text:
Calculate molar mass of a compound from a given molecular mass
The mass of one molecule of a compound is $2.19\times10^{-22}\ \mathrm g$. What is the molar mass of the compound?
My attempt:
Using the formula moles=mass/molar mass, I found how to calculate molar mass using the formula molar mass=moles/mass.
Therefore, I was under the impression that, surely, $6.02\times10^{23}$ divided by the mass of the compound would have resulted in the molar mass. However, the answer was wrong.

You know how many molecules of anything are in one mole of that particular anything, so start from there.

@ToddMinehardt Well I know that I have to divide the amount of moles of this compound over its mass to get to the molar mass. I have tried to get the number of moles through two different but haven't been able to because the compound hasn't been specified , but came to the wrong result twice.

You might want to include the steps you took towards solving this problem  in your post. There is a homework policy for this site which states (in part) that a good-faith effort has to be made by the OP in order to keep the question open. I answered your query based on the effort you detailed in the comment (above).

Okay, I'll do that now.

To get the molar mass of your mystery compound, make use of the fact that there are $N_{\rm A}$ (Avogadro's number) of molecules of it in one mole of it.
To solve the problem, we multiply the molecular mass (given) by Avogadro's number (known) to get the mass of one mole of the mystery compound:
$${\rm molar \;mass} = N_{\rm A}\times 2.19\times 10^{-22}\,{\rm g} = (6.022\times 10^{23}\,{\rm mol}^{-1}) \times (2.19\times 10^{-22}\,{\rm g}) = 131.88\;{{\rm g}\over{\rm mol}}$$

Oh I see now. Instead of multiplying Avogadro's number by the mass of the compound, I was dividing the two.


text:
Hydrogen bonding
It is mentioned in my textbook:

"Hydrogen atoms of water molecule create strong electrical field due to their small sizes"

How is the size related to electrical field and its strength?
A little touch of physics will be greatly appreciated.

The magnitude of the electric field is basically given by Coulomb's law. In SI units, the electric field $E$ at the distance $r$ from the charge $Q$ is
$$ E = \frac{Q }{4\pi \epsilon_0 r^2} $$
The sentence only says the simple point that if $r$ (distance from the charge, in this case the proton i.e. nucleus) which is squared in the denominator is short, then the electric field will be large (strong) because it's the inverse square law (involving indirect proportionality).

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text:
I drink city water, city has lead pipe, how is this safe?
6 years ago, I needed to replace the main water shut off valve to my house.  This required the city to turn off the water at the street. Incompetent city employee drives key through water pipe. Plumbers fix pipe, I realize that pipe is lead.
Yearly, the city (90% of which was built before 1950) sends out a water quality report, which reports undetectable amounts of lead.  That is interesting since probably 90% of the city has lead pipe.
Several questions :

How is lead pipe safe from leaching lead into water?
Do cities usually treat the water to prevent lead from leaching into the water?
What is the form of harmful lead (e.g. is it in aqueous solution, lead oxide, or forming  ligands with some other molecule)?

Interesting fact: plumbers are called so because of the latin name for lead, plumbum.

In light of the recent events in Flint Michigan, this issue has come to the forefront (see this article from the American Chemical Society's Chemical  & Engineering News and see this article). To summarize:
Active Measures to prevent lead corrosion:

Maintain high pH. This makes lead carbonates less soluble in water.
Add orthophosphates which create relatively insoluble layer on the inside of the pipes.

Without these treatments, the addition of chloride disinfectants aggravates the corrosion problem by forming lead (II) chlorides (which is water soluble).
As mentioned by Jaroslav, the time that water sits in the system also plays a factor.

Lead pips are a common problem. Usually, it is safe to drink water from it.

So long as water is flowing, there will not be unsafe (or even detectable) levels of lead in the water. - True remark by  @Some_Guy.

The problems accrue when the pipeline is not used for longer time. The lead concentration arises and can reach harmful levels. There is a possibility to install filters to remove lead from the water. This site is quite useful and provide more information.

Lead is contaminating water if the pipeline is made of lead.
I am not aware of any method that can prevent lead leaching.
Over longer time oxides can be formed:

$\ce{2Pb(s) + O2(g) + 2H2O(l) -> 2 Pb(OH)2(s)}$
and other:
$\ce{PbCO3}$ or $\ce{Pb(CO3)2^2-}$
Elementary lead does not dissolve in water under normal conditions.
Lead compounds are generally soluble in soft, slightly acidic water.
Lead binds to sulphur ($\ce{S2-}$), or to phosphor ($\ce{PO4^3-}$).
Reference here.

The key point here is that, so long as water is flowing, there will not be unsafe (or even detectable) levels of lead in the water.  I can't speak for where OP lives, but in my area, anywhere that has lead piping has a periodic test for lead, and it never comes up positive.

That right. I will update the answer.

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text:
Atmospheric carbon removal: asphalt from CO2?
In reading a text about $\ce{CO2}$ recycling and sequestration of various kinds, I came to think about storing atmospheric $\ce{CO2}$ as asphalt. Given cheap energy as in the case of Iceland, and improved atmospheric carbon capture technologies, could this be feasible? Is there a name for the reverse cracking process?

Organic revers process to cracking is called Photosynthesis.

Well, I was thinking of building very long carbon chains, that's not photosynthesis, is it? Photosynthesis would just be the first step, right?

An alternative to photosynthesis will be great breakthrough.

Reducing CO2 back to alkanes (not neseserily using light energy) isn't viable as we're burning fuel to get energy - so what one country would be producing energy and the other losing it to reverse it?

Trading in negative emissions would make carbon removal itself a business, under cap- and trade systems. Under carbon tax systems the tax could be made symmetric around zero, so that removal is subsidized by the same amount that an emission of the same size would cost. Energy for the process might come from nuclear or renewables.

See this article http://www.sciencedaily.com/releases/2015/08/150819083117.htm
They have an aike idea.

I just realized the word I was looking for is polymerization.


text:
IUPAC name, phthalic anhydride
Could someone help me please (preferably talk me through not just provide the answer) with the IUPAC name of phthalic anhydride?  I'm sorry for not posting attempts but I'm afraid I have absolutely no idea how to begin here...

The preferred IUPAC name (PIN) of phthalic anhydride is 2-benzofuran-1,3-dione.
Please add details to isolate your problem and to narrow the answer set.

Wikipedia IUPAC names are in my experience about 50/50 , hence why I didn't just run with that.  If that is indeed the correct name, how is it derived?

Your approach is correct.
The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-65.7.7 Cyclic anhydrides
P-65.7.7.1 Cyclic anhydrides formed from two acid groups attached to the same parent hydride structure are named in two ways:
(1) as heterocyclic pseudoketones;
(2) by changing the class term ‘acid’ to ‘anhydride’ in the systematic or retained name of the dibasic acid.
Method (1) generates preferred IUPAC names

Using Method (1), the first step is naming the heterocyclic component as follows.

P-25.2.2.4 Heteromonocyclic components fused to a benzene ring
Unless listed as a retained name (…), a benzene ring fused to a heteromonocycle of five or more members (a benzoheterocycle) is named by placing the locant(s) indicating the position(s) of the heteroatom(s) at the front of the name consisting of the fusion prefix ‘benzo’ followed by a retained name, a Hantzsch-Widman systematic name, or a name formed by skeletal replacement (‘a’) nomenclature (…). The locants cited correspond to the full bicyclic structure. As in Hantzsch-Widman names, locants are placed in the order corresponding to the order of citation of the heteroatoms in the heterocyclic component. The locant ‘1’ is always assigned to the atom of the heterocyclic component next to a fusion atom. Heteroatoms are allocated lowest locants as a set, without regard to kind; if there is a choice, lowest locants are assigned in accordance with the seniority of the ‘a’ prefixes (…). In general nomenclature locants may be omitted when the name is unambiguous; for preferred IUPAC names locants must be cited. The letter ‘o’ of the ‘benzo’ prefix is elided when followed by a vowel. Indicated hydrogen is placed at the front of the name, when required.
(…)

Accordingly, the name ‘benzofuran’ consists of the prefix ‘benzo’ followed by the retained name ‘furan’.

The locant ‘1’ is assigned to the atom of the heterocyclic component next to a fusion atom. The locant ‘2’ is assigned to the heteroatom and it is placed at the front of the name.

Finally, the corresponding pseudoketone is named using the suffix ‘one’ in accordance with rules expressed for ketones. Thus, the preferred IUPAC name (PIN) of phthalic anhydride is ‘2-benzofuran-1,3-dione’.

Benzofuran is named by the position of the oxygen, hence 2-Benzofuran.  (below is 1-benzofuran)

The =O s are then added on the 1 and 3 carbons, hence, 1-3-dione
Compare, cyclopentan-1,3-dione

The end result is
2-benzofuran-1,3-dione

How about 1,3-dioxo-2-benzofuran?

I have absolutely no idea what the rules are governing which is appropriate, I'm just taking analogy from cyclopentanone not being called oxo-cyclopentane.  But yeah, I have no idea :/

i think it doesn't matter, coz IUPAC nomenclature is designed such that each name refers to exactly one compound, but not necessarily the other way round.


text:
Calculating pH: Weak Acid, Strong Base
This is a question that had me puzzled for quite a while.  I feel that information is missing.

A weak acid, $\ce{HA}$, $K_\textrm{a} = \pu{1.0E-4 M}$, is titrated with $\ce{NaOH}$.  The concentration of $\ce{NaA}$ at the equivalence point is $\pu{0.010 M}$.  What is the $\mathrm{pH}$ at the equivalence point?
a) 4.0
b) 6.0
c) 7.0
d) 8.0
e) 11.0

The correct answer is :

 d) 8.0

The answer is fairly easy to guess, as the $\mathrm{pH}$ needs to be greater than 7, but 11 is much too basic.  However, I'm not sure how I would be able to figure that out from the information given?  I've tried using the Henderson-Hasselbalch equation, but I can't figure out the relationship between $\ce{[NaA]}$ or $\ce{[A-]}$ and $\ce{[HA]}$.

At the equivalence point, you have an aqueous solution of weak base $\ce{Na}A$ at the concentration of $c$. The pH of the solution is given by the equation: $$\mathrm{pH}=\frac{1}{2}(\mathrm{p}K_a +\mathrm{p}K_w -\mathrm{p}c)$$
In your case: $\mathrm{pH}=\frac{1}{2}(4.0 +14 -2.0)=8.0$
Now, if you are not familiar with the above equation, you can write the reaction that could happen in the solution; the reaction of  $\ce{Na}A$, completely dissociated, with water:
$$A^- +\ce{ H2O <=> OH- +H}A$$
The constant of this equilibrium is: $$K=\frac{K_w}{K_a}= 10^{-10}$$
On the other hand,  you have:

$$K=\frac{x^2}{c-x}=\frac{x^2}{1.0\times10^{-2}-x}\approx\frac{x^2}{1.0\times10^{-2}} =1.0\times10^{-10}$$
Where $x$ is the concentration of ion hydroxide and the weak acid $\ce{H}A$.
We have, $x=1.0\times10^{-6}\mathrm{M}$.
$$[\ce{H+_\mathrm{(aq)}}]=\frac{10^{-14}}{1.0\times10^{-6}}\mathrm{M}$$
 $$\mathrm{pH}=8.0$$

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text:
Calculating [HI] given Kc, [H2], and [I2] for the reaction 2HI <-> H2+I2

In the reaction in equilibrium
$$\ce{2HI <=> H2 + I2}$$
The value $K_c$ is $0.0198$ at $\pu{721 K}$. Calculate $\ce{[HI]}$ at $\pu{721 K}$ if $\ce{[H2]} = \pu{0.0120 M}$ and $\ce{[I2]} = \pu{0.0150 M}$.

We use an ICE box. Since $K < 1$, the equilibrium is to the left. We are spending products to create reagents, so...
$$\begin{bmatrix}&\ce{[HI]} & \ce{[H2]} & \ce{[I2]}\\
\text{Initial} & 0 & 0.0120 & 0.0150\\
\text{Change} & +2x & -x & -x\\
\text{Equilibrium} & 2x & 0.0120 - x & 0.0150 - x
\end{bmatrix}$$
The equilibrium is given by
$$0.0198 = \frac{(0.0120-x)(0.0150-x)}{(2x)^2}$$
Solving yields:
$$x = 0.01907$$
Therefore
$$\ce{[HI]} = \pu{0.0381 M}$$
But the options are

$\pu{0.00211M}$
$\pu{0.133M}$
$\pu{0.0334M}$
$\pu{0.006567M}$
$\pu{0.0953M}$

While it is pretty close to $\pu{0.0334M}$, I can't see how could I have made my calculation any more accurate (or if they were the ones who were inaccurate).
Was my procedure correct?

For the record, at the temperature presented, the single equilibrium reaction cited is but one of many reactions. Missing are those involving the hydrogen atom and also the iodine atom, see https://aip.scitation.org/doi/10.1063/1.1730174 .

You're close. Using the expression for $K_c$, you just need to solve for $[\ce{HI}]$:
$$K_c = {[\ce{H2}][\ce{I2}]\over[\ce{HI}]^{2}}$$
$$0.0198 = {(\pu{0.012 M})(\pu{0.015 M})\over[\ce{HI}]^{2}}$$
$$[\ce{HI}] = \sqrt{{(\pu{0.012 M})(\pu{0.015 M})\over 0.0198}} = \pu{0.0953 M}$$

Well I can see that this method is way better. Do you know what went wrong with my particular procedure, though?

The ICE method takes you from initial conditions to equilibrium (final) conditions: in your problem, above, we're already at equilibrium, so the business with subtracting $x$ from each product concentration is not called for, as you're given those concentrations (0.012 M and 0.015 M). You might think of this problem as just the final step in the ICE procedure, perhaps, where the Change has already happened.

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text:
IUPAC nomenclature: putting the number before or in the middle of the name
Which is preferred according to "strict" IUPAC:
2-pentanone or pentan-2-one ?

The name “2-pentanone” and “pentan-2-one” describe the same compound. The name “2-pentanone” is in accordance with obsolete IUPAC Recommendations. The relevant rule of the IUPAC Nomenclature was changed in the 1993 IUPAC Recommendations:

R-0.1.2 Position of locants
Locants (numerals and/or letters) are placed immediately before the part of the name to which they relate, except in the case of traditional contracted forms (see R-2.5).

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.3.2 Position of locants
Locants (numerals and/or letters) are placed immediately before that part of the name to which they relate, except in the case of the traditional contracted names when locants are placed at the front of their names.

Examples:

hex-2-ene (not 2-hexene)
cyclohex-2-en-1-ol (not 2-cyclohexen-1-ol)
pentan-2-one (not 2-pentanone)
(propan-2-yl)benzene

Examples for traditional contracted names with locants at the front of the name:

2-naphthyl (not naphth-2-yl), the preferred name is naphthalen-2-yl
2-pyridyl (not pyrid-2-yl), the preferred name is pyridin-2-yl

so in that case would the IUPAC names for MIBK and https://en.wikipedia.org/wiki/N-Vinylpyrrolidone
be 4-Methyl-2-pentanone
and
N-ethenyl pyrrolid-2-one respectively ?   Whenever I lookup IUPAC names online they seem to be pretty all over the place with this convention :/

@Some_Guy The preferred IUPAC names are “4-methylpentan-2-one” and “1-ethenylpyrrolidin-2-one”. The locants “2-” are placed immediately before the suffix “-one” to which they relate.

Thanks.  While we're here, why is N-Ryl not correct?

@Some_Guy The locant “N-” is not necessarily wrong; for pyrrolidin, however, the locant “1-” is preferred. Anyway, if you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context.

@Some_Guy Comments are not meant for follow-up questions. If you have a new question, ask it. If it's too trivial to be a question, maybe it's not a question that would require a lot of thinking or research.


text:
Find the stoichiometric coefficient of a reactant given Kp, Kc, and the stoichiometric coefficient of the product

Have an hypothetical equilibrium reaction:
$$\ce{2 A (s) + x B (g) <=> 3 C (g)}$$
With the constants
\begin{align}
K_\mathrm{p} &= 0.0105 \\
K_\mathrm{c} &= 0.45
\end{align}
both at $\pu{250 ^\circ C}$. What is the value of $x$?

Since $\ce{A}$ is solid, I guess we can ignore it, right?
$$\ce{x B(g) <=> 3 C (g)}$$
Using $K_\mathrm{c}$:
$$0.0105 = \frac{\left[\ce{C}\right]^3}{\left[\ce{B}\right]^x}$$
Then
$$\left[\ce{B}\right]^x = \frac{\left[\ce{C}\right]^3}{0.0105}$$
But to solve this, I need the concentration of $\ce{C}$. How can I find it out?

Using the relation of $K_\mathrm{p}$ to $K_\mathrm{c}$:
$$\frac{K_\mathrm{p}}{K_\mathrm{c}} = (RT)^{\Delta n}$$
where $\Delta n$ is the change in the number of moles of gas in the reaction (here, $\Delta n = 3 - x$). Plugging in (using $R = \pu{0.0821 L atm mol-1 K-1}$ and $T = \pu{523.15 K}$) we get:
$$\frac{0.0105}{0.45} = 0.0233 = 42.950615^{(3-x)}$$
To solve we take the natural logarithm of both sides:
$$\ln{(0.0233)} = (3-x) \ln{(42.950615)}$$
$$\frac{\ln{(0.0233)}}{\ln{(42.950615)}} = -1 = (3 - x) \implies x = 4$$
And so $x = 4$. We can check to see if our answer is correct by plugging the calculated value for $x$ into the second expression from the top of this post:
$$0.0233 \buildrel{?}\over{=} 42.950615^{-1}$$
which does turn out to be correct.


text:
Why are there only 7 types of unit cells and 14 types of Bravais lattices?
Why are there only 7 types of unit cells and 14 types of Bravais lattices?
I was reading about solid-state chemistry for the first time and this limitation made no sense to me.
I tried to do the math and realized that there could be many more possibilities. Usually, the standard unit cell is described on the basis of whether the sides are at 90 degrees to each other and whether the sides are equal or not. However, any angle other than 90 could be chosen as the reference angle. So why 90?
Is it because only these 14 systems are found in nature and there is no need to study others?
Moreover, why are there only 4 sided shapes or hexagonal shapes?
Why not pentagonal or trigonal?
Any reference from any site or textbook would be appreciated.

Because three dimensional space is so constricted, maybe?  If we could live in more dimensions there would be room for more choices.  But, we can't.

All quotes will be from Solid State Physics by Ashcroft and Mermin.
Bravais Lattice:

A fundamental concept in the description of any crystalline solid is that of the Bravais lattice, which specifies the periodic array in which the repeated units of the crystal are arranged.  The units themselves may be single atoms, groups of atoms, molecules, ions, etc., but the Bravais lattice summarizes only the geometry of the underlying periodic structure, regardless of what the actual units may be."

Primitive Unit Cell:

A volume of space that, when translated through all the vectors in a Bravais lattice, just fills all of space without either overlapping itself or leaving voids is called a primitive cell or primitive unit cell of the lattice.

Unit Cell; Conventional Unit Cell:

One can fill space up with nonprimitive unit cells (known simply as unit cells or conventional unit cells). A unit cell is a region that just fills space without any overlapping when translated through some subset of the vectors of a Bravais lattice. The conventional unit cell is generally chosen to be bigger than the primitive cell and to have the required symmetry.

Crystal Structure:

A physical crystal can be described by giving its underlying Bravais lattice, together with a description of the arrangement of atoms, molecules, ions, etc. within a particular primitive cell.

So, one comes up with 14 Bravais lattices from symmetry considerations, divided into 7 crystal systems (cubic, tetragonal, orthorhombic,monoclinic, triclinic, trigonal, and hexagonal). This comes solely by enumerating the ways in which a periodic array of points can exist in 3 dimensions.
Now, what is on those points is a unit cell, which will itself have some symmetry. Thus, the combination of Bravais lattice and unit cell symmetry can again be enumerated and one comes up with 230 space groups.
Now for some of your related questions:
All cubic-related Bravais lattices will have 90 degree angles because they are based on cubic symmetry.  The trigonal Bravais lattice has no 90 degree angles, but isn't talked about much in more basic textbooks because, well, it looks weird.
Why no pentagonal unit cells? Well, because you can't fill space with a 5-fold symmetric Bravais lattice. Quasicrystals, while they have 5-fold symmetry, are a tiling through space that does not obey the rules for a Bravais lattice.

can you please explain the last line of the definition of Braviais lattices? Also another thing I asked ,"Is it because only these 14 systems are found in nature "?

@KaranSingh - an example. Take a simple cubic Bravais lattice. The periodic structure is the set of points that make up the vertices of a bunch of cubes stacked up to fill space.  If you look at it from any direction, that is what it looks like.  Now, at each point place a sphere (unit cell). It really looks no different. OK, now place an ellipsoidal solid (aka American football) on each point, with the long axis of each aligned in the x-direction.  You have the same Bravais lattice, but a different space group, because the unit on the lattice has reduced symmetry.

I think the short answer is "these 14 systems are the only ones possible." You can prove using math that there cannot be another.

Indeed - group theory is a wonderful thing for chemists and physicists to learn. It certainly was a big part of Wigner's work.

Thanks @GeoffHutchison.But  can't any more structures be produced forcibly(read artificially)?

@KaranSingh All periodic structures have to obey the mathematical rules of group theory. It isn't about what exists in nature but about the logical constraints on symmetry. Having said that, nature and man have produced structures that don't play by those rules but these are strictly called aperiodic as they don't have strict long range repetition but are a little irregular. Some have close to 5-fold symmetry (exact 5-fold symmetry isn't possible in 3D). These are the 3D analog of 2D things like Penrose Tiles which break the rules of 2D tessellation.

How does the trigonal lattice look weird?  It'a just a cube stretched or squeezed along one body diagonal.

@JonCuster I can't get how 14 Bravais lattices and unit cell generate $230$ space groups. Bravais lattice is just a periodic array of points (the geometric canvas that motifs are hooked into). Unit cell is arbitrary so for a given lattice we can have in general more than 1 unit cell. How the 230 spaces groups are derived? I can't understand if the space group or point group and the symmetry operations that include refers to the lattice (or unit cell in case of point group) or the crystal structure.

The seven crystal systems and 14 Bravais lattices (and 230 space groups) are all that are theoretically possible. Nobody has any doubt about that, and I believe it is mathematically proven.
If you want to follow the rule, that a crystall is formed by endless translational symmetry of a unit cell, then the only possibilites to start from are two structures: parallelepipeds and hexagonal prisms.
You can always, and only, except for hexagonal prisms, fill space by stacking parallelepipeds in all three directions. Rhombohedral, cubic, trigonal etc. are all special cases of the "triclinic" unit cell with higher symmetry, it is obvious that there are not endlessly more options that are more symmetric. Those make up for six of the seven crystal systems, and hexagonal is the special case making up the seventh.
The Bravais lattices come from unit cells which have an internal symmetry. You could go without these by describing them with one of the less symmetric crystal systems, but the rule is to assign the crystal system with highest symmetry. There are again not so many possibilities to have an internal symmetry, so this only makes 14 Bravais lattices out of the 7 crystal systems.
The paradigm is not to think of ways to make the system endlessly more complicated, but to start from the most odd system that is able to fill space, and think of the (limited) possibilities to make it more simple (i.e. symmetric).

The crystal structure and symmetry depends upon lattice parameter a, b, c and angles $\alpha, \beta$ and $\gamma$. During repetition of these combination it repeats one of the 14 Bravais lattice. Hence, no further combination is possible. That is why we have only 7 crystal system and 14 Bravais lattices

Bravais lattices have translational symmetry. As you mentioned all the possible translational symmetries are categorized into 14 types. Some of the lattices with extra symmetries (in addition to translational symmetry) are placed in certain types. For example, the cubic crystal system with three subcategories have the most degrees of symmetry. Any other lattice that doesn't have an extra symmetry is under Triclinic. Therefore, you can not have any other type. Because all the possible translational symmetries with extra point symmetries are categorized and the rest are in Triclinic type. Noting is left outside.

There are point groups I.e. combination of symmetry elements. There could be many combination of symmetry elements possible. But only those are permitted which somehow incorporate well with translation symmetry. In three dimensional space, based on above restrictions,only 32 point groups are permissible. There 32 permissible point groups require 14 bravais lattices which are groups into seven systems.


text:
What results from combining monopotassium phosphate, magnesium sulfate and calcium nitrate?
If I mix monopotassium phosphate, magnesium sulfate, and calcium nitrate together in aqueous solution, what chemicals will result?
Are these safe to combine (i.e. any toxic gases, etc.)?
I thought this answer about just magnesium sulfate and calcium nitrate was insightful.

Based on the linked question, I assume you're talking about mixing them in aqueous solution?

You won't get any gases.


text:
What are the intermolecular and intramolecular forces between Hg molecules/ atoms in liquid mercury?

Mercury is the only metal that is liquid at room temperature.

I know that metallic bonding exists between metal atoms but my knowledge is limited to metallic bonding found in solids. Will there be metallic bonding between $\ce{Hg}$ atoms in the liquid phase?

Moreover, there will be London dispersion forces between the $\ce{Hg}$ molecules.

Does mercury liquid only have metallic bonding as the intramolecular force and London dispersion force as the intermolecular force?
If liquid mercury (with both metallic bonding and LDF) is compared with a hypothetical liquid that has hydrogen bonding along with LDF, will the boiling point of mercury be higher than the aforementioned hypothetical liquid?

related http://chemistry.stackexchange.com/questions/3996/do-metallic-bonds-contain-london-dispersion-forces

Mercury is, at best, a fair conductor of heat and electricity.  See this earlier answer for an explanation of mercury's physical properties.

@ron I understand. Can I conclude that liquid mercury only has LDF as intermolecular force based on these links?

@Peregrine  Yes.


text:
Get K for a weak acid equilibrium when given the ionization percentage

There is some weak acid with a concentration of $0.0284$ M, which is
  ionized by $6.70$%. What is the value of $K_\mathrm a$?

So we got
$$\ce{HA <=> H+ + A-}$$
The formula for the ionization percentage is $\frac{[\ce{H+}]}{[\ce{HA}]}\cdot100$. So since the value is $6.70$ we can solve and get that
$$[\ce{H+}] = 1.90\cdot10^{-3}$$
So now I got concentration of $\ce{H+}$ ions. However, as far as I understand, to calculate $K_\mathrm a$ I need
$$K_\mathrm a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$
I am missing the value of $[\ce{A-}]$. How do I proceed from here?

Think about the relationship between $\ce{[H+]}$ and $\ce{[A^{-}]}$.

For every mole of $\ce{HA}$ that dissociates, you get equimolar amounts of $\ce{H+}$ and $\ce{A-}$. Therefore:
$$[\ce{H+}] = [\ce{A-}] = 1.9\times 10^{-3}\,\mathrm{M}$$
and
$$[\ce{HA}] = (0.0284\,\mathrm{M}) - (1.9\times 10^{-3}\,\mathrm{M}) = 0.0265\,\mathrm{M}$$
and so

 $$K_{\mathrm a} = {(1.9\times 10^{-3})^{2}\over 0.0265} = 1.36\times 10^{-4}$$

where the units of molarity are omitted because they cancel due to the implicit presence of $\ce{H2O}$ in the system (which would otherwise show in the denominator).


text:
How to find the amount of moles given mass percentage data

A hydrated aluminium sulphate, $\ce{Al2(SO4)3.xH2O}$, contains 8.10% of aluminium by mass. Find the value of x.

My attempt: (Assuming 100g sample)

Calculate the amount of aluminium using the formula $\mathrm{amount = \frac{mass}{molar~mass}}$, yielding $\mathrm{0.300~mol}~\ce{Al}$ having a mass of $\mathrm{8.10~g}$.

Determine the amount of sulphate ($\ce{SO4^2-}$) by recognising a ratio between the aluminium and sulphate. 2 moles of aluminium to 3 moles of sulphate, 2:3.

Using the ratio, I determined the amount to be $\mathrm{0.450~mol}~\ce{SO4^2-}$, which has a mass of $\mathrm{43.2~g}$.

I can now calculate the mass of the anhydrous compound, aluminium sulphate ($\ce{Al2(SO4)3}$), to be $\mathrm{51.3~g}$, leaving $\mathrm{48.7~g}$ of anhydrous substance, or $\ce{H2O}$.

Using the figure of $\mathrm{48.7~g}$, I can calculate the amount of $\ce{H2O}$ using the formula $\mathrm{amount = \frac{mass}{molar~mass}}$ which produced a figure of $\mathrm{2.71~mol}~\ce{H2O}$.

By using the principle of relative amounts, I find that x=9.

The answer is incorrect. Where did I go wrong?

Basis: 100 g
$m_{\ce{Al}}=8.10\ \mathrm g$
 $n_{\ce{Al}}=\frac{8.10\ \mathrm g}{27\ \mathrm{g/mol}}=0.3\ \mathrm{mol}$
 $n_{\ce{Al2(SO4)3}}=0.5 \times n_{\ce{Al}}=0.15\ \mathrm{mol}$
 $m_{\ce{Al2(SO4)3}}=0.15\ \mathrm{mol} \times 342.15\ \mathrm{g/mol}=51.3\ \mathrm g$
$m_{\ce{H2O}}=100\ \mathrm g-51.3\ \mathrm g=48.65\ \mathrm g$
$n_{\ce{H2O}}=\frac{48.65\ \mathrm g}{18\ \mathrm{g/mol}}=2.7\ \mathrm{mol}$
$x=\frac{2.7\ \mathrm{mol}}{0.15\ \mathrm{mol}}=18$


text:
How to determine the strongest oxyacid of chlorine?

Which is the strongest acid here?

$\ce{HClO}$
$\ce{HClO2}$
$\ce{HClO3}$
$\ce{HClO4}$

I found this question, didn't really get the first answer, but this one in particular seemed interesting.
Apparently, you calculate a $\mathrm{p}K_\mathrm{a}$ for each substance.
According to this website.
$$\mathrm{p}K_\mathrm{a} = -\log_{10}K_\mathrm{a}$$
I need to find out $K_\mathrm{a}$ then. From what I understand $K_\mathrm{a}$ is obtained from dividing the concentrations of products by the concentrations of reagents. However, how am I supposed to do so if I am not given any equilibrium reactions?

Evans' and Bordwell's $p\mathrm{K_a}$ tables have a large amount of data on different compounds. Other sources are also available but these are two of the best that I know.

http://chemistry.stackexchange.com/questions/39451/how-to-assess-the-strength-of-the-acid

In such questions you're supposed to know certain rules of acidity. You can find some of them on this page. In your case fourth rule, concerning oxoacids, is relevant and as you can see there, it's described on this very case. More precise formulation is so called Pauling's Rule described here. In general the more oxo groups connected to central atom, the more acidic compound.
Related: How are Pauling's rules applied to Nitric acid?


text:
Is Hexafluorocyclohexane the "Most Polar" Small Molecule?
Earlier this year, there was a lot of attention when all-cis 1,2,3,4,5,6-hexafluorocyclohexane was synthesized*:

Notably, C&E News quoted the lead author:

This compound is remarkable for being the most polar nonionic compound now known to exist

Now, the original paper reports a molecular dipole moment of $\pu{6.2 D}$, which is extremely high for an alkane. It also claims the:

unusual property of a facially polarized ring in organic chemistry

Certainly having one side of the cyclohexane ring completely substituted with fluorines is unusual, so I agree with the facial-polarity property being interesting.
But there are many small molecules (e.g., under 20-30 atoms) with large dipole moments, e.g., p-nitroaniline:

A quick PM7 calculation gives the predicted dipole moment of $\pu{7.93\pm1 D}$. So I dispute the claim from C&E News - nitroaniline is clearly more polar and non-ionic.
So my question is this: What's the largest dipole moment for a known non-ionic organic molecule under 30 atoms?
*: O'Hagan, et. al, "All-cis 1,2,3,4,5,6-hexafluorocyclohexane is a facially polarized cyclohexane" Nature Chemistry 7, 483–488 (2015)

I'll indicate that at the moment, I have no idea what the answer is, but I suspect it's highly conjugated.

Also, if someone can't easily find or compute the dipole moment of their candidate, please send me a message and I'll help. Others probably can, as well, if I'm not around.

Your calculated dipole moment of 7.93 +/- 1 D is notably higher than the experimental dipole moment for p-nitroaniline which is variously reported between  5.5-6.3 D.

@ron indeed, although IIRC, it's on the higher end. In any case, I gave the quick (perhaps poorly chosen) example that D-$\pi$-A is likely to yield higher dipole moments. I'll find a better example later.

I don't think it's a bad example.  In fact, if you replace the benzene with several ethylene linkages and the nitro with some type of CO, you get the molecule (poorly) described here (p. 396) which the author says has a dipole moment of 17.7 D

@Geoff Did you see, that the 6.2 D were calculated? A molecular dipole value of 6.2 D for 1 was calculated at the M11/6-311G(2d,p) theory level using natural bondorbital (NBO) analysis with NBO 6.0.

Merocyanine-type dyes, in particular spyropyrans can have extremely high dipole moments. Random googling turns up cases with dipoles supposedly as large as 47 D. Not sure whether they clear your "non-ionic" requirement, though, as they straddle the line between neutral molecules and zwitterions. You may be interested in articles by Frank Würthner, who has studied this class of materials for organic electronics.

Are zwitterions with less than 30 atoms allowed or forbidden? (cc @Martin - マーチン)

@andselisk - if you think it's a stable ground-state, zwitterion, sure.

Henry Rzepa calculated a Dipole Moment of 31.5!

At first, I thought, that those 6.2 D of O’Hagan et al. have been measured somehow but it as later described, they calculated it (1):

A molecular dipole value of 6.2 D for 1 was calculated at the M11/6-311G(2d,p) theory level using natural bond orbital (NBO) analysis with NBO 6.0.

I was using your given database for a little search, starting with nitroaniline, and came up with a few small molecules with higher dipole moments. Those seem to be PM7 calculated values and are shown in the following table:
\begin{array}{lll}
\hline
\text{Name} & \text{Dipole moment in D} & \text{Source}\\
\hline
\text{4-Nitroacetanilide} & 8.36 \pm 1.08 & (2)\\
\textit N\text,\textit N\text{-Dimethyl-4-nitroaniline} & 9.53 \pm 1.08 & (3)\\
\textit P\text ,\textit P\text{-Dimethyl-}\textit N\text{-(4-nitrophenyl)phosphinic amide} & 10.89 \pm 1.08 & (4)\\
\hline
\end{array}
As there were some comments, “it seems that this level of calculation does not provide accurate estimates of dipole moments”, I did some further calculations on levels that might provide better estimates.
There is a quite new paper of Hickey and Rowley (5) which says:

CCSD, MP2, or hybrid DFT methods using the aug-cc-pVTZ basis set are all able to predict dipole moments with RMSD errors in the 0.12–0.13 D range and polarizabilities with RMSD errors in the 0.30–0.38 Å³ range.

Within the paper, there is later on a comparison between cc-pVTZ, Sadlej-pVTZ and aug-cc-pVTZ:

The Sadlej basis set yields results that are improved over the cc-pVTZ basis set and have effectively the same accuracy as the aug-cc-pVTZ basis set.

That’s why I chose Sadlej’s basis set at first … but I wanted to look if it could not be easier hence faster. Therefore I also chose the def2-SVPD and looked, what an improvement a def2-TZVPD single point calculation on this geometry would yield.
I use ORCA 3.0.3 for calculations – the command line is like the following:
! B3LYP Sadlej-pVTZ TightSCF TightOpt

Table 1: "Pure" dipole moments (in Debye) straight from the calculations
\begin{array}{lllll}
\hline
 & \text{def2-SVPD} & \text{def2-TZVPD} & \text{Sadlej-pVTZ} & \text{PM6}\\
\hline
\text{4-Nitroacetanilide} & 9.18261 & 9.00380 & 9.07922 & 8.6738\\
\textit N\text ,\textit N\text{-Dimethyl…} & 7.39418 & 7.29188 & 7.30862 & 9.4209\\
\textit P\text ,\textit P\text{-Dimethyl…} & 8.58913 & 8.44430 & 8.51591 & 11.8352\\
\text{all-}\textit{cis }\ce{C6H6F6} & 6.17422 & 5.97013 & 6.04794 & 5.6298\\
\hline
\end{array}
What I did then, was to think of how to produce values that could be properly compared to the experimental values from the paper. As they said, with their methods they can yield RMSDs of around 0.12 D. So I took their values for the Sadlej basis set and made my own calculations on their test molecules for the two def2 basis sets (and also for PM6 with Gaussian 09 Rev. A.02).
Table 2: Some analytical values to compare with the publication values; the units are in Debye
\begin{array}{lllll}
\hline
 & \text{def2-SVPD} & \text{def2-TZVPD} & \text{Sadlej-pVTZ} & \text{PM6}\\
\hline
\text{MAE} & 0.08 & 0.08 & 0.08 & 0.40\\
\text{RMSD} & 0.13 & 0.12 & 0.12 & 0.61\\
\hline
\end{array}
So for my two ways, the MAE and the RMSD are not too far away from the triple zeta values but are calculated way faster. Now I have made a linear regression between the experimental and the calculated dipole moments from Hickey and Rowley’s test molecules to be able – as said before – to calculate proper error ranges at a given probability.
Figure 1: Regression functions for the four tested methods (blue), prediction intervals at 95% (gray); from top left to bottom right: def2-SVPD, def2-TZVPD//def2-SVPD, Sadlej-pVTZ, PM6

Now let’s turn the “pure” values from Table 1 into somehow more real values:
\begin{align}
\text{def2-SVPD} \to \text{exp}&: -0.0097439128 + 0.9763275\ x\\
\text{def2-TZVPD//def2-SVPD} \to \text{exp}&: -0.0045050428 + 0.99095691\ x\\
\text{Sadlej-pVTZ} \to \text{exp}&: -0.012922547 + 0.98812009\ x\\
\text{PM6} \to \text{exp}&: -0.16057241 + 0.89545049\ x\\
\end{align}
Prediction Intervals (for the error):
$$\small\begin{align}
\text{def2-SVPD} \to \text{exp}&: 0.24473369 \sqrt{1.0217391 + 0.010348758 (x-1.4590217)^2}\\
\text{def2-TZVPD//def2-SVPD} \to \text{exp}&: 0.23719452 \sqrt{1.0217391 + 0.010656662 (x-1.4321957)^2}\\
\text{Sadlej-pVTZ} \to \text{exp}&: 0.23333979 \sqrt{1.0217391 + 0.010593477 (x-1.4448261)^2}\\
\text{PM6} \to \text{exp}&: 0.98446665 \sqrt{1.0217391 + 0.0097476423 (x-1.7592391)^2}
\end{align}$$
Table 3: Converted dipole moments (in Debye) from Table 1 including error ranges at 95% probability
\begin{array}{llll}
\hline
 & \text{def2-SVPD} & \text{def2-TZVPD} & \text{Sadlej-pVTZ} & \text{PM6}\\
\hline
\text{4-Nitroacetanilide}  & 8.96 \pm 0.31 & 8.92 \pm 0.30 & 8.96 \pm 0.30 & 7.61 \pm 1.20 \\
\textit N\text ,\textit N\text{-Dimethyl…}     & 7.21 \pm 0.29 & 7.22 \pm 0.28 & 7.21 \pm 0.27 & 8.28 \pm 1.24 \\
\textit P\text ,\textit P\text{-Dimethyl…}     & 8.38 \pm 0.30 & 8.36 \pm 0.29 & 8.40 \pm 0.29 & 10.44 \pm 1.40 \\
\text{all-}\textit{cis }\ce{C6H6F6} & 6.02 \pm 0.27 & 5.91 \pm 0.26 & 5.96 \pm 0.26 & 4.88 \pm 1.06 \\
\hline
\end{array}
What can be seen from this now? The smaller B3LYP/def2-SVPD method yields nearly the same values as the more expensive Sadlej version. Putting the def2-TZVPD single point on top of them, yields slightly deeper results with actually no accuracy improvements. PM6 is no real good option as the error ranges are way higher, with values of about $1-1.4$ D.
In the end, the dipole moment of all-cis 1,2,3,4,5,6-hexafluorocyclohexane is probably somewhere around $6$ D but there are probably many other molecules with higher dipole moments … “my” three molecules shown above go at least up to somewhere around $9$ D.
As long as no experimental values are available, everything here is only a good guess – but you will know this too good.

Sources:

O’Hagan et al., Nature Chemistry, 2015, 7, 483–488
http://pqr.pitt.edu/mol/NQRLPDFELNCFHW-UHFFFAOYSA-N
http://pqr.pitt.edu/mol/QJAIOCKFIORVFU-UHFFFAOYSA-N
http://pqr.pitt.edu/mol/FICBIFYYTQXSEY-UHFFFAOYSA-N
A. L. Hickey, C. N. Rowley, J. Phys. Chem. A, 2014, 118 (20), 3678–3687

As I mentioned above, it seems that this level of calculation does not provide accurate estimates of dipole moments.  For example,  the experimental dipole moment for N,N-Dimethyl-4-nitroaniline is reported as 6.0+/- 0.4 D (dioxane, ref).

But then at least the error range covers 6.2 D. And this made it a good starting point to look if the others or similar systems might have higher dipole moments.

@pH13 Did you consider replacing the benzene parent structures with azulene (which itself has a significant permanent dipole moment), e.g. N,N-dimethyl-2-nitroazulen-6-amine?

@Loong: N,N-dimethyl-2-nitroazulen-6-amine has a calculated dipole moment of $\mathrm{(12.15 \pm 0.36)\ D}$ (at the B3LYP/def2-TZVPD//B3LYP/def2-SVPD level)

At first I checked common organic compounds and solvents in CRC Handbook [1, 9-59] that have dipole moment above $\pu{4.5 D}$:
\begin{array}{llrr}
\hline
\text{Name} & \text{Formula} & \text{Atom count} & \mu, \pu{D} \\
\hline
\text{Thiophenecarbonitrile} & \ce{C5H3NS} & 10 & 4.59(2) \\
\text{3-Methyl-2-butenenitrile} & \ce{C5H7N} & 13 & 4.61(13)\\
\text{Ethyl cyanate} & \ce{C3H5NO} & 10 & 4.72(9) \\
\text{Ethylene carbonate liq.} & \ce{C3H4O3} & 10 & 4.9 \\
\text{Propylene carbonate liq.} & \ce{C4H6O3} & 13 & 4.9 \\
\text{2-Nitroanisole liq.} & \ce{C7H7NO3} & 18 & 5.0 \\
\text{Hexamethylphosphoric triamide liq.} & \ce{C6H18N3OP} & 29 & 5.5 \\
\text{2-Pyridinecarbonitrile} & \ce{C6H4N2} & 12 & 5.78(11) \\
\text{Hydrogen cyanide trimer} & \ce{C3H3N3} & 9 & 10.6 \\
\hline
\end{array}
Nothing special or really interesting except for $\ce{HCN}$ trimer, which is not really organic. Time to look at something more exotic.

Conjugated double bonds
German paper "Spektroskopische Untersuchungen an Merocyaninen I.) UV- Und NMR-Spektren von Malodinitril-Substituierten Vinylogen Säureamiden" [2] presents a series of malodinitrile-substituted vinylogous acidic amides of the following formula
$$\ce{Me2N-[CH=CH]_n-CH=C(CN)2} \qquad (n \in [0, 4])$$
and the following experimentally established ground-state ($S_0$) dipole moments:
\begin{array}{rlrr}
\hline
n & \text{Formula} & \text{Atom count} & \mu, \pu{D} \\
\hline
0 & \ce{C6H7N3} & 16 & 8.6 \\
1 & \ce{C8H9N3} & 20 & 11.4 \\
2 & \ce{C10H11N3} & 24 & 13.5 \\
3 & \ce{C12H13N3} & 28 & 15.1 \\
4 & \ce{C14H15N3} & 32 & 16.3 \\
\hline
\end{array}

Aromatic compounds
From the recent paper "Hexasubstituted Benzenes with Ultrastrong Dipole Moments" [3] there are a few more candidates:

2,2‐diheptyl‐1,3‐dihydro‐1,3‐benzodiazole‐4,5,6,7‐tetracarbonitrile
$\ce{C25H32N6}$ (63 atoms).
$\mu_\mathrm{calcd} = \pu{12.7 D}$ (level-of-theory: DFT-B3LYP; basis set: 6-311 ++G(d,p));
$\mu_\mathrm{exp} = \pu{12.2\pm0.3 D}$ (dielectric permittivity measurements in DMAc).

2,2‐dimethyl‐1,3‐dihydro‐1,3‐benzodiazole‐4,5,6,7‐tetracarbonitrile
$\ce{C13H8N6}$ (27 atoms).
$\mu_\mathrm{calcd} = \pu{12.3 D}$ (level-of-theory: DFT-B3LYP; basis set: aug-cc-pVTZ);
$\mu_\mathrm{exp} = \pu{10.9\pm0.3 D}$ (dielectric permittivity measurements in THF).

Zwitterions
As stated in a handbook "Solvents and Solvent Effects in Organic Chemistry" [4, p. 79], the highest dipole moments among solvents found to date are probably zwitterionic 3-alkyl-1,2,-oxadiazolium-5-olates (sydnones). For example, 4-ethyl-3-(1-propyl)sydnone $\ce{C7H12N2O2}$ (23 atoms) has a dipole moment of $\pu{10.7 D}$ [5].

Another zwitterion and multifunctional dye, Brooker's merocyanine $\ce{C14H13NO}$ (29 atoms), has ground state dipole moment of $\pu{26 D}$ in trichloromethane, and which overall greatly depends on solvent polarity [4, 6].

In 2021 Henry Rzepa proposed another potential candidate in his blog, cycloprop‐2‐en‐1‐ylium‐1‐ylethyne (7 atoms), with the calculated dipole moment $\pu{11.9 D}$ in chloroform.

Bibliography

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.
Scheibe, P.; Schneider, S.; Dörr, F.; Daltrozzo, E. Berichte der Bunsengesellschaft für physikalische Chemie 1976, 80 (7), 630–638 DOI: 10.1002/bbpc.19760800711.
Wudarczyk, J.; Papamokos, G.; Margaritis, V.; Schollmeyer, D.; Hinkel, F.; Baumgarten, M.; Floudas, G.; Müllen, K. Angewandte Chemie International Edition 2016, 55 (9), 3220–3223 DOI: 10.1002/anie.201508249.
Reichardt, C.; Welton, T. Solvents and Solvent Effects in Organic Chemistry, 4th ed.; Wiley-VCH: Weinheim, Germany, 2011.
Handa, M.; Kataoka, M.; Wakaumi, M.; Sasaki, Y. BCSJ 1997, 70 (2), 315–320 DOI: 10.1246/bcsj.70.315.
Morley, J. O.; Morley, R. M.; Docherty, R.; Charlton, M. H. J. Am. Chem. Soc. 1997, 119 (42), 10192–10202 DOI: 10.1021/ja971477m.

Calicene ($4.66~\pu{D}$, 14 atoms) and its derivatives are also good candidates.

For Hexaphenylcalicene, a hydrocarbon with 74 atoms, a dipole moment of $6.3~\pu{D}$ has been reported.[1] 2,3-dicyano-5,6-diphenylcalicene (34 atoms) has a dipole moment of $14.3~\pu{D}$.[2]
It would be interesting to see how two conjugate double bonds between the rings would influence the dipole moment

References

E.D. Bergman, I. Agranat, Chem. Commun, 512 (1965).
Houben-Weyl, Methods of Organic Chemistry, Vol. E 17d, 4th Edition, p. 2967.

Isn't 2,3-dicyano-5,6-diphenylcalicene a zwitterion? I'm not sure how Houben-Weyl present this compound, but in original German paper it is pretty much drawn as a zwitterion. Also, feel free to substitute ref. 2 with this paper, as the table with dipole moments in Houben-Weyl how I see it on GoogleBooks is a complete replica of the one in this paper:)

Personally, I would not call it a zwitterion, but  as long as there's  no official definition of the term, it seems to be a matter of taste. 2. The original article is behind a paywall. Many people (including me) don't have access to it. So I prefer to cite the publicly available pieces of information.


text:
Is white phosphorus really a cool burning fuel?
About 11 months ago, I posted a question on this SE site on cool flames.
One of the answers mentioned white phosphorus. However, a few weeks earlier, I played a game called Spec Ops: The Lines, in which the protagonist uses white phosphorus laden mortar munition, horribly burning dozens of soldiers and civilians. I was confused that the answer mentioned WP because of that moment from the game, and further googled it to find that WP munitions are actually heavily regulated, and using them against civilians is considered a warcrime. To me, that doesn't really seem compatible with "cool burning" at a temperature you can hold your hand in.
It's not my intent to cause this answer to be downvoted, but I'm confused whether the answer is factually accurate. According to Google, it burns at 5,000 degrees Fahrenheit, but that's the munitions.

In answer you linked there was probably misunderstanding with ignition temp. and flame temp.

@Mithoron Also perhaps a difference between burning small amounts of phosphorus vapor and burning the actual stuff.  Solid chunks of white phosphorus are truly nasty weapons for the reasons mentioned.  It's allowed for use as a smoke producer, but the Israeli army in particular continues to use it as an incendiary weapon.

Burning white phosphorus sticks to skin and causes horrible burns. It's also quite toxic. Highly recommended to stay away...


text:
Bond angles in BrF5
If in an octahedral structure there is one lone pair placed on any two of the axial position then will the lone pair distort all the 90 degree angles due to repulsion from lone pair or the angles will remain same? I found one question regarding $\ce{BrF5}$.

You are right! You asked the question and you replied simultaneously.
According to this website, the structure of bromine pentafluoride confirmed your claims.

Consider the figure given in the previous answer:
Due to a lone pair of $\ce{e-}$ and thus, due to lp-bp repulsions, all equatorial fluorine atoms will move upwards but still remain in a plane. Now, the $\ce{Br}$ will be out of that plane. This will cause the $\ce{F_{axial}-Br-F_{equatorial}}$ angle to reduce. I think $\ce{F_{equatorial}-Br-F_{equatorial}}$ angle would remain $\mathrm{90^\circ}$.
According to this website, the experimental data are:
$$\ce{F_{axial}-Br-F_{equatorial}}\mathrm{~=84.8^\circ}$$
$$\ce{F_{equatorial}-Br-F_{equatorial}}\mathrm{~=89.529^\circ}$$
I am wondering, where is the difference of $~\mathrm{360^\circ -4\times89.529^\circ=1.884^\circ}~$?
One more thing, why it is written equitorial not equatorial in this website?
Edit: I found another source$^1$ according to which the equatorial bond angles are at $\mathrm{90^\circ}$.

Reference:

Page 291, University Chemistry, 4th Edition, Bruce M. Mahan/ Rollie J. Meyers, Pearson (ISBN 978-81-317-2957-1)

It might be a typo or poor proofreading. I found that it's a noun: "republic of equitorial guinea".


text:
Why is HCl used in hydrolysis of sucrose?
I am aware that $\ce{HCl}$ can be used to help speed up the hydrolysis of sucrose but I haven't really been able to find out why. I've read somewhere that the glycosidic link in sucrose is broken off and that the $\ce{H+}$ ions from the $\ce{HCl}$ is responsible for this.
I've also come across the ether link during my research and would like to know how this plays a part in the hydrolysis of sucrose and if the $\ce{HCl}$ interacts with it.

related http://chemistry.stackexchange.com/questions/34650/how-many-ml-of-1-0m-hcl-would-be-required-to-completely-decompose-sucrose-into-g

https://www.quora.com/What-are-the-effects-of-HCL-on-starch

It’s not an ether linkage, it’s an acetal linkage. And also: that is exactly the mechanism if you already found it. Protonate the linking oxygen, break up of the complex. Whether $\ce{HCl}$ or any other (sufficiently strong) acid is used is irrelevant.

The mechanism really isn't much more complicated than what you describe.

The $\ce{H+}$ binds to the acetal linkage. Water then binds to the fructose ring, breaking its bond with the acetal linking $\ce{O}$. Finally, the water molecule on the fructose loses a proton, restoring the catalyst. As Jan mentioned in the comments, the specific acid used is irrelevant so long as it is strong enough to sufficiently acidify the solution.
(Credit to Khan Academy for the image)

The answers that have been provided so far are correct in one sense, but they skip over the key to your question. The key to why a reaction is catalyzed by a substance has to do with the mechanism, in particular the rate determining step. The answers above do not show that step or discuss it, but rather they show a step which precedes it, namely the protonation of the bridge oxygen. It is true that this step is important, and that it involves the hydrogen ion. But it is the following C-O bond cleavage step that is more likely the rate determining step.
For a detailed answer to your question see this post which references a publication that solved the mechanism details:
Hydrolysis of Sucrose over Sn1Ac or Sn2Ac
Based on that reference, I do not believe @Tyberius is correct in saying, “Water then binds to the fructose ring, breaking its bond with the acetal linking O”. Rather, water only comes in after the oxocarbonium ion has been formed in the C-O bond cleavage step. The detailed answer to your question is that the presence of an acid makes that C-O bond cleavage step happen quicker.

I don't think you can say they "solved the detailed mechanism". There are multiple feasible pathways, and computational studies show the energetic differences are quite subtle, see https://chemistry.stackexchange.com/a/80353

@KarstenTheis You make a good point. It depends how one interprets the word "solved", and I did not mean to imply that no other solutions were possible. I fully agree that other solutions are possible, but for the purpose of my answer I was not focusing on those issues but rather the key question related to the original post; namely, the C-O bond cleavage step. By the way, I am an author on that paper and you will see that in it we specifically mention other possible solutions.

Congrats on the paper! I enjoyed exploring how you managed to distinguish between mechanistic hypotheses. No matter whether SN1 or SN2, and no matter which glycosidic bond ends up being cleaved, the reaction starts with a protonation step, and that is why the rate depends on the pH. Even though it is not the rate determining step, the hydronium concentration is expected to appear in the rate law because it is a fast equilibrium generating a species that is a reactant in the rate determining step.

@KarstenTheis Yes, I agree concerning H+. To observe the initial products revealing the mechanism, finding the exact right [H+] was the key. It turned out to be the same pH used all the time in the commercial hydrolysis of sucrose. Although noted in ref 34, what's not mentioned is that it took over a year to figure that out ;)


text:
How to control and vanish termiticide/pesticide fumes from room
I had some wood work done in my room and termiticide (UNIBAN-20 (Chlorpyriphos-20% EC) was used so that there is no termites attack in future. The stupid carpenter was using this product for the first time and it had to be diluted 10 times in kerosene oil before use which he didn't do. He poured the concentrated solution of this termiticide.
Now we have very strong smell and poisonous fumes (the head starts spinning within seconds) in the room.
How can I neutralize the smell and get rid of the poisonous fumes? It has been 13 days now.
I have exhaust fan switched on but the smell is still there in the room.

I think you should seek professional advice to solve this problem. It doesn't sound like the sort of thing that you want to mess around with.

@bon: Thanks for your response! Who should i approach?

As stated by bon in the comments, this is probably something you should contact someone who is actually supposed to be able to answer this question about. The fumes might just be an annoyance, or they could be highly and immediately dangerous.
One place to start is with the manufacturer. Indeed, the manufacturer is probably the best place to go in general with chemical safety concerns. This chemical seems to be produced by a number of firms, so I was not able to locate a specific one, but a google search shows a number of possible contacts. This MSDS provides contact information for a Dow subsidiary. I suspect they will be willing to help, but if they refuse to because they were not the manufacturers of your specific product, you will need to ask the carpenter for the documentation. In a number of locations, the law specifies he must be able to provide an MSDS.
Alternatively, you might wish to contact your local fire department (non-emergency line). They can probably point you to someone local whose job it is to handle potential chemical contamination.
I will note that the MSDS I cited does not suggest extreme danger. However, it is still best to contact a producer or specialist of some sort.


text:
What is Enthalpy?
At school, they teach the following two formulas: $$\Delta H = -mc\Delta T$$ and $$q= mc\Delta T$$ What i am not sure is that my teacher says that q and $\Delta H$ are the same thing but to think of q as a magnitude and hence it is always positive. However what I am confused about is how can q always be positive? What if the change in temperature is negative? Also, what is the difference between q and $\Delta H$. My current understanding is that they are the same thing and they measure the change in heat energy. Is this right?

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Enthalpy is thought as the total potential energy of a system. The actual thermodynamic equation for enthalpy is: $$\Delta H = \Delta U + \Delta {(pV)}$$ where $\Delta U$ is the change in internal energy. Internal energy can be thought as the total amount of heat energy there is in a system. Hence $\Delta U$ can be thought as the total energy that is required to make a system, while $\Delta {(pV)}$ can be thought as how much energy is required to make space for the system.
Q on the other hand is actually a different thing to enthalpy. Q refers to the heat transfer of a system. Note that this means that q can be positive or negative (if heat energy moves from the system to the surroundings) and hence it is not always positive. The correct formula for heat transfer is: $$q = \Delta U -w $$ So actually, q doesn't always have to equal to enthalpy change. However at a school level it is taught that they equal the same thing. This is because of we assume that pressure is constant (which is a common assumption made as usually most experiments have a constant pressure), enthalpy becomes: $$\Delta H = \Delta U + p\Delta V$$ Also since $w = p\Delta V$, enthalpy becomes: $$\Delta H = \Delta U - w = q$$ Hence, enthalpy and q equal the same values only when pressure is constant.
The equation that your teacher has given you is actually an equation for heat transfer, not for enthalpy change. However because enthalpy equals heat transfer in most cases, it is commonly said that enthalpy change also equals to that equation. However the one mistake is that there shouldn't be a minus in the equation for enthalpy change. The equation for heat transfer and enthalpy should both be the same.


text:
Physical Interpretation of Gibbs Free Energy
I know that $\Delta G = \Delta H - T\Delta S$ and that if $\Delta G$ less than $0$ a reaction will be spontaneous. What I am wondering is what is the actually physical interpretation of Gibbs free energy. I mean, if the Gibbs free energy of a particular reaction $20\ \mathrm{J\ mol^{-1}}$, what does that actually mean?

I'd be careful with the word 'spontaneous' - as an example, graphite has the lowest free energy of the carbon solid phases at room temperature and pressure.  Graphite is about 2.9 kJ/mol more stable than diamond.  Yet, diamonds don't mysteriously turn in to chunks of graphite - you need to get over a nucleation barrier and that barrier is pretty high.

Gibbs free energy is a  measure of the total non-expansionary work that a system can do. So in the case of your example, that system has the potential to do 20 joules of non-expansionary work per a mole of substance. Non-expansionary work refers to work other than pressure-volume work, such as electrical work. Hence not only is Gibbs free energy is important to tell whether a reaction will be spontaneous or not, it is also important to engineers as they tell how much non-expansionary work a system, such as motor engine, can do.
Gibbs free energy is really derived is from the Clausius inequality which is: $$dS\geq \frac{dq}{T}$$ Basically is states that in a reaction, the change in entropy must always be equal (in the case of an ideal gas) or greater than the transfer of heat energy divided by temperature. Therefore, for a reaction to be spontaneous, the following relation must be satisfied: $$dS-\frac{dq}{T}\geq0$$ Now, lets assume that the pressure is constant so we can simplify this relation.
Constant Pressure
Enthalpy is given by: $$\Delta H_p = \Delta U + p\Delta V$$
and internal energy by: $$\Delta U = q + w$$
Now by substituting $\Delta U$ in the expression for enthalpy we get: $$\Delta H _p= q + w + p\Delta V$$ Now, since w = $-p\Delta V$, the above equation simplified so that: $$\Delta H_p = q$$ So, now in the Clausius inequality, we can substitute dq with dH and multiply by $T$ to get: $$TdS \geq dH$$ Now by integrating both sides and putting all the variables on one side, we get the question for gibbs free energy: $$\Delta G = \Delta H - T\Delta S$$ and hence this is why $\Delta G$ has to be negative for a reaction to be spontaneous

Thanks for the answer. Also, I am also interested in the derivation of the formula. Could you please show me the derivation for Gibbs free energy also.

@benzene The Gibbs free energy can be derived by doing two subsequent Legendre transforms on the formula for U... Is that enough to get you started?

What are Legendre transforms?

@benzene Eh I think it's easier to say that Gibbs energy is defined to be $G = H - TS$. That's what most texts would write. At const $T$, $\Delta G=\Delta H-\Delta (TS)=\Delta H-T\Delta S$. The reason why we choose this definition of $G$ is because it's convenient to have the intensive variables $p$ and $T$ as the differentials: for a reversible process $dG=V,dp-S,dT$, at const. $p$ and $T$, $dp=dT=0$ so $dG=0$, which is the condition for equilibrium. Legendre transformations are just another way of saying the same thing. https://en.wikipedia.org/wiki/Legendre_transformation#Thermodynamics

@benzene Consider if we stuck with the internal energy $dU = T,dS - p,dV$. The condition for equilibrium would then be $dU = 0$ at const volume (which is still okay) and const entropy - but there's no way you can make sure a system stays at const entropy. To convert this to something usable, you need to do the Legendre transformation - since $T$ and $S$ are a pair, you just add $TS$ to $U$. You could stop here to get the Helmholtz free energy $A$, but for us, doing one more step and subtracting $pV$ to get $G$ makes things even easier since we like dealing with const $p$, not const $V$.

I'm going to concentrate on the $T\,S$ term (which you likely find the most mysterious) and hopefully give a little more physical intuition.
The $T S$ term arises roughly from the energy that is needed to "fill up" the rotational, vibrational, translational and otherwise distractional thermal energies of the constituents of a system. Simplistically, you can kind of think of its being related to the idea that you must use some of the energy released to make sure that the reaction products are "filled up" with heat so that they are at the same temperature as the reactants. So the $T S$ term is related to, but not the same as, the notion of heat capacity: let's look at this a bit further.
Why can't we get at all the energy $\Delta H$? Well, actually we can in certain contrived circumstances: let's look at these circumstances to find out why in a meaningful definition we must leave behind an amount $T\,\Delta S$. Let's think of the burning of hydrogen:
$$\ce{H2 + 1/2 O2 -> H2O};\quad\Delta H \approx 143\:{\rm kJ\,mol^{-1}}\tag{1}$$
This is a highly exothermic reaction, and also one of the reactions of choice if you want to throw three astronauts, fifty tons of kit and about a twentieth of the most advanced-economy-in-the-world’s-1960s GDP at the Moon.
The thing about one mole of $\ce{H2O}$ is that it can soak up less heat than the mole of $\ce{H2}$ and half a mole of $\ce{O2}$; naively this would seem to say that we can get more heat than the enthalpy change $\Delta H$, but we'll see this is not so. We imagine a thought experiment, where we have  a gigantic array of enormous heat pads (all individually like an equilibrium “outside world") representing all temperatures between absolute zero and  $T_0$ with a very fine temperature spacing $\Delta T$ between them. On my darker days I find myself imagining an experimental kit that looks eerily like a huge pallet on wheels of mortuary shelves, sliding in and out as needed by the experimenter! We bring the reactants into contact with the first heat pad, which is at a temperature $T_1 = T_0 - \Delta T$ a teeny-tiny bit cooler than $T_0$  thus reversibly drawing some heat $\Delta Q(T_1)$  off into the heat pad. Next, we bring the reactants into contact with the second heat pad at temperature $T_2 = T_0 - 2\,\Delta T$, thus reversibly drawing heat $\Delta Q(T_2)$ off into that heat pad. We keep cooling then shifting to the next lowest heat pad until we have visited all the heat pads and thus sucked all the heat off into our heat pads: see my sketch below:

Now the reactants are at absolute zero. There is no heat needed to "fill them up" to their temperature, so we can extract all the enthalpy $\Delta H$ from the reaction as useful work. Let's imagine we can put this work aside in some ultrafuturistic perfect capacitor, or some such lossless storage for the time being.
Now we must heat our reaction products back up to standard temperature, so that we know what we can get out of our reaction if the conditions do not change. So, we simply do the reverse, as sketched below:

Notice that I said that $\ce{H2O}$ soaks up less heat than the reactants. This means that, as we heat the products back up to standard temperature, we take from the heat pads less heat in warming up the water than we put into them in cooling the reactants down.
So far, so good. We have gotten all the work $\Delta H$ out into our ultracapacitor without losing any! And we're back to our beginning conditions, or so it seems! What's the catch?
The experimental apparatus that let us pull this trick off is NOT back at its beginning state. We have left heat in the heat pads. We have thus degraded them: they have warmed up ever so slightly and so cannot be used indefinitely to repeatedly do this trick. If we tried to do the trick too many times, eventually the heat pads would be at ambient temperature and would not work any more.
So we haven’t reckoned the free energy at the standard conditions, rather we have simply calculated the free energy $\Delta H$ available in the presence of our unrealistic heat sink array. To restore the system to its beginning state and calculate what work we could get if there were no heat sink array here, we must take away the nett heat flows we added to all the heat pads and send them into the outside World at temperature $T_0$. This is the only "fair" measure, because it represents something that we could do with arbitrarily large quantities of reactants without the presence of artificial, finite resources like heat pad arrays.
But the outside World at $T_0$ is warmer than any of the heat pads, so of course this heat transfer can’t happen spontaneously, simply by dent of Carnot’s statement of the second law!
We must therefore bring in a reversible heat pump and use some of our work $\Delta H$ to pump this heat into the outside world to restore standard conditions: we would connect an ideal reversible heat pump to each of the heat pads in turn and restore them to their beginning conditions, as sketched below:

This part of the work that we use to run the heat pump and restore all the heat pads, if you do all the maths, is exactly the $T\,\Delta S$ term.
The above can also cast some light on an endothermic reaction. In such a reaction, we imagine having an energy bank that we can borrow from temporarily. After we have brought the products back up to temperature, we find we have both borrowed $-\Delta H$ from the energy bank and put less heat back into the heat pads than we took from them. So heat can now flow spontaneously from the environment to the heat pads to restore their beginning state, because the heat pads are all at a lower temperature than the environment. As this heat flows, we can use a reversible heat engine to extract work from the heat flowing down the gradient. This work is, again, $-T\,\Delta S$, which is a positive work gotten from the heat flowing down the temperature gradient. The  $-T\,\Delta S$ can be so positive that we can pay back the $\Delta H$ we borrowed and have some left over. If so, we have an endothermic reaction, and a nett free energy: this energy coming from the heat flowing spontaneously inwards from the environment to fill the higher entropy products (higher than the entropy of the reactants). A good thought experiment here is to imagine towing an iceberg into warmer latitudes, and then putting it into contact with the outside world through heat engines which convert a fraction $1-T_\mathrm i/T_\mathrm o$ of the heat flowing into work.
Take heed that, in the above, I have implicitly assumed the Nernst Heat Postulate -the not quite correct third law of thermodynamics - see my answer here for more details. For the present discussion, this approximate law is well good enough.


text:
Why is Cubic closed packing called "cubic"?
How is Cubic Closed Packing "cubic" and also how is hexagonal closed packing "hexagonal"?
I have been thinking over this for a few hours but I am unable to visualise.

I understand that when placing the third layer on top of the second layer ,$2$ situations arise:
If the third layer is placed on the tetrahedral voids it forms an A-B-A-B... type structure which is called hcp (why?)
and If the third layer is placed on the octahedral voids it forms an A-B-C-A-B-C... type structure which is called ccp (why?)

Start with the hexagonal A-B-A-B structure.  It is hexagonal because, choosing one atom in a layer, it is surrounded by 6 equivalent atoms. So, you have a 6-fold symmetry axis, hence the 'hex-'. The fact that it is close-packed can be gotten at mathematically (or at least shown that it is closer-packed than other crystal structures).
Cubic close packing is, at least in most materials science and metallurgy circles, much more commonly called face centered cubic. The 'standard' unit cell is a cube of atoms with additional atoms at the center of each face.  It is somewhat harder to visualize how this is related to close-packed planes, although there are many pictures available on-line (one, chosen somewhat randomly, is from Florida State). The point is that the stacking direction of the close-packed planes are along the 111 body diagonal of the cube.
Rotate the cube so that it is standing on one point with the opposite point on the body diagonal pointing up. Your close packed planes are:
(1) the bottom point, (2) the top point, (3) the three corner atoms closest to the bottom point as well as the face atoms closest to the bottom atom, (4) the three corner atoms closest to the top atom as well as the face atoms closest to the top atom. It may take some squinting...

RE: "Cubic close packing is, at least in most materials science and metallurgy circles, much more commonly called face centered cubic." That just isn't right. There is simple cubic (SC), body centered cubic (BCC) and face center cubic (FCC). BCC and FCC are two different packing arrangements with the same packing fraction.  BCC can't be called FCC.

@MaxW - yes, there are different cubic systems. However, only one is close packed (fcc).

HUH?!? BCC has the same packing fraction as FCC $\approx 74%$

@MaxW - No, that is not correct. The packing fraction of fcc is about 0.74, bcc is only about 0.68, and simple cubic is about 0.52. This is a standard solid state physics homework problem (say, Ashcroft & Mermin chapter 4).

Dah... I confused BCC with hexagonal close packing (HCP) which does have the same packing fraction as FCC. You are right too on the point that only FCC is called cubic close packing. // Thanks for keeping me honest...


text:
Cinnamaldehyde listed as (2E)-3-phenylprop-2-enal , but is the first 2 redundant?
Wikipedia lists the molecule as being named "(2E)-3-phenylprop-2-enal", however it would appear that, as there is only 1 double bond which has a possibility of stereoisomerism, the 2 is redundant.
Would the correct IUPAC name not in fact be E-3-phenylprop-2-enal?

The preferred IUPAC name (PIN) for cinnamaldehyde is (2E)-3-phenylprop-2-enal.

With regard to numbering of locants that describe the position of stereogenic units, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-91.3 NAMING OF STEREOISOMERS
(…)
  In preferred IUPAC names, stereodescriptors are placed immediately at the front of the part of the name to which they relate. They are placed at the front of the complete name when related to the parent structure; they are cited in parentheses followed by a hyphen. When they relate to substituent groups, they are cited at the front of the corresponding prefix. They are preceded by a numerical or letter locant to describe the position of the stereogenic unit when such locants are present; general rules of numbering are applied (see P-14.4).
  (…)

Accordingly, all stereodiscriptors have to be preceded by a locant.
In case of the PIN for cinnamaldehyde, the stereodiscriptor ‘E’ is preceded by the numerical locant ‘2’.
In general nomenclature, however, locants are usually omitted when there is no ambiguity.
Nevertheless, the current IUPAC recommendations are prescriptive about when omission of locants is permissible in PINs. Locants are omitted in PINs only in the exceptional cases that are explicitly mentioned in the IUPAC recommendations. None of these cases applies to the locant of the stereodiscriptor in (2E)-3-phenylprop-2-enal.
As a rule of thumb, if any locants are essential for defining the structure, then all locants must be cited in the PIN.
Note that the current recommendations given in the 2013 Blue Book incorporate a change from earlier recommendations given in the 1993 Guide. By way of comparison, the following examples are explicitly mentioned in the 1993 Recommendations as well as in the 2013 Recommendations:

(2E)-but-2-ene (PIN, 2013 Recommendations)
(E)-but-2-ene (1993 Recommendations)

(1Z)-1,2-dibromo-1-chloro-2-iodoethene (PIN, 2013 Recommendations)
(Z)-1,2-dibromo-1-chloro-2-iodoethene (1993 Recommendations)

(2Z)-2-methylbut-2-enoic acid (PIN, 2013 Recommendations)
(Z)-2-methylbut-2-enoic acid (1993 Recommendations)

(2E)-but-2-enedioic acid (PIN, 2013 Recommendations)
(E)-but-2-enedioic acid (1993 Recommendations)

This is pretty surprising, but very informative, thanks.  I'm surprised that even something like a but-2-ene needs a number now, I wonder why they made that decision...

The 1993 IUPAC "Blue Book" is the easiest to find online thanks to ACDLabs. In this version, numbers are only used with stereochemical descriptors if there is more than one:
(E)-phenylprop-2-enal is the correct systematic name for cinnamaldehyde.
The most recent edition is the 2002 edition, which I do not have access to. The 2002 edition may contain a different recommendation.

The current IUPAC Recommendations are from 2013.


text:
Equation for reaction of tin and iodine to produce tin(IV) iodide

Tin(IV) iodide is prepared by direct combination of the elements.
Add 2.00 g of granulated tin to a solution of 6.35 g of iodine. Write an equation for this reaction.

Attempt:
1. Determine the amount of tin using moles = mass/molar mass. This yields a result of 0.017 mol $\ce{Sn}$.
2. Determine the amount of iodine using moles = mass/molar mass. This yields a result of 0.5 mol $\ce{I}$.
3. Determine the ratio of tin to iodine. Using relative amounts, for every 1 mol $\ce{Sn}$ there are 3 mol $\ce{I}$, so the ratio is 1:3. Because iodine is a diatomic molecule, I will adjust accordingly, so the ratio becomes 2 mol $\ce{Sn}$ to 3 mol $\ce{I2}$ or 2:6.
4. Construct the formula:
$$\ce{2Sn + 3I2 -> SnI4}$$
with excess tin and iodine.
Is my procedure correct?

I think you are overcomplicating it. Your equation doesn't balance so it can't be correct. Surely the correct answer is just $\ce{Sn + 2I2 -> SnI4}$.

@ToddMinehardt That was my question (I couldn't remember my account), but I don't feel like I received a satisfactory answer (I don't know if you remember, but I couldn't leave a comment for some reason).

I still don't see why you are attempting to calculate the equation using the masses given. The balanced chemical equation is independent of the amount of each substance which is added and in this case is easily determined as $\ce{Sn + 2I2 -> SnI4}$. All the masses tell you is that excess tin is being added.

@bon Okay, so the equation is Sn + 2I2 --> SnI4. But how do you conclude that there is excess tin?

As you stated, the ratio of tin to iodine added is 1:1.5 but the ratio of tin to iodine in the equation is 1:2 so there is excess tin.

The balanced chemical equation is independent of the amounts of each reactant added. In this case we can easily deduce the equation to be: $$\ce{Sn + 2I2 -> SnI4}$$
What the given masses tell us is that there is excess tin being added. We can see this because the $\ce{Sn:I2}$ ratio added is $1:1.5$ whereas the stoichiometric ratio is $1:2$ and so there is less iodine than is required to completely react with the added tin - or in other words there is an excess of tin.

The procedure is correct, be careful with this:

The amount of iodine is $\pu{0.050 mol}$, but the molecular amount $\ce{I2}$ is $\pu{0.025 mol}$.

For the ratio use the amount of point 1 and 2:

$$ \frac{\text{Amount of Sn}}{\text{Amount of }\ce{I2}} = \frac{\pu{0.017 mol}}{\pu{0.025 mol}} \approx \frac{2}{3}$$

Don't confuse! This equation is not well balanced. Stoichiometry is inherent of the reaction and is based on the mass conservation law; amount of substance tells you which reactant is limiting and which one is in excess.

$$
\ce{\underset{\pu{0.017 mol}}{Sn} + \underset{\pu{0.025 mol}}{2I2} <=> \underset{\pu{? mol}}{SnI4}}
$$
Now with this you can tell which one is on excess and for how much.


text:
Why is C=C launching electrophilic attack on Bromine instead of the other way round?
So during my lecture on reaction of Alkenes with Halogens, my lecturer used the example of $\ce{Br2}$ gas reacting with $\ce{CH2CH2}$. He mentioned that the $\ce{C=C}$ launches an electrophilic attack on the electropositive Bromine atom (the one that is nearer to $\ce{CH2CH2}$) and I'm wondering why is the $\ce{C=C}$ launching the electrophilic attack? Shouldn't it be done by the Bromine gas since the $\delta^+ \ce{Br}$ is electrophilic and needs the electrons so it attacks the double bond which is most electron dense?

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That's seems wrong. The bromine molecule is the electrophile and the double bond is the nucleophile and I would normally describe it as an electrophilic attack by the bromine on the double bond.

If it just happened once, I am guessing it was a slip of the tongue by your lecturer, it happens... I have said much stupider stuff than that in my lessons :/ In any case, as bon said, $\ce{Br_2}$ is the electrophile and $\ce{C2H4}$ the nucleophile.


text:
Combining amphetamine and MAOIs
Could combining amphetamine and a (possibly selective) MAOI (MonoAmine Oxidase Inhibitor) increase the amphetamines potency without serious toxic side effects? If so by what factor, and would this be profitable for clandestine amphetamine producers to lower production costs?
I guess this is most at home at the Chemistry Stackexchange. I'm looking into this matter from a medicinal chemist point of view.


text:
Does Cute Poison actually work?
For those of you that have watched Season 1 of Prison Break (TV Series), in the "Cute Poison" Episode Micheal Scofield combined $\ce{CuSO4}$ (Copper Sulfate) and $\ce{H3PO4}$ (Phosphoric Acid) to weaken the metal. Is it true?

Perhaps you could give some more detail about the situation. What was the metal? What happened when the solution was added to it?

Copper sulfate is an ingredient in root killer, flushed down toilets when sewer lines are too close to large trees. If the prison was older and had large establish trees growing near outgoing sewer lines the might have need for copper sulfate. The chemical would probably be kept in a maintenance or cleaning cabinet.

The Wikipedia page on "Cute Poison" episode says that it wasn't anhydrous copper (II) phosphate, but Gypsum.
I haven't seen the episode, and the Wikipedia page's reaction seems to be the backwards reaction of yours. But I imagine yours is the one the pro/antagonist is gonna use to escape the prison; as

Gypsum is commonly found and accessible:

Gypsum is a soft sulfate mineral composed of calcium sulfate dihydrate, with the chemical formula $\ce{CaSO4·2H2O}$. It can be used as a fertilizer, is the main constituent in many forms of plaster and in blackboard chalk, and is widely mined. A massive fine-grained white or lightly tinted variety of gypsum, called alabaster, has been used for sculpture by many cultures including Ancient Egypt, Mesopotamia, Ancient Rome, Byzantine empire and the Nottingham alabasters of medieval England.

Calcium phosphate's solubility will decrease with temperature increase and it will precipitate, leaving you with a solution of sulfuric acid.

It's possible for a double displacement reaction to occur in aqueous medium (with a spark):
$$\ce{2H3PO4(aq) + 3CaSO4(aq)·2H2O(l) \leftrightharpoons 3H2SO4(aq) + Ca3(PO4)2(aq) + 6H2O(l)}$$
$\ce{H2SO4}$ is sulfuric acid. It's a very strong acid in water, a diprotic acid. Its $p {\rm K_a}$s are −3 and 1.99 according to Wikipedia.

Its corrosiveness on other materials, like metals, living tissues or even stones, can be mainly ascribed to its strong acidic nature and, if concentrated, strong dehydrating and oxidizing properties.

Later in the same article, you can read how it reacts with different material. It can weaken metal pretty fast and easily, as 3 moles of it will be present in any 7 moles of products, which is high above the dilute concentration (arbitrarily) defined in the Wikipedia page.
Copper sulfate also reacts with phosphoric acid and forms sulfuric acid, but it's hardly accessible in prison, I reckon.


text:
Finding the density of nitrogen at 0 degrees Celsius and 1 atmosphere

The density of nitrogen at $\pu{0 ^\circ C}$ and $\pu{1 atm}$ is most nearly equal to which of the following quantities?
(a) $\pu{0.001 g/L}$; (b) $\pu{0.01 g/L}$; (c) $\pu{0.1 g/L}$;
(d) $\pu{1 g/L}$; (e) $\pu{10 g/L}$.

The answer is (d), $\pu{1 g L-1}$.
I think this is how you do it, but please correct me if I'm missing something. I didn't end up using the standard molar volume, so this might be incorrect.
$$pV = nRT$$
We know the pressure is $\pu{1 atm}$, the temperature is about $\pu{273 K}$ and $R = 0.08$.
Solving for $\frac {n}{V}$, we find $\frac {n}{V} = \frac {1}{22}$.  Now, this is in moles per litre, so by multiplying by the molar mass of nitrogen (approximately $\pu{28 g mol-1}$), we find that we have about $\pu{1 g}$ per litre as the density.

"… the following quantities" – These are not "quantities", they are "quantity values",
"values of a quantity", or simply "values".

Your approach is correct. But use units whenever needed and follow correct significant figures.

@FadedGiant  According to my reading of IUPAC's defintion of "quantity" (https://goldbook.iupac.org/terms/view/Q04982), the HW question's use of the term is acceptable:  "The term quantity may refer to a quantity in a general sense, for example length, mass, or to a particular quantity, for example length of a particular rod, mass of a specified object".  Hence quantity could also refer to the density of a particular sample of gas, which is precisely what the HW question is listing.

Also, "quantity value" and "value" are problematic, since if you take quantity in the first sense IUPAC uses then for, say, "0.01 g/L" the quantity would be "density", the value would be "0.01", and the units would be "g/L".  See, for instance, how NIST distinguishes quantity, value, and units here: https://webbook.nist.gov/cgi/cbook.cgi?ID=C24203369&Mask=40 .  Alas, this is inconsistent with how NIST uses "value of quant." here, which seems along the lines you're suggesting:  https://www.nist.gov/pml/special-publication-811/nist-guide-si-chapter-7-rules-and-style-conventions-expressing-values

Using the ideal gas law
\begin{align}
  pV &= nRT, &
  \frac nV &= \frac{p}{RT},
\end{align}
where
\begin{align}
  T &= \pu{273 K},\\
  p &= \pu{1 atm},\\
  R &= \pu{0.0821 L atm mol-1 K-1},\\
  M(\ce{N2}) &= \pu{28.0 g mol-1},
\end{align}
we get
\begin{align}
  \frac{n}{V} &=
    \frac{\pu{1 atm}}{\pu{0.0821 L atm//mol K} \times\pu{273 K}},\\
  \frac{n}{V} &= \pu{0.0446 mol//L},
\end{align}
and finally
$$
\rho(\ce{N2})
  = \pu{0.0446  mol//L} \times \pu{28  g//mol}
  = \pu{1.2  g//L}.
$$
Therefore answer (d) is correct.


text:
Regarding the mechanism for biological activity of tyramine, amphetamine and ephedrine
Graham Patrick explains the mechanism for the activity of tyramine, amphetamine and ephedrine in his book Introduction to Medicinal Chemistry as follows:

Some amines such as tyramine , amphetamine , and ephedrine closely resemble noradrenaline in structure and are transported into the nerve cell by noradrenaline’s transport proteins.
  Once in the cell, they are taken up into the vesicles. Because these
  amines are competing with noradrenaline for transport proteins,
  noradrenaline is more slowly reabsorbed into the nerve cells.
  Moreover, as the foreign amines are transported into the nerve cell,
  noradrenaline is transported out by those same transport proteins.
  Both of these facts mean that more noradrenaline is available to
  interact with its receptors. Therefore, amphetamines and similar
  amines have an indirect agonist effect on the adrenergic system.

I have two question concerning this explanation:

When these foreign amines have been taken up into the vesicles how do they proceed? Do they re-enter the synaptic cleft by exocytosis, so that they can re-exercise their negative influence on noradrenaline re-uptake? How are they eventually cleared from the synaptic area?
Why do the noradrenaline re-uptake proteins transport noradrenaline out of the axonal terminal under the influence of these foreign amines?

Well, it's probably on-topic but imo it would be better to ask that on Biology SE

Question: When these foreign amines have been taken up into the vesicles how do they proceed?...Do they re-enter the synaptic cleft by exocytosis...How are they eventually cleared from the synaptic area?
Here is the mode of action of these agents (indirectly acting sympathomimetics):

The most
important drugs in the indirectly acting sympathomimetic amine
category are tyramine, amphetamine and ephedrine, which are
structurally related to noradrenaline.
Amphetamine enters the nerve terminal via the noradrenaline
transporter (NET) and enters synaptic vesicles via the vesicular
monoamine transporter (VMAT), in exchange for noradrenaline
(NA), which accumulates in the cytosol.
Some of the NA is degraded
by monoamine oxidase (MAO) within the nerve terminal and some escapes,
in exchange for amphetamine via the noradrenaline transporter, to act
on postsynaptic receptors. Amphetamine also reduces NA reuptake via
the transporter, so enhancing the action of the released NA.

Exocytosis is not involved in the release process, so their actions do not require the presence of Ca2+. They are not completely specific in their actions, and act partly by a direct effect on adrenoceptors, partly by inhibiting NET (thereby enhancing the effect of the released noradrenaline) and partly by inhibiting MAO.
These agents are cleared by metabolism via oxidative deamination with monoamine oxidase(MAO) enzyme, however  are poor substrates for MAO and, thus, show a prolonged duration of action.
Amphetamine is an example of a drug metabolized through oxidative deamination:

Question: Why do the noradrenaline re-uptake proteins transport noradrenaline out of the axonal terminal under the influence of these foreign amines?
I am not really sure about the “why” part but instead the “how” part AFAICT, this is more of a displacement mechanism:
These drugs, do block reuptake 1(NET) releasing stored transmitter from noradrenergic nerve endings and they displace vesicular norepinephrine into the cytoplasm. The norepinephrine is then transported out of the nerve terminal by the reverse action of reuptake 1 and activates adrenoceptors.
So essentially the drugs increase the availability of noradrenaline to stimulate adrenergic receptors by:
i. releasing or displacing NE (noradrenaline) from sympathetic nerve endings,
ii. blocking the transport of NE into sympathetic neurons or
iii. blocking the metabolizing enzymes, MAO or COMT.
How this helps
References

Rang and Dale’s Pharmacology 8th Edition. Section 2: Chemical Mediators

Medical Pharmacology at a Glance M. JNeal: Drugs acting on the sympathetic system

Lippincott Illustrated Reviews: Pharmacology 6th Edition: Adrenergic Agonists

Goodman and Gilman Manual of Pharmacology and Therapeutics: Agents Acting at the Neuromuscular Junction and Autonomic Ganglia

Principles of Clinical Pharmacology 2nd Ed Atkinson etal
Basic and Clinical Pharmacology B. Katsung et al.


text:
Why the bond angle of PH3 is lesser that that of PF3?
We can explain why the bond angle of $\ce{NF3}$ (102°29') is lesser than $\ce{NH3}$ (107°48') by the VSEPR theory, since lone pair lone pair repulsion is greater than lone pair bond pair repulsion. Then for $\ce{PH3}$ and $\ce{PF3}$, also, it is expected that the bond angle of $\ce{PF3}$ will be smaller. But I found that the thing is just reverse. Why is this so? Is it an exception?

I don't see how VSEPR can explain the bond angles in $\ce{NF3}$ and $\ce{NH3}$. Take a look at this question for the correct reason.

related: Bent's rule. See http://chemistry.stackexchange.com/q/15671/4945 and http://chemistry.stackexchange.com/q/15620/4945

The answer can not be determined using VSEPR theory. The number of lone pairs in both compounds is the same. The answer lies in electronegativity.

In $\ce{NH3}$ nitrogen is more electronegative than hydrogen and therefore it will pull the electrons towards it. Therefore the electrons will be closer to each other hence more repulsion.
But in $\ce{NF3}$ fluorine will pull electrons towards itself therefore the electrons are further comparatively and hence repulsion will be less and so is the bond angle.

But in case of $\ce{PF3}$ and $\ce{PH3}$  back bonding is possible in $\ce{PF3}$ therefore it has a larger bond angle .

What type of backbonding are you envisaging?

Fluorine is also larger than hydrogen so sterics have a contribution.

Can you please explain why back bonding makes the repulsion increase in $PH_3$?

Actually it forms a partial double bond between P and all the three flourine .

Phosphorus does not access d-orbitals. $-1$

The reason it doesnt access d orbitals is because its already having full octet so as the F orbital hence therefore d orbitals are not needed to be accessed Sir @Jan ?

VSEPR works by accident in the cases of $\ce{NH3}$ and $\ce{NF3}$. In reality, there are much more things to consider as shown in this answer. All four compounds should have bond angles of $90^\circ$ if there were no other effects present.
For both nitrogen compounds, the effects are the short $\ce{N-X}$ bonds which lead to steric clash of the three substituents. Thus, the bond angle is broadened from the original, hypothetical $90^\circ$ to whereever it may seem comfortable to the molecule. Since fluorines are larger atoms than hydrogens are, the $\ce{N-F}$ bond is longer than the $\ce{N-H}$ bond ($137~\mathrm{pm}$ versus $102~\mathrm{pm}$) and therefore the atoms can agree on a smaller bond angle (less crowding).
For phosphorous, we can initially assume that the bond angle of $\ce{PH3}$ is close enough to $90^\circ$ for no hybridisation to be necessary (semi-proven by this answer of Martin) and likewise for $\ce{PF3}$. The bond lengths are $142~\mathrm{pm}$ and $156~\mathrm{pm}$, respectively. The difference in bond lengths is only half of that of the nitrogen compounds ($14$ versus $35~\mathrm{pm}$). Therefore, we can assume the bond length to be mainly dictated by the central phosphorous atom, while the bond angles must be dictated more by the outer atoms. Since again fluorine is larger than hydrogen, this time the better way to stabilise the molecule is not to lengthen the bonds but to expand the angle.

Whats the reason to conclude "dictated by the central phosphorous atom, while the bond angles must be dictated more by the outer atoms. " Isnt bond length greater due to size of F greater than H? P remains same for both

In PH3 and PF3 bond angle of PF3 is greater as in PF3 back bonding takes place.
Back bonding is possible in PF3 as P has vacant d orbital(as its atomic no. is 15 therefor its electronic configuration is 1s(2e) 2s(2e) 2p(6e) 3s(2e) 3p(3e)
this shows it has vacant 3d orbital which can perform back bonding with fluorine as fluorine has pair of electrons.
as backbonding takes place the lonepair of P will be used in doublebond formation with F. and obviously if their is no lonepair then their is no one to repell. there for more bond angle
also in PH3 no backbonding can take place as hydrogen has no pair ofelectron(e) for backbonding to take place.
May this solution help you.:)

That's not true - significant usage of d-orbitals in p-block element's compounds is outdated theory

lol  this is the logic outdated theory all concepts of chemistry are years old and you creating your own theories I would say go for it

Chemistry marches on see for ex. http://chemistry.stackexchange.com/a/18544/9961 for newer description, as this was experimentally disproved.


text:
Uniform circular motion of electrons?
Can we use the uniform circular motion formulae to calculate the velocity  of an electron along with the formula of radius of nth Bohr orbit? And using that can we calculate the kinetic energy of an electron?

No - get circular motion for atomic orbitals out of your head out of your head right now.  Bohr's model is an (early) stepping stone from classical to quantum, and persisting in thinking in that fashion will lead you to horribly wrong places.

Yes Sir you are right ..!

http://physics.stackexchange.com/questions/20187/how-fast-do-electrons-travel-in-an-atomic-orbital

http://chemistry.stackexchange.com/questions/26501/how-fast-do-electrons-move-around-the-nucleus

Well yes you can if you assume the Bohr model is correct. However in reality, it is a little bit more complicated than that. In reality everything have both particle and wave like properties, which builds the foundation of quantum mechanics. The lighter the object, the more wave like properties it exhibits and the heavier the object, the more particle like properties it exhibits. So a ball actually has wave light properties but it is too heavy for it to be notices. Meanwhile a light wave actually has particle properties, however it is too light to be noticed. Electrons on the other hand turn out to a mass between these two extremes, hence both its particle and wave like properties are able to be noticed.Hence it is important to stop thinking electrons as small point particles.
Another important concept is the Heisenberg Uncertainty Principle which is the following formula: $$\Delta x \Delta p \geq \hbar$$ This means that the uncertainty of an object's position multiplied by the uncertainity in an object's momentum at best can only be equal to the constant value: $\hbar$ which equals $1.054*10^{-31}$ This equation can be applied to electrons. Hence, the important thing that this equation tells us about electrons in an atom is that rather electrons moving in a predictable circular orbit, they actually travel in any path that they want and we actually can't precisely determine position of the electron in an atom. Hence the quantum mechanical model of the atom actually consists of orbitals, rather than orbits.
Orbitals is the area which an electron is in 90% of the time. There are 4 main type of oribitals, s, p, d and f shown below.

So, finding the velocity of an electron is little more complicated than using simple classical mechanic equations. However its velocity can be determined in simplified conditions using quantum mechanics.