Are matrices a good way to do stoichiometry?

Sure, most of the equation-balancing Javascript calculators on the web use that method. It's overkill to balance a single equation that way, but if you're automating the process or if you have a lot of equations to balance and check it can be convenient.

If you search on "balancing equations matrix" you'll find many simple matrix inversion  approaches for balancing equations that fail if you don't have a single unique solution.  Here's one that's more general:

L. R. Thorne, An Innovative Approach to Balancing Chemical-Reaction Equations: A
Simplified Matrix-Inversion Technique for Determining The Matrix Null
Space, Chem. Educator 2010, 15, 304–308

The basic idea is that coefficients in balanced equations are found in the nullspace of the formula composition matrix (that is, the coefficients are solutions $\bf x$ of the matrix equation $\bf C x = 0$, if $\bf C$ is the formula composition matrix). For example, if I want to find a balanced equation that relates $\rm HCl$, $\rm K_2Cr_2O_7$, $\rm FeCl_2$, $\rm CrCl_3$, $\rm H_2O$, $\rm FeCl_3$, and $\rm KCl$, I can build a formula composition table

$$\begin{matrix} & \ce{HCl} & \ce{K2Cr2O7} & \ce{FeCl2} & \ce{CrCl3} & \ce{H2O} & \ce{FeCl3} & \ce{KCl}\cr H & 1 & 0 & 0 & 0 & 2 & 0 & 0\cr Cl & 1 & 0 & 2 & 3 & 0 & 3 & 1\cr K & 0 & 2 & 0 & 0 & 0 & 0 & 1\cr Cr & 0 & 2 & 0 & 1 & 0 & 0 & 0\cr O & 0 & 7 & 0 & 0 & 1 & 0 & 0\cr Fe & 0 & 0 & 1 & 0 & 0 & 1 & 0\end{matrix}$$

I can find the nullspace vectors using the multistep procedure outlined in that paper, or I can do it very simply in Maxima:

C : matrix([1,0,0,0,2,0,0],
           [1,0,2,3,0,3,1],
           [0,2,0,0,0,0,1],
           [0,2,0,1,0,0,0],
           [0,7,0,0,1,0,0],
           [0,0,1,0,0,1,0])$
nullspace(C);

$$\mathrm{span}\left( \begin{bmatrix}14\cr 1\cr 6\cr -2\cr -7\cr -6\cr -2\end{bmatrix}\right)$$

which corresponds to the equation

$$\ce{ 2CrCl3 + 7H2O + 6FeCl3 + 2KCl\ =\ 14HCl + K2Cr2O7 + 6FeCl2}$$

As chemists, we'd write that equation reversed,

$$\ce{ 14HCl + K2Cr2O7 + 6FeCl2\ =\ 2CrCl3 + 7H2O + 6FeCl3 + 2KCl }$$

Here are a few more nails you can pound with your linear algebra hammer: you can determine the minimum number of chemical equations required in a reaction mechanism, or discover how many equations in a reaction mechanism are stoichiometrically independent. For example, consider the following halogen/hydrogen reactions:

$$\begin{array}{rcl}
\rm H_2 + X_2&\xrightarrow{k_1}&\rm 2HX\\
\rm H_2 + X_2&\xrightarrow{k_2}&\rm 2X + H_2\\
\rm H_2 + X_2&\xrightarrow{k_3}&\rm 2H + X_2\\
\rm H + X_2&\xrightarrow{k_4}&\rm X + HX\\
\rm H_2 + X&\xrightarrow{k_5}&\rm H + HX\\
\rm HX + H&\xrightarrow{k_6}&\rm X + H_2\\
\rm H_2 + HX&\xrightarrow{k_7}&\rm X_2 + H_2\\
\rm 2HX &\xrightarrow{k_8}&\rm X_2 + H_2\\
\end{array}$$

Let's make a formula matrix for the five species in the reactions above:

$$\begin{pmatrix}{H}_{2}\cr H\cr HX\cr X\cr {X}_{2}\end{pmatrix}=\begin{pmatrix}2 & 0\cr 1 & 0\cr 1 & 1\cr 0 & 1\cr 0 & 2\end{pmatrix}\begin{pmatrix}H\cr X\end{pmatrix}$$

Any set of $S-C$ stoichiometrically independent equations is a "complete set" of reactions, where $S$ is the number of species and $C$ is the number of independent equations for the conservation of elements (components). There are five species ($S=5$) and the rank of the formula matrix is 2, so the system has two components ($C=2$). There must be $S - C = 5 - 2 = 3$ stoichiometrically independent reactions for the mechanism to be complete.

Do the reactions above constitute a complete set? We can write a stoichiometric matrix like so:

$$\begin{pmatrix}-1 & 0 & 2 & 0 & -1\cr 0 & 0 & 0 & 2 & -1\cr -1 & 2 & 0 & 0 & 0\cr 0 & -1 & 1 & 1 & -1\cr -1 & 1 & 1 & -1 & 0\cr 1 & -1 & -1 & 1 & 0\cr 0 & 0 & 0 & -2 & 1\cr 1 & 0 & -2 & 0 & 1\end{pmatrix} \begin{pmatrix}{H}_{2}\cr H\cr HX\cr X\cr {X}_{2}\end{pmatrix} = \begin{pmatrix}0\cr 0\cr 0\cr 0\cr 0\cr 0 \cr 0 \cr 0\end{pmatrix}$$

The rank of the stoichiometric matrix is 3, so the eight equations above are equivalent to three independent equations. The mechanism is complete. Whether it's correct is of course a different matter. Stoichiometric constraints don't determine a mechanism; they limit the number of reactions in a particular reaction channel, but you can have more than one channel (more than one mechanism running side-by-side).

Why is this useful---why would you want to know what the number of independent stoichiometric equations ($R$) are? Well, according to Gibb's rule of stoichiometry, $R \le S-C$. If you have $R < S-C$, that tells you that you have a restricted equilibrium: the components must be constrained by conditions other than stoichiometric equations. The restricted equilibrium composition will depend on which species are present initially in that case. There are also experimental design implications, since in a single experiment you can get only $S-C$ reaction rates. You may have to figure out how to favor one channel over another in separate experiments to get all of the rates.

TL;DR: If you're looking for applications of linear algebra in chemistry, chemical kinetics is one playground; there are many others.


Are nucleophilic substitutions reversible?

In general, yes, $\mathrm{S_N2}$ reactions are reversible. But not in this particular case.

Rates of $\mathrm{S_N2}$ reactions depend on several factors: the nucleophile, the leaving group, the alkyl group undergoing substitution, and so on. In the context of your question, the leaving group ability is quite possibly the most important. Iodide, $\ce{I-}$, is an excellent leaving group. Fluorine, $\ce{F-}$, is an extremely poor leaving group. Again, there are several reasons which dictate leaving group ability, but at the simplest level, it has to do with the stability of the negative charge on the leaving group. This broadly correlates with basicity: iodide is hardly basic at all (its conjugate acid, $\ce{HI}$, is a very strong acid) and is quite happy to exist as $\ce{I-}$. Conversely, fluoride is reasonably basic (its conjugate acid, $\ce{HF}$, is a weak acid) and is not all that happy existing as $\ce{F-}$.

Using this logic you will find that hydroxide ion $\ce{OH-}$ is a very poor leaving group. Consequently, it's almost impossible to get ethanol (or other alcohols) to directly react in an $\ce{S_N2}$ fashion. So in the particular case you have raised, no, the reaction is effectively irreversible.*

The way around the leaving group problem is to convert the hydroxyl group to something that leaves better. Because alcohols are common, lots of people have devoted effort into making sophisticated ways of doing this. But by far the simplest way is to simply protonate it:

$$\ce{ROH + H+ <=> ROH2+}$$

The leaving group now is not hydroxide $\ce{OH-}$; it's now water $\ce{H2O}$, which is a much, much better leaving group. Consequently, if you treat alcohols with $\ce{HCl}$, you have a much better chance of getting an alkyl chloride, as opposed to if you were to just add chloride ion (in the form of e.g. $\ce{NaCl}$). The $\ce{H+}$ protonates the alcohol, and the $\ce{Cl-}$ does the actual substitution.

For an example of a reversible $\mathrm{S_N2}$ reaction, the best example is one where the nucleophile is the same as the leaving group. For example, consider a chiral alkyl iodide $\ce{R1R_2R_3CI}$. Let's say we have an enantiomerically pure sample of this alkyl iodide. Since $\mathrm{S_N2}$ reactions occur with inversion, if we treat it with sodium iodide, can we effect an $\mathrm{S_N2}$ reaction which will convert it entirely to the opposite enantiomer ?

The answer is no, and that's precisely because the reverse reaction will also occur. Both the forward and reverse reactions have the same nucleophile and the same leaving group, so in this case there aren't any considerations which would make one less effective than the other.

* "effectively" because technically speaking every reaction is reversible.


One-step oxidation of hex-1-ene to hex-2-en-1-ol

You're supposed to turn an alkene into an an allylic alcohol in one step. The "classical" reagent for this conversion is selenium dioxide, $\ce{SeO2}$. Have a look at some older references, such as J. Org. Chem., 1979, 44, 4683-4689.
DOI

As kindly pointed out by orthocresol, this elegant method will unfortunately furnish the the wrong allylic alcohol. Whether in situ isomerization is possible (allylic hydroxylation in acidic medium) is feasible can't be decided until I have another coffee ;)

(Please note that a procedure using singlet oxygen will need two steps: the Schenck ene reaction furnishes a hydroperoxide which you would have to reduce.)


Is 1.5M NaOH a strong enough base to deprotonate benzyl alcohol?

Only a very small fraction of the benzyl alcohol ($\mathrm{p}K_\mathrm{a} = 15.4$) would be removed by sodium hydroxide. Acetic acid is about 10 orders of magnitude more acidic than benzyl alcohol with a $\mathrm{p}K_\mathrm{a}$ of $4.8$, so it is completely deprotinated to form sodium acetate and water.

For the purposes of your experiment you can assume that there is effectively no reaction between benzyl alcohol and sodium hydroxide.


Thermodynamics: Is Isolated system possible?

There are no 100% isolated systems - even the universe itself may not be truly isolated (depending on how you define universe).

However, in practice, many systems are "isolated enough" that we can treat them as conserving mass and energy, especially over short time frames or small distances.

For example, we often treat a well-insulated and closed reaction vessel as isolated when doing thermodynamic calculations, and this approximation works very well for fast reactions.

Another example where this works is an adiabatic process. Many processes can be successfully modeled as adiabatic even though they happen in poorly-insulated containers - for example, the rapid expansion or compression of a gas.

Another example is the use of small sub-domains of a non-isolated system, which over short time scales behave as isolated systems themselves. For example, in finite element methods (FEM), a system is broken up into very small sub-domains. Each one acts as a small system in which mass and energy can transfer across the boundaries. Locally, an element is an open system, and over long time scales the entire system is open, but on short time scales, the sets of domains that share boundaries are treated as isolated. If energy leaves one domain, it enters the neighboring domain - it doesn't completely leave the system. In effect, you can model an open system as a large collection of isolated systems, with the outermost having open boundaries.

As you can see, even though the assumption of an isolated system is not really true, you can find many, many places where it is "true enough" that we can use it practically. In fact, it would be very hard to do anything in thermodynamics without it!


Is there an error on Wikipedia's page on characteristic rotational temperature of HF?

The units of the inertial constant I are reported in $\pu{amu \cdot Å^2}$. One unified atomic mass unit (or Dalton) is approximately $\pu{1.660e-27 kg}$. Using the correct formula in MATLAB:
>> amu=1e-3/6.022e23;
>> h=6.626e-34; kb=1.3806e-23; tr=@(I) ((h/(2*pi))^2)/(2*I*kb*amu*1e-20);
>> tr(0.86)
ans =

 2.820249699307124e+01

Using instead $\pu{I= 0.80 amuÅ^2}$ from the NIST website returns a value closer to that on the Wikipedia table.


Do dark lines on porcelain dishes indicate usage of aluminium for the color?

If the two bowls look different although treated similar, the material must differ.

Your bowls are almost fully covered with a glaze - except for the ring on the bottom on which the ceramic was standing while fired in an oven.

Here, water can enter the ceramic core. If the ceramic material of one bowl is more porous, water can be taken up easier. Upon heating, as in a dishwasher, the water absorbed expands. The glaze can't withstand the tension build up and breaks.

This results in the cracks that you observe.


Awkward. It just so happens that the renovations guy is right. The term you need to google is "smokeweb".

The ionisation process of burning (particularly) synthetic materials like plastics, rubbers and polymers, creates charged smoke particles which can attract towards certain surfaces, and attract to each other creating chains, and ultimately web-like formations. They form under certain conditions called wet smoke, which involves low heat, or smoldering type environments. Smoke webs are typically sticky and pungent and will smudge easily and show up in particular areas where there is limited air flows, typically corners and edges of ceilings, in curtain folds, cabinets etc.

Spider webs will burn up very quickly in a real fire.


Total number of stereoisomers of 2,3,5,6-tetrachlorooctane

If you have studied combinatorics, and if you know the fundamental principal of counting, then this is basically making a four letter word with two letters- R and S. Each place has two choices, and you have four spaces to fill, so you get 2×2×2×2= 16 optical isomers for this molecule. Since the terminal alkyl groups are different, so there is no chance of a meso-compound. And as Ivan rightly stated in the comment, to list the isomers, make all possible combinations of R and S(you have missed a few as pointed out by porphyrin in the comments)


The H atom is neither oxidized nor reduced in the redox semi-equation. It remains at +1. Only Manganese reacts, as it is reduced from +7 to +2. The redox couple should be written :  MnO4-/Mn2+, without mentioning Hydrogen

And MnO4 does not lose electrons, as you say. On the contrary, MnO4 fixes or consumes 5 electrons in its reduction to Mn2+.


Yes, the formal charge at such atoms would be different.  Such atoms, typically found at the at the edge of some crystalline or regular solid, are known as "defects" in the solid.  Because they have an incomplete outer shell and lack a stable bonding configuration (bonding vacancies), they usually exist as a radical, anion or cation.  They define an area of extremely high reactivity on the solid's surface and mark the spot where reactions often begin.

Edit: references to charges at defect sites

this link covers inorganic systems, see p.5 for some nice pictures.

this link covers graphene systems, from the "Conclusions" section; "We have performed scanning gate microscopy on graphene and observed such charge inhomogeneity and puddles due to extrinsic doping from possible sources such as metal contacts, graphene edges and surface residues."


Why is absorbance the negative logarithm of transmittance?

Absorbance is useful because it is additive. That is, it's absorbance which is used in Beer's law:

$$A = \epsilon \cdot c \cdot l$$

While you can certainly make a version of this law which uses transmittance instead of absorbance, the math becomes much less straightforward.

The explanation of why absorbance is additive falls out of the derivation of Beer's law. The official derivation relies on calculus, but you can get a sense of things by thinking of the solution not as a single thing, but as a succession of "slabs":

Take a single thin slab of a colored material. This absorbs a certain amount of light at a given frequency. But it doesn't absorb a fixed amount of light. Instead, each photon that passes through has a certain probability of encountering an absorbant molecule and being absorbed. The more light, the more absorbed. For a given slab under typical conditions, we can define the percent transmittance (%T), which is the fraction of light that goes through that material.

But what happens if you start varying the width of the material. Or equivalently, if you stack several of such slabs together. The amount of light going through one slab would be %T. The amount going through two slabs would be %T times %T. (For example, if the slab had a %T of 50%, then the first slab would absorb half the light. The second slab would only "see" half the original light, and would then absorb half of that light. So the amount being transmitted out the back of the second slab would be only 0.5*0.5 = 25% of the total amount.) Three slabs gives you %T * %T * %T. Generally, for N slabs, you'd get only $(\%T)^N$ of the light coming through. (Note the necessity of converting from percentages to fractions to do the calculation here.)

Okay, but how about varying concentration? Well, since the chance that a photon is absorbed is proportional to the chance it interacts with an absorbing molecule, then if you double the concentration, you double the number of interactions that could happen. If you work out the math, it turns out that (in dilute solutions) doubling the concentration is equivalent to just doubling the path length - you effectively have two "slabs", just compressed into the space of a single slab. The effect is the same as increasing the path length. So for a solution at twice the concentration, you's have $(\%T)^2$ the transmittance.

Note the use of exponents. This is all because when you combine two things which absorb, you have to multiply their percent transmittance. Exponents are hard to work with. So instead, we do a trick which can convert mulexponentstiplication into addition: take a logarithm. If we work with logarithms of the transmittance, all of those multiplications are now additions, and instead of using exponents for path length and concentration, we multiply by each. Taking the negative log is just another trick to make things work out reasonably: it's more convenient to have the measurement move in the same direction as concentration and path length (such that more absorbance means a larger concentration), so we stick a negative sign on the result of the logarithm to make things easier.

That's basically it. You certainly could do all your calculations with transmittances, but the equations and reasoning gets more complicated. If you use absorbances, you get the nice Beer-Lambert law, which states the concentration and path length are proportional to absorbance.


How to determine the boiling point of substances?

While it may be difficult to determine the exact boiling point of a substance, many factors play to compare if a substance has a higher boiling point than the others.

1) Intermolecular force. Notice that it is "intermolecular force" instead of
"intramolecular force". The reason being when you boil something you want to transform the substance from liquid state into gas state, and that can only be achieved by weakening the connection of molecules, not by breaking the intramolecular bond of a molecule. There are generally three types of intermolecular force : London dispersion force [weakest], dipole-dipole, and hydrogen bonds [strongest]. The stronger the force, the larger the amount of energy needed to break off the connection between the molecules, thus the boiling point is higher.

London dispersion force (LDF/Van der Waals) usually occurs when there exists no significant dipole in the molecule (propane, hexane)
Dipole-dipole occurs when there exist an observable amount of charge difference between atoms in a molecule. This can be seen in the case of hydrogen chloride. The chlorine atom is more electronegative, thus it attracts the more electropositive hydrogens of other molecules too (but doesn't bond with them).
The hydrogen bond is the strongest one. It occurs if a molecule has a hydrogen atom bonded to either F (fluorine), O (oxygen), or N (nitrogen). These atoms are highly electronegative. This happens in the case of water $H_2O$. I reccomend you to think why the hydrogen bond is the strongest and Van der Waals (London dispersion force) is the weakest intermolecular force.

2). Molecular mass. Suppose you have two substances, both of which happen to interact through LDF. Another thing to consider is the mass of the molecule. If the molecule is bigger, then the surface area is larger, which results in a bigger LDF. This leads us to the conclusion that the mass of the molecule is proportional to the boiling point.

3). Branches. In alkanes (consisting of C and H only), generally a straight chain alkane has a higher boiling point than similar branched alkanes due to the surface area between two adjacent molecules. This would be harder if the alkane has branches. Compare : 2,3-dimethylbutane and hexane. Both have 6 carbons, but the boiling point of 2,3-dimethylbutane is 331.15 K whereas hexane is 341.15 K.

Take a look at your case. $H_2O$ has oxygen in it, so hydrogen bond. Although the electronegativity of hydrogen, selenium, sulfur, and tellurium are around 0.1 - 0,48 difference, it doesn't really contribute much to the overall dipole of the molecule. So these three must interact with LDF. We can see that the molecular mass of sulfur < selenium < tellurium (S < Se < Te). From what we know, the bigger the mass, the larger the boiling point.

From this point we can tell that :
$H_2S < H_2Se < H_2Te < H_2O$

And a quick Google search will show us that the boiling point of these molecules are :
$H_2S (-60) < H_2Se(-41,25) < H_2Te(-2.2) < H_2O$ (100)


How do I calculate the order of a reaction given slope?

As mentioned, you don't want a direct answer so I am just going to ask you some question to think about, that hopefully help in getting you to the answer. Please reply in the comments whenever something is unclear to you!

You're expecting first order in both reactants, so if you would write down the reaction rate equation, what does that look like?

With the equation you should have one unknown only: $k$ (or two that make up $k$: $k_0$ and one other, which?, if you are interested in the temperature effect as well).

Now calculate $k$ for each reaction separately. Are they the same? Do you expect them to be? If they're not, should you average them?


Which of the following statements is true about the trend down group 7?

The answer is D, the melting points increase. This is absolutely true (source for values):
\begin{array}{lrr}
\text{Halogen} & \text{Melting point}/^\circ\mathrm{C}& \text{Boiling point}/^\circ\mathrm{C}\\\hline
\text{fluorine}   & -220  & -188  \\
\text{chlorine}  & -101  & -35   \\
\text{bromine}  & -7.2  & 58.8  \\
\text{iodine}   & 114   & 184   \\
\text{astatine}  & 302   & 337   \\\hline
\end{array}

And the boiling point increases, too, so answer C is definitely wrong.

Another obviously wrong answer is B, the atoms get smaller. With increasing main quantum number the atoms obviously have to become bigger since the main electron density moves further away from the nucleus. In numbers:
\begin{array}{lrr}
\text{Halogen} & \text{Covalent radius}/\mathrm{pm}& \text{Ionic radius }\ce{(X^{-})}/\mathrm{pm}\\\hline
\text{fluorine} &  71  & 133  \\
\text{chlorine}   &  99  & 181  \\
\text{bromine}  & 114  & 196  \\
\text{iodine}   & 133  & 220  \\
\text{astatine}  & 150  &      \\\hline
\end{array}

For answers A and E it cannot be unambiguously answered, because reactivity is not a well defined concept, see "What is reactivity really, and can it be quantified?"  for example. The IUPAC goldbook states, that reactivity is a kinetic property. It goes on to describe, that it can only be absolutely used in a given context or in reference to another system. However, it is also often used in expressing general trends, which is the case when looking at this question.
Given this context, you can find the following portion on UC Davis ChemWiki:

  Reactivity of Elements: decreases down the group
  The reactivities of the halogens decrease down the group ( At < I < Br < Cl < F). This is due to the fact that atomic radius increases in size with an increase of electronic energy levels. This lessens the attraction for valence electrons of other atoms, decreasing reactivity. This decrease also occurs because electronegativity decreases down a group; therefore, there is less electron "pulling." In addition, there is a decrease in oxidizing ability down the group.

In this context, answer A is wrong. However, I would not consider E to be strictly true either. Have a look at the electron affinities, and you will find, that fluorine behaves anomalous. This was also discussed in "Why does chlorine have a higher electron affinity than fluorine?"
However the general statement will usually be given as: "Electron affinity decreases down the group."
\begin{array}{lc}
\text{Halogen} & \text{Electron Affinity}/\mathrm{kJ\cdot mol^{-1}}\\\hline
\text{fluorine}  &  -328.0  \\
\text{chlorine} &  -349.0  \\
\text{bromine}   &  -324.6  \\
\text{iodine}    &  -295.2  \\
\text{astatine}  &  -270.1  \\\hline
\end{array}
The same trend can be observed with bond enthalpies of the $\ce{X2}$ series (source):

This trend is broken for the hydrogen halides $\ce{HX {(g)}}$.

Your assumption is therefore correct, maybe you have already done the same analysis. However, we can only definitely state and proof, that answer D must be correct, while answer E might be correct depending on the reference system.


How can a group be both a good nucleophile and a good leaving group?

I'm not entirely sure of this, comments appreciated

Leaving group tendency is a thermodynamic variable. So, it depends more on the equilibrium and energies of the reaction. We calculate it from the basicity of the group (weak base -> more stable while solvated -> better leaving group).

Now, iodide is a weaker base than $\ce{OH-}$, since its conjugate acid $\ce{HI}$ is a stronger acid than $\ce{H2O}$. So, iodine is a better leaving group.

On the other hand, nucleophilic tendency is basically "how willing to sacrifice a lone pair is the group?".  It's more dependent on forces, and less on energy, as it is a kinetic phenomenon. The outer shell electrons in $\ce{I-}$ are more loosely bonded to it than those in oxygen in $\ce{OH-}$. So, it is a better nucleophile.

Another way to look at it would be via the "accessibility" of the group in solution. This basically attempts to explain the difference between a nucleophile and a leaving group. In a leaving group, you have to "give up" the ion to the solution. This means that solvation doesn't really matter here -- once you've given it up, it's gone. The process will have a similar rate for different ions. However, when looking at the reverse process, you need to check how solvated the ion is. If the ion is surrounded by water molecules (like $\ce{F-}$, it will be much harder for the electrophile to "extract" it. On the other hand, an ion like $\ce{I-}$ is less solvated, and thus more "accessible".


How does a volcano for a science fair work?

The food color-dyed water is just there to add volume and color.

The household chemicals used are acetic acid (vinegar, $\ce{CH3COOH}$) and sodium bicarbonate (baking soda, $\ce{NaHCO3}$).

The acetic acid and sodium bicarbonate react to form carbon dioxide gas ($\ce{CO2}$), dissolved sodium acetate ($\ce{CH3COONa}$) and water ($\ce{H2O}$).  It is the liberation of $\ce{CO2}$ gas that causes the foaming and bubbling.

The chemical reaction is written as:

$$\ce{CH3COOH(l) + NaHCO3(s) -> CH3COO-(aq) + Na+(aq) + CO2(g) + H2O(l)}$$

Where:
l = liquid
s = solid
aq = aqueous (dissolved)
g = gas (vapor)


Does 11.2 dm3 of nitrogen consist of 0.5 moles at STP?

You are right that 0.5 mol of $\ce{N2}$ is 11.2 L, but read the question carefully. Option B says that 11.2 L of $\ce{N2}$ contains 0.5 mol of atoms, which is wrong, it contains 11.2 L of $\ce{N2}$ contains 0.5 mol of molecules. It seems to be a trick answer. The first option is correct as 0.5 mol of $\ce{N2}$ has a mass of 14 grams.


The pH scale is merely a way of keeping track of the concentration a species (the hydrogen ion $\ce{H^+}$, $pH = -log\  a_{H^+} \simeq -log\  [H^+]$) in a medium (usually liquid, not necessarily aqueous). You can define arbitrary scales for other species, too. Analytical chemists are more familiar with scales such as pCl, with $pCl = -log\  a_{Cl^-} \simeq -log\  [Cl^-]$, and that's fine, but they're much rarer. The reason chemistry pays so much attention to hydrogen ions specifically is because the pH scale keeps track of what is arguably the most reactive species in all of chemistry (one could make a case for alpha particles, i.e. bare helium-4 nuclei $\ce{^4_2 He^{2+}}$ instead, but hydrogen ions are nevertheless comparable in strength and far more common).

If you stop to think about it, a (non-solvated) hydrogen ion, $\ce{H^+_{(g)}}$ is a bare proton. Let me explain why this is an important realization. In chemistry, we talk a lot about how electrons get shoved around when different molecules meet (also known as chemical reactions), and the exact way in the electrons shuffle about is dependent on complicated electromagnetism at the quantum level. However, to a simple but enlightening approximation, one can imagine how substances behave based on how much electrical charge they concentrate in a region; simply, concentrated charges will display stronger electromagnetic effects.

In most of chemistry, we deal with charges of order unity relative to the fundamental electrical charge ($e=1.6022 \times 10^{-19}\ C$) spread in the volume of a few atoms (of order $10^{-3}\  nm^3$). The resulting charge density can be found by dividing the charge by the volume. Now here's the thing - the nucleus of an atom is tiny compared to the size of the entire atom (with its electrons). A lone hydrogen atom has a radius of approximately $0.1\ nm$, but a lone hydrogen nucleus has a radius of about $1\ fm = 0.000001\ nm$, a difference of five orders of magnitude. The volume difference is thus fifteen orders of magnitude. Therefore, the bare nucleus which comprises a $\ce{H^+}$ ion has a charge density 15 orders of magnitude greater than a similarly-charged atomic ion, for example $\ce{Li^+}$. In effect, this means that a bare $\ce{H^+}$ has such a huge positive charge density that it will tear opposite charges (electrons) from absolutely anything. Helium and neon are notoriously unreactive, but both react quite happily (and rather exothermically) with gaseous $\ce{H^+}$ forming hydrohelium(+) and hydroneon(+) as the respective products.

So why doesn't water (or any ready source of $\ce{H^+}$ ions, i.e. protic substance) immediately dissolve absolutely everything it touches? It so happens that, in condensed phases, whatever molecules close to a hydrogen ion are so attracted that they effectively encase the hydrogen ions into solvated hydrogen ions, forming a shell which protects everything else from the aggressive reactivity of the hydrogen ion. For example, in water, $\ce{H^+}$ is actually not a very correct description of reality. Rather, one finds $\ce{H(H_2O)_{n}^+}$ (often alluded to by formulae such as $\ce{H3O^+}$, $\ce{H5O2^+}$, $\ce{H9O4^+}$ and others), or more concisely (but less distinguishably) $\ce{H^+_{(aq)}}$. After being surrounded by water molecules, the hydrogen ion loses almost all of its reactivity, as the original charge is now effectively dispersed over several molecules, greatly dimishing the charge density of the solvated hydrogen ion compared to the non-solvated hydrogen ion. Even after the massive handicap, solvated hydrogen ions are still quite reactive. Thus, it is important to keep track of their amount, to figure out if certain materials are capable of withstanding the onslaught.

After all I said, it seems like it would be a good idea to minimize the amount of $\ce{H^+}$ in water, which can be done by going to high values in the pH scale. However, by doing so you begin to significantly increase the amount of hydroxide ions $\ce{OH^-}$ present, in order to obey the autodissociation reaction equilibrium of water. In some sense, hydrated $\ce{OH^-}$ ions in solution are about as reactive as solvated $\ce{H^+}$ (even though non-solvated $\ce{H^+}$ is much, much more reactive than non-solvated $\ce{OH^-}$). Thus, going to either end of the pH scale can be a bad idea. One could say that water solutions are least reactive* when their pH is close to 7 (including pure water), as one minimizes the sum of $\ce{H^+}$ and $\ce{OH^-}$ present.

* Least reactive in the specific sense that if a material resists acidic aqueous solutions or basic aqueous solutions, then it must resist pure water and neutral aqueous solutions, assuming that whatever else anything else in the water incapable of altering the pH value is also unreactive; a pH 7 sodium chloride solution for example can be quite corrosive to metals, but not because of the water itself.


emulsion - why does it make a differnce if you have have more/less water/oil

The two combinations (10/1 and 1/10) of oil and water will give different emulsions: the first will be water-in-oil and the second will be oil-in-water. The difference lies in the continuous phase; the minor phase will be tiny droplets, with a size depending on how much energy you put into shearing it. Neither emulsion will be particularly stable; that's why we need emulsifiers: to stabilize the emulsion, and also to make it easier in the first place.

The yolk of an egg is primarily fatty (that's why it tastes so good). It's so fatty that it does not absorb the egg white, which is 90% water and about 10% protein. The yolk does contain some hydrophilic groups - but not enough to emulsify the albumen. But the yolk can emulsify/stabilize a small amount of water, and the yolk will disperse easily in (some) oils. So the yolk, spread out in oil, can stabilize a smaller amount of water. But you can't get the yolk well dispersed in water: it agglomerates quickly. And extra oil just makes the destabilization quicker and more obvious.


You have already correctly identified the parent structure of the compound as bicyclo[2.2.2]octane.
The most important simplified criteria for the numbering for such structures are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

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

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The compound that is given in the question doesn’t have any principal characteristic group that is expressed as a suffix. Thus, Rule (c) is not relevant in this case.
The compound also doesn’t have any multiple bonds. Thus, also Rule (e) is not relevant in this case.
Next, low locants are assigned to the substituents that are expressed as prefixes according to Rule (f). Note that all substituents are considered together in a series of increasing numerical order. Therefore, the correct name is 5-bromo-2-chloro-7-fluorobicyclo[2.2.2]octane rather than ‘2-bromo-5-chloro-8-fluorobicyclo[2.2.2]octane’ since the locant set ‘2,5,7’ is lower than ‘2,5,8’.

Only if there is still a choice, a low locant would be given first to the bromo substituent in accordance with Rule (g).


What happens to gunpowder when it gets wet?

In general, putting explosives in water will not prevent them from exploding, because most explosives already contain all of the necessary reactants to produce an explosive chemical reaction.  They do not need to draw oxygen from the air, and the reaction happens much too fast for the water to act as a heat sink.

Traditional "black powder" is an exception; it doesn't explode when wet.  I don't know exactly why this is, but two plausible reasons are:

Unlike most other explosives, black powder is shock-insensitive; it has to be initiated with a flame or spark.  Water can easily soak up all the heat energy from a small flame, like the one you get from a percussion cap.  This is the same as how it's harder to set fire to a living tree than a dry log, but if the fire gets hot enough it will still catch.
Black powder is a "low" explosive, meaning that the chemical reaction happens relatively slowly (for an explosion), which gives the water more time to absorb heat energy and quench the reaction.

Worse, black powder can be permanently ruined by water.  This is because it's a mixture of saltpeter (potassium nitrate, $\ce{KNO3}$), sulfur, and charcoal, and saltpeter is water-soluble.  If you get it wet enough, the saltpeter can be completely washed out of the mixture.  This doesn't always happen; the 19th century industrial manufacturing process for black powder involved deliberately dampening it, both to make it less dangerous during manufacture, and to make saltpeter crystals form inside each particle of charcoal.  But they were careful to use only just enough water.


Products which adhere to human nail while still being removable/reusable?

Try flour and water paste or washable school pva glue. Start with the flour. It would be an interesting experiment to try a 50/50 mix of both.
Another possibility is rubber cement (art gum). This will be quicker drying.


Capillary action only works with glass tubes?

Few things in chemistry are truly unique. The tube does not have to be glass; it only has to be wettable, which glass typically is and plastic typically isn't.

In the tubes made of non-wettable materials, capillary action works the other way: the liquid will go down rather than up, akin to Hg on the picture provided.

 (source)

So it goes.


does water's polarity help explain its unusual phase properties?

Not just its polarity, but its ability to form hydrogen bonds, together with the size and shape of its molecules.

Hydrogen bonds give water an unusually high boiling point for a molecule its size; many molecules of similar size that can't hydrogen-bond are gases at room temperature. They also give water a relatively high heat capacity and high enthalpies of fusion and vaporization. These latter two properties mean that ice has to absorb more heat to melt and liquid water has to absorb more heat to boil than would otherwise be necessary without hydrogen bonding.

Water's shape lets it form 4 tetrahedrally arranged hydrogen bonds. This lets it hydrogen-bond into networks of chair-shaped hexagonal rings in ice. The voids in the center of the rings make ice less dense than water. This is unusual; the solid form of a compound is most often denser than the liquid form.


Difference between a dimethyl and ethyl substituent groups

If an organic compound has dimethyl in its name than that means that it has 2 methyl groups. If they are on the same carbon they form a V in the bond line structure. An ethyl group on the other hand  forms a 2 carbon unbranched zigzag in the bond line structure instead of the V that a dimethyl group forms in the bond line structure.

For the ethyl group it is standard on all bond line structures to represent it as a 2 carbon unbranched zigzag. For the dimethyl group it is standard to represent it as a V in all bond line structures.

As to how we know this it is because an ethyl group coming off of a carbon looks like propane if you just look at that carbon and the ethyl group. In fact some people in the early years of organic chemistry might have called propane ethylmethane before the prop prefix was used for alkanes. However if this was the case some of them would call propane dimethylmethane. This was because they were looking at different carbons as the central atom. If you look from one of the ends it looks like an ethyl group sticking off of methane. If you look at the center it looks like a dimethyl group sticking off of methane.


Why are most glow-in-the-dark things green?

The human eye is most sensitive in green wavelengths. Therefore, if one were to put the same amount of light energy into different wavelengths, the green portion of the spectrum would appear brightest, though it contained no more energy than the rest.

If a phosphor or glow-stick manufacturer is seeking the "greatest bang for the buck", i.e. the most light for the least amount of chemical, green is a good choice.

There is a variety of phosphor colors, such as these listed in Wikipedia:

Red: Yttrium oxide-sulfide activated with europium is a red phosphor used for CRT's.
Yellow: Zinc cadmium sulfide (Zn,Cd)S:Ag, emits yellow light.
Green: Zinc sulfide with copper, ZnS:Cu, emits green light, ~530 nm. It was widely used in CRT oscilloscopes and radar displays because of its long persistence and simple chemistry. It is known as phosphor P31.
Blue: Zinc sulfide with few ppm of silver, ZnS:Ag,emits blue light with peaking at 450 nm. It is known as phosphor P22B and, unlike ZnS:Cu, has a very short persistence, so is suitable for slit-photography CRT's.

For a more complete list of phosphors and their colors, see this useful PDF or Phosphor Technology.

Many stores carry a variety of inexpensive chemiluminescent products, such as glow-in-the-dark necklaces and bracelets, in colors ranging from red through violet.


How well do d-block electrons actually shield?

Let us read Keeler and Wothers's Chemical Structure and Reactivity carefully. (It's really a great book. I wish I had used it in the past.) We excerpt the relevant paragraphs that detail the points you mention, and highlight particularly important statements regarding screening. All page numbers are taken from the second edition.

Point 2, on Slater's rules.

[Energy ordering:] (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) ...

If the electron being considered is in an ns or np orbital, then electrons in the next lowest shell (i.e. that with (n-1)) each contribute 0.85 to $S$. Those electrons in lower shells (i.e. (n-2) and lower) contribute 1.00 to $S$.

If the electron being considered is in an nd or nf orbital, all electrons below it in energy contribute 1.00 to $S$ [the shielding constant]. (p. 262)

Point 3, on Z$_\text{eff}$ and d-block contraction.

...there is a downwards kink in the [s-block] orbital energy on moving from the 3s to the 4s [orbitals]. [...] This difference is due to the filling of the 3d orbitals between calcium and gallium. The effects of all the extra protons in the nucleus is to cause an additional lowering in the energy of the valence orbitals for the p-block elements in Period 4. However, the effect is not really drastic: the 4s and 4p orbitals in gallium do not seem to have experienced the full effect of the ten extra protons. We interpret this by saying that the effect of the extra protons on the orbital energies of the 4s and 4p has, to a large extent, been cancelled out by the electrons that have been added in the 3d orbitals.
We could rephrase all of this by looking at the problem from the perspective of effective nuclear charges rather than orbital energies. The increase in Z$_\text{eff}$ for gallium is slightly greater than might have been expected by comparison with boron and aluminium because of the filling of the d-block. (p. 266)

Point 4, on orbital energies.

...between scandium (Sc) and copper (Cu) electrons are being added to 3d orbitals, and such electrons form an effective screen for the 4s electrons. To put it another way, as far as the 4s electrons are concerned, on moving from one element to the next between scandium and copper, the effect of the extra proton is largely cancelled out by the addition of the electron to the lower 3d orbital. Consequently the energy of the 4s falls rather slowly. (p. 261)
The outer electron in potassium occupies the 4s AO, rather than the 3d. This is because the 4s orbital penetrates to the nucleus more effectively than does the 3d, resulting in the energy of the 4s being lower than that of the 3d. [...] ...both the 3s and the 3p have subsidiary maxima close in to the nucleus, but the first maximum for the 3s is very much closer in. The 3d has no such subsidiary maxima, and therefore is much less penetrating than 3s or 3p.
...the electrons in the 4s do not screen the 3d particularly well since much of the electron density from the 4s is further out from the nucleus than that from the 3d.
From scandium onwards the energies of both the 4s and 3d AOs drop steadily: this is simply the result of the increase in nuclear charge not being quite offset by the increase in electron-electron repulsion. Put another way, the electrons in the 3d sub-shell do not screen one another particularly well. Although both AOs fall in energy, the 4s falls less steeply, which can be explained by noting that this AO is quite well screened by the 3d electrons, not least as these are in a lower shell. (p. 627)

We see that Keeler and Wothers are at least self-consistent regarding points 2, 3, and 4: 3d electrons are effective at shielding 4s and 4p electrons and not effective at shielding 3d electrons. I believe, therefore, that the contradiction lies in point 1, which should be corrected as follows:

d-block electrons shield poorly s or p electrons in the same shell

I suspect the statement that d-block electrons shield poorly is taught in reference to the concept of penetration. If we look at plots of radial probability density for the 3s, 3p, and 3d orbitals, we might naively conclude, on the basis of mean distance from nucleus, that 3d electrons are closest to the nucleus, which means that they should shield 3s and 3p electrons well. In reality, because of penetration, this is not the case. Thus we conclude that 3d electrons shield (s or p electrons in the same shell) poorly, and the misconception arises if we try to extend this poor shielding to 4s electrons and the like.


Bromination of alkanes with Br2 will not occur in a dark environment. In the presence of light, multiple halogenations can occur given enough time.

For more information, including the full mechanism, check out: https://en.wikipedia.org/wiki/Free-radical_halogenation


The locant set ‘1,2,4,6’ is not lower than ‘2,1,5,3’.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the lowest locant set is defined as follows:

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
  (…)

Especially note that the locants are cited in order of increasing value when they are compared.

Thus, the locant set ‘2,1,5,3’ is regarded as ‘1,2,3,5’, which is lower than ‘1,2,4,6’.

Therefore, the correct name is 2-bromo-1-chloro-5-fluoro-3-iodobenzene.


How does salt and water promote corrosion?

I think a tantalising clue can be found in the principle of cathodic protection used in industry.

Cathodic protection is used in the shipping industry to prevent iron/steel hulls from rusting. Essentially you make your hull into a cathode by connecting a (typically) zinc anode to the hull. How do we know the ship's hull is the cathode and zinc the anode?

And zinc in fact?

Lets go back to the standard reduction potential tables. The following reaction has a reduction potential of $\ce{-0.7618V}$:

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

Whilst the following has a reduction potential of $\ce{-0.44V}$:

$$ \ce {Fe^2+(aq)} + 2{e}  \ce{->} \ce{Fe(s)}  $$

Clearly the rusting process works in the opposite direction (and thus so do the above equations) which just means we reverse the voltage to get the oxidation value.

A higher reversed voltage means that that process has a greater tendency to occur, which means that zinc here is preferentially oxidised over iron. Hence if we connect an iron hull to a zinc rod, the zinc acts as the anode whilst the hull is the cathode i.e any $\ce{Fe^2+}$ directly formed by reaction with sea-water is pushed back to iron metal by the production of $\ce{Zn^2+}$, a process which liberates two electrons.

This should suggest that some electrolytic process contributes to salt water affecting the oxidation of iron. The most basic explanation is that the sea water acts as an electrolyte, which facilitates this process much more easily than, say, distilled water.

What does this mean however when we don't have an electrochemical cell as above, and only have a single electrode i.e. the ship's hull?

We might like to look up the standard reduction potentials for both chloride and iron, however we quickly see that this isn't the case; both chloride and iron metal have to lose electrons in our case, which isn't going to form an electrochemical reaction.

It's an interesting case that bleach (alkali salts of chlorite, $\ce{ClO-}$)undergoes the following reaction with iron:

$$ \ce {Fe(s)} + \ce{ClO^-(aq)} +\ce{H2O}  \ce{->} \ce{Fe^2+(aq)} + \ce{HClO2(g)} +\ce{2H+(aq)}  $$ which has a favourable overall voltage of around $\ce{2.11V}$.

So we've seen that chloride doesn't directly react with iron, but chlorite does. This is a possible route for corrosion, although it's not obvious by which natural process chlorite comes about.

Another point to note is that seawater just doesn't contain sodium chloride, but myriad other ions which can contribute to this corrosion process,especially sulphate.

$$ \ce {SO4^2-(aq)} + \ce{4H^+(aq)} + \ce{Fe(s)}  \ce{->} \ce{Fe^2+(aq)} + \ce{SO_2(g)} + \ce{2H2O(g)} $$ which occurs with a total voltage of $\ce{0.61V}$, which tells us that the chlorite process is much more favourable thermodynamically.

So in conclusion, there are many different reactions that can utilise seawater to oxidise iron—it's not incredibly clear which ones do contribute, and which ones don't, but the above provides the means for some educated guesswork, so to speak.

As a final point, iron can oxidise in air without water, it just takes a long time. The presence of an electrolyte will speed up the process however. Corrosion also needn't be a reaction with oxygen either, but any process which makes iron ions from iron metal.


What indicators can be used when substituting iodine with other halogens in the Winkler method?

It will likely not work with other halides. As far as I understood the winkle method, the oxygen is first fixated with $\ce{Mn^2+}$ in basic conditions:
$$\begin{align}
\ce{Mn^2+ (aq) + 2 OH- (aq)&-> Mn(OH)2 (aq)}\tag{1}\label{alkalinemanganese}\\
\ce{2Mn(OH)2 (aq) + O2 (aq) &-> 2 MnO(OH)2 (s) v}\tag{2}\label{bindingoxygen}\\
\end{align}$$

The sample is acidified and iodide is added. This is oxidised to $\ce{I2}$, which forms a complex with remaining iodide.
$$\begin{align}
\ce{MnO(OH)2 (s) + H+ (aq) &-> Mn(OH)3 (aq)}\tag{3}\label{acidicmanganese}\\
\ce{2 Mn(OH)3 (aq) + 2 I- (aq) + 6 H+ (aq)
&-> 2 Mn^2+ (aq) + I2 (aq) + 6 H2O}\tag4\label{iodide-ox}\\
\ce{I2 (aq) + I- (aq) &<=> I3- (aq)}\tag5\label{triiodide}\\
\end{align}$$

The $\ce{I3-}$ forms the indication complex with the starch, which can then be titrated with thiosulfate solution. When the colour is gone you have successfully converted all of the prepared $\ce{MnO(OH)2}$ back to manganese(II) ions.
$$\ce{I3- (aq) + 2 S2O3^2- (aq) -> 3 I- (aq) + S4O6^2- (aq)}\tag6\label{triiodide-reduction}$$

The reason why this works is because you have the equilibrium $\eqref{triiodide}$ which produces the  triiodide ion, which is relatively stable in aqueous solution. Iod by itself does not dissolve in water (much). This complex again forms a very colourful complex with starch, the switch can easily be detected by the human eye.

Polyatomic ions of other halogens are not unheard of (ref. Trichloride ion (c.f. triiodide ion) and therein), but they are a lot less stable. I am not aware that they form a complex with starch so they won't be that easily detected.

Another aspect, I don't want to go much deeper here, is that the redox reactions necessary $(\ref{iodide-ox},\ref{triiodide-reduction})$ for this procedure to work might not work at all with other halides.

The original method was proposed in the 19th century and is still applied today. It is a quite accurate and robust method. There is not really a reason to tamper with it.


Effect of specific heat capacity on heat loss

Assume you add a fixed amount of heat to a substance held in an isothermal bath, so that as soon as you start heating it, the substance gives off heat to the surroundings. As heat is added the substance increases in temperature to a maximum. When the heater is turned off, the substance continues to dissipate heat until it is again in equilibrium with the surroundings.
If the rate at which heat is added is much faster than that of heat dissipation and the heat capacity is approximately constant, then the maximum temperature will be approximately
$$T_{\mathrm{max}}=T_{\mathrm{ini}} + \frac{Q}{C_p}$$
where Q is the total amount of heat added. As expected, a substance with a smaller heat capacity will become hotter.
Now consider the rate of heat dissipation. The rate of heat dissipation can be expressed with the heat equation (in one dimension) as
$$\frac{dQ}{dt} = -k \frac{\partial T}{\partial x}$$
where k is the thermal conductivity.
Therefore if k is the same for two substances, then the rate at which heat is released will differ according to the heat capacity, and the lower heat capacity substance will release heat more quickly (since temperature gradients will be greater in that substance).
This can also be argued based on the value of the thermal diffusivity $\alpha$, which is inversely proportional to the heat capacity $c_p$:
$$\alpha = \frac{k}{\rho c_p}$$
As the Wikipedia explains,

In a sense, thermal diffusivity is the measure of thermal inertia.[6] In a substance with high thermal diffusivity, heat moves rapidly through it because the substance conducts heat quickly relative to its volumetric heat capacity or 'thermal bulk'.

Since the substance with smaller $c_p$ will have a larger thermal diffusivity (assuming all else constant) it will also lose heat more quickly.
This can of course be expected to result in greater uncertainty in a measurement of the heat absorbed by the substance.


Very shortly after a radioactive element decays in a molecule, you could consider that for a brief moment, it does indeed form a molecule with the new element in the same place as its originating radioisotope. So yes, after the decay of $HTO$ you could get $HeOH^+$ via the unusual reaction $TOH → HeOH^+ + e^- + \bar\nu_e$. Of course, though, this is a very unstable compound and it would immediately participate in further chemical reactions, very likely releasing the helium atom at some point. Even if the molecule were relatively stable after having one of its atoms transmuted, the molecule might still decompose as the nuclear decay releases a lot of energy, which can tear the comparatively fragile molecule to pieces.

A curious tidbit is that, in fact, the decay of tritiated compounds has been considered as a means to produce legitimate helium compounds, given that directly reacting helium with other substances has been fruitless other than for very trivial species, such as $HeH^+$. The problem is that tritium has a rather long half life of about 11 years, so it would be quite hard to produce reasonable quantities of helium compounds in a reasonable timespan, along with the likelihood of the helium compound being very poorly bound and the emission of vast quantities of nuclear decay energy which could easily decompose the fleeting substance.


There's something called the "solvent leveling effect". In short, there's a lowest pKa in a particular solvent, based on the basicity of the conjugate base.

In water, you're limited by $\ce{OH-}$. So given the same concentration of $\ce{HI}$ and $\ce{HCl}$ in water, they will indeed have the same pH.

In other solvents, you may have a difference between acids, depending on the pKa.


The molecular equation for this titration is $$\ce{K2Cr2O7 + 6Fe(NH4)2(SO4)2 + 7H2SO4 ->
3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 6(NH4)2SO4 + 7H2O}$$

The complete ionic equation is $$\ce{2K+ + Cr2O7^2- + 6Fe^2+ + 12NH4+ + 19SO4^2- + 14H+ ->
6Fe3+ + 14SO4^2- + 2Cr^3+  + 2K+  + 12NH4^+  + 7H2O}$$

When you cross out everybody who shows up on both sides in the same form, you get the net ionic equation for this titration: $$\ce{Cr2O7^2- + 6Fe^2+ + 14H+ -> 6Fe^3+ + 2Cr^3+ + 7H2O}$$

Just as the hydrogens in $\ce{H2SO4}$ become $\ce{2H+}$ so the potassiums in  $\ce{K2Cr2O7}$ become $\ce{2K+}$ and they don't change on the other side of the equation where they form $\ce{K2SO4}$.


Why does hydrogen burn with a pale blue flame while its emission spectral lines are red in colour?

It is a very interesting question, but comparing a combustion spectrum with an atomic emission one is like comparing apples and oranges. A flame is a luminous gas phase chemical reaction where the hydrogen atoms are combining with oxygen atoms. It is a chemiluminescence phenomenon. A discharge tube emission is an atomic emission.
You posted a picture of a hydrogen spectrum from a discharge tube which consists of discrete lines in the UV, visible, and IR regions. In the same way, the hydrogen-air flame has an emission in the ultraviolet, visible and in the IR. The blue flame is not a line spectrum, as you might speculate. It is a continuum (broad band), which indicates that this is due to molecular emission in the flames, not from hydrogen atoms! This is in contrast with the alkali metals whose compounds easily atomize in the flame.
Also note that what may appear to us as a single color, like blue, is rarely a single wavelength.
There is a beautiful article by R.W. Schefer , W.D. Kulatilaka, B.D. Patterson, and T.B. Settersten on Visible emission of hydrogen flames, Combustion and Flame 156 (2009) 1234–1241. Go to Google Scholar and find it; it is open access on Google.
Look at their experiment: They view the "blue" hydrogen flames by various optical filters which block or pass certain wavelengths. See many colors are present in the "blue" flame of hydrogen. In fact most of the emission is in the ultraviolet region.

Direct flame luminosity photographs in a laminar, diffusion H2-jet flame. (a) Unfiltered, f /2.4 aperture; (b) short-wavelength pass filter with 550 nm cutoff wavelength,
f /2.4 aperture; (c) long-wavelength pass filter with 530 nm cutoff wavelength. Images taken at f /2.4 aperture with 90-ms exposure time. Fuel jet exit velocity is 47 m/s
with a coflow air velocity of 0.57 m/s. Reynolds number = 837.
Most of the emission is in the UV and infrared as you can see from the spike near 300 nm and intense band after 700 nm. Most people cannot see beyond 700 nm. The infrared emission is due to excited water molecule or perhaps its fragments.

Note: The above spectrum consists of two parts, UV on the left, and visible extending from $\pu{350 nm}$ to $\pu{850 nm}$ on the right hand side, recorded separately. For this figure, the later was amplified by a factor of 6.5 with respect to the OH in the ultraviolet.


Why is cysteine more acidic than serine?

Good question. First let's recognize that thiols (-SH) are more acidic functional groups than the corresponding alcohols (-OH) (some rationale here). So if we were talking about pKr we would see around 8 for cysteine's thiol side chain and around 15 for serine's alcohol side chain.

Now, to address your question about the carboxyl acidity, pK1, of these compounds. Serine's OH side chain has the more electronegative atom than cysteine's SH group, so we would indeed expect a stronger electron-withdrawing inductive effect in serine, possibly increasing the carboxyl acidity. However, the O atom is two carbons away from the carboxyl group, so this effect may be minimal to negligible... Not to mention of course the empirical data you provided implies that cysteine has the more acidic hydroxyl, not serine.

So I'm just speculating without looking at the actual experiments used to establish the pK1s, but it might be possible that when attempting to measure the pK1 of cysteine's carboxyl, the acidity of its thiol may have slightly skewed the measurement to make the carboxyl appear more acidic. I'm not sure if this was controlled for in the pK1 determination, but either chemically or computationally removing the contribution of the thiol to cysteine's acidity might result in a slight discrepancy compared to what it "really" is.

Additionally, one might imagine that the acidic thiol of could intramolecularly protonate cysteine's amino group, creating a proton-labile positive charge (-NH3^+) quite close to the carboxyl, and this might serve to stabilize the negative charge of the carboxylate conjugate base (-COO^-) thus increasing the carboxyl group's acidity. This effect would be more prominent in cysteine (due to the acidity of the thiol) than in serine (whose hydroxyl would be less likely to protonate the amine to stabilize the carboxylate). So this is a possible (but perhaps oversimplified) mechanistic rationale of why cysteine's carboxyl is more acidic than serine's.

As for the amino group, being more often protonated by the side chain in cysteine, it may also appear to be more acidic in serine.

This is just conjecture, for a complete understanding I would recommend looking at some experiments in the literature to see how they handle determining pK1 and pK2 when an acidic (or basic) side chain is present. It's possible that common practice is to eliminate the side chain's effect by computational or chemical methods, so without examining the primary sources one cannot be certain exactly what the reported pKa values actually represent.


Does Nucleophilic substitution require water to happen?

Keeping other parameters fixed, an elevated temperature may enhance the rate of elimination such the product of elimination is more dominant than the one by substitution.  But there is not a "critical temperature" above which only elimination will occur every time, rather than the structure of the substrate and of the reagent equally influence this.

For the other part, transesterifications as one example work better in absence of water.  See, the oxygen present in the water molecule, with highly concentrated electron density, it may act as a nucleophile, too -- a rather hard one (HASB princible) -- leading to parasitic side reactions.


Reaction between alumina and aqueous sodium hydroxide, which one's right?

Quoting from chemguide (emphasis mine):

  With hot concentrated sodium hydroxide solution, aluminium oxide reacts to give a solution of sodium tetrahydroxoaluminate.

  $$\ce{Al2O3(s) + 2NaOH(aq) + 3H2O(l) -> 2NaAl(OH)4}$$

  Note: You may find all sorts of other formulae given for the product from this reaction. These range from $\ce{NaAlO2}$ (which is a dehydrated form of the one in the equation) to $\ce{Na3Al(OH)6}$ (which is a different product altogether).

  What you actually get will depend on things like the temperature and the concentration of the sodium hydroxide solution. In any case, the truth is almost certainly a lot more complicated than any of these.

  The version I am using is perfectly acceptable and is consistent with the aluminium chemistry you will find elsewhere on the site.

You can see that products of the reaction depends on the concentration of sodium hydroxide (Basic pH) but  other factors like temperature and pressure also come into play.

If you do not add water and only change the concentration of sodium hydroxide, product will also be different:

$$
\begin{align}
\ce{Al2O3 + 6 NaOH &-> 2 Al(OH)3 + 3 Na2O}\\
\ce{Al2O3 + 2NaOH &->[\pu{900 .. 1100 ^\circ C}] 2 NaAlO2 + H2O}
\end{align}
$$

Sodium aluminate is generally $\ce{NaAlO2}$ but its aqueous solution contains many other species like $\ce{[Al(OH)4]−}$, $\ce{[Al(OH)6]^3-}$ or perhaps the $\ce{[Al(H2O)2(OH)4]−}$. They are all considered to be aluminate species. So, if you are getting sodium tetrahydroxoaluminate or sodium hexahydroxoaluminate, they are no different compounds but are all hydrated form of sodium aluminate (addition of water + change in the concentration of sodium hydroxide).


Effect of pressure on gas solubility in liquids

The pressure underwater increases by roughly 1 bar every 10 meters, that is, ambient pressure is doubled every 10 meters compared to standard atmospheric pressure.
When you breathe air underwater using a SCUBA tank, your lungs expand against the surrounding ambient pressure in order to increase their volume. As the lungs expand they fill with gas which props up the lungs (balances pressure within the lungs with the surrounding pressure). The SCUBA device adjusts the pressure of the delivered air so that it balances the ambient pressure. The partial pressure of the component gases in the gas mixture inhaled from the SCUBA tank is not the same at the water surface and when a diver is submerged. That means the pressure of the air in the diver's lungs is not ~1 bar, it can be far greater, within limitations imposed by the potential toxicity of high blood concentrations of oxygen and nitrogen.


Firstly, we should realise that sanitizers should not be considered as pure alcohols, alcohols are indeed a major component of it but many other important components are also present. A typical sanitizer contains these things -

Isopropyl alcohol/ Ethyl alcohol, Glycerin, Water, Carbomer(stabilizer) and some other components mostly related to cosmetics.

Now, i will answer your questions one by one

After applying sanitizers, can they catch fire?

Yes, there is 60%-70% alcohol content which is highly flammable. Organic compounds like alcohols, alkanes etc have very high enthalpy of combustion ($\Delta_\mathrm{c}H^\circ$ = -2000 kJ/mol for Isopropyl alcohol) and they can evaporate very easily due to the heat of our hands($\Delta_\mathrm{vap}H^\circ$= 40kJ/mol for isopropyl alcohol).

As soon as we apply alcohol, it evaporates so why is it still flammable?

This is actually wrong, as I earlier mentioned, that sanitizers are not only alcohols, in fact, the 'carbomer' present acts as a stabilizer to emulsify/dissolving all the components in it. These stabilizers prevent the alcohol to evaporate immediately and allow it to disinfect our hands fully. Also, after applying sanitizers, many of you might feel hands becoming sticky it is because of the presence of Glycerin and other cosmetic products used in it which act like a moisturizer. So generally it takes about 30-40 seconds for all(mostly) the alcohol to evaporate and sustaining a risk to be inflammable for a long time.
A thing to keep in mind is that the residue (mostly Glycerin) is also flammable(not that much, it's flammability rating is one) so after the alcohol has evaporated, you aren't fully safe from catching fire.
If your are working in kitchen, there is no actual need of using sanitizers as you can always wash hands in sink with soap and water.
I hope this answer cleared some of your doubts, stay safe :).


Which of the following compounds were involved in the titration?

Yeah,  you just gotta do that.

Assume the variables $x$, $y$, $z$. You need three equations to get the values.

First equation: Add all three and equate to $\pu{0.3148g}$.

Second equation: In the titration involving methyl orange, due to it's pH range being 3.1 to 4.4, changes color as soon as carbonates and bicarbonates are neutralized. That will give you the second equation.

Third equation: Consider all three bases for phenolphthalein titration, since phenolphthalein changes color at 8.3 to 10.0.This gives the third equation.

Solve all the three, and enjoy the efforts of your hard work.


Is carbon dioxide more acidic than carbon monoxide?

Carbon dioxide is a much stronger Lewis acid than carbon monoxide. In carbon dioxide, the carbon has a much higher oxidation state (formally, $+4$ vs. the $+2$ of carbon monoxide), and the two oxygen atoms heavily withdraw electron density from the carbon atom, leaving a very large partial positive charge ($\delta^{+}$). This is rigorously confirmed by electrostatic potential maps and calculated orbital coefficients. As a consequence, the electron-deficient carbon of $\ce{CO2}$ is highly electrophilic, and reacts with a variety of nucleophiles (e.g., Grignard reagents, organolithium reagents, etc.).

Carbon monoxide, on the other hand, actually contains substantial electron density at the carbon atom (in spite of the bonds being generally polarized towards the oxygen atom), which possesses a fairly localized and active lone pair. Consequently, a few notable exceptions aside, carbon monoxide mostly functions as a Lewis base.


What is the significance of Law of Multiple Proportions?

These laws all date back from the very early days of chemistry and are only taught at school to give pupils a way to grasp parts of the subject more easily. Except for the law of conservation of mass, which some professors/supervisors love stating frequently, they are hardly mentioned in universities outside of science history classes.

That said, as I understood them, they refer to slightly different cases. The law of definite proportions should be understood to read:

  No matter how much mass of two elements will react with each other, the product will contain the masses of its elements in a certain ratio and this ratio only.

Taking an easy example, if I reacted $46~\mathrm{g}$ of sodium and $16~\mathrm{g}$ of oxygen, I would get $62~\mathrm{g}$ of $\ce{Na2O}$ where the ratio of sodium to oxygen is $46:16$. If I had instead reacted $92~\mathrm{g}$ of sodium with an unlimited amount of oxygen, I would have gotten $124~\mathrm{g}$ of the same product which contains sodium and oxygen in a $46:16$ ratio. No matter how much sodium I let react with how much oxygen, the product will contain the definite proportion of sodium to oxygen $46:16$.*

The law of multiple proportions now states that some elements are special. For example, iron can react with chlorine to give two different chlorides: $\ce{FeCl2}$ and $\ce{FeCl3}$. For each of the two you can, as you realised set up a law of definite proportions which often gives you a rather weird rational ratio. However, when comparing those two with each other, you find that the mass of chlorine in $\ce{FeCl3}$ is 1.5 times as much as the mass of chlorine in $\ce{FeCl2}$ — corresponding to a $3:2$ ratio. And it is these very small (usually single digit) whole numbers that are actually meant.

Taking a different example, the relative atomic masses of sulphur and chlorine are $32.07$ and $35.45$, respectively. So a sulphur dichloride would have mass ratios of $32.07:70.9$ — far from any small number, but still a rational ratio. Sulphur tetrachloride exists, too, with another pretty unclean ratio. However, the ratio of the chloride contents is $1:2$ — isn’t that a nice simple ratio?

*: This example is not technically correct. Burning sodium gives $\ce{Na2O2}$, sodium peroxide. But I wanted nice and whole numbers and oxygen and sodium have nice and whole mass numbers.


Will sodium thiosulfate react with nitrate and/or nitrite ions

First, typically only high concentrations of chloride will interfere with the colorimetric test...I assume you are using NitraVer or something similar? If you aren't using seawater (i.e., $\ce{Cl-} \gt \pu{100 mg/L}$) then you will be okay. But, let's say you are using seawater. There is an easy solution, simply make your standards in an aqueous matrix containing the same amount of chloride. (I hope you are making standards or have a check standard and are not just blindly trusting the UV-Vis result...)

Second, thiosulfate is a reductant and thus could potentially reduce nitrate and nitrite. However, your most powerful answer will be your own tests - you need to do control experiments. Make some nitrate and nitrite solutions of known concentrations and then mix them with thiosulfate at concentrations you would use. Check the nitrate/nitrite concentrations after mixing and determine if thiosulfate reduced them.


I am confused about the stability of carbanion III and IV in the image

I think in 3, carbanion is resonance stabilised. While in 4, C=C-C=O are in conjugation, thus will not stabilise carbanion.


GAMESS and Gaussian : what is the maximum efficient number of supported nodes?

Is there any criteria for selection of best number of CPUs for optimum
  resource usage?

The criteria is basically the same as for choosing the level of theory: benchmarking. Parallel scaling depends on many factors: the level of theory, the size of the molecule, the machine architecture, etc., so there is no golden rule.

Though from my experience I can say that Gaussian 09 single point energy, geometry optimisation and frequencies calculations:

scale well for HF & DFT up to 16 physical cores;
scale well for MP2 up to 8 physical cores;
almost do not scale for CI & CC.

Also note that Gaussian 09 is about 2 times faster for almost any kind of calculations than the previous version (Gaussian 03) and doesn't suffer from some awkwardly strange memory effects (see e.g. here).


How can hydrogen and oxygen coexist in high temperature? Where did the oxygen come from?

I'll answer the question in general of how Hydrogen and Oxygen can coexist in metastable mixtures at higher temperatures, though without hazarding a guess for your specific case. Like you and others, I'd be very surprised if your bulk gas is actually well mixed at $\pu{900 ^{\circ}C}$ without exploding! It's far enough beyond the autoignition temperature, that an explosive reaction seems inevitable. However, metastable combinations of methane, hydrogen and oxygen in the $\pu{550 ^{\circ}C}$ range have been reported in the literature.

Hydrogen can be a surprisingly picky burner. I was told by a colleague who specialized in rocket propulsion that in between Hydrogen's upper and lower explosive limits are a host of regions where it strongly resists ignition, even above its reported autoignition temperature of ~$\pu{570 ^{\circ}C}$. According to her, the reaction between hydrogen and oxygen is quite complex with several important intermediates. Having concentrations in these non-explosive regions apparently allows the hydrogen-oxygen mix to be metastable for long periods. This would of course not extend to long time periods at $\pu{900 ^{\circ}C}$!

More in my territory, there's also many chemicals which inhibit the slow combustion (non-explosive) of hydrogen and oxygen up through the $\pu{500 ^{\circ}C}$s. The most common is $\ce{KCl}$, which strongly inhibits the reaction of oxygen and hydrogen in non-explosive conditions allowing a metastable mixture to exist. Walls of vessels can also similarly inhibit reaction. The reaction mechanism apparently stems from $\ce{KCl}$ and other surfaces' abilities to consume reaction intermediates of the combustion such as monatomic Hydrogen:

$$\ce{KCl + H^. -> K^. + HCl}$$
$$\ce{SiO2 + H^. -> SiO(OH)}$$
or direct adsorption on metal surfaces like nickel, palladium, etc.

After doing a bit of digging, methane is apparently a superb inhibitor of hydrogen and oxygen combustion. The effect is explained in a 1959 paper by R. Baldwin, et al. (see following reference). The paper explains that when methane is above a sharp critical concentration it drastically raises the autoignition temperature of hydrogen's explosive reaction and almost completely eliminates slow combustion in non-explosive conditions. In other words, when enough methane is present, the hydrogen and oxygen mixture either does nothing or explodes completely. From data taken in systems between $\pu{450\! -\! 550 ^{\circ}C}$, there was a very clear relationship of methane concentration vs. explosive limit taking the form of (values approximate):

Explosive Limit (in $\pu{K}$) = $\frac{\pu{3350K}}{2.3-\log_{10}(X_\ce{CH4})}$

where $X_\ce{CH4}$ is the mole fraction of methane, studied in the 0.01 - 0.1 range (significantly lower than yours). For this particular heuristic, the mole fraction of hydrogen was 0.28, oxygen 0.14, and the balance was nitrogen. The paper suggests the mechanism is non-equilibrium formaldehyde formation.

As a final note, please be very cautious when handling or recreating these mixtures. Just because the hydrogen isn't burning, doesn't mean it can't explode. A lab down the hall from mine was fortunate to loose only a wall to a hydrogen evolving system which ran fine for months and then, quite suddenly, decided to explode.

Best of luck learning more about your system! It sounds fascinating.

Reference:

R. R. Baldwin, N. S. Corney, R. W. Walker, “The inhibition of the hydrogen + oxygen reaction by methane,” Trans. Faraday Soc. 1960, 56, 802–814 (DOI:10.1039/TF9605600802).


Long linear molecule with alternating elements?

An example of a linear structure is given by cycopentadienyl lithium, whose structure is explored in this question.  There is some covalent interaction in this polymeric structure, as in most organolithium compounds, but the bonding between lithium and cyclopentadienyl groups is highly polar.


Two factors may be at play. (1) While bacteria and molds do indeed grow on sugar, a very high concentration of sugar inhibits growth. So fruit preserves, syrup, honey, molasses etc. do not need to be refrigerated and generally do not require other preservatives to be added. Soft drinks contain less sugar and are more susceptible to spoilage, hence the use of preservatives. I would definitely refrigerate a soft drink after opening it, if not consumed immediately. Acidity also helps inhibit growth (e.g. sweet pickles).

(2) Another factor is that bacteria and molds require water for growth, so foods such as dried fruits, that contain a moderate level of sugar but only a small amount of water, may have a long shelf life.


Why does glass shatter when heat is applied to it?

This is not a chemical process.

Objects typically expand when heated. Think in railway profiles in the summer, think in loosening a stuck nut by heating it with a blowtorch.

However, when the material is under stress and a bad heat conductor, unevenly heating will result in cracking.

While it is a pity for the cup, the breaking of glassware to thousands of rounded (!) pieces upon uneven or punctual (mechanical) impact may be exactly what you want under different circumstances. Think in tempered (toughed) glass used in windows of railroad passenger cars.

On a side note, there's the Bologna Bottle. In the making, these bottles are slowly cooled on the outside - but very rapidly on the inside, which maximizes the stress.

As a consequence, you might use the bottle as a hammer to drive a nail into something, but beware if you drop something in the bottle. Here's a video to demonstrate the effect. (Be warned that there's a lot of screaming kids in it.)


Toilet bowls are made of ceramics which can contain feldspar. Toilet bowls also often have a feldspar glaze. There is a study that disolved Cr(VI) can be removed by feldspar.

Studies on the removal of Cr(VI) from waste-water by feldspar
http://onlinelibrary.wiley.com/doi/10.1002/jctb.280530204/abstract

The IMA currently recognizes 90 Cr bearing minerals. 16 minerals if you only count those that contain also Si. Feldspar can be either K, Na or Ca-Feldspar. Of those 16 minerals 6 contain Na, 2 K and 5 Ca. Reading those papers might even narrow the choices down to a couple of minerals that can form in standard-toilet-bowl-conditions.

Mineral database that lets you select minerals by their chemistry:
http://rruff.info/ima/

I think that unless you are willing to donate your toilet bowl to science that the answer won't get any more precise than this.

Edit:
Cu would be the favorite choice for a blue color mineral (Azurite, ...). There is also a study about the removal of copper by feldspar:

Experimental studies of the interaction of aqueous metal cations with mineral substrates: Lead, cadmium, and copper with perthitic feldspar, muscovite, and biotite
http://www.sciencedirect.com/science/article/pii/S0016703797001178

Again there are 603 Cu-minerals. Cu+Si=32 minerals. With the common Feldspar cations your left with about 10 each. I don't think the minerals could be recognized visually. The favorite method would be x-ray diffraction of a powder sample of the blue stuff.


Why is the cathode reaction required in a fuel cell?

Why do we need two half reactions, one at the anode and another at the cathode?

The whole thing eventually grinds to a halt without the other reaction. The electrons generated at the anode travel through the circuit to the cathode. What happens then? Something needs to happen to the electron.

Additionally, what happens to that proton? As the concentration of protons increases, what will balance the charge that is produced in the cell, given that the electron is elsewhere?

We can imagine that we can set up a cathode in which the reverse reaction of the anode occurs:

Cathode: $\ce{H2 -> 2H+ + 2e-}$
Anode: $\ce{2H+ +2e- -> H2}$

However, in this case, there is overall no net reaction, which means that $\Delta G =0$ and since $\Delta G = nFE$, then the cell potential $E=0$. We need a different reaction for the cathode to consume electrons.

For example:

$$\ce{O2 + 2H2O +4e- -> 4OH-}$$

Since this reaction produces hydroxide anions, then we have something for the proton to do:

$$\ce{H+ + OH- -> H2O}$$

Now we have the following combination of reactions:

Anode: $\ce{H2 -> 2H+ + 2e-}$
Cathode: $\ce{O2 + 2H2O +4e- -> 4OH-}$
Neutralization $\ce{H+ + OH- -> H2O}$
Net: $\ce{2H2 + O2 -> 2H2O}$

This net reaction is exergonic, and thus produces a potential.


Is there an example where interpreting reaction coordinate as time leads to an incorrect conclusion?

Transition states are very short lived species, generally lasting less than a picosecond. At the single molecule level, crossing a transition state is a very fast event: it goes from reactants to products in less than a picosecond.

Some reactions are very slow because the probability of crossing the transition state is very low, but never because they spend a long time at the transition state. Thus, interpreting reaction coordinate as time is incorrect.

In a slow reaction, a small number of molecules will react early, but the majority will spend a long time in the reactants state, waiting to cross the transition state. If one could take a static picture of an ensemble of molecules, some would be in the reactants state, some would be in the products state, and it would be almost impossible to see any molecule at the transition state.


How to create liquid carbon, and what kinds of properties does it have?

It's possible to create liquid carbon, but it's extremely hard. As you can see on the Carbon phase diagram  if you want to obtain "easely" liquid carbon you must do it at $\ce{1 GPa}$ and $\ce{5000 K}$.

And for the properties I really don't know but carbon follows the octet rule and it might be very reactive as an electrophile because it will just have four electrons around it.

I hope it can help you a bit, i'm seeking in some data-bases but there are not really things to answer your question.

Have a nice day.


Soda drinks are carbonated meaning that carbon dioxide is dissolve into it. When carbon dioxide dissolves, the following reaction occurs: $$\ce{CO2(g) + H2O(l)~ \leftrightharpoons ~H2CO3(aq)}$$ As you can see, carbonic acid (a weak acid) is formed. It this which is responsible for the 'fizziness' of the drink.

Note that this reaction is in equilibrium, meaning that not all of the carbon dioxide dissolve to form carbonic acid, but only a certain percentage actually dissolves. Usually, the percentage of carbon dioxide that is dissolved is constant, however this can be changed by doing certain things to the system.

For example, by decreasing the pressure the equilibrium will shift to the left, meaning that the percentage of carbon dioxide that is dissolved will decrease. This will result in an increase of gaseous carbon dioxide in the bottle. This reason for this is explained by the Le Chatelier's principle. You can also imagine it as a piston pushing the gas into the liquid as shown in the below image. Image (a) is at low pressure while image (b) is at high pressure.

Normally in soda bottles, the pressure inside the bottle is relatively high to ensure that more carbon dioxide dissolves into the drink. So when you open a soda can you decrease the pressure. This results in more carbon dioxide gas to produced. It is the release of this gas that causes that pfffftt sound.

Now when you shake the bottle, due to turbulence, small bubbles containing carbon dioxide form.The formation of these small bubbles allow even more carbon dioxide gas to form. So if you were to shake a soda can and then open it soon after, the soda will jump up quite violently as there is now a lot of carbon dioxide being released.

However if you were to shake a soda can and then wait a while before opening it, it most likely just create the normal pfffftt sound. This is because you give the system some time to go back to its original equilibrium position, meaning that some of the carbon dioxide gas will redissolve back into the solution .


Why is the name 'propylene oxide' preferred for epoxypropane?

why is 'propylene oxide' the preferred name for epoxypropane?

Propylene oxide is not the "preferred" IUPAC name (PIN): Wikipedia lists the PIN as 2-methyloxirane, which is in line with the 2013 IUPAC recommendations.

  why is epoxypropane the only isomer encompassed by the designation 'propylene oxide'?

It's more of a tradition than anything else. Note how the two ring carbons in propylene oxide are the same as the two doubly bonded carbons in propylene: this is not a coincidence. A very common way of synthesising an epoxide is to oxidise the corresponding alkene (you can think of it as simply adding an oxygen atom across the double bond), and so "alkene oxide" has become a rather conventional way of referring to an epoxide (see also https://en.wikipedia.org/wiki/Styrene_oxide for another example).

So, the meaning of "propylene oxide" is not really "an oxidation product of propylene". It should pretty much be read as "the epoxide derived from propylene", or "the epoxide formed when you add an oxygen atom across the double bond of propylene".

In the strict sense of oxidation/reduction being addition/loss of oxygen, oxetane is an oxidised form of propylene, but there is no simple chemical oxidant that will perform this transformation. So, calling oxetane "propylene oxide" would never have - and has never - caught on.


Why do helium balloons expand in volume as they go higher?

I didn't know that balloons expanded during the fly because of thermodynamics, and I didn't know how high they can fly, but a rapid search tells that a partially unfilled regular balloon can fly until an altitude of around $\pu{25 km}$.

Now, $\pu{25 km}$ means that it reaches the first part of the stratosphere, with temperatures of $\pu{-60 ^\circ C}$, that gradually increase until $\pu{0 ^\circ C}$ at $\pu{50 km}$.
As for the pressure, it goes from around $\pu{40 mmHg}$ to $<\pu{1 mmHg}$ in the range $25\text{–}\pu{50 km}$.

If you try a $pV=nRT$ calculation on these data, you see that the gas is already at around 10 times its initial volume with $\pu{40 mmHg}$ pressure and a $\pu{213 K}$ temperature, and that at the $\pu{50 km}$ point the volume is increased 700 times!

Also: while the trend of the pressure is quite logical, that of temperature is caused by complex interations (eg: sun rays that heat particles). You can find this image quite interesting:


Are there substances that cannot undergo combustion?

It depends a little on your definition of combustion.

Most combustion we see involves things reacting with the oxygen in air. Most combustion reactions are a self-sustaining exothermic reaction of some substance with oxygen gas. But atmospheric oxygen is relatively dilute (only about 21% of air is oxygen) and this limits what will burn sustainably in air (also, at slightly lower concentrations many commonly flammable things won't stay lit). The requirement for things to be combustible in air is, crudely, a reaction path with oxygen that releases energy.

Examples include many types of compound. Hydrogen reacts with oxygen to give water; hydrocarbons react to give carbon dioxide and water. Many metals react to give metal oxides (eg magnesium, spectacularly when lit; iron, only when hot and finely divided).

Many things won't burn in air for two reasons: there are activation barriers to a potential oxidation meaning a potential reaction can't be sustained (aluminium has a strong oxide layer on its surface protecting the reactive metal from further oxidation, for example); other compounds are already highly oxidised an no further oxidation can happen (eg silicon dioxide, a major ingredient of sand and glass).

But there are many more things that will combust in pure oxygen as Apollo 1 discovered, tragically. And as many demonstrations show spectacularly (see demo here). But highly oxidised things like silica can't be oxidised further and won't burn not matter how concentrated the oxygen is.

But oxygen isn't the only gas that can support exothermic reactions. Rather more things will burn in fluorine, for example. Even relatively unreactive substances can support combustion if there is an available reaction path that releases enough energy and a reaction can get started. One of the best known examples is that magnesium will burn in carbon dioxide (the reaction producing magnesium oxide releases energy) and even nitrogen (producing magnesium nitride). See the explanation here and the video here.

So, to return to the original question, are there any substances that won't combust? Whatever the definition of combustion is the answer is still yes. Noble gases, for example, though some can be coerced into reacting under extreme circumstances, never combust or support combustion.

If we restrict our definition to combust in air, many substances won't combust though many of them might do so under different circumstances.


Sigma and pi symbols in molecular orbital theory

These symbols are for bonding orbitals. A covalent bond is formed when two orbitals in different atoms hybridize. $\sigma$ and $\pi$ refer to two different configurations in which this can happen.

In a $\sigma$ bond, two $s$ orbitals or two $p$ orbitals oriented towards each other join into one orbital, with the electron density concentrated between the two atoms.

In a $\pi$ bond, two $p$ orbitals oriented perpendicular to the bond axis merge, and, in the resulting orbital, electron density is concentrated above and below the bond axis.


Can baking soda or citric acid help clean a pot with burned-in milk rice?

The hotter the water, the more likely it will be to remove burnt-on starch. Not only does $\ce{NaHCO3}$ raise the boiling point of the water, but it also helps saponify fats (actually, "washing soda", $\ce{Na2CO3}$ would be better).

However, you still need to do a lot of scrubbing afterwards to remove the carbonized layer.


Getting the wrong pH value, according to data and online calculator

As you can see from the point that $n_\ce{OH} > n_\ce{AcOH}$, there will be an excess of hydroxide ions. What you want to calculate is the pH for this amount of hydroxide ions within the total volume at this point.

Thus the calculation looks like:
\begin{align}
[\ce{OH-}] &= \frac{n_\ce{NaOH}-n_\ce{AcOH}}{V_\text{a} + V_\text{b}}\\
\Leftrightarrow \mathrm{pOH} &= -\lg ([\ce{OH-}])\\
\Leftrightarrow \mathrm{pH} &= 14+\lg ([\ce{OH-}])
\end{align}

Your error is based on a rounding error ... please don't round intermediate result as all rounding errors will sum up. You just need to round the final results to a proper amount of decimal places.

The correct calculation:

$$\mathrm{pH} = 14 + \lg \left(\frac{0.0024768 - 0.00246}{0.025 + 0.0258}\right) = 10.5194 \approx 10.52$$

vs. your calculation:

$$\mathrm{pH} = 14 + \lg \left(\frac{0.0025 - 0.00246}{0.025 + 0.0258}\right) = 10.8962 \approx 10.90$$


What is the IUPAC nomenclature of alkane with large numbers of C atoms (more than 200)?

The definitive answer is, of course, in the rules established by the IUPAC’s “Commission on nomenclature in organic chemistry”. The reference you are looking for is:

  “Extension of Rules A-1.1 and A-2.5 Concerning Numerical Terms Used in Organic Chemical Nomenclature”, Pure Appl. Chem., 1986, 58, 1693-1696

which can be found here as a PDF and here as an HTML version. In particular:

  NT-1.1 - The fundamental numerical terms for use in hydrocarbon names or as multiplying prefixes for simple features are given in the following list:

  \begin{array}{|c:c|c:c|c:c|c:c|}\hline
    1&\rm \text{mono- or hen-} &10 &\rm \text{deca-}&100&\rm \text{hecta-}& 1000&\rm \text{kilia-} \\\hline
    2&\rm \text{di- or do-}&20&\rm \text{icosa-}&200&\rm \text{dicta-}&2000&\rm \text{dilia-}\\\hline
    3&\rm \text{tri-}& 30&\rm \text{triaconta-}&300&\rm \text{tricta-}&3000&\rm \text{trilia-} \\\hline
    4&\rm \text{tetra-}&40&\rm \text{tetraconta-}&400&\rm \text{tetracta-}&4000&\rm \text{tetralia-} \\\hline
    5&\rm \text{penta-}&50&\rm \text{pentaconta-}&500&\rm \text{pentacta-}&5000&\rm \text{pentalia-}\\\hline
    6&\rm \text{hexa-}&60&\rm \text{hexaconta-}&600&\rm \text{hexacta-}&6000&\rm \text{hexalia-}\\\hline
    7&\rm \text{hepta-}&70&\rm \text{heptaconta-}&700&\rm \text{heptacta-}&7000&\rm \text{heptalia-}\\\hline
    8&\rm \text{octa-}&80&\rm \text{octaconta-}&800&\rm \text{ octacta-}&8000&\rm \text{octalia-}\\\hline
    9&\rm \text{nona-}&90&\rm \text{nonaconta-}&900&\rm \text{nonacta-}&9000&\rm \text{nonalia-}\\\hline
\end{array}

So, $\ce{C_205H_412}$ is the molecular formula of pentadictane. $\ce{C_7547H_15096}$ is heptatetracontapentactaheptaliane.