text:
How does epoxy stick to stuff?
According to Wikipedia what gives epoxy its holding power is apparently "ionic" interactions between the epoxy and the surface.
But I thought that epoxy was overall neutral when cured. How can it make ionic bonds with the surface it is bonded to? Or are there unprotonated alkoxides left in the epoxy which then make the bonds with the surface?

Good question... since epoxy adheres better to many metals than to aliphatic plastics, e.g. polyethylene, is it perhaps forming a metallic compound interlayer?

How anything adheres to anything is a [ahem], sticky subject.  The particular wiki paragraph you mention has no citation, and my bias is that this statement is in correct.  I would expect there are polar interactions between the epoxy and the surfaces, between the surface bonded epoxy and the bulk epoxy.  However, I don't have library access anymore to verify.

I can't say I've done much research, but this link looks interesting from an actual resin/adhesive company.

To sum it up, the adhesive property of an epoxy is the result of inter-molecular forces and the formation of mechanical bonding.

Adhesives cure when the small resin molecules (mers) join together to form extremely large molecules known as polymers.
Some adhesives are used only with specific substrates because of the cure time needed to allow for good bonding (both with the substrate and for polymer formation.)
I had trouble finding information saying that all epoxies use ionic bonding, and it  seems to me that most epoxies will actually just utilize intermolecular forces such as

Ion-induced dipole forces
Ion-dipole forces
Van der Waals forces
Hydrogen bonding

An important first step in applying an epoxy is the wetting of the surface and then the polymerization of the epoxy. The ability to wet is related to viscosity and the inter-molecular forces. If there is a low viscosity and strong attraction between the substrate and the adhesive then we can witness the formation of mechanical bonding. Below is an explanation:

In order for an adhesive to bond  two surfaces, there must be several types of interaction between the adhesive and both substrates. The first type of interaction is that the adhesive must wet the substrate, meaning that the adhesive must spread itself out into a film that covers the substrate surface. In order for this to happen, the adhesive must have a low enough viscosity so that it will flow. Because viscosity is temperature dependent, the application of a cold adhesive to a substrate, or the application of an adhesive to a cold substrate, may result in poor wetting. Another factor that affects wetting is the relative strengths of cohesive forces (between like molecules, such as two adhesive molecules) and those of adhesive forces (between unlike molecules, such as an adhesive molecule and a substrate molecule). If the cohesive forces among adhesive molecules are weaker than the adhesive forces between the adhesive molecules and the substrate surface, then the adhesive molecules will spread out over the substrate and wet its surface. An adhesive that has a relatively low viscosity and is able to wet the substrate surface will flow into any tiny cracks or pores on the substrate surface, thus promoting what is known as mechanical bonding. Mechanical bonding increases the strength of an adhesive bond and, as a result, a forced separation of the two substrate surfaces is more apt to tear the substrate surfaces.
Mechanical bonding is one of several ways that an adhesive bonds substrates. All surfaces, except those that are highly polished, have pores. If the adhesive flows into these pores and then polymerizes, a mechanical bond is formed. It is similar to placing a wick into liquid candle wax. Once the wax solidifies the wick can not be easily removed. A mechanical bond has formed.

Quotation and information from this answer were sourced here.


text:
Finding new pH after NaOH added to buffer solution
I have a buffer containing 0.2 M of the acid $\ce{HA}$, and 0.15 M of its conjugate base $\ce{A-}$, with a pH of 3.35. I need to find the pH after 0.0015 mol of $\ce{NaOH}$ is added to 0.5 L of the solution.
I started by converting everything to moles, writing out an equation for $\ce{HA}$ and $\ce{OH-}$ reacting to make $\ce{H2O}$ and $\ce{A-}$, getting 0.0985 mol of $\ce{HA}$ left after all the $\ce{OH-}$ was used up, and 0.015 mol of $\ce{A-}$ produced.
At that point I got a bit stuck. I tried doing things a bunch of different ways after that, such as trying to calculate $\mathrm{p}K\mathrm{_a}$ by using the Henderson-Hasselbalch equation, but I'm not getting anywhere near the answer. I calculated a $\mathrm{p}K\mathrm{_a}$ of 3.47. I added the amount of $\ce{A-}$ made when $\ce{HA}$ reacts with $\ce{OH-}$ to the original moles of $\ce{A-}$ in the buffer (I calculated 0.18 M $\ce{A-}$), and tried putting that, the leftover moles of $\ce{HA}$, and my $\mathrm{p}K\mathrm{_a}$ value into the equation, but that wasn't right either.
What am I doing wrong?

did you account for the change in volume?

Hmm no. Would I just assume a base of 1L originally for the plan buffer solution? I don't have a given initial amount.

you said you have 0.5L of the buffer at the start

Ok, right, yea, sorry. So short of googling density of NaOH which would be odd since the book doens't usually give you questions where you'd have too look up something outside of the book, how would I figure out the change in volume? In the other questions it gave me a volume of NaOH for each thing.

what is the final pH supposed to be?

The new pH is supposed to be 3.37

Start by calculating $\mathrm{p}K_\mathrm{a}$ using the Henderson–Hasselbalch equation and the initial data given. $$\mathrm{p}K_\mathrm{a} = 3.35 - \log\frac{0.15}{0.2} = 3.47$$
Calculate the amount of substance of $\ce{HA}$ and $\ce{A- }$ after addition of $\ce{NaOH}$. $$n_{\ce{HA}} = 0.15\times0.5 - 0.0015 = 0.0765$$ $$n_{\ce{A- }} = 0.2\times0.5 + 0.0015 = 0.0985$$
Calculate the final $\mathrm{pH}$. The final volume is not needed because it is the same for both concentrations and so it cancels in the equation. $$\mathrm{pH} = 3.47 + \log\frac{0.0985}{0.0765} = 3.58$$

the Henderson Hasselbach equation is actually
pH = pKa + log[A-]/[HA]
so using original conditions
3.35 = pKa + log[0.15/0.2]
Therefore pKa is 3.22
The rest of the equation is correct

source: Stack Exchange

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text:
Can neutralization be reversed?
So far, maybe it's just that my course isn't advanced enough, but I have never seen anything called 'neutralization constant'.
Any such reaction involving an acid and a base will be complete no matter the strength of either reactant.
Buffer works the same way. If the reaction is reversible then the buffet is not effective.

I imagine by "neutralization constant" you mean the inverse of the autoionization constant. For water that would mean $\ce{H^+ + OH^- -> H2O},\ \ K=1.0 \times 10^{14}$. This massive constant is why even weak acids or bases completely dissociate and react when faced against a strong base or acid, respectively.

The reverse reaction of neutralization is called hydrolysis ( X salt + H2O = acid + base )

But that's if you start with water and salt

Hmm then i guess i don't understand the question.

I tried again to understand the question hahah and i hope this time I'll help.It's not that any of these reactions will be complete and it depends on the strength of either reactant.Neutralisation between an acid and a base either of which  or both are weak will be a reversible reaction which means that it is not complete, on the other hand strong acid a bases in neutralisation will be complete and it will be a ireversible reaction.

It doesn't matter if you start with pure salt - reaction is the same.

In general under normal conditions of temperature and pressure the reverse reaction of the neutralization is not possible because of their (acids & bases)  tendency (release of H+ ion & OH- ion) to form salt water due to which there is no any constant so called "neutralization constant"
But as for the consideration, the reverse of the neutralization reaction is called hydrolysis. In a hydrolysis reaction a salt reacts with water to yield the acid or base:
BA + H2O → HA + BOH

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text:
What is the significance of an organic compound's M peak?
Background
On the topic of identifying an organic compounds using peaks generated in Mass Spectrometry, a rule of thumb expressed in the UK educational textbook, CGP, is that that "the M peak is the one with the second highest mass/charge ratio".
Question
What is the significance of this M peak anyway?

The largest peak is representative of the most stable isotope produced, referred to as the bass peak.

Cheers @LukeAutrey!

"M" stands for molecular ion.  It is often represented as $\ce{M, M^{+} or M^{+~.}}$. In the mass spec experiment, an electron is knocked out of some of the sample molecules.  The peak at M/q (q=1) represents the molecular weight of the entire molecule, less an electron, before it fragments to smaller ions.  From precise measurement of the molecular ion (e.g. 82.34567890) you can determine the exact molecular formula of your sample compound.  This can help greatly with sample identification.


text:
Why would the 2+ molecular ion of butane have a mass spectrum peak at 29?
Question 7 of Edexcel's June 2014 Unit 2 exam paper reads:

In a mass spectrum of butane, C4H10, where would a peak be seen for
  the molecular ion if it had a charge of 2+?

The answer is 29.
My question is: why?

The mass spectrometer observes the mass to charge ratio (m/z). Most of the time, small organic molecules like butane will be singly charged by the ionizer, which means that the mass can be read off directly since z =1.
If butane was somehow ionized to the +2 state, then the mass spec would observe 58/2=29, which is the mass of butane (58 g/mol) divided by the charge (+2).
This becomes particularly important for larger compounds with multiple easily ionizable groups, such as peptides. A series of peaks corresponding to different charge states can be used to identify the compound of interest.


text:
Why does hot steam produce MgO instead of Mg(OH)2?
Background research

Magnesium hydroxide exists naturally as brucite (source: wiki)
Passing hot steam over magnesium produces magnesium oxide (source: Edexcel's June 2014 Unit 2 exam (CH02_01): question 18)

Question
Why doesn't magnesium hydroxide form when treated with hot steam, instead of magnesium oxide?

There are two steps in the process:
$\ce{Mg + H2O -> MgO + H2}$
and
$\ce{MgO + H2O <-> Mg(OH)2}$
The equilibrium point in the second step is dependent on the temperature. See file Equilibrium Assemblages for more information.

The chemical equation has to be balanced

$\ce{Mg(OH)2}$ decomposes at high temps into $\ce{MgO}$ and $\ce{H2O}$. To be exact, 350 C


text:
Thermometer fluid is separated, defective?
I bought a thermometer from Cole-Parmer having kerosene as the fluid, and the fluid is broken up. The fluid in the bottom part goes up to about 10 °C even though it is 20 °C in the room, then there is a gap followed by fluid between 23 °C and 25 °C, then there is no fluid going all the way to the top where there is another small stretch of fluid. There is a short black rubber sleeve on the top of thermometer.
Does this mean the thermometer is defective or am I using it incorrectly, not “calibrating” it or something?

No, it is not defective, just slightly damaged, possibly from shipping. You should call customer service at Cole-Palmer and see if they will replace (or repair) the thermometer.
If not, there are 2 ways to fix it. I haven't tried either one, so I don't know how well they work.
If there is a bulb at the top and the thermometer is low range (under 150°C), then you would heat the thermometer in an upright position until all the bubbles are pushed up into the bulb.
The other method is to cool the thermometer down until all the liquid is in the bottom bulb. The second method may not work if the liquid freezes before all the liquid has moved into the bulb.


text:
Radial Distribution Function for a Hard Sphere Gas in the Percus-Yevick Approximation - Is There a Quick Way to Evaluate it Numerically?
The question is as in the title.
I know that for a one-component hard-sphere gas, there is an analytic solution in the $\vec{k}$-space (PRA, 5, 939). I know that to obtain the correct solution in $r$-space I just need to do something like a Fast Fourier Transform. The problem is that in doing so I can only obtain values for the radial function on a grid, and problem arises as to the magnitude of errors on the grid values, the errors in interpolating between the grid values, and so on.
I have to apologize for the fact that I rarely do fast fourier transforms, and even less estimate their errors.
Is there a standard way (e.g fitting formula) to do this? Thanks.

I seem to find a reference using density expansion (PRA, 43, 5418; "radial distribution function for hard spheres") for a one-component gas and a reference for a multi-component gas, (Journal of Chemical Physics, 108, 3683; "Structure of multi-component hard-sphere mixtures"). Other better references are much appreciated.

Indeed there are many references which claimed to give a formula for the evaluation of the radial distribution function for a hard-sphere gas, but many of them contains typos, such that they probably will not give the correct answer. One of them is the ref. given in the previous post (PRA, 43, 5418; "radial distribution function for hard spheres").

One ref. which seemed to give the correct result is, Molecular Physics, 1975, 30, 1505; "Hard sphere correlation functions in the Percus-Yevick approximation". I would like to stress that the radial distribution function $g(r)$, the total correlation function $h(r)$, as well as the correlation function $c(r)$ are all discontinuous across the hard-sphere diameter $d$, such that a naive (numerical) fourier inversion may or may not work.

While it is certainly true that $h(r)$ and $c(r)$ are discontinuous across $d$, their difference $h(r)-c(r)$ is not, as can be seen from the Ornstein-Zernike equation. This implies that the Fourier transform of $h(r)-c(r)$ is more well-behaved, and should succumb to a numerical Fourier inversion.


text:
Determining the Average Bond Enthalpy for the C-F bond
Given the following data, how would you work out the average bond enthalpy for $\ce{C-F}$ bond. I've tried setting up the chemical equations and applying Hess's Law, but that's not getting me anywhere.
$\Delta H_\mathrm f^\circ(\ce{CF4(g)})=-680~\mathrm{kJ~mol^{-1}}$
Bond enthalpy, $\ce{F2(g)}=+158~\mathrm{kJ~mol^{-1}}$
$$\ce{C(s) -> C(g)}\quad \Delta H=+715~\mathrm{kJ~mol^{-1}}$$
EDIT: These are the equations I used:
$$\begin{align}
\ce{C(s) + 2F2(g) &-> CF4(g)}\\[6pt]
\ce{F2(g) &-> 2F-(g)}\\[6pt]
\ce{C(s) &-> C(g)}
\end{align}$$

Welcome to Chemistry.SE! Did you take the stoiochiometry for $\ce{C + 2F2 -> CF4}$ into account?

@KlausWarzecha Yes, but I still couldn't get an answer. Am I taking the right approach by using Hess's Law?

Using Hess's Law is fine! Did you consider that you have 4 $\ce{C-F}$ bonds?

Your approach to use Hess's Law is reasonable!
\[\Delta H_r = -680 - (715 + 2\cdot158) = -1711\ \mathrm{kJ\cdot mol^{-1}}\]
That's the enthalpy for $\ce{CF4}$ - a molecule with four $\ce{C-F}$ bonds.
The average $\ce{C-F}$ bond enthalpy is smaller:
\[\frac{1711}{4}\ \mathrm{kJ\cdot mol^{-1}} \approx  427\ \mathrm{kJ\cdot mol^{-1}}\]


text:
Can metals have a net negative charge
Normal metals like sodium or Calcium have a positive charge as $\ce{Na}^+$ or $\ce{Ca}^{2+}$.
Transition metals have a loot of variable oxidation states.
Yesterday I read about Iridium in Wikipedia and it stated that

a +9 oxidation state has been found in $\ce{IrO_4^+}$

which is astounding for me...
So i was thinking can metals have a net negative charged ion also?
According to Google :

Ions: An atom or group of atoms gain an electric charge by gaining or losing an electron, usually through bonding. Cation: An atom loses electrons and is positive ex: $\ce{Ca^2^+}$. Anion: An atom gains electrons and is negative ex: $\ce{Cl-}$. Metals usually form cations while nonmetals usially  form anions. [source: chemwiki UCDavis]

Since it is written that metals USUALLY have positive charges which means there may be conditions or compounds when they have negative charges too. Is it possible?
Also can noble gases form ions too?

Addressing your noble gas ion question, they have been known for over 50 years to make compounds with formal positive charges on the noble gas atoms, though discrete noble gas cations (i.e. $\ce{Ng^{x+}}$) are not known in compounds.

More interestingly, regarding the possibility of noble gas anions, all noble gases up to radon have an endothermic electron affinity, and could not even in principle stabilize a negative charge on them. However, due to relativistic effects, the superheavy noble gas ununoctium may be able to form an isolated $\ce{Uuo^-}$ anion, though it would presumably be quite reactive. Whether a compound can be made with a $\ce{Uuo^-}$ anion is not known.

related: http://chemistry.stackexchange.com/questions/6923/bonds-between-metals-and-non-metals/6939#6939, http://chemistry.stackexchange.com/questions/26527/except-pure-alloys-are-there-any-compounds-with-more-metal-elements-proportion/26548#26548

@Nocolau there is no longer "ununoctium".  It's now oganesson, Og.

Search “Alkalide” for information on things like $\ce{Na-}$ and other first column negative states

@nicolau "discrete noble gas cations are not known in compounds." We are closer than one might think with radon-fluorine compounds, where a complex ion $\ce{RnF^+}$ is evident. See https://en.wikipedia.org/wiki/Radon.

These species usually do not exist in nature, but they can be synthesized.

Silver has been reduced in liquid ammonia to give $\ce{Ag-}$.
A lot of anionic metal carbonyl complexes $\ce{M(CO)_{n}^{m-}}$ have been synthesized:

-1

$\ce{[V(CO)6]-}$, $\ce{[Nb(CO)6]-}$, $\ce{[Ta(CO)6]-}$, $\ce{[Mn(CO)5]-}$, $\ce{[Ir(CO)4]-}$, $\ce{[Co(CO)4]-}$, $\ce{[Rh(CO)4]-}$

-2

$\ce{[Ti(CO)6]^{2-}}$, $\ce{[Zr(CO)6]^{2-}}$, $\ce{[Hf(CO)6]^{2-}}$, $\ce{[Fe(CO)4]^{2-}}$, $\ce{[Ru(CO)4]^{2-}}$, $\ce{[Os(CO)4]^{2-}}$

-3

$\ce{[V(CO)5]^{3-}}$, $\ce{[Nb(CO)5]^{3-}}$, $\ce{[Ta(CO)5]^{3-}}$, $\ce{[Mn(CO)4]^{3-}}$, $\ce{[Re(CO)4]^{3-}}$

-4

$\ce{[Cr(CO)4]^{4-}}$, $\ce{[Mo(CO)4]^{4-}}$

There's a review by John E. Ellis: Highly Reduced Metal Carbonyl Anions: Synthesis, Characteri's zation, and Chemical Properties on these compounds.

Edit
While Christian Aichinger's comment concerning $\pi$-backbonding in metal carbonyls is correct, it is important to realize that $\ce{CO}$ is a neutral ligand!
With other words: The Mond process: $\ce{Ni(s) + 4CO -> Ni(CO)4}$, using nickel powder and carbon monoxide, is not a redox reaction and the oxidation state of the metal still is $0$ in the metal carbonyl.

The metal carbonyls have pretty much the same old positive oxidation state if we remove those carbonyls... But the Ag- is good ....

@NeilRoy Unless I got you completely wrong: Did you take into account that carbon monoxide is a neutral ligand?

Another noteworthy example would be CsAu which contains gold in the -1 oxidation state.

I don't fully agree with @NeilRoy, but even though CO is neutral, it can pull electron density away from the metal via π backbonding ($ \ce{ M^{-}-C=O <-> M=C-O^{-} } $). I think it's helpful to remember that oxidation states are a helpful formalism to keep track of electrons, but they don't always describe the real electron distribution very well.

Very interesting examples of metal anions are the alkalides $\ce{Na^-}$, $\ce{K^-}$, $\ce{Rb^-}$, and $\ce{Cs^-}$. Yes that's right, monoanions of the familiar alkali metals! Who said old dogs can't learn new tricks?

@NicolauSakerNeto Change your comment into an answer

Or the answer here.

Question about alkalides: https://chemistry.stackexchange.com/questions/53906/alkalide-compounds

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text:
How adding one more electron increases the ionization energy?
Elements of group 6A, compared to 5A, require less ionization energy due to the paired electrons of 6A.
But a question arises: why does group 7A, compared to group 6A, require more ionization energy in spite of the paired electrons in group 7a too.
Regardless of the general trend that the ionization energy increases as we go from left to right in periods, why didn't we apply the same procedure as the first comparison?

It's not just adding one more electron. Another proton is also added to the nucleus. The overall trend across a row is driven by shielding effects.
In general, as you go across a row, the shielding effect is fairly small compared to the effect of increasing nuclear charge.
As you note, there is a slight deviation in this trend between 5A (i.e., $\ce{N}$) and 6A (i.e., $\ce{O}$) because the extra electron in oxygen is paired. This increases the electron-electron repulsion, so the ionization energy decreases slightly.
As you note, $\ce{F}$ (7A) resumes the overall trend of increasing ionization energy driven by the attractive effects of the higher effective nuclear charge $Z_{eff}$.

Elements of group 6A (also called group 16) require less ionization energy than group 5A (group 15) elements because group 5A elements have a half filled p-orbital.
Group 7A (group 17) DOES NOT have paired electrons. Ionization energy of group 7A elements is more than that of group 6A elements due to higher effective nuclear charge on group 7A elements.

but Group 7A (17) has paired electrons in the first and the second orbitals and an unpaired electron in the last orbital, of the three orbitals of the subshell p !!!

@user14840 Elements are said to have paired electrons when all the electrons are paired. In this case 1 electron is still unpaired.

I am not aware of such a definition of paired electrons and I think it is a poor choice. There would be no way to describe complexes between high and low spin.

@Martin I agree with you on the issue of distinguishing between high spin and low spin complexes. However in this case we're talking about native elements and their stability and ionization energies, and I feel in this case while referring to paired electrons, it is generally meant all electrons are paired.


text:
Slow acidification process
I'm trying to find a substitute of Glucono delta-lactone (GDL). This food additive is usually used to produce silk tofu. Because of its slow hydrolysis the pH decreases very smoothly, which results in a tofu(coagulated soy milk protein) very soft. At the same time it is neutral in flavour, which is a great feature for a food additive. Now, my problem is that I'm trying to find an alternative because GDL is not allowed in organic food (in EU). Does anyone know any possible alternative compatible with the allowed additives (Annex III in:
http://eur-lex.europa.eu/LexUriServ/LexUriServ.do?uri=OJ:L:2008:250:0001:0084:EN:PDF)
Or, how can I know the reaction rates of these acidifiers? For example, does Lactic acid hydrolyses slowly?
Note: When I refer to fast or slow, my time reference is a minute (fast) or hours (slow).

Lactic acid does not hydrolyse. It quickly dissociates as acids are used to.


text:
Calculate species concentration from first-order kinetic reactions
I am trying to determine the concentration of products from the kinetic reactions of wood pyrolysis. The kinetic scheme separates the wood into cellulose, hemicellulose, and lignin. For example, the kinetic reactions for just the cellulose are shown below:

The $G1$ and $G2$ terms represent a group of chemical species, $LVG$ represents levoglucosan, and the $K$ terms represent the kinetic rate constants for each reaction.
I am using Python (see code below) to plot the concentration profiles and mass balance of the system. As shown in the plot below, the total mass remains constant (as it should). The plot represents each component as a percent of the original wood concentration which is given as $kg/m^3$. My next step is to determine the concentration ($kg/m^3$) of each chemical species. From the plot, I have the amount of Group 1 ($G1$) produced but I would like to determine the concentrations for $H_2O$ and $Char$. For example, $5\, H_2O + 6\, Char = G1$.
How can I determine the concentration (as $kg/m^3$) of each individual species in Groups $G1$ and $G2$ ?

Python function for the cellulose reactions (see below).
Note that the activation energy units are in $\frac{kcal}{kmol}$ but the concentration of wood provided to the system is a mass basis represented as the density of the wood $\frac{kg}{m^3}$. However, when using the appropriate units for $R$ the rate constant $K$ is in units of $\frac{1}{s}$ where s is seconds. The amount of cellulose provided to the reactions is assumed to be 50% of the original wood as $0.5 * 700 \frac{kg}{m^3}$. The concentrations returned from the cellulose function cell are in terms of $kg/m^3$. The plot (see above) displays the percent of the concentration relative to the original concentration of wood. For example, cellulose in the plot is calculated as $\frac{\rho_{cell}}{\rho_{wood}}*100$ where $\rho$ is the concentration in terms of $kg/m^3$.
def cell(T, pw, dt, p):
    """
    T = temperature, K
    pw = vector of initial wood concentration, kg/m^3
    dt = time step, s
    p = total number of time steps
    """

    # array to store species concentrations as a density, kg/m^3
    # row = chemical species
    # column = concentration at time step
    spec = np.zeros([16, p])

    # initial cellulose concentration in wood, kg/m^3
    spec[0] = pw*0.5

    R = 1.987   # universal gas constant, kcal/kmol*K

    # A = pre-factor (1/s) and E = activation energy (kcal/kmol)
    A1 = 4e13;  E1 = 45000  # CELL -> CELLA
    A2 = 0.5e9; E2 = 29000  # CELLA -> products
    A3 = 1.8;   E3 = 10000  # CELLA -> LVG
    A4 = 4e7;   E4 = 31000  # CELL -> 5*H2O + 6*Char

    # reaction rate constant for each reaction, 1/s
    K1 = A1 * np.exp(-E1 / (R * T))         # CELL -> CELLA
    K2 = A2 * np.exp(-E2 / (R * T))         # CELLA -> G2
    K3 = A3 * T * np.exp(-E3 / (R * T))     # CELLA -> LVG
    K4 = A4 * np.exp(-E4 / (R * T))         # CELL -> G1

    # concentrations for each chemical species, kg/m^3
    for i in range(1, p):
        r1 = K1 * spec[0, i-1]
        r2 = K2 * spec[1, i-1]
        r3 = K3 * spec[1, i-1]
        r4 = K4 * spec[0, i-1]
        spec[0, i] = spec[0, i-1] - (r1+r4)*dt          # CELL
        spec[1, i] = spec[1, i-1] + r1*dt - (r2+r3)*dt  # CELLA
        spec[2, i] = spec[2, i-1] + r4*dt               # G1
        spec[3, i] = spec[3, i-1] + r2*dt               # G2
        spec[4, i] = spec[4, i-1] + r3*dt               # LVG

    # return species array concentrations, kg/m^3
    return spec

Python script to plot the reactions from the above cellulose function:
# Modules
#------------------------------------------------------------------------------

import numpy as np
import matplotlib.pyplot as py
import funcRanzi as kn

py.close('all')

# Parameters from Papadikis 2010a
#------------------------------------------------------------------------------

rhow = 700  # density of wood, kg/m^3
Tinf = 773  # ambient temp, K

# Initial Calculations
#------------------------------------------------------------------------------

dt = 0.01   # time step, delta t
tmax = 4   # max time, s
t = np.linspace(0, tmax, num=tmax/dt)   # time vector
p = len(t)  # total number of time steps

# Calculate Concentrations of Chemical Species
#------------------------------------------------------------------------------

# vectors for wood, gas, tar, char concentrations as a density, kg/m^3
pw = np.zeros(len(t))   # wood
pg = np.zeros(len(t))   # gas
pt = np.zeros(len(t))   # tar
pc = np.zeros(len(t))   # char

pw[:] = rhow    # initial wood concentration as density

# array of chemical species concentrations as a density, kg/m^3
spec = kn.cell(Tinf, pw, dt, p)

# concentration as percent relative to original wood, %
cell = spec[0]/rhow*100     # CELL
cella = spec[1]/rhow*100    # CELLA
g1 = spec[2]/rhow*100       # G1
g2 = spec[3]/rhow*100       # G2
lvg = spec[4]/rhow*100      # LVG

total = cell + cella + g1 + g2 + lvg

# Plot Results
#------------------------------------------------------------------------------

py.rcParams['xtick.major.pad'] = 8
py.rcParams['ytick.major.pad'] = 8
py.rcParams['lines.linewidth'] = 2
py.rcParams['axes.grid'] = True

py.figure(1)
py.plot(t, cell, label='cell')
py.plot(t, cella, label='cella')
py.plot(t, g1, label='g1')
py.plot(t, g2, label='g2')
py.plot(t, lvg, label='lvg')
py.plot(t, total, label='total')
py.title('Cellulose Reactions at T = {} K'.format(Tinf))
py.xlabel('Time (s)')
py.ylabel('Conversion (% dry basis)')
py.legend(loc='best', numpoints=1)
py.show()

According to the diagram you wrote, g1 and g2 are just stoichiometric groupings of certain components.  Because there is no modeled back-reaction from those components to the rest of the system, you can find the component amounts by simple stoichiometry.
Asssuming the "stoichiometric" coefficients in the groupings are meant as mass coefficients, then the calculations would be:
py.plot(t, 6*g1/(6+5)+0.61g2/(sum([0.8, 0.2, 0.1, 0.25, 0.3, 0.21,...etc...]), label='char')
py.plot(t, 5*g1/(6+5)+0.83g2/(sum([0.8, 0.2, 0.1, 0.25, 0.3, 0.21,...etc...]), label='water')
py.plot(t, 0.88g2/(sum([0.8, 0.2, 0.1, 0.25, 0.3, 0.21,...etc...]), label='HAA')
py.plot(t, 0.2g2/sum([0.8, 0.2, 0.1, 0.25, 0.3, 0.21,...etc...]), label='glyox')

etc.

Gavin, I'm not following.  Each unit of G1 produced contains 5 units of water, according to your diagram.  You could say there are 5 moles of water per mole of G1, or 5 molecules of water per "molecule" of G1, or whatever. You could also think of G1 being a molecule that is converted to 5 H2O  +  6 char at infinite speed relative to the other system kinetics.  Either way, unless I am misunderstanding what you mean by your diagram,   G1 = 5 H2O  +  6 char.

The only other possibility I can think of is that the system units are currently in mass, but you want to find moles of water?  In that case you would have to adjust by the formula weight of water.

You would still calculate the mass balance the same way.  Or, if you wanted, you could calculate it as total = cell + cella + char + water + HAA + .... + lvg, that is, with all the components expanded.  But you can't count G1, G2, or any of their components more than once.  So if you include G1, don't include water and char, etc.

Keep in mind that the products returned from my cell function are in $kg/m^3$. So cell, cella, g1, g2, and lvg returned from the function are in units of $kg/m^3$ not moles. The total for the system should always equal 0.5700 kg/m^3 (350 kg/m^3). I'm confused on how 5H2O + 6Char = G1 if G1 should be some amount as kg/m^3. If H2O is 5g1 then it would be greater than g1.

I agree there is some inconsistency then.  What is the meaning of the coefficiencts "5", "6", and "0.61", etc. in your diagram?  You could interpret them as mass coefficients, in which case you would have to calculate water at 5/11g1  + 0.83g2/sum(g2coeffs) and char as 6/11g1 + 0.61g2/sum(g2coeffs), and so on.  Or you could interpret the coefficients as molar coefficients, in which case to find mass you would have to multiply by FWs.  For example, 185/11g1/(18+12)  + 0.83*g2/sum(g2coeffs)*18/sum(g2FWs).

If total = cell + cella + g1 + g2 + lvg then total = cell + cella + (5*g1 + 6*g1) + g2 + lvg does not give the same number as the previous approach.

Gavin, I see now as I hope my preceding comment made clear.  You need to decide what "units" the coefficients of g1 and g2 are in.  If they are in mass units, then you just normalize them so the sum is one.  I'll update my answer.

If those coefficients are molar coefficients, though, you need to first weight each coefficient by the formula weight of the corresponding compound before normalizing.

Ah ha, I see what you are saying now. But one last thing regarding the updated answer: For g2 the char should be 0.61*g2/(0.8+0.2+...+0.02+0.61) and water should be 0.83*g2/(0.8+0.2+...+0.02+0.61)?

Yep, just re-edited.  Thanks for being so patient!  In my mind this problem is a good example why people invented moles...

Let us continue this discussion in chat.


text:
Why does the second electron affinity has an opposite sign of the first one?
Many first electron affinities are positive, indicating a favourable process, but the corresponding second electron affinities are negative. For example, the first and second electron affinities of oxygen are $+141$ and $-780~\mathrm{kJ~mol^{-1}}$ respectively. Why is this so?

Many non-metals like those in Group 16 and 17 have first electron affinities that are negative.  In other words, the following reaction is exothermic and energy is released.
$$\ce{X + e^{−} -> X^{−} + ~energy}$$
By gaining an electron the group 17 atoms attain a complete octet of electrons and become stabilized.  By gaining an electron the Group 16 atoms are just one electron away from having a filled shell.  Further in both the Group 16 and 17 cases, the protons in the nucleus are not fully screened by the outer electrons, so adding one more electron to form an anion is electrostatically favorable.
However, if we try to add a second electron the reaction becomes endothermic; adding a second electron is unfavorable and energy must be added (endothermic).
$$\ce{X^{-} + e^{−} + ~energy~ -> X^{2−}}$$
It is difficult to add a second electron primarily because of the increased electron-electron repulsion that occurs with the second electron.
Since adding the first electron is exothermic, while adding the second electron is endothermic, the sign reverses when we measure the energy added or released by these two processes.

OK, but why basically the exothermic and endothermic reactions occur with different type of elements? In other words, how can we know,for example, that that elements of group 17 releases energy whereas others absorb???
I couldn't understand the relationship basically

Elements that form stable negative ions (like oxygen, fluorine, etc.), and especially those that achieve a stable octet upon negative ion formation (fluorine, chlorine, etc.), will always have exothermic (negative) first electron affinities.  Most other elements, especially those that can achieve an octet by losing one or two electrons, don't want to gain an electron (that moves them further away from the stable octet configuration), so when we "force" a metal to add an electron it is an endothermic reaction.

You will find it easier if you write down the electronic configurations of each element in a period or group and then check out their electron affinity. There is a good trend in most and you will find quite interesting things about why various elements' electron affinities differ! For example, the trend is slightly changed in case of caesium due to the variation in energies of the subshells. And when you move to the right side of the periodic table, the group 16 has a rather chaotic trend I would say but everything has an explanation...


text:
Where to find data for Gibbs energy, enthalpy, and entropy?
Where can you find the enthalpy and Gibbs free energy of formation and entropy of compounds either as a data file or web API? Technically, I only need two since I can figure out the other one from two of them.

NIST webbook does have a lot of data, though they are not in any kind of an API form as far as I know.
http://webbook.nist.gov/chemistry/

There's no API, but there are some tools to "scrape" the NIST site, e.g. this for UV/Vis

This is a good answer, but isn't it incomplete? For instance, I cannot find the enthalpy of formation for solid water in the NIST webbook. Am I just being silly, or does this exclude a lot of very useful (and well known) information?

NIST is the best place to turn for lots of data.  However, more easily parsed, smaller datasets are available in a couple of other locations.

The CHNOSz package in R has thermochemical data for a variety of species, mostly inorganic.  Their database is referenced back to the chemical literature.  See an answer I gave to an old question for an example of how to use it.
Robert Alberty's Mathematica package for biochemical thermodynamics has a lot of built in thermochemical data for common biochemical compounds -- relative to a biochemical standard state of constant pH.
Water-specific databases.  In case you're only interested in water, ice, and steam, there are Excel plugins that let you get thermochemical data by calling functions inside Excel.
There appears to be a Python library called thermopy which contains a burcat() module in which some thermochemical property data lives.  I haven't tried it out though.
The database in that python module seems to be this one.
For biochemical reaction systems, there is also eQuilibrator which has a web interface.

Free Energy and Enthalpy Database (FREED)
This has proved extremely valuable (at least for me) and I think that others can also benefit from this.

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


text:
Ratio of copper(II) ions to EDTA?

What is the ratio of copper(II) ions to EDTA in a $\ce{[Cu(EDTA)]^{2-}}$ complex?

I know that EDTA forms up to 6 coordinate bonds with the central atom, but is that what they are asking for with "ratio"?
According to this website, it seems that copper actually bonds with all 6 free electron pairs.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website. And, "homework" tag is for questions when you want a more "teacher" approach. It doesn't mean "homework" literally.

They're asking for the ration of ions i.e. for each $Cu^{2+}$ ion, how many $EDTA^{6-}$ are there.

So all that would be is 1:1, that's it?

Yes that's right.

@BinaryGeek EDTA has a maximum of a -4 charge. It can never reach -6.

@LDC3 Yeah, I wrote that by mistake. Thanks for telling me.

The ratio of Cu to EDTA is the number of copper ions per EDTA ion. In this case there is one EDTA ion for every copper ion so the ratio is 1:1.
Copper bonds with six electron pairs on EDTA to form the complex shown below where the central metal species is copper(II):


text:
Green and red crystals of [Ni(Ph2EtP)2Br2]

From the reaction of $\ce{NiBr2}$ and $\ce{Ph2EtP}$, it is possible to isolate green and red crystals of $\ce{[Ni(Ph2EtP)2Br2]}$?
A) coordination number and geometries of the complexes?
B) possible isomeric structures for the green and red crystals.

I know that d orbitals effect the color but I don't know how to apply it to isomers.

The coordination number of $\ce{[Ni(Ph2EtP)2Br2]}$ is four;
thus, the complex may be tetrahedral

or square planar

(figures taken from here)
The complex $\ce{[Ni(Ph2EtP)2Br2]}$ contains $\ce{Ni^2+}$, which has eight d electrons.
When filled with eight electrons, the tetrahedral structure is paramagnetic, and the energy difference between the highest occupied and the lowest unoccupied orbital is relatively small (4.4 Dq).
The corresponding square-planar structure is diamagnetic, and the energy difference between the highest occupied and the lowest unoccupied orbital is relatively large (10 Dq).
Therefore, the square planar complex absorbs light of higher energy (lower wavelength) than the tetrahedral complex.
The green crystals absorb red light, whereas the red crystals absorb green light. Thus, the red crystals absorb light of higher energy than the green crystals. Therefore, the red crystals correspond to the square-planar complex and the green crystals correspond to the tetrahedral complex.
Furthermore, since the complex $\ce{[Ni(Ph2EtP)2Br2]}$ contains two different ligands, there are cis and trans isomers of the square-planar structure.

It's a well known experiment (J. Am. Chem. Soc. 1970, 92, 2691). Of course, it's possible to isolate green and red crystals $\ce{[Ni(Ph2EtP)2Br2]}$. The coordination number for both complexes is four.
The geometry of the complex with  red crystals is square planar. It's diamagnetic. Hybridization of $\ce{Ni^2+}$ is $\ce{dsp^2}$ The geometry of the complex with green crystals is tetrahedral. It's paramagnetic.
Hybridization of $\ce{Ni^2+}$ is $\ce{sp^3}$
Electron configuration of $\ce{Ni^2+}$: $\ce{[Ar] 4s^0 3d^8}$.
We can explain the formation of the red crystals as follows:

On the other hand, we can explain the formation of the green crystals as follows:

As for possible isomers, there are two isomers (cis and trans) for the square planar complex since it has two kinds of ligands.


text:
Why should a complexometric titration be kept neutral?
In my next lab, I'll be doing a complexometric titration of a copper solution with EDTA and a murexide indicator. The lab guide advises to keep the pH level nearly neutral by adding a diluted $\ce{NH3}$ solution. Why is that necessary?
I understand that during the formation of the copper complex $\ce{[Cu(EDTA)]^{2-}}$, the solution becomes more acidic. Do we need to add the basic solution in order to make sure the murexide does not change color due to the decreasing pH level? Or is it simply so that we can revert the color change (of the formation of the copper complex) and continue the titration until the equivalence point is reached (and all copper ions are in complex form)?

The colour of murexide in aqueous solution does indeed depend on the pH: blue in alkaline, red in acidic solution.

Your last guess is exact. This is a classical method, the equivalent point is seen by our eyes, so in our interest is for the complex to be formed so we can recognize the colour, and we should add $\ce{NH3}$ so the colour won't change immediately after the equivalent point.

I am still having problems understanding exactly how the ammonia changes the color of the solution. Is it because the murexide acts simultaneously as a pH indicator and an indicator for the metal complex? Adding ammonia does not change the copper complexes that have already been formed, right?

Yes, it only keeps it that way for a while so we can recognize the colour easily, so the acid doesn't dissociate and change the colour immediately.


text:
Mechanism of styrene dimerization

I am struggling to find a mechanism for the above reaction.
So far I have suggested that the proton adds to one of the reacting aromatic species leaving the resonance stabilized secondary cation. This then adds to the ortho position on the second reactant (since alkyl groups are activating). I then get a positively charged intermediate.
For a simpler reaction I would simply remove the proton but I wonder whether the reaction from this step to the product is concerted? I couldn't see an obvious step so I deprotonated and again could find no obvious way to continue apart from a rather elaborate proton transfer from a carbon that does not possess the most acidic proton.

You're right that the first step involves protonation of the terminal carbon in the  styrene double to yield a resonance stabilized benzyl carbocation.  This carbocation then adds to another styrene molecule in the exact same fashion, reacting at the terminus of the styrene double bond to again yield a resonance stabilized benzyl carbocation.  This latter carbocation then alkylates itself (like a Friedel-Crafts alkylation).  Loss of a proton from this final carbocation yields the indicated product.

The mechanism depicted above is most likely correct but please do not push arrows this way. All the arrows are backwards and suggest electrons originating from carbocations which are most definitely sinks and not sources.


text:
What is the mechanism of chlorosulfonation of benzene?

The above reaction looks very similar to a conventional sulfonation reaction but if the reaction pathway is analogous to sulfonation I do not see how the $\ce{SO_2Cl^+}$ would form. Surely $\ce{Cl^-}$ would leave rather than $\ce{H_2O}$ in the formation of the electrophile? Please help me find a mechanism.
Also, I would like to approach questions, such as this one, like I would a puzzle rather than calling upon my memory. Is there a series of steps that I should follow? (i.e: 1. Identify electro/nucleophile, 2. Identify acidic protons ...).

Rob, IMO this is a difficult problem, so don't feel bad about not "seeing" the solution right off.
Chlorosulfonic acid (or chlorosulfuric acid) is both difficult and hazardous to work with so it's reactions aren't discussed all that frequently.  It can also give rise to different products at different temperatures.  At higher temperatures it generates $\ce{SO3}$ which will behave as an electrophile and react with aromatic compounds to produce sulfonation products.

image source
At lower temperatures (your case), chlorosulfonic acid reacts by way of the following equilibrium (who knew this?)
$$\ce{3 ClSO2(OH) <=> SO2Cl^{+} + 2 SO3Cl^{-} + H3O^{+}}$$
The equilibrium generates $\ce{SO2Cl^{+}}$ which is the electrophile of interest in your reaction.  $\ce{SO2Cl^{+}}$ will react with an aromatic nucleus to produce the corresponding sulfonyl chloride by way of a typical electrophilic aromatic substitution reaction.
If you'd like to read more about the complexities of reactions involving chlorosulfonic acid see here starting around page 11.
The difficult part here was identifying the source of the electrophile. Looking at the final product you get an idea of what the electrophile is, but how is it produced?  Sometimes you have to do a little googling to sort things out.
Again, IMO this was a tough one.

The $\ce{OH}$-group is protonated by another molecule of chlorosulphuric acid, creating a cation. This cation can be attacked by benzene under the right conditions. $\ce{H2O}$ is not a better leaving group than $\ce{Cl-}$ (look at the charge, they mostly tell a lot about the ability of the leaving group, though not always). A famous example of this reaction is of course the famous old synthesis of saccharin (Clayden et al. p. 644 & p. 645) giving the widely used $\ce{TsCl}$ as byproduct:

Is the ortho product the major product?

The answers provided so far are actually incorrect.
First, the benzene reacts with chlorosulfonic acid to produce benzenesulfonic acid and HCl.  The chloride acts as a leaving group, not the OH.  Second, the benzenesulfonic acid reacts with a second equivalent of chlorosulfonic acid to produce benzenesulfonyl chloride and sulfuric acid.  Thus, in net you need two equivalents of chlorosulfonic acid to convert benzene to benzenesulfonyl chloride.
Edit update:
I should add that the arrow-pushing mechanism of sulfonation with SO3, provided above by Ron, is actually incorrect too.  Modern research has shown that the traditional organic chemisty mechanism of electrophilic aromatic substitution, in this case, is actually not true.  The intermediate (sometimes call the sigma complex or the Wheland intermediate) in which SO3 is bound to the arene leaving a positive charge on the arene, actually does not exist.  Instead, modern computational research, as well as kinetics studies that are much older, indicate a termolecular pathway.  Typically, three molecules are involved.  In concentrated SO3 or oleum, two molecules of SO3 form a transition state with the arene... it is a concerted mechanism.  In sulfuric acid, the termolecular complex involves the arene, sulfuric acid, and SO3.  In fact, sulfonations with higher SO3 concentrations (e.g. oleum) actually don't produce the sulfonic acid product; they first produce sulfonic anhydrides which are later hydrolyzed during work up.
For whoever downvoted me, go look up the various papers by Cerfontain for the 2nd order in SO3 data, and look up various recent computation papers (Shi, Computational Theoretical Chemistry, 2017, 1112, 111; or, Schaefer, 2016, Acc. Chem Res., 49, 1191).  You will see that the simplified world of organic chemistry's golden age is unfortunately not perfect or correct.
The mechanism provided for chlorosulfonation in which ClSO2+ is generated... this is very hand wavy.  You should at least provide a reference justifying this sort of thing.  It's hard to believe that equilibrium (if even real) out-competes the equilibrium of ClSO3H with SO3 and HCl.
Response to Ron:
For the chlorosulfonic acid mechanism, the ref your provided is a review book which links to a journal from the Russian Journal of Organic Chem from 1975.  I unfortunately can only access the abstract because that year does not appear to be electronically uploaded and I don't have access to the physical copy of the journal.  Anyways, the abstract indicates kinetics that are third order in chlorosulfonic acid.  This is great and pretty neat, but does not necessarily support the crazy ion-pair mechanism that is proposed.  A lot of these crazy old mechanisms from the 'golden era' of organic chemistry (40s to 80s approx) are often wrong.  Good data kinetic data was often obtained, but the mechanisms are often later found (often with computational support) to be imprecise.
One of the best examples is SO3, which you have listed above.  The mechanism of sulfonation with sulfuric acid varies depending on conditions due to the wide variety of species present (see the ref you listed; the chlorosulfonic acid book).  But, no one does sulfonations this way, because you need forcing conditions to drive the reaction to completion; people do them in highly concentrated sulfuric acid or with oleum.  Under these conditions SO3 is often the sulfonating agent, or pyrosulfuric acid is, which is what you get when you react sulfuric acid with SO3 (e.g. oleum).  Under these realistic conditions, the reaction is typically termolecular.  Please read the refs I listed.  The intermediate you listed above is actually too far uphill energetically, and modern academics have abandoned it.  This is why I am extremely doubtful of your listed chlorosulfonation mechanism, and why the SO3 mechanism you listed is not currently supported.

Please start reading around p. 11 in the link provided in my answer.  It provides data and references supporting the two mechanisms I described.

response was too long to fit in the edit box; please see response in original post above


text:
How to calculate the enthalpy of formation of hydrazine from tabulated values?

Estimate the standard enthalpy of formation of hydrazine, $\ce{N2H4(g)}$, from the following data.
\begin{align}
\Delta{}H_\mathrm{B}(\ce{H2}) &= 436~\mathrm{kJ/mol}\\
\Delta{}H_\mathrm{B}(\ce{N2}) &= 944~\mathrm{kJ/mol}\\
\Delta{}H_\mathrm{B}(\ce{N–N}) &= 163~\mathrm{kJ/mol}\\
\Delta{}H_\mathrm{B}(\ce{N–H}) &= 388~\mathrm{kJ/mol}\\
\end{align}

While I know that $\Delta{}H_\mathrm{B}(\ce{N–N})$ refers to the energy released from breaking that bond, I'm not sure what $\Delta{}H_\mathrm{B}(\ce{H2})$ or $\Delta{}H_\mathrm{B}(\ce{N2})$ refers to. Could someone please help me understand?

In order to "make" hydrazine, $\ce{H2N-NH2}$ from hydrogen and nitrogen atoms, you have to

break $\ce{N2}$ once, giving 2 nitrogen atoms
break $\ce{H2}$ twice, giving 4 hydrogen atoms
form one $\ce{N-N}$ bond
form four $\ce{N-H}$ bonds

You have all the enthalpies that you need. Just pay attention to the coefficients and the signs.

Standard Enthalpies of Reaction
The magnitude of  $\Delta H_\mathrm f^\circ$  is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, minus the sum of the standard enthalpies of formation of the reactants, also multiplied by their coefficients:
Here is the table of enthalpies required to break bounds for the different molecular assemblies.
\begin{array}{lcccc}\hline
\textbf{Bounds} & \ce{H-H} &\ce{N-H}&\ce{N-N} &\ce{N#N} \\ \hline
 \Delta H_\mathrm B=-\Delta H_\mathrm f^\circ & 436 & 388 & 163 & 944 \\\hline
\end{array}
This table gives the amount of energy in $\pu{kJ mol-1}$ required for breaking chemical bounds. Thus it is called $-\Delta H_\mathrm B^\circ$ and means the opposite of forming chemical bounds $\Delta H_\mathrm f^\circ$.
The sign convention for $\Delta H_\mathrm f^\circ$ is the same as for any enthalpy change: $\Delta H_\mathrm f^\circ\lt0$  if heat is released when elements combine to form a compound and   $\Delta H_\mathrm f^\circ\gt0$   if heat is absorbed according to the usual conventions.

Producing $\ce{N2(g)}$ requires that: given $\ce{2N}$, produce $\ce{N2}$, 1 time,  so it means

$$\Delta H_\mathrm f^\circ[\ce{N2}]=\pu{-944 kJ mol-1}$$

Producing two $\ce{H2(g)}$ requires that: given $\ce{4 H}$, produce $\ce{H2}$, 2 times, so it means

$$2\Delta H_\mathrm f^\circ[\ce{H2}]=\pu{-872 kJ mol-1}$$

Producing $\ce{N2H4(g)}$ from $\ce{N2H4(l)}$ requires absorbing $$\Delta_\mathrm{vap} H^\circ=\pu{+44.7 kJ mol-1}$$

Reacting $\ce{N2H4(g)}$ requires that: to produce $\ce{4 H}$ and $\ce{2 N}$, the experimenter has to destroy 4 bounds $\ce{N-H}$ and he has to destroy 1 bound $\ce{N-N}$, so it means

$$\Delta H_\mathrm f^\circ[\ce{N2H4}]=\pu{-1715 kJ mol-1}$$
Finally, by making the sum of the energy needed to produce hydrogen and nitrogen, minus the energy got from the reaction of the hydrazine molecule, we get a standard enthalpy of reaction:
$$\Delta_\mathrm rH^\circ=\pu{-56.3 kJ mol-1}$$
proving that the decomposition of hydrazine is exothermic.


text:
How do I go about balancing these equations?
I have to use oxidization number approach.
$\ce{Co^2+ (aq) + H2SO3(aq) -> Co^3+(aq) + S2O3^2- (aq)}$
This is my approach so far:
The change in Co is 1 electron, the change is S is 2 electrons, so CO must be multiplied by two
$\ce{2 Co^2 + H2SO3(aq) -> 2 Co^3+ + S2O3^2-}$
I balance the O's, then balance the H's and I get
$\ce{2 Co^2+ + 2H2SO3(aq) + 2H^+ -> 2 Co^3+ + S2O3^2- + 3H2O(l)}$
I did it the half reaction way as well, I got the same answer but the coefficients on the Co's were 4. I suspect my problem is my assumptions about the oxidization number change at the beginning.
Thanks in advance

Also it is in acidic conditions

You've got two sulphurs...

You need to balance charges as well.

I agree with you, the right equilibration of the redox reaction is the following:
$$\ce{4 Co^2+(aq) + 2H2SO3(aq) + 2H^+ (aq)-> 4 Co^3+(aq) + S2O3^2- (aq) + 3H2O(l)}$$
The problem is  in your assumptions about the oxidization number change at the beginning, and more precisely about sulfur.
The oxidation number of  sulfur in $\ce{H2SO3}$ is four. But in thiosulfate ion, you have two inequivalent sulfur atoms:
Please see:
 Oxidation states of the sulfur atoms in the thiosulfate ion
The oxidation number is $\ce{+IV}$ for the central one, while it's zero for the terminal one.
In sum, the variation in the oxidation number for sulfure is four, not two.
(As if you have two sulfur atoms with oxidation number of four in $\ce{2H2SO3}$. One of them changes its oxidation number to zero. while the other one stays intact).

The problem with your answer is that the charges do not balance. There is a net charge of +6 on the left hand side and +4 on the right hand side. This can be solved by doubling both of the Co coefficients giving a net charge of +10 on each side.


text:
Why are the masses of atoms less than the sum of their subatomic particles?
The mass of carbon-12 is $\pu{12 u}$ by definition. However, one carbon-12 atom comprises 6 neutrons (each weighing $\pu{1.0087 u}$), 6 protons (each weighing $\pu{1.0072 u}$), and 6 electrons (each weighing $\pu{0.0005 u}$), which all add up to $\pu{12.0894 u}$.
Where does the $0.7\%$ difference in mass come from?

In very simple and short terms: In bound states, the mass of the involved particles changes, i.e. mass defect. The loss of bonding energy translates to a loss of mass, basically demonstrated by $E=m\cdot\mathcal{c}^2$.

Exactly that mass difference is used in nuclear reactions to "generate" energy: for light nuclei (smaller than iron), energy is released by fusing them to heavier ones; for large nuclei (larger than iron), it is released by splitting them.

This is due to the mass-energy equivalence and a phenomenon called binding energy.
Forming a nucleus releases energy because the nucleons are falling into a potential energy well. Due to Einstein's mass energy equivalence this results in the mass of the new nucleus being less than that of the particles that formed it.
The binding energy of carbon-12 is quoted on Wikipedia as $\pu{92.162 MeV}$. Therefore we can estimate the mass defect of a carbon-12 atom, $\Delta m$, using $ E = (\Delta m)c^2$:
$$\Delta m = \frac{(\pu{92.162 \times 10^6 eV})(\pu{1.6022 \times 10^-19 J eV-1})}{(\pu{2.9979 \times 10^8 m s-1})^2} = \pu{1.6430 \times 10^-28 kg} = \pu{0.098943 u}$$
The difference in the mass of carbon-12 to the mass of its constituent particles is $\pu{0.08940 u}$, so we can see that our calculation is a reasonable estimate of the mass defect. The slight difference is due to other more complicated factors that I have not taken into account, but it still illustrates the main reason for the mass defect.

The small difference should be binding of the electrons to the nucleus, though I cannot see the value above (yellow boxes)


text:
Finding out the concentration of oxygen in water
I read in a biology course that you can determine the concentration of oxygen in water by adding some $\ce{NaOH}$ and then a small amount of $\ce{MgCl2}$. They say that a dark coloured precipitate is formed.
I can't see a way of writing this reaction. Can anyone help me?

Probably MnCl2 - manganese chloride not magnesium.

It is definitely not magnesium, but manganese.

This is the old Winkler titration, published as
Die Bestimmung des im Wasser gelösten Sauerstoffes by Ludwig Wilhelm Winkler in Ber. Dtsch. Chem. Ges., 1888, 21, 2843-2854 (SRC)
How does it work?

Add $\ce{MnSO4}$ to your water sample, then $\ce{NaOH}$ and $\ce{KI}$.

Under alkaline conditions, oxygen dissolved in your water sample oxidizes $\ce{Mn(II)}$ to $\ce{Mn(IV)}$.
\[\ce{2Mn^{2+} + O2 + 4 OH- -> MnO(OH)2 v}\]
This is the brownish precipitate mentioned in the question.

Now, the sample is acidified (Winkler used hydrochloric acid) and the solution becomes yellowish due to the iodine formed. Under these conditions, $\ce{Mn(IV)}$ oxidizes iodide:

\[\ce{Mn(IV) + 2 I- -> Mn^{2+} + I2}\]

The amount of iodine formed, which repesents the aount of oxygen in the water sample, can be determined by titration with thiosulfate solution.


text:
Reverse Friedel-Crafts mechanism

I need to find a mechanism for the above reaction but I cannot make much progress. I do have something in mind but it doesn't seem entirely plausible: First the tert-butyl group leaves (this seems wrong but tert-butyl cation is stable and thus a reasonable leaving group maybe? Also the negatively charged ring that it would leave behind would be stabilised by the EWG Bromines). The cation is then attacked by Cl and the ring is attacked by a proton (one product formed). Then a Friedel-Craft alkylation to make the other product (hence the excess of benzene to promote only mono-substitution). I doubt that is right, purely because of the first step. Can you help?

You are close. Think about reordering the steps you've described to avoid any negatively charged intermediates. Under acidic conditions only positively charged and neutral species will be present.

I am sorry could you tell me the first step please. I am really stuck. Why only positive charge in acidic conditions?

The Friedel-Crafts alkylation is reversible (see here under "Friedel–Crafts dealkylation"). So first protonate the ring and then dealkylate

This is really a reverse Friedel-Crafts followed by a normal Friedel-Crafts. So whatever the last step of a FC alkylation is will be the first step of this reaction.

This reaction proceeds through a reverse Friedel-Crafts alkylation followed by a forward Friedel-Crafts alkylation. The last step of Friedel-Crafts alkylation is deprotonation, which means that the first step of the reverse process will be protonation. Next, the carbon-carbon bond breaks, with 1,2,4,5,-tetrabromobenzene acting as a leaving group. The resulting t-butyl cation can undergo the typical electrophilic aromatic substitution steps, attack by an aromatic ring (benzene here) and deprotonation to regain aromaticity.
One might wonder why aluminum chloride is present in the reaction mixture, since we can write a reasonable mechanism that excludes it. Likely it is present in case chloride (from HCl) traps it to form t-butyl chloride. Aluminum chloride will catalyze the reformation of t-butyl cation.

Having come back to this question again, I realized that the second species that you have drawn in your mechanism seems massively unstable (cation on carbon with bromine substituent). I can't see another way of doing this so I don't doubt the mechanism. However, why is this possible? Also, is Acylation reversible too?

@RobChem Inductively bromine is electron withdrawing (destabilizing cation) but it's also electron donating by resonance (stabilizing cation), so I think it's a reasonable intermediate. Theoretically, FC acylation is reversible too. I'm not aware of any examples though.

I guess the aluminium chloride is necessary so the chloride anion has a place to go.


text:
Looking for a compound
TL;DR;
I need a compound that is harmeless to the human body, odorless, tasteless, causes some kind of visible reaction (urine coloration perhaps) AND can be bought without medical recomendation.
LONG VERSION;
Food stealing in my company fridge has become a serious problem. It's not only a recurring issue but a targeted one... our thief has been targeting "high" value, particularly coconut water. HR seems to ignore our complains... so I decided to take matters in my own hands... since I have access to timesheets, I thought that if I can scare out thief to leave work or skip it, I can narrow-it-down/point-him-out because we have about 30 people in our office...

Urine typically is yellow, but in combination with blue dyes, a green colour can be obtained.
Noteworthy compounds are methylene blue, thymol, Cimetidine (antacid), or Mitoxanthrone (used in cancer therapy).
I've mentioned these just for the sake of completeness - I think it is irresponsible and stupid to spike food with anything that doesn't belong there.
Hiding a webcam in a shelf and running zoneminder probably isn't legal either and will get you in trouble with the privacy officer, but at least that doesn't mess around with the health of anybody else.

for the sake of completeness I do intend to label my coconut water bottle with a big "DO NOT EAT. SCIENCE EXPERIMENT UNDER DEVELOPMENT" label... wich i'm sure the little thief will ignore...

You could use asparagus extract. That would attract your nose's attention.


text:
Why is the decarboxylation of 2-hydroxybenzoic acid (and related compounds) done 'under nitrogen'?
I was reading Cannabinoid Chemistry - 6.2 Extraction & pondered over the following:

Cannabis must be dried be(fore) it is extracted, because it is not possible to remove more than 50% of the cannabinoids from fresh material. THC-Acid is difficult to extract. If you plant(sic) to convert the THCA to THC, the plant material should be thoroughly decarboxylated by heating it under nitrogen at 105℃ for 1 hour before performing a solvent extraction.

Why is decarboxylation done 'under nitrogen'?  My first guess would be that at 105℃ the plant begins to oxidize (but that is just my naive guess).
I've also sourced this graph:

Effect of heating time and temperature on the THC content of an n-hexane
marihuana extract after heating on the glass surface in an open reactor.  Journal of Chromatography 520 (1990).
Which suggests that 122℃ for 27-30 minutes will be even more effective for decarboxylation, but if the sample begins to oxidize at 105℃ it seems even more important to get the right mixture of gases around it at 122℃.

I took the liberty to broaden the title of your question, since it is a problem of other 2-hydroxy benzoic acid derivatives and phenolic compounds in general.

[…] it seems even more important to get the right mixture of gases around it at 122℃

No, it's only important to keep the oxygen away while you are heating the sample.
Have a look at the structures: the thermal decarboxylation of the THCA isn't different from the decarboxylation of salicylic acid.
But with the carboxyl group or without, (upon heating,) phenols are easily oxidized by air.
In principle, the sample could be heated under vacuum too. In this case, you might want to use a cooling trap to collect volatile material. But usually, one does that anyway to protect the vacuum pump.


text:
How do you identify reducing / non-reducing sugar by looking at structure?
Identifying reducing / non-reducing sugar been confusing me for a while now , I know that reducing sugar contain aldehyde or ketone group . It's easy to identify them in monosaccharides but this becomes confusing in case of disaccharides!

I fail to see carbonyl group here, How can I identify reducing / non-reducing sugar by looking at structure in picture? can anyone help me out?

Many sugars are drawn in the cyclic, closed form where the carbonyl group has been converted to a hemiacetal.

Once you realize that a hemiacetal can equilibrate with a carbonyl (e.g. it is a carbonyl in disguise), identification of reducing sugars becomes easier.
Here is an example from Wikipedia, it is Maltose, the same as your third compound

See the aldehyde in the open isomer drawn on the right; see how it can close to form a hemiacetal on the left.

The way I always heard what that (in a sugar) was that the anomeric C atom was the one bound to two oxygens.  And if one of those two oxygens was part of an OH group, then the sugar was a reducing sugar.


text:
Why are 720 nm LEDs so uncommon?
Disclaimer: Potentially more of a physics question, but this area is muddy.
I'm trying to find 720 nm LEDs but they're proving to be very uncommon. In fact, there seems to be a gap between 660 nm up to ~730 nm. I've only been able to find one supplier of 720 nm LEDs (data sheet here). These LEDs also tend to be very low powered compared to their counterparts in the nearby range. For example, a typical 690 nm LED outputs 690 mW, whereas the 720 nm LED only ouputs 170 mW, and that's the strongest one I could find (which is no longer carried).
My gut feeling says this has to do with the material being used in the LED (in this case GaAlAs). Perhaps it's very difficult to dope the material in such a way that it emits 720 nm light? I've already noticed that these LEDs are prone to burning out. Is the material just not very good for this purpose? Hopefully someone with more insight into this topic could shed some light (heh) on the situation.

This belongs to physics subsection.

I think it is on topic here, too, but you still might have a better chance for an answer at physics.se. If you don't get replies, you should think about migrating it.

This isn't a real answer to the underlying problem, but it seems that EPIGAP has some, although I haven't seen anything with an output > 35 mW.

I like the question however, and a great read for you @wes34499 is why the blue LED got a nobel prize in physics in 2014.

+1 Muddy subject. ;-) This area of physics is so chemist-ry that most physicist despair. Of course you also don't find many chemists who know their way around semiconductors. Nobody does solid-state chemistry any more. I blame the soft-matter hype. :-D

I suspect the answer is that AlGaAs is an alloy with a bandgap between that of AlAs - 2.16 (which is indirect) and GaAs - 1.42 eV (which is direct). If there is enough aluminum, the resulting semiconductor is an indirect bandgap semiconductor, which means that to emit a photon, it must be coupled with a lattice phonon, greatly reducing the probability of photon emission, which would make a AlGaAs LED of short enough wavelength rather inefficient at emitting light, instead producing a lot of heat rather than light.


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Why does the OAc group leave in the first step of Oxymercuration?

The above picture shows the first step of Oxymercuration. The electrons attack the Mercury atom but the OAc group leaves. Why is this? Hg is a transition metal with a number of available oxidation states, why cant it just accept the electrons? For similar reactions (like glycol cleavage using periodate or lead tetra acetate) how is it possible to tell whether a group will leave (retaining oxidation number) or the oxidation state changes?


text:
Why do we sometimes write reactants above the reaction arrow?
We write:
\[\ce{Alcohol ->[\ce{KMnO4}] Acid}\]
The alcohol forms a carboxylic acid, i.e. gains an $\ce{O}$ from $\ce{KMnO4}$. This means $\ce{KMnO4}$ is a reactant! Then why is it written above the arrow?
What else do we write above the arrow?

If we only care about what happens to alcohol it's useful notation.

Btw, if you're interested in using $\LaTeX$ for math and equations and really want really sharp looking chemical equations, have a look at all the options provided by the mhchem package. Check out this and the original documentation at CTAN. You will love it!

Things written on arrows are chemicals that are part of the reaction in some way, but not one of the reactants or products. In this case, the $\ce{KMnO4}$ is your oxidizing agent, which can be swapped out for any other strong oxidizer, and is not an inherent part of the reaction.
Other things you could see on the arrow would be $\ce{H2O}$ or Diethyl Ether, both of which are solvents. Other things include $h\nu$ and $\Delta T$, both of which are conditions under which the reaction takes place. Lastly the most common thing you'll see are catalysts such as $\ce{Pt}$. As you  can see, all of these things are necessary to the reaction, but aren't really what we care about in the reaction/can be substituted for other, similar chemicals.

but why cant we call it a reactant. also is it a reagent ? Can we call KMnO4 a reagent?

Does KMnO4 provide oxygen atom to acid?

@KlausWarzecha Is KMnO4 a reagent here? Thanks very much wes3449 ..

@Princess Yes, it does provide the oxygen and calling it a reagent is fine. Wow, two answers in one :)

@Princess, a reagent is the same as a reactant. Just two different ways of saying the same thing. If you want more info about the full reaction, you should look for the full reaction mechanism.

Do we write reagent name at top of arrow?

If you want to write a proper balanced chemical equation then you do not write reagents on the arrow!  Catalysts, energy sources, environmental conditions, yes...

So where do I write KMnO4? @User58220

On the left, with the other reagents...  See equations in http://www.chemguide.co.uk/organicprops/acids/preparation.html

@wes3449 In case of light and heat are these reactants?


text:
Can paper burn without oxygen or air?
If you put a paper inside an oven vacuum packed and go beyond $250~\mathrm{^\circ C}$, will it burn? There's no oxidizer. If not, what will happen?

https://en.wikipedia.org/wiki/Pyrolysis

No, the paper will not burn without oxygen being present.
Paper is made primarily of cellulose which is a polymer of glucose.

If you heat paper in a vacuum the cellulose simply decomposes to $\ce{H2O}$, $\ce{CO2}$, $\ce{CO}$ and carbon.  As the paper decomposes it will "char" or turn brown to black as the cellulose polymer degrades.  Here is a link to an abstract describing the process in more detail.

Depending on the precise temperature and duration of heating, a lot of interesting degradation products can form before things wind up as water, CO2, CO, or C.  For example pyrolysis oil is made by heating biomass -- which like paper contains a lot of cellulose.

In addition to @CurtF. comment about yielding liquids you may use, do not forget the wood gas generators especially popular during WWII (e.g., https://en.wikipedia.org/wiki/Alternative_fuel_vehicle#Wood_gas), and more in detail -- including post WWII examples, https://en.wikipedia.org/wiki/Wood_gas_generator.  Albeit feed by wood, I speculate paper could in principle undergo the same as wood (albeit with less efficiency than wood).

It will decompose into other compounds. This is how charcloth is made. It is called dry distillation or pyrolysis.

The title poses, "Can paper burn without oxygen or air?" A little sneaky pedantry here from an old-school chemist - No Oxygen present, no air either, nor any mention of a vacuum. This allows us the presence of other strong oxidisers such as Chlorine and other halogens which will support combustion, which is what most people mean by "burn". As an aside, I well remember a chemistry practical demonstration where my teacher plunged a spoon of Phosphorus into a cylinder of pure chlorine to see a pale green flame spontaneusly appear, accompanied by copious white smoke (Phoshorus pentachloride) being emitted. So the answer is yes, but only in very uncommon enviroments, but which are found in the chemical industry where the Fire Service need to be aware of potential fire hazards.

If you put a paper inside an oven vacuum packed and go beyond $250~\mathrm{^\circ C}$, will it burn? There's no oxidizer. If not, what will happen?

Probably not.
As other answers point out. However, paper is a complex and varied material and some varieties called paper could contain trace quantities of materials that upon heating, could produce oxidizing species. So a hard, firm "No" may not be true in all cases.

Can paper burn without oxygen or air?

Yes!
You can watch iron, sulfur, coal, and cotton spontaneously ignite when exposed to fluorine gas in the video Fluorine - Periodic Table of Videos
I'd say cotton would be a reasonable proxy for paper here. Not only do these materials burn in fluorine, combustion is spontaneous!

Incidentally you can look up individual elements on Periodic Table of Videos at http://www.periodicvideos.com/


text:
Sequential Kinetics
I was solving some problems on chemical kinetics, then I was just struck at some ques of sequential reaction kinetics.
Given reactions $$\ce{O_3 + Cl \rightarrow O_2 + ClO ~~~ k_1=5.2 \times 10^9~Lmol^{-1}s^{-1}}$$
and  $$\ce{ClO + O \rightarrow O_2 + Cl ~~~ ~~~  k_2=2.6 \times 10^{10} ~Lmol^{-1}s^{-1}}$$
So, which value is closest to rate const of overall net reaction? $$\ce{O_3 + O \rightarrow 2O_2}$$
and the answer given was $=5.2 \times 10^9 ~\mathrm{Lmol^{-1}s^{-1}}$
Therefore I am confused, how did they find the rate const of the net reaction?

The rate constant of the overall reaction is determined by the rate constant of the RDS, Rate Determining Step.
The RDS is the one with the lowest rate constant. Hence in this case the overall rate of the reaction is $5.2 \times 10^9 Lmol^{-1}s^{-1}$


text:
Formation of O₂, Water, and Electricity from Hydroxide?
The Clark Type Oxygen Electrode converts water, oxygen, and electrons into hydroxide via a hydrogen peroxide intermediate:
\[\ce{O2 + 4e^{−} + 2 H_2O -> 4 OH^{−}}\]
Which is the sum of:
\[
\begin{align*}
\ce{2H2O +O2 + 2e- &-> H2O2 + OH- \\
H2O2 + 2e- &-> 2 OH-}
\end{align*}
\]
Can this be reversed?

Yes, the overall reaction can be reversed.  The reverse reaction is called the oxygen evolution reaction or simply water oxidation.  Depending on the pH, you could write it as either
$$\ce{2 H2O -> O2 + 4 H+ + 4e-}$$ or
$$\ce{4 OH- -> O2 + 2 H2O + 4e-}$$
The reverse reaction happens when water is electrolyzed to make hydrogen at one electrode and oxygen at the other.  You may have seen that demo at one time or another -- here's a Youtube version.
Oxygen is a thermodynamically powerful oxidant, which is another way of saying that it is easily reduced.  This means that the water oxidation reaction requires energy input, about 1.23 V at the chemical reference state of pH 0.  But it usually takes even more energy to effectively reverse the reaction -- notice they used a 9 V battery in the Youtube demo, so about 8 of those volts are wasted in the demo.  Developing new electrode materials that can carry out the water oxidation reaction with less energy loss is a major area of fundamental research in chemistry today.
You also said the Clark electrode works with $\ce{H2O2}$ as an intermediate.  Depending on the details of the particular materials used, in the water oxidation reaction hydrogen peroxide may or may not be an intermediate.  Working out the fundamental reaction steps that happen at electrode surfaces can be very difficult.


text:
What is the minimum number of given sets of information required to complete an ICE chart for an equilibrium reaction?
I'm considering creating a program that automatically completes an ICE chart/table when the minimum given information is provided. The reason I'm posting my question here is because I need to know the possible cases of minimum information required before I can even start the program.
I know you can complete the table if you're given these following sets of information:

K and initial concentrations
K and final concentrations
K and initial concentrations of reactants and final concentrations of products

I also believe there are a ton more, but I'm not sure how to complete the list.

I'm not thinking about Ksp and Ka. Only just Keq.

Alright I think you have the basics down, first of all I have a couple pointers for you since I have done many (many) RICE charts. What I would do if I were you? Since I have some experience in programming I would just use molarities of the initial and final.
You must make the user input the data, say if the user was given moles then make him/her convert it to molarity.
I think you have it pretty much down, you need either the initial concentrations and K or find the K through using the initial concentrations of the products and reactants or find the initial using the product and K.
I mean then again I would work through making this first part down because those are the essentials (atleast what I've learned) and then work on making updates to that program.
Since I do have programming experience I would just use if statements to do the entire thing. If this is available then solve for this.
Don't worry about the entire thing yet. Otherwise I would have to say you have pretty much everything down.
NOTE: This is not about Ka, Ksp, Kb, just for Keq. Because then you would have to work with the pH and pOH more equations.


text:
Can we really beat the speed of light?
This is a rather physics question but I wanted to get an insight from a chemistry perspective. I heard from some that you can beat the speed or go faster than it but is that really true. Physicists as we all know look at the theoretical possibilities but what is your insight from a chemistry-ish view.
Here and here for more information.

If we can then how much more faster is it? I heard one view into this that goes along the lines of:
If you distorted space and moved it then you would be light-years away from where you started. This seemed very dumb at first but as I thought about it is this anyway possible.

What exactly do you mean by "chemistry-ish view"? I don't think chemistry has much to do with this sort of thing other than that chemistry has strong ties to physics.

Exactly thats what I mean that chemistry that has strong ties with physics. Is there anyway to explain this whole beating light in a chemistry way?

Chemistry by its very definition is mostly concerned with matter on the level of protons and electrons. There are relativistic effects involved that are also of interest to chemists but finding a way to travel faster than the speed of light wouldn't change chemistry in my opinion. I can't imagine new reactions becoming possible or anything since you would have to "cheat" anyway (like maybe distorting space) if at all possible to travel faster than the speed of light.

In the words of Dr. Kit Bowen, "Chemistry is simply condensed phase physics."

@LordStryker Or as a friend of mine put it: "Chemistry is the nice part of physics." :)

Well on Physics SE there's tag faster-than-light with 252 questions: http://physics.stackexchange.com/questions/tagged/faster-than-light How this question could be not off-topic here?

Chemistry examines the individual "blocks" of matter and their properties, similarities and differences. Physics mainly deals with what behavior all of matter shares, and passing the speed of light is one. So, this question is off-topic here.

I'm voting to close this question as off-topic because it belongs to physics.SE.

I think the question is on-topic.  The OP clearly wants a chemist point of view.  If this happens to coincide with physics then so be it.

You aren't going to find a magical way to beat the speed of light in chemistry. Compared to contemporary physics, chemistry is a small subset which operates on a very mild energy scale, with modest distances and time intervals involved. If there were some process so crazy as to break relativity at this level, then major discrepancies from theory and experiment would abound in astrophysics and particle physics, and we would have detected such by now.

There are a number of things that exceed the speed of light:

Particles in a medium such as water can enter faster than light [in that medium] but are slowed and release energy as Cherenkov radiation.
The phase velocity of a wave can exceed that of light [in that medium].
An arbitrary point, such as the point on a pair of scissors where blades,
meet, can move faster than light.
Most puzzling to me: entangled particles. Einstein called this "spooky action at a distance". Although each particle moves at C or less, information might be shared between them in an indefinitely short time; see superluminal communication. Entanglement does apply to chemistry, as do other odd quantum effects; e.g. the detection of single photons with silver halide photography as they form an interference pattern in the double-slit experiment.

Regarding entanglement, your own link says that it cannot be used to transmit information faster than $c$: "It is not possible, however, to use this effect to transmit classical information at faster-than-light speeds[10]"

From classical special relativity it can be shown that anything (particle, object, electromagnetic wave, Z-wave, a dishwasher, anything) that travels faster than the speed of light (in vacuum) would violate the causality principle. In relation to chemistry, this means more qualitatively that the second law of thermodynamics would be violated. Hence I would argue that from a chemistry viewpoint traveling faster than the speed of light is impossible.


text:
Why the magnitude of dipole moment influences on FTIR % transmittance?
In 3:50 of this video two FTIR spectras are shown. He explains that the % transmittance (double bond in his example) depends on the dipole moment, $\mu = \delta \cdot d$.
I understand that the absorbance center frequency depends on $\delta, d$, but why the percentage of transmittance depends on $\delta, d$? A private case of this would be a symmetric bond ($\delta=0$): why the transmittance is 100%?

Light has both oscillating electric and magnetic fields.  In the common UV-Visible-IR spectroscopy experiments it is the interaction of the molecule with the  electric field that is being observed. A molecule is very small compared to the size of a beam of light, so it is fair to view the molecule as being immersed in or surrounded by the light.  In other words, it is fair to view the molecule as being immersed in a uniform electric field.
The strength of the interaction ($I_s$) of the light's electric field ($E$) with the molecule is dependent upon both the strength of the light's electric field and the vector description of the molecule's electric field.  The molecule's electric field is described by the spatial arrangement of the molecule's electrons.  The spatial arrangement of the molecule's electrons is described by the dipole moment ($\mu$), a vector sum.  Hence,
$$I_s = E \cdot \mu $$
The absorption intensity of a spectral transition (the area under the absorption curve - usually it correlates with the peak height) is given by the same equation except that rather than use the molecular dipole moment $\mu$, one uses what is called the transition dipole moment.  Rather than consider the ground state distribution of electrons around the molecule, one must use the rearranged distribution of electrons around the molecule that occurs when the molecule absorbs light.
If the transition dipole moment is always similar to the ground state dipole moment, then what the video says is correct.  This may be true for larger molecules when we are looking at an absorption that excites a $\ce{C-H}$ or $\ce{C-C}$ bond, or something similar.  In such cases, the small perturbation in electron density around the specific bond probably doesn't radically change the overall electron distribution around the entire molecule.  In smaller molecules like $\ce{CO}$ it seems possible to me that light absorption could make a significant change to the overall electron distribution and that consequently the ground state dipole moment might differ substantially from the transition dipole moment.

So basically what you're saying is that higher (transition) dipole moment causes, according to some unknown principle, more interaction with incident light. Why would that be the case?

No, I'm saying something else.  The transition dipole moment may be larger or smaller.  The two electric fields interact in a well-understood manner.  Simplifying for a moment, the interaction of 2 point charges (attract, or repel) is really the interaction of their respective electric fields and the equations to describe this are well known.  The interaction of a non-symmetric electric field generated by all of the electrons in a molecule with the electric field associated with a light wave is also understood,

but much more complex, in a mathematical sense, but these are not unknown principles.  These two fields may interact more or less strongly than the electric field generated by the light interacted with the ground state molecular electric field.  If the interaction during the transition process is larger then the absorption is more intense than if the interaction is smaller.

but why higher dipole means more interaction (=more absorbance)? or: why zero dipole means no absorbance?

As the equation above says, the bigger the molecule's  electric field (transition dipole moment), the bigger the interaction with the lights electric field and therefore the bigger the absorption (the product of the two)

..but why? what's the physical principle behind this interaction?

The magnitude of the interaction between 2 electric fields.


text:
Spin coat of Graphene oxide on glass slides
I would like to spin coat Graphene oxide on the glass slides and intend to have a thickness of 100nm/0.1micrometre. May I know what is the amount of Graphene oxide to be poured onto the glass slides, spin time, spin speed and time for it to develop to obtain the optimized and desired thickness of 100nm?

The procedure will vary with the size and shape of the glass, the concentration and viscosity of your graphene oxide dispersion, the solvent you're using, the wettability of your glass slide, and so on. You'll need to figure out the parameters yourself by varying the spinning conditions, measuring the resulting film thicknesses via profilometry/AFM/cross-sectional SEM, and then constructing spin curves (examples).


text:
Acidity order of acetone, methyl acetate and N,N-dimethylacetamide

Why is the acidity order true?
  $$\ce{CH3COCH3} > \ce{CH3CO2CH3} > \ce{CH3CON(CH3)2}$$

I deduced that in methyl acetate and N,N-dimethylacetamide there is a −I effect of the methoxy group, $\ce{N(CH3)2}$  and a +R effect, too so the attached keto-carbon develops a positive charge and the conjugate base becomes unstable due to the +R effect of methoxy and $\ce{N(CH3)2}$ groups.
Is my logic right?

Which protons are we considering the acidity of?

@bon Although OP should clarify, I think its the hydrogen atoms attached to the carbon atoms. Those are the only ones common in the 3 compounds.

The functional group in the second compound is an ester. I don't think its right to view it as a methoxy group attached to a keto-carbon.

The increased acidity of the protons on the alpha-carbon in these compounds, compared to the acidity of the protons in methane, is due to stabilization of the resultant negative charge through resonance with the adjacent carbonyl.  In the top line of the following figure we show a ketone where the proton has been removed.  We can draw 2 resonance structures to describe this carbanion.  In structure II the negative charge has been delocalized onto the electronegative oxygen atom - a very favorable and stabilizing situation.

To whatever degree we remove the carbonyl's ability to resonance stabilize that adjacent charge we will decrease the acidity of the alpha-protons.
In the case of both the ester (III) and amide (VI) we can draw a third resonance structure (V and VIII respectively).  To whatever extent these resonance structures contribute to the overall description of the ester and the amide, the carbonyl will be less available to stabilize the negative charge through structures IV and VII.
It turns out that resonance structures V and VIII are significant contributors, with the amide structure VIII having a positive charge on nitrogen being more important than resonance structure V which places the positive charge on the more electronegative oxygen (less favorable).  In fact, resonance structure VIII is such a significant contributor that restricted rotation occurs about the $\ce{C-N}$ bond and we can typically observe two different R group signals in the proton-nmr spectrum of amides.
So as stated earlier, as the contribution of the third resonance structure increases, the contribution of the second structure (the structure that stabilizes our carbanion) will decrease making our compound less acidic.  The ketone doesn't have a third structure so it will be the most acidic structure. Structure VIII is more important than structure V, so the ester will be more acidic than the amide.  Therefore, the overall order of decreasing acidity is
$$\ce{CH3COCH3 > CH3CO2CH3 > CH3CON(CH3)2}$$
just as you noted.


text:
How is it that fructose has a different metabolic pathway than glucose but yet glucose is converted to fructose?
Fructose is described to have a different metabolic pathway (a more fat-inducing one) than glucose (see: http://healthyeating.sfgate.com/difference-between-sucrose-glucose-fructose-8704.html) as it can only be metabolized by the liver and does not impact insulin.
What I don't understand, is that in glycolysis, a paramount step of metabolizing glucose, it is very quickly converted to fructose-6-phosphate. Why does the body want to convert glucose into fructose?

The metabolic pathway you are talking about is how fructose is converted into energy and how its concentration in the blood is regulated. It is indeed true that blood fructose level does not affect blood insulin level, and this is why it is more "fat-inducing", as it cannot be effectively regulated like glucose.
It is also true that in glycolysis, glucose is converted to fructose-6-phosphate. However, this should not be confused with fructose itself. The fructose phosphate that fructolysis produced is fructose-1-phosphate, not fructose-6-phosphate.

It still feels strange because it is essentially a fructose molecule (although an isomer) so it must possess some of the same properties?

@BjarniJohannsson in biochemistry, structure matters. Enzymes and receptors depend on the molecular structure to recognize and differentiate substances. Also, the "properties" of fructose that make them "fat-inducing" requires it to be in the blood, and glycolysis is entirely carried out in the cytosol, and there likely isn't any membrane protein that allows fructose-6-phosphate to exit the cell as it is an essential intermediate product in cellular respiration.


text:
Is this a possible explanation as to why sigma bonds are stronger than pi bonds?
Sigma bonds form due to the axial overlap of two orbitals whereas pi bonds form due to their lateral overlapping. Since the former is more effective, we can say that sigma is stronger than pi. We mean that the orbitals overlap to a further extent in sigma bonds and not as much in pi bonds(...or at least that is what I infer). To me this sounds like a circular proof/explanation, so I thought up another way to understand it.
In a sigma bond, the electron cloud of one atom is attracted to the nucleus of the other atom's nucleus.This happens because of the electrostatic forces of attraction between the positively charged protons in the nucleus of one atom and the negatively charged electron cloud of the other atom. The attractive force causes the electron clouds to the atoms to overlap, despite the fact that the electron clouds should repel each other.  However, the electrons start to repel each other after a certain extent of overlapping has occurred. (Because the repulsive forces between the two nuclei and the electron clouds negate the attractive forces)
In a pi bond, there is no positively charged nucleus that would attract the electrons, so there must be another factor favoring the bond formation.
Since I haven't studied the quantum mechanics of bonding yet, my best guess is that the pi bond (or lateral overlapping) occurs because of the pairing of electrons.
So this means that sigma bonds are stronger because of electrostatic forces, which consequently causes more effective overlapping than pi bonds.
Is my reasoning correct to some (if any) extent?

when you go through perturbation theory you will result in what is known as the klopman-Salem equation. In its simplified form this equation has two terms a Coulombic interaction and an orbital term. \begin{equation} \Delta E = -\frac{Q^2}{R}+\frac{[c_1c_2\beta]^2}{E_{Homo}-E_{Lumo}}\end{equation} this latter term is what we can look at. The c terms are orbital  coefficients and beta is a resonance integral (basically how well the orbitals overlap) the denominator is the difference between the orbitals energies. Therefore a large gap means less contribution.

From the orbital term. This is a good way to understand bonding. So we have the coulomb part. An overlap part. And an energy difference part.

Look at the directionality of the overlap in the 2 cases.  In a sigma bond the orbitals overlap is very effective as it is head-on.  Pi overlap of orbitals is less effective because it is lateral.

Thanks so much @AngusTheMan and I hope I understand ur first comment when it's finally taught

...and sorry, @ron, but I have heard that reason at least a hundred times this year and the purpose of this post is to find an alternative explanation, but thanks for the consideration

Related: http://chemistry.stackexchange.com/questions/6244/why-are-sigma-bonds-stronger-than-pi-bonds

@Martin Related but definitely not a duplicate.  And I would agree with your opinion about the accepted answer.

This is a very short answer and it might as well have just been a comment. It is probably also tempting in it's simplicity. For starters, let's assume, that nuclei are point charges.
When two atoms form bonds, all the electrons experience the attractive force of the nuclei, of both of them. This is independent of the type of orbital the electrons occupy.
In a σ orbital, the path of maximum electron density is between the nuclei. It is also where the electronic field of the nuclei is strongest. For a π orbital the situation is different.
Let's take a step back and consider an atom. The s orbital is spherical symmetric. The maximum electron density is (formally) in the nucleus. For the p orbital, there is a node at the nucleus. The maximum electron density is further away from the opposite charge, causing attraction to be lower. The energy of the orbital will be higher.
Similarly we can see that for π orbitals. The node of these kind of orbitals is exactly where the field of the nuclei is strongest. Hence the attraction is lowered, the energy raised, because the maximum electronic density is further away from the nuclei.
In very short terms: In σ orbitals the maximum electron density is on the bonding axis, while in π (and δ) orbitals it is off the bonding axis.

The answer to this question is simple:
Since sigma bonds are the first bond that is formed b/w 2 atoms, its formed by head-on overlapping of orbital, this kind of overlapping gives us a strong bond as compared to sideways overlapping(which happens in case of pi bonds).
Also, If you observe the two ways orbitals arrange themselves in the two bonds.....in sigma bonds the orbitals are closest to the 2 nuclei as compared to the orbitals of pi bond.
Being closer to the nuclei makes them strong as each nucleus pulls the e- of the opposite atoms towards itself and the orbitals overlap.
In a more relaxed language:
In sigma they come straight at each other whereas in pi they bend themselves towards each other.
Also, another imp. point: In sigma the 2 orbitals simply overlap whereas in pi bond the 2 have to bend themselves towards each other and that takes energy. Since pi bonds require certain extra energy for formations this can also be another reason why they are weaker.

Please don't use unnecessary abbreviations like "e-", "imp.", "b/w". Never heard of this "bending energy" theory for π bonds, too. Do you have a reputable source?