How to (numerically) model a phosphoric acid titration curve

Much easier is to calculate the inverse function $[\ce{Na+}]=f([\ce{H+}],K_\mathrm{a1},K_\mathrm{a2},K_\mathrm{a3})$.
Calculate fractions of respective phosphate forms as the function of $\ce{[H+]}$ and acidity constants.
First, calculate the common denominator $CD$:
$$a_1 = [\ce{H+}]^3$$
$$a_2 = [\ce{H+}]^2 \cdot K_\mathrm{a1} = a_1 \cdot \frac {K_\mathrm{a1}}{[\ce{H+}]}$$
$$a_3 = [\ce{H+}] \cdot K_\mathrm{a1} \cdot K_\mathrm{a2} = a_2 \cdot \frac {K_\mathrm{a2}}{[\ce{H+}]}$$
$$a_4 =  K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot K_\mathrm{a3} = a_3 \cdot \frac {K_\mathrm{a3}}{[\ce{H+}]}$$
$$CD = a_1 + a_2 + a_3 + a_4$$
.. and then reuse it and its additive terms(*):
$$[\ce{H3A}]=c_0 \cdot \frac {a_1}{CD}$$
$$[\ce{H2A-}]=c_0 \cdot \frac {a_2}{CD}$$
$$[\ce{HA^2-}]=c_0 \cdot \frac {a_3}{CD}$$
$$[\ce{A^3-}]=c_0 \cdot \frac {a_4}{CD}$$
$c_0$ is the total phosphate concentration. From the fractions, you can get $\ce{[Na+]}$:
$$\ce{[Na+]} = \ce{[H2A-]} + 2 \cdot \ce{[HA^2-]} + 3 \cdot \ce{[A^3-]} + \frac{K_\mathrm{w}}{[\ce{H+}]} - \ce{[H+]}$$

There is one important thing to consider: Titration dilutes the acid.
If  there is the volume $V_0$ of $\ce{H3PO4}$ with molar concentration $c_\mathrm{0,init}$, and if we spend the volume $V_1$ of $\ce{NaOH}$ solution of concentration $c_1$, then:
$$[\ce{Na+}]=c_1 \cdot \frac {V_1}{V_0 + V_1 }$$
$$c_0=c_\mathrm{0,init} \cdot \frac {V_0}{V_0 + V_1}$$
So perhaps the best approach can be:

There are given concentrations of respective solutions
For given volumes of respective solutions calculate $c_0$ and $\ce{[Na+]}$
For given $c_0$ and $\ce{[Na+]}$, calculate $\ce{[H+]}$ by finding the solution of the equation below ( $f([\ce{H+}]) =  0$ ) by numerical methods, like Regular falsi, or Newton one.

$$ \ce{[H2A-]} + 2 \cdot \ce{[HA^2-]} + 3 \cdot \ce{[A^3-]} + \frac{K_\mathrm{w}}{[\ce{H+}]} - \ce{[H+]} - \ce{[Na+]} = 0$$
For $y_1=f(x_1)$, $y_2=f(x_2)$, $y_1 \cdot y_2 \lt 0$, there is the Regula falsi iteration:
$$x_3 =  \frac{x_2 \cdot y_1 - x_1 \cdot y_2  }{  y_1 - y_2}$$

(*) Note that the formulas are quite common, I have just rearrange them. They should be able to be found on internet. I have once derived them myself, but not this time, because I remember them.


If a body absorbs a quantity of heat $q$ its temperature will normally rise by a value $\Delta T$. The average heat capacity  over this temperature range is defined as $C_{av}\equiv q/\Delta T$. The instantaneous heat capacity at temperature $T$ is $C\equiv dq/dT$. This definition is not exact enough, however, until the path of heating is specified. From the first law $q$ could be replaced by $dU+dw$ so it is necessary to fix another parameter which is $V$ or $p$ giving rise to the definitions you quote.
Note (a) that at a phase change $C$ becomes infinity because heat is absorbed with no change in temperature. (b) As the heat capacity is the gradient of internal energy with temperature it describes how the internal energy increases. For example when $C$ becomes constant at high temperatures the molecules are absorbing heat a constant and maximum 'rate' with increase in $T$. (c) Using statistical mechanics the heat capacity can be calculated from a knowledge of a molecule's energy levels.


What is the chemical name of the compound H2S4O?

Section IR-5.4 of IUPAC’s Nomenclature of Inorganic Compounds (Red Book) of 2005 lists the rules that shall apply to naming inorganic compounds by a generalised stoichiometric name which does not carry any information about the compound’s structure. These rules dictate that:

The constituents of the compound to be named are divided into formally electropositive and formally electronegative constituents. There must be at least one electropositive and one electronegative constituent. Cations are electropositive and anions electronegative, by definition. Electropositive elements occur later in Table VI than electronegative elements by convention.
In principle, the division into electropositive and electronegative constituents is arbitrary if the compound contains more than two elements. In practice, however, there is often no problem in deciding where the division lies.
Section IR-5.4.1

In our case, oxygen must be considered electronegative and hydrogen must be considered electropositive since they have the highest and lowest electronegativity, respectively.
Within the classes of electronegative or electropositive substituents, ordering is strictly alphabetical. Electropositives come first, electronegatives last.
Section IR-5.4.2.1 allows the use of multiplicative prefixes as is the case with binary compounds. Further information such as charge, oxidation numbers or structure is not at our disposal.
Therefore, according to the principles laid out, we have two choices to name $\ce{H2S4O}$, depending on whether we wish to consider sulphur electropositive or electronegative:

Dihydrogen tetrasulphur oxide

Dihydrogen oxide tetrasulphide

To the best of my knowledge, this compound does not exist. But the principles of nomenclature are designed in a way to allow the naming of all compounds whether synthesised, hypothesised or fancily drawn on paper.


Gibbs free energy for the evaporation of water

The average heat capacity $C_p$ of liquid water and water vapor over the range between $\pu{25^\circ C}$and $\pu{100^\circ C}$ are, respectively $4.18\ \pu{kJmol^{-1}K^{-1}}$ and $1.72\ \pu{kJmol^{-1}K^{-1}}$.  So the enthalpies of liquid water and water vapor at $\pu{100^\circ C}$ are:
$H_l(\pu{100^\circ C})=-285.8+(0.018)(4.18)(75)=-280.2\ \pu{kJmol^{-1}}$
$H_g(\pu{100^\circ C})=-241.8+(0.018)(1.72)(75)=-239.5\ \pu{kJmol^{-1}}$
So, at $\pu{100^\circ C}$, $\Delta H=40.7\ \pu{kJmol^{-1}}$
The entropies of liquid water and water vapor at $\pu{100^\circ C}$ are:
$S_l(\pu{100^\circ C})=0.06996+(0.018)(4.18)\ln{(373.15/298.15)}=0.08684\  \pu{kJmol^{-1}K^{-1}}$
$S_g(\pu{100^\circ C})=0.1887+(0.018)(1.72)\ln{(373.15/298.15)}=0.1956\  \pu{kJmol^{-1}K^{-1}}$
So, at $\pu{100^\circ C}$, $\Delta S = 0.1088\ \pu{kJmol^{-1}}$ and $T\Delta S=40.6\ \pu{kJmol^{-1}}$ So the change in Gibbs free energy between saturated liquid and saturated vapor at $\pu{100^\circ C}$ and 1 atm is $$\Delta G=40.7-40.6=0.1\ \pu{kJmol^{-1}}$$ which, to within roundoff error is zero (as expected).
The Spoiler answer is obviously incorrect.


What does an equal sign mean in chemical equations?

The IUPAC Green Book (Quantities,
Units and Symbols in Physical Chemistry, Third Edition, 2007) reads:

  2.10.1 Other symbols and conventions in chemistry
  [...]
  (iv) Equations for chemical reactions

  (a) On a microscopic level the reaction equation represents an
  elementary reaction [...] A single arrow is used to connect reactants
  and products in an elementary reaction. An equal sign is used for the
  "net" reaction, the result of a set of elementary reactions. [...]

  $\ce{H + Br2 -> HBr + Br} \quad$ one elementary step in HBr formation
  $\ce{H2 + Br2 = 2 HBr} \quad$ the sum of several such elementary steps

  (b) On a macroscopic level, different
  symbols are used connecting the reactants and products in the reaction
  equation, with the following meanings:

  $\ce{H2 + Br2 = 2 HBr}\quad$ stoichiometric equation
  $\ce{H2 + Br2 -> 2 HBr}\quad$ net forward reaction
  $\ce{H2 + Br2 <--> 2 HBr}\quad$ reaction, both directions
  $\ce{H2 + Br2 <=> 2 HBr}\quad$ equilibrium

  The two-sided arrow $\ce{<->}$ should not be used
  for reactions to avoid confusion with resonance structures [...]

  2.12.1 Other symbols, terms, and conventions used in chemical kinetics
  [...]
  (ii) Composite mechanisms
  A reaction that involves more than one elementary reaction is said to occur by a composite mechanism. The terms complex mechanism, indirect mechanism, and stepwise mechanism are also commonly used. Special types of mechanisms include chain-reaction mechanisms, catalytic reaction mechanisms, etc.

    Examples
    A simple mechanism is composed of forward and reverse reactions
    $\quad\quad\quad\quad\quad\quad\ce{A -> B + C}$
    $\quad\quad\quad\quad\ce{B + C -> A}$
    It is in this particular case conventional to write these in one line
    $\quad\quad\quad\quad\quad\quad\ce{A <--> B + C}$

  However, it is useful in kinetics to distinguish this from a net reaction, which is written either with two one-sided arrows or an "equal" sign
  $\quad\quad\ce{A <=> B + C}$
  $\quad\quad\ce{A =  B + C}$

  When one combines a composite mechanism to obtain a net reaction, one should not use the simple arrow in the resulting equation.
  $\quad$Example
  $$\begin{align}
\ce{A &\to B + C}\quad\text{ unimolecular elementary reaction} \\
\ce{B + C &\to D +E}\quad\text{ bimolecular elementary reaction} \\
\hline
\ce{A &= D + E}\quad\text{net reaction (no elementary reaction,no molecularity)}
\end{align}$$


How do I create a transparent liquid for ferrofluid display?

There's a really good video on YouTube by Nilered/Nileblue on ferrofluids. In it he suspends magnitite with oleic acid (main component of olive oil) as a surfactant and uses Kerosene (Paraffin Oil) as a carrier.
There's a PTFE 'dry lubricant' sold under the WD-40 brand name that might be able to prevent the glass from staining as it creates a thin coat of PTFE on the glass which is highly inert. Just spray the piece of glass with the PTFE Spray and wait for it to dry (~7mins). [The PTFE spray has a myriad of other uses too, you can put it on any surface you want to have low friction, to be lubricated, an unreactive surface, dust repellant surface, etc].
This might be a way around the staining issue you are getting. Or at least point you in the right direction. Good luck!


A compound S follows the Lambert-Beer law if the absorbance $E_\lambda$ at a particular wavelength $\lambda$ is proportional to the concentration $c$ of S.

\[E_\lambda = \left(\frac{I_0}{I} \right) = \epsilon\cdot c \cdot d \]

A "deviation" is observed when the real concentration $c$ is different from the expected/assumed.

This may happen if

S is an acid or a base:

variation of the pH leads to (de)protonation and forms another species with a completely different absorption spectrum, or at least a different $\epsilon$ at the wavelength under observation

S undergoes a reaction to S' (with different optical properties) at elevated temperatures (or change of pH)
Decrease of temperature or change of pH leads to the formation of a turbid solution from which light is scattered into the detector


Why do acids favor the oxidizer? And bases the reducer?

The effect that can be observed by adding acids to a solution to increase conductivity is known as the Grotthuss mechanism. Protons have a much higher mobility in solution ($3.62\cdot10^{-3}~\mathrm{cm^2~V^{-1}~s^{-1}}$) than any other cation, basically allowing it to tunnel through a solution. (A similar effect can be noted for hydroxyl anions.)
Basically this allows a much faster charge exchange through hydrogen bond networks, also resulting in a higher cell voltage.

(But then again I am not really sure I understood your question correctly.)


I left a bottle of water in the fridge for the past three months. I took it out today. There was a block of ice in it. How?

The piping for refrigerant circulation is in the fridge walls, so obviously it's colder near the back as there is no heat from the front door reaching that area and heat exchange with compressor/coolant circuit is at maximum.

As for the partially frozen water, this happens to mineral water and water containing any salts in general. As water freezes out, the salinity increases, resulting in freezing point depression for remaining solution. These processes are the basis for fractional freezing.


Asking about definitions of word "sugar" in biochemistry

Absolutely correct!

  N-Linked oligosaccharides are initially synthesized as lipid-linked
  precursors. The lipid component in this process is dolichol, a
  long-chain polyisoprenol of 14 to 24 isoprene units (17–21 units in
  animals and 14–24 units in fungi and plants; isoprene units are C5
  units with the carbon skeleton of isoprene , which is linked to the
  oligosaccharide precursor via a pyrophosphate bridge.

Dolichols apparently anchor the growing oligosaccharide to the endoplasmic reticulum membrane. These compounds have strong hydrophobic interactions with membrane lipids, anchoring the attached sugars to the membrane, where they participate in sugar-transfer reactions.

The pathway of dolichol-PP-oligosaccharide synthesis involves stepwise addition of monosaccharide units to the growing glycolipid by specific glycosyltransferases to form a common “core” structure:

Dolichol pyrophosphate glycoside. The carbohydrate precursors of N-linked glycosides are synthesized as dolichol pyrophosphate glycosides. Dolichols are longchain polyisoprenols (n = 14–24) in which the α-isoprene unit is saturated.

References

Voet and Voet Biochemistry
Lehninger Principles of Biochemistry


The way to get a handle on why gases are all similar is to think about the relationships between how the particles of the gas behave and the macro properties temperature and pressure.

The temperature is a measure of the average kinetic energy of the particles. The pressure is a consequence of the particles of the gas bouncing off the walls of the vessel the gas is contained in or the surfaces in contact with the gas. In a gas made of heavy particles the average speed of the particles will be much lower than for a gas made of light particles. When you do the calculations about the effect at a surface (where the pressure is a result of the particles bouncing elastically off the surface) the effects of the mass cancel out because the heavier particles travel more slowly at the same temperature. Overall this means that, for an ideal gas, the mass of the particles doesn't matter for the pressure and all gases behave the same regardless of the particles they are made from.


Determining change in entropy and enthalpy during mixing of gases

Your equation for $\Delta S_{system}$ is incorrect.  The correct equation is $\Delta S_{system}=\int{\frac{dq_{rev}}{T}}$.  I hope you understand the difference, and the fact that the subscript "rev" refers exclusively to a reversible path between the initial and final states of the system.  To get that, you need to forget entirely about the actual irreversible path, and instead focus on the two end states.  You then have to devise a reversible path between these same two states so that you can determine the value of the integral.  For your problem, this recipe will yield a positive value for $\Delta S_{system}.$

Regarding the enthalpy change, you are aware that the enthalpy of mixing of ideal gases is zero, correct?


Why is the groundstate electronic degeneracy of monoatomic O(g) 5?

S = 1, L = 1, so (S+1)(L+1) = 6

You are using here not just alternative physics, but also alternative math. For, even if the formula were right, you should get $2 \cdot 2 = 4$ and not $6$.

Anyway, the right formula for the degeneracy is $(2S + 1)(2L + 1)$ which being used for the ground $^1D$ state of oxygen atom (i.e. for $S = 0, L=2$) indeed gives $1 \cdot 5 = 5$.


If a sugar is nonreducing, does that mean it doesn't ionize in water?

I believe the confusion lies with the term Reducing sugar. From my perspective sucrose is not an ionic compound, it is covalently bonded. I believe that this should indicate to you that it does not "ionize" or the term that I was taught "dissociate" since it has nothing to dissociate from.

To quote wikipedia:

  "The van 't Hoff factor is the ratio between the actual concentration
  of particles produced when the substance is dissolved, and the
  concentration of a substance as calculated from its mass."

Since it does not "dissociate," I would say that the answer is no, it does not "ionize," and therefore has a Van't Hoff factor of 1 when mixed with pure water.

Like anything with electrons, they can be lost. This takes place in our bodies through more complicated reactions than I understand completely. I believe however, according to wikipedia, that sucrose has a Glycosidic bond which when broken, will separate sucrose into smaller sugars, if this takes place in your solution then I believe it will have a Van't Hoff factor > 1.


Balancing complex haloform redox reaction

As you noted, you need to find the limiting reactant. This is most of the time done by taking the one with a 1:1 ratio of educt:product.

So in this case, the $\ce{ROCH_3}$. So you take that once. Now you see immediately that you need 2 $\ce{Na}$ on the right side, so you can put a 2 in front of the $\ce{NaOH}$. You have two $\ce{X}$ on the left, but 4 on the right, so put a 2 in front of the left one. Now count all your atoms. Subtract them from each other:

You have $a * \ce{H}$ on the left and $b * \ce{H}$ on the right. if $a-b=0$ its balanced. Do that for every reactant and you're done.

There are of course more strictly formulated methods, but in my time I noticed that a straightforward approach with a little thinking can be applied most universally without having to memorize some formulas.


How does chlorine form more than 1 bond?

The order of bonding, and so the valence state of Cl in $\ce{ClO_{x}-}, x>1$ compounds is very debatable. Generally, two models exist.

$\ce{Cl}$ atom, just like $\ce{S, P}$ and some others has unoccupied $d$-orbitals in the valence shell. It is possible to move some electrons from $p$-orbitals to $d$-orbitals, producing half-occupied orbitals that can participate in covalent bond formation
$\ce{O}$ atom has 6 electrons at the valence shell with four orbitals. It is possible to move one electron from chlorine atom to oxygen atom to form $\ce{O^{\dot{-} }}$ ion, that can than form a bond with newly formed half-filled orbital of chlorine. However, being electronegative and having high positive charge, chlorine partially pulls back the electron pair of newly formed bond.

Currently the second option is in favor of theoreticians, both by electron population analysis and relatively high charges in the oxochloric compounds, however, in school-level chemistry people traditionally resort to the first option. On the other hand, the second approach is the only option to adequately describe nitric acid and nitrates as nitrogen does not have $d$-orbitals on the valence shell.

Now, the stability of oxochloric compounds is a completely different matter. All of them are not very stable compounds, easily freeing their oxygen. Still, the process has significant activation energy, the state in between is even less favorable then oxochloric molecule. both oxygen and chlorine have electron-rich shells that repel potential reactants. Perchlorate ion is especially stable, at its surface is completely covered by outer negative shells of oxygen, so electrons cannot reach positively charged chlorine.


How to formally calculate the surface density of layered materials such as graphene?

The C-C length in graphene is l = 0.142nm and the area of a hexagon can be calculated with the formula:

$ A = \frac{3\sqrt{3}}{2} l^2 =  0.0523nm^2$

In each hexagon, there are 2 full carbon atoms(1/3*6) so the surface density of one single layer is:

$ S_d = \frac{2*massCarbon}{A}  =  \frac{2*1.994 × 10^{-26}Kg}{0.0523× 10^{-18}m^2} = 76.26 × 10^{-8} Kg/m^2 = 7.63× 10^{-8} g/cm^2$

If you are considering 2, 3, etc. layers than the surface density it's twice, three times etc. the surface density of the single layer.

Additional note: The distance between layers is h = 0.335 nm and therefore its density can be calculated as:

$ d = \frac{S_d}{h}  =  \frac{7.63 × 10^{-8} g/cm^2 }{0.335× 10^{-7}cm} = 2.28 g/cm^3  $

This is very close to the experimental value I have found online says that the density of graphene is $2.267 g/cm^3$


What is the molar mass of a compound with 0.5 atoms in its molecular formula?

You should use 0.5 lithium atoms in the molecular mass. Alternatively you can multiply the formula indices by two to obtain $\ce{LiFe_2(PO_4)_2}$. This formula describes exactly the same compound but it's much more readable, for instance is now clear that the substance contains one $\ce{Li(I)}$, one $\ce{Fe(II)}$, one $\ce{Fe(III)}$ and two phosphate anions per $2n$ electrons transfered.

Here the value of $n$ is multiplied by 2 because you also have your stoichometry multiplied by 2. If you use $\ce{Li_{0.5}Fe(PO_4)}$ to calculate $M$ then the value of $n$ stays unchanged.


Calculate the amount of energy released when 500 cm³ of methane gas at STP reacts with excess air according to the equation:

How you find the moles from volume? Did you assume $500\ \mathrm{cm^{3}}$ is $500\ \mathrm g$? This is wrong, because the density of $\ce{CH4}$ is not $1\ \mathrm{g/cm^{3}}$. You have to use another way to calculate moles of the gas. Think about ideal gas law or how much volume for STP gas.


Sign of work done by the system in first law of thermodynamics equation

I always got lost in remembering the rules for the signs, so I just ignore them and think about the two parts from the standpoint of the system (cylinder):

If heat enters the system, then it gets hotter so that contribution is positive. This part's fairly intuitive.

If I let the cylinder expand, it's relaxing to a less compressed state and so loses energy, which means the work contribution is negative. In order to compress the cylinder I have to push on it, adding some of my own energy to the system, so the work contribution is positive.

In your case, the change in energy from the heating (positive), and loss of energy through expansion (negative) is:
$$ \Delta E = 3600kJ - 800kJ $$

Edit: The problem itself is ambiguous regarding whether it's expansion of contraction. My answer assumes expansion since that's what would naturally happen upon heating a gas cylinder. Gaurang's answer is correct if you read it as contraction.


What would happen if I combined this polymer and this metal (assuming I can)

You can combine materials in two ways: chemically and mechanically.
On average, if you want to the desired physical properties to add up, you want to go mechanical way, not the chemical one.

Once the reaction takes place, the there is absolutely no reason to assume the product preserves properties of both precursors. It might sometimes, but more often than not it's a completely different beast.
Speaking of retaining physical properties in terms of logic and addition operation, in chemistry both $\text{true} + \text{true} = \text{false}$ and $2 + 3 = -42$ are possible.
Mixing water and sodium won't result in liquid metal.

What you probably want to look into instead is mechanical combination, or, more precisely, composite materials such as fiberglass or carbon plastics alternative where titanium would act as a 2D or 3D supporting mesh for sorbothane, though not reacting with it chemically.
Such poly-phase materials are already used for monolithic composite armor plates.


Latent heat of vaporization/evaporation:  It's the amount of heat required for liquid ---> gas phase change of a substance at it's Boiling point.

Not exactly. It's the amount of heat required for liquid ---> gas phase change of a substance at the particular temperature, there is usually and implicitly the substance boiling point. But generally, it is defined for any temperature.

Is it equivalent to only the amount of latent heat of vaporization or is it plus the amount of energy required to raise it's temperature upto it's boiling point.?

Neither, neither. It is between these 2 values.
The latent heat of water evaporation at temperature $T_2 \lt T_1$ (let we take $T_1$ as the boiling point ) is equal to the  latent heat of vaporization at temperature $T_1$
plus the heat required to warm up water  from $T_2$ to $T_1$
minus the heat released to cool down vapour from $T_1$ to $T_2$.
$$\Delta H_{\mathrm{evap},T_2} = \Delta H_{\mathrm{evap},T_1} + C_\mathrm{p,water}\cdot (T_1 - T_2) + C_\mathrm{p,vapour}\cdot (T_2 - T_1)$$
where $\Delta H$ are molar latent evaporation heats at constant pressure ( called molar enthalpies of evaporation) at respective temperatures,
$C_\mathrm{p}$ are respecive molar heat capacities of water and vapour ( for simplicity considered constant to avoid integrals).
It is application of the Hess' law, which is application of the energy conservation law. It says the energy change depends only on the initial and final state, not on the way how is the final state reached. If it were not so, we could easily construct perpetuum mobile by switching between 2 states by 2 different ways.
So getting A->B directly, or A->C->D->B is equivalent, where

A is cold water
B is cold vapour
C is hot water
D is hot vapour

Generally, the latent heat of evaporation increases with decreasing temperature, so you need more energy to evaporate water at lower temperature. At water critical temperature near $374 ^{\circ}\mathrm{C}$ the properties of liquid and gaseous water  converge to the same values and the latent heat of evaporation converges to zero.
If water at $50^{\circ}\mathrm{C}$ is provided by latent heat of evaporation at $50^{\circ}\mathrm{C}$, then it will evaporate. If water at $50^{\circ}\mathrm{C}$ is provided by latent heat of evaporation at $100^{\circ}\mathrm{C}$, then some water will not evaporate. Because the latent heat of evaporation is bigger at $50^{\circ}\mathrm{C}$, compared to $100^{\circ}\mathrm{C}$.
Technically, it is different question, as evaporation at ambient temperature takes energy from the environment, while you have to supply energy for the boiling.
Note that physicists/chemists prefer to talk about vapour, a gaseous phase of substance, that is not considered as a "permanent" gas. Steam is rather a technical/engineering/everydays term, covering pure water vapour, mixtures of vapour with gases like air, or the same with an aerosol of water or ice particles. See water phase diagram. Water vapour exists even at at mesosphere at altitude near 80 km at temperatures almost $-100^{\circ}\mathrm{C}$. Also, ice can sublimate directly into vapour, without melting.


What is the least dense liquid under normal conditions?

Isopentane $\ce{C5H12}$ has the density of $0.6201~\mathrm{g\,cm^{-3}}$ at $20~\mathrm{^\circ C}$ [1, p. 3-330].

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; CRC Press, 2017; Vol. 97. ISBN 978-1-4987-5429-3.


How were the first molecular formulas determined?

Lavoisier burned hydrogen and oxygen together and found that water was produced.  Dalton incorporated this information into his thinking that the most stable arrangement of any two atoms involved a 1:1 combination, therefore water would be "OH".

In slightly more recent times, but prior to modern-day spectroscopy, chemists would chemically chop molecules into identifiable fragments.  They would also combust the molecule and get a C,H,N,O ratio.  Putting all of this together they would (often) come up with remarkably accurate molecular formulas.


Energy loss in the process for exothermic reactions

No.  If means that to maintain the temperature of the products equal to the temperature of the reactants, you need to remove heat, and, form the first law, if you do this, then $\Delta E$ is negative.  However, if you don't remove the heat, the temperature of the products will be higher than those of the reactants, and $\Delta E$ of the system will then be zero (as required by the first law).


Whether environments are the same or not is based on symmetry considerations and not the number of hydrogens attached to the adjacent carbon atom.

Molecules and atoms within a molecule can be of three types:

identical - they can be interconverted by simple rotations or translations.
enantiotopic - they can be interconverted by reflection in a mirror or about a center of symmetry
diastereotopic - they cannot be interconverted.

Molecules or atoms within a molecule that are found to be identical are said to be in the same environment.  Molecules or atoms that are identical, or in the same environment, must be identical with respect to all chemical and physical properties.  These properties would include chemical reactivity, spectroscopic environment, etc.

In benzophenone, I can rotate the benzene ring about the $\ce{C5-C7}$ bond. This rotation interconverts $\ce{C4}$ and $\ce{C6}$.  If I told you to look away and I performed this rotation (and we removed the numeric subscripts), there is no way that you could tell if the original carbon or the other carbon was now in the $\ce{C4}$ location.  These two carbons (and their attached hydrogens) are therefore said to be equivalent by the above definition.  The same applies to $\ce{C1}$ and $\ce{C3}$, they are also equivalent to each other, or in the same environment.

I can also rotate the molecule around the $\ce{C7=O}$ carbonyl axis.  If I do this I can interconvert $\ce{C4, C6, C8, C_{10}}$.  This same rotation also (separately) interconverts $\ce{C1, C3, C_{11}, C_{13}}$, and also (separately) $\ce{C2}$ and $\ce{C_{12}}$.

Therefore, these 3 sets of carbons

$\ce{C4, C6, C8, C_{10}}$
$\ce{C1, C3, C_{11}, C_{13}}$
$\ce{C2, C_{12}}$

are different (diastereotopic) from one another (I can't interconvert $\ce{C4}$ and $\ce{C_{12}}$, for example).  So we have 3 different chemical environments, but within each environment there are several identical atoms. The same methodology and reasoning applies to the hydrogens attached to these carbons.  They also exist in 3 different environments, and within each of these 3 environments there are several equivalent hydrogens.


How to convert CO2 ppm volume to ppm weight?

You can't determine the concentration of CO2 in the water just from measuring the concentration in the air above the water.

The solubility of CO2 is extremely dependent upon pH. See: http://ion.chem.usu.edu/~sbialkow/Classes/3650/CO2%20Solubility/DissolvedCO2.html

There is no indication you are considering pH at all.

Solubility of CO2 also depends upon temperature.

Also, you need to consider whether or not the phases are in equilibrium.


Electron Configuration of an element with no. of electrons > 18

But, according to the 2n2 rule, the M-shell can hold upto 18 e-. So, doesn't this mean that the e- configuration of Ca can also be written as 2,8,9,1 or say, 2,8,7,3 or something?

Experimentally, the ground state electron configuration is 2, 8, 8, 2, so there is no other correct way to write it.  2,8,9,1 and 2,8,7,3 would be incorrect.

Really the only way to know is through experiment or possibly very high level calculations that include relativity and quantum electrodynamics.  As atomic number increases, the problem becomes more and more complex.  It isn't necessarily true that one configuration accurately describes the ground state.

For example for Lawrencium (element 103), for which no experimental determination has been made, you will see some periodic tables that say it has a 7p electron (8 electrons in the P shell and 3 electrons in the Q shell) and some that say it has a 6d electron (9 electrons in the P shell and 2 electrons in the Q shell).

See The low-lying level structure of atomic lawrencium (Z=103): energies and absorption rates to appreciate how difficult it is to really calculate such information.


Which is more stable out of the given carbocation?

$\ce{C-D}$ bonds are stronger. Hyperconjugation is thus more easily seen  with hydrogen rather than deuterium.   Hyperconjugation> Inductive effects at stabilization.I believe you can reason the rest out by yourself.  To see why $\ce{C-D}$ bonds are stronger.

See this


Why the DMM (multimeter) register negative resistance when measure a drop of hydrochloric acid

The multimeter probe tips are dissolving and setting up a galvanic cell that augments the current flow thru the ohmmeter.
The probe tips may be of the same metal, but the ohmmeter voltage will favor the dissolution of one over the other, and the cell thus formed will assist the applied voltage.
Quickly switching to voltmeter (or milliammeter) mode will show a cell voltage (current).
Using the ohmmeter to measure resistance of a non-corrosive solution (e.g., NaHCO3) will show some (positive) resistance.
Inert electrodes (Pt, Au) will show a positive resistance.


Suggestions for 200-600 nm range transmitting polymers/materials

No plastic of any practically useful thickness$^\dagger$ has $80\%$ transparency down to $200~\mathrm{nm}$. Period.

Below is a UV transmission chart for various materials from a marketing brochure published by BrandTech, a manufacturer of disposable UV/Vis cuvettes (click image to enlarge):

The chart only goes down to $210~\mathrm{nm}$; extrapolating the trend, not even quartz is $80\%$ transparent at $200~\mathrm{nm}$, but it's probably a decent choice for your purposes as long as it sufficiently meets your other criteria.

Beyond that, fluorite $\left(\ce{CaF2}\right)$ is the only specific material I know of with better UV transparency; it appears to have better than $90\%$ transparency down to around $200~\mathrm{nm}$ (figure is from ThorLabs online catalog):

A chart at that ThorLabs link indicates two additional candidate materials: $\ce{MgF2}$ and sapphire. As with quartz, the choice among these will boil down to how well they meet your other criteria.

$^\dagger$ As a caveat to the above: If a very thin plastic film ($\leq 100~\mu \mathrm{m}$ or so) is acceptable, a specialty material might give sufficient transmittance. You would have to convince the manufacturer of such a material to make it for you, however, which would probably not be trivial, or cheap.


What is melting? Which bonds do we break to melt something?

This is a good question. Unfortunately there are several criteria by which chemists indentify whether a process is melting or not. One of them is called the Lindemann Criteria which says:

"Crystals are considered to melt, when the vibrational amplitude becomes half of the interatomic spacing in the crystal lattice."

What does it mean? Normally, at temperatures greater than 0K, atoms possess kinetic energy. Infact, temperature itself is a measure of the kinetic energy of the constituent atoms. The kinetic energy of an atom is related to the temperature as:

$$\text{Kinetic Energy}=\frac{3}{2}k\text{T}$$

where $k$ is a constant called Boltzmann Constant, with a value of $1.38×10^{-23}\text{J/mol}$. Atoms in a solid are characterized by their fixed mean positions. They only vibrate within a certain boundary around them. Since they have kinetic energy (and velocity as well), they tend to leave their current location, but repulsive forces from the other atoms push it back to its original position. In this way, you could consider atomic bonds like "little springs".

The extent to which the atom displaces itself from the mean location is called it's vibrational amplitude. As we increase temperature, the atoms will have more velocity, and can consequently displace itself further from the mean position. Lindemann defines the melting temperature as the temperature at which the amplitude becomes half of the space between two adjacent atoms of the crystal.

Yet another criteria for defining the melting point is the Born criteria that says:

"Crystals are considered to melt, when the shear modulus approaches zero"

You might wonder, what is this shear modulus? It's infact, a measure of how much tangential stress a solid object can handle, and the tangential strain caused by it. Tangential forces on an object are forces that act parallel to the surface of the object, hence the name. See the picture below that shows a tangential force applied on an originally cuboidal object:

Tangential (or shearing) stress is defined as:

$$\sigma _{shear}=\frac{F_{shear}}{A}$$

where $F_{shear}$ is the tangential (shearing) force, and $A$ denotes area of surface in question.

Considering the picture above, shearing strain would be defined as:

$$\epsilon _{shear}=\frac{\Delta x}{h}$$

There exists a unique proportionality between the stress applied and strain produced, and is given by:

$$\frac{\sigma _{shear}}{\epsilon _{shear}}=S$$

where $S$ is called the shear modulus. Its a constant for a given material and a given temperature.

In a physical sense, liquids are considered as substances that cannot withstand a tangential stress. When a tangential stress is applied, liquids simply keep on increasing the strain, even under small stress. This itself is a useful way to consider when a solid melts, when its shear modulus becomes zero!

Now, let's consider how we consider melting for the different examples that you have asked for:

Diamond: Diamond has a structure of covalent bonds arranged tetrahedrally to each carbon. Each bond has it's characteristic bond length. Applying Lindemann condition over here would be considered advisable , so that melting point is where vibrations of carbon atoms is half of the $\ce{C-C}$ bond distance. Ultimately some of the $\ce{C-C}$ would break and the molten mixture would largely consist of radicals of varying sizes, that permit free movement between themselves.

Graphite: While each layer of graphite is only weakly bonded to the other layer by van der Waals' forces, each layer (called graphene) itself is a quite a large molecule. You can consider a liquid as a large number of small molecules, so clearly graphene doesn't look like this. To melt it, you would have to break some of the $\ce{C-C}$ bonds so that we can produce small molecules that can move freely. We apply Lindemann condition in this case. The molten mixture would be similar to that of diamond.

Branched polymers: Remember how I told to consider liquids as small molecules that can move freely sliding among themselves? We consider the same thing over here. The branch bonds are comparatively weaker than the rest of the bonds in the polymer, and these are the ones that will break when heated. As far as bond-breakings are concerned, we use the Lindemann condition. While the unbranched polymers are big molecules, they are still small enough to exhibit liquid character at the high temperature required to break the branched bonds. The molten mixture would consist of the straight-chain polymers in radical forms.


If this is a real sample, and not a provided coursework spectrum, then the best recommendation is to run a HSQC and/or HMBC. This will provide complete data to allow this assignment fully, and in much the same time as it takes to run a DEPT135. I don't understand the urge to run DEPTS any more - I haven't run one for years. A phase edited HSQC will give you all of the information that a DEPT can provide, as well as connectivities between proton and carbons. Anyway, off my soapbox.

Below is a simulation of this molecule, which has reasonable alignment with your peaks.

A methyl centre trans to a alkyl chain will be downfield shifted to an equivalent methyl group trans to a proton. So 19 will be downfield shifted from 20, and be fairly similar in chemical shift to 16. Simulations have their limitations, and carbon shifts are solvent dependent to +/- a few ppm at least. To avoid any uncertainty, run HSQC and HMBC. These should be part of your normal procedure for assigning structures.


The question: The heats of combustion of $\ce{C_xH_y}$, $\ce{C}$, and $\ce{H2}$ are $a, b,$ and $c$ $\pu{cal mol-1}$, respectively. What will be the heat of formation of $\ce{C_xH_y}$?

We have three combustion reactions to consider:
$$\ce{C + O2 -> CO2} \quad \Delta H_{\mathrm{comp}(\ce{CO2})} = b \tag{1}$$
$$\ce{H2 + 1/2 O2 -> H2O}  \quad \Delta H_{\mathrm{comp}(\ce{H2})} = c \tag{2}$$
$$\ce{C_xH_y + ($x$ +y/4) O2 -> x CO2 + y/2 H2O}  \quad \Delta H_{\mathrm{comp}(\ce{C_xH_y})} = a \tag{3}$$
Assuming the given answer is correct, I have to tell this straight: The wrong doing in this question is giving all three heat of combustion values as their numerical values and not mentioning it in the question (thus, solver has to assume it). That, I called being tricky at this level of education. For example, if the question is a multiple choice, then it may include both $-\left(x b+\frac{y c}{2}-a\right) \ \pu{cal mol-1}$ and $+\left(x b+\frac{y c}{2}-a\right) \ \pu{cal mol-1}$ as chices to pick as an answer. Too bad for students.
Let's assume $\Delta H_{\mathrm{comp}(\ce{CO2})}$ is $-b$ in the equation $(1)$, and so on. Now we can rewrite above equations as:
$$\ce{C + O2 -> CO2} \quad \Delta H_{\mathrm{comp}(\ce{CO2})} = -b \tag{4}$$
$$\ce{H2 + 1/2 O2 -> H2O}  \quad \Delta H_{\mathrm{comp}(\ce{H2})} = -c \tag{5}$$
$$\ce{C_xH_y + ($x$ +y/4) O2 -> x CO2 + y/2 H2O}  \quad \Delta H_{\mathrm{comp}(\ce{C_xH_y})} = -a \tag{6}$$
From the equation $(4)$:
$$\Delta H_{\mathrm{comp}(\ce{CO2})} = \Delta H_{f(\ce{CO2})} - \left(\Delta H_{f(\ce{C})}  + \Delta H_{f(\ce{O2})}  \right) = \Delta H_{f(\ce{CO2})} - \left(0  + 0  \right) = -b\\ \therefore \ \Delta H_{f(\ce{CO2})}  = -b \tag7$$
Simillarly, from the equation $(5)$:
$$\ \Delta H_{f(\ce{H2O})}  = -c, \tag8$$
and from the equation $(6)$ :
$$\Delta H_{\mathrm{comp}(\ce{C_xH_y})} = \left(x\Delta H_{f(\ce{CO2})}  + \frac{y}{2}\Delta H_{f(\ce{H2O})}  \right) - \Delta H_{f(\ce{C_xH_y})} = \left(-xb - \frac{yc}{2}\right) - \Delta H_{f(\ce{C_xH_y})} = -a$$
$$\ \Delta H_{f(\ce{C_xH_y})}  = +a - \left(xb + \frac{yc}{2}\right) \tag9$$
The equation $(9)$ is the correct answer.


Does acid anhydride give a positive iodoform test?

According to this source, both acid halides and anhydrides can form enols reasonably readily and can be halogenated. I'd imagine, for anhydrides, this to be related to the potential for intramolecular hydrogen bonding in the enol form (as shown below):

This allows halogation at the alpha position by treatmeant with the halogen. I'd also imagine this process can be catalysed in anhydrous acidic conditions, avoiding hydrolysis.
The same cannot be said for base-facilitated enolate formation. As user55119 implied, the highly reactive $C=O$ is likely to be attacked before the alpha hydrogens (or even if deprotonation occurs first, proceed through an E1cB pathway due to the presence of a good leaving group; implied by Clayden Organic Chemistry), decomposing the anhydride instead of forming the enolate.
Enolate formation in basic conditions is usually required for multiple halogenation, and generally three substitutions are required to form a suitable leaving group for the haloform reaction to proceed. This is why it is a test for a methyl group next to a $C=O$ (as Mathew has said); as there are three alpha hydrogens.
Considering basic conditions are not an option, only mono-halogenation is feasible (perhaps even in acidic conditions) and the haloform test is made virtually impossible.


Why the concentration with respect to time equation of a first-order reaction does not depend on the stoichiometric coefficient of reactant?

The key point which you will study in advanced classes, is the concept of elementary reactions. When a given reaction is an elementary reaction, the stoichiometric coefficients become the power of the rate law or in better words, the reaction order corresponds to the stoichiometric coefficients.

Definition of elementary reaction: An elementary reaction is a chemical reaction in which one or more chemical species react directly to form products in a single reaction step and with a single transition state (more details on elementary reactions).

Rate of reactions are determined by experiments. We never write the rate law based on the balanced equation. Compare this with equilibrium constant where coefficients are important. Even $\ce{2 A -> B}$ may be proceeding in several elementary steps. The observed rate may be the effect of several combined elementary reactions.

In those kinetic problems like yours, the writer will tell you the order of the reaction. Don't proceed to write the rate law on the basis of balanced equation.

Reaction rate and writing the rate law are two different things. For a reaction

$$\ce{a A + b B → p P + q Q}$$

Reaction is written as

$$r = -\frac{1}{a}\frac{\mathrm d[\ce{A}]}{\mathrm dt} = -\frac{1}{b}\frac{\mathrm d[\ce{B}]}{\mathrm dt} = \frac{1}{p}\frac{\mathrm d[\ce{P}]}{\mathrm dt} = \frac{1}{q}\frac{\mathrm d[\ce{Q}]}{\mathrm dt}$$

but we cannot write the rate law i.e. you do not know what would be the order of the reaction in general.

The relation above only shows how the rate of disappearance of reactants and rate of appearance of products are related. It should not be used anywhere else. Your book answer is correct.


Why does the temperature of a tyre rise when it is pumped up rapidly?

Once the pressure in the tire has risen a little, the volume of the tire doesn't change much any more.  But, you are increasing the number of moles of air inside the tire when you pump air into it.  The new air pumped into the tire compresses the gas that was already in there previously and raises its pressure.  So the new gas is doing work on the previous gas to compress it  adiabatically and approximately reversibly.  This causes the internal energy of the gas and its temperature to increase.


How to calculate volume for NaOH and drano max gel (includes dilution and pH, pOH)

Yes.

If you have for drano $C_1 = \pu{5 mol/L}$ and $V_1 = \pu{0.1 L}$ that is $\pu{0.5 mol}$, and for $\ce{NaOH}$ $C_2 = \pu{0.01 mol/L}$ and $C_2 = x$. Using formula for dilution:

$$C_1V_1 = C_2V_2 \quad \implies \quad V_2 = \frac{C_1V_1}{C_2}$$

General tip: With encountering questions like these, always think about what units you have (mol/L, L etc.) and what units you want to go to. Most of the time, it gets clear when you just look at the units.


Bond Type of Lead: Metallic or Network Covalent?

As you proceed down Group 14 the metallic bonding structure tends to become more stable than the tetrahedrally bonded structure we see in diamond, Si and Ge.  Tin is actually a transitional case.  At typical room temperature it is metallic (white tin) but upon chilling it transforms to the tetrahedral allotrope (gray tin).  By the time we get to lead we have fully gone over to the metallic phase.

Because the tetrahedral bonding structure requires each atom to contribute four valence electrons, this tetrahedral --> metallic transition may be seen as a form of the "inert pair" effect that emerges in heavy atoms.


Why is the formation of slacked lime accompanied by a hissing sound and the formation of bubbles?

Good eye for the inconsistency.  Cultivate that, it will serve you well.  But this is an apparent, rather than a real inconsistency.  The reaction is VERY exothermic.  Check out how they make quicklime to find out how much heat it takes.

Where the water touches the quicklime, enough heat is evolved to boil a little bit of the water.  Hence the bubbles and hiss.


Value of the frequency factor in the Arrhenius equation

The Arrhenius constant, A, does not have to necessarily be the same for two reactions taking place at the same temperature, due to the fact that it depends on the order of a reaction as well as the frequency with which successful collisions will occur during a reaction.

As for the second part, as k ∝ x-Ea, for 2k to be a larger value, Ea for 2k would have to be less than the one for k, i.e. e1 < e2 if the activation energy for 2k is e1 and that for k is e2.


How is the central carbon in 3‐bromopentane‐2,4‐diol chiral?

Carbons #2 and #4 are chiral centres themselves. There are four possible diastereoisomers:

If C2 and C4 have different configurations, which is the case in molecules A and D, then C3 is termed a pseudoasymmetric centre and labelled with a small 'r' or 's'. These diastereomers are meso compounds: although they contain chiral centres, they are not chiral, because they have an internal plane of symmetry.
On the other hand, in molecules B and C, C2 and C4 have the same configuration and C3 is not a chiral centre (and not a pseudoasymmetric centre either). These diastereomers are chiral, though: the chirality comes from C2 and C4.


Why do increasing concentrations of zinc chloride not linearly decrease the freezing point?

You've reached the other side of the eutectic.

See, that graph of lowering freezing temperature is always a part of a bigger picture.

Once you are past that minimum, it is no longer freezing point that you are measuring. Instead, it is the point where the solution becomes saturated and $\ce{ZnCl2*6H2O}$ (or whatever hydrate is stable at those conditions) starts crystallizing.

You'd see a similar pattern with pretty much any soluble salt. Zinc chloride, if anything, is good in that it dissolves immensely well, so the graph is wide, and its features can be seen more clearly.


Synthesis of a bicyclo[2.2.2]octane from benzoic acid

Supposed that the synthesis does involve a Diels-Alder reaction, it is nowhere written that it is the last step in the sequence.

Take another look at your product and imagine a $\ce{C=C}$ double bond somewhere. Does it now look more like a Diels-Alder product? The $\ce{C=C}$ can be turned to $\ce{C-C}$ using... (You figure that out).

Work your path back from here and try to identify diene and dienophile.

Don't look until you've figured out the diene ;)

   Convert acid to ester, 2 x $\ce{MeMgBr}$, Diels-Alder reaction with vinylmethyl ether, hydrogenate double bond. You didn't cheat, did you? ;)


My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'.

That is not a problem because the tendency of the metal to lose or gain electrons has no direct bearing on the reactivity of the metal or the reduction potential. You are thinking of this on an "per atom of metal" basis, but the reduction potential is on an "per electron" basis. One way to see that is that the cell potential does not change when you double the coeffients in an equation:

$$\ce{Ag+ + 1/2Mg -> Ag + 1/2 Mg^2+}$$

has the same cell potential as

$$\ce{2Ag+ + Mg -> 2Ag +  Mg^2+}.$$

(This is saying something like the potential of a D cell battery is the same as that of a AAA cell battery, even though the D cell battery is much bigger). On the other hand, the Gibbs energy of reaction would be twice as large for the second reaction compared to the first. One way to think about this is that the work the second reaction would do (if this is set up as a voltaic cell) is twice that of the first reaction because two electrons are transferred instead of one (across the same cell potential).

  How has the voltage helped us determine the difference in the tendencies of these two metals?

What we really want to know is which of the two metals would be reduced and which would be oxidized. For that, we need to know the potential (or in your picture, the excess or deficiency of electrons in the respective electrodes). This will tell you the direction of the reaction, and the cell potential when connecting the two electrodes and the two electrolytes to allow the reaction to proceed.

You can measure the potential at next to no current, so the stoichiometry of the reaction does not matter at this point - you just have to consider the charge separation that already has happened at the electrodes before you connect them.


How can you calculate the tendency of one compound to become another?

Depending on the constraints you impose on the system, there are different criteria for spontaneity, however, the commonly running theme is that it is related to minimizing some state function. For example, as a simple example from classical mechanics, the time evolution of the system is such that the potential energy of the mechanical bodies is minimized.
When we do a chemical reaction, we usually do it at constant temperature and pressure. In this case, the evolution of the system is determined by a potential function known as Gibb's free energy. If there is a state in which the system which reduces Gibb's free energy, then the system will evolve to reach that state. When I say evolve I mean to have processes such as chemical reactions.
I had actually written a very detailed answer about this on physics stackexchange, check it out for more information (here)

Refer these notes by MIT for the criteria for spontaneity for different constraints :
MIT ocw notes
Discussion about constant P and T chemical reactions:
Here


For the process to go forward, the entropy of the universe has to increase. For this specific case, the reaction is exothermic and the entropy of reaction (entropy change of the system) is positive. If all of the enthalpy is released as heat, the entropy of the universe will increase (the entropy of the surrounding increases because of the heat transfer, and the entropy of the system increases because molecules are more dispersed in the gas phase).

If there is a mix of work and heat exchange, you just have to make sure the entropy of the universe still increases. In this case, the work possible exceeds the enthalpy change. When you do maximal work, you have heat transfer into the system. The entropy decrease of the surroundings is offset by the entropy increase in the system.

So in this case $$|\text{work}| > |\Delta H|$$

For a picture, see https://chemistry.stackexchange.com/a/112958 scenario 2.

  Cool, so can you substitute work, W and heat, Q into the equation and write W=Q−TΔS to directly show that ΔS>0?

No, where there is work, enthalpy change is no longer equal to heat.

Also, you have to specify which ΔS you are talking about. If it is the entropy change in the system (due to the reaction) $\Delta S_\text{sys}$, the equation below is a variation of the second law (the left-hand side is equal to the entropy increase of the universe times temperature):

$$T \Delta S_\text{sys} - q_\text{rev} > 0$$

And if you want to relate work to enthalpy, you use the first law:

$$\Delta H = q + w$$

and insert it into equation above to get:

$$T \Delta S_\text{sys} + w - \Delta H > 0$$

or for maximal work with a bit of rearrangement,

$$  w_\text{max} = \Delta H - T\Delta S_\text{sys}$$

So you can say that if the work you get out of a reaction is larger than the enthalpy change, the entropy of the reaction has to be positive.


Is methanol really more acidic than water?

First, water's $\ce{pK_a}$ in water is $\ce{14}$ as explained here. This means that $\ce{H2O}$ is slightly dissociated in liquid form, such that $\ce{[H+] = [HO-] = 10^{-7}}$.

Second, I would say that caring too much about which compound is the best acid is a bit like arguing by definition. Acidity is a fuzzy concept used as a shorthand for deeper meaning. This means there is little point arguing about the shorthand when we have the actual properties it synthesizes.

In this case (sou rce):

$$
    \begin{array} {|c|c|c|l|}
    \hline
    \ce{solvent} & \ce{pK_a\ of\ water} & \ce{pK_a\ of\ methanol} & \ce{Interpretation} \\
    \hline
    \ce{water} & 14.0 & 15.5 & \ce{water\ is\ more\ dissociated\ than\ methanol} \\
    \hline
    \ce{DMSO} & 31.4 & 29.0 & \ce{methanol\ is\ more\ dissociated\ than\ water}\\
    \hline
    \end{array}
$$
So you can argue (as in the answer you linked) that methanol is more acidic than water in the abstract sense, because $\ce{DMSO}$ doesn't stabilize the anions by hydrogen bonding and so is closer to the "ideal" case (whatever that may be).

But what's the point ? You already have the $\ce{pK_a}$ values, any practical question you might have about the concentrations of theses solutes in $\ce{H2O}$ and $\ce{DMSO}$ is already answerable.


Why polymers have a range over which they melt instead of having a specific melting temperature?

No. Although we need to be careful in defining what is and isn't a "polymer". The nature of (nearly all) polymers is that they are (covalently) bonded (in at least one dimension) over large (relative to atomic diameters) distances. The nature of a liquid is that is exhibits no long range order, no long range "connectivity". There are, as far as I know, no polymer crystals, in their most ordered configurations they are quasi- or semi-crystalline, having disordered zones as well as often but not always having zones of some regularity/order. Often the degree of crystallinity of a polymer is determined by its processing history (temperature, pressure, and stresses applied to it), but even the most crystalline of polymers is not fully crystalline. What this means is that the transition between solid and liquid phases is not sharp; the disengagement of the polymer chains from one another takes place gradually with no sharp/distinct difference between the solid phase and the liquid phase. Perhaps a better way of saying that is that the polymer becomes more and more amorphous but continues to have some long-range order as each chain's motion relative to its neighbors increases. see https://en.wikipedia.org/wiki/Reptation and https://en.wikipedia.org/wiki/Amorphous_solid. Note that the presence of polymer  chains (assuming linear or nearly linear polymers) of varying MW increases the gradual transition from amorphous solid to polymer melt. If the amount of oligiomer is large enough, it may act as a solvent for the larger chains. Note also that the theory of glass to liquid and glass to solid phase transitions is an active area (not yet "settled science").


Which combinations of orbitals lead to pi bonds according to Valence Bond Theory?

Here are the requirements for a $\pi$ bond:

Each contributing orbital has to have exactly one nodal plane containing the bond axis.  There can be other nodal planes that might cut through the axis, but don't contain it so those planes don't count.  Two nodal planes containing the axis, as with a $d_{xy}$ orbital, also fail because that's too many.

The nodal planes identified in (1) that contain the bond axis must line up, so they end up as a single nodal plane going through the bond.  Draw, let us say, the $p_x-p_x$ pair, which is easy to see it's a $\pi$ bond, to see what I mean.

If you sketch the orbital combinations for each of the nine given cases you see immediately that three fail test (1) (the $s$ orbital is no-go for a nodal plane, and $p_z$ has its nodal plane oriented wrong), but the others pass both tests (1) and (2) and they are the six combinations identified by the book.
The hidden lesson here is that $d$ orbitals can form $\pi$ bonds too.  You will see them do so when you study transition-metal complexes.


The gate of a FET works much like the grid of a radio tube (valve). The gate is insulated from source and drain; the only current flow (if the insulator were perfect) is to charge the intrinsic gate capacitance. Even a MOSFET has some leakage, but it's pico- or femtoamperes. See this discussion on capacitance and leakage.


Can 4-pyridones be synthesised analogously to 2-pyridones?

A reliable and old reference on the synthesis of 1-methyl-2-pyridone (Org. Synth. 1935, 15, 41, DOI) does not mention the formation of 1-methyl-4-pyridone as a side product and I doubt that it and can be formed that way.

The synthesis of 1-methyl-2-pyridone starts by reacting equimolar amounts of dry pyridine and dimethyl sulfate to form the N-methyl pyridinium salt. Then, the mixture is diluted with water and solutions of $\ce{NaOH}$ and $\ce{K3[Fe(CN)6]}$ are simultaneously added.

This seems to do the trick: the $\beta$-amino alcohol formed by adding $\ce{OH-}$ ortho to the nitrogen atom is immediately oxidized, which prevents the opening of the pyridinium salt.

For the synthesis of 1-alkyl-4-pyridones, a somewhat classical approach would be to react the corresponding 4-pyranone with a primary amine under more or less drastic conditions, as shown in Can. J. Chem., 1976, 54, 3377.

Altenatively, Grignard reaction of a 2-substituted 4-benzyloxypyridine-N-oxide, followed by debenzylation/methylation with methyl triflate and subsequent treatment with $\ce{NaOH}$ will yield the desired a 1,2,5-trisubstituted 4-pyridone (DOI).


It is entirely arbitrary whether you call it an organic compound or not, though most would not.

The distinction you make that organic compounds should be found in living things is not a useful criterion. Moreover you are wrong that carbon dioxide isn't: it is both made and used by living things. Animals make it when they metabolise sugars to release energy; plants consume it when they build more complex organic molecules through photosynthesis. In fact most organic molecules are, ultimately, derived from $\ce{CO2}$.

Even more importantly most molecules considered organic are neither made by nor are found in living things. Chemists make new carbon compounds all the time (tens of millions in the history of chemistry) and most have never been made by animals or plants.

The organic/inorganic terminology is mostly very simple: covalent compounds containing carbon are organic. The only fuzzy area is around very simple molecules like $\ce{CO2}$ where the distinction doesn't matter much. So we would not normally think of diamond or silicon carbide as organic. But we might (though many would not) call calcium carbide organic because it contains a $\ce{C2}$ unit with a carbon-carbon triple bond.

However since the terminology is mostly very obvious and also somewhat arbitrary, it isn't worth much argument to sort out those very simple but awkward edge cases.


What is the typical range for the molar absorptivity of paint?

Well, I cannot say much from the above. Estimation of molar absorptivity experimentally would involve some measurements with different concentrations. I think single measurement may not give that information.

$TiO_{2}$ if present in paint then it is only one component ( ~10-40% or more maybe). Other components may affect the molar absorptivity too.
 If you want to find molar absorptivity of paint (with several components combined) you can just measure paint highly diluted in a solvent  and measure UV-spectra at different known concentrations and use the Beer-lambert law.

For molar absorptivity of only $TiO_{2}$ I have different proposal.

If you have some
$TiO_{2}$ then you can weigh 5 different amounts of it. Then dissolve it in acid (sulphuric acid or mixture with hydrochloric acid or you can check yourself which is better ) to make 5 solutions (should be dilute solution ). You know the concentration for these solutions.

Then, measure absorption with UV spectrometer.
Finally, use the Beer-Lambert law to get the molar-absorptivity based on the absorption peak (~350 nm).

Anyway this is a weird idea. (I am not sure if absorption of acid species, say sulphate ions will affect this measurement.)
And if you really try, let us know.


Nitrate is a strong oxidant which oxidizes the iodide to iodine.

$$\ce{NO_3^- + 2I^- + 2H^+\rightarrow NO_2^- + I_2 \uparrow + H2O}$$

Note that the oxidation number of the nitrogen atom in $\ce{NO3-}$ is $+V$ and in $\ce{NO2-}$ it's $+III$, so over all, we have a reduction equation of:

$$\ce{NO_3^- + 2e^- + 2H^+\rightarrow NO_2^- + H2O}$$

On the oxidation side we want to form $\ce{I_2}$ out of $\ce{I^-}$, so the oxidation equation is:

$$\ce{2I^- \rightarrow I_2 + 2e^-}$$


How is it possible to quantify the strength of a Lewis acid/base?

$$\ce{A~  + :\!\!B -> \ \ A\!\leftarrow:\!\!B}$$

$$K_f = \dfrac{\ce{[A\!\leftarrow:\!\!B]}}{\ce{[A][B]}}$$

See sections 4.1 and 4.2 in the following reference for more information:

http://goicoechea.chem.ox.ac.uk/teaching/abse/abse_lecture2.pdf

But $K_f$ is specific to an acid-base pair.

A more universal scale is formed by the Gutmann–Beckett method by observing the phosphorous NMR spectrum of triethylphosphine oxide as it interacts with a sample lewis acid.

An Acceptor Number (AN) scale is formed with AN=0 corresponding to triethylphosphine oxide in hexane (P chemical shift of 41), and AN = 2.21 (d-41), where d is the chemical shift of triethylphosphine oxide interacting with the sample. The greater AN is, the more acidic the compound.


Think of the planetary system which inspired Bohr to think about his atomic model. The Earth is rotating around the Sun in a fixed orbit. What keeps the Earth rotating around the Sun? The mechanics of the planetary motion, and other electrical phenomena were very well understood in Bohr's time, so much so that by late 1880 to early 1900 a physics professor advised his student, who would become a Nobel laureate later, not to study physics because there is nothing left to do in physics

...here might still be a little dust or bubble at one or the other angle to check and classify, but the system as a whole is quite secure, and theoretical physics is noticeably closer to that degree of perfection that geometry has had for centuries..
"Max Planck: Wege zur Physikalischen Erkenntnis."

So classical electrodynamics (motion of charges) predicted in Bohr's time that a moving negatively charged body should spiral into the positively charged proton. Obviously this was not happening. Secondly, a moving charge should emit electromagnetic radiation because the great Maxwell had already established that.

Bohr had a moment of enlightenment. He just said let us say that the electron does not spiral into the nucleus but stays in a fixed orbit nor it emits radiation. Lo and behold, that assumption helped him explain most the hydrogen atomic spectrum in the UV, Visible, and in the near infrared. This was a big big feat! That is why when you read Bohr's theory, his assumptions are called Bohr's postulates.

What is a postulate in logic as per a dictionary's definition? "A fundamental principle, presupposition, or condition, esp. one assumed as the basis of a discipline or theory; (also) a proposition that is (or is claimed should be) taken as granted; esp. one (to be) used as a basis for reasoning or discussion, a premise."

"Taken for granted" is the keyword, which means when you are reading my theory, accept my assumptions that they are true-don't question them.


The "sugar" you refer to is sucrose, which is a disaccharide (a fancy term we use for a type of sugar, composed of two other bits of sugar called monosaccharides; which in this case, are glucose and fructose). I felt I ought to have brought up the sucrose specification, since when you talk chemistry, "sugar" can refer to a whole lot of things.
Anyways have a look at this diagram:

Observe carefully, and you'll see that there are 8 -OH groups (hydroxyl groups) in a molecule of sucrose. Now why did I direct your attention to these particular groups?  It's because they're capable of taking part in a fascinating phenomenon called hydrogen bonding (I believe you already know a bit about it, so I won't go into details).
Now when you add a solute to water, that is capable of H-bonding, it sorta causes all the water molecules and solute molecules to 'clump' together. The greater the amount of H-bonding the solute displays with water, the greater the tendency for all those interaction to put up some resistance to the flow of water. This ability to 'resist flow', that is inherent to every real fluid (to different degrees of course), is what you call viscosity.
Now 8 hydroxyl groups, can result in some pretty extensive H-bonding. To give you a rough idea as to how much that contributes to the viscosity of the sugar solution, think of glycerine (glycerol). Rub some of it between your fingers, it's pretty viscous right? Now the thing is, a glycerol molecule has only 3 hydroxyl groups. So imagine how much more viscous/sticky the sugar solution ought to be, when it's got nearly 3 times as many hydroxyl groups! [No, I'm not saying H-bonding is the only factor you should consider]. Of course, the more concentrated the sugar solution, the more the sucrose molecules and hence more the H-bonding; so it'll get stickier and stickier.
Now another thing you ought to consider is the geometry of the sucrose molecule. Now it isn't as important as the H-bonding bit, but it's still noteworthy. What you must know is that the sucrose molecule is not a planar molecule. It isn't some 2-dimensional structure. It's got all sorts of groups sticking out above and below the plane of the carbon rings. Now at low concentrations this isn't really an awfully important thing, but the story changes when you're dealing with concentrated solutions. That's when the 'jagged' physique of the sucrose molecule comes into play. It serves to compound the effect of H-bonding and further obstructs flow.
Want an analogy for that bit on the importance of the molecule's geometry?
Imagine you've got this large sack of coins. Dump them onto the floor and they spread out so well, that you wouldn't be wrong if you said they flowed. But suppose you had nothing better to do, and decided to glue a couple of screws on both faces of each coin. Put all your coins back in the sack, and proceed to dump out the whole lot. You can easily picture how well this 'modification' obstructs the flow of the coins, and probably now, you can appreciate the importance of molecular geometry in this regard better.


Why does shaking a match put the fire out?

Combustion of small materials, such as a match or birthday candle, actually involve the release of volatile vapours, which themselves burn. It is not the solid material that burns. There needs to be a minimum amount of volatile material present in this combustion zone (just above the burning match) for the ignition to occur. As the combustion process continues, heat is given off, and more volatile materials are released, which in turn continues the combustion cycle. Now, if you shake a match or blow on a candle, you rapidly disperse these volatile fuels from the combustion zone, and there is no longer sufficient fuel to ignite. It is effectively removing the fuel component of the fire triangle, for just a brief moment.

Large fires can reignite because there is sufficient heat left in the fuel to further release volatile fuels, which can either self-ignite or ignite through the presence of embers.

Blowing gently on a small wood fire increase the oxygen content into the combustion zone, without dispersing the fuel. Similarly, if you experiment with a small enough wood fire, you will see that blowing on different parts of the fire will have a different outcome: blowing on the base of the fuels will increase oxygen content, and not affect volatile fuels. Blowing on the top of the fire (where the flame starts to burn) will probably put the fire out.

Will this put out a small paper fire? That will depend on the heat retained by the burning fuel. A single piece of A4 paper if shaken hard enough will extinguish. A ream of A4 paper that has burned half way down the page will be put out this way, but could easily reignite if the paper pages are pulled apart to allow oxygen into the released volatile fuels.

Generally, forest fires are accelerated by strong winds. Winds affect forest fires in a number of ways, and increase the rate of spread significantly. This is a topic for a whole other question.


What an aqueous solution really contains?

Uh... I think the insight you need is about chemical equilibrium.
An aqueous solution is exactly the mixture formed when something dissolved in water. This is easy to understand, but here is a more difficult question: what's the component of this mixure?
To answer this is not easy.
In a glass of pure water, there are still a little $\ce {H+}$ (to say presicely, $\ce {H3O+}$) and $\ce {OH-}$. This is because sometimes proton ($\ce {H+}$) will "jump" from one $\ce {H2O}$ molecular to another one.
$\ce {2H2O <=> H3O+ + OH-}$.
(Since we usually do not use the expression $\ce {H3O+}$, this process can be rewrited as $\ce {H2O <=> H+ + OH-}$ to avoid disruption.)
This phenomenon (called the autoionization of the water) is very weak. The concentrations of $\ce{H+}$ and $\ce{OH-}$ in "pure water" is $\ce{[H+]=[OH^{-}]=}1.0 \times 10^{-7} M$ when the temperature is 25°C (298K).
Then you put something in this glass of water. As other people mentioned, most of covalent compounds, such as glucose, will not break apart in water since the intermolecular force (covalent bond) do not break. Water moleculars may surround those covalent compounds like a cage to make them dissolved. So there won't be other things.
But sometimes it breaks: when acid compounds are dissolved into water, like $\ce {HCl}$ and $\ce {H2SO4}$, they will give their hydrogen atoms as proton to water:
$\ce {HCl + H2O -> H3O+ + Cl-}$
$\ce {H2SO4 + H2O -> H3O+ + HSO4-}$
This is because they are more acidic than $\ce {H3O+}$, here "acid" means "who gives proton(s)", while "base" means "who accepts proton(s)" (This kind of definion is called as Brønsted-Lowry theory, and as you see, in autoionization of water, one of the $\ce {H2O}$ is acid and another is base).
Noticed that acid like $\ce {HCl}$ and $\ce {H2SO4}$ will nearly definitely give their $\ce {H+}$ to water compounds (they are too “willing” to do it), we call those acid as strong acid. But some acid are not: protons sometimes jump back.
For example, $\ce{HSO4-}$ will meet an equilibrium:
$\ce{HSO4- + H2O <=> H3O+ + SO4^2-}$
This kind of acid will be called as weak acid. In an aqueous solution of a weak acid, most of the weak acid compounds keep unchanged.
Yet we haven't mentioned what you concerned: how about aqueous solution of ionic compounds? They usually break their intermolecular forces (ionic bond) via hydration, i.e. combination with water compound. Water compound have some charges on them, so they will attract ions and break them apart. Especially, we will see bases we are familiar with -- those who will ionized as cations and $\ce {OH-}$ -- first.
In fact, the structure of hydrated cation will be complex ("$\ce {H3O+}$" is also not the actual situation), so we usually just write them as unhydrated form, such as $\ce {Fe^2+}$ or $\ce {Na+}$. As you know, there are weak bases and strong bases.
So bases like $\ce {NaOH}$ are quite "willing" to accept protons:
$\ce {NaOH -> Na+ + OH-}$
But $\ce {Fe(OH)3}$ may not, and it also meet an equilibrium:
$\ce {Fe(OH)3 <=> Fe^3+ + 3OH-}$
Again, in an aqueous solution of a weak acid, most of the weak base sturcture keep unchanged. And for this reason, weak bases are usually insoluable in water, since water "have no good idea" to deal with the whole relatively large ion sturcture. But in fact, the word "insoluable" means that there is another equilibrium like this:
$\ce {Fe(OH)3_{(aq)} <=> Fe(OH)3_{(s)}}$
and the solid form dominates.
So here we meet the hardest part: what about adding some salts into water, like $\ce {MgCl2}$ you mentioned?
So far you have learned about ionization of the salt, and you know most of salt will completely ionized. But how about those ions?
A balance like $\ce {Mg(OH)2 <=> Mg^2+ + 2OH-}$ shows that $\ce {Mg(OH)2}$ is a  weak base. But if we see it as $\ce {Mg^2+ + 2OH- <=> Mg(OH)2}$ it means $\ce {Mg^2+}$ is a weak acid. And if you remember I have said in autoionization of water, the water compound can be both base and acid (we say this kind of substances are amphoteric). Then we can know that there will be a new equilibrium like this:
$\ce {Mg^2+ + 2H2O <=> Mg(OH)2 + 2H+}$
(We usually call this as hydrolyzation of salt.)
And because $\ce {Mg^2+}$ is a weak base, you may expect that there will be not a lot $\ce {Mg(OH)2}$ and it's still soluable. Also there will be not a lot $\ce {H+}$ so the solution is just a bit more acidic then water.

Addition: You can also think this comes from these two balances:

$\ce{H2O <=> H+ + OH-}$ (autoionization).
$\ce{Mg(OH)2 <=> Mg^2+ + 2OH-}$. in further calculation you may meet in future,  this kind of idea can make you avoid some mistakes.

So you may expect in a $1M$ aqueous solution of $\ce {MgCl2}$, there will be a lot of $\ce {Cl-}$ ($2M$) and $\ce {Mg^2+}$ (a bit less than $2M$), a little $\ce {H+}$ and a little $\ce {Mg(OH)2}$, and there will be $\ce {OH-}$  which is negligible in the most case. (Of course there is very much water!)
While you can see in the case of $\ce {NaCl}$ there won't be $\ce {NaOH}$ and $\ce {HCl}$ compounds in the solution: because $\ce {NaOH}$ and $\ce {HCl}$ are "strong", there won't be a hydrolyzation equilibrium. But yes, you can say there are $\ce {H+}$ and $\ce {OH-}$, as you have known, there are $1.0 \times 10^{-7} M$ of them. However, we usually not concern about this. And as you see, all those $\ce {H+}$ and $\ce {OH-}$ in an aqueous solution, in effect, come from water compounds. (This is why there will be "strong" acid and base: because they are all giving/receiving protons from water, even if they actually have different ability, compared to water, they are behaving undistinguishably. This is called as leveling effect of solvent.)
There are a lot of things I haven't mentioned here, because I think you are not able to calculate something about equilibrium, and maybe you are not familiar with thermodynamic idea yet. But here I just hope what I said is interesting enough for you.
Well, a simple conclusion:

To find out what will be there in an aqueous solution, you may consider:

If the solute ionize: Is it a covalent compound? Is it acid?
If yes, whether the ions are weak acid or base (in Brønsted-Lowry theory)? Remember, a cation which can be formed from a weak base will be weak acid; while an anion which can be formed from a weak acid will be weak base. (Or you may use an awful word: they are conjugate acid base pair!)

I think these two step are enough for you for the time being.


How does pH level change the concentration of dissolved oxygen (DO) in sea water?

The equilibrium of dissolved-to-gas-phase oxygen is not affected by the aqueous hydronium ion concentration ($\ce{H3O+}$), and thus the pH of water does not affect dissolved oxygen concetrations of sea water.


What makes DNA/Protein in a gel run in straight lines?

What you are trying to achieve in the electrophoresis experiment is a homogenous electric field. If the electric field points downward homogenously, then the greatest force will be exercised in a downwards direction which will lead to a downwards movement of your charged sample.

While not being exactly equivalent, you can show a similar feature (homogenous field) with a horseshoe magnet. If you place a piece of paper or cardboard over the top of it and then sprinkle iron filings onto that, they should arrange themselves in a linear fashion between the magnet’s poles.


Do hydrogen atoms not bound to carbon also appear in 1H NMR spectra?

Yes, with a but.

First the yes: they do appear, albeit at slightly different chemical shifts as, of course, the environment around a proton in $\ce{-CH2\!\;\!-}$ is slightly different to the electronic environment around the proton in $\ce{-NH2}$. The $\ce{N-H}$, $\ce{O-H}$, and $\ce{C-H}$ bonds all have slightly different electronegativies.

Now for the but: the thing that more significantly distinguishes $\ce{N-H}$ and $\ce{O-H}$ from $\ce{C-H}$ is their lability in protic solvents. Exchange of labile protons in the sample molecule with deuterons (i.e. deuterium ions, $\ce{^2H+}$) in the solvent can can remove the proton peak from the spectrum (this exchange is an equilibrium process, but as the solvent is in large excess the equilibrium lies over towards full deuteration of the sample's labile groups. This can actually be useful in identifying peaks: first run a sample in, for example, $\ce{CDCl3}$, then add a couple of drops of $\ce{D2O}$ to your NMR tube, give it a shake, and re-run the spectrum -- your labile protons will have vanished.

Edit 1: Furthermore, as @user1573870 points out in the comments, exchange of protons (exhanging protons for protons, rather than exchanging protons for deuterons),  on "the NMR timescale" can lead to broadening, and the resulting peak can be, while technically present, very difficult to spot. I'd suggest anyone doing NMR do background reading on that, because it's both useful and interesting (and I need to do some, because I'm out of practice with NMR).

Edit 2:To directly answer @Chewie's question whether "we have to count or omit them in the integration of the spectrum", you only count each of the peaks that are present. Using the example of ethanol, as asked in the comments, $\ce{CH3CH2OH}$ should give three peaks:

The $\ce{CH3}-$ peak should integrate to show three protons,
The $\ce{-CH2}-$ peak should integrate to show two protons, and
The $\ce{-OH}$ peak should integrate to show one proton.

If due to exchange with the deuterated solvent the $\ce{-OH}$ peak is not present, or if the $\ce{-OH}$ peak is 'there' but so broad as to be invisible to you (or so broad that it is difficult to integrate accurately), then you would only give the integrals for the other two peaks, and those two peaks should give integrals with a ratio equivalent to their respective number of protons, i.e. 3:2. If the $\ce{-OH}$ peak is not there, its proton does not just get thrown in with one of the other peaks/integrals (unless, sneaky proton, it just happened to have the same integral as another peak, and was hidden behind it, but for something as simple as $\ce{CH3CH2OH}$ that is unlikely).

The punchiest way to say all of thise would be that one does not "integrate spectra"; one integrates peaks.


Solubility in homogenous mixtures of solvents

The complication comes from solubility of the solvents with each other whilst simultaneously having some solvation power of a solute. I have made a compound before that was so water soluble that it caused sodium chloride to precipitate out. A real example was of an organic compound that had 1%w/v solubility in ethanol but around 6.5%w/v in toluene. In a 50:50 mixture,  the compound had solubility of 13%w/v. A completely non-linear result! The solubility was even higher in 90:10 toluene-ethanol. So, the complex nature of three way interactions mean it would be very difficult to predict these solubility graphs just using two data points of the two pure solvents.
If we asked the same question of boiling points, we would point straightaway to azeotropes. So, analogously, other properties exhibit this behaviour. However,  in this instance the nature of the solute as a third component makes it too complicated to predict without having the wealth of property data available for all components. If this is known and you have the software, you may have a chance.


Mathematically balancing chemical equation yields weird results

As @Ivan Neretin and @Tyberius pointed out, the negative coefficient of a reactant can be flipped to be positive by moving the reactant to the products' side.

So, in my case, the equations become
$$\begin{align}
\ce{11C7H6O3 + C4H6O3 &-> 9C9H8O4}\\
\ce{8C4H6O3 &-> 2C7H6O3 + 9C2H4O2}
\end{align}$$

Now, since the original equation $\ce{C7H6O3 + C4H6O3 -> C9H8O4 + C2H4O2}$ consists of 2 independent equations which can be separately balanced and combined in an arbitrary proportion. See  More than one way of balancing a chemical equation for a more detailed explanation.


According to Martin Chaplin's Water Dissociation and pH:

  In ice, where the local hydrogen bonding rarely breaks to separate the constantly forming and re-associating ions, the dissociation constant is much lower (for example at $-4~\mathrm{^\circ C}$, $K_\mathrm{w} = 2 \times 10^{-20}~\mathrm{mol^2~L^{-2}}$).

So $[\ce{H+}] = 1.4 \times 10^{-10}~\mathrm{mol\ L^{-1}} \Longrightarrow \mathrm{p[\ce{H+}]} = 9.9$

For more information see Self-Dissociation and Protonic Charge Transport in Water and Ice Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences Vol. 247, No. 1251 (Oct. 21, 1958), pp. 505-533

This is a review article by Nobel Prize winner Manfred Eigen , after whom hydrated $\ce{H3O+}$ is sometimes referred to as the Eigen Ion.


"Nitrogen (gas) is physisorbed over iron (metal) at room temperature": Is this for real?

I think you are misinterpreting the course book issued by your school since it never mentions what happens at room temperature.
For me, none of the books is (seems) wrong, because they do not contradict each other. I will mention each affirmation including the textbook which said it (Levine, or Your book):

Levine says: N2 is physically adsorbed on any solid at low temperature. This agrees with your text saying N2 is physically adsorbed at $\pu{83 K}$ on Fe.
Levine says: N2 is chemisorbed at room temperature on Fe. Your book does not mention about room temperature. No contradiction here.
Your book says N2 is chemisorbed at $\pu{773 K}$. Levine does not mention this temperature, but if it already occurred at room temperature it will also occur at higher.

The only 'gray area' is at which temperature does N2 get chemisorbed (room temperature or not?). In the article of de Boer et al. mentioned by @Linear Christmas above, they measure chemisorption of N2 even at $\pu{23.9 ^\circ C}$.
They clearly state:

Hitherto it has been reported in literature$^{1,2}$ that an iron catalyst starts to chemisorb nitrogen at a measurable rate at about $\pu{200 ^\circ C}$. The chemisorption activity of our intensely reduced sample was considerably higher; already at $\pu{0 ^\circ C}$, nitrogen was chemisorbed at a measurable rate.

Reference:
J. J. F. Scholten, P. Zwietering, J. A. Konvalinka, J. H. de Boer, "Chemisorption of nitrogen on iron catalysts in connection with ammonia synthesis. Part 1.—The kinetics of the adsorption and desorption of nitrogen," Trans. Faraday Soc. 1959, 55, 2166-2179 (https://doi.org/10.1039/TF9595502166).


Product of the reaction between chalcone and sodium methoxide in methanol

The first step is called Michael addition. Have a look at this wikipedia page.

  The Michael reaction or Michael addition is the nucleophilic addition of a carbanion or another nucleophile to an α,β-unsaturated carbonyl compound. A newer definition, proposed by Kohler, is the 1,4-addition of a doubly stabilized carbon nucleophile to an α,β-unsaturated carbonyl compound.

Of course, in your case, the nucleophile is $\ce{OCH3-}$. Perform 1,4-addition by adding it to the double bond at the beta position (with respect to the carbonyl), and you'll expect an enolate ion, which looks like this -

Finally protonate the $\ce{O-}$ and tautomerise to get your product as @Zenix already stated.

For the second step, check out this link.

References: Michael reaction, wikipedia || Ketone halogenation, wikipedia || Enol, wikipedia


What is number of 1H NMR signals for toluene?

In general, the aromatic region of any $\ce{^1H-NMR}$ spectrum would display a complex coupling pattern even if for mono substituted nucleus such as toluene. However, your $\ce{^1H-NMR}$ spectrum is taken by using a low field NMR instruments such as 60 MHz, you'd not able to get one with good resolution to show these patterns (other factors involve in having broad multiplets instead of resolved peaks are low quality calibration, RF instability, etc.). The following spectrum is taken using a high resolution NMR instrument. The $\delta~7.70-7.70$ region was expanded to show the coupling pattern of 5 aromatic hydrogens. In a low-field instrument, those peaks would overlap to give you a broad multiplet. Read this article to learn about basic principles of coupling patterns.


Why is ice sticky between −20 to −8 degrees Celsius but not below −30 °C?

I think I remember that episode. Basically, crystals of water in kerosene were not a problem during the whole flight (at minus thirty something) but crystals were too sticky when the temperature raised to -20°C and prevented the fuel to go to the engines just a few millimeters before entering the "warming cell" which was supposed to prevent any such blockage.

Crystals at low temperature (let's say -35°C) will be quite hard and quite big, even if some molecules of the fuel will be trapped in the crystals of ice (because there is much more fuel than water). If the temperature goes up to a certain level (let's say -20°C), crystals will break where there was molecules from the fuel. As a consequence, the big crystals will break into very small crystals. Ice crystals do not like their environment (i.e. the fuel) because they cannot find any stabilizing effect with it, but they love other ice crystals. Moreover, the overall surface greatly rises when the size of the crystals diminishes. As a consequence, they will try to stay together and form a "slush".

In that accident, it was not the only factor. The other two problems were that acceleration increased the flux of kerosene (so more small crystals of ice were available to create a slush), and there was a misconception in the heat exchanger which was aimed at avoiding water crystals.


Diffusion-controlled reactions, activation-controlled reactions and rate equation

To add a little to what @geo has written. In all types of bimolecular reactions in the liquid phase, the reactants diffuse together; they have to, there is no other way for them to meet. The reaction is diffusion controlled if the activation barrier is so small that the first time the reactants meet, reaction is effectively instantaneous. This means that the barrier is only a few times average thermal energy (3RT/2) and that reaction occurs before the species can diffuse apart again.

In an activation controlled reaction the species repeatedly meet by diffusion but only occasionally$^*$ do they acquire enough energy from the surrounding solvent molecules that, at that instant, have enough energy to surmount the activation barrier.  Once they have reached  the barrier top the energy fluctuations that caused the barrier to be surmounted in the first place must cease after the transition state is crossed and products are then formed. (Recall that molecules in solution are continually jostling about and, from the Boltzmann (exponential) distribution, there is always a small chance that a molecule has, by random fluctuations, and for a short time, many times the average energy ).
$^*$By 'occasionally' I mean perhaps 1 in a million or even tens of millions of encounters.


The method of production of sea salt usually involves bulk evaporation of seawater. This allows the salt to crystallise and due to it's bigger size it is sold without much processing. On the other hand, the table salt has much smaller size and is processed a lot more than we could assume before it is sold.
This could be one reason why sea salt has coarse texture.
See also: Maxine Builder and Maxine Builder. What's the Difference Between Kosher Salt and Sea Salt? (via the Internet Archive)
I was interested to answer this question because I have always thought that both the salts are just the same. The tiny size of salt grains just seems less coarse but it essentially is having the same texture is what I would believe.
But then I looked at sugar as well. The tiny grains of sugar or a little bigger sugar diamonds seem smooth compared to the rough textured bigger blocks of sugar. And so I was convinced that it was due to it being processed. But then think of powdered sugar? That appears to be smooth.


Identify products of Propiophenone using nmr

As this is clearly a homework type question, this answer will just give you some helpful pointers to head you in the right direction, and leave you to piece together the answer.

My usual approach for these types of problems is to have a look at the NMR data first, and ignore all the IR data. We might use that later for double checking.
From both spectra, it is possible to identify the following spin systems:

-CH2CH3
-C6H5 (monosubstituted benzene).

We also know we must have a -OC(O)- group in there also. That satisfies our molecular formula.

Looking at Isomer C first:

CH2 group at 2.7, indicative of it being adjacent to the carbonyl of an ester
Ortho aromatics at 7.0  and upfield of the meta, indicative of substitution with an electron donating (or aromatic activating) group, Ar-O-
I'll let you piece together the structure of C.

Looking at Isomer D:

CH2 group at 4.4, indicative of being adjacent to a heteroatom, so -O-CH2
Ortho protons at 8.0  and downfield of the meta , indicative of substitution with an electron withdrawing (or aromatic deactivating) group; Ar-C(O)-
I'll let you put this structure together also.

And what about the IR data? We know the following, which is consistent with what we propose from NMR:

Al-C(O)-O-Ar 1740-1770cm-1
Ar-C(O)-O-Al 1715-1730cm-1

The general rule with IR absorptions is that conjugation lowers the frequency - I'll leave you to rationalise this with some fancy arrow-pushing from the aromatic ring to the carbonyl group.


Why does HF generate a carbocation with an alkene

However I felt like the bond strength of $\ce{HF}$ is so high that it couldn't be possible to cleave it for the donation of $\ce{H+}$

HF has the highest single bond energy and hence you were right in expecting that it may not dissociate.However in concentrated solutions, due to the formation of stable hydrogen bonded complex $\ce{[HF_2]-}$, the reaction is pushed in the forward direction making it  a resonably strong acid (in fact strongest among corresponding concentrated solutions of all the hydrogen halides).So, you could generate a carbocation with that.However, as generally dilute solutions are used which are economically feasible, more stronger mineral acids in dilute solutions  like $\ce{HCl, HBr}$ and $\ce{HI}$ are used.
EDIT:
As suggested by Oscar Lanzi in the comments, acids like $\ce{HCl, HBr}$ and $\ce{HI}$ are safer to handle when compared to $\ce{HF}$ as the reactions of $\ce{HF}$ are associated with highly negative enthalpy changes [highly exothermic]


Why can't element with paired electron have colour

Your assumption is wrong. There are elements with paired electrons which are coloured.

Colour is a product of electronic transitions in atoms or molecules. If those transitions match the energy of a photon of visible light, then the substance will be coloured (though there are some other rules about how intense the colour will actually be).

Some elements do show colour. Halogens, for example, exist as diatomic molecules in their native state and are all coloured. Chlorine is a greenish gas and bromine is an intensely red liquid. Sulphur is usually found as a yellow solid. All have paired electrons.


Yes, superheating can happen in some alloys. Whether it occurs at all, and its extent if it occurs, is very specific to each alloy (and heating rate dependent, obviously). It also depends on things like sample geometry (particles, solid, etc.) and size. This paper reports delays in melting by up to 100 K for some Al-Pb and Cu-Pb alloys.


Do lone pairs on substituents (e.g. in aniline) count towards Hückel's rule?

It is only the electrons from atoms in the ring that count when applying Huckel's rule. Electrons from substituents on the ring are only cross-conjugated with the aromatic π-system. When you think of it in terms of perturbational molecular orbital theory the substituent electrons take the part of a perturbation for the ring's aromatic system.

The reason for only using the ring electrons traces back to the origin of aromaticity. Take a look at the π-orbital energy levels of benzene (figure taken from p 36 of Fleming, Molecular Orbitals and Organic Chemical Reactions, Reference Edition):

Here, $\alpha$ is the energy of an isolated p-orbital and $\beta$ is the stabilization energy of the π-bond in ethylene. You can see that for benzene you get the best stabilization if you have 6 π electrons ($4n + 2$ for $n=1$) in the system because then you fill the stabilized MOs $\psi_1$, $\psi_2$ and $\psi_3$ which are lower in energy than an isolated p-orbital and gain the maximum stabilization energy possible. If you'd have more electrons in the system you would have to fill the destabilized MOs $\psi^{*}_{4}$, $\psi^{*}_{5}$ and $\psi^{*}_{6}$ which would cost a lot of energy because of the large energy gap between the bonding and antibonding MOs.

Now, what happens if you add a cross-conjugating substituent to the ring. This can be seen in the (Huckel-)MO-diagram of styrene (Fleming, p 71).

The main effects of the substituent are that the degeneracies of some orbitals are lifted and that the orbital energies shift a bit, especially those of HOMO and LUMO are of interest. In the case of styrene the HOMO lies higher in energy and the LUMO lies lower in energy compared to benzene. Now there are 4 stabilized energy levels instead of 3 - any electron that occupies one of those levels will lead to a stabilization of the molecule. But that is nice and well as you get 2 electrons from the substituent which will want to occupy one of the 4 levels and that leaves you with 3 levels that have to be occupied by ring electrons. Thus, your system will be most stable if the π-system of the ring contains 6 electrons, exactly like in benzene. Any more than 6 electrons will lead to destabilization as there is still quite a big HOMO–LUMO gap (even though it is smaller than in benzene).

The argument I made for styrene will remain valid for other molecules with different substituents (like aniline). The MO scheme of aniline will look a little different but the essential features should be the same.


Assignment of the 13C NMR of 2-heptanone

For open chain alkanes, chemical shifts in 13C are dependent on substituents at the α, β and γ positions. α substituent effects are largely correlated to electronegativity. β substituents cause a shift of similar magnitude, but are not related to electronegativity. In fact a number of studies have shown an inverse relationship to electronegativity, but this is not consistent. The origins of the β effect are unclear. The downfield shift is larger for primary/secondary carbons than for tertiary/quaternary (hence the larger downfield shift for C-5 which has 2 β carbons). The γ effect is small and has a dependence on sterics of the bonding pathway.

The chemical shifts for straight chain alkanes can be calculated quite accurately using additive tables (such as in Pretsch, or used by Chemdraw etc). As a β substituent, an alkane -CH2- has a shift increment of +9.4, versus -C(O)- having an increment of +0.5. Hence your difference of about 8ppm for carbons 3 and 5.

From Chemdraw

Node     Shift    Base + Inc.   Comment (ppm rel. to TMS)
CH2     22.9         -2.3      aliphatic
                     18.2      2 alpha -C
                      0.5      1 beta -C(=O)-C
                      9.4      1 beta -C
                     -2.5      1 gamma -C
                     -0.4      general corrections
CH2     31.3         -2.3      aliphatic
                     18.2      2 alpha -C
                     18.8      2 beta -C
                     -2.7      1 gamma -C(=O)-C
                     -0.7      general corrections


How does hydraulic equilibrium work on the particulate level?

The membrane has plenty of holes that are a little bit bigger than small molecules like water. Bigger molecules or bigger ions like usual solutes cannot go through these holes. So water has the tendency to cross the membrane both ways. When a water molecule hits a hole in the membrane from the pure water side, it can cross it. Nothing prevents it. When a solute molecule hits a hole of the same membrane, it blocks the passage for a couple of microseconds. No water can go through this hole from the solution side of the membrane. So on average, more water can cross the holes of the membrane in the direction water to solution, then backwards.
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