Strong activation of phenoxide ion compared to others

I assume by your notes that your teacher was mentioning electrophilic aromatic substitution reactions. These are two-step reactions.  The product-determining step (whether it is ortho-, meta- or para-) in the substitution mechanism is the first step, which is also the slowest (high activation energy), and hence the rate determining step (see [energy Diagram] below).  The second step is relatively faster, because positively charged intermediate ion loose a proton to get back the resonance stabilized aromatic ring again. Accordingly, the rate of the reaction is roughly depend on the influence of each of the given substituents to stabilize the benzenonium intermediate (positively charged adduct). You can best see it by looking at the capability of each substituent to delocalize the positive charge within the benzenonium intermediate.

Each given substituent ($\ce{-OH, -NH2, -NHR, -NR2}$) except for the phenolate ion has a lone pair. Thus, each can share it with the benzenonium nucleus to reduce the positive charge, reducing the energy of intermediate. However, phenolate ion ($\ce{ph-O^-}$) has full negative charge, which can neutralize benzenonium intermediate completely, to give least energy among all substituents considered here.


Why is H2O a weaker acid than acetylacetone? Shouldn't a hydrogen connected to an oxygen be more acidic than a hydrogen connected to a carbon atom?

Your instructor is right, up to a certain point. There are factors that are more important (the atom to which it is bound) and factors that are less important (resonance, etc.), and it's good to be able to order these relative factors. The devil is in the details, though. If your resonance stabilisation is strong enough (and in this case it is) then it can overcome the more important factor.
Before starting, let me point out that CH's are less acidic than OH's, but this isn't actually to do with electronegativity. If it was about electronegativity, then OH's should be more acidic than SH's as O is more electronegative than S, but the opposite is true. Same applies to HF and HCl: F is more electronegative but HF is actually a weak acid. It has more to do with the X–H single bond strength than anything, as a weaker bond is easier to break (homolytically). The X–H heterolytic cleavage, relevant for an acid–base reaction, is sort of like a homolytic cleavage, plus an electron transfer from H to X. The electronegativity comes into play in the electron transfer step, but it's just not so important a consideration, compared to the bond strength which describes the homolytic cleavage.

I'm giving pKa's to the nearest 5(ish) in this discussion. The exact numbers don't matter too much, we're more interested in broader trends.
If you compare water and methane, their pKa's are ~15 and ~50 respectively, which is a difference of about 35 units. That's roughly how much the adjacent-atom factor counts for, when comparing C and O.
If you then compare acetone and methane, pKa's are ~20 and ~50 respectively, which is a difference of about 30 units. Not quite as big as the previous one, which was 35.
Acetylacetone, though, has two carbonyl groups, that's twice as many as in acetone. Now, resonance stabilisation doesn't scale linearly in that fashion and hence the additional stabilisation over acetone is only ~10 units. However, it's still enough to bring the pKa down to ~10, which is more acidic than water.


Abbreviations in electronic configurations

I would recommend following the convention used by NIST which is to list the subshells in order of principle quantum number $(n),$ and in order of angular momentum quantum number $(\ell)$ for a given $n.$
See also ELECTRONIC STRUCTURE OF THE ELEMENTS (PDF).
The question refers to “rules on how to fill the orbitals”, but I don't think there are any rules, just memorization aids.
See The low-lying level structure of atomic lawrencium $(Z = 103):$ energies and absorption rates for an sample of how difficult it is to calculate the ground state configuration of an atom.  The only rule is quantum electrodynamics.
Also, the configurations are really just designations or leading configurations, with the actually ground state being a mix of configuration.  See for example the above lawrencium reference or A critical compilation of energy levels and spectral lines of neutral boron, where it explained “Configuration and
term labels have little physical meaning for highly mixed levels”. Boron ground state is $95\%$ $\mathrm{(1s)^2(2s)^2(2p)}$ and $4\%$ $\mathrm{(1s)^2(2p)^3}.$  Lawrencium is calculated to be $86\%$ $\mathrm{(7s)^2(7p)}$ and $6\%$ $\mathrm{(6d)(7s)(7p)}.$


Does the temperature during ice melting change?

Here is a naive interpretation, that may help you understand this phenomena.
In the presence of salt, water is attracted and "wants" to be adsorbed around $\ce{Na+}$ and $\ce{Cl-}$ ions. But this operation requires some energy for water molecules to separate the positive and negative ions. This is why the dissolution of $\ce{NaCl}$ in water is endothermic.
Now if you put salt in contact with ice, water in ice "wants" also to be attached to surrounding ions. But it cannot move. It first have to be a liquid. And there is no heat in the surroundings to make ice melt. Nature has found a way. Every piece of ice is divided in two parts. One part gives heat to the second part, that melts, makes water to dissolve the salt. But the remaining part has lost heat. So it gets cooler and cooler. The temperature of the remaining ice decreases from $0$°C to about -$20$°C.


I think the term combined "$\ce{SO3}$" is nothing but $\ce{SO3}$ of $\ce{(SO3 + H2O)}$ and free $\ce{SO3}$ means that is $\ce{SO3}$ in oleum sample.

And value of % leveling of oleum is from (100% to 122.5%).


The reaction for complete combustion of acetylene is
$$\rm 2C_2H_2 + 5O_2 \rightarrow 4 CO_2 + 2H_2O$$

...you'll get complete combustion when acetylene and oxygen are mixed in a 2:5 mole (or volume) ratio. If you have a higher ratio than that, combustion will be incomplete. The incomplete combustion products will appear as soot and carbon monoxide.

Air is 1/5th O$_2$, so you need 5 times as much air as you'd need pure O$_2$. The optimal acetylene/air ratio should be about 2:25.

Look at the acetylene/air ratios for each tube. The gases are both at the same temperature and pressure, so the volume ratio is equal to the mole ratio. Any ratio that's greater than 2/25 will probably produce soot, and the higher the ratio, the more soot you'd expect. Any tube that contains an appreciable amount of oxygen mixed in with the acetylene will burn quickly.

The full tube burned slow because oxygen needed to mix with acetylene at the mouth of the tube. The little bit of acetylene that burned there produced soot, because the acetylene/air ratio was probably much higher than 2/25.

The tubes that were 3/4 and 1/2 full had 3:1 and 1:1 acetylene/air ratios. Both ratios are fuel-rich, so soot was produced (you'd expect more soot with the 3/4 full tube than the 1/2 full tube). Both tubes burned quickly because there was a fair amount of O$_2$ mixed with the acetylene.

The tube that was 1/10 full had a 1:9 acetylene/air ratio. 1/9 is just a bit higher than 2/25, so this tube was fuel-rich, too, but much less so than any of the other tubes. Very little soot would be expected; maybe none, if more air was drawn into the tube as the acetylene burned.


What's the biggest organic molecule that could have a smell?

I'll quote from$\ce{^{[1]}}$:

  The general requirements for an odorant are that it should be
  volatile, hydrophobic and have a molecular weight less than
  approximately 300 daltons. Ohloff (1994) has stated that the largest
  known odorant is a labdane with a molecular weight of 296. The first
  two requirements make physical sense, for the molecule has to reach
  the nose and may need to cross membranes. The size requirement appears
  to be a biological constraint. To be sure, vapor pressure (volatility)
  falls rapidly with molecular size, but that cannot be the reason why
  larger molecules have no smell, since some of the strongest odorants
  (e.g. some steroids) are large molecules. In addition, the cut-off is
  very sharp indeed: for example, substitution of the slightly larger
  silicon atom for a carbon in a benzenoid musk causes it to become
  odorless (Wrobel and Wannagat, 1982d).

  A further indication that the size limit has something to do with the
  chemoreception mechanism comes from the fact that specific anosmias
  become more frequent as molecular size increases. At the “ragged edge”
  of the size limit, subjects become anosmic to large numbers of
  molecules. An informal poll among perfumers, for example has elicited
  the fact that most of them are completely anosmic to one or more musks
  (e.g. Galaxolide® mw 244.38) or, less commonly, ambergris odorants
  such as Ambrox®, or the larger esters of salicylic acid.

  One can probably infer from this that the receptors cannot accommodate
  molecules larger than a certain size, and that this size is
  genetically determined (Whissel-Buechy and Amoore, 1973) and varies
  from individual to individual.

References

$\ce{[1]}$: "Structure-odor relationships: a modern perspective", by Luca Turin (Dept of Physiology, University College London, UK) and Fumiko Yoshii (Graduate School of Science and Technology, Niigata University, Japan), which appears as chapter 13 of: Handbook of Olfaction and Gustation. Richard L. Doty (ed.). 2nd ed., Marcel Dekker, 2003.

$\ce{[2]}$: Ohloff, G. (1994) Scent and fragrances: the fascination of odors and their chemical perspectives [Berlin, Springer].

$\ce{[3]}$: Whissell-Buechy D. and Amoore J.E.(1973) Odour-blindness to musk: simple recessive inheritance Nature Sep 21;245 (5421):157-8.


I would suggest a Frigorific Mixture of calcium chloride (often sold as DampRid) and Ice in a cooler.  You will need quite a bit of each and to moniter at least twice a day, but it wont break the bank and can reach the desired temperature.


Lattice enthalpy is given by the Born-Lande equation:
$${\displaystyle E=-{\frac {N_{A}Mz^{+}z^{-}e^{2}}{4\pi \epsilon _{0}r_{0}}}\left(1-{\frac {1}{n}}\right)}$$
This seems very complicated, but removing some constants, we can reduce it to say:
$$E \propto \frac {z^+z^-}{r_0}$$
here, $z^+$ and $z^-$ are charge on cation and anion and $r_0$ is distance to closest ion.

Coming to your question, In the sulphate series of Alkaline earth metals, the charge remains the same for all members: what changes is the distance to closest ion, which is smaller for small ions such as $\ce{Be^2+}$, thereby causing their lattice energies to be high.

The decrease in lattice enthalpy may not neccessarily mean a decrease in ionic character. If two lattices have similar values of $r_0$, then a higher lattice enthalpy for one of them may imply larger ionic character due to large charges on either of the atoms.


Formic Acid vs Formica - Which came first?

The name 'formic acid' has been derived from the name 'formica' as it is derived literally from ants. Wikipedia clearly states:

  The word "formic" comes from the Latin word for ant, formica,
  referring to its early isolation by the distillation of ant bodies.

If you type 'formic acid etymology' in Google, it will give two links that proves the above statement:

  1791 (in formic acid), literally "from ants," coined from Latin
  formica "ant". The acid first was obtained in a fairly pure form in 1749 by German chemist Andreas Sigismund Marggraf (1709-1782), who
  prepared it by distilling red ants.(etymoline)

  As early as the fifteenth century, some alchemists and naturalists
  were aware that ant hills gave off an acidic vapor. In 1671, English
  naturalist John Ray became the first person to describe the isolation
  of formic acid by the distillation of large numbers of ants. These
  insects secrete the substance for attack and defense purposes. Thus
  the name "formic acid" was coined from the Latin word for ant,
  formica.(newworldencyclopedia)


What's the least common element in the universe?

(edited based on @JonCuster's comment) The rarest known elements are most probably the man-made heavy elements. These isotopes (elements) are very unstable. They most probably also formed in supernovas, but their natural formation is minimal, and due to their short lifetime, they do not accumulate either.
As Jon added, most probably heavier elements are also formed in supernovas, however, I am not aware of any of them observed. The reason for that is 1) again, short lifetime prevents any large accumulation, 2) under the formation conditions they are not atoms, but bare nuclei /plasma, so one cannot observe their atomic spectrum.


Why is it that the bigger the side-chain is, its hydrogen atoms tend to be more equatorial?

It's not exactly clear from your question what distances you consider, but I have taken snapshots of a methylcyclohexane with the methyl group in the equatorial and axial positions:

      (equatorial)

    (axial)

The distance of the methyl $\ce{H}$ atoms to the nearest cyclohexane $\ce{H}$ atoms are displayed, and you'll see that indeed the axial position has a shorted $\ce{H–H}$ distance of ~ 2 Å, while equatorial position has longer $\ce{H–H}$ distances.

Also, as others have remarked in the comments: only the groups (and $\ce{H}$ atoms) directly on the cyclohexane ring are said to be in axial or equatorial position. Atoms of side chains do not follow this terminology, in particular for side chains that can rotate freely as it means nothing there.


Can hydrofluoric acid etch glass at a pH of 7.0?

In $\ce{SiO2}$ etching, there are two active fluorine-containing species, $\ce{HF}$ and $\ce{HF2-}$.

The rate of etching at 25 degrees C in Angstroms per second is:

$$9.66[\ce{HF2-}] + 2.50 [\ce{HF}] -0.14$$

according to A Study of the Dissolution of SiO2 in Acidic Fluoride Solutions J. Electrochem. Soc. vol. 118, pp. 1772-1775.

The relavent equilibrium constants at 25 degrees C and ionic strength of 1M are

$$\frac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]} = 0.0013M $$

$$\frac{[\ce{HF}][\ce{F-}]}{[\ce{HF2-}]} = 0.104M $$

Because of the -0.14 term in the rate equation, only at enormously high F- concentration would there be etching at pH 7 (such as in this situation).

This study was concerning high (~1M) concentrations of fluorine species, so it might not be completely relevant for low concentrations or very long time periods.


Is there anything akin to molecular chirality for crystalline?

There are some crystal structures where the nature of the crystal structure creates chirality and therefore optical activity.

A trivial example (important because of its history: it was involved in the discovery of optical activity and molecular chirality) is tartaric acid. But the crystal is chiral because the molecule is chiral so that might not be what yyou were looking for.

But there are examples of crystals where the structure as a whole is chiral but the building blocks are not. Quartz, for example, is formed from a network of linked SiO4 tetrahedra (not molecules but a giant interlinked network where each oxygen is shared between two silicons). The overall crystal structure, though, is chiral due to small shifts in the rotation of the linked tetrahedra (see wikipedia for a detailed explanation).

I think this is a crystal analogy to molecular chirality.

Only a few crystal structures can do this and the ones you use as examples can't.


Role of phosphoric acid in redox titrations

I don't know why your professor is unnecessarily creating an example which has no real value! No one would do potentiometric titrations of iron in the presence of phosphate.

(i) Potentiometric titrations have nothing to do with color or need masking.
(ii) One would need to do a preliminary separation of iron via ion exchange resin.


Where does the energy of formation of covalent bonds come from?

Your view of bonding is a little simplified and confused. For the details it is worth reading the great answer referred to in some comments already: Fundamental forces behind covalent bonding .

I'll stick with a simpler view that focusses just on the energy involved and not the mechanism of chemical bonding. Firstly, you are wrong that boding two hydrogen atoms needs an input of energy. It actually releases energy. This is a thermodynamic view of the reaction. This tells us whether there is more or less energy in the product of the reaction being described than in the chemicals being combined to create the product.

But a lot of chemical reactions are thermodynamically favourable (that is the reaction releases energy) don't happen easily (this is probably not true for atomic hydrogen which should react very easily to form H2). But there are plenty of other reactions that release energy according to thermodynamics but that don't happen easily. For example, elemental carbon plus oxygen will release energy if converted to CO2. But neither the diamond in jewellery or the coal in your fireplace (which is basically impure carbon) reacts easily to give carbon dioxide despite being in an environment with a lot of Oxygen gas in it. Here the problem is that room temperature oxygen molecules and carbon just don't have enough energy to kick off the process that creates the product. This is where we have to look at the kinetic view of the reaction rather than just the thermodynamics. Although the overall reaction is favourable according to thermodynamics, we need to add energy (firelighters for coal fires, a blowtorch or diamond rings) to the mixture to give the atoms in the reactants enough energy to overcome the kinetic barriers to the reaction. Think of having to climb a tall mountain to get to a lower valley on the other side. Or think of giving oxygen molecules enough energy to break carbon atoms apart from their normal solid structure.

Once coal or diamond are warm enough, the reaction will happen spontaneously and you have either destroyed your engagement ring or you have a nice coal fire. In this case you can feel the energy released from the reaction as a pleasant warming of your room.

Back to the question. The energy involved in a reaction is an accounting of the amount of energy used in the bonding of the products versus the reactants. The energy released in reactions such as burning carbon goes into the environment as heat. Other reactions happen spontaneously (try dropping a small lump of sodium into water). Some reactions require a tiny amount of encouragement (oxygen and hydrogen in certain concentrations will react explosively to form water when encouraged with the tiniest spark).

We can drive some reactions the other way using inputs of energy. For example we can electrolyse water into a mixture of hydrogen and oxygen. This requires us to add energy to the system (about 286kJ/mol) because the products are in a higher energy state then the reactants.

If you just focus on the thermodynamics of a reaction the numbers (kcal or kJ/mol) tell you how much the energy difference between the products and reactants is. But many reactions that should release energy require some input to kick off the reaction. This is confusing, but chemistry wouldn't be interesting and we wouldn't exist if thermodynamics were the only thing driving reactions.

Also energy is not created or destroyed here, just moved around. You can put energy in to a system to drive a reaction or the environment gets energy back from reactions that release it.

So, some reactions require energy to drive them because the products have higher energy, some release energy because the products have lower energy but many release energy but, confusingly,  require some energy input to kick off the actual reaction process.


definitely $\ce{O-}$

In $\ce{O-}$ 7 outer electrons are tied to nuclei with 6 positive charges (number 8 minus 2 electons of inner shell), while in $\ce{F-}$ 8 outer electrons are tied to nuclei with 7 charges. The difference, however, should be quite small.


Quantum mechanical explanation for Le Chatelier's principle?

The main thing that classical and quantum statistical mechanics have in common is probability.  In classical stat mech the equilibrium value of an observable is given by
$$\langle A \rangle = \sum_i^N p_iA_i$$
where $A_i$ is the value of $A$ in microstate $i$ and $p_i$ is the probability of microstate $i$.  The expression for $p_i$ at equilibrium depends on the ensemble (microcanonical, canonical, etc).  One way to define equilibrium is that the probabilities are constant
$$\frac{\partial p_i(t)}{\partial t}=0$$
So if one perturbs the system so that this equation doesn't hold the $p_i$'s will change until it does and Le Chatelier's principle is a heuristic way of predicting whether the individual $p_i$'s increase or decrease.

In quantum stat mech the equilibrium value of an observable is given by
$$ \langle A \rangle = \tfrac{1}{N} \sum_\alpha^N \langle \Psi^{\alpha}| \hat{A} |\Psi^{\alpha}\rangle$$
where $\Psi^{\alpha}$ is the wavefunction for microstate $\alpha$

The wavefunctions are typically expanded in an orthonormal basis ($\phi$'s)
$$ \langle A \rangle = \tfrac{1}{N} \sum_{k,l} \rho_{l,k}\langle \phi_l| \hat{A} |\phi_k\rangle $$
where the density matrix is can be defined by a density operator
$$ \rho_{l,k} = \langle \phi_k| \hat{\rho} |\phi_l\rangle $$

The equilibrium is then defined as
$$\frac{\partial \hat{\rho}(t)}{\partial t}=0$$

So if one perturbs the system so that this equation doesn't hold the $\hat{\rho}$ will change until it does and Le Chatelier's principle is a heuristic way of predicting whether the individual $\Psi^{\alpha}$ contributions to  $\hat{\rho}$ increase of decrease.

The time dependence of $\hat{\rho}$ can be calculated by the time-dependent Schrödinger equation.  So Le Chatelier's principle is a heuristic way of predicting roughly how the time-dependent wavefunctions will evolve at the simulation approaches equilibrium.

To get a more physical picture one can introduce the Born-Oppenheimer approximation in which case the numerical simulation of the time-dependent Schrödinger equation becomes a classical MD simulation (with a QM energy function) from which the time dependence of the probabilities $p_i$ can be computed and Le Chatelier's principle is a heuristic way of predicting whether the individual $p_i$'s increase or decrease.


Has the molecular diatomic oxygen trianion been observed or predicted?

According to "Small Multiply Charged Anions as Building Blocks in
Chemistry" Acc. Chem. Res. 1996, 29, 497-502

  it is concluded that all isolated diatomic dianions or
  other multiply charged diatomic anions are not stable
  to loss of an extra electron or fragmentation in their
  ground electronic states

So even $\ce{O2^{2-}}$, let alone $\ce{O2^{3-}}$ cannot exist in isolation.


The usual commercially available burettes are standardizes in the international standard ISO 385:2005 Laboratory glassware – Burettes. This standard includes small burettes with capacities of 1, 2, 5, 10 and 25 ml. The very small versions are typically advertised as “micro burettes”.

$$
\begin{array}{lllll}
\hline
\text{Nominal capacity} & \text{Graduation divisions} & \text{Accuracy limits} & & \text{Max. length}\\
& & \text{Class A and AS} & \text{Class B} \\
\mathrm{ml} & \mathrm{ml} & \pm\ \mathrm{ml} & \pm\ \mathrm{ml} & \mathrm{mm}\\
\hline
1 & 0.01 & 0.006 & 0.01 & 575 \\
2 & 0.01 & 0.01 & 0.02 & 650 \\
5 & 0.01 & 0.01 & 0.02 & 820 \\
5 & 0.02 & 0.01 & 0.02 & 800 \\
10 & 0.02 & 0.02 & 0.05 & 820 \\
10 & 0.05 & 0.03 & 0.05 & 620 \\
25 & 0.05 & 0.03 & 0.05 & 820 \\
25 & 0.10 & 0.05 & 0.10 & 620 \\
50 & 0.10 & 0.05 & 0.10 & 820 \\
100 & 0.20 & 0.10 & 0.20 & 870 \\
\hline
\end{array}$$


If we have 2 variables
with absolute uncertainties(+)
$A \pm \sigma_A$ and $B \pm \sigma_B$

(+)  expressed as standard deviations, but applicable to confidence intervals as well.

Then $C = A \pm B$ has uncertainty

$$\sigma_C = \sqrt{{\sigma_A}^2 + {\sigma_B}^2}$$

For $C=A/B$ or $C=A\cdot B$,

$$\sigma_C = C \cdot \sqrt{\left(\frac {\sigma_A}{A}\right)^2+ \left(\frac {\sigma_B}{B}\right)^2 }$$

Let $A$ is content of cassein in 100 ml of milk.
Let $B$ is the actual volume of used milk.
Let $C=A\cdot B$ is the mass of  cassein in the volume of milk used in analysis.

Then
$\sigma_C$
is the estimation of the error of the amount of cassein (known as the result error)
$\sigma_A$
is the estimation of the  error of the cassein mass in 100 ml of  milk (unknown)
$\sigma_B$
is the estimation of the  error of the milk volume  (1 ml )

The error introduced by scales can be neglected, or introduced as additive variable $D=0$ with absolute error estimation $\sigma_D=0.001~ \text{g}$.

Another, the biggest  factor is the used analytical method with its bias $E$ and error $\sigma_E$.

$$\begin{align}
C &= D + E + A\cdot B \\
\sigma_{AB} &= A\cdot B \cdot \sqrt{\left(\frac {\sigma_A}{A}\right)^2+ \left(\frac {\sigma_B}{B}\right)^2 }\\
\sigma_C &= \sqrt{{\sigma_D}^2 + {\sigma_E}^2 + {\sigma_{AB}}^2}\\
\sigma_{AB} &= 2.4 \cdot \sqrt{\left(\frac {\sigma_A}{2.4}\right)^2+ \left(\frac {1}{100}\right)^2 }\\
0.13 &= \sqrt{{0.001}^2 +{\sigma_E}^2 + {\sigma_{AB}}^2}\\
\end{align}$$


Reduction of amides with lithium aluminium hydride

Not according to the commonly accepted mechanism here

The first step is

  The nucleophilic H from the hydride reagent adds to the electrophilic C in the polar carbonyl group of the ester.

Note: the pKa of primary amides is in the 22-25 range so not close to alcohols.


I think Ivan Neretin has given the right answer.

Any alloy between Sn and Pb melts at relatively low temperatures. You stated that the compound you fabricated did not even melt at 2300 °F. I have been looking at the stable lead oxides and tin oxides. The only compound of this set that has a melting temperature above 2300 °F is tin dioxide at 2970 °F.

Please note that I cannot be certain of this, as I only checked a small set of compounds I found to be the best candidates. It could be the case that you have created something more complex. (beautifully phrased by DrMoishe Pippik as a 'toxic miasma' :-) )


What is the name of the formula ΔE = q + w?

This is a statement of the first law of thermodynamics. The first law says that the energy of an isolated system is conserved. An isolated system is one in which neither mass nor energy is exchanged with the surroundings. This equation shows the two means by which a closed system (meaning a system in which mass is conserved but energy is not) can exchange energy with its surroundings - by heat, or by work.

$$
\Delta E=q+w
$$

All energy transfers can be described as either heat flow or as work done on or by a system, and therefore the total amount of heat energy that goes in or out ($q$), plus the total amount of work energy that goes in or out ($w$), must be equal to the total internal energy change of the system ($\Delta E$).

This equation is particularly important in chemical thermodynamics because we have no way of measuring the total internal energy of a real system directly - all that we can measure is the amount of heat and work that goes in or out of the system, and therefore the change in internal energy.


a) You are wrong. $l=2$ simply means $\ce{d}$-orbitals, thus, only $\ce{3d}$ is the right answer.

b) Assuming that the meaning of the question is to specify subshells that can contain two electrons in the same spin state, i.e. both with $m_s = +1/2$ or both with $m_s = -1/2$, $\ce{s}$-orbitals should indeed be excluded.


How can I make my yeast experiment more chemistry-oriented?

It sounds like a perfect chemistry experiment to me. It sounds good enough to even be one of my AP chemistry labs. As an objective for the lab, you could try to determine the amount of CO$_2$ gas produced per amount of sugar you use, (with all other ingredients in excess). I'm pretty sure the equation for fermentation is C$_6$H$_{12}$O$_6$ → 2 C$_2$H$_5$OH + 2 CO$_2$ (although note that common granulated sugar has the equation C$_{12}$H$_{22}$O$_{11}$ and decays into two glucose molecules when combined with water).Why not test a certain amount of sugar (like .1 moles) and see if you yield the theoretical .2 moles of CO2. You could use the ideal gas laws (PV=nRT) to determine the moles of CO2 product you yielded. Compare what you yielded to the theoretical amount you should've yielded. If you yield less, then you could include in your report/conclusion including why you think your yield was short.

The lab would involve gas laws, chemical equations, and percent yield, all of which are chemistry topics/principles. What is your current level of chemistry experience? If you haven't learned of anything I mentioned, I can help explain it better and point you in the right direction if you'd like.


What is the percentage of d-tartaric acid in a mixture of d and l isomers?

Your calculation is correct.

You can check your result since you know that pure (+)-tartaric acid has a specific rotation of $+12^\circ$ and that pure (−)-tartaric acid has a specific rotation of $-12^\circ$. Therefore, a mixture of $75\ \%$ (+)-tartaric acid and $25\ \%$ (−)-tartaric acid has a specific rotation of
$$0.75\times(+12^\circ)+0.25\times(-12^\circ)=6^\circ$$

A similar exercise can be found in Hart, Hadad, Craine, Hart Organic Chemistry – A Short Course: Google Books

The corresponding solution can be found here: Google Books


What is the most stable structure amongst the keto-enol tautomers of pentane-2,4-dione?

The compound in option (c) is highly stable because it forms a 6 membered ring by tautomerism and that ring is highly stable because all bonds in that ring have developed partial double character. Whereas in option (d) there is ring formation but only two bonds of that ring have partial double character. This makes (c) more stable than (d).


Finding initial pressure in a chemical equilibrium

If the initial pressure of $\ce{CO}$ and $\ce{H_2O}$ are equal, then let $p_{\ce{H_2O}}=p_{\ce{CO}}=x$.
Since their total is 1.5 bar, i.e. $\ce{p_{H_2O} + p_{CO}}=1.5$
Can you solve this equation?

Edit: This is under the assumption that there was originally no carbon dioxide or hydrogen, I have not yet managed to generalise it to any case.
It is given that $K=4$. From the equation for the equilibrium constant, $K=\frac{\ce{[CO2][H2]}}{\ce{[CO][H2O]}}$.
It is also given that initially, $\ce{p_{H2O} = p_{\ce{CO}}}$; from the ideal gas law,
$PV=nRT \implies P=\frac{nRT}{V}=kC  \tag{1}$
where $k$ is some constant and C is concentration.
So initially, $[\ce{H2O}]=[\ce{CO}]=y$, where $y$ is a constant. Let the concentration consumed by the equilibrium shifting to the right be x.
At equilibrium, $\frac{\ce{[CO2][H_2]}}{[\ce{CO}][\ce{H2O}]}=\frac{x^2}{(y-x)^2}=4$.
Solving this equation gives $x=\frac{2y}{3}$.
Now then, if you look back at (1), pressure is proportional to concentration. Try to use that relationship to solve the problem.

 Now, from $(1)$, we see that pressure is directly proportional to concentration. We have worked out from our algebra that: $[\ce{CO2}]=[\ce{H2}]=\frac{2y}{3}$, therefore $[\ce{CO}]=[\ce{H2O}]=\frac{y}{3}$.

So then $p_{\ce{H2}}= p_{\ce{CO2}} = 2p_{\ce{CO}} = 2p_{\ce{H2O}}$ at equilibrium.
So from substitution into Martin's equation, $\ce{p_{H_2O} + p_{CO} + p_{\ce{CO2}} + p_{\ce{H2}}}=1.5$, we find that $p_{\ce{H2O}} = p_{\ce{CO}} = \pu{0.25 bar}$.
But they are now only a third of their original concentration, and so their pressure has also decreased by a factor of 3. So their original pressure is $\pu{0.75 bar}$ each.


Fraction of concentration of benzene in air water

You don't need the original concentration or any of the variables you are missing. Henry's Law gives you everything you need. The usual formulation of Henry's Law at equilibrium:

$$c_{i,s}=k_{H,i,s}c_{i,s}$$

Where $c_{i,air}$ is the concentration of solute $i$ in the vapor above the solution, $k_{H,i,s}$ is the Henry's Law constant for solute $i$ in solvent $s$, and $c_{i,sol}$ is the concentration of $i$ remaining in solution.

We can rearrange Henry's Law to give you what you need - a ratio of the amount of benzene in the air to the amount in the water:

$$k_{H,i,s}=\frac{c_{i,air}}{c_{i,sol}}$$

The other snippet of information you need is the relationship between moles and concentration. There is a reason you were given volumes.

$$n_{air}=c_{air}*V_{air}$$
$$n_{sol}=c_{sol}*V_{sol}$$

You only need the ratio of $\dfrac{n_{air}}{n_{sol}}$, so you really only have one unknown. It's algebra time!

$$K_{H,i,s}=\frac{c_{i,air}}{c_{i,sol}}=\dfrac{\dfrac{n_{air}}{V_{air}}}{\dfrac{n_{sol}}{V_{sol}}}$$

$$\dfrac{K_{H,i,s}V_{air}}{V_{sol}}=\dfrac{n_{air}}{n_{sol}}$$


Depending on who you ask you will receive a different concept to define macromolecules. These concepts typically arise from the sub-field of those who made the definition, so a polymer chemist is more inclined to consider repeating monomers of exactly one type since that is what plastics are. At the same time, the branch of macromolecular chemistry usually deals with systems that are much smaller than your average multi-domain protein.

The definition most can agree on (to a certain extent) is that macromolecules must be ‘larger’ than ‘small molecules’ (note the pairs of inverted commas). In this definition, it is perfectly reasonable for lipids to be grouped as macromolecules as they are much larger than most secondary metabolites. However, it is true that they are not made up of subunits linearly connected (whether repeating or not) so one may not consider them biopolymers.


How to remove salt from powdered soup mix?

Peel and thickly slice several potatoes and cook these in your soup. Remove them when they have softened. They will absorb the salt while cooking.

You can use this method if you accidently add too much salt to stews or sauces (normally this becomes apparent when reducing.)

The removed potatoes can be discarded or used in another meal, maybe mashed.


Nature of bond in oxyacids of phosphorus

In general, I'm not sure double bond and coordinate covalent bond are mutually exclusive labels and depending on the specific chemical you're looking at, they might both apply.

My Reasoning:

Coordinate bonds are bonds where both electrons were donated by one atom. Usually, the concept of a double bond is used in the context of valence bond theory and coordinate bond is used in the context of crystal field theory. There is evidence for pi bonding in coordinate complexes (i.e. back bonding) leading to what could be called a double bond in valence bond theory.
If you look at bond length data you can usually determine if there is a double bond. (Or if you have some of the acid and access to an IR, you can do this yourself.) The CRC should contain reference bond lengths.

I 'm not an expert in phosphorous chemistry, but these could be helpful:

Structural calculations for many compounds, including a few acids
Study on structures of phosphates, not acids


Flow rate of the fuel consisting of ethanol and water burning in air of a given flow rate

I think you overcomplicate the problem. There are many variables you don't need and many components that are duds (nitrogen, water, pressure). What important is that we burn ethanol:

$$\ce{C2H5OH + 3 O2 -> 2 CO2 + 3 H2O}$$

so that the ratio between ethanol and oxygen is $1:3$. I have taken chemistry engineering classes long time ago and I don't remember exact notations, so I'd rather denote molar flow rate with an n-dot notation:

$$\dot{n} = \frac{\mathrm{d}n}{\mathrm{d}t}$$

where $n$ – amount; $t$ – time. Since fuel is only 90% alcohol, then

$$\dot{n}(\text{fuel}) = \frac{\dot{n}(\ce{C2H5OH})}{x(\ce{C2H5OH})}$$

where $x$ denotes molar fraction. From the combustion reaction

$$\dot{n}(\ce{C2H5OH}) = \frac{\dot{n}(\ce{O2})}{3}$$

and for the air the following is fulfilled:

$$\dot{n}(\ce{O2}) = x(\ce{O2})\frac{\dot{n}(\text{air})}{1 + α}$$

where $α$ is the partial excess of air. Now we can put everything together and plug in the numbers:

$$
\begin{align}
\dot{n}(\text{fuel}) &= \frac{x(\ce{O2})\dot{n}(\text{air})}{3x(\ce{C2H5OH})(1 + α)}\\
  &= \frac{0.21\cdot\pu{100 mol s-1}}{3\cdot 0.9\cdot (1 + 0.2)}\\
  &= \pu{6.48 mol s-1}
\end{align}
$$

One hell of a burner.


The mathematical condition for a complete complexation reaction

Ok, to start let's just check the given answer of 24.3 mL.

The total solution is 34.3 ml (10 mL Ag + 24.3 ml NH3).

We'll assume that all 0.001 moles of silver are as species $\ce{[Ag(NH3)2^+]}$ so:

$\ce{[Ag(NH3)2^+]} \approx (\dfrac{10}{34.3})0.100 = 0.0291 $

$\ce{[NH3]_{initial} =}(\dfrac{24.3}{34.3})0.100 = 0.0708$

We'll assume $\ce{NH4^+}$ and $\ce{[Ag(NH3)^+]}$ are both negligible so:

$\ce{[NH3] =} 0.0708 - 2(0.0291) = 0.0126$

Now from the equation

$\dfrac{\beta_2}{\beta_1}= 10^{3.9} = 7.94 \times 10^3 = \dfrac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}}$

$\ce{[Ag(NH3)^+] =} \dfrac{[\ce{Ag(NH3)2^+}]}{(7.94 \times 10^3)\ce{[NH3]}} = \dfrac{0.0291}{(7.94 \times 10^3)(0.0126)} = 2.91 \times 10^{-4}$

$\ce{[Ag^+] =} \dfrac{[\ce{Ag(NH3)^+}]}{(2.00E3)\ce{[NH3]}} = \dfrac{2.91 \times 10^{-4}}{(2.00E3)(0.0126)} = 1.15 \times 10^{-5}$

fraction of total Ag as $\ce{Ag(NH3)^+}$

$\dfrac{2.91 \times 10^{-4}}{0.0291} = 0.01 $

fraction of total Ag as $\ce{Ag^+}$

$\dfrac{1.15 \times 10^{-5}}{0.0291} = 3.97  \times 10^{-4}  $

Limits on ratios

Given the two equations:

$\mathrm{K}_1= 10^{3.3} = \frac{\ce{[Ag(NH3)+]}}{\ce{[Ag+][NH3]}}$

$\mathrm{K}_2 = 10^{7.2} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag+][NH3]^2}}$

then we can derive a third

$\mathrm{K}_3 = 10^{7.2 - 3.3} = 10^{3.9} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}}$

It is possible to solve for some lower limits on the ratios assuming that the $\ce{[NH3]}$ can at most be 0.1 M. So:

$\frac{\ce{[Ag+]}}{\ce{[Ag(NH3)+]}} \geqslant \frac{1}{\mathrm{K}_1\ce{[NH3]}} = 5.01 \times 10^{-3}$

$\frac{\ce{[Ag+]}}{\ce{[Ag(NH3)2^+]}} \geqslant \frac{1}{\mathrm{K}_2\ce{[NH3]^2}} = 6.31 \times 10^{-6}$

$\frac{\ce{[Ag(NH3)+]}}{\ce{[Ag(NH3)2^+]}} \geqslant \frac{1}{\mathrm{K}_3\ce{[NH3]}} = 1.26 \times 10^{-3}$

So the absolute upper limit on the % of the silver as $\ce{[Ag(NH3)2^+]}$ is 99.874%.

Quick and dirty calculations...

To check this I brute forced some calcs in a spreadsheet.

The calculations below ignores the reactions below (alkali solutions do not produce appreciable amounts of silver hydroxide due to the favorable energetics for the reaction to the oxide):
$$ \ce{NH3 + H2O <-> NH4^+ + OH^{-}}$$
$$\ce{Ag^+ + OH^{-} <-> AgOH(s)}$$
$$\ce{AgOH(s) -> Ag2O(s) + H2O} \text{ (pK = 2.875)}$$

 NH3    [Ag(NH3)2+] [NH3]tot    [NH3]       [Ag(NH3)+]  [Ag+]       mol(Ag)     F*
 20.1   3.32E-02    6.68E-02    3.32E-04    1.26E-02    1.90E-02    1.950E-03   4.87E-01
 20.2   3.31E-02    6.69E-02    6.62E-04    6.30E-03    4.75E-03    1.334E-03   2.50E-01
 20.3   3.30E-02    6.70E-02    9.90E-04    4.20E-03    2.12E-03    1.191E-03   1.61E-01
 20.4   3.29E-02    6.71E-02    1.32E-03    3.15E-03    1.20E-03    1.132E-03   1.17E-01
 20.5   3.28E-02    6.72E-02    1.64E-03    2.52E-03    7.68E-04    1.100E-03   9.11E-02
 20.6   3.27E-02    6.73E-02    1.96E-03    2.10E-03    5.35E-04    1.081E-03   7.46E-02
 20.7   3.26E-02    6.74E-02    2.28E-03    1.80E-03    3.95E-04    1.067E-03   6.31E-02
 20.8   3.25E-02    6.75E-02    2.60E-03    1.57E-03    3.03E-04    1.058E-03   5.47E-02
 20.9   3.24E-02    6.76E-02    2.91E-03    1.40E-03    2.40E-04    1.051E-03   4.82E-02
 21.0   3.23E-02    6.77E-02    3.23E-03    1.26E-03    1.95E-04    1.045E-03   4.31E-02
 21.1   3.22E-02    6.78E-02    3.54E-03    1.14E-03    1.62E-04    1.041E-03   3.91E-02
 21.2   3.21E-02    6.79E-02    3.85E-03    1.05E-03    1.36E-04    1.037E-03   3.57E-02
 21.3   3.19E-02    6.81E-02    4.15E-03    9.69E-04    1.17E-04    1.034E-03   3.29E-02
 21.4   3.18E-02    6.82E-02    4.46E-03    9.00E-04    1.01E-04    1.031E-03   3.05E-02
 21.5   3.17E-02    6.83E-02    4.76E-03    8.40E-04    8.82E-05    1.029E-03   2.84E-02
 21.6   3.16E-02    6.84E-02    5.06E-03    7.87E-04    7.77E-05    1.027E-03   2.66E-02
 21.7   3.15E-02    6.85E-02    5.36E-03    7.41E-04    6.91E-05    1.026E-03   2.50E-02
 21.8   3.14E-02    6.86E-02    5.66E-03    7.00E-04    6.18E-05    1.024E-03   2.36E-02
 21.9   3.13E-02    6.87E-02    5.96E-03    6.63E-04    5.56E-05    1.023E-03   2.24E-02
 22.0   3.13E-02    6.88E-02    6.25E-03    6.30E-04    5.04E-05    1.022E-03   2.13E-02
 22.1   3.12E-02    6.88E-02    6.54E-03    6.00E-04    4.58E-05    1.021E-03   2.03E-02
 22.2   3.11E-02    6.89E-02    6.83E-03    5.72E-04    4.19E-05    1.020E-03   1.94E-02
 22.3   3.10E-02    6.90E-02    7.12E-03    5.48E-04    3.85E-05    1.019E-03   1.86E-02
 22.4   3.09E-02    6.91E-02    7.41E-03    5.25E-04    3.54E-05    1.018E-03   1.78E-02
 22.5   3.08E-02    6.92E-02    7.69E-03    5.04E-04    3.27E-05    1.017E-03   1.71E-02
 22.6   3.07E-02    6.93E-02    7.98E-03    4.84E-04    3.04E-05    1.017E-03   1.65E-02
 22.7   3.06E-02    6.94E-02    8.26E-03    4.66E-04    2.82E-05    1.016E-03   1.59E-02
 22.8   3.05E-02    6.95E-02    8.54E-03    4.50E-04    2.63E-05    1.016E-03   1.54E-02
 22.9   3.04E-02    6.96E-02    8.81E-03    4.34E-04    2.46E-05    1.015E-03   1.49E-02
 23.0   3.03E-02    6.97E-02    9.09E-03    4.20E-04    2.31E-05    1.015E-03   1.44E-02
 23.1   3.02E-02    6.98E-02    9.37E-03    4.06E-04    2.17E-05    1.014E-03   1.40E-02
 23.2   3.01E-02    6.99E-02    9.64E-03    3.94E-04    2.04E-05    1.014E-03   1.36E-02
 23.3   3.00E-02    7.00E-02    9.91E-03    3.82E-04    1.93E-05    1.013E-03   1.32E-02
 23.4   2.99E-02    7.01E-02    1.02E-02    3.70E-04    1.82E-05    1.013E-03   1.28E-02
 23.5   2.99E-02    7.01E-02    1.04E-02    3.60E-04    1.72E-05    1.013E-03   1.25E-02
 23.6   2.98E-02    7.02E-02    1.07E-02    3.50E-04    1.63E-05    1.012E-03   1.22E-02
 23.7   2.97E-02    7.03E-02    1.10E-02    3.40E-04    1.55E-05    1.012E-03   1.19E-02
 23.8   2.96E-02    7.04E-02    1.12E-02    3.31E-04    1.47E-05    1.012E-03   1.16E-02
 23.9   2.95E-02    7.05E-02    1.15E-02    3.23E-04    1.40E-05    1.011E-03   1.13E-02
 24.0   2.94E-02    7.06E-02    1.18E-02    3.15E-04    1.34E-05    1.011E-03   1.10E-02
 24.1   2.93E-02    7.07E-02    1.20E-02    3.07E-04    1.28E-05    1.011E-03   1.08E-02
 24.2   2.92E-02    7.08E-02    1.23E-02    3.00E-04    1.22E-05    1.011E-03   1.06E-02
 24.3   2.92E-02    7.08E-02    1.25E-02    2.93E-04    1.17E-05    1.010E-03   1.03E-02
 24.4   2.91E-02    7.09E-02    1.28E-02    2.86E-04    1.12E-05    1.010E-03   1.01E-02
 24.5   2.90E-02    7.10E-02    1.30E-02    2.80E-04    1.07E-05    1.010E-03   9.93E-03
 24.6   2.89E-02    7.11E-02    1.33E-02    2.74E-04    1.03E-05    1.010E-03   9.73E-03
 24.7   2.88E-02    7.12E-02    1.35E-02    2.68E-04    9.89E-06    1.010E-03   9.55E-03
 24.8   2.87E-02    7.13E-02    1.38E-02    2.62E-04    9.51E-06    1.009E-03   9.37E-03
 24.9   2.87E-02    7.13E-02    1.40E-02    2.57E-04    9.15E-06    1.009E-03   9.20E-03
 25.0   2.86E-02    7.14E-02    1.43E-02    2.52E-04    8.82E-06    1.009E-03   9.04E-03

(1) In the table the column [Ag(NH3)2+] assumes that the 1 millimole of silver is all $\ce{Ag(NH3)_2^+}$. The value is expressed as a concentration so it falls due to the dilution of the extra $\ce{NH3}$. In the column mol(Ag) all the silver species are totaled. The value should be 1.00E-03 moles or 1 mM Ag. Values significantly above that indicate that the calculations are in error. (Note that the way the calculations are done the mM Ag will always be too high.)

(2) The column [NH3]tot is the concentration of all NH3 (mMol NH3)/(total volume)

(3) The column [NH3] is the "free" ammonia. It assumes ammonia is either free $\ce{NH3}$ or $\ce{Ag(NH3)_2^+}$ and thus it ignores the single amine complex $\ce{Ag(NH3)^+}$. It calculates free ammonia as [NH3]tot - 2*$[\ce{Ag(NH3)_2^+}]$

(4) The column labeled F* is the fraction of silver as $\ce{Ag(NH3)^+}$ and $\ce{Ag^+}$. Thus the real decision is what fraction is low enough since the fraction continues to fall as excess ammonia is added.

Conclusion

The choice of the "end point" is arbitrary...

mL        end point
----      -----------------------------------
21.4      97% [Ag(NH3)2^+]
21.6      [Ag^+] 10% of [Ag(NH3)^+]
22.1      98% [Ag(NH3)2^+]
24.3      [Ag(NH3)^+] 1% of [Ag(NH3)2^+]
24.4      99% [Ag(NH3)2^+]
24.6      [Ag^+] = 1*10^-5


Health Risks Associated with Working in Chemistry Labs

I would recommend looking at the MSDS information for anything that you work with specifically and/or researching the long term health effects of those specific substances


Why is the mass of 1 mole of an atom/molecule equal to its gram atomic/molecular mass?(proof)

$\def\p{\mathrm{p^+}}\def\n{\mathrm{n}}$
Mass of one proton = $m_\p$ and mass of one neutron = $m_\n$. (Masses are in unified atomic mass unit-$\mathrm{u}$)

C-12 isotope consists of 6 $\p$ and 6 $\n$. Therefore, mass of C-12 isotope is equal to 6 $m_\p$ + 6 $m_\n$. Since $m_\p$≈ $m_\n$, the mass of C-12 can be replaced by 12 $m_\p$. (electrons are not taken into account since they have negligible mass)

1 $\mathrm{u}$ is defined as 1/12 of the atomic mass of C-12 isotope.

⟹1 $\mathrm{u}$ = 1/12 × (6 $m_\p$ + 6 $m_\n$)

⟹ 1 $\mathrm{u}$ ≈  1/12 ×12 $m_\p$

⟹ 1 $\mathrm{u}$ ≈  1 $m_\p$   ...equation(1)

Now,

12 g of C-12 has $\pu{N_A}$ particles and constitutes 1 mole of it. (according to the definition of 1 mole)

⟹ 1 mole of (6 $\p$ + 6 $\n$) have a mass of 12 g.

⟹ 1 mole of (12 $\p$) have a mass of 12 g.

⟹ 1 mole of $\p$ have a mass of 1 g

⟹ 1 mole of $\p$ × 1 $m_\p$ = 1 g

⟹ 1 mole of $\p$ × 1 $\mathrm{u}$ = 1 g

⟹ 1 mole of $n$ $\p$ × 1 $\mathrm{u}$ = $n$ g

Therefore, mass of 1 mole of $n$ $\p$ is equal to $n$ g

Now, the final part,

Let an atom have $x$ $\p$  and $y$ $\n$ . The atom can be considered to be $(x+y)$ $\p$  for our purposes.

1 mole of $(x+y)$ $\p$ have a mass of $(x+y)$ g.

Thus the mass of 1 mole of atoms of an element is equal to its gram atomic mass.

Molecular mass of an element or a compound is equal to the sum of the atomic masses of each type of it's constituent atoms.
Thus the mass of 1 mole of molecules of an element or a compound is equal to the gram molecular mass.


What are the effects of storing drinking water in a copper container?

As you state, copper storage containers have long been used to prevent growth of pathogens in potable water.

As for the danger of copper toxicity, it depends on the pH and water temperature. Between pH 8 and 11, about 1 ppm of copper dissolves, so if you're consuming ~5 l/day or about 5,000 g/day of water, and if the water has had sufficient time to become saturated, then you'd get ~5 mg/day of copper, exceeding suggested safe limits for long-term exposure.

That said, I think that, if the water pH is between 8 and 11, and if it is refrigerated (e.g. 4 C), and if it has been standing quiescently for no more than eight hours, it would not have time to be saturated with copper, so it probably would be safe to drink if that were the only source of copper in your diet.

Given all those mealy-mouthed caveats, then, you might at least check the pH of the tap water, and use a different container if it's acid. You would not want to be like the drinkers from Yudono spring, unless you're really into Heavy Metal.


Why isn't Mg deposited on the cathode after electrolysis?

You are in aqueous solution. Everywhere where there is an aqueous solution, you have a certain concentration of a further cation: $\ce{H+}$. It has a potential of $\pu{\pm0 V}$ by definition, thereby being nobler than $\ce{Mg^2+}$.

Thus, hydrogen gas is ‘deposited’ on the cathode first and bubbles away while the solution gets removed.

It is not possible to deposit considerably less noble metals on the cathode in normal, aqueous solutions.


pH of CH3COOH + HCl solution (shouldn't be a buffer)

Or does some reaction happen that I'm not aware of?

There are three reactions, and you are probably aware of all of them:

$$\tag{1}\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$

$$\tag{2}\ce{HCl(aq) -> H+(aq) + Cl-(aq)}$$

$$\tag{3}\ce{Ch3COOH(aq) <=> H+(aq) + CH3COO-(aq)}$$

The concentration of $\ce{H+}$ in those three reactions is the same because it all happens in the same phase. So what happens in reactions (1) and (2) influences the equilibrium of reaction (3).

  Can I simply find the [H+] released by the reaction of CH3COOH with water and add them to the [H+] released by HCl, add the two concentrations and then calculate pH that way?

No you can't because if you go back to reaction (3), you are no longer at equilibrium. To avoid trying to fix one equilibrium while messing up another, there are two strategies:

a) Put everything into a system of equations, and solve them in one swoop (preferred method if you are using an equation solver).

b) Start with the major species ignoring minor species and reactions that don't influence major species much. Then, move on to those other reactions and the minor species. This is preferred when you have to do the calculations on paper and don't need the exact solution.

The approximate answer

After mixing and ignoring all acid dissociation reactions, the concentrations are the following:

c(acetic acid) = 50 / 125 * 0.3 M = 0.12 M

c(hydrochloric acid) = 75 / 125 * 0.2 = 0.12 M

So here are the steps:

Let hydrochloric acid dissociate and get hydrogen ion concentration and pH
Check whether acetic acid dissociates appreciably
Check whether water dissociated appreciably

For the first step, we get [H+] = 0.12 M, and pH = 0.92. These are tentative because we did not let the other acids (water and acetic acid) dissociate.

For the second step, we are not at equilibrium yet (no acetate yet). However, the pH is very low compared to the pKa of acetic acid, so unless the pH changes a lot, it won't dissociate much. Let's try to calculate the acetate concentration assuming that the change in acetic acid and hydrogen ion concentration is negligible.

$$ [\text{acetate}] = K_a * [\text{acetic acid}] / \ce{[H+]} = \pu{1.8e−5} * 0.12 / 0.12$$

So we are not making a big mistake if we say the concentrations of acetic acid and hydrogen ions did not change much due to reaction 2. If we want, we can update the hydrogen ion concentration from 0.12 M to 0.120018 M.

For the third step, we do what we always do when the pH is substantially acidic. We just calculate the hydroxide concentration assuming the hydrogen ion concentration is not much affected by water dissociation. It comes out as $\pu{8.3e-14}$. If we want, we can update the hydrogen ion concentration from 0.120018 M to 0.120018000000013 M and a pH of 0.92075. (None of this makes sense because we only had 2 significant figures for the dissociation constant of acetic acid.)

Because we assumed a hydrogen ion concentration of 0.12 M for the second step and 0.120018 M for the third, even though this is not quite true, reactions (2) and (3) are not quite at equilibrium for the concentrations we calculated. The good thing is that in most cases, that does not matter.


There are several ways to determine oxidation number/state. The most useful method for organic (covalent) compounds uses the following three rules, and works for most elements engaged in covalent bonding (just replace carbon with some other element):

Every bond between a carbon atom and a more electronegative atom increases the oxidation number of that carbon atom by 1.
Every bond between a carbon atom and a less electronegative atom (hydrogen is most common) decreases the oxidation number of that carbon atom by 1.
Every bond between a carbon atom and another carbon atom has no effect on the oxidation number of the carbon atom (although we could get more complicated since carbon atoms with different "hybridizations" have different electronegativities).

This approach is equivalent to assigning all of the electrons in a covalent bond to the more electronegative atom for purposes of determining oxidation number. This page on Master Organic Chemistry has a number of examples.

Here's an example from the Wikipedia article on the matter, using acetic acid:

Atom            Bonds to   Contribution to Ox#
H (methyl)       C         +1 (2.5 > 2.1)
          TOTAL Ox#        +1
C (methyl)       H         -1 (2.1 < 2.5)
                 H         -1
                 H         -1
                 C         +0 (same atom)
          TOTAL Ox#        -3
C (carboxyl)     O         +1 (3.5 > 2.5)
                 O         +1 (double bond counts as 2 bonds)
                 O         +1
                 C         +0
          TOTAL Ox#        +3
O (carbonyl)     C         -1 (2.5 < 3.5)
                 C         -1
          TOTAL Ox#        -2
O (hydroxyl)     C         -1
                 H         -1 (2.1 < 3.5)
          TOTAL Ox#        -2
H (hydroxyl)     O         +1 (3.5 > 2.1)
          TOTAL Ox#        +1


Silver:

Silver sulfide precipitate is black.  You are pobably confusing a precipitate in a test for silver, with actual silver.  The precipitate in a test will be a compound of silver, such as silver sulphide which is black.

Nanophase metal particles are optically different from bulk in many cases, and can be black, but this doesn't seem to be what the question is referring to.


What interesting things can be seen in spectra between 400-700nm?

Absorbance spectrum of colored solutions may be boring for a "sight seeing" student, as you call yourself. Emission spectrum is far more interesting for visual learners. Anything luminous, which you can see with your eyes, should have a spectrum in the visible range. This includes the sky, moon, a red hot stove, fluorescent lamps, computer screen, flames, bulb, LED etc.

  The visible spectrum (400-700nm) tends to give a very blurry spectral
  picture

I gather you are talking about absorbance spectrum of solutions but this is not universally true. If you ever had a chance to try holmium oxide "solution" placed in a spectrometer, its absorbance spectrum has very sharp features.

Emission spectrum on the other hand is far more interesting. One can see very sharp emission lines even with a CD spectroscope. All you have to look at sodium street lamps, white fluorescent bulbs or the moon (one can see very weak solar spectrum with dark lines).

If are looking for low budget "instruments" I would suggest that you play with a CD or DVD as a diffraction grating rather than using a lab based absorbance measuring instruments. That is just a black box.

https://hackaday.io/project/28529-portable-dvd-spectroscope


Galvanic Electrochemical Cell Without half cells

This differs from the lemon galvanic cell (aka, 'battery') in several important ways: 1) Mg is used as the anode, rather than Zn, 2) there is no salt bridge, unlike in the the lemon cell, where the lemon serves as a crude 'salt bridge', and 3) the electrotyle is 2 M HCl rather than citric acid (and other stuff) in lemon juice. An open circuit voltage will be produced, easily measured with a digital multimeter in voltage input mode. A short circuit current will be produced, easily measured with a digital multimeter in current input mode. It may be possible to light a red LED, since they have low threshold voltage and low current requirements. I suspect it would light the red LED, though feebly.

But regardless of whether or not there is an external connection of the Mg and Cu electrodes, there will be a rapid redox reaction at the Mg electrode: $$\ce{Mg(s) + 2 H^+ <=> Mg^{2+} + H_2 (g)}$$

This unwanted competing redox reaction is the same as the net reaction obtained from the two relevant half cell reactions and it justs wastes some of the available energy as heat in the cell's vessel. So some electrical power is available by connecting an electrical load (e.g., red LED or resistor) between the two electrodes. But the parasitic oxidation of the Mg anode will continue until the electrode is consumed or the HCl is almost entirely consumed. (Magnesium will very slowly react with boiling water.)

Notes: 1. Clean copper, i.e., without surface compounds or so-called 'patina', does not react with 2 M HCl at room temperature. So the 2 M HCl does not contain any $\ce{Cu^+}$ or $\ce{Cu^{2+}}$ ions. 2. The reaction of Zn with 2 M HCl is considerably slower than is the reaction of Mg with 2 M HCl.

For more information, and at a much deeper level, see K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", Journal of Chemical Education, 95 (2018) 1801-1810.


Is most of the weight you lose actually released as carbon dioxide when you exhale?

Let's start with:

  How much is heat energy?

2000 kilocalories per day, considering $E=mc^2$, is
$9.3 \times 10^{-8}$ grams

  is all the carbon released later when you exhale?

No.  Brainstorming, the others ways would include: feces, urine, perspiration, flatulence, shedding skin, hair and nails, mucous from nose and mouth, for women: menstruation, vaginal discharge, lactation, and giving birth, and for men ejaculation.

According to The Value of Human Feces:

Per day the amount of carbon excreted in feces is 21 grams and in urine is 7 grams.

According to The Composition of Human Perspiration Proc R Soc Med. 1934 May; 27(7): 839–848, a typical amount of perspiration a day is 700mL, and per 100mL the main carbon containing compounds are 20 mg urea, 20 mg glucose and 5mg amino acids.  So clearly less than 1 gram of carbon loss per day from perspiration. If we were considering loss of salts instead of just carbon, that would be about 3 grams per day from perspiration.

For menstruation, blood loss is about 40mL per cycle, and blood is 92% water, so maybe 1 gram of carbon on a heavy day of a period.

Skin loss per day is 1.5 grams according to Cells for life, cells for an instant.

A highly controlled study having a person in a chamber for 24 hours and analyzing the air found 653g of CO2 or 178g carbon "A Respiration Calorimeter with Appliances for the Direct Determination of Oxygen" Carnigie Institution of Washington Publication 42.

So, for a person that is not lactating or giving birth, I would estimate about 30 grams per day of carbon loss is other than from exhaling carbon dioxide.  Therefore:

About 86% of carbon loss is through exhaling carbon dioxide.

However, this will certainly vary depending upon the amount of indigestible carbon such as fiber, in ones diet.

Also, if mass loss through nitrogen and phosphorous (but not hydrogen and oxygen) were also considered the loss through breathing would come down to about 80%.  Inclusion of salts would bring this percentage down further.

The sum of water from feces, urine and respiration is about 1.8 kg per day and the total mass emission of the body is 2.6kg, according to the Carnigie source above.


Arsenic bound proteins confirmatory test

You may wish to give this suggestion a try, courtesy of the literature on constructing nano-particles.

First, dissolve the protein samples in say ethanol or other flammable solvents.

Next, place it into an alcohol lamp with a wick composed preferably from glass fiber, but one may be able to employ woven cotton or a cotton string.

Lastly, light the lamp and place a bowl over the lamp to capture/condense escaping particles.

My recollection is that the target metal of interest, is slowly draw through the wick and thermally broken down. Still looking for the source, but here is a related article.

This is clearly relates to the final step in the Marsh Test, where the arsenic presence is distinctive.

Note, this proposed procedure, if successful, avoids the conversion of a sample of a compound into deadly AsH3 gas per a nascent hydrogen pathway, which is perhaps not required here, as you already have constructed (via an alcohol lamp), a path to 'burn' the As(III).


Clarification required on whether to use $C_v$ or $C_p$ in this question on finding entropy change

The correct form can be identified as follows. Start from the combined 1st and 2nd laws:

$$\begin{aligned} d S & = \frac{dU}{T} +  \frac{p}{T}dV \end{aligned}$$

For an ideal gas

$$\begin{aligned} d U &= nC_V dT \\ \frac{p}{T} &= \frac{nR}{V} \end{aligned}$$

such that

$$\begin{aligned} d S & = nC_V \frac{dT}{T} +  nR \frac{dV}{V} \end{aligned}$$

Integrating

$$\begin{aligned} \Delta S & = nC_V \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{V_2}{V_1} \right) \end{aligned}$$

Substitute in the ideal gas law in the right-hand side (note $n$ is constant)

$$\begin{aligned} \Delta S & = nC_V \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{P_1 T_2}{P_2 T_1} \right) \\ & = n(C_V +R) \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{P_1 }{P_2 } \right) \\ & = nC_p \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{P_1 }{P_2 } \right) \end{aligned}$$


Why Li can't make it's Peroxide when heated with more O2?

It has to do with the small size and high charge density of the lithium ion, which effectively polarizes any peroxide ion that might form and thus pulls it apart.  See http://www.chemguide.co.uk/inorganic/group1/reacto2.html.


The use of HBr with peroxide leads to an anti-Markonikov's product(known as the peroxide effect).The carbons are both substituted once, but not equally! One has a methyl group while the other side has a phenyl group. The chemical basis for the Markovnikov's  rule is the formation of a a stable carbocation during the addition process.

So, with just HBr a nomal Markovnikov's addition takes place i.e the Br group gets attached adjacent to the phenyl group while the H on the next carbon(as a benzylic carbocation is more stable).Final product: Ph-CHBr-CH2-CH3.
With the peroxide however , the opposite happens leading to the formation if Ph-CH2-CHBr-CH3.


Can dissolved salts (or any soluble chemicals, for that matter) 'bump' each other out of solubility?

Yes, this effect is termed salting out for non-ionic compounds and the common ion effect for ionic ones.

Note that in specific cases, the converse may hold true. For example, "the solubility of silver chloride in water is reduced if a solution of sodium chloride is added to a suspension of silver chloride in water (op cit)." However, adding more chloride re-dissolves the precipitate due to the formation of complex ions:

$\ce{AgCl(s) + Cl- (aq) -> AgCl2- (aq)}$


How is the potential energy between two atoms measured?

The morse potential (equation see e.g. wikipedia) basically contains the bond dissociation energy, a "force constant" and the bond length at ground state. It does not give any measurable reality, but is just a mathematical model describing (approximating) the same.

The lower part of the curve (usually the ground state and perhaps the first two vibrationally excited states) are approximated by the harmonic oscillator, the energies of which you can get from  IR or Raman spectroscopy. The harmonic oscillator (= an ideal spring) has only one parameter, the force constant.

The bond lengths you can get either directly from microwave rotational spectroscopy, or from the rotational fine structure of you IR spectrum. They give you the reduced mass of the structure, which gives you the bond length if you know the absolute mass of the two atoms.

For the math and actual application, I refer you to any standard physical chemistry textbook.


What is sprayed on chromatography paper when developing?

It depends on what you will use the chromatography paper for. Different testable substances will require different mediums. However, for a plain-simple chromatography without staining/substance identification, aluminium oxide (or silica) coated on a plate is used, however paper can also be used.


If this is an energy drink for humans then the active are either soluble in water or ethanol. The standard factory method is to use supercritical CO2 extraction. Check the method for production of instant coffee. Briefly, they mix coffee beans with CO2 at 40C (above critical temperature and under very high pressure. Then they filter it and drop the pressure. Components precipitate out. You can drop it to certain value, filter to second chamber, drop pressure slightly, filter move to third chamber, drop pressure to 1 Atm. In this case second chamber can have a single component. In case of coffee caffeine precipitates in second chamber. Now mix stuff from first and third chamber to get decaf.

Ethanol extraction works ok and is used by pharmacists. It is hard to concentrate it however. And ethanol is an active ingredient on its own.

Water extraction is the dirtiest way mainly because it is difficult to concentrate your material.


Why is it possible to form bromoalkanes but not iodoalkanes from free radical substitution?

Mechanistic Summary

Initiation: Involves homolytic cleavage of the $\ce{X-X}$ bond to generate $\ce{X^.}$ radicals. Initiation is endothermic, and the energy requirement is usually met by thermal treatment or photoirradiation.
Propagation I: Abstraction of a $\ce{H}$ atom by the $\ce{X^.}$ radical, to produce a $\ce{R-H2C^.}$ radical. This transformation usually takes place via a $[\ce{R-H2C...H...X}]^\ddagger$ transition state. Also, this process is endothermic
Propagation II: This stage involves the abstraction of $\ce{X}$ atom from $\ce{X-X}$ by the the $\ce{\ce{R-C^.}}$ to prduce $\ce{R-H2C-X}$. This process occusrs via a transition state that looks so something like this: $[\ce{R-H2C^....X...X}]^\ddagger$. This process, if it is sufficiently exothermic, compensates for the previous endothermic step and the reaction proceeds.

Termination: Involves radical-radical recombination.

Now let us look at the energetics of the reaction. The first propagation step is exothermic for fluorine and this explains its extreme reactivity. For the subsequent halogens, it becomes increasingly endothermic from chlorine to iodine. This trend can be explained on the basis of decreasing bond strengths of $\ce{H-X}$ bond. (Data given in the table above)

Now let me compare to extremes, fluorination of methane and iodination of methane. As one can observe, the first propagation step for fluorination is not only exothermic, but it also has a small activation barrier.

Thus in the transition state, $[\ce{H3C...H...F}]^\ddagger$, the $\ce{H-CH3}$ distance is only slightly greater than that in $\ce{CH4}$, and $\ce{H-CH3}$ bond is weaker than the $\ce{H-F}$ bond. Thus only a slight shift of the hydrogen atom towards the fluorine radical is sufficient to break the carbon-hydrogen bond and form the fluorine-hydrogen bond. This kind of a transition state is called an early transition state, and is typical of fast,exothermic processes.

In contrast, iodination displays a late transition state which is typical of slow,endothermic processes.The activation barrier for the first propagation step is relatively high, in fact it is almost the same as the endothermicity of the reaction. The transition state is not reached till the the carbon-hydrogen bond is almost fully broken, and the hydrogen-iodine bond is fully formed.

These rules about transition states are called Hammond's Postulates, and I encourage you to read about them.

Now, indeed the reason why iodination doesn't proceed is that even though final step is exothermic it doesn't release sufficient energy to compensate for the extremely endothermic first step. Thus, the net reaction is endothermic.

However, it is not the exothermicity of $\ce{H-X}$ bond formation that drives the halogenation reaction, rather it is the energy released upon $\ce{-C-X}$ bond formation (in the second propagation step) that ensures reaction spontaneity. (this can be seen in the data presented in the table)

This is why bromination, or for that matter chlorination can take place.

My apologies, the question asked about iodethane and bromoethane, however, the examples I took up was that of iodomethane (not the first time I have misread a question in my rush to post an answer). However, the ideas presented here still apply

References: Data table, and reaction profile taken from Organic Chemistry: Structure and Function by Vollhardt&Schore.


What's happening at the beginning of a weak acid titration?

Unfortunately this is not correct.

  $$\ce{CH3COOH~(aq) + H2O(~l) <=> CH3COO- ~(aq) + H3O+~(aq)}$$

If you have a pure acetic acid solution the equation you state governs the pH of the solution.

If you are now adding a strong base like $\ce{{}^{-}OH}$, you are removing the hydronium ion from the right hand side of the equation. According to Le Châtelier, the equilibrium adapts and shifts more to the right, i.e. more acetic acid dissociates. As you remove $\ce{H3O+}$ the pH increases.
Another way of looking at this is, that you remove acetic acid from the left hand side (similar to what you stated).
$$\ce{H3CCOOH + {}^{-}OH <=> H3CCOO- + H2O}$$
According to Le Châtelier, the equilibrium (in the first equation) has to shift to the left, to regenerate more acetic acid, removing once again $\ce{H3O+}$ and the pH increases.

In any case you choose to look at, you increase the concentration of acetate in the solution and another equilibrium comes into play.
$$\ce{H3CCOO- + H2O <=> H3CCOOH + {}^{-}OH}$$
This equilibrium is essentially responsible for raising the pH.


https://www.youtube.com/watch?v=B0E4PX3e3RE It feels somewhat weird to answer my own question but I think this video describes exactly what I wanted. As it supports, when heat is applied to a piece of wood, some bonds of the molecules that make up wood, break and thus different compounds are formed. These compounds are not held back by some force and so they are released in the air. When these compounds meet atmospheric oxygen, under heat (=energy), they burn and thus more heat is released along with carbon dioxide and water. That stage can be described as ignition. Finally, this produced heat is able to preserve the fire.


I haven't dived into your math, but I think there are 3 things that are confusing:

intermediate calculations, even though you weren't asked to do any;
missing units and inconsistent notations (e.g. base 10 is not always shown);
ambiguous phrase "atoms occupy a volume of $2\times 10^{-4}$."

I always find it cleaner and more productive to derive the formula first, and only then plug in the numbers minding the proper units. As for the volume issue, I tend to believe that $2\times 10^{-4}$ means the fraction of the volume occupied by gas atoms and it is not the absolute value expressed in cubic centimeters. In fact, in this case you get the correct answer and you don't need to use gas volume ($\pu{230 cm3}$) at all.

Check this out. Radius of an atom $r_\mathrm{a}$ is indeed can be found from the volume of an atom $V_\mathrm{a}$

$$r_\mathrm{a} = \sqrt[3]{\frac{3V_\mathrm{a}}{4π}}.$$

At the same time

$$V_\mathrm{a} = \frac{V_\mathrm{atot}}{N}$$

where $V_\mathrm{atot}$ – volume of all gas atoms depicted with spheres; $N$ – number of the atoms that can be found via Avogadro's number $N_\mathrm{A}$ and amount $n$:

$$N = nN_\mathrm{A}.$$

Finally, you are allowed to use ideal gas law to determine the amount $n$:

$$n=\frac{pV}{RT}$$

where $p$ – pressure; $V$ – volume of the gas (macro scale); $R$ – the molar gas constant; $T$ – temperature. Now, lets's put all the things together assuming that  $V_\mathrm{atot}/V = 2\cdot 10^{-4}$:

$$
\begin{align}
r_\mathrm{a} &= \sqrt[3]{\frac{3V_\mathrm{atot}RT}{4πpVN_\mathrm{A}}}\\
  &= \sqrt[3]{\frac{3\cdot2\cdot10^{-4}\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{300 K}}{4\cdot 3.14\cdot\pu{101325 Pa}\cdot\pu{6.022e23 mol-1}}}\\
  &= \pu{1.25e-10 m}
\end{align}
$$

Now that's the value that actually makes sense:)


How do I know crystal structures from formula?

I'll just point out the direction in which you could look for an answer.

In general case, it should be almost impossible to determine the crystal structure by looking at the chemical composition of a substance.

You are, however, asked to explain why different fluorides adopt different structure types, which is a different task. Perhaps, you should take into account the difference (sorry) in the ionic radii of Mg and Ca. Some structures prefer "fat" cations, some - "lean" ones. Most of the time, it's just plain geometry combined with experimental observations. I know the explanation is primitive, and ions are not just some rigid spheres, but the "ionic radii" approach works in many cases.

For many crystal structures there are ranges of ionic radii ratios in which the particular structure is more or less stable. Some of these ratios are so important that they have proper names, such as Goldschmidt tolerance factor for perovskites. Returning to your example, this page lists the cation/anion radii ratios for fluorite (0.73 or above) and rutile (between 0.41 and 0.73). You can check these ratios against the dimensions of ions in magnesium and calcium fluorites.

For the reference - this Shannon table hosted by Imperial College London is an excellent source of the values of ionic radii in oxides and halides. Just one more thing: the reference ratios are likely to be designed to work with so-called Shannon Ionic Radii, not the less-common Shannon Crystal Radii.


How to measure and transfer sticky solids

Ideally, you want your sticky solid to already be in a Schlenk flask or similar. If it is unfortunately still in a normal bottle, take the weight of a Schlenk flask, fill in enough of the sticky solid and re-weigh. Note the difference, this is what you have in your flask.

Assuming that the sticky solid has a low enough vapour pressure (spoilers: practically all solids do), evacuate the Schlenk flask under high vacuum and backfill with argon (or nitrogen, if that is your inert gas of choice. Add a known quantity of your (dry, stored over molecular sieve and under inert atmosphere below a septum) solvent. Ideally, you either want an amount of solvent that gives you an easy-to-handle concentration (e.g. $10~\mathrm{M}$) or an amount so that what you need is in an even volume of liquid (e.g. $1~\mathrm{ml}$).

Finally, transfer the solution into your reaction flask.

If you need the sticky solid in different solvents and it is air-stable, you can evaporate the dry solvent at the rotavap. If not, keep it in solution under argon (or nitrogen). If the latter is the case, do your colleagues a favour and choose a standard concentration, not a good filling amount for exactly your reaction.


How to find the ionization constant of an unknown weak acid?

Look at the answer given here and modify it:
How to set up equation for buffer reaction?

In your case the concentration of base $c_B \approx 0$ and $[\ce{OH^-}] = K_\mathrm{w}/[\ce{H^+}] \approx 0$, so that the equation reduces to $$K_\mathrm{a} = \frac{[\ce{H^+}]^2}{c_\mathrm{a}-[\ce{H^+}]} \approx \frac{[\ce{H^+}]^2}{c_\mathrm{a}},$$
as $c_\mathrm{a}$ is so large.


Solubility and heat release of ytterbium (III) sulfate in water

When the solubility of a salt decreases with temperature, this indicates that the enthalpy of dissolution is exothermic, i.e. heat is released during this process. This is the case when the lattice energy, which is at least required to separate the ions by breaking up the crystal lattice, is smaller than the energy which is released when the ions are solvated. The solubility curve of ytterbium(III) sulfate therefore proves the statement (B).


Can I engrave borosilicate glassware without compromising its structural integrity?

Any scratch weakens glass somewhat. However, a vibratory engraver is very likely to shatter the vessel.
You can label glass permanently by etching with warm concentrated $\ce{NaOH}$ or room-tempeerature $\ce{HF}$, perhaps with a stencil. Use great caution with $\ce{NaOH}$, and even more caution with $\ce{HF}$. Both are very caustic, and $\ce{HF}$ can release toxic fumes and cause deep tissue damage.
There are commercial etching creams with HF that might be less likely to splash.
Note that etching weakens glass slightly, but less so then engraving.


Calculating the ionic strength for an aqueous solution containing potassium ferricyanide and sodium sulfate


why does a pot of boiling water quiet down after it reaches boiling point but is still roiling?

Oh, you've come to the right place.  Many years ago I asked this question as a kid on a TV write-in programme called Don't Ask Me.  The TV people invited me on the show and built a series of demonstrations to explain.

The first thing that happens when you warm up a cold pot of water is that the pan bottom gets to boiling temperature.  The water starts to boil.  Liquids boil when their vapour pressure equals atmospheric pressure.  Otherwise bubbles could not form in the the liquid.  So, the pan bottom gets hot, bubbles form, but then they move up into cooler water and implode as the vapour pressure drops.  This makes a sharp snapping sound.  Lots and lots of these imploding bubbles make a roaring sound.

As convection currents then start to mix up the water, it becomes more uniform in temperature, and the bubbles actually rise to the top of the liquid and burst.  The implosions stop and you get the softer bubbling sound.

Hope this answers your question!


Why can sulfur have an expanded octet but not oxygen?

sulfur has 2 electrons in its 3s subshell , and 4 in 3p subshell. On absorbing energy the paired electrons get excited to to the next higher energy level which is the 3d subshell. So after exciting it's electrons six unpaired electrons are present which can take part in bonding via overlapping of orbitals. Other compounds can also be explained similarly. Also the d- orbitals contribute to only two bonds out of six bonds( which would have been meant in the paper by Reed and Weinhold (1986) ). Oxygen has no empty orbitals with it . It only possess orbitals upto 2p which contains 4 electrons. So it cannot expand its octet by exiting it's electrons.


Understanding the unit cell of Chalcopyrite vs its formula

Some of the Copper and Iron lie on the faces of the unit cell.

Copper atoms

The 8 Copper atoms at the corners all represent the same Copper atom so that means only 13 - 7 = 6 Copper atoms.
Then there are an additional 2 repeats on the top and bottom faces and front and back faces bringing it to 4 Copper atoms.

Iron atoms

The Iron atoms similarly have a pair repeat on the left and right faces, the top and bottom faces and the front and back (10 - 3 = 7).
Additionally the 4 Iron atoms in the centre line (from left to right) represent the same atom so thats 7 - 3 = 4 Iron atoms.

This is how the formula and the unit cell are consistent with each other.


It is a large, soft, Lewis acid. One can use protic acid and water to promote the same reaction, but the product distribution can sometimes be different.
 In the oxymercuration reaction Hg can form a pi complex with alkenes or alkynes, and these "bridged" cations will not undergo rearrangments to the extent that those produced by protic acids might.
 Bottom line - they're both essentially performing the same function in this reaction - activating the pi bond towards attack - but the oxymercuration is milder.


Identification of a compound C10H12O with 1H and 13C NMR data

Like you said, the 2 doublets at around $\pu{7 ppm}$ belong to a para-substituted benzene. The three single-proton multiplets around $\pu{5-6 ppm}$ belong probably to a single substituted alkene. A singlet and a doublet between $3$ and $\pu{4 ppm}$ belong to $\ce{CH3-{}}$ and $\ce{-CH2-{}}$ groups, respectively (Note that the signal that belongs to the $\ce{-CH2-{}}$ group is a doublet, indicating that it is adjacent to a carbon with only a single proton, so it is probably connected to the alkene).

The absorption maximum at $\pu{275 nm}$ gives us a hint that the disubstituted phenyl ring of our compound is not conjugated to the $\ce{C=C}$ bond.

Concerning the oxygen atom: it is definitely neither a carbonyl group (indicated by the absence of an intense peak at $\gt \pu{180 ppm}$), nor a hydroxy group. So it probably is an ether.

So far, we have two possible candidates:

Allyl para-tolyl ether
1-allyl-4-methoxybenzene

The chemical shift of the $\ce{CH3}$-group is shifted downfield compared to the $\ce{CH3}$-shift in toluene (where it is around $\pu{2.3 ppm}$, while this compound has $\gt \pu{3.7ppm}$), and thus more likely belongs to $\ce{CH3-O-Ar}$ protons. So, our compound is 1-allyl-4-methoxybenzene.

However, the multiplets of the vinylic protons in the $\ce{^1H}$-NMR spectrum don't fit this compound well. The allyl group should give two dd signals for the terminal protons and a ddt for another vinylic proton. But the $J_2$ constant for terminal protons is very small ($\lt \pu{2 Hz}$) and can't be seen on some old NMRs, what would lead to the simple two doublets.

Why do we see dddd instead of ddt for $\ce{R-CH=CR2}$ proton? I'm not sure, but maybe this is because the $\ce{-CH2-{}}$ protons are diastereotopic. Then it makes sense.


Maximum possible hypochlorite concentration from electrolysis?

A little bit of Wikipedia research reveals that

  Sodium hypochlorite can also be obtained as a crystalline pentahydrate $\ce{NaOCl·5 H2O}$, which is not explosive and is much more stable than the anhydrous compound.5 The formula is sometimes given as $\ce{2NaOCl·10H2O}$. The transparent light greenish yellow orthorombic[10][11] crystals contain 44% NaOCl by weight and melt at 25−27°C. The compound decomposes rapidly at room temperature, so it must be kept under refrigeration. At lower temperatures, however, it is quite stable: reportedly only 1% decomposition after 360 days at 7°C.6[12]

So, it looks like 44% w/v would be an upper bound to any solution of NaOCl in water, regardless of whether it was obtained via electrochlorination or not. This corresponds to approximately 43.1% active chlorine.

Also, note that that the term "active chlorine" measures overall oxidative capability, not necessarily hypochlorite concentration — hypochlorites can easily decompose into chlorates without altering the greater solution's overall oxidative capabilities
$$\ce{3OCl- (aq) -> ClO3- (aq) + 2Cl- (aq)}$$
as demonstrated via their molar-equivalent iodometries
$$\ce{3OCl- (aq) + 6I- (aq) + 6H+ (aq) -> 3I2(s) + 3Cl- (aq) + 3H2O (\ell) \\ ClO3- (aq) + 6I- (aq) + 6H+ (aq) -> 3I2(s) + Cl- (aq) + 3H2O (\ell)}$$
so more accurate measurements of hypochlorite concentration may likely require additional techniques (such as spectroscopy, for example) as well.

Finally, due to the unstable nature of solutions with high [NaOCl(aq)] and their natural tendency to
$$\ce{2OCl- (aq) -> 2Cl- (aq) + O2 (g)}$$
maximizing [NaOCl(aq)] solely via electrolysis/electrochlorination will likely require a dedicated setup that simultaneously:

Limits atmospheric exposure around the anode to prevent loss of chlorine gas while allowing hydrogen gas to escape by the cathode (or, alternatively,
reduce atmospheric oxygen to water)
Keeps the solution at low temperatures (to limit various other decomposition reactions)
Maintains some external "tending" voltage (to preserve the Nernst equilibrium)

For practical applications, this could be a very unwieldy form of storage!


Why does calculating formal charges help us draw correct Lewis structures?

Formal charges are helpful for two reasons:

at lower levels, you use them to show your instructor that you can properly count electrons and ‘keep the books’. You would need to realise that each oxygen has an additional electron and therefore a formal negative charge while sulfur is lacking two electrons and therefore has a formal positive charge. Adding up the formal charges should give the charge of the molecule.
Since all lone pairs are typically drawn at this level, lacking formal charges can easily be rederived.

at higher levels, lone pairs are typically not drawn unless they are important for some reason (e.g. if they take part in a resonance mechanism discussed at this very moment). Therefore, the only information on the number of electrons around an atom we can extract from the formula is the formal charge.
If at higher levels, you draw a tetravalent sulfur atom without formal charges, we must assume that it is actually a formally neutral sulfur and that there is an additional lone pair. This, of course, makes a big difference; compare the electronic states of sulfur in $\ce{SO4^2-}$ and $\ce{SF4}$.


Acidity of naphthol vs hydroxynaphthoquinone

For naphtol both the acidic and the basic forms have many resonance structures along the two fused benzene cycles. However, in the case of 2-hydroxy-1,4-naphthoquinone, the basic conjugated form creates novel resonance structures that engage the double bond of the quinone cycle as seen in the written resonance structure in the question. The novel resonance structures of the basic forms will stabilize the basic form.
Therefore, the hydroxyl group in 2-hydroxy-1,4-naphthoquinone has higher tendency to lose the hydrogen ion than the hydroxyl group in naphtol. This explains the higher acidity of 2-hydroxy-1,4-naphthoquinone in comparison with naphtol.


Can nuclear shielding affect relaxation time?

I'd say yes, but not by any detectable amount. In some relaxation mechanisms (e.g. dipolar relaxation, a major relaxation pathway for $\ce{^1H}$), the rate of relaxation varies with $\langle B_\mathrm{loc}^2\rangle$, i.e. the magnitude of the random fields that the spin experiences due to the magnetic momenta of all the other spins. The magnitude of any field that a nucleus experiences will be attenuated by the induced magnetic field arising from its own electrons, so it stands to reason that – all else being equal – a more shielded nucleus will have a smaller $\langle B_\mathrm{loc}^2\rangle$, and a longer relaxation time, than a less shielded nucleus. This applies to both $T_1$ and $T_2$.

Bear in mind, however, the most obvious effect of nuclear shielding, which is to change the resonance frequency of the nucleus. The spread of frequencies is on the order of $10^{-6}$ relative to the strength of the external magnetic field. If we crudely estimate that the spread in $T_1$ and $T_2$ arising from sheilding differences is also on the ppm level, that corresponds to ~µs levels of difference, which I strongly doubt is detectable.

In any case, for two protons that have different chemical environments, there would be other factors which affect the relaxation times to a much greater extent.


Is it safe to inhale O2 from electrolysis of a pH 13.7 KOH solution?

No. Untreated and unpurified, the gas coming directly out of your reaction vessel would be unsafe to inhale. It is likely that water vapour containing traces of KOH would be present, and this poses a significant safety risk to the person breathing this vapour. You could test this yourself by measuring the pH of the headspace above your reaction vessel. Hold a piece of indicator paper here for a period, and it will slowly show a change.


Why does fructose reduce Tollen's reagent and Fehling's solution?

The presence of the hydrogen atom in the aldehyde group gives reducing properties to the aldehyde molecule. Aldehydes are easily oxidized by mild oxidizing agents such as Tollens’ reagent ($\ce{Ag+}$ in aqueous ammonia), Fehling’s reagent ($\ce{Cu^2+}$ complexed with tartrate), or Benedict’s reagent ($\ce{Cu^2+}$ complexed with citrate). Ketones do not have such reducing properties and are not oxidized under similar conditions. Therefore, Tollens’ reagent, Fehling’s reagent, or Benedict’s reagent can be used to distinguish aldehydes from ketones.

However, fructose reduces such reagents even though it contains no aldehyde group. Reduction occurs because the reagents are basic solutions and fructose is readily isomerized to a mixture of aldoses (glucose and mannose) under basic conditions.

Under basic conditions, the proton alpha to the carbonyl group (aldehyde or ketone) is reversibly removed. If the resulting enolate ion reprotonates on the oxygen, an enediol intermediate results. Thus, the base-catalyzed enediol rearrangemet can move the carbonyl group up and down the chain.

Furthermore, in the enolate ion, the affected carbon atom is no longer asymmetric. Reprotonation can occur on either face of the enolate, giving either configuration. Thus, the base-catalyzed epimerization results in a mixture of stereoisomers.

Therefore, under strongly basic conditions, the combination of enediol rearrangements and epimerization leads to a complex mixture of sugars (see also: Lobry de Bruyn–van Ekenstein transformation).


The esterification of carboxylic acids with tertiary alcohols usually occurs by an SN1 mechanism wherein the oxygen of the alcohol is protonated and a tertiary carbocation is formed. It in turn reacts with the carboxylic acid to form the tertiary ester. The hydroxyl group in citric acid, while tertiary, is alpha to a carboxyl group. A cation alpha to a carboxyl group (-CO2H) is a destabilized entity. This mechanism would be precluded. Traditional Fischer esterification is likely sluggish owing to steric hinderance.


What are the Waves Modeling when Referring to the Atomic Orbitals

In a one-dimensional system (I choose this to keep things as simple as possible) the state of the system is described by the wave function $\psi = \psi(x, t)$, which depends on the coordinate $x$ of the particle at time $t$. Wave functions in general are complex functions of real variables. The symbol $\psi^{∗}(x, t)$ denotes the complex conjugate of $\psi(x, t)$. The quantity $P(x, t) = \psi^{∗}(x, t) \psi(x, t) \mathrm{d}x = | \psi^{∗}(x, t) |^{2} \mathrm{d}x$ gives the probability that at time $t$ the $x$ coordinate of the particle lies in the small interval $[x, x + \mathrm{d}x]$. The probability of the particle being in the interval $(a, b)$ on the $x$ axis is given by $\int_{a}^{b} | \psi^{∗}(x, t) |^{2} \mathrm{d}x$. The probabilistic interpretation of the wave function was proposed by Max Born. By analogy with the formula: mass = density $\times$ volume, the quantity $\psi^{∗}(x, t) \psi(x, t)$ is called the probability density that a particle at time $t$ has position $x$.

Edit: Atomic orbitals

For atoms you have a system that is spherically symmetric. Thus, it is most conveniently expressed in spherical coordinates instead of cartesian ones. The wave function solving the time-independent Schroedinger equation is then expressed in terms of the radial distance $r$, the polar angle $\theta$ and the azimuthal angle $\varphi$.
This wave function $\Psi(r, \theta, \varphi)$ can be split into a radial part $R_{n,\ell}(r)$ and an angular part $Y_{\ell,m} (\theta, \varphi )$, so that $\Psi(r, \theta, \varphi) = R_{n,\ell}(r) Y_{\ell,m} (\theta, \varphi )$, where $n$, $\ell$ and $m$ are the Principal, Azimuthal and Magnetic quantum number, respectively, and the functions $R_{n,\ell}(r)$ are the solutions of the radial Schroedinger equation

\begin{equation}
\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} }  - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R_{n,\ell}(r) = 0
\end{equation}

with the nuclear charge $Z$ and the mass of an electron $m_{\mathrm{e}}$. Those radial wave functions are the part of the atomic wave functions $\Psi$ that determines their "extent" and their radial nodes, i.e. the nodes that grow with the Principal quantum number $n$: for example a $1\mathrm{s}$ orbital has no radial nodes (it is a simple sphere without any sign-change in the wavefunction) while the $2\mathrm{s}$ orbital has 1 radial node (it is a smaller sphere wrapped in a bigger spherical shell whereby the inner sphere has a different sign than the outer spherical shell such that you get one nodal plane between inner and outer part). The angular wave functions $Y_{\ell,m} (\theta, \varphi )$ are the so called Spherical Harmonics. They determine the angular momentum of the atomic wave function $\Psi$ and its characteristic shape, i.e. $\mathrm{s}$ orbitals are spheres while $\mathrm{p}$ orbitals are dumbbell-shaped (side note: the Spherical Harmonics are complex functions which are rather hard to visualize; the shapes that every chemist is used to comes from the usage of the so called real Spherical Harmonics, which are just linear combinations of the complex-valued Spherical Harmonics).


Is it true that heavy water is not blue?

Based on your description, I may have found the article you originally saw, or at least one very similar.

Researchers from Dartmouth College published a paper$\mathrm{^1}$ in which they report, among other things, the results of viewing sunlit white paper through two 3 meter lengths of plexiglass; one filled with $\ce{H2O}$ and one with $\ce{D2O}$. Sure enough, because of the lower frequency of the maximum absorption of $\ce{D2O}$ in the red to near IR wavelengths, the blue color that is characteristic of $\ce{H2O}$ is far less pronounced in $\ce{D2O}$. This website is based on the published paper and additionally shows a photograph of the blue colored $\ce{H2O}$ on the left with the far less colored $\ce{D2O}$ on the right:

1) "Why is Water Blue", Charles L. Braun and Sergei N. Smirnov, J. Chem. Edu., 1993, 70(8), 612


What compressible rubber like gasket material is the least permeable to CO2?

Viton is a trade name for a fluoroelastomer type. There are different grades which have different properties. I've had a Viton seal be deformed in a vessel when using tetrahydrofuran as a solvent, whereas Kalrez does not. I've seen different Viton grades again not have great compatibility with acetone. So do check out the compatibility of each supplier's product.
We have two seals that we use for $\ce{CO2}$ applications. EPDM and PTFE.  These both cope with 100 bar pressure.


Can two electron occupy the same spatial spot in a statistical way?

Electrons do not occupy places. Other than that, you are mostly right. Indeed, $\psi(r_1,r_2)$ must vanish at $r_1=r_2$ because of the spin-statistics theorem, if not for other reason. Upd. It was brought to my attention that this reasoning is not quite right; see the comment below. Still, I would expect at least a downward-pointing cusp, much like the upward-pointing cusp found at the nucleus, only the other way around.

This, however, is not easy to visualize. We're not used to imagining multidimensional surfaces. And if we try to focus on one electron at a time, then we're back with the one-electron approximation, and the accurate multiparticle $\psi$ is lost to us.

BTW, the argument of infinite repulsion does not work. After all, $\psi$ doesn't vanish near the nucleus; wouldn't that imply infinite attraction? No, it would not. You integrate over the entire space and come up with some figure for energy, which is surely finite.


I have heard that electrons absorb or eject photons when transitioning from one orbital to another. Is this correct?

Not exactly.  The atom as a whole emits or absorbs the photon.  There is no reason to single out the electron versus the nucleus in such transitions.

  Can atomic nuclei eject photons?

Yes there are two ways a nucleus in particular (as opposed to an atom taken as a whole) can emit photons.

First, nuclei can exist in excited states. When an excited nucleus transitions to the ground state, a photon in the gamma or X-ray range of the electromagnetic spectrum is emitted.  Electrons do not have an analogous mechanism for emitting photons, because electrons are fundamental particles, while nuclei are not.

Second, if a nucleus has a nuclear spin (S), then in a magnetic field there will exist  2S + 1 quantized energy states of the nuclei.  Transitions of nuclei between these states emit or absorb photons, the energy of which depends upon the strength of the field, for example in the radio frequency range.  This is the basis of NMR and MRI.  The electron has an analogous mechanism for emitting photons, known as ESR or EPR.


Implication of negative order on kinetics

What does this $[\ce{O2}]^{-1}$ part mean?

It means the higher the concentration of oxygen, the lower the rate. The two-step mechanism you show relies on oxygen atoms (radicals) to react with ozone in the second step. Dioxygen acts as a radical scavenger (first reaction, reverse direction). So if dioxygen concentrations are high, the reaction will not proceed via this mechanism.

When ozone is produced as part of smog in the lower atmosphere (i.e. the air we breathe), this mechanism is not expected to help lower the ozone concentration. The equilibrium of the first reaction would be too far on the side of the reactants.

In the higher layers of the atmosphere, however, this sequence is more feasible. Concentrations of all species are extremely low, so bimolecular reactions are slow as well. In fact, in the image linked below, concentrations are given as molecule per mL (as apposed to mol / L) and rate a similarly expressed as molecules reacting per time unit.

https://www.ems.psu.edu/~brune/m532/meteo532_ch7_stratospheric_chemistry_files/image018.jpg

What actually happens in the ozone layer is a bit more complicated than the two step process you show, see for example https://www.ems.psu.edu/~brune/m532/meteo532_ch7_stratospheric_chemistry.htm

  For example, initially when there is no O2, the rate should be ∞; how is that possible? Or is it normal and I should accept it?

See the excellent comment by Nicolau Saker Neto. Another way to say this is "the molecules don't know the concentration of species, but they still do their thing without worrying about division by zero".


I don't think this question has an answer.  First, muscle soreness is not caused by lactic acid.   Second, distance runners run aerobically, meaning very little lactic acid would be generated by running in the first place.  (Sprinters are the exception here.)  Third, even if lactic acid buildup was a problem (which it isn't), both the cellular cytoplasm and human bloodstream are very well-buffered.  The cytoplasm is primarily buffered by potassium glutamate and the circulation by bicarbonate (created from respiration).  Production of "lactic acid" would really be production of lactate salts.


Free Radical Reaction of Methane and Chlorine

OK, let's focus on the stoichiometry of your last 4 equations:

$\ce{2 CH4 + 2 Cl2 -> 2 HCL + 2 CH3 + 2 Cl~~~~}$  this looks OK

$\ce{2 Cl + 2 CH3 -> 2 CH3Cl~~~~~~~~~~~}$ this looks OK

$\ce{2 CH3Cl + 2 CH3Cl -> 1 C2H6 + 1 Cl2~~~~}$ this should be $\ce{1 CH3Cl + 1 CH3Cl -> 1 C2H6 + 1 Cl2}$

$\ce{2 HCl -> 2 H2 + 2 Cl2~~~~~~~}$ this should be $\ce{2 HCl -> 1 H2 + 1 Cl2}$

I'm not sure I agree with all of your chemical reactions, but at least now you don't wind up with more chlorine then you started with.


Can MCM-41 (mesoporous materials) be used for ion exchange?

Ion-exchange resins need to have charged groups that can bind your ions of interest. If a certain ion-exchanger will work for your problem depends on what kind of ions you want you to exchange or separate. The pores must be larger than your molecules of interest. E.g. polystyrene resins which are fine for water desalination (to remove e.g. calcium and magnesium ions) won't work well for separating proteins, as their pores are too small for biomolecules to enter the beads.