Extracting urea from urine What is a process for the extraction of urea from urine? These are the steps I have come up with:

Add 2-propanol to get rid of any bacteria present. (The 2-propanol likely contained some sodium chloride.)
Precipitate out the proteins and neutralize any acids present using sodium carbonate. (At least, that is what appears to happen upon addition.)
Crystallize out the sodium and potassium chlorides, as urea is much more soluble than those salts.

At the second step, three layers formed: a top clear yellow layer (which I assume was 2-propanol), a middle very cloudy yellow layer (which quickly disappeared into the bottom layer) and a bottom slightly cloudy yellow layer (which I assume was water). The bottom two layers contained particles floating around.

Then, I assume that the water layer remaining contains mostly urea. However, how do I get rid of everything else (urobilin, etc.)?

The end goal is to generate ammonia from the urea by adding some calcium hydroxide. However, it would also be nice to get rid of other impurities which do not impede this reaction.

First, from Boundless.com (Chemical Composition of Urine), "Urine is an aqueous solution of greater than 95% water, with the remaining constituents, in order of decreasing concentration urea 9.3 g/L, chloride 1.87 g/L, sodium 1.17 g/L, potassium 0.750 g/L, creatinine 0.670 g/L and other dissolved ions, inorganic and organic compounds (proteins, hormones, metabolites)."

Urine color is generally a result of the degradation of the heme molecule from blood into the urinary products, called bilirubin, urobilinogen, urochrome, urobilin and uroerythrin. In essence, they are all tetrapyrroles amounting to something on the order of 1mg per 100ml (or roughly 0.001% of urine)... so this is a very minor part of urine. Also, it should be noted that over 3,000 chemicals have been found in urine, so there's something to consider.

Activated carbon filtration might be able to be used to preferentially remove the tetrapyrroles and some other organic molecules while allowing urea to pass (for the most part).

There are various forms of activated carbon (of course), and so the amount of urea absorption (by carbon) will be variable; the adsorption capacity of the activated carbon for urea will (likely) be different for activated carbon from different sources. Absorbance of urea from one report was about 1.69 mg urea/g carbon, where the urea concentration was 100 mg/L (see picture below). At greater concentrations (like 930mg urea per liter), lesser amounts of acivated carbon will become saturated; in other words the activated carbon will not be able to absorb more urea, but it might be able to effectively remove the tetrapyrroles or some other minor stuff.

But I think that the first thing to do is deal with the concentration. Urea is only about 0.93% of urine. So you only have about 0.93g urea per 100ml of urine.

To concentrate urea: boil off the water (which will also kill bacteria). Urea doesn't boil. The melting point for pure urea is 132.7 °C and urea will begin to decompose at around 150 °C. The residue (which should look creamy) will have a relatively high urea concentration which can then be extracted with (warm/hot 60 °C) ethanol or diacetone alcohol. Salt and protein are significantly less soluble in alcohol(s), so they should hopefully be (mostly) removed at this step.

The solubility of urea in 95% alcohol (100 g/L), absolute ethanol (50 g/L), and methanol (166 g/L) at 20 °C (room temperature) sounds pretty fair. And from TECHNICAL MEMORANDUM No. 13/M/67 (Investigations of Crystal Modification of Urea) "in the case of diacetone alcohol the solubility [of urea] ranged from 2 per cent at 20 °C to about 10 per cent at 100 °C, offering possible development of a process yielding free-flowing crystalline urea."

The solubility of salt (at 25 °C) in ethanol (0.65 g/L) and methanol (14.9 g/L) is significantly lower.

Now, the solubility of a hodgepodge of stuff including tetrapyrroles is something that you will have to experiment with. But, many similar compounds are readily soluble in acetone and tetrachloroethylene (often had as brake cleaner). Urea should not be highly soluble in these two solvents, but I'm a afraid I can't find a reliable solubility listed online, and I happen to be out of acetone right now. Cleaning an alcoholic extract of boiled urine with tetrachloroethylene might require using a separatory funnel.

On a historical note, (around 1727) Herman Boerhaave was able to extract urea from urine by working with the concentated-by-boiling residue (having the consistency of "fresh cream"). After filter pressing the residue, he allowed it to dry for a year... until he obtained a solid mass with an unknown oily substance (on top).  He discarded the oil and rinsed the crystals (filtride) with cold water to remove salt, leaving urea. He then recrystallized the urea from hot water. Previously (in 1669), Hennig Brand discovered the philosophers stone (phosphorus) by cooking the bejeezus out of (lots of) urine.

Personally, instead of waiting a year, I would take a chance on extracting urea from the cream. Also, I read today that boiling urine causes irreparable stink-damage to a home; so get a hot plate and an exension cord and boil it outside. Another recipe (as if I had nitric acid to spare), calls for reacting concentrated urea (say 100g in 100ml water) to make urea nitrate. The addition of nitric acid is exothermic and urea nitrate is explosive, so make sure the solution doesn't boil off (ice bath)... and urea nitrate is precipitated in the freezer. Then after you have clean urea nitrate, you can neutralize the acid under alcohol to recover urea. I would reserve this (nitric acid) method for the final work up, using relatively clean urea.

Also, here's a couple of pay-walled articles that might be useful to you if you really want to do this: Solubility of urea in water-alcohol mixtures and Urea in the history of organic chemistry: Isolation from natural sources


Why is Polonium the only metal to crystallize forming Simple Cubic Cell? I have read that Polonium is the only metal that crystallizes forming SCC. I searched for a concrete explanation but all they were saying is the stability of Polonium at STP. I can't really comprehend their explanation.

The inorganic crystal structure database (ICSD), by specifying that
      the structure contains only a single element, you can reduce the
      number of hits to 55. Structure examples are Cr, Po, Mn, Sb, Se, P,
      Ca, Li, As. Some are specific phases occurring at different
      temperature or pressures.

So Po is not the ONLY metal to have a simple cubic structure, but
the other cases require special conditions.

  The origin of the stabilized simple-cubic (SC) structure in Po is
  explored by using the first-principles band calculations. We have
  found that the prime origin is the inherent strong spin-orbit (SO)
  interaction in Po, which suppresses the Peierls-type structural
  instability, as usually occurs in p-bonded systems. Based on the
  systematic analysis of electronic structures, charge densities, Fermi
  surfaces, and susceptibilities of Se, Te, and Po, we have proven that
  the stable crystal structure in VIA elements is determined by the
  competition between the SO splitting and the crystal-field splitting
  induced by the low-symmetry structural transition. Our study suggests
  that the large SO interaction would suppress the Peierls instability
  which is generally expected to occur in one-dimensional conductors.

  In fact, Po becomes airborne with ease: if a sample is heated in air
  to 55 °C (131 °F), 50% of it is vaporized in 45 hours to form diatomic
  Po2 molecules. (The melting point of polonium is 254 °C and its
  boiling point is 962 °C.)

Putting it all together, Po is in the column with Se and Te, and just to the right of Bi, none of which are typical of "normal" metals. The formation of $\ce{Po2}$ molecules at near ambient temperatures suggests that the outer electrons of Po form bonds of a highly covalent nature.

But it is indeed unusual and fascinating, and if Po were not 5000 times more radioactive than radium, I might be tempted to do a little chemistry.

References

Quora:
"What are examples of simple cubic crystal structure?" Gervais Chapuis, Nov 3, 2016
B. I. Min, J. H. Shim, Min Sik Park, Kyoo Kim, S. K. Kwon, and S. J. Youn, Origin of the stabilized simple-cubic structure in polonium: Spin-orbit interaction versus Peierls instability, Phys. Rev. B,  73, 132102 – Published 7 April 2006
https://journals.aps.org/prb/abstract/10.1103/PhysRevB.73.132102
https://en.wikipedia.org/wiki/Polonium


percent dissociation Would these answers be correct?
. Of 100% of the NaCL molecules dissociated, the Van't Hoff factor i, would be equal to 2 for NaCl. We assumed it equaled two when we figured out Kb of the NaCl solution.
a. Now we want it find the percent dissociation of the NaCl molecules by finding i using the equation below (use Kb = 0.512℃/m and your calculated molarity of NaCl solution ) show all work
 i = ( ▵Tb) / Kb x m)
i experimental = 1.8 [ {1.8/2} * 100% = 90% dissociation]    i theoretical = 2 (100% dissociation)

my answer - % dissociation = (1.74/2) *100% = 86.9%

b. . Do you feel comfortable in making the assumption that the NaCl in your experiment was completely dissociated? Why or why not?
I do not feel comfortable in making the assumption that that the NaCl in my experiment was completely dissociate because it lead to a 16.6% difference in my experiment

Thank you

This is one of those questions where I think you'll need to know more chemistry than you probably have been taught.

The freezing point depression formula is only usable (i.e. linear) for "dilute" solutions. As the concentrations of solutes become larger, the relationship becomes non-linear.

So when $\ce{NaCl}$ is dissolved in water all the sodium and chloride ions are truly dissociated. There are no "molecules" of $\ce{NaCl}$ floating around in the aqueous solution.

Imagine the solution this way. Instead of the water molecules bouncing around in the solution more or less randomly, the positive charged $\ce{Na^+}$ ion attracts the more negative oxygen atom of the water molecule leaving the positive hydrogen atoms on the outside of this first "shell" of ligands. Now a second "shell" of water molecules will be attracted to the first (Oxygen to hydrogens) and so on. Now each shell has more and more water molecules but the binding energy decreases as the shell number increases. So the gist is that there is an exponential decrease in the ordering of the water molecules. Ultimately the ordering becomes random again as in the bulk of the water.

Now in dilute solutions these spheres of water around each ion are separated by so much water that they never interact. But as the solution becomes more concentrated then these spheres of water do start to interact. The interaction is measured by what is known as an activity coefficient.


How is this Steampunk patina being applied? Can anyone explain what is being deposited in this video @ 11:40?

It's Steampunk patina technique to make things look old and decrepit.

  "The mixture consists of: 50 g of 30% vinegar, 20 g of salt and 50 g of water. I used a 5 volt and 2 amp power supply. Connect the positive terminal to the part, and the negative terminal is lowered into the water.  Time of blackening of the metal somewhere in 10 minutes"

And the cathode is stainless steel. It wipes off with a scourer. So specifically,

What is being deposited?
Is the deposit related directly to the cathode material?
What happens if I change the cathode to say normal steel /copper wire?

I'm not a chemist, so I won't understand anything clever or those molecular spider diagrams...

I haven't found much in literature about this process, we can try to find the solution reasoning, but some experiments should helpful.

What is being deposited?

If he connects the positive terminal to the object it means that the object is the anode. So something is reduced on the anode.  Copper(II) oxide quite common oxide used for balckening. How can we obtain this by reduction? I would say that it could be the reduction of copper acetate to copper(II) oxide.

So where the copper acetate come from?
In fact, to the cathode there is also a copper wire wrapped around the steel, it could be that the copper ions are produced from that wire on the cathode:

$$\ce{Cu + 2H^{+} -> Cu++ + H2 }$$

Vinegard contains acetic acid and reacts with copper:
$$\ce{2CH3COO- +Cu++ -> Cu(CH3COO- )2}$$

Or directly from the acid attack of acetic acid:
$$\ce{2CH3COO- + 2H+ + Cu ->Cu(CH3COO- )2 +H2}$$

What could happen is the following (decribed here):
$$\ce{Cu(CH3COO- )2 +H2 -> (CH3COO- )2CuH2}$$
$$\ce{(CH3COO- )2CuH2 + Cu(CH3COO- )2 +H2O -> Cu2O +4CH3COOH}$$

Is the deposit related directly to the cathode material?

If this is true there is no relationship with the steel used at the cathode.

What happens if I change the cathode to say normal steel /copper wire?

The cathode may oxidize. If the copper wrapped around the steel has a part on the process, the process could not work if you don't use copper.

Do we not understand the role vinegar and some salt plays?

Salt is probably used to increase the conductivity of the solution.
The vinegard could be used for different reasons, but as  I said before it is probably used for producing copper acetate.

Why is this simple process open to debate and conjecture?

This is an excerpt from an article written at the New Zeland Institute of chemistry:

  Electroplating is both an art and a science. Although based on several
  technologies and sciences, including chemistry, physics, chemical and
  electrical engineering, metallurgy, and perhaps others, it retains in
  some ways the aspects of an art, in which experience is the only
  teacher. In fact, of course, all the sciences have elements of art
  which can be learned only by experience; all the reading of textbooks
  on chemistry will not produce a chemist. No text on electroplating
  will produce an expert electroplater; there is no substitute for
  experience and what is somewhat inelegantly termed know-how.

Is not easy to predict the role of each component if the experiment is poorly described, there are many variables.


Why is the calomel electrode widely preferred despite its toxicity? The following pagargraph is from the Wikipedia page of calomel:

  Mercurous chloride is employed extensively in electrochemistry, taking
  advantage of the ease of its oxidation and reduction reactions. [...] Over the past 50 years, it has been superseded by the
  silver/silver chloride (Ag/AgCl) electrode. Although the mercury
  electrodes have been widely abandoned due to the dangerous nature of
  mercury, many chemists believe they are still more accurate and are
  not dangerous as long as they are handled properly.

Yes, the calomel electrode is a breakthrough in electrochemical industry due to its ease of its oxidation and reduction reactions. But, the nature of mercury is such that every chemist fears it. There has been cases of mercury poisoning and deaths related to it. Despite its toxicity, electrochemist prefer the SCE
over the silver/silver cloride electrode. Why?

Due to its toxicity, the SCE has almost been replaced by the silver/silver chloride electrode. But still, the SCE is used as a reference electrode. Why? Why is the silver/silver chloride electrode not as good as the SCE?

Also, what are the safety measures/protocols undertaken by electrochemists while performing experiments using the calomel electrode?

[…] Why and how chemist are still able to perform experiments using Calomel electrodes despite its toxicity? […]

I mostly used non-aqueous silver electrodes as a reference for my measurements in acetonitrile.

But I wouldn't be afraid to use a saturated calomel electrode (SCE) when necessary. We have recently discussed the toxicity of mercury metal and I think it is largely overrated.

As far as $\ce{Hg2Cl2}$ (calomel), the other mercury component in the SCE is concerned, have a look as the wikipedia article that you have cited yourself. The solubility is extremely low ($\mathrm{2.3\ mg\cdot l^{-1}}$) and the ${\mathrm{LD_{50}}}$ is $\mathrm{210\ mg\cdot kg^{-1}}$ (rat, oral uptake). This isn't overly dramatic either, when compared to the infamous potassium cyanide (${\mathrm{LD_{50}}}$ $\mathrm{5\ mg\cdot kg^{-1}}$, rat, oral uptake).

Actually, calomel had medical uses :)

In an article on Venereal disease prevention, the author writes:

  […] the absence of therapeutic modalities spurred the United States Armed Services to utilize Credé's solution for urethral irrigation and locally applied calomel ointment for prophylactic purposes in World War I […]

As far as the toxicity is concerned, both mercury metal and mercury(I)chloride are by no means comparable with dimethylmercury.


Why is adiabatic process isentropic? Adiabatic process is isentropic, but I can't get it why is it so. If we go by statistical method
$$S = K\ln W,$$
where $W$ is the thermodynamic probability, and if we look into adiabatic expansion the volume increases, we can say that the thermodynamic probability increases, and hence the entropy of the system should increase.
Moreover, if we look into isothermal expansion of an ideal gas in vacuum, there would be no work done, and so no heat would enter or move out of the system. But since there is a change in volume, we can write
$$∆S = nR\ln\frac{V_2}{V_1}.$$

You left out the adjective “reversible.”  The entropy change for an adiabatic reversible process is zero.  In an adiabatic reversible process, the temperature also changes.


Colour of metal complexes with the same type and number of ligands Crystal Field Theory says that because of d-orbital splitting caused by different ligands we have 2 different energy levels for d-orbitals.

It also says that transition metals are coloured because photons excite electrons, and when they have the crystal field splitting energy through a specific wavelength of light, the electrons to go to the higher energy d-orbital.

When electrons come back down they emit the same wavelength of light and that's the colour we see.

But then why is $\ce{[Ni(H2O)6]^2+}$ green and $\ce{[Cu(H2O)_6]^2+}
$ blue? They have the same type and number of ligands.

When electrons come back down they emit the same wavelength of light and that's the colour we see.

I think this is a major misconception held by students. The color of these transition metal complexes is due to the absorption of light in the visible region. Very crudely, imagine you have six colors in the rainbow, if you remove a certain color portion from that rainbow, the transmitted light is no longer white to our eyes. For example, copper (II) absorbs red, that is why it appears blue to us. Yes, the electron has to relax back to its ground state, it does so non-radiatively.

The spacing (∆o) in the energy level determines the wavelength which will be absorbed.


Are all hydrolysis reactions non-redox reactions? Are all hydrolysis reactions always non redox reactions? For example:
$$\ce{N2O3 + 2H2O -> 2HNO2}$$
The main reason I say that this statement is correct is because oxygen and hydrogen generally show oxidation states of −2 and +1 respectively when bonded to other atoms.
This means that any change in oxidation state needs to take place between the nitrogen atoms involved in the molecular structure and this is very rarely seen.
Hence most (if not all) hydrolysis reactions are non-redox in nature. Is there a mistake in my understanding?

Hydrolysis refers to any compound reacting with water, including cases where the water acts as an oxidizing or reducing agent.  Magnesium silicide, for instance, reduces water leaving both hydrogen and oxygen from the water in negative oxidation states:
$$\ce{Mg2\overset{-4}{Si} + 4\overset{+1}{H}_2O -> 2Mg(OH)2 + \overset{+4}{Si}\overset{-1}{H}_4}$$


Why does using NaI as a nucleophile and acetone as solvent lead to an SN2 reaction mechanism instead of SN1? Is I- not considered a weak nucleophile? In a reaction with a secondary leaving group, why does using $\ce{NaI}$ with acetone as a solvent lead to an $\mathrm{S_N2}$ reaction mechanism instead of $\mathrm{S_N1}$? I thought that halogens were weak nucleophiles, and only really led to $\mathrm{S_N1}$ reactions?

As Jan wrote, there isn't a hard and fast rule encompassing every mechanism, so there may not be a definitive answer to the question. Nevertheless, I found an interesting site that talked a little bit about the Finkelstein reaction:

$$\ce{NaI + RX -> NaX + RI}\qquad\ce{X} = \ce{Cl}, \ce{Br}$$

This is in acetone solvent (I don't know how to format it along the arrow).

If you go by Henry Rzepa's blog, the solvation of sodium halides by acetone leads to a solvent shell that isn't well "fit" for a chloride or bromide anion. The writer discusses how the gaps between solvated sodium ions is just large enough to fit an iodide anion, which means the anion has minimal distance to the sodium atoms around it, and therefore a minimized potential energy. Bromide and chloride are too small, leaving a gap of empty space around them, and the distance to sodium cations is larger than optimal. It becomes energetically more favorable for the sodium and halide ions to simply fall out of solution than arrange themselves in a solvated cage. This drives the reaction forward in accordance with Le Chatelier's principle.

While this relates to the solubility of the halide salts, maybe it can also serve to speculate on the mechanism of the Finkelstein reaction. In the vicinity of the solvated $\ce{NaI}$ cage, there isn't a favorable space to release bromide or chloride into, so the iodine attack is forced to proceed through $\mathrm{S_N2}$.

This is far from rigorous research, so I'd take it with a grain of "salt" as a correct answer, but thought it might be interesting to think about.


Reactivity of Alkanes, Alkenes, and Alkynes In my textbook, it says that alkanes are generally the least reactive of the three, alkenes are slightly more reactive, and alkynes are even more reactive. However, alkanes have single bonds, alkenes have double bonds, and alkynes have triple bonds. How can the molecule with triple bonds be the most reactive if triple bonds require the most energy to break?

Maybe it will help if you look at reactivity in terms of ability to attract electrons instead of bond enthalpies. Alkanes are $sp^3$ hybridized, and hence have $25$% $s$ orbital character and $75$% p character. $s$ orbitals are closer to the nucleus and thus have a 'contracting' effect on the hybrid orbital. Greater the $s$ character, 'smaller' the hybrid orbital. This means that electrons are more closely packed. Alkanes have only 25% $s$ character, hence the hybrid orbitals are comparatively larger, and the effective nuclear charge on outermost electrons is less. Alkenes ($sp^2$) and alkynes ($sp$) have 33.3% and 50% $s$ character respectively. Thus their hybrid orbitals are SMALLER; i.e. effective nuclear charge is more. This means that it is easier to accept electrons, as now the effective strength of nucleus is more. Hence, alkynes can easily accept electrons, followed by alkenes and then alkanes. Thus, alkynes are most reactive, followed by alkenes and alkanes. Even though increase in $s$ character increases bond enthalpy, but it also increases the electron withdrawing capacity; and it is the latter which predominates.

(In fact, in general, greater the $s$ character, more the electronegativity and more the acidic nature).


Dissociation constant of molecular dinitrogen as a function of temperature? I'm looking for the dissociation constant of molecular dinitrogen ($\ce{N2}$) as a function of temperature,

$$\ce{N2 <=>[$K$] 2N} \quad K=\frac{[\ce{N}]^2}{[\ce{N2}]}$$

Casas & Bermelo (1978) reported three data points for nitrogen (I couldn't access their references):

$$K_{298}=2\cdot 10^{-160},\ K_{2000}=10^{-3},\ K_{20000}=1.29\cdot 10^5$$

Are these the accepted values today? Is there a more complete database online?
Would it be correct to fit an exponential function using these data points to extract $K$ for other temperatures?

Update: If you plot the $\log(K_p)$ vs temperature it does not give a straight line over such a large temperature range so you are unlikely to be successful.

However, you can estimate $K_p$ as the molecular properties of dinitrogen are known so you can calculate the partition functions Z and so the equilibrium constant at any temperature. You need the vibrational frequency (and possibly anharmonicity), rotational constant and dissociation energy. Work out the partition functions for translation, rotation & vibration for dinitrogen and translation for N atoms and that is all you need.

The equations needed are given in most phys. chem. textbooks and specialist books on statistical mechanics. At a standard state of $1$ bar pressure
$$k_P= \frac{Z_{trans}(N)^2}{NZ_{trans}(N_2)Z_{rot}(N_2)Z_{vib}(N_2)}\exp(-E_0/(kT))$$

The experimental values at low temperature are likely to have the biggest error since dinitrogen is such a stable molecule, and it probably does not matter much how small the $K_p$ is its always tiny at low temperatures. Using the method above I calculated values as $2.4 10^{-160}, ~ 8.4 10^{-19}$ and $ 62400$ at the three temperatures you quote (using the dissociation energy as 9.756 eV).


Are bisulfates and bicarbonates soluble in water? Generally, are bisulfates and bicarbonates soluble in water?
Specifically, are calcium hydrogen carbonate and magnesium hydrogen carbonate soluble in water?

According to it's Wikipedia page, calcium bicarbonate is not a known solid compound, but it does exist in solution, presumably prepared by dissolving calcium carbonate and bubbling carbon dioxide through the solution until bicarbonate dominates the following pH-dependent disproportionation:

$$\ce{2HCO3- <=> H2O + CO2 + CO3^2-}$$

It apparently has a solubility of 16.6 g per 100 mL of water at room temperature. which is about a 1 M solution.

Magnesium bicarbonate similarly exists only in aqueous solution, though it is much less soluble at 0.077 g / 100 mL.

As solids, these salts probably fall victim to the following equilibrium( $\ce{M}$ is a generic divalent metal:

$$\ce{M(HCO3)2 <=> H2O + MCO3 + CO2(g)}$$

The carbon dioxide escapes, preventing the system from ever staying at equilibrium.


Is Mg(OH)2 more insoluble than MgCO3? In removing temporary hardness by boiling, $\ce{Mg(HCO3)}$ is converted to $\ce{Mg(OH)2}$ but not $\ce{Mg(CO3)}$. The answer that I found in our coursebook was that it has a higher solubility product, but if its solubility product is higher, then it should be more soluble than $\ce{Mg(CO3)}$. Can you tell me the correct reason?

I finally found the answer.

In temporary hardness, we have to remove $\ce{Mg^2+}$ ions by precipitating it. In $\ce{Mg(OH)2}$, $\ce{Mg^2+}$ ions in water are present in the concentration that is cube root of the solubility product, but in $\ce{Mg(CO3)}$ it is the square root of the solubility product. So, despite $\mathrm{K_{sp}}$ of $\ce{Mg(OH)2}$ being higher than $\ce{Mg(CO3)}$, $\ce{Mg^2+}$ ions in $\ce{Mg(OH)2}$ are comparatively less than $\ce{Mg^2+}$ ions in $\ce{Mg(CO3)}$.


Can the electrophile attack at the position of the NH2 group in nitration of aniline? In the nitration of aniline, we observe the formation of a meta-substituted product (47%) almost as much as a para (52%).

However, can't the electrophile attack at the position where $\ce{-NH2}$ is already attached, so that after the attack, it will be stabilised by the mesomeric (+) effect of $\ce{-NH2}$? Maybe just after some time, the $\ce{-NH2}$ group will become $\ce{-NH3+}$ (acidic medium) and $\ce{NH3}$ would depart as a leaving group?

Maybe just after some time, the $\ce{-NH2}$ group will become $\ce{−NH3+}$ (acidic medium) and $\ce{NH3}$ would depart as a leaving group?

No! You are confusing the leaving group oxonium ion with the ammonium ion. Simply put it, $\ce{R-NH2}$ does not leave as $\ce{R+}$ and $\ce{NH3}$, as you might expect from the hydroxyl group leaving mechanism.

But that does not mean there is no other way to make an $\ce{-NH2}$ group (which is a very bad leaving group) leave. You can instead make it leave as dinitrogen gas!

The way this works is by the reaction of $\ce{R-NH2}$ with $\ce{HNO2}$. If $\ce{R}$ is an:

aromatic group: a diazonium ion, $\ce{R-N2+}$, will be formed (stable at ice cold temperature). Further reactions such as Balz-Schiemann, Gattermann, Sandmeyer, etc. may be employed to replace the $\ce{-N2+}$ with another group, and make it leave as dinitrogen gas.
alkyl group: $\ce{R-N2+}$ will be formed but it will, being highly unstable (given that it does not enjoy the resonance stabilization of an aromatic ring), instantly decompose to release dinitrogen gas and form a classical carbocation $\ce{R+}$. This carbocation may rearrange, if necessary, and then receive an attack by a nucleophile.

This is how you make the bad leaving group, $\ce{-NH2}$, leave any aromatic ring or an alkyl group.


Why does Mn (II) behave as "hard"? Why, within HSAB, does $\ce{Mn^{2+}}$ behave as a hard acid?

If we go across the third block, then all metals to the left of manganese form soft divalent cations (e.g. $\ce{Ti^{2+}}$, $\ce{V^{2+}}$, $\ce{Cr^{2+}}$) and all to the right are borderline (e.g. $\ce{Fe^{2+}}$, $\ce{Co^{2+}}$, $\ce{Ni^{2+}}$, $\ce{Cu^{2+}}$, $\ce{Zn^{2+}}$).

An example of the effect of its "hardness" on reactivity is that $\ce{Mn^{2+}}$ forms more stable complexes with oxalate ($\beta=10^{3.8}$) than with ethylenediamine ($\beta=10^{2.2}$).

It has to do with the fact that $\ce{Mn^{2+}}$ is a high-spin $d^5$ ion (in most cases) which is afforded special stability, as all frontier $d$ orbitals are half-filled. This reduces affinity for more covalent (soft) interactions that would involve disrupting the $d^5$ state by addition of electrons. So, the character is more ionic.

Editing after some additional searching: This also explains why ions such as $\ce{Fe^{3+}}$ are also hard. However, it seems to contradict in cases like organocuprates, where it would seem that the formation of $\ce{Cu(I)}$ would result in a similarly stabilized $d^{10}$ ion (same with silver). In these cases, is it simply because polarizability of transition metals in that region outweigh the $d^{10}$ stabilization?


Why does oxygen have less negative electron gain enthalpy than sulphur? Oxygen has less negative electron gain enthalpy than sulphur.

This statement is given in my book with a short reason:

  due to compact size of oxygen

I'm unable to understand why. I would appreciate any worthy explanation for this.

It is because Oxygen is first element of Oxygen family in its table and Electron gain enthaply decreasing from top to bottom in the group.

This exemption happens because of the reduced 2p orbital of Oxygen, there is enormous interelectronic r between the electrons in 2p orbitals thus Oxygen has less proclivity for an additional electron and the EGE of Oxygen is more positive than Sulphur.


Exhaust gas purification I'm more asking this because I'm not sure how far science has come.
Pretty much, I've been thinking about the car exhaust system lately. The two main components that lead to hazards from cars are $\ce{CO}$ and $\ce{NO_{$x$}}$. Carbon monoxide is hazardous because it thrives to become $\ce{CO2}$ and robs living beings of the oxygen in their bloodstream to become carbon dioxide. And the nitrogen oxides are hazardous only as that compound, but nitrogen and oxygen are quite harmless as $\ce{N2}$ and $\ce{O2}$.

Would it be possible to separate the oxygen from $\ce{NO_{$x$}}$ and somehow give that to the $\ce{CO}$, and any left over oxygen simply becoming $\ce{O2}$?

This all might be a stretch, but I have an active imagination and like seeking out answers to these questions.

This pretty much what happens (among other processes) in a three-way catalyst of a car:

$$\ce{2CO + 2 NO ->[\mathrm{cat}] 2CO2 +N2}$$


Baking soda bath to deacidify newsprint? I have some oil-pastel sketches on newsprint I would like to treat for framing (and longevity). My plan is to rinse them in a deacidifying bath, then use wheat paste to wet-mount them to heavyweight pH-neutral 100% rag paper. And finally, to frame them with UV-protective glass recommended for art.

My question is this: Is there any reason not to use baking soda for the deacidifying bath?

I have experimented with it, and it seems to work great. I used a garden-soil pH test solution before and after, and sure enough, the treated paper shows no sign of acidity. The wheat paste bond seems just as good with the treated newsprint as with the untreated newsprint. The oil pastel colors seem unaffected (not surprising, oil pastel being very water-resistant).

Is there anything about sodium bicarbonate that might, over time, cause a visible effect on the paper?

It seems that adding bicarbonate is how acid-free paper is made, though more modern paper is neutralized by the use of chalk as filler. The wiki page says that magnesium or calcium bicarbonate are commonly used, but that may be just due to cost rather than any specific issues with using sodium bicarbonate. Source


What does “to the left” mean? However, the concentration of hydronium and hydroxide ions will be very, very small; in fact, the equilibrium that's established in solution lies so far to the left, that only 18 in 1010 molecules of water will undergo autoionization.

What does "to the left" mean in the extract above? Does it mean the left side of the chemical equation?

Yes, "to the left" refers to the left side of an equilibrium expression.

Traditionally, the autoionization of water is written as

$$\ce{2H2O <=> H3O+ + OH-}$$

When we talk about equilibrium lying "to the left", it means that the educt/reactant is favored, i.e. more $\ce{H2O}$ than $\ce{H3O+}$ or $\ce{OH-}$.

Conversely, an equilibrium that lies "to the right" is one that favors products. In this case, it would favor the production of $\ce{H3O+}$ and $\ce{OH-}$ over $\ce{H2O}$,


Why would the loss of H₂ to space mean that the Earth's geochemistry would be based on acid/base reactions? I am reading a paper[1] that contains the line:

  Most of the $\ce{H2}$ in the Earth's mantle escaped to space early in the Earth's history; consequently, the overwhelming majority of the abiotic geochemical reactions are based on acid/base chemistry, i.e., transfers of protons without electrons.

The connection between the loss of $\ce{H2}$ and acid/base chemistry is not clear to me.

Why would the loss of $\ce{H2}$ to space make acid/base chemistry the default?

Falkowski, P. G., T. Fenchel, and E. F. Delong. (2008) The microbial engines that drive Earth's biogeochemical cycles. Science 320:1034-1038.

The statement you cite contains a reference to “How Earth’s atmosphere evolved to an oxic state: A status report”, D. C. Catling and M. W. Claire, Earth and Planetary Science Letters 237 (2005) 1–20. This paper in turns give some more details on this assertion, which I'll summarize below.

It all turns around the question of how reducing the prebiotic atmosphere was. The paper I cited above contains this quote:

  before life existed the atmosphere contained mainly $\ce{N2}$, with negligible $\ce{O2}$, probably more $\ce{CO2}$ than today, and only minor levels of reducing gases such as $\ce{H2}$.

So, in the absence of widely available gases such as oxidative $\ce{O2}$ or reducing $\ce{H2}$, oxygen was mainly present as $\ce{H2O}$ and $\ce{CO2}$. This means redox reactions are not big players in the overall chemical scheme of the periodic atmosphere, and acide-base reactions dominate (by default).


Which is more stable — a carbanion on a vinylic position or a carbanion on a benzylic position? I am getting a dilemma over this. Should the negative charge on the benzylic position participate in resonance with the benzene ring and make it more stable or is it the electronegativity of the $\ce{sp^2}$ hybridized carbon of the vinylic group that will make the carbanion more stable on it ?

One of the more accurate methods to measure carbanion stabilities is negative ion photoelectron spectroscopy.  In this process, a carbanion is placed (generated) in the gas phase (in a high vacuum).  The gas-phase carbanion is then irradiated with light.  Each frequency of light provides a specific energy ($E=h\nu$).  The frequency of the light is varied (and the energy supplied changes as a consequence of the change in frequency) until enough energy is supplied to knocked out of one of the carbanion electrons (this is called photodetachment).

$$\ce{CH3^{-} + hν -> CH3• +~~ e^{-} +~ KE_{electron}}$$

The kinetic energy of the photodetached electron is then experimentally measured and the electron affinity ($EA$) of the carbanion is given by

$$EA = hν ~~– ~~KE_{electron}$$

The $EA$ is the energy required to remove one electron from the carbanion. The larger the electron affinity, the more tightly the electron was held.  Stable carbanions hold their electron more tightly than less stable carbanions. The electron affinity of the allyl anion was measured as 0.481 eV.  For the vinylic carbanion, the EA was measured as 0.667 eV.

The fact that the vinyl carbanion is more stable than the allyl carbanion suggests that hybridization effects outweigh resonance effects in terms of stabilizing these carbanions.  That sounds reasonable and is what I would have guessed.

reference for EA measurements


Can we make a rusting battery Since in rusting (oxidation) of Iron, transfer of 4 electrons takes place is it possible to use this reaction under catalytic conditions to create a simple electric cell, even if it just gave a millivolt of potential.

$\ce {Fe + H2O -> FeOH3 + H2}$

Considering that we have complete control over the surroundings, like volume expansion, disposing hydrogen and other such factors.

Sure! A nickel-iron battery uses the oxidation of iron at the negative plate to produce about half the voltage. Though invented by Jungner, it's often called an Edison battery. See http://en.wikipedia.org/wiki/Nickel%E2%80%93iron_battery.

There are very low power semiconductor devices which could be powered by an iron battery. You could make three or four small cells from iron nails and a more electronegative element, e.g. lead, to power a digital watch; even without optimization to remove layers of hydrogen and rust, it should operate for months.

For some other low power ideas, see https://www.enocean.com/en/energy-harvesting/.


Diffusion of CO2 and O2 in Water/Atmosphere System Consider the following experiment:

One reservoir of pure water is interfacing the atmosphere at standard condition.
Both are at rest and at the same temperature.
Both have no internal gradients (ie: constant concentrations, temperature, ...)
The interface between them is a perfect flat surface (for sake of simplicity).
The dissolved $\ce{CO2}$ in the water has a concentration somewhere between 0-50ppm.
The dissolved $\ce{O2}$ in the water has a concentration somewhere between 0-11ppm.

Questions
in order of importance

How to mathematically model the $\ce{CO2}$ and $\ce{O2}$ rate of change per unit area (flux) in the water and in the atmosphere considering only diffusion?
Suppose the atmosphere is at constant velocity relative to the water reservoir. How this would change the mathematical model above?

Notes: Analytical methods would be appreciated instead of 'heavy' computer simulations.

Such questions are relatively standard in discussing transport phenomena, so I'll suggest Bird, Stewart, and Lightfoot's Transport Phenomena as a good reference.

We assume a flat plane of contact infinite in extent. We assume that the bulk atmosphere has a constant concentration $c_A$ of some species of interest, and that the bulk liquid has a constant concentration $c_B$ of that species. The species is only allowed to mix between the two bulk phases for times $t > 0$. Then the problem reduces to the one-dimensional diffusion equation (Fick's first and second laws) $$\partial_t c(z,t) = D\,\partial_z^2 c(z,t)$$ subject to the initial condition $$c(z,0) = \frac{c_A+c_B}{2} + \frac{c_B-c_A}{2}\text{sgn}(z),$$ and the boundary conditions $$c(-\infty, t) = c_A, \quad c(+\infty, t) = c_B,$$ where $D$ is the species' diffusion constant and $\text{sgn}(z)$ is the signum function.

For ease of solution, make the substitutions $$\bar{c}(z,t) := c(z,t) - \frac{c_A+c_B}{2}, \quad \Delta c := \frac{c_B-c_A}{2}, \quad \phi(z,t) := \frac{\bar{c}(z,t)}{\Delta c},$$ which yields $$\partial_t\phi(z,t) = D\,\partial_z^2 \phi(z,t), \quad \phi(z,0) = \text{sgn}(z), \quad \phi(\pm\infty, t) = \pm1.$$ By definition $\phi$ is a dimensionless quantity, so it must depend on the variables relevant in the problem in a dimensionless form, for otherwise $\phi$ would have units. The relevant variables in our problem are the diffusion constant $D$, the length $z$, and the time $t$.

The only dimensionless variable that arises from this set of quantities is $\eta := z/\sqrt{Dt}$, so we know $\phi(z,t) = \phi(\eta)$. Substituting this result into our diffusion equation and applying the chain rule yields the ordinary differential equation $$-\frac{\eta}{2}\phi'(\eta) = \phi''(\eta).$$ Integrating twice yields the result $$\phi(\eta) = \alpha + \beta\int_0^\eta\text{d}\eta'\,\exp\left(-\frac{\eta'^2}{4}\right),$$ and it remains to determine the constants $\alpha$ and $\beta$ from our boundary conditions. Actually, for standardization, we will first take $\eta' \to 2\eta'$ and $2\eta \to \eta$, so $$\phi(\eta) = \alpha + 2\beta\int_0^\eta\text{d}\eta'\,\exp(-\eta'^2).$$ By symmetry $\alpha = 0$, and by explicit evaluation $\beta = 1/\sqrt{\pi}$, and hence we have $$\phi(\eta) = \text{erf}(\eta), \quad \text{or} \quad \phi(z,t) = \text{erf}\left(\frac{z}{\sqrt{4Dt}}\right).$$ Going back and untangling all our substitutions, we end up with $$c(z,t) = \frac{c_A+c_B}{2}+\frac{c_B-c_A}{2}\text{erf}\left(\frac{z}{\sqrt{4Dt}}\right).$$ The flux then follows as $$J(z,t) = -D\,\partial_zc(z,t) = -D\,\frac{c_B-c_A}{\sqrt{4\pi Dt}}\exp\left(-\frac{z^2}{4Dt}\right),$$ and we are done.

Regarding your second question, I don't believe the movement of one phase relative to the other will have any effect. As we've seen above, the xy-plane essentially drops out of the picture, and you won't be introducing any further gradients in chemical potential by translating one phase relative to the other, since the concentration of each phase is a function of $z$ only. Of course, if you had a finite area of contact, then things would be different. Perhaps see BSL 18.5, in this case.

In addition, keep in mind that there's important chemistry that's being neglected in this simplistic model. For example, the dissolution of carbon dioxide in water follows not just from diffusion but also from reaction with water, forming bicarbonate ion and related species in solution. This model doesn't account for that.


How to correctly prepare a standard solution? Having just spoken to my teacher about the creation of a standard solution, and she suggested one way of doing it, however a previous chemistry teacher had told us a different way of doing it. My question is, (of the below methods) which is the "correct" way of doing it, and what are the pros/cons of doing one over the other? If you're interested in the full procedure of making this/any other standard solution (with the names of compounds, it is page 24 here).

Method 1:

The solid was weighed accurately in a weighing bottle, and then transferred to the volumetric flask, and then the bottle is re-weighed to calculate the amount of solid transferred to the flask.

Method 2:

Rather than re-weighing the solid in the weighing bottle, the weighing bottle is rinsed with DI water, and then poured into the volumetric flask. (this is repeated a few times).

If I'm not including enough detail in the above methods please let me know and I'll try to expand. For reference the 2 methods are Part 1, step 3 of the above link.

The reason you are using a weighing glass is because it is a lot smaller and it has a much smaller mass than a volumetric flask. These glasses usually are more stable, less affected by temperature, and pressure changes, or drafts in the laboratory. The smaller surface also tends to not accumulate much dirt due to static, your hands, etc.

If you were using method one, then you could skip the part with the weighing glass altogether. This method suffers from getting a stable reading as due to the size it is more affected by the conditions in the laboratory.
Method two is certainly the better method, because the total error tends to be much smaller. When you are using such equipment, then your concentrations are typically smaller than 1 mol/L and you are working with small quantities of solid substances. Continuously rinsing out the weighing vessel will ensure to transfer almost all of your solute.
If you were to produce a saturated solution, then the mass of the solute would not matter in the first place.

If your compound does not readily dissolve then the attempted concentration is too high to be used for analytical chemistry and you should rather start from a saturated solution. All primary standards are easily dissolved. If you are not dissolving a primary standard, then your should check the concentration against one.

Therefore, method one can be used if accuracy is of a minor concern (leave out the weighing glass) because it has to be determined again with a primary standard. Method two should be used when preparing accurately concentrated solutions of primary standards.


What is the net ionic equation for rection of liquid bromide with sodium iodide crystals? I have this problem,

Liquid bromine is added to a container of sodium iodide crystals

I interpreted it with this molecular formula $\ce{Br_{2(l)} + 2NaI_{(s)} \rightarrow 2NaBr_{(s)} + I_{2(g)}}$.

From my understanding, since none of the products nor any of the reactants is aqueous, there shouldn't be a net ionic equation. However, my friend has proceeded with the problem as follows:

$\ce{Br_2 + 2Na^+ + 2I^- \rightarrow 2Na^+ + 2Br^- + I_2}$

$\ce{Br_2 + 2I^- \rightarrow 2Br^- + I_2}$.

Could somebody explain which one of us is correct and why?

Technically you are correct. How can there be any ions when the reactants are not in aqueous medium ? But here it is a matter of convention and maybe an illustration of how to write the ionic equations from molecular ones. Likewise, if you have done balancing of redox reactions , you might notice half of the reactions given to balance are completely impossible according to the laws of chemical combination. You'll always be at a loss when you overthink in chemistry, even if your argument is right.  Taking it as just a convention would be a better idea .

Hope this helped !


What is the transparent film that covers water-soluble dishwasher tablets? These dishwasher tablets (Finish Powerball Shine) seem to be covered in plastic!

But that can't possibly be true since there's no unwrapping, and they're completely soluble in water.

What is the film made of?

It is made using Polyvinyl Alcohol (Mowiflex material).


Effect of +M and +R effect in basicity of 4-methoxy aniline We know that methoxy group exerts a +M effect, as a result of which basicity should increase even after resonance as +M effect is considered to be stronger than resonance effect.
So by this reason 4-methoxy aniline must be more  basic than cyclohexane amine, but this is not true.
I am not able to understand what factor should make cyclohexane amine more basic than 4-methoxy aniline.

Since you mentioned the $\mathrm +M$ effect, I'm assuming you're familiar with resonance and aromaticity.
This is a structure of cyclohexane amine:

Here, the lone pair of electrons present on the Nitrogen atom make it a Lewis Base (an electron donor).
This is a structure of 4-methoxy aniline:

Here the lone pair of Nitrogen is delocalized due to resonance. Hence it's not readily available for donation. That's why 4-methoxy aniline is a weaker base.


The source for the N-N bond dissociation energy of 240 kJ/mol in the table Authoring an exam question, I made the mistake of taking a bond dissociation energy from a table I found on the internet. It comes in different styles, but is always labeled Table 7.1. It gives a bond dissociation energy of $\pu{240 kJ/mol}$ for the - admittedly uncommon - nitrogen:nitrogen single bond. A student was puzzled that the enthalpy of formation for hydrazine, estimated based on this value, was so far off from the experimental value; the student then found a value for the bond dissociation energy of $\pu{159 kJ/mol}$ in a more rigorous search.

The General Chemistry textbooks available to me from home have values close to $\pu{159 kJ/mol}$ (OpenStax Chemistry: $\pu{160 kJ/mol}$; Kotz 6th ed (2006): $\pu{163 kJ/mol}$; Radel (1994): $\pu{170 kJ/mol}$). I am not worried about the small differences - it depends which set of molecules containing $\ce{N-N}$ bonds was used, and a substructure search of commercially available compounds shows that there is quite a variety, with many having partial double-bond character (the following sample are 3 from the 100 or so just from Sigma Aldrich):

My question is: Where did Table 7.1 originate, and where did the value for $\ce{N-N}$ originate?

Here is a selection of links to variations of table 7.1, with the first example as picture:

https://wps.prenhall.com/wps/media/objects/4678/4790506/ch07_02.htm

https://www.chegg.com/homework-help/questions-and-answers/use-bond-dissociation-energies-table-calculate-approximate-h-kilojoules-reaction-acethylen-q27780023

https://www.chegg.com/homework-help/questions-and-answers/average-bond-dissociation-energies-d-kj-mol-h-h-4362-h-c-410-h-f-570a-h-ci-432a-h-br-3662--q36015212

https://cdn.clutchprep.com/problem_images/65691-36fd337a4e3e837c.png

Your question about the source of Table 7.1 is correctly answered by the andselisk's comment. The value of $\pu{240 kJ\:mol-1}$ for $\ce{N-N}$ bond is probably initiated from Encyclopedia of Inorganic Chemistry, which has listed bond energies of wide variety of bonds and their respective bond lengths. It has given two values for $D_\circ$ of $\ce{N-N}$ bond: ~$\pu{167 kJ\:mol-1}$ for $\ce{N2H4}$ in general and $\pu{247\pm 13 kJ\:mol-1}$ for $\ce{H2N-NH2}$ in particular.

The reason is given in the very first paragraph of the article. In their words:

  There are two different ways to define the bond energy even for the simplest diatomic molecule (see the diagram below). $\Delta D_{\circ'}$ is the dissociation energy measured from the very bottom of the potential energy well. However, the molecule possesses zero-point energy and thus, the experimentally measured dissociation energy $\Delta D_{\circ}$ is somewhat less:

In poly atomic compound there are other considerations. For example, the mean bond energy of $\ce{N-H}$ is the average of three bond energies associated with sequential fission of the three $\ce{N-H}$ bonds in ammonia ($\ce{NH3}$). These three values are different from each other and also different from their average, the mean bond energy of $\ce{N-H}$. Using this mean bond energy of $\ce{N-H}$ in $\ce{NH3}$ and the heat of atomization of $\ce{H2N-NH2}$, a value of $\pu{159 kJ\:mol-1}$ has been found for the $\ce{N-N}$ single bond energy. Then again, using the mean bond energy of $\ce{N-F}$ in $\ce{NF3}$ and the heat of atomization of $\ce{F2N-NF2}$, a value of $\pu{172 kJ\:mol-1}$ has been found for the $\ce{N-N}$ single bond energy. These two values are in somewhat good agreement. However, the dissociation of the two molecules are quite different:

$$\ce{H2N-NH2 -> 2 H2N^.} \qquad \pu{247 kJ\:mol-1}$$
$$\ce{F2N-NF2 -> 2 F2N^.} \qquad \pu{88 kJ\:mol-1}$$

Therefore, it is warned that care should be exercised in using any values of the bond energies.

Source:

R. Bruce King (Editor in chief), "Bond Energies," Encyclopedia of Inorganic Chemistry; 2nd Edition, John Wiley & Sons, Ltd.: New York, NY, 2005 (ISBN: 978-0-470-86078-6). Bond Energies is available online: https://doi.org/10.1002/0470862106.id098


Why does pyridine make such a big change from SNi to SN2? When an alcohol is treated with SOCl2 alone, retention through SNi takes place, whereas adding pyridine makes the reaction follow SN2 mechanism and inversion takes place. What concept lies behind this occurrence?

Both cases share, that the reaction of the alcohol and the thionyl chloride yields an alkyl chlorosulfite under liberation of HCl. Then

Either the C-O bond of the alkyl chlorosulfite brakes, a (tight) ion pair of carbo-cation and chlorosulfit anion is formed.  The later decomposes into $\ce{SO2}$ and $\ce{Cl-}$, the latter reacting with the carbo-cation.  This yields the rentention of configuration observed.

.OR.

In presence of pyridine, HCl generated does not leave the reaction mixture (as above), but reacts with pyridine to yield pyridinium chloride.  Pyridinium chloride is able to react along SN2 on the alkyl chlorosulfite.  Seeing this variant from the beginning till the SN2 of the alkyl chlorosulfite yields the inversion of configuration observed.


Why does the potential difference of a current drop, when silver nitrate is added to deionised water? Semi-related to Why do some salt ions in solution conduct electricity better than others?

After conducting more experiments, using deionised water as an analyte only, I found the potential difference decreases as more silver nitrate is added. This goes against what I was lead to believe that it is actually disruption in forms of "pure" water that actually conduct electricity (i.e. pure water should not technically conduct electricity at all).

This graph shows how the potential difference drops, from my findings. The last data points plotted are at 1cm3 of silver nitrate added; the interval is 0.05cm3.

So, why is it that I see a massive drop in potential when even the slightest (0.50cm3) amount of silver nitrate is added?

The potential of an electrode in the absence or paucity of redox reactions at the interface is not predictable using the Nernst equation. This situation is known as an unpoised electrode and the electrode's measured potential is related to the formation of an electrical double layer at the interface, which behaves very much like a capacitor.

If you try to measure the potential of an electrode that won't easily oxidize or reduce in deionized water, due to the very low concentration of ions, establishing the double layer is slow and easily disrupted. You can think of it as having a large solution resistance in series with the double layer capacitance. When there are few ions, the resistance is very high. On that same note, how good the voltage measurement instrument is becomes important. Any current that a voltmeter draws will induce a large voltage at the electrode (as there is a large resistance to bringing ions to the interface, excess charge at the electrode is not easily balanced)—a voltmeter with a 10 nA input bias current may induce 100–200 mV of error in a very resistive solution, so it takes very good instruments to deal with these types of measurements.

This error is proportional to the solution resistance and even adding a tiny amount of ions to deionized water will dramatically reduce the resistance. I'm guessing that repeating the experiment with a supporting electrolyte will prevent this effect.

The other trouble here is that neither of your electrodes can be assumed to maintain a fixed potential in your solution. The potential of the silver electrode should vary according to the Nernst equation once you have an appreciable quantity of silver ions in solution, but the graphite electrode is not in equilibrium with any components of your solution and its potential is free to fluctuate as the solution composition changes. Without a supporting electrolyte, adding the silver nitrate may have a large effect on the potential of the graphite electrode.


Failed esterification of 1-butanol and ethanoic acid I did an esterification experiment of 1-butanol and ethanoic acid.

30 mL of 1-butanol was added followed by 30 mL of ethanoic acid, and then 1.5 mL of concentrated sulphuric acid and some boiling chips.

The mixture was colorless at first. However, it turned brown during the reflux of esterification immediately after heating the solution, and the solution had a burnt fruity smell.

I suspect this is because I didn't swirl the mixture before heating. Also, before heating, the bottom of the round bottom-flask felt warm, and I suspect this could be the reaction between the concentrated sulphuric acid and ethanoic acid.

One thing that I can't seem to understand is, how did the color of the solution turn brown? Could it be that the product of the reaction between sulphuric acid and ethanoic acid reacted with the 1-butanol to produce a colored solution?

P.s: I couldn't obtain any pure ester after the distillation of the ester as the temperature didn't rise above 100 °C.

I don't think the brown color necessarily means that the reaction failed. It could be due to some contaminant in your reagents but in a small quantity that wouldn't cause the reaction to fail. The fruity smell could very well be due to your product see: n-butyl acetate has a sweet banana smell. I would take the distillate you collected (or skip the distillation altogether) and try an aqueous work-up: Add very carefully aqueous $\ce{Na2CO3}$ until basic $\ce{pH}$ and then extract this mixture with ether. This way any ester product will be in the the organic layer and sulfuric/acetic acid will stay in the aqueous together with most of the alcohol.

To finish the work-up dry the ether layers over $\ce{Na2SO4}$, filter and remove the ether (unless you have access to a Buchi evaporator, the safest way is to leave it in an open flask/beaker overnight). You can further purify the final product by distillation.

You might improve the yield if you cool the reaction mixture in an ice bath before adding sulfuric acid


Is beryllium an alkaline earth metal? For context: I got a question asking, "Which of the following alkaline earth metals do not give flame colour?". I quickly marked $\ce{Be}$ and $\ce{Mg}$ and got negative marks.

The following is a quote from my book (emphasis in the original):

  The elements of Group 2 include beryllium, magnesium, calcium, strontium, barium and radium. These elements with the exception of beryllium are commonly known as the alkaline earth metals. These are so called because their oxides and hydroxides are alkaline in nature and these metal oxides are found in the earth's crust.

As you can see, it's stated that beryllium is not an alkaline earth metal!

Is beryllium an alkaline earth metal or not?

There is some disagreement in usage among authors, but IUPAC standard nomenclature approves calling beryllium an alkaline earth metal, as explained on page 51 of IUPAC's last Red Book.

In fact, all the elements belonging to group 2, $\ce{Be,Mg,Ca,Sr,Ba,Ra}$, are called alkaline earth metals with IUPAC's approval.

Other common traditional names approved by IUPAC are alkali metals for the elements of group 1 except hydrogen, i.e. $\ce{Li,Na,K,Rb,Cs,Fr}$; halogens for $\ce{F,Cl,Br,I,At}$ in group 17,whose only member excluded from such a designation is $\ce{Uus}$ (now $\ce{Ts}$), and noble gases for all the elements of group 18 except $\ce{Uuo}$ (now $\ce{Og}$), i.e. $\ce{He,Ne,Ar,Kr,Xe,Rn}$.


Why do alkali metals have an exothermic electron affinity? Alkali metals should have positive electron gain enthalpy as they are electropositive elements and also there atomic size is big in their periods so they should be reluctant to take electrons but they show negative electron gain enthalpies. Why?

I think you mean to ask about the electron affinity of the alkali metals. That is, the enthalpy gain or lost when adding an electron to an atom.

While the electron affinity for alkali metals are much smaller than the halogens, they are slightly positive. That is, quoting from Wikipedia:

  Electron capture for almost all non-noble gas atoms involves the release of energy and thus are exothermic.

The catch, is that a positive electron affinity (Eea) refers to an exo-thermic process:

  Confusion arises in mistaking Eea for a change in energy, ΔE, in which case the positive values listed in tables would be for an endo- not exo-thermic process.

You refer to the alkali metals as "electropositive" -- in this case, you're referring to their electronegativity relative to, say, halogens. Yes, a sodium atom is likely to become a sodium cation and give up an electron. But the electron configuration is $3s^1$ and consequently, can also gain an electron and become the more stable $3s^2$ state.

It's more likely that a sodium atom will lose an electron, and become a cation, but the electron affinity reflects the modest gain in stability as $3s^2$ after attaching an extra electron.


Boiling chips for concentrating sulfuric acid I am concentrating sulphuric acid. What should I use as boiling chips? I have tried broken mug pieces, the solution became cloudy, and I have tried broken glass (retrieved from a drinking class, not crystal) and the solution became yellow.

Clean porcelain or glass shouldn't react with concentrated sulfuric acid.
For boiling chips you want to use homogeneous material with a substantial/developed surface:

Porous unglazed ceramic.
Raschig rings, which are also widely used industrially for the similar distillation purposes, are probably the ideal choice.
An alternative for a consumer market would be porous ceramic rings for aquarium filters, which is basically the same thing (stick with the white ones).

The majority of glass types (borosilicate glass, quartz, …).
Broken pieces of chemical glassware or light bulbs would do.
For higher efficiency, use glass capillaries or sintered glass pieces.

Potential cause of cloudy solution when you used mug pieces might be the glaze on top of porcelain or the porcelain itself if it hasn't been properly thermally treated (remaining binder), and, of course, porcelain dust particles leftovers may float around forming a suspension.
Yellow color of sulfuric acid when you used glass pieces is likely an indicator that acid came in contact with organics (e.g. oil, grease).
Pure colorless glass won't cause this coloration.
On a side note, there is no need to use grease at all as sulfuric acid itself is quite viscous and can serve as a sealant for the joints (if you are using any).
Notes, advises

If the liquid has been cooled down, make sure to add at least one new chip before heating again to the boiling point (prevents superheating).
Avoid using boiling chips for vacuum distillation.
Instead, use mechanical stirring or bubble inert gas (argon, nitrogen) through a submerged glass capillary.
Boiling chips don't work under vacuum.
Never add boiling chips to a hot solution as it will result in rapid vaporization and the liquid is going to boil over or bump.
Keeping a substantial amount of sodium carbonate nearby might come in handy in neutralizing spilled acid, if there is any.


Is there a physical explanation why benzene's pi molecular orbitals in a Frost circle are degenerate but pyridine's are not? From here, the Frost Circles for both benzene and pyridine:

The $\pi$-MOs formed by 6 $\pi$-AOs of an aromatic ring are:

Listed with respect to the ordering in the above image, these MOs in benzene's $D_{6h}$ symmetry correspond to the group representations:

$\pi_6=B_{2g}$

$\pi_4 / \pi_5=E_{2u}$

$\pi_2 / \pi_3=E_{1g}$

$\pi_1=A_{2u}$

The $D_{6h}$ symmetry of benzene is reduced to $C_{2v}$ in pyridine. I believe the pyridine MOs transform in $C_{2v}$ symmetry as:

$\pi_6=B_2$

$\pi_5=B_2$

$\pi_4=A_2$

$\pi_3=A_2$

$\pi_2=B_2$

$\pi_1=B_2$

Onto my question:

Because the doubly degenerate $E$ representation does not exist in the $C_{2v}$ symmetry of pyridine, the corresponding $\pi$ orbitals are no longer degenerate. I could just stop there and accept that symmetry dictates degeneracy and there's nothing more for me to understand, but is there a way to dig a level deeper and connect the degeneracy to a physical explanation?

For example, I can imagine nodes breaking the extended conjugation of the $\pi$-atomic orbitals, which would explain why the $\pi$-MO with zero nodes is the lowest energy bonding orbital, and the $\pi$-MO with the most nodes the highest energy antibonding orbital. In this way, I can understand why the pyridine MOs climb in energy with increasing nodes. However, if it were solely about nodes and conjugation, then the benzene $\pi_2 / \pi_3$ and $\pi_4 / \pi_5$ MOs should not be degenerate; the one with more nodes would be higher lying in energy.

Or maybe the $\pi_3$ MO in pyridine is higher in energy because it has no $\pi$-atomic orbital situated on the nitrogen atom—the node is located on the nitrogen. Nitrogen being more electronegative than carbon, there should be a lower potential with some electron density localized in an atomic orbital on nitrogen. Therefore, the $\pi_2$ MO would lie lower in energy than its $\pi_3$ counterpart because nitrogen possesses an atomic orbital in the $\pi_2$ MO.

Is there a physical explanation why these orbitals in benzene are degenerate, and vice versa why the corresponding orbitals in pyridine are not?

My first thoughts are along the lines of your second-last paragraph. Here is a quick sketch of how to formalise it. From simple Hückel theory, you can obtain the coefficients of the AOs in the MOs:

$$|\psi_i\rangle = \sum_a c_{n,i} |n\rangle$$

where $c_{n,i}$ denotes the coefficient of AO $|n\rangle$ in the MO $|\psi_i\rangle$.

In Hückel theory the main difference between benzene and pyridine is that one atomic orbital is shifted in energy, so one of the $\alpha$ terms turns into, say, $\gamma$. Without loss of generality we will label the nitrogen with the index $n = 1$. In theory, the resonance integrals $\beta$ involving the nitrogen should change as well, but to a first approximation we assume they are the same.

In this case therefore the Hamiltonian matrix, when cast in the AO basis, changes from $\mathbf{H}_0$ to $\mathbf{H}$:

$$\mathbf{H}_0 = \begin{pmatrix}
\alpha & \beta & 0 & 0 & 0 & \beta \\
\beta & \alpha & \beta & 0 & 0 & 0 \\
0 & \beta & \alpha & \beta & 0 & 0 \\
0 & 0 & \beta & \alpha & \beta & 0 \\
0 & 0 & 0 & \beta & \alpha & \beta \\
\beta & 0 & 0 & 0 & \beta & \alpha
\end{pmatrix} \qquad \longrightarrow \qquad \mathbf{H} = \begin{pmatrix}
\gamma & \beta & 0 & 0 & 0 & \beta \\
\beta & \alpha & \beta & 0 & 0 & 0 \\
0 & \beta & \alpha & \beta & 0 & 0 \\
0 & 0 & \beta & \alpha & \beta & 0 \\
0 & 0 & 0 & \beta & \alpha & \beta \\
\beta & 0 & 0 & 0 & \beta & \alpha
\end{pmatrix}$$

Most of the Hamiltonian is unchanged, so we could consider this within the framework of perturbation theory, where we have a perturbed Hamiltonian $\hat{H} = \hat{H}_0 + \hat{V}$. The relevant perturbation is therefore

$$\mathbf{V} = \begin{pmatrix}
\gamma - \alpha & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0
\end{pmatrix}$$

or to condense this back into operator form, $\hat{V} = (\gamma - \alpha)|1\rangle\langle 1|$.

The proper mathematical analysis is slightly annoying because the (unperturbed) eigenstates that we're interested in looking at are degenerate, which then necessitates that you cast the matrix $\hat{V}$ in the basis of the degenerate states of interest. Here, the two states we're interested in are $|\psi_2\rangle$ and $|\psi_3\rangle$. The application of Hückel theory to benzene gives the formulae for these unperturbed states as

$$\begin{align}
|\psi_2\rangle &= \frac{\sqrt 3}{3}|1\rangle + \frac{\sqrt 3}{6}|2\rangle - \frac{\sqrt 3}{6}|3\rangle - \frac{\sqrt 3}{3}|4\rangle - \frac{\sqrt 3}{6}|5\rangle + \frac{\sqrt 3}{6}|6\rangle \\
|\psi_3\rangle &= \frac{1}{2}|2\rangle + \frac{1}{2}|3\rangle - \frac{1}{2}|5\rangle - \frac{1}{2}|6\rangle
\end{align}$$

Note that this choice of coefficients is not unique, as the two states are degenerate: any linear combination of these two MOs is also a permissible MO with the same energy. In any case, the relevant matrix elements of $\hat{V}$ in the new basis $\{|\psi_2\rangle,|\psi_3\rangle\}$ are given by

$$V_{i-1,j-1} = \langle \psi_i | \hat{V} | \psi_j \rangle$$

and if you assume that $\langle n | n' \rangle = \delta_{nn'}$ (i.e. the AOs have no overlap with each other, which is a key assumption of simple Hückel theory anyway), then it becomes very simple:

$$\mathbf{V} = \begin{pmatrix} \frac{\gamma-\alpha}{3} & 0 \\ 0 & 0\end{pmatrix}$$

Perturbation theory tells us that the eigenvectors of $\hat{V}$ are the "states which are stable towards the perturbation". What this means is that, since the degeneracy is lifted by the perturbation, you also remove the free choice of having any linear combinations be a permissible MO: the coefficients of AOs in the new MOs must be uniquely determined (ignoring the phase factor). Since $\hat{V}$ is diagonal in the basis $\{|\psi_2\rangle,|\psi_3\rangle\}$, it conveniently means that both $|\psi_2\rangle$ and $|\psi_3\rangle$ are the eigenvectors of $\hat{V}$.

More interesting here is the first-order corrections to the eigenvalues of $\hat{H}_0$, which are simply the eigenvalues of $\hat{V}$. In this case, since $\hat{V}$ is diagonal, the eigenvalues are just $(\gamma-\alpha)/3$ and $0$: that is to say, the MO $\psi_2$ is shifted in energy by $(\gamma-\alpha)/3$ and the MO $\psi_3$ is unshifted in energy.


What type of intermolecular forces will dominate Diethyl ether? Is Diethyl ether (also known as ethyl ether) a polar molecule? What type of intermolecular forces dominate it? Dipole-Dipole Interactions, London Dispersion Forces or Hydrogen Bonding? Please Explain.

As you have already figured out, diethyl ether is a small dipole due to the electronegativity difference between carbon and oxygen, which can also be seen in this electron density plot. As both dipole moment vectors point roughly in the same direction, they do not cancel each other out like in the case of carbon dioxide, so the molecule has a net dipole moment and is therefore (weakly) polar.

London dispersion forces, which result from short-lived dipoles induced by fluctuations in the electron shell of molecules, are also present. They are responsible for the weak interactions between the alkyl chains, like in other alkanes and non-polar molecules. The permanent dipoles can also interact with each other, in the form of intermolecular attractive interactions between centers with $\delta+$ and $\delta-$ partial charge (and likewise repulsion between centers of equal partial charge).

Hydrogen bonding between C-H groups and oxygens, although possible in principle (see comments to your question) and, for example, found in protein structures, can be neglected in this case due to the low number of donors and the weakness of the interaction (< 17 kJ/mol). Nevertheless, the ether oxygen can act as a hydrogen bond acceptor in combination with better hydrogen bond donors like OH groups, which explains why diethyl ether is sparingly soluble in water despite its low polarity (69 g/l at 20 °C (source)). On the other hand, the absence of hydrogen bonding between ether molecules becomes apparent in the much lower boiling point of diethyl ether (34.6 °C, source) compared to water.


Please explain the shapes of the orbitals For a lot of years, I had been believing that sphere was the most stable 3-dimensional shape. But after coming across the p,d and f-orbitals, I am unable to comprehend the fact that these orbitals have such crude shapes. Can we prove that these shapes(dumb bell and crossed dumb bell) are the regions in which the electrons would be most stable (No wave functions please. I am only in high school)?

P.S.: An intuitive answer would be much appreciated than a mathematical one.

P.S.S: This is not a duplicate question. I don't want to know how the shapes are like that. I know it is due to the probability functions, I want to know if it is just the rule of nature, or can we show it is because of stability.

I'm guess that you have read about orbital shapes in the wikipedia article, or done a google search on the term.

In general, such "orbitals" shown are typically calculated for a lone electron, not any "real" multi-electron configuration.

Think of an orbital like a single loop of cotton fiber wound into a cotton ball. The planetary model of the electron-nucleus pair indicates that the electron is a solid ball following the fiber. The quantum model of the electron-nucleus pair indicates that the electron has no fixed position, but that it is essentially at all places on the loop at the same time.

So how big is the cotton ball in diameter? It has no limiting diameter, it depends on how hard you squeeze! So if I catch a Hydrogen atom with my magic tweezers, there is a finite probability according to the Schrödinger wave equation that its electron can be in orbit even further away than the dwarf planet Pluto, or inside the nucleus! So we we talk of an the "size" of an ion it is somewhat of an artificial abstraction, and the Schrödinger wave equation can't be "the gospel truth." (I don't mean to badly disparage the Schrödinger wave equation for it is very useful.)

Not to leaving you totally confused, the size of an ions does have some real physical significance. For instance consider table salt, $\ce{NaCl}$. This is really $\ce{NaCl}$. Using x-ray diffraction we can measure how close the $\ce{NaCl}$ are, so we know their "size."

Starting with the Aufbau principle for the atoms, chemists can predict the electron structure of an atom, say carbon. Given the structure of bonds in the atoms, say carbon and hydrogen, then chemists can predict how carbon and hydrogen will bond to form molecules. The fact that chemists can make predictions about molecular structure is the "proof" that the models work.

Howver weird things do have with real electron orbitals. For instance if there was a single "normal electron configuration" for chromium, then there would only be one chromium chloride. However chemists have synthesized three.

Chromium(II) chloride, also known as chromous chloride.
Chromium(III) chloride, also known as chromic chloride or chromium
trichloride
Chromium(IV) chloride

So there must be weird configurations that are stabilized by some sort of "hybridization." As another example all the C-H bonds in methane, $\ce{CH4}$, are the same because the four carbon orbitals (one 2s, and three 2p) orbitals of the carbon atom hybridize into four $sp^3$ orbitals that are equivalent.

To put such orbital shapes into perspective, a chemist thinks a unicorn is simply a hybrid of a horse and a rhinoceros. So a model helps making predictions, but you can't have a fixated belief that the model is the whole
truth.

Another example would be showing you a picture of a blueberry pie. Would just such a picture "explain" a blueberry pie? In order for the picture to have meaning you must have some underlying knowledge.

You know in general what a pie is.
You have know that pies contain fruit pieces, so this kind is made
from blueberries.
You know that pies are sweet.


What type of compound/element might be in these glowing tiles I saw these tiles and these as well while surfing the net, they obiviously glow when their electrons are exited to a higher state and then de-exite over a long period to give off visible light.

My question is which chemical might get exited by visible light and them emit back visible light for a long time.

Glow in the dark materials are materials that exhibit phosphorescence. The most common materials used for this effect are zinc sulfide and strontium aluminate, doped with various elements that affect the colour of luminescence.

Phosphorescence is similar to fluorescence, where an electron is excited to a higher energy state, but instead of dropping back immediately by emitting a photon, an intersystem crossing event occurs where the electron's spin flips resulting in a triplet state (S=0 to S=1 in the figure). Because the electron has the same spin as the other electron it left behind in the ground state, the transition to the ground state is forbidden and thus occurs much more slowly. In practice, a given material usually exhibits both fluorescence and phosphorescence. Materials that are very phosphorescent have a high probability of intersystem crossing so that when light is shined on them, a large population builds up in triplet states that will release light slowly. What makes a given material have a high intersystem crossing probability is pretty complicated, but it tends to be higher in molecules with large atoms—even things like fused silica glasses are phosphorescent, but very weakly.

Source


Potassium sulfate as a byproduct of manganese heptoxide synthesis? Manganese heptoxide may be  formed by the action of concentrated sulfuric acid on potassium permanganate (ref):

$$ \ce{2KMnO4 + 2H2SO4 -> Mn2O7 + H2O + 2KHSO4} $$

Could the amount of potassium permaganate as a reactant in the reaction be increased to yield potassium sulfate as a product instead of potassium bisulfate as shown?

The reaction you mentioned shows the formation of manganese heptoxide and potassium bisulfate. But if you alter the stoichiometry of the reaction, i.e. change the moles of the reactant, you will obviously get different products like manganese(II or III)sulfate, manganese oxide, potasium sulfate etc. For this, the reaction conditions should be changed.

$$\ce{2KMnO4 + 2H2SO4 ⟶ Mn2O7 + H2O + 2KHSO4}$$

  Potassium permanganate react with sulfuric acid to produce
  manganese(VII) oxide, potassium bisulfate and water. Potassium
  permanganate - solid. Sulfuric acid - 98% solution. The reaction
  proceeds at room temperature. Impurities: manganese(IV) oxide
  $\ce{MnO2}$(source)

$$\ce{2KMnO4 + H2SO4 → Mn2O7 + K2SO4 + H2O}$$

  Potassium permanganate react with sulfuric acid to produce
  manganese(VII) oxide, potassium sulfate and water.(source)

$$\ce{4KMnO4 + 6H2SO4 → 4MnSO4 + 2K2SO4 + 5O2 + 6H2O}$$

  Potassium permanganate react with sulfuric acid to produce
  manganese(II) sulfate, potassium sulfate, oxygen and water. Sulfuric
  acid - 60% solution. Impurities: ozone $\ce{O3}$.(source)

$$\ce{2KMnO4 + 4H2SO4 ->[70-75°C] Mn2(SO4)3 + K2SO4 + 4H2O + 2O2}$$

  Potassium permanganate react with sulfuric acid to produce
  manganese(III) sulfate, potassium sulfate, water and oxygen. Sulfuric
  acid - 98% solution. This reaction takes place at a temperature of
  70-75°C.(source)


Does pure iron not rust? In my school science textbook, written more than twenty years ago, it said that pure iron does not rust. Accompanying was a photograph of an ancient iron statue situated outside, which had not rusted.

Is this correct?

The purity of the iron is less important here than the conditions of the atmosphere and other surroundings.  Rusting of iron involves the formation of hydrated oxides and so requires the presence of moisture, either as water vapor or as liquid.  In a dry environment iron is indeed more resistant to corrosion because it forms anhydrous oxides which, as long as moisture is kept away, do provide some passivation.


Conjugate Theory and Redox I'm trying to apply conjugate theory, which I can apply very well to acids/bases, to redox. Can you verify my logic, which I have broken down below?

1) I know that solid sodium is a strong reducer.

2) I know this because solid sodium when placed in water reacts violently, forming, among other products, sodium ion.

3) Reducers are oxidized. Reducers lose electrons. Correct; $\ce{Na^{+}}$ is the product of placing solid sodium in water.

4) So in this half-reaction: $\ce{Na<=>Na^{+} +e^-}$, sodium is the reducer. $\ce{Na^{+}}$ is the conjugate oxidizer.

5) Oxidizers are reduced. Oxidizers gain electrons.

6) Because sodium is a strong reducer, which means it has a strong tendency to be oxidized, or a strong tendency to lose electrons, its conjugate oxidizer must be weak. Oxidizers are reduced; oxidizers gain electrons. The potential for $\ce{Na^{+}}$ to gain the electron that $\ce{Na}$ just lost must be small, or $\ce{Na}$ would not be a good reducer.

7) Therefore, a strong reducer's conjugate oxidizer must be weak.

8) Likewise, a strong oxidizer's conjugate reducer must be weak.

9) Likewise, a weak reducer's conjugate oxidizer must be strong.

10) Likewise, a weak oxidizer's conjugate reducer must be strong.

Also, if anyone could point me to a good reference on gaining insight into redox, tha would be great!

And while we're on the topic of redox, what's the mechanism for this reaction? Are both bonds cleaved homolytically? I see how the nucleophilic chlorines attack the electrophilic hydrogens but how's the bond broken?

$\ce{H_2 + Cl_2 ->2HCl}$

Your question covers two different topics:

Redox properties can be well described with redox potentials attributed to a given reaction. It is essentially the same concept that you are talking about: we assign a number (conveniently it is an actual measurable potential) to an electrode reaction which includes an oxidized and reduced form of the species we are talking about. This number, the redox potential, can show you how strong oxidizer/reducer is the corresponding corresponding form, and just as you guess, strong oxidizers generally have a weak reducer pair.
The $\ce{HCl}$ formation reaction is generally a radical chain reaction, triggered e.g. by light, and not a redox reaction. The key here is to split the $\ce{Cl2}$ (e.g photo-chemically), then the obtained free radicals induce a chain-reaction with elementary steps like
$\ce{Cl + H2 -> HCl + H}$ or
$\ce{H + Cl2 -> HCl + Cl}$
The two gas mixed is more or less inert in total darkness.


Determining ratio of competing fragmentation processes in mass spectrometry In a photoionization experiment, three photons of wavelength 266 nm were absorbed by a neutral molecule of 6-methyl-2-heptanone. If the kinetic energy of the electron removed during ionization is 1.4 eV, what is the ratio of the simple cleavage to the rearrangement produced fragment ions? The following plot describes the fragmentation of the molecular ion of 6-methyl-2-heptanone. The IE of this compound is 9.10 eV.

My intuition would just be to use the value of 9.10 eV provided in order to look at the rates of cleavage to rearrangement in order to determine ratios. I'm not sure how the 1.4 eV and the 266 nm photons come into play.

Your three photons represent the total energy added to the system:

$$E_\mathrm{photon} = 3 \cdot \frac{hc}{\lambda} = \pu{13.98 eV}$$

That, less the sum of ionization energy and kinetic energy of the removed electron, leaves the change of the internal energy of the cation:

$$E_\mathrm{int} = \pu{13.98 eV} - \pu{9.10 eV} - \pu{1.4 eV} = \pu{3.48 eV}$$

Finding the value of $\log k$ for both curves at this energy will give the relative rates of simple cleavage vs rearrangement.


Is the electrical conductivity only measurable in aqueous solutions? I have been looking for other ways to measure it, but all I've found is through aqueous solutions. So I want to know if there's another way to do it, or why it isn't possible through other means.

Did you look into ionic liquids? Also in ionic complexes like poly (ethylene oxide)? For instance, look into these references:

P. V. Wright, Electrical conductivity in ionic complexes of poly(ethylene oxide), Polymer International, 1975, 7(5), 319-327 (https://doi.org/10.1002/pi.4980070505).
Q.-G. Zhang, S.-S. Sun, S. Pitula, Q.-S. Liu, U. Welz-Biermann, J.-J. Zhang, Electrical conductivity of solutions of ionic liquids with methanol, ethanol, acetonitrile, and propylene carbonate, J. Chem. Eng. Data, 2011, 56(12), 4659–4664 (https://pubs.acs.org/doi/abs/10.1021/je200616t).


What causes energy peaks during a molecular dynamics energy minimization? I'm minimizing the energy of a protein with NAMD using the fixedatom feature. I put my protein in a water box, and I allow all atoms other than the protein backbone to be unconstrained. After running the minimization, I used VMD's NAMD Plot for NAMD's log file and the TOTAL3 total energy option, which resulted in some energy peaks that weren't present until this step:

Are those peaks normal to have? If not, what causes them?

Here is the NAMD log file. The NAMD configuration file used for the minimization is:

#############################################################
## JOB DESCRIPTION                                         ##
#############################################################

# Minimization and Equilibration of
# Human Pi GST in a Water Box
# N, V, E simulation

#############################################################
## ADJUSTABLE PARAMETERS                                   ##
#############################################################

structure          ionized.psf
coordinates        ionized_h_free_min.pdb

set temperature    310

firsttimestep      0

# output
set output              Cry1_side_free
outputname              $output
dcdfile                 ${output}.dcd
xstFile                 ${output}.xst
dcdfreq                 50
xstFreq                 50

binaryoutput            no
binaryrestart           no
outputEnergies          100
restartfreq             1000

#############################################################
## SIMULATION PARAMETERS                                   ##
#############################################################

# Input
paraTypeCharmm      on
parameters          /usr/local/lib/vmd/plugins/noarch/tcl/readcharmmpar1.3/par_all27_prot_lipid_na.inp
temperature         $temperature

# Force-Field Parameters
exclude             scaled1-4
1-4scaling          1.0
cutoff              12.
switching           on
switchdist          10.
pairlistdist        14.0

# Integrator Parameters
timestep            1.0  ;# 2fs/step
#rigidBonds          all  ;# needed for 2fs steps
nonbondedFreq       1
fullElectFrequency  2
stepspercycle       10

# Constant Temperature Control
langevin            on    ;# do langevin dynamics
langevinDamping     5     ;# damping coefficient (gamma) of 5/ps
langevinTemp        $temperature
langevinHydrogen    off    ;# don't couple langevin bath to hydrogens

# Periodic Boundary Conditions
cellBasisVector1     -75.1629982   0.0   0.0
cellBasisVector2     0.0    -78.31999779 0.0
cellBasisVector3     0.0    0.0  -98.59599972
cellOrigin          -12.303732872009277 19.206361770629883 -36.690704345703125

wrapAll             on

# PME (for full-system periodic electrostatics)
PME                 yes
PMEGridSizeX        80
PMEGridSizeY        80
PMEGridSizeZ        100
dcdUnitCell     yes
# mobile atom selection:
# protein
fixedAtoms              on
fixedAtomsFile          fixedatoms_backbone.pdb
fixedAtomsCol           B

#############################################################
## EXECUTION SCRIPT                                        ##
#############################################################

# Minimization
minimize            10000
reinitvels          $temperature

At the top of your config file, you imply that you want to perform an NVE calculation, but you’re actually doing NVT, because you have the Langevin thermostat turned on. Langevin works by giving the system a random “kick”, which is probably spiking the energy.


Is there an abundance threshold of a 'synthetic' element for it to be considered natural? I always (possible naively) thought that elements are either natural or synthetic, either one or the other.  But, according to the Wikipedia page about Synthetic Elements, it states that:

  All elements with atomic numbers 1 through 98 are naturally occurring at least in trace quantities, but the following elements are usually produced through synthesis. Except for Francium, they were all discovered through synthesis before being found in nature.

These are shown in pink in the image below:

Image attribution: "Periodic table, good SVG" by Incnis Mrsi - Own work. Licensed under CC BY-SA 3.0 via Wikimedia Commons.

I understand that the amounts of these elements exist naturally in trace quantities, not necessarily on Earth, for example, according to the Jefferson Lab page about promethium:

  Promethium does not occur naturally on earth, although it has been detected in the spectrum of a star in the constellation Andromeda.

This leads me to my question:

Is there a natural abundance threshold of a 'synthetic' element for it to be considered natural?

Is there a natural abundance threshold of a 'synthetic' element for it
  to be considered natural?

$\mathrm{> 0}$

In other words, any element that can be detected outside of the laboratory (e.g. in the natural environment) is a naturally occurring element.  This would include the 80 stable elements along with the 18 radioactive elements with very long half-lives, and the shorter-lived radioactive elements produced upon their decay.

Therefore there are 98 naturally occurring elements; all of the others are synthetic, artificial, man-made, whatever you choose to call them.

This Wikipedia article provides nice background, especially the following sections:

the introductory section, especially the last paragraph therein
Occurrence and origin on Earth
List of the 118 known chemical elements - Occurrence


Would there be need for periodic DFT if we are able to get all the forcefields of all the elements in the periodic table in ReaxFF? Since ReaxFF combines the advantages of DFT and Classical MD, would there be any need for periodic DFT calculations if we can have all the forcefields of all the elements of all the possible systems?

Quoting from the article The ReaxFF reactive force-field: development, applications and future directions

"Such empirical methods, including reactive force-field (ReaxFF),1 trade accuracy for lower computational expense, making it possible to reach simulation scales that are orders of magnitude beyond what is tractable for QM."

So yes, if you want accuracy, especially for something which the approximations made in generating a ReaxFF force field are not good, you are going to have to get out the Quantum Mechanics.


How to measure saturated salinity and mineral levels for flaked sea salt? I am a sea salt enthusiast looking to make the perfect flakey sea salt. I start with sea water and boil it down to a thick brine. Then with less heat I reduce the brine more slowly and inverted salt flake pyramids are made. I am looking for tools/instruments that will help me measure/understand the content of the brine (mineral content and salinity). Sometimes I get perfect flakey salt pyramids like the photo attached and sometimes I get I get more chunky results. The temperature of the brine is always the same but I get different results at times. This leads me to think that its the content of minerals in the saturated brine solution.

Are there chemistry tools/instruments I can use to understand the actual chemical/mineral makeup of the brine that is turning to salt.

Why not use salt (NaCl) bought from a store, or even better at a defined purity grade from a chemical vendor?

When you use sea salt, you get lots of other ions but more importantly you get colloidal particles, organic substances, and other compounds that will affect the crystallization. Chunky results sounds to me like a colloid particle that started crystallization in all directions at once.

If you want perfect crystals, do it perfectly, i.e. with a pure solution, in a clean and new glassware or polished stainless container. If you want perfect crystals you have to reduce the new-age-ness of it ;)

As for understanding, nobody fully understands crystallization, but a lot of things are known. There are books on this topic, check amazon and search for "crystallization". As for instruments, x-ray fluoroscopy and diffraction wouldn't be amiss but you will get very far with an optical microscope, I know I can watch crystals through a microscope for hours and still find something interesting. Barring XRF, XRD and microscope, and within a household budget there are salinity measurement probes and conductance measurement probes which will be able to tell you something of the brine such as it is.


Origin of "ortho", "meta", "para" Definition

The ortho position refers to the two adjacent positions on a benzene ring.

The meta position refers to the positions separated by one carbon atom on a benzene ring.

The para position refers to the opposite position (separated by two carbon atoms) on a benzene ring.

Background

The dictionary.com pages for ortho, meta, and para state that they are independent from the prefixes ortho- (correct), meta- (after; beyond; a higher level), para- (alongside) respectively.

However, I suspect that the meta position is just after the ortho position, so it is named meta ("after" in Greek).

Question

How did these words become affiliated with the positions on a benzene ring?

The designation of an ortho-, meta-, or para-substitution pattern was coined when spectroscopic analysis was just starting (Robert Bunsen, for example).  Some of the spectroscopic techniques to assign substitution patterns (IR-, and more importantly, NMR spectroscopy) were simply not available.  To quote wikipedia's entry about origines of arene substitution pattern:

  The prefixes ortho, meta, and para are all derived from Greek, meaning
  correct, following, and beside, respectively. The relationship to the
  current meaning is perhaps not obvious. The ortho description was
  historically used to designate the original compound, and an isomer was
  often called the meta compound. For instance, the trivial names
  orthophosphoric acid and trimetaphosphoric acid have nothing to do with
  aromatics at all. Likewise, the description para was reserved for just
  closely related compounds. Thus Berzelius originally called the racemic
  form of aspartic acid paraaspartic acid (another obsolete term: racemic
  acid) in 1830. The use of the prefixes ortho, meta and para to distinguish
  isomers of di-substituted aromatic rings starts with Wilhelm Körner in
  1867, although he applied ortho prefix to a 1,4 isomer and the meta prefix
  to a 1,2-isomer. It was the German chemist Karl Gräbe who, in 1869,
  first used the prefixes ortho-, meta-, para- to denote specific relative
  locations of the substituents on a di-substituted aromatic ring (viz,
  naphthalene). In 1870, the German chemist Viktor Meyer first applied
  Gräbe's nomenclature to benzene. The current nomenclature was
  introduced by the Chemical Society in 1879.

Besides many among the chemists are familiar of this designation, it is limited to describe the relationship of two substituents of a benzene.  Hence frequently chemical suppliers and data bases prefer to describe the position of them by numbers.  IUPAC's Blue Book still allows the use of o, m, and p (rule A-12.3), yet the consent moves into the direction of a consistent numbering that is more adaptable in terms of number of (potentially different) substituents and their relative arrangement to (hetero)aromatic nuclei.

To illustrate this, think of examples like lutidine (systematically 2,6-dimethylpyridine), a frequently used mild, sterically hindered base that does not fit into the other system.

In the end, we've to learn understand both dialects + some of non-systematic names, too.


How do I calculate the cost of a single Aluminum atom? I am doing a Chemistry assignment in which I was given a piece of aluminum foil which I had to measure and use for my calculations. I am supposed to calculate the cost of a single aluminum atom using the given data from my teacher and the data which I have measured from my piece of aluminum foil. I understand the basic concept of how I am supposed to do this, but I am having some trouble. Here is the data:

Data given by teacher:
  Cost of a roll           $5.53
  Roll size                30.5 cm x 30.5 m
  Density of Al            2.70 g/cm^3
  Molar Mass Al            26.98 g/mol

Data measured from my piece of Aluminum foil:
  Mass     1.35 g
  Size     18.6 cm x 12.5 cm

I have determined a way to do this, but I am having trouble with it. First of all, I have determined the amount in moles which is $\mathrm{0.0500~mol}$. Then I multiplied the amount in moles with Avogadro's constant ($\mathrm{6.02 \cdot 10^{23}~mol^{-1}}$) to determine the number of aluminum atoms in the piece of aluminum foil I received which is $\mathrm{3.01 \cdot 10^{22}}~\text{atoms}$. Now I just have to find the cost of a single aluminum atom, but this is where I am stuck. I cannot use the cost of the aluminum foil roll to divide it by the number of atoms since I was given a piece of aluminum foil so I need to find the cost of the piece of aluminum foil and then divide it by the number of atoms.

So basically my question is: How do I find the cost of the piece of aluminum foil? If I can find the cost then I can divide it by the number of atoms to determine the cost of a single aluminum atom.

However, if there is a way to find the number of atoms in the roll of aluminum foil, I could then use the cost of the aluminum roll to find the cost of a single aluminum atom. Please explain if there is a way to do this using the data given by the teacher.

You can compute the area of aluminum foil in the roll as:

$$30.5\;\rm{cm}\cdot 3050\;\rm{cm} = 93025\;\rm{cm}^{2}$$

You have measured the mass per square centimeter as:

$${1.35\;\rm{g}\over 18.6\;\rm{cm}\cdot 12.5\;\rm{cm}} = {1.35\;\rm{g}\over 232.5\;\rm{cm}^{2}}$$

You have been given the cost of the roll (and thus the cost per area):

$${\$5.53\over 93025\;\rm{cm}^{2}}$$

Using those numbers, the molar mass for aluminum, and Avogadro's constant (here, it's just the number of atoms so I designate it as such):

$$\rm{cost\;per\;atom} = \left({\$5.53\over 93025\;\rm{cm}^{2}}\right)
\left({232.5\;\rm{cm}^{2}\over 1.35\;\rm{g}}\right)
\left({26.98\;\rm{g}\over 1\;\rm{mol}}\right)
\left({1\;\rm{mol}\over 6.022\cdot 10^{23}\;\rm{atoms}}\right)
$$
$$\rm{cost\;per\;atom} = \$4.59\cdot 10^{-25}$$