Why don't polar and non-polar compounds dissolve each other? Dipoles can also be induced in polar and non polar compounds, then why don't they dissolve?

Very simply, you explain the reason for this solubility rule by taking in consideration the energy requirements for the breaking of intermolecular forces between the molecules in the solute and the solvent.

Note: this is only a simplified explanation as it also depends on other factors such as change in entropy

Here is some background information on intermolecular forces. In non-polar substances, there are dispersion forces between each molecule. These dispersion forces are relatively weak and hence only require little energy to break them. In polar substances, there are dipole dipole and hydrogen bonding (depending on the substance) between each molecule. These forces are much stronger than dispersion forces and require more energy to break.

Now lets consider the following cases:

Non-polar Solute and Solvent

For the solute to dissolve, the dispersion forces between the molecules in the solute and solvent need to break. This only requires very little energy. However when the solute dissolves into the solvent, they are able to be make dispersion forces with each other. The making of these forces releases very little energy. So simply put, very little energy is required to break the forces and very little energy released when making the forces. Hence overall everything balances out and the process occurs.

Non-polar Solute and Polar Solvent (and vice versa)

For the solute to dissolve into the solvent, both dispersion forces and dipole dipole forces are broken which require large amount of energy. However the molecules in the solute and solvent are only able to make dispersion forces with each other (as they aren't both polar). This only releases very little energy. Therefore overall, more energy is required than released and hence the process won't happen.

Polar Solute and Solvent

For the solute to dissolve into the solvent, dipole dipole forces are broken which require large amount of energy. However when they do dissolve, the molecules in the solute and solvent are able to form dipole dipole forces which releases large amount of energy. Therefore overall, everything balances out and the process occurs.


Why is tin-112 stable, but indium-112 radioactive? According to Wikipedia, tin-112 is a stable nuclide, but indium-112 is radioactive (half-life only 14 minutes). What is the explanation for this?

For isobars (i.e. nuclides with a constant mass number $A$), the binding energy $E$ as a function of the atomic number $Z$ describes a parabola, the so-called valley of β-stability (see also Weizsäcker’s formula or semi-empirical mass formula).

$$E = a \cdot Z^2 + b \cdot Z + c \pm d/A^{3/4}$$

The most stable nuclides lie at the bottom of the valley (but not necessarily exactly at the minimum of the parabola). Isobars with lower atomic numbers $Z$ are unstable to $\beta^-$ decay. Isobars with higher atomic numbers $Z$ are unstable to $\beta^+$ decay or electron capture. Therefore, at least one of two adjacent isobars must be radioactive (see Mattauch isobar rule).

Thus, since Sn-112 is stable, we expect In-112 to be radioactive.

The last term $(\pm d/A^{3/4})$ gives rise to three different parabola depending on whether the nuclides are even-$A$ (with even $Z$ and even $N$), even-$A$ (with odd $Z$ and odd $N$), or odd-$A$ (with even $Z$ and odd $N$, or with odd $Z$ and even $N$).
In the last case, in which the mass number $A$ is odd, the result is a single parabola with one stable nuclide at the bottom.
However, for even mass numbers $A$, we find two parabolas; the odd-odd-curve always lies above the even-even curve. Hence, it is possible that two or even three stable isobars exist.

The case with two parabolas also applies to your example. The stable nuclide Sn-112 has 50 protons and 62 neutrons (even-even). The radioactive nuclide In-112 has 49 protons and 63 neutrons (odd-odd). The next neighbour, Cd-112 has 48 protons and 64 neutrons (even-even) and is stable.


Is Saliva an acid or base? I have a debate with some friends who believe that saliva is just basic, but with the enzymes which make them reactive with objects in our mouth, shouldn't it be an acid in this case?

I can count four reasons that say saliva pH isn't the same among humans:

As people age, the composition of saliva changes in their body; this is easily explainable with the fact that the body isn't performing the actions the same way as you grow up or become old.
People have different eating habits; this will put the body in impression of producing especial kinds of enzymes and reduce\increase the production of certain factors in the saliva.
People aren't the same when it gets to taking care of the insides of their mouth. You can be brushing your teeth twice as much as a close friend of yours does.
Human digestive system isn't always performing its duties consistently and without "ups" and "downs" during the day, which has an important effect on enzymes produced.

And a sea of other reasons....

But, if we are to answer this question:

  Saliva is a watery substance located in the mouths of animals, secreted by the salivary glands. Human saliva is 99.5% water, while the other 0.5% consists of electrolytes, mucus, glycoproteins, enzymes, and antibacterial compounds such as secretory IgA and lysozyme. Saliva, wikipedia

I'll also say that according to here, a healthy range of pH of saliva varies between 6.5 and 7.5 during the day, but is optimally between 6.5 and 6.8.

Source

Thus, saliva is a very weak acid, or sometimes a very weak base. Considering something like that an acid or a base is like saying that the water we drink is a base and the water of the raindrops that fall from the sky of a peaceful country is acidic.

Saliva is really important for it's being a buffer.


A Curious (definitely wrong) result of Absolute Entropy The Gibbs free energy is defined as
$$G=H-TS$$
Differentiating both sides
$$dG=dH-TdS-SdT$$
For an adiabatic reversible expansion of an ideal gas, we would have $dG=0$, and $dS=0$.
Thus we can write, $dH=SdT$
Also, $dH=n C_p dT$.
Finally I reach the strange result that $S=nC_p$. How can I define absolute entropy so simply. Something is definitely very wrong here. Please correct me.

The Gibbs free energy change is zero in the case of reversible processes carried out at constant temperature and pressure, but that isn't the case if these conditions are not observed.
As a demonstration consider an isothermal reversible expansion of an ideal gas.
Since the temperature is constant, the free energy change is given by
$$\Delta G = \Delta H - T \Delta S$$
Because the gas is ideal and T constant, $\Delta U = 0$, and from the second law of thermodynamics it follows that
$$\Delta S = \frac{q_\text{rev}}{T}= -\frac{w_\text{rev}}{T} = \frac{\int_{V_1}^{V_2}p dV}{T} = \frac{\int_{V_1}^{V_2} \frac{nRT}{V} dV}{T} = nR\log \left( \frac{V_2}{V_1} \right)$$
It also follows that
$$\Delta H = \Delta U + \Delta (pV) = 0 + \Delta (nRT) = RT \Delta n = 0$$
since the system is closed.
But then
$$\Delta G = - nRT\log \left( \frac{V_2}{V_1} \right) \neq 0$$
The free energy change is not zero because the pressure changed during the process.
For the reversible adiabatic expansion of an ideal gas
$$\begin{align}  \Delta S &= 0  \\ \Delta H &= \int_{T_1}^{T_2} C_p dT  = C_p \Delta T \\ \Delta G &= \int_{P_1}^{P_2} V dP - \int_{T_1}^{T_2} S dT =  (V_1P_1^{C_V/C_P})\left(\frac{C_P}{C_P-C_V}\right)( P_2^{1-C_V/C_P}-P_1^{1-C_V/C_P}) - S\Delta T  \end{align}$$
The integral over P in the expression for $\Delta G$ can be evaluated after substituting the following expression for the volume in terms of pressure:
$$V=(V_1P_1^{C_V/C_P}) P^{-C_V/C_P}$$


Changing State of Alumina Oxide I intend to bring solid Alumina Oxide into a liquid state, for the purpose of pouring the substance into a mold.  Subsequently, I will allow it to cool with an ambient air temperature of 70f, slowly allowing it to return it to a solid state.  Melting will be performed with a gas mixture.

My question is, will the Alumina Oxide return to the original molecular structure as is was prior to melting it from a solid state?

Thank you for your time and consideration.

Rob

First of all, there is no such thing as "alumina oxide". Alumina is aluminium oxide. Melting alumina is pretty hard, its melting point is above 2000 degrees Celsius. It's so high that people use it as a container (crucible) for melting other stuff. Such high temperatures are not easily achieved without melting a bunch of stuff around it.

Let's assume that you did manage to melt it, maybe using one of those fancy levitation laser techniques. Your  mold has to be heat resistant and inert so it will not melt or react with the molten alumina. Graphite is usually a good one.

Now, you were asking about the structure. Assuming that you started with regular crystalline alumina (aka corundum), then yes: the resulting material after cooling in air will be the same corundum. Amorphous alumina (glass) will only form by extremely rapid cooling, for example dropping a very small amount to ice cold water. In all normal circumstances it will be crystalline.


When a molecule loses hydrogen atoms, does it become oxidised? I had the following question on a bio test covering cell respiration and glycolysis, but I'm not sure if I agree with the answer.

  When a molecule loses hydrogen atoms (as opposed to hydrogen ions), it becomes
  B) oxidised

According to my textbook (Principles of Life, 2nd Edition), oxidation is defined as

  Relative loss of electrons in a chemical reaction; either outright removal to form an ion, or the sharing of electrons with substances having a greater affinity for them, such as oxygen. Most oxidations, including biologial ones, are associated with the liberation of energy.

Hydrogen atoms have one proton and one electron in their typical state. If a molecule is losing hydrogen atoms, then its net charge wouldn't change. For example, if a (hypothetical) molecule had 7 protons and 6 electrons, it would have a charge of $+1$. If it then lost a hydrogen atom, it would now have 6 protons and 5 electrons, which still yields a net charge of $+1$. Am I missing some critical piece of information, or is this still considered oxidation?

This phrasing is only true for organic molecules. If for example, sodium hydride loses hydrogen, the sodium ion will get reduced. But since you seem to come from a biochemical background, this simplification is okay since you will be dealing with organic molecules primarily.

The idea behind that statement is that hydrogen atoms in biomolecules are typically only bound to carbon, nitrogen, oxygen and sulfur. Hydrogen is less electronegative than all these elements and therefore any $\ce{X-H}$ bond will be polarised towards $\ce{X}$; the non-hydrogen atom.

When determining oxidation states, bonds are formally cleaved heterolytically in such a way that the electronegative partner gets both electrons. Then, the electrons on the formal atomic ions created this way are counted and subtracted from the number the compound should have. Bonds between the same element are cleaved homolytically and the same procedure applied. Thus, if we take ethene ($\ce{C2H4}$, structure see below) the $\ce{C-H}$ bonding electrons are formally attributed to carbon entirely while the $\ce{C=C}$ double bond is split equally. Doing this, each carbon atom ends up with six electrons, formally, and therefore a formal charge of $2-$ or an oxidation state of $\mathrm{-II}$. If we did the same for ethyne ($\ce{C2H2}$) — a compound with two hydrogen atoms removed — we arrive at five formal electrons and thus an oxidation state of $\mathrm{-I}$. Therefore, going from ethene to ethyne is an oxidation.

$$\begin{align}\ce{H2C=&CH2} & \ce{H-C&#C-H}\\
\text{ethe}&\text{ne} &\text{eth}&\text{yne}\end{align}$$

Having gotten the theory out of the way, what does this mean for the simplified rule? Well, as I mentioned any $\ce{X-H}$ bond is polarised away from the hydrogen and thus effectively adds an additional electron to $\ce{X}$. Carbon, which is the most common bonding partner of hydrogen, is less electronegative than all the other typical elements of biomolecules, and thus will formally lose electrons to them. When removing hydrogen from a compound chemically, we will always remove at least one $\ce{C-H}$ bond in favour of a $\ce{C-C, C-O, C-N}$ or $\ce{C-S}$ bond. Thus, the carbon atom will formally lose electrons and we will have an oxidation.

This can also be rationalised in another way to determine oxidation states: the additive method. Starting from a neutral compound, assign each hydrogen atom a $\mathrm{+I}$, each oxygen $\mathrm{-II}$. Add that up and invert the sign; this is the oxidation state of all the carbons (assuming a molecule of the formula $\ce{C_mH_nO_x}$). Divide by the number of carbons to get an average oxidation state of carbon.

$$\chi(\ce{C}) = -\frac{n - 2x}{m} = \frac{2x - n}{m}$$

As you see, each hydrogen atom in a compound enters into the formula with a negative sign. Thus, removing a hydrogen must increase the average oxidation state of the carbons. Unfortunately, there is no trivial way to extend this formula to compounds containing oxygen and nitrogen or oxygen and sulfur.


Trans-hydrogenation of alkenes Both this and that question seem to indicate that the hydrogenation of alkenes to alkanes or alkynes to alkenes is always a syn process.

However since hydrogenation is not a concerted process, I do not see why we could not have a non-negligible (although small) portion of the trans product.

Following this reaction mechanism for syn additions :

before the penultimate step (the addition of the second hydrogen), can't the molecule simply rotate around the $\ce{C-C}$ axis (I expect this should be possible with small enough alkenes) ?

The Wikibooks article on alkenes does indicate that typically the hydrogenation of alkenes leads to only $98\%$ of the syn product, but I have been unable to find more precise information elsewhere.

This link has a good summary of hydrogenation. Syn addition is very much preferred because of the way the molecule 'adheres' to the surface. As you note, however, there are cases where anti-addition is observed - particularly for sterically hindered molecules such as 1,2-dimethylcyclohex-1-ene which shows up to 30% of the trans/anti-addition product (which also so happens to be the thermodynamic product).

The reason we see the syn-addition more often is because the Hydrogen atom attack happens really quickly (it's the kinetic product). If you pump more heat energy into the reaction though, you could expect to see more trans product - but this is dangerous as hydrogen is explosive.


Why is aluminium(III) considered a hard acid if it forms covalently bonded compounds like aluminium oxide? If something is a hard acid, it tends to form ionic bonds as it is polarizing. So why is the aluminium ion a hard acid when it forms bonds with covalent character. A similar argument follows for why carbon dioxide is considered a hard acid?

Short summary - HSAB theory has it's limits of applicability, and this is a good example.

More recent discussion of acid-base theory center on the frontier orbital concept, e.g. W. B. Jensen, The Lewis Acid-Base Concepts (1980) pp. 112-155.

In this concept, an "acid-base" reaction creates new HOMO-LUMO orbitals of the product. So it's about energy matching between the acid (empty LUMO) and base (filled HOMO) (and also symmetry and orbital overlap). It's still very much a Lewis picture, since you have the electron pair on the base and an electron acceptor on the acid.

For example, if we react $\ce{NH3 + H+ -> NH4+}$ we combine the HOMO of $\ce{NH3}$ (i.e., a lone pair) with an empty LUMO on $\ce{H+}$ to create the adduct.

IMHO, this is a nice concept, since the orbital energy match between Al and O is good, you get covalent character to the resulting product. (Now you also get ionic character, but that's a different question.)


Only one unique adiabatic path between two states My Textbook says:

Two states A and B can never lie both on a reversible as will as irreversible adiabatic path. There lies only one unique adiabatic path linkage between two states A and B.

I don't understand what the book means. Why cannot there be both reversible and irreversible adiabatic paths between two states? And if only one path is possible, then how do we know which one?
What I understood is: In adiabatic process: ΔU=W. Now since ΔU is state function, it will mean that Wirreversible=Wreversible, which is not possible. But mathematical proof aside, is there any example/physical explanation to this?

For an infinitesimal adiabatic process $dS_\mathrm{surroundings} = 0$ since no heat is exchanged with the surroundings. But $dS_\mathrm{system} = 0$  is only true for a reversible process, since only then  $dS_\mathrm{system} = dq/T$. For an irreversible process $dS_\mathrm{system} \neq dq/T =0$. Therefore
$$dS_{\mathrm{irrev}}\neq dS_{\mathrm{rev}}=0$$
This means that the state accessed with a reversible process through an integral change $\int dS_{\mathrm{rev}}=0$ cannot be accessed directly through an irreversible adiabatic process, for which $\Delta S = \int dS_{\mathrm{irrev}} \neq 0$.
Now since S is a state function, the terminal states of the two processes must therefore differ (QED).
As regards the second part (how do we know which is the reversible), you can use the approach just outlined to conclude that for the reversible process $\Delta S =0$.


Kolbe's electrolysis method for getting alkanes It involves the electrolysis of "Sodium and Potassium salts of Carboxylic Acids". Why only Sodium and Potassium? Will some other metal like Li, Mg, Zn etc. not work? As long as the element is a better reducing agent than Hydrogen, won't our destination be attained? What is the role of the metal here?

The metal does not play any role other than to form soluble salts here. The metal cations move to the cathode while carboxylate anions move to the anode where oxidation to a $\ce{R-COO·}$ radical takes place. These radicals decompose to $\ce{CO2}$ and $\ce{R·}$ radicals. Two of latter combine to form a C-C bond.


Testing for nitrate ions with NaNO3 in Devarda's test When we use Devarda's test on $\ce{NaNO3}$ solution (we add $\ce{NaOH}$ and $\ce{Al}$ powder, heat the mixture, and produce ammonia), what is the equation of the process?

I understand that $\ce{Al}$ reduces $\ce{NO3-}$ to $\ce{NH4+}$ then $\ce{NH4+}$ and $\ce{OH-}$ have a reversible reaction to produce $\ce{NH3}$, and water, but I'm not sure how to summarize this in an equation. In particular, what happens to the aluminium?

Quoting from Svelha G., Vogel's Qualitative Inorganic Analysis,

  A solution of nitrate ion is boiled with zinc dust/aluminium powder
  and sodium hydroxide solution to get positive result. Excellent result
  is obtained by the use of Devarda's alloy (45% Al, 5% Zn, 50% Cu) but
  the ammonium ions must be removed by boiling the solution with sodium
  hydroxide solution or almost evaporating it to dryness before addition
  to metal.

  The overall reaction is:

  $$\ce{3NO3- + 8Al + 5OH- + 18H2O ->3NH3 ^ + 8[Al(OH)4]-}$$


Why does free silicon react with water? Free silicon is not found in nature (in water, it might be present in the form $\ce{Si(OH)_{4}}$. Some compounds of silicon, such as silane ($\ce{SiH_{4}}$) will react spontaneously in an aqueous environment, while the analogous methane won't.

Is there an explanation for the tendency of silicon compounds to react with water?

Is there an explanation for the tendency of silicon compounds to react
  with water?

Reaction with water, oxygen, etc., the answer is the same and has to do with the strength of the $\ce{Si-O}$ bond. Let's compare the following two reactions:

$$\ce{CH4 + \frac{1}{2}O2 -> CH3OH}$$
$$\ce{SiH4 + \frac{1}{2}O2 -> SiH3OH}$$

In both reactions we break an $\ce{X-H}$ bond and an $\ce{O-O}$ bond, while we make an $\ce{X-O}$ bond and an $\ce{O-H}$ bond.  The $\ce{O-O}$ bond breaking and the $\ce{O-H}$ bond making are common to both reactions and therefore cancel out.  Let's compare the only differences, the $\ce{X-H}$ bond breaking and the $\ce{X-O}$ bond making.

The following Table contains the bond energies for these bonds.

\begin{array}{|c|c|c|c|} \hline
\ce{X} & \text{X-H Bond Strength} & \text{X-O Bond Strength} & \text{Overall Change}  \\ & \text{(kcal/mol)} & \text{(kcal/mol)} & \text{(kcal/mol)}\\ \hline
\ce{C} & 99 & 86 & -13\\ \hline
\ce{Si} & 75 & 110 & 35\\ \hline
\end{array}

Looking at the numbers it becomes clear that the $\ce{Si-H}$ bond is a lot weaker than the $\ce{C-H}$ bond, and the $\ce{Si-O}$ bond is a lot stronger than the $\ce{C-O}$ bond.  Consequently, the reaction involving silicon is more exothermic than the analogous reaction involving carbon.  This greater stability in the $\ce{Si-O}$ product is reflected in the transition state leading to it.  It lowers the energy of the transition state and consequently the reaction involving silicon is much more facile than the same reaction involving carbon.


Will heating diatomic oxygen enough break the O=O bonds? The bond enthalpy associated with a $\ce{O=O}$ double bond is equal to 495 kJ/mol. Does that mean that adding enough kinetic energy in the form of heat will eventually cause the bonds to break and create monoatomic oxygen?

Yes, adding at least 495 kJ/mol of kinetic energy one way or another (thermally, photochemically by irradiation with photons of that energy, sonication, etc.) will cause $\ce{O2}$ to dissociate into monatomic oxygen.

$$\ce{O2 ->C[energy]\ 2O}$$


What is a "symmetrical product" of an irreducible representation with itself? In their classic paper outlining the Jahn–Teller theorem (Proc. R. Soc. Lond. A 1937, 161 (905), 220–235), the authors wrote "where $[\phi^2]$ denotes the representation of the symmetrical product of $\phi$ with itself". Reading this I'm thinking about a regular product of representation (scalar product with each characters).

But in the next paragraph there is this equality for the $D_\mathrm{4h}$ group:

$$[\mathrm{E_g^2}] = [\mathrm{E_u^2}] = \mathrm{A_{1g} + B_{1g} + B_{2g}}$$

Obviously something is wrong, as the direct product is a 4-dimensional representation, but the right-hand side is 3-dimensional.

I would like to add that there is a reference of Weyl 1928 unfortunately this article is in German and I cannot find anything in it.

What is the correct way to carry out this direct product?

In your case you need the equations for the characters as there are symmetric and antisymmetric parts in the product. Your question gives just the symmetric part and the antisymmetric part is $A_2$ normally written as $[A_2]$ to indicate this.

The characters are found using $\chi^+ =\left([\chi(R)]^2+ \chi(R^2) \right)/2$ for the symmetrical part and $\chi^- =\left([\chi(R)]^2- \chi(R^2) \right)/2$ for the antisymmetric part. Using the point group table you need to work out the square of all the operations for $E_g$ this gives $\chi(R^2)$ (e.g. $(C_2)^2 \to E$) and then square the operations which gives $[\chi(R)]^2$, i.e. $[E]^2\to 4$. From the list of characters you then identify the symmetry species, either directly by inspection (as for the antisymmetric in this case), or by forming the irreps from the reducible representation.


why are lp-lp repulsions greater than bp-lp and bp-bp? My teacher told me that lone pair and lone pair are closer together hence the repulsion is greater and they take up more space, but I do not understand? can anyone explain it to me?

So, from Wikipedia article on VSEPR theory we read:

  The overall geometry is further refined by distinguishing between
  bonding and nonbonding electron pairs. The bonding electron pair
  shared in a sigma bond with an adjacent atom lies further from the
  central atom than a nonbonding (lone) pair of that atom, which is held
  close to its positively charged nucleus. VSEPR theory therefore views
  repulsion by the lone pair to be greater than the repulsion by a
  bonding pair. As such, when a molecule has 2 interactions with
  different degrees of repulsion, VSEPR theory predicts the structure
  where lone pairs occupy positions that allow them to experience less
  repulsion. Lone pair-lone pair (lp-lp) repulsions are considered
  stronger than lone pair-bonding pair (lp-bp) repulsions, which in turn
  are considered stronger than bonding pair-bonding pair (bp-bp)
  repulsions, distinctions that then guide decisions about overall
  geometry when 2 or more non-equivalent positions are possible.

As I said in my comment, the key point is given in the second sentence: the bonding pair shared in a bond lies further from the atom than a nonbonding pair of that atom which is held close to its positively charged nucleus. Consequently, lone pairs are closer to each other than any other combination of pairs (lone pair-bonding pair and bonding pair-bonding pair), and thus, repell each other more strongly.


What facts are being used in this atomic structure question? I wish to understand how to solve this atomic structure question:

  For the ground state, the electron in the H-atom has an angular
  momentum = $\hslash$ , according to the simple Bohr model. Angular
  momentum is a vector and hence there will be infinitely many
  orbits with the vector pointing in all possible directions. In
  actuality, this is not true, because:

Bohr model gives incorrect values of angular
momentum.
only one of these would have a minimum energy.
angular momentum must be in the direction of spin of electron.
electrons go around only in horizontal orbits.

Actually, we have been taught Bohr orbits a lot in class, with formulae for everything (atomic radius/electron orbital speed/various energy levels/etc.)  and also basics of the quantum mechanical model. However, we didn't go into such a great depth of Bohr orbits as asked in the above question.

Specifically, I don't know of any reasonable arguments against points 1,2,3. Point 4 is evidently wrong based on the quantum mech model.

  How do I develop any claim for/against the points 1,2,3? What are the facts involved here?

The only correct answer is #1.

In actuality, the ground state of the hydrogen atom has zero angular momentum.

Regarding #2 and #4, there is no preferred plane in which one point should orbit another point.  By symmetry, they are all equivalent.

Regarding #3, there is no angular momentum in the ground state, and in a state whether there is angular momentum, it doesn't need to be in the same direction as the electron spin angular momentum.


Regarding the use of dilute sulphuric acid in this titration I did a titration today for the first time.
A solution of Potassium Permanganate of unknown molarity was titrated against a solution of ammonium iron II sulphate of known molarity.
I have to say, seeing the purple solution turn clear was a lot of fun.
Dilute sulphuric acid solution was used twice during this experiment. First to prepare the ammonium iron II sulphate and sulphuric acid solution, and then 20 ml was added to each sample before titration.
Why did these have to be added ?
Also, could I have used pure sulphuric acid ? Why did it have to be diluted ?
I think the first usage provided the H+ for the reaction to occur but I am not sure. Would pure sulphuric acid not be more efficient ?

A little bit of sulphuric acid is used when preparing iron(II) solution. This acid prevents the possible hydrolysis of $\ce{Fe^{2+}}$ ion according to : $$\ce{Fe^{2+} + 2 H_2O <=> [Fe(OH)_2(s) + 2 H^+}$$ In case this reaction occurs, the added sulphuric acid will react and destroy the precipitate $\ce{Fe(OH)_2}$ according to the inverse reaction : $$\ce{Fe(OH)_2 + 2 H^+ -> Fe^{2+} + H_2O}$$
Now, during the titration, the ion $\ce{H^+}$ is needed to carry out the reaction of permanganate on the ion $\ce{Fe^{2+}}$ . Hopefully you know this equation.
And concentrated sulphuric acid is not wanted, because it reacts strongly with water just during its dissolution. It produces a lot of heat.


How to determine the products of a redox reaction I'm doing a paper on water treatment methods, and in one section, I'm describing how a nearby water treatment plant is removing Trichloroethylene ($\ce{C2HCl3}$) from water by oxidizing it using Ozone ($\ce{O3}$).

However, I can't find any documents confirming the products of this reaction, so I'm going to need to determine them on my own, but I'm not sure how. I'm fairly certain the equation would be formatted as

$$\ce{C2HCl3(aq) + O3(g) -> H2O(l) + }$$

but I'm not sure of anything past that.

I did some research on this topic (water treatment for TCE) and I come out three answer which is by using peroxide, ozone and permanganate. Since you asked for ozone first, so I will give that first.

What are you missing in the reaction is that water is the reactant of this reaction instead as a product. Here's the equation:

$$\ce{C_2HCl_3 + O_3 + H_2O -> 2CO_2 + 3HCl}$$

For the reaction involved ozone, The reaction will be more effective in an acidic medium and it must perform in a enclosed delivery pipes.

Second, The reaction involving peroxide. Hydrogen peroxide react with TCE will also produce carbon dioxide and hydrogen chloride but water is a product here. Here is the equation:

$$\ce{C_2HCl_3 + 3H_2O_2 -> 2CO_2 + 2H_2O + 3HCl}$$

Same as the ozone reaction, this reaction is more effective in an acidic medium.

Last one is the reaction involving permanganate. We will be using potassium permanganate but you can use sodium permanganate, calcium permanganate or magnesium permanganate. It will react with TCE to produce carbon dioxide, manganese (IV) oxide, metal chloride and hydrogen chloride. Here is the reaction:

$$\ce{C_2HCl_3 + 2KMnO_4 -> 2CO_2 + 2MnO_2 + 2KCl + HCl}$$

This reaction is a bit different compared with the other two. It could be effective in acidic or basic medium between the PH range of 3.5 to 12.

Hope this helps you.


Electrode Potential Half Reactions Does flipping a reaction as written change the sign of the value of E or is it a completely different value?

For example if I have the value for the half reaction $\ce{Cl2 + 2e- -> 2Cl-}$ but need the value for the opposite reaction ($\ce{2Cl- -> Cl2 + 2e-}$), do I just put a negative sign and treat it like $\Delta H$ values?

Yes, you are correct - just flip the sign. As quoted from the venerable UC Davis ChemWiki:

  To determine oxidation electrodes, the reduction equation can simply be flipped and its potential changed from positive to negative (and vice versa).

They give the value for your (forward) reaction as +1.358 V, so for the reaction in question, you'll use -1.358 V.


What effect does Baking Soda and Citric Acid have on a soft drink? In chemistry, we are doing a project where we have to make our own carbonated soda. The ingredients our group is using are:

Baking Soda ($\ce{ NaHCO3}$)
Citric Acid ($\ce{C6H8O7}$)
Flavoring powder (ex. KoolAid flavoring
    powder)
Sugar
Water

Using stoichiometry, our group has already calculated how much citric acid we need based on the 1 gram of baking soda that we are deciding to putting in: we have to use 0.76 g citric acid based on our calculations.

My question is, what effect does baking soda have and what effect does citric acid have on the soda? Does baking soda make it "bubbly"? Does citric acid make it more concentrated?

Actually if we care just about the reaction between Baking Soda ($\ce{NaHCO3}$) and Citric Acid ($\ce{C6H8O7}$) the following reaction should ensure :

$$\ce{3NaHCO3(aq) + C6H8O7(aq) -> 3 H2O (l) + 3 CO2(g) + Na3C6H5O7(aq)}$$

So basically the reaction should produce water, carbon dioxide and trisodium citrate (as known as E331 : a food additive)


Potassium Atom can act as reducing agent, yet ion cannot? I am currently looking through a revision sheet, however it has caught me on a question 'Explain why a potassium atom can act as a reducing agent yet a potassium ion cannot'.

I have done brief searches, however none have given the answer I'm looking for. So please, why is this?

Also, if the explanation doesn't already include it; is this only the case with Potassium, and how is this to be determined? (How is it known whether it is referring to a cation or anion?)

Thanks,

The ability of K to act as a reducing agent is mainly due to its electronic structure.
$K = [Ar] 4s^1$ . In this case, such atom tends to donate his $4s^1$ electron (acting as a reducing agent). This is due to the gain in energy that produces to have its octet complete and stay in the cationic form $K^{+}=[Ar]$ which is quite more stable that the pure element one.

This is common to all elements in the first column of the periodic table, just because all of them have a quite similar electronic structure, just with more electrons on their inner orbitals.


How does % purity on household products work? For example, toilet bowl cleaner is 9.5% HCl. Does this mean that for every 90.5 moles of H2O (or other stuff) there is 9.5 moles of HCl? Does it mean that for every 90.5 g of H2O there are 9.5 g of HCl?

For example, toilet bowl cleaner is 9.5% HCl.

First off, do you see "w/v" mentioned after those figures, i.e- $\ce{HCl}$ 9.5% (w/v)?

  Does this mean that for every 90.5 moles of H2O (or other stuff) there is 9.5 moles of HCl? Does it mean that for every 90.5 g of H2O there are 9.5 g of HCl?

Even if you don't, I'd still expect the concentration to be expressed in the form: $x$ grams of $\ce{HCl}$ for every $y$ units of water. As for whether water's mass or volume's considered, it would depend on the regulations for such products (which I'm unfamiliar with).

However, I wouldn't be surprised if it did turn out to be a w/v quantity (which is quite common in pharmaceutics).

  Kinda weird that we don't know if this is weight or molarity.

Like I said, I'm going to throw my weight behind "weight", since it's far more common than molarity outside of a lab (come to think of it, I've never seen a household product's ingredient concentration expressed in molars).

Not exactly important, but since you mentioned "molarity", here's a handy formula to convert something in % w/v to molars:

$$\text{Molarity} = \frac{{10}{(w/v)}{(d)}}{M}$$

Where $\ce{d}$ is the density of the solution, $\ce{M}$ is the molar mass of the solute and $\ce{w/v}$ is the mass concentration of the solute.


What does the "s" mean in s-cis and s-trans conformations I am learning about conjugated unsaturated systems, and came upon s-cis and s-trans conformations in conjugated dienes.

So what does the "s" mean in s-cis and s-trans conformations?

The s stands for single, as in single bond. It means the double bonds are cis or trans to each other based around the single bond.

It is a common enough expression to search for, if more information is required.


synthesizable functional groups I was told in this question: https://chemistry.stackexchange.com/questions/4789/synthesizability-and-energy-minimization-of-a-structure/4791#4791 to

[Count the] "number of accessible functional groups (alkenes, amines, alcohols, esters, etc) which make it easier to prepare, 5) number of inaccessible functional groups (aromatic rings, polycyclic systems, quaternary carbons, complex or long alkyl groups),"

But when I googled accessible functional groups and inaccessible functional groups no pages had a longer list. I found a SMARTS file which lists some common functional groups. Could someone help me determine which are readily synthesizable?

Thank you

What you are asking amounts to the collected knowledge, wisdom, and intuition of the organic chemistry community. Your best bet will be to find an organic chemist, preferably one with a graduate degree, to work with you on this project. Said organic chemist can apply his/her knowledge/wisdom/intuition to each functional group in less time than it would take you to look up the answer. Doing so frees you to work on the algorithm you describe in your other question.

The beginning of the answer to this question takes me two semesters to get through with my introductory organic chemistry students, and that is just for the common functional groups. After covering the basics of organic structure and reactivity, most texts have 15-20 chapters organized by functional group containing the reactions of and syntheses of said functional groups.

The common functional groups covered in most introductory sequences:

alkanes
cycloalkanes
alkyl halides
alkenes
conjugated dienes and other unsaturated systems
alkynes
aromatic ring systems
alcohols
glycols and other diols
phenols
ethers
epoxides
aldehydes
ketones
carboxylic acids
esters
anhydrides
acyl halides
amides
amines
nitriles
thiols
sulfides
acetals
imines
enamines
heterocycles
nitro compounds
diazo compounds
carbohydrates
amino acids
nucleic acids

If you want to know about every functional group, you will need an advanced, probably graduate level, synthesis course. However, even I have not heard of some of the designations on your list, like semithiocarbazide.

  Some good ideas on where to start, if you choose not to consult with an organic chemist.

Check an undergraduate level organic chemistry book out of your local library and get some sense of the subject. In your case, it does not look like you need to know how the reactions work, just that they exist, so you can use the text as a reference source.
Sign up for a free MOOC in organic chemistry. Again, you just want to know that the reactions exist, so you can pick and choose the content from the course.
Acquire a copy of one of the seminal books in organic chemistry, March's Advanced Organic Chemistry. This book is not for the fainthearted but it gives good treatment to the functional groups, including the less common ones.
Find (do not buy - these books are not worth the price tag) a copy of either Comprehensive Organic Transformations: A Guide to Functional Group Preparation by Richard Larock (which is now 14 years old) or the set of Comprehensive Organic Functional Group Transformations, which appears to be many volumes (now 9-10 years old).


What is the difference in the effect of DTT (Dithiotreitol) and SDS (sodium dodecyl sulfate) for PAGE? In class, we have studied that when running a polyacrylamide gel with our protein sample to separate the proteins by molecular size, we can use several techniques, among which we have denaturing PAGE and SDS-PAGE. Basically, we focused on the latter, in which SDS is a detergent added to negatively charge all proteins in the sample (at a constant mass:charge ratio), as well as denature them. Then, the professor mentioned that we also add reagents such as DTT or beta-mercaptoethanol to reduce the disulphide bonds in between proteins.

My question is: if SDS already denatures proteins (so we have primary structure only), why would we be interested in adding DTT?

Proteins are thought of as long polyamide chains. This long chain of amino acids is known as the primary structure. Due to hydrogen bonding of the backbone amide groups, a protein chain will organise most of itself locally either as an α-helix or a β-sheet; which one depends on the angles to the α carbon. These secondary structure elements then cluster to form a tertiary structure in which hydrophobic residues typically cluster together, aromatic residues may form π stacks, positively and negatively charged residues come together to build up ionic interactions, cysteins come together in disulphide bridges and hydrogen bonds are also formed where applicable. This last step is repeated if a protein consists of multiple subunits to form the multi-protein complex’s quarternary structure.

Almost all of the structure elements I mentioned that explain tertiary structures are intermolecular, meaning that they do not constitute a chemical bond between the connected fragments. Therefore, these structure elements are easily broken by the action of SDS whose aliphatic chain inserts into these tertiary structures, interacts with them and practically captures them on its inside.

The only exception to the former paragraph are disulphide bridges between cysteins. These are actually a chemical bond conneting the two fragments. In a simplified reaction scheme, the synthesis of a disulphide bridge looks like this:

$$\ce{\underset{\text{cystein}}{2 R-S-H} ->[][- 2e- \ -2 H+] \underset {\text{disulphide bridge}}{\ce{R-S-S-R}}}$$

Chemically, it is an oxidation of two cysteins.
Therefore, these disulphide bridges cannot be broken by the action of SDS alone. It takes a reductive agent such as DTT to resupply the electrons to reform the two cysteins. If no reducing agent is present (but there are disulphides in the protein structure), the disulphide bridge will remain intact and the protein will move not through the PA gel as it should according to its size since it is cyclic.


Hyperconjugation in bridged carbon systems While studying about hyperconjugation, I came across the following compound which posseses one α-hydrogen (with respect to the $\mathrm{sp^2}$ hybridised carbon atom with an empty $\mathrm{p}$-orbital), however, apparently does not show a hyperconjugation effect.

This led me to draw an ‘assumption’ that an alpha hydrogen on a bridged carbon atom may not get involved in hyperconjugation when placed next to a carbocation, alkene, free radical, or an arene. Is this true? If so, then why?

Hyperconjugation is indeed very difficult to practically in cases involving bridgehead protons. The underlying reason is essentially the same that also explains why these protons cannot be enolised if there is a carbonyl group adjacent to them. This is best explained by drawing out the compound in a three-dimensional manner, which I have done for you in the figure below.

Figure 1: approximate three-dimensional representation of the cation in question; cation represented by the p orbital.

If you look at figure 1 closely, you can see that the empty p orbital and the bridgehead hydrogen σ-bond are almost at a $90^\circ$ angle towards each other. For hyperconjugation, you want approximately parallel arrangements, so this obviously fails.

However, you should always be aware that this type of compounds, essentially a norbonyl cation, is essentially the example species for non-classical cations. In this case, instead of a hyperconjugation involving the $\ce{C-H}$ bond one essentially expects the $\ce{C-C}$ bond of the ‘back half’ of the six-membered ring to provide that hyperconjugation. For a more accurate and detailed explanation, please consult the Wikipedia page or the textbook of your choice.


Why does osmolarity take into account both penetrating and non-penetrating solutes? According to Wikipedia:

… osmolarity takes into account the total concentration of penetrating solutes and non-penetrating solutes, whereas tonicity takes into account the total concentration of non-freely penetrating solutes only.

But if osmolarity has units of $\pu{Osm L^-1},$ shouldn't it include only non-penetrating solutes, because the penetrating solutes won't contribute to the osmoles?

Collected and elaborated info from the chat with the OP:
To summarize the osmolality and tonicity difference, elaborated in the below link:

Osmolality takes into account all solutes.
Tonicity excludes solutes able to pass the semipermeable membrane.

So isotonic solutions need not to have the same osmolality.
The osmotic pressure is caused by decreased water activity, what does not depend on if solute is or is not able to pass the membrane. By other words, osmotic pressure and related osmolality  is property of solvent, not solutes, similarly as kryoscopic and ebulioscopic effects, all belonging to the class of colligative properties( depending on solute molality, but not on solute nature )
Primary is osmolality and osmotic pressure difference it causes. The difference then causes water osmotic flow.
When the osmolality is getting about balanced, then there is progressively coming into the play the difference in tonicity.
The same osmolality but higher tonicity means there is lower osmolality and therefore indirectly concentration of permeable solutes. This means there is concentration gradient powering the diffusion flow toward more tonic side. It is taking water with it, keeping osmolality balanced.
As result, volume of more tonic side increases even with the same osmolality.
At certain circumstances, the residual osmotic pressure may be compensated by the permeable solute concentration gradient, keeping status quo.


Recharge battery using heat A galvanic cell works by having two electrochemical equilibriums which have a different potential.

If you (theoretically) had a galvanic cell with two electrodes which have almost identical potentials and are sensitive to heat (at the anode the oxidation is exothermic and at the cathode the reduction is), connected the two half cells with a diode contrary to the normal current and heated the whole thing up, shouldn't it be possible to reverse the current and recharge the cell?

Please excuse language mistakes I might have made; I am not a native English speaker and therefore don't know all technical terms in English.

An electrochemical cell works on the principle that the Gibbs free energy gets converted into electrical energy by making the electron transfer indirect. This is done by itself to make the 'heat energy' that would normally be released in a redox reaction into 'electrical energy' so ideally heat shouldn't be lost.

Also you say you want to charge the cell by heating it so as to make the reaction go in the opposite direction. But for a spontaneous cell reaction, increasing the temperature makes the Gibbs free energy more negative and $\Delta G^\circ = -nFE^\circ$ ($E^\circ$ is standard cell potential).

So the $E^\circ$ value increases, and so does the concentration (because of backwards reaction), so the cell doesn't actually get charged up.

Ideally just charge it by passing current in the other direction.


Carbonic acid in water If we add a weak acid such as carbonic acid in water it weakly dissociates to form bicarbonate and hydrogen ions . Now water already has some hydrogen ions, having a ph=7.

So now indirectly we added hydrogen ions and base (bicarbonate ) ions to water which already has a hydrogen ion concentration.  The resulting solution is acidic . This means that in the addition of equimolar hydrogen ions and bicarbonate ions( base) hydrogen ions dominate.

The only reason which I can think of is this: lets say there were 1000 hydrogen ions in water, now addition of 100 hydrogen ions and 100 bicarbonate ions will cause a net increase in hydrogen ion concentration because bicarbonate wont be able to counteract the increase in hydrogen ions because bicarbonate, when acting as a base, will be in equilibrium with carbonic acid. I.e. 100 bicarbonate ions won be able to neutralize 100 hydrogen ions present in the solution, and because of their equilibrium they will ionize only say 50 and overall, and there will be an increase of 50 hydrogen ions.

Is my explanation correct?

To start off, here is the equilibrium we are talking about so we can refer to it later:

$$\ce{H2CO3 <--> H+ + HCO3-}$$

Since the pH of the solution is $\mathrm{-log[H+]}$, the pH of pure water will decrease upon the addition of carbonic acid.

Where I think you are getting confused is regarding the action of the bicarbonate that is produced. As this is an equilibrium reaction, bicarbonate will indeed act as a base, which is the reaction from right to left above. But lets say a molecule of bicarbonate reacts with hydrogen ion to reform carbonic acid. That just means another molecule of carbonic acid will dissociate to take it's place. This is all just a way of saying that the reaction shown above is an equilibrium, such that both the left to right and right to left reactions are occurring. But in the end there is never any net hydroxide produced or hydrogen ions consumed by this reaction, as there is always plenty carbonic acid available to counteract the basic action of the bicarbonate.

I hope I understood your question correctly. Feel free to use the comments below to ask for any clarifications.


Which is more acidic HOF or HOCl? My attempt to the question:
We know that more polar the $\ce{O-H}$ bond, the more acidic is the compound. According to electronegativity, the $\ce{O-H}$ bond of $\ce{HOF}$ should be more polar than $\ce{O-H}$ bond of $\ce{HOCl}$ since electronegativity order is $\ce{F} > \ce{Cl}$. But we should not miss another factor! That is the backbonding of  2p electrons (lone pairs) of oxygen with the 3d vacant orbital of chlorine. This is not possible in $\ce{HOF}$ due to the absence of vacant orbitals. Backbonding reduces (maybe, I do not know) electron density in oxygen and therefore polarity of $\ce{O-H}$ bond increases.

If we compare the HOF and HOCl
1) In HOF , the oxygen atom cannot do backbonding since fluorine has no vacant orbital while in HOCl it can do hydrogen bonding it can do backbonding which reduces electron density on oxygen by a slight extent( Remember 2pπ-3pπ backbonding is not very much effective due to size difference of 2p and 3p orbitals) . But yeah this factor makes HOCl bond more polar .
But there are two more important factors-:

2[)INDUCTIVE EFFECT]-  among oxygen and fluorine , fluorine is more electronegative , so it partially pulls the sigma bonded electrons towards itself giving a  delta+ charge to oxygen .
While in HOCl , O is more electronegative than Chlorine so the opposite mechanism happens and O gets a slight negative charge.
So according to this HOF has increased bond polarity than HOCl.
3) This factor is the most important and sufficient to answer almost any acid comparison . Compare the stability of the ion formed after releasing H+ ( formally called as the conjugate base of the acid).
So the question left is simply the comparison between OF- and OCl- ,
 The first two factors again work here but another important factor is the 2p-2p electron repulsions( due to smalls ize of sigma bond the electrons in the orbitals of O and F repel each other. This factor  in itself is quiet dominating) present in OF- makes OF- less stable than OCl-

So according to me HOCl must be more acidic than HOF.

Another important thing i wanted to draw attention to is that HOF is not the common acid that can be kept in laboratory. It is quite reactive and reacts vigorously even at 0°C , measuring its acidity is quite diffcult . I also searched online but couldn't find any data concerning Ka values for ionisation of HOF.
I hope this helps.


Reaction of urea and thiourea with nitrous acid I was told that urea on reaction with nitrous acid gives nitrogen gas, carbon dioxide and water. While on the other hand, thiourea on reaction with nitrous acid gives $\ce{H+}$, thiocyanate ion and water.

I tried looking for the mechanism and thought that this would be related:

I carried out the above procedure on both the nitrogens of urea and was able to obtain the products of the first reaction enlisted above. Now, my questions are:

• Isn't the nucleophilicity of the lone pairs of urea's nitrogens negligible, and any reaction should proceed via attack by oxygen's lone pairs? (I thought that this was similar to amidines and amides)

• Why are the products of the reactions with urea and thiourea different, as the carbonyl oxygen, or sulphur doesn't seem involved in the reaction according to me

• More importantly, is this even the correct mechanism?

As implied in your link, the reaction of urea and nitrous acid will follow the mechanism you posted, repeating a second time with the nitrogen on the remaining amide. As for why the nitrogen is more nucleophilic, I can speculate, but I don't want to be wrong.

The difference for thiourea is that sulfur can also act as a nucleophile instead of the nitrogen in the thioamide. The product is dependent on the pH. You can read more about it here. Here is the mechanism:

That's a thiourea that I squeezed in on the far right as a product from the strong acid. Sorry for the mess.


Surface Functional Groups/Notation I'm reasonably unfamiliar with chemical notation, and recently I've come across something I'm not sure how to interpret.  In this article, the authors refer to the "surface groups" denoted by $\ce{#Ti-F}$ and $\ce{#Ti-OH}$.

Similarly, this article refers to surface functional groups $\ce{#Si-O-}$ and $\ce{#Si-OH}$.

To my (limited) knowledge, "$\ce{#}$" denotes a triple bond.  If this is the case, what am I to take from this?  Are they referring to a (triple) dangling bond, or is this some notation I'm not familiar with to denote surface species?  The first page of the first article does state that "$\ce{#Ti-X}$" refers to surface species, but I'm finding it difficult to believe the notation was chosen arbitrarily.

I appreciate any explanation.  As I've said, I am a novice with chemistry and chemical notation.

The notation used does not necessarily represent a triple bond - for example, using one of the examples that you have provided:

  $\equiv\text{Si}-\text{OH}$

Which is for the functional group silanol, that is defined by the $\ce{Si-O-H}$ connectivity, it's chemical structure is

Image source

Using this example, the hydrogens represented by the  ≡ symbol can be replaced by other chemicals forming other members of the silanol functional group, for example, Trimethylsilanol:


What would be a stable way to hold lithium hydride powder? I currently have 100 grams of lithium hydride powder that needs to be held in a solid form. Initially, I was considering using cement, however, the moisture require converts the lithium hydride to lithium hydroxide. What would another moisture-free agent be that I could use for this task?

If it helps, the lithium hydride and the solidifying agent are to be used for neutron moderation.

If the LiH needs to be exposed to ambient conditions, it should be enclosed is a moisture-impermeable container, e.g., a glass ampoule which is closed off. Glass can be blown quite thin - but then you need to be quite careful. As DrMoishe suggested, PE or PP would be good, but I'm thinking of it as a closed tube, not as a binder.


Formating chemical equations for proteins binding in hugely complex configurations In a previous question I asked about a simple case, I'm now trying to write equations for proteins that bind multivalently. In the simplest case proteins $A_1$ and $A_2$ can bind to up to two $L$ proteins. This gives a massive chemical equation:
`

where $(AL_1AL_2)_r$ is a tetramer where $A_1$ and $A_2$ are both bound to both of the $A$ proteins in a ringlike structure. But this diagram misses out one possible state, $A_1LA_2$ which is connected in this way:

.

I can't see a way of joining this extra state to the above chemical equation without arrows crossing each other.

What is the best way of formatting these kinds of complex protein reaction equations?

Generally speaking, for these sorts of complex reactions, you want to go to more than two dimensions. (At least for the initial bit.)

For each potential binding reaction you have a different dimension. That is, for $L$ binding to $A_1$ on site 1, you go horizontally. For $L$ binding to $A_2$ on site 1, you go vertically. For $L$ binding to $A_1$ on site 2, you go in the Z dimension, and then for $L$ binding to $A_2$ on site 2, you go in the 4th dimension.

Obviously, you can't put an actual 4D representation on paper. In fact, you can't put a 3D representation on paper. But if you're lucky, you can do a bit of topological rearrangement of the network and then do a projection of your network onto a 2D plane.

For example, if you have three potential binding interactions, each independent, then your system can be represented as the corners and edges of a cube. You can then represent this system by any of the standard projections of a 3D cube into 2D space.

For four potential binding interactions, it's the the corners and edges of a tesseract. Projecting a tesseract onto 2D space is much harder, in that there's no way to do it without lines crossing. But depending on what you want to represent, you might be able to adequately mark the crossings, or add "jumps" between two sub-graphs. (That is, have two more sub-graphs and have a systematic way of going from nodes on one graph to the node on the other graph with implicit reaction edges.)

For an arbitrary reaction graph, there's no guarantee that the crossing number of the graph is going to be zero -- that is, there's no guarantee that you'll be able to represent the network on a 2D page without crossings or jumps. Often the best you can do is attempt to be systematic about things, and then add crossings/jumps as appropriate.


Regarding reversible phase transition of a real gas According to change in entropy for a reversible phase transition, say from liquid to gas at constant pressure $p$ and temperature $T$
: $\Delta S =\frac{\int dq_p}{T} $  which comes out to be $\Delta S =\frac{\Delta H_{vap}}{T_b}$. This means that this process is spontaneous since Entropy change is positive. But the Gibb's Energy change for this process is $0$. Why is that?
Change in entropy for this process suggests that this is a spontaneous process but Gibb's Energy change suggests it is not. Why does irregularity occur?

$$\Delta S_\mathrm{sys} =\frac{\Delta H_\mathrm{vap}}{T_b}$$
is the entropy change of the system.
$$\frac {\Delta G}{T_\mathrm{b}} = \Delta S_\mathrm{tot} = \Delta S_\mathrm{ext} + \Delta S_\mathrm{sys} = \Delta S_\mathrm{ext} + \frac{\Delta H_\mathrm{vap}}{T_b} = 0$$
is proportional to the total entropy change.


Why does gum get harder to chew after a while? My Sister asked me this:

  "Why is it that when we chew gum,  it is soft and mushy to begin with , but slowly gets firmer and firmer like after 20 mins of chewing?"

I think  it is because when we initially chew gum it has plenty  of sugar causing rapid salivation and hence mushiness and the rubber polymer is very flexible to begin with and slowly looses its elasticity hence becomes firmer.

Is this the reason?

What is the chemistry involved ?

I would be surprised if there was reticulation of the gum's polymers upon contact with your saliva. According to this manufacturer, plasticizers are used, and the plasticizers appear to be fats. This would mean that, upon chewing, the plasticizers would be removed from the gum, rendering it unpleasantly harder.

Plasticizers are used to render a polymer more pliable, more elastic, than it naturally is. This could get its own question, imo.


Propylene Glycol Degradation due to Sunlight I had a translucent container of propylene glycol exposed to sunlight.  When I went to use it, I noticed it has a strong "acetone" like odor.  Someone said it was a photo-oxidation reaction of the glycol into ketones or aldehydes.  What happened and will the odor get worse even though it is is now removed from sunlight?  Do I need to worry about these ketones or aldehydes in the glycol?

The thermal degradation of propylene glycol and its uses in e-cigarettes has been a discussed in a previous question here on Chem.SE.

  Upon complete oxidation and degradation, propylene glycol will fully
  decompose to carbon dioxide and oxygen. However, if there is partial
  oxidation, you will get some intermediate species - formaldehyde or
  acetaldehyde.(1st answer)

  Propylene glycol does not decompose on heating, it simply boils. In
  theory, under certain conditions (pressure/encapsulation/catalysts) it
  could dehydrate to either acetone or Propionaldehyde, probably the
  former.(2nd answer)

Also, this link shows the decomposition products of propylene glycol which are all acid based:

  The main degradation product of propylene glycol was lactic acid (2-hydroxypropionic
  acid), with formic, acetic, and oxalic acids as minor products.

So, mainly degradation of propylene glycol in sunlight would yield a bunch of acids and small amounts of acetone. Worrying about the odors is a personal medical question and is off-topic in Chem.SE.


Stable natural helium hydride? Reading the transcript of the Royal Society of Chemistry podcast Helium Hydride, they state that helium hydride is possibly the most ancient compound to form in the Universe.  They make the assertion that despite helium hydride being most stable as an ion, but:

  The neutral molecule can briefly exist, but only as an excimer, where the excess electron has enough energy to keep it at a higher level.

Under what natural conditions could a stable neutral helium hydride molecule exist?

As explained in Emission Spectra of Bound Helium Hydride Physical Review Letters vol. 77 pages 2941-2944:

  The ground state [of neutral HeH] is repulsive, as shown by
  numerous experiments and calculations.

but excited states of HeH that are bound exist (as first experimentally observed by this paper, but previously calculated).

Given that only excited states are bound, it don't think it proper to say HeH stably exists under any conditions natural or otherwise.

The paper does state:

  Because of the abundance of He and $\ce{H_2}$ in stellar atmospheres
  and interstellar space, HeH is expected to
  be formed there. Therefore our data may be useful to
  find the HeH lines there.

However, that it exists does not mean it stably exists.


Why do the melting points of Group 15 elements increase upto Arsenic but then decrease upto Bismuth? The boiling points of group 15 elements increase on going down the group (or, as size increases) but the same is not true for the melting points. The melting points increase from $\ce{N}$ to $\ce{As}$ and then decrease from $\ce{As}$ to $\ce{Bi}$. Why is this the case?

A lower melting point is an indication of a lesser organised solid structure. So the question is actually:

  What in the structures of arsenic and phosphorus make them less prone to melting?

Let’s start off by separating nitrogen from the entire group. It is a prime example of a diatomic gas such as hydrogen with a very inert $\ce{N#N}$ triple bond. As throughout the periodic table, multiple bonds with s and p orbitals between atoms of the same element are magnitudes more stable in the second period than anywhere else due to much better p-orbital overlap (smaller size, smaller interatomic distance). Therefore, we can consider phosphorus, arsenic, antimony and bismuth more or less as similar while nitrogen is an odd one out.

Phosphorus has four stable allotropes with vastly different structures. The white phosphorus allotrope, $\ce{P4}$ molecules, basically falls into the same class as dinitrogen and melts at around 40 degrees. Red, purple and black phosphorus build up network-like structures in two dimensions, either in tube-like chains (red) or as ‘wavy graphite’ (black). Only the black modification is clearly defined. Red phosphorus has melting points of around $600~\mathrm{^\circ C}$ while black phosphorus sublimes in vacuum at $400~\mathrm{^\circ C}$ — probably meaning that its melting point at standard pressure is much higher, maybe even higher than arsenic’s sublimation point. I found no references to liquid phosphorus so I have no idea whether the bonds of the network structure need be broken to liquidify or not.

Figure 1: crystal structure of black phosphorus, taken from Wikipedia.

Henceforth, we have established a structure element which we will be sticking to with only moderate modifications. The transition phosphorus — arsenic — antimony — bismuth is more or less the non-metal—metal transition. What explicitly changes is the distance between the sheets as shown in table 1.

$$\textbf{Table 1: }\text{Comparison of bond lengths and inter-sheet contact distances for pnictogens}$$
$$\begin{array}{lrrrr}
\hline
\text{ } & \ce{P_{black}} & \ce{As} & \ce{Sb} & \ce{Bi} \\
\hline
\ce{E-E}\text{-bond/pm} & 223.0 & 251.7 & 290.8 & 307.1 \\
\ce{E\bond{...}E}\text{-contact/pm} & - & 311.9 & 335.4 & 352.8 \\
\text{d}_\text{contact}/\text{d}_\text{bond} & \approx 1.5 & 1.239 & 1.153 & 1.149 \\
\hline
\end{array}$$

So while the phosphorus structure can be described well as consisting of semi-isolated sheets stacked above each other; when we have reached bismuth, these sheets are much closer to an overall three-dimensional network. Therefore, bismuth atoms are in a distorted octahedric environment as other metals would be. Arsenic, where the sheets are still well separated, exhibits almost metallic condictivity perpendicular to the sheets.

Putting this together and taking the extreme points, we can say:

The phosphorus structure is highly organised in that two directions are strictly discernable. Melting along one of these is no problem (perpendicular to the sheets) but melting within a sheet is a huge problem as $\ce{P-P}$ bonds would need to be broken.
The bismuth structure no longer has a true distinction of directions; all three axes are more or less equal. Therefore, it takes similar energies to break bonds in one direction as it does to break them in another direction meaning the overall process is easiert.
Furthermore, liquid phases are generally those that display a high local ordering but a low long-range ordering. The high local ordering can be preserved in bismuth even if the long-range ordering present in the solid state is lost. Conversely, the local ordering is phosphorus is a lot stricter and cannot easily be transferred into a state where long-range ordering is low.
Finally, note that non-metals whose solid phase does not consist of small, well defined molecules generally have higher melting points than metals. (Compare carbon’s sublimation point of c. $3600~\mathrm{^\circ C}$.)


Can you color a flame and what would the effects be? I read this question (Flame temperature for cooking food) and came up with the following questions:

Could companies color their flames so that products would heat quicker?
How would they do this?
Could you use this to make an infrared or ultraviolet flame? If
you could, how hot would they be?

Unfortunately not, I'm afraid.

I think you might have gotten cause and effect mixed up here. Blue flames are hotter than yellow flames not because of their color, but because hotter flames generate blue light and cooler ones generate yellow/red light.

To take an example of this kind of problem, I know that very red tomatoes tend to be tastier than very pale tomatoes. Does this mean that we could make tastier tomatoes by injecting food coloring into them? Well, probably not, because it's not the color that causes the flavor. Rather, tomatoes that generate more color tend to also generate more things like sugar and flavors, and so flavorful tomatoes tend to have more color.

With flames, it's even more black-and-white because while its possible to have a pale tomato that happens to taste good, the color of the flame is set by the universe: unless you add colors yourself, it is impossible to have a blue flame that is the same color as a yellow one. The color of the flame is determined by something called "blackbody radiation," and if you're curious about why flames are colored the way they are, it's definitely something to take a look at.


What is role of palladium in hydrogenation reactions? Why is palladium used in hydrogenation reactions? Is there any theory about the role of palladium and other catalysts in those reactions?

The general catalysis cycle of hydrogenation reaction is pretty simple and AFAIK well documented. It consists of for reactions, all well documented in metalloorganic chemistry.

Hydrogen activation. It comes in many flavors, but idea is the same: $\ce{H-H}$ bond is broken and instead $\ce{M-H}$ bonds are formed.
Alkene bonding with a metal center
Alkene insertion in $\ce{M-H}$ bond, forming $\ce{R-M}$ alkyl bond
Elimination of alkane from $\ce{R-M}$ and $\ce{M-H}$ fragments.

This general process can occur on mononuclear or polynuclear complexes and on metal surface. However, when selecting particular metal for the process it is important that the intermediate particles were neither too stable, or they will cover catalyst completely and remain there, nor too unstable or the process won't happen. It so happens, than Ni, Pd, Pt are in the sweet spot with Pd being one of the best. They are not, however, the only hydrogenation catalysts in existence, they are simply the ones with high activity and simple composition. For example, $\ce{Ni}$-promoted $\ce{WS2}$ catalystalso has significant hydrogenation activity.


What is hydrogen-rich water? My question is: Is hydrogen-rich water true? I am aware of the alkaline water. I have read several articles in the Internet regarding this kind of water but I cannot seem to be convinced. Wikipedia isn't even talking about it.

I am puzzled with how can a water be enriched with hydrogen. It's definitely $\ce{H2O}$ and everybody knows it. It's not water anymore if it's not $\ce{H2O}$.

This link is one of the examples of websites that I read about hydrogen-rich water.

Hydrogen water is a solution of molecular hydrogen $\ce{H2}$ in water, which shouldn't be confused with acidic or basic solutions.  The first scientific reference to hydrogen water appears to be in the journal, Nature Medicine which can be accessed here if you have a personal or institutional subscription to the journal. The claim is that molecular hydrogen can serve as a mild reducing agent and therefore has antioxidant properties; namely, it can reduce hydroxyl radicals:
$$\ce{H2 +2 OH\cdot -> 2H2O}$$
The article has generated a considerable amount of research, with well over 300 journal articles citing this work originally published in 2007.  There seems to be some merit to the claim although in my brief scan of the abstracts, there does not appear to be a large body of evidence suggesting that commercial products are beneficial.  One such article, which is publicly accessible is here.  As mentioned in another answer, the commercial product approach to generating hydrogen water appears to be the oxidation of magnesium in water, which generates $\ce{Mg^2+}$ ions.  As with any other nutritional supplement, I would be very wary of entering this type of treatment without consulting a professional.  Magnesium is an important nutrient, however at high doses it does show signs of toxicity.


Electric field effect on reactions What would be the effect of putting a solution in a strong electric field? Would the reactions in the solution be affected?

I'm thinking yes – lots (if not all) – chemistry is dependent on charge (partial/real charges) and by applying an electric field, we'd be affecting the very property of the molecules/ions that drives reactions.

Has this been researched?

Would the reactions in the solution be affected?

Yes, they can be.  Here are some nice examples.  Electric fields also exert a significant effect on electron transfer reactions.  Basically you would expect charged molecules to be more affected than polar molecules, which should be more affected than non-polar, but polarizable molecules.  It's just a question of how big a field you can produce and how polarized and long-lived are the various entities (starting materials, intermediates, etc.) that exist along the reaction coordinate.


Using Dalton's law of partial pressures A container of volume $\pu{4.0 L}$ contains nitrogen at $\pu{25 °C}$ and $\pu{803 kPa}.$ This container is connected by a valve to another container of volume $\pu{10.0 L}$ that contains argon at $\pu{25 °C}$ and $\pu{47.2 kPa}.$ We now open the valve so that the two gases are mixing.
Calculate the partial pressure of each gas after the mixing and the total pressure of the mixture.

Using the Dalton's law of partial pressures, isn't the partial pressure of each gas exactly $\pu{803 kPa}$ and $\pu{47.2 kPa}$ and the total pressure $\pu{803 kPa} + \pu{47.2 kPa}?$
What am I missing here?

Dalton's law of partial pressures means that the total pressure in the final volume $V=V_1+V_2$ is equal to the sum of the partial pressures P(N2) + P(Ar) in this volume. However, the partial pressure of N2 in the total volume $V$ is not the same as the pressure of N2 in its initial volume $V_1$. You have to account for the expansion. This probably means you also need to use the ideal gas law for that.


Trend in ionisation potential in period 2 Carbon has valence electronic configuration of $2s^2$ $2p^2$. Beryllium has configuration $2s^2$.So carbon should have less ionisation potential than Beryllium as it is easier to remove electrons from $2p$ subshell than fully filled $2s$ subshell.

But I found that in JD Lee it is mentioned that carbon has more I.P. than beryllium. How can it be explained?

Emperically speaking, Be is on the "metallic" side of the table. Lower IE than non-metals is obviously expected.

A quick glance at the $IE_1$ trends for period II tells us that like the OP noted, Be certainly has an abnormally high $IE_1$, deviating significantly from the expected plot, due to the fully filled configuration.

However, Carbon (Z=6) has a higher nuclear charge(=more protons) than Be (Z=4), a smaller radius and hence a stronger hold on its valence electrons.

Even Nitrogen displays an exception, due to half-filled $2p$ subshell, as evident in the plot.

Just because of a tiny bump in the plot, expecting N and Be to have higher IE than all the elements of the periodic table is clearly an exaggeration.


Can we produce tritium from helium 3? If yes how? I want to know the best and economic way of the manufacture of tritium from helium-3

From my knowledge, helium-3 is actually a by-product of tritium decaying. Tritium is usually manufactured industrially by bombarding lithium-6 nuclei with neutrons in a nuclear reactor, and is currently the cheapest way to manufacture tritium.


Do metallic bonds contain London dispersion forces? On our chemistry exam, a question asks "Which force is present between Fe particles? a) Van der Waals b) Metallic"

At first it seemed obvious that it was metallic; however, upon closer reading, it seemed like a trick question. First of all, there's no such thing as a metallic force. In addition, I'm not sure if Van der Waals forces exist for metallic bonds. If they do exist, then the answer was definitely "Van der Waals".

So my question is, would Van der Waals/London dispersion forces exist between particles of iron?

If by "Fe particles" you mean distinct particles (as opposed to the atoms inside a continuous Fe crystal structure), then the answer would be VdW/London forces.

From wikipedia:

This is the only attractive intermolecular force present between neutral atoms

Which, in a way, can be applied to particles or particle surfaces as well, as long as they're not charged.

Whereas, if you are talking about a continuous iron crystal structure, and by particles you mean atoms within that structure, then you have a metal-type bond between them caused by delocalization of electrons into the conductive band. See wikipedia for further details.


Which reaction is this? In my book there is a reaction that is

  $\ce{HCCl2-CH2F}$ in presence of $\ce{RO^- ROH}$ gives $\ce{CCl2=CH2}$

I could not understand my type of reaction it is?

And why does $\ce{F-}$ leave whe $\ce{Cl-}$ should leave as it is a better leaving group?

It appears to be an e1cb mechanism, where formation of the carbanion on the dichloro end is the rate limiting step. This appears to be a far more acidic proton than the ones on the fluorine-bearing carbon. Once the carbanion is formed, it rapidly eliminates fluoride ion to give the stable dichloroethylene product, rather than eliminating chloride (admittedly a better leaving group) to give the carbene, a higher energy product.


Why is the amino acid cysteine classified as polar? Cysteine amino acid has an embedded sulfur group in its side chain. Looking at the electronegativity difference of hydrogen and sulfur, it can be considered a non-polar side chain because the electronegativity difference is less than 0.5. This makes it non-polar but still we put it in polar category, why?

[...] we still put it in polar category

If you search for amino acid classification, you will find that there is no agreement on how to classify cysteine.
What you should know is that cysteine has different roles in proteins. In cysteine proteases, it acts as a nucleophile (and its surrounding often makes it more prone to deprotonation compared to a netural aqueous environment). Cysteines that form disulfide bonds often are at or near the surface of the protein; once the disulfide is formed, you have a quite hydrophobic group. Protonated cysteine is incapable of making conventional hydrogen bonds, and the electronegativity of carbon and sulfur are quite similar. This explains why methionine, the other sulfur-containing amino acid, is classified as hydrophobic.

[...] , why?

The classification of amino acids is sometimes used as proxy of the tendency of amino acid side chains to be located on the surface or in the interior of a protein. Similarly confusing, proline is sometimes classified as polar amino acid; however, its side chain is made exclusively of carbon and hydrogen atoms. If you study where proline is mostly located with a folded protein, you find it is on the surface. This is not because proline is polar but because it is a helix and strand "breaker", so it is often found in turns, which are mostly on the surface of the protein.
Also, you should know that there is no all-or-nothing classification of every amino acid side chain. Tyrosine, threonine, even lysine have hydrophobic parts while being capable of making hydrogen bonds. Often, the environment is just right, with hydrophobic parts interacting with other hydrophobic parts, and all hydrogen bond potential satisfied, either by interactions with other side chains, with main chain or with solvent.
Why is cysteine not classified in a more consistent manner?
Here is an excellent starting point to answer this question, starting with the abstract:

Cysteine (Cys) is an enigmatic amino acid residue. Although one of the least abundant, it often occurs in functional sites of proteins. Whereas free Cys is a polar amino acid, Cys in proteins is often buried and its classification on the hydrophobicity scale is ambiguous. We hypothesized that deviation of Cys residues from the properties of a free amino acid is due to their reactivity and addressed this possibility by examining Cys in large protein structure datasets. Compared to other amino acids, Cys was characterized by the most extreme conservation pattern, with the majority of Cys being either highly conserved or poorly conserved. In addition, clustering of Cys with another Cys residue was associated with high conservation, whereas exposure of Cys on protein surface with low conservation. Moreover, although clustered Cys behaved as polar residues, isolated Cys was the most buried residue of all, in disagreement with known physico-chemical properties of Cys. Thus, anomalous hydrophobic behavior and conservation pattern of Cys can be explained by elimination, during evolution, of isolated Cys from protein surface and clustering of other Cys residues. These findings indicate that Cys abundance is governed by Cys function in protein rather than by the sheer chemical and physical properties of the free amino acid, and suggest that high tendency of Cys to be functionally active can considerably limit its abundance on protein surface.


What happens to the pressure of gas when only one molecule of it is placed in a very large container? We say that pressure due to gas is constant throughout the container. Pressure is created by molecules present in it. When there are many molecules I agree that they can apply the same pressure all over.

But when there are very few or just one molecule in the container, how can that single molecule create a uniform pressure in the container. I mean pressure will be created (say in a manometer) when  a molecule applies force on it (Hg). If we take a very large container and place just one molecule and also keep two manometers at the two extremes. Then how can that single molecule create the same pressure or apply the same force on both the manometers at the same time (container is very large)?

What happens to the pressure of gas when only one molecule of it is placed in a very large container?

I think you have forgotten what pressure represents in terms of the kinetic theory of gases. When studying gas kinetics, you start by understanding the behavior of a single molecule/atom moving around a container and colliding with the surface and bouncing in random directions. Then you add a few more molecules/atoms so that now there are collisions and as long as the mean free path of the gas is less than the dimensions of your vessel you still have motion that acts in random diffusion-like behavior. Once sufficiently many atoms have been added that a given amount of area experiences a regular number of collisions per unit of time, then you have something that can be given as an average rather than a count of collisions. This is what pressure represents, an average force on a surface per unit area of molecular collisions on that surface. The instantaneous force can change for a given spot nanosecond to nanosecond, but averaging all spots over a span of microseconds will give an average force per area can be resolved which we call pressure.

The problem here that gives you a misunderstanding is that you have taken a behavior of an aggregation of many molecules and tried to apply it to a single molecule. This approach will disappoint. But to summarize and actually answer the question, pressure cannot describe a system of only one gas molecule, and is therefore undefined.


Does vinegar neutralize the strongly basic compounds in cement? I have heard that washing one's hands with vinegar is effective in neutralizing the strong bases in cement and is more effective than water.

I want to know how cement burns and what bases are involved in this reaction.

I have tried googling "cement burns" and found websites explaining the importance of washing one's hands with vinegar as a preventative measure. I have tried searching PubMed but I've found mostly case reports that do not address how chemical burns actually happen.

If anyone knows of some good papers or books that cover the topic of chemical burns in greater detail than what I've found; please let me know.

If you look for Portland Cement safety issues, it is stated that cement is highly alkaline so it can cause alkaline burn and presumably since vinegar is acetic acid, it neutralises the alkaline effects of cement as tested here and the burns you suffered are probably related to alkaline burn. You can look at this article section 6.5 for some details on alkaline burns, which was suggested by @Cornelius or perhaps the mentioned chapter in this book (Harchelroad FP, Rottinghaus DM. Chemical burns. In: Tintinalli JE, Kelen GD, Stapczynski JS, Ma OJ, Cline DM, eds. Emergency Medicine: A Comprehensive Study Guide. 6th ed. New York, NY: McGraw-Hill; 2004: chap 200.).


Intuition for electrode potential If I connect a zinc half cell (Left hand side) and a copper half cell (right hand side), I get an E cell value of 1.1V taken from the voltmeter. If I replace the voltmeter with a light bulb, I will get light and heat energy of the bulb.

I learned that the above set up is equivalent to adding Zinc to copper sulphate solution. However, in this case, there is no wire for the electron to flow...

MY QUESTION:

1.Then, where does the energy go when I add Zinc to copper sulphate solution (that in the first scenario I converted to electrical energy)? Is it released as heat?

2.As the electrons flow from the Zinc half cell to the copper half cell through the light bulb, will the voltage between the electrodes decrease as the charges on the electrodes are supposed to decrease. Then what would the formula for total energy released be?

Please correct me if any of my concepts are wrong. Thanks in advance.

Ad 1) I am not sure what do you mean by "equivalent". But you add Zinc in a copper sulphate solution, the "energy" will be consumed through cementation.

to a recation Cu$^2$$^+$(aq) + Zn(s) → Cu(s) + Zn$^2$$^+$(aq)

Ad 2) Please see Equilibrium Constant of an Electrochemical Cell Reaction Example Problem where you can find ansewer to your second question.


Why does solid phase peptide synthesis (SPPS) always proceed from the C to the N terminus? Is it just for historical reasons? Because the first person who did it decided to attach the C-terminus to the solid support?

Or are there chemical reasons why it would be impossible to proceed in the opposite direction?

The reasons are

The use of an acylating agent in excess (e. g. acetic anhydride or N-acetyl imidazole) drives the reaction forward.
UNCAs (urethane-type protected  N-carboxy anhydrides (e. g. Fmoc, Z, Boc, ...)) can be used. They prevent racemization of the amino acid, that it is attached to.

But there are (despite of ribosomes themselves) also instances, where N -> C directionality was applied:

http://onlinelibrary.wiley.com/doi/10.1002/anie.199710991/abstract
http://www.ncbi.nlm.nih.gov/pubmed/10891165


Did the finding of the electron from cathode rays/ oil drop experiment at least temporarily throw Dalton's Atomic theory "out the window"? Because chemists learned that the atom was divisible, did Dalton's theory get any criticism or did they just adjust the theory to fit the new idea?

Dalton's theory was about how atoms behaved: the observation of Cathode rays provided part of the explanation of why they behaved like this

The key elements of Dalton's theory were observational. He explained how the idea of atoms could explain chemical reactions and many observational facts about them. This was an important breakthrough and explained far more than previous ideas. But it didn't provide much of an explanation of why they behaved that way.

The observation of cathode rays was one of the first steps in providing the why not a fundamental challenge to Dalton's actual theory. Chemists didn't get a (mostly) complete explanation of what atoms behave the way they do or of some of the characteristic features of atoms until the discover of the neutron in the 1930s. Why, for example, do atomic weights not always follow the atomic numbers (or the pattern of chemical behaviour observed across the periodic table)?

We now understand that chemical properties are determined by the electrons in atoms and the number of electrons depends on the number of nuclear protons. But the mass also includes neutrons which occasionally alters the apparent periodic relationship between atomic mass and chemical properties.

None of this is a fundamental challenge to Dalton's ideas. All of it is a deeper explanation of why the observed properties and behaviours of atoms in Dalton's theory work as they do.

So we didn't throw Dalton's ideas out. We refined them with a more fundamental theory that explained why they behave like that.

This contrasts with other changes in our understanding of chemistry. For example, we discarded the old phlogiston theory of the explanation for combustion because observations didn't refine it but completely refuted it.


Carbon numbering in carbohydrates Carbons have been numbered differently in two different pictures. What are the rules of carbon numbering?

Actually, the numbering is the same, at least for the fructose portion of the molecule (which is the right-hand monosaccharide in your sucrose image - the one on the left is glucose).

The key point in carbohydrate numbering is to look for the carbon which is an aldehyde or a ketone (the carbonyl carbon). This will be the end of the monosaccharide which is numbered lowest. For aldoses, this carbon would be carbon 1. For ketoses, it's the lowest number that can be reached in the linear, ring-open form (e.g. for fructose, it's 2). The rest of the carbons are numbered accordingly.

When you have a linear Fischer projection (like the "D-fructose (linear)" images on the second image) the carbonyl carbon should be obvious. In the ring form, it's less obvious.

Instead of looking for an aldehyde or a ketone in the cyclic form, you're looking for an acetal or hemiacetal. That is, you're looking for a carbon with multiple bonds to oxygens. In the first image, this is the carbon which is making the oxygen bridge between the two monomers (the one on the left of the furanose) - the oxygen of the bridge is one oxygen, and the oxygen of the furanose ring is the other. In the second picture, it's the carbon on the far right. You have the ring oxygen of the furanose, and then a free hydroxyl oxygen. Both of these carbons are numbered "2", and if you imagine rotating the furanose ring 180 degrees to flip over that acetal position, you'll see the other carbons match up as well.

(One thing that might be tripping you up is the directionality of the 1 carbon with respect to the 2 carbon. Keep in mind that the hemiacetal form can interconvert with the ring-open carbonyl form, and the "anomeric position" of the monosaccharide is free to invert in chirality - that is, it can readily flip between "up" and "down". It's only when it's in the bonded acetal form that it's locked into either the alpha or beta configuration.)


What to do about a un-dissolved substance? To exterminate a coackroach infestation, I bought solid boric acid (in powder form). Rather than awaiting the cockroaches' encounter with the boric acid, I thought about spraying boric acid onto affected areas or cockroaches. I only took high school chemistry, but remembered solubility so checked its solubility in water. I then compared it with the solubility of other compounds, to see if this were an effective idea:

Boric acid: At 100 Celsius, 27.53 g/100 g of water.
Sodium chloride: At 100 Celsius, 38.99 g/100 g of water.

This looks fine, so I poured boiling water into a bottle with the boric acid powder. I shook it and saw the boric acid dissolve. Yet two weeks later, the boric acid seems to have reversed dissolution. Now, it's separated from the water and sits at the bottom of the water.
Can substances do this? What should I do now? I'd need a pot to boil the mixture over the stove. But then, in the interest of health and safety, I couldn't use this pot anymore, so don't want to sacrifice one.
Update: With thanks to and in view of the instructive answers, would it help to refill the bottle, to benefit from the low solubility of boric acid at room temperature? There's only around 10% of water left, with the boric acid sitting at the bottom.

I think what you forgot is that the solubility of boric acid (or sodium chloride.. or most solids) is lower at room temperature.

So when you dissolved the boric acid in boiling water, there's no problem at first. As it cooled, you will have a supersaturated solution until eventually the boric acid will recrystallize or crash out.

I'd create a new mixture using room temperature water and add only a small amount of boric acid (as mentioned by @K_P, about 5g / 100 mL of water). Let it dissolve slowly while you mix.

Then if you spray, you shouldn't leave large amounts of boric acid around.

Updated Answer

Yes, if there's only ~10% water, you will want to add more (room temperature) water. This will dissolve more boric acid. Stir/shake to dissolve as much as possible, wait a bit and then you can spray again.

As mentioned, you don't want to add water to a hot acid. In this case, adding more room temperature water to a room temperature acid will probably produce heat, but add slowly and carefully.


Why does boiled water have a different taste as compared to cool water? I have noticed it frequently and decided to check with some of my peers and friends about this and they too have felt such a thing.
The taste comparison occurs after the boiled water is cooled down to room temperature.
My question is: does water even have a taste?
And whether heating it may change the taste.

The changed taste that stays after cooling the heated water

Hot (boiling) water has a much lower capacity for dissolving gases. When boiling water, $\ce{CO2}$ is removed and through the equilibria behind ($\ce{CO2 + H2O <=> H2CO3}$ and $\ce{H2CO3 <=> H+ + HCO3-}$) so is $\ce{HCO3-}$.

Basically, this removes  $\ce{CO2}$, overall carbonic acid species and also acid equivalents: the resulting water will be more basic (less acidic).

$\ce{CO2}$, $\ce{H2CO3}$ and $\ce{HCO3-}$ are not only present in carbonated (sparkling, fizzy) water but also in tap water. For example, this analysis sheet of the Munich municipal water supply cites 300 mg/l $\ce{HCO3-}$, 15 mg/l $\ce{CO2}$ and 0.5 mg/l $\ce{CO3^2-}$ and a slightly basic pH.

It is known that mammals have $\ce{CO2}$ receptors in their acid taste cells, and this is thought to contribute to the water taste. (I don't have the full text, but Zocchi et al: The cellular mechanism for water detection in the mammalian taste system, Nature Neuroscience volume 20, pages 927–933 (2017) looks promising.)

BTW, two further candidates for that bland/flat/stale taste are
deionized water and reverse osmosis drinking water.

Warm vs. cold water taste

There will be further changes with temperature, including acid base equilibria and pH, partial pressure of volatile molecules and so on between cold and warm water. And this also applies to our sensing in the taste cells. Put this together with taste/smell sensory input being subject to further processing in our brain to produce the taste/smell impression, and I think it is very hard to compare taste/smell of cool and hot things.


Why are the melting and boiling points of the group 1 metals lower than the group 2 metals in the same period? I thought it was because the group 1 metals are smaller than the group 2 metals but the answer sheet says it's because there are more valence electrons and a stronger positive charge in group 2 metals. But why does it matter if there's more valence electrons and a stronger positive charge?

In the photo, you can see that for Group $2$ metals, the cation is doubly positively charged, making the attraction stronger.

Credits

The photo is from Figure 12.20, Section 12.5, Principles of General Chemistry (v. 1.0).


What happens when ferric chloride is dissolved in water? Ferric chloride can be prepared by dissolving hematite in hydrochloric acid:

$$\ce{Fe2O3 + 6HCl -> 2FeCl3 + 3H2O }\tag{1}$$

But the resulting ferric chloride will be exposed from its formation to water.

After investigating a little, I found that anhydrous ferric chloride, when dissolved in water, hydrolysed and forms ferric hydroxide:

$$\ce{FeCl3 + 3H2O -> Fe(OH)3 + 3HCl }\tag{2}$$

But isn't it supposed that, when dissolved in water, hexahydrate is formed? What really happens, hydrolyses or hydrates?

Iron(III) chloride indeed hydrolyses in water, or to be precise, $\ce{Fe^3+}$ does assuring acidic medium.
Hydrolysis isn't completely suppressed even in strongly acidic solutions.
In the absence of other factors, what happens exactly depends on the concentration of chloride and temperature.

In diluted cold solution a typical formation of metal aqua ions takes place:

$$
\begin{align}
\ce{FeCl3 + 4 H2O &<=> [Fe(H2O)4Cl2]+ + Cl-}\tag{R1.1}\\
\ce{[Fe(H2O)4Cl2]+ + H2O &<=> [Fe(H2O)5Cl]^2+ + Cl-} &\quad \mathrm{p}K_\mathrm{a} &= 0.30\tag{R1.2}\\
\ce{[Fe(H2O)5Cl]^2+ + H2O &<=> [Fe(H2O)6]^3+ + Cl-} &\quad \mathrm{p}K_\mathrm{a} &= 1.48\tag{R1.3}\\
\ce{[Fe(H2O)5Cl]^2+ + H2O &<=> [Fe(H2O)4(OH)Cl]+ + H3O+} &\quad \mathrm{p}K_\mathrm{a} &= 3.05\tag{R1.4}\\
\end{align}
$$

In concentrated cold solution aquacomplex is also accompanied by  tetrachloroferrate(III):

$$\ce{2 FeCl3 + 4 H2O <=> [Fe(H2O)4Cl2)]+ + [FeCl4]-}\tag{R2}$$

In boiling-hot solution iron(III) chloride decomposes to hydroxochloride

$$\ce{FeCl3 + 2 H2O ->[Δ] Fe(OH)2Cl(s) + 2 HCl}\tag{R3}$$

which is partially hydrolyzed further to colloidal iron(III) hydroxide ($\ce{Fe(OH)2Cl}$ is still more soluble than $\ce{Fe(OH)3}$).

Finally treatment with steam results in iron(III) oxide:

$$\ce{2 FeCl3 + 3 H2O ->[\pu{350 - 500 °C}] Fe2O3 + 6 HCl}\tag{R4}$$

Reactions were adopted from [1, pp. 420–421].

References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


Why does a dynamic equilibrium exist? We know that at equilibrium Gibbs free energy is minimum. We also know that at equilibrium both forward and reverse reactions occur simultaneously, and we also know that for a reaction to be spontaneous Gibbs free energy should decrease in that direction. So if at equilibrium we have minimum Gibbs energy then both forward and backward reactions will be non spontaneous, since they are characterised by an increase in Gibbs free energy, therefore there should be no forward and backward reactions and hence, no dynamic equilibrium. Then why does a dynamic equilibrium exist?

To understand the confusion here you need to take two different points of view about what is happening in a reaction: the molecular view and the statistical view.

The statistical view ("Gibbs free energy is a minimum") is an average of all the molecular activity in the reaction. It describes the net, bulk behaviour of the reaction. At equilibrium, there is no net direction of change. But this tell us little about what is happening to the individual molecules or atoms in the reaction.

The molecular (or atomic) view is about what is happening with the individual things that make up the reaction. At any temperature above absolute zero there will be plenty of things happening at the molecular level. There is plenty of energy around and the molecules will bang into each other a lot. Moreover, not all the components will have the same energy (some molecules are slow moving, others fast: the distribution of energies will follow a Boltzmann distribution). Sometimes there is enough energy for the products of the reaction to fall apart; sometimes there is enough energy for the reactions to react to give the product. Equilibrium is reached when, at a given temperature, the rates of both reactions are the same not when everything stops at the molecular level. Only when we look at the overall averages in the bulk reaction do we see no net change: at the molecular level things are always happening.

It is worth noting that the molecular view gives more insight than the purely thermodynamic view which only considered bulk averages. Some reactions at the molecular level happen fast, others slowly. Often in chemistry it is the kinetics that determine the product not the thermodynamics and thinking about the molecular point of view often explains the observed situation in the bulk.

It is quite important and insightful in chemistry to be able to switch from one view to the other as you will get a much richer interpretation of what happens in real reactions if you can.