Titration of barium(II) hydroxide by hydrochloric acid $\pu{25 ml}$ of $\pu{0.024 M}$ $\ce{HCl}$ is titrated into a volumetric flask which initially contains $\pu{45 ml}$ of $\pu{0.034 M}$ $\ce{Ba(OH)2}$. Calculate the $\mathrm{pH}$.

This is how I calculated.

$$\ce{Ba(OH)2 + 2HCl -> BaCl2 + 2 H2O}$$

Moles leftover / excess moles will determine the $\mathrm{pH}$ because the others that have reacted will neutralise each other.

So calculating moles of $\ce{Ba(OH)2}$ and $\ce{HCl}$ in their respective volume
$C=\frac{n}{v}$, hence $n= c \times v $.

Moles of $\ce{Ba(OH)2} = \pu{0.034M} \times \frac{45}{1000} = 1.53 \times 10^{-3}$

Moles of $\ce{HCl} = 0.024 \times \frac{25}{1000} = 6 \times 10^{-4}$ but $2\times6\times10^{-4}$ since it’s $\ce{2HCl}$ which is $1.2 \times 10^{-3}$.

Now to find leftover moles, you’ll have to subtract as such
${1.53} \times 10^{-3} - 1.2 \times 10^{-3} = 3.3 \times 10^{-4}$, this is the excess moles of $\ce{Ba(OH)2}$.

Looking at solution of excess $\ce{Ba(OH)2}$

$\ce{Ba(OH)2  -> Ba^2+ + 2 OH-}$

Since $\ce{Ba(OH)2}$ : $\ce{2OH-}$, moles of $\ce{OH-} = 3.3 \times 10^{-4} \times 2 = 6.6 \times 10^{-4}$ moles of $\ce{OH-}$

So, concentration $= \frac{6.6 \times 10^{-4}}{\frac{25+45}{1000}} = \frac{6.6 \times 10^{-4}}{0.07} = 0.009..$

$\mathrm{pOH} = -\log \ce{[OH]} = -\log[0.009] = 2.026$

Since $\mathrm{pH} + \mathrm{pOH} = 14$, $\mathrm{pH} = 14 - \mathrm{pOH}$
So $\mathrm{pH} = 14 - 2.026 = 11.97$

HOWEVER, the book is saying $\mathrm{pH} = 12.55$. I have done this problem multiple times and cannot figure it out.

Your solution is correct up to the point you assumed that you can double the concentration of hydrochloric acid.
Unfortunately, this is wrong and not at all how stoichiometry works.
Let's focus on what's important.
In the nutshell, we are dealing with a typical neutralization reaction:

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

and note that $\ce{BaCl2}$ as a salt of a strong acid and a strong base won't undergo hydrolysis and, as a consequence, won't have a noticeable impact on $\mathrm{pH}$.

As you [semi-]correctly calculated, there is an excess of a base ($n_0$ values refer to the initial amounts):

$$
\begin{align}
n_0(\ce{H3O+}) &= n(\ce{HCl}) \\
  &= c(\ce{HCl})\cdot V(\ce{HCl}) \\
  &= \pu{2.4e-2 mol L-1}\cdot\pu{2.5e-2 L} \\
  &= \pu{6e-4 mol}
\end{align}
$$

$$
\begin{align}
n_0(\ce{OH-}) &= 2\cdot n(\ce{Ba(OH)2}) \\
  &= 2\cdot c(\ce{Ba(OH)2})\cdot V(\ce{Ba(OH)2}) \\
  &= 2\cdot\pu{3.4e-2 mol L-1}\cdot\pu{4.5e-2 L} \\
  &= \pu{3.06e-3 mol}
\end{align}
$$

One can see that there is a tremendous excess of hydroxide of the following amount:

$$
\begin{align}
n(\ce{OH-}) &= n_0(\ce{OH-}) - n_0(\ce{H3O+}) \\
  &= \pu{3.06e-3 mol} - \pu{6e-4 mol} \\
  &= \pu{2.46e-3 mol}
\end{align}
$$

and the leftover concentration is

$$
\begin{align}
c(\ce{OH-}) &= \frac{n(\ce{OH-})}{V(\ce{HCl}) + V(\ce{Ba(OH)2})} \\
  &= \frac{\pu{2.46e-3 mol}}{\pu{2.5e-2 L} + \pu{4.5e-2 L}} \\
  &= \pu{3.51e-2 mol L-1}
\end{align}
$$

Finally, as you rightfully suggested (use the correct value for $c(\ce{OH-})$, however):

$$
\begin{align}
\mathrm{pH} &= 14 - \mathrm{pOH} \\
  &= 14 + \log(\pu{3.51e-2}) \\
  &= 12.55
\end{align}
$$

Notes

Using the term "moles of X" is incorrect.
The proper terminology here is the amount of substance.
Never try to substitute a physical quantity with its unit.
Watch out for the proper capitalization, which is actually a big deal in natural sciences.
$\ce{HCl}$ is a formula for hydrogen chloride, whereas $\ce{HCL}$ is not and may denote "hydrogen-carbon-ligand" or be an abbreviation.
Same goes for math operators, including $\mathrm{pH}$ and $\mathrm{pOH}$: $\ce{Ph}$ is phenyl and $\ce{PH}$ can be read as "phosphor-hydrogen", or be yet another abbreviation.


Is nifedipine a weak base or weak acid? I was wondering whether someone could explain me if nifedipine is a weak base or a weak acid.
I barely can find any information about nifedipine's physicochemical properties in studies, however some suggested that it is a weak acid. If it is a weak acid, which hydrogen atom is dissociable and what would the pKa of that group be?

Thanks in advanced.

Nifedipine's structure, as shown on wikipedia:

Nifedipine is a weak acid.

$pK_b = 10.07$
$pK_a = 3.93$
$pK_a = 4.3$


Why do properties of colloidal solution depend only on dispersion medium's physical state? According to a past year JEE question, properties of colloidal solution depends only on dispersion medium's physical state and not on nature of dispersed phase or dispersion medium, or temperature of system. I cannot find any further information on this subject. I would like some clearer guidance on this. Would someone please help?

The assumption that properties of colloidal dispersion only depends on the medium is completely incorrect. The vague part of the question is "properties" in the question, because the examiner is not clear about what type of properties he/she is talking about. It is not uncommon to find open ended questions in public exams. The properties of colloidal solutions heavily depend on the temperature (since their viscosity is temperature dependent as well as flow rate dependent). Similarly, the size of colloidal particle controls the optical properties of the dispersion. Similarly the surface charge on the particles determines the sign/magnitude of the zeta potential. If you want to read further, read DLVO theory (named after four persons) on the stability of the colloids from Wikipedia. One should not worry about an incorrect question. If it were a true-false question, one can simply mark it false.


Does relative amount of lithium and cobalt oxide change in recharged battery? There is a question on my assignment about stoichiometry. I spent 2 whole days searching and review this topic but still clueless.

Reaction in lithium ion battery
$$
\ce{Li(s) + CoO2(s) → LiCoO2(s)} \, .
$$
When the battery is recharged, electrical energy will reverse the reaction. How do the relatives amounts of lithium and cobalt oxide change as the battery is recharged and how?

About maximum $20~\%$ (as I remember, perhaps more now) of the $\ce{Li}$ goes (as $\ce{Li+}$) out of the structure of $\ce{LiCoO2}$, through the elctrolyte, and into the electrode on the other side of the battery. That electrode consists mainly of graphite, which intercalates the $\ce{Li}$ in it's layers in a quasi-metallic form.

So it is $\ce{Li_{1-x}CoO2 + x~Li}$ after charging. $\ce{LiCoO2}$ has a special structure that does not break down when some $\ce{Li}$ is missing, except if you keep a charged battery at over $\mathrm{50~^\circ C}$ or so.

By the way, the related wikipedia article tells you all this. Two days of researching? ;-)


What is the cause of hygroscopy? Hygroscopy is the ability of a substance to attract and hold water molecules from the surrounding environment.

Why are some compounds hygroscopic and others are not? What is it about their structures that allows them to hold water? Is it due to the atoms being arranged in a certain structure due to bonding?

How can you tell if a compound is more hygroscopic than another?

Ions are hygroscopic because they form very stable, low energy hydrates, which are thermodynamically favorable over water vapor. I.E. salt attracts water vapor because the hydrated salt is lower energy than the water vapor. Other very porous materials are hygrosopic due to capillary action. Water vapor is attracted to the surface initially due to a variety of reasons, then the narrow channels in the porous material cause the water to spread out to lower their energy.


How does zinc become oxidized when it's not in direct contact with copper ions in a galvanic cell? Whenever a galvanic cell is explained in textbooks/videos, it usually starts by showing that if you put a piece of $\ce{Zn}$ in a $\ce{Cu^{2+}}$ ion solution, the $\ce{Zn}$ will become oxidized and transfer $\ce{2e-}$ to the $\ce{Cu^{2+}}$ ion, so the $\ce{Cu^{2+}}$ becomes a neutral $\ce{Cu}$ metal.

I understand this, but in a galvanic cell, the $\ce{Zn}$ never actually touches the $\ce{Cu^{2+}}$ ions, the $\ce{Zn}$ and $\ce{Cu}$ are only connected with wires. I don't understand how the $\ce{Zn}$ can be oxidized and transfer ions through the current.

Zinc electrode is anode and copper electrode is cathode. The two electrodes are connected by a electrically conducting wire. At anode, $\ce{Zn}$ atoms are oxidised to $\ce{Zn^2+}$ which go into the solution, and the electrons are left on the metal strip.

The electrons produced in the oxidation half-reaction at anode are transferred through the wire to $\ce{Cu}$ electrode where they are consumed. The $\ce{Cu^2+}$ ions are reduced to metallic $\ce{Cu}$ which coats on the $\ce{Cu}$ strip.

Thus, oxidation and reduction half reactions occur at separate electrodes and the current flows through the wire. A salt bridge maintains neutrality in both the compartments by a flow of ions.

The overall cell reaction can be given as:

$$\ce{Zn (s) + Cu^2+ -> Zn^2+ + Cu (s)}$$


Why does mutarotation occur only in specific media? I was reading a book on biomolecules, and read that muta-rotation in hexoses, can happen only in media like water, which can act as an acid and as a base. Why is this so?

Essentially, it is just the formation of the open chain form, and then back to the other cyclic anomer, right? What role does the medium have in this process?

Essentially, it is just the formation of the open chain form, and then back to the other cyclic anomer, right?

Precisely, mutarotation involves interconversion of the α-and β-forms (cyclic anomers) of the monosaccharide with intermediate formation of the linear aldehyde (or ketone).

The mutarotation of glucose is a general acid- base-catalysed reaction which involves two proton transfers. Studies have proposed that these proton transfers are either carried out in a stepwise or concerted manner, with the latter alternative possibly incorporating several water molecules in a cyclic transition state.

Two possible mechanisms of mutatoration have been investigated:

The first involves an intramolecular proton transfer i.e acid-catalyzed reaction.
The second uses one solvent water molecule to assist proton transfer i.e water-catalyzed process.

There is some experimental  evidence1 for the solvent participation in the mutarotation reaction that appears from the studies of primary solvent effects on the rates of glucose and tetramethylglucose observed in $\ce{H2O + D2O}$ mixtures. In addition the absence of intramolecular proton transfer can be deduced from the small values found for the mutarotation rate of glucose in organic solvents.

  Eigen1 proposed that the two proton transfers could proceed
  in aqueous solution via a pathway involving two or more water
  molecules in a cyclic hydrogen-bonded transition state, where the
  intramolecular proton transfer does not take place in aqueous
  solution.

Previous studies2 have shown that the reaction order with respect to water is zero for an acid-catalyzed reaction involving a stepwise mechanism, however in aqueous solution without added catalyst, water molecules must act as an acid or base and it is possible that the mutarotation reaction does proceed via a cyclic-concerted mechanism involving about three water molecules.

  The results obtained in experimental work in the  newer
  studies1 have shown that mutarotation is a
  solvent-assisted process that obeys pseudo first-order kinetics. Besides that, the first water molecule from the solvent acts as a
  catalyst, and plays a decisive role, since it completely changes the
  mutarotation mechanism.

So a solvent is necessary, the evidence been seen  in proton transfers in  reactions that do occur via cyclic concerted mechanism as proposed in these studies.

References:

A theoretical study of glucose mutarotation in aqueous solution. A. M. Silva et al. / Carbohydrate Research 341 (2006) 1029-1040. Carbohydrate Research 341 (2006) 1029-1040.
The Mutarotation of Glucose in Dimethylsulfoxide and Water Mixtures. Ballash and Robertson CAN. J. CHEM VOL. 51. 1973.
CM Garrett, Grisham Biochemistry


Why does chlorine undergo disproportionation in alkaline media? One reaction used to form sodium hypochlorite (A type of bleach) is as follows:

$\ce{2NaOH(aq) +Cl2(g)->NaOCl(aq) + NaCl(aq) + H2O(l)}$

Provided the sodium hydroxide is dissolved in some substance.

I know that the sodium hydroxide dissociates to form the $\ce{Na+}$ and $\ce{OH-}$ ions, but I do not know why the $\ce{Cl2}$ breaks down into 2 separate atoms instead of just staying the way it is. My questions are: what are the driving forces behind this reaction and why does the $\ce{Cl2}$ break down? I see how it happens, I just don't know why.

At constant pressure and temperature, the spontaneity of all chemical reactions is controlled by the difference in Gibbs free energy ($\Delta G$), as described by the following equation:

$\Delta G = \Delta H - T\Delta S$

Where $\Delta H$ is the enthalpy change of reaction, $\Delta S$ is the entropy change, and $T$ is the absolute temperature. For a reaction to take place spontaneously, the free energy change must be negative (i.e., $\Delta G < 0$). In the specific reaction you're asking about, it's very probable that the change in entropy is negative, since a gas is being consumed and an equal total number of moles products are being formed as moles of reactants, so that the end result is an increase in order with respect to the reaction mixture. However, the formation of products results in a net release of heat (i.e., $\Delta H$ is negative), which increases the entropy of the surroundings sufficiently, such that, overall, entropy is increased in a global sense.

Ultimately, what drives this reactions forward is its exothermic nature (under standard conditions). In this reaction, the formation of stable chemical bonds releases energy as heat. If one were to attempt this reaction at very high temperatures, it would eventually fail, as the increase of the $T\Delta S$ term would at some point dwarf the enthalpy change.

There are other factors at work as well, including chemical equilibrium. By Le Chatelier's principle, the process can be made favorable by, e.g., maintaining a continual excess of reactants, or removing products formed from the reaction mixture as it proceeds.

Edit: Here's one source which indicates 348cal/g energy released, to put a specific number on the process. You could also calculate the enthalpy change using tables of standard thermodynamic data.


How to treat Magnesium powder and ribbon I am a master student of nuclear physics. Due to some measurements I'd like to make, I have to use Magnesium as a beam target. Searching through the net to see it's properties I found that it can be easily oxidized. In addition I've read that it's highly flammable.

I have in hand two forms of Magnesium:powder and ribbon. In order to make the targets I have two alternatives.

Use the powder to make highly pressured pellets. I thought of flushing a bottle with Argon or Nitrogen, making the pellet as fast as possible using this procedure(which involves acetone) in the atmosphere and then after the pellet is formed put it in the Argon or Nitrogen bottle.
Use the ribbon as is. The only problem is that it not wide enough, so I will have to make a spiral of this ribbon around something(I don't yet know what the "something" will be, so any idea is welcome!). The ribbon I have has a width of 3mm but I want a width of 6mm. So I thought of spinning the ribbon around an item, to form a spiral.

The goal is to make the target as safely as possible and avoid Magnesium oxidizing, because I don't want to have Oxygen in my target. So my question has three parts.

How to safely treat magnesium powder/ribbon so as to avoid any possible accident and preventing it from oxidizing?
Which of the above would be the most effective way to make the target?
How to store the target until the experiment will be held?

Note that after the target is made, it will go to an evaporator, in order to evaporate some gold. Any advice, help or direction is more than welcome!

Magnesium powder and ribbon are relatively stable in air at normal conditions, probably because of thin protective layer of oxide. However, ribbon can burn in air with blinding light and powder gives self-burning (i.e. not needing external oxygen sources like air) mixtures with many common substances, some of them explosive. Moreover, burning magnesium may continue to burn in $\ce{CO_2}$ atmosphere, under quartz sand and especially under water (producing hydrogen, that can form explosive mixture). The only common material that can safely extinguish burning magnesium is iron powder. Surprising enough, being stable in cold water for short time, magnesium readily react with methanol (and probably ethanol), chlorinated hydrocarbons and hot water.

The good thing is that magnesium ribbon is not easy to lit up, and powder is almost impossible to lit up when on surface of some massive object (common gas lighter flame lit magnesium ribbon after 15-60 seconds of heating)

Magnesium must not be heated on air or under nitrogen. Period. Technical grade argon probably contain significant amount of nitrogen, so it should not be used either.

Magnesium theoretically can be melted under eutectic mixture of $\ce{NaCl}$-$\ce{KCl}$-$\ce{CaCl_2}$, but by doing so without experienced supervision you ask troubles. Still, theoretically speaking, if you find an iron (not silicate, porcelain or corundum) crucible you probably can melt magnesium in it under mentioned mixture of dry salts, let it crystallize and then carve it out.

However, magnesium alloy sheets are available on the market, so maybe you can find something and cut needed form from commercially available sheets. Try this first before doing anything requiring heating magnesium. I have doubts, that you will be able to press good tablets from magnesium powder, so if commercially available sheets and maybe some magnesium compounds (oxide, sulfide and others) won't not work for you, you probably will end with melting magnesium. I hope you won't need it, it is dangerous.


electrolysis of aqueous zinc nitrate Can anyone explain why in the electrolysis of aqueous solutions such as zinc nitrate, the oxidation  half equation shows hydroxide ions forming oxygen and water -   can this half equation be balanced producing oxygen and hydrogen ions - if not why not?

You mean, why we write $\ce{4OH- -> O2 + 2H2O + 4e-}$ and not $\ce{4H2O -> O2 + 4H+ + 4e-}$?

Indeed, you may find both these reactions in the table of standard electrode potentials (at $E^0=+0.401$ and $+1.229$, correspondingly; also, note the reversed direction). In some sense, the second equation may even be considered more appropriate, since $\ce{H+}$ is a more abundant species in the solution than $\ce{OH-}$ (zinc salts would produce slightly acidic solutions due to hydrolysis).

But the first reaction was chosen not without reason. See, when we electrolyze something, it is anions that rush to the anode to get oxidized. Water molecules are neutral; electric field would not drag them in either direction. So this might be the reason why the equation is written this way, even if the amount of $\ce{OH-}$ is minuscule and the overall reaction looks more like the second one.


Why do the names of most chemical elements end with -um or -ium? Why do the names of most chemical elements end with -um or -ium for both primordial and synthetic elements?

To expand on @BelieveInvis's answer -- in the early 19th century, when the Royal Society was really in the swing of things, the dominant language of scholarship was still Latin. Since Latin didn't have words for the new metallic elements, new words were coined from the existing terms for the substances and given Latinate endings.

From the OED's entry on -ium:

The Latin names of metals were in -um, e.g. aurum, argentum, ferrum; the names of sodium, potassium, and magnesium, derived from soda, potassa or potash, and magnesia, were given by Davy in 1807, with the derivative form -ium; and although some of the later metals have received names in -um, the general form is in -ium, as in cadmium, iridium, lithium, osmium, palladium, rhodium, titanium, uranium; in conformity with which aluminum has been altered to aluminium.

So, I think after that, other elements were simply given the suffix to fit the generally useful naming scheme, and then, metal names which were already in common use kept their common language names (e.g. gold as opposed to aurum) simply by force of usage.


Sodium + water reaction, noxious fumes? I was reacting pure sodium and tap water, and at one point I put some splattered sodium back into the water. I observed fumes that reacted with my respiratory system, producing a burning sensation.

I read that it is possible to produce sodium oxide fumes from sodium hydroxide at high temperatures (the sodium did burn at the end, producing an orange flame).

My question is what could be the chemical composition of these fumes?

I agree.  This is a dangerous reaction and should not be tried without proper equipment, including a lab hood and fire extinguishing equipment.

The fumes could be "just" droplets of sodium hydroxide solution.  I've seen similar fumes involving hydrochloric acid, just by adding water to rinse out traces of concentrated acid (in a hood, of course).  They may be carried by increased evaporation even without boiling and without need for a further "chemical reaction".


Designing a device to separate water from water-ethanol mixture using dry ice baths I've been trying to design a device to increase the percentage of ethanol and water, or to at least concentrate the water-alcohol mixture without using distillation.

The way I'm trying to do this is by using a dry ice bath (dry ice+propanol). I then insert a stainless steel vessel in the bath with the said water-alcohol mixture (between 30-60% ethanol to begin with). Going by the ethanol-water phase diagram and the temperature of the ice bath (-78C), I should, in theory, be able to concentrate the mixture up to ~80% ethanol.

Where this goes wrong is that instead of cleanly separating into water ice and concentrated water-ethanol mixture, what I get is a kind of ice-alcohol slushy.

Any suggestions on how to get a clean separation and get ice that I can scoop out?

Thanks!

One suggestion: try cooling the alcohol-water mixture more slowly. Slower crystal growth may produce larger ice crystals that occlude less of the ethanol and are easier to separate by filtration. How to do it? Place the steel beaker in the outer bath before adding dry ice to the outer bath, then add the dry ice incrementally to gradually bring the temperature down to where you first observe ice crystals, then keep the temperature constant for a while as crystals form. Then lower the temperature a degree or two and repeat the cycle. Probably better not to stir the inner mixture. It will probably help to use a thermometer to carefully monitor the temperature of the alcohol mixture.


Why does CO2 lowers the pH of water below 7? Ok so I understand that having a pH below 7 is considered to be acidic and I also understand that part of the definition of an acid is that it gives off H+ when dissolved in water(even though not all acids necessarily need to give off H+ when dissolved in water)

But I was the rather confused when I came upon this ( when they refer to the 2 gases they are talking about CO and CO2)

What I dont understand is how the equation came to be ? ( the product side of the equation)

And how would It have looked if CO was dissolved in H2O, would it have still lowered the pH??

Carbon Dioxide (CO2) readily dissolve in water and form Carbonic Acid (i.e H2CO3 (aq) )

This is the formation of bonds.

Then Carbonic Acid (i.e H2CO3  (aq)  ) dissociate in water as follows.

So water gets H+ ions, so that cause water acidic.

The following shows dissociation of Carbonic Acid (i.e H2CO3 (aq) ) more clearly.

Carbon Monoxide (CO) do not readily react with water in room temperature in standard conditions. So it do not dissolve in water.


Reaction of Silicon Dioxide with Hydrofluoric Acid I learnt that $\ce{SiO2}$(Silicon dioxide) doesn't react with any acid except $\ce{HF}$. So what is special about $\ce{HF}$?
Why does only $\ce{HF}$ reacts with $\ce{SiO2}$ even though $\ce{HF}$ has a very high bond energy compared to other acids in that group?

It is surprising that Silicon halides behave differently with water according to the choice of the halogen fluor or chlorine. The following reactions are known to happen : $$\ce{SiO_2 + 4 HF -> SiF_4 + 2 H_2O} ......... (1)$$
$$\ce{SiCl_4 + 2 H_2O -> SiO_2 + 4 HCl}........ (2)$$ These equations are exactly the opposite from one another. Why does reaction ($1$) not proceed in the other way like ($2$) ? I don't know, but I would like present a personal interpretation. Maybe it is due to the fact that reaction ($1$) proceeds in two steps, with the intermediate formation of $\ce{H_2SiF_6}$. Maybe ($1$) proceeds like that $$\ce{SiO_2 + 6 HF -> H_2SiF_6 + 2 H_2O}$$ $$\ce{H_2SiF_6 <=> SiF_4(g) + 2 HF } $$$\ce{H_2SiF_6}$ is well known to be formed if $\ce{HF}$ is in excess. And of course HCl cannot form $\ce{H_2SiCl_6}$, because the chlorine atom is too big to act as a ligand this octahedral structure. $\ce{H_2SiCl_6}$ does not exist.
This is a personal idea. I would be pleased to hear about any comment.
Sorry for presenting these lines as an Answer. Alas the text is much too long for a comment...


What are the reaction products in my dishwasher from a steel colander? In my dishwasher I noticed that a steel colander and a coffee maker (the Moka Pot) both became discolored after being in the dishwasher. The metal is clearly reacting with something. They aren't made of aluminum (I don't think) and I've never seen this reaction before when I do dishes in the sink.
To clarify: the metal on my coffee pot and the colander was relatively smooth and shiny, and then after they were in the dishwasher they became grayer, and the texture was clearly rougher. That said, i was able to remove the discoloration with a Scotch-brite plastic pot scrubber sponge or Brillo pad, with a little elbow grease.
So, the chemicals I saw on the back of my dishwashing liquid were:
NaCl, Na$_{2}$SO$_{4}$, Citric Acid, Na$_{2}$CO$_{3}$, Sodium percarbonate, sodium silicate and a couple of organics (it says enzymes, but I don't know how much of that is marketing hype). I looked up the rinse aid (it's Finish brand) and I came up with alcohol ethoxylate, sodium polycarboxylate, sodium cumene sulfonate and tetrasodium EDTA, Methylisothiazolinone and methylchloroisothiazolinone, and CI Acid Blue dye.
I do not know the xact metal composition of my colander or coffee pot, though I suspect it's is stainless steel I don't know which particular type and what the exact mix of carbon, chromium, etc. is. I can say that the colander is one that is quite old; it's age is measured in decades. The coffeepot not so much; I got that maybe a decade ago? At least that. Both are made of a gray steel that has a slight "rough" feel; it's not the super-shiny modern variety. The Colander in particular seems a type of metal I recall being quite common with cheaper kitchen stuff in the 70s and 80s.
In any case, I wanted to know what the following might be:
-- The reaction products of the dishwashing chemicals dissolved in the water; clearly there is something going on
-- If I need be concerned about toxicity, since this is something that touches food. I would assume there is less of an issue if I scrub the discoloration off. And it's not likely to be in concentrations that are harmful one-off, but why ingest weird stuff if you don't have to and all that.
Any insights are appreciated; this is something I was rather curious about (and may make a good demo for a chemistry class).
ETA: In the images you can see the darker gray on the metal -- that was the discoloration.
ETA II: apropos of blacksmith37 's comment/ answer I checked if there was any magnetism and the fridge magnet does not stick so it's either aluminum or a steel that is non-magnetic.

enter image description here

I am confident both are aluminum ; Steel would be magnetic, austenitic stainless is non-magnetic ,but normal relatively shiny or bright in consumer products. All steels are about 3 X more dense than aluminum if you could judge the density. Aluminum is subject to corrosion in solutions of salts, acids ,and bases. But I couldn't pick one of your components as the most likely cause. That is, I expect it is corrosion of aluminum you are seeing. I have had a very similar colander  but washed it by hand because it was fairly large to put in a dish washer.


Why are acids and bases are defined as both donors and acceptors of species in the acid-base theories? So, a Brønsted-Lowry acid is a proton donor, and a Brønsted-Lowry base is a proton acceptor. On the other hand, a Lewis acid is an electron pair acceptor, and a Lewis base is an electron pair donor.

Why is it that in these acid-base theories, acids and bases are defined as both donors and acceptors of species? Or is there a part of acid-base theory I don't quite understand?

These are primarily old theories that were kind of evidence or observation based. These are not completely correct theories as there are may loop holes in them. For eg. many types of compounds with significant acidic and basic characters cannot be predicted by these theories. Also, to a certain extent the Bronsted and Lewis concepts are complementary.Wikipedia says - ""

  A Lewis base is often a Brønsted–Lowry base as it can donate a pair of electrons to H+; the proton is a Lewis acid as it can accept a pair of electrons. The conjugate base of a Brønsted–Lowry acid is also a Lewis base as loss of H+ from the acid leaves those electrons which were used for the A—H bond as a lone pair on the conjugate base.

and

  In 1923, Lewis wrote An acid substance is one which can employ an electron lone pair from another molecule in completing the stable group of one of its own atoms. The Brønsted–Lowry acid–base theory was published in the same year. The two theories are distinct but complementary.

Basically, proton is a $H^+$ ion and we all know from our simple understanding that compounds that give $H^+$ or proton are acids and those that accept $H^+$ or proton are bases.


Dipole Moment of Normal Water vs Heavy Water The question is in the title itself.

My guess: Higher for Normal water.

My Reasoning: Internet told me Deuterium is more electronegative than Protium. So,  there should be less difference between the electronegativity values of $\ce{D}$ and $\ce{O}$. Consequently, the dipole moment for $\ce{D2O}$ should be lower.

Is this a valid reasoning or should I take into account factors like lesser vibration amplitude of $\ce{D}$ than $\ce{H}$ that forms more stable bonds.

There is no measurable difference.

HOH 1.8546 ± 0.0006 D

DOD 1.8558 ± 0.0021 D

Dipole moment of water from Stark measurements of H2O, HDO, and D2O J. Chem. Phys. 59, 2254 (1973)


Inter-classification of organic chemistry functional groups (carboxyl vs alcohol & ether vs ester) I have a few questions about whether organic functional groups can be classified as another.

First, can carboxylic acids be also be considered alcohols (they have $\ce{-OH}$)? If not, can they at least behave like alcohols?

Similarly, can esters ($\ce{RCOO-R'}$) be considered ethers because the $\ce{R'}$ group is linked to the other half of the ester via an $\ce{-O -}$ linkage? In whichever case, what are some chemical characteristics that distinguish ethers and esters?

Would it make sense to classify a pistol as a type of pipe? The barrel of a pistol is a cylindrical metal tube, after all.

Clearly, no. The other key defining components of a pistol (stock, trigger, etc.) combine with the cylindrical metal tube to radically change its function and behavior.

Similarly, even though a carboxyl group does contain an $\ce{-OH}$ group bonded to a carbon atom, its behavior is sufficiently different from that of an alcohol that it makes no sense to group them together. A parallel argument holds for ethers and esters as well, as ably noted by MaxW.


lanthanide contraction "The Lanthanide Contraction refers to the fact that the 5s and 5p orbitals penetrate the 4f sub-shell so the 4f orbital is not shielded from the increasing nuclear change."

First, 5s and 5p are filled before 4f, so if the atom wants to give an electron out, which energy shell does the electron come out of? How 5s and 5p penetrate 4f? Isn't 4f outer than 5s and 5p?

According to Aufbau principle, for multi-electron atoms, in the ground state of an atom, an electron enters the orbital of lowest energy first and subsequent electrons are fed in the order of increasing energies.

For multi-electron atoms, the following sequence is observed for orbitals in the increasing order of energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s.

Here the numbers 1, 2, 3, 4.. indicates shell number. It may be remembered that, though filling of electrons takes place according energy of orbitals, when electrons are removed from the neutral atom, electrons with highest value of n are removed first. So, electrons from 5p or 5s is removed first than from 4f.

To understand lantanide contraction, lets consider an example. If you consider Pd (of 5th period) and Pt (of 6th period), you would expect Pt to have relatively large radii than Pd. In practice, they have roughly same radii. This is because, Pd has 4d electrons and Pt has 5d and 4f electrons. Due to diffused shaped of 4f electrons, they can't shield 5d electrons of Pt. Thus, contracting the size of Pt roughly same as Pd.

Considering your last question, no 4f subshell lies closer to the nuclei than  5s and 5p. For more info, I would suggest you to read this page: Lanthanide Contraction


Stability of isomers of N4O I have googled and found that the structure of $\ce{N4O}$ is linear with a positive and negative charge on 2 nitrogen atoms. But, the following structure seems more stable because it fulfils octet for all elements, the structure has no charge and it is resonance stabilised:

Why is this not the actual structure of $\ce{N4O}$? Are there any destabilizing effects that outweigh the stabilizing effects in this structure? The structure doesn't seem to have much angle strain. Where am I wrong?

From The Nitrosyl Azide Potential Energy Hypersurface: A
High-Energy-Density Boom or Bust? J. Am. Chem. Soc. 1996, 118, 4860-4870 :

  [Abstract] ...While the ring isomer is predicted to be the most stable structure on the
  hypersurface, the barrier to dissociation is most likely between 1 and 2 kcal mol-1 (including zero-point vibrational
  energy [ZPVE], the existence of any barrier becomes questionable) making isolation theoretically possible but
  experimentally difficult. This small barrier also detracts from the attractiveness of the N4O ring structure as a high energy-density material. The trans-chain isomer, however, lies in an energy valley with higher sides, consistent
  with its previous experimental observation.

  [full text]...In an ideal five-membered ring
  with 6 $\pi$ electrons, the $\pi$ electrons would be distributed evenly
  among all the bonds. In the present case, the highly electronegative
  O atom prefers to keep electrons around itself, leading
  to a partial negative charge on the oxygen. Energetically, there
  is a certain degree of stability associated with the ring isomer,
  although not on the order of common aromatic systems. The
  ring isomer is predicted to be at most (DZP CISD) 20.9 kcal
  mol-1 more stable than the trans-chain isomer; however, this
  value decreases to 13.2 kcal mol-1 with TZ2P CCSD.
  The ring isomer TS to dissociation into N2 and N2O is shown
  in Figure 5. ... Energetically, the barrier to dissociation
  is at most 15.3 kcal mol-1 (DZP CISD) and drops lower
  with improvements in both basis set and correlation scheme.
  In going from a DZP to a TZ2P basis set, for example, this
  barrier drops by 5.2 kcal mol-1 for CISD and 4.3 kcal mol-1
  for CCSD. Assuming a similar trend in moving from DZP
  CCSD(T) to TZ2P CCSD(T), the ring dissociation barrier is
  expected to drop to 1-2 kcal mol-1 with the addition of f-type
  functions possibly making it even lower. A barrier of this size
  lies below the ZPVE, throwing doubt on the existence of the
  N4O ring isomer.

So thermodynamically, yes, the ring form is the lowest energy isomer.  However, all the isomers are unstable to decomposition and the linear isomer is in a deeper potential energy well than the ring isomer. Therefore, the linear isomer is easier to observe experimentally for kinetic reasons.

There may be additional information in Theoretical study on structures and stabilities of N4X (X = O, S, Se, Te) series International Journal of Quantum Chemistry Volume 109, pages 226–235.


Do conjugated double bonds take priority over a +1 formal charge? Do conjugated double bonds take priority over a +1 formal charge?

The first one is what I solved for (ignore the hexagon that has no double bonds attached to it), and the one below it is the answer from the solutions.

Why did they move the conjugated bonds around first? This ends up causing the formal charge to move in the opposite direction than mine. Does it matter for this problem?

Given the symmetry of this benzyl cation, the "direction" does not matter. I have put direction in quotes because there actually is no such thing. Electron pushing arrows are a helpful tool (at the very least for accounting purposes), but it is not what is happening in this molecule (or most other cases, either). What you have here is a resonance structure, meaning that the true situation is not well modeled by a single Lewis structure, thus we use several, and the true situation is somewhere in between the structures with varying contributions from the individual ones. So it does not matter how one arrives at these structures, it is important that they be reasonable and do contribute.

One could argue that the answer given is more extensive than your attempt, but I would consider both equal. The reactivity of the molecule is not dominated by the aromatic ring, thus placing extra emphasis on it seems superfluous.


What are the height and width of the large and small nodes of the sp3 hybridized orbitals of carbon and silicon? I will make an effort to interpret answers based on wave functions, but I am a senior pursuing a bachelor's degree in Biology, so I don't have the background in quantum physics for a purely mathematical answer to make intuitive sense.

I would appreciate any partial answers which cover only carbon or only silicon, as well as partial answers that give the height and width of only the small node or only the large node of the sp3 orbital. I am asking about the width of the large and small nodes at their widest points, not at their center of electron density.

I am aware the sp3 hybrid orbitals only form when these atoms are bonded to four other atoms, so there are no empirical observations showing the height of sp3 orbitals from isolated atoms (for this question, assume a tetrahedral lattice composed purely of carbon or silicon). That said, any estimates based on halving bond lengths should account for orbital overlap in these bonds so that the height of the orbital from each bonding atom is not underestimated.

Pieces of information that I would find very useful as an addendum to answering my main question are estimates of the electric potential energy, kinetic energy, and total energy (constant) of an electron as it travels in the largest ellipse contained within the sp3 orbital of either a carbon or silicon atom in a lattice – including the large and small nodes, and assuming this ellipse does not move from one bonded atom to another.

Thank you!

EDIT to update: Having taken organic chemistry I and II, I am also aware that orbitals are simply the 95% confidence interval for finding an electron. I am also aware that the common understanding of electron movement in orbitals is that they can move between states of equal energy without occupying all of the states in between. In my question about an electron moving in an elliptical path, I am asking for answers to assume that the electron is following such a path, and that this path is a planar ellipse which corresponds to a cross-section of the outer edge of the 95% confidence interval. Smaller elliptical paths would have lower energy, and I'm asking this because it is the highest energy path an electron would take in the orbital if it were following a planar elliptical path.

So, after I get past the shock and awe that you are asking us to describe a fictional object,

this page has some pictures which will be suitable for you to construct visualization: ].

I eyeball it as about 4 times larger in all three dimensions for the 2s/2p level.  The radial distribution function for 3s/3p has its highest density about 2.2 times further away from nucleus as the 2s/2p level
]


Is it possible for a liquid chemical to have a pH value lower than its aqueous solution? During the QC check of one of the products that we make as a company, the product had a $\mathrm{pH}$ of $10.44$ on one $\mathrm{pH}$ meter and $10.20$ on the other. (The product is a liquid, so it's straightforward to immerse $\mathrm{pH}$ probes to take the measurements.)

For QC of the product, it is required to make a $1\%$ solution using demineralised water and to take a $\mathrm{pH}$ reading.

The first $\mathrm{pH}$ meter read $11.72$ for this $1\%$ solution, and the second meter read $11.41$.

How is it possible for the raw product's $\mathrm{pH}$ to be lower than the $1\%$ solution's $\mathrm{pH}$? The product is supposed to be basic and not acidic at all.

The concept of pH tends to break down when water is not greatly in excess of the solute (e.g. if your product is a highly concentrated amine base). The apparent pH of your product, when measured using a pH meter or an indicator, may be relatively low (seemingly less basic) if the solution has a less polar environment than that of pure water--because ionization will be less favorable.

If no water is present at all, the product might still give a pH reading with a meter or indicator, but the term "pH" in this case would be irrelevant.

It always helps to talk specifics with chemistry, so I am curious as to what the ingredients of the product are.


Confusion regarding mechanism of dehydration of alcohols 1)Today I came across a problem which asks to find the order of dehydration of 2-hydroxycyclohexanone(1), 3-hydroxycyclohexanone(2),4-hydroxycyclohexanone(3). The answer is 2>3>1 But I thought answer should be 2>1>3 cause in 2 and 3 the product would be conjugated and hence more stable.
2)My second confusion is regarding the pathway taken by such hydroxyketones. Do they follow E1 pathway and form carbocation or do they take some other pathway?

I have provided what I think but I am not sure whether the same mechanism will be followed or not.

I think your mechanism is reasonable.
Coming to rate of dehydration,
In first case Alpha hydroxy ketone will have the least rate of dehydration because a carbocation is formed which is destabilized by Ketone group and even notice there is hydrogen bonding in reactant making reactant more stable and thus rate of dehydration is decreased in first case.
In second case, though carbonation is not stabilized,  here breaking of H will be easier because the H are more acidic being next to electron withdrawing group , so 2 has highest rate .
In third case, since carbocation is not so unstable and even the H is not much acidic. So  3 has moderate rate of dehydration.
Therefore,  rate of dehydration:2>3>1


Why are not all metals included in the reactivity series Why aren't all metals included in the activity series ? For example why isn't molybdenum included ? Now we know that hydrogen displaces those metals from their oxides that are less reactive than iron. But how do I know whether Hydrogen would also displace Mo if its not there in the reactivity series ?

It's a good question: "reactivity" may vary depending on concentration, temperature etc. but should generally closely approximate the electromotive (galvanic) series.

That said, as @Karl mentions, reactivity series I've seen usually are incomplete, often leaving out transition metals with multiple valences such as Mo and W. To fill in the missing species, look at a good table of the electromotive series.

The omission of so many metals is downright shocking, revolting. One would think a current table would have included more metals; these reactivity series seem like works in transition.


Proving a thermodynamic relation between $(\partial H/\partial T)_p$ and $(\partial U/\partial T)_V$ (This question is taken from Problem 1.1(b) of the book Chemical Thermodynamics: Principles and Applications.)$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$

  Prove $$\pd HTp = \pd UTV + \left[ V - \pd HpT \right] \pd pTV. \tag{1}$$

First, I expressed $H$ as  $U + PV$, and took the partial derivative against $T$, keeping pressure constant:

$$\pd HTp = \pd UTp + p\pd VTp \tag{2}$$

Expressing $U$ as a function of $(T, V)$:

$$\pd UTp = \pd UTV + \pd UVT \pd VTp \tag{3}$$

Using the well-known identity that $$\pd pVT \pd TpV \pd VTp = -1, \tag{4}$$

$$\pd UVT \pd VTp = -\pd UpT \pd pTV = \left[ V - \pd UpT - V\pd ppT \right]\pd pTV \tag{5}$$

This, of course, means that

$$\pd HTp = \pd UTV + \left[ V - \pd HpT \right]\pd pTV + p\pd VTp \tag{6}$$

Based on my working, there is an additional $$p\pd VTp$$ term. Is my working incorrect? I can’t find my mistake.$\endgroup$

$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$

I would start from

$$ dH = \pd HTp dT + \pd HpT dp$$

This gives rise to

$$ \pd HTV = \pd HTp  + \pd HpT \pd pTV$$

in which you can recognize some components of the solution. Since

$$dH = dU +PdV + VdP $$

so that

$$ \pd HTV = \pd UTV  + V \pd pTV$$

it follows that

$$ \pd UTV  + V \pd pTV = \pd HTp  + \pd HpT \pd pTV$$

which is readily rearranged to obtain the desired equation.
$\endgroup$


Nomenclature of organic molecules with more than one functional group I need some help naming compounds with more than one functional group based on their priority order. Here are some examples:

A 4-membered carbon chain with an aldehyde at one end and a ketone right next to the aldehyde
A 4-membered carbon chain with an aldehyde at one end, and the ketone and aldehyde having 1 carbon in between.
A 5-membered carbon chain with a $\ce{C=C}$ double bond at carbon number 2  and a $\ce{C#C}$ triple bond at $\ce{C}_4$ (maybe it is the reverse)

I think that for the one with a double bond and a triple bond, the triple bond gets higher priority and so it would be pent-4-en-2-yne but I am not sure about it.

There are some priority rules about naming compounds with more than 1 functional group.

Aldehyde has higher priority than ketone (I don't know why but I think it is because of aldehyde always take place at the end of carbon chain)
Alkene has higher priority than alkyne, because -ene suffix comes before -yne

So the answer for your question is:

2-oxobutanal
3-oxobutanal
pent-1-en-4-yne


Molecular Assembly Program I'm not totally sure which Stack Exchange community this would go in, but Chemistry seemed best.
I've been thinking for a long time now, that it would be cool to have a program where you could design your own atomic models, from the sub-atomic scale and up, then test them in a real situation. You could create anything from simple molecular structures to...anything.
I've looked around online, but there aren't any programs that you can build your own atomic structure. Anyone know if there are any out there like this?

EDIT: I've tried a few other programs, including Avogadro, this, and another one by PHET.

You  might want to try BioVia's (formerly Accelrys) Discovery Studio Visualizer. This is free software and the "gateway drug" to their sophisticated suite of molecular modeling and chem/bio informatics software. It is a step above the various drawing programs out there and lets the user generate realistic 3-D models of an infinite variety of chemical structures. http://accelrys.com/products/discovery-studio/visualization-download.php


2 problems with elimination-addition of halobenzene using NH2- When converting chlorobenzene to aniline, I drew the following mechanism and the book says it is correct, but there are 2 questions that I do not understand:

In the 2nd step, I am thinking that it is a nuclophilic attack, since $\ce{NH2^-}$ is a good nucleophile. However, I do not understand what makes the $\ce{C#C}$ bond in benzyne electrophilic. Doesn't a triple bond have high electron density and render it more nucleophilic instead? How can the reaction happen?

In the 3rd step, I don't know why proton transfer happens here. Isn't forming $\ce{NH2-}$ from $\ce{NH3}$ not thermodynamically viable as $\ce{NH2-}$ is a strong base?

My thought on (2): forming $\ce{NH2-}$ ion from $\ce{NH3}$ is more stable than leaving the carbanion (the aniline with an extra electron) there, but I am not sure about whether this is a correct explanation.

1. Electrophilicity of the triple bond
Hard-soft acid-base theory ties in with this question. The general trends in nucleophile-electrophile reactivity show that hard (highly charge dense) nucleophiles have a tendency to react well with hard electrophiles, and soft (low charge density) nucleophiles with soft electrophiles. This relationship is due to two competing factors; electrostatics and overlap of frontier orbitals. Hard-hard interactions lend their stability to electrostatic attraction, while soft-soft interactions rely on a good HOMO-LUMO overlap.
A π-bond is generally considered a soft species that doesn't reply on electrostatics very strongly. Good soft nucleophiles have relatively high HOMO's (highest occupied molecular orbitals) and good soft electrophiles have relatively low LUMO's (lowest unoccupied molecular orbitals). The third bond in benzyne is abnormal in that it doesn't involve overlap of 2 unhybridised p-orbitals, but 2 hybridised $sp^2$ orbitals that aren't completely parallel. From Clayden Organic Chemistry:

This will, as expected, result in a very weak orbital overlap. As a result, the split in energy between the resulting bonding and antibonding orbital is very small, giving a very low LUMO which can easily accept a nucleophile's electrons. From here:

This is the source of the triple bond's electrophilicity.
2. The final protonation step
(Image from Clayden Organic Chemistry)

Your assumption is correct that the anion is very unstable. According to this source, , benzene has a $\mathrm{p}K_\mathrm{a}$ of 43. This is due to the negative charge being situated in an $\pu{sp^2}$ orbital, which is highly charge dense, with no delocalisation into the ring (since it is not a part of the π-system). Ammonia has a $\mathrm{p}K_\mathrm{a}$ of 38, so its conjugate base, $\ce{NH2^-}$, is an extremely good base. This value is still lower than benzene's, so the benzene anion is still less stable. By a quick calculation, this gives the reaction above an equilibrium constant of roughly $\pu{10^5}$, heavily in favour of ammonia deprotonation.


Why does the alkaline hydrolysis of benzamide yield ammonia? Nucleophillic addition to the carbonyl group is governed by the rule that a less basic group is displaced by a more basic group since $\ce{NH2^-}$ is more basic than $\ce{OH-}$ the reaction should not proceed forward but it does why so?

If the reaction were simply having the form

$\ce{RC(O)NH2 + OH^- <=> RC(O)OH + NH2^-}$

then your reasoning might well apply.  But, in the real world, the carboxylic acid $\ce{RC(O)OH}$ deprotonates (more than the amide $\ce{RC(O)NH2}$) and the proposed, strongly basic amide ion of course protonates to ammonia.  These proton transfer reactions push the displacement forward giving the true reaction

$\ce{RC(O)NH2 + OH^- -> RC(O)O- + NH3}$


Why is carboxylic acid hard to enolize? Why do we need PBr to help enolize a carboxylic acid. My textbook doesn't give a reason for this at all. Does it have to do something with the acidity of carboxylic acid. Thanks.

Think about the pKa of the carboxylic acid protons compared to the pKa of the enolisable proton (hint: the acid is more acidic).

Once the carboxylic acid is turned into the carboxylate, enolisation is not possible, as the carbonyl cannot simultaneously conjugate with the carbon and the oxygen.

The use of phosphorus tribromide is known as the Hell-Vollhard-Zelinski reaction, it allows carboxylic acids to be directly brominated at the alpha position without initial protection of the carboxylic acid via in situ formation of the acyl bromide which hydrolyses upon workup.


What role does sulfuric acid play in the titration of oxalic acid and potassium permanganate? The titration is between oxalic acid and potassium permanganate, with $\ce{H2SO4}$ added to oxalic acid. My question is what role other than a dehydrating agent and maybe catalysing does the $\ce{H2SO4}$ play in this titration.

The question came to me while reading a book (Advanced Chemistry by Philip Matthews - Cambridge university press), I was reading about titrations and this question came to me, but it isn't mentioned in the book.

The used analytical reaction of purple permanganate to colourless $\ce{Mn^2+}$ occurs under acidic conditions:

$$\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O}$$

Under neutral conditions, permanganate would be reduced to dark brown manganese(IV) oxide:

$$\ce{MnO4- + 4H+ + 3e- <=> MnO2 + 2H2O}$$

Therefore, sulfuric acid is added to make the solution acidic. Actually, excess acid is required since $\ce{H+}$ is consumed during the reaction (8 mol $\ce{H+}$ per 1 mol $\ce{MnO4-}$).

Sulfuric acid is used because it is stable towards oxidation; whereas, for example, hydrochloric acid would be oxidized to chlorine by permanganate.


What chemical compounds were used in CRT phosphors? In the old days computer monitors and televisions were built from cathode ray tubes (CRTs) where the image was generated by an electron beam hitting phosphors which emitted light approximating the three primary colours.

What were the phosphors made from and what were the key reasons why the particular substances were chosen?

You should ask yourself: "For which application?". In fact there are different types of "phosphors" (most of them have no "Phosphorus" inside) for different applications and different compound were used during the evolution of CRT monitor. Some kind of phosphors may have advantages in terms of:

Efficiency of the luminescence induced.
Cheaper price.
Longer life.
Better color.
Shorter decay time.

Note however that in many application longer decay times are prefer because allow to the user to see transient signal one of this application is radar in this case is used the GB phosphors (it usually used the WTDS symbol to refer to a specific compound. In Phosphor Handbook
 by Shigeo Shionoya,William M. Yen,Hajime Yamamoto p. 607 you can find the relative compound in this case is  $(Zn,Cd)S:Cu + (Ca,Mg)SiO_3:Ti)$) that has a time decay between 100 ms and 1 s.

If you are  interested in RGB monitor and you want a list of the phosphors manly used.
I've copied the the following table from Practical Applications of Phosphors by William M. Yen,Shigeo Shionoya,Hajime Yamamoto p. 232 showing how phosphors compound for every RGB channel changed during time.

\begin{array}{c c c  c}\hline
&Blue  & Green  & Red \\ \hline
1               & ZnS:Ag^{2+}   &  Zn_2SiO_4:Mn^{2+}   & Zn_3(PO_4)_2:Mn^{2+}\\
2 & ZnS:Ag^{2+} &  (Zn,Cd)S:Ag    &  (Zn,Cd)S:Ag  \\
3& ZnS:Ag^{2+} & (Zn,Cd)S:Ag & YVO_4:Eu^{3+}\\
4 & ZnS:Ag^{2+} &   (Zn,Cd)S:Cu,Al & Y_2O_2S:Eu^{3+}or\space Y_2O_3:Eu^{3+} \\
Current & ZnS:Ag^{2+}       & ZnS:Au,Cu,Al  \space or \space ZnS:CU,Al    &Y_2O_2S:Eu^{3+}    \\ \hline
 \hline
\end{array}


Analytical gradient of the Hartree-Fock nuclear-electronic repulsion term The analytical gradient of the Hartree-Fock energy is given by (Szabo & Ostlund eq. C.16):

$$ \frac{\partial E}{\partial X_{A}} = \sum_{ij}{P_{ji} \frac{\partial H_{ij}}{\partial X_{A}}} + \frac{1}{2} \sum_{ijkl}{P_{ji}P_{kl} \frac{\partial (ij|kl)}{\partial X_{A}}} - \sum_{ij}{Q_{ji} \frac{\partial S_{ij}}{\partial X_{A}}} + \frac{\partial V_{NN}}{\partial X_{A}} $$

Of course, the Born-Oppenheimer approximation means that the nuclear repulsion term is easy to handle, and the derivative is given as:

$$ \frac{\partial V_{NN}}{\partial X_{A}} = Z_{A} \sum_{B} \frac{Z_{B} (X_{B} - X_{A})}{R_{AB}^{3}}$$

The remaining terms require computing the derivatives of the Gaussian functions (depending upon the molecular integral evaluation scheme used). However, in Szabo & Ostlund the nuclear-electronic attraction term is provided as:

$$ \frac{\partial V_{Ne}}{\partial X_{A}} = -Z_{A} \sum_{i} \frac{X_{i} - X_{A}}{r_{iA}^{3}}$$

In the Hartree-Fock equations, the molecular Hamiltonian is formed as $H = T + V_{Ne}$. Presuming that the gradient of the molecular Hamiltonian is formed the same way, how can I treat this formulation of the analytical gradient with only a scalar value for the nuclear attraction term, when I must take into account the density matrix? Is this solution to the nuclear attraction gradient of no use in calculating the gradient of the energy?

  [1] A. Szabo and N. S. Ostlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Dover Publications Inc., New York, pp. 441-442.

There are a few points to discuss:

Since there are $3N$ possible $\{X_{A}\}$, each term where an $X_{A}$ appears will result in $3N$ matrix elements. In your first equation, that will be $S$, $T$, $V$, and the 2-electron contribution, which should really be rewritten into a Fock matrix term $F$ where the ket is precontracted with the density.
There is nothing wrong with this solution, though it is a little strange because it's in the MO basis, and we normally prefer to work in the AO basis to avoid any unnecessary transformations. I'm going to drop the $Ne$ from here on out since $V$ is understood in this context to only be the one-electron nuclear attraction term.
I think you're asking what happens to the density matrix. The derivative of $V$ expands into multiple parts, as we will see below. I will reference a bunch of equations that use $h$, where $\left( h = H = H^{\text{core}} \right) \equiv T_{e} + V_{Ne}$, and differentiating into a sum should hopefully be clear. What follows should work for any real-valued one-electron operator. I will use the book by Yamaguchi with liberties taken. In many places I use $p,q$ rather than $i,j$, as these indices run over all MOs, not just occupied ones.

Expanding $\frac{\partial V_{ij}}{\partial X_{A}}$ using the product rule gives

$$
\begin{align*}
\frac{\partial V_{ij}}{\partial X_{A}} &= \frac{\partial}{\partial X_{A}} \left( \sum_{\mu\nu}^{\text{AO}} C_{\mu i} C_{\nu j} V_{\mu\nu} \right) \tag{3.80} \\
&= \sum_{\mu\nu}^{\text{AO}} \left( \frac{\partial C_{\mu i}}{\partial X_{A}} C_{\nu j} V_{\mu\nu} + C_{\mu i} \frac{\partial C_{\nu j}}{\partial X_{A}} V_{\mu\nu} + C_{\mu i} C_{\nu j} \frac{\partial V_{\mu\nu}}{\partial X_{A}} \right), \label{3.81}\tag{3.81}
\end{align*}
$$

where the third/last term is the true AO integral derivative, and the first two terms, the MO coefficient derivatives, come from differentiating the density matrix.

The simplest term to write out, but the most difficult to implement, is the AO integral derivative. I will use $\mu,\nu$ rather than $\chi_{\mu},\chi_{\nu}$, so they refer to both AO basis functions and their matrix indices.

$$
\begin{align*}
\frac{\partial V_{\mu\nu}}{\partial X_{A}} &= \frac{\partial}{\partial X_{A}} \left< \mu | \hat{V} | \nu \right> \tag{3.24} \\
&= \left< \frac{\partial \mu}{\partial X_{A}} | \hat{V} | \nu \right> + \left< \mu | \frac{\partial \hat{V}}{\partial X_{A}} | \nu \right> + \left< \mu | \hat{V} | \frac{\partial \nu}{\partial X_{A}} \right> \label{3.25}\tag{3.25}
\end{align*}
$$

What you have presented,

$$
\frac{\partial \hat{V}}{\partial X_{A}} = -Z_{A} \sum_{i} \frac{X_{i} - X_{A}}{r_{iA}^{3}},
$$

is only the derivative of the operator. The basis function derivatives are also required, as the AOs in this case depend on the nuclear positions. Contrast this with an electric field perturbation, where only the derivative of the operator is required, or magnetic field derivatives, where basis functions may be perturbation-dependent if gauge-including atomic orbitals (GIAOs) are used. Also important is that the index $i$ here refers to an electron, not an occupied MO. This is an electric field integral.

Now for the derivative of the MO coefficients/density matrix. Clearly they don't appear in the final HF gradient expression. They disappear through the magic of Wigner's $2n + 1$ rule. From page 25:

  When the wavefunction is determined up to the $n$th order, the expectation value (electronic energy) of the the system is resolved, according to the results of perturbation theory, up to the $(2n+1)$st order. This principle is called Wigner's $2n+1$ theorem [29-31].

More explicitly, we have the zeroth-order wavefunction, so we must be able to calculate the first-order correction to the energy. Worded differently, any first derivative of the energy must be easily calculated without differentiating MO coefficients, which is only required for second derivatives, such as the molecular Hessian or the dipole polarizability.

First, rewrite the MO coefficient derivatives

$$
\frac{\partial C_{\mu i}}{\partial X_{A}} = \sum_{m}^{\text{MO}} U_{mi}^{X_{A}} C_{\mu m}, \tag{3.7}
$$

where the index $m$ runs over all occupied and unoccupied/virtual MOs. The same goes for $p,q$. I've replaced $a$ from the text with $X_{A}$, but this holds for any general perturbation (such as $\lambda$, pick your favorite unused index). The key insight is that we can write the effect of a perturbation on the MO coefficients as the contraction of the unmodified MO coeffcients with a unitary matrix describing single-particle excitations from occupied to virtual MOs, as well as deexcitations from virtual to occupied MOs. In matrix form, this is

$$
\mathbf{C}^{(X_{A})} = \mathbf{C}^{(0)} \left( \mathbf{U}^{(X_{A})} \right)^{T},
$$

where it is usually easiest to have the dimension of $\mathbf{U}$ be $[N_{\text{orb}}, N_{\text{orb}}]$, with only the occ-virt and virt-occ blocks being non-zero. The problem now becomes solving for $\mathbf{U}$ and eliminating it from the final gradient expression. I will walk through parts of section 4.3 to show how this is done.

Given the energy expression for an RHF wavefunction,

$$
E = 2 \sum_{i}^{\text{d.o.}} h_{ii} + \sum_{ij}^{\text{d.o.}} \left[ 2(ii|jj) - (ij|ij) \right], \tag{4.1}
$$

the first derivative with respect to $X_{A}$ is

$$
\frac{\partial E_{\text{elec}}}{\partial X_{A}} = 2 \sum_{i}^{\text{d.o.}} h_{ii}^{X_{A}} + \sum_{ij}^{\text{d.o.}} \left[ 2(ii|jj)^{X_{A}} - (ij|ij)^{X_{A}} \right] + 4 \sum_{m}^{\text{all}} \sum_{i}^{\text{d.o.}} U_{mi}^{X_{A}} F_{im}, \label{4.16}\tag{4.16}
$$

where the Fock matrix is defined as

$$
\begin{align*}
F_{pq} &= h_{pq} + \sum_{k}^{\text{d.o.}} \left[ 2(pq|kk) - (pk|qk) \right] \tag{4.6} \\
&= h_{pq} + 2J_{pq} - K_{pq},
\end{align*}
$$

and I've introduced the Coulomb and exchange matrices $\mathbf{J}$ and $\mathbf{K}$ as well. I skipped all the steps in expanding the first derivative, and $\eqref{4.16}$ is what results upon collecting terms with $\mathbf{U}$.

Using the RHF variational conditions, the Fock matrix from a converged calculation is diagonal in the MO basis, corresponding to the MO energies

$$
F_{pq} = \delta_{pq} \epsilon_{pq}, \tag{4.7}
$$

so $\eqref{4.16}$ simplifies to

$$
\frac{\partial E_{\text{elec}}}{\partial X_{A}} = 2 \sum_{i}^{\text{d.o.}} h_{ii}^{X_{A}} + \sum_{ij}^{\text{d.o.}} \left[ 2(ii|jj)^{X_{A}} - (ij|ij)^{X_{A}} \right] + 4 \sum_{m}^{\text{all}} \sum_{i}^{\text{d.o.}} U_{mi}^{X_{A}} \epsilon_{im}, \tag{4.17}
$$

which can be further simplified as

$$
\frac{\partial E_{\text{elec}}}{\partial X_{A}} = 2 \sum_{i}^{\text{d.o.}} h_{ii}^{X_{A}} + \sum_{ij}^{\text{d.o.}} \left[ 2(ii|jj)^{X_{A}} - (ij|ij)^{X_{A}} \right] + 4 \sum_{i}^{\text{d.o.}} U_{ii}^{X_{A}} \epsilon_{ii}. \tag{actual 4.17}
$$

Now we use one of the most important tricks in quantum chemistry. Given the orthonormality of the MOs,

$$
S_{pq} = \delta_{pq}, \tag{3.44}
$$

we must have

$$
\frac{\partial S_{pq}}{\partial X_{A}} \overset{!}{=} 0. \tag{3.45}
$$

This is where not using general notation in the above is a bit confusing because it seems to conflict with your original expression, but remember that similar to $\eqref{3.81}/\eqref{3.25}$, this is in fact multiple terms: two for the basis functions, and two for the MO coefficients, giving

$$
\begin{align*}
\frac{\partial S_{pq}}{\partial X_{A}} &= \sum_{\mu\nu}^{\text{AO}} C_{\mu p} C_{\mu q} \frac{\partial S_{\mu\nu}}{\partial X_{A}} + \sum_{m}^{\text{all}} \left( U_{mp}^{X_{A}} S_{mq} + U_{mq}^{X_{A}} S_{pm} \right) \tag{3.40 + 3.43} \\
                                       &= S_{pq}^{X_{A}} + \sum_{m}^{\text{all}} \left( U_{mp}^{X_{A}} S_{mq} + U_{mq}^{X_{A}} S_{pm} \right). \tag{3.43}
\end{align*}
$$

The sum over all MOs can be eliminated by reusing the orthonormality condition, so in the first term $m \overset{!}{=} q$ and for the second term $m \overset{!}{=} p$, and the overlap matrix in the MO basis is unity for those terms, giving

$$
\frac{\partial S_{pq}}{\partial X_{A}} = S_{pq}^{X_{A}} + U_{qp}^{X_{A}} + U_{pq}^{X_{A}} \overset{!}{=} 0. \tag{3.46}
$$

Recognizing that we only need diagonal terms, this can be rewritten as

$$
U_{pp}^{X_{A}} = -\frac{1}{2} S_{pp}^{X_{A}}, \tag{4.20}
$$

which is then plugged back into the first derivative expression to give

$$
\begin{align*}
\frac{\partial E_{\text{elec}}}{\partial X_{A}} &= 2 \sum_{i}^{\text{d.o.}} h_{ii}^{X_{A}} + \sum_{ij}^{\text{d.o.}} \left[ 2(ii|jj)^{X_{A}} - (ij|ij)^{X_{A}} \right] + 4 \sum_{i}^{\text{d.o.}} \left( -\frac{1}{2} S_{ii}^{X_{A}} \right) \epsilon_{ii} \\
                                                &= 2 \sum_{i}^{\text{d.o.}} h_{ii}^{X_{A}} + \sum_{ij}^{\text{d.o.}} \left[ 2(ii|jj)^{X_{A}} - (ij|ij)^{X_{A}} \right] - 2 \sum_{i}^{\text{d.o.}} S_{ii}^{X_{A}} \epsilon_{ii}. \tag{4.21}
\end{align*}
$$

Rewrite the last term

$$
\begin{align*}
\sum_{i}^{\text{d.o.}} S_{ii}^{X_{A}} \epsilon_{ii} &= \sum_{i}^{\text{d.o.}} \sum_{\mu\nu}^{\text{AO}} C_{\mu i} C_{\mu i} \frac{\partial S_{\mu\nu}}{\partial X_{A}} \epsilon_{ii} \\
                                                    &= \sum_{i}^{\text{d.o.}} \sum_{\mu\nu}^{\text{AO}} C_{\mu i} C_{\mu i} \epsilon_{ii} \frac{\partial S_{\mu\nu}}{\partial X_{A}} \\
                                                    &= \sum_{\mu\nu}^{\text{AO}} W_{\mu\nu} \frac{\partial S_{\mu\nu}}{\partial X_{A}} \tag{4.24}
\end{align*}
$$

to use the energy-weighted density matrix $\mathbf{W}$, which in your expression is called $\mathbf{Q}$.

Again, the elimination of the $\mathbf{U}$ matrix is one of the most important results in quantum chemistry, as it means the coupled-perturbed SCF equations do not need to be solved for first derivatives of SCF wavefunctions. This is why you do not see any density or MO coefficient derivatives in the gradient expression.

References

Yamaguchi, Yukio; Goddard, John D.; Osamura, Yoshihiro; Schaefer III, Henry F. A New Dimension to Quantum Chemistry: Analytic Derivative Methods in Ab Initio Molecular Electronic Structure Theory (International Series of Monographs on Chemistry); Oxford University Press: 1994.
Epstein, Saul T. General Remainder Theorem. J. Chem. Phys. 1968, 48, 4725.
Epstein, Saul T. Constraints and the $V^{2n+1}$ theorem. Chem. Phys. Lett. 1980, 70, 311.

If you don't have access to the Yamaguchi book, the seminal article on HF derivatives contains a condensed version of this derivation, but be aware of typos.

Pople, John A.; Raghavachari, Krishnan; Schlegel, H. Bernhard (Berny); Binkley, J. Stephen. Derivative Studies in Hartree-Fock and Møller-Plesset Theories. Int. J. Quantum Chem. 1979, 16 (num. S13), 225-241.


Can a small amount of scandium be added to gold, platinum or palladium to create a wedding ring? The number 21 has a special meaning to us, so we would like to make wedding rings with some scandium in it (element 21). I know scandium can be alloyed with aluminium or titanium, but can it also be alloyed with say gold, platinum or palladium without detrimental effects?

The go-to place for alloy information is ASM International's Alloy Phase Diagram Database (providing your institution has access).

Scandium-Gold: as noted in the comment by @MaxW, you can get several atomic percent (but less than 10 at.%) Sc in Au. There exist 6 compounds across the diagram which you would want to avoid most likely.

Scandium-Platinum: likely up to about 10 at.% of Sc in Pt. At least several compounds to avoid.

Scandium-Palladium: also order of 10 at.% of Sc. 5 compounds to avoid.

All the compounds should be avoided because (1) they are higher in scandium content, and (2) they are mostly line compounds and unlikely to be malleable.

So, yes, you can get scandium in to a gold-platinum-palladium mix. Find a good jeweler and they can melt it in for you. Scandium is not all that expensive (at least compared to the cost of getting the jeweler to do a custom alloy).


Why polyethylene (PE), wool (polyamide) and cotton (cellulose) burnt with different smells, while they just product CO2 (and N2)? Both PE and cotton produce CO2, but they distinctly have different smells while being burnt (Burning PE has smell of candle wax, while burning cotton’s smell is like burning leaves). Wool produces CO2 and N2 which is odorless, but yet has smell of hair burnt.
  Moreover, they also differ in burning process (PE and wool burns slowly, while cotton burns rapidly). So why there are such differences when their products are similar?

Things are more complex than you suppose

There are two key reasons why these compounds burn with different smells.

The first is that your assumption about what the substances contain is wrong. Biological fibres contain plenty other than carbon and hydrogen. Often they will have lots of nitrogen and sulphur and various other elements and these often produce a variety of pungent chemicals that provide strong odours even in low concentrations. Plus artificial polymers will contain small amounts of other ingredients to modify their properties and these will often contain elements that give more complex combustion products some of which may be smelly.

The second is that burning these compounds in normal conditions is far from a perfect reaction yielding just the obvious oxidation products like carbon dioxide (which would be odourless). Even pure PE (and most of the PE you get will not be pure as it will contain various fillers and additives with other elements in it) will not burn cleanly to produce water and carbon dioxide). In normal conditions the burning will produce a variety of hydrocarbons and particles some of which will have odours. This is fairly obvious from the soot produced (which you would not get at all if the burning was completely efficient in producing carbon dioxide). The variety of those "imperfect" combustion products will be very different for different starting materials. So even compounds with roughly the same chemical composition (but different structure) may burn to give different combustion side products and, hence, will smell different.


Can ZnO act as a reducing agent for noble metals like Pd? Palladium ions (e.g., $\ce{Pd^2+}$) can be reduced to $\ce{Pd}$ by more reactive metals such as zinc. Metal oxides such as $\ce{ZnO}$ are very stable but can its atoms in the oxide structure still be displaced by Pd?
I know the answer is probably "No" but in one of my samples $\ce{ZnO}$ nanoparticles were removed from the solution after I added $\ce{Pd^2+}$ solution to the system and $\ce{Pd}$ nanoparticles were formed. The solvent was methanol. After washing the solution and drying I did XPS and no trace of zinc was observed but saw $\ce{Pd}$ peaks.

Zinc chemistry is overwhelmingly that of the +2 oxidation state, it does not go higher. For ZnO to reduce Pd+2, the Zn has to go up from +2 which it cannot. If you are getting reduction of Pd+2, Zn is not responsible.


Why can't oxygen in furan be sp-hybridized? From the Chem LibreTexts article on Aromaticity and the Huckel $4n + 2$ Rule:

So far, you have encountered many carbon homocyclic rings, but compounds with elements other than carbon in the ring can also be aromatic, as long as they fulfill the criteria for aromaticity. These molecules are called heterocyclic compounds because they contain one or more different atoms other than carbon in the ring. A common example is furan, which contains an oxygen atom. We know that all carbons in furan are $\mathrm{sp^2}$ hybridized. But is the oxygen atom $\mathrm{sp^2}$ hybridized? The oxygen has at least one lone electron pair and is attached to an $\mathrm{sp^2}$ hybridized atom, so it is $\mathrm{sp^2}$ hybridized as well.

Is this a general pattern - if an adjacent atom is $\mathrm{sp^2}$-hybridized, and if there's at least one electron pair, then our atom it $\mathrm{sp^2}$-hybridized as well?
Why can't oxygen in furan be $\mathrm{sp}$-hybridized? Is this only because the overall structure will not be consistent, because sp-hybridization will make the $\ce{C-O-C}$ line straight, and thus the $\ce{O-C-C}$ angles would have to be lesser than the carbons' $\mathrm{sp^2}$-hybridization would allow?

Is this only because the overall structure will not be consistent, because sp-hybridization will make the COC line straight, and thus the OCC angles would have to be lesser than the carbons' sp2-hybridization would allow?

Yes, this is the correct answer. A very common misconception is that hybridisation can be used to predict the geometry, or that hybridisation somehow involves an energy cost associated with 'promoting' electrons into the hybrid orbitals. This is entirely wrong.

Hybridisation is always determined by geometry. You can only assign hybridisation states to an atom if you already know its geometry, based on some experimental or theoretical evidence. The geometry of the oxygen in furan is trigonal planar and therefore the hybridisation must be $sp^2$.


Why is fat nonpolar? Why is fat nonpolar while water is polar?

In water there is the $\ce{H-O}$ bond and the $\ce{O}$ is more negative, we have
$\text{dipole = charge}* \text{distance}$

A fat molecule is like
$\ce{H-O-C-C-C-C-C-C...}$
but the carbon atoms don't matter, the polarity is still ${q*d}$, as in the water molecule, so they should have the same amount of polarity.

So why do people say fat is a nonpolar molecule?

The polarity of a single bond might indeed be the same for some different molecules. When speaking of polarity of molecules however, the whole molecule should be taken into account, since the macro properties are dictated by all the parts of a molecule.

Regarding the fatty acids: The 'head' of such a molecule is indeed polar, and in specific cases this means that the interactions and reactivity will be like polar molecules. This effect however is only contained to the head of the molecule. Fatty acids as a whole are still considered non-polar because this type of molecule will not be dissolved in water. Note that the main reason for this is not due to the interactions of water with the fatty acid. The fatty acid is just as 'happy' interacting with water as it is with other fatty acid chains. Water, on the other end, does not 'want' to interact with fatty acids, since it will have to rearrange around the fatty acid molecule to ensure the same number of hydrogen bonds are made. This is an entropic cost, that makes the total Gibbs Free Energy change unfavorable.

The result is that water molecules do not rearrange the hydrogen bonds with other water molecules and the fatty acids interact with each other. Due to this effect, water not being willing to rearrange hydrogen bonds, fatty acids are considered non-polar.

Hope this has helped, and if I used to much terminology or jargon, please let me know!


How to calculate the mass of magnesium necessary to produce a certain volume of hydrogen gas? How many grams of magnesium metal are required to produce 44.8 liters of hydrogen gas at STP according to the chemical equation shown below? (Answer: $\pu{48.6 g}$)
$$\ce{Mg(s) + 2 HCl(aq) -> MgCl2(aq) + H2(g)}$$

What I have done:
Amount of hydrogen to be produced= 44.8/22.4 = 2 moles of $\ce{H}$ = 2.02 grams of $\ce{H}$.
Amount of magnesium metal needed $$=\frac{\pu{2.02 g} (\text{of}~\ce{H})\cdot \pu{1 mol} (\text{of}~\ce{H2})\cdot \pu{1 mol} (\text{of}~\ce{Mg})\cdot \pu{24.31 g} (\text{of}~\ce{Mg})}{\pu{2.02 g} (\text{of}~\ce{H2})\cdot \pu{1 mol} (\text{of}~\ce{H2})}$$
I'm not getting the correct answer, can someone please explain how to do this using dimensional analysis? Please just tell me only how to do this, do not show it.

It is very helpful when you start with using proper notation, i.e. $V(\ce{H2})=44.8~\mathrm{L}$. Proper notation like $M$ for molecular mass, $m$ for mass, and $n$ for amount of substance will help you follow through your thinking.
When do the calculations, always carry the units with every quantity you use.

It is necessary to know a few principle formulae, for example that molecular mass and mass are connected through the amount of substance: $$M = \frac{m}{n}$$
There are a few more you will need and with their repeated use you will get to know them easily.

When attacking this problem, find out how the quantities relate to each other. In this example it is: If I want to produce one mole of hydrogen gas, how many moles of magnesium do I need?

   one, because $n(\ce{Mg})\ce{->}n(\ce{H2})$

A key principle (and asumption) in this exercise is the ideal gas law,
$$pV=n\mathcal{R}T.$$
You need it to calculate the amount of substance of hydrogen that should be produced.
(You have indirectly used it by dividing the given value by the molar volume of an ideal gas at STP.)

   In this case about two moles of hydrogen gas shall be produced, i.e.
 $\displaystyle n(\ce{H2})=\frac{pV(\ce{H2})}{\mathcal{R}T}\approx2~\mathrm{mol}$
 (Old definition of STP: $p=1~\mathrm{atm}$ and $T=298.15~\mathrm{K}$)

Now you also know the amount of substance of magnesium you need.

   two moles, i.e. $n(\ce{Mg}=2~\mathrm{mol}$

You just have to use the molecular mass of magnesium to find the mass of magnesium you need with the above connection.


Should β-mercaptoethanol and anything that touches it be handled only under fume hoods for safety aspects? Background: I recently started in a lab that regularly uses β-mercaptoethanol (βME) for protein purification. I know that the chemical is an irritant for the lungs if inhaled (mucus membranes), damaging if in contact with eyes, toxic if in contact with skin. However, I have had projects in different labs that handle this chemical differently. Some strictly only open it in the fume-hood, have designed waste in the fume-hood for any pipette tips that touch βME and always discard any protective gloves that they used while handling this chemical directly after use – so not to spread residual βME all over the lab. This also goes for any buffers that have βME in it.
In the lab that I currently work in, they open the concentrated bottle of βME in the fume hood, but they don't care too much about discarding pipette tips in designated waste nor change gloves after use (and I know that the gloves smell like βME after opening the bottle). In addition, they do not work with lammeli buffer in the fume hood (they use 4 × lammeli for SDS-PAGE). However, from the safety manual of lammeli-buffers — which also contains βME (and you can immidiately smell that) it seems to be just as hazardous as the concentrated βME.
I don't have hypochondria when it comes to lab safety, but I would like to develop a rational understanding of how dangerous this is common chemical is. In addition I believe that the fridge which we store βME in smells quite strongly of βME. Should the bottle be stored in a secondary container (plastic bag) in addition?
Question: How do people who commonly use this chemical treat it during their everyday lab routines? How hazards is it to regularly inhale trances of βME from e.g. the fridge or when people work with lammeli buffers on the lab bench? What do people think about discarding their gloves after working with it, before they start touches other lab equipment?

β-Mercaptoethanol is not that dangerous a chemical. Yes, you should not drink it, you shouldn’t dip your bare fingers in it, you shouldn’t inhale tons of it and you shouldn’t use it to rub yourself down. But other than that it is hardly more dangerous than most other thiols. Simple methanol is probably more toxic and is also routinely handled.

The thing about β-mercaptoethanol is that it  stinks like hell.  And it has the tendency to stick around stinking for quite some time. Like with most thio-compounds, you can smell it long before it becomes seriously dangerous; your sense of smell is a warning against being too risky. You know yourself that the gloves stink if you used the bottle, that the pipette tips stink after pipetting it and that any flasks that contained it tend to stink for quite a while afterwards.

And that is the reason why some labs rigorously segregate it from other chemicals; have dedicated wastes for pipette tips and gloves all kept under a fume it. It is all simply meant to stop the stench as early as possible, to keep it confined in a single fume hood and keep most of the lab air smelling better. The safety aspect is secondary. If you develop a routine of changing equipment every time after handling β-mercaptoethanol, you are less likely to forget to take off your glove and the lab has less smell to deal with.

Therefore, you would be doing your lab colleagues a favour by keeping the bottle confined within its own small plastic bag; but it’s only an olfactory favour, not that much a health favour.


UV-Transparent Plastic? I am looking for a type of plastic, preferably in the form of a sheet or film, that is weatherproof, durable, flexible, and both transparent and UV-transparent. Most films I've seen so far are treated to be UV-resistant, where can I find untreated plastics for purchase?
Planning to use this plastic in order to cover a solar panel, without inhibiting energy output.

Did you consider PMMA?

A 3 mm Plexiglas G UVT layer seems to have a transmittance > 80% at $\lambda = 300~\mathrm{nm}$ and above.


Is beryllium difluoride covalent or ionic? My textbook  says that despite the large electronegativity difference $\ce{BeF2}$ is covalent since the beryllium ion will have too much charge density and it will attract the fluorine electron cloud and therefore forms polar covalent bonds.

I looked it up on Wikipedia and found out that it has a crystal lattice,  is soluble in water, and has a quite high melting point.

What else does one need to call a compound ionic? Am I or the textbook right?

There is a very relevant article by Gillespie Covalent and Ionic Molecules: Why Are BeF2 and AlF3 High Melting Point Solids whereas BF3 and SiF4 Are Gases? J. Chem. Educ., 1998, 75 (7), p 923.

According to the article, the charge on Be is +1.81 and the charges on the Fs are -0.91. (citing to his earlier article Reinterpretation of the Lengths of Bonds to Fluorine in Terms of an Almost Ionic Model Inorg. Chem., 1997, 36 (14), pp 3022–3030)

So while nothing is completely covalent or ionic, BeF2 is extremely toward ionic.


Why does Fe(CO)₄ adopt a tetrahedral, as opposed to square planar, geometry? Why does $\ce{Fe(CO)4}$ adopt a tetrahedral, as opposed to square planar, geometry? Here's my analysis:

Factors for square planar:

coordination: 4
$\ce{Fe(0)}$: d8
$\ce{CO}$: strong field ligand (gives large d-splitting)

Factors for tetrahedral:

oxidation state: 0 (gives weak d-splitting)
$\ce{CO}$: relatively bulky (?) ligand

Other thoughts:

The general guidelines for determining whether or not a row 1 transition metal complex is square planar or tetrahedral (given it  has 4-coordination and d8 electrons), are the strength and bulkiness of the ligand. Generally, strong field ligands like $\ce{CO}$ and $\ce{CN^{-}}$ gives square planar complexes. In fact, $\ce{[Ni(CN)4]^{2-}}$ is square planar. Thus, the only way I could explain why $\ce{Fe(CO)4}$ would be tetrahedral would be because of its low oxidation state. Is this correct?

The hypothetical $\ce{Fe(CO)4}$ complex is isoelectronic to the observed $\ce{[Ni(CN)4]^{2-}}$ complex and therefore would also be square planar.[1,2] Both are 16 electron complexes, i.e. one of the valence orbitals of the central atom remains unoccupied. Your assessment for $\ce{Fe^{\pm0}}$ as $\mathrm{d^8}$ is absolutely correct, also that carbonyl is a good ligand causing a strong field and hence a strong splitting. In a tetrahedral complex there would be triply degenerated HOMO, which would only be occupied by four electrons. This arrangement cannot be more stable than a planar complex. However, the iron tetracarbonyl complex as a singular entity has not been observed (to my current knowledge). The observed structures for the compound are various dimers and trimers.[3]

There is also a neutral iron carbonyl complex, but it binds five instead of only four ligands. Therefore $\ce{Fe(CO)5}$ is a trigonal bipyramidal 18 electron complex.[4]

The observed $\ce{[Fe(CO)4]^{2-}}$ complex[5] is isoelectronic to $\ce{Ni(CO)4}$ which is a tetrahedral 18 electron complex.[1,4] Without going into more detail, the iron in this complex is of course $\ce{Fe^{-II}}$ and therefore $\mathrm{d^{10}}$. A square planar arrangement would no longer be more stable and a complex with a higher symmetry and more degeneracy is preferable.

References

J. Demuynck, and A. Veillard, Theor. chim. acta,
1973, 28 (3), 241-265.
ChemSpider.com
Raymond K. Sheline, J. Am. Chem. Soc., 1951, 73 (4), 1615–1618
Dario Braga, Fabrizia Grepioni, and A. Guy Orpen, Organometallics, 1993, 12 (4), 1481–1483.
Henry B. Chin, and Robert Bau, J. Am. Chem. Soc., 1976, 98 (9), 2434–2439.

Note
Carbonyl is nowhere close to a bulky ligand. Apart from halides and hydrides it might as well be the smallest possible ligand.


Biochem Book age and suggestions I'm taking chem classes, and am interested in learning a some biochem on the side, on my own. I've bee looking for a decent book, though I'm on a bit of a tight budget so not brand shiny new ones upwards of a hundred. Hopefully. Obviously I don't want something ancient just for reduced price, so two questions. How many years back is ok to use as a cutoff for publication date, and any suggestions?

I would suggest the Lehninger Principles of Biochemistry, which is one of the best in my opinion since it covers most of the key aspects in biochemistry.
It is quite expensive to buy as new, but since it is quite common you can maybe borrow it from a library. In this way you can get an impression about the book and eventually decide if it is worth buying it or not. If you are interested in having it I would buy a secondhand one and I would say it does not matter if it is not the last version, the previous one should be good as well.


Role of intermediate in determining the exothermic or endothermic nature of reaction I was given a question to answer:

Which of the following reactions is “not” exothermic?

$\ce{CH3-Cl + CH3-CH3 -> CH4 + CH3-CH2-Cl}$
$\ce{CH3-Cl + (CH3)3-CH -> CH4 + (CH3)3C-Cl}$
$\ce{CH3-Cl + CH2=CH-CH3 -> CH4 + CH2=CH-CH2-Cl}$
$\ce{CH3-Cl + CH2=CH2 -> CH4 + CH2=CH-Cl}$

Now, what my teacher told was that; the 4th reaction proceeds through a very unstable free radical as compared to the other three reactions. Therefore the 4th reaction must not be exothermic.
But I have a doubt that stability of an intermediate in any reaction can only give us an idea of the activation energy required to reach the intermediate from the reactant via a transition state. Whereas the stability of product (with respect to the reactant) determines whether a reaction is endothermic or exothermic.
So how do we check here product stability to determine exothermic or endothermic nature of the reactions?
Or if something is wrong in what I think then please correct it.

A typical method for determining whether a reaction is endo or exothermic is to compare the heats of formation ($\Delta_f H$) of the products to those of the reactants. If we do not have measured values for the enthalpies of formation, we can get a fairly good estimate by summing up typical values for the bond enthalpies of the bonds in the compounds.
When using bond enthalpies, we can simplify the calculation by considering only those bonds that are broken or formed during the reaction, since the other bonds appear in both the products and reactants and so cancel out when we look at the difference.
Thus,
$$\textrm{Enthalpy of reaction = }\Delta_r H = \textrm{[sum of enthalpies of bonds broken]} - \textrm{[sum of enthalpies of bonds formed]}$$
[Here I'm using the standard convention that bond enthalpies are reported for the breaking of bonds and so are positive values.]
Comparing the four reactions given, we can simplify even further by noting that all four reactions involve breaking of a $\ce{H3C-Cl}$ bond and formation of a $\ce{H3C-H}$ bond. Since those occur in all four reactions, we can ignore them for the purpose of comparison.
We now have reduced the problem to a simple comparison of the change of an $\ce{R-H}$ bond to an $\ce{R-Cl}$ bond.
You are correct that it does not matter what intermediates might occur on the reaction path. The radical stability is relevent, however, because the bond enthalpy is defined as the change in enthalpy needed for homolytic bond cleavage, ie
$\ce{A-B -> A. + B.}$
Since the products are radicals, the stability of the radicals gives us information about the bond enthalpy. It is for this reason that comparing the stability of radical "intermediates" is a way to tell which of your given reactions is not exothermic.


In which direction does imidazole affect the pH? I'm reading something about nickel affinity chromatography and it talks about using imidazole to affect pH. In which direction does it change the pH?

Imidazole is an amphoteric molecule and can therefore act as an acid and a base.
 (courtesy of wikipedia)
It's acidity is given with $\mathrm{p}K_a=14.5$ for the proton as $\ce{N~1}$. The basic site is the lone pair at $\ce{N~3}$. Its basicity is given through the following equilibrium: $$\ce{Im + H+ <=> IMH+}$$ with a $\mathrm{p}K_a=7.05$.
(source)
You can obtain more information about this nice little molecule on its wikipedia page.

So in conclusion, it very much depends on the solution with which you are working, which effect of imidazole takes over.


How do I determine the temperature at which a dissolved substance evaporates? I would like to understand the temperature at which a substance will vaporize when dissolved in a liquid. I have researched this online for hours, but haven't found a conclusive answer. Is it the boiling point of the dissolved substance? I'm attempting to find the temperature at which caffeine vaporizes when dissolved in water or other vegetable glycerin. Many thanks for your help.

First, it is a good idea to clarify the meaning of the word vaporize from here:

  (Physics / General Physics) to change or cause to change into vapour or into the gaseous state
  to evaporate or disappear or cause to evaporate or disappear, esp suddenly
  (Physics / General Physics) to destroy or be destroyed by being turned into a gas as a result of extreme heat (for example, generated
  by a nuclear explosion)

Chemists should be careful when using this word because it can be used to describe evaporation, sublimation, and boiling.  Note the third definition, which suggests that the vaporization of matter typically involves a large amount of energy.

Since the mixture you refer to in your question is in a liquid state, I am going to focus on liquid boiling.  (Note, we can say that boiling is the rapid vaporization of a liquid.)  The boiling point of a liquid is the point at which the vapor pressure of the liquid is equal to the surrounding pressure.  The boiling point of a liquid is affected by intermolecular forces such as hydrogen bonding, dispersion and dipole-dipole interactions.

When a solute is dissolved in a pure liquid (your example of caffeine dissolved in water), these intermolecular forces are altered.  The end result is an elevation of the boiling point.  This change can be predicted quite accurately using concepts typically covered in the first year of a college-level general chemistry class.

So, we know now that a mixture of caffeine and water will have a boiling point higher than pure water; however what happens as the solution continues to boil?  I ignored an important point in the previous paragraph; boiling point elevation due to addition of a solute will only occur if the solute is nonvolatile. Caffeine is such a solute.  Therefore, as you boi your mixture, the water will be turned into a gas, but the caffeine will be left behind.  The volume of solvent is decreasing; therefore the solution concentration is increasing.  Once the caffeine solubility limit is reached, it will start to crystallize.

When you are left with a mound of caffeine you can continue to heat it.  As you heat it above about $230\,^{\circ}\mathrm{C}$ it will start to sublime, or in more general terms, vaporize.

For those interested, the experimental procedure for the extraction and purification of caffeine from tea leaves is described at an undergraduate level here.


Make paper from hair? Hair is fibrous... Is there a good reason, why we don't recycle our own hair (and clothing, what have you) into paper-products? (rather than cutting down trees)?

The diameter of human hair is typically 100 microns (from 17 to 181 microns: https://hypertextbook.com/facts/1999/BrianLey.shtml ). The diameter of a cellulose fiber is about 10 microns ( https://en.wikipedia.org/wiki/Paper ), and a cellulose molecule is much smaller (up to about 10,000 glucose units long: https://en.wikipedia.org/wiki/Cellulose_fiber ).

While it should be possible to make a coarse papery product from hair, something is needed to make the hair fibers stick together without being sticky in the final product. Think how obnoxious it would be if you wash your hair, then dry it and it turns into cardboard!

Paper from cellulose fibers sticks together when it dries because there are so many hydroxyl groups on every cellulose molecule and fiber - no extra adhesive needed. It can be made quite thin - commercial paper is available as thin as 60 microns ( https://hypertextbook.com/facts/2001/JuliaSherlis.shtml ).

Hair, however, is long enough, and frequently curly enough, to be able to be woven into yarn or thread, and then into thick 2-dimensional products. The thickness of the yarn or thread gives the product more strength.


Does sulfur have various allotropes at room temperature? Recently I had a question in my chemistry test that said:

  Which of the following nonmetals does not have multiple allotropes at room temperature?
  A) Oxygen
  B) Hydrogen
  C) Sulfur
  D) Phosphorus

The answer was apparently hydrogen, but can't sulfur also be a correct answer? I am judging this just because in my chemistry book it says that rhombic crystalline form exists below $\pu{96^\circ C}$. Monoclinic exists above $\pu{96^\circ C}$. Are there any allotropes of sulfur at room temperature?

"Exists" is an awfully tricky word: it can be understood in a thermodynamic sense or in a kinetic (=everyday) sense, and the meanings are sometimes quite different. Ditto for "stable".

Thermodynamically, there exists only one stable allotrope of any element at any given conditions. All other allotropes are not allowed to exist and must eventually transform to that one. This is true, but the problem is that the transformation may take forever.

Now, your book on sulfur was using this meaning, but your test was not. It was meant to be understood in a kinetic sense, where multiple allotropes may coexist all right. In particular, the high-temperature monoclinic sulfur can be cooled down to room temperature and still remain monoclinic. I can attest to that; I used to hold it with my own hands, as did numerous other chemists. That's what people call metastability.

There is also plastic sulfur and other forms, but that's another story. And as you already know, there is oxygen vs ozone, and white phosphorus vs red, so hydrogen is really the only element here that does not have multiple allotropes.

So it goes.