Changes in which factors affect both the rate and the rate constant of a first order reaction?

If you see the rate equation for the Arrhenius equation:
$$
k = A e^{-E_a/(R T)}
$$
you can see that the rate constant increases for an increase in temperature (and as activation energy increases, the rate constant decreases.

Now, since the rate constant increases, this implies that the rate of reaction increases with temperature.

Regarding concentration: you are correct in that concentration affects the rate of reaction. However, a rate constant does not change according to concentration. You can observe this in the Arrhenius equation, where A is independent of the concentration of the substrate. The fact that the rate increases does not imply the rate constant changes.

Possibly more clearly: it is implied that increasing the rate constant will increase the rate of reaction. However, increasing the rate does not increase the rate constant.

An increase in temperature increases the rate constant and hence the rate.

An increase in concentration increases the rate but not the rate constant.


Does Victor Meyer test distinguish alcohols and alkyl halides?

Aaron,

You could be correct if your alkyl halides to be ditinguished are 1,2,3 degree alkyl IODIDES.
Because we ultimately are converting 1,2,3 degree alcohols into IODIDES in step-I; followed by reactions with NaNO2/HCl and then NaOH.

@Pritt Balagopal

He had a valid point, when he said "alkyl halides and alcohols are said to react in the same way with the reagents used for Victor Meyer test"


No reaction occurs with the reagents you have stated at room temperature. Maybe with different conditions (eg. basic conditions) the haloform reaction that Ivan commented would occur.


What reagents change the time of a chemical clock?

First, a review of the underlying chemistry per an educational source:

A clock can be formed from molecules that react at a rate from the time the chemicals are mixed to the time the reaction occurs.
Explanation: There are two forms of iodine:
Element form (Iodine): I2 + starch -------> blue
Ion form (Iodide): I- + starch --------> colorless
There are two reactions taking place simultaneously in the solution:
#1. 2I- + H2O2 ---(slow)-------> I2 + H2O
#2. I2 + Vitamin C ----(fast)----> I (colorless)
In Reaction # 1 iodide ions react with hydrogen peroxide to produce iodine element which is blue in the presence of starch. BUT, BUT......
In Reaction # 2 The Vitamin C is immediately reacting with any iodine formed in reaction # 1. The net result, at least for part of the time is that the solution remains colorless with excess of iodide ions being present.
Now after a short time as the reactions keep proceeding in this fashion, the Vitamin C gets gradually used up. The Vitamin C creates a clock reaction (1 - 2 minutes) and once it is used up, the solution turns blue, because now the iodine element and starch are present.

So, to answer the question: "which reagents in the experiment are responsible for changing the time before the colour change in the clock", the answer is H2O2 and Vitamin C. Now, while the rate-controlling step, associated with the slowest reaction is H2O2 (creating the colourful I2 highlighted by starch), some Ascorbic acid must be present to turn the solution clear again.
On the question:"Can you please clarify which reagents cause the increase/decrease in time of the chemical clock reaction?", the strength of the H2O2 (try changing from 3% to 10%) and the pH of the hydrogen peroxide (try adding alkaline Washing Soda Na2CO3 to the H2O2), both I would likely expect to have impacts on the Vitamin Clock timing.


What is the product of the reaction of diol with tosyl chloride followed by lithium aluminum hydride?

There are two replaceable $\ce{H}$s in the starting material, but only 1 equiv of $\ce{TsCl}$ is given. Thus, primary $\ce{-OH}$ will win the battle due to steric reasons, and therefore, B represent the intermediate product X. The second reaction is firstly acid-base reaction between secondary $\ce{-OH}$ and $\ce{H-}$, and then, nucleophilic substitution ($\mathrm{S_N}$2) reaction to replace $\ce{-OTs}$ group by $\ce{H-}$. That leaves D as the product Y, after acid workup.


Preparation of alcohols from Monosubstituted cyclic ethers

Yes, Girgnard reagents can act as nucleophiles in the ring opening of epoxides. Nucleophilic attack occurs much faster at the less hindered carbon, so the product of this reaction will predominate. The reaction results in the formation of an alkoxide, and does not form the alcohol until treated with a protic acid. The mechanism for Grignard addition to epoxypropane (methyl oxirane) is:

$\hspace{4cm}$

For 2-methyl-2,3-epoxybutane, a more hindered epoxide, the addition still occurs at the less hindered carbon:

$\hspace{3.8cm}$


How should I construct the orbitals of dilithium using only pen and paper?

When combining orbitals, your first approximation should be to relate similar ones with similar ones. In this case, on each lithium (or nitrogen or fluorine) atom we have:

1 s-orbital
1 p-orbital in the direction of bonding
2 p-orbitals perpendicular to the bonding axis.

We should treat these separately initially, see what the results give us, and then think if we should combine them.

Realise that the point group of the molecule is $D_\mathrm{\infty h}$. However, that is a bad point group to work with. Instead, assume $D_\mathrm{2h}$ and correlate your results back to $D_\mathrm{\infty h}$ according to a table originally presented here:

$$\begin{array}{cc}
\hline
{\text{representation in }D_\mathrm{2h}} & {\text{representation in }D_\mathrm{\infty h}}\\
\hline
A_\mathrm{g} & \Sigma_\mathrm{g}^+\\
B_\mathrm{1g} & \Sigma_\mathrm{g}^-\\
B_\mathrm{2g} + B_\mathrm{3g} & \Pi_\mathrm{g}\\
A_\mathrm{g} + B_\mathrm{1g} & \Delta_\mathrm{g}\\
B_\mathrm{1u} & \Sigma_\mathrm{u}^+\\
A_\mathrm{u} & \Sigma_\mathrm{u}^-\\
B_\mathrm{2u} + B_\mathrm{3u} & \Pi_\mathrm{u}\\
A_\mathrm{u} + B_\mathrm{1u} & \Delta_\mathrm{u}\\
\hline\end{array}$$

Ignore the difference between $\Sigma^+$ and $\Sigma^-$ for now. What is relevant is that both the two s-orbitals and the two p-orbitals along the bond axis transform as $A_\mathrm{g} + B_\mathrm{1u}$ in $D_\mathrm{2h}$, and thus as $\Sigma_\mathrm{g}^+ + \Sigma_\mathrm{u}^+$ in $D_\mathrm{\infty h}$. Which means that we expect a bonding and an antibonding orbital from both — no surprises here. So initially, assume (as it looks you tried to do) a pair of $\unicode[Times]{x3c3}$ and $\unicode[Times]{x3c3}^*$ orbitals for both.

Since the energy difference between s- and p-orbitals is large enough (you can and should always assume that) the antibonding $\unicode[Times]{x3c3}_\mathrm{s}$ orbital should still be lower than the bonding $\unicode[Times]{x3c3}_\mathrm{p}$ orbital. These two now have a different symmetry and thus cannot mix productively.

Fill in your electrons. You have two, which will end up in the lowest $\unicode[Times]{x3c3}_\mathrm{s}$ orbital.

What can still mix is $\unicode[Times]{x3c3}_\mathrm{s}$ with $\unicode[Times]{x3c3}_\mathrm{p}$ and their corresponding antibonding orbitals. This will lead to the fidgety task of determining how much stabilising interaction can still be found. Assume the $\unicode[Times]{x3c3}_\mathrm{s}$ to be stabilised slightly by $\unicode[Times]{x3c3}_\mathrm{p}$, because the former is populated while the latter is empty. That should give the $\unicode[Times]{x3c3}_\mathrm{s}$ orbital a slight look as if it derived from sp hybrid orbitals. (Only weakly though!) Note however, that this secondary mixing can be ignored at first approximation.

Also note that any of this mixing must result in symmetric or fully antisymmetric orbitals with respect to the plane of symmetry perpendicular to the bond axis ($\sigma_\mathrm{h}$).

Taking all this and reviewing your picture:

Your lowest orbital looks good. A fully bonding orbital of the s-type with very slight hints of a p-orbital mixing in.
Orbital 2 should look much more like an antibonding $\ce{H2}$ orbital. Since both antibonding orbitals are unpopulated, mixing is unproductive. If you must, very slightly add a bit of p-contribution on the inner side of the atoms.
Orbital 3 should look almost like orbital 1, but clearly with a much larger p-look; so have two nodal planes, one through each atom. Once you get to and past diboron, this is the primary bonding orbital.
Orbital 4 is technically good. However, as per the initial approximation both lobes on each atom should be equal, and only due to the secondary mixing should the inner lobes be slightly smaller. Draw it, as you may be accustomed to $\mathrm{(sp^3)^*}$ hybrid orbitals.

If you want a shortcut, remember to consider s and p separately. Two s-orbitals mix as you know it from hydrogen, creating a very big spherish blurb and an antibonding double-sphere. Two p-orbitals mix to give something like your orbitals I and II (bonding and antibonding). The two s should be lower than both the p.

In a second step, you can mix the two bonding orbitals (and the two antibonding orbitals; but not bonding with antibonding) to add a little p-contribution to your s-type orbitals. However, the lower should clearly always remain an s-type (with a little p mixed in).


There are no lone pairs in $\ce{As}$ after it forms a bond with each of the five $\ce{F}$.

Hence, the number of lone pairs of electron around the central atom ($\ce{As}$) is zero, and the number of sigma bonds the central atom forms is five (a single bond has one $\sigma$-bond only).

Hence the hybridization number of $\ce{As}$ in $\ce{AsF5}$ is $0+5=5$ and it follows $\ce{sp^3d}$ hybridization geometry with no change in actual shape because of absence of lone pairs.

So, it has a trigonal bipyramidal structure: two $\ce{F}$ ions in the axial positions and three others in equitorial positions.


Why doesn't hydrogen displace metals from their oxides?

If magnesium reacts with water you get magnesium oxide and hydrogen.

True. So you can't put out a magnesium fire with water. It just burns hotter.

$$ \ce{Mg(s) + H2O ->T[aqueous] MgO(s) + H2(g) ^}$$

    But hydrogen is generally a lot more reactive than magnesium, so isn't this backwards?

No, in the above reaction you're reacting water with magnesium. The hydrogen is given off as a gas, though a tiny amount will stay dissolved in the water.

    So why does it actually happen the other way round?

It will under suitable conditions. So with heat and under a pressurized flow  of hydrogen gas the reverse reaction will occur.

$$ \ce{ MgO(s) + H2(g) ->T[heat, pressure] Mg(s) + H2O(g) ^} $$


How to dissolve beeswax on the underside of a porous wood structure

Esters are good for dissolving wax, but use a long carbon chain that is not flammable.
Isopropyl myristate is sold online to people that want to make their own cosmetics. It is relatively cheap. It will leave an oil film that can be removed with liquid detergent in warm water.

For thick wax, warm with a blow dryer and scrape with a plastic spatula. Then try above.


Choice of doping elements in semiconductors

Look at the II-VI options from a chemical perspective.

For silicon, Group III boron p-doping would be replaced by Group II beryllium doping, and Group V phosphorus n-doping would be replaced by Group VI sulfur doping.

The beryllium would tend to act like a Lewis base that donates a pair of electrons to some nearby silicon atom, forming a stronger and more localized bond than is possible with the odd electron count created by boron doping. Similarly, the sulfur would tend to act like a Lewis acid and form a localized sulfide bond with some nearby silicon atom, versus the odd electron count created by phosphorus.

So, the main reason for picking elements immediately "next door" (off by one electron) is to avoid introducing paired electrons, which on average will bond more strongly and more locally via a Lewis base-acid process.

That said, there's enough complexity going on there that it's still possible that you could get II-VI or maybe even I-VII doping to work in the right situations.


Which of the following has higher enol content?

The two main factors affecting the stability of the enol forms of (A) and (B) are

Hydrogen bonding
Aromaticity

In the enol form of both compounds, hydrogen bonding is present:

But the resonance structure of enol form of (B) is aromatic. It contains $6$ $\pi$-electrons:

The enol form of (A) lacks this stability. Hence, the enol content of (B) is more than that of (A)


There is one possibility with glycerol With density 1.261 g/cm 3.Here the exact numerical value of mass is greater than that of volume of glycerol in kg and litres respectively.Moreover glycerol is soluble in water and vice versa.


Back titration of excess HCl used for dissolving limestone

It is generally a good idea to approach problems in the order that sub-questions are asked in. There is often a paedagogic reasoning in that order that can help you approach the answer.

In this case, the key point is that the limestone is impure $\ce{CaCO3}$, and that thus we do not know how much of the acid was initially neutralised with $\ce{CO2}$ evolution. Approaching the question from this side, as you did, is not very helpful, since we are missing an important value to calculate from.

What we do know, however, is the final end of the experiment: that $8.50~\mathrm{ml}$ of $0.1~\mathrm{M}\ \ce{NaOH}$ were able to neutralise a certain portion of the remaining hydrochloric acid. You should think of this as being a titration.

Once you have realised that, it makes sense why the first question asks for the amount of substance in the $20.0~\mathrm{ml}$ aliquot — it is the one you can actually calculate quickly, and it will lead you to the entire solution.

Thus, your plan of action is:

find the concentration or amount of $\ce{HCl}$ in the $20.0~\mathrm{ml}$ portion by known formulae used for titration problems (your work so far shows you can do this);

   We have $\ce{HCl + NaOH -> NaCl + H2O}$, so
 $$\begin{align} n_\ce{HCl} = n_\ce{NaOH} &= (\pu{8.50 mL})(\pu{0.1 M}) \\ &= \pu{0.00085 mol} \end{align}$$

thencefrom calculate the concentration or amount of $\ce{HCl}$ in the entire $200~\mathrm{ml}$ sample from similar formulae;

   A $\pu{20 mL}$ portion of the $\ce{HCl}$ simply has one-tenth of the original amount of $\ce{HCl}$, so in the entire $\pu{200 mL}$, we have:
 $$n_\ce{HCl} = (\pu{0.00085 mol}) \times \frac{\pu{200 mL}}{\pu{20 mL}} = \pu{0.0085 mol}$$

Next, decide how much hydrochloric acid was lost in the dissolution of the limestone.

   At the start, we had a grand total of
 $$\begin{align} n_{\ce{HCl},\mathrm{tot}} &= (\pu{15.0 mL})(\pu{1.4 M}) \\ &= \pu{0.021 mol} \end{align}$$
 which you already calculated. After reacting with the limestone, we were left with only $\pu{0.0085 mol}$ of that. So, the amount that reacted is simply:
 $$n_{\ce{HCl},\mathrm{reacted}} = \pu{0.021 mol} - \pu{0.0085 mol} = \pu{0.0125 mol}.$$


What is the non-ideal equivalent of mass action kinetics?

According to Catalytic reaction rates in thermodynamically non-ideal systems Journal of Molecular Catalysis A: Chemical 163 (2000) 189–204 (citing to kinetics textbooks) non-ideal kinetics needs to consider the activity of the transition state in addition to the activities of the reactants:

For $\ce{A + B ->}$

rate $=\frac{k_BT}{h}K^{\ddagger}\frac{a_Aa_B}{\gamma_\ddagger}=k_0\frac{\gamma_A\gamma_B}{\gamma_\ddagger}C_AC_B $

Where $K^{\ddagger}$ is the equilibrium constant for the formation of the transition state, $a$ is activity, $\gamma$ is activity coefficient, $C$ is concentration, $k_B$ is Blotzmann constant, $T$ is temperature, $h$ is Planck's constant and $k_0$ is the thermodynamically ideal rate constant.

Another reference to look at for more info is The status of transition-state theory in non-ideal solutions and application of Kirkwood–Buff theory to the transition state J. Chem. Soc., Faraday Trans. 2, 1986,82, 1297-1303.


Urethane foam from isocyanates, commonly sold at hardware stores for insulation.

Click links for general information and for its chemistry.


What are some good examples of rate equations for a math class?

How thoughtful of you to include chemistry in your differential equations course! We appreciate your effort, especially going the extra mile to make it realistic.

I hope you've seen the rate equation page on Wikipedia as it contains a good deal of the mathematics in several examples of reactions. You should also be interested in the reaction order page (you can find examples here of reactions with fraction or negative order with respect to some reagents).

Let me do a quick recap on basic chemical kinetics theory. Consider the general reaction equation:

$$\ce{a\ A + b\ B + c\ C + d\ D + ... \longrightarrow w\ W + x\ X + y\ Y + z\ Z + ...}$$

, where uppercase letters indicate different molecules in a gas or solution, and lowercase letters indicate the stoichiometric coefficients (negative for the reactants, positive for the products). Naively, we can assume the reaction happens because, while the molecules are jostling about, at some point all of the reactants ($a$ molecules of $\ce{A}$, $b$ molecules of $\ce{B}$, $c$ molecules of $\ce{C}$, etc.) will bump into each other all at the same time and undergo reaction. If you think of collisions as independent events with a probability of happening proportional to the concentration of each species, then it isn't too hard to understand that the frequency with which reactions happens is proportional to $[A]^a[B]^b[C]^c[D]^d...$, and so the rate equation for the reaction would be:

$$r=k[A]^a[B]^b[C]^c[D]^d...$$

, where r is the reaction rate and k is a proportionality constant. The consumption or production rate for any species $\Gamma$ with stoichiometric coefficient $\gamma$ is trivially related to the reaction rate by:

$$\frac{d\Gamma}{dt}=\gamma r$$

However, experimentally things aren't so simple (fortunately for nature, unfortunately for our minds). A reaction that requires the simultaneous collision of many molecules would be highly unlikely, meaning that it would happen very, very slowly. Even so, experimentally we observe many reactions that involve a large amount of molecules. For example, the combustion of cyclohexane in air is formally given by the equation:

$$\ce{C6H12 + 9 O2 → 6CO2 + 6 H2O}$$

If the reaction really required ten molecules to bump into eachother at the same time with sufficient energy and in the right geometry, then this reaction probably could not happen in air, as oxygen would be too rarefied to compensate the extremely low proportionality constant k for a single-step reaction. In actuality, a gaseous mix of cyclohexane and air can react so fast as to cause an explosion, converting all the reactants into products in a miniscule fraction of a second. Clearly something is wrong.

It turns out the assumption that reactions occur in one step is incorrect in general. Usually, there is more than one step involved, and indeed there generally is more than one path from reactants to products. The speed at which a reaction occurs is really described by considering all paths at once, with all their steps, and adding their contributions (like a start and end point in a complex tree diagram). In general this is pretty difficult and unnecessarily complex. It is often a reasonable approximation to model the reaction by selecting only the fastest route from reactants to products, and the reaction rate is bottlenecked by the slowest step of the fastest route, which becomes responsible for determining the rate equation.

Now, since the reaction rate depends on selecting the fastest route, you shouldn't expect any of the steps in it to contain some very huge bottleneck, such as requiring six molecules to react simultaneously; there would likely be a slightly different route, perhaps with more steps, but in which each step doesn't involve so many molecules interacting at once, and hence being a faster route. Because of this, it turns out that most of the fastest reaction pathways involve slowest steps that very rarely depend on more than two molecules at once. Most reactions which can be modeled simply like this therefore have rate equations of the type $r=k[A]$, $r=k[A]^2$, $r=k[A][B]$ or even $r=k$. Few reactions rates are of the third order, of the type  $r=k[A]^3$,  $r=k[A]^2[B]$,  $r=k[A][B][C]$ or similar. I don't know of any reaction that goes as $r=k[A]^3$ specifically. Fourth order reaction rates are so rare that they are the focus of research when found. I don't expect anyone to know a fifth order reaction.

At the end of the day, kinetic theory is hard enough that reaction kinetics end up being determined simply by parameter fitting from experiments. Even if the underlying reaction mechanism is not completely understood, if a fit of the type $r=k[A]^{3/2}[B]^{-1}$ happens to be good, then so be it. Reactions which have been studied in depth are significantly more complex, such as the misleading simplicity of the reaction between hydrogen and bromine.

Edit: I don't know why I didn't search for a fifth order reaction rate! Turns out some exist, such as this one. The reaction rate is of the type $r=k[A][B]^4$, so you could have an example where $\frac{d[B]}{dt}=k[B]^4$.


How will the degree of conjugation effect the light emitted? (glow stick reaction)

The initial chemiluminescent (CL) species transfers its excited state energy to the dye. This depends on the overlap of the CL emission spectrum with the absorption spectrum of the dye and their spatial separation, i.e. better spectral overlap and higher concentration more transfer and more dye emission. (see Forster energy transfer for mechanism).  The dyes used have 80-100% fluorescence quantum yield, so v. high chance of emission. The intensity you experience will also depend on how your eyes respond to different colours.

The energy gap in the dyes does not greatly affect their fluorescence efficiency, this is primarily governed by interactions within the dye to produce triplet states which drain energy out of the fluorescent singlet state.


Is this enough information to solve this problem?

You need to use the Henderson-Hasselbach equation:

$$\pu{pH = pK_a + \log\frac{[Base]}{[Acid]}}$$

The $\pu{pK_a}$ of $\ce{HNO2}= 3.15$

Therefore,

$$10^{\pu{pH-pK_a}} = \pu{\frac{[Base]}{[Acid]}}$$
$$\pu{10^{3.35-3.15} = \frac{[Base]}{[Acid]}}$$
$$\pu{1.58 = \frac{[Base]}{[Acid]}}$$

Implies that the numerator must be $1.58$ times that of the denominator.

The base is $\ce{NaNO2}$; the acid is $\ce{HNO2}$. The base is two times as concentrated as the acid.

This means, for every $\pu{mL}$ of acid you use, you need $\pu{1.58/ 2 = 0.79 mL}$ of base.

If,

$$V_T = V_A + V_B$$

And,

$$V_B = 0.79V_A$$

Then,

$$\pu{300 mL} = V_A + 0.79V_A  \Rightarrow V_A = \pu{167.60 mL}$$


Where did amino acids get their one-letter codes?

Tryptophan

For instance, is there a reason 'W' specifically was chosen for tryptophan (other than the fact that 'T' was taken)?

Once you have assigned the other 19 amino acids, there are only 7 letters of the alphabet left: B, J, O, U, W, X, and Z. (Certainly not a nice Scrabble hand to have!)
If one wants to use a letter found within the name of the amino acid, the only available letter would be O. However, the usage of U and O was historically discouraged because these letters could be easily confused with other letters (U with V; O with G, Q, C, D, and the number 0).
It turns out that the choice was made because W is a very fat letter and was reminiscent of the indole ring system present in tryptophan (the only amino acid to contain a bicyclic system). (See below for source.)

What happened to the other six letters, then?
All 26 letters of the alphabet now find use as a one-letter code for amino acids or various combinations thereof.
After their discoveries, selenocysteine and pyrrolysine (the latter only found in bacteria) were assigned U and O respectively. Furthermore, B is used to represent aspartic acid OR asparagine; J is used to represent leucine OR isoleucine; Z is used to represent glutamic acid OR glutamine; and X is used to represent an unknown amino acid.
I believe B and Z find use because, in protein sequencing, acid hydrolysis is often used to break peptide bonds. This has the undesirable side-effect of hydrolysing the amide groups in asparagine/glutamine, leading to the formation of aspartic/glutamic acids, which means that one cannot tell exactly which amino acid it was at the start. J is used in NMR spectroscopy where isoleucine and leucine are difficult to distinguish.

Why did they choose the letters they did?
As far as I am aware the usage of one-letter symbols is adopted by both IUPAC and IUB (since 1991, IUBMB) in their joint 1983 recommendations on "Nomenclature and Symbolism for Amino Acids and Peptides".1 ankit7540's concise summary of the historical development already mentioned these recommendations.
In particular, Section 3AA-21.2 "The Code Symbols" has a description of why the letters were chosen. This document is probably the most authoritative stance on the matter. The rationale is mostly in line with Jan's answer:

Initial letters of the names of the amino acids were chosen where there was no ambiguity. There are six such cases: cysteine, histidine. isoleucine, methionine, serine, and valine. All the other amino acids share the initial letters A, G, L, P or T, so arbitrary assignments were made. These letters were assigned to the most frequently occurring and structurally most simple of the amino acids with these initials, alanine (A), glycine (G), leucine (L), proline (P) and threonine (T).
Other assignments were made on the basis of associations that might be helpful in remembering the code, e.g. the phonetic associations of F for phenylalanine and R for arginine. For tryptophan the double ring of the molecule is associated with the bulky letter W. The letters N and Q were assigned to asparagine and glutamine respectively; D and E to aspartic and glutamic acids respectively. K and Y were chosen for the two remaining amino acids, lysine and tyrosine, because, of the few remaining letters, they were close alphabetically to the initial letters of the names. U and O were avoided because U is easily confused with V in handwritten material, and O with G, Q, C and D in imperfect computer print-outs, and also with zero. J was avoided because it is absent from several languages.
Two other symbols are often necessary for partly determined sequences, so B was assigned to aspartic acid or asparagine when these have not been distinguished; Z was similarly assigned to glutamic acid or glutamine. X means that the identity of an amino acid is undetermined, or that the amino acid is atypical.

One can only hypothesise what they meant by "associations that might be helpful in remembering the code" in the case of N/Q/D/E. My best guess is:

D and E were possibly chosen for aspartic and glutamic acids because they were the only consecutive pair of letters left, emphasising their chemical similarity. Aspartic acid is shorter than glutamic acid by one methylene group (CH2), so it gets the earlier letter D.
Glutamine sounds like Q-tamine. If you don't think it sounds similar, repeat it 50 times until you do.
AsparagiNe was assigned N.

Reference

IUPAC-IUB Joint Commission on Biochemical Nomenclature. Nomenclature and Symbolism for Amino Acids and Peptides: Recommendations 1983. FEBS J. 1984, 138 (1), 9–37. DOI: 10.1111/j.1432-1033.1984.tb07877.x. A HTML version (perhaps more user-friendly) can be found at this address.


Heat produced when magnesium is reacted with acid

It can help to imagine the experimental setup.

Imagine a thermally insulated reaction container, let's call it calorimeter, that is filled with 50 ml 1 M $\ce{HCl}$ ( = a solution of hydrogen chloride gas in water).

We have to measure a temperature change ($\Delta{} T$), that comes from an exothermic reaction. Where would we best measure this temperature change? Our reaction medium apparently is a good choice. Therefore, there's most likely a thermometer (insulated from the outside) in good thermal contact with the water.

Now we let the magnesium strip fall into the reaction medium (using some clever mechanic) while our calorimeter is closed), let it react and measure the temperature change until the temperature reaches some final (maximum) value.

As you wrote, the temperature change ($ \Delta{} T$) is +7 °C.

I don't see that an initial temperature was given, but that's not important. We have the $ \Delta{} T$.

Now we need a specific factor that correlates the heat liberated in the reaction with the temperature change in a specific medium.

Here, the specific heat capacity of water comes into play.

There you go.


It is such a beautiful picture, perhaps good for a chemistry book cover. It is a similar phenomenon as what happens in a so-called chemical garden. Water seems to be creeping up to the nail head, and helping in oxidation of iron salts to insoluble hydroxides. Finally the gelatinous membranes act like semi-permeable membranc in the osmosis like experiment.

The figure shows several salt crystals soaked in a slightly alkaline medium.

Ref: From Chemical Gardens to Chemobrionics,Chemical Reviews, 2015, 115(16)
DOI: 10.1021/acs.chemrev.5b00014


In a saturated solution, why is there ongoing dissolution and crystallisation?

Because equilibrium is dynamic not static

There are many things in the world of chemistry which could be simpler but are not. Equilibrium is one of them.

There are few chemical processes where equilibrium is equivalent to the situation where everything just stops. Equilibrium normally occurs when the rates of processes going forward are the same as the rates going backward, not when things stop happening. In this case the rate of dissolution is the same as the rate of crystallisation.

On a molecular level things are very dynamic. Molecules in the solution are hitting crystals all the time. Molecules in the crystal are vibrating and being hit by molecules of the solvent: sometimes the molecules in the crystal have enough energy to escape from the crystal and go into solution. Equilibrium is the point where those two reactions have equal rates, not where all the activity stops.

We know this is true in real solutions from several experiments. We can, for example, label the molecules in the crystal with radioactive versions of one of their atoms (eg carbon 14). Even if we start with a saturated non-radioactive solution and add radioactive crystals of the same compound, the radioactivity will spread to the solution (see this question Proof of Dynamic Nature of Equilibrium).

The key thing is that all equilibria in chemistry are dynamic. The major difference in many cases is how dynamic. Some happen very slowly, others very quickly. But the position of the equilibrium only has a weak relationship to the speed of the equilibrium.


How to approximate the pH of a box of water molecules with one proton?

For a rough, back of the envelope, sanity check sort of calculation, your second approach is more appropriate than the first. (Though check your value for Avogadro's number.) This is something that should jump out at you if you keep track of units in your calculation. The $\ce{[H+]}$ in the equation should be measured in moles of hydrogen ion per liter, so your calculation needs to result in an inserted value that's in moles of hydrogen ion per liter. You can also rule out the first equation by performing a thought experiment: if the size of the box doesn't factor into the equation, does it make sense that the pH wouldn't change if the box size changes? (No matter how much the size changes?)

However, the statement that pH is equal to the negative log of the hydrogen ion concentration is strictly speaking only true for dilute aqueous solutions near standard temperature and pressure. For concentrated acids/bases, solutions which are not primarily water, or for pressure and temperatures well outside standard, things get more contentious. There's a number of different interpretations of what "pH" means, and each have their advantages/disadvantages, depending on the situation.

One of the big considerations here is that "concentration" is not really really the thermodynamically relevant parameter when figuring out how chemical systems behave. Instead, it's the compound activity which is typically the more important quantity to keep track of. As it happens, the thermodynamic activity of a compound is normally proportional to the concentration in dilute solutions, so by convention we set unit activity to be equivalent to 1M solution.* This means the distinction between concentration and activity is normally ignored/glossed over in most introductory chemistry courses, but make no mistake, it's the activity of the compounds which should be used in things like the Henderson–Hasselbalch equation, not the concentration.

*(This normalization by 1 mol/L is the reason we can take the log and not run up against any strange results with dimensional analysis - it's technically a dimensionless value inside the log: the actual concentation, divided by the reference concentration of 1 M.)

The reason this matters for you is that you're dealing with a situation which isn't quite the same as a dilute aqueous solution. Nominally, it sounds like you're close. Your water density is around standard water density, and the amount of hydrogen ions are close to 1 mole per litre. But there are other considerations to take in to account. For example, what temperature and pressure are you using? Extreme values can change the hydrogen ion activity. Also, you didn't mention anything about the presence of a counter-ion. The unbalanced electrical charge might affect the predicted activity as well. The other issue is that just a single hydrogen ion will result in some pretty extensive fluctuations in the system as the system changes state. The "pH" (or hydrogen ion activity) of the system may actually exhibit time dependence, which might be important for the system you're modeling.

If you're truly interested in what the equivalent pH of the solution is, you should look into simulation techniques which attempt to actually gauge the chemical activity of the hydrogen ion under the conditions you're interested in. -- That is, figure out why you want to know the "pH" of the system, and look for more direct measures of that.


We essentially deal with two different trends here and I think the answer lies within analyzing these trends. First we have that the ionization energy obviously increases with increased nuclear charge/atomic number. Secondly we have that with a higher magnetic quantum number (the s-, p-, d- and f-orbitals) the energy of the orbital increases and therefore that the ionization energy decreases. The decrease in ionization energy that the higher magnetic quantum number introduces is larger than the increase that the single extra positive charge introduces.

We should also take Hund's rule in account, which is favorable for a lower ionization energy of $\ce{Be}$ compared to $\ce{B}$. This factor is however relatively small compared to the above ones.

To verify some of these trends yourself, you can take a look at this table of ionization energies.


Why are there no dihedral mirror planes in the D3h point group?

The σ$_v$ planes in a molecule are a direct result of the C$_n$ main axis i.e. they are interchangeable if you perform the C$_n$ operation on them. On the other hand σ$_d$, which are called dihedral planes bisect the dihedral angles between members of the σ$_v$'s set. Therefore molecules with odd C$_n$ such as BH$_3$ wont have any σ$_d$ planes; only n (3 for BH$_3$) σ$_v$'s. In contrast a molecule such as benzene which has a C$_6$ main axis will have 3 σ$_v$'s (the ones including the carbon atoms) and 3 σ$_d$'s which are the ones passing between the carbon atoms. Of course it doesnt really matter which set is called dihedral and which is called vertical, it could be the other way round as well. Ref: Chemical applications of group theory by Cotton.


How to calculate the volume of oxygen in a reaction between hydrogen peroxide and bleach using this apparatus

This is how it works in my head:

If instead of peroxide in the syringe, you just had $\pu{20cm^{3}}$ of air, then injecting the air will cause a reading of $\pu{20cm^{3}}$ to happen.  The water gets pushed out of the way.

Injecting peroxide would do the same thing (approximately).  After the reaction, there is still saltwater and leftover reactants taking up space.  The oxygen produced will take up additional space (and push the water away) because it is a gas and it wants to expand.


If you are using water alone, then sodium lead alloy reacts as sodium would but, with the lead added, it's much safer.  But you can have a wider variety of options if, as the comments suggest, you use a dilute acid like vinegar (reacting that, for instance, with zinc).  Wear eye protection no matter what you do.


Identifying the gas given ratio of its specific heats

I'd go about such a question looking at the easiest thing first: ideal gases. So, what do you know about the relations between $C_{p}$ and $C_{v}$ (both being the molar heat capacities at constant pressure and constant volume, respectively) for ideal gases? For one thing, it is well known that

\begin{equation}
  C_{p} - C_{v} = R
\end{equation}

with $R$ is the universal gas constant. You can rearrange this equation for the ratio you are after:

\begin{equation}
  \frac{C_{p} - C_{v}}{C_{v}} = \frac{R}{C_{v}} \\
  \frac{C_{p}}{C_{v}} - 1 = \frac{R}{C_{v}} \\
  \frac{C_{p}}{C_{v}} = \frac{R}{C_{v}} + 1 \ .
\end{equation}

Now, you need to find $C_{v}$. Here, the degrees of freedom that you mentioned come into the game, since the equation that helps us out is the equipartition law, stating that each excited quadratic degree of freedom contributes $\frac{1}{2} R T$ to the molar internal energy $U$

\begin{equation}
  U = \frac{1}{2} R T (f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}})
\end{equation}

with $f_{\text{trans}}$, $f_{\text{rot}}$ and $f_{\text{vib}}$ being the quadratic degrees of freedom for translation, rotation and vibration respectively. It is important to note that only excited degrees of freedom contribute to $U$. Which ones are active depends heavily on temperatur. Vibrations are usually not excited at room temperatur (there are exceptions, e.g. iodine) - they usually need temperatures in the order of $1000 \, \text{K}$ but that varies greatly. Rotations are much easier to excite - I know of no example where they aren't excited at room temperature. Translations take very little energy to excite and can always be considered active. If all degrees of freedom are active you get: For linear molecules $f_{\text{trans}} = 3$, $f_{\text{rot}} = 2$ and $f_{\text{vib}} = 3N - 5$, where $N$ is the number of atoms in the molecule. For non-linear molecules $f_{\text{trans}} = 3$, $f_{\text{rot}} = 3$ and $f_{\text{vib}} = 3N - 6$.
Since $C_{v} = \frac{dU}{dT}$ you get

\begin{equation}
  C_{v} = \frac{1}{2} R (f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}}) \ .
\end{equation}

Substituting this into the equation for the ratio between $C_{p}$ and $C_{v}$ yields

\begin{equation}
  \frac{C_{p}}{C_{v}} = \frac{2}{(f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}})} + 1 \ .
\end{equation}

Now, you only have to know how many degrees of freedom each of the molecules has and plug them into the equation.


Difference between Coupled Cluster and Full CI

Full Coupled Cluster (FCC) vs. Full Configuration Interaction (FCI)

The main theoretical difference is the way excitations are used. "Excitation" refers to putting one or more electrons in higher orbitals than the "reference" calculation, which is often a Hartree-Fock calculation, would. CI uses a "simple", linear excitation operator, whereas CC uses a more complicated, exponential one. One can think of CI as a sum of excitations while CC results in a product of excitations.
Symbolic equations for the wave function ($\Psi$) ansatz of either method read:
$$
\Psi^{\text{CI}} = (1 +C_1+C_2+C_3+...) \Psi^{\text{HF}} \\
\Psi^{\text{CC}} = \exp(T_1+T_2+T_3+...) \Psi^{\text{HF}} \\
= (1 +T_1+T_2+T_3+\frac{1}{2}T_1^2+\frac{1}{2}T_2^2+T_2T_1+\frac{1}{2}T_3^2+...+\frac{1}{6}T_1^3+...) \Psi^{\text{HF}}
$$
In CI, the excitation operator is often denoted $C$ and the prefactor of determinants are called CI coefficients, whereas CC uses $T$ and "amplitudes". In either, the subscript refers to the number of excited electrons.
For "full" calculations, that is, calculations that take into account every possible excitation, the results of CC and CI will be identical.

Full calculations are too expensive for all but the smallest systems, because the cost rises as $O(N!)$, where $N$ is some measure of the system size involving electron number and basis set size. For this reason, one typically decides to "truncate" the excitations to lower the cost.

Single-reference Truncated CC vs. Single-reference Truncated CI (this is what CC and CI commonly refer to)

Truncation first involves collecting the excitations into "classes" of excitations. Within a class, one gets similar cost and hopes for similar contributions. Thus, one may sort of gauge the cost and accuracy of the calculation. Standard ways of doing this have evolved, and this is what CCSD, CCSD(T), CISD etc. refer to. CASSCF/RASSCF can be categorized like this, as well.

With truncation, another difference between CI and CC arises: CC is no longer variational and CI is no longer size-extensize. Variationality is the property that the resulting energy has the true energy (of the given basis set) as a lower bound - one approaches the energy monotonically from above as more excitations become involved. Size-extensivity is the property that two identical systems at infinite separation from each other yield twice the energy of a single system. The latter property is thought of as the more important one, mainly because of its implications for closer, interacting systems. Variationality is also less of a concern because finite basis sets must be used, thus, one is typically far above the true energy anyway.

Another technical difference is that truncated CC and CI lead to different equations, for which different methods of solving have been developed. Overall, CC implementation is, or at least used to be, more difficult. Properties other than the total energy can also be more difficult to implement.

Scaling of CC and CI

Basically, when truncation is performed at similar levels, the two methods scale the same, but the prefactor may be different - this also depends on the quality of the implementation. For reasons stated above, CC has been worked on more intensively since, say, the mid 1990s. So, CISD and CCSD scale similarly, CISDT and CCSDT would as well. Bottom line: CCSD(T) will still be affordable for molecules where full CI would require specialized implementation on a dedicated supercomputer.


How do foundries prevent zinc from boiling away when alloyed with Aluminum?

When alloys are made by mixing molten metals (actually an alloy only need contain one metal and at least one other compound, metal or not) the metals only need to be heated to their melting point, not all the way to their boiling point. In the example you've given, the melting point of aluminum is $\pu{660^oC}$, which is $\pu{247^oC}$ below the boiling point of zinc, so the volatilization of zinc is negligible under these conditions.

However, the issue you bring up does present problems in other cases. For example this article states the following:

  One difficulty in making alloys is that metals have different melting points. Thus copper melts at $\pu{1,083^oC}$, while zinc melts at $\pu{419^oC}$ and boils at $\pu{907^oC}$. So, in making brass, if we just put pieces of copper and zinc in a crucible and heated them above $\pu{1,083^oC}$, both the metals would certainly melt. But at that high temperature the liquid zinc would also boil away and the vapour would oxidize in the air. The method adopted in this case is to heat first the metal having the higher melting point, namely the copper. When this is molten, the solid zinc is added and is quickly dissolved in the liquid copper before very much zinc has boiled away. Even so, in the making of brass, allowance has to be made for unavoidable zinc loss which amounts to about one part in twenty of the zinc. Consequently, in weighing out the metals previous to alloying, an extra quantity of zinc has to be added.

Summary, TL;DR:
In your example of aluminum and zinc, each metal melts well below either of their boiling points so that loss via volatilization is not a problem. There are cases however, such as alloying copper and zinc, where the boiling point of one metal is lower than the melting point of the other. One way to minimize (but not eliminate) the loss of the more volatile metal is to quickly dissolve it in the high-melting metal and then cool the solution. Although this does not eliminate losses due to volatilization, it can greatly reduce the problem. And actually, since alloys are frequently composed of predominantly one metal, it is not uncommon to dissolve the lesser components into the primary component as a matter of practice anyway.

I hope the example I gave addresses your question. Please don't hesitate to ask for any clarifications in the comments below.


What is the pH of a 5M solution of hydrochloric acid?

As explained in pH Paradoxes: Demonstrating That It Is Not True
That pH ≡ -log[H+] Journal of Chemical Education Vol. 83 pages 752-757.

pH = -log (hydrogen ion activity).

and "pH of 7.6 M HCl is about -1.85 (not -0.88)"

So how far off is -log (5) = -0.69 from the real answer?

From this table of activity coefficients 5m HCl has an activity coefficient of 2.38, so activity is 11.9, which yields pH = -1.1.

Unfortunately the table is in terms of molality rather than molarity.  5M HCl is about 5.5m, and activity coefficient is dramaticly increasing with concentration (see table I of the pH paradox reference) so the true pH deviates somewhat further from -0.69, but pH = -1.1 is a reasonable estimate.


Actually arabinohexose got its name from the family of six carbon membered mono saccharide(i.e. aldohexose) and the first name arabinose is taken from the another popular name i.e. gum arabica because it was first extracted from it!


Why is the cis isomer of 1,3-dimethylcyclobutane more stable than the trans isomer?

The four carbon atoms in cyclobutane are not coplanar, as that would lead to a large degree of unfavourable eclipsing interactions. So, it is not a flat square; it adopts a puckered conformation. One of the carbon atoms makes a $25^\circ$ angle with the plane formed by the other three carbons, and this mitigates some of the eclipsing interactions, at the cost of a small increase in angle strain. The conformation is commonly called the "butterfly", and two equivalent puckered conformations interconvert rapidly. Here's a visual reference to help you

Pictured above is an unsubstituted cyclobutane.

Take a closer look at the figure on the left, and notice how there would be some transannular interaction between the two $\ce{H'}$ hydrogens at $\ce{C-1}$ and $\ce{C-3}$, marked in $\color{red}{\text{red}}$. This is bad. In the figure on the right, the $\ce{H'}$ hydrogens are no longer axial, but equatorial, and this is better.

In short, placing substituents in an equatorial position is better than placing them in an axial position.

Now, imagine what the cis and trans isomers would look like for 1,3-dimethylcyclobutane.

In the cis isomer both $\ce{H'}$ on $\ce{C-1}$ and $\ce{C-3}$ would be replaced by $\ce{-CH3}$ groups (since they are on the same side). Of course there will be considerable repulsions if they are arranged in the conformation on the left, which is why ring flip takes place and we will predominantly get the conformation on the right with both the methyl groups in equatorial positions. This is ideal.

The trans isomer would be like replacing one $\color{red}{\text{red}}$ $\ce{H'}$ and one $\color{blue}{\text{blue}}$ $\ce{H}$ at $\ce{C-1}$ and $\ce{C-3}$ with methyl groups. No matter what you do, you can only get one of the methyl groups at the equatorial position, but never both.


Why is the sum of two inexact differentials exact?

There's not much to do, as you nearly spotted the difference by yourself: $U$ is a state function, thus is path-independent. In other terms, one can estimate it at any moment only with an initial and a final value (and express it in terms of its variables only). But with $Q$ and $W$, it's different: these are path functions (or process functions) consequently depending on the path. Hence, their differential is not something one might express depending only of its variables. (these are not "real" functions for mathematicians...)

Is the sum of two path-dependent functions a path-independent function?

The sum of inexact differentials is an exact differential in that particular case. In other cases, it depends on the units involved: for instance, with entropy, we have $\mathrm{d}S = \delta Q_\mathrm{rev} / T$. So, no typical case here!

what is the difference between a partial derivative and an inexact derivative?

These are absolutely not the same! A partial derivative is a differentiation of a multi-variable function regarding one of its variables only, while what we have here is an inexact differential, thus not a derivative as 1) no variable is specified here, and 2) an inexact differential can be of a function depending on only 1 variable.


Reactions of Grignard Reagent: 1,2 vs 1,4 addition

This is not a 1,4 addition.  It is a 1,2 addition across a carbon-carbon double bond.  This mode of reaction is unusual for Grignard reagents, but here a highly stabilized "double benzylic" anion is formed and the competitive reaction of adding to the carbonyl group is sterically hindered.


In an article by Katharina C. Kress, Martin Kaller, Kirill V. Axenov, Stefan Tussetschläger, and Sabine Laschat, published in
Beilstein J. Org. Chem., 2012, 8, 371-378 (DOI), this procedure is described. From the Supporting Information, I read that the authors obtained the product in 72% yield after

  column chromatography on silica gel (hexanes/EtOAc, 1:1)

I wouldn't be astonished if that's as good as it can get.


If you just use hydroxide, then the initiator is water.


Using a Linear Combination of Atomic Orbitals (LCAO) will not exactly satisfy the Schrödinger Equation / be an eigenfunction of the Hamiltonian
We can write down the Hamiltonian for the $\ce{H2+}$ cation with bond length $2R$, nuclei at $\vec{R_1}$ and $\vec{R_2}$, using the Born-Oppenheimer approximation:
$$
H\Psi(\vec{r}) = \left[-\frac{\nabla^2 }{2} - \frac{1}{\vert\vec{r}-\vec{R_1}\vert} - \frac{1}{\vert\vec{r}-\vec{R_2}\vert} \right]\Psi(\vec{r}) = E \Psi(\vec{r})
$$
If you feed in a linear combination of 2 $1s$ orbitals into $H\Psi$ you get a mess that is not a multiple of the linear combination you put in - i.e. a linear combination of 1s orbitals is not the exact solution.
In this case, because we only have one electron, the Schrödinger Equation can be solved to arbitrary accuracy using a variety of numerical techniques for second order partial differential equations in 3D; such as solution in an basis of functions or on a grid in space.
Solving in Practice
A suitable infinite basis of functions can describe all possible functions in 3D.  Two different examples are spherical harmonics with suitable radial components (Slater type orbitals) such as the $1s$ orbital, or Gaussian functions with suitable angular components (Gaussian type orbitals more commonly used in computational chemistry today).
With a infinite number of functions we would have an exact solution, but with only the first few terms in a suitable series we can be remarkably accurate.
In practical electronic structure theory, this means 3-30 functions per atom with suitable chosen parameters, centering the functions on atomic nuclei, as that's where the electrons will be mostly due to attraction to nuclei.
Variational Principle
Seeing as using a finite number of orbitals will not give an eigenfunction of the Hamiltonian, and certainly not the lowest energy ground state, we need another way to confirm a solution.
Instead we solve the Schrödinger equation "projectively".  Hamiltonians are hermitian operators.  One of the properties of hermitian operators is that their eigenfunctions are  orthogonal.  What this means is that for a Hamiltonian $H$ and two exact solutions $\Psi_i$ and $\Psi_j$ the integral:
$$
<\Psi_i\vert H\vert\Psi_j> = \iiint_{\text{all space}} \Psi_i^* H \Psi_j \;\text{d}V = \cases{E_i \text{ if } \Psi_i = \Psi_j\\0 \text{ otherwise}}
$$
Taking the integral of two wavefunctions, one multiplied by the complex conjugate of the other, is often referred to as projecting one wavefunction onto another.
Another property of the Hamiltonian operator is that the value of the integral with some generalised wavefunction $\Psi$:
$<\Psi\vert H\vert\Psi>$, the "expectation value" of the energy of $\Psi$, is minimised by the exact ground state wavefunction.  This property is know as the variational principle.  Hence in self consistent field/Hartree Fock calculations, we solve equations to minimise that integral of our LCAO wavefunctions by varying the coefficents to get the best LCAO approximation to the true wavefunction.


Your initial compound weights 1g, and it is only made of C and H. The products are CO2 and H2O. In other words (the reaction is not balanced, but for our purpose we don't care):

CxHy + O2 --> CO2 + H2O

From the grams of CO2 you can get back to the moles (and therefore grams) of C in the initial compound, while from the grams of water you can get to the moles (and grams) of H.

moles of C = 1.42 g CO2 x $\frac{\text{1 mol CO2}}{\text{44.0 g CO2}}$  x $\frac{\text{1 mol C}}{\text{1 mol CO2}}$ = 0.0323 mol C
You then multiply by the molar mass of carbon, which is 12 g/mol, and you get 0.388 g.

We do the same thing for hydrogen:

moles of C = 0.872 g H2O x $\frac{\text{1 mol H2O}}{\text{18.0 g H2O}}$  x $\frac{\text{2 mol H}}{\text{1 mol H2O}}$  = 0.0969 mol H = 0.0989 g

Note that we did the same thing twice, but since we have two moles of H in water we multiplied by 2 in the second equation.
Now we just have to find the percentage:
You get

$\frac{\text{0.388 g}}{\text{1 g sample}}$ *100 % =  38.8 % C

$\frac{\text{0.0989 g}}{\text{1 g sample}}$ *100 % =  9.89 % H

I hope this helps!


There are several Inconels , the basic one is 600. Because the Inco and Huntington divisions are now gone or acquired ,it would be difficult to get literature . But they produced copious amounts , you may be able to find some ; I have about a 6" stack of brochures I have been too lazy to discard. American Society for Metals Handbooks should be easy to find ( there are several issues). ASM is a common and comprehensive source. What do you mean "Flare" ?

The 600 Inconels do not age harden; The 700 Inconels do age harden. All have good strength and oxidation resistance at high temperatures . Some useful above 2000 F. If you are talking about a rocket ,I guess you do not need blade and vane alloys ( for turbines). 800 and 900 are Incoloys  ; 400 and 500 are Monels
Addenda
Looking at SpaceX info they are using something like Inconel 750 ( formerly X , and X750 ), so it is age hardened for strength due to pressure in the engine. Supposedly they have a modification named SX 500 . Apparently by "flare" they means rapid oxidation because of the oxygen in the engine , but they do not explain.


Determine the ΔH for the reaction in kJ/mol of ethanol?


Which formula to use to find the pH of a solution?

From reading your question it is obvious that you are confused. When I studied chemistry in college in the 1970's I don't remember the Henderson-Hasselbach equation being a famous equation. It seems like it is now being used as some magic can opener.

You need to go back to the chemistry and figure out the chemistry first, then do the math.

Ok, let's start by noting that the pKa of acetic acid is 4.76 from Wikipedia. There are two significant figures in the mantissa so a mathematical solution good to 1% will be adequate. Also we are dealing with solutions on the order of $10^{-3}$ molar. This allows us to make two very important assumptions which significantly simply the mathematics.

First we assume that we can ignore the autoionization of water.
Second we assume that since acetic acid is a weak acid the solution ends up being essentially 0.004 molar sodium acetate and 0.001 molar acetic acid.

Because we have excess acetic acid we know that the solution will be acidic. If we wanted a more exact solution we'd ignore the second assumption and use the $K_\mathrm{a}$ expression

$K_\mathrm{a} = \dfrac{\ce{[H^+][Ac^-]}}{\ce{[HAc]}}$

to solve the problem more exactly. Letting $\ce{[H^+]} = x$ then $\ce{[Ac^-]} = 0.004 + x$ and $\ce{[HAc^-]} = 0.001 - x$. This requires solving a quadratic equation which is doable, but mathematically messier.

Now back to using our two assumptions we can go to the Henderson–Hasselbalch equation and solve the problem with a simple formula.

$\mathrm{pH} = \mathrm{pKa} + \log{\dfrac{\ce{[Ac^-]}}{\ce{[HAc]}}} = 4.76 +  \log{\dfrac{0.004}{0.001}} = 5.36$

This means that the $\ce{[H^+]} = 4.4 \times 10^{-6}$.

So our assumption that the final solution ends up being essentially 0.004 molar sodium acetate and 0.001 molar acetic acid is reasonable.
The assumption that the autocatalysis of water can be ignored is a bit dicier.


How to find the average half life of radioactive nuclide which undergoes two different decays?

But I am getting the correct answer only if I take $\lambda_2$ as $2$ times the given decay constant for the second reaction. Is this because of $\underline{2}A$ on the reactant side instead of $A$?

Yes.

The way I tend to approach half-life problems is to recast them as the relevant kinetic differential equations.  Assuming the first-order kinetics as stated in the problem, the contribution of reaction 1 is
$$
\left(\frac{\mathrm{d}[A]}{\mathrm{d}t}\right)_1 = -\lambda_1[A],
$$
and that of reaction 2 is†
$$
\left(\frac{\mathrm{d}[A]}{\mathrm{d}t}\right)_2 = -2\lambda_2[A].
$$

Implicit in the above is the assumption that the first-order decay constants apply to the stoichiometry of the reactions as-written. As you correctly note, that extra factor of $2$ in the differential equation for reaction 2 is because two $A$ nuclei are involved per "unit" of reaction progression.

The total rate of loss of $A$ is the sum of the above:
$$
\frac{\mathrm{d}[A]}{\mathrm{d}t}
= -\lambda_1[A] -2\lambda_2[A]
= -\left(\lambda_1 +  2\lambda_2\right)[A].
$$

This is a straightforward first-order ODE, readily solved for the overall half-life by standard methods.

† While it seems quite unusual to me to have a second-order reaction ($2A\longrightarrow B$) with a first-order rate constant (units of $\mathrm{s}^{-1}$), for the sake of the problem I'm content to roll with it. I'm sure they cast the problem this way so that the calculus was straightforward.

As well, simultaneously having $A \longrightarrow B$ and $2A \longrightarrow B$ processes would seem to violate all kinds of conservation laws--the problem makes no physical sense as a result. Plus, I know of no second-order nuclear process that occurs outside of a particle accelerator. $2A \longrightarrow C$ would have been much less preposterous, even if implausible in the context of low-energy radioactive decay.


Enantiotopic hydrogens in cyclopropanone

Cyclopropanone is a reactive species, as such there aren't many experimental reports in which it is characterised, however some (possibly questionable data) does exist:

  Cyclopropanone is a highly reactive compound. It is stable for a few days at liquid nitrogen temperature but rapid polymerisation takes place above o or even below if traces of water are present. The reaction is highly exothermic.

  Ref: Recueil des Travaux Chimiques des Pays-Bas, 1966, 11, 1170. DOI:10.1002/recl.19660851113

.

  The NMR spectrum  (neat) shows a singlet  at 1.72 p.p.m.  relative  to  tetramethylsilane.

  Ref: Recueil des Travaux Chimiques des Pays-Bas, 1966, 11, 1170. DOI:10.1002/recl.196608511

It appears that in the absence of a chiral environment, all of the protons are in a single environment at 1.72 ppm (Hans Reich quotes a single peak at 1.65 ppm) which is what we'd expect (the molecule is essentially planar, and all of the protons are equivalent).

As you point out in your question, replacing any one of the protons (hypothetically) would yield two enantiomers, which, in the presence of a chiral environment, should be distinguishable. It can therefore be concluded that in a chiral environment, the protons should appear to be enantiotopic.*

* there are two slight caveats to this. Firstly, the compound would still be unstable and prone to polymerising, therefore there is no way of actually testing this. Secondly, the chiral environment must be able to interact with the molecule, for things like chiral alcohols, lanthanide shift reagents are (/used to be) commonly used, but I'm less convinced how you put a ketone into a chiral environment.


What's with the weird definition for the Calorie?

A number of the Chemistry texts I've used or examined for adoption have indeed confused heat and thermal energy. They sometime refer to a body possessing heat. But Physics books are pretty unaminous that heat is the transfer of thermal energy and not a state property of matter.

I've never seen, in texts I've used, heat restricted strictly to the interface between the participants in the exchange. It refers to the total amount of thermal energy transferred between two bodies. It can also be used to describe how much energy was lost by one of the bodies in the exchange and how much was gained by the other. One way to reconcile this with your interface statement is that once the surface of a body warms due to contact with a hotter body, the higher temperature at the surface relative to the rest of the body will continue to cause heat to flow within the body.

As to the choice of temperature, it shouldn't near zero since this is where a phase change occurs.

Edit. In response to your edit, #2. Search for a table of "specific heat of water". Notice the specific heat is not constant as temperature varies.
I'll hazard a guess on the 14.5-15.5 part. Think 15 degrees. But, since an interval is required, let's begin a half degree before and end an equal amount after. As to why 15 degrees, hmmm. One answer to a recent question on the Physics forum says this is roughly room temperature. Pretty cold for me but maybe true when and where this was defined. (Sorry for no link, I'm not that fascile on iPad.)


Is there a difference between potassium manganate and permanganate?

No, they are somewhat different. $\ce{K2MnO4}$ is potassium manganate and $\ce{KMnO4}$ is potassium permanganate. Notice the two potassium for each $\ce{MnO4}$.

The Roman numerals mark the oxydation state. So, manganate will have a (VI) and permanganate will have a (VII).


You are drawing very nice $\mathrm{p}_x$, $\mathrm{p}_y$ and $\mathrm{p}_z$ orbitals onto that piece of paper. But if you take an isolated atom, which direction is $x$? Which is $y$? The coordinate system you’re seeing the atom in is just something we planted over its top because humans like things nice and organised. It is entirely arbitrary.

By the way, the direct calculations do not give $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals (strictly speaking not even $\mathrm{p}_z$. They give you $\mathrm{p}_0$, $\mathrm{p}_1$ and $\mathrm{p}_{-1}$ (the indices referring to the value of $m_l$, the ‘magnetical quantum number’). In spherical coordinates, which are used because it makes the calculations easier, only the $z$ axis is semi-defined by values $\theta = 0$. Because $\psi(\mathrm{p}_0) = f(r, \theta)$ (the wavefunction of one of the p-orbitals is independent of $\phi$), we say that this represents the $\mathrm{p}_z$ orbital. $\mathrm{p}_1$ and $\mathrm{p}_{-1}$ (the direct solutions of the Hamiltonian) cannot even be mapped to directions as they involve complex numbers; only via linear combinations can we transform them into two real orbitals pointing in an arbitrary $x$ and an arbitrary $z$ direction.

Things change if you introduce an external magnetic field. The atom’s $z$ axis will now align with the magnetic field in some way, giving your system one preferred direction. One, not three. $x$ and $y$ are still essentially undefined. However, in a magnetic field it would make a difference which orbitals the electrons populate.

You might also be interested in checking out the Jahn-Teller effect which occurs in transition metal chemistry to avoid filling some of all degenerate orbitals. In a nutshell it is energetically better for certain complexes to distort themself from a perfect octahedron to create distinct energy levels that can be fully populated, half populated or unpopulated.


Order of second ionization energy for cations

You are correct about $\ce{Li^+}$: the second electron to be extracted would be one of the two extremely tightly bound $\text{1s}$ electrons, which explains the very high second ionisation energy of that ion.

As for the comparison $\ce{Be^+}$ to $\ce{C^+}$, the main effect here is effective nuclear charge, $Z+1$, which in the case of $\ce{Be^+}$ is $5$ and in the case of $\ce{C^+}$ is $7$.

Equivalent electrons (same or similar orbitals) are more tightly bound to the nucleus, hence harder to extract, for higher $Z+1$. This results in a trend of higher second ionisation energies for higher $Z$, across a period.

This is a general trend across each period.

  What is the general steps of rules to solve such kinds of questions?

Chemguide is a good place to start.


Why do unsaturated hydrocarbons burn with a sooty flame while saturated hydrocarbons burn with a clear flame when burnt in the air?

Your observation that unsaturated hydrocarbons burn with a sooty flame is not always true and depends on the circumstances. There are more factors at play than the nature of the hydrocarbon that is burning.

The most important reason why flames are sometimes smoky is that the combustion is incomplete. This means that not enough oxygen was present to fully oxidise all the carbon to carbon dioxide or carbon monoxide gases (and the amount of CO will vary with the ratio as well as, given enough oxygen, it should also be completely combusted to carbon dioxide). If there is not enough oxygen in the flame there are many competing side reactions that generate sooty resides (the reactions are pretty complex). Some pyrolysis reactions of hydrocarbons may occur before combustion itself (if the mixing in the flame is less than perfect) and these may contribute to the generation of sooty products which may be harder to burn than the original hydrocarbon.

Incomplete combustion is as much a function of the environment of burning as it is the thing that burns. Some environments don't enable enough oxygen to get into the gas mixture to drive complete combustion and the reactions can easily leave a lot of sooty emissions. A candle, for example, burns with a smoky flame despite being made from mostly saturated long chain hydrocarbons. It does so because a candle wick doesn't do a good job of mixing air with the vapours of the burning hydrocarbon (this is deliberate as the prime role of candles is to provide light and this comes from incandescent particles of soot in the flame: a pure flame burning efficiently would provide little light by itself).

Even well controlled environment such as engines can't completely avoid side reactions. But Diesel engines (which use mostly saturated hydrocarbons) produce more sooty particles than petrol (gasoline) engines though petrol engines have far more unsaturated fuels. The exact reasons are complex and depend on the precise environment inside the engine.

In well controlled environment where the oxygen-fuel ration is set carefully most hydrocarbons burn cleanly with little soot. Gas-based cookers (using mostly saturated methane, ethane, propane or butane) are very clean because they control the gas mix very precisely (use the wrong burner and they may well be sooty!). Welding and cutting torches can happily use the highly unsaturated acetylene with a clean, non-sooty, flame (the oxygen-acetylene flame is one of the hottest achievable) because they control the gas mix very carefully leaving little room for side reactions.

Uncontrolled flames might well show your general observation that unsaturated fuels burn more sootily, but this is as much a product of the environment as it is the product of the hydrocarbon. It is hard, without a carefully controlled experiment, to tell whether the general observation is true at all.


What is some advance reference books about alloy that gives a lot historical background and applies a lot of theoretical equation or formulas

The classic would be Porter and Easterling's "Phase Transformations in Metals and Alloys", apparently still available. This should give you a good basis for understanding the thermodynamics of binary (and higher) systems.

For a deeper look at how computational thermodynamics is done, like CALPHAD, the Lukas, Fries, and Sundman book "Computational Thermodynamics" might do, but that has zero historical background.

You might also try asking this in Physics using the materials-science tag.


There is no restriction that soap be natural or unnatural. Soap are salts and detergents are not necessarily salts. The latter is a very broad term. Soap has a narrow meaning.
IUPAC clearly defines soap as

"A salt of a fatty acid, saturated or unsaturated, containing at least
eight carbon atoms or a mixture of such salts."

Just note that not all soaps are used for cleaning nor there is any condition that they should be soluble in water. Insoluble soaps of metallic salts serve as lubricants.
Similarly,
IUPAC defines a detergent as

"A surfactant (or a mixture containing one or more surfactants) having
cleaning properties in dilute solutions (soaps are surfactants and
detergents)"

Just like banana is a fruit but not all fruits are bananas, in the same way, all water soluble soaps are detergents and surfactants. Not all detergents are soaps.
Visit the IUPAC Gold Book and search the entries detergents and soap.


Option (D) is the answer. This may be predicted by redistribution of hybrid orbital theory or simply "Bent's Rule".
Clearly, during movement of the lone pair from one sp2 hybrid orbital to the other(to interconnect between the cis and trans isomers), there has to be an intermidiate transition state, during flipping when the lone pair resides in an unhybridized p-orbital. However the activation energy to achieve this intermidiate transition state is low for option (D).
Between N and H (as in the question), the electronegativity of N is higher than H. This pulls the shared electron pair in the N-H bond towards the N atom. In order to stabilize the high electron density, there is a redistribution of hybridization observed. The s character of the sp2 hybrid orbital of N (forming bond with the H ) increases ( eg from sp2 to sp1.8). This causes a corresponding decrease of s character of the other sp2 hybrid orbitals of N, such as the one containing the lone pair. The decrease in s character is associated with a a corresponding increase in p character of the lone pair containing orbital( eg from sp2 to sp2.3). Since it is easier to achieve transition from sp2.3 to unhybridized p orbital (having 100% p character), in the transition state it is easier to mediate between geometric isomer, possibly even existing in equilibrium.
For further reading see this answer:
Why are oxime geometrical isomers stable?


The structure of the triiodide ion places a negative formal charge on the central iodine atom.

No it doesn’t. The two resonance structures that describe the four-electron three-centre bond put the negative formal charge on the outer iodines ($\ce{1/2-}$ each).

That said, polarity is usually defined as having a non-zero dipole moment. The dipole moment’s vector must display the same symmetry as the entire molecule. Since the molecule is linear and both $\ce{I\bond{...}I}$ distances are equal, its point group is $D_{\infty \mathrm{h}}$ which includes $i$ which means the overall dipole moment must be zero.

The charge it carries does not matter. Any single-atom ion also has zero dipole moment and would thus be called non-polar. ‘Non-polar’ does not mean ‘free of electrostatic interaction’.


A nomenclature note: conventionally, in chemistry "a metal" is an element (such as a transition metal or alkali metal). If what you're talking about is a substance that behaves as a metal, that's typically referred to as metallic, like metallic glasses.

Therefore, I would rephrase your question as "Can metallic ammonium exist?" - in fact, this has been a question for quite awhile. From what I've read, there's a theoretical basis that it should be possible at given temperatures and pressures, but it hasn't been experimentally observed. The papers I was reading were from the 50s-70s, so I'll keep looking for more recent information.

Solvated electrons are slightly related, in that there is a "free" electron, and this was in fact originally documented in liquid ammonia.


Fuel cell potential for the electrolysis of ammonia

You can see my detailed response on the sign of electrode potentials here:
Still taught to reverse oxidation half cells in electrochemistry?

To summarize, electrochemists all over the world have decided to quote all electrode potentials as reduction potentials. This tug of war of signs between the so-called European convention and American convention of electrochemistry for more than a century, resulted in this decision. Basically, the current convention is the same as that of Ostwald.


Which etching process is better, photochemical or electrochemical?

While I have no expertise in this subject I did a web search for "advantages acid etching has over electroetching" which led to the following which I copy verbatim, specifically the part that seems relevant to your comments:

  Electro-etching offers a great alternative to etching with acid, but
  as with everything, it also has some disadvantages:

  • A feature of electro-etching is that it will bite isolated lines deeper than those that are close together, or cross hatched lines.
  Also it tends to bite harder in the external areas than at the center
  of plates (edge-effect).
  In addition, electrolytic etching bites harder on the edges of wide
  lines and edges of resists of large areas of open bite. This is more
  noticeable in prolonged bites. However, all these drawbacks can be
  compensated for by using a cathode grid in the electro-etching unit
  instead of a cathode plate.


Why is the principal energy of an electron lower for excited electrons in a higher energy state?

Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~\mathrm{eV}
$$
which is the negative of the energy required to remove an electron from the ground state of a hydrogen atom.

If we increase $n$ to say $n=2$, then we have,

$$
E=-E_0/4=-3.4~\mathrm{eV}
$$
which is a larger number than for $n=1$. Don't let the minus sign confuse you.

This is a very common source of confusion when these equations are seen for the first time. The confusion often stems from the fact that we are free to choose the zero of energy wherever we would like. So, in this case, zero energy corresponds to the case where the electron and nucleus are infinitely separated which is the $n\rightarrow\infty$ limit. So, more negative numbers correspond to lower energies and more tightly bound electrons.


By what mechanism do acids deprotect primary silyl ethers?

Fluoride-mediated deprotection of silyl ethers proceeds through a pentavalent silicon pathway:[1]

This mechanism proceeds the same for both primary and secondary silyl ethers, and there's no reason the mechanism for acid-catalyzed deprotection would be any different. Though I couldn't find a mechanism proposed in literature, it seems likely it would proceed through a similar pentavalent intermediate (Nu = nucleophile):

Steric effects are what drive the chemoselectivity of these reactions. Usually the differences in steric environments around the carbinol carbons is enough, but the use of different silyl groups on the two alcohols can further increase yield of the desired product. These steric effects are particularly pronounced in the following transformation, in which a primary TBS ether is selectively deprotected in the presence of a primary TBDPS ether and a secondary TIPS ether:[2]

Reaction conditions can also have a major effect on the chemoselectivity of desilylation reactions. For example, when HF•pyr is used in a THF-pyridine mixture, selective removal of the primary TBS ether in a protected triol was achieved in 84%. After oxidation of the newly deprotected alcohol to a carboxylic acid, the secondary TBS ether was cleaved using HF•pyr without excess pyridine, leading to lactonisation:[3]

Nelson and Crouch have written a review on the selective deprotection of specific silyl ethers in a compound with multiple silyl protecting groups.[4]

References

Kim, S.-K. Chitin and Chitosan Derivatives: Advances in Drug Discovery and Developments; CRC Press, 2013; p 80.
Larivée, A.; Unger, J. B.; Thomas, M.; Wirtz, C.; Dubost, C.; Handa, S.; Fürstner, A. The Leiodolide B Puzzle. Angew. Chem. Int. Ed. 2011, 50 (1), 304–309 DOI: 10.1002/anie.201005850.
Körner, M.; Hiersemann, M. Enantioselective Synthesis of the C8−C20 Segment of Curvicollide C. Org. Lett. 2007, 9 (24), 4979–4982 DOI: 10.1021/ol702092h.
Nelson, T. D.; Crouch, R. D. Selective Deprotection of Silyl Ethers. Synthesis 1996, 1996 (09), 1031–1069 DOI: 10.1055/s-1996-4350.


How to derive these general formulae for number of stereoisomers of a compound with a possible plane of symmetry?

I managed to crack the formula for optical isomers with odd chiral centers, so I'll share my attempt here. Hopefully others may innovate on it and post solutions for other formulae.

Pseudo-chiral carbon atoms - an introduction
The Gold Book defines pseudo-chiral/pseudo-asymmetric carbon atom as:

a tetrahedrally coordinated carbon atom bonded to four different entities, two and only two of which have the same constitution but opposite chirality sense.

This implies that, in your case:

If chiral carbons 2 and 4 both have configuration R (or both S), then the central carbon 3 will be achiral/symmetric, because now "two and only two of its groups which have the same constitution" will have the same chirality sense instead. (Your approach by "plane of symmetry" is wrong. Find more details on this question)
Hence, there can be two stereoisomers (r and s) possible on the 3rd carbon due to its pesudochirality. But, there will be only one if both substituents on left and right have the same optical configurations.

Building up an intuition by manual counting
For optical isomers with odd number of chiral centers and similar ends, you can guess that, if there are $n$ chiral centers, then the middle ($\frac{n+1}2$-th) carbon atom will be pseudo-chiral. To build up an intuition, we'll manually count optical isomers for $n=3$ and $n=5$:
Case $n=3$
Take the example of pentane-2,3,4-triol itself. We find four (=$2^{n-1}$) isomers:
$$
\begin{array}{|c|c|c|}\hline
\text{C2}&\text{C3}&\text{C4}\\\hline
R&-&R\\\hline
S&-&S\\\hline
R&S&R\\\hline
R&S&S\\\hline
\end{array}
$$
As expected from the relevant formula, we find that the first two ($=2^\frac{n-1}2$) are meso compounds, and the remaining two ($=2^{n-1}-2^\frac{n-1}2$) are enantiomers.
Case $n=5$
Take the example of heptane-2,3,4,5,6-pentol:

We expect $16~(=2^{n-1})$ isomers, with the C4 carbon being pseudo-chiral. To avoid a really large table, we observe that the number of meso isomers is easily countable (<< number of enantiomers). Here is a table of those four (=$2^\frac{n-1}2$) meso isomers:
$$
\begin{array}{|c|c|c|c|c|c|}\hline
\text{C2}&\text{C3}&\text{C4}&\text{C5}&\text{C6}\\\hline
R&R&-&R&R\\\hline
R&S&-&S&R\\\hline
S&R&-&R&S\\\hline
S&S&-&S&S\\\hline
\end{array}
$$
Note that the total optical isomers are given by $2^{n-1}$ isomers (more on that below). Hence, the number of enantiomers is easily $12(=2^{n-1}-2^\frac{n-1}2)$.

A formula for the number of meso isomers
As you must have observed from the table, the sequence of optical configurations, when read from the fourth carbon atom, is exactly the same towards both left and right. In other words, if we fix an arbitrary permutation for the optical configurations of the carbon atoms on the left (say RSS), then we will get only one unique permutation of the optical configurations on the right (SSR).
We know that each carbon on the left has two choices (R or S), and there are $\frac{n-1}{2}$ carbon atoms on the left. Hence, the total number of permutations will be $2\times2\times2\cdots\frac{n-1}{2}\text{ times}=2^\frac{n-1}{2}$.
Since, our description ("the sequence of optical configurations, when read from the fourth carbon atom, is exactly the same on both left and right") describes meso isomers, we have hence counted the number of meso isomers, which is $2^\frac{n-1}{2}$.

A formula for the number of total isomers
We note that there are $n$ chiral carbons (including that pseudo chiral carbon). Again, each chiral carbon has $2$ choices. Hence, the maximum possible number of optical isomers is $2\times2\times2\cdots n\text{ times}=2^n$. This is the maximum possible, not the actual total number of isomers, which is much lower.
The reduction in number of isomers is because the string of optical configurations reads exactly the same from either terminal carbon. Example: RSsRS is the same as SRsSR. This happens because the compound has "similar ends"
Hence, each permutation has been over counted exactly twice. Thus, the actual total number of isomers is half of the maximum possible, and is  $=\frac{2^n}2=2^{n-1}$.

Conclusion
Hence, we have derived that, if 'n' (number of chiral centers) is odd for a compound with similar ends, then:

$\text{Number of meso isomers} = 2^{(n-1)/2}$
$\text{Total number of optical isomers} = 2^{n-1}$
$\text{Number of enantiomers} = 2^{n-1}-2^{(n-1)/2}$


Na+ or Ca2+ . Which ion has the larger radius?

The thing is that you need to know the coordination environment in the first place as ionic radii are C.N.-dependent. From Shannon's canonical paper "Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides"[1]:

\begin{array}{cc}
\hline
\text{Ion} & \text{C.N.} & \text{C.R., Å} & \text{I.R., Å} \\
\hline
\ce{Na+} & 4 & 1.13 & 0.99 \\
& 5 & 1.14 & 1.00 \\
& 6 & 1.16 & 1.02 \\
& 7 & 1.26 & 1.12 \\
& 8 & 1.32 & 1.18 \\
& 9 & 1.38 & 1.24 \\
& 12 & 1.53 & 1.39 \\
\hline
\ce{Ca^2+} & 6 & 1.14 & 1.00 \\
& 7 & 1.20 & 1.06 \\
& 8 & 1.26 & 1.12 \\
& 9 & 1.32 & 1.18 \\
& 10 & 1.37 & 1.23 \\
& 12 & 1.48 & 1.34 \\
\hline
\end{array}

where C.N. - coordination number; C.R. - crystal radius; I.R. - ionic radius. The data is also available online for free. Comparing similar coordination environments, it's easier to notice that overall $\ce{Na+}$ is indeed larger than $\ce{Ca^2+}$:

Reference

Shannon, R. D. Acta Cryst A, 1976, 32 (5), 751–767 DOI: 10.1107/S0567739476001551.


Simulating H1 NMR spectra using chemical shift/multiplicity data?

I am not aware of any software that can transform a spectrum into a molecule. But if you are asking for look up database have a look at Spectral Database for
Organic Compounds SDBS

Here you can input simple data as you asked for like ppm, and then get suggested moleulces as output.


How does one determine the relative basicity of a nitrogen in a molecule?

Use a combination of

the exclusion principle: sort out the unlikely centres with low electron density and
paper work to figure out those nitrogen atoms for which the additional positive charge can be distributed over a larger part of the molecule without moving the proton.

In the case of caffeine (1) the amide and the imide are rather unlikely centres for protonation! You have probably already ruled out the amide nitrogen in the benzamide case (your second molecule) in favour of the aniline site.

From an initial number of four possible centres in 1, only two remain.

Now, let's see whether the positive charge resulting from protonation can be stabilized.

This looks fair, but creates a quarternary ammonium ion for which no obvious charge distribution seems to be available.

Let's examine the last possibility:

Would you agree that this looks much more promising?


Why do gummy bears explode when added to hot potassium chlorate?

Potassium chlorate is a source of oxygen. After heating, it decomposes to $\ce{O2}$ and $\ce{KCl}$:
$$\ce{4 KClO3 → KCl + 3 KClO4}$$
$$\ce{KClO4 → KCl + 2O2}$$

The gummy bear is mainly composed of sugar and other carbohydrates. Those carbohydrates will react with oxygen, combustion occurs. For example, glucose will react in this manner:
$$\ce{6O2 + C6H12O6 -> 6CO2 + 6H2O}$$

If there is any material present which does not burn, such as $\ce{H2O}$, the temperature will not rise as high. For gummy bears the reaction works spectacularly because they are mainly carbohydrates (>70%).

An apple, for example, has only ~13% carbohydrates – unless you dry it, of course. On the other hand, this video on YouTube is an example of how sugar itself reacts violently with potassium chlorate.


What is the standard way of reporting/interpreting measurement uncertainty in scientific publications?

In my experience, this varies greatly with your field of study.

In organic chemistry, I have yet to experience proper uncertainty reporting in journal articles or lab reports with the notable exception of physical organic chemistry. Once in my undergraduate years, I noted while writing a report that $130~\mathrm{mmol}$ where actually $1.3 \times 10^2~\mathrm{mmol}$ and wrote it that way. Okay, I should have written $0.13~\mathrm{mol}$ in a more correct notation to signify the significant digits. But the way the TA marked the report clearly told me ‘I don’t give a damn about correct error margins or significant digits, just God damn write $130~\mathrm{mmol}$!’ Lesson learnt from that anecdote: synthetic organic chemistry does not care about minor details such as significant figures or uncertainty.

The closer you get to physical chemistry, the more your measurements actually represent data values to be cited in a table and thus the more exact the values determined become, the more people care about reporting error margins and uncertainty. And funnily enough, in biology this is also done much more regularly. I suspect that there are different underlying reasons: One side (physical chemistry/physics) does not want the negative outcome of ‘you pretended your results are more accurate than they are!’, while the other side, biology, knows that it is intrinsically dealing with very error-prone systems that require large repetition counts which can then be analysed statistically.

Although synthetic organic chemistry practically also deals with rather error prone systems, either the idea is that there are too many uncontrollable variables or the idea is $n=1$ (always!); in any case, error bars or uncertainty are deemed superfluous.


What is the fate of sulfur in cysteine when it participates in gluconeogenesis?

Cysteine is a dispensable amino acid required for synthesis of protein and non-protein compounds. These non-protein compounds include pyruate (hence acetyl coenzyme A), taurine, sulfate, and glutatione (GSH). The fate of the cycteine is hence sketched in following diagram (Ref.1):

As evidence from the diagram, the sulfur atom in cysteine has been converted to sulfate ion ($\ce{SO4^2-}$) through $\ce{HS-}$ ion. This fact has been previously proved experimentally (Ref.2). Accordingly, the first major step in cysteine catabolism in mammalian tissues is the oxidation of cysteine to cysteine sulfinic acid catalyzes by cysteine dioxygenase (CDO). The reaction mechanism of this non-heme iron dioxygenase has been recently reviewed (Ref.3).

References:

G. Courtney-Martin, P. B. Pencharz, “Chapter 19: Sulfur Amino Acids Metabolism From Protein Synthesis to Glutathione,“ In The Molecular Nutrition of Amino Acids and Proteins; Dominique Dardevet, Ed.; Academic Press, Elsevier Inc.: London, United Kingdom, 2016, pp. 265-286 (https://doi.org/10.1016/C2014-0-02227-7).
Catherine L. Weinstein, Rudy H. Haschemeyer, Owen W. Griffith, “In vivo studies of cysteine metabolism. Use of D-cysteinesulfinate, a novel cysteinesulfinate decarboxylase inhibitor, to probe taurine and pyruvate synthesis,” The Journal of Biological Chemistry 1988, 263(32), 16568-16579 (http://www.jbc.org/content/263/32/16568.full.pdf+html).
Crisjoe A. Josepha, Michael J. Maroney, “Cysteine dioxygenase: structure and mechanism,” Chem. Commun. 2007, (32), 3338-3349 (DOI: 10.1039/B702158E).


Making three membered ring using different reactants

I think that both reactions involve carbene as common intermediate which then is the species involved in the subsequent cyclopropanation reaction.
These are the two equations which lead to the formation of the carbene:
$$\ce{KOH + CHCl3 ->  :CCl2 + KCl + H2O}$$
$$\ce{Cl3CCO2Na ->  :CCl2 + NaCl + CO2}$$
Subsequently, the alkene undergoes cyclopropanation by dichlorocarbene.
In the second case, heat is required in order to promote decarboxylation, which is entropically favored.


How do I visualise bicyclo[4.4.1]undeca-1,3,5,7,9-pentaene?

This compound is 1,6-methano[10]annulene, and a classic example of an unusual compound displaying aromaticity.

Aromaticity requires the pi-system to lie in a plane. One might speculate that [10]annulene (shown below) would be aromatic, because it contains a cyclic pi-system of 10 electrons, which appears to be planar. However, because the hydrogens indicated are trying to occupy the same space, the compound is forced out of planarity, and the compound it non-aromatic.

Bianchi and coworkers reported the crystal structure of 1,6-methano[10]annulene. In this compound, the hydrogens that clash in [10]annulene are replaced by a methylene (CH2) group. From directly above, the compound looks basically like naphthalene, with a little extra space in the middle. More interestingly, from the side, it's clear that the atoms of the 10-membered ring are nearly planar, with that methylene bridge oriented perpendicular to the plane.

Reference: Structure of 1,6-methano[10]annulene, R. Bianchi, T. Pilati and M. Simonetta, Acta Cryst. (1980). B36, 3146-3148, https://doi.org/10.1107/S0567740880011089


Is the hybridization only related to atomic orbitals but not to the molecular orbitals?

Hybridisation is a concept devised by man to ease explanation of chemical observations. It is not a 1:1 equivalent of anything seen in molecules but rather a way of rationalising what one has seen — one that is as mathematically correct as molecular orbitals (that span the entire molecule) are.

There are two fundamentally different approaches on how to describe ‘bonds’. The first one is the molecular orbital approach. It considers electrons that can freely move around the entire molecule, in a so-called potential energy space: the potential energy derives from the (negatively charged) electron being close to (a series of positively charged) nuclei which attract each other. As you may remember from your very first chemistry classes ever, the mass of even a proton is $2000$ times that of an electron, so a carbon nucleus is some $24000$ times heavier than an electron. According to Newton, the nuclei have a lot more inertia than the electrons do, so it is reasonable to assume that whichever fixed positions the nuclei take, all electrons can then assume favourable movement patterns.

Quantum mechanics, and further quantum chemistry, give us methods to calculate what movement patterns these electrons might assume — although technically the equation is not solveable, because it also must contain electron-electron interactions, there are good approximations out there. The wave functions that one can derive from all this information correspond to molecular orbitals you might have seen before: They abide by the molecule’s symmetry (will always fit into one irreduceable representation of the molecule’s symmetry point group) but extend over the entire molecule (even the orbitals representing atomic core electrons can be thought of being delocalised somewhat).

Using the quantum mechanical method of calculation (operators) you get an energy for each molecular orbital. Thereby you have ordered your molecular orbitals by energy (approximately equalling the number of nodal planes) — this gives you HOMO and LUMO that you might have heard of before.

To reduce this back into ‘bonds’, we need to use mathematical tricks. Simply spoken: All wave functions can be treated as vectors, and vectors can be linear combinated to give other vectors — meaning we take $0.5$ times orbital $\Psi_1$, $0.01$ times orbital $\Psi_2$ and so on for each molecular orbital, add them up, divide them by a normalisation factor, and we get a new orbital that has a different shape (and a new energy). If we do this in a sensible way, we can manage to reduce molecular orbitals to those that only ‘span’ across one bond — because we have cancelled out all wave functional contributions from other parts of the molecule. This would correspond to your typical $\sigma$ and $\pi$ type bonding, nonbonding and anti-bonding orbitals.

The second method is the hybridisation approach. This time, you take the general shape of the molecule and consider each atom by itself. For each atom, you can assume a hybridisation based on its geometry: If there are three other atoms in a triangular shape around it, all sharing the same plane with the atom in question, we can assume it to be sp² hybridised.

Once we have done this, we need to take a look at neighbouring atoms to create ‘bonds’ again. Usually, there will be two hybrid orbitals ‘pointing’ at each other, ready to create a bond. Using the same technique we have to create the hybrid orbitals, we can mix these two hybrid orbitals to create another orbital. This orbital is not considered to be of a new sp-hybrid type, but rather gets a new type of name: It can be $\sigma$, $\pi$ or $\delta$ (to name the most common ones). These Greek letters closely resemble s, p and d, and the bond orbitals in question share a common symmetric trait with the corresponding atomic orbitals: $\sigma$ orbitals can be rotated freely around the bond axis, $\pi$ orbitals have one and $\delta$ orbitals two nodal planes along the bond axis (the bond axis is completely part of the nodal plane).

If we did this correctly, we will arrive at orbitals of similar shape and energy as with the above method, only we are much, much faster, which is why this method is often taught in schools.

Onto the question of how do the $\ce{C-C}$ (and $\ce{C-H}$) bonds differ in $\ce{C2H4}$ and $\ce{C2H2}$:
You just saw how we decided what orbitals we want to combine to form a bond. In $\ce{C2H4}$, we are combining a carbon’s sp² orbital with a hydrogen’s s-orbital to create a $\sigma$ bond. For simplicity, we can label this bond $\sigma_{\ce{sp^2-s}}$. Similarly for $\ce{C2H2}$, only that this one has to be labelled $\sigma_{\ce{sp-s}}$. You see that you created bonds with different hybrid orbitals and therefore the bonds will have different ‘shapes’ and energies. This is what is meant with

  The $\ce{C#C}$ bond of ethyne ($\ce{HCCH}$) has higher s character ($50\,\%$) than the $\ce{C=C}$ bond of ethylene ($\ce{H2CCH2}$) ($33\,\%$)

Technically, this is short for:

  The hybrid atomic orbitals of the carbons of ethyne have a higher s character ($50\,\%$) than those of ethene ($33\,\%$), so the $\sigma_{\ce{C-H}}$ bonds will retain that property somewhat.

Long story short: No, we cannot say that a $\sigma$ bond has sp² or sp³ character; we can (and should) only say that the atomic orbital of one side of that bond has this character.

And: It is wrong to assume $\ce{sp^2 + s} = \ce{2 sp}$, because that would recentre a p-orbital from the carbon to the hydrogen.