text:
How could you predict degenerate pi molecular orbitals just knowing what symmetry point group a molecule belongs to?
If you were to apply the Huckel method to some molecules which belonged to, say, $C_{2v}, D_{2h}, $ and $ D_{4h}$, which ones would you expect to have degeneracies?
I'm not sure if this is mixing up math jargon with chemistry jargon here, but I know the Mulliken symbol E is doubly degenerate and T is triply degenerate. So would the $D_{4h}$ molecule have degeneracies because it has the irreducible representations $E_g$ and $E_u$?

This might be interesting for you.

Which orbital symmetries you can get in a molecule always depends on the molecule’s point group. Only for those point groups which have degenerate irreducible representations can you get degenerate orbitals; so out of your examples only the $D_{4h}$ molecule could. This is entirely independent of Hückel’s method to determine energy levels.
But just because your molecule has the irreducible representations, doesn’t mean that a $\pi$ orbital has to have that degeneracy. Instead, it again depends on what you get when following symmetry rules.
You can calculate whether you will get $\pi$ degeneracies by drawing your molecule of interest, including all the p atomic orbitals and determining the molecule’s point group. Once you have that, for each symmetry operation, check how many p orbitals would be transformed onto themselves by said operation (and note whether with or without a phase change). Add that up. Once you’ve done that, you basically have the representation of the p-orbitals of the molecule that you can reduce to irreducible representations, finding out how many discrete and degenerate energy levels you have (albeit not including energy values).

Agreed. In some cases, it could be easier than this. Let's say you have a central atom (e.g, $\ce{Pt}$). You can then look in the character table to see the representations of the p and d orbitals and know if there's a degenerate orbital set for those atomic orbitals. Generally, though you need to perform the symmetry analysis you describe.


text:
Propose a chemical formula for the white solid that forms during the initial stages of the reaction of Sn with benzyl chloride
In my inorganic class, we carried out the synthesis of chlorotribenzyltin, $\ce{SnCl(CH2C6H5)3}$.

The synthesis begins with a reflux I'll describe below:

In a hood, place 2.0 g (17 mmol) of 325 mesh tin powder, 4.0 mL of $\ce{H2O}$ and 6.0 mL (6.6 g, 52 mmol) of benzyl chloride in a 50 mL round-bottom flask. After the reaction mixture has been heated to reflux for 3 hours*, remove the heat and cool the flask in ice. The lower liquid phase should solidify to form a white solid.
*The white solids that appeared shortly after refluxing commenced should have largely disappeared, although small amounts of white or gray solids on the sides of the flask are usually still present. The reaction mixture at this point should consist of two liquid phases, plus a small amount of unreacted tin.

So the question the textbook asks is

Propose a chemical formula for the white solid that forms during the initial stages of the reaction of Sn with benzyl chloride.

I have a number of ideas as to what this could be:

The chloride bonded to the tin forming some kind of salt like
$$\ce{SnCl2}$$

The reaction did some kind of partial reaction creating something like
$$\ce{SnCl2R2}$$

The reaction went too far and created
$$\ce{SnR4}$$ and $$\ce{SnCl4}$$ and keeping them in reflux redistributed the ligands  based on stoichiometric amounts like
$$\ce{SnCl4 +3SnR4->4SnClR3}$$

This last one seems most likely based on supporting evidence from the text:

$\ce{SnCl4}$ reactions with Grignard reagents to give organotin compounds:
$$\ce{SnCl4 +4RMgX->SnR4 +4MgXCl}$$ ...
(it goes on to say)...the product, chlorotribenzyltin $\ce{SnCl(CH2C6H5)3}$, is a colorless solid that is stable in air for several days. In some respects, the reaction is analogous to the preparation of Grignard reagents, $\ce{RMgX}$ by treatment of magnesium metal with organic halides.

Is my reasoning for choice 3 solid? Is there something I'm not thinking of?

Turned in this report and my professor commented with his thoughts
$$\ce{Sn + 2RCl \rightarrow R2SnCl2 \rightarrow R2Sn }\text{ (white solid)} \ce{+ SnCl2 \rightarrow  R3SnCl}$$
but I didn't get the question wrong. So unless if someone contributes another answer, I'll accept this as the answer.

Apart from the stoichiometry not being quite right, Holleman and Wiberg agree with that.

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text:
What is the role of pH in azo coupling reaction of diazonium with phenol and aniline?
Why is it so that azo coupling with phenol needs basic pH (9–10) and with aniline it needs acidic pH (4–5). How does pH participate in the mechanism? I have searched the internet for quite a while nowhere does it explain the actual ‘role’ of the pH. Any light on this matter would be helpful.

Deprotonation of the phenol and protonation of aniline result in species that easily react with a diazonium cation in the intended manner.
Let's have a look at the species involved. On the one hand, there is the diazonium cation:

Deprotonation of phenol yields phenolate, for which a resonance structure with a negative charge in para position to the substituent can be written:

We conclude: Deprotonation of the phenol means activation!

In the case of aniline, the lone pair on the nitrogen atom is the preferred centre for the reaction with the diazonium cation:

This isn't what we want! Protonation of aniline avoids this side reaction.
We conclude: Protonation of aniline means "masking" the wrong centre!

In the case of phenol, Why lone pairs of oxygen are not the preferred center for reaction with diazonium cation??

If aniline is protonated, then won't the benzene ring be deactivated in aniline? NH3+ is supposed to be a strong deactivating group. This deactivation should then prevent azo coupling because azo coupling reactions need highly activated groups.

Klaus, could you add to your post to answer the comments? They're good questions, and making a new post for them seems unnecessary.

what is the role of Ph in the answer given?

@PranoyDe orthocresol's answer here answers why HCl is being used to form the azobenzene.

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text:
What are the spectroscopic signatures that differentiate triple state water from water at 1 atm?
According to the Wikipedia article Triple point of water, the triple point of water occurs at

The single combination of pressure and temperature at which liquid water, solid ice, and water vapor can coexist in a stable equilibrium occurs at exactly 273.16 K (0.01 °C; 32.02 °F) and a partial vapor pressure of 611.657 pascals (6.11657 mbar; 0.00603659 atm).

Shown in the illustration below (from the Wikipedia article):

Crucially, to put this into a context, the 611.73 pascals quoted in the article corresponds to the

minimum pressure at which liquid water can exist. At pressures below the triple point (as in outer space), solid ice when heated at constant pressure is converted directly into water vapor in a process known as sublimation.

Potentially useful information to determine if an exoplanet can possibly have water (if all other conditions are suitable).
What are the spectroscopic signatures that differentiate triple state water from water at 1 atm?

The triple point of water is point of dynamic equilibrium between three phases of water. Any small change in Pressure or Temperature will lead to one phase regions or two phase lines on the phase diagram.
The triple point is not a state of matter rather a combination/mixture of three phases . So i very much doubt we can get any unique spectroscopic signatures . We would get a distorted result as this point has three different phases.
Also Emission and Absorption spectra depends on  temperature and pressure.

Great answer (+1), a question though - would the distorted result be 'characteristic' enough to be differentiated from water at 1 atm?

water will have specific peaks at 1 atm and room(or experimental) temperature. But when we have three phases together like this , we will be  seeing random peaks (i think). So we can get the spectra of the individual phases by spatial and spectral seperation of components.


text:
Why does a reduced compound become a better electron donor?
When a compound gains an electron, why are they more likely to give them up (i.e. become stronger reducing agents)?

Simple electrostatic repulsion can be used to explain this. For example, if you have $\ce{O^2-}$, there are 8 protons in the nucleus, but 10 electrons, so there is a net negative charge of 2. Like charges repel, pushing the electrons away.
BTW, compounds rarely "gain an electron", the reactive species are usually radicals.


text:
Air contains 20 % O2 by volume. How many cm³ of air will be required for oxidation of 100 cm³ of acetylene?

Air contains 20 % $\ce{O2}$ by volume. How many cm³ of air will be required
for oxidation of 100 cm³ of acetylene?
(a) 1064 cm³
(b) 212.8 cm³
(c) 500 cm³
(d) 1250 cm³

Answer: (d)
My solution:
Equation: $\ce{2C2H2  +5O2 -> 4CO2 +2H2O}$
From stoichiometric calculations:

2 moles of $\ce{C2H2}$ will react with 5 moles of oxygen
1 $\ce{C2H5} \rightarrow $ 5/2 moles of $\ce{O2}$
$\pu{22.4L}$ $\ce{C2H2} \rightarrow 5/2\times \pu{22.4L}$ of $\ce{O2}$
1 volume of air $\rightarrow$ 0.2 (20/100) volume of oxygen

Now the only thing I don’t know is how to convert the volume into liters so that I can find the correct answer.

Let's begin with your answer:
$2\ \mathrm{mol}$ $\ce{C2H2}$ will react with $5\ \mathrm{mol}$ of oxygen.
$1\ \mathrm{mol}$ $\ce{C2H2}$ will react with $5/2\ \mathrm{mol}$ of $\ce{O2}$.
If we suppose that the gases are under standard conditions, then
$22.4\ \mathrm{L}$ $\ce{C2H2}$ will react with  $\frac{5}{2} \times 22.4\ \mathrm{L  } ~ \ce{O2}$.
This means that:
$0.1\ \mathrm{L}$ $\ce{C2H2}$ will react with $\frac{5}{2} \times 0.1\ \mathrm{L}$ $\ce{O2}$.
$0.1\ \mathrm{L}$ $\ce{C2H2}$ will react with $\frac{5}{2} \times 0.1\times 5\ \mathrm{L}$ of air.
So, the volume of air is $1.25\ \mathrm{L}= 1250\ \mathrm{cm^3}$

What I suggest is we can use Gay Lussac's law of combining volumes of gases. According to this law, 2 volumes of acetylene reacts with 5 volumes of oxygen. From the balanced equation, the ratio between volumes of gaseous reactants and products are in the ratio 2:5:4.
2 volumes of $\ce{C2H2}$ requires 5 volumes of $\ce{O2}$.
The volume of $\ce{O2}$ used for $\pu{100 cm3}$ of $\ce{C2H2} = (5/2) \times 100  = \pu{250 cm3}$.
But air contains only 20% of $\ce{O2}$, i.e. $0.2$ of $\ce{O2}$.
Thus, the amount of air required to get $\pu{250 cm3}$ of oxygen $= 250×100÷20 = \pu{1250 cm3}$

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text:
Oxygen Preventing the Formation of Large Organic Compounds?
My textbook tells me that when carbon bonds with oxygen, the electrons are too close to the oxygen, and therefore they cannot just string together and form large molecules like glucose. It goes on to say that because, during photosynthesis, carbon becomes bonded to less electronegative atoms like hydrogen, it becomes more capable of forming larger compounds. Why is that?

Glucose is a large organomolecule? Since when? (But of course, being large is relative)

Might be referring to glucose polymers.

@MARamezani Much larger than carbon dioxide anyway.

If we take a look at glucose we can see there are oxygen atoms in it, just with extra presence of the hydrogen.
The assumption :

carbon bonds with oxygen, the electrons are too close to the oxygen, and therefore they cannot just string together and form large molecules like glucose.

is wrong.
Take a look at Polymeric carbon oxides

Carbon suboxide spontaneously polymerizes at room temperature into a carbon-oxygen polymer, with 3:2 carbon:oxygen atomic ratio. The polymer is believed to be a linear chain of fused six-membered lactone rings, with a continuous carbon backbone of alternating single and double bonds. Physical measurements indicate that the mean number of units per molecule is about 5–6, depending on the formation temperature.
Carbon monoxide compressed to 5 GPa in a diamond anvil cell yields a somewhat similar reddish polymer with a slightly higher oxygen content, which is metastable at room conditions. It is believed that $CO$ disproportionates in the cell to a mixture of $\ce{CO2}$ and $\ce{C3O2}$ ; the latter forms a polymer similar to the one described above (but with a more irregular structure), that traps some of the $\ce{CO2}$ in its matrix.
Another carbon-oxygen polymer, with C:O ratio 5:1 or higher, is the classical graphite oxide and its single-sheet version graphene oxide.


text:
High or low spin complex of Zn2+ and Ti3+
The zinc ion in its +2 state can show no more pairing, since all the electrons are already paired - so in any complex it forms it will have the same magnetic moment as it does right now. $\ce{Ti}$ in its +3 state will show both outer and inner orbital complexes without any electrons pairing.
So while calling these complexes high spin or low spin complexes is not useful, since there is no possibility of pairing involved in either - but all the same, won't they be called high spin complexes, since no lowering of magnetic moment occurs?
My teacher said, however, that since $\ce{Zn}$ in +2 state forms only outer orbital complexes, its complexes will be called high spin - and $\ce{Ti}$ in its +2 state similarly when it forms inner orbital complexes (which it does without pairing), will be called low spin because the complex is inner orbital. But shouldn't such a complex of $\ce{Ti}$ be a high spin complex? (I'm saying this in reference to VBT).

Are you familiar with ligand field stabilisation energy? This approach answers your question. Interaction of the central metal ion with the surrounding ligands causes splitting of the $d$ orbitals into an essentially upper and lower energy level, the form of the splitting is rather complex and depends directly upon the symmetry point group that the ligand complex belongs to.
Either way, high spin complexes only form if the reduction in inter electron repulsion energy is greater than the energy splitting of the orbitals, with the definition of high spin being a none aufbau filling of the split orbitals. It does not make sense to describe the electronic structure of a $\ce{Zn^2+}$ ion in a complex as high or low spin, there is only one configuration available as there are 10 $d$ electrons present, and the 5 $d$ orbitals split to form 5 new orbitals of varying degeneracies (dependent upon the point group of the complex and the symmetry of the coordination environment).
$\ce{Ti^2+}$ has 2 valence d electrons available to be involved in bonding with the ligand environment, however there is no coordination environment that I am aware of (The splitting would have to be extremely small and other thermodynamic factors such as the ionic contribution to the Gibbs energy of complex formation would mean that the complex would never be observed anyway - such a low splitting would only be the result of very poor overlap which would be caused by large separation between the central ion and the ligands, resulting in a lower ionic term) that causes splitting where the magnitude of the splitting is less than the repulsion term invoked when both electrons occupy the same molecular orbital, be it degenerate or not.
The $\ce{Ti^3+}$ ion has only 1 $d$ electron, it will definitely reside in the lowest available molecular orbital. However, the first excited states, ignoring hyperfine splitting, of both titanium ions will be high spin - the magnitude of the splitting usually corresponds to wavelengths of light in the visible region of the spectrum, hence many transition metal complexes are coloured.

I didn't understand why you said Ti forms high spin complexes in +3 state. Are you talking about the splitting according to the Crystal Field Theory? I think the d electron will go to the lowest orbital - though this really isn't non aufbau filling, since no pairing occurs - I think it should be high spin. High spin means complexes in which no pairing occurs and magnetic moment of the metal ion doesn't reduce on complex formation - am I right? This is what I learnt.


text:
Why do complex ions not emit light from de-excitations of electrons?
I understand that complex ions are coloured due to d-orbital splitting which results in electrons being able to absorb wavelengths of visible light and become excited to the higher energy state meaning the transmitted light is coloured. However I don't see why the electron would not simply de-excite and emit the same wavelength of light that it absorbed which would result in no visible difference. Is there some reason why the same wavelength is not emitted or perhaps the electron does not de-excite at all?

The fate of electronically excited states is not immediately obvious.
First of all the rate at which the electron emits a photon and falls down a level is dependent upon the energy difference between the two levels, it is only an expectation value of a macroscopic observable, on an individual atomic scale it is impossible to predict when the electron in the upper level will emit a photon, however multiple instances of the same measurement will tend towards the expectation value. The direct result of this is that there is a path difference between the emitted photon and the 'rest' of the photons that were not absorbed by the complex and the result is a different interference between photons in the transmitted light compared to the incident light and hence a new colour is observed.
There is another fate of excited electrons, the Franck-Condon principle invokes the very plausible assumption that the nuclei do not respond immediately (at least on the time scale of electronic transitions) to the new equilibrium bond position that is determined by the new electronic state of the complex. This means the excited electronic state will most likely not be in the lowest vibrational state of the new equilibrium bond configuration, the splitting of vibrational energy levels is orders of magnitude lower than those of electronic transitions and hence the same law that governs electronic emission applies, the electron is much more likely to fall down these vibrational energy levels first, without emitting photons of visible light.
The electron will then be expected to, by the time it is occupying the lowest vibrational state of the excited electronic state, invoke a transition down to the ground state energy level. The energy of the excited state is described in a basic form by
$$H=E^*+v$$
where $H$ is the energy of the absorbed photon, $E^*$ is the energy of the excited state in the lowest vibrational mode and $v$ is the contribution from the initial vibrational state of the excited complex.


text:
For protease, lipase, and amylase, include an image of the chemical, and the chemical that it operates
I have no idea how to find a picture for these enzyems. My teacher did not explain the chemical reaction and I am really lost. Please help!

Welcome to the site. The chemical reaction? What chemical reaction are you looking for? Also, have you tried Google for finding an image?

Your question is too broad; protease is a family of enzymes which  performs protein digestion aka proteolysis; same goes to lipase and amylase for lipids and sugars respectively. These enzymes/proteins have a complex chemical structure. If you want to see the structure of proteins or enzymes I would recommend RCSB PDB database. You can search for protein structure by name,PDB ID, sequence etc. in this database. It also gives detailed information about that protein.
For example a search for Serine proteases ( proteases that have serine bonded at the active site) will give you 458 strutural hits. There are different parameters such as organism, taxonomy etc available to refine your search

An image of HIV-1 Protease

D30N tethered HIV-1 protease dimer/saquinavir complex (PDB ID=4Q5M)


text:
How to determine the value for n in the Gibbs free energy and redox potential equation?
I am having a problem with this equation for redox potentials
$$\Delta G = -nFE_\mathrm{cell}$$
In this equation I never am totally sure about what the value of $n$ should be, for example for the reaction shown below, would the $n$ be 2 electrons or 1 electron? Personally, I think it should be 2 electrons because that is the number of mol of electron under the simplest whole number ratio.
\begin{align}
\ce{cyt $c$(Fe^3+) + e- &-> cyt $c$(Fe^2+)}\\
\ce{NADH &-> NAD+ + H+ + 2e- }\\
\end{align}
I think this is a different matter than in the already answered question "Does the relationship equation between standard cell potential and equilibrium constant violate potential's intensive properties?" Because in that Nernst equation the change in $n$ would be balanced by the change in the equilibrium constant. I assume that you are insinuating that there is one of the variables here which will correct for the increase in $n$. I don't see which would, as $E_\mathrm{cell}$ should be independent of quantity and $F$ is a constant.

Do you think your question is similar to this one, Matthieu?

Personally, I think it should be 2 electrons because that is the
  number of mol of electron under the simplest whole number ratio.

Yes, it is $\ce{2e-}$.
Assuming that the two reactions provided above are half-reactions, the $n$ has to be $2$.
$E_\mathrm{cell}$ is independent of quantity because it is an intensive property. It does not depend on the amount of matter present. The $n$ accounts for the number of moles of $\ce{e-}$, this is used to calculate $\Delta G$, which in turn depends on the amount of matter present.
Overall, one must understand that $E_\mathrm{cell}$ is just the difference in cell potentials of two half-reactions. Increasing or decreasing the amount of both does not matter.

This statement that $n$ "has to be 2" is not correct. By no means does $n$ have to be 2, it can be anything OP so chooses.

The accepted answer is not correct, the number $n$ is not equal to 2 simply because 2 is the lowest common multiple of 2 and 1. Furthermore $n$ is not the "number of moles of electrons", which would have units of $\pu{mol}$; $n$ is in fact dimensionless.

In fact the number $n$ can be anything you like, and it depends on how you combine the two half-equations.
To use a different (not very realistic, but I don't care) example consider the half-reactions (data from Wikipedia)
$$\begin{align}
\ce{Na+(aq) + e- &-> Na(s)} & E_1 &= \pu{-2.71 V} \tag{1} \\
\ce{Br2 (l) + 2e- &-> 2Br-(aq)} & E_2 &= \pu{+1.07 V} \tag{2} \end{align}$$
We have $E_\mathrm{cell} = E_2 - E_1 = \pu{+3.78 V}$. To construct a full reaction for this cell the "usual" procedure taught is to take (1), multiply it by 2, and subtract it from (2).
$$\ce{Br2(l) + 2Na(s) -> 2Na+(aq) + 2Br-(aq)} \tag{3}$$
In this case, we have $n = 2$ as there are two electrons being transferred in this reaction, from bromine to sodium. Equivalently, you can think of (3) as being the sum of two half-reactions, each featuring the gain or loss of two electrons:
$$\begin{align}
\ce{2Na(s) &-> 2Na+(aq) + 2e-} \tag{4} \\
\ce{Br2(l) + 2e- &-> 2Br-(aq)} \tag{5}
\end{align}$$
Note that the stoichiometric coefficient of $\ce{e-}$ - the number in front of the electrons - is 2, for both half-equations.

However, you could easily construct a different full reaction, simply differing from (3) by a multiplicative factor. There's no reason why you cannot, for example, divide (2) by 2, then subtract (1) to get:
$$\ce{1/2 Br2(l) + Na(s) -> Na+(aq) + Br-(aq)} \tag{3*}$$
and here you should only have $n = 1$, because there is only one electron being transferred here, from bromine to sodium. Again you can decompose this into its half-reactions:
$$\begin{align}
\ce{Na(s) &-> Na+(aq) + e-} \tag{4*} \\
\ce{1/2Br2(l) + e- &-> Br-(aq)} \tag{5*}
\end{align}$$
and the stoichiometric coefficient of $\ce{e-}$ here is simply 1.

The bottom line is that

$E_\mathrm{cell}$ doesn't change with the stoichiometric coefficients.
$n$ does, and hence $\Delta G = -nFE_\mathrm{cell}$ does. Your choice of the full cell reaction determines the value of $n$ you work with.
The best definition of $n$ is the stoichiometric coefficient of the electrons in the half-reactions into which the cell reaction can be divided (Atkins & de Paula, Physical Chemistry).


text:
Crystallization issue
For solid-liquid mixtures, cooling crystallization process can be used to separate the solute substances from each other. Assume that we have two salts that have very close solubility in a certain solvent which we are going to use. First, we dissolve the salts in the solvent while heating. After a point, we let the mixture cool down and observe the crystallization. My question is, how can we do this experiment accurately if they have really close crystallization point that they would nearly crystallize at the same time?

You could do as Louis Pasteur did to separate the chiral isomers of ammonium sodium tartrate... select crystals by shape.

I might suggest that changing the solvent would be a good idea.

If you seed the supersaturated solution with a few crystals of just one salt, that one will likely crystallize out faster, affording a degree of separation.

The best result would be obtained by using a solvent that one of the product salts is more soluble in than the other salt.

Given the constraints of the question, you are screwed, don't even bother.

Under the right conditions, crystallization can be a very effective means of obtaining a pure solid form of the solute. Unfortunately, in your case one of those conditions is violated; you have a second solute with similar solubility characteristics.  You didn't quantify the solubilities of the solutes, but the phrase "very close solubility in a certain solvent" tells me that under the conditions of your system, crystallization will not be an effective means of obtaining a pure solid form of the solute.
One option, as mentioned in a comment to your question, is to choose a different solvent. Selecting the right solvent / solvent mix can be a bit of an art and will also require some research and experimentation on your part.  Because you didn't discuss the specific solvents or solutes used in your system, it is not really possible to give more specific advice on solvent selection.
Another option is to choose a different means of solute separation, like some type of chromatography. Again, not knowing anything about your compounds makes it difficult/impossible to give further advice regarding this technique.


text:
Why wash a sample before titration?
In the awwa notebook for $\ce{Na2SiF6}$ determination, why it is required to wash the product sample in an alcoholised $\ce{KCl}$ solution before titration?

I think we need a bit more info to be able to answer this question. In general, you wash your sample to get rid of something unwanted. What this might be in your case, nobody can tell you without more details about the experiment setup.

It is true that this is not an academic question that can be answer with a notebook from high school.

Here the summerised protocole 1. Wheight precise amount of sample to analyse, 2. Suspend sample in enough solution of KCl/carbonate/ethanol/water to keep the solution alcaline, 3. Filter solution in a paper filter, 4. Rince sediment with a solution of KCl/ethanol/water untill filtrate gets a neutral pH, 5. Dissolve filtered sediment in distilled water under heat, 6. When temperature of solution reach 80 C, start titrate with NaOH, 7. As solution is boiling, end titration when the solution is alcaline per phenilphtalein, 8. Amount of NaOH used correlate with amount of Na2SiF6 in sample.Thanks

Your protocol gives you the answer: to make sure your sediment is neutral. After all, you titrate with NaOH afterwards, so any wrong starting point in terms of pH would skew your result.


text:
Why is energy released as electrons move?
Why is it that energy is released from electrons when they move from one atom to a more electronegative one?

The previous answer is correct. I just wanted to add that atoms, electrons and generally matter, tries to ''organise'' itself in order to adopt a state, where it would have the lowest energy possible.
An electronegative atom tends to attract electrons and has the ability of handling their charge better. Thus when an electron moves from a less electronegative atom to a more electronegative one, it is like moving from a state of high energy to a state of low energy. This is also described as realease of energy.

Due to difference in electronegativities the atoms have different binding energies and this difference is radiated in some form. If you're talking about a transition in a single atom the difference is the binding energies of different orbital/orbits/sheels.


text:
Charge consistency in fragment qm/mm methods
I'm calculating the multipole moments on several fragments of a molecule. What are some procedures to ensure that the charges etc are representative of the complete molecule? How do I ensure I'm not getting an overall charge in a neutral molecule?

Draw a Lewis Dot Structure. This may help figure out the position of $e-$ and thus help you find the charge of a fragment.

That might help if we were dealing with integer charge, but any sub part of a molecule most of the time doesn't have that. For instance in water the O might have a charge of -0.35 and H1 might be 0.15 and H2 might be 0.2 . I'm looking for a somewhat mathematical rigorous procedure to do this. For instance if the water described above is the true result  and my fragments are each atom If my O is -0.4 H1 0.1 and H2 0.15, how do I approach the true result?

I really don't understand what you are asking. Do you think, that you will lose or gain an electron during a calculation? If you set up a calculation, you specifically set the parameters to a fixed number of electrons and nuclei. The program just finds the best orbitals where to put the electrons around the nuclei. So the second case in your comment is just impossible. It would be helpful, if you could describe your methodology more precisely, maybe I am interpreting it wrong.

I'm performing a Distributed Multipole Analysis on a protein.

If you're doing any sort of computational run (either quantum chemistry, solid state, or molecular dynamics), you have to set the charge on the overall system.
You said in your comments that you're doing a distributed multipole analysis across multiple fragments of a protein.
In this case, one of the constraints on the calculation is exactly that the overall charge must match that specified. The program won't arbitrarily change the charge to be an anion or a cation if you specified a neutral system.
Now, that's not to say that the individual fragments won't have fractional or integer charges. Depending on the method, it's entirely reasonable that there's some amount of partial charge transfer between fragments.
It's just that there's absolutely no way that the sum of the charges will be different than the constraint.


text:
Proportion of gases filled in cylinder of sea-divers
In my book it is given that earlier cylinder were filled with 2% oxygen and 98% helium. Nowadays it is filled with 11.7% helium, 56.2% nitrogen and 32.1% oxygen.
There are few questions which come up in my mind

Earlier 2% oxygen is changed to 32%. So was 2% oxygen enough for huaman?
98% helium is reduced to 11.7% why it is so?
I don't get reason why helium and nitrogen are filled?

From this source

Helium is used in scuba tanks to dilute oxygen.  Why helium?  First, it is chemically inert, that is to say, it does not participate in chemical reactions.  Second, helium is much less soluble in water than many other gases, such as nitrogen.  The low solubility means it does not enter the blood stream, even under pressures commonly experienced by deep sea divers.
  Under water, a scuba diver is subjected to added pressure.  In shallow water the pressure difference is not so great, and one can simply use compressed air to breathe.  However, at a depth of 300 feet, a diver experiences nearly 10 times normal atmospheric pressure.  Under these conditions, breathing compressed air can be harmful.  The increased pressure requires an inert gas to be used to dilute oxygen.
Oxygen is essential for breathing, above and beneath the surface.  However, one can get too much of it.  Under high pressures, greater amounts of oxygen enter into the bloodstream.  If the concentration becomes too high, oxygen poisoning can result, with symptoms that include confusion, impaired vision and nausea.  So, under high pressures, oxygen must be diluted with some other gas.
Nitrogen is very abundant and relatively inert, but it cannot be used to dilute oxygen in scuba tanks.  At the elevated pressures experienced by divers, larger amounts of nitrogen will dissolve in the blood causing nitrogen narcosis.  Dissolved nitrogen also can cause a painful condition, called "the bends" if a diver makes too rapid an ascent.  As the pressure decreases, the dissolved gas can form bubbles that can stop circulation in capillaries and damage the nervous system.

The wikipedia article on breathing gas says

Helium (He) is an inert gas that is less narcotic than nitrogen at equivalent pressure (in fact there is no evidence for any narcosis from helium at all), so it is more suitable for deeper dives than nitrogen. Helium is equally able to cause decompression sickness. At high pressures, helium also causes high-pressure nervous syndrome, which is a CNS irritation syndrome which is in some ways opposite to narcosis.
  Helium fills typically cost ten times more than an equivalent air fill.
Equivalent air depth is used to estimate the decompression requirements of a nitrox (oxygen/nitrogen) mixture. Equivalent narcotic depth is used to estimate the narcotic potency of trimix (oxygen/helium/nitrogen mixture). Many divers find that the level of narcosis caused by a 30 m (100 ft) dive, whilst breathing air, is a comfortable maximum

I suspect earlier dives were not so deep , as the suits, storage and general technology were probably lacking. So a 2% was probably enough for the shallow dives.
Helium is a good diluting gas , but i think the new mixture was adopted to reduce costs as helium is very expensive.

Hint:
Try to use Henry's Law. According to which solubility of gases increases with increase in pressure. If we go deeper downwards, pressure increases and when we come to the surface, pressure decreases, so there might be that soem buubles are created in the blood which may be fatal. So air cylinders are made keeping this in mind.


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Is there a difference between hydrated oxides and hydroxides?
My guess is that they are different.
I think hydrated oxides would have a lattice with the cation and oxide anions as lattice points and molecules of water trapped in interstitial spaces, while hydroxides would have the cation and hydroxide anions as lattice points with no water molecules in interstitial spaces.

I think that when we talk about bases, we immediatly have to do with hydroxides because they contain the hydroxide group OH-. Hydrated oxides on the other hand can be only simple oxides who contain also Hydrogen. I didn't find anything on the internet so this is just a thought.

$\ce{Mn(OH)2}$ is a hydroxide while $\ce{MnO(OH)2}$ is a hydrated oxide.

Maybe hydrated oxides don't even have a lattice; that's not important. What is important is their non-stoichiometric composition.

This formular $\ce{[Zr4(OH)8 *16 H2O] ^{8+} }$ is a "white gel called zirconium hydroxide but generally considered to be a hydrous oxide"
from Nielsen 2005

zydrate comes in a little glass vial


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Reaction of secondary alcohol with ammoniacal silver nitrate (Tollens' reagent)

What are the products of the reaction of 2-butanol with silver nitrate?

For the moment, I am a bit stumped as I only have $\ce{AgNO3}$ as an oxidizing agent to form a carboxylic acid from an aldehyde. Additionally, the lack of $\ce{H2O}$ indicates that this is not the reaction I should be looking for. The only reaction that seems feasible to me is an SN2 mechanism where the nitrate anion acts a nucleophile and the hydroxide ion as the leaving group.
Here is my proposed mechanism much in the style of Tollen's reaction with an aldehyde.

Do you know the product of the reaction? Or the reaction conditions, solvent?

@Jori, No they only gave this information, but in our course we have only seen $\ce{AgNO3}$ as an oxidizing agent for aldehydes. I guess it would be safe to assume that the solvent was water but even then I am not sure how the reaction would proceed...

@Matthieu Kratz Are you are referring to ammoniacal silver nitrate (Tollens' Reagent)?

@bon, yup referring to tollen's reagent, but the form in which were given it was a mix silver nitrate and ammonia/ water. I can think of a mechanism that would work on ketones, but I have never seen it used in such a way.

Your question does not make complete sense. In the comments you clarified that it is ammoniacal silver nitrate, but then in the question image you had still only written $\ce{AgNO3}$. IIRC ammoniacal water is important for Tollen's reagent to function. Please clarify that the initial question image was incomplete, otherwise your question body and question title are currently mismatching.

This question refers to the use of Tollens' Reagent (ammoniacal silver nitrate). The oxidising species here is the diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$.
Tollens' reagent is usually used as a test for aldehydes, with which it gives a characteristic silver mirror:

It is often said that Tollens' reagent does not react with alcohols. However, from personal experience, gentle heating of a primary alcohol with Tollens' will cause a small amount of oxidation and result in a fine black precipitate. Again, in my experience, secondary alcohols (such as in the question) do not react with Tollens' reagent and so I would predict that no reaction will take place here.
EDIT: Your proposed mechanism certainly looks plausible but whether it will take place in reality I can't say for certain (my experience of Tollens' reagent says no).

Thanks for answering, I guess they made a mistake in their question but could you check out the mechanism that I suggest in my post once I edit it, as the question seems to require a product.

This is only based on my experience. I certainly don't profess to be an expert on the subject and so there might be some other reaction which I am not aware of.

Thanks for answering anyway, but I wouldn't be suprised if one of the examiners made a mistake when writing the question because our chemistry deparment ain't always the best...


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Why are hollow cathode lamps used for atomic absorption spectroscopy?
In 1955, it was proposed that hollow cathode lamps are good emission sources for monochromatic radiation, and thus for atomic absorption spectroscopy (AAS).  In looking into the topic, I've read that "continuum radiation is not preferred in AAS." Why not?

Atomic absorption lines are very narrow.
For the Beer-Lambert law to be applicable, the bandwidth of the source should be narrow in comparison with the width of the absorption peak. Otherwise, the signal-to-noise ratio and the slope of the calibration curve would be low; the resulting sensitivity would be poor.
Continuous radiation sources use monochromators; however, even good monochromators have effective bandwidths that are significantly greater than the width of atomic absorption lines.
Therefore, the use of line sources with bandwidths even narrower than the width of absorption peaks is preferred.

One of the strengths of atomic absorption and other ionizing elemental analysis techniques using optical detection (ICP-OES is the main other one I know of) is that the optical absorption and emission properties at particular wavelengths are highly element-specific.
Despite this, the reasons for desiring monochromatic, element-specific AAS light sources are as follows:

Most detectors used in AAS / OES are broad-spectrum: they report just about any light that strikes them within their respective ranges of sensitivity, which usually span several hundred nanometers.
While the "fingerprint" wavelengths of each element are highly specific, they often sit at wavelengths close to those of other elements.
In AAS, any light that is passed through the sample has the potential to be absorbed by anything present in the sample.

Therefore, the use of a continuum light source in AAS would obliterate the analytical specificity afforded by the technique: anything present in the sample that absorbs light in the region of overlap of (a) the emission wavelength range of the light source and (b) the wavelength range of sensitivity of the detector would be recorded in a given measurement.

Yes, thanks but I need a more specific answer than the oliteration of analytical specificity by the continuum light source, for example: how does it have to do with the optical width ? Optical width of the absorbtion line is small, 10 times less than the optical width of the monochromator. Continuum sources in tiny spectral regions can affort not so much energy. I don't clearly understand these relations above I read in my book.


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Endothermic Conversion of Liquid to Gas
When a liquid (in my case I used butane) under high pressure is released into the air, it becomes gaseous, and the container seems to get colder. Why is it that turning a liquid to a gas requires energy when it was simply pressure that was keeping it a liquid? Where does it get this energy? Would the gas formed be colder, or no?

Somewhat related http://chemistry.stackexchange.com/questions/14273/liquefying-butane-in-a-freezer

This is probably a bit of a simplistic explanation.
Due to the butane being released from the from the container, the pressure in the container decreases, which in turn, decreases the temperature of the container according to the Pressure-Temperature Law (also known as the Admonton Law), that states for an ideal gas:
$$P  \propto  T$$
The boiling point of butane is normally −1 °C at normal pressures – so in typical use at room temperature, it should have already vapourised. But, due to it being under pressure, it remains a liquid, undergoing evaporation in the container. So, butane has a vapour pressure that increases with temperature, as seen below:

Image source: Wikipedia, who sourced it from the CRC Handbook of Chemistry and Physics 44th ed
Even though butane (like all gases) is not an ideal gas, we can still surmise that if you drop the pressure (by releasing the gas), then the temperature of the gas also drops, especially considering the combined pressure due to that from the container and vapour pressure is rapidly released = rapid depressurisation = rapid loss of temperature, which is felt on the outside of the container.

Thank you for that, Loong - could not figure out how to do the degrees symbol

The butane I released was collected in another container. Would the gas I collected also be cold?


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Is chirality a physical or a chemical property?
In my opinion, it should be both, because it will affect the way in which a molecule will interact with plane polarized light (physical), but also affect chemical interactions such as receptor binding.
On the other hand, consider $\ce{CHDBrCH3}$, which is chiral, as there 4 distinct groups attached to the central carbon. This would insinuate that chirality is a purely physical property, as despite deuterium having the same chemical properties as hydrogen, we are able to form a chiral molecule.

I would not consider chirality as a property--either physical or chemical--it is part of the definition of a molecule's structure, which in turn gives rise to the properties it exhibits (rotation of polarized light, melting point, reactivity with X, etc.). In your example, the fact that hydrogen and deuterium are almost (but not entirely) chemically equivalent will make the substance behave almost the same as its dihydrogen analog, and the two mirror image isomers nearly indistinguishable by physical or chemical means.


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Increase in Potential Energy When Bonds Are Broken
Why is it that the potential energy of a system increases when the bonds present in it are broken? (i.e., when a solid turns into a liquid.) I thought the potential energy of the system was mostly contained in these bonds? If these bonds are broken, then why does the system’s potential energy increase?

First, changing a solid into a liquid does not involve breaking bonds in the normal sense of the word, i.e. covalent bonds. Mostly intermolecular interactions are broken as heat is put in to melt something. These may be hydrogen bonds e.g. water, $\pi-\pi$ interaction e.g. benzene or dispersion as in cyclohexane and similar non polar but (slightly) polarisable molecules.
Starting with free atoms, when these are brought together and a molecule is formed then energy is released to the surroundings and a stable molecule is formed, say, H$_2$, or HCl etc. The figure shows the potential energy plotted vs atom separation. The energy is zero when the atoms are separated at infinity; in reality about 10 bond lengths is equivalent to infinity.

To dissociate the molecules, potential energy has to be added as shown by the dissociation energy. Any energy above this is released as kinetic energy of the atoms.
The heats of formation of different types of molecules have a wide range of values. Ignoring vibrational energy, which is small compared to bond enthalpy, this is in effect moving from the total minima in the potential wells of reactants (of the whole molecule not just a bond) to the minima in the products. This may involve absorption or release of heat to the surroundings and so may involve gain or release of potential energy overall. It just depends what the reaction is. (In calculating the Free energy changes entropy changes have to be considered also.)

It is not true that "potential energy of a system increases when its bonds are broken." For example, there is less potential (chemical) energy when the bonds of acetylene are broken.
In an exothermic reaction, energy is released (often as heat) as bonds form. For example, heating ammonium chloride (requiring input of thermal energy), it decomposes to ammonia an hydrogen chloride; when these reactants recombine, they release that energy.
$\ce{NH4Cl <=> NH3 + HCl}$

"For example, there is less potential (chemical) energy when the bonds of acetylene are broken." Isn't breaking of individual bonds always endothermic - so that the potential energy of the particles will rise? Of course, the net energy of the particles can decrease through bond formations, but that's another matter.


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Why does an acid have a low pH value?
When study biology in class, I don't know how my teacher get to the topics on pH value. So, she says that pH stands for percentage hydrogen. Then, I thought there is something strange about it cause acid has a higher percentage of hydrogen then base. But why does an acid have a low pH value? Is there any mistake on the pH definition or my understanding?

Just a fact: The $p$ in front of something as in $pH$, literally means, "take the negative log of." This is why we have those cool identities, $pH+pOH=14$, and using log rules we get $[H^+][OH^-]=10^{-14}$. If you have $pABCD$, it means $-\log (ABCD)$.

Actually in pH the p is for potenz or power of hydrogen. The concentration of proton $(\rm H^+\equiv H_3O^+)$ in water is often $10^{-3},10^{-5},10^{-14}$ which is very tedious to write, so pH is defined as $3,5,14$ for these solutions. Or:
$$\rm [H^+]=10^{-pH}\iff \log_{10}[H^+]=-pH\iff pH=-\log_{10}[H^+]$$
Note that pH scale and proton concentration works in reverse direction, one increases so the other decreases.
The pH Scale at normal room temperature is from 0 to 14 (debatable) and 7 is neutral (no excess proton or no deficiency of proton) and 0-7 is acidic (excess proton) wheras 7-14 is basic (deficiency of proton). This is defined because purest form of water at this temperature has pH 7 and pH more than this (and hence concentration of proton less than this) is caused by bases which take up proton and and pH less than this (and hence concentration of proton more than this) is caused by acids which release proton.

I wouldn't say it's "debatable" whether the pH scale in water goes from 0 to 14 at room temperature; confining aqueous pH to those limits is objectively incorrect. There are many examples of aqueous solutions with pH lower than 0 or higher than 14. They're just not something the average person will come across because they of course present significant safety hazards. I make a point of bringing attention to it because it's a common misunderstanding. See, for example, this Chem.SE question.

Also, the p in pH is not potenz. There hasn't been an accurate definition of  it, it is very debatable.

@Asker123 you would like to see this

Mais, non! C'est ne pas potenz, mais le puissance Hydrogen. http://www.sciencehq.com/chemistry/ph-scale-puissance-of-hydrogen.html

I am very old and maybe something changed : pH ( nothing to do with % ) used to be " negative the log ( base 10) of the hydrogen ion concentration in g-moles per liter of water" . Because it is a negative, the smaller the pH the greater the number on hydrogen ions. So pH of 1 means a lot of H ions and pH 13 means much fewer H ions.


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Dichromate ion: Tetrahedral or bent?
I am currently studying about the vespr theory and I'm stuck on an ion which is dichromate (VI) ion $\ce{Cr2O7^2-}$. So this is what I am thinking about:
1) If I make the oxygen atom at the middle the centre atom, it is a bent.
2) If I make the chromium atom thecentre atom, it is a tetrahedral.
So, my problem is that whether dichromate (VI) ion bent or tetrahedral?

Well, the overall shape of the molecule doesn't get anything close to bent and/or tetrahedral. I suggest doing some readings about hybridization and MO basics, rather than sticking to most-beloved, most-loathed VSEPR.

I am going to start reading about hybridization after Vespr

@MARamezani The VSEPR model should be digested in a very careful way - it predicts some molecular shapes correctly for the entirely wrong reasons. And for those reasons it is also often very wrong.

Both, Bent about Oxygen and tetrahedral about chromium:
$\hskip2.5in$

So, how do I call it? Tetrahedral bent?

@Simon-Nail-It nothing nothing I suppose


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Thermodynamics and Coffee
Heyo, I'm sure you all know of the experiment where you take a mixture of ice and water, plop some salt in there and the temperature of the solution progressively drops. The salt increases the entropy of the liquid state, shifting somewhat the ice/water equilibrium towards water. As ice is converted to liquid water, the endothermicity of this process leads to a decrease in the temperature of the solution.
Now, I was wondering if the reverse would ever be possible, say you were to have a thermos flask which would be sealed. It would contain Hot Tea and have a slot where one could pour in salt grains without releasing steam. By the logic in the previous question, won't some of the water vapor move to the liquid state (exothermic), resulting in the heatin up of your already hot tea. If possible would this change even be measurable or useful in any sort of way?

You're exactly right: the phenomenon is called boiling-point elevation:

Boiling-point elevation describes the phenomenon that the boiling point of a liquid (a solvent) will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent. [...] It is an effect of the dilution of the solvent in the presence of a solute. It is a phenomenon that happens for all solutes in all solutions, even in ideal solutions, and does not depend on any specific solute-solvent interactions. The boiling point elevation happens both when the solute is an electrolyte, such as various salts, and a nonelectrolyte. In thermodynamic terms, the origin of the boiling point elevation is entropy and can be explained in terms of the vapor pressure or chemical potential of the solvent. In both cases, the explanation depends on the fact that many solutes are only present in the liquid phase and do not enter into the gas phase (except at extremely high temperatures).

The empirical method for computing boiling point elevation for a given solute is described here. In short:

$\Delta T_b = \sum_i{n_iK_{b,i}m_i}$

Here, $\Delta T_b$ is the observed increase in boiling point, and $K_{b,i}$ and $m_i$ are the boiling point elevation constant and the molality (moles of solute per kilogram of solvent), respectively, of solute $i$.  The $n_i$ factor is the number of ions into which solute $i$ dissociates upon dissolution (e.g., $n_{sucrose} = 1$ and $n_{\ce{NaCl}} = 2$).
Your best bet for a full thermodynamic description is a college or graduate level textbook on the topic. I don't know offhand of any that specifically address boiling-point elevation, though.  The Wikipedia page on colligative properties actually has a bit of thermodynamic detail, providing the expression for computing the boiling point elevation constant from the enthalpy of vaporization $(\Delta H_{\mathrm{vap}})$ and other physical parameters:

$K_{b,i} = \frac{RM_iT_b^2}{\Delta H_{\mathrm{vap}}}$

where $R$, $M$, and $T_b$ are the ideal gas constant, the formula weight of solute $i$, and the boiling point of pure solvent, respectively.

Thanks for responding, but do you know if there is a method of calculating exactly how much a given amount of salt would cause the temperature to increase? I have a rough idea that would involve the gibbs free energy relating to the equilibrium constant and the enthalpy change of condensation...


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cupola based hydrocarbons
My favorite groups of shapes are cupolas. Are there hydrocarbons in the shape of these similar to how cubane is to a cube? If so how stable are they, and could you please give some general information about each one.

They would be highly unstable due to the inverted tetrahedral geometry of the carbons in the top face. Generally, a polyhedron can form the basis of a stable hydrocarbon if it has at most 3 edges for each corner. A big caveat is that the hydrocarbon will distort the shape if the angle strain is too great. I found that out in my computational research on prismanes.

So one does exist, the diagonal cupola exists.

Yes, that would be prismane

If one forgets about the correct mathematical notation for a moment:

a cupola is like a bowl
a bowl is a fragment of a spherical object

Molecules of this shape are found in the field of fullerenes (bucky balls).
One of the most simple bowl-shaped molecules is sumanene. If you like to play a bit with the structure, please fetch sumanene.cml from pastebin and open it in Avogadro.

The shape is more obvious in the following animation.

The structure of two extended bowl-shaped molecules has been published in
J. Am. Chem. Soc., 2011, 133, 16319–16321 (
DOI), and although they are not true cupolas, you might find them interesting anyway.
I've uploaded the structures bowl-01.cml and bowl-02.cml as CML files.


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Basic character of amines
I learnt that basic character is directly proportional to the positive inductive effect and inversely proportional to steric hindrance

Based on the above mentioned properties how is $\ce{(CH3)2NH}$ is more
  basic than $\ce{CH3NH2}$ while $\ce{(CH3)3N}$ is less basic than the
  secondary amine $\ce{(CH3)2NH}$?

Welcome to Chemistry SE! This looks like a homework question, please show your work into this, so that we are not doing the homework for you. :)

Actually to convey it to you..this is a MCQ question that i came across in a competitive exam :) the answer as i have mentioned above to my knowledge is (CH3)2NH i just don't understand as to why it has to be the secondary amine that is more basic than the tertiary amine or primary amine :)

@Asker123 I believe he did show sufficient work into it. He explained what he knows and if he didn't know anything about the solvation effect there is not much more he can say. I feel like this should be re-opened as it is a very valid question and one I have been wondering about myself

what we expect here would be the tertiary amine to have the greatest basicity and the least would be the primary amine.
But through experiments it has proven that secondary amines to have the greatest basicity. the order of basicity is as follows.
NH3 < primary amine < secondary amine
The basicity of the tertiary amine is approx. equal to that of primary amine.
The reasons behind this is:

Steric hindrance
Solvation of ions

3 + inductive effect
When the alkyl group attached increases the basicity reduces. It is because the size of the alkyl group is so large that it prevents an attack from Hydrogen ion. So according to this theory decreasing order would be
Primary amine > Secondary amine > Tertiary amine
But there's another thing you've got to consider. That is the solvation of ions. Amines dissolve in water to form protonated amines thus increasing the number of  possibilities of hydrogen bonding. More the H bondings, greater the dissolution in water. Higher the H bonding, greater is their stability and greater is their basicity
Experiments have indicated that primary amines are the ones who have more H bonds and tertiary amines are the least.
So by combining all those reasons scientists have shown that basicity of amines vary as follows.
NH3 < primary amine ~ tertiary amine < secondary amine

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Least evil way to make soap?
I'm not a chemist but something of a tinkerer and adult boy scout.  Currently if you want to make soap, you need fat and lye.  Fat I'm good with, it's easy to get and you can even get it from plants. Lye usually requires wood ash though and that I would rather avoid, if possible. If not possible and burning something is mandatory, then I want the biggest bang for my buck. That means I want to get the most lye I possibly can with the least pollutant release, especially greenhouse gasses and we shouldn't be thinking of just wood. Anything that will get us some low impact soap in the end. We may even want to replace soap as an idea with something else that you can make at home and wash humans, clothing and dishes with.

By far the largest source of lye nowadays is the electrolysis of concentrated solutions of $\ce{NaCl}$ or $\ce{KCl}$ in water, the famous and incredibly industrially important chloralkali process, producing millions of tons of $\ce{NaOH}$ and $\ce{KOH}$ every year. While electrolysis is an energy-intensive process, presumably the electricity could be supplied by a low-carbon energy source such as solar or wind power.
"Replacing soap", in the sense of halting the usage of surfactants, is not a good idea and likely not even possible. Surfactants are too useful a class of substances. In fact, soaps are likely the most environmentally friendly surfactants, given that they're based on a simple hydrolysis and acid-base reaction with common biomolecules, fatty acids and triglycerides. Other surfactants such as alkylbenzenesulphonates (dishwashing liquid), tetraalkylammonium salts (hair conditioner) and perfluoroalcoxylates, among others, can be toxic to aquatic lifeforms and take longer to degrade in the environment. The excessive usage and improper disposal of surfactants is a recognized environmental problem, but tackling soaps won't really get you far in solving the issue.

OK. Interesting. So what it sounds like is that there is a means by which you can get lye from salt IF you can figure out what to do with the Chlorine. Is that right?

The chlorine produced by the chloralkali process is hardly an unwanted byproduct; while dangerous, chlorine gas is a very important chemical feedstock, and ultimately has many, many uses, in the most diverse areas of manufacturing.


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Why does pyrite form cubic crystals?
Pyrite ($\ce{FeS2}$) forms cubic crystals, like these ones:

I know that the crystal structure is primitive cubic, but I don’t see how cubic structure on the molecular level translates to cubic structure on much larger scales. If you are given some number of cubes, there are many different ways to assemble the cubes face to face, and only one of these ways is to create a large cube. So why is it that the pyrite in this rock forms nearly perfect cubes?

How and why crystals have the 3D shape that they have is an active field of research. There isn't an adequate modeling theory as of yet. The mineral calcite has more than 300 crystal forms and thousands of crystal variations. http://www.galleries.com/calcite

Actually the is only one way how those little cubes can be fit together: the whole idea of unit cell is based on the assumption that the whole lattice is built up those units shifted according to translational symmetry.

Pyrite can actually come in a range of crystalline morphologies, as seen below:

Image source: Modelling nanoscale $\ce{FeS2}$ formation in sulfur rich conditions (J. Mater. Chem. 2009, 19 (21), 3389), who state that the shape is dependent on the concentration of sulphur during its formation.
Modelling of cubic pyrite crystals, as in the picture in your question, are explained in the article Modeling the Shape of Ions in Pyrite-Type Crystals (Crystals 2014, 4 (3), 390–403.), modelling the following structure:

From the caption, the blue spheres represent $\ce{Fe}$ and the ellipsoids represent $\ce{S}$. The shapes of the constituent chemicals used in this diagram is based on their modelling, specifically one of the conclusions the author makes is that:

It turns out for $p$ valence shell ions that an isotropic ionic radius only occurs on cubic lattice sites. For all other site symmetries, however, two or three radial parameters will apply. Appropriate geometric shapes for $p$ valence shell ions are thus given by ellipsoids instead of spheres.

The $p$ valence ion in this case being sulphide.  With this model in mind, the explanation provided in the caption for this diagram states:

The mesh inscribed into the sulfur ions now reveals the ellipsoidal compression along <111> directions. In this model, the number of contact points of $\ce{S}$ and $\ce{Fe}$ ions become four and six, respectively, and the ellipsoidal deformation is concluded to enable a stable packing.

The molecular shape does not necessarily mean that the pyrite crystal has a cubic shape, however if the conditions are right (pressure and temperature), and the composition is 'correct' - if the formation is in a iron rich environment (as per the first reference), then this molecular shape could express as a cubic crystal.

I don't see how this answers the question. I am not asking why does pyrite have primitive cubic structure on the molecular level. I am asking why cubic structure on the molecular level implies cubic structure on larger scales.

Note carefully that the dodecahedral form is not a regular dodecahedron.  For an example of a truly regular dodecahedralform see here.

Coincidence.
While pyrite crystallises in spacegroup $\mathrm{Pa\bar{3}}$, i.e., a cubic spacegroup, there is no guarantee that the unit cell shape is resembled in  the macroscopic crystal shape. The faces of the crystal do however always represent a plane that can be described by some Miller indices of the crystal lattice. And indeed, there are multiple alternative crystal habits for pyrite (eg dodecahedral, see Wikipedia article and santiago's answer), while retaining the same internal structure.
Which habit forms in practice, is mainly determined by the crystallisation conditions. These have to be determined empirically, as there is no way to predict them.

It's not coincidence obviosly because the probability that you arrange say a million cubes face-to-face and end up with a perfect cubes is basically 0. Still no one has managed to answer my question.

There is exactly one way to arrange the unit cells ("cubes")  - otherwise you would not have a crystal. This makes the internal symmetry. What I am saying is, is that this single internal structure can lead to to multiple appearances of the macroscopic crystal - and that there is no way of accurately predicting what that would look like ("coincidence"), even though it puts some limits to it.

Kinetics.
The crystal grows with the same speed in three perpendicular directions, by slowly starting new planes in each direction, which then fill much faster.
Why?
Quickly filling new planes minimises surface energy, so that seems logical. That a cubic space group has no preferred growth direction is not as compellent, but fits the bill in case of pyrite.
(Be careful: This rationalises the fact, but predictions, like the other answers say, are difficult.)

This is more what I'm getting at. It must be energetically favorable in some way to have a growing cube. But this seems hand wavy. What about a sphere?

A sphere with diameter x made out of cubic building blocks has very nearly the same surface area as a cube with sides of length x, but significantly less volume. That is very unfavourable. You only get spheres out of an amorphous mass, never crystalline material.

All completely pure primitive cubic crystalline compounds will form macroscopic, perfect cubes. Deviations in macroscopic form are the result of impurities somewhere in the molecular lattice, otherwise a low cohesive energy such that the chunks of material don't stay together well (e.g. by electrostatics). Perfectly pure $\ce{NaCl}$ forms cubes likewise, but smaller ones, because the cohesive energy is lower (if I'm not mistaken). Thus the energetic affinity of a material for itself (over available impurities) increases probability of perfect macroscopic formation that models the crystal unit-cell. Most compounds attract foreign species though, so nicely perfect cubes of any cubic-structure chemical are hard to find -- even of pyrite, as mentioned in the other answer. Lots of materials form surface oxides before expanding a macroscopic version of the unit cell; others bond with similar ionic "pieces" -- e.g. contamination of $\ce{K+}$ in $\ce{NaCl}$ to form chunks of rock salt, $\ce{KCl}$.
$\ce{NaCl}$ crystal is used as an optic barrier for samples in infrared spectrometers because when pure it can be shaped linearly in any dimension to form any desired (semi-)planar shape with atomic precision. Effectively you can remove one atomic layer at a time because it's a pure two-component cubic macro-system.
I remember when I traveled to the Dead Sea in Israel -- there were really nice, big (like 1-inch), perfectly formed salt cubes on the shores. I have some pyrite at home too. It's cubic but some chunks form out in different proportions and directions due to impurities/chemical environment.
To enhance the point -- even more complex but still cubic unit-cell materials make macro-cubes, if pure of foreign material, e.g. MOF-5 and others:


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What prevents weak acids from dissociating completely?
What prevents weak acids from dissociating completely in water?

I'm voting to close this question as off-topic because a lack of evidence of research effort.

possible duplicate of Why are weak acids weak?

To give a straight answer: intramolecular forces.
Take two examples, $\ce{HCl}$ and $\ce{HF}$. We all know that $\ce{HCl}$ is a strong acid and it dissociates completely, while $\ce{HF}$ doesn't. This is because of the strength of Fluorine and Chlorine differs. Fluorine is strongly bonded to Hydrogen so less of it dissociates.
Another reason is position which is mostly found in oxoacids.

I'll format it later, don't have time.

Ok so the $HF$ intramolecular forces are strong, but this doesn't explain why some $HF$ molecules dissociate and some don't.

I answered that part for you!

HF is almost completely dissociated, but fluoride binds hydronium.

It seems that your fundamental misunderstanding is that some molecules will dissociate and some will not.  Who is to say that they do not dissociate?  In fact, virtually every single molecule will dissociate in solution, BUT here's the catch: the same holds for protonation of the conjugate base.  In reality, these molecules are all constantly dissociating and reprotonating, and the relative stability of the acid and conjugate base, represented by equilibrium constants, determines the rate of the forward and backward reactions, which determines where the forward and backward reactions proceed at the same rate.  This is the fundamental concept of equilibrium in a reversible process, that the system is constantly in flux, but there is a point where the overall concentrations remain constant.
What we term "strong acids" are simply acids with pKa values so low that they completely dissociate and any time a conjugate base is reprotonated it almost immediately dissociates again.  There are also classes of acids that are "near strong acids" and "weak acids" with increasing pKa values.

The Acid (HA), which calls as weak, will produce an ion (A-) by dissociation. According to Bronsted-Lowry acid-base theory, the produced ion (A-), which is called as Conjugate base of acid (HA), is unstable when its respective acid (HA) is weaker one.
Because of the instability, the conjugate base (A-) involves in backward directive reaction to form its parental acid (HA) which is in non-dissociated form. In other words, there could be equilibrium in between non-dissociated acid (HA) (reactant in forward direction because of the ability to donate H+ ion) and conjugate base (A-) (reactant in backward direction because of the instability).

That is the reason there is no complete dissociation occurs in weak acids.
    Apart from this, there could be the situation in which the weaker one is comparative term. I mean, in the presence of strong acid (H2SO4) the weak acid (CH3COOH) is in the state of non-dissociation form that is because of ‘Common-ion effect’. In this case also there is no complete dissociation from weak acid.

according to ostwald dilution law, there is relationship between dissociation constant and degree of dissociation of weak electrolyte. Since, weak acids have very small dissociation constant ('k' value ) , so they do not donate all of its hydrogen ion (H+).That's why , they do not dissociate completely.


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Why does opened wine develop the sour taste and smell of vinegar?
An opened bottle of wine developed into a sour taste and smell of vinegar. Why?

It's acidic bacterial waste from contamination.

See https://en.wikipedia.org/wiki/Acetobacter_aceti

The ethanol (alcohol content) of the wine, is oxidised, by the oxygen in the air into ethanoic acid. Vinegar is dilute acetic acid which is where it gets its smell.
The reaction happens in to stages,

The ethanol is partially oxidised to ethanal
The ethanal is further oxidised to ethanoic acid

Ethanol will not just get oxidized by air, not commonly. It's typically digested by bacteria in aerobic fermentation, either by contamination or unfinished fermentation. You can make red wine vinegar at homme by just leaving wine in a wide bowl or pan and adding some live vinegar.

Or oxidation involves first other compounds present in the wine. Saying that ethanol is oxidised by atmospheric oxygen, though practically true, is a rather poor answer in chemistry. Ethanol let to air evaporates without being oxidised. OP should rather refer to the comment by Stagg C. and this one of mine. I don't know the details, but the formation of some peroxides should happen first, than these oxidize the alcohol.

authors: Владислав Динул


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Smith Hieftje correction method
I completely don't understand how this method works. My book says that this method works by the principle of self-absorption which occurs by the emitted radiation of hollow cathode lamp when it works on high intensity of electricity, due to which we have more atoms who lead to an extension of emission lines. This is how self-absorption occurs.  Does anyone have any proper explanation for this?
What does this self-absorption actually mean?
How does that lead to a correction?

The Smith-Hieftje background correction is basically another means of "chopping" the signal. The amplifier locks onto the chopper frequency to cancel out the background noise. "Locking on" essentially means that the amplifier will measure the difference between the two signals.
So one way to chop the "light" from the emission tube would be to insert a "fan blade" to mechanically chop the light on and off.
The Smith-Hieftje background correction is a bit more complicated. Basically what happens is that this emission tube has two parts. A heater to vaporize the source of the atoms in the tube and an electron beam to stimulate an excited state. In this type of emission tube the heater is pulsed high and low. When the heater is low it produces enough metals atoms so that the atomic line emission is created. But when the heater is on high heat then it vaporizes so many atoms that the vapor cloud of metal atoms self-quenches the emitted atomic line emission. The high density cloud of atoms also means that the atoms bump into one another. So instead of the atoms in the cloud all having about the same kinetic energy they have a range of kinetic energies which results in line broadening.

High intensity of electricity added to the source ( $\ce{HCL}$ ), gives us more atoms, and when we have more atoms then the chances for self-absorption are higher. One atom will get excited and then turning back will emit radiation, the other atom from the same radiation will absorb that emission. In the beginning, the instrument measures the intensities of radiation at normal intensity of electricity. After that, it makes the measurement on higher electricity. The difference between them, should give us the background absorption.

I'll stast by explaining what a HCL is and how it works, and then move on to background correction.
A hollow cathode lamp (HCL) is a hollow glass tube filled with low pressure of some noble gas (Neon or Argon), inside the glass tube there is a cathode and an anode. The cathode is made with the metal you are interested in measuring.
The lamp works as follows: A potential difference is applied between the cathode and the anode, such that the cathode has a lower potential than the anode. The potential must be high enough to ionize part of the gas that fills the tube, the positively charged cations accelerate towards the cathode, while the electrons fly towards the anode. When the cations hit the cathode some of the metal atoms in its surface detach, afterwards, this detached atoms become excited due to collisions with nearby atoms and release this excitation energy as photons (light). The photons emitted are, in principle of different single specific wavelenghts depending on the energy difference between the excited and base states, however, there are some phenomena (mostly the fact that there are other nearby atoms that slightly shift the energy levels, and the Doppler effect) that contribute to the single wavelenghts being instead small intervals of wavelenght. This is called "Line Broadening". This line broadening is more apparent if there is a higher concentration of atoms in the vicinity of the emitting atom, or if temperature is higher.
If a higher potential difference is applied between the cathode and the anode, then a higher amount of atoms is emitted from the cathode, because cations impact the cathode with a higher kinetic energy. This high concentration of atoms means two things: First, its very likely that the emitted radiation will be reabsorbed by some nearby atoms, this has the effect of lowering the intensity of the "lines" emitted by the HCL. Second, a high concentration of atoms and higher temperature (atoms collide with higher kinetic energy, wich implies higher temperature) will mean that the lines will broaden considerably (a link to an image is provided at the end of the answer)
The Smith-Hieftje background correction works as follows: First the absorption by the sample of the lines emitted by the HCL is measured normally by the detector. This includes the absorption by the analyte and the background absortion due to matrix effects, etc. Then a pulse of high voltage is applied to the HCL, so that the emission lines are broadened and the center of the lines is self-absorbed by the atoms in the lamp as described previously.
Because the analyte can only absorb the wavelenght that was already self absorbed by the HCL, the analyte won't absorb anything. The matrix and other impurities, however, will absorb the wavelenghts of the broadened line (background absorption typically has the form of wide absortion bands, not single lines, so taking out a single frequency doesn't affect it much, think continuum background correction with deuterium lamps).
Now one has both, the value of the absortion of the analyte + background and of the background. The only thing left to do is substract them, which is done automatically. This process is repeated periodically, during the duration of the measurement (after the pulse of voltage is made, a small interval of time has to pass so that the detached metal atoms in the HCL can deposit themselves again in the surface of the HCL and don't interfere with subsequent measurements).
Some sources:
www.agilent.com/cs/library/applications/a-aa14.pdf (hollow cathode lamps)
https://books.google.com.ar/books?id=eTz6AwAAQBAJ&pg=PA478&lpg=PA478&dq=Smith-Hieftje+background+correction&source=bl&ots=Rst6iNgPM6&sig=2JOZBeSiYOutOOzSuZg2RNqSNls&hl=en&sa=X&ved=0ahUKEwizjPSi2Z3LAhXLh5AKHYL4Cn4Q6AEIUDAI#v=onepage&q=Smith-Hieftje%20background%20correction&f=false (a neat book, not specifically on AAS but has a nice explanation of background correction, check page 478 for an image of a self-absorbed, broadened emission line).

All good answers, and I am way late, so I will just post two figures (from optical calculus simulations) that help illustrate how the Stan Smith and Gary Hieftje background correction works. First, the HCL and absorber spectral profiles for Ca at the usual detection wavelength:

First, note the shift due to the source and absorbers being at different temperatures. Second, the HCL profile is narrower. Third, it is OK that the HCL in not monochromatic or perfectly aligned with the absorber profile.
The next figure shows the HCL profile and the absorber profiles at low HCL current and at high HCL current:

The self-reversal and broadening of the HCL band is clear. So now the background correction concept should be clear, as nicely discussed in the previous answers.


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Alkalinity in Water
If you add Alum - $\ce{Al2(SO4)3}$, to a clay water sample which contains $\ce{Ca(HCO3)2}$ - due to the water's alkalinity, the pH of this water sample would drop as Alum drops the pH levels. How would you counteract this imbalance. As in, what could you add to the sample to keep the pH the same as it was before the addition of Alum.

If you've lost your original browser cookie, you can use the Contact Us link at the bottom to recoup your account.  Then you can edit your post without having to leave an answer.

Just to make sure: aluminium sulfate - Al2(SO4)3 is not considered as "Alum". Alums are double sulfate salts, like KAl(SO4)2.

One thing I can suggest are buffers which maintain the pH of a solution within a certain limit when either acid or bases are added. These often are based around the concept of weak acids or bases. See this.

authors: يقينى بالله يقينى


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When will organometallics act as base or nucleophile?
I would like to ask if the hydrogen next to carbonyl group is acidic,and organometallics are strong base, when will it act as a base and extracts the acidic proton and when will it attack the carbonyl carbon?
Also I would also like to ask, for bases when will it attack the acidic proton to form a enolate and when will it attack the carbonyl carbon directly to form an hemiacetal/hemiketals?

Grignard is a base as well as an nucleophile. When there's a strong electron deficient group it'll attack there, when there' a very acidic group it'll attack there. For other bases it depends upon their relative basicity and nucleophilicity. BAsicity is thermodynamically related while nucleophiicity is kinetically decide. They're silent about each other. You need to practice and gather examples and an experience too :D

I like this answer! Especially the delineation between thermodynamics and kinetics,

@Dissenter happy to hear a word from from you since a long time :D

I will add that acid-base reactions tend to be kinetically favored over other reactions.

@Dissenter since formation of water has a equilibrium constant $10^{14}$ :D


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Understanding FHF- molecular orbital diagram?
I need help understanding the molecule FHF-'s molecular orbital diagram:
So I know that "ag" refers to the 2pz and 2s orbitals of fluorine atoms...but I am not sure why "ag" is labeled for the topmost orbital level? I don't really know what that is...but it's interacting with both hydrogen's 1s orbital and fluorine's 2p orbital, so does it imply that it's either antibonding or bonding orbital? And the one right below the topmost orbital--would that be the antibonding sigma 2pz orbital?
Thanks, I'd appreciate the help!

So we are treating the two fluorines as ligand groups on the right hand side and we introduce a Hydrogen on the left. We observe that each fluorine has a 2S, 2px, 2py and 2pz. If we take the 2pz as being along the internuclear axis then we assign it with sigma symmetry, while the x and y p orbitals have pi symmetry.
It is the 2pz that can interact with the Hydrogens 1S because it has the correct symmetry. It turns out that the symmetry of that orbital is ag. This creates as you say the bonding and anti bonding pair with the H's 1S. It is important to note that the bonding and anti bonding pair are of the same symmetry.
The hydrogen only has the 1S orbital (symmetry ag) and no others, so every other orbital of a symmetry that is not ag is non-bonding since it has no respective partner in the FHF environment.
With regard to the assignation of symmetry to an orbital you must have an understanding of SALCs (symmetry adapted linear combinations of molecular orbitals), if you perform a projection operator method then you will generate the symmetrys ... however this is a bit advanced and not something expected at undergraduate level (to my knowledge).
I hope that helps :)


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Chemical based Water Purifier
I am currently researching as too which is better for purifying a water sample. I have obviously tried the standard Google search for 'Aluminium Sulphate or Ferric Chloride' and have come up with lots of results. The only problem, is that most of these results are based more around peoples opinions or psuedo-scientific observations. So, to the question:
What chemical reasons are there for choosing Aluminium Sulphate over Ferric Chloride to refine a water sample?

On addition of inorganic coagulants like aluminium sulfate(alum) or Iron salts causes several chemical and physical interaction take place among particles.
This inorganic coagulants neutralize the negative charge on particles. Also metal hydroxide precipitates of the aluminum and iron salts ions begin to form.
This precipitates combine to form larger particles under natural processes such as Brownian motion
Aluminum hydroxides are formed within a fairly narrow pH range, typically: 5.5 to about 7.7. Iron (III) hydroxides can form over a larger pH range including pH levels lower than are effective for alum, typically: 5.0 to 8.5

Source wikipedia

Can you explain the Brownian motion?

@BobDylan I have added link and also updated source link. Please check out both the link.


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Peroxide effect in reagents apart from HBr and CHCl₃
Is the peroxide effect observed only with addition of hydrogen bromide and chloroform? If so, why not with other reagents in electrophilic addition on alkenes?

What is the "Peroxide effect"?

@Jori  anti markovnikov  rule

Is the peroxide effect observed only with addition of hydrogen bromide
and chloroform?

No, the "peroxide effect" or "anti-Markovnikov" addition of a free radical to an unsaturated bond is observed with other compounds too, $\ce{CCl4}$ is another  example.
Also, just to be clear, these reactions involve the addition of radicals ($\ce{X^{.}}$), not electrophiles, to alkenes (or alkynes).
Let's examine the radical addition of $\ce{HX}$ to an olefin.  I'll use a general method that you can later use to examine any radical addition to an unsaturated bond.
There are two propagation steps that are key to our analysis:
$$\ce{X. + H2C=CH2 -> X-H2C-CH2. ~~(1)}$$
$$\ce{HX + X-H2C-CH2. -> X-H2C-CH2-H + X. ~~(2)}$$
If two halogen radicals combine, or if a halogen radical and a halo-alkyl radical (formed in the propagation steps) combine, the reaction is terminated. The propagation steps are in competition with the termination steps. If the propagation steps are endothermic, then they compete less effectively with the termination steps and the radical addition will not proceed in a useful manner.
In propagation step (1) we are breaking a pi bond and making a $\ce{C-X}$ bond, let's look at the thermodynamics of this  step.
\begin{array}{|c|c|c|c|} \hline
\ce{C-X} & \text{Bond Strength} & \text{Pi Bond Strength} & \text{Overall}  \\ & \text{(kcal/mol)} & \text{(kcal/mol)} & \text{(kcal/mol)}\\ \hline
\ce{C-F} & 116 & 66 & -50\\ \hline
\ce{C-Cl} & 81 & 66 & -15\\ \hline
\ce{C-Br} & 68 & 66 & -2\\ \hline
\ce{C-I} & 51 & 66 & 15\\ \hline
\end{array}
We see that this step is exothermic for the first 3 hydrogen halides, but is endothermic for the $\ce{HI}$ case.  Already we see that hydrogen iodide will not add to olefins via a radical pathway at room temperature.
In propagation step (2) we are breaking an $\ce{HX}$ bond and making a $\ce{C-H}$ bond.  Let's perform a similar thermodynamic analysis of this step.
\begin{array}{|c|c|c|c|} \hline
\ce{H-X} & \text{Bond Strength} & \ce{C-H}~ \text{Bond Strength} & \text{Overall}  \\ & \text{(kcal/mol)} & \text{(kcal/mol)} & \text{(kcal/mol)}\\ \hline
\ce{H-F} & 135 & 99 & 36\\ \hline
\ce{H-Cl} & 103 & 99 & 4\\ \hline
\ce{H-Br} & 88 & 99 & -11\\ \hline
\ce{H-I} & 71 & 99 & -28\\ \hline
\end{array}
(source for bond strengths)
Both propagation steps (1) and (2) must occur for the reaction to proceed.  Only in the case of the addition of $\ce{HBr}$ is the reaction exothermic for both steps.  For the hydrogen halide series, only $\ce{HBr}$ will add to an olefin by a free radical mechanism.

Your answer is wrong because Markonikov addition $\ne$ Peroxide Effect. And I would be happy if you could provide some source for CCl4's addition

Plus CCl4 is not an electrophile.

Ok Now I can agree, but still I learnt that Peroxide -anti markonikov, please check it again.

Sorry but you gave wrong lectures, I am really sorry but please see where it is written "Mechanisms that avoid the carbocation intermediate may react through other mechanisms that are regioselective, against what Markovnikov's rule predicts, such as free radical addition. Such reactions are said to be anti-Markovnikov" I hope you don't argue before checking, sorry.

@ADG You are correct, it shoulkd be anti-Markovnikv.  I have corrected the answer.  Free radicals are generated and a free radical addition to a double bond will follow when peroxides are used to initiate a reaction. Such reactions result in an anti-Markovnikov addition pattern to the double bond. Therefore terms like "peroxide" or "anti-Markovnikov" are often used to describe them. None of these reactions involve electrophiles, they involve free radicals. Here is a link to an example of a free radical addition of $\ce{CCl4}$   to a double bond. Look about half-way down the page.

What has the fact the reaction is exothermic or not to do with whether it occurs? I thought it's under kinetic control, so shouldn't it occur just if it is fast enough (if it is endothermic one only has to supply the required heat at a constant temperature)?

I added a paragraph to explain this, basically the propagation steps and the termination steps are in competition.  If the propagation steps are endothermic they do not successfully compete with chain termination and the radical addition pathway is shut down.

How exactly does the endothermic nature of the step affect how it competes?

Endothermic means uphill, energy is required for the reaction to occur.  It will not proceed spontaneously.

Is the peroxide effect observed only with addition of hydrogen bromide
and chloroform?

No, there are in fact a few more of them that I know. For example, $\ce{CHBr3}$, $\ce{CCl4}$ and $\ce{CBrCl3}$.

Reagent
positive part
negative part

$\ce{CHCl3}$
$\ce{H+}$
$\ce{CCl3^-}$

$\ce{CHBr3}$
$\ce{H+}$
$\ce{CCl3^-}$

$\ce{CCl4}$
$\ce{Cl+}$
$\ce{CCl3^-}$

$\ce{CBrCl3}$
$\ce{Br+}$
$\ce{CCl3^-}$

Some useful examples:
$$\ce{R-CH=CH2 ->[\ce{H-CCl3}][R-O-O-R]R-CH2-CH2-CCl3}$$
$$\ce{R-CH=CH2 ->[\ce{Br-CCl3}][R-O-O-R]R-CHBr-CH2-CCl3}$$

Note:

Apart from using H₂O₂, peroxide effect (or Kharasch addition)
is also observed when other peroxides like R₂O₂ or (RCO)₂O₂ are used.
Wikipedia's article on Morris S. Kharasch is worth reading to understand peroxide effect.

Welcome to CH SE site!
Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.
Suitable formatting can be done according to formatting math/chem expressions/equations.

Thanks @poutnik, but i think I have jotted down all of the things given in the screenshot, so it must contribute to a better search result...

You could for the table easily use the ---|---|---  SE Markdown syntax.

In addition to @Poutnik 's comment, I don't understand what meaning five (six counting the background, which is for some reason not transparent/white) different colors are supposed to convey. In scientific writing every mean of formatting signifies something. Random colors only distract. Also, please refrain from abbreviations such as "+ve"/"-ve", they are illiterate, confusing and the latter doesn't even make semantic sense since you've used a hyphen, not a minus (cf. -,−).

@andselisk well, it is just a screenshot from my class notes... And my organic chemistry teacher likes to make colorful eye-catching notes... Btw, i am removing the image.


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Determining enthalpy of neutralization and heat capacity of calorimeter by overlying with straight lines or "eye-o-metric method"
Below is a model example of how to interpret gathered data. Temperature on y-axis, time on x-axis.
We are supposed to read  (temperature change by neutralization) and  (temperature change by resistive heating - to determine heat capacity of calorimeter) by overlaying the selected area with straight lines.
What I don't get is the way the straight lines are put the way they are. Maybe the bottom one: we expect the temperature drop caused by heating the calorimeter will continue by this straight fashion so we must take it into account in subsequent temp change reading.
But what about the upper line? Why read the upper value above the graph and not on its local maximum?

Ah, I understand now. It's of analogous reason to the first case. By adding heat during a time interval we expect that some of it will be absorbed by the calorimeter.


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Electronic oxygen concentration sensor
I bought an electronic meter which measures the concentration of oxygen in air. But I think the sensor has failed because it is too old, as it had a crusty substance coming from it.but I can no longer find spare sensors for it.
How might this sensor have worked? I presume the property of some substance was measured; this property changes with the concentration of oxygen.
Perhaps one of these:

Its resistance changed as air was passed through some sort of chemical, and the resistance measured to determine the oxygen concentration.
Or perhaps its opacity to light measured with an LED and an appropriate optic sensor.
Or the volume of the substance changed, or some other property, which ultimately varies the capacitance.

What substance might this have been? I want to make my own device - I don't intend to fix this one, so I don't need to identify the substance used in this particular sensor.

I cannot say as to your specific sensor, but most oxygen sensors are electrochemical sensors built around concentration cells.
Take a look at the following diagram of a simple zirconia-based oxygen probe from Wikipedia. This type of sensor is very like the one in automobiles.

In the middle of the sensor is a gas permeable zirconia $(\ce{ZrO2})$ membrane between two gas permeable electrodes. The membrane and electrodes form the electrochemical cell. In many cases, an electrochemical cell will use a reference electrode with a standard reaction going on for comparison. In this sensor, the standard reaction is the same one that is going on at the other electrode. The difference is that at the standard electrode, the partial pressure ("concentration") of oxygen is known and at the other electrode it is unknown.

But wait, how does that work?

Let's take a generic electrochemical process, where $\ce{M}$ in the following equation represents something generic that we can oxidize:
$$\ce{M(s) + O2(g) -> MO2(s)}$$
This cell has a standard potential $E^\circ$. The actual measured potential, however, will depend on the partial pressure of oxygen $P_{\ce{O2}}$ because the standard potential is defined at $P_{\ce{O2}} = 1\ \text{atm}$. The relationship that gives us the actual electrode potential at some other oxygen level is the Nernst equation;
$$E=E^\circ - \frac{RT}{nF}\ln{Q}$$
Where:

$R$ is the ideal gas constant
$T$ is the temperature
$n$ is the number of electrons transferred (4 in my example)
$F$ is the Faraday constant
$Q$ is the reaction quotient, in this case $Q=\frac{1}{P_{\ce{O2}}}$

If $P_{\ce{O2}}$ is large, then $E$ is small and vice versa.
Now, we have two electrodes, our unknown and our reference, each with their own potential:
$$E_{ref}=E^\circ - \frac{RT}{nF}\ln{\frac{1}{P_{\ce{O2},ref}}}$$
$$E_{unk}=E^\circ - \frac{RT}{nF}\ln{\frac{1}{P_{\ce{O2},unk}}}$$
The voltmeter in the device measures the difference between these two potentials and does some maths to calculate back $P_{\ce{O2},unk}$ and then convert it to concentration or percent. Note that it does not even matter what $\ce{M}$ and $\ce{MO2}$ were, since they are not in the equation and even $E^\circ$ cancels out.
$$E_{unk}-E_{ref} = \frac{RT}{nF}\ln{\frac{P_{\ce{O2},ref}}{P_{\ce{O2},unk}}}$$

What substance might this have been?

If your sensor was this kind of sensor, the crusty mess is probably zirconia.

authors: حمزة إبراهيم


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Why most excreta are brown?
This might sound funny but my friend asked me why most - if not every - faeces are of a brown or brownish color?
Why not grey or slightly bluish or green for example?
Is there any simple scientific explanation for this?
Thanks

The brown color of faeces is mainly due to the presence of bile and bilirubin.  Bilirubin is a breakdown product of haemoglobin, which is destroyed by the body as part of red-blood-cell turnover.  The half-life of a human red blood cell is about 120 days.  Faeces may be green if bilirubin is not being produced (although it has never happened to me)

This question more likely belongs to Biology SE

@TomD you should consider adding this as an answer!

@TomD: I don't think green - green would be the primary decomposition products of haemoglobin decomposition which are excreted from the bile. Sometimes these can go back to the stomach and are vomited with greenish color, but  when they pass the colon especially the large intestine they get oxidized to the brown coloured product. If the bile gets stuck from some liver infection or stones or the like and no billirubin enters the colon the faeces appear yellowish to white - not green.

@Franky_GTH.  Biliverdin, a key breakdown product of heme, is certainly green, and NADPH-linked biliverdin reductase plays a key role in converting this green catabolic intermediate to bilirubin (which is excreted).

@TomD: No objections against that! The green color of the bile excretions come from that. Just the brown color of the faeces is not bilirubin its according to what I know decomposition products of bilirubin.


text:
Why so differance in alkali halide's crystal forms?
I always wonder that though Both $\ce{NaCl}$ and $\ce{CsCl}$ are alkali halides, why they adopt different crystal forms, FCC and BCC respectively? and I have a doubt about $\ce{CsCl}$ structure that Why the $\ce{Cs+}$ or $\ce{Cl-}$ ions are not located at voids/lattice points as like as $\ce{NaCl}$ structure?

Perhaps you have heard of the following terminology: "rock salt", "cesium chloride", "zinc blende" structures? If so, you may recall that in this case, the main factor dictating structure is "size" or atomic radius ratio, rather than group. For example, in the case of the rock salt structure, it is most likely that there is "a cation/anion radius ratio of 0.414 to 0.732". As such, you could consider the radius of cesium in comparison to sodium: significantly larger. Thus, it adopts the cesium chloride configuration, which is preferred when "ions are of roughly the same size ".