Is it possible to chemically treat grass to make it digestible?

The indigestible component of grasses is cellulose, which is polymerized glucose.

Yes, it is possible to hydrolyze the cellulose to glucose, which is of course digestible.

The hydrolysis can be catalyzed enzymatically with a cellulase, or using acid hydrolysis.

The process of hydrolyzing cellulose is often discussed in the context of so called cellulosic ethanol, as the glucose is then fermented to ethanol subsequent to hydrolysis.


Reaction of cyclohexane-1,2,4-triol with acetone

As long as you have a proton source in your solution (hint: all the time) these acetals will equilibrate back and forth until they find the thermodynamic hole. So you will never observe a hemiketal like the one you drew; rather, it will always move forward with the elimination of water (there is only little water around if acetone is the solvent, so liberation of water is entropically favourable) to the formation of the $\ce{O{,}O}$-ketal.

Note that you can form two different ketals if the stereochemistry of the hydroxy groups is correct: If they are 1,3-syn, the six-membered cyclic ketal can also be formed.

Which one you end up with will likely depend solely on the reaction conditions. Kinetically, the five-membered ring is favoured. Thermodynamically, I assume the six-membered ring to be more stable, if it is able to find a conformation with little steric stress.


You should treat everything as ideal gas. Then you calculate the volume change from the air itself, using ideal gas law and pressure being constant (atmospheric, maybe). Calculate also the number of moles of air in the syringe, it will come handy soon.

The remaining change in volume is then attributed to the solvent, again using ideal gas law, you put in the given volume, temperature, pressure and obtain number of moles, which turned into vapor.
Having this, you know number of moles of solvent and number of moles of air, so again assuming everything ideal (Raoult's law), you know the partial pressures (i.e. vapor pressure).

For added fancines, you can take the decrease of the liquid volume into account.

Side note: While using ideal gas and ideal mixture of ideal gases approximation, all quantities are nicely additive, so don't be afraid.


Reactivity of alkaline earth metals towards water

Do not forget solubility effects involving the hydroxide product.  Magnesium is plenty electropositive enough to decompose water, only to passivate with a hydroxide that has limited solubility (you can detect the presence of some reaction by decanting the liquid and then adding phenolphthalein, or by carefully looking for a few hydrogen bubbles during the reaction).  Put magnesium in methanol, which is less protic than water; the methoxide is less passivating and you can actually bring the alcohol to a boil!  See https://www.google.com/url?sa=t&source=web&rct=j&url=https://m.youtube.com/watch%3Fv%3DNMfs3e9OdZQ&ved=0ahUKEwjlnLOxpPDWAhVDKiYKHVFwB5wQwqsBCCgwAA&usg=AOvVaw2TaAZjVq0m8EHL-JluhxGn.


Can we determine the presence of carbon by simply heating or charring the substance?

Not all carbon-containing chemical char. For example, sodium carbonate, $\ce{Na2CO3}$ just produces $\ce{CO2}$. Sodium carbonate might be viewed as inorganic, though.

An organic chemical that behaves in a similar way is hexamethylenetetramine, or hexamine, $\ce{(CH2)6N4}$. On heating, it decomposes primarily into formaldehyde, $\ce{H-CHO}$, and ammonia, $\ce{NH3}$, though there are traces of other gases such as $\ce{HCN}$ and $\ce{CO}$. For that reason, hexamine is used as a smokeless fuel (no char left behind, as the saying goes).


What is the product when Lithium solid is dropped in water?

I think the major error you made is applying what you call the "Metal + Water = Metal Oxide" rule.  This may apply to some metals, but not all.  Specifically, Lithium is a Group 1 metal, the so-called Alkali Metals.  This means that Group 1 metals tend not to form oxides with water, but rather form hydroxides with water.  That is, the metal ions bond with hydroxide ion, $\ce{OH-}$, which has a -1 charge.  If you have ever seen this reaction (lots of YouTube videos if you have not done the reaction yourself), you will see that lithium in water produces lots of gas.  And in some videos, you will see that this gas can be set on fire!  Given the elements involved, that's a good indication that the gas is hydrogen, a very flammable gas.

So the most likely equation for lithium in water forms an hydroxide of lithium along with hydrogen gas.  Is that enough information for you to propose a different equation than what you answered?


Entropy and Equilibrium Constant: Chemical Thermodynamics

I agree it's not a duplicate because, while you are asking about the same homework problem discussed in the other question you linked, the issue you are having is different.  Hence, because the last answer didn't address your specific confusion, you are asking a different question about that same homework problem.

Here's where you've got things mixed up:  Whether a reaction is exothermic or endothermic depends on the inherent nature of the reactants and products, not the equilibrium constant.  I.e., $\Delta H^{o}_{rxn}$ is the enthalpy change for a complete conversion of reactants to products at standard state, irrespective of the equilibrium constant.  It's not, as I believe you were thinking, the enthalpy change to go from pure reactants to the equilibrium mixture. Hence if you could, say, change $K_{eq}$ purely by tweaking $\Delta S^{o}_{rxn}$, then $\Delta H^{o}_{rxn}$ wouldn't change just because you've changed $K_{eq}$.

So B is correct.   If you raise T (and here we're assuming $\Delta H^{o}_{rxn}$ is independent of T), and $\Delta H^{o}_{rxn} > 0$, the reason the reaction shifts to the right is this: In calculating $\Delta S^{o}_{surr}$, we have to divide $q_{rev, surr} = q_{surr}$ by T, and  T is now higher.  Hence, for the same heat flow from the surroundings (which is given by $-\Delta H^{o}_{rxn}$), the resulting unfavorable entropy change in the surroundings is less.

[I've used standard state here, which means it's carried out at a constant external pressure of 1 bar, but what I've written applies at any constant external pressure.]


Is there an alternative wet chemical method or electrochemical method to obtain precipitation of iron, nickel and silicon powder?

Both metal powders can be preapred electrochemically.

Iron dust is prepared using solution containing $\ce{FeCl3}$ with addition of $\ce{AlCl3}$ to make the solution more conductive. The process occurs at elevated temperature 60°C. Unfortunately exact process parameters are hard to find.  But following graph could help.

Nickel could be prepared electrochemically too and can be prepared by dissolving of Ni-Al alloy in hydroxide -Raney_nickel. The description of patent is here.

This is link to a study book PRÁŠKOVÁ METALURGIE, K. Skotnicová, M. Kursa
, unfortunately only in czech.


What is thermal plasma in-flight oxidation technique?

Original paper where this method (and, I guess, the term itself) has been introduced was written in Japanese in 1982 [1]. In 1997 there was a great overview published in English by Toyonobu Yoshida, author of the original paper, with a chapter (2.3) devoted to this particular method:

  Processes used to manufacture materials by injecting a solid, liquid, or gas into a thermal plasma to use its high enthalpy and reactivity are referred to as in-flight plasma processes (IFP processes).

Due to the nature of the method which utilizes plasma jet with reactants generated via various torch-types, there is actually a flow of chemicals reacting while they "fly", of course, in the plasma area.

Various plasma torches provide various times-of-flight, temperatures and velocities. For example, the direct current jet torch has the following characteristics:

  An argon plasma of about $\pu{80 kW}$ that has been generated in air reportedly produces a temperature of about $\pu{11600 K}$ and a velocity of about $\pu{410 m/s}$... The reaction time for reactants injected near the anode at high temperatures is about $\pu{400 \mu s}$ and the gaseous raw materials are
  mainly used for the synthesis of fine powders... The reaction time of the reactants is estimated to be $\pu{1 ms}$.

Bibliography

Yoshida, T.; Akashi, K. Tetsu-to-Hagane 1982, 68 (10), 1498–1502. DOI 10.2355/tetsutohagane1955.68.10_1498.
Ultra-fine particles: exploratory science and technology; Hayashi, C., Uyeda, R., Tasaki, A., Eds.; Materials science and process technology series; Noyes Publications: Westwood, N.J, 1997. ISBN 978-0-8155-1404-6.


Strictly speaking you might have a point there. But it is probably more difficult to judge.

I was unable to unearth an official recommendation by the IUPAC, so I unfortunately have resort to Wikipedia ([1] Kristi Lew (2009). Acids and Bases. Essential Chemistry. New York: Chelsea House Publishing.):

  Homogeneous is a term in physical chemistry and material science that refers to substances and mixtures which are in a single phase. This is in contrast to a substance that is heterogeneous.[1] The definition of homogeneous strongly depends on the context used. Generally it refers to smooth variation of properties with no discontinuities or jumps.

If we talk about all the air around the world, it is certainly not uniform. There are concentration differences across certain areas. When we consider a vector into space, the total concentration will decrease along this vector. As we know, we will pass certain areas with different chemical compositions, for example the ozone layer.
But all these changes will be gradually; there are no discontinuities. In this sense one would call air homogeneous, even though it does not fit in the definition of phase.

Additionally, if you take an arbitrary (volume) unit of air, all at the same temperature and pressure, and you allow it to establish an equilibrated state, then yes, this will not only be homogeneous, it will also form one phase.
The above linked article further considers air as a homogeneous mixture. It rambles on with considering it a gaseous solution of oxygen, carbon dioxide, etc. in nitrogen. This is technically incorrect, since the gaseous state of matter is specifically excluded from the IUPAC definition of solution. Also a solution can never be at the same time a mixture, as the former can be separated by mechanical means, while the latter cannot.

The problem here is obviously the lack of a unique and rigorous definition for homogeneity. For whatever purpose when it really matters, one should probably use a more robust definition like phase, which is essentially what your second statement describes.

Depending on the context you might even obtain a homogeneous mixture of water and oil - the question is then more or less how stable such a suspension is. After a while it will certainly separate in two phases.


Effects of evaporating the solvent of a supersaturated solution

Two caveats:

If you sufficiently disturb a supersaturated solution, the solute will spontaneously come out solution.  So you would need to heat it without sufficiently disturbing it.

The solubilities of some solutes decrease with temperature.  So let's assume your solute's solubility increases with temperature (as is the case for most, but not all, solids dissolved in a liquid).

So if we allow for the above then, as you heat the solution, you will eventually reach at point at which it is no longer supersaturated.
Indeed, the reverse of this is how supersaturated solutions are typically prepared:  A solution is prepared at a high temperature with a concentration of dissolved solid just below the saturation point (you don't want any undissolved solid left, since that will act as a nucleation site and thus prevent supersaturation).  Then it is carefully cooled.  So, by heating it sufficiently, you would simply be moving it back towards the starting point.
Finally, someone might argue that, if the solution is supersaturated near the boiling point, it might not be possible to heat it up sufficiently to get below the saturation concentration.  I suppose that's possible in theory, but I don't know how you would prepare such a system, since it couldn't be created using the typical cooling process.


How dangerous is an industrial scale styrene spill?

According to this safety data sheet (SDS), styrene monomer has a Category 3 respiratory classification, where 1 would be most serious. Category 3 is defined "substances produce transient (short duration or temporary) target organ effects."

Add to that these hazard, precautions and carcinogenicity statements:

H335: May cause respiratory irritation.
H372: Causes damage to organs through prolonged or repeated exposure if inhaled.
P260 Do not breathe dust/fume/gas/mist/vapor/spray.
NTP Reasonably anticipated to be a human carcinogen

Read the cited SDS or others for statement on environmental precautions and cleanup. Frankly, they're a bit vague to me, "If the product contaminates rivers and lakes or drains inform respective authorities."


How can off-diagonal Hamiltonian matrix elements be nonzero when the overlap matrix is diagonal?

I believe the confusion lies in some ambiguity in the use of the term overlap.

For the $S_{AB}=0$ approximation of Hückel theory, overlap is used in the specific mathematical sense of the "overlap integral":

$$
S_{AB} = \left<\varphi_A|\varphi_B\right> =
\int_{\vec r}{\varphi_A^*\!\left(\vec r\right)\varphi_B\!\left(\vec r\right)\mathrm{d}\vec r}
$$

However, this does not mean that at each point in $\mathbb{R}^3$ only one of $\varphi_A$ or $\varphi_B$ is allowed to bear a non-zero value.  In fact, at most points in $\mathbb R^3$, $\varphi_A$ and $\varphi_B$ will both bear non-zero values. Since $\varphi_A$ and $\varphi_B$ will in general have "co-located regions of non-zero values," the two functions can be described qualitatively as overlapping, in the same sense as in your second quote. It is the overlap of the functions in this more general, qualitative sense, that allows $H_{AB}=\left<\varphi_A^*\middle|\hat H\middle|\varphi_B\right>\neq0$.

(To note,  in order for Hückel's approximation to be exact, the orbitals would have to be selected such that the specific integral $S_{AB}$ over all $\mathbb R^3$ equals zero$^\ddagger$. I believe Hückel did not actually do this, but instead just coerced to zero any nonzero $S_{AB}$.)

To help explain more explicitly why this is, I'll first write out the integral implied by the Dirac notation for $H_{AB}$:

$$
H_{AB}=\left<\varphi_A^*\middle|\hat H\middle|\varphi_B\right>=
\int_{\vec r}{\varphi_A^*\!\left(\vec r\right)\hat H\!\left[\varphi_B\!\left(\vec r\right)\right]\mathrm{d}\vec r}
$$

Then, I'll introduce two (real-valued) one-dimensional functions defined on the closed domain $x=[0,1]$ to use for demonstration purposes:

$$
\varphi_A\!\left(x\right) = \sin\!\left(2\pi x\right) \\
\varphi_B\!\left(x\right) = \sin\!\left(3\pi x\right)
$$

$\hspace{10mm}$

It is straightforward to show that $S_{AB}=0$ for these two functions.  One way is to plug the overlap integral into Wolfram Alpha. Another option is to do it by hand.  In the below figure, the green curve is a plot of the product $\varphi_A\!\left(x\right)\cdot\varphi_B\!\left(x\right)$, and the orange curve is a plot of the cumulative integral of this product:

$$
\int_0^x{\varphi_A\!\left(x'\right)\cdot\varphi_B\!\left(x'\right)\mathrm{d}x'}
$$

$\hspace{10mm}$

As can be seen, the integral over the entire domain is zero, satisfying $S_{AB}=0$.

Now, let's introduce a simple "Hamiltonian"$^\dagger$:

$$
\hat H = x
$$

The next figure is a replicate of the first one above, with $\hat H\!\left[\varphi_B\right]$ also plotted as the dashed red curve:

$\hspace{10mm}$

As in the second figure above, one last plot shows the local value of $\varphi_A\!\left(x\right)\hat H\!\left[\varphi_B\!\left(x\right)\right]$ along with the cumulative integral:

$$
\int_0^x{\varphi_A\!\left(x'\right)\cdot\left[x'\cdot\varphi_B\!\left(x'\right)\right]\mathrm{d}x'}
$$

$\hspace{10mm}$

The value of $H_{AB}$ here is clearly not equal to zero.  (Wolfram Alpha concurs.) Thus, it is absolutely possible to have $H_{AB} \neq S_{AB} = 0$.

To note, the principle exhibited here on a closed domain in $\mathbb R^1$ applies equally well to orbitals on an infinite domain in $\mathbb R^3$. Note also, if I were actually performing a quantum mechanical calculation here, I would have been a lot more careful about normalizing my wavefunctions (viz., ensuring $S_{ii}=1$).

$^\ddagger$Numerous methods are known for generating orthogonalized orbitals with $S_{AB}=0$ strictly, from a general set of orbitals where $S_{AB}\neq0$ occurs for some non-null subset of the $\left(\!A,\!B\right)$. To my knowledge, the most commonly used orthogonalization for atomic orbitals is Löwdin's method (see this PDF). The Gram-Schmidt orthogonalization (usually in its "modified" form) is used in a variety of other areas of quantum chemistry. Wikipedia also has a pretty solid article on orthogonalization methods in general.

$^\dagger$You'd never use this as a Hamiltonian operator. Further, I'm not sure if it technically even is a Hamiltonian operator. (I think it'd work as a dipole moment operator?) But, whatever, leave me alone, it's for purposes of illustration.


All strong acids dissociate in water: $\ce{HX + H2O -> H3O+ + X-}$.
Metals reacts with hydronium ions, not with a particular acid, e.g. like $\ce{M(s) + 2 H3O+ -> M^2+ + H2(g) + H2O}$.
( With nitric acid, it is more complicated due its reduction to $\ce{NO}$ or $\ce{NO2}$, but it is not important now. )
Metallic cations and acid anions are not paired with a particular counter ion, until they start precipitating. They all form together a "ionic soup".
As a rule of thumb, the guidance is the salt solubility. But in case of mixed anions and/or cations and non stoichiometric cation/anion ratios, more important is solubility product
$$K_\mathrm{sp}=[\ce{M^n+}]^{m} \cdot [\ce{X^m-}]^{n}$$
The salts that would reach their solubility product would start precipitate,
but then hard-to-control kinetic factors of crystallization initiation and propagation take part.
The excess of the counter ion decreases solubility of the salt, compared to salt solubility in water. It is similar effect as gravimetric precipitation of very little soluble salts, where excess of the counter ion causes much lower ion residual concentration, compared to being just in water.
Other ions may alternate salt solubility by side reactions  ( if applies ) with the considered ions, or by affecting activity coeeficients - see below.

For the case 1, sulphates would have the minor effect of decreasing somewhat the activity conefficients that therefore somewhat increased solubility of the phosphate. But it is general aspect of ions presence.
For the case 2: It depends on if there is formed a double salt like LiCdPO4 or not.

In particular specific cases, there is possibility of forming double salts, where precipitate consists of integrated more than 1 cation or anion. E.g. crystallization from solution containing $\ce{Fe^2+, NH4+, SO4^2-}$ may lead to crystallization of "Mohr salt" $\ce{(NH4)2Fe^{II}(SO4)2 \cdot 6 H2O}$. Similarly for alums with $\ce{K}$ and $\ce{Al/Cr/Fe^{III}}$ as  $\ce{M^{I}M^{III}(SO4)2 \cdot 12 H2O}$.

During rapid crystallization there occurs also coprecipitation:

There are three main mechanisms of coprecipitation: inclusion, occlusion, and adsorption. An inclusion occurs when the impurity occupies a lattice site in the crystal structure of the carrier, resulting in a crystallographic defect; this can happen when the ionic radius and charge of the impurity are similar to those of the carrier. An adsorbate is an impurity that is weakly bound (adsorbed) to the surface of the precipitate. An occlusion occurs when an adsorbed impurity gets physically trapped inside the crystal as it grows.

In the case of rather concentrated solutions of well soluble salts, there is complication we should consider the ion activities  instead:
$$K_\mathrm{sp}=a_{\ce{M^n+}}^{m} \cdot a_{\ce{X^m-}}^{n}$$
Acticity coefficients for diluted solutions can be calculated by Debye-Huckel theory, but for concentrated solutions it is hard to theoretically predict them. We can use concentrations for rough predictions, but the numbers may not fit well.


Covalent and ionic bonds are best thought of as opposite ends of a spectrum, where electrons are shared evenly (as in the case of a symmetric molecule like N2) or unevenly (like HF).

That being said, bonds with more ionic character are generally considered stronger due to the significant electostatic contribution to bonding (the cationic and anionic components experience attraction).

This principle is the core concept of Pauling Electronegativities, where the degree to which electrons in a bond are "attracted" to one atom over the other (that is, the degree to which a bond is ionic) is ultimately derived from experimental bond dissociation energies (i.e. bond strengths).


Difference between voltaic cell in chemistry and battery in physics

I think you've confused the flow of electrons with the flow of current. Before it was discovered that electricity is the result of the flow of negatively charged electrons it was proposed that electricity result from the flow of a positive charge. We called this current and this convention has remained to this day.

So current is effectively the flow of positive charge not the flow of electrons, which are negatively charged. So current flows from the cathode to the anode, but electrons flow from the anode to the cathode. This is true in both physics and chemistry.


Alcohol or Carbonyl - Which is more reactive towards PCl5?

Reaction of alcohols with PCl5 and PCl3 (for reaction with alcohol)
Mechanism for conversion of ketone to dichloride with phosphorus pentachloride
(for reaction with ketone)
Now as per the question as it is not given in excess so it will react with any one of the alcohol or ketone. It's quite hard to tell which one of them will react first because both of them show totally different kinds of reactions. Like ketones prefer neucleophile attack and alcohols prefer substitution and elimination.  But if you look at the mechanism quite closely you would find all the favourability of the reaction depends on the lone pair donation tendency of oxygen.  Now as you know, in sp³ hybridization state the s -character is nearly 25% and in sp² hybridization state s-character is nearly 33% and as s-character increases electronegativity increases
Why a higher s character increases a carbon atom's electronegativity?
and the tendency to hold the electron pair increases so for ketone it is quite difficult to donate it's electron pair to PCl5. So in that perspective I can say alcohol is quite more reactive than ketone in this case.


Cahn-Ingold Prelog rules concerning tie breaking on second level

look, all i can say is that $ D > C > A $

why ?

Lets compare A and C
$ OCH$3 and $ H $ are common to both. one has CH3 and the other has CH2F. of course, C > A.

compare A and D in the same way and find that D > A

compare D and C now.
there is clear proof that D > C because it has F in two places while C has H in those two exact positions.
thus it is seen that D > C > A

now see B and C
in B : after O we have CH3 and after C we have { H , H , F }
in C : after O we have H and after C we have { H , H , O }

looking at second layer after oxygens : B > C
looking at second layer after carbons : C > B.
thus even i am confused when it comes to B
but definitely D > C > A


Difference between intercalated lithium and metallic lithium (Li-ion batteries)

Intercalation is the addition of lithium ions into a host material without significantly changing the host's structure. In my research, I will often prepare test cells where the cathode is an intercalation material, like $\ce{CoO2}$, which can accomodate (or "intercalate") layers of lithium ions between the oxygen layers in the crystal lattice to form $\ce{LiCoO2}$. It's not that there are layers of cathode material surrounding lithium, but rather there's only 1 material, but lithium ions can move into and out of it.

$$ \ce{CoO2 + Li+ + e^- -> LiCoO2} $$

I mostly use lithium metal for the anode because it's easy to work with and has a stable potential, making it a good reference electrode:

$$ \ce{Li^+ + e^- -> Li_{(s)}} $$

In this setup we have an intercalation cathode ($\ce{LiCoO2}$), and a lithium metal anode. The major downside is that when you cycle lithium metal, it forms dendrites which can puncture the separator and short-circuit the cell, leading to catastrophic failure. To solve this problem, commercial cells use graphite anodes, which can intercalate lithium ions:

$$ \ce{C6 + Li+ + e- -> LiC6} $$

This solves the dendrite problem, but makes the battery less efficient. Since dendrite formation is only a problem with repeated cycling, non-rechargeable ("primary") batteries just use lithium metal instead. Lots of researchers have tried to solve "the dendrite problem" for decades, with little success.

The diagram below shows a lithium ion battery with an intercalation cathode on the right: the green parts are lithium, the blue parts are transition metals (Ni, Co, etc.) and the red spheres are oxygen. The anode on the left is the graphite intercalation anode, but you could replace it with a metallic lithium anode.


what is the basic factor which determines the order of ionization energies of K,Li,Ca,S?

Looked up ionization energies (kJ/mol) from Wikipedia

K  = 418.8
Li = 520.2
Ca = 589.8
S = 999.6

so the book answer is correct...

Ok, here is best hand waving explanation that I can give. The general trend...

Li and K are both alkali metals. As the atomic number goes up the shells get further away from the nucleus and the inner electrons shield the valence electron. So in order of increasing ionization energy

Li > Na > K > Rb > Cs > Fr

So to start we easily get Li > K.

Ca and K are in the same period in the periodic table. Generally as you go up in atomic number in a period the ionization energy will increase. So Ca > K.

Sulfur is a non-metal in the upper right hand of the periodic table so it should have the highest ionization energy.

So we're down to either S > Ca > Li > K or S > Li > Ca > K.

  To me giving either of these answers wouldn't be embarrassing.

If you go to the Bohr model of the hydrogen atom then energy of the $n$th shell is

$\mathrm{E}_n  \propto \dfrac{Z^2}{n^2} $

If we assume that the 2s and 3s electrons are perfectly shielded by all the electrons in the lower shells, then

$ \mathrm{E_{Li}} \propto \dfrac{1}{2^2} = 0.25$

The two Ca 3s electrons being equal the charge is 2

$ \mathrm{E_{Ca}} \propto \dfrac{2^2}{4^2} = 0.25$

So I would have guessed S > Li > Ca > K

Now I know that is wrong. I know that the "problem" is with the shielding of the electrons. Below is a section of the periodic table with ionization energies listed.

I just don't know how to explain the "right" answer without hand waving about shielding.

I don't know how you'd rationalize all the relationships between the group 1 and 2 elements.

If you assume the Bohr model as noted above then you end up with this which correctly predicts the trend down a group but overestimates shielding of the inner electrons.


Can I perform electroplating with sodium carbonate and copper?

You can try the electrodeposition in a solution prepared by mixing baking soda with vinegar and the copper sulfate. Wear glove and protection glasses. The aluminium must be cleaned before plating. Please let me know the result.

Acidic copper plating on aluminium will cause evolution of hydrogen, so less acidic solution is better. On the other hand you need to have the solution conductive and the baking soda could help.


What commonly availible salts are strong dessicants?

Have you looked into silica gel or zeolites? Both are non-toxic, and thus likely obtainable by private persons. Zeolites are used in self-refrigerating beer kegs.


How do I find the final temperature of a calorimeter using specific heat capacities?

All you did is essentially right, your only mistake is in the last step, as LDC3 already pointed out in the comments. However, I am encouraging you to use units all the way and when dealing with thermodynamics use Kelvin instead of Celsius.
\begin{align}
Q &= mc\Delta T\\
\end{align}
Now you can form the equations for each of the problem, while substituting $\Delta T$ with a temperature range, being $x$ the final temperature the whole system will end up on. Also note, that the iron will be cooled down, while the water will be heated. (I am using a different approach than you.
\begin{align}
Q_\mathrm{loss} &= m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}[T(\ce{Fe})-x]\\
Q_\mathrm{gain} &= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}[x-T(\ce{H2O})]\\
\end{align}

The transferred heat has to equal $$Q_\mathrm{gain} = Q_\mathrm{loss}$$

With this you can solve for $x$.
\begin{align}
m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}[T(\ce{Fe})-x] &= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}[x-T(\ce{H2O})]\\
%m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe})-m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}x &= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}x-m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})\\
% m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})&= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}x+m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}x\\
% m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})&= [m(\ce{H2O})\cdot{}c(\ce{H2O})+m(\ce{Fe})\cdot{}c(\ce{Fe})]\cdot{}x\\
x &=\frac{m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})}{ m(\ce{H2O})\cdot{}c(\ce{H2O})+m(\ce{Fe})\cdot{}c(\ce{Fe})}\\
%
x &=\frac{30~\mathrm{g}\cdot{}0.449~\mathrm{\frac{J}{gK}}\cdot{}417~\mathrm{K} +40~\mathrm{g}\cdot{}4.184~\mathrm{\frac{J}{gK}}\cdot{}293~\mathrm{K}}{ 40~\mathrm{g}\cdot{}4.184~\mathrm{\frac{J}{gK}}+30~\mathrm{g}\cdot{}0.449~\mathrm{\frac{J}{gK}}}\\
x &= \frac{5616.99~\mathrm{J}+49036.48~\mathrm{J}}{167.36~\mathrm{\frac{J}{K}}+13.47~\mathrm{\frac{J}{K}}}\\
x &= \frac{54653.47}{180.83}~\mathrm{K} =302.24~\mathrm{K}\\
x &\approx 29~\mathrm{^\circ{}C}
\end{align}


Is this chemical name wrong? 2-bromobutane vs 1-bromobutane

The numbering of locants in the question is correct. In acyclic hydrocarbons, the longest chain is numbered from one end to the other.

The corresponding subsection of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-15.1.7.1.4 The longest chain is numbered from one end to the other by arabic numbers, the direction being chosen so as to give the lower locants to the substituent groups (side chains) [see P-14.4(f)]. (…)


What is θ in Bent's rule equation (%s - character)?

TL;DR: $\theta$ is the angle between any two equivalent bonds and using  Coulson's theorem for anything beyond the second period is out of its scope.

First things first, Coulson's theorem (the generalization of said formula) cannot be applied in cases of spndm. Therefore using $\ce{PCl5}$ as a comparison is not the right idea. Also, the formula that you've stated is only when finding the bond angle between two equivalent bonds as would be proved below.
This equation is supposedly a simplification of Coulson's theorem that states,
$$1 +\lambda_1\lambda_2\cos(\theta) = 0$$
Here, $\lambda_1$ stands for the hybridization index of one bonded orbital, $\lambda_2$ denotes the hybridization index of the second bonded orbital and $\theta$ signifies the bond angle between the two orbitals. However, this rule as said above can only be applied to spn hybridisation compounds where $\sqrt{n}$ is the hybridisation index. Its formal proof is given in Wikipedia if you are curious and want to find out.
The simplified idea behind Coulson's theorem (as I understand it), is as follows.
Each set of equivalent bonds is made up of a linear combination between s and p orbitals. This is what hybridization is. Given any pair of bonds, we can find the bond angle between the two through their hybridization indexes ($\lambda_1$ and $\lambda_2$) and vice versa. Let's take the example of $\ce{CH4}$,
$\ce{CH4}$ has four equivalent $\ce{C-H}$ bonds and we know that the bond angle is $109.5^\circ$and so according to Coulson's theorem, the hybridisation of the $\ce{C=H}$ bond can be found to be,
$$1+\lambda_\ce{C-H}\cdot\lambda_\ce{C-H}\cos(\theta_\ce{H-C-H}) = 0 $$
$$1 +\lambda_\ce{C-H}^2\cos(\theta_\ce{H-C-H})  = 0 $$
\begin{align}
\therefore \lambda_\ce{C-H}^2 &= \frac{-1}{\cos(109.5^\circ)} \\
&\approx 3
\end{align}
Therefore, the $\ce{C-H}$ bonds are sp3 in nature.

Proof of $\cos \theta = \frac{s}{s-1}$
From Coulson's theorem, we have,
$$1 +\lambda_1\lambda_2\cos(\theta) = 0$$
Now, if both bonds are equivalent in nature, we get $\lambda_1 = \lambda_2 = \lambda$, therefore,
$$1 + \lambda^2\cos(\theta) = 0$$
This implies,
$$\cos \theta = \frac{-1}{\lambda^2}$$
Now for such a molecule the hybridisation would be sp$\lambda^2$ and so percentage s-character would be
$$\text{%s character} = \frac{1}{1 + \lambda^2}$$
Solving for $\lambda^2$, we get:
$$\lambda^2 = \frac{1-s}{s}$$
Substituting this in the equation for $\cos \theta$, we get:
$$\cos \theta = \frac{s}{s-1}$$
Note: An assumption that I took here is that the bonds are equivalent. If this is in fact true, then this rule can only be used to find the bond angle between any two equivalent bonds and hence in the case of $\ce{PCl5}$, only the axial-axial ($\ce{P-Cl}$) bonds or equatorial-equatorial $\ce{(P-Cl)}$ bonds can be compared. θ then is the angle between any two equivalent bonds.


$n$-Factor is basically a method to find out a relationship between the compound and what it is equivalent to in terms of acidic nature or basic nature. Note that the two hydrogen atoms in $\ce{H2SO4}$ are both attached to oxygen. But in $\ce{H3PO3}$ two of the three hydrogens are attached to oxygen and the other hydrogen is directly attached to phosphorous.

Note that only those hydrogens attached to highly electronegative elements can be donated easily in case of inorganic acids. One could explain better by using the concept of stabler structures with the $\ce{H}$ removed from $\ce{O}$ than from $\ce{P}$ by resonance. But for now know it as a thumb rule.

Therefore, since $\ce{H3PO3}$ has only two $\ce{H}$ attached to $\ce{O},$ its $n$-factor is 2 and not 3.


I think something like potassium 2-methylpropan-2-olate and 3-bromopentane in $\ce{SO(CH3)2}$ should do the trick.


Creating an efficient synthetic route to this elastomeric block copolymer

How about starting with a difunctional isocyanate, adding hydrazine in step 1 and polyethylene glycol in step 2. I would call it an addition polymerization. Your structure does not show the number of repeating units, but that could be largely controlled by varying the monomer proportions.


The "fishy" odor that you're familiar with is brought about by a whole bunch of compounds, and not any single one. Then again, if we were to narrow this down a bit, we could say that simple nitrogen compounds are the main culprits.

But suppose we want to blame only a single compound for the delightfully pungent odor of rotting fish, and we couldn't be bothered with the nuances of olfactory physiology or food chemistry to go through a list of compounds running into the hundreds (if not thousands). In that case, we can pin the blame on trimethylamine (TMA).

With regard to fish, the TMA is a result of the degradation of trimethylamine oxide (TMAO) present in the body fluids (blood, lymph, excreta) of the fish.

Just so you don't get the wrong idea, TMAO is present in the body fluids of all vertebrates ̶I̶ ̶c̶a̶n̶ ̶t̶h̶i̶n̶k̶ ̶o̶f̶, not just fish (albeit in much lower concentrations, which is why you don't mistake ground chicken for tuna).

Now why is TMAO more abundant in fish?

Ever tried to pickle a cucumber (or anything for that matter)? Notice that the cucumber shrivels up over time? It's losing water to the pickling fluid by a process known as exosmosis: the water is drawn out from a less concentrated environment (cucumber's tissue) to a more concentrated one (pickling fluid).

 Disclaimer: As user @lly pointed out in the comments, this particular bit has the potential to cause a bit of confusion. The inside of a cell (the cytoplasm I mean) is technically an aqueous colloidal solution (the pickling fluid is an aqueous solution as well). In the context of aqueous solutions, "concentration" is an indication of how much solute is present, not water. So in this context, if something is more concentrated, it has less water in it

Now the same thing is going to want to happen to a fish in salt water, which isn't so nice from the fish's perspective.

To counter the outflow of water, a fish has a significantly higher concentration of the familiar metabolite urea in its body fluids (as compared to terrestrial vertebrates). This establishes a situation where the fish's body fluids have roughly the same "concentration" as the salt water it swims in (the more correct term here would be "osmolarity", in place of "concentration")

That urea, if left to its own devices, will seek to destabilize  macromolecular structures and inhibit cellular functions; the TMAO present alongside the urea helps counteract those detrimental effects.

Of course, going by this logic, saltwater fish should have a much stronger "fish odor" than freshwater fish (which is certainly the case).

Lets not forget the classic kitchen tip: Rubbing fish with something acidic like vinegar, b̶a̶t̶t̶e̶r̶y̶ ̶a̶c̶i̶d̶ or lemon juice helps reduce the odor (the TMA degrades under acidic conditions).

Also, some of you might know that a fish-like smell is observed in rancid food. That, apparently, is the result of the oxidation of omega-3 fatty acids.


Which allotropes of sulfur exist naturally?

The alpha and gamma form of cyclo-S8 are the only forms to occur naturally. Besides the reference provided in the question, you can also find more information in the wikipedia article of "sulfur allotropes".


How might one reduce the concentration of bubbles in their ice, from ice cube trays?

IIRC, the solubility of virtually all gases in water decreases with temperature. This is the reason why heating water will result in bubble nucleation, and this can be exploited to produce very clear ice by boiling water, perhaps several times, before putting it into the trays.


Nuclear zero-point energy and quantum delocalization effects of water molecules

So, what was actually said in the paper1 with respect to AIMD simulations of water is the following:

  Nuclear zero-point energy and quantum delocalization effects are neglected making classical water overbound.

To understand the meaning of this statement we first need to know how AIMD works. Ab Initio Molecular Dynamics (AIMD) is a molecular dynamics simulation in which potential energy surfaces for nuclear motion are obtained from first principles (ab initio) rather than represented by a force field. The quantum chemical calculation of potential energy surfaces are done in the Born–Oppenheimer approximation in which nuclei are considered to be stationary classical particles.

As its very name signifies, it is, of course, just an approximation. It ignores that nuclei move and aren't perfectly localized in space. The nuclear motion can be approximately separated into vibrational and rotational the first of which energetically dominates the second. Taking the vibrational motion on the nuclei into account results in some additional so-called zero-point vibrational energy that rises the total energy. Other quantum effects (tunneling, delocalization) can be also taking into account for the nuclei which results in some further changes in energy.

Altogether, the more and more quantum effects are taking into account, the closer are the results of the simulations to the physical reality. AIMD simulations ignore few quantum effects, and thus, do not quite match the reality. For instance, water tends to be overbound in such simulations comparing to what we have in the real world. That is, essentially, the meaning of the statement quoted above.

1) A.D. Hammerich, V. Buch, F. Mohamed, Ab initio simulations of sulfuric acid solutions, Chemical Physics Letters, 2008, 460, 423-431, http://dx.doi.org/10.1016/j.cplett.2008.06.053.


I used to work in an environmental water testing laboratory.  The regulatory standard method for testing for lead in water uses an Atomic Absorption (AA) spectrophotometer.

It seems that the issue with Flint's water appears to be that the new source of water is corrosive, the water has not been treated for corrosion, and pipes in Flint are commonly made out of lead.  So if your pipes at home are made of lead and are corroding, you may have a problem that the lab at the water treatment plant was not able to test for.

If you are curious about checking your water at home, you can obtain a water testing kit from a local hardware store.  These kits come with test strips that allow you to check your water for things like nitrates, pH, alkalinity, hardness etc.  Make sure your kit allows you to test for lead.


Can Isopropyl β-D-1-thiogalactopyranoside be degraded into either 1-deoxy-galactose or just galactose?

I'm not sure if I understood your question correctly, and my thought is based on this assumed reaction:

Your indication of a resolution of about 1.8 Å reads like the data were collected by Cu K$\alpha$ radiation, rather than by Mo K$\alpha$.  If sulfur was split off from your molecule, than (even more with softer Cu than Mo X-rays) this should have decreased a lot the contribution of anomalous scattering in the X-ray data, as CHO's atomic scattering factors -- in comparsion -- are mainly described by $f'$, and not so much about $f''$. What about looking on value .and. esd of the Flack parameter of your model for your new product, compared to a "typical / reference" model of IPTG?
(By the way, Howard Flack passed away just at the beginning of this year.)


What surface treatment can turn nickel black?

The article provided by Karl has answered the question with electroplating method, which is not preferred but this is the only answer I have.

The method is based on MIL-P-18317. The electroplating solution consists of $\ce{NiSO4, Ni(NH4)2(SO4)2, ZnSO4}$ and $\ce{NaSCN}$

The black nickle coating are $\ce{NiS}$ and $\ce{ZnS}$.
$\ce{NiS}$ and $\ce{ZnS}$ are deposited by $\ce{M^2+ + H2S -> MS + 2 H+}$.
$\ce{H2S}$ is generated by $\ce{SCN- + 3H+ + 2e- -> H2S + HCN}$

note: Is reducing thiocyanate a standard method to generate sulfide in electroplating ? I know soluble sulfide salts smells horrible, but it is worth the trouble ? not mentioning the cyanide byproduct, in acidic solution.

Metal $\ce{Ni}$ and $\ce{Zn}$ are also deposited $\ce{M(NH3)2^2+ + 2e- -> M + 2NH3}$.
The role of $\ce{NH4+}$ is to generate $\ce{M(NH3)2^2+}$ prior to metal deposition.


Does the van der Waals force not affect the alkali metal 's trend of decreasing melting points down the table?

Because, in the condensed phases, you don't have molecules of metallic elements.  You have, as a first approximation, a "rice pudding" structure where cations are embedded (like raisins) in a "pudding" of delocalized valence electrons (metallic bonding).  The main cohesive force in the condensed metals is then electrostatic attraction between the opposing charges of the cations and electrons.  In alkali metals where each atom breaks up into the same charge cation (+1 charge) and produces the same number of delocalized electrons (one per atom), this electrostatic attraction goes down with larger atoms.


Do electron donating groups make a molecule nucleophilic?

Sometimes.

Sure, electron donating groups can make a molecule more nucleophilic by virtue of increasing the negative charge density on the nucleophilic atom.

However, nucleophilicity is a kinetic property, and having a lot of electron donating groups attached can significantly affect the ability of the nucleophile to approach the electrophile. This reduces nucleophilicity. An example would be tert-butoxide. Yes, oxygen with a negative charge is generally nucleophilic, and yes, the tert-butyl group is an inductive donor, but tert-butoxide is not very nucleophilic because of steric hinderance.

  One group, the alkyl group in this example, is electron donating in $\ce{CH3CH2O-}$. Will the molecule be nucleophlic?

Maybe. Nucleophilicity also depends on the solvent. A charged ion such as ethoxide ion would be well-solvated in a polar solvent, such as water, and less well-solvated in a non-polar solvent, such as hexane.

Solvation impacts reaction kinetics, and in turn, impacts nucleophilicity. Consider the halides for example. In polar protic solvents, iodide ion is the strongest nucleophile. Protic solvents are not able to solvate iodide ion as well as they can solvate the fluoride ion, which has a much higher negative charge density. In polar aprotic solvents, however, fluoride ion is generally a stronger nucleophile than iodide ion because fluoride ion is not as strongly solvated and it also has a great negative charge density.

In sum, nucleophilicity is not an inherent property. It depends on a variety of factors - both external and internal to whatever molecule or ion that you are considering.


Determining the value and units of the rate constant

Alright, so in the reaction

$$\ce{2NOBr(g) -> 2NO(g) + Br2}$$

the rate law is

$$\text{rate} = k[\ce{NOBr}]^x$$

where $k$ is some constant and $x$ is the order of the reaction in respect to $\ce{NOBr}$.

By seeing how the initial rate changes when we change the concentration of $\ce{NOBr}$, we can determine the value of $x$. We can use any two of the three. I'm going to use the first and third trials. If we divide them we get

$$\frac{\text{rate 3}}{\text{rate 1}} = \frac{k[\ce{NOBr}]_3^x}{k[\ce{NOBr}]_1^x}$$

$$\frac{2.43\times10^2}{1.08\times10^2} = \frac{0.900^x}{0.600^x}$$

The $k$'s cancel out.

$$2.25 = 1.5^x$$

$$x = 2$$

The rate is second order in respect to $\ce{NOBr}$, and the rate law is written $\text{rate} = k[\ce{NOBr}]^2$. If you double the concentration, the rate will quadruple.

$$\text{rate before doubling concentration} = k[\ce{NOBr}]^2$$

$$\begin{align}\text{rate after concentration} &= \left(2[\ce{NOBr}]\right)^2\\
&= 2^2[\ce{NOBr}]^2\\
&= 4[\ce{NOBr}]^2\\
&= 4 \times \text{rate before doubling concentration}\end{align}$$

A tripling of the concentration will increase the rate by a factor of nine, a quadrupling of the concentration increases the rate by a factor of 16, and so on.


What is the maximum covalency of oxygen?

We know the maximum covalent is at least 3 with such species as  $\ce{H3O+}$.  Also the pyrylium ion (https://en.m.wikipedia.org/wiki/Pyrylium_salt) has an oxygen covalency of 3 by forming a pi bond for the third bond.
Can we get 4-coordination?  As mentioned by @Rob, yes.   @Rob's comment states that basic beryllium acetate appears to have 4-coordinate, covalently bound oxygen.
And it check out.  Basic beryllium acetate, $\ce {Be4O(C2H3O2)6}$, has the four-coordinate oxygen in the center, then four beryllium atoms coordinated to this oxygen at the vertices of a tetrahedron, then an acetate group bridging each edge of the tetrahedron.  In addition to beryllium, zinc forms a similarly structured basic acetate, which can be obtained by heating normal zinc acetate (https://en.wikipedia.org/wiki/Zinc_acetate).
Might the central oxygen be mostly ionically bonded?  Magnesium and heavier alkaline earth metals, which form more strongly ionic bonds with oxygen, are not known to form such a basic acetate structure.  This corroborates the hypothesis that covalent bonding of the central, four-coordinate oxygen is involved.


Can electrons switch orbitals within a shell?

The short answer to this is: according to our current understanding of quantum mechanics, we can't tell.

And now for the long answer:

First off, if you haven't already seen Richard Feynman's 60-second primer to the scientific method, you should definitely watch it now. It's short. I promise I won't go anywhere while you're watching it.

...

Alright. So the way science works is to test ideas by experiment. This means that in order to distinguish between which of two ideas is right, we'll need to come up with some difference in their behavior. This can sometimes be problematic in science because the difference in behavior between two models can be incredibly small. To take just one example, it wasn't until the 20th century that an apparent error in the orbit of Mercury was confirmed, which was eventually explained by relativity. Before that, if you had given someone the theory of relativity and the Newtonian theory of gravity and asked them to tell you which one was right, they would have shrugged their shoulders. There simply wasn't enough high-quality evidence to decide between one or the other.

On the other hand, what if there's a nightmare situation in which we simply can't tell the difference no matter how good our instruments are? Imagine that I give you a small cube which lights up when you shake it, but I've used magic so that no matter what experiments you try to use to figure out what's inside of the cube, you can't get any information. You can try to use x-rays to image the inside, or weigh it, or spin it around to see if the insides are unevenly distributed, or look at it with an IR camera...but every single time, the magic renders your results useless.

Now suppose I asked you how the box worked. Without being able to get any info, you're stuck! No matter how clever you are in making your guess or how many guesses you come up with, you can't really say anything about any of them, because you're being magically prevented from getting information about what's in the box.

This is a very important idea, so I want to make sure you really understand it. Without some sort of actual behavioral difference that can be looked at, science has no way of distinguishing between alternate ideas.1

Now, quantum mechanics is sometimes a lot like our magic box (to pretty much anyone who tries to study it, it is a magic box). Our current understanding of QM says that we have this wavefunction that tells us what the system is doing. We can't actually look at the wavefunction. The best we can do is to use some operators on it, which will give us the values of certain properties (don't worry if this all sounds very mysterious, it takes years of study to understand some of this stuff).

So how can we see if the wavefunction matches what we think it is? Well, we can look at what we should get from the operators, and what we actually do get from measuring it. For instance, by applying something called the Hamiltonian operator to the wavefunction, we get the total energy of the system. We can then measure the total energy of the system and see if theory and experiment agree. We can do that a whole bunch of times, measuring the momentum, and kinetic energy, and the average position of the system. If we do it enough times, we can say "eh, it's probably correct" and then move on to the next problem.

Now, to get back to your original question: can electrons ever switch places? Well, our current understanding of quantum mechanics is that, when electrons switch places, the wavefunction changes sign.2 "Aha!" I hear you say. "So we can watch the wavefunction and see if it changes sign, and if we ever see it flip, we can say that the electrons switched places!"

Well...not really. Because we can't look at the wavefunction directly, we can only measure things through the operators. And here's where the nasty trick comes into play: for all operators that we can actually observe, there won't be any change if the wavefunction flips its sign.

We're stuck! Just like our magic box, we can't get any meaningful information about what's going on inside, and so we can't really tell if the electrons switched places. Even if they did, it would look exactly the same to us (no matter how good our instruments are, we cannot see a difference, because there literally isn't one). It's an undecidable problem!

Now, if you're anything like me, this is a really crummy and disappointing answer. It took me a long time to accept the fact that there are some things in science that we just can't decide. So let's get a little creative: what would have to change in science for us to be able to tell?

First, and most directly, we would need an alternate theory to QM that allows us to distinguish between the two cases. Seeing how incredibly complex current theories are, I doubt that will happen anytime soon (though I could be pleasantly surprised). The second is that we would need some way to measure the wavefunction directly, instead of looking at it through operators. I've heard whispers of groups making progress on that front, though I honestly don't know if there's any truth to that or not. One way or the other, I don't expect we'll be able to answer this question tomorrow, or anytime relatively soon.

Some time back, I heard about a very clever paper that had to deal with this problem. There was an effect that could either depend on the temperature or density of water. Of course, you can't control the temperature and density of water independently (the stuff expands when you heat it up), and so everyone had just assumed that temperature was responsible. This group realized that, if you supercool water, it takes on densities similar to those found above 0C. By carefully supercooling the water and measuring this effect, they found that at similar densities of water, the effect was the exact same, even when the temperature differed by over 50C. It was density that was responsible all along!
This is one of the most mysterious things about all of physics. Even after two years of studying quantum mechanics, I have yet to find anyone who can offer a satisfying explanation for why fermions (half-integer spin particles) should flip their wavefunction upon exchange of (ostensibly identical) particles. If you know of a good read on the subject, please put it in the comments and I'll add it here!


How does H's ionization energy relate to its transition energy (Bohr's Model)?

Ionization energy of some arbitrary atom is not related whatsoever to its transition energy. Also, there are many different transitions (even in a hydrogen atom), each with its own specific energy. It is just that this particular transition of this particular atom happens to have the energy of 3/4 its ionization energy, one of the few things the Bohr model managed to get right. You can't generalize that to other atoms and other transitions.


Phase rule for a pure substance in vapor-liquid region

Note that the phase rule is qualitative. It says nothing about the particular amounts. It simply enumerates the theoretical number of independent parameters needed to determine amounts and compositions for the given number of components and phases.
The definition of a system, like total amounts of components, does not count among degrees of freedom.
The number of degrees of freedom of a system is the number of independent parameters, that are needed for the given system to determine its state.
For the given container volume and the substance amount, a given temperature determines the equilibrium pressure and vice versa. So there is 1 degree of freedom.
Amounts of particular  phases are determined by the temperature as the  single degree of freedom.


How to calculate the percentage dissociation of phosgene using the pressure equilibrium constant?

Here goes a possible answer, I tried to describe it step by step so you can see your (and my) possible errors. The first part of your calculation was conceptually good, but you make the calculation wrong.

As one of the comments say, $$K_p=\frac{(342-254)*(352-254)}{254}=33.95$$

So you have $K_p$. Then

\begin{array}{lcccc}
 & \ce{CO &+& Cl_2 &<=> &COCl_2}\\
\text{Initial pressure of substance} &0 && 0 && A\\
\text{Final pressure of substance}   & x && x && A-x\\~\\
\end{array}

With the condition that $x + x + A-x=760$, we have $A+x=760$.

Using our first calculation:

$$K_p=\frac{(x)*(x)}{A-x}=33.95$$
(Kp is the same at same $T$)

$$K_p=\frac{x^2}{760-2x}=33.95$$
here we get x=130.229, then:
dissociation of $A= 1-\frac{(A-x)}{A}$
we get dissociation of A=1-0793=0.207

or 20.7% (0.01% flew away)


Which are the Geometrical isomers of the complex [Cr(NH3)2Cl2(en)]+?Is it one or two?

You have an octahedral complex where two of the adjacent positions are occupied by the ethylenediamine ligand. Let's make these equatorial for a clearer picture.

There are only two positions in the remaining four positions that are trans (C and D). You can choose to put two chlorides there, two ammines there, or one of each. Those are the 3 geometric isomers.
If the trans ligands are the same, then there is a plane of symmetry that goes through C-M-D and bisects the ethylene diamine ligand, so the complex will not be chiral. But if the two ligands are different in positions C and D, then there will not be a plane of symmetry and the complex will be chiral. You can draw them out to see what they are.
To summarize, there are three geometric isomers, one of which is chiral. If you have trouble with them types of problems, sometimes, it's helpful to just consider the cis/trans or mer/fac relationships between groups of ligands.


What is considered room temperature in Celsius?

Considering you've read Wikipedia article on room temperature (r.t.), you probably already have an idea that there is no uniformly defined value due to the subjective matter the term is defined upon.
There are numerous standards across industries defining temperature ranges, most of which lead to the value for r.t. of $\pu{(20 ± 5) °C}.$

In natural sciences there are several well-defined standard conditions describing the combination of exact values for temperature, pressure and humidity.
Room temperature is not among these, so ideally, an exact value must be specified alongside, if relevant, e.g. "the parameter has been determined at r.t. $(\pu{20 °C})$".

German Wikipedia article presents an observation that in physical chemistry and statistical physics in European region the typical value taken for r.t. is $\pu{20 °C},$ whereas in U.S.A. the majority of textbooks refer to r.t. as to $\pu{25 °C}.$

In nuclear chemistry, thermal neutron energy $E_0 = \pu{25.26 meV}$ is fixed as characteristic energy at r.t., which is a reference point for tabulated parameters such as neutron cross section.
Using $E = k_\mathrm{B}T$ relation, an exact value for r.t. can be determined, which is usually taken as $\pu{20.43 °C}$ (see e.g. [1, p.  11-173] and references therein).

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.


1,3- and 1,4- shifts don't really happen. It's just too far away. If you have a double bond at 4 or 5 carbons away, maybe you can do a ring closure or form a non-classical carbocation. The cation you've drawn could however undergo a sequence hydride shifts. Each hydride shift would move the cation closer to the methyl group, until the 3rd shift places the cation on the tertiary carbon. You'd need to out-compete any kind of elimination or cation reaction with a nucleophile...


large wavelength range and list of x/y numerical data points of elements

I don't know about the IR spectra of metals. Certainly metals will have vibrational spectra, but I'm sure where to find it (if NIST does not have).

I can tell you how to get x/y data from NIST, if they have the spectrum.

For example, for benzoic acid. At the bottom of the page, there is a link to download the sepctrum in JCAMP-DX format. The .JDX file format is really a space-delimited text file containing x/y data with some markup at the front for the JCAMP plugin.

See all that lovely x/y data? Force the .JDX file open in your favorite text editor, like notepad. Do not use a word processor - they will add additional markup. Do not use a spreadsheet program yet. Spreadsheets can open .JDX files, but they won't always choose the most helpful delineation. Once open in notebook, resave the file as a text (ascii) file. Make sure you type in the .txt to change the extension (alternately, you can just change the extension to .txt using any other means you may feel comfortable with).

Now... open your favorite spreadsheet program, and use it to open your shiny new text file. You may have to tell the program to look for "all files" (MS Excel is bad about this). You should get some dialog/options on how to open the file. Choose "Delimited" if its an option, and when it comes time to choose your delimiters, make sure your choice includes "space":

Now you should have a file with the x/y data. The first column is wavenumber (at least for IR) and the second is transmission ($\frac{I_{\text{out}}}{I_{\text{in}}}$). Some of these files have more than one column of y-data. When I graphed each column from the benzoic acid file, I got the same spectrum (almost).


Is there any reason why halogenation of substituted alkanes wouldn't occur through free radical substitution?

In principle, this reaction should be possible. However, the intermediately formed 2-hydroxy-2-propyl radical is not stabilized by the OH substituent, as the alternative mesomeric structure with C=OH bond and unpaired electron at O would violate the octet rule. 2-bromo-2-propanol should still be the major product, but there are also a lot of possible side products like multiply brominated compounds. Other substituents which stabilize an adjacent radical (by conjugation, for example) will increase the selectivity of the reaction. The possibility of reaction of the bromine radicals with the functional groups and sensitivity of the substrate to the reaction conditions (UV radiation etc.) also need to be considered. For the selective halogenation of certain substrates, often more suitable reagents than $\ce{Br2}/ h\nu$ are available, for example, NBS/AIBN for allylic bromination.


Is activation energy temperature-independent?

There is a simple (some might say simplistic) way to get an intuition about this and it involves thinking at the molecular or atomic level rather than about the bulk properties of the reaction (the thermodynamic view).

For a simple reaction where molecule A has to bang into molecule B to create molecule C, the reaction will only happen if the amount of energy involved in the collision is large enough. The amount of energy involved depends on the nature of the reaction which is determined by the electronic structures of A and B. Within limits, those electronic structures don't vary with temperature and a given reaction will only occur when there is enough energy in the collision for the reaction to happen.

What changes with temperature isn't the energy required for the reaction to occur, it is the number of molecules that have enough energy to overcome the barrier. At higher temperatures more molecules have that energy, but the amount of energy required doesn't change. The amount of energy require to initiate the reaction is roughly equivalent to the activation energy. It can be calculated by looking at the way the overall rate of the reaction changes with temperature. So the rate of reaction changes with temperature but not the barrier to the reaction happening.

This view is a little simplified but it gives the right intuition about the problem.


Why does basicity go down along the group 15 hydrides?

The very idea of $\ce{sp^3}$ hybridization you are referring to is pretty much not there, starting from $\ce{PH3}$. Look at its bond angles, they are almost exactly $90^\circ$. This means that phosphorus contributes to the $\ce{P-H}$ bonds almost exclusively with its $\ce p$ orbitals, and the lone pair just sits there hidden on $\ce s$ orbital, rather than stick out as $\ce{sp^3}$.

The same is even more true for the following hydrides.

Also, with nitrogen being the most electronegative of all, its partial negative charge in $\ce{NH3}$ attracts $\ce{H+}$, which is not the case for $\ce{PH3}$ and the rest.


Why are MD (Molecular Dynamics) simulations performed at room temperature for the purposes of studying proteins?

Another reason why people don't bother to use 310K versus 300K is because any time you are using MD, you have an approximate force field describing the dynamics of the system which means you have an approximate phase diagram for the system. When comparing MD and experiment, it is a reasonable question if you really want to compare properties at the same temperature or compare properties at "equivalent" points on the phase diagram. As an example, some models of water which are very widely used have a phase diagram which is drastically different from the real phase diagram of water (i.e. they boil at 400K and other weird things) and you will see studies which describe the phase diagram of DFT water or TIP4P water so that you can make comparisons based on the temperature and on the phase diagram.

So, this is to say, if you submit a paper where you do MD at 300K when studying an enzyme in solution, it is unlikely you will get complaints that this is a biomolecule so it should be at 310K because even comparing to properties of the actual enzyme at 310K isn't a direct comparison since you aren't at the same locations on the respective phase diagrams. What could be more relevant is to make comparisons to the real world across a range of temperatures so that you can see if the trends are the same rather than just a single point.

Another less physical reason is that people have already been doing simulations at 300K, so you can't really get complaints about a simulation at 300K since everyone has already been using 300K. This is actually very common in science.


Current prices are about 34 € (500 g) for 1-bromobutane and 32 € (500 mL) for 1-butanol in decent lab quality.

The most cost efficient method is ordering 1-bromobutane from a chemical supplier of your choice!

On a technical scale, one would probably react 1-butene with anhydrous hydrogen bromide under radical conditions, using peroxides or UV light to initiate homolytic cleavage of the $\ce{H-Br}$ bond.


Which is the cheapest plastic that retains its structure at 85 °C or higher temperature?

The melting point is not going to give you the answer you want.  To properly select a polymer you need to look at the glass transition temperature $(T_\mathrm{g})$.  This is the temperature at which a material loses crystallinity and starts to more easily flow.  This parameter varies widely with thermal history and internal stress of a polymer material and can be raised by annealing.

Based on UV resistance, I would exclude any olefin or vinyl polymers as well as ABS.

For water resistance, I would exclude cellulose polymers and acetals.

Based on the temperature requirement I would say polystyrene ($T_\mathrm{g} = \pu{100^\circ C}$), acrylics $(T_\mathrm{g} = 85 - \pu{165^\circ C})$, polycarbonate ($T_\mathrm{g} = \pu{135^\circ C}$), or PPE/PPO ($T_\mathrm{g} = \pu{215^\circ C}$) have sufficiently high $T_\mathrm{g}$'s to remain as acceptable for the mentioned specifications.


Why does bond length decrease with increased basis set quality for SCF methods?

I think that your rationale is right. I would think it in terms of energy instead to make a little more precise. Thinking about a diatomic molecule, the nuclei-nuclei repulsion function of distance does not change with the basis set used. So the difference is strictly in terms of electronic part. At large distance the atoms behave more similar to an isolated atom, so few spherical harmonics are enough (for a relative description). As atoms get closer it is needed more flexibility to represent the very different probability density that there is in the space region located between atoms. With a smaller basis set this wave function is approximated poorer. If so, it is expected that the energy rises in such region if one consider the variational principle.

Here I calculated for the oxygen molecule in the region close to equilibrium distance (about 1.2 A) the difference in electronic energy between HF/STO-3G and HF/6-31* basis set:

The behavior close to 1.4A should be just an artifact in convergence (I did not take care about that).

Sadly I never found a discussion about that in the literature. Up to my understanding there is not to much that can be said about that.


Copper Sulfate Pentahydrate experiment, help please?

Frankly you're going about this all wrong. First you want to exploit the difference in solubility of the salt in hot water versus cold water.

Copper sulfate's solubility is 32 g/100ml water at 20 C, and 114 g/100ml water at 100 C.

2-3 drops of sulfuric acid is all that is needed. You're just trying to prevent the formation of $\ce{Cu(OH)2}$ or $\ce{CuCO3}$.

  I'm not sure about battery acid. I assume that you're using new battery acid, not actually from an old battery. So the sulfuric acid will be a little less concentrated. Use 5-6 drops instead of 2 or 3.

After you make a saturated solution of the copper sulfate (aim for 100g /100ml water) in hot water, let it cool slowly and don't stir at all. Stirring will promote nucleation of many small crystals.

To get a really large crystal you'd make a saturated solution at room temperature, filter it, and add one of these seed crystals. You'd then let the water evaporate over weeks very slowly.


These two sources both put bone at a hardness of 5:

http://www.chacha.com/question/how-hard-is-bone-according-to-moh's-hardness-scale

https://answers.yahoo.com/question/index?qid=20110310200841AABwtMj

Whether they are trustworthy is questionable though, so take it as you will.

This source put brass at 3 and iron at 4.5:

http://www.jewelrynotes.com/the-mohs-scale-of-hardness-for-metals-why-it-is-important/

and this image puts brass at 4 and iron at 4-5 (Similar to 4.5):
http://patentimages.storage.googleapis.com/WO2001048807A1/imgf000009_0001.png

While these different sources seem to have conflicting data, I think it would be safe to assume that Brass is the softest of these three materials, Iron comes second, and Bone is the hardest.

Edit: In the description of that monster, the adjective used is 'strong'. You may want to consider how much force each of these materials can withstand instead of how hard they each are :)


Is there a broader nomenclature convention to explain why the ‘a’ in ‘chlorophyll a’ is italicized?

This is only a partial answer to your question.

The suffixes a and b to the name chlorophyll were most likely introduced by Richard Willstätter and Arthur Stoll. They are widely used in their book UNTERSUCHUNGEN ÜBER CHLOROPHYLL (Julius Springer, Berlin, 1913). Here, however, the suffixes are not yet italicized and appear as normal text.


What is the difference between the KLMN and SPDF methods of finding electronic configuration?

The KLMN(OP) method is based on electron shells, with the labels KLMN(OP) being derived from an experiment in which the spectroscopist wanted to leave room for lower energy transitions in case there were any.

K denotes the first shell (or energy level), L the second shell, M, the third shell, and so on. In other words, the KLMN(OP) notation only indicates the number of electrons an atom has with each principal quantum number ($n$).

The SPDF notation subdivides each shell into its subshells. For further information about the other quantum numbers, especially $l$, which defines the subshell, see the accepted answer to this question: What do the quantum numbers actually signify?

  The K shell can hold two electrons: $n=1,\ l=0$

When $l=0$, we have an s subshell, which has one orbital $m_l=0$, with room for two electrons.

  The L shell can hold 8 elections: $n=2,\ l=0,1$

When $l=1$, we have a p subshell, which has three orbitals $m_l=-1,0,+1$, with room for 6 electrons. The L shell also has an s subshell.

  The M shell can hold 18 electrons $n=3,\ l=0,1,2$

When $l=2$, we have a d subshell, which has 5 orbitals $m_l=-2,-1,0,+1,+2$, with room for 10 electrons. The M shell also has s and p subshells.

  The N shell can hold 32 electrons! $n=4,\ l=0,1,2,3$

When $l=3$, we have an f subshell, which has 7 orbitals $m_l=-3,-2,-1,0,+1,+2,+3$, with room for 14 electrons. The N shell also has s, p, and d subshells.

This table here summarizes the relationship between the two notations for all elements. Here is an example for scandium, $\ce{Sc}$:

Scandium has 21 electrons ($Z=21$). Its electron configuration in KLMN notation is

$$\begin{array}{cccc}
\mathrm{K} & \mathrm{L} & \mathrm{M} & \mathrm{N} \\
2 & 8 & 9 & 2
\end{array}$$

Its SPDF notation (based on the aufbau principle) is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2}$.


Old question, but I thought I'd add to it:

In your example, the first undissolved solid is indeed gold. It's not easily dissolved by $\ce{HNO3}$ alone, but requires a mixture of $\ce{HNO3}$ and $\ce{HCl}$ (aqua regia). $\ce{Ag}$ and $\ce{Cu}$ would dissolve by hot $\ce{HNO3}$. The precipitate formed by addition of $\ce{HCl}$ is indeed $\ce{AgCl}$, so you're on the right track.

A brief list from experience:
[At room temperature in aqueous solution]

$\bullet \ce{Au2S}$: insoluble (quite sure this is the precipitate formed in basic solution when sodium metabisulfite is added to $\ce{HAuCl4}$. Either that or $\ce{Au2S3}$).

$\bullet \ce{AuCl3}$: highly soluble (can be formed by dehydrating and evaporating $\ce{HCl}$ from $\ce{HAuCl4}$, which is the product of $\ce{Au}$ dissolved in aqua regia).

$\bullet \ce{KAu(CN)4}$: soluble

$\bullet \ce{KAu(CN)2}$: soluble

$\bullet \ce{AgCl}$: insoluble

$\bullet \ce{CuCl2}$: highly soluble

$\bullet \ce{CuCl}$: soluble, but harder to synthesize than $\ce{CuCl2}$

$\bullet \ce{(Au/Ag/Cu)(NO3)_x}$: soluble ($x = 1,2$)

I think the overall result here is that gold, silver, and copper salts generally follow one of the primary trends of solubility: that is, that $\ce{NO3-}$ salts are always soluble regardless of cation, and $\ce{Cl-}$ are usually soluble except in cases $\ce{Ag+}$, $\ce{Pb^{2+}}$ and $\ce{Hg+}$.


There is sodium and potassium in wood.  When the wood burns some of the sodium and potassium become bound to oxygen.  The oxides become hydroxides when exposed to water.


Ok. Finally found some time to look up references.

The first piece of glassware is a bubbler. See:
https://www.chemglass.com/product_view.asp?pnr=AF-0513

You use this to help keep positive pressure in your inert atmosphere system. You need to fill it with mineral oil or mercury and you can vent your sealed system though this. This allows you to keep piping in say nitrogen or argon and keep positive pressure without blowing anything up.

The second piece of glassware is a fritted filter used for inert atmosphere chemistry. See:
https://www.chemglass.com/product_view.asp?pnr=AF-0540

The side arm allows you to equilize the pressure between the two sides (otherwise the liquid may not flow from top to bottom). There are ground glass joints on both ends so that you can create a sealed system free from water or oxygen.


Diffuse functions for Ahlrichs/Karlsruhe basis sets in Gaussian?

If they are not in Gaussian yet, there won't be keyword to include these diffuse functions, but you can add them manually to your jobs.

You can find the basis set specification for def2-TZVPD for the atoms you are interested from the Basis Set Exchange. From there, you can either just insert the diffuse functions after your molecule specifications using the keyword Extrabasis or just insert the whole basis manually using the gen keyword where you would normally put the name of the basis set.


How can I find the stoichiometry of a reaction from the volumes of the reactants and products given?

In a problem like this I think that the best way to answer the question is to set up the equations and try all the answers. However, in this case we can infer from the text that $B$ need to be the limiting reacting, as all of it reacts but some of $A$ is left unreacted. The only answers providing this are answer b and answer d. However, the latter contains a minus sign which doesn't makes sense chemically, so let's focus on the former.

We can compute the moles involved in the reaction using the ideal gas equation $PV = nRT$. Let's assume the volume is $1 \mathrm{L}$, we can calculate the amount of moles for both $A$ and $B$ (as they have the same volume and are present in the same conditions):

$$n = \dfrac{PV}{RT} = 40.88 \mathrm{mmol}$$

If the stoichiometry is the one proposed in answer b ($\ce{A + 2B -> 2C}$)
then the whole quantitity of $B$ will react with half the quantity of $A$ to give the same amount of moles of compound $C$, which will be $n = 40.88 \mathrm{mmol}$. Half of $A$ will be left unreacted, which is $n = 20.44 \mathrm{mmol}$. Using the aforementioned equation we can now compute their volumes:

$$V = \dfrac{nRT}{P}$$

Which is $V_a = 1 \mathrm{L}$ and $V_c = 2 \mathrm{L}$, which in total is $V_{tot} = 3 \mathrm{L}$, exactly three times more than the original volume.


How can you inhibit crystallisation of hot superconcentrated magnesium sulphate

First, your question involves inorganic chemistry, not organic, unless you're talking about the unmentioned gelling agent.

Second: Impurities serve as a source of nucleation, and stirring increases crystallization. If you leave the supersaturated solution in a clean container and allow it to cool slowly, e.g. in a Styrofoam cooler chest, it may take a long time to form crystals. If you use some thickener, such as PEG or gelatin, it might take longer yet. Left undisturbed, though, supersaturated $\ce{MgSO4}$ can persist for years. There are videos of it crystallizing after a seed is added.


How will my studies in mathematical statistics be applied to and assist me with problems in chemistry?

Most definitely yes.

I took multiple statistics classes as an undergraduate and they have helped me considerably. (I can argue fairly that it helped me get tenure.)

A few suggestions from my own work:

Statistical design of multivariate experiments: Many scientists think you should test one variable at a time. No. I taught a grad student (in another group) about optimization theory and proper design and three months later, a great paper came out.
Data mining and exploration: Many chemistry data sets are extremely complicated, with cross-correlated factors. Understanding how to draw empirical inferences, find the most important factors, etc. is enormously helpful across a wide range of chemistry.
Principal component analysis: An emerging area in spectroscopy and imaging is taking complicated spectra and performing PCA to help with quantifying chemical components in mixtures and following everything from cancer growth to materials design.
Multivariate regression and ANOVA: My group has created numerous quick statistical screens using multivariate surrogates for complex simulations or experiments.
Cross-validated analysis: Again, the understanding of training and test sets, bootstrapping, etc. is very useful.

Now, you also mention SDEs and Markov processes. These are crucial in chemical kinetics, including areas like reaction-diffusion systems. Monte Carlo methods are also highly used in studying dynamical processes.

The list would be enormous. Basically, I think it's a great skill to have and you'll use it throughout your career.


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

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

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


Can single molecules of C and O2 react in isolation, and if so how will momentum be conserved?

$\ce{C + O2}$ is awfully complicated, so let's just pretend you've asked this:
In a single act of the reaction $\ce{H. + H .-> H2}$, how is momentum conserved?
That's a legitimate concern all right. After all, we are taught that this reaction does happen instantly, once given a chance, and that's in fact true. Also, we know that it releases a lot of heat. Now, heat is nothing but the motion of molecules; how does energy convert to the motion of one molecule as a result of one single reaction act?
It doesn't. The conservation of momentum forbids that, just as you reasoned. Chemical reactions are collective phenomena. Nobody cares about a single molecule.
Now what really happens to a single molecule which has just formed as a result of the mentioned reaction? That's really simple: the molecule is vibrating wildly, ready to break apart. Is has just enough energy to do so. It will do so half of the times, or maybe more often. But that doesn't matter. What matters is that sometimes the vibrating molecule will hit another molecule and sent both of them flying away in opposite directions, thus releasing a part of its energy and becoming more or less stable.
An emission of a photon is also an option, but that's another story.
So it goes.


How to name molecules with a hydroxyl group in a side chain?

The proposed name 2-(hydroxymethyl)propane-1,3-diol is correct.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) includes three different methods for systematic names of alcohols. The preferred names are formed using the suffix ‘ol’ and (if necessary) the prefix ‘hydroxy’.

  P-63.1.2 Systematic names of alcohols, phenols, enols, and ynols

  Hydroxy compounds are named in three ways:

  (1) substitutively, using the suffix ‘ol’ and the prefix ‘hydroxy’. The presence of several ‘ol’ characteristic groups is denoted by the numerical multiplying prefixes ‘di’, ‘tri’, etc.; the final letter ‘a’ in a multiplying prefix is elided before the suffix ‘ol’. Rule P-44 is applied when a principal chain or a senior ring system must be chosen. When there is a choice for numbering, the starting point and the direction of numbering of a compound are chosen so as to give lowest locants to the ‘ol’ suffixes;

  (2) by functional class nomenclature and the class term ‘alcohol’;

  (3) as assemblies of identical units by multiplicative nomenclature when the conditions for its use are fulfilled (see P-51.3)

  Method (1) generates preferred IUPAC names. Names as assemblies of identical units, method (3), are preferred to those that are formed by simple substitution (P-51.1.5).

The first step is the selection of the principal chain, which will serve as the parent structure. The relevant general criteria for the choice of a principal chain in such compounds are:

greater number of groups that are expressed as suffix
longest chain
lower locants for groups that are expressed as suffix

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.

  The senior ring, ring system, or principal chain:

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

For the compound that is given in the question, the principal characteristic group is $\ce{-OH}$, which is expressed as suffix ‘ol’. Whichever way you look at it, the principal chain can only have two of these groups. Therefore, the principal chain is a propane; the parent structure including the substituents corresponding to the principal characteristic group is propane-1,3-diol.

The compound that is given in the question actually has a branched chain structure. Therefore, the next step is the identification and naming of side chains, which will be treated as prefixes. In this case, the resulting name is 2-methylpropane-1,3-diol.

The third $\ce{-OH}$ group cannot be expressed as suffix ‘ol’ since it is not situated on the principal chain. Therefore, it has to be indicated by the prefix ‘hydroxy’.

  P-63.1.4 Substitutive nomenclature, prefix mode

  Hydroxy groups are indicated by the prefix ‘hydroxy’ when:

  (1) a group having priority for citation as the principal characteristic group is present; or

  (2) a hydroxy group cannot be denoted by a suffix.

In this case, the hydroxy group is placed on the methyl side chain, which results in the compound substituent group ‘hydroxymethyl’.

  P-29.4 COMPOUND SUBSTITUENT GROUPS

  P-29.4.1 Compound substituted substituent groups

  A compound substituted substituent group is formed by substituting one or more simple substituents into another simple substituent that is considered as the principal chain. The choice of the principal chain is fully discussed in Section P-44.3. The first criterion to be applied is that the principal chain is the longest chain; it will be applied in the following examples of acyclic compound substituent groups.

  (…)

Parentheses are used around such compound prefixes in order to avoid ambiguity.

  P-16.5.1 Parentheses (also called curves or round brackets)

  P-16.5.1.1 Parentheses are used around compound (see P-29.1.2) and complex (see P-29.1.3) prefixes; after the multiplicative prefixes ‘bis’, ‘tris’, etc.; to enclose a multiplied parent compound consisting of a parent hydride and a substituent suffix or a multiplied substituent prefix even though preceded by a basic numerical prefix, such as ‘di’, ‘tri’, etc.

Therefore, the complete systematic name for the compound that is given in the question is 2-(hydroxymethyl)propane-1,3-diol.


Relative strength of intermolecular forces

I would say that your analysis of first, second and fourth one is correct, as H-bonds are stronger than dipole-dipole interactions. However, you made a mistake while filtering compounds having dipole-dipole interaction. The $1$ and $3$, both have dipoles!
The $\ce{C=O}$ bonds in $\ce{CO2}$ are polar and hence they possess dipoles, but unlike a water molecule, $\ce{CO2}$ shape is linear, and therefore the equal dipoles cancel each other, resulting in a non-polar molecule as seen in the following figure,

That's the reason $\ce{CO2}$ possess dipole-dipole interactions whereas $\ce{F2}$ doesn't as it has only one bond which is non-polar!
Image taken from courses.lumenlearning.com: Polar Molecules


Why does N-phenylhydroxylamine give Fehling's test?

Quoting from "REDUCTION OF AROMATIC NITRO COMPOUNDS TO N-PHENYLHYDROXYLAMINES - A Comparative Study of the Yields Obtained when
Aromatic Nitro Compounds are Reduced, under Similar Conditions, to N-Phenylhydroxyiaminee by William Henry James"

  N-Phenylhydroxylamine is unstable. It is readily affected by light,
  especially white light. It is also affected by moist air, being
  readily oxidized to nitrosobenzene. It has strong reducing power,
  reducing almost quantitatively Fehling's Reagent to cuprous oxide, and
  causing the formation of a silver mirror when warmed with an
  aramoniacal solution of silver oxide.

So, I guess that its instability and its reducing power are the cause of the effect explained.

Reference

James, William Henry. Reduction of aromatic nitro compounds to N-phenylhydroxylamines. Thesis (M.A.), Boston University, 1932. Boston University Libraries OpenBU


What is the physical state of a single atom?

This answer has been written so it hopefully can be understood by people that do not have a degree. If something is not correct, let me know, but be aware that throwing around fancy words will not help anyone understanding this any better. Consider the other posts here concerning liquids and solids, I don't want to be redundant and repeat what has been said already.

The answer is a bit complicated. One would intuitively say no, but that doesn't tell the whole story.

Take the interstellar medium. We have about 1 atom per cm³ in there and it is called a "gas". Now that's what I would call an isolated atom in the gas phase.

An atom is an object that belongs into the realms of quantum mechanics. A solid, liquid or gas is something that belongs into our classical world. If you zoom in close enough, that means if we look at a problem microscopically, words like "solid" or "liquid" no longer make sense. If you now begin to zoom out, the quantum effects vanish. This is sometimes called quantum decoherence and we enter the world of classical physics where solids and and liquids exist again.

So I would argue it all depends on how closely you look at the problem. Isolated atoms in a large volume? A gas if you ask me. But if you have to zoom in to talk about an individual atom within a larger compound consisting of many more atoms, let's say one atom within a piece of iron, you no longer can say that this single atom is "solid".

Think of it like this:

Can a single person have a political system?

Yes, why not. I can proclaim my own nation and be the only person in it. But mostly, a nation is something that many people form together. If you look at one individual person, you are in the realm of psychology (or biology). If you look at an entire state, you are in the realm of (for example) politics. If you zoom in at one individual person, you can no longer say that person has a certain political system. But if you zoom out far enough, you suddenly get a political system. Zooming out of course is a bit more abstract here, but, as I said, this is the realm of psychology/politics ...


How does Fermi's golden rule yield a specific rate?

I'm fairly sure the states $\require{\begingroup} \begingroup \newcommand{\ket}[1]{|#1\rangle} \newcommand{\braket}[1]{\langle #1 \rangle} \ket{i}$ and $\ket{f}$ are eigenstates of the unperturbed Hamiltonian $\hat{H}_0$. Therefore, by virtue of the Hermiticity of $\hat{H}_0$, they are necessarily orthonormal

$$\braket{i|f} = \delta_{ij}$$

therefore if we define $\hat{V'} = \hat{V} + k$ where $\hat{V}$ is the perturbation Hamiltonian and  $k$ is some arbitrary constant

$$\begin{align}
\braket{i|\hat{V'}|f} &= \braket{i|\hat{V}|f} + \braket{i|k|f} \\
&= \braket{i|\hat{V}|f} + k\braket{i|f} \\
&= \braket{i|\hat{V}|f}
\end{align}$$

if $\ket{i}$ and $\ket{f} \endgroup$ are different states. If they are the same state, then you are calculating the transition probability from one state to itself - not a very useful concept.


Why is only the carbonate anion used to precipitate Barium?

Actually, the textbook should remove this question. You should discuss this with your teacher. It will be a service to future students. The heavier elements of group II elements form insoluble sulfates, phosphates and carbonates.

Now carbonates display another interesting phenomenon. If you keep bubbling carbon dioxide in into a group II carbonate solution (or better say suspension), the group II ions form water soluble bicarbonates. It was a classical experiment of "old school" chemistry. Bubble carbon dioxide into a clear solution of lime, it would become cloudy, if you continue that, it becomes clear again.

Another reason, which may be valid is that you may wish to re-dissolve precipitated barium carbonate after filtration. That would be easy to do so by adding a small amount of strong acid. However re-dissolving barium sulfate or phosphate is very difficult.


Relation between various bond angles of Nitrogen compounds

$\ce{NH3}$ and $\ce{NF3}$ are easier to analyse and predict since there is only one set of atoms around the nitrogen — not three groups of four or three groups of seven as for trimethyl- and triethylamine, respectively.

Note that you did not (but should!) include bond lengths in your analysis. It is important to remember them. For ease, here they are as taken from Wikipedia (along with the bonding angles):

$$\textbf{Table 1: Bond lengths and angles}\\\textbf{of nitrogen compounds}\\
\begin{array}{lcc}
\hline
\text{compound} & d(\ce{N-X}) & \angle (\ce{X-N-X})\\
\hline
\ce{NH3} & 102~\mathrm{pm} & 107.8^\circ\\
\ce{NF3} & 137~\mathrm{pm} & 102.5^\circ\\
\ce{NMe3} & 146~\mathrm{pm} & 110.9^\circ\\
\hline
\end{array}$$

Starting with the difference between $\ce{NH3}$ and $\ce{NF3}$ — oh wait, I already answered that once. The gist is both compounds would love a $90^\circ$ bond angle but cannot due to too small bond lengths — the outer atoms would clash into each other, figuratively speaking. Therefore, the bond angles are expanded (at a loss of energy) to reduce this steric strain (gain of energy) up to an optimal sweet spot. Since the $\ce{N-H}$ bond is shorter than $\ce{N-F}$, the fluorines are comfortable with a slightly smaller bond angle as they are further away from each other (even though they are larger).

The difference when going from fluorine atoms to methyl groups (or ethyl groups) — as highlighted above — is the additional set of atoms. These cause further crowding, even though the slightly larger carbon atom (with respect to fluorine) allows for an even longer bond length. The steric strain of the additional hydrogens means that the bond angle must be extended even further to accomodate for all these.

Not much changes when going from trimethyl to triethylamine — except that the steric strain is larger still, and that therefore the bond angle is probably increased slightly again. One source I found including a crystal structure including triethylamine did not, however, give notably different bond angles when compared to trimethylamine, so this point may be moot. Note, however, that the angles reported are already larger than the perfect tetrahedral angle, indicating that the lone pair has an even lower s-proportion which would destabilise the overall molecule. Only by reducing steric strain can this low s-proportion be explained in any way.

As I already commented on the question: Back-bonding will not happen since there are no accessable empty orbitals on nitrogen that could accept a back-bond. Try drawing a back-bond in a Lewis formula without exceeding nitrogen’s octet to see that it is impossible.


How to clean a copper pan from black burned fat chemically?

The crust is a giant mess of hydrocarbons that have been burned, caramelized, and congealed into the surface of the pan.

I would recommend that you try the steel sponge to get some of the crust off, you may find that as you start breaking into the top layers of crust that it will fall apart. Another strategy would be to put the pan in the oven to the highest heat that is safe for the cookware and this might cause the pan to expand enough to create cracks in the crust or release it from the micro-cracks that were already in the pan surface. As a last resort you will have to chisel it off slowly.


Will alcohol or soap damage plastic or rubber?

Will soap damage plastic or rubber?

Probably not. Most plastics and rubbers are resistant to aqueous solutions and ionic compounds (including most surfactants and soaps).

  Will alcohol damage plastic or rubber?

Maybe. There are two problems, in that neither alcohol nor plastic refer to specific compounds, but are rather generic terms encompassing hundreds of compounds each. The number of possible combinations is endless. As Leonardo suggests, the best approach is either to look in the manual or experiment in a small inconspicuous part of the product.

However, assuming that by alcohol you either mean ethanol, which is the drinking kind of alcohol, or isopropanol, which is the alcohol in many cleaning products, then I can give you some advice based on common plastic types. Both are similar enough in their solvent properties that we can generalize.

Alcohol will damage some plastics, but not all. I will list my answers by the resin code, or "recycling symbol", found on most plastic items:

Poly(ethylene terephthalate), PET or PETE - PET is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening due to the dissolution of plasticizers.
High-Density polyethylene, HDPE - HDPE is resistant to most things.
Poly(vinyl chloride), PVC - PVC is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening.
Low-Density polyethylene, LDPE - LDPE is resistant to most things.
Polypropylene, PP - PP is resistant to most things.
Polystyrene PS - PS is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening.
This stands for "other", but the most common is polycarbonate, which is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening.

Rubber - most rubbers are probably resistant to alcohols, but prolonged exposure will cause loss of elasticity