text:
The differences between integrals evaluations of contracted an uncontracted basis functions
I wanna to evaluate molecular integrals (overlap, kinetic and etc) using Gaussian basis functions. To do so, there are two approaches: contracted and uncontracted basis functions in which the former means the primitive Gaussian functions ($G_1$ and $G_2$) are summed to form contracted ones and finally by summing the contracted functions  ($\phi_1$ and $\phi_2$) using appropriate coefficients, the final molecular orbital is formed, for two contracted basis functions and three primitives we have:
$$
\psi=c_1\phi_1+c_2\phi_2
$$
$$
\phi_1=d_{11}G_{11}+d_{12}G_{12}+d_{13}G_{13}
$$
$$
\phi_2=d_{21}G_{21}+d_{22}G_{22}+d_{23}G_{23}
$$
$$
G_{11}=e^{-\alpha_{11}r^2}
$$
where the exponents are given as a 2x3 matrix (generally exponents matrix in contraction basis functions approach is a [number of electrons]x[number of primitive Gaussians] matrix and for uncontracted functions approache is a [number of electrons]x[number of basis functions] matrix).
In this way the final overlap matrix for example is
$$
\begin{bmatrix}
c_1c_1\phi_1\phi_1 && c_1c_2\phi_1\phi_2\\
c_2c_1\phi_2\phi_1 && c_2c_2\phi_2\phi_2\\
\end{bmatrix}
$$
For simulating total wave function using two uncontracted basis functions, we have
$$
\psi=c_1\phi_1+c_2\phi_2
$$
where it's not necessary to use any summation to obtain $\phi_1$ and $\phi_2$, instead we can construct them directly by knowing Gaussian exponents:
$$
\phi_1=e^{-\alpha_{11}r^2}
$$
Again the overlap matrix has a form like above matrix.
Are the mentioned explanations true? How should we apply these differences in a Mathematica code?

A lot of stuff is going on in this question. So for clarity let's make this answer very explicit. We can try to consider the two cases of using an uncontracted basis and a contracted basis to clarify the differences. As a case let's look at STO-$2$G for $\mathrm{H_2}$. In the STO-$2$G basis set $\mathrm{H_2}$ would normally be described by two basis functions each consisting of two primitives.
Uncontracted case
If we use the STO-$2$G basis set in the case of using uncontracted Gaussian basis functions, each hydrogen will be described by two basis functions. Our overlap matrix will therefore be of the form:
$$ S=
\begin{bmatrix}
\left<\phi_1^\mathrm{H_1}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_1^\mathrm{H_1}\left|\phi_2^\mathrm{H_1}\right.\right>  &
\left<\phi_1^\mathrm{H_1}\left|\phi_3^\mathrm{H_2}\right.\right>  &
\left<\phi_1^\mathrm{H_1}\left|\phi_4^\mathrm{H_2}\right.\right> \\
\left<\phi_2^\mathrm{H_1}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_2^\mathrm{H_1}\left|\phi_2^\mathrm{H_1}\right.\right>  &
\left<\phi_2^\mathrm{H_1}\left|\phi_3^\mathrm{H_2}\right.\right>  &
\left<\phi_2^\mathrm{H_1}\left|\phi_4^\mathrm{H_2}\right.\right>  \\
\left<\phi_3^\mathrm{H_2}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_3^\mathrm{H_2}\left|\phi_2^\mathrm{H_1}\right.\right>  &
\left<\phi_3^\mathrm{H_2}\left|\phi_3^\mathrm{H_2}\right.\right>  &
\left<\phi_3^\mathrm{H_2}\left|\phi_4^\mathrm{H_2}\right.\right>  \\
\left<\phi_4^\mathrm{H_2}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_4^\mathrm{H_2}\left|\phi_2^\mathrm{H_1}\right.\right>  &
\left<\phi_4^\mathrm{H_2}\left|\phi_3^\mathrm{H_2}\right.\right>  &
\left<\phi_4^\mathrm{H_2}\left|\phi_4^\mathrm{H_2}\right.\right>
\end{bmatrix}
$$
In this notation we have that:
$$\left<\phi_i^\mathrm{H_k}\left|\phi_j^\mathrm{H_l}\right.\right> =
N_iN_j\int_{-\infty}^\infty P_i\exp\left(-\alpha_ir_i^2\right)P_j\exp\left(-\alpha_jr_j^2\right) d\tau$$
With $P$ being a function of the angular momentum, in cartesian coordinates:
$$P=x^ly^mz^n$$
It can be seen that $P=1$ for s-functions. As we can see we have 4 basis functions in total (two on each hydrogen). You wrote something about, you would expect that in the case of uncontracted that we would have ""[number of electrons]x[number of basis functions] matrix"". As we can see this is not the case. Right now our matrix is in the atomic basis. When we solve the Hartree-Fock equations and find the coefficients that will take us to a molecular basis, we will see that the orbitals that are occupied in the atomic orbital basis, will have a contribution to occupied orbitals in the molecular orbital basis, and are therefore still important.
Contracted case
Lets now take a look at how the STO-$2$G basis set will be used in the case of contracted basis functions. Now our overlap matrix will take the following form:
$$ S=
\begin{bmatrix}
\left<\phi_1^\mathrm{H_1}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_1^\mathrm{H_1}\left|\phi_2^\mathrm{H_2}\right.\right> \\
\left<\phi_2^\mathrm{H_2}\left|\phi_1^\mathrm{H_1}\right.\right>  &
\left<\phi_2^\mathrm{H_2}\left|\phi_2^\mathrm{H_2}\right.\right>
\end{bmatrix}
$$
At first glance the overlap matrix look just like the one in the uncontracted case, but smaller. What has changed is how the elements are defined:
$$\left<\phi_i^\mathrm{H_k}\left|\phi_j^\mathrm{H_l}\right.\right> =
\sum_I\sum_J c_Ic_JN_IN_J\int_{-\infty}^\infty P_I\exp\left(-\alpha_Ir_I^2\right)P_J\exp\left(-\alpha_Jr_J^2\right) d\tau$$
We can see that we now sum over all the primitives included in the contraction. To make it more clear, let's write out how the second element in the overlap matrix looks like. Now remember that each of the basis functions consists of two primitives:
$$
\begin{eqnarray}
\left<\phi_1^\mathrm{H_1}\left|\phi_2^\mathrm{H_2}\right.\right>
&=& c_{11}c_{21}N_{11}N_{21}\int_{-\infty}^\infty \exp\left(-\alpha_{11}r_{11}^2\right)\exp\left(-\alpha_{21}r_{21}^2\right) d\tau \\
&+&c_{11}c_{22}N_{11}N_{22}\int_{-\infty}^\infty \exp\left(-\alpha_{11}r_{11}^2\right)\exp\left(-\alpha_{22}r_{22}^2\right) d\tau \\
&+&c_{12}c_{21}N_{12}N_{21}\int_{-\infty}^\infty \exp\left(-\alpha_{12}r_{12}^2\right)\exp\left(-\alpha_{21}r_{21}^2\right) d\tau \\
&+&c_{12}c_{22}N_{12}N_{22}\int_{-\infty}^\infty\exp\left(-\alpha_{12}r_{12}^2\right)\exp\left(-\alpha_{22}r_{22}^2\right) d\tau
\end{eqnarray}
$$
In this notation the first number refers to the basis function and the second refers to the primitive. As we can see each element of the overlap matrix in the contracted basis is made from four integrals. Even though our overlap matrix is four times as small compared to that of the uncontracted basis, we still have to evaluate the same number of primitive integrals. In this case we have investigated here, where we used STO-$2$G, we have no unoccupied orbitals. In general, this will not be the case, most basis sets (also when used in the contracted basis) will have unoccupied orbitals.
Other comments
In your question I can see that you mention molecular orbitals. Just to be clear. When we do the integrals we are in the atomic orbital basis. The molecular orbital basis is the one we find when we solve the Hartree-Fock equations. I.e. the molecular orbitals have nothing to do with uncontracted vs contrancted basis functions, though we will get more molecular orbitals if we use uncontracted basis functions, because we will get the same number of molecular orbitals as we had atomic orbitals.
Regarding how to implement this in Mathematica I have no idea. Though as my understanding of how Mathematica is ought to be used, I would recommend that you would implement your integral code in a language like Python. For inspiration you can take a look at this guide, which implements all the integrals with contracted basis functions.

Many Thanks for your detailed answer. I think somewhere you misunderstood my mean, what is the exponents matrix in both cases? is that other than a 2x4 matrix?

@NSR I see. How you store the exponents in your code is pretty much free choice. You can store them in a matrix if you want. In the first case (uncontracted), you need to exponents for every integral (one exponent for each Gaussian). In the second case with STO$2$G you need two exponents for the first basis function, and two exponents for the second basis functions, i.e. you will need two vectors of length four.

Yes, however, we can combine two vectors of length four and build a 2x4 matrix again, so finally, we encounter with the same exponent matrices, right?


text:
Nomenclature of common naming of alkynes vs alkenes; both -ene?
Is there a difference between common naming of alkenes vs alkynes? Why is Ethylene the common name for an alkene and acetylene for an alkyne? Isn’t acet- the common form of naming ethyl groups anyway so why is ethyl used in a common name?

Related Q: Naming system -“ylene”

The official IUPAC name for ethyl with a double bond is ethene, but this is a relatively recent change.
Before the official name (as an exception) was ethylene. I’m not sure about propene/propylene but am fairly confident the official IUPAC name for propyl double bond is propene and not propylene. However, these compounds are often referred to with the unofficial name as per tradition.
For alkynes, the IUPAC name is, for ethyl, the name is ethyne.
Don’t remember to specify the location of the double/triple bond (ex: 2 - ethene)
For alkenes, the formula is $\ce{C_nH2_{n+2}}$ and for alkynes is $\ce{C_nH_{2n}}$.

To be precise, the non-systematic name “ethylene” was still retained in the 1979 IUPAC recommendations but was no longer recommended in the 1993 recommendations.

Is there a 2-ethene? For acyclic cmpds: alkanes, CnH2n+2; alkenes, CnH2n; alkynes, CnH2n-2.

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text:
Thiol prefixes for nomenclature
I've seen multiple prefixes for naming thiols: mercapto-, sulfanyl-, and -thio. Is there a rule for when specific ones are used? And are they used for IUPAC and/or common naming?

Mercapto- and sulfanyl- are both prefixes for use with the $\ce{-SH}$ group (known as a thiol), while thio- is used to denote the "sulfur" equivalent of an oxygen-containing functional group.
More details from the IUPAC Blue Book:

Mercapto- is mostly abandoned, and only used as a prefix for -SH
groups. This quote is from the older edition of the Blue Book. "When -SH is not the principal group, the prefix "mercapto-"
is placed before the name of the parent compound to denote an
unsubstituted -SH group."
Sulfanyl- "In these [1993] recommendations,
the prefix "sulfanyl-" is preferred to "mercapto-" which was used in
previous editions of the IUPAC Nomenclature of Organic Chemistry."
Thio- "In nearly all its applications "thio" denotes replacement
of oxygen by sulfur."

Please note that the links point to obsolete IUPAC recommendations from 1979 and 1993, respectively. However, that does not necessarily mean that names formed according old recommendations are no longer in accordance with current recommendations.

Yeah I forgot to mention that, thanks for adding the clarification.

If the $\ce{-SH}$ group is the principal characteristic group, the compound is named using the using the suffix ‘thiol’ (e.g. propane-2-thiol).
In presence of another characteristic group having seniority for citation as a suffix, the $\ce{-SH}$ group is cited using the prefix ‘sulfanyl’ (e.g. 3-sulfanylpropanoic acid).
The prefix ‘mercapto’ for ($\ce{-SH}$) is no longer recommended.
The use of the functional replacement prefix ‘thio’ is no longer used to replace oxygen atoms in alcohols etc. (e.g. benzenethiol, not ‘thiophenol’).
The corresponding rule in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-63.1.5 Sulfur, selenium, and tellurium analogues of hydroxy compounds
Sulfur, selenium, and tellurium analogues of hydroxy compounds are named substitutively using the suffixes ‘thiol’, ‘selenol’, and ‘tellurol’, and the prefixes ‘sulfanyl’, ‘selanyl’, and ‘tellanyl’, respectively; the presence of several of the same kind of ‘ol’ characteristic groups is denoted by the numerical multiplying prefixes ‘di’, ‘tri’, etc. The prefixes ‘mercapto’ ($\ce{-SH}$), and ‘hydroseleno’ or selenyl ($\ce{-SeH}$), etc. are no longer recommended.
Functional class nomenclature is not used.
(…)

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text:
Lyophilization Advantage
I've been reading about freeze drying and how it increases the shelf life of certain foods and is overall a good preserving technique.
My question is with regard to the actual process. I understand that freeze-drying involves freezing the actual material and then proceeding to dry it by placing it in a vacuum. This way the ice is turned into vapour without going into any liquid phase. My question is why is avoiding the liquid phase necessary and beneficial? When it comes to vaccines, foods, and polymer storage for example, why is it advantageous? I don't intuitively understand the reasoning behind why lyophilization is so favourable.

I don't think the point of freeze drying is to avoid liquid water -- I mean, there's liquid water already in the substance to be dried -- but to avoid heat.  Ordinarily, to dry something out to preserve it, you need to heat it to speed up the evaporation of water, certainly if you want to get rid of the water quickly (which you would want to do in any commercial setting, where time = money).  But heating causes all kinds of changes to foods, and while some foods are fine with it, others  suffer degradations of perceived quality, e.g. undesirable changes in color and texture.   The point of freeze-drying is that you can get rid of the water at a low temperature.


text:
Meanings of isopropyl, n-butyl, isobutyl, s-butyl and t-butyl and differences between them
I am having a hard time learning organic chemistry because I cannot understand the meanings of isopropyl, n-butyl, isobutyl, s-butyl and t-butyl. There is no proper explanation given in my book about them. I have also read pages on the internet but I ended up confusing myself more.
My questions:

What are they? Are they alkyl groups (substituents) like methyl and ethyl or just a portion of the molecule (not a substituent)?
How was their name obtained? (I believe it was obtained on the basis of their structure and i would like to know on what basis)?
How are the names for such type of groups for higher alkanes obtained?
Are these names a part of IUPAC nomenclature?

Related: Are common names of substituents accepted by IUPAC?

(1) Yes, they're alkyl substituents.  But that doesn't mean they have to have been literally added to some other molecule.  There's nothing wrong with speaking about a part of a molecule as a "substituent," even if that exact piece was never actually literally substituted on the main chain in a chemical reaction.  Think of it as a hypothetical substituent.
(2) You get the names this way:
(a) Use the parent hydrocarbon and change "-ane" to "-yl."  So a 4 carbon substituent becomes "butyl" from "butane."  The 3-carbon substituent becomes "propyl" from "propane."
(b) There are two isomers of butane.  The one with 4 carbons in a row is called "n-butane" for "normal" butane.  The other has 3 carbons in a row, with the 4th branching off, this is called "isobutane" from a Greek prefix "iso" meaning "same as," which is the same prefix in "isomer."  Of course it's not the same molecule, but it has the same chemical formula.
(c) There are two ways to attach the group derived from n-butane to a main chain.  You could attach it from one of the end carbons, in which case we call it "n-butyl."  Or you could attach it from one of the inner carbons, in which case we call it "sec-butyl" because the attachment point is a secondary carbon (instead of a primary carbon as it is in n-butyl).
(d) There are also two ways to attach the group derived from isobutane. You can attach it from one of the end carbons, in which case we call it an "isobutyl" group, or you can attach it from the center carbon, in which case it's "t-buty" or "tert-butyl" for "tertiary" butyl because the attachment carbon is a tertiary carbon.
There are only two groups you can derive from propane: the group can either be attached from one of the end carbons, in which case it's a "propyl" group, or it can be attached from the middle carbon, in which case it's an "isopropyl" group.  (There is no isopropane parent, so this is a little inconsistent.)
(3) There are a few names for bigger alkyl substituents, but they aren't as often used I believe, because it starts to get even more complex and idiosyncratic, which is why...
(4) ..these are not the IUPAC names for these alkyl substituents.
You can certainly swear a vow to always use IUPAC names, and many of us have done so when we first learned organic.  But with time comes familiarity with these little guys, who are more compact and roll off the tongue faster than their systematic equivalents, and you should not be surprised if after some years (if you stick with it) you find yourself using them all the time just because it's convenient.


text:
How to name binary (inorganic) compounds given their chemical formula, and vice-versa?
How can I deduce the name of a binary (inorganic) compound like $\ce{B2F2}$ or $\ce{N2O4}$, from the formula?
And how can I convert a chemical name like diboron difluoride to the chemical formula?
I don't see why we do not need the numeric prefixes in the cases of lithium oxide and magnesium chloride but we do with dinitrogen tetroxide and iodine monochloride.
Please limit the answer to simple, binary, ionic and covalent compounds, and exclude hydrocarbons.

Prerequisites
If you're uncomfortable with any of the following, please first head over to the corresponding links before continuing.

A chemical symbol is a shorthand representation of the name of an element, for example, $\ce{N}$ for nitrogen, and $\ce{Na}$ for sodium. More details on the Wikipedia page.
Polyatomic anions/Radicals: anions with more than one element, like nitrate ($\ce{NO3-}$) or sulfate ($\ce{SO4^2-}$). More details on the Wikipedia page.
Oxidation state: an integer or decimal number assigned to an element in a chemical species. It is a tool that helps us do nomenclature easily. Read a detailed introduction here.
Ionic and covalent compounds: You must understand what ionic and covalent compounds are. You must also know the few elementary examples of each. For example, you should know that $\ce{N2O4}$ would be a covalent compound, while $\ce{NaCl}$ would be ionic. Here's an introduction by LibreTexts if you need a refresher.

Introduction
There are two separate cases here for ionic and covalent compounds.[1]. Some of the logic overlaps, but some does not, and care must be taken to avoid using one's rules for another.

Naming an ionic compound from its formula

Name the cation and then the anion. ("sodium sulfate", but not "sulfate sodium"). (How to identify the cations and anions in an ionic compound? The cation is usually a metal and the more electropositive element, while the anion is a non-metal and more electronegative element)
Keep everything lowercase (details) ("sodium sulfate", but not "Sodium Sulfate" or "sodium Sulfate" or "Sodium sulfate")
Do not apply numeric prefixes to either the cation or the anion. ("sodium sulfate", but not "disodium sulfate" or "sodium monosulfate")
Remember the suffixes for anions. The single atom anions of group 15, 16, and 17 have the suffix ending "-ide". Examples: fluoride, oxide and phosphide. However, polyatomic anions generally do not end in an -ide. See the Appendix for details.
Transition metal cations (and some p-block metals) also need their oxidation-state to be specified in parentheses, as they exhibit variable oxidation states.  Hence, "mercury(II) chloride", but not "mercury chloride". Notice that the oxidation state is denoted in capital Roman numerals.[2]

Let's take an example: $\ce{MgCl2}$. We can identify it's composed of magnesium and chlorine. Thus, the name would be magnesium chloride – in all lowercase and with the cation (metal) preceding the anion (non-metal). There is no need to denote the oxidation state as magnesium isn't a transition metal.
Examples:
$\ce{CsF}$, $\ce{NaNO3}$, $\ce{Al2(SO4)3}$, $\ce{FeCl2}$, $\ce{Pt3(PO4)4}$
Answers: (hover to peek!)

 caesium fluoride, sodium nitrate, aluminium sulfate, iron(II) chloride or ferrous chloride, platinum(IV) phosphate

Writing the formula for an ionic compound from its name

First write the cation and then the anion ($\ce{KBr}$ but not $\ce{BrK}$)
Use parentheses to denote multiple polyatomic anions ($\ce{Mg3(PO4)2}$ but not $\ce{Mg3PO42}$). Avoid them for a single polyatomic anion ($\ce{Na2SO4}$ but not $\ce{Na2(SO4)}$)
Use the trick of cross-transferring oxidation states between the ions to determine the count of each cation and anion involved. This trick works because the formula should denote an electrically neutral species. (you can verify this mathematically) So, to write the formula for aluminium sulfate:

Note that:

In case both oxidation states have a common factor, you should cancel them. Example: magnesium oxide:
.
drop the number 1. Hence, potassium sulfide is written as $\ce{K2S}$ but not $\ce{K2S1}$.

Examples:
potassium phosphate, calcium chloride, magnesium cyanide
Answers: (hover to peek!)

 $\ce{K3PO4, CaCl2, Mg(CN)2}$

Naming a covalent compound from its formula

Name the less electronegative element and then the more electronegative element ("carbon monoxide" but not "oxygen monocarbide")
Apply suffixes to anions in exactly the same way as we did for ionic compounds. ("sulfur hexafluoride" but not "sulfur fluoride") Also, keep the name all lowercase as before.
Use numerical prefixes.[3] Hence "dinitrogen tetroxide" for $\ce{N2O4}$ but not "nitrogen oxide". Here's the first ten suffixes: $$\begin{array}{|c|c|} \hline
\mathrm{prefix} & \mathrm{number}\\ \hline
\mathrm{mon(o)} & 1 \\
\mathrm{di} & 2 \\
\mathrm{tri} & 3 \\
\mathrm{tetr(a)} & 4 \\
\mathrm{pent(a)} & 5 \\
\mathrm{hex(a)} & 6 \\
\mathrm{hept(a)} & 7 \\
\mathrm{oct(a)} & 8 \\
\mathrm{non(a)} & 9 \\
\mathrm{dec(a)} & 10 \\
\hline
\end{array}$$ Note that:

the letter in parentheses is dropped when the anion begins with a vowel ("tetroxide", but not "tetraoxide")
we do not specify the mono- prefix for the first element ("carbon tetrachloride" but not "monocarbon tetrachloride")
we can construct similar numerical prefixes for larger counts by using the Greek system. Example: "undeca" for 11. Look at a larger list on Wikipedia.

Let's take an example: $\ce{P4O6}$. In one molecule of $\ce{P4O6}$, we can see that it has four atoms of phosphorus and six atoms of oxygen. So, its name should be of the form "Xphosphorus Yoxide". The table of numerical prefixes tells us that we should use the prefix "tetra" for nitrogen and "hexa" for oxygen. Since the name "oxygen" begins with a vowel, we shall write "hexoxide" (and not "hexaoxide"). Thus, the correct name for this chemical formula is "tetraphosphorus hexoxide".
Examples:
$\ce{CO2}$, $\ce{P4O10}$, $\ce{N2O}$, $\ce{N2S5}$
Answers: (hover to peek!)

 carbon dioxide, tetraphosphorus decoxide, dinitrogen monoxide, dinitrogen pentasulfide

Writing the formula for a covalent compound from its name
This is much simpler from the previous case, rather it follows an exactly reverse process.
We'll just take a look at two examples:

boron triiodide: we see that its formula would be: $\ce{B_$x$I_$y$}$. Now, $x=1$ since boron lacks any numerical prefix, and $y=3$ due to the "tri" prefix. Hence, the formula is $\ce{BI3}$.
dinitrogen pentoxide: we see that its formula would be: $\ce{N_$x$O_$y$}$. Now, $x=2$ and $y=5$ due to the "di" and "pent(a)" prefixes. Hence, the formula is $\ce{N2O5}$.

Examples:
hydrogen bromide, iodine heptafluoride, diboron hexahydride
Answers: (hover to peek!)

 $\ce{HBr}$, $\ce{IF7}$, $\ce{B2H6}$

Naming of hydrogen compounds
Several hydrogen compounds have non-systematic names that have persisted through the years in standard nomenclature and are now well recognized. Note that for alkali metals and alkali earth metal hydrides, the name is simply "metal + hydride", like sodium hydride, on the basis of rules described earlier.
However, covalent hydrides have varying names. There are not much rules around them, but rather "patterns" that the following list would show you:

Group 13: $\ce{BH3}$: borane, $\ce{AlH3}$: alumane, $\ce{GaH3}$: gallane, etc. Pattern: half of the element name suffixed by "-ane".
Group 14: $\ce{CH4}$: methane, $\ce{SiH4}$: silane, $\ce{GeH4}$: germane. Pattern: half of the element name suffixed by "-ane". Methane is a remnant of organic chemistry, and is a part of nomenclature of hydrocarbons.
Group 15: $\ce{NH3}$: ammonia, $\ce{PH3}$: phosphine, $\ce{AsH3}$: arsine, $\ce{SbH3}$: stibine, $\ce{BiH3}$: bismuthine. Pattern: half of the element name suffixed by "-ine". "Ammonia"'s origins, though, are a bit more interesting (read the Wikipedia page).
Group 16: $\ce{H2O}$: water, $\ce{H2S}$: hydrogen sulfide, $\ce{H2Se}$: hydrogen selenide, $\ce{H2Te}$: hydrogen telluride. Note that while "oxidane", "sulfane", "selane", etc. are acceptable, in the case of oxidane, that name is only used when denoting water derivatives. Read Do chemists refer to water as "dihydrogen monoxide"? and Does water have a chemical name?.
Group 17: they are simply named as "hydrogen halide". Example: "hydrogen bromide" $\ce{(HBr)}$. It is not named "hydrogen monobromide", as you may believe from earlier examples, and that it because hydrogen can only form a single halide with the halogens (see Note 3 below).

The curious students would also notice that – when writing chemical formulae – for groups 13, 14, 15, the hydrogen atom is written after the group atom $\ce{(NH3)}$, while in groups 16 and 17, it is written before the group atom $\ce{(H2O)}$. See this answer to understand why it is so.

Notes:

It is true that no compound is 100% ionic or 100% covalent. The above rules for "ionic compounds", thus, apply to compounds with dominant ionic character. Similarly, rules for "covalent compounds" apply to compounds with dominant covalent character.
Another way to write such a name is to use an "-ic" suffix for a higher oxidation state and "-ous" suffix for a lower oxidation state. As in "mercuric chloride" for "mercury (II) chloride". But, according to the IUPAC Recommendations 2005, it is no longer acceptable.
Why are numerical prefixes necessary? Note that in covalent compounds, a given anion can be present in more than one count. For example, the $\ce{F}$ in $\ce{BrF_x}$ can be present as $\ce{BrF}$, $\ce{BrF3}$, or even $\ce{BrF5}$. Hence, it is necessary to indicate if it is a monofluoride, a trifluoride, or a pentafluoride.
Several covalent compounds and ionic compounds have their own unique, common/trivial names. These names do not follow any systematic rules and are based on the source (location/discoverer) or properties of that compound. Learning these common names and their formulae comes by experience over time. Examples include: diborane $\ce{(B2H6)}$, Epsom salt $\ce{(MgSO4)}$, oil of vitriol (sulfuric acid), etc. Read more on the Wikipedia page.

Appendix: polyatomic anion names
Apart from the regular -ide suffix for monoatomic anions, there are some more suffixes for the polyatomic anions.

most pseudohalides retain the "-ide" suffix. (except cyanate derivatives)
the other two suffixes are "-ate" and "-ite". "-ate" is used when the central atom is in a higher oxidation state, whereas "-ite" is used for the lower oxidation state.
the "per" prefix is used when the central  halogen atom is in its highest oxidation state, while "hypo" is used when it is in its lowest oxidation state.

A few examples would make these rules clearer:

Sulfate $(\ce{SO4^2-})$ and sulfite $(\ce{SO3^2-})$: both are sulfur and oxygen anions, but in sulfate, sulfur has higher oxidation state as compared to that in sulfite. Similar case applies to nitrate, nitrite and phosphate, phosphite.
Hypochlorite $\ce{(ClO^-)}$, chlorite $\ce{(ClO2^-)}$, chlorate $\ce{(ClO3^-)}$, perchlorate $\ce{(ClO4^-)}$: have chlorine in increasing oxidation numbers. Similar series exist for other halogens (except fluorine).

Of course, there are many more polyatomic anions in chemistry, but these ions are relevant to our discussion at an elementary level.

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text:
Why lattice constant (unit cell) is always a parallelepiped?
Wikipedia defines lattice constant as physical dimension of unit cells in a crystal lattice.
Unit cell is defined as:

The unit cell is defined as the smallest repeating unit having the
  full symmetry of the crystal structure. The geometry of the unit
  cell is defined as a parallelepiped.

Why that elementary pattern that repeats itself in crystals is always a parallelepiped? In real life such elementary repeating unit/part/block is very often not a parallelepiped at all!
Is this some abstraction and a "bounding parallelepiped" is meant - the one into which definitely fits whatever real elementary repeating pattern is?

What cases do you think at not described by a parallelepipied? Are you confusing the lattice and the basis?

It is not the only one but the norm for chemists/biochemists, solid state physicists define the primitive cell differently  as the Wigner-Seitz cell.  This has  a rather polygonal look. Choose a victim point to be the centre. Connect this point to other nearby points with lines, then draw perpendicular lines at the midpoints. These latter intersect to form the primitive cell.

A unit cell is always a parallelepiped because this is the definition of a unit cell. In case when you don't have a repeating unit of this kind (which might indeed be the case in real life), you don't have a crystal, also by definition. You might be dealing with a quasicrystal or something.
3D periodicity is what defines a crystal. If you have it, this means you have three non-coplanar periodic vectors, and these inevitably define a parallelepiped.

For a 3D-lattice, due to the periodicity of the system all the information necessary to encode it can be encoded in a space described by three vectors and a set of operations/symmetries that tells you how those cells get sewn together to tile the plane.
In linear algebra speak, the rank of the matrix made from composing the lattice vectors is equivalent to the dimension of the space they span. Any fewer vectors, and there isn't enough information to map to a specific system (does not span the lattice.) Any more vectors, and there is excess information (each vector can be written as a linear combination of the other vectors, and thus encodes superfluous information that can be trimmed down.) The symmetries encode the rest of the information and just describe any other restrictions to the configuration.
This is generalizable to lattices and parallelepipeds in n-dimensions, as lattices are just an example of a free abelian group over some topological space (here, a vector space) and parallelepipeds are the "fundamental regions" of lattice groups acting on Euclidean topological spaces.


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Comparing size of ions lithium, gallium ions
How is the size of $\ce{Li+}$ ($\pu{0.68}$) larger than $\ce{Ga^3+}$ ($0.62$)?
This is very difficult for me to understand  why they say that size of Li+ with 3 protons and just 2 electrons should be considerably more than that of $\ce{Ga^3+}$. I understand the poor shielding effect of $\ce{3d^{10}}$ electrons but it isn't obvious to me why it will make its size smaller than $\ce{Li+}$ which has just one shell with 2 electrons and more effective nuclear charge. Where is my logic wrong?

Because removing three electrons from Ga leaves a much stronger net positive charge, pulling in the orbitals, than pulling one from Li.

We removed remove the 1 electron from 4p1 and then remove 2 electrons from 4s2 these three electrons where the ones on which the nucleus pulled on with more force but now they are not there.

@drmoishepippik but what about Ga3+ containing 3 shells and Li only one?

Could you link the source of the image?

@holgerfielder https://chemistry.stackexchange.com/questions/98166/comparing-size-of-ions?noredirect=1#comment181545_98166

@Jasmine I have removed the image because most of it was irrelevant to the discussion. Still, you need to link to the webpage from where you obtained the values. Note that the image was missing the distance units too (is it Angstrom or picometers?)

@gaurangtandon I meant to relate the size. I will take care next time.

That is impressive.  What you may need to bear in mind is that as the nuclear charge increases, all the electron orbitals move in closer to the nucleus.  You may have been assuming, consciously or unconsciously, that the electron shells appear at a relatively fixed distance from the nucleus -- that is, that the n=1 shell is always at about the same distance from the nucleus, the n=2 shell is always at about the same additional distance, and so on.
Not the case!  For example, if you take a look at the quantum mechanical model of the hydrogen atom and solve it for a "hydrogen-like" atom with atomic number Z and one electron (so, He+, Li++, Be+3,....,U+92 ha ha) then you will find that the average orbital radius of the electron scales like 1/Z.  So if we ignored electron-electron repulsion entirely, the radius of each shell would decrease like 1/Z as Z increases.  That's quite a contraction.  It would mean that the n=1 shell in carbon, say, would be 1/6 the size of what it is in hydrogen.  In the case of Ga, Z=31, the n=1 shell would be 1/31 the size it is in H!  Each new shell would contract similarly.
So in the absence of electron-electron repulsion we would expect the size of atoms to shrink drastically as Z increases, going across the Periodic Table, and bumping up slightly at the beginning of each row with the addition of a new shell.  Of course, real atoms do have electron-electron repulsion, lots of it, so the effect is considerably muted.
Nevertheless, the size of atoms does shrink considerably going from left to right in the Periodic Table, and it's worth remembering this is because all of the electron shells shrink, not just the valence shell.

This surely helps. I didnt know that the size radius by 1/Z and i think it may even lead to repulsion between the electrons in 2 consecutive shells


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Regularity of electron configuration for heavier atoms
Edit: Is there structure in the nucleus that might be responsible for part of the regularity of the (ground-state) Aufbau process? The periodic addition of electrons is more than one might naively predict from Coulomb forces alone--so is this an intrinsic property of electrons or does the nucleus somehow mediate or influence the Aufbau process?

This is a very naive question.
Only after making graphs of electron counts by orbital for elements in sequence and seeing the symmetry of some of the heavier (and possibly non-realizable) elements (below is 170 or so, from the Wiki electron configuration data page) did it occur to me that the picture seems to imply a lot about the nucleus that I do not recall from general chemistry.
In particular for heavier elements, despite some irregularity along the way, it seems to imply that there are mirroring forces inside the nucleus that govern the buildup of electrons and that in some sense the nucleus is at least as complex as the rest of the atom. Not only are the electrons balanced in terms of charge numerically, but there is seemingly some structural (for lack of a better word) equivalence as well.
If someone could give a qualitative synopsis of what is known about this and/or correct any misconceptions that would be great. An introductory reference would also be of interest.

What are you asking really? This pic looks like illustration of aufbau principle, with numbers of electrons fitting into subshells. There are reasons to think there are also nuclear shells but quite different.

@Mithoron: I have inserted a proposed edit which reflects the original question but hopefully more clearly elicits the (IMO good) answer.

Yes, the nucleus has a complex structure, too.  The most basic evidence for this comes in the existence of patterns in which nuclei are most stable, which manifests itself in the decay lifetime and to some extent in the cosmic abundance of isotopes.  I don't think there is any reflection of it in the electronic structure: from the point of view of the electrons, the nucleus is pretty much just a point positive charge.  The patterns in the electron configuration you are noticing are due to the Pauli exclusion principle and the various ways quantum mechanical particles can fit into a spherical potential well.   You can find a description in any basic general chemistry textbook, and the math will be developed in any undergraduate physical chemistry textbook.
You can get started on nuclear structure by googling "nuclear shell structure."  That will lead you to very basic introductions, e.g. on Wikipedia or Hyperphysics, and you can decide where to go from there.  I don't know any textbooks that treat it simply, because the assumption is that if you are interested in the nuclear physics of nuclei you are already very familiar with the quantum mechanics of electrons.

This answers my question. Having looked at "nuclear shell structure" it seems a little surprising that electrons see only a point charge and that there is no connection between the structure in the nucleus and that of the electrons but none is implied in the article so you must be right--thanks.

Oh there are certainly small corrections to the energy levels and orbitals  of the electrons due  to the finite size of the nucleus, and the distribution of charge you assume it to have. They're just very small because the nucleus is about 10^-4 to 10^-5 times as big as the atom.  So it's not exactly a point charge, but it's pretty close.


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Reacting NaOH in solution with O3
I use common air to produce ozone. However the gas after the ozone generator does contain $\ce{NO2}$ gas, which I have to remove for the subsequent steps. I tried to remove the $\ce{NO2}$ with a stainless steel gas wash bottle filled with distilled water. This works fine, however the distilled water will subsequently get charged with $\ce{NO2}$ and turn to dilute nitric acid.
After a few days of operation, I have changed the water and tested its acidity. The $\mathrm{pH}$ was around 0.4 - it was possible to etch a copper pipe with it. So the formation of this acid is not very good for the stainless steel gas wash bottle. I have decided to add 2 more stainless steel gas wash bottles and fill the first one with $\ce{H2O + NaOH}$, in order to neutralize the $\ce{HNO3}$. The other two bottles are for capturing the remaining traces of $\ce{NO2}$.
Bottle 1: 5 liter $\ce{H2O}$ + $\pu{15g}$ $\ce{NaOH}$
Bottle 2: 5 liter $\ce{H2O}$
Bottle 3: 5 liter $\ce{H2O}$
Now I have done some ozone measurements with and without these gas wash bottles. The result was as follows:

It seemed as if $\ce{O3}$ would pass through pure $\ce{H2O}$ without any reaction once the solution is saturated and the saturation seems to happen almost immediately.
However the solution with the $\ce{NaOH}$ seems to react the first 2-3 minutes and then suddenly stop reacting, which is actually good, because it does not use up ozone, however I have no clue what is happening in detail.

Here are some charts from the measurement. Upper chart: test of the $\ce{H2O + NaOH}$ solution. Lower chart: test of the $\ce{H2O}$. Note: the lower the value in the chart, the higher the ozone concentration.
The duration of each test was about 5 minutes. I did measure the $\mathrm{pH}$ of the $\ce{NaOH}$ containing solution and it basically did not change due to the ozone treatment. (It was around 12.35 before and after ozone bubbling...)

Any idea what is happening there and why?

You could oxidize some hydroxide ions, to wit:
$\ce{OH^- + O3 -> HO2^- + O2}$
Such an oxygen transfer is favored by strongly basic solutions because $\ce{HO2^-}$ is a weaker base ($\mathrm{p}K_\mathrm{a}$ hydrogen peroxide = 11.75) than $\ce{OH^-}$, as well as ozone being a more powerful oxidizer than diatomic oxygen.


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What occurs when stannous fluoride is combined with baking soda in the oral cavity?
I have googled this question in many forms over many months repeatedly.
For some background, when stannous fluoride is combined with water, it forms precipitates and/or oxidizes, rendering it less effective in the oral cavity.  Crest Pro-Health, as an example, adds a stabilizer, a gluconate, to maintain stability of the stannous fluoride. (If anyone can answer the question of how the gluconate stabilizes the stannous fluoride, that would be great too.)
My main question: What is the effect (via the chemical reactions) of combining a sodium bicarbonate/hydrogen peroxide solution with stannous fluoride in the oral cavity?  And how is that effect altered if hydrogen peroxide is not used?
According to me, the high pH of the baking soda neutralizes the acidity of the stannous fluoride, thus nullifying the effects of $\ce{SnF2}$. I also feel that the hydrogen peroxide oxidizes the stannous fluoride - and there may be some precipitates. But I am not sure.
I am asking these questions because I want to determine whether I can use a baking soda/peroxide toothpaste followed by an $\ce{SnF2}$/hexametaphosphate toothpaste.


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Chemical Kinetics- Net Number of Particles
I have a box of 5 types of particles, $A$, $B$, $C$, $x$ and $y$. They react in the following way.
$\ce{$A$ + $B$ -> $C$ + $x$}$
and
$\ce{$C$ + $y$ -> $A$ + $B$}$
If only the first reaction was happening then $N(C) = k \cdot N(A)N(B)$. Here $N(I)$ is the number of particles of $I$ in the box.
Can I write the total number of $C$ in the box as $\text{number of $C$ created} - \text{number of $C$ destroyed} = N(C) = k \cdot N(A)N(B) - k' \cdot N(C)N(y)$?

Not the total number but instead the number /time which is the rate $dN_c/dt$ . These equations then have to be integrated knowing the initial number present. Normally we do not use numbers in a box but equivalently and more conveniently  concentration. In your scheme only x is present at long times given that A, B and y are present initially.

Actually, the rate law for the species $\ce{C}$ is written as:
$$\frac{\mathrm{d}\ce{C}}{\mathrm{d}t} = k \ce{A} \ce{B} - k' \ce{C} \ce{y}$$
If you integrate this, you would get an equation in terms of the number of particles, as you require. However, this is a very typical integral, and there would be a number of other arbitrary constants, which would make the final result not equal to the one you've shown.
What you've written is a mistake on replacing the rate of formation of $\ce{C}$ with its number of particles itself. Both are not equivalent.


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Are dopant ions on a crystal lattice free or bonded?
This question seems simple but I don't seem to find many answers right away for this.
When you make a crystal doped with ions (like a luminescent phosphor) and the ions then become a defect in the crystal structure and still have emission lines similar to the free ion, is that ion really free? Or is it actually in some kind of bond?
I mean, if it did make a ionic or molecular bond, it wouldn't be a free ion anymore, the energy levels and the probabilities for the electron transitions would be very different, but the crystal has a luminescence that's similar to the free ion.
Of course there will be interactions, the electron transitions won't have the same probabilities of the free ion (you do see some lines of emission been strengthened, weakened  and/or shifted when you use a same ion in different crystals), but it's quite similar in the overall.
Saying the kind of defect an atom is on a lattice (like interstitial or substutional) only tells me about the geometry and array of the atoms on the lattice.
But I want to know about the chemical bonds they're making (or not).
Are they just "trapped" inside the structure by electrostatic repulsion without making an actual bond with any molecule around them?
Thanks!

It depends on lattice, dopant and doping procedure. There isn't any one answer.

The dopants do not act like free ions. However, the difference between energy levels on the dopant atom can be similar to a free atom/ion split by a crystal field (depicting the neighboring lattice points as point charges).

Would it be possible to summarize the effects of "lattice, dopant and doping procedure" in a short answer? This could be a helpful question with some long term value. @MaxW (cc other closevoters)


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Is cyclohexane saturated or unsaturated?
I have this confusion that cyclohexane is of form $\ce{C_{n}H_{2n}}$ and so it must be an unsaturated compound, but also it does not have a double or triple bond so it must be saturated.

You are right in that its formula matches CnH2n. You are also right in that it has no multiple bonds. If these facts lead to a contradiction, then probably your definition of saturated compounds has a contradiction in itself.

Your definition of saturated is flawed and only applies to acyclic alkanes

Cyclo comes from the Greek word 'kyklos' which means circle.
The cyclohexane molecule is in the form of a circle therefore it can have the formula $\ce{C_nH_{2n}}$ and still be saturated.

As what is mentioned in the cpmment by Ivan and Waylander, your definition of saturated compounds is flawed.
So let just set the definition straight first. The definition of saturated compounds are compounds that the carbon atoms are bonded singly with each other. (This definition will work for now in our example.)
So based on the definition above and the image provided by CKA, we can clearly so all the carbon atoms are bonded singly with each other and thus prove that it is a cyclohexane is a saturated compound.
But how can a compound be saturated but have the general formula $\ce{C_{n}H_{2n}}$ you might ask?
The key thing here is that $\ce{C_{n}H_{2n}}$ is unsaturated when is acyclic or non-cyclic such as n-heptene, metylbut-1-ene, 2,3- dimetylpent-1-ene etc. These compounds have straight or branched carbon chain (and the carbon are not formed into a ring or a loop).
We can try prove that a cyclic saturated compound has a general formula of $\ce{C_{n}H_{2n}}$.
Let take an acyclic compound with formula $\ce{C_{n}H_{2n + 2}}$ (which will kinda look like this $\ce{CH3(CH2)_{n-2}CH3}$).
Then, we will remove one hydrogen atom from both end of the carbon chain (which will kinda become like this $\ce{CH2(CH2)_{n-2}CH2}$), the formula will be $\ce{C_{n}H_{2n + 2 - 2} -> C_{n}H_{2n}}$.
Lastly, since the carbon atom at both end is still capable to have one more single bond, we will connect the single carbon atom which the last one and thus make a loop of carbon chain/a cyclic compound. Since we are just connecting the carbon atoms and not adding or removing any atoms from it, the formula will still remains as $\ce{C_{n}H_{2n}}$.


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In Reimer-Tiemann reaction why does phenol attack the carbene from ortho position?
From the Wikipedia article on Reimer-Tiemann reation:

In step 5, why doesn't the oxygen attack the carbene. The way I see it: upon attacking the carbene from ortho position, the mechanism proceeds via a non aromatic intermediate. Oxygen bearing a negative charge is a good nucleophile too.
So I proceeded with oxygen acting as an nucleophile and came up with this mechanism:

The final product is not a major product (I wasn't even able to find it listed as a product after a cursory google search), so why does phenoxide attack the carbene from ortho position.

EDIT
The product I'm asking about has not even been mentioned in the linked question.

Related: Products of Reimer–Tiemann reaction of 4-methylphenol

Due to continuous conjugation of negative charge in phenoxide ion, the lone pair of electrons are present more at ortho position rather than at para position + carbene is highly electron deficient compound

First of all, in the phenoxide ion, the lone pair on the oxygen atom doesn't remain fully available, it remains in continuous conjugation with the double bond of the benzene ring. In the resonance hybrid, there is also electron density on the Carbon atom in the ortho and para position,and also definitely on the oxygen atom itself. But, Carbon being less electronegative acts as a better nucleophile in this case rather than Oxygen.  So, attack from the Carbon atom rather than Oxygen atom is more probable.  More over, attack from the ortho position, though creates a chance of Steric crowding, but the intermidiate generated from the attack by ortho-position creates an ordered transition state.The metal ion(say, $\ce{Na+}$) can easily interact with the electronegative $\ce{O}$ and $\ce{Cl}$ atoms, and can increase the stability of the transition state. Here is a picture of the intermidiate and a 3D view of it,
This ordered transition state will not appear in case of attack by the Oxygen atom. This is also another reason why that transition state is more favoured, and the reaction rate through that step is higher than any other possible steps.

what you have shown here needs a citation. Regardless I don't see why chlorine should have any interaction with the $\ce{Na+}$ ion. It doesn't have any charge on it. ("Backbonding" is not possible there is evidence that it forms a cyclopropyl dichloro ring)

@Soumik But the wikipedia states that ortho product is major even in case of KOH. Can you please shed some light on it?

Another way of rationalising is through the use of the Hard-Soft Acid-Base principle. The dichlorocarbene generated is a soft electrophilic species while the oxyanion form of the phenoxide is a hard nucleophilic species. The nucleophilic $\pi$ bond is much softer and thus, it is much more favourable for it to be the attacking nucleophile (soft-soft interaction) rather than the negatively-charged oxygen atom (hard-soft interaction).


text:
Why are negative enthalpies becoming more positive described as decreasing?
This website and my lecturer have describe negative enthalpies that are becoming less exothermic as decreasing. A relevant diagram from the linked to website to illustrate what I mean:

I see the enthalpy of hydration is decreasing in magnitude but it is becoming more positive so why is it said to decrease? Is this a convention of some sort?

It is used this way because the values are closer to zero, i.e. the relative magnitude is smaller ignoring the sign because the signs are the same.  But it can be confusing: what if the hydration was +322, then one would have to say it had increased.

@porphyrin If I'm not mistaken, hydration enthalpies are always exothermic, right? Likewise for lattice formation enthalpy. And thank you for your comment, this makes more sense now.

I was just giving an example irrespective of being correct chemically :)


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What is the solublity of water in liquid carbon dioxide at low temperature?
The solubility of water in liquid carbon dioxide has been investigated by Hosmer W. Stone in 1943 from
-29 to about 22 °C.
Is there any research done on the solubility of water in liquid $\ce{CO2}$ at still lower temperatures ?

The lower the temperature and higher the pressure, the more soluble CO2 is in water.  Pepsi and Coke have staudiesd this for years.

You have it backwards. The OP wanted solubility of water in liquid CO2 and actual data, not just a generalization.


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Repeating unit of poly lactic acid (PLA)
Which repeating unit for poly lactic acid (PLA) is correct, picture 1 or picture 2?
 and
I read somewhere that in polymerization, acids give $\ce{OH}$ and alcohol gives $\ce{H}$, creating water. According to this explanation, picture number 2 is correct but I have seen on websites like Wikipedia that picture 1 is correct. So, which is the correct picture and why?

In fact, both pictures show the same polylactic acid, only different disconnection points have been chosen:

The red disconnection (or your picture 1) is preferred because you can immediately see that it is a polyester.
Edit:
Your teacher is right in that, during the polymerization, the $\ce{-COOH}$ group loses $\ce {-OH}$ and the $\ce{-OH}$ group loses $\ce {-H}$ to form $\ce{H2O}$.
Therefore, your teacher prefers this arrangement:

If we draw it the other way, it are the wrong $\ce{-OH}$ and $\ce{-H}$ at the end of the chain, in the sense that it are not the ones that will give water:

However, note also that in either case, the part between brackets is not formed by identical fragments.
The atoms are indeed the same, but since not all are connected to the same atoms, they are not all chemically equivalent. There are different functional groups.
There are acid, alcohol and ester groups; mainly ester of course.
So in reality, we are writing $\ce{[...]_n}$, but those $n$ are not all the same and thus it is not correct.
If you want to depict the whole chain including the polymer chain and the ends that can continue polymerizing, the most correct way is to draw the end monomers explicitly.
At that point, you can place the brackets where you want, since the structures are absolutely equivalent.

Thanks for your answer, but my teacher don't accept picture 1 , he said that in polymerization,the acid gives OH and the alcohol give H so he chose the picture 2 because in this picture if you want extend repeating unit , you must add OH to acid (acid part in picture 2 without OH) and H to alcohol (alcoholic part in picture 2 without H) , so he said that pic 2 is correct.I said to him that both picture are correct but he don't agree with . on your opinion how can I convince him that both pictures are correct?If you have any resources , please write them,then I show the resources to my teacher.


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Why does SYBR green emit more signal when bound to DNA?
Principle of SYBR Green

SYBR Green I (SG) is an asymmetrical cyanine dye used as a nucleic
  acid stain in molecular biology.

  (wiki)

This molecule is commonly used in qPCR to meassure the amount of double stranded DNA. The fluorescent dye SYBR Green I binds to the minor groove of the DNA double helix. In solution, the unbound dye exhibits very little fluorescence (left figure below), however, fluorescence is greatly enhanced upon DNA-binding (right figure below).

(image source)

Question
Although its common use I cannot figure out why SYBR green fluorescence more when bound to DNA.

*I first doubted whether I should have posted this on biology stackexchange due to its relevance in this field, however I think this question covers way more chemistry than biology.

In dyes such as this rotation about bonds that can be involved in the conjugated part of the molecule contribute to radiationless transitions causing the excited state to transfer its energy to  a triplet or back to the ground state rather than fluorescing. When intercalated between base pairs as you suggest in your diagram, the molecule is held more rigidly and so radiationless transfer is reduced and fluorescence increased.
In molecules with a dipole it may additionally be the case that in a polar solvent the stabilisation due to the solvent polarity brings the initially formed excited state closer to a crossing to a charge-transfer state and this non-radiative process will lower fluorescence yield. Intercalated in the DNA the polarity will be lower than in water and again this will result in more fluorescence.  If your molecule changes wavelength in polar vs non-polar solvents then this is indicative of this process.


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Electroplating diamond grit
I want to do electroplating with diamond grit and make profiling, grinding and cutting disks. (see the attached photos)
I want to know the process to make the machine or the company whom is selling machinery to do it.
UPDATE :
My question is not about how to shape metal like this, I know how to do it.
My question is that, how can I electroplate diamond grit on the surface of these profiling wheels?

One cannot electroplate with diamonds, other methods have to be used to cover things with them.

Well, you can electroplate from a bath containing diamond bits, encapsulating some diamond in the film. It is a pretty standard industrial process.

@JonCuster Can you elaborate on that?

@JonCuster Please give me some documentations if you have. Any source or tutorial about it?

Googling 'electroplating diamond grit' produces a wealth of sites. Since it is an industrial process, likely each firm has their own 'magic recipe', but you could start at, say, https://www.dpcanada.com/en/articles/plating-brazing

@JonCuster I have done the googling a lot and have read your shared link many times, None of the sources on internet has referred to details of the process, Until now, I know how to plate nickle on the surface of work piece but the problem is plating diamond grit with nickle together on the surface of work piece, one guy has told I should suspend diamond grit in electroplating bath but had not give more detail about how to do that!

Yes, likely a problem for such processes.  Each company has their trade secrets.

Tou can make the form blanks and send them to norton abrasives and they can diamond electrolyte them for you. It doesnt cost as much as you would think, and when they are worn out just send them back. If you go on thier website they have all of the info in printable form in the diamond abrasives catalog. It covers dimensions for blanks to be undersized before coating. I am a toolmaker we use them for fine radii on punches and other tooling. Hope this helps. James

Maybe these links would help:
US3046204A - Method for making diamond tools - Google Patents
https://patents.google.com › patent
The Basics: Electroplating & Vacuum Brazing - Diamond ...
https://www.dpcanada.com › articles › pl...
US4079552A - Diamond bonding process - Google Patents
https://www.google.com › patents
https://www.jstage.jst.go.jp/article/matertrans/43/10/43_10_2604/_article/-char/ja/
https://www.sciencedirect.com/science/article/pii/S0263436808001649
https://patents.google.com/patent/US6286498B1/en


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What do 1T and 2H mean in front of compounds?
I have seen 1T and 2H used as prefixes for metallic compounds in literature, but I am having trouble finding what they mean. For example, 1T-HfS2 or 2H-HfS2.

These are notations that are used in identifying polytypes in layered compounds.
2H indicates 2 layers per Hexagonal unit cell (H for hexagonal).
1T indicates 1 layer per Trigonal unit cell (T for trigonal).
At least, that's my understanding.

This is the correct answer. Quite common in describing SiC crystals.


text:
What is the effect of temperature changes on rate of chemisorption?
In chemisorption, if temperature increases, then after some time why does its rate decrease? I tried to explain it by Le Chatelier's principle but I don't believe it's valid as chemisorption is irreversible, so there won't be any equilibrium.
I also wonder what effect would the decrease of the temperature have on chemisorption. Going by the above pattern I believe it should be the reverse but I am not sure.

If the surface is of finite extent, when full no more molecules can be adsorbed. In the simplest (Langmuir) model, the rate of adsorption is proportional to the fraction of the surface that is empty.


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Why are highly branched alcohols solid?
Usually branching (or decrease in surface area) leads to increased volatility. But in case of alcohols, my book states that:

The higher alcohols (butanols to decanols) are somewhat viscous, and some of the highly branched isomers are solid at room temperature.

Here the decerease in surface area did not increase the volatility; on the contrary, volatility has considerably reduced.
What is the reason of this anomaly? Or is this even an anomaly?

https://chemistry.stackexchange.com/questions/65933/why-is-the-melting-point-of-tert-butyl-alcohol-140-c-higher-than-that-of-sec-bu/65983#65983

Hydrogen bonding might break the simple model that works for hydrocarbons.

Well, if you're talking about melting points, the simple argument that more compact molecules should have higher mp because of increased dispersion forces can easily fail.  The problem is that melting is a much more complicated phase transition than boiling, with significant entropic effects coming from considerations of packing and entanglement -- especially relevant for floppy large molecules -- so that arguments based purely on the enthalpic changes due to intermolecular forces can easily be insufficient.


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How to synthesize a material at 2000atmospheres?
I am trying to combine various elments and at high pressure. How to achieve these pressures for a gas and in quantities of at least 1 cubic centimeter.

With a compressor. You can do a little plug reactor like a column for a HPLC and make your reaction inside. I know that for polyethylene reaction takes place in plug reactors which are about one kilometer long and the pressure inside is around 2500 to 3500 atm! :)

@Hexacoordinate-C Good Idea. I am considering using high temperatures as well. Maybe this is a problem for plug reactors?

I will try to give some more information knowing what you added in the comments. As you didn't precise which temperature I can't answer you in a very precise way.
However as I already said you can try to figure out how a reactor for polyethylene's production is made. If you're able to have access to some engineering literature, then you should have a look to tubular and autoclave reactors. If not you will have to contact someone or a company to explain what you want and get maybe an answer. In any case I doubt that having a reactor able to handle both ultra high pressure and (ultra) high temperature could be cheaper. The main reason is the savety. I bet that if you need such high pressure and temperature, you might use some dangerous chemicals and if not you have the savety and security problems of high pressure and temperature.
I found something that may be interesting for you here. You can achieve a pressure of 2.000 bars at a maximum temperature of 427 °C. On these reactors you can go up to 10.000 bars but with a lower temperature.
I would advise you to contact them they might have some solutions for your problem.


text:
Bond order analysis in molpro
I'm trying to compute bond orders in a molecule, let's consider S2 as an example. I'm not entirely sure what's the right way to do that, but I assume NBO (Natural Bond Orbital) should give me what I need. For calculations I'm using MOLPRO. I've tried to launch it with UCCSD(T)-F12 as follows:
 symmetry, nosym
 geometry = {
 S1
 S2, S1, bond
 }
 bond = 3.60529338 BOHR
 basis = cc-pVTZ-F12
 set, spin = 0

 hf
 uccsd(t)-f12a, scale_trip = 1
 nbo, keep_wbi = 1

It seemed to work and gave me the following output:
  - - - - - - - - - - - - - - - - - - - - - - - -
        NBO  OCCUPATION       LABEL     CENTERS
  - - - - - - - - - - - - - - - - - - - - - - - -
         1    2.000000         CR       1
         2    1.999827         CR       1
         3    1.999918         CR       1
         4    1.999989         CR       1
         5    1.999848         CR       1
         6    2.000000         CR       2
         7    1.999827         CR       2
         8    1.999918         CR       2
         9    1.999989         CR       2
        10    1.999848         CR       2
        11    1.999151         LP       1
        12    1.989845         LP       1
        13    1.999151         LP       2
        14    1.989845         LP       2
        15    2.000000         BD       1    2
        16    2.000000         BD       1    2

CR seems to stand for core, LP - lone pair, BD - bonding. From this I can conclude the bond order is 2, since we have two bonding orbitals with 2 electrons on each. However, the spin state was set to 0 here, whereas it's supposed to be 2 (2 unpaired electrons). When I tried to set spin to 2, the same calculation failed:
No suitable structure found!
 ? Error
 ? The problem occurs in nbo_transform

Can anyone suggest what is wrong here? Are there any other ways to compute bond order?
The full outputs: spin=0; spin=2

You could try the many methods by Multiwfn for bond orders.


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What do numbers in parenthesis after elements respresent?
I have seen elements with a number in parentheses after them in literature, but cannot find what it means. What does the number in parentheses after an element represent, such as in W(100) or Mo(110)?
https://aip.scitation.org/doi/abs/10.1063/1.1727919
https://www.sciencedirect.com/science/article/pii/0039602880900886

Something like this? https://aip.scitation.org/doi/abs/10.1063/1.1727919 "The interaction of hydrogen with a single‐crystal tungsten (100) surface has been studied"

It refers to the crystalline plane which forms the surface.

These are called Miller indices and they identify a particular surface of a crystal.  (You can form different faces by cutting the crystal in different ways.)  When you see something like "Mo(110)" it means "the 110 surface of a crystal of molybdenum."

An explanation of what miller indices are would probably help a bit (even if you don't get too technical).


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What are the functional group names of halogenoalkanes?
The functional group name of other organic compounds is easily findable, but that of the halogenoalkanes (or haloalkanes) is not. This makes me wonder if such a name exists for the halogenoalkanes or if they are an exception.
Are there such names? If yes, what are they for the different possible halogens present in a given compound?
Clarification: With functional group name I mean hydroxyl for the alcohols, phenyl for arenes, etc..

They don't have a functional group name similar to that of hydroxyl/phenyl/etc. Instead, you've to name them as if they were an alkyl branch - "haloalkyl" - and that too within parentheses.

Have a look at this simple comparison:

Notice that both 2,3-dimethylbutyl and chloromethyl are both within parentheses, whereas phenyl isn't. Therefore don't mistake "haloalkyl" to be a short name as "phenyl" is. There simply is no such short name, and you've to name it as an alkyl chain instead.


text:
Why is SF6 inert towards hydrolysis?
Why is $\ce{SF6}$ inert towards hydrolysis? Would not $\ce{H - F}$ bonds be created in water which is much stable and has a high negative enthalpy of formation. This hydrogen bond making must disintegrate the whole compound.

Related: https://chemistry.stackexchange.com/questions/42705/why-is-sulfur-hexafluoride-more-stable-than-selenium-or-tellurium-hexafluoride/42707#42707

In short its kinetically stabilised, attack at S is close to impossible due to steric hindrance and rigidity of FSF angles. Thermodynamically with water it is very unstable with respect to HF$_{aq}$ and H$_2$SO$_4$. Creepy ...

Hello I was wondering if the fluorine atoms would make hydrogen bonds with water molecules, since the HF bond is very stable this bonding must release enough energy to break the SF bonds,but I suppose I'm wrong, can you tell me why?

The group electronegativity (see Huheey) of -SF$_5$ is about 3.0, that means F in F-SF$_5$ is not very strongly polarized (F is anyway quite hard), so its a bad hydrogen bond acceptor, thus the F-S bond is not really substantially weakend. The only "dangerous" attack would be a nucleophilic one at sulphur by water lone pairs, but that is sterically hindered.


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Finite probability from discrete distribution and zero probability from continuous distribution in Maxwell-Boltzmann statistics
I have difficulty understanding a certain concept with the derivation of the Maxwell-Boltzmann distribution $f(v)$ function from Boltzmann statistics. The derivation starts with the Boltzmann statistics formula that gives the probability for an energy $E_x$:
$$p(E_x) = \frac{\exp(-E_x/kT)}{\sum_{i=0}^\infty \exp(-E_i/kT)}$$
From this formula the Maxwell-Boltzmann distribution function is derived:
$$f(v) = \left(\frac{m}{2\pi kT}\right)^{3/2} \cdot \exp\left({\frac{mv^2}{2kT}}\right)$$
I understand that this formula gives the probability density of a particle with velocity $v + \mathrm{d}v$. The reason that the formula gives the probability density for velocities between $v$ and $v + \mathrm{d}v$ is because velocity and thus energy are considered continuous, and therefore just 1 specific value of energy or velocity would have a probability of 0. Therefore, a range of speeds must be taken to give a non-zero probability.

If one specific value of energy or velocity should give a probability of 0, how come the formula for the Boltzmann statistics gives a non zero probability for a specific value of energy $E_x$?
And how can one reason a formula that considers a specific value of energy or velocity to be zero from a formula that doesn't? Is there a way to explain this?

I don't find this to be a duplicate of my other question. My other question states a misconception about translational energy being continuous in general. This question states how the derivation is done between two formulas, one considering energy to be continuous and the other doesn't.

Read here: https://en.wikipedia.org/wiki/Borel%E2%80%93Kolmogorov_paradox

Possible duplicate of Derivation of Maxwell–Boltzmann statistics

@Mithoron his question 2 seems to boil down to "how can there be a finite number of permutations of a continuous continuous value". This question boils down to "how can we say a given energy has zero probability even though the Boltzmann equation gives a nonzero probability". They both in part stem from the same misconception about how to treat probability densities, but I don't think they are the same question.

What is the probability of picking a real number in the interval $\ce{[0,1]}$?

The probability of any one specific velocity is not zero, it's just very, very small.  Infinitismal, or tending towards zero, is how we usually put it.  The reason is because the spacing between allowed translational states is inversely proportional to the size of the container, and when you are talking about a macroscopic container (a bottle of gas, say, or beaker of liquid) then from the atomic point of view this is gargantuan and the spacing between allowed translational energies very, very small.  (If you want to get a feel for this, look up the allowed energies for a particle in a box.  You'll see that the size of the box goes in the  denominator and as the box gets bigger the spacing between energy levels gets smaller and smaller.  You can amuse yourself by calculating the exact difference between the 1st and 2nd energy levels for, say, an H atom in a box 1cm x 1 cm x 1cm.  It's really small!)
So small that to do the sum over states in your top formula is a gigantic pain, quite challenging.  But if you convert it to an integral, the integral can be done pretty easily.  So that's what we do.  We replace a huge sum, over 10^20 terms or more, each of size 10^-20 or less, with an integral.  Think of it as the reverse of when you learned integral calculus, and in order to figure out an integral you replaced it with a large sum -- and let the size of each term go to zero while the number of terms went to infinity.
Here, you are taking a sum which has a very large number of terms, each of which is very small, and replacing it with an integral.  It's not exact, but it's very good, and it becomes perfect in the limit of an infinite-sized container, and anyway the errors for ordinary-sized containers are ridiculously small, unmeasureable.

Thanks for your answer. I have two questions if you don't mind.
1. With "the sum over states", do you mean the sum over each state's probability?
2. Regarding my top formula in my OP, if you fill in a value of say 5 for the energy Ex, would the formula then give the probability for particles having an energy of exactly 5.00000000...... or would it give the probability of particles with energy that is within a certain range of 5? If it's the latter, how large is that range regarding my top formula then?

@JohnnyGui 2. The numerator in the probability is for a single $E_x$ only. If you wanted to consider some energy "width", you would need to add up all probabilities, one for each possible energy, within that width. 1. "Sum over states" is referring to the denominator, which is the partition function. The article says it well:

@JohnnyGui "In classical mechanics, the position and momentum variables of a particle can vary continuously, so the set of microstates is actually uncountable. In classical statistical mechanics, it is rather inaccurate to express the partition function as a sum of discrete terms. In this case we must describe the partition function using an integral rather than a sum."

Got it. Does this mean that, clascially, we can NOT use the first formula in my OP to calculate probabilities (without being too inaccurate)?

@JohnnyGui like pentavalentcarbon said, in "classical statistics" the concept of number of microstates cannot be defined at all; the statistical weight is simply the volume in phase space $\Delta p \Delta q$. In a "quasi-classical" sense, given $s$ degrees of freedom, we can write down the number of states $\Delta \Gamma = \frac{\Delta p \Delta q}{(2 \pi \hbar)^s}$

@JohnnyGui That is correct. Because of that, the formula is a bit poor: for quantum systems, or those with a finite/countable number of states, we do perform that summation to calculate the partition function. It isn't infinite. Say, for a single boson that can occupy 3 different non-degenerate levels, $i\in{0,1,2}$. The formula is still incredibly useful, just not for classical behavior.

@JohnnyGui Also, a technicality of the site: if you want me to read a comment, you'll have to ping me (at sign, then username), because I'm not the author of the answer.

As has been mentioned already, your concern is valid: from a mathematical perspective, the probability that a continuous random variable $E$ takes a specific value $E = \epsilon$ is indeed zero. How do we rectify this issue?
(Note that the probability distribution $P$ and the probability density $f$ mentioned here and in the problem statement refer to the velocity vector $\mathbf{V}$ and not the energy $E$, so they should be parametrized with respect to $\mathbf{V}$ and not to $E$. With regards to notation, we will take the upper-case variables $\mathbf{V}$ and $E$ to be random variables and the lower-case variables $\mathbf{v}$ and $\epsilon$ to be specific values that those variables can take.)

Correspondence principle. Assume independence in $x$, $y$, and $z$, so that $$P(\mathbf{V}=\mathbf{v}) = P_1(V_x = v_x)P_1(V_y = v_y)P_1(V_z = v_z),$$ where $P_1$ is the one-dimensional probability distribution of velocities. Let our system exist in a finite cubic box of volume $L^3$, and consider the limit as $L \to \infty$. We can partition all of phase space so that each point of the probability distribution $P(\mathbf{V}=\mathbf{v})$ is associated with cubes of identical volume in phase space. As $L \to \infty$, the number of points grows (the quantum of translational energy becomes smaller), and the volume associated with each point shrinks. In the continuum limit we can replace each point by its associated infinitesimal cube; i.e., we substitute $$P(\mathbf{V}=\mathbf{v}) \to P(\mathbf{v}-\boldsymbol{\delta} < \mathbf{V} < \mathbf{v}+\boldsymbol{\delta}) \approx f(\mathbf{v})\times(2\delta)^3$$ for $\boldsymbol{\delta}$ a constant vector with every element $\delta$, where $2\delta$ is the spacing between neighboring points. The purpose of this entire construction is to show that $P(\mathbf{V}=\mathbf{v}) \propto f(\mathbf{v}) \propto e^{-\beta mv^2/2}$, for then the proportionality constant can be recovered by a simple normalization.
Classical statistical mechanics. An important result from this subject is simply that $$f(\mathbf{r},\mathbf{p}) \propto \exp(-\beta\mathcal{H}(\mathbf{r},\mathbf{p})),$$ which yields the desired probability density directly. It's hard for me to elaborate, because I haven't found a satisfactory classical proof of this result. An interesting approach is here, but it has its own problems.

In either case, it's formally inappropriate to go from a discrete to continuous model without likewise going from a probability distribution to a probability density. Nevertheless, the physical motivation and ideas behind such a switch should be rather clear.

I have a hard time understanding how it is concluded that P(V = v) is proportional to exp(-Bmv^2/2) and I can't find a satisfactory source that explains this. For example, where does the pi come from in the distribution function f(v) when treating the distribution classically?

@JohnnyGui, reading over my original response again, part 1. seems possibly incorrect (and at least quite confusing) and probably needs to be modified, so I wouldn't put too much stock into it. The $\pi$ comes from normalization of the probability density.

Thanks for your response. Is there a way to explain what the normalization entails?

@JohnnyGui, the normalization condition is that $\int\text{d}v,f(v) = 1$. Since we know $f(v) = Ne^{\beta\epsilon}$ for a constant $N$, the normalization condition is an equation by which $N$ can be identified.

This helped me understand it better, thanks. One other question, if f(E) is the probability distribution density of the energylevels, does that mean that f(E) is basically the derivative of p(Ex) (the derivative of the first equation in my OP, but now considering Ex as continuous)?

@JohnnyGui, no. Your first equation deals with the discrete probability $P(E=\epsilon)$ of being in a certain state, which would correspond to a probability density $p(E=\epsilon)$ in a continuous system. The probability density is the derivative of the cumulative distribution function, which is the probability $p(E<\epsilon)$.

This makes sense to me. I have 2 questions to make sure I understand this: 1. If I integrate the probability density function all up to a certain energy level Ex, then this would give me the cumulative probability p(E < Ex), correct? 2. To calculate the probability of a certain interval p(Ex to Ex + dE ), then I would have to integrate the probability density function all up to Ex and then substract this integration from the integration of the probability density function all up to Ex + dE, correct? (I can't type the epsilon sign in a comment, so I have substituted it with Ex)

@JohnnyGui, yes and yes. But note that for your second question you're essentially just calculating $$\int_\epsilon^{\epsilon+\text{d}\epsilon}\text{d}E,p(E) \approx p(\epsilon),\text{d}\epsilon.$$

Thanks for verifying. I blacked out in my second question and your integration does indeed make sense to me. Reading other sources, I noticed that the continuous probability distribution f(v) (the second equation in my OP) is derived by reasoning the probability of finding particle within infinitesimal momentum increments in 3 dimensions that form an imaginary shell of a sphere. Does that mean that one can not use the discrete probability function in order to derive a continuous density function, and that it has to be derived from scratsch using other parameters?

@JohnnyGui, you can, kind of. Your first equation is an equation over microstates. Classical statistical mechanics works on phase space. The semiclassical association between microstates and phase space is to partition phase space up into little hypercubes of side length $h$, and treat each little hypercube as a microstate. Then we have $P(\omega) \approx f(v) h^3$, where $\omega$ denotes a microstate, and also $P(\omega) \propto e^{-\beta\epsilon(x,v)}$ from your first equation, so then $f(v) \propto e^{-\beta\epsilon(x,v)}$. And this comment should really have been my answer.

Thanks a lot for this. I have one last question if you don't mind. I don't quite get how a probability density f(v) multiplied by a unit of volume (h^3) would give a discrete probability P(w). Is there a way to explain this?

@JohnnyGui, note that $h$ is small. Then $$P(v_x'\in(v_x,v_x+h),v_y'\in(v_y,v_y+h),v_z'\in(v_z,v_z+h)) = \int\text{d}\mathbf{v}' f(\mathbf{v}') \approx f(\mathbf{v}) h^3.$$ We claim that this phase space cube is associated with a particular microstate, so their probability must likewise be equal.


text:
Intermolecular interaction between aniline and dichloromethane
Short background
As it is known there are interactions between:

Charges
Charge and dipole
Hydrogen bonding
van der Waals (VDW) forces

From stronger (1) to weaker (4).
VDW forces are divided into dipole-dipole, dipole-induced dipole, induced-induced dipole; also from stronger to weaker.
The problem
Reading about solutions I found interesting the intermolecular forces involved on aniline and dichloromethane mixing. Below both structrures are shown:
Aniline
 Dichloromethane
Question
What is the most important intermolecular force between those molecules and why? It might be interesting to see an AIM analysis of aniline, but I have no idea how to do that. Also, I'm not sure about what experiment could give valuable information on this case.
My attempt
Those previous interactions aren't mutually exclusive. $\ce{-NH2}$ might be a negative density region of the molecule (inductive attraction), but it is also donor by resonance, so the ring has partially a negative charge density. In consecuence, it is not easy to see the dipole. Dichloromethane, on the contrary, is an easy-to-see dipole.
It might result that London dispersion is the most relevant interaction here, but I can't very well justify it.

According to this NIST publication (caution: link leads directly to a biggish PDF file) the dipole moment in the gas phase of dichloromethane and aniline are quite similar (1.6 and 1.5 D respectively).  These are respectable dipole moments, and neither molecule is very big, so offhand I would say an argument can be made that the dipole-dipole interaction is most important, particularly because it's orienting.
Note by the way that rating dispersion forces as the weakest should be done cautiously. It works for small molecules, but since dispersion forces scale with the number of electrons in a molecule and no other interaction does, once you get above a fairly modest size of molecule dispersion forces (along with excluded volume forces) dominate.  That's why eicosane is a solid at room temperature while water is a liquid.

Very interesting, do you find it easy to see the dipole direction on aniline? I'm in trouble with that...

Nope.  That's why I looked it up.  For what it's worth, I could rationalize it by arguing that the entire C6-N system has some pi delocalization going on because it lowers the energy of the pi electrons, to be able to spread out over all 7 atoms.  But that would necessarily mean the N has a partial + and the ring a partial - because the lp is not always on the N, roughly speaking.  Hence, a dipole.  But that's all ex post facto hand waving.  One would have to do a quantum chemistry calculation to know whether it checks out.

From a VB point of view it would suggest a contribution from the resonance structure with NH2+ on one end (with a double bond to C1) and a carbanion at the para position of the ring.  My o-chem is rusty, but isn't that res str the reason NH2 is para (as well as ortho) directing for electrophilic substitution?

Cool, thanks :) What is VB? Yes you are right with substtutions, that could be seen in some calculations also, I suppose. So the dipole is possibly positive on N and negative on para... interesting.

Sorry, VB = "valence bond theory, " I mean if you are trying to explain the electronic structure by some combination of Lewis structures.


text:
Molar heat of combustion with different fuels

I'm not sure exactly how to attack this question, I'm sure that you have to use equations such as
$$Q=mc\,\Delta T$$
$$n=\frac{M}{m}$$
But I don't know where to use them. Any help would be appreciated
I just attempted the question, I'm not sure if this is the correct working.
Assuming 1 mol of each substance
Ethyne:
$$\begin{align}n&=\frac{M}{m}\\[6pt]
1&=\frac{26.038\ \mathrm{g/mol}}{m}\\[6pt]
m&=26.038\ \mathrm g\end{align}$$
producing 1630 kJ
Hexane:
Same working as above so:
$$m=86.178\ \mathrm g$$
producing 1300 kJ
Ethane:
$$m=30.07\ \mathrm g$$
producing 1560 kJ
Simplifying each of the mass and energy relationships for each substance gives:
Ethyne: 1 gram for 62.6 kJ
Hexane: 1 gram for 15.09 kJ
Ethane: 1 gram for 51.89 kJ
This would imply that ethyne would be the best fuel, as it uses the least amount of mass and the greatest amount of energy is produced.
Not sure if this is correct. Any tips would be great.

I think you are on the right track and already have the answer (which is the highest energy density per unit mass). Also this is an exemplary way to pose questions like this on this site as you showed your working (many similar questions get closed because people don't bother to show their thoughts).

This question is asking which substance has the greatest heat density, or energy released per mass combusted (something like J/g).
$\ce{C2H2}$ has a mass of $26.04\ \mathrm{g/mol}$. Since the $\Delta H$ in its combustion is $1\,630\ \mathrm{kJ/mol}$, you will be looking at a heat density of $\frac{1\,630\ \mathrm{kJ/mol}}{26.04\ \mathrm{g/mol}} = 62.6\ \mathrm{\frac{kJ}{g}}.$
Similarly, for $\ce{C6H14}$, you get $\frac{1\,300\ \mathrm{kJ/mol}}{86.178\ \mathrm{g/mol}} = 15.1\ \mathrm{\frac{kJ}{g}}.$
For $\ce{C2H6}$, you get $\frac{1\,560\ \mathrm{kJ/mol}}{30.07\ \mathrm{g/mol}} = 51.9\ \mathrm{\frac{kJ}{g}}.$
It looks like you were on the right track.


text:
Understanding elementary rate laws from a probability stand point
I am learning about chemical kinetics and dynamics and as I understand for a general chemical reaction
$$\ce{aA + bB -> cC + dD} $$ whose reaction rate, r, can be described by an elementary rate law can be written as follows: $$r= k[\ce{A]^a[\ce{B}]^b} $$
What I do not seem to conceptually understand is if you look at a chemical reaction from a collision theory standpoint, would it not be true that as one reactant that is in a higher stoichiometric ratio relative to the other, the probabilty that the reactants collide (assuming any collision gives a reaction) would be inversely proportional to the stoichiometry of the reactant. For example in the following:
$$\ce {A + 2B -> C} $$
You basically need 2 mols of B for every mol of A for a successful reaction to produce C, so if the probability that A successfully collides with a mol of B is $\propto$ to $1/z$, then the probability that A successfully collides with TWO mols of B  $\propto 1/z^2$ which is lower, so why does a reaction rate, which is $\propto$ to the probability of a collision, with an elementary rate law $\propto$ to a concentration raised to an exponent?

Are you acquainted with the Gillespie algorithm?

$k$ is different for different reactions, see my comment to one of the answers.

Let's divide the reaction volume into V little boxes and randomly throw ${\rm N}_A$ molecules of A into it and ${\rm N}_B$ molecules of B.  The probability $p_A$ that a given box has a molecule of A in it is ${\rm N}_A$/V, in the limit that $V >> {\rm N}_A$, i.e. an ideal dilute solution, and likewise the probability $p_B$ that it has a molecule of B in it is ${\rm N}_B/V$.  The probability it has two molecules of B in it is ${p_B}^2$, again in the limit that $V >> {\rm N}_B$ and ${\rm N}_B >> 1$.
Consequently, the probability all three atoms are close to one another is $({\rm N}_A/V)({\rm N}_B/V)^2$, which is proportional to the terms in your rate law.  (This isn't a derivation, by the way, but only a rationalization I hope is appealing to intuition.  I've swept all kinds of detail and qualification under the rug.)
Where your intuition may be leading you astray is in the assumption that squaring the molarity of B makes the RHS of your rate law bigger, and that doesn't match up with your intuition that it should be smaller if it involves a less likely event (a trimolecular collision).  But that is not the case.  Squaring numbers only makes them bigger if they are greater than 1, below 1 it makes them smaller.  And of course the prefactor will be different, too.  What the squaring really does is change the dependence of the rate law on changes in B concentration. It says the rate law is much more sensitive to changes in B concentration than A concentration.  And, given that you need two B molecules to collide with each A molecules, that should be intuitively compelling.

If all concentrations are 1 M, you might think reactions have the same rate no matter what the rate law. However, the rate constant $k$ depends on the reaction. For termolecular reactions (three particles colliding), the expectation is that the rate constant is “supersmall” compared to dimolecular reactions.


text:
What are the predominant acids in sphagnum peat moss?
I'm trying to figure out what the predominant acidic substances in peat moss are. Peat moss is said to have a pH of approximately 4.0. So, I'm curious what is making it acidic.
I know not all acids bind to various elements in the same ways. So, I'm trying to find out which acids (and/or acidic salts) are present in peat moss in order to aid me in my research to discover approximately how long it might take to acidify soils with it compared with the traditional methods of adding various forms of sulfur to soil. You don't need to tell me how long it takes, by any means. I can do more research after I find out which acids or acidic salts are in peat moss, in what levels.
I'm not familiar with this particular SE site's customs. So, if this is the wrong place to ask, or if I used the wrong tag(s), feel free to let me know. I figured people might be more likely to know here than on Gardening and Landscaping, though.

This 1992 article by Gagnon and Glime suggests that the ability of Sphagnum to lower pH depends on the presence of cations in the surrounding environment and is not (contrary to prevalent belief) due to organic acids in the plant itself. Here is a more recent article from New Phytologist amplifying that cation exchange is responsible for lowering of pH.
If you want to acidify soil using sphagnum/peat--assuming the authors are right--it's more important what cations you put into the soil. The sphagnum will then exchange them for $H^+,$ lowering the pH of the soil.
This does not mean there aren't organic acids in moss. You would have to look in specialized journals to determine what acids are found in the plant, but they are not necessarily very acidic and are, according to both articles, not the cause of acidity of the soil associated with the plants.

By "in the plant" do you mean in the peat moss?

@Shule: Yes I mean in the peat moss.


text:
Why is it a reduction from a ketone to an alcohol?
Why is it a reduction from a ketone to an alcohol (using $\ce{LiAlH4}$)? I'm not able to figure out the oxidation state of carbon.

Related: https://chemistry.stackexchange.com/questions/70488/oxidation-state-of-propene/70504#70504

As you can read from the link mentioned in the comment to the question, oxidation number is not a useful concept when dealing with organic compounds.
So we won't be determining oxidation number of atoms to know whether oxidation or reduction has happened.
We can follow the basic definition of oxidation and reduction to determine which one of the two is taking olace:

Oxidation is gain of oxygen or loss of hydrogen.

Reduction is loss of oxygen or gain of hydrogen.

So in case of conversion of a ketone into a 2° alcohol, two atoms of hydrogen is add for one carbonyl group.
And since hydrogen atoms are being added to the molecule (here, ketone), we can conclude that the molecule has been reduced.

Like Ivan already states in Oxidation State Of Propene, there is limited use of oxidation states in organic chemistry. However, you can still apply the schemes that I have written about in Electronegativity Considerations in Assigning Oxidation States.
In the following reaction I use $\ce{R_{1,2}}$ in place of alkyl moieties.

A case can be made for the convention that alkyl moieties have an overall oxidation state of zero.
After that you'll assign $-2$ to the oxygen in the ketone; and since the molecule is neutral the carbon has to have the oxidation state of $+2$.
In the alcohol you'll also assign $-2$ to the oxygen atom. Hydrogen is the least electronegative element and it can give away one electron, leaving it with an oxidation state of $+1$. The hydroxyl moiety therefore has an overall oxidation state of $-1$ affecting the carbon. Since there is another hydrogen the overall oxidation state of carbon is $0$.
The carbon formally goes from $+2$ to $0$, i.e. it gains electrons, therefore it is reduced.
Note that oxidation states are not actual (partial) charges. Below you can see the partial charges of acetone and 2-propanol optimised on the DF-B97D3/def2-SVP level of theory. You might note that they are quite different from the oxidation states.


text:
Polyamides Reaction mechanism
I already know that polyamides are made through condensation polymerisation with a carboxylic acid and an amine. But how exactly does the mechanism work?
If the reaction takes place in acid for nucleophilic acyl substitution, the amine group would get an extra proton and as a result the nucleophilic acyl substitution can't happen, like it would for an carboxylic acid and an alcohol to form polyesters.

Commercially the diacid/diamine mixture is heated (>250C) to overcome the amine protonation and remove water or by formation of the di-acyl chloride followed by reaction with the diamine. Reference here 1.
Nylon 6 is also formed by adding a little NaH to caprolactam, mechanism here 2


text:
Comparing the pKb of piperazine and methylpiperazine

In comparison to piperazine, the inductive effect will cause the $\mathrm{p}K_\mathrm{b}$ of methylpiperazine to be:

The answer given is that the $\mathrm{p}K_\mathrm{b}$ will be lower due to electron donation by the methyl substituent. I understand that the methyl group stabilizes through the inductive effect by releasing electron density. However, wouldn't this make the $\mathrm{p}K_\mathrm{b}$ higher (a weaker base) since it's more stable and less likely to react?

In terms of the logic employed by your answer key
Whether an inductive effect is stabilizing depends on the nature of the molecule and reaction you're thinking about.
You're right to think of the methyl group as inductively release electron density into the ring. However, this inductive release from the methyl group is actually increasing the electron density in the piperazine ring, which in turn makes the N lone pair higher in energy and more reactive, thus making it a stronger base. However!
What is actually the case in real life
I was able to find a detailed study of $\mathrm{p}K_\mathrm{a}$ values of protonated piperazines and methylpiperazines, and assuming $\mathrm{p}K_\mathrm{b} = 14 - \mathrm{p}K_\mathrm{a}$.
Then at 298 K...

$\mathrm{p}K_\mathrm{b}\text{(piperazine)} = 4.27$
$\mathrm{p}K_\mathrm{b}\text{(N-methylpiperazine)} = 4.86$
$\mathrm{p}K_\mathrm{b}\text{(N,N'-dimethylpiperazine)} = 5.62$
$\mathrm{p}K_\mathrm{b}\text{(2-methylpiperazine)} = 4.43$

This tells us that your answer key is actually wrong, and that in real life sterics may be playing a role in the basicity of the N atom. This data is in line with the fact that secondary bases tend to be slightly stronger than tertiary ones. For example, the

$\mathrm{p}K_\mathrm{b}(\text{diethylamine)} = 3.02$
$\mathrm{p}K_\mathrm{b}(\text{triethylamine)} = 3.25$