Why do lipids in a cell hydrolyze into glycerol and fatty acids? Most Lipids in a cell have ester bonds (made up of glycerol and fatty acids).

I have learned in my chemistry class that we can hydrolyze an ester by simply adding water to it.

$$\ce{CH3COOCH2CH3 + H2O <=>[H^+ (aq)] CH3COOH + EtOH}$$

Then why do lipids which  have an ester bond don't hydrolyze inside a cell which is 70 percent  water? While searching throught web I have found this link which explains of different ways to hydrolyze an ester. In that it explains about how to hydrolyze esters using dilute acids like HCl. In our stomach HCl is present so logically HCl should destroy the ester in stomach lining (which are made up of cells) instantaneonusly, but this is not happening. Why?

$%edit$

Organisms control rates of chemical reactions in two ways. First, by controlling the conditions, and second, by controlling the location of reagents.

Conditions

Hydrolysis of esters is very slow at normal physiological conditions: 37 oC and pH 7.4. A mixture of water and simple esters like ethyl acetate remains stable for years. At higher or lower pH, the reaction becomes faster. As you mention, the HCl in the stomach would definitely increase the rate of hydrolysis. The reaction is also accelerated by enzymes, and organisms use enzymes for all intentional ester hydrolysis.

Location

For a reaction to occur, the water molecule and the acid catalyst needs to get to the ester. In cell membranes, that ester functional group is not on the surface of the membrance; the ionic phosphate part is. The ester group is just below the surface, in the hydrophobic interior of the membrane. Neither water, nor acid can penetrate into this hydrophobic region.

In the stomach, there is an extra layer of protection: mucus, which prevents the stomach acid from reaching the cells.


Variation of angle in GeX2 molecules where X is an halogen I'm wondering what is the exact reasoning to explain why in $\ce{GeX_2}$ the angle $\widehat{XGeX}$ is smaller when $\ce{X}=\ce{F}$ than when $\ce{X}=\ce{I}$.

I ask this question because I had an exams some days ago and I failed in this question and for me my reasoning is not false.

What I said is that

  From $\ce{F}$ to $\ce{I}$ the difference of
  electronegativity between the germanium and the halogen is
  decreasing. Then the electrons in the $\ce{Ge-X}$ bond are more near
  $\ce{X}$ when $\ce{X}=\ce{F}$ than when $\ce{X}=\ce{I}$. Then the two
  bonds repulse themselves more for $\ce{GeI_2}$ than for $\ce{GeF_2}$ and so the angle increase from $\ce{F}$ to
  $\ce{I}$.

Why my reasoning is absolutely false, and what could be the perfect one?

As for all atoms of period 3 and higher, we can consider germanium to be essentially unhybridised and the bonds being formed entirely by $\unicode[Times]{x3c3}_\mathrm{p-p}$ overlaps. Therefore, we expect a perfect $90^\circ$ angle for all three compounds.

However, while fluorine is small, the size of the halogens increases from fluorine to iodine. The larger the halogen the more space it requires. Since the central atom remains the same size (and, let’s approximate, is roughly the size of bromine) it cannot offer more space for a strict $90^\circ$ angle without lengthening (= weakening) the bonds which would be unfavourable. It is less unfavourable to widen the angle to generate more space. This requires the s-orbital to hybridise slightly (less favourable) but keeps the desired shorter $\ce{Ge-X}$ bond length (more favourable than anything else in this consideration). Not the bonding electrons repulse each other, but the steric stress of the halogen’s lone pairs causes them to move apart. The same effect can be observed in $\ce{NH3}$ versus $\ce{PH3}$ or $\ce{PH3}$ versus $\ce{PF3}$.

Note the following: If you have two bonds from a central atom which are each described by a bonding and an antibonding orbital, then these two sets of orbitals will always be orthogonal to each other (the integral of overlap is $0$). Thus they will be indifferent to each other, not repulsive.


Why are the two CH2 groups in allene perpendicular to each other? Allene, $\ce{H2C=C=CH2}$, has two $\ce{=CH2}$ groups perpendicular to each other. This is the same as hydrogen peroxide's $\ce{H}$ atoms. Is there any reason why they adopt the perpendicular arrangement, when the VSEPR model allows both the perpendicular and planar arrangements for the groups on the ends? I don't think it's a steric effect, because the hydrogen atoms/methylene groups are facing opposite directions and far apart.

No, VSEPR theory does not allow for a coplanar arrangement. Let us consider allene and its orbitals:

Source. Green indicates $\mathrm{p}$ orbitals; blue $\mathrm{sp^2}$; and red $\mathrm{s}$.

Hybridization

The terminal carbons are $\mathrm{sp^2}$ hybridized, and form three $\sigma$-bonds each. This means that each terminal carbon has one unhybridized $\mathrm{p}$-orbital.
The central carbon atom is $\mathrm{sp}$ hybridized, and forms two $\sigma$-bonds. This means it has two unhybridized $\mathrm{p}$-orbitals. For our sake, we will call these two orbitals $\mathrm{p_x}$ and $\mathrm{p_y}$. These orbitals are perpendicular to one another.

$\mathrm{\pi}$-Bonds

A $\pi$-bond is the result of the sharing of two $\mathrm{p}$-electrons, and it is particularly important for us to consider them as belonging to a particular region of space. In order for the central carbon's $\mathrm{p_x}$-electron to form a $\pi$-bond with one of the terminal carbons, it must be with another $\mathrm{p_x}$-electron. If the two did not share the same spacial orientation no bond could form. Since the $\mathrm{x}$-dimension is occupied by the electrons of a $\pi$-bond between the central carbon and one of the terminal carbons, the only way for the central carbon to form an additional $\pi$-bond is by sharing its $\mathrm{p_y}$-electron with the other terminal carbon. Hence, the two terminal carbons are necessarily perpendicular to one another.


Textbook problem with molar mass of carbon dioxide $\ce{K2CO3}$ and $\ce{HCl}$ react to produce $\ce{CO2}$. In an empty flask ( m = $\pu{85.431 g}$ ) we put in the produced gas and the mass of the flask is now $\pu{85.510 g}$. After that we fill the flask with water. Volume of the flask filled with water is $\pu{122 ml}$. Pressure is $\pu{101325 Pa}$, temperature of air is $\pu{300 K}$. Density of water is $\pu{0.99893 g cm-3}$, and the molar mass of air is $\pu{28.8 g mol-1}$. Find the molar mass of carbon dioxide.

I have calculated the mass of $\ce{CO2}$, using the mass of the flask filled with gas and subtracting the mass of the empty flask. I have figured out that somehow through $pV = nRT$ I am supposed to get to $n$ and take into consideration that $n$ is not only the number of moles of $\ce{CO2}$ but also of $\ce{H2O}$ and air.
What I don't understand is the volume of water used to fill the flask. In that volume do we only have $\ce{H2O}$ or is that volume of 122 mL also the volume of air + $\ce{CO2}$ + $\ce{H2O}$?
Any hint is appreciated.

The problem is that, when the flask is empty, it weighs $85.431\ \mathrm g$. But this weight is the weight of the empty flask plus the air included. So the first thing to do is to calculate the weight of the air, and the corresponding number of moles of air in the flask before adding $\ce{CO2}$. The volume of the air is the same as the volume of water when it is filled with water, namely $122\ \mathrm{mL}$.
The amount of air is $$n = pV/RT = \frac{122\times10^{-6}\ \mathrm{m^3}\times101325\ \mathrm{Pa}}{8.316\times300\ \mathrm{J/mol}} = 4.956\times10^{-3}\ \ \mathrm{mol}$$ The mass of this air is : $$m = 4.956\times10^{-3}\ \mathrm{mol}\times28.8\ \mathrm{g/mol} = 0.1427\ \mathrm g$$ So the mass $m_0$ of the empty flask without air is : $$m_0 = 85.431\ \mathrm g - 0.1427\ \mathrm g = 85.2883\ \mathrm g$$
A s consequence, the mass of the $\ce{CO2}$ in the flask is : $$m(\ce{CO2}) = 85.510\ \mathrm g - 85.2883\ \mathrm g = 0.2227\ \mathrm g$$
The molar mass of this $\ce{CO2}$ is: $$M = \frac{m}{n} = \frac{0.2227\ \mathrm g}{4.956\times10^{-3}\ \mathrm{mol}}= 44.8\ \mathrm{g/mol}$$
This is nearly the expected value for the molar mass of $\ce{CO2}$. If you take into account the fact that the volume $122\ \mathrm{mL}$ is known with a precision $\pm1\ \mathrm{mL}$, or $\pm1\ \%$, the final result will be defined with an uncertainty of $\pm1\ \mathrm{g/mol}$. It is: $$M(\ce{CO2}) = 44.8\ \mathrm{g/mol} \pm 1\ \mathrm{g/mol}$$


If a gas absorbs heat, then how can its temperature remain constant? My book has this line:

  In an isothermal expansion, the gas absorbs heat and does work

But, if a gas absorbs heat, then how can its temperature remain constant? (as the process is isothermal)

From the first law, $\Delta U=Q-W$.  If the process is isothermal, then $\Delta U=0$ (assuming an ideal gas), so Q = W.  The heat energy is consumed by your gas doing work, so there is no change in internal energy (temperature).


Do we lose significant figures with digital instruments? With digital scales, we get an answer to the hundredths or thousandths place, but we do not have the ability to estimate "between the lines".  Is the real significance of a measurement different when it comes from a digital scale, or thermometer or other device than when it is read with the possibility of visual interpolation?

As an example, I might have a digital scale that gives a reading of 25.36 grams.  It could be anywhere between 25.355 and 25.365 grams.  If I read a sliding scale to 25.36 grams, the 6 is an estimate and the true value could range by a few hundredths of a gram while the digital scale at most would have a range of +/- 5 thousandths of a gram.

I realize that you won't have a sliding scale that can match a digital scale anyway, but the same could occur with digital thermometers reading to 0.1 degrees and indicating a range of true value of +/-0.05 while a reported value to the nearest 0.1 degree on a thermometer could be several tenths of a degree off.   So is there any acknowledgement that the reported values have different meanings with regard to significance?

As far as I know the visual interpolation of analogous measurements is bad scientific practice anyway.
We assume that rounding errors work in both directions and compensate each other in the long run. On the other hand visual interpolation by a human can introduce systematic errors.

But putting this discussion away one has to note that even if you visually interpolate, you still digitalise your data (just with a finer grid). By digitalising I do not mean that you write them into a computer, but the abstract mapping of continous data points of $\mathbb{R}$ into a finite subset.

In the end it just is a question of how fine the grid is you are measuring with. And yes, if you don't introduce systematic errors a visual interpolation makes your grid a little bit finer.
On the other hand switching to electronic devices usually gives you orders of magnitude better results.

Some examples:

Let's assume that someone wants to measure lengths of $\mu$m with a normal ruler that has 1 mm spacing.
It is intuitively clear that this won't work.
(Even if you can interpolate.)

Now let's look at another example that was brought up in the comments by @Zhe. The problem that every number in your computer is an element of a finite set. This introduces rounding errors, when saving real numbers (real as in $\mathbb{R}$).
If you execute the following python code or something similar in your language of choice, you will see this rounding error.

import sys
print(sys.float_info.epsilon)
# for python2:
# print sys.float_info.epsilon

On my 64bit machine this is in the order of $10^{-16}$. The standard deviation of the meter is in the order of $10^{-11}$ (wikipedia).
Floating point noise will rarely be a problem, when dealing with experimental lengths.

This kind of analysation has to be done for every experimental measurement device you are using. Regardless of electronics vs. non electronic measurements. The advantage of "old analogous" measurements is that you can intuitively see from the scale how fine the grid is, that you are working with.


Sodium Peroxoborate reacting with Sulfuric Acid and Potassium Permanganate This is one of the reactions I had to undertake in a lab. I think the reason for the test in the first place is to test for the peroxoborate anion in the solution.

From my understanding what's happening is that sodium peroxoborate will liberate $\ce{H2O2}$ when reacted with sulfuric acid. Adding the $\ce{KMnO4}$, the hydrogen will react to reduce $\ce{Mn(VII) ->Mn(II)} $ and form $\ce{O}$

I'm having difficulty interpreting this into balanced chemical equations. This is what I've tried but not sure if it's correct

$$\ce{Na2[B2(O2)2(OH)4] + H2SO4 ->Na2SO4 +B2O3 +2H2O2 +H2O}$$

This equation shows the liberated $\ce{H2O2}$ but not sure how to write a reaction now where $\ce{KMnO4}$ is added.

Any help would be much appreciated. I've been strugging with this for a while and there isn't much information online.

Well $\ce{KMnO4} $ and $\ce{H2O2}$ react as follows:

$$\ce{2KMnO4 +3H2O2 -> 2MnO2  + 2KOH + 2H2O + 3O2}$$

Adding the two reactions together I get:

$$\ce{Na2[B2(O2)2(OH)4] + H2SO4 + 2KMnO4 + H2O2 ->Na2SO4 + 2H3BO3 + 3O2 + 2MnO2 + 2KOH}$$


What does H2SO4 do to YAS? I have been reading VOGEL'S qualitative analysis and I came across the reagent Yellow ammonium sulphide. It's written here that it is used to distinguish Group 2B elements. But my question is that how will addition of H2so4 affect YAS itself?

Answer given is that it turns the cloudless solution into white /yellow but how does it exactly do this ?what is the reaction?

Ammonium sulphide is $\ce{(NH_4)_2S}$; it is colorless. Yellow ammonium sulphide contains some sulfur $\ce{S}$ dissolved in the preceding solution. So the anion  $\ce{ S^{2-}}$ is mixed with some disulphide $\ce{S_2^{2-}}$, which is yellow. If you add an acid to this solution, you produce the reaction $$\ce{S_2^{2-} + 2 H^+ -> H_2S_2}$$ or$$\ce{S_2^{2-} + 2 H_3O^+ -> H_2S_2  + H_2O}$$
But $\ce{H_2S_2}$ is thermodynamically unstable with respect to disproportionation. It gets slowly decomposed according to $$\ce{H_2S_2 ->  H_2S + S}$$The sulfur S produced here is not soluble into water, and will make the solution cloudy. If there is enough disulphide, the sulphur will make a precipitate of elementary sulphur, which may be pale yellow or white.

Ref.: N.N. Greenwood A. Earnshaw, Chemistry of the Elements, A. Wheaton & Co, Exeter, 1986, p.805 - 807.


Orbital Electronegativity Considerations in Resonance Structures After consulting three organic chemistry texts (Wade, Klein, and Brown), only one actually clearly delineates a few guidelines for picking the most significant resonance structure. The others mention the same factors (maybe less one or two). So in sum, the important factors to consider are:

1) A filled valance shell is preferable to a partially filled valance shell.

2) The more covalent bonds the better.

3) Least separation of oppositely signed charges. I'm thinking this is a Coulombic effect. Some books imply formal charge through the term "charge" but what about partial charges (which are much more reflective of true charges than formal charges. Which one should I really be looking at? Least separation of opposite formal charges or opposite partial charges?

4) Negative charges should generally be assigned to the more electronegative elements; the opposite is true for electropositive elements. Again, formal charge misleads - consider the ammonium ion (positive formal charge on nitrogen; negative partial charge on nitrogen). Again, one book implies formal charge but really, isn't looking at partial charges a bit more informative?

5) Unmentioned factor in any of the three textbooks: What about hybridization? If we have a molecule with two nitrogens, one $\ce{sp}$ hybridized and one $\ce{sp^2}$ hybridized but otherwise identical, shouldn't we also take into account "orbital electronegativity" - i.e. the ability of hybrid orbitals with higher s-character to better stabilize negative charge (and conversely higher p-character better stabilizes positive charge)?

6) Unmentioned factor in any of the three textbooks: Induction? Shouldn't induction also play a role in resonance stabilization? Resonance delocalizes electron(s); can't induction further stabilize electrons? debunked.

As far as hybridization goes, the hybridization of all of the atoms must be the same in all resonance structures. In considering resonance structures, none of the nuclei should be moved in space. Since hybridization is designed to explain the arrangement of nuclei/electrons in space, hybridization cannot change between resonance structures, otherwise nuclei may be forced to occupy new positions.

In practice, this means that certain atoms may appear to be sp3 hybridized but with consideration of resonance structures actually should be treated as sp2. This often means that a lone pair is not in an sp3 orbital as expected, but an unhybridized p-orbital.


greater bond energy, more stable? I am very confused about bond energy. In my book, it says "the most stable bond is the bond with the highest bond energy". However, later on it mentions that " A system with low energy is a stable system. Thus, a bond is formed when electron energy level is the lowest".
So, does high energy means more stability or less stability?

Look at it in this way. Consider the energy to break a bond to be arbitrarily assumed to be at $\pu{0 J/mol}$. Now, there are two bonds, $\ce{A-B}$ and $\ce{C-D}$. Let's say they have energies of $\pu{-242 kJ/mol}$ and $\pu{-378 kJ/mol}$. Clearly, the $\ce{C-D}$ bond has the lower energy.

Now, you can say that a stronger bond needs more energy to break it. So, this means that you need to supply $\pu{242 kJ/mol}$ of energy to break $\ce{A-B}$ and $\pu{378 kJ/mol}$ of energy to break $\ce{C-D}$. You can definitely see that $\ce{C-D}$ needs more energy to break it, and to bring the energy to $\pu{ 0 J/mol}$.

Hope this answer helps you.


Systematic IUPAC names for beta-glucose and lactic acid What are the IUPAC systematic name for $\beta$-glucose and lactic acid (as in the monomers used in polymerization for PLA)?

For lactic acid, I found this:

  2-hydroxypropanoic acid

Is this correct?

For $\beta$-glucose, I have not find anything conclusive so far. I did find on this site (Carbohydrates) that "carbohydrates have been given non-systematic names". Is this true, and does that imply that "$\beta$-glucose" is simply $\beta$-glucose?

A compound that can be isolated from a natural source is commonly given a trivial name. These trivial names are usually related to the biological origin of the material. When the full structure is known, a systematic name may be generated in accordance with the IUPAC recommendations for nomenclature of organic chemistry; however, this name may be too cumbersome (for example for repeated use in the text of scientific publication).

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name (PIN) for lactic acid is the systematic name 2-hydroxypropanoic acid. Nevertheless, the name lactic acid is retained for use in general nomenclature.

  P-65.1.1.2.3 The names citric acid, lactic acid, glyceric acid, pyruvic acid, and tartaric acid, related to natural products, are also retained; no substitution is recommended, but the formation of salts and esters is allowed.

For carbohydrates such as glucose, the current IUPAC recommendations do not identify a PIN. The systematic name for the β-pyranose form of ᴅ-glucose is (2R,3R,4S,5S,6R)-6-(hydroxymethyl)oxane-2,3,4,5-tetrol. Since this name may be too cumbersome, a semisystematic name may be formed based on the retained name ᴅ-glucose as follows.

  P-102.3.4 Cyclic forms of monosaccharides

  Most monosaccharides exist as cyclic hemiacetals or hemiketals. Two aspects of the internal cyclisation must be examined: first, the size of the ring, and secondly, the conformation of the newly created chirality center.

  P-102.3.4.1 Ring size

  Out of the various possible heterocyclic ring sizes resulting from hemiacetal or hemiketal formation, those with five and six members, including an oxygen atom, prevail and are discussed in this Section. Their names are based on those of the parent heterocycles furan and pyran, respectively. Names are formed by including the terms ‘furan’ and ‘pyran’ before the ending ‘ose’ in the name of a sugar. (…)

Thus, the retained name ᴅ-glucose is changed to ᴅ-glucopyranose to indicate the cyclic form having a six-membered ring.

  P-102.3.4.2 Anomeric forms; the stereodescriptors ‘α’ and ‘β’

  P-102.3.4.2.1 In the cyclic form, the configuration of the newly created chirality center ‘C-1’ must be expressed. This center is called the ‘anomeric center’. The two stereosiomers are called ‘anomers’; they are designated by the stereodescriptors ‘α’ and ‘β’ according to the configurational relationship between the anomeric center and the so called ‘reference center’.

  P-102.3.4.2.2 Configurations ‘α’ and ‘β’ for monosaccharides

  The anomeric reference center in a monosaccharide having a retained name is the configurational atom as defined in P-102.3.3. (…)
  The anomeric stereodescriptor ‘α’ or ‘β’, followed by a hyphen, is placed immediately before the configurational stereodescriptor ‘ᴅ’ or ‘ʟ’ of the carbohydrate name.

Therefore, the resulting name for the β-pyranose form of ᴅ-glucose is β-ᴅ-glucopyranose.


Why isoelectronic species give similar spectral lines? In my perspective those species should give similar spectra those which have same energy of different energy levels as the transition would yeild same energy photon
But still the answer to this question is always the isoelectronic species
Not the ions of the element

You are completely forgetting the nuclear charge of isoelectronic series. Their valence shell configuration is the same but the electrons experience a different attraction in each. As you stated

$\ce{Mg^{2+}}$ and $\ce{Al^{3+}}$ give series of similar spectral lines
(this was the question of my exam i.e.NEET)

Yes the spectrum pattern is similar but not identical. Atomic spectrum and that of ionized atoms is like a fingerprint, otherwise identifying the elements in the stars will become difficult and perhaps impossible. There is certainly some overlap in atomic spectrum but not identical as your question title suggests. A quick search of Google shows that this non-sensical question is quite popular in local exams. It is wrong. Isoelectronic species will not have an identical matching spectrum. Even the wavelengths are different.
Search this title in Google Scholar to see the wavelengths of isoelectronic series, where it available free of charge. None match with neutral sodium spectrum.
Wavelengths and levels of the Na I isoelectronic sequence from K IX through Mn XV
Journal of the Optical Society of America Vol. 66, Issue 9, pp. 899-904 (1976)


What will happen to the entropy and free energy of the gasses when the partition is removed? Consider a container of volume $ 5.0$  L that is divided into two compartments of
    equal size. In the left compartment there is nitrogen at $1.0$ $atm$ and $25 °C$; in the
    right compartment there is hydrogen at the same temperature and pressure. What
    will happen when the partition is removed?

  $A) $The entropy decreases, and the free energy decreases.
  $B)$ The entropy increases, and the free energy decreases.
  $C) $The entropy increases, and the free energy increases.
  $D) $The entropy decreases, and the free energy increases.

Logic tells that upon removing the partition, randomness increases and hence entropy increases. I am confused about free energy. First law of thermodynamics has to be applied , I think. But I can't seem to get the right direction.

A Spontaneous process is characterized by an increase in the total entropy (for both system and surroundings).

Spontaneous processes are characterized by a decrease in free energy (analogous to the decrease in gravitational potential energy occurring for a ball rolling downhill).


In a galvanic cell, why do electrons move through the wire from zinc to copper? I understand that there is aqueous $\ce{Cu^2+}$ in this scenario which pulls on the electrons from the zinc causing the electrons to move through the wire, but how does the electronegativity of $\ce{Cu^2+}$ pull on the electrons from the $\ce{Zn}$ when they aren't "connected".

What I mean by that is the $\ce{Zn}$ metal and the wire are only connected to the $\ce{Cu}$ metal and not the aqueous $\ce{Cu^2+}$ solution. This would make more sense to me if the $\ce{Zn}$ metal was connected to $\ce{Cu^2+}$ solution via a wire rather than being connected to the $\ce{Cu}$ metal.

If we consider the classical Daniel cell
$$\ce{Zn|Zn^2+||Cu^2+|Cu|}\text{(wiring  to Zn)}$$

then due higher tendency of zinc ions to    leave the metal, getting hydrated in solution, the zinc electrode obtains more negative electrostatic potential due extra electrons, compared to cupper electrode.

As electrodes are galvanically connected, electrons flow from more negative potential of the $\ce{Zn}$ electrode toward more positive $\ce{Cu}$ electrode, increasing potential of the former and decreasing the potential of the latter.

The $\ce{Cu}$ electrode, with the potential lower then its equilibrium potential, reduces $\ce{Cu^2+}$ to $\ce{Cu}$.

The $\ce{Zn}$ electrode gets the higher than equilibrium potential, as electrodes leave it by the wire, so more $\ce{Zn^2+}$ ions are transferred to $\ce{Zn^2+}$ salt solution, to approach the equilibrium potential.

Ions in both solutions, connected by diaphragm or salt bridge, migrate along the potential and concentration gradients, caused by unequal potentials and electrode reactions.

The whole cycle is closed.


Any possible additives to a drain cleaner I use as a source of NaOH? It is in pellets. And the box says main ingredient $\ce{NaOH}$ at least $30\%$ (I hope it's more). The box also says "aluminium catalyst". My goal was to take these pellets and dissolve them in water so that I would make a $1\textrm{M}$ solution of $\ce{NaOH}$.

I set everything up and most of it refused to dissolve in the solution. And the solubility for $\ce{NaOH}$ is pretty high. So I added some heat.This made almost everything slowly dissolve. Another thing was these small aluminium bits seemingly reacting and going up and down in the solution. I'm not sure if there was any gas released, but the solution definitely was not still.

I filtered it to remove the small aluminium balls. Any ideas If I have a relatively pure solution of $\ce{NaOH}$, or is it full of some other drain cleaner components. It is transparent-whiteish-little-yellowish. Filter kinda yellowish too.

What could these other components be?

Thanks

The product you've described contains the following ingredients (per the ingredient data sheet on the manufacturing company website):

Sodium hydroxide
Sodium chloride
Aluminium
Mineral oil

One possibility is that the aluminium is there to provide a foaming action, as its oxidation in the presence of aqueous sodium hydroxide liberates $\ce{H2}$ gas, per the following reaction (source):

$$\ce{2Al + 2NaOH + 2H2O -> 2NaAlO2 + 3H2}$$

The aluminium and sodium hydroxide would need to be kept separate and relatively anhydrous in the bottle to avoid reaction, so I wonder if the purpose of the mineral oil (which is hydrophobic) is to act as a moisture barrier and prevent mixing. If this is indeed the reaction that occurs, then the hydrated aluminium salts present are going to be at least somewhat soluble in solution, so I think it's improbable that what you have is pure sodium hydroxide. That said, if the quantity of aluminium is indeed small, this might not be a problem for your purposes.


Best reagents state for air free chemistry. Choosing halide for grignard synthesis I am planning to synthesize a grignard and I have a choice between the iodo or chloro aryl halide. The iodo is a solid and the chloro is borderline. For my previous reaction I melted the chloro and then tranfered it using a flushed sringe. As an alternative I could just weigh out the iodo and then pull vacuum on it a few times before running my reaction. Is there a better option between the two or they are just the same?

The chloride can be a bit slow to react, particularly if the aryl halide is hindered. You might need a small amount of iodine to get the reaction started. (Maybe you ran into this already?)

Once you form the Grignard, they should be identical for most applications.

Personally, I would just go with iodide because the chloride is probably a bit less reactive, and frankly, weighing out substances that are almost solid is extremely annoying. Strongly prefer the actual solid.

The main reason I would elect the chloride over the iodide in this case is if the cost of the chloride were much lower, and even then only if the reactivity weren't seriously impeded by using the chloride.


Gibbs Free Energy and Equlibrium Constant I have some problems understanding the relationship between $\Delta$$G°$ and $K$. For example, in the reaction of $$\ce{N2 + H2 <=> 2NH3}$$ where $\Delta$$G°= - 33.5$ kJ/mol.

$\Delta$$G°=- 33.5$ kJ/mol is obtained from the $\sum G^°_{\text{f, products}} – \sum G^°_{\text{f, reactants}}$
. This means that the complete conversion of 1 molar of $N_2$ gas and 1 molar of $H_2$ gas gives 2 molars of $NH_3$ gives -33.5 kJ/mol of free energy. This is assuming that the reaction goes fully to completion, yet in reality it doesn't.

From this formula below

$\Delta{G^°} = –RT \ln K_p \tag{5-6}$

$K_p$ can be obtained as $7*   10^5$, assuming at 298K and 1 atm.  My question is, since the complete reaction corresponds to  $\Delta$$G°= - 33.5$ kJ/mol of free energy, why should there be $K$ in the first place? $- 33.5$ kJ/mol tells me this value is for the complete reaction, so why should we insert this value into the equation just to get $K$ which tells me otherwise that the reaction doesn't go to completion? It doesn't make sense to me? In fact, shouldn't have K be infinity since the value that we had inserted is one obtained for a complete reaction?

Also, from this equation $\Delta{G} = \Delta{G^°} + RT \ln Q \tag{5-5}$

does having 1 mol of $N_2$ , 1 mol of $H_2$ and 1 mols $NH_3$ of in the reaction correspond to $- 33.5$ kJ/mol, since that makes $Q = 1$? And since $Q$ slowly reaches the value of $K$,  $\Delta{G} = 0$, then  $- 33.5$ kJ of free energy must have corresponded from the initial state to the equlibrium state. Yet again to me this doesn't make sense, because my understanding is that $- 33.5$ kJ is a value obtained only for the ideal scenario that the reaction goes to completion, and not from its initial standard states to equilibrium.

How does this all work out?

Q is supposed to be calculated in terms of partial pressures, not moles.  $\Delta G^0$ is based on a reversible process you have devised to take the pure reactants in stoichiometric proportions and 1 atm (actually, 1 bar) and 298 K and convert them to pure products in stoichiometric proportions, also at 1 atm. and 298 K.


How to determine the purity of lactic acid with titration? $\pu{25 mL}$ of $\pu{0.5 \%(w/v)}$ lactic acid ($\ce{C3H6O3}$, molecular mass $\pu{90.1 g mol-1}$, $\mathrm{p}K_{\mathrm{a}} = 3.86$ was neutralised by $\pu{13.1 mL}$ of $\pu{0.1010 M}$ monobasic base. Estimate the end point $\mathrm{pH}$ and calculate the amount of $\ce{C3H6O3}$ in % of the initial one.

This question was part of my final paper. I can't remember the exact words used in the paper for the % part, but I think it is asking for the % purity. I am not sure how to do it.

I calculated the $\mathrm{pH}$ using the formula $\mathrm{pH} = 0.5 (\mathrm{p}K_\mathrm{a} - \log M)$ but I think it is wrong. How is this doen correctly?

At the end point, you have a solution of sodium lactate (weak base). The concentration of sodium lactate is given by the equation:
$$
C' =
\frac{C_\mathrm{b}\, V_\mathrm{b,eq}}{V_\mathrm{a} + V_\mathrm{b, eq}},
$$
where $C_\mathrm{b}$ is the concentration of the base, $V_\mathrm{b, eq}$ is the volume of the base at the end point and $V_\mathrm{a}$ is the volume of lactic acid.
$$
C' =
\frac{\pu{0.1010 mol L-1} \times \pu{13.1 mL}}{\pu{25 mL} + \pu{13.1 mL}}
 = \pu{0.03473 mol L-1}
$$
The $\ce{pH}$ of the solution at the end point is given by the equation:
\begin{align}
  \mathrm{pH}
    &= \frac{1}{2}(\mathrm{p}K_\mathrm{w} + \mathrm{p}K_\mathrm{a}
      - \mathrm{p}C')\\
    &= \frac{1}{2}(14 + 3.86 + \log 0.03473) = 8.20
\end{align}

To calculate the purity percentage of lactic acid, we calculate the amount of substance of pure lactic acid: At the end point
$$n_\mathrm{a} = n_\mathrm{b}.$$
This means:
$$
n_\mathrm{a}
  = C_\mathrm{b} V_\mathrm{b}
  = \pu{0.1010 mol L-1}\times \pu{13.1E-3 L}
  = \pu{0.001323 mol}
$$
The mass of pure lactic acid is:
$$
m
  = \pu{0.001323 mol} \times \pu{90.1 g mol-1}
  = \pu{0.1192 g}
$$
Or the mass of $\pu{0.5 \%(w/v)}$ lactic acid is:
$$
m'
  = \frac{\pu{0.5 \%(w/v)} \times \pu{25 mL}}{100\%} = \pu{0.125 g}
$$
The purity percentage of lactic acid $P\%$:
$$P\% = \frac{m\times 100\%}{m'}=95.36\%$$


Does crystal field splitting energy Δ increase going down a group of transition metals if the ligands are weak field? Going down a group, the metal orbitals are more diffuse so there is greter
overlap between the metal and the ligand orbitals.
If there are no π-interactions, i.e. with σ-donor ligands like $\ce{NH3}$, then the $\mathrm{e_g}$ MOs are split further with more metal-ligand overlap, and $\Delta_o$ increases.
If there are π-acceptor interactions, then the $\mathrm{t_{2g}}$ MOs are split further with more metal-ligand overlap, but $\Delta_o$ still increases as it is the gap between the bonding $\mathrm{t_{2g}}$ and antibonding $\mathrm{e_g^*}$.
However with π-donor ligands ($\ce{Cl-}$, $\ce{Br-}$ etc), surely more overlap would cause greater splitting of the $\mathrm{t_{2g}}$ orbitals, which would decrease $\Delta_o$, as it is the energy difference between the antibonding $\mathrm{t_{2g}^*}$ and $\mathrm{e_g^*}$.

So, does $\Delta_o$ still increase going down a group if the ligands are π-donors?

I see where you are coming from, but you forgot that even with π-donor ligands, there is still a σ-bonding present in the complex. In the 4d/5d metals, the $\mathrm{e_g}^*$ orbital is pushed up by a lot due to the σ-type overlap.

The $\mathrm{e_g}$ overlap is always stronger than $\mathrm{t_{2g}}$ overlap because the metal $\mathrm{e_g}$ orbitals point directly at the ligands ($\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$), whereas $\mathrm{t_{2g}}$ orbitals point in between the ligands ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$).

So, yes, the $\mathrm{t_{2g}^*}$ orbital is pushed up higher by the π-donor, but the $\mathrm{e_g^*}$ orbital is pushed up even higher. Thus, for example, the complex $\ce{[RuCl6]^2-}$ has a $(\mathrm{t_{2g}})^4$ configuration (i.e. low spin) even though $\ce{Cl-}$ is a π-donor ligand.


Why are rates of reaction faster for ionic compounds than for covalent compounds? Why is it that ionic reactions (i.e., reactions involving ionic compounds) are faster than covalent reactions (reactions involving covalent compounds), even though the ionic bond is stronger than the covalent bond and hence more time will be required to break the bond?

Not all ionic reactions are not faster, i.e. have a larger rate constant than those between neutral species, however, at similar activation energies there are effects that only ionic reactions suffer. There are two considerations, the dielectric constant of the solvent used, water, hexane, ethanol, acetone etc., and the relative charges on the ions, similar or opposite as well as their value, $\pm1,\pm 2 $ etc.

For slow reactions, that is not diffusion controlled, (see below) the 'primary kinetic salt effect' is important. This shows that the rate constant can increase or decrease depending on the ionic strength of the solution and the charge on the ions. It is based on the Debye-Huckel theory of ions in solution which relates activity coefficients in dilute solution to the ionic strength.

The rate constant for a bimolecular reaction  has the form
$$ \log_{10}\left(\frac{k_2}{k_2^0}\right) = 2\alpha z_az_b\sqrt I$$
where I is the ionic strength, the z the charges on the two ions, $(-1,+1),(-1,-1) $ etc. and $\alpha \approx 1.6 \cdot 10^{-2} \pu{ m^{3/2} mol^{-1/2}}$ is a constant originating in Debye-Huckel theory. At low ionic strength ($ \lt 0.5 \pu{mol dm^{-3}}$ ) the rate constants increase or decrease linearly with $\sqrt I$.

The effect of increasing the dielectric constant (relative permittivity) is to lower the rate constant, thus rate constants are lower in water ($\epsilon = 78$) than in cyclohexane ($ \epsilon = 2.2$) for the same reaction. This is due to the fact that, at a given distance, in a high dielectric solvent the electric field of the ion is greatly reduced compared to its value in a low dielectric solvent. Thus in a high dielectric the ions are far more isolated from one another and so rate constants are only affected when ions come in relatively close contact. Conversely in low dielectric solvents the ions 'feel' the presence of one another at a large separation, this can considerably increase or decrease the rate constant depending on relative charges.

The rate constant is given by
$$ \ln(k_2) =\ln(k_2^{\infty}) - \frac{z_az_bq^2 }{4\pi \epsilon _0 \epsilon d_{ab}k_BT}$$
where $k_2^{\infty}$ is the limiting rate constant for a solvent of infinite dielectric constant. The permittivity of free space is $\epsilon _0$, the relative permittivity $\epsilon$, q is the charge on the electron, $d_{ab}$ the separation of the ions at reaction and $k_B$ the Boltzmann constant. A plot of the rate constant vs. $1/\epsilon$ yields a reasonably straight line  for all but the lowest $\epsilon$ values.

When a reaction's activation energy  is low, less than $\approx 10 \pu{kJmol^{-1}}$ the reactants react so rapidly on contact that the rate constant in determined by how quickly they can approach one another in solution; the rate constant then becomes diffusion controlled and thus depends on the solution viscosity.

Using the Stokes-Einstein equation and assuming equal sized molecules the rate constant has the simple form
$$ k_2= \frac{8RT}{3\eta}$$
where $\eta$ is the solvent viscosity. Water has a value $\approx 1\cdot 10^{-3} $ Pa s or $1$ centipoise (cP) at room temperature. In fluid solution such as hexane the rate diffusion controlled constant is typically $10^{10}$ to $10^{11} \pu{dm^3mol^{-1}s^{-1}}$.

The effects of ions changes the diffusion controlled rate constant for the reasons described above and is now
$$ k_2= \frac{8RT}{3\eta}\left(\frac{x}{e^x -1}\right)$$
where
$$ x=\frac{z_az_bq^2 }{4\pi \epsilon _0 \epsilon d_{ab}kT}$$
and is the term used before.

When the ionic charges are the same and the dielectric constant small the rate constant is greatly reduced, and conversely it is enhanced when the charges are opposite in sign to one another. In high dielectric solvents these changes are greatly reduced. These effects are similar to those described above.


Is change of pH of water with respect to temperature linear? Only recently I learned that the pH of water varies with temperature, from about 7.47 at 0 °C to 6.14 at 100 °C.

My question is, first, is there a geometric explanation for this, in the sense that water can perhaps arrange itself in a way that partial bonding of oxygen to a third hydrogen can happen without simultaneously producing hydroxide ions to balance things out?

And--related--is the spread of pH above enough to create a detailed profile for the variation of pH with temperature? If so is it linear?

To summarize... in aqueous solutions:

Basic means $\ce{[OH−] > [H+]}$
Acidic means that $\ce{[H+] > [OH-]}$
Neutral means $\ce{[OH−] = [H+]}$

There is a for the Wikipedia article for the autodissociation of water. The  chemical equation for the autodissociation is:
$$\ce{H2O <=> H^+ + OH^-}$$
The mathematical relationship between $K_d^t$, $\ce{[H^+]}$, $\ce{[OH^-]}$ and $t$ is: $$K_d^t =\ce{[H+][OH-]}\quad\text{for }t > 32\text{ C}$$
Note that $K_d^t$ changes with temperature $t$, but for pure water at any temperature, $t$,  the water will be neutral, and $\ce{[OH−] = [H+]}$. The temperature dependency of $K_d^t$ also means that pH + pOH is not always 14.

The Wikipedia article shows that the relationship between $K_d^t$ and temperature is a curve. (The curve shown is a bit odd in that water boils at 100 C, so higher temperatures must be taken under pressure.)

Also regardless of temperature the $\text{pH}$ of an aqueous solution is defined as $\text{pH} = -\log{\ce{[H^+]}}$ so the pH does not depend on $\ce{[OH^-]}$.


Relation between constant-pressure and constant-volume heat capacities: Cp - Cv = nR For an ideal gas, we have

$$C_p - C_V = nR$$

where
$C_p$ is heat capacity at constant pressure,
$C_V$ is heat capacity at constant volume,
$n$ is amount of substance, and
$R=N_\mathrm A\cdot k_\mathrm B=8.314\,462\,618\,153\,24\ \mathrm{J\ mol^{-1}\ K^{-1}}$[source] is the molar gas constant.

How can I prove this?

Preliminaries

Consider $U = U(V,T, p)$. However, assuming that it is possible to write an equation of state of the form $p = f(V,T)$,  I don't  have to explicitly address  the $p$ dependence of $U$, and I can write the following differential:

$$\mathrm{d}U = \underbrace{\left ( \frac{\partial U}{\partial V} \right)_T}_{\pi_T} \mathrm{d}V  + \underbrace{\left ( \frac{\partial U}{\partial T} \right)_V}_{C_v} \mathrm{d}T \tag{1}$$

so one writes,

$$ \mathrm{d}U = \pi_T \mathrm{d}V + C_v\mathrm{d}T \tag{2}$$

also, for an ideal gas, the internal pressure ($\pi_T$) $= 0$

Additionally for changes in internal energy at constant pressure,

$$ \left (\frac{\partial U}{\partial T}\right)_p = \pi_T \left (\frac{\partial V}{\partial T}\right)_p +C_v \tag{3}$$

Here, I take the opportunity to define two new quantities, namely $\alpha$ and $\kappa_T$ (the expansion coefficient and the isothermal compressibility respectively)

$$\alpha = \frac{1}{V}\left (\frac{\partial V}{\partial T}\right)_p$$

and, $$\kappa_T = \frac{-1}{V}\left (\frac{\partial V}{\partial p}\right)_T$$

Rewriting (3) using $\alpha$

$$ \left (\frac{\partial U}{\partial T}\right)_p = \alpha \pi_T V +C_v \tag{4} $$

Now, although the constant volume heat-capacity is defined as $C_v = \left (\frac{\partial U}{\partial T} \right)_v$, if we let $\pi_T = 0$ in equation (4), we get $C_v = \left (\frac{\partial U}{\partial T} \right)_p$

This holds true for a perfect gas, and one can quickly obtain the desired relation at this stage.

Derivation: Difference between constant volume and constant pressure heat capacities for a perfect gas

Consider, $$C_p - C_v  = \overbrace{\left (\frac{\partial H}{\partial T} \right)_p}^{\text{definition of} \ C_p} - \overbrace{\left (\frac{\partial U}{\partial T} \right)_v}^{\text{definition of} \ C_v} \tag{5}$$

Introducing, $H = U+ pV = U+nRT$, and exploiting $C_v = \left (\frac{\partial U}{\partial T} \right)_v =  \left (\frac{\partial U}{\partial T} \right)_p$, equation (5) yields:

$$ C_p - C_v = \left (\frac{\partial (U+nRT)}{\partial T} \right )_p - \left (\frac{\partial U}{\partial T} \right )_p = nR$$

Derivation: Difference between constant volume and constant pressure heat capacities (general case)

However, as my contribution to this discussion I would like to derive a relation between heat capacities that is universally true for any substance, not just a perfect gas. So let's return to equation (5):

$$C_p - C_v  = \overbrace{\left (\frac{\partial H}{\partial T} \right)_p}^{\text{definition of} \ C_p} - \overbrace{\left (\frac{\partial U}{\partial T} \right)_v}^{\text{definition of} \ C_v} \tag{5}$$

Here, we substitute $H = U + pV $ and obtain,

$$ C_p -C_v = \overbrace{\left( \frac{\partial U}{\partial T} \right)_p}^{\text{evaluated in} \ (4)}+ \left( \frac{\partial (pV)}{\partial T} \right)_p - C_v \tag{6}$$

The first partial derivative was already taken care of in equation (4). For the second one, since the derivative is to be evaluated at constant pressure, we can do the following

$$\left( \frac{\partial (pV)}{\partial T} \right)_p = p \overbrace{\left( \frac{\partial V}{\partial T} \right)_p}^{\alpha V}$$

Putting all of this together, one obtains

$$C_p -C_v = \alpha \pi_T V + \alpha pV = \alpha(p+ \pi_T)V \tag{7}$$

At this stage, I will make use of the following relation (derived in additional comments) $$\pi_T = T \left (\frac{\partial p}{\partial T}\right )_v - p$$ After substituting this into (7) we get:

$$ C_p -C_v = \alpha T V \left( \frac{\partial p}{\partial T} \right)_V \tag{8}$$

I wish to transform the last remaining partial derivative, and to do so I consider $ V = V(T,p) $ which yields the following differential

$$ \mathrm{d}V = \left( \frac{\partial V}{\partial T} \right)_p \mathrm{d}T + \left( \frac{\partial V}{\partial p} \right)_T \mathrm{d}p $$

At constant volume, $\mathrm{d}V = 0$ so one gets,
$$\left( \frac{\partial V}{\partial T} \right)_p \mathrm{d}T  = - \left( \frac{\partial V}{\partial p} \right)_T \mathrm{d}p$$

$$\overbrace{ \left( \frac{\partial V}{\partial T} \right)_p}^{\alpha V} = \overbrace{-\left( \frac{\partial V}{\partial p} \right)_T}^{\kappa_T V} \left( \frac{\partial p}{\partial T} \right)_V$$

Note: One can avoid all of this work, and simply invoke the Euler Chain Rule

Rearranging,  $$\left( \frac{\partial p}{\partial T} \right)_V = \frac{\alpha}{\kappa_T}$$

We can finally substitute this into (8) to get

$$C_p -C_v = \frac{\alpha^2 TV}{\kappa_T} \tag{9}$$

This is true for any substance, not just a perfect gas. Now, for a perfect gas $ pV = nRT$ holds true, and thus $\alpha = \frac{1}{T}$ and $\kappa_T = \frac{1}{p}$. Making these substitutions into (9), we get our desired result

$$C_p -C_v = nR $$

Additional Comments

This might seem like an unnecessarily complex, and not to mention convoluted way to get to the desired result (especially, in light of a much simpler method presented by @orthocresol), however, I think the deriving the expression for a general case first, and then reducing it to the special case is illuminating. Moreover, in spirit and approach, it is not that far from what @orthocresol did.
Physical Significance of certain terms/quantities

$\pi_T$ is called the internal pressure (it has the dimensions of pressure) and is a consequence of the interactions between molecules. For an ideal gas it is necessarily zero.
$\alpha$ i.e the expansion coefficient is the fractional change in volume that accompanies a rise in temperature. A large volume of $\alpha$ implies that the sample responds very strongly to changes in temperature.
Similarly, $\kappa_T$ is a measures of the response to a change in pressure. The negative sign insures that $\kappa_T$ is a positive quantity, because a pressure increase causes a decrease in volume ($\mathrm{d}V$ is negative)
Since equation (9) holds true for any substance, for solids and liquids one might be tempted to say $C_p \approxeq C_V$ because $\alpha$ is small for solids and liquids. However, one must be careful because $\kappa_T$ can be small as well, which makes the fraction $\frac{\alpha^2}{\kappa_T}$ large. In other words, even though a little work has to be done to push back the atmosphere when a solid expands, a great deal of work will go into pulling the atoms apart.

Supplementary Derivation: For a system where, $N$ doesn't change the fundamental equation of thermodynamics is:

$$\mathrm{d}U = T\mathrm{d}S -p\mathrm{d}V$$

This seems to suggest that, $U = U(S,V) $. Thus, one can write the following differential and after comparing to the one above can equate $T$ and $-p$ (as indicated by annotations) with the partial derivatives given below:

$$\mathrm{d}U = \underbrace{\left ( \frac{\partial U}{\partial S}\right)_V}_{T} \mathrm{d}S + \underbrace{\left ( \frac{\partial U}{\partial V}\right)_S}_{-p}\mathrm{d}V$$

Moreover, dividing both sides of the fundamental equation by $\mathrm{d}V$ (yeah, I know), and imposing constraint of constant temperature) we can manipulate it into the following form:

$$\overbrace{\left( \frac{\partial U}{\partial V} \right)_T}^{\pi_T} = \overbrace{\left ( \frac{\partial U}{\partial S}\right)_V}^{T}\left ( \frac{\partial S}{\partial V}\right)_T - \overbrace{\left ( \frac{\partial U}{\partial V}\right)_S}^{-p}$$

Thus, we have $$ \pi_T = T\left ( \frac{\partial S}{\partial V}\right)_T - p$$

Invoking, the Maxwell Relation

$$ \left ( \frac{\partial S}{\partial V}\right)_T = \left ( \frac{\partial p}{\partial T}\right)_V$$

one finally gets, $$\pi_T = T \left (\frac{\partial p}{\partial T}\right )_v - p$$


Why oxygen is enclosed in parenthesis in some chemical reaction? While going through my chemistry text book I came through the following reaction.

$\ce{RCH2OH + (O) -> RCHO + H2O}$

In the above reaction why oxygen atom is enclosed in parenthesis ?

It represents oxygen from an oxidising agent, as opposed to from diatomic oxygen. It's normally done in square brackets like this:
$$\ce{CH3CH2OH + 2[O] → CH3COOH + H2O}$$
Is the reaction between ethanol and potassium manganate (VII) to form ethanoic acid. The oxygen in the reaction is from the permanganate, not from diatomic oxygen. It kind of simplifies the reaction, so you only see the reaction of what you want -the ethanol and oxygen - instead of the whole reaction with the manganate there. (Manganate is an oxidising agent).

In your reaction the alcohol:
$\ce{RCH2OH}$ is reduced by the oxygen to form an aldehyde and water, using oxygen from an oxidising agent - the [O].


Calculating the mass in g of a molecule If I have the amount of the molecule in moles, how do I calculate the mass of an individual molecule?

For example, if the molecule was water?

The "amount of molecules in moles", such as 7 moles, provides no information concerning the mass of an individual molecule.

This is like asking, "if I have the amount of eggs in dozens, how do I calculate the mass of an egg?"


Why does electron flow from anode? The above is a picture of Galvanic cell. Zinc anode is placed in Zinc sulfate solution so we have ions as Zn2+ and SO42-. On the cathode side we have the same situation we have Cu2+ and SO42-. My question is why the zinc rod is getting converted into zinc ions and copper ions are getting converted into atoms and get deposited on copper rod. Why is electron flowing from anode to cathode?

The semi-reaction of reduction for Zn is :

$Zn^{2+} +2e^- \rightarrow  Zn \quad E^\circ =-0.763$

The semi-reaction of reduction for Cu is :

$Cu^{2+} +2e^- \rightarrow  Cu \quad E^\circ= 0.337$

The standard redox potential for Cu is more positive thus it is reduced and Zn is oxidized.
Electrons flow from a negative to a positive potential. They flow towards the cathode because their potential V, at the anode is more negative, thus they have an higher potential energy: $U_e = V*e$

note that $e$ the charge of the electron is negative as well as V.


What are the practical differences between the various DMSO oxidations (Swern, etc.)? Recently, I was reading a paper about a total synthesis of (+)-rubriflordilactone A (the article is open access: Angew. Chem. Int. Ed. 2015, 54 (43), 12618–12621). In two different steps where an alcohol -> aldehyde oxidation was required, the authors favoured using the Parikh–Doering reaction (DMSO activated with pyridine-sulfur trioxide complex):

In practice, is there a reason why the Parikh–Doering might be favoured over the Swern (DMSO activation with oxalyl chloride), or is it simply a case of "we just got better yields with it"?

I checked in Kurti–Czako. No hints were offered, but they also did list two more Swern variants, namely Pfitzner–Moffatt (activation with a carbodiimide) and Albright–Goldman (activation with an acid anhydride). On top of that, as Jan mentioned in a comment, there is also the Corey–Kim oxidation where N-chlorosuccinimide is used to generate the chlorosulfonium salt from DMS (instead of DMSO).

In general: are there benefits or drawbacks that are specific to each variant that would influence a synthetic chemist's decision, and if so, what are they?

tl;dr:
Swern and Parikh–Doering are by far the most practically useful and widely used oxidations. Whilst both have slight differences, the reality is that most research groups have a preference. This may be because they just have the reagents on hand (distilled oxalyl chloride vs good quality sulfur trioxide–pyridine), or because someone had a bad experience with one over the other.
The Anderson group (in the post you linked) commonly uses the Parikh–Doering conditions, whilst other groups appear to favour the Swern conditions using oxalyl chloride in place of sulfur trioxide–pyridine.

A brief overview of various DMSO oxidations and their drawbacks:
One thing to mention about all of these DMSO based reactions is the formation of stoichiometric dimethylsulfide (stench) as a by-product, which increasingly makes them avoided, especially by process chemists. That said, the reagents are cheap (compared to a Dess–Martin, for example), and the reaction is generally very reliable and amenable to use on moderate scales.  The DMSO and triethylamine must also be distilled for optimal results (which can be a drawback, especially if large scales are needed).
1. Moffatt oxidation
Reagents: DMSO, DCC, PPA.
Reference: Pfitzner, K.; E.; Moffatt, J. G. A New and Selective Oxidation of Alcohols. J. Am. Chem. Soc. 1963, 85 (19), 3027–3028. DOI: 10.1021/ja00902a036.
The Moffatt oxidation was one of the earliest examples of a DMSO-mediated oxidation. Two issues generally exist. Firstly, a strong acid (polyphosphoric acid, TFA, etc.) is required to get the reaction going, and, although the reaction becomes basic as it proceeds, the initial conditions using polyphosphoric acid/similar quite often have unwanted side effects (stripping of protecting groups, from experience). Secondly, and more importantly from a scale point of view is the stoichiometric N,N'-dicyclohexylurea produced as a by-product (similar to the Steglich esterification).
Due to the difficult-to-remove stoichiometric by-products, and the harsher conditions, Moffatt oxidations have largely been forgotten/ignored. Reagents such as EDCI have been introduced, however the high cost relative to oxalyl chloride/sulfur trioxide–pyridine have limited the appeal of Moffatt oxidations.
2. Albright–Goldman oxidation
Reagents: DMSO, acid anhydride
Reference: Albright, J. D.; Goldman, L. Dimethyl Sulfoxide-Acid Anhydride Mixtures. New Reagents for Oxidation of Alcohols. J. Am. Chem. Soc. 1965, 87 (18), 4214–4216. DOI: 10.1021/ja01096a055.
The Albright–Goldman oxidation commonly uses acetic anhydride, and goes at room temperature - although with significant reaction times which can give side reactions time to occur.
The biggest drawback, and reason why the Albright–Goldman is rarely used, is due to a competing Pummerer reaction (which is often significant), along with protection of the alcohol as its acetate ester if unhindered.
Some improvements have been made using TFAA in place of acetic anhydride. However, mixing TFAA and DMSO causes a violent exotherm unless the temperature is carefully and cautiously controlled.
3. Parikh–Doering
Reagents: DMSO, sulfur trioxide–pyridine, triethylamine
Reference: Parikh, J. R.; Doering, W. v. E. Sulfur trioxide in the oxidation of alcohols by dimethyl sulfoxide. J. Am. Chem. Soc. 1967, 89 (21), 5505–5507. DOI: 10.1021/ja00997a067.
Along with the Swern (see below), the Parikh–Doering is one of the most widely used DMSO based oxidations, with many examples in the synthetic literature.
The Parikh–Doering is able to be ran at mild temperatures (0 °C to room temperature), and the by-products are easily removed (crashing out and filtration). One advantage to the Parikh–Doering over the Swern is the fact that sulfur trioxide–pyridine is a bench stable solid, compared to oxalyl chloride which is a liquid that must be distilled and stored in the freezer to prevent degradation.
4. Swern oxidation
Reagents: DMSO, oxalyl chloride, triethylamine
Reference: Omura, K.; Swern, D. Oxidation of alcohols by “activated” dimethyl sulfoxide. a preparative, steric and mechanistic study. Tetrahedron 1978, 34 (11), 1651–1660. DOI: 10.1016/0040-4020(78)80197-5.
The Swern is the newest of the widely used DMSO oxidations, and often represents the best combination of reliability, cost of reagents, and ease of reaction.
Unlike the Parikh–Doering, the initial stages of the reaction must be ran at −78 °C to avoid unwanted decomposition. However, the conditions are equally as mild. One advantage over the Parikh–Doering is that oxalyl chloride decomposes into gaseous by-products, meaning that the only thing needed to be removed is the hydrochloride salt of triethylamine. Swern reactions can often be filtered, concentrated, and telescoped into the subsequent reaction without attrition in yield.

References:
The majority of the chemical information here comes from Comprehensive Organic Synthesis, Vol. 7. A whole chapter is dedicated to DMSO based oxidations and their examples in total synthesis: see Lee, T. V. Oxidation Adjacent to Oxygen of Alcohols by Activated DMSO Methods. DOI: 10.1016/B978-0-08-052349-1.00191-8.


Routes of formation of esters with highest yield Esters can be formed in two ways according to the syllabus I'm following:

(1) Mixing an alcohol with a carboxylic acid in the presence of sulfuric acid; and

(2) Mixing an alcohol with an acid chloride.

My question is:

Which method produces the highest yield of esters?

I think method (2) does because the $\ce{Cl-}$ will attach with the $\ce{-OH}$ in alcohol. However, I'm not exactly sure of the reason. Can someone explain it clearly with full details please?

Option 2 generally gives higher yields.

This is because option 1 (Fischer esterification) usually results in an equilibrium mixture which contains significant amounts of products and reactants, limiting the yield. This is mitigated by using a stoichiometric amount of a concentrated acid catalyst, such as sulfuric acid, which is also a strong dehydrating agent, and therefore removes the water from the equilibrium as it forms, favouring the products. Additionally, the reaction mixture can be heated to evaporate the water, or you can use a large excess of one of the reactants, typically the alcohol because it also acts as the solvent.

Conversely, option 2, more easily gives high yields because the reaction is more or less complete, with very little reactants left at the end. This is because the reaction produces hydrogen chloride gas instead of water, which is lost from the reaction mixture, driving the reaction forward. The downside of this option is that acid chlorides are generally more dangerous than carboxylic acids and are harder to prepare, or more expensive to buy. Aditionally, hydrogen chloride gas is produced which must be disposed of safely.

Another, often used method, is esterification using acid anhydrides. These are intermediate in reactivity between carboxylic acids and acid chlorides and are easier to use than acid chlorides, but still provide excellent yields. A classic example of this is the acetylation of salycylic acid with acetic anhydride to produce Aspirin.


How can we do amide coupling reaction in water? Does anyone know if it is possible or if there are any protocols for amide coupling in water (and not a normal polar aprotic solvent like DMSO, DCM, ACN, etc.)?

A search on the phrase "Amide Formation in Water" produces a good selection of hits.
Recent papers such as this and this offer conditions.
From the abstract of 1

A general and environmentally responsible method for the formation of amide/peptide bonds in an aqueous micellar medium is described. Use of uronium salt (1-cyano-2-ethoxy-2-oxoethylidenaminooxy)dimethylaminomorpholinocarbenium hexafluorophosphate (COMU) as a coupling reagent, 2,6-lutidine, and TPGS-750-M represents mild conditions associated with these valuable types of couplings


How do organic chemistry mechanisms become accepted? Now I have been learning chemistry for five years. I remember when I started organic chemistry, it was fun to draw arrows between molecules to show, as if in a mathematical demonstration, how the reactions occurred. In every lesson I had, teachers explained to us how a specific reaction (for example the Shapiro reaction) occurs step by step, explaining the chemistry of each group in each intermediate as if things were obvious (you know how teachers are).

But I've been wondering for some weeks now how does a mechanism come to be considered as accepted or still discussed?

If they use some, what kind of spectrometry techniques are used to measure the amount of each intermediate? If not how do they proceed? Do they use computational chemistry? Because for example for a reaction such like a $\mathrm{S_N2}$ it doesn't look too tricky to find how it works, whereas for Fries rearrangement (I don't know if the mechanism is considered as accepted or not) it seems to be more tricky.

(Ref.)

So can you explain the methods (at least the most used) to confirm a mechanism? I am aware that "confirm" does not mean that we are 100% sure, but rather that it is simply the best we have found so far.

Great question!
When I was teaching, Anslyn and Dougherty was a decent text for this. Here are some general comments:

First, please note that you cannot be sure about a mechanism. That's the real killer. You can devise experiments that are consistent with the mechanism but because you cannot devise and run all possible experiments, you can never be sure that your mechanism is correct.

It only takes one good experiment to refute a mechanism. If it's inconsistent with your proposed mechanism, and you're unable to reconcile the differences, then your mechanism is wrong (or incomplete at best).

Writing mechanisms for new reactions is hard. Good thing we have a whole slew of existing reactions that people already have established (highly probable, but not 100% guaranteed) mechanisms for.

Computational chemistry is pretty awesome now and provides some really good insights into how a specific reaction takes place. It doesn't always capture all relevant factors so you need to be careful. Like any tool, it can be used incorrectly.

The types of reactions you run really depend heavily on the kind of reaction you're studying. Here are some typical ones:

Labeling -- very good for complex rearrangements
Kinetics (including kinetic isotope effects) -- good for figuring out rate-determining steps
Stereochemistry -- Good for figuring out if steps are concerted (see this example mechanism I wrote for a different question)
Capturing intermediates -- This can be pretty useful but some species that you capture aren't involved in the reaction, so be careful.
Substitution effects and LFER studies -- Great for determining if charge build-up is accounted for in your mechanism

For named reactions, the Kurti-Czako book generally has seminal references if you want to actually dig through the literature for experiments.
For your specific reaction, what do we think the rate-determining step is? Probably addition into the acylium? You could try to capture the acylium intermediate.
You could run the reaction with reactants that have two labelled oxygens and reactants that have no labelled oxygens. Do they mix? If not, it's fully intramolecular. Otherwise, there's an intermolecular component and the mechanism as written is incomplete.
A quick Google search suggests that the boron trichloride mediated version has been studied via proton, deuterium, and boron NMR. I didn't follow up on this, but there's clearly some depth here.
When I was T.A.ing for Greg Fu, he really liked to use an example with the von Richter reaction. I might be able to find those references...


How do I use the priority list of substituent groups to decide the parent chain for this compound? I know that the -OH group has a higher priority than the -Cl group, but does that mean I should favour including it in my parent chain even if there's another chain that has a lower locant on the first substituent group?

My intuition is that the chain with the lower locant should be favoured. Is my reasoning in line with IUPAC rules?

The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)

So, yes, the priority list of substituent groups has to be considered first in order to decide the parent chain for this compound according to Rule P-44.1.1. However, you have already done that in both of your suggested examples: The principal characteristic group that is expressed as a suffix is the acid group ($\ce{-COOH}$). Therefore, the parent structure is a butanoic acid.
Looking at your two examples, your actual question is about which of the two possible butanoic acids is the principal chain: 3-chloro-2-(2-hydroxyethyl)butanoic acid or 2-(1-chloroethyl)-4-hydroxybutanoic acid. Thus, your actual question is about how to identify the principal chain after you have already considered the priority order of functional groups.
Both possible chains have the same length; therefore, the next rule (P-44.3.2) cannot be used to make a decision.
Both possible chains don’t have any multiple bonds; therefore, also Rule P-44.4.1 (a and b) doesn’t help.
In both possible chains, the principal characteristic group ($\ce{-COOH}$) is located at the start of the chain; thus, also Rule P-44.4.1 (h) doesn’t make any difference.
The number of substituents expressed as prefixes is the same (two) in both cases: chloro and 2-hydroxyethyl compared to chloroethyl and hydroxy; therefore, Rule P-45.2.1 cannot decide which chain is the principal one.
The decision is finally made using Rule P-45.2.2 since the locants et ‘2,3’ in 3-chloro-2-(2-hydroxyethyl)butanoic acid is lower than ‘2,4’ in 2-(1-chloroethyl)-4-hydroxybutanoic acid.
Therefore, the preferred IUPAC name for the compound that is given in the question is 3-chloro-2-(2-hydroxyethyl)butanoic acid.


13C NMR spectrum only showing solvent This is my first time taking a $\ce{{}^{13}C-NMR}$ spectrum and while I'm aware the signal to noise ratio is much lower compared to $\ce{{}^{1}H-NMR}$, I didn't realize it was quite this bad. I took a spectrum and I only see the $\ce{CDCl3}$ triplet. What concentration range is generally appropriate given say 256 scans (10 minutes on our machines?), or is it rare to see any good results after without drastically increasing the scan time?

Spectrum for reference (taken on a 300 MHz Varian Mercury):

There are several reasons why $\mathrm{^{13}C}$ experiments are far less sensitive than proton experiments.

The natural abundance for $\mathrm{^{13}C}$ is only 1.1%, while $\mathrm{^{1}H}$ has a natural abundance of 99,9%. This difference alone means that if you don't label your sample with $\mathrm{^{13}C}$, the spectra will be around 90 times less sensitive than proton spectra due to the lower natural abundance alone.

$\mathrm{^{13}C}$ also has a four times lower gyromagnetic ration than $\mathrm{^{1}H}$, which also lowers the sensitivity. But the way the experiment is usually measured, you get magnetization transfer from directly-attached protons to the carbons, which increases the sensitivity.

$\mathrm{^{13}C}$ also relaxes more slowly, which means that the time between experiments is longer and you can measure fewer scans than for proton experiments in the same time.

Another factor that is important is which kind of probe your spectrometer is using. For an inverse probe like e.g. a Bruker TXI, the inner coil is the proton coil, which means it's especially suited for proton experiments. For $\mathrm{^{13}C}$ experiments an observe probe like a BBO is more sensitive, as the carbon is measured on the inner coil. If you have multiple spectrometers, you should ask which one is better suited for $\mathrm{^{13}C}$ experiments.

If you have a very small amount of sample, a Shigemi tube allows you to measure in half the sample volume (200-250 $\mu$l) compared to standard tubes, you can achieve a higher concentration that way. They're pretty expensive though and they have to be matched to the solvent, so you would need a CDCl3 tube.

And there are various problems that could drastically reduce your sensitivity for $\mathrm{^{13}C}$, the most important ones if we disregard hardware problems, are proper tuning and matching for $\mathrm{^{13}C}$, and the correct pulse length for $\mathrm{^{13}C}$.

I've avoided your actual question, as the required concentration depends on quite a few factors. I'd recommend you to ask the person responsible for the spectrometers, as they should be able to give you a better estimate based on the actual hardware you have.


Why does this elimination reaction occur? I was confused by why the following elimination reaction hits only one of the hydrogens instead of both isomers that are possible. Please help me out!

Conjugated dienes are significantly favorable to non-conjugated dienes (because of delocalization and hybridization, etc.), so this will be the major product. It is not to say that the other reaction cannot happen, but that a combination of factors leads to that product being major. All of chemistry occurs in an equilibrium of some degree, and the reaction above will become most stable in conjugated diene form.


Are modern newspaper inks a health risk? It's very common in India to see newspapers being used to absorb oils from oily snacks, or simply for temporary food packaging. Some articles say that covering food items in newspaper is a health risk. They also mention that using newspapers for these purposes are strictly disallowed. For example:

Eating food wrapped in newspapers can cause cancer!

  Newspapers and cardboard boxes used for packaged foods are made of recycled paper that may be contaminated with harmful chemicals like diisobutyl phthalate and di-n-butyl phthalate which can cause digestive problems and also lead to severe toxicity.
  Recycled paper also has printing ink residues trapped from previous prints. These trapped residues have found to contain hormone disruptors like benzophenones and mineral oils.

Dangers of eating newsprint

  Canned food and drinks, newspapers used for packing food, cosmetics like eye makeup, hair dye, sindhoor, lead-based paints, ceramic glazes, industrial emissions, exposure at battery units, diety-making units, and tea are all potent media for lead poisoning.

Healthy eating : Watch out for poisonous packaging

  Exposure to a class of organic chemicals called arylamines, such as benzidine, 2-Naphthylamine and 4-Aminobiphenyl, is associated with high risks of bladder and lung cancer. Apart from these, printing inks also contain colorants, pigments, binders, additives and photo-initiators (used for speeding up the drying process of the ink), which have harmful effects.

I am highly inclined to believe these articles and therefore I strongly advise against using newspapers as absorbent paper and packaging paper.

But I read some articles that say that the West has now implemented strict regulations and therefore modern newspaper inks are safe.

So I've come here to clear my confusion. Are they safe now?

What chemicals do modern newsprint and the ink contain exactly?
Are they still toxic?

Before anybody may wrap some food in a newspaper, it needs to be printed.

Given that workers in print shops are the first that are exposed to potentially dangerous compounds, this is a matter occupational safety their national regulations.

In Germany, the BG ETEM provides an extensive exclusion list for inks and solvents. Compounds covered by this list are not be be used!

In order to avoid an endless recital of compound names or CAS numbers, the exclusion list is based of GHS hazard statements and includes (among others):

materials with CMR properties:
H340, H341, H350, H350i, H351, H360, H360F, H360D, H360FD, H360Fd, H360Df
materials with specific organ toxicity, acute toxicity and other health risks:
H370, H372, H300, H310, H330, H301, H311, H331, EUH 029, EUH 031, EUH 032

Summary

Does this mean that German newspaper are safe for handle in production and daily use? Yes, unless printing houses violate existing regulations.

Is the same true for Indian newspapers? I have no idea, since I am not familiar with the legal situation in India.


Potassium thiocyanate and iron(II) acetate I am trying to find the net-ionic equation for the reaction of potassium thiocyanate and iron(II) acetate.

It seems reasonable to me that the molecular equation is

$$\ce{2KSCN + Fe(CH3COO)2 -> 2CH3COOK + Fe(SCN)2}$$

which would give the net-ionic form

$$\ce{SCN- + Fe^2+ -> Fe(SCN)2}$$

My chemistry teacher says the correct form is

$$\ce{4SCN- + Fe^2+ -> Fe(SCN)4^2-}$$

First I don't know what molecular equation would give rise to this. Second, why in the world is my guess, which is simpler, wrong and this one correct?

As best I can tell, your answer is correct. I can't find any reference for iron(II) thiocyanate in the form your teacher says is correct, and several references that say $\ce{Fe(SCN)2}$ is the correct form.

For precipitation reactions, the charge will never change. You should always assume that the charges in the product will balance. There might be some cases where the product has an unexpected formula that you can't predict based on charges alone, but I can't think of any off the top of my head, and you certainly shouldn't be expected to know them at this stage.

edit:

This is the best reference I could find, it's a google books link to the CRC handbook


1-propanamine to cyclopropane? I was solving some questions on amines when I happened to see the below question:

  propan-1-amine + ($\ce{NaNO2 + HCl}$)= a mixture of products
  (originally, the mixture in brackets on the reactant side was just
  written as $\ce{HNO2}$)

  The mixture contains:

  a.    1-Propanol

  b.    2-Propanol

  c.    Cyclopropane

  d.    All of these

The answer happened to be "d. All of these". I could understand the formation of 1-propanol and 2-propanol but not cyclopropane. The answer given in the textbook wasn't clear and I couldn't understand. It said that the carbocation intermediate had something to do with it which I couldn't follow.

Can someone explain what is behind all this?

Diazotisation of primary amines gives the diazonium salt which decomposes to the primary cation and $\ce{N2}$. From this, in aq. $\ce{HCl}$, you directly get the primary alcohol, primary chloride and primary nitro by immediate trapping of the primary cation by ions in solution and the terminal alkene by 1,2 elimination.
A cation shift to give the secondary cation also occurs to give secondary alcohol and secondary chloride.
The cyclopropane arises from a 1,3-elimination of the primary cation.

pics from reference 1

References

Amines with Nitrous Acid - chem.libretexts.org
Nitrosation, diazotisation, and deamination by J. H. Ridd  Q. Rev. Chem. Soc., 1961, 15, 418-441 DOI: https://doi.org/10.1039/QR9611500418


Are all highly shock sensitive chemicals highly temperature sensitive too? I found that some highly shock sensitive explosives like lead azide, mercury fulminate have a relatively high auto-ignition temperature. I want to know if it's a general property or not and why so? Doesn't shock sensitivity means that the molecular bonds can be broken easily by supplying only a small amount of energy?

Yes -- but only to a degree

I assume you are talking primarily about explosives. All explosives are sensitive to both shock and heat to some degree. They are also sensitive to friction and spark. In most jurisdictions, as part of the regulatory approval process for new explosives the initiation sensitivity to each of these stimuli is determined with standardised tests.

In general, the lower the activation energy of the decomposition reaction, the greater the sensitivity to all stimuli. However, there is no simple relationship between sensitivity to different stimuli. Comparing two explosives, one may be more sensitive to impact than the other, but less sensitive to heat; a third may be less sensitive to either heat or impact, yet very sensitive to friction.

These sensitivities are not only a function of the chemical nature of the explosive, but also its physical form, purity, and the other ingredients in a composition. For example – not surprisingly! – addition of oily or greasy ingredients, even in small amounts, tends to reduce friction sensitivity. Most famously, liquid nitroglycerine is extremely shock sensitive, but is greatly desensitised by adsorption into a porous medium.

Again, for some important crystalline primary explosives the heat sensitivity is more or less fixed, but the shock sensitivity depends on the particle size distribution.


Why are lithides not known? In the last few decades, many alkalides - anions of alkali metals - have been synthesised. The most famous is undoubtedly that of sodium: $\ce{[Na(\text{2.2.2-cryptand})]+Na-}$, but the alkalides $\ce{K-}$, $\ce{Rb-}$, and $\ce{Cs-}$ are all known. However, $\ce{Li-}$ is not.

James Dye writes in a 1984 review1 that for the product of the reaction

$$\ce{M(s) + N(s) + L(s) -> [ML]+N-(s)}$$

to be thermodynamically stabilised ($\ce{M}$, $\ce{N}$ are metals, and $\ce{L}$ is the macrocyclic ligand), several criteria must be met:

  Small lattice energies for $\ce{L(s)}$, $\ce{M(s)}$, and $\ce{N(s)}$ so that the enthalpies of sublimation will not be too large.
  Low ionization energy of $\ce{M}$.
  High electron affinity of $\ce{N}$.
  Large complexation energy of $\ce{M+}$ by $\ce{L}$.
  Large lattice energy of $\ce{[ML+]N-}$ (which depends mainly on the interionic separation).

From these given factors, I suppose the only possible explanation is the larger heat of sublimation of lithium. However, lithium is still known to form electrides in solution - compounds of the form $\ce{[ML+]e-}$. In order for these to be formed, the sublimation energy of $\ce{Li}$ still has to be overcome.

Is there a thermodynamic reason why lithides, $\ce{Li-}$, have not yet been made?

Reference

Dye, J. L. Electrides, negatively charged metal ions, and related phenomena. Prog. Inorg. Chem. 1984, 32, 327–441. DOI: 10.1002/9780470166338.ch4.

One barrier is the greater tendency of lithium to form partially covalent bonds than heavier alkali metals.  This is a good thing for organic chemists, since this stabilizes organolithium compounds and enhances the solubility of lithium chloride as a chloride ion source (see this answer).  But it's not so hot for forming lithides.  Covalent bonding to more electronegative elements, meaning to practically all other elements in this case, oxidizes the intended lithide ion.  In effect the cryptand in an intended lithide is converted into one of those organolithium compounds.

Linear Christmas, in the comments to the question, hints at a way to beat this tendency.  Two of the references given there, https://doi.org/10.1021/ja056314+ and https://doi.org/10.1002/cphc.201200805, involve theoretical investigations of lithides built with pyrrole-based cryptands.  Such cryptands resist covalent bonding to lithide (or other nucleophilic) anions because this would destroy the aromatic of the ring and require electrons to overlap into a high-lying, strongly antibonding orbital.  These compounds, or counterparts with heavier alkalide ions, draw interest for their expected nonlinear optical (NLO) properties, making them synthetic targets.  So, stay tuned.


Which of the four balloons will shrink? Question:

Four gas balloons A, B, C, D of equal volumes containing $\ce{H2}$, $\ce{N2O}$, $\ce{CO}$, $\ce{CO2}$ respectively prickled with a needle and immsersed in a tank containing $\ce{CO2}$. Which of them will shrink after some time?
a) A
b) B
c) C
d) Both A and C

My attempt: I think that the ballons filled with the gases which will react with $\ce{CO2}$ will be the ones that will shrink. Now $\ce{H2}$ reacts with $\ce{CO2}$ to form $\ce{CH4}$, so the amount of $\ce{H2}$ will decrease and the balloon will shrink. Also I know $\ce{CO}$ cannot react with $\ce{CO2}$, so the answer has to be (a). I need hep determining the correct solution.

The relative molecular mass ($M_r$) of the gas determines the rate of diffusion of the gases from a region of higher concentration to a region of lower concentration.

The smaller the $M_r$ of the gas, the greater the rate of diffusion (movement of gas molecules).

  $M_r$ of $\ce{CO2}$ = 40

$M_r$ of $\ce{H2}$ = 2 (Balloon A)
Since hydrogen gas has a smaller $M_r$ than carbon dioxide gas, more hydrogen gas diffuse out of the balloon than carbon dioxide gas into the balloon per unit time. Hence, this balloon will shrink.

$M_r$ of $\ce{N2O}$ = 44  (Balloon B)
Since nitrous oxide gas has a greater $M_r$ than carbon dioxide gas, more carbon dioxide gas diffuse into the balloon than nitrous oxide gas out of the balloon per unit time. Hence, this balloon will inflate

$M_r$ of $\ce{CO}$ = 28  (Balloon C)
This balloon will shrink (same logic as $\ce{H2}$).

$M_r$ of $\ce{CO2}$ = 40  (Balloon D)
Since the balloon and the tank both contains $\ce{CO2}$ (same $M_r$ of gases), there is no net diffusion of gases in/out of the balloon, thus the balloon will not inflate/shrink (theoretically).


Would nodal planes of participating Atomic orbitals be nodal planes of molecular orbital? As I take sigma 2p antibonding molecular orbital as an example

Would it have 3 or 1 nodal planes. The two supposed nodal planes pass through the centres of the two atoms and are same as in the original atomic p orbitals. By LCAO A and B, I suspect that there will be some electron density due to the wave function of B where A is zero. Is there a fault in my logic?

For pi there is no problem, there are 1 nodal planes for pi bonding mo and 3 for antibonding mo as there is a common nodal plane to both MO.

AFAIK: The nodal planes are where the wavefunction sign inverses. If you drew the colors according to the sign value then you would see that for antibonding sigma MO there is a virtual nodal plane, since the sign changes between + and - between the two centre orbital lodes. Even if they by some magic grew larger, they can never touch, they interfere destructively. There will always be a plane between them where the wavefunction returns 0. (when the two spheres meet they will form a planar interface. Imagine soap bubbles.)

I found this, it is more descriptive; https://socratic.org/questions/how-can-i-draw-antibonding-orbitals


Partial pressure of gases in a container with a piston I've been having problems with the following problem.

Figure 1.

A container with a piston like that shown in Figure 1 is filled with $0.10$ $mol$ of Ar and $1.00$ $mol$ of water (liquid and vapor). The temperature of the entire container is $87 °C$ and the total pressure is $1$ $bar$. It is assumed that the volume of the liquid or cup is negligible with respect to that of the gas, and that the temperature of water and Ar is always maintained at $87 °C$. In this initial situation, the partial pressure of Ar is $A$ $bar$, the partial pressure of water vapor is $B$ $bar$, the volume of the gas is $C$, and the number of moles of liquid water is $D$ $mol$.

The piston is then pulled back quickly to fix the gas volume to $15.8$ $L$. At the moment when the piston
is pulled, the partial pressure of Ar decreases to $E$ $bar$ and the partial pressure of water vapor
decreases to $F$ $bar$. In this situation, the water will boil, because the total pressure of the gas is
lower than the saturated vapor pressure of water at $87 °C$, which is $0.6226$ $bar$. The partial pressure of water vapor increases
due to the boiling until the boiling eventually stops. When the boiling stops, the number of moles of
liquid water is $G$ $mol$. After that, evaporation proceeds until equilibrium is reached. At
equilibrium, the number of moles of liquid water is $H$ $mol$.

Calculate the appropriate numerical values for blanks $A$ - $H$.

The reason I am having trouble solving the problem is the fact that there seems to be too many unknown variables: the initial volume of the container $(V_0)$, the partial pressures of Ar $(p_{Ar,1})$ and water $(p_w)$ and the number of moles of liquid water $(n_{w,l})$. I've come up with the following formulas but I am unable to get any further.
$\ce{p_{total}=p_{Ar,1} + p_{water}=1 bar}$
$\ce{n_{total}=n_{Ar,1} + n_{water,g}=0.10 mol + (1 mol-n_{water,l})}$
$\ce{p_{Ar,1}=\frac{n_{Ar}*R*T}{V_0}}$
$\ce{p_{w,1}=\frac{n_{water}*R*T}{V_0}}$
$\ce{V_0=\frac{n_{total}*R*T}{p_{total}}}$

$B=\pu{0.6226 bar}$ as the water-water vapour system would be in equilibrium.
Now,
$A+B= \pu{1 bar}\\
A= \pu{0.3774 bar}$
For initial volume of the container,
$
P_{\ce{Ar}_i}V_i=n_{\ce{Ar}_i}RT\\
V_i=C=\frac{n_{\ce{Ar}_i}RT}{P_{\ce{Ar}_i}}\\
C=\pu{\frac{0.1 \times 0.08314 \times 360}{0.3774} L}\\
C=\pu{7.93 L}\\
$
For initial moles of water vapour,
$
P_{\ce{H2O_{(v)}}_i}V_i=n_{\ce{H2O_{(v)}_i}}RT\\
n_{\ce{H2O_{(v)}_i}}=\frac{P_{\ce{H2O_{(v)}_i}}V_i}{RT}\\
n_{\ce{H2O_{(v)}_i}}=\pu{\frac{0.6226 \times 7.93}{360 \times 0.08314} L}\\
n_{\ce{H2O_{(v)}_i}}=\pu{0.165 mol}\\
$
Now,
$
n_{\ce{H2O_{(v)}_i}} + n_{\ce{H2O_{(l)}_i}} =\pu{1 mol}\\
n_{\ce{H2O_{(l)}_i}} = D = \pu{ (1 - n_{\ce{H2O_{(v)}_i}}) mol}\\
D = \pu{ (1 - 0.165) mol}\\
D = \pu{ 9.835 mol}\\
$
Now as the volume is doubled by Boyle's law we can say that the partial pressure of argon and water vapour will decrease to half of original.
Thus,
$
E=\frac{A}{2}=\pu{0.1887 bar}\\
F=\frac{B}{2}=\pu{0.3113 bar}\\
$
Now as the partial pressure of water vapour is less than the vapour pressure of water at that temperature, water will evaporate to make it equal.
Now as the temperature is same thus the vapour pressure of water is also same as initial. Thus only change in system is it's volume which was doubled. So we can conclude that twice amount of water vapour as compared to initial conditions will be required to attain this pressure.
Thus,
$
n_{\ce{H2O_{(v)}_f}}=2 \times n_{\ce{H2O_{(v)}_f}}\\
n_{\ce{H2O_{(v)}_f}}=\pu{0.33 mol}\\
$
But as the total moles of water is conserved,
$
G=H=\pu{(1-n_{\ce{H2O_{(v)}_f}}) mol}\\
G=H=\pu{0.67 mol}\\
$


Does the iodoform reaction form hydrogen iodide or water? I am currently learning about the iodoform reaction, i.e. when CH3COR compounds react with iodine and sodium hydroxide to form iodoform.

My course book and Chemguide both says that the reaction is as follows:

$$ \mathrm{CH_3COR + 3I_2 +4OH^- \rightarrow  CHI_3 + RCOO^- +3I^- +3H_2O}$$

However, this picture from Wikipedia suggests that hydrogen iodide is formed instead of water:

Which one is correct?

Your balanced equation is correct, both water and iodide (and iodide in water will be in equilibrium with hydrogen iodide) are formed in the iodoform reaction.

$$ \mathrm{CH_3COR + 3I_2 +4OH^- \rightarrow  CHI_3 + RCOO^- +3I^- +3H_2O}$$

The Wikipedia article that you mention, could have been written a bit clearer.  Specifically, there are two key steps in the iodoform reaction.  First, there is formation of the tri-iodo ketone and water is generated in this step.

In a subsequent step, the carbonyl undergoes an SN2 reaction to generate iodoform.

The Wikipedia drawing in your post seems to focus on the mechanism of the second step (3 to 4b).  The first step (2 to 3) is just illustrated in a cursory fashion and the formation of water is not included in the figure.


Hot and cold air interaction in a sealed environment What would happen if air heated to 1200 degree's Fahrenheit in a sealed environment was injected with colder air?
No specific pressure defined yet just a general idea of potential interaction between the two.

Edit
My apologies, I meant within a sealed environment like an engine cylinder. I know that the hot air would naturally expand and push a piston but how would the injection of cold air alter this, if at all?
For example purposes we can assume the cylinder is iron while the piston is aluminum which is common design today.

In order to work out this problem, several parameter values are required to define the initial state, preferably

amount $n_1$, volume $V_1$, or mass $m_1$ of the hot air in the cylinder
initial temperature $T_1$ of the hot air
initial pressure $p_1$ of the hot air
amount $n_2$, volume $V_2$, or mass $m_2$ of the injected cold air
initial temperature $T_2$ of the cold air
initial pressure $p_2$ of the cold air

Unfortunately, only the initial temperature of the hot air is defined as $T_1 = 1200\ ^\circ\mathrm F$, which is $T_1 \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$ in proper units (the use of the unit degree Fahrenheit is deprecated). This leaves us with five unknown variables, which is certainly too many for a vivid analytical description.

Nevertheless, we may investigate this problem by guessing reasonable parameter values as follows. By way of example, we arbitrarily assume that equal masses of hot and cold air $(m_1 = m_2 = 1\ \mathrm{kg})$ are mixed. The assumed temperature of the cold air sample is equal to room temperature, i.e. $T_2 = 68\ ^\circ\mathrm F = 20\ ^\circ\mathrm C \approx 293\ \mathrm K$, whereas the temperature of the hot air sample is given as $T_1 = 1200\ ^\circ\mathrm F \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$. The assumed pressures of the air samples is equal to normal atmospheric pressure, i.e. $p_1 = p_2 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$. These conditions define the initial state and also allow to determine other interesting properties using tabulated reference values for air, e.g. density $\rho$, Volume $V$, specific internal energy $u$, and internal energy $U$.

1. Hot air sample

temperature $T_1 = 1200\ ^\circ\mathrm F \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$
pressure $p_1 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$
mass $m_1 = 1\ \mathrm{kg}$
density $\rho_1 = 0.3827\ \mathrm{kg/m^3}$
volume $V_1 = \frac{m_1}{\rho_1} = 2.613\ \mathrm{m^3}$
specific internal energy $u_1 = 819.4\ \mathrm{kJ/kg}$
internal energy $U_1 = u_1 \cdot m_1 = 819.4\ \mathrm{kJ}$

2. Cold air sample

temperature $T_2 = 68\ ^\circ\mathrm F = 20\ ^\circ\mathrm C \approx 293\ \mathrm K$
pressure $p_2 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$
mass $m_2 = 1\ \mathrm{kg}$
density $\rho_2 = 1.2046\ \mathrm{kg/m^3}$
volume $V_2 = \frac{m_2}{\rho_2} = 0.830\ \mathrm{m^3}$
specific internal energy $u_2 = 335.3\ \mathrm{kJ/kg}$
internal energy $U_2 = u_2 \cdot m_2 = 335.3\ \mathrm{kJ}$

3. Mixture at constant volume

The hot air sample is confined by a piston in a cylinder. The cold air sample is added, thus allowing the two air samples to mix. The total mass $m_3$ is given by
$$\begin{align}
m_3 &= m_1 + m_2 \\
&= 1\ \mathrm{kg} + 1\ \mathrm{kg} \\
&= 2\ \mathrm{kg}
\end{align}$$

For simplicity’s sake, we ignore any work $W$ performed by or on the system (e.g. the work that is required to inject the cold air into the cylinder). Furthermore, we assume that the process is adiabatic, i.e. no heat $Q$ is transferred into or out of the system. Therefore, in accordance with the first law of thermodynamics, the total internal energy $U_3$ is given by
$$\begin{align}
U_3 &= U_1 + U_2 \\
&= 819.4\ \mathrm{kJ} + 335.3\ \mathrm{kJ} \\
&= 1154.7\ \mathrm{kJ}
\end{align}$$
and the new specific internal energy $u_3$ is
$$\begin{align}
u_3 &= \frac{U_3}{m_3} \\
&= \frac{1154.7\ \mathrm{kJ}}{2\ \mathrm{kg}} \\
&= 577.3\ \mathrm{kJ/kg}
\end{align}$$

At constant volume $V$, i.e.
$$\begin{align}
V_3 &= V_1 \\
&= 2.613\ \mathrm{m^3}
\end{align}$$
the new density is
$$\begin{align}
\rho_3 &= \frac{m_3}{V_3} \\
&= \frac{2\ \mathrm{kg}}{2.613\ \mathrm{m^3}} \\
&= 0.7654\ \mathrm{kg/m^3}
\end{align}$$

The new temperature $T_3$ and pressure $p_3$ can be determined from the density $\rho_3$ and the specific enthalpy $h_3$ using tabulated reference values:
$$\begin{align}
T_3 &= 658\ ^\circ\mathrm F \approx 348\ ^\circ\mathrm C \approx 621\ \mathrm K \\
p_3 &= 1.37\ \mathrm{bar} = 0.137\ \mathrm{MPa}
\end{align}$$

4. Mixture at constant pressure (ignoring work)

Allowing the system to expand to normal atmospheric pressure by moving the piston, the new pressure $p_4$ is
$$p_4 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$$

We know that the mass $m$ of the air in the closed system does not change, i.e.
$$\begin{align}
m_4 &= m_3 \\
&= 2\ \mathrm{kg}
\end{align}$$
and we still (incorrectly) assume that no work $W$ is performed by or on the system and that no heat $Q$ is transferred into or out of the system. Therefore,
$$\begin{align}
U_4 &= U_3 \\
&= 1154.7\ \mathrm{kJ}
\end{align}$$
and thus
$$\begin{align}
u_4 &= u_3 \\
&= 577.3\ \mathrm{kJ/kg}
\end{align}$$

The new density $\rho_4$ can be determined from the pressure $p_4$ and the specific internal energy $u_4$ using tabulated reference values:
$$\rho_4 = 0.5681\ \mathrm{kg/m^3}$$

Thus, the new volume $V_4$ is
$$\begin{align}
V_4 &= \frac{m_4}{\rho_4} \\
&= \frac{2\ \mathrm{kg}}{0.5681\ \mathrm{kg/m^3}} \\
&= 3.521\ \mathrm{m^3}
\end{align}$$

The volume change $\Delta V$ during expansion is
$$\begin{align}
\Delta V &= V_4 - V_3 \\
&= 3.521\ \mathrm{m^3} - 2.613\ \mathrm{m^3} \\
&= 0.908\ \mathrm{m^3}
\end{align}$$

The corresponding pressure-volume work $W$ can be calculated assuming an external pressure of $p_\mathrm{ex} = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$.
$$\begin{align}
W &= p_\mathrm{ex} \cdot \Delta V \\
&= 101\,325\ \mathrm{Pa} \times 0.908\ \mathrm{m^3} \\
&= 92.0\ \mathrm{kJ}
\end{align}$$

The amount of energy transferred to the surroundings as work $W$ significantly reduces the internal energy $U$ of the system. We ignored this fact when we assumed that $W_4 = 0$ and $U_4 = U_3$. We can eliminate this error by calculating a new state as follows.

5. Mixture at constant pressure (including work)

If the internal energy is reduced (due to energy transfer to the surroundings as work) to
$$U_5 = 1087.8\ \mathrm{kJ}$$
and thus the specific internal energy is reduced to
$$\begin{align}
u_5 &= \frac{U_5}{m_5} \\
&= \frac{1087.8\ \mathrm{kJ}}{2\ \mathrm{kg}} \\
&= 543.9\ \mathrm{kJ/kg}
\end{align}$$
the new density $\rho_5$ after expansion to normal atmospheric pressure
$$p_5 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$$
can be determined from the pressure $p_5$ and the specific internal energy $u_4$ using tabulated reference values as
$$\rho_5 = 0.6111\ \mathrm{kg/m^3}$$

Thus, the new volume $V_5$ is
$$\begin{align}
V_5 &= \frac{m_5}{\rho_5} \\
&= \frac{2\ \mathrm{kg}}{0.6111\ \mathrm{kg/m^3}} \\
&= 3.273\ \mathrm{m^3}
\end{align}$$
the volume change $\Delta V$ during expansion is
$$\begin{align}
\Delta V &= V_5 - V_3 \\
&= 3.273\ \mathrm{m^3} - 2.613\ \mathrm{m^3} \\
&= 0.660\ \mathrm{m^3}
\end{align}$$
and the corresponding pressure-volume work $W_5$ is
$$\begin{align}
W_5 &= p_\mathrm{ex} \cdot \Delta V \\
&= 101\,325\ \mathrm{Pa} \times 0.660\ \mathrm{m^3} \\
&= 66.9\ \mathrm{kJ}
\end{align}$$

We find that
$$\begin{align}
U_5 &= U_3 - W_5 \\
1087.8\ \mathrm{kJ} &= 1154.7\ \mathrm{kJ} - 66.9\ \mathrm{kJ} \\
\end{align}$$
This result is in agreement with the first law of thermodynamics, which states that the change in the internal energy $U$ of a closed system is equal to the amount of heat $Q$ supplied to the system (which is $Q = 0$ in this case), minus the amount of work $W$ done by the system on its surroundings.

$$\small
\begin{array}{llllll}
\hline
\text{Quantity} & \text{Symbol} & \text{Unit} & \text{State 1} & \text{State 2} & \text{State 3} & \text{State 4} & \text{State 5} \\
\hline
\text{Temperature} & T & ^\circ\mathrm F & 1200 & 68 & 658 & 658 & 580 \\
& & ^\circ\mathrm C & 649 & 20 & 348 & 348 & 304 \\
& & \mathrm K & 922 & 293 & 621 & 621 & 577 \\
\text{Pressure} & p & \mathrm{bar} & 1.01325 & 1.01325 & 1.37 & 1.01325 & 1.01325 \\
& & \mathrm{MPa} & 0.101325 & 0.101325 & 0.137 & 0.101325 & 0.101325 \\
\text{Mass} & m & \mathrm{kg} & 1 & 1 & 2 & 2 & 2 \\
\text{Density} & \rho & \mathrm{kg/m^3} & 0.3827 & 1.2046 & 0.7654 & 0.5681 & 0.6111 \\
\text{Volume} & V & \mathrm{m^3} & 2.613 & 0.830 & 2.613 & 3.521 & 3.273 \\
\text{Specific internal energy} & u & \mathrm{kJ/kg} & 819.4 & 335.3 & 577.3 & 577.3 & 543.9 \\
\text{Internal energy} & U & \mathrm{kJ} & 819.4 & 335.3 & 1154.7 & 1154.7 & 1087.8 \\
\text{Work} & W & \mathrm{kJ} & \text{n/a} & \text{n/a} & 0 & 0 & 66.9 \\
\hline
\end{array}$$