forming a new bond later.

Is the strength/length of the bond what determines the amount of energy needed to break it? I'm also unsure of what the strength or length of a bond is based on.

A: The strength of a bond is the same as the amount of energy needed to break it (in the absence of other reactants). Bond length tends to correlate inversely with bond strength, as for example in going from a single to a double to a triple bond between carbon atoms. Bond strength depends on a number of factors--bond strength tends to be greater between two atoms of about the same size. Bonds between two electronegative atoms like oxygen or chlorine are weaker than a bond between two carbon atoms.

Am I correct in thinking that bond order is about the number of shared electrons in a covalent bond? But what about ionic bonds? Is it the amount of electrons "taken" by the more electronegative atom?

A: You are correct here in your understanding of bond order. But the term "bond order" does not apply to so-called ionic bonds -- for example in sodium chloride -- where an electron is completely donated from sodium to chlorine and does not participate in keeping the atoms together. Rather, it is the electrostatic attraction between the sodium and chloride ions that keep them together.


Will the total equilibrium pressure increase in gas phase reaction PCl5(g) = PCl3(g) + Cl2(g) after chlorine is removed? Will the total pressure be greater than the initial equilibrium pressure if all $\ce{Cl2}$ is removed and the following reaction is allowed to re-equilibrate?
$$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$$
Assume constant temperature and volume.

I know that when $\ce{Cl2}$ is removed, the reaction will go forward and I also know expressions of $K_\text{p}$ and $K_\text{c}$.  I also know Le Chateliers Principle. I have thought about it for a long time but have not made any progress.

First it is necessary to understand that due to the stoichiometry of the reaction, if you move the equilibrium to the products side, pressure will be increased. This is based on the fact that 1 mole of gas produces 2 moles of gas. In other words the number of molecules in the system is increased. By removing all of the $\ce{Cl2}$ you force the reaction to move to the products side (Le Chatelier's Principle). Thus more $\ce{PCl3}$ will be produced along with $\ce{Cl2}$. Eventually you will end up with more atoms inside the system and that will lead to increase of the pressure.


What would happen if Hofmann bromamide reaction is carried out in Br2 and KOD? What would happen if Hofmann bromamide reaction is carried out in Br2 and KOD ?

I already know the mechanism the question is the water molecules that is attacked by the isocyanide is it already present in the solution or is the one formed within the reaction in situ?, secondly If we are given the reaction is carried out in Br2 and KOD then do i have to assume that it is actually  KOD in D2O because my teacher says there would be no crossover product that if the reaction is carried out in aforesaid conditions the product would be RND2but i think if the water molecule in the reaction mechanism are  the ones generated in situ then there is a fair chance of other products like RNHD or even RNH2.

So the main question are :-

1) Are the water molecules the ones generated in situ ?

2) Is KOD equivalent to KOD in D2O ?

3) Would RNHD and RNH2 also form in the given conditions ?

This isn't the most well-thought-out scenario, because if you dissolve $\ce{KOD}$ in $\ce{H2O}$ all the deuteriums will be lost immediately anyway.

$$\ce{KOD + H2O <=>> KOH + HOD}$$

Assuming that the amount of $\ce{KOD}$ is much less than the amount of $\ce{H2O}$ (after all $\ce{H2O}$ is the solvent) then the position of this equilibrium will lie very, very far to the right. So, at the end of the day, if you are using $\ce{H2O}$ as the solvent, the use of $\ce{KOD}$ won't have any effect. If you want to see any difference in deuteration, you have to at least use $\ce{KOD/D2O}$. However, there is a second problem with this:

If you dissolve an amine in $\ce{D2O}$, then all the amine hydrogens will be very rapidly replaced with deuteriums. It does not even matter mechanistically what the pathway is. Regardless of whether the initial product is $\ce{RNH2}$, $\ce{RNHD}$, or $\ce{RND2}$, as long as there is $\ce{D2O}$ in the mixture, all three will become $\ce{RND2}$. The same can be said of dissolving them in $\ce{H2O}$.

To answer your questions:

(1) No. In general, protonation can and will occur by any solvent molecule that happens to be close by. If you are running the reaction in $\ce{D2O}$ solvent, then there is much, much more $\ce{D2O}$ than there will ever be $\ce{H2O}$ produced. So, protonation by $\ce{H2O}$ is statistically close to impossible.

(2) None of us can answer that, as we cannot presume what your teacher was thinking of when they set the question.

(3) Depends on what your solvent is, see comments above on dissolving amines in $\ce{D2O}$ or $\ce{H2O}$.


pH of dilute acids in hexane Consider a strong acid $\ce{HA}$. In a $\pu{1000 mL}$ of hexane (assume it is unable to accept a proton) we "dissolve" (this can mean forming a suspension with vigorous stirring - otherwise we can use a polar solvent which cannot accept a proton) $\pu{1e-30 mol}$ of the acid along with enough water to remove all the protons.

Now we have extremely dilute solution of $\ce{H+}$ where autoprotolysis is insignificant compared to the volume of solvent, so according to $\mathrm{pH} = -\log[\ce{H+}]$, pH should be extremely high. However, we are still using an acid, so pH should not be too high. What accounts for the difference?

pH is defined for water solutions; using it elsewhere is either impossible or leads to contradictory results like yours. Who said pH should not be too high? Well, you might argue that high pH implies alkaline solution, does it not? No, it doesn't. Alkaline solutions are (roughly speaking) those with high concentration of $\ce{OH-}$. In water this means the same thing, because $\ce{[H+]\cdot[OH-]=10^{-14}}$, so low $\ce{[H+]}$ means high $\ce{[OH-]}$ and vice versa. In hexane, not really.

To determine pKa values which are way beyond the physically possible range of pH, we extend it with things like Hammet's $H_0$, and thus build a ladder of measures suitable for a much wider range. You may call it "effective pH" or like that, but remember: it has little to do with literal concentration of $\ce{H+}$, and your hypothetical solution of $\ce{10^{-30}M\; HCl}$ in hexane is in fact neither acidic nor basic.


How does P4O10 make sense? I'm trying to understand the gains phosphorus has from forming $\ce{P4O10}$. The molecular structure is such that every oxygen atom forms two bonds (either two with one phosphorus atom, or one with two), while every phosphorus atom forms five bonds.

I'm clear about the gains the oxygen has from this -- it fills its outer shell to the octet rule, mimicking neon effectively. But what about phosphorus? We're talking about a covalent bond, so forming five bonds would leave it with five extra electrons, which is two more than it needs to reach an optimal configuration.

I remember reading somewhere (possibly on this site) that ionic bonds can be regarded as an extreme form of covalent bonds, where the difference in electronegativity is so great, that the metal gives its extra electrons to the non-metal. Can we assume that the difference in this case is also large enough, such that phosphorus effectively 'gives' away five electrons, therefore reaching neon too? Or am I oversimplifying it?

"ionic bonds can be regarded as an extreme form of covalent bonds." They "can be regarded" as magic, too. What's your point? In what specific context is such regard useful? (I'm not questioning that there may be some specific context that you could regard them that way, rather I AM saying that it is not generally useful.)
 The octet rule is used to predict valence.  Obviously H doesn't obey it. It's quite useful for the chemistry of C,N, O, and F.  But consider ClF5. The Rule is not going to get you there. Nor is it useful, imho, for P4O10 --- but there are ways to bend the meme to "explain" the deviancy. see for example the discussion of hypervalent compounds in https://en.wikipedia.org/wiki/Octet_rule#Hypervalent_molecules. Once you get away from the 2nd & 3rd Period and the 13th - 17th Groups, it shouldn't be relied upon. So, its applicability is limited, but fortunately it's applicability is for a lot of elements which are interesting for us in our environment here on Earth. Since d-orbital (and f-) contributions to hybridizations is generally small, the octet rule is generally useful. What sort of confidence should you have in using it to predict valence? Well, there's two ways to answer this: 1. as applicable to a student learning basic chemistry or 2. as applicable to experts who have in addition to the rule a vast literature on the subject. It is useful for the first and nearly useless for the 2nd.


Decarboxylation product of bicyclic compound What is the decarboxylation product of the following species:

It may seem straightforward but the product enolate involves a double bond with one end at the bridgehead carbon which is impossible/very unstable. What is the correct product and how is the product reached mechanistically.

${\beta}$-keto acids typically decarboxylate very smoothly at temperatures between 100-200°C. The mechanism involves a cyclic, 6-membered transition state that first produces an enol intermediate (see figure).

However in the case of your bicyclo[2.2.1] ${\beta}$-keto acid, a bridgehead double bond would result.  As you correctly point out, such a structure would be very high energy.  Therefore this compound does not decarboxylate at normal temperatures (reference, see p. 218, scheme 6.53).

I'm sure if you heated it hot enough something would happen.  It may decarboxylate via a free radical mechanism. The resultant bridgehead free radical could fragment or dimerize.  Alternately, and my preferred guess, would have the starting compound, in equilibrium with its non-bridgehead enol tautomer, undergo a reverse Diels-Alder reaction as pictured above.


How to rationalise the difference in halogen bonding? Can anyone explain the difference in halogen bonding to me. I understand the explanation in (c). It appears to me, that accepting that would contradict the answers to parts (a), (b), and (c).
Is the reason for part (a) because the first molecule does not have resonance and for (b) the $\ce{F3C-I}$ molecule has a larger inductive effect? If so, that would not make sense for (c) to be correct.

A halogen bond can, quite analogous to a hydrogen bond, be considered a mostly electrostatic interaction with some covalent interaction. Hence the halogen bond acceptor is the electron donating molecule, i.e. a Lewis base. The halogen (bond) takes the role as the Lewis acid.

In (b) you compare $\ce{H3C-I}$ and $\ce{F3C-I}$. Fluorine is more electronegative than hydrogen, and the statement $$\angle(\ce{H-C-H})>\angle(\ce{F-C-F})$$ holds. The carbon directs mor $\ce{p}$ orbital character into the carbon fluorine bonds than in the carbon hydrogen bonds. As a consequence the $\ce{s}$ orbital character in the carbon iodine bond (from the carbon) is higher in $\ce{F3C-I}$ compared to $\ce{H3C-I}$. That causes the bond to be slightly shorter and having more electron density at the carbon. This is also making the iodine a little more electron deficient and a better electron acceptor, i.e. a stronger acid (see also Bent's Rule). The electronic interaction will therefore also be slightly stronger.
This observed effect is also often called inductive effect, which like you correctly stated is larger in $\ce{F3C-I}$.

In (c) you are comparing the Lewis bases $\ce{NMe3}$ and $\ce{OMe2}$ acting towards $\ce{F3C-I}$ (or $\ce{HC3-I}$, the reasoning would remain the same). Oxygen is more electronegative than nitrogen, hence the lone pair orbitals of oxygen are more compact, i.e. the maximum electron density is closer to the oxygen nucleus. This causes dimethylether a weaker electron donor. Hence the the oxygen iodine interaction in $\ce{F3C-I\bond{~}OMe2}$ can be expected to be weaker than the corresponding nitrogen iodine interaction in $\ce{F3C-I\bond{~}NMe3}$. (Just like it is stated in your sheet.)
Additionally, since methyl groups are usually slightly electron donating, and trimethylamine has one more methyl group than dimethylether, it can be expected to be an even better electron donor, stronger base.

Resonance does not contribute to these bonding situations in a significant way. (Looking at this from a valence bond perspective would certainly exceed this format and it is not necessary for a good model explanation.)

Last a word of caution. Oxygen (in neutral molecules) can in most cases not be described as $\ce{sp^{3}}$ hybridised. Symmetry considerations and it's electronegativity are usually the reason for this. It is better described as $\ce{sp}$ hybridised in terminal position (like in ketones), or $\ce{sp^{2}}$ hybridised in bridging positions (like in ethers).


What is the physical significance of the universal gas constant R? While studying work done in isothermal processes I was told that the universal gas constant $R$ is the work done by the gas per mole per kelvin. I am sure I am missing something basic but I didn't find anything beyond that statement when I looked for an explanation.
This is what I found on Wikipedia:

The physical significance of $R$ is work per degree per mole.

In doing what, is 8.314 joules of work done by one mole of an ideal gas?

It may be helpful to look at a related value $k_\mathrm{B}$, the Boltzmann constant, which is widely used in thermodynamics.
These two are related by $R = k_\mathrm{B}N_\mathrm{A}$, allowing the ideal gas law to also be written:
$$PV = Nk_\mathrm{B}T$$ where $N$ is the number of particles, as opposed to the number of moles.
The units are $\pu{J\cdot K^{-1}}$.
It's a proportionality between energy and temperature. In the ideal gas law, it relates temperature on the right to the product of pressure and volume, which is an energy.
This general idea is frequently used in thermodynamics, as you will see factors of the form $\exp{(-E/k_\mathrm{B}T)}$, where the $k_\mathrm{B}$ allows the exponent here to be unitless.
As examples:

Planck's law, where the energy is in the form of quantum energy level spacing:
$$B_{\nu}(\nu, T) = \frac{2h\nu^{3}}{c^{2}}\frac{1}{\exp\left(\frac{h\nu}{k_{\mathrm{B}}T}\right) - 1}$$
Maxwell-Boltzmann distribution, where the energy refers to the kinetic energy of gas molecules:
$$f(v) = \left(\frac{m}{2\pi k_{\mathrm{B}}T}\right)^{\frac{3}{2}}4\pi v^{2}\exp\left(-\frac{mv^{2}}{2k_{\mathrm{B}}T}\right)$$


Oxygen reduction - beginner level I am embarrassed to ask but I don't get something really basic.

When molecular oxygen and hydrogen react to produce water, we say the oxygen has been reduced and the hydrogen has been oxidised. Reduction, I understand, is the gain of electrons.

So I look at the oxygen atoms prior to the reaction and it seems to me each has six of its own plus two more that are shared.

And I look at the oxygen in the water molecule and it has... six of its own and two that are shared.

So in what sense has it gained electrons?

Of course, before posting this question I have searched and read for several hours. All I can find is talk of half-reactions. The clue here seems to be that the oxygen is ionised to $\ce{O_2-}$ prior to reacting with $\ce{2H+}$. But this just confuses me. Maybe this intermediate step does involve reduced oxygen but the end point is still the same number of electrons as we started with, so if it does get reduced it must be oxidised again immediately afterwards, for no net effect.

Where am I going wrong?

This is a really good question which I've never given much thought about.
I'll try to explain the reason as much as I know it. There may be a better explanation.

The reaction can be represented as -
$$\ce{2H_2 + O_2 <=> 2H_2O}$$

Now in O2 you have a structure like
$$\ce{O\dbond O}$$

And in H2O you have a structure like (ignoring the tetrahedral hybridisation and bent shape)

$$\ce{H\sbond O\sbond H}$$

We know that Oxygen is second most electronegative element after fluorine, which is because of it's small size and strong effective nuclear pull towards electrons.

Hydrogen on the other hand, is not as electronegative as Oxygen even after having a small size, because of very less effective nuclear pull.

When a bond between H and O is formed, O pulls the shared pair of electrons towards itself because it is more electronegative than H. This is the "gain" of electrons. The atoms do not actually take up electrons (when it is covalent bond) but pull it towards themselves, thus getting reduced. You will not find this "gain" or pull of electrons in the O-O bond because they have equal pulls, hence 0 oxidation state. But in water, Oxygen has +1 oxidation state.

Similarly, for hydrogen, it pushes the electrons towards Oxygen in water. Not exactly losing the electrons but density of electrons around Hydrogen atom decrease due to this push by hydrogen and pull by Oxygen. Thus it has "lost" it's electron, becoming positively charged which wasn't the case when it was in H2
Also, you'll notice that when more bonds are formed, there is more change in oxidation state of the atom.

Hope this helped.


Separating boric acid from sodium chloride I want to create some boric acid with borax in the following reaction:

$$\ce{Na2B4O7·10H2O + 2 HCl → 4 BH3O3 + 2 NaCl + 5 H2O}$$

However as you can see, the leftover solution will contain a mixture of boric acid, sodium chloride and water. Both chemicals are soluble in cold water so neither will want to immediately crash out of solution.

My question is, how can I remove the excess water and sodium chloride to leave myself with some powdered/dry boric acid crystals?

I know I can probably just use some liquid ammonia, but then the boric acid is only slightly soluble in it leaving much of it behind. I could perhaps use glycerin also, however I am limited to how much I currently have.

Here is a potential solution that shouldn't even require heating. Note that I have not tried this myself.

Boric acid is soluble in lower alcohols like ethyl alcohol and isopropyl alcohol.
The solubility of sodium chloride in ethyl alcohol at room temperature is about 0.05%$^1$.

Given these properties, I suggest dissolving the dried boric acid / sodium chloride in a minimal amount of cold 95% denatured ethyl alcohol or 99% isopropyl alcohol (available at most pharmacies).

The boric acid should dissolve readily and the sodium chloride should not, so you can just filter it and let it dry to a fairly sodium chloride free boric acid solid.

To make the sodium chloride even less soluble you could cool the solution in a freezer prior to filtering.  If you notice any significant increase in solids that means boric acid is coming out of solution and you'll want to abort this step.

1) Journal of Chemical and Engineering Data, Vol. 50, No. 1, 2005


When blue vitriol is dissolved in water when CuSO4.5H2O is dissolved in water, what physical and chemical effects happens to its water of crystallization?
While solving a stoichiometric problem the ques was that 1litre of CuSO4.5H2O was dissolved in 1 liter water and the concentration of Cu2+ ions were asked. When calculating molarity the volume which those 5 water of crystallization were also taken along with the solvent water.
I had the doubt that if the 5H2O is attached to CuSO4.5H2O then how can we include that also while calculating volume of solvent (for denominator of molarity). Those 5 water molecules are not free, they are bonded then why do we consider it?

When a sample of $\ce{CuSO4⋅5H2O}$ is dissolved in water, the $5$ molecules water are mixed with the water used as solvant, and cannot be distinguished any more from the other molecules of water. It does not really matter, as if you want to calculate the molarity, you never dissolve your sample of $\ce{CuSO4·5H2O}$ in exactly $1$ litre water. No ! You dissolve it in a volume of water big enough to get one liter solution, whatever the exact amount of water used for this dissolution. It should be a little less than one liter water. The only important parameter is the final volume of the solution, namely one liter.


Copper electrode in a Daniel cell Suppose I have a beaker containing $\ce{CuSO4}$ solution and I dip $\ce{Cu}$ rod in it. What reactions are going to take place? Is the $\ce{Cu}$ rod going to acquire any positive charge?

I am asking this question to clear my doubt  regarding a Daniell cell as to why $\ce{Cu}$ rod used in the cell should be positively charged. I don't get it, though $\ce{Zn}$ rod dipped in $\ce{ZnSO4}$ solution acquires negative charge due to oxidation of $\ce{Zn}$ to $\ce{Zn^2+}$ is clear to me, but how can $\ce{Cu}$ rod dipped in $\ce{CuSO4}$ be positively charged. I suppose it to be neutral.

What is the answer to this conceptual question?

If viewing these as charged objects, we're using a physics/electrostatics viewpoint, not usually done in electrochem.  Or at best, it's a topic for postdoc courses in surface nano-chemistry (yet I think if it were discussed in undergrad classes, it would clear up much student misconception!)  The metal/electrolyte interface forms a capacitor.  For this reason we must deal with actual potentials, NOT the shifted values shown in tables of Standard Electrode Potentials.

You're right to be confused.  As with the zinc rod, the copper rod will also acquire a negative charge.

From a physics standpoint, the two electrodes in contact are 1) the electrolyte and 2) the metal.  A half-cell then has two conductive electrodes, not one.  A complete electrolysis cell has four.  In a half-cell, a genuine voltage appears between the metal and the electrolyte.  The voltage is not undefined or somehow abstract or "ghostly."  It's just difficult to measure.  A simple voltmeter cannot do it.

These concepts go back over a century, to when chemistry and physics weren't fully separate.  Alessandro Volta and Michael Faraday studied electrolysis from this viewpoint.  I've seen "single" electrode potentials discussed in 1930s undergrad textbooks, but not in the modern ones.  (Well, Bockris electrochem has it.)

Search: Galvani Potential, also Volta Potential, both found at surfaces and junctions between differing materials.  Perhaps also look for Helmholtz layer as well as Helmholtz Double Layer.

.

Whenever metal first contacts electrolyte, a brief oxidation reaction takes place which establishes a static potential appearing across the double-layer located at the metal surface.  The net reaction halts once these potentials appear.  The copper is then left with a negative charge, same as the zinc, with the electrolyte having net positive charge.

However, if either a salt-bridge or a common electrolyte is connecting the half-cells, the zinc and copper rods will always acquire two different amounts of negative surface-charge, leading to a well-known potential-difference measured between them.

For working with these physics/nanochem concepts, it's useful examine the SHE reference electrode.  The relative potential-difference of the Standard Hydrogen Electrode is roughly 4.44V, not zero.  This is the Galvani potential between metal and electrolyte, also called "Absolute electrode potential," although that term comes from chemistry, and a physicist would insist that this is a genuine potential-difference, not "absolute."  (The two electrodes after all are the conductive electrolyte on one side, and conductive metal on the other!)  If the SHE potential-difference is 4.44V, then all the values in the table of Electrodes need to be shifted by that amount.  In that case, the polarity of every electrode is negative.  Zinc and copper will both end up with the same polarity.  Copper is not positive-charged.

However, once these concepts are grasped, there's no major need for them, since measurements in electrochemistry are always biased by the inherent Galvani potential of various Reference Electrodes (SHE, or SCE, or Ag/AgCl.)  If we ignore electrostatics considerations, and deny that the electrolyte is itself a conductive electrode, then we can take the reference electrode potential-difference to be zero.  In that case, voltage measurements become trivially easy.  No vibrating non-contact probes needed, no quadrant electrometers or Faraday-cup charge-detectors.

But in that case, surface nanochemistry questions such as @Sharad 's cannot be answered unless we first discuss the zero SHE-potential being a "convenient lie," and the table of Standard Electrode Potentials must be shifted by 4.44V to reflect electrostatic reality.  (All the potentials in that table have the same polarity.  Oxidizing of solid metal always removes positive ions, even for the metals listed on the table "below" the SHE's zero-voltage entry.)

See also: we already learned all this in undergrad classes, right?


Devitrification vs glass transition temperature This graph comes from a so-called "differential thermal analysis". It is studying the effects of temperature on glass.

I don't understand what the difference between "glass transition" and "devitrification" is, and why one would be endothermic and the other would be exothermic.

I'm having trouble analysing these types of graphs. Does anyone know of a simple source online that I could check out?

This graph comes from a so-called "differential thermal analysis". It is studying the effects of temperature on glass.

Differential thermal analysis (DTA) is a method that heats two sample pans, one with a sample and one empty as a reference so that the effects of the sample pan can be negated. The heating is performed at a constant heat flow and the temperature of the sample is compared to the reference. Any exo- or endothermic process will add or subtract additional heat to the material and cause a spike in temperature upward or downward.

  I dont understand what the difference between "glass transition" and "devitrification" is, and why one would be endothermic and the other would be exothermic.

Glass transition is the temperature where bonds of a glass can relax and stress can be anneal out of a sample. It is neither endothermic or exothermic but the specific heat capacity increases as the atoms now have additional degrees of freedom.
Devitrification is where enough is put into the bonds that the atoms may rearrange and form crystalline structures from their vitreous state (devitrification). Crystallizing is an exothermic process because the atoms become more ordered and minimally space compared to a vitreous state, this reduces the strain in the bonds, releasing energy.


In which country is it allowed to practice non-clandestine amateur chemistry? I travel a bit so am flexible where I live, and have taken a recent interest in chemistry to try to gain an understanding of physical reality. I find learning from textbooks to be good, but want to have the comfort of applying a lot of this knowledge in my own lab. To see and feel how it works first hand. From my understanding, the west has a huge paranoia with clandestine chemistry, which is of no interest to me. However, it seems a vast quantity of common reagents and catalysts in organic chemistry books are banned in western countries due to this.

Would someone know how it's possible to practice amateur chemistry from home, without fear of accidentally breaking the law by using some reagent I'm not supposed to? And if not, what would be the best way to gain hands-on experience like this, where I can have complete freedom to experiment with what I want to do? I find at university I'm limited to lab sessions, and if one day I gain employment with a lab I'd no doubt not be free to conduct my own experiments. I wish to emphasize, I'm financially independent from my chemical studies, so purchasing the equipment isn't the issue. It's the legality I'm concerned about.

Is there an option to do this, to pursue amateur knowledge in chemistry, or is it illegal? Is there a country in which it isn't illegal?

Your question is basically too broad to be answered in full detail. I suggest that if you want a more detailed answer about a certain part, you should open a question for that matter separately. That being said, I'll try to summarize my own experience as a hobby chemist.

First of all, are you actually studying chemistry? IMHO, if this is within your possibilities, it is the best way to gain experience and also do your your own research. You will have supervision and the best equipment available. You will get the space to do research in your own areas mostly during your graduate studies, but here (The Netherlands) you will also get some space as an undergraduate.

Consider the following points (in no particular order):

You say that finance is not a problem, but do not underestimate the price of all the chemicals and equipment you will need to do any bit of serious chemistry. Scales, vacuum pumps, glassware, solvents, a dedicated fridge, a specially constructed cabinet to store flammables, a fume hood with explosion proof suction, etc (and I'm not even mentioning NMR, mass spectrometers, IR, all of which are available to you as student). To run any decent lab from your backyard be prepared to invest several thousands of dollars (or euros). This can be done of course a lot cheaper, but then you will be limited in your possibilities or you will have to exchange safety for money.
Almost every chemical you will ever need when doing a normal synthesis (i.e. no explosive making or pharmaceuticals) is legally available in most countries (I heard however that some states in the US are quite notorious w.r.t. hobby chemistry, like Texas). You are even able to obtain chemicals that are clearly dangerous and should not be sold to amateurs, like osmium tetroxide, sodium cyanide (not everywhere), gaseous hydrogen chloride in large 110L pressurized tanks. There are a few exceptions:

Common precursors and preprecursors to illicit drugs in large quantities. Most notably phenylacetone and derivatives, piperonal, piperidine, methylamine, benzaldehyde, nitroethane, etc.
Chemicals that can be used to manufacture explosives in large quantities. Examples include: nitromethane and ammonium nitrate.
Chemicals that are severely dangerous to the environment or can be easily abused to poison people, although oddly enough some of the most hazardous are freely available. Examples are: elemental mercury, a lot of (organo) arsenic and other heavy metal compounds, a lot of radioactive compounds, some cyanides (may be available though from mining suppliers), etc.

Note: you are able to obtain most of the above mentioned chemicals (also in larger quantities) if you start a business and have it legitimately registered to the Chamber of Commerce. This business does not actually have to do anything, it can function as a facade. It still won't work for pharmaceutical precursors or very large quantities of chemicals however, but for a hobby chemist that is mostly no problem.
Note 2: At least here in The Netherlands some pharmaceutical precursors and preprecursors can be legally owned and synthesized, they are just not available commercially because of the high change for abuse. These chemicals however (e.g. phenylacetone) can be used in many other legal synthesis and hence it might be useful to synthesize them yourself.
Be prepared to spend a lot of time on this hobby, both studying theory and doing practical work. Independently doing (serious) chemistry is not easy and you will have to rethink and adjust your setup a lot of times. Also the possibility of failure or accidents is higher without supervision. It is not entirely without risk.


How to calculate the depression in freezing point of a solution in which there are multiple solutes being added? In just about every problem I've encountered earlier, all such questions describe the process for calculating the depression in freezing point for a solution wherein there is only one solute. However, on a take-home exam, we were asked to solve the following problem:

What is the freezing point of a solution containing $\pu{0.2 mol}$ glucose and $\pu{0.2 mol}$ $\ce{KCl}$ in $\pu{250 g}$ $\ce{H2O}$?

My thinking is to use the general formula
$$\Delta T_\mathrm{f} = imK_\mathrm{f}.$$
What I'm concerned about is how to reconcile the multiple molalities, as both glucose and $\ce{KCl}$ have the same molality, but ought we add them together? Multiply them? Just a little confused.
Also, he gave us a set of “useful equations”, but on the sheet he did not specify a van 't Hoff factor for $\ce{KCl}.$ This is confusing to me because I'm pretty sure that $\ce{KCl}$ is a strong electrolyte and ought to have one. Should we just assume $i = 2$ since there are two species $(\ce{K+}$ and $\ce{Cl-})$ we are sure to get afterwards?

Freezing point depression is a colligative property, well defined by this Wikipedia article (emphasis mine):

In chemistry, colligative properties are properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present.

This means the problem is the same regardless of the number of different solvent species. You simply add them up to get the total solute concentration and calculate the freezing point depression in the same way you would if they were all the same species.  As you mentioned, you do need to keep track of dissociating species like $\ce{KCl},$ i.e. $\pu{1 mol}$ $\ce{KCl}$ dissociates into $\pu{2 mol}$ of total solutes.


Do ions still "conduct electricity" if a physical connection made by aqueous electrolytes between the anode and the cathode is nonexistent? I always had the misconception that when people say ions conduct electricity, what they actually mean is a connection is made between the cathode and the anode by the ions and electrons can flow from the cathode to the anode through this connection.
However, I recently learned that ions conduct electricity because chemical reactions are happening around the electrodes.
Taking the example of the electrolysis of sodium chloride water solution. At the cathode, $Na^+$ captures electron ($Na^+ + e^- \rightarrow Na$); at the anode, $Cl^-$ releases electron ($2Cl^- \rightarrow Cl_2 + 2e^-$). And this is because the battery takes electrons from the anode and feeds electrons to the cathode, and positively charged anode attracts anions and negatively charged cathode attracts cations. That is to say electrons do not flow from cathode to anode but are given to and taken from the ions.
I still feel confused after knowing this explanation because it seems to me that the anode and the cathode do not need to be connected by aqueous electrolytes to "conduct electricity". Suppose the cathode and the anode are placed into different solutions in different containers, anode still attracts anions and cathode still attracts cations. So that there should still be electricity flowing through the circuit.
Some people say that a "closed circuit" is required for ions to conduct electricity. If that is the case then what am I missing here?

If a cathode (negative pole) is placed alone in an ionic solution (without any anode in the solution), it will attract cations from the solution, and discharge maybe one or two or one thousand of these cations. This will last maybe 1 nanosecond, and then it will stop. The reason is that this number of atoms is extremely small, so small that the neutral atoms which are deposited on the cathode cannot be seen or weighed. And then the cathodic discharge will stop, because the whole solution has lost positive charges (cations), but has not lost any negative charges (anions). As a consequence, the whole solution becomes negatively charged. This negative charge increases with the time and prevents the cathode from bringing new electrons necessary for discharging further cations. The electrolysis will stop. The electrode is said to be "polarized".


When recovering Copper from Copper Chloride via Aluminum, why can't we use Sodium instead to make Sodium Chloride? About a week ago in my Middle School Science Class, we used an aluminum washer and some copper chloride, which we had gotten from creating circuit boards, in a closed system to recover the copper (if it was possible). We resulted in getting some copper flakes, but as a consequence the copper was replaced by aluminum in the chloride solution. So technically it is still a waste material that can't be used in everyday life, despite having recovered the copper.

So I got thinking: If Salt, or Sodium Chloride, is Sodium and Chlorine, and is much more useful than Aluminum Chloride or Copper Chloride since we need it in our everyday diets, what prevents us from using it to recover copper from chlorine?

I asked my teacher what she thought about the subject, and she explained that apparently sodium is very reactive when not bonded to anything and so is hard to come by naturally without extracting it from something else. I gathered that Aluminum, as reactive as it is, is apparently more common to find, or cheaper to manufacture, than sodium is. This was an unsure answer, as the costs and waste produced when manufacturing aluminum vs. manufacturing sodium were not looked up or explained.

However, in an earlier lesson we had learned that manufacturing aluminum produces a lot of hazardous waste, and so does not have a great impact on the environment to make. I do not know about sodium and its costs and waste production.

With that information provided, and a consideration of the environment at hand, would it be better to consistently use sodium in the recovery of copper from copper chloride than aluminum? Consider all waste produced from manufacturing and all costs to create when answering, please.

Thank you.

As your teacher mentioned, sodium is a very reactive material. She mentioned that it is hard to obtain it naturally as a result. However, this isn't quite the reason as to why sodium isn't used for recovery.

If we were to only consider the cost in energy, aluminum would be a poor choice.
The main process for production of aluminum metal requires a current of 400 kiloamperes and temperatures of 950 C. In contrast, the main process for sodium metal production requires a current of 30 kiloamperes and temperatures of 600 C.

The main reason would be because sodium is too reactive to be used. The sodium metal would react with the water and form sodium hydroxide. This reaction would be too fast and violent that you wouldn't be able to recover copper.
So your teacher was right in pointing out that sodium is too reactive.


Production of acetone from egg shells and vinegar I learned from these wiki pages that acetone can be produced from calcium acetate, which can be produced by the reaction of eggshells and vinegar. However, I couldn't find the required temperature. Would it be possible to produce acetone vapours by simply heating calcium acetate over a natural gas flame (~1500 degrees C)?

Calcium acetate decomposes starting at around 160 degrees Celsius, according to Wikipedia. So yes, it is definitely possible to do so in a gas flame.


How is antioxidant activity of carotenoids determined? I am doing some research regarding carotenoids and beta-carotene. After extracting the sample, I have not been able to find a way to measure antioxidant activity of beta-carotene. I have $\pu{0.1mM}$ of DPPH (2,2-diphenyl-1-picrylhydrazyl) solution dissolved in methanol, but I just read that you cannot use DPPH to measure beta-carotene free radical inhibition. Is this true? How should I go about doing this?

As Buck Thorn and I commented elsewhere, the antioxidant activity of carotenoids has been determined using $\ce{DPPH}$ radical ($\ce{DPPH^.}$)(Refs 1 & 2). According to Ref.1, antioxidant activities of the isolated α- and β-carotene have been analyzed in five different concentrations of 0.001, 0.005, 0.01, 0.05, and 0.1% ($w/v$) using several assays: The metal chelation, $\ce{DPPH}$ radical scavenging, and superoxide scavenging activities. All of these assays have been showed almost similar results in terms of high activity at lowest concentrations. Thus, it is not true that $\ce{DPPH}$ radical assay cannot be use to measure β-carotene free radical inhibition.

For latest information on $\ce{DPPH}$ radical assay, read Ref.3. In general, following is a typical procedure for the assay:

  The $\ce{DPPH}$ assaying of $\beta$-carotene can be performed using known antioxident (e.g., Trolox, ascorbic acid, etc.) as reference compound. The assay can be carried out in either alcoholic media (e.g., methanol or ethanol) or in an aprotic solvent (e.g., acetonitrile). To the degassed solutions ($\pu{2 mL}$) of antioxidant (with a final concentration from $0$ to $\pu{60 μM}$), were added $\pu{2 mL}$ of $\pu{140 μM}$ $\ce{DPPH^.}$ stock solution (70 μM $\ce{DPPH^.}$ final concentration). The solution was stirred at room temperature in the absence of light, and, after $\pu{60 min}$, the absorbance was measured at $\pu{517 nm}$ using an UV-vis spectrometer. The working $\ce{DPPH^.}$ concentration was calculated from the absorbance response at $\pu{517 nm}$ of the initial solution in the absence of antioxidant. The $\ce{DPPH^.}$ scavenging percentage or inhibition ($I\%$) was calculated from the measured absorbance using the following expression:

  $$I\% = \frac{A-A_0}{A_0}\times 100$$
  Where $A$ and $A_0$ represent the absorbance at a given antioxidant concentration and the absorbance in the absence of given antioxidant, respectively.

References:

Lars Müller, Kati Fröhlich, Volker Böhm, “Comparative antioxidant activities of carotenoids measured by ferric reducing antioxidant power (FRAP), ABTS bleaching assay (αTEAC), DPPH assay and peroxyl radical scavenging assay,” Food Chem. 2011, 129(1), 139−148 (https://doi.org/10.1016/j.foodchem.2011.04.045).
Surashree Sen Gupta1, Mahua Ghosh, “In Vitro Antioxidative Evaluation of $\alpha$- and $\beta$-Carotene, Isolated from Crude Palm Oil,” Journal of Analytical Methods in Chemistry 2013, Article ID 351671 (10 pages) (http://dx.doi.org/10.1155/2013/351671).
Om P. Sharma, Tej K. Bhat, “DPPH antioxidant assay revisited,” Food Chem. 2009, 113(4), 1202−1205 (https://doi.org/10.1016/j.foodchem.2008.08.008).
The most recent reference for $\ce{DPPH}$ assay: Magda M. Becker, Gilvanda S. Nunes, Danilo B. Ribeiro, Francisco E. P. S. Silva, Gaëlle Catanante, Jean-Louis Marty, “Determination of the Antioxidant Capacity of Red Fruits by Miniaturized Spectrophotometry Assays,” Journal of the Brazilian Chemical Society 2019, 30(5), 1108-1114 (http://dx.doi.org/10.21577/0103-5053.20190003).


Difference between calcination and pyrolysis What is the difference between calcination and pyrolysis?

I understand they both involve heat treatment but each under a different type of atmosphere, which leads to some sort of reduction reaction? What are the differences in underlying mechanisms?

Essentially calcination is decomposition at high temperature with inorganic reactants such as calcium carbonate in making cement.
$$\ce{CaCO3 ->[\Delta ] CaO + CO2}$$
Pyrolysis is decomposition or oxidation at high temperatures of organic materials such as cellulose in making charcoal.
$$\ce{(C6H12O6)_{n} ->[\Delta] C7H4O + C_{x}H_{y} + H2O}$$
where $\ce{ C_{x}H_{y}}$ is volatilized byproducts.


Composition of arcanite (potassium sulfate) agricultural grade. Potassium content of random drawn sample? I bought potassium sulfate from this source:
Potassium sulfate granulate
I'm using it as a potassium additive in hydroponics. The composition is listed as K2O = 50%, S = 18%. This equals about 41.5% K (50* (mass of K / mass of K2O) Calculated from mass percentages (mass of K2 / K2SO4) potassium sulfate should contain about 44.9% K and and 18% S. So based on this calculations I concluded that the potassium content is within the margin of error tolerable for fertilizer application. (About 3.4% for potassium)
However after further reading I came across an article which suggested that raw materials should be as homogeneous as possible because otherwise a sample drawn randomly could have huge variations in the content of the chemicals involved. How high could the deviation be with respect to potassium?f I think the source material is arcanite
EDIT: The security data sheet states that the purity of the granulate is guaranteed to be between 90-99%

The deviation can be in range from negligible to huge, depending on the level of nonhomogeneity and locality of the sample.
Note that routine analysis of potentially non-homogenous sources involve sophisticated sampling and sample averaging schema, so the result is representing number.


Powdering aluminum chemically? I need a fine aluminum powder for many reactions/experiments. Could I dissolve the metal in an acid and then drop it out of solution using a base, or vice versa? What would be the best chemicals/process to produce a high yield of aluminum powder?

There is no simple and affordable way to obtain fine aluminium powder chemically. As permeakra said, in the absence of a ball mill, it's just easier to buy some.

Note that a finely dispersed aluminum powder obtained commercially is always going to have low level of impurities such as $\ce{Fe, Mn, Si, Mn}$ (from the grinder's balls) and traces of wax/fats on the surface (added by manufacturer on purpose to prevent adhesion and separate particles from each other).

As for the chemical/physicochemical methods, there is a great overview included in the US5885321A patent, among which there are the following:

exploding of aluminum wire in a vacuum by a high electric current (particles' size varies a lot);
condensation of vaporized aluminum in a current of inert gas (very energy-inefficient);
decomposition of aluminium carbonyls (there is no experimental evidence for any of these to exist);
reduction of halides in solution (issue with separation of the particles from the salts);
CVD of aluminium thin films by decomposition of tertiary amine complex of aluminum hydride (complex multistep process with low yield and expensive and volatile precursors);
decomposition of the dialkylaluminum hydride in $\ce{H2}$ atmosphere at elevated temperatures (up to $\pu{260 °C}$) with a catalyst ($\ce{Ti, Zr, Hf, Ln}$) and a tertiary amine (no particle size control, questionable efficiency);
decomposition of diethylhydridoaluminum or diisobutylhydridoaluminum in diisopropyl ether or triethylamine at $\pu{90 °C}$ to $\pu{185 °C}$ with titanium isopropoxide catalyst.

The patent itself proposes

  decomposing alane-adducts in organic solvents under an inert atmosphere to provide highly uniform particles selectably sized from about $\pu{65 nm}$ to about $\pu{500 nm}$

where adducts are

  trialkyl amines and tetramethylethylenediamine, ethers and other aromatic amines;

and conditions such as

  atmospheric pressure and at temperatures as low as $\pu{50 °C}$ with xylene solvent, ... , toluene, dioxane, and tetramethylethylenediamine.

As a catalyst, titanium halide, amide and alkoxide are used. Authors also provide a complete table of the systems tested:

References

Higa, K. T.; Johnson, C. E.; Hollins, R. A. Preparation of Fine Aluminum Powders by Solution Methods. US5885321A, March 23, 1999. (PDF)


Does oil break down condoms? It seems that every website on sexual health advises against using oil-based lubricants with condoms. It is claimed that "oil breaks down latex". One source claimed that a latex condom completely breaks down in only 60 seconds. It made me curious, so I made an experiment.

I took a piece of rubber latex condom and soaked it into regular canola oil I found in the kitchen. I checked the condom after 1 minute, nothing changed. So I let it soak for about 5 minutes more, and then 5 hours more, still nothing. It was able to hold a large amount of water without leaking or breaking.

So I am wondering, is it really true that oil degrades latex? What sort of chemical reaction is supposed to happen? What properties of the latex material and the oil influence this reaction?

First off, may I say that I applaud your decision to test this through an experiment. It's rare to see that than I would like.

Now, on to the matter at hand. It's fairly well known from industrial chemistry that non-polar solvents degrade latex quite heavily.

I work with latex seals a lot, and the hexanes we use routinely break the seals down in under a day. Of course, if you're lubricating your condoms with hexanes, you're a) an idiot or b) absolutely insane.

A paper I managed to find suggests that there really isn't too much direct data on condoms, and it muses that the warnings might have arisen from industry, where nonpolar solvents decidedly do degrade latex.

To find out, they did a burst experiment with condoms that had been treated with various oils. Glycerol and Vaseline-treated condoms showed a very, very minor decrease in strength, while mineral oil/baby oil-treated ones burst at less than 10% of the volume of an untreated condom.

They also found that 10 month-old condoms have half the burst volume of 1-month old ones, so you could argue that using 1-month-old condoms that have been slathered in Vaseline is still much safer than using older ones.

As for the actual chemistry of the weakening, I honestly don't know. If I were to hazard a guess, I would note that the latex looks like a bunch of ethylenes glued together,

so my guess would be that the solvents get between the chains and force them apart, weakening them. For this to happen, the solvent must be nonpolar, but still small enough to slip between the chains of the polymer.

That's probably why vaseline and canola oil don't have much of an effect---they're just too big to fit between the chains. Again though, I don't know for sure, so don't quote me on this last paragraph.


1,2-dibromopropane reaction with OH and NaNH2 We have the reaction of the alkyl halide $\ce{CH3-CHBr-CH2Br}$ with 1) $\ce{OH-}$ 2) $\ce{NaNH2}$ The answer is propyne.

I am unable to understand why we have performed an elimination reaction here. Why can't it be a substitution reaction instead? Also, how do we decide which two carbon atoms would lose their hydrogen atoms? (as in elimination, there are two choices for losing the $\ce{H}$ atom - from left and right carbon respectively)

Substitution, meaning displacement reactions are Sn1 or Sn2. These reactions are dependent on the leaving group, nucleophilicity vs base of attacking group, and solvent. Currently you have two good leaving groups (Br) and strong bases. Strong bases (have localized electron density, no resonance etc.) favor proton attack due to partial + charge on H, no spatial constraints. So the protons are attacked preferentially. It would be much easier sterically and kinetically compared to either Sn2 (recall l, these also requires specific orientation of attack) or Sn1. These reactions will occur but at much much small amounts, such that on time of reaction they are minimal.
The reason for the -yne product is because there are two adjacent leaving groups. Once the first comes off (which is aided by partial resonance by the other Br), the electron densities around the double bond, the large radii of Br, along with the positive charge character on Br due to resonance, it easier to remove the next Br, but this requires a strong base. The amine is definitely strong enough, but so is the OH ( less so due to higher electronegativity of Oxygen which doesn't want to share its electrons so to speak )
Because you end up with a -yne it doesn't matter which Br is removed first, regarding the final regiostereochemistry.


Does deprotonation of a hexaaquaion continue until neutral charge I learned that deprotonation is the process in which a proton is lost from ligand $\ce{H2O}$ molecules surrounding a central metal ion (I am referring to hexaaqua ions). Also, the process continues until the charge of the complex ion is neutralized.

My question:
Why does it only continue until the charge is neutralized, why don't all the six ligand water molecules lose a proton?

Let's look at an example: Hexaaqua aluminium, or $\ce{[Al(OH2)6]^{3+}}$

It is acidic and will undergo deprotonation of a ligand water molecule spontaneously:
$$ \ce{[Al(OH2)6]^{3+} <=> H+ + [Al(OH)(OH2)5]^{2+}} $$

You can even get it to deprotonate twice:
$$ \ce{[Al(OH)(OH2)5]^{2+} <=> H+ + [Al(OH)2(OH2)4]^{+}} $$

But, as indicated, both reactions are equilibrium reactions. So if the concentration of protons in solution is high enough, the hydroxo ligands will be reprotonated and become neutral aqua ligands.

If you add some hydroxide to the solution, you can precipitate aluminium hydroxide, since it is not soluble in water. While this might be a driving force for the deprotonation, it doesn't seem to be the case here.

So we have shown that the process does not necessarily continue until the charge of the complex ion is neutralized.

Now, concerning your question why this doesn't happen until all aqua ligands have lost a proton:

Again, a part of the solution is the equilibrium idea brought forward in the upper part. But there's another way to think about it: If you have a negatively charged ion, you also need a positively charged counterion. The only counterion that exists in the solution of water and the aluminium compounds is the freshly abstracted proton, which happily will bind back to the hydroxoligand.

There are also $\pi$-donor effects that are stronger in the hydroxo than the aqua ligand. So it might be favourable to have no more than three hydroxo ligands in an octahedral complex so no two $\pi$-donors are across each other.


How many transition elements are there? I searched the internet and most of sites say there are 38 transition metals. But our textbook says that group 12 or the group of zinc can not be counted as transition metals, which already puts the number to 37.
Moreover, Sc, La and Ac also can't be counted as transition metals. That puts the number to 34. Y being in the same group as Sc should also be cancelled as transition metals making the number 33.
These are all my thoughts and haven't been varied by any reliable sources. And also it's based on our textbook. I would like to know what is wrong with these thoughts.
Note: The book considers thorium a d block element. Thus, according to the book the number of d block elements is 41.

Shakespeare apparently said "What’s in a name? That which we call a rose, by any other word would smell as sweet." Rule no. 1, don't trust the internet until and unless you have some idea about the credentials of the website (and this comes with experience).

The usage of the term transition metal is nothing but historical. You can stick to IUPACs definition (https://goldbook.iupac.org/terms/view/T06456)

  An element whose atom has an incomplete d sub-shell, or which can give rise to cations with an incomplete d sub-shell.

By that zinc should not be a transition element. The definition has nothing to do with paired electrons as you wrote in the comments.

Look at the history of word, this is from 18th century from the unabridged OED.

  transition element  n. Chemistry  [after Russian perexodnyj èlement (see transitional element n. at transitional adj. and n. Compounds)] (originally, now historical) any of the nine metallic elements forming group VIII of Mendeleev's periodic table; (now) any of a large class of metallic elements occupying the central block of the periodic table (groups IV b–VIII, I b, and II b, or 4–12, the d-block), which are characterized by partly filled d orbitals, commonly show variable valency and an ability to form coordination compounds, and form many coloured compounds; often extended to include also those elements having partly filled f orbitals (the lanthanides and actinides).


Why isn't 4-hydroxybenzoic acid formed along with salicylic acid in Kolbe's reaction? I've read that in Kolbe's reaction of phenol, i.e. reaction of phenol $(\ce {C6H5OH})$ with $\ce{NaOH}$ forming sodium phenoxide $(\ce {C6H5O^{-}Na^{+}})$, followed by reaction with $\ce{CO_2}$ and acidification; salicylic acid (2-hydroxybenzoic acid ) is formed, but when $\ce{KOH}$ is used instead of $\ce{NaOH}$, 4-hydroxybenzoic acid is formed.

I don't understand why two different products are formed in the same reaction using very similar bases. Moreover, there is electrophilic aromatic substitution on the phenoxide ion in both reactions and  $\ce{O^{-}}$ being an ortho-para directing group must form both products in both the reactions.

I don't understand why two different products are formed in the same reaction using very similar bases.

If the bases make a a difference, there must be a difference between them.
If the anion is the same, the cation must make the difference.

The difference between $\ce{Na+}$ and $\ce{K+}$ lies in the ionic radii of 0.95 and 1.33 $\mathring{A}$, respectively.
It is believed that $\ce{Na+}$ serves as an anchor and guides $\ce{CO2}$ into the ortho position during the formation of the Wheland intermediate. In the latter, $\ce{Na+}$ is chelated between the oxygen atoms of carboxylate and the phenolate. Apparently, this is very effective.
This does not mean that the phenolate doesn't attack $\ce{CO2}$ via C-4 at all. However, the formation of a Wheland intermediate (= the rate-determining step) is normally (and definitely in the case of the para attack) reversible.
It is open to question if full reversibility can be claimed for the $\ce{Na+}$ directed ortho attack under modest reaction conditions.
It is however likely that the back reaction (= cleavage Wheland intermediate) will stronger come into play at elevated temperatures. Consequently, some 4-hydroxybenzoic acid will be formed then.
With the bigger $\ce{K+}$, the anchoring and chelating effect is seemingly not possible, $\ce{K+}$ doesn't fit and the para direction effect of the phenolate prevails.


Predicting the solution with the highest boiling point I am facing problem with the following question:

Predict which solution will have the highest boiling point in increasing order?

$1~\%$ solution of $\ce{C6H12O6}$

$1~\%$ solution of $\ce{NaCl}$

$1~\%$ solution of $\ce{ZnSO4}$

$1~\%$ solution of  $\ce{(NH2)2CO}$

I know that more is the van ’t Hoff factor more will be the elevation in boiling point and hence the solution will have more boiling point. Hence 1. should have the lowest boiling point followed by 4. But I am not sure about options 2. and 3.

The first aspect to consider is the van't Hoff factor. Thus #1 is first and #4 is last. Since both #2 and #3 have the same factor of 2, we must consider other properties.

A major part of BP is how many molecules of solvent are exposed on the surface layer. Therefore, the ions with the greater diameter should take up more surface area, decreasing the number of solvent molecules available to 'escape'.. decreasing the vapor pressure and increasing the BP elevation.
My guess is that the polyatomic ion is quite large in comparison and the deciding factor.

from lowest BP to highest
 #1,#2, #3, #4


Can sodium dithionite reduce the nitro group of 2-nitrobenzoic acid? Can sodium dithionite reduce o-nitrobenzoic acid to anthranilic acid under basic conditions with mild reflux?

If so, how could I separate anthranilic acid from the reaction mixture?

Yes I believe it can. Dithionite is a reliable reducing agent for aromatic nitro groups. The reaction of dithionite produces an acid solution which will need to be neutralised before isolation. Anthranilic acid is not very water soluble so an extraction with e.g. chloroform should separate the product cleanly. Tetrahedron Lett. 2013, 54 (21), 2600–2603 discusses dithionite reduction of o-nitrobenzamides and refers to other papers that have used dithionite on similar substrates.


How to test a sample of cigarette tobacco for its nicotine content? I was wondering how one would go about testing samples of different cigarette tobaccos for their nicotine content. Is there a kit of some kind available online to do this?

Unfortunately, nicotine analysis is prohibitively complicated and expensive for a home kit. The presence of other alkaloids similar to nicotine pretty much necessitates that somewhat expensive equipment (i.e. a gas chromatograph) is used to first separate out the nicotine prior to analysis (typically by mass spectroscopy).

The only practical means for a layperson to test tobacco products for nicotine content is to out-source the work to a GC-MS equipped laboratory that routinely performs such analysis.


What role does glycerin or sugar play in bubble formula? From: http://bubbleblowers.com/homemade.html

  Exploratorium Bubble Formula
  from the Exploratorium web site
  2/3 cup Joy dishwashing soap
  1 gallon water
  2 to 3 tablespoons of glycerin (available in pharmacies or chemical supply houses)

  Cyndi's Bubble Recipes from the Nathan's Wish web site 1/2 cup of
  dishwashing liquid (Dawn or Joy) 2 cups of water 2 teaspoons of sugar

What role does glycerin or sugar play in bubble formula?  Are sugar or glycerin better than each other in some way w.r.t this formula?

From:

https://www.scientificamerican.com/article/bring-science-home-best-bubbles/

  The solution with only water and detergent probably made smaller,
  shorter-lived bubbles compared to the solutions with glycerin or corn
  syrup. Bubbles burst when the layer of water molecules between the
  detergent molecules evaporates. But Glycerin and corn syrup form weak
  bonds with the water molecules and slow down the evaporation process,
  thus improving the life span and durability of the bubble. Glycerin
  makes stronger, longer-lasting bubbles, but corn syrup is often
  substituted in bubble solutions because it is cheaper. Can you find
  other substances around the house that can be added to water to make a
  bubble solution? If you try products such as shampoo or liquid hand
  soap, you can check their ingredients to see what might be helping to
  make the bubbles form.


Is it possible to have a ferrocenol? I'm specifically talking about having a hydroxy- group on the ferrocene ring.

Is it simply impossible because the ring is anionic?
I haven't seen any hydroxyferrocene molecules available online. (unless they are a methanol group)

Yes we can have hydroxyferrocene.  See here for a brief summary description of possible synthesis and a nucleophile substitution reaction forming alkoxyferrocenes.


Effect of temperature in the free radical polymerization of acrylic acid I am working with the model of a free radical polymerization process in a tubular reactor. In the simulations, I observe that by keeping the feed flow of the initiator (sodium persulfate) and the monomer (acrylic acid) constant and just by increasing the jacket temperature, the concentration of the monomer (for example of the product) increases, even though the reaction is exothermic.
I have used the method of moments to model the chain length distribution of polymers. Change of viscosity has not been modeled.

My question is: Is the increase in the monmoer concentration by the increase of temperature, an expected behavior in polymerization? If so, do you know some literature that I can learn more about this?

Yes, this is expected.

With higher temperature, there is a larger concentration of radicals in the polymerization mixture, and initially the polymerization goes faster than at lower temperatures.  This leads to a faster increase in viscosity, which slows the ability of molecules to diffuse in solution.  This means it is more difficult for unreacted monomer to find growing radicals, and for unreacted initiator to find monomer.  Radicals are highly reactive so when they can't react the have a habit of engaging in termination reactions which are nonproductive regarding conversion.

This is normally called autoacceleration, and is fairly well described in Odian's 'Principles of Polymerization'.  Generally, you do not need to heat radical polymerizations beyond the bare minimum to decompose your thermal initiator.


Find molar concentration of sodium carbonate dissolved in water The question is

  A solution of $\ce{Na2CO3}$ is made by dissolving $5.3~\mathrm{g}$ of the powder in enough water to make $250~\mathrm{mL}$ of solution. Find the molar concentration.

I have done the following but I don't know why it's not working.

  Molar mass: $106~\mathrm{g}$

  $5.3~\mathrm{g} / 106 = 0.05\ \mathrm{mol}$

  $0.05\ \mathrm{mol} / 250\ \mathrm{mL} = 0.0002\ \mathrm{mL}$

However, the answer is $0.2\ \mathrm{mol/L}$ but I got $0.0002$. It does become $0.2$ when I multiply by $1000$ but the answer is already in $\mathrm{mL}$. How do I arrive at the correct answer?

Use units! Consistently, properly, all the time!

You have units all over the place sometimes fitting the quantity you are calculating, most often not. If you approached the problem with correct use of units, there would be no question.

First, the molar mass.

$$M(\ce{Na2CO3}) = 105.99~\mathbf{\frac{g}{mol}}$$

Grams are not equal to grams per mole. Then, let’s do the rest without losing or inventing units as we go along. (I bolded the units for you.)

$$n(\ce{Na2CO3}) = \frac{m(\ce{Na2CO3})}{M(\ce{Na2CO3})} = \frac{5.3~\mathbf{g}}{105.99~\mathbf{\frac{g}{mol}}} = 0.05~\mathbf{mol}$$

$$c(\ce{Na2CO3}) = \frac{n(\ce{Na2CO3})}{V} = \frac{0.05~\mathbf{mol}}{250~\mathbf{ml}} = \frac{0.05~\mathbf{mol}}{0.250~\mathbf{l}} = 0.2~\mathbf{\frac{mol}{l}}$$

Different versions of jumbling around the powers of 10 (especially in the millilitre part) will get you the same result if done correctly. It really bears repeating.

Use units! Consistently, properly, all the time!


Binary Molecular Compounds of Nonmetals How does one determine the formula for two non metal elements?

Such as Nitrogen and Oxygen.

I look at the charges with Nitrogen having -3 and Oxygen having -2. How would I get to zero?

How do you build a formula based on those two elements.

Why does it equal $\ce{N2O5}$         ?

Nonmetals form covalent bonds with other nonmetals, not ionic bonds. This means you can't use ionic bonding rules to figure out what the structure will be. In fact, there is no way to predict the structure - all you could do is propose Lewis structures for various combinations to predict what it could be.

This is also why there is a special set of nomenclature rules for molecular compounds - it's not enough to just give the names of the elements, you also have to specify the number of each atom.

In your example, if all you are given is N and O, you would have no way of predicting the formula. It could be any one of maybe half a dozen compounds (maybe more). To end up with
$\ce{N2O5}$, you would need to know the name of the compound first.

For binary molecular compounds, the naming rules are pretty easy. First, you need to know the prefixes:

1: mono-
2: di-
3: tri-
4: tetra-
5: penta-
6: hexa-
7: hepta-
8: octa-
9: nona-
10: deca-

Then, for each element (starting with the least electronegative one - the one furthest to the left on the periodic table) you just use the prefix in front of the element name. $\ce{N2O5}$ would be called dinitrogen pentoxide.

There are two small details. The first is that we don't use "mono" when there is only one atom of the first element. This means that $\ce{CO}$ is carbon monoxide rather than monocarbon monoxide. The second is that when an element starts with a vowel and the prefix ends in a vowel, we drop the prefix's ending vowel. This means that we say pentoxide instead of pentaoxide.

To find the formula given the name, you just reverse this process.

$\ce{NO}$ - nitrogen monoxide

$\ce{N2O}$ - dinitrogen monoxide

... you get the idea.


Cis and trans substituents on rings different stereodescriptors The attached two compounds are diastereomers. They can be differentiated by cis/trans descriptors. Is there another way of describing them as cis/trans seems to be outdated?

EDIT: How to determine the r/s configuration of the pseudochiral C atoms in this case? The usual rule is that R precedes S, but here two identical groups are bounded to the same pseudochiral atom. How to assign r/s in this case?

Edit: strongly related to your previous question How to determine the configuration of C atom bounded to two groups with same configuration? and Yolo123's question that was suggested to you Number of diastereomer pairs of 1,3-dichloro-1,2,3-triphenylpropane where the nomenclature was explained in detail by Jerrepierre. The same situation applies here.

Although the cis/trans nomenclature might seem 'outdated', it is still by far the easiest way to describe disubstituted ring systems, and is commonplace in both conversation between organic chemists, and in the organic literature when such relationships are introduced into a molecule.

As these ring systems possess some symmetry, the (R) and (S) descriptions commonly used are not applicable, but (r) and (s) may be assigned to carbons ('pseudo asymmetric'), based on the IUPAC 2013 recommendations for describing achiral cyclic compounds.

I've included the assignments for the two molecules you presented in the image below. In essence (taken from March's Organic Chemistry):

  The small letters used for the psuedoasymmetric center are assigned using established rules. [...] The descriptors “r” and “s” are used to denote such centers; they are assigned in accordance with Sequence Rule 5, taking into consideration that “(R)” has precedence over “(S)” in the order of priority.

This additional example, taken directly from the IUPAC blue book 2013 may clarify the reasoning behind assignment of (r) and (s):

Its worth noting that this nomenclature only applies when you can find symmetry, in many cases this is non-existent, and normal (R) and (S) nomenclature can be applied.


Entropy change of surroundings and system The entropy change of the surroundings can be calculated by the equation $dS_{sur}=\frac{dq}{T_{sur}}$ regardless of the path (irreversible or reversible).

The argument is that because the surroundings may be approximated as either constant volume or constant pressure, the heat absorbed/released by it is equal to the internal energy or enthalpy change, respectively. As both are state functions, it does not matter if the path taken was reversible or not, the changes would be the same for both, so $dU_{sur,rev}=dU_{sur,irrev}$ and $dH_{sur,rev}=dH_{sur,irrev}$. However, $dH=dq$ for constant pressure, and as per definition $dS=\frac{dq_{rev}}{T}$, the initial equation follows.

This is somewhat the way my book derives the formula. It seems to be inconsistent to me because it equals $\frac{dq_{rev}}{T}=\frac{dq_{irrev}}{T}$, which some pages after the book says that is not the case and uses the fact that $dq_{rev}\geq dq_{irrev}$ to derive the Clausius inequality.
I do understand the Clausius inequality and its derivation, however I do not understand why is it that it is not used to calculate the entropy change of the surrounds, only the system. The arguments for its derivation seem to apply to the first case as well.

Similar question that has no answer: Entropy change of surroundings

In evaluating the entropy change of the surroundings, you need to first look exclusively at its initial state and its final state, before and following the irreversible process.  The only difference between the initial state and the final state is a difference in enthalpy of the surroundings, equal to amount of irreversible heat that flowed to the surroundings.  To get the entropy change for the surroundings, you next need to separate it entirely from the original system, and subject it to a new reversible process while in contact with its own new surroundings (i.e., a second set of surroundings).  No matter what new (alternative) reversible process you devise, you will always end up with the same enthalpy change and same beginning and ending pressures, and its entropy change will always be the same, and equal to the irreversible heat it received in the original process divided by its absolute temperature.

On the other hand, for the original system, if you follow the same procedure (separating it from the original surroundings and putting it into contact with a third surroundings), the reversible process you devise will give you a different value for the integral of $dq/T_{boundary}$ than the integral of $dq/T_{boundary}$ for the original irreversible process.


Where do the electrons come from to make the carbon stable during bombardment of alpha particles on beryllium When beryllium is bombarded with alpha particles, where do the electrons come from to make the carbon stable?

Source: https://en.wikipedia.org/wiki/Triple-alpha_process

When beryllium is bombarded with alpha particles, where do the electrons come from to make the carbon stable?

Nowhere.  This is a nuclear reaction, not a chemical one.  Charge is conserved and no extra electrons were supplied.

The carbon nucleus is stable with respect to nuclear decay.

Chemical stability relates only to the behavior of electrons with respect to atoms.  It has no place in this nuclear  reaction.

Eventually the carbon nucleus may attract additional electrons from some external source, but that's nothing to do with stability of the nucleus.  The nucleus does not require any electrons to be stable.


Polymers with no molecular mass distribution? What sort of ideal conditions are required - theoretically - to conduct polymerization reactions to make all the macromolecules end up with exactly the same molecular mass, just like the products in normal chemical reactions? How do living organisms manage to synthesize macromolecules, for instance proteins, different than humans do in the lab?

Living polymerizations can afford nearly zero polymer polydispersivity.  You can go anion (Kraton thermoplastic elastomers - and tell us about the one silicon atom in each polymer molecule), free radical (TEMPO or PROXYL capping of the free radical end, thermally activated), redox (copper halide), olefin metathesis polymerization (ROMP and ADMET), and other clevernesses.

Proteins are made from a DNA template via RNA, one AA residue at a time. There is no dispersivity.  Post-synthesis modification, e.g., glycosylation in erythropoietin, can be polydisperse.


How to list elements according to their radius? For example Ag, Cd, Au Can you check the accuracy of these sentences I've created? I've asked a similar question before, but now I want to find out the real rule. Pleas tell me if my interpretation is correct: Cd, Ag , Au (increasing ray order). This happens because although we know that period and group effect the radius length, most important one is the group. So firstly we should list the elements according to their group and then list them in the same group... For example Zn, Cd (same group), then Pd, Pt, (same group), then Ph, ir (same group). From our lesson we know that from Cd to Hg the ray doesn't change but I think when we have to list them we should consider that it grows a little. Are these statements right? If not, please help me clear the concept. I'm not a native speaker of english, so I may not express correctly in some occasions

This is the radius general trend on the periodic table:

The main reason that explains how it changes is the nuclear effective charge experimented by the electrons when atracted by the nucleos. Briefly speaking, the more electrons you add as you go down a group (in different energy levels), the lower the nuclear effective charge that the electrons will experiment, so the electron denisity will be less "compresed", resulting in a larger radius.
Going from the right to the left, you add one by one, a new electron that experiments almost the same nuclear effective charge, so it will be atracted with a similar force, and also, you are adding another proton to the nucleus, so the overall atraction will be higher, the electron density will decrease and the atomic radius will decrease.


How do I get irreducible representations? I'm currently trying to do a problem where it asks me to "show that the irreducible representations spanned by the atomic x-, y-, and z- coordinates is independent of the choice of local coordinate system. Use the diagrams below."

I know the general process is to determine the irreps for the separate coordinates. So, when I attempted to do the first one, by doing the symmetry operations to the x vectors from the D3h character table. However, it did not work as well. So now I don't know where to go with it. Please help and thank you.

Have a look at the method given in this answer. In your example subject each vector on each atom to the operation in the point group, e.g rotation, reflection etc. Count 0 if the vector is moved else count 1 if it remains unchanged and -1 if it is unchanged but just points in the other direction. Now you have the reducible representation. Use the method suggested above to get the irreps.

If the angles are not 180 then it is necessary to use a rotation matrix and determine the trace (sum of diagonal elements). (This is explained in a good symmetry text book such as R. Carter, 'Molecular Symmetry and Group Theory'. Of course this has all been worked out and so it is necessary only to use a formula for $C_n$ and $S_n$ operations.

These are  $\displaystyle \chi(C_n)=1+2\cos\left(\frac{2\pi}{n}\right) $ for $n$ fold rotations and $\displaystyle \chi(S_n)=-1+2\cos\left(\frac{2\pi}{n}\right)  $ for rotation reflections, e.g. $\chi(S_3)=-2\;; \chi(C_5)=(1+\sqrt{5})/2$,

Excepting rotation, for  each atom each operation contributes an amount as shown below.

$$\begin{array}{ccc}
\text{operation} & E & \sigma & i  \\
\chi(n) & 3 & 1 & -3
\end{array}$$