text:
How do I know crystal structures from formula?

Give an explanation why $\ce{MgF2}$ and $\ce{CaF2}$ adopt different structure types.

For example, the structure of calcium fluoride is 'fluorite' and the structure of magnesium fluoride is 'rutile'. I am struggling, just from looking at the formulae,  on how I determine this?

You don't. That being said, a post factum explanation is somewhat possible.

Just to note that it's a bit misleading to say that calcium fluoride adopts the fluorite structure. Calcium fluoride is fluorite, and the structure was named after the mineral.

I'll just point out the direction in which you could look for an answer.
In general case, it should be almost impossible to determine the crystal structure by looking at the chemical composition of a substance.
You are, however, asked to explain why different fluorides adopt different structure types, which is a different task. Perhaps, you should take into account the difference (sorry) in the ionic radii of Mg and Ca. Some structures prefer "fat" cations, some - "lean" ones. Most of the time, it's just plain geometry combined with experimental observations. I know the explanation is primitive, and ions are not just some rigid spheres, but the "ionic radii" approach works in many cases.
For many crystal structures there are ranges of ionic radii ratios in which the particular structure is more or less stable. Some of these ratios are so important that they have proper names, such as Goldschmidt tolerance factor for perovskites. Returning to your example, this page lists the cation/anion radii ratios for fluorite (0.73 or above) and rutile (between 0.41 and 0.73). You can check these ratios against the dimensions of ions in magnesium and calcium fluorites.
For the reference - this Shannon table hosted by Imperial College London is an excellent source of the values of ionic radii in oxides and halides. Just one more thing: the reference ratios are likely to be designed to work with so-called Shannon Ionic Radii, not the less-common Shannon Crystal Radii.

Not really a complete answer, rather an addition to the given answer (which I think you should accept). As of now, it's possible to predict existence of certain compounds, mostly binary and ternary inorganic ionic solids, at the given temperature and pressure, as well as their crystal structure.
There are many computer programs, among which USPEX is arguably the most feature-rich and promoted. It uses Oganov-Glass evolutionary algorithm and works best for network solids (including particles and clusters). Fairly recently there also were some advancements for predicting crystal structures of small organic molecules.

Suggestion:  The blind tests moderated by the CCDC you mention advanced further than to the fourth round your link points to (cf. https://www.ccdc.cam.ac.uk/Community/initiatives/cspblindtests/).  Which is why I'm going to replace the link set by you by the one pointing to the sixth one  -- equally published as open access article on Acta Cryst B.

@Buttonwood Good catch, thank you for the edit!

You knew better about USPEX, and me a bit about the contests.  It's one of the enjoyable ideas of SE, bringing the knowledge of multiple people into one forum, an exchange.

You can treat it as a "generic problem solving" type of question. @voffch gave some specific ways to do this.
You're asked why they form different structures, but don't have to say anything about their actual structures. This question sounds like something from an introductory undergrad general/inorganic course or high school honors gen chem course. Typical material for such a course discusses ionic structures in terms of sphere packing/ionic radius (keywords: fcc, bcc, hcp) and covers ionic radius in terms of electronegativity and shielding.
You're given:

Mg(II) and Ca(II) are different cations
Their fluorides form different structures (e.g. the anion is the same)
(you should know) Mg and Ca pretty much always are in +2

So the question is implicitly "Of the differences between Mg and Ca cations, which differences make their fluorides have different structures?."
The specific mention of F suggests there's something special about it too. F is pretty "extreme" all the way in its corner so it was probably chosen as a "superlative," e.g. "most electronegative," "smallest, "biggest occupational hazard" or something like that.
At this point, you can just compile a list of all the things you've learned about metal/halide ions over the relevant time period and pick ones that affect geometry. Also consider what you've been taught about crystal structures.
It could play out like this:

The course covered elemental trends in ionic radius, and reasons why (electronegativity, shielding)
The course covered simple ionic structures (sphere packing)
Sphere packing is only affected by radius
Mg, Ca have different radii. Both are "big"
F is "small"
The reason the structures are different is because Mg, Ca have different radii
(Only if the course covered this) Mg(II) is [bigger/smaller] than Ca(II) because [electronegativity, orbitals, etc.]

By way of explanation, you can explain radius trends + draw a sketch of the different packings, approximately to scale.
If your course covered different material-- Compare Mg, Ca in terms of the stuff your course covered.


text:
What are Technetium's physical stress characteristics?
I've looked at periodic and data tables from several sources, but none give mechanical / engineering property data on Technetium.  Why is this?

Tc is unstable ans is certainly not a constructing material, but it doesn't mean there is no data for it. What mechanical properties are you interested in, exactly?

Elastic moduli of polycrystal Technetium: https://www.sciencedirect.com/science/article/pii/0022508870900391

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text:
Simple soda–lime decarboxylation
This is my homework question I tried it hard the question is as follow:

I attempted it as follow:

According to my understanding as acid base reaction are fastest the carbanion will gain a $\ce{H+}$. From the water formed and will the product as I made, but the answer is completely different:

Please guide me where I am wrong.

You are wrong in thinking that the anion of the chloroalkene will simply pick up a proton. These reactions are run at a high temperature ao there is plenty of energy around for forming unstable species

You have a leaving group beta to anion. You would have no doubt about what will happen if it were on an alkane.

You are correct in the first step - the decarboxylation gives the anion of the chloroalkene.
It is important to remember that such soda lime reactions are run at high temperature and CaO is a powerful drying agent, so there will be no water present to quench the anion. What happens next would be obvious if you were dealing with an alkane; there is a leaving group beta to an anion so there would be an elimination with loss of Cl-. This is what happens; Cl- leaves to give the triple bond (again remember that this is a high temperature reaction so plenty of energy for forming unstable species). This is clearly unstable within a [6] ring so it dimerises by a 2+2 process to give the product drawn.

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text:
Properties of Water
Water is a substance characterised by properties such as high boiling point and its ability to act as a universal solvent. So why do we say that “the smallest unit of water is a water molecule”? Single molecules cannot possess such properties. Am I missing something here?

You need a lot more water molecules until you have water. A drop of 10,000 will show a huge difference e.g. in vapour pressure. Even a billion would still be quite strongly influenced by it's surface, and not show the ideal bulk water properties.

Not all water is defined the same. "Water" defines a molecule - $H_2O$ - but in our everyday parlance also a bulk substance which is a collection of such molecules. Perhaps it would help to say "the smallest unit of a volume of water molecules is a single water molecule". Not very enlightening perhaps.

Water is a substance. Would it perhaps be more correct to say that a water molecule is the smallest unit that defines the properties of the substance water?

@J.Smith I don't think that's going to solve your question, because you still need to define "properties". The properties in question can depend on how many water molecules you have.

The problem here is that conventional usage of "the smallest unit of water is a water molecule" leaves out a lot of qualifications. The sentence ignores the qualification that some properties are intrinsic to the molecule but others only appear when there are a lot of molecules together. Every molecule in bulk water has the same mass, geometry and electric dipole but many of the bulk properties only appear when there are millions of the molecules (since they depend on how the molecules interact).

No, you aren't missing anything. Indeed, one molecule doesn't have any boiling point, neither high nor low. In fact, you've just stumbled upon one of the holes deliberately left in the early definitions. It is a very common thing in the textbooks on natural sciences: first we give the kids a simplified definition, so they don't have to read a ton of prerequisites before they start reading the book, and then they kinda already know what a molecule is, so why bother returning to it.
If we were to invent a more-or-less realistic definition of a molecule in simple terms, it would go like this: "the smallest unit of matter such that the matter retains its properties when split to these units and then allowed to settle down".
Indeed, say we have a glass of water, and we magically split it to individual molecules and then step aside. The molecules will assemble back into liquid water; probably it will be all over the place, but who cares, it is liquid water all the same, with unchanged boiling point and stuff.
Now what if we magically split a sample of some matter to smaller units, like atoms? Well, as soon as we step aside, those will assemble back to molecules, but not necessarily the same molecules. They may well be different, and so will be the physical properties of the sample.
This definition is not without its flaws either, but that's another story.

Many molecules won't survive being taken into vacuum. ;-)  Imo trying to define real world stuff like molecules in terms of abstract entities like "smallest unit of matter" is generally a bit of a philosophical exercise in futility. You get to statements which are often flawed, true in best case, but all pretty useless in gaining further understanding.

The word "water" can be used to mean different things. In everyday language we usually use "water" to mean a large collection of water molecules, which we associate with such phenomena as boiling, freezing, solubilization of salts, etc. We can conjure a different definition of "water", namely, a single water molecule, with molecular formula $\ce{H2O}$. These are not the same thing. In a way, one is a subset of the other.
The solvation capabilities and boiling point of water are thermodynamic properties, usually (but not only) defined for macroscopic ("bulk") quantities of water under equilibrium conditions. The boiling point defines a temperature at which a transition occurs between the liquid and the gas states of a substance. The boiling point is a property of the ensemble, the entire collection of molecules. It is important to emphasize that when chemists discuss properties such as boiling points they are pretty strict. Such properties depend on the pressure, amongst other things. Even the geometry of the water volume can influence the experimentally observed boiling point. Chemical thermodynamic properties belong by definition to a collection of molecules under well defined equilibrium (constant) conditions. Thermodynamicists observed that properties like solubility and phase transition temperatures reach limiting values provided you have sufficient molecules, that is, above a limit dealing with bulk matter. In fact, a thermodynamic limit can be defined by observing how properties change with the amount of a substance.
It doesn't make much sense to attribute a bulk property such as a boiling point to a single water molecule, but you can say that a single molecule of water has intrinsic molecular properties that lead reproducibly, when the molecule aggregates with others of its kind, to specific macroscopic properties.
On a historical note, properties such as the boiling point were defined quantitatively as the foundations of thermodynamics were layed down and a formal molecular theory of matter began to be developed, the latter involving amongst other things statement of Avogadro's hypothesis and, later, advances in spectroscopy.

Thanks for the response Try Hard. Does that mean that we cannot define a single molecule of water to be the substance “water”? Since water is characterised by the aforementioned thermodynamic properties.

Water is divisible down to the molecular level. It exhibits properties (thermodynamic or possibly other) that may depend on the number of such molecules and their interactions.

Then let me ask this question: What properties characterise a substance? Are they thermodynamic properties or something else? What am I missing here?

The word "substance" typically implies a macroscopic ensemble, a large collection of molecules. A thermodynamic description might be adequate, assuming the substance is in equilibrium, which for a pure homogeneous liquid such as water would mean stating what temperature and pressure it is at. Then you can proceed to list all of the properties you might be interested, extensive (such as volume, enthalpy of formation, etc) or intensive (molar properties such as molar heat capacities, thermal expansion coefficient, etc).

With thermodynamic properties you would not be speaking of one water molecule, but of the properties of the ensemble of molecules. But of course properties can also include spectroscopic properties that are impossible to explain without an atomic/molecular/QM description of matter. When you start dealing with certain spectroscopic properties, you move away from the aggregate properties and shift into the realm of the molecule. The aggregate properties become "perturbations" on molecular properties that are similar in an ensemble and in a single molecule. Often, but not always.

the smallest unit of water is a water molecule

To say that a molecule is the smallest unit of a substance that shares its properties is not a good definition, and has the same problems as defining an atom as the smallest unit that has all the properties of a given element.
For the definition of an atom, you can salvage that definition by specifying exactly what properties are retained. For example, the electronic transitions involving inner electrons will be the same for an isolate atom and for different allotropes of the element (and you can use that to identify atoms from their spectrum even if they are extraterrestrial).
Similarly, you can specify a small set of properties that a bulk sample of a pure molecular substance has in common with a single one of the molecules it contains: Composition, bond lengths and angles (with some minor variations), electronic spectra (more or less).
Of course, if your bulk sample is in the gas phase, there will be more properties that are the same, so pure water in the gas phase or water in air will share most properties with a single isolated water molecule.


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Does 1 mole have the same molar mass?
I have tried to look it up, and find answers, it really intrigues me!

One mole of nitrogen does not have the same mass as one mole of oxygen...

The question doesn't make much sense. It's like asking "Does 1 metre have the same colour?"

A mole of any substance is defined as Avogadro's number of molecules of that substance. So every mole has the same number of molecules. Since the masses of molecules differ, the mass of a mole of any substance will be dependent on the mass of one molecule of that substance. That is,
(Mass of one mole) = (Avogadro's number) x (the mass of one molecule).


text:
How does di-isobutylaluminium hydride ( DiBAl-H ) reduces carboxylic acid
$\text{DiBAl-H}$ reduces by electrophilic attack and provides hydrogen by $\text{H}^-$ transfer. Unlike $\text{LiAlH}_4$ it does not directly provide $\text{H}^-$ ion. So it should not undergo acid base reaction with carboxylic acid. But my friend argued that it undergoes acid base reaction. I can't find any reaction mechanism for reduction of carboxylic acid with $\text{DiBAl-H}$, can someone propose a mechanism for it?

Reduction of esters of carboxylic acids directly to aldehydes is of great synthetic interest. Zakharkin and Khorlina in 1962 (Ref.1) has shown that, diisobutylaluminum hydride (DIBAL-H) could be very useful in reducing esters to aldehyde in high yields at low temperatures (about $\pu{-70 ^oC}$). They also found that solvents affect reduction significantly. For example, the reduction reaction in toluene or hexane proceeds to give corresponding aldehydes in yields l0-15% more than those in ether as the solvent. Previously, in 1959, Miller et al. (Ref.2) found that DIBAL-H is a more selective reagent than lithium aluminum hydride in the reduction of nitriles to aldehydes, and it can also be used to reduce benzoic acid to give 72% yields of benzyl alcohol (they also showed esters can be reduced to corresponding alcohols). The conditions and yields of these references are summarized in this page. In Ref.2, during acid reduction, Miller et al. have used 3 equivalents of DIBAL-H and observed that a gas (hydrogen) evaluation upon the addition of the hydride to benzoic acid until one mole of the reducing agent had reacted. Thereafter, the addition of more DIBAL-H gave no gas. Thus, the first reaction is, presumably an acid-base reaction. However, DIBAL-H here should be acting as a Lewis-acid instead of a base, as illustrated in scheme 1:

The ist mono-adduct, then further reacted with second DIBAL-H to give di-adduct, which upon acid hydrolysis would give expected aldehyde (Scheme 2):

References:

L. I. Zakharkin, I. M. Khorlina, “Reduction of esters of carboxylic acids directly into aldehydes with diisobutylaluminum hydride,” Tetrahedron Letters 1962, (14), 619-620 (https://doi.org/10.1016/S0040-4039(00)70918-X).
A. E. G. Miller, J. W. Biss, and L. H. Schwartzman, “Reductions with Dialkylaluminum Hydrides,” J. Org. Chem. 1959, 24(5), 627–630 (DOI: 10.1021/jo01087a013).

At lower temperatures (around -78oC), carboxylic acids (with 2 equiv. of DIBAL-H) is reduced by DIBAL-H

What does aqueous base have to do with the question? Two equivalents of DIBAL-H may be required. One as a base; one as a reductant. The bis-aluminum complex (lower right corner) must be stable prior to work up at low temperature.

@Sarahjane this does not answer the initial part of my question. Further this mechanism does not seem to be correct. 1 eq. of DiBAl-H can reduce ester, here's the mechanism (
https://i.stack.imgur.com/1Dwzk.jpg ). So why does carboxylate Ion require 2 eq. of DiBAl-H inspite being more electron rich?


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Combination of several unit symbols when finding new pressure from ideal gas law

As a car is driven, the tires heat up and the volume and the pressure change. Calculate the new pressure of a 35.0 L tire initially at 235 kPa that expands to 36.5 L when it heats up from 15.0 °C to 43.0 °C. (4 marks)

$P_1 = \pu{235 kPa}$
$P_2 = ?$
$V_1 = \pu{35.0 L}$
$V_2 = \pu{36.5 L}$
$T_1 = \pu{15.0 °C} = 15.0 + 273 = \pu{288 K}$
$T_2 = \pu{43.0 °C} = 43.0 + 273 = \pu{316 K}$
The number of moles of gas in the balloon is not changing, so $n_1 = n_2$.
$$P_1V_1/n_1T_1 = P_2V_2/n_2T_2$$
(since $n_1 = n_2$ we can cancel them from both sides)
$$P_1V_1/T_1 = P_2V_2/T_2$$
(cross multuple to allow for isolation of $P_2$)
$$P_1V_1T_2 = P_2V_2T_1$$
(divide both sides by $V_2T_1$ to isolate $P_2$)
$$P_2 = P_1V_1T_2/V_2T_1$$
(now plug in values)
$$\begin{align}
P_2 &= (\pu{235 kPa})(\pu{35.0 L})(\pu{316 K})/(\pu{36.5 L})(\pu{288 K})\\
  &= 2 599 100 / 10 512 \\
  &= \pu{247 kPa}
\end{align}$$
The new pressure of the tire would be 247 kPa.
I like keeping track of the units throughout solving, but not sure how to do that with three units...
Can I write $P_2 = 2599100\cdot\pu{kPa/L/K}/10512\cdot\pu{L/K}$ (L and K cancel out and leave me with kPa), or is it recommended not to record the units in that step and only write the final unit with the final answer?

Section 5.1 Unit symbols in SI Brochure [1, p. 130] has an answer:

In forming products and quotients of unit symbols the normal rules of algebraic multiplication or division apply. Multiplication must be indicated by a space or a half-high (centred) dot (⋅), since otherwise some prefixes could be misinterpreted as a unit symbol. Division is indicated by a horizontal line, by a solidus (oblique stroke, /) or by negative exponents. When several unit symbols are combined, care should be
  taken to avoid ambiguities, for example by using brackets or negative exponents. A solidus must not be used more than once in a given expression without brackets to remove ambiguities.
$\pu{m kg/(s3 A)}$,
  or $\pu{m kg s–3 A–1}$,
but not $\pu{m kg/s3/A}$,
nor $\pu{m kg/s3 A}$.

I personally prefer to use powers of the units — subjectively, this makes the expressions clearer and also reserves a slash "/" for separating the symbol of the quantity from the units on plots and in the table headers. However, in your case you don't need any of that. According to the ideal gas law
$$pV = nRT$$
and
$$n_iR = \text{const} \quad \implies \quad \frac{p_iV_i}{T_i} = \text{const},$$
so you have a simple proportion
$$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \quad \implies \quad p_2 = \frac{V_1T_2}{V_2T_1}p_1.$$
Obviously, there is no need to juggle around with the units as all of them except for the pressure units are cancelled out (note how I also grouped homogeneous quantities within the fraction one under another, and wrote pressure separately for clarity):
$$p_2 = \frac{V_1T_2}{V_2T_1}p_1 = \frac{\pu{35.0 L}\cdot\pu{316 K}}{\pu{36.5 L}\cdot\pu{288 K}}\cdot\pu{235 kPa} = \pu{247 kPa}.$$
And yes, generally it is considered a good practice not to do any intermediate calculations unless you are explicitly asked so, and solve the problem algebraically in the first place.
References

BIPM. Le Système International d’unitès / The International System of Units (“The SI Brochure”), 8th ed.; Bureau international des poids et mesures: Sèvres, 2006.


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How could the ideal gas law be discovered from experiments on real gases?
The gas laws, namely Boyle's Law, Charles' Law, Avogadro's Law and Gay-Lussac's Law, are all experimental laws. Combining these laws, we get the ideal gas law $pV=nRT$. Also, "real life" gases do not exactly follow this law, so there are more laws for "real life" gases: van der Waals' law, Dieterici equation, etc., which approximately describe these laws within certain boundaries of the gas parameters: pressure $p$, volume $V$ and temperature $T$.
But there seems to be an apparent logical flaw: the ideal gas law $pV=nRT$  was found by experimenting on "real life" gases, but these "real life" gases do not follow the ideal gas law. How could this be the case?

The simple answer is: because while real matter may not be perfectly ideal, it comes pretty dang close.

The same way Newton and Copernicus discovered simple laws of gravitation, even though real gravity is very slightly different (general relativity, and we still don't have a theory of quantum gravity).  The difference is negligible for systems they could observe, so their model did fit the data.  For gases it's easier to create systems there the difference is non-negligible, but still.

The experiments done used to derive the gas laws (boyle, charles, etc) were done in conditons where real gases behave approximately ideally (very low pressures, for example). Thus, they were able to approximate the behavior of ideal gases.

You also have to understand the historical progression towards the ideal gas law. Boyle published his work in 1662. Charles formulated the original law in his unpublished work from the 1780s. Avogadro hypothesized in 1811 that two given samples of an ideal gas, of the same volume and at the same temperature and pressure, contain the same number of molecules.

Nice that @MaxW made us think of 200 y in between. Sometimes we do not get the real flow of things.  Not chemistry, but a genial intuition such as Olbert paradix seems a stupid assertion if state to someone born in the second half of the last century, for instance.

The ideal gas law is a very good approximation of how gases behave most of the time
There is no logical flaw in the laws. Most gases most of the time behave in a way that is close to the ideal gas equation. And, as long as you recognise the times they don't, the equation is good description of the way they behave.
The ideal gas equations assume that the molecules making up the gas occupy no volume; they have no attractive forces between them and their interactions consists entirely of elastic collisions.
These rules can't explain, for example, why gases ever liquefy (this requires attractive forces). But most of the common gases that were used to develop the laws in the first place (normal atmospheric gases like oxygen or nitrogen) are usually observed far from the point where they do liquefy.
As for the volume taken up by the molecules of the gas, consider this. A mole of liquid nitrogen occupies about 35mL and this is a fair approximation of the volume occupied by the molecules. at STP that same mole of gas occupies a volume of about 22,701mL or about 650 times as much. So, at least to a couple of decimal places, the volume occupied by the nitrogen molecules is negligible.
The point is that for gases not close to the point where they liquefy (and few components of the atmosphere are), the ideal gas laws are a very good approximation for how gases behave and that is what we observe in experiments on them. The fancy and more sophisticated equations describing them are only really required when the gas gets close to liquefaction.

Veering slightly into history-of-science with this one, but were measurements back in those days even accurate enough to detect deviations from ideal behaviour?

@orthocresol I'm not sure it is a good generalisation to say that in the "old days" scientists were less accurate in their measurements. More importantly, just how big are the deviations from ideality for typical atmospheric gases in room-temperature laboratories?

I don't know, and am rather too lazy to calculate a ballpark figure; but I was just curious as to whether technology several centuries ago was good enough to detect them. Maybe 'accurate' was a poor choice of word; substitute 'precise' for it if you prefer. (I'm not disagreeing with anything you said!)

Please note that, at STP according to current IUPAC recommendations, one mole of nitrogen actually has a volume of 22.701 l. The mentioned value of 22.4 l corresponds to the old definition of STP.

@Loong Sorry. I've lost the will to keep up. But I will alter the text to match.

I think it's worth mentioning that the "degree of ideality" is a literal experimental parameter called the compressibility factor, which is equal to exactly 1 for an ideal gas and deviates from unity for a real gas. However, it is clear that the value is typically quite close to 1 for a wide range of common gasses in reasonable conditions.

In the last paragraph, the parenthetical note is a "garden path sentence".  Why are most components of the atmosphere not at room temperature?  Are they hotter, colder, or what?  Oh--few of the components of the atmosphere are close to the point of liquefaction at room temperature.  Perhaps the note would be better as simply "(i.e. most of the components of the atmosphere, at commonly available temperatures)".

You must consider this:

The question whether a physical system follows a particular law is not a "yes or no" question. There is always an error when you compare what you measure with what the law predicts. The error can be at the 17th digit, but it's still there. Let me quote a very insightful passage by H. Jeffreys about this:

It is not true that observed results agree exactly with the predictions made by the laws actually used. The most that the laws do is to predict a variation that accounts for the greater part of the observed variation; it never accounts
  for the whole. The balance is called 'error' and usually quickly  forgotten or altogether disregarded in physical writings, but its existence compels us to say that the laws of applied mathematics do not express the whole of the variation. Their justification cannot be exact mathematical agreement, but only a partial one depending on what fraction of the observed variation in one quantity is accounted for by the variations of the others. [...] A physical law, for practical use, cannot be merely a statement of exact predictions; if it was it would invariably be wrong and would be rejected at the next trial. Quantitative prediction must always be prediction within a margin of uncertainty; the amount of this margin will be different in different cases[...]
The existence of errors of observation seems to have escaped the attention of many philosophers that have discussed the uncertainty principle; this is perhaps because they tend to get their notions of physics from popular writings, and not from works on the combination of observations.

(Theory of Probability, 3rd ed. 1983, §I.1.1, pp. 13–14).

The error can be different in different ranges of the quantities you're measuring.
When trying to guess a physical law, scientists often consider the simplest mathematical expression that's not too far from the data.

The "ideal gas law" is very accurate in the range where the temperature $T$ is high, the pressure $p$ is low, and the mass density $1/V$ (inverse volume) is low. In these ranges the law is very realistic. If you 3D-plot various measurements of $(p,V,T)$ for a fixed amount of a real gas, you'll see that a curved surface given by $pV/T=\text{const}$ fits the points having large $T$, low $p$, low $1/V$ very accurately. "Real" gases are very "ideal" in those ranges.
Moreover, the law involves simple multiplication and division: $pV/T$, so it's one of the first a scientist would try to fit the points.
Finally, the law wasn't just suggested by fitting, but also by more general physical and philosophical principles and beliefs. Take a look at
S. G. Brush: The Kind of Motion We Call Heat (2 volumes, North-Holland 1986)
for a very insightful presentation of the history of this and other laws.

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I like @pglpm  answer, but can't comment on it.
I would like to mention that the technology of those years was not precise enough for very exact measurements, and not good enough for measurements at "boundary" conditions (where differences from the ideal gas law increase greatly in magnitude), so the experimental/measurements errors would hide any deviation from the "ideal gas law".

Suppose you have a crate, and when you put one apple in the crate, the crate + apple weighs 1100 grams. The crate plus two apples weighs 1200 grams. The crate plus three apples weighs 1300 grams. Do you have any guesses as to what one, two, and three apples would weigh without the crate?
The gas laws are more complicated, but it's the same basic principle. If you hypothesize that $pV$ is equal to some interaction between an "ideal" function of $nRT$ and some complicating factor from real gasses, it's not too difficult to take a bunch of data points of different $p$, $V$, $n$, and $T$ for different gasses, and isolate out the "ideal gas component" from the rest of the effects, especially since you can take extreme cases where the ideal gas component dominates.
It's made easier by simply applying intuition to the situation. It's logical that the volume should be proportional to number of molecules. Increasing the temperature should also increase the volume. Increasing the pressure should decrease the volume. Much of the the ideal gas law can be worked out from first principles, and experiments then verify the models (as well as providing the value of $R$).


text:
Melting point of unsymmetrical dimethylhydrazine vs 1,2-dimethylhydrazine
Unsymmetrical dimethylhydrazine ($\ce{H2NN(CH3)2}$) has a melting point of −57 °C, but its isomer, 1,2-dimethylhydrazine ($\ce{(CH3NH)2}$) has a melting point of −9 °C. That looks to me like a significant difference, so there must be some mechanism which makes these two compounds extremely different in this respect. However, I'm unable to figure out why that difference exists; could you please explain that?
These are the structures of the compounds (from wikimedia commons):
Unsymmetrical dimethylhydrazine:

1,2-dimethylhydrazine:

Here're some of the arguments which I considered:

The packing in crystal structures could be 'different' in a significant way. This could potentially be supported by the fact that the difference between their boiling points is much lower: unsymmetrical dimethylhydrazine has a boiling point of 64 °C while 1,2-dimethylhydrazine's is 87 °C. However, this is an extremely vague argument; it could be strengthened by some details regarding how the different packings contribute to the melting points.
Hydrogen bonding should be a pretty significant factor to consider while analyzing these compounds. I suspected that steric hindrance may allow only one $\ce{N}$ atom for hydrogen bonding in unsymmetrical dimethylhydrazine. However, this would suggest that 1,2-dimethylhydrazine should have a lower melting point, which is clearly not the case.

We had a discussion about these two compounds' melting points in the Physics SE chat a while back, so I thought I'd pop over here and ask.

hint: symmetry.

It is difficult to answer without the exact crystal structure of both compounds (which I didn't even try to get), but the C2-symmetry of 1,2-dimethylhydrazine (1,2-DMH) is really stinking, compared to unsymmetrical dimethylhydrazine. If you consider the plane "H–N–N–H" in 1,2-DMH, it is obvious that close molecule can form a sheet just by hydrogen-bonds. It is not the alpha and omega of the crystal structure, but I am not surprised at all by the melting points you gave.

According to the Carnelley’s Rule:

That of two or more isomeric compounds, those whose
  atoms are the more symmetrically and the more compactly
  arranged melt higher than those in which the atomic
  arrangement is asymmetrical or in the form of long chains

If we attend to the structure of the 1,1-dimethylhydrazine, one can see that the steric hindrance due to the methyls groups inhibits the formation of H-bond in the central nitrogen atom for this isomer, in contrast to the 1,2-dimethylhydrazine which both nitrogen atoms are capable to form H-bonding.
For the thermodynamics point of view, the melting point of a solid is related to the enthalpy and entropy of melting by the following expression:
$$T_{\text{m}}=\frac{\Delta_{\text{m}} H}{\Delta_{\text{m}}S}$$
where the subscript "$\text{m}$" stands for "melting". Is difficult to estimate the impact of both enthalpy and entropy without further information than the molecular structure, but in most of the cases the process of melting is driving by enthalpic factors; therefore if the symmetric molecule can make more H-bonds than the unsymmetrical it's reasonable to assume that the symmetric molecule has a higher enthalpy of fusion and consequently a higher melting temperature provided the entropic effects are ignored.
References and further reading:
Effect of molecular symmetry on melting temperature and solubility
Melting Point and Molecular Symmetry
Carnelley’s Rule and the Prediction of Melting Point

Melting points of related compounds: hydrazine, 2 °C; monomethylhydrazine, −52 °C (Wikipedia).
Three of the four compounds have an $\ce{NH2}$ group, so this is not a deciding factor. Hydrogen bonding may be present without being a factor in the different melting points.
The common feature in the two high-melting compounds is a symmetrical N-N bond; an unsymmetrical bond is common to both low-melting compounds.
Hydrocarbons can display a similar trend in melting points without significant hydrogen bonding influences: n-pentane ($\ce{CH3CH2CH2CH2CH3}$) melts at −130 °C while the extremely symmetrical neopentane ($\ce{C(CH3)4}$) melts at −16.5 °C.


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Finding major product when bromonium compound is acted upon by a nucleophile
The original question is:-

The major product obtained in the following conversion is:-

(Note that Me here means methyl group)
The initial step would be the formation of a bromonium intermediate. Which would then be attacked by a nucleophile $\ce{CH_3OH}$ as it is present in excess in solvent.
What I think is, $\ce{CH_3OH}$ would attack the carbon atom close to ketone, because that’s the carbon best able to stabilize positive charge due to resonance with keto group. This is what I think will happen

But the proposed answer was

Can anyone explain me why how and why this compound is formed?

https://chemistry.meta.stackexchange.com/questions/3044/hidden-points-of-editing-you-probably-didnt-know?r=SearchResults&s=1|27.6568

The bromonium ion may be attacked by the non-carbonyl oxygen of the acetate forming a 6-membered ring. The 6-membered ring is then opened by SN2 displacement with methanol. Basically, a neighboring group effect.

I'd like to point out that the ketone would actually disfavor an attack at its alpha position:

Placing two positive charges on adjacent carbon atoms is much less favorable than not, especially when there is an oxygen perfectly poised to help stabilize this positive charge, as user55119 pointed out.

That makes sense.... Will methanol attack the same carbon(carbon atom away from keto group) if acetate is absent?

Yes, it should.

I like the ringo's point, but literature evidence does not support the classic cyclic bromonium ion mechanism. Effects of a carbonyl group on the ring opening of a neighboring bromonium ion in some bromination of reactions of $\alpha,\beta$-unsaturated ketones have been studied. The authors have used methyl acrylate and methyl crotonates (cis- and trans-isomers) in this study. In addition to halogens (used in preliminary studied), they have mainly used brominr chloride ($\ce{BrCl}$) for the electrophilic addition under ionic and radical conditions (Ref.1). For example, under ionic conditions, addition of $\ce{BrCl}$ to methyl acrylate ($\ce{CH2=CHC(O)OCH3}$) gave primarily anti-Markownikov product ($\beta$-attack). The products ratio is 83% $\ce{CH2(Cl)CH(Br)C(O)OCH3}$ (beta attack) and 17% $\ce{CH2(Br)CH(Cl)C(O)OCH3}$ ($\alpha$-attack). However, under radical conditions (ultraviolet irradiation), methyl acrylate has yielded 100% of $\ce{CH2(Br)CH(Cl)C(O)OCH3}$.
Then, during another project on aldehyde and ketones, directed by the same principal investigator (Ref.2), it was found that some simple $\alpha,\beta$-unsaturated aldehydes and ketones do not react with halogens by the expected attack on the $\ce{C=C}$ bond. They proposed a mechanism, which agrees with the  formation of products obtained. The simplest version using $\ce{BrCl}$ to show the $\beta$-attack is depicted in Scheme 1:

Based on above suggested mechanism and Ringo's prediction, following mechanism is depicted (Scheme 2):

References:

V. L. Heasley, D. W. Spaite, D. F. Shellhamer, G. E. Heasley, “Considerations on the Effect of a Carbonyl Group on the Ring Opening of a Neighboring Bromonium Ion,” J. Org. Chem. 1979, 44(15), 2608-2611 (DOI: 10.1021/jo01329a003).
V. L. Heasley, D. F. Shellhamer, T. L. Carter, D. E. Gipe, R. K. Gipe, R. C. Green, J. Hordeen, T. D. Rempel, D. H. Spaite, G. E. Heasley, “Reaction of halogens with some $\alpha,\beta$-unsaturated aldehydes and ketones,” Tetrahedron Letters 1981, 22(26), 2467-2470 (https://doi.org/10.1016/S0040-4039(01)92934-X).


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Why are carbons of Inositol chiral centers?
I was asked to draw all possible stereoisomers of inositol (1,2,3,4,5,6-cyclohexanehexol). To obtain the answer I had to assume that all six carbons of the molecule are asymmetric, which (bearing in mind possible molecular symmetry) results on the following nine stereoisomers:[1]

The problem is that I fail to see why these carbons can be considered chiral centers. Given:[2]

The most common cause of chirality in an organic molecule, although not the only one, is the presence of a carbon atom bonded to four different groups [...]. These carbons are now named chiral centers [...]

The chirality of the molecule is not the reason of my doubts, I can see that the only chiral isomers are D- and L- chiro-inositol. Since all carbons in the ring possess a hydroxyl group, can they not be considered to be bonded to equivalent substituents, making them symmetric? Is the possibility of the -OH groups to be above or below the plane of the ring what makes them asymmetric?

From Wikipedia
McMurry, John. Organic Chemistry. CENGAGE Learning, 2008. 7th edition.

Seven of the nine inositols are achiral. Any carbons lying in a plane of symmetry are in achiral environments (r and s). Those carbons not in a plane of symmetry are in chiral environments (R and S). I Googled "inositols chirality" and found this: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/inositol.html

When we count different substituents to determine chiral centers, we have to include the relative locations of atoms as well as stoichiometry.
Let's look at the top carbon atom in the D-chiro form.  If you go clockwise around the ring you encounter five CHOH groups, and if you go counterclockwise you get the same five CHOH groups.  The substituent atomic formulas look the same.  But:

in the clockwise direction you see the hydroxyl group attached to the first carbon from the top is on the same side of the ring as the hydroxyl group on the top carbon itself.  Whereas:
in the counterclockwise direction you see the hydroxyl group attached to the first carbon from the top is on the opposite side of the ring as the hydroxyl group on the top carbon itself.

That difference is enough to count as two different substituents even though the formulas in terms of atoms are the same.  So you have a chiral center, and the CIP convention even identifies the orientation of the top carbon as S.  Similarly any of the other ring carbons can be a chiral center.
Contrast this with the scyllo form, for instance, where the substitution on any carbon is truly fully symmetric including stereochemistry.  So no chiral centers there.  Most of the other forms listed above do have chiral centers by the method above, but end up being meso compounds.  so only the forms labeled "chiro" actually have a net molecular chirality.

Consider 2,3-butanediol.

This compound has 2 stereocenters. The meso version of this compound is achiral. However, there are still two stereo centers. As drawn, the left one is (R) and the right one is (S). It is precisely because these two stereocenters are identical but opposite in configuration that leads the molecule to be meso.
A carbon center is asymmetric if it has 4 different substituents regardless of whether or not the whole molecule is chiral. The ring substitution in inositol is such that the centers around the ring are not identical (including when you look at stereochemistry--for example, (R) has higher priority than (S)). This of course isn't enough to help you easily make stereochemical assignments for all of the centers of inositol, but hopefully, this allows you to understand why the centers are still chiral centers.
As an aside, there are a few centers that cannot be chiral because they lie on a mirror plane. Interestingly though, changing the configuration of the center produces a different molecule. These centers are called pseudochiral, and we label them with lower case r and s.

I offer a different perspective to your question about the inositols and how the chirality of the "top" carbons in D-chiro-inositol 2 (C2; S-configuration) and scyllo-inositol 3 (C2; r-configuration) are determined.
Generic inositol 1 is numbered the same way as D-chiro-inositol and scyllo-inositol. The method for assigning the CIP descriptor for each inositol at C2 will be illustrated. Starting from C2 in structure 1, the counterclockwise path C2-->C3-->C4-->C5 is colored blue while the clockwise path C2-->C1-->C6-->C5 is in red.
Starting with D-chiro-inositol 2, temporary descriptors (Ro/So) at all carbons except C2 are assigned. At each of these carbons the hydroxyl group has the top priority while hydrogen has the lowest priority. C2 is taken as the next to highest priority group, no matter how far the carbon in question is from C2.1 The temporary assignments for D-chiro-inositol are illustrated in structure 2. To determine whether the blue or red path has the higher priority, step 1 is to compare the assignments of C3/C4 (So/Ro) with C1/C6 (So/Ro). No decision can be made at this point. In step 2, one compares C3/C5 (So/So) in blue with C1/C5 (So/Ro) in red. Because "like" descriptors (RR or SS) take precedent over "unlike" descriptors (RS or SR), whether they are temporary or not. The priorities for C2 are OH>SoSo>SoRo>H. C2 is of the S-configuration. Digraph 4 illustrates this point as well. Because D-chiro-inositol is chiral, each of the carbons is both stereogenic and chirotopic. Each carbon must be analyzed independently as was done for C2.  Notice that the temporary assignments (Ro/So) have no relationship to the final assignments (R/S). continued

Scyllo-inositol 3 is achiral having three planes of symmetry passing through C1/C4, C2/C5 and C3/C6. All carbons are equivalent; assign one, you have assigned all six. After assigning temporary descriptors and proceeding all the way to step 3 (see rules next to structure 1), no decision can be made. C2 and all other carbons are pseudoasymmetric (stereogenic and achirotopic). The default position is to use the temporary assignments at C1 and C3 (see digraph 5). CIP Rule 5 states R>S (and Ro>So). The priorities for C2 are OH>Ro>So>H. C2 is of the r-configuration. Lower case r/s descriptors are reserved for pseudoasymmetric carbons.
The two inositols discussed here have descriptors that are either R/S or r/s. This need not be the case. For example, muco-inositol has four chirotopic (R/S) carbons and two achirotopic (r/s) carbons.
1) This method is a simplification of a more complex methodology.

OK, I'll bite.  How is the orientation around the top carbon atom in the scyllo form determined when there is obviously a mirror plane going right through that carbon atom?  (And one mirror plane or another through all the other carbon atoms too.)

@Oscar Lanzi: Because of the plane of symmetry thru C2/C5, the red/blue paths are indistinguishable. C1 is Ro while C3 is So. Ro > So. Priorities OH>Ro>So>H. You can do the same exercise for anyone of the scyllo carbons because they are all equivalent. C2 in D-chiro is S and not R. See the colored picture. Don't fret. It took me a long time to understand this stuff.


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Nail polish chemistry question
When I read my ingredients on my nail polish remover, it has water as its first ingredient.  Why does nail polish have water, yet the warnings still say its flammable?

I'm voting to close this question as off-topic because ingredients of healthcare (and other) products are listed in descending order of percent composition and, more to the point, mixtures can still be flammable even if water is the first component listed.

If you've ever held a flame to rum or other high proof alcoholic drink, you'll understand why.

Mixtures of water with flammable liquids can themselves be flammable, largely due to the fact that the flammable liquids generally are more volatile than water, (ie they evaporate more easily.) As a result, the air above a solution of nail polish will have a lot of molecules of the flammable liquid in it. (That's why you can smell them easily). If you have a flame near the liquid, those molecules in the air above it can combust. The heat generated will evaporate more liquid, which will fuel the fire. Some of the heat is lost to vaporizing water, which doesn't burn, but if the fraction of the total liquid that is flammable is high enough, the combustion continues.
In nail polish, the flammable liquid is usually acetone or ethyl acetate.

Flammability (technically, inflammability, but I guess people are confused by this term) is a function of some vapor. Yes, water is not flammable, but the flammable liquid, in the liquid state, also doesn't burn. It's specifically the vapor that is flammable. Usually, there is some threshold concentration, reflected in the partial pressure of the flammable gas, that is required to ignite the material. Once it ignites, the heat generated by the combustion reaction promotes combustion of the adjacent vapor and heats the liquid enough to produce more gas.
If you mix water and your flammable solvent (assuming they're miscible, which they kind of have to be), you have a solution. Raoult's law says that the partial vapor pressure of the gas above the solution is proportional to the mole faction of the liquid in the solution. In this case, it means that even if the flammable liquid is diluted with water, there's still vapor above the liquid. And though the pressure may be less, it may be high enough that it's still highly flammable.


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Dependence of primary alcohol oxidation on presence of water and "strong" or "weak" oxidants
It is known that the oxidation of primary alcohols can result in aldehydes or carboxylic acids. The product obtained from the reaction is described as dependent on reagents and conditions;[1]

"Primary alcohols are oxidized to either aldehydes or carboxylic acids, depending on the reagents chosen and the conditions used."

A distinction is typically made between "weak" (eg. PCC) and "strong" (eg. chromic acid) oxidants[2], the former being capable of giving an aldehyde product while the latter carries out the oxidation to carboxylic acids. However, the accepted mechanism for the reaction shows that the oxidation always results on an aldehyde, which can then be further oxidized to a carboxylic acid through an hemiacetalic aldehyde hydrate forming in the presence of water.[1][3]

"Most other commonly used oxidizing agents [...] oxidize primary alcohols directly to carboxylic acids. An aldehyde is involved as an intermediate in this reaction but can’t usually be isolated because it is further oxidized too rapidly."

This indicates that it is the presence of water in the reaction medium which determines the product of the oxidation. Thus, "weak" oxidants could still yield a carboxylic acid in aqueous solutions (in fact, they are always noted to be used in anhydrous conditions) and "strong" oxidants will result on an aldehyde when used with organic solvants. If so, I fail to understand the classification of "weak" and "strong".
Furthermore, some sources, like Wikipedia, seem to imply the existence of a direct oxidation from primary alcohol to carboxylic acid, but I've been unable to find such a mechanism in bibliography, always ending up with the hydrate intermediary method.
To summarize:

Is the distinction between "strong" and "weak" oxidants meaningful or does the presence of water always determine the reaction product?
Does a method for the direct oxidation of primary alcohols to carboxylic acids without an aldehyde hydrate exist?

McMurry, John. Organic Chemistry. Brooks Cole, 2011. 8th edition.
Alcohol Oxidation: “Strong” and “Weak” Oxidants
Reagent Friday: Chromic Acid, H2CrO4

"Weak" and "strong" in the examples are used because the weak oxidant stops at an aldehyde while the strong oxidant goes on to the carboxylic acid. Water is the critical factor. "Weak" PCC in the presence of water can give carboxylic acids. PS: PCC oxidation of primary alcohols under anhydrous conditions can produce esters via the hemiacetal of the aldehyde formed from unoxidized alcohol.

I'm afraid I still don't understand the difference. If "weak" oxidants can behave as "strong" oxidants, then why use those terms at all? Is it that a single equivalent of "strong" oxidant is enough to obtain a carboxylic acid, while a "weak" one requires two equivalents?

Forget about "weak" and "strong". The oxidation of a primary alcohol with Cr(VI) reagents goes through the aldehyde, then hydration, then it goes on to the carboxylic acid. Without water, the rxn stops at the aldehyde. Look up Collins oxidation and PCC oxidation.

After looking up more information and taking the comments of user55119 into account, I've decided to post an answer myself, hoping to clearly address the main question points.

The determinant factor of the products obtained from an oxidation reaction of primary alcohols is the presence of water in the reaction medium, as it allows the formation of an aldehyde hydrate which can be further oxidized. Thus, the distinction between "weak" and "strong" oxidants is not necessarily informative (and could even be misleading) and emphasis should be placed in the reaction mechanism alongside the reaction medium.
The oxidation of primary alcohols with Cr(VI) reagents takes place through the mechanism depicted in the question or slight variations of it (depending on the reagent chosen). Carboxylic acids are obtained only through an aldehyde intermediate.

Further reading:
McMurry, John. Organic Chemistry. 8th edition.
MasterOrganicChemistry


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Why is it vital that atoms in the same sample are all given the same charge? Mass spectrometry
When being tested in the spectrometer why is it vital that the atoms in the same sample are given the same electrical charge e.g all are c+1 or all are +2 etc

As far as I know, the degree of ionization of particles in mass spectrometry depends on the energy supplied. In other words, it's tunable. However, you would want to use your electron beam (let's say it's electric ionization) as an analytical tool, not a sledgehammer. It makes the MS result analysis more feasible if the ionization energy is just enough to remove as few electrons as possible. Simplifies the spectrum, so to speak. Is that what you're asking about?

This is a misconception that during ionization, each ion must get a +1 or +2 charge. It is desirable to have +1 so m/z corresponds to the molecular mass of the ion or its fragment. The state of ionization heavily depends on what you are using to ionize your target analyte. For example if you use electron impact, you will see fragments mostly with +1 charge. However in other techniques called matrix assisted laser desorption (MALDI), the state charge can be as high as +12 and so on. They are detected as such. In electrospray ionization, +2 is also common.
There is no way a mass spectrometer "knows" a priori the state of charge of an ion. All charge ratios are detected. However it can distinguish the sign. It depends on the user whether negative or positively charged ions reach the detector. Some ions are better detected in positive mode and some in negative mode.

In mass spectrometry, the position of each peak you observe depends on the value m/z (i.e. mass over charge). If you have only z=1, then the position will depend only on the mass (m/1 = m).(1)
In a mass spectrometer, you will generate a mixture of charges at first but ions with a charge greater than 1 are filtered out.
(1) Obviously, it also works with a -1 charge.


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Naming this organic compound
What’s the name of the following compound?
$$\ce{CH3C(p-ClC6H4)2CH(Br)CH3}$$
My teacher said that she strongly thinks( with another teacher) that it should be 3-bromo-2,2-bis-(4-chlorophenyl)butane, though many other books and other students (similar to me) stated that it should be 2-bromo-3,3-bis-(4-chlorophenyl)butane.
She asked us try to find it out. Any organic chemistry experts out there?

I think your teacher is correct, numbering of carbons is done acording to the lowest locant rule.

The red numbering gives you : 2, 3, 3
The green numbering gives you : 2, 2, 3
We see that the first point of difference between the two numberings occurs at second number, so the green numbering is preferred.
Consequently, the correct name should be : 3-Bromo-2,2-bis(4-chlorophenyl)butane


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Molecular weight of NiMnGaFe compund problem
I have a big problems calculating molar/molecular etc masses, because in my case I have a monocrystal, co I cannot imagine it as a molecule and my compound is made of Ni-Mn-Ga-Fe. I know the mass and the atomic and the mass percents and I need to get molecular weight.
In the atomic percents I have: 50% of Nickel, 25% of Manganese, 17,5% of Gallium and 7,5% of Iron. So the ratio (of number of atoms) is 2:1:0,7:0,3 and I can find the atomic masses of all these elements but what is going to be the molecular mass?
I am sorry for probably trivial question (I am really dumb in this area unfortunately and I was searching on the internet for answers but every time the structure of molecule was known and this is a crystal)

:/ It's an alloy not a compound, so there's no molecular or even molar mass.

I was thinking that but I needed to calculated the Bohr magneton f. u. from emu, so I need to search for different equation, thank you

At least now I know for sure.

Seems it's not even a single phase https://www.sciencedirect.com/science/article/abs/pii/S0925838818321406

This alloy has too much of iron so after Laue diffraction it looked as a polycrystal but my samples with less of iron are single phase and succesfuly oriented. Making monocrystal is just harder so there are plenty of articles about polycrystals. The magnetization measurement however were succesful for all.

https://www.researchgate.net/post/How_do_you_convert_emu_g_to_Ub_mewB_per_TM_atom I found the aswer here.


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Average Molar Weight
We have a mixture of two gases, say nitrogen and hydrogen.
To calculate the average molar weight M , the formula says:
M (average) = (hydrogen pressure/100).M(H2) + ( 1 - nitrogen pressure/100).M(N2)
What I haven't understood is that why do we have pressure in the formula?
What I don't find logical as well is that the unit on the left side isn't the same as that on the right side of the equation.

I suppose the $100$ is not unitless, but implies to be $\pu{100 kPa}$.
Then both equation sides have the same dimension and average molar mass is a weighed average of gas molar masses.

Other option could be to use a molar or volume fraction.

Thank you :) I have read on wiki  (https://en.wikipedia.org/wiki/Mole_fraction) as well that : In a mixture of ideal gases, the mole fraction can be expressed as the ratio of partial pressure to total pressure of the mixture. But I still need to know why :)

It comes as consequence of the ideal gas laws and the ideal gas state equation $$pV=nRT$$

You are correct that the pressure units should cancel. However your formula
M (average) = (hydrogen pressure/100).M(H2) + ( 1 - nitrogen pressure/100).M(N2)
doesn't make any sense to me.
If we assume that the 100 is for 100 kPa, and that the total pressure of the gas is 100 kPa, then the formula should be:
M (average) = (hydrogen pressure/100).M(H2) + (nitrogen pressure/100).M(N2)


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Does propylene glycol dissolve plastic?
According to Médecins Sans Frontières propylene glycol can dissolve plastic:

As propylene glycol can dissolve plastic... (Source).

Is this claim true? This seems rather spurious to me. If it matters, for a specific "plastic" I am particularly interested in evaluating the claim in regards to polypropylene and polyethylene as both are materials common for making syringes.

The kind of plastic almost always matters, I doubt this source meant PE/PP - they aren't particularly soluble in solvents.

I understand that maybe based on the medical source of the question this might be downvoted, but otherwise I don't really understand. If a presumably reputable source makes a farfetched statement and someone posts it here I think that is a good question. Perhaps meta addresses why this is not a worthy question?

Thank you, Mithoron and @NightWriter for the comment. I am asking this question because of a question on Medical Sciences Beta. If you read more, you will realize that I actually doubt the claim and am asking this question as the claim comes from a medically reputable source.

@Mithoron From what I could find, plastic syringes are made from PE/PP. Of course, please correct me if I am wrong.

As propylene glycol can dissolve plastic...

This claim is absurd. Propylene glycol cannot dissolve poly(olefin) polymers especially polypropylene or polyethylene. It is possible for glycols to cause paneling of containers made to such, but not dissolve; and certainly wouldn't happen in the amount of time it takes to perform an injection.

Out of curiosity, would you say that poly(olefin) polymers could leach out from the syringe into the propylene glycol?

@BarryHarrison No, though there could be unreacted monomer, but far below any concentration of concern. Think about how unclean it is to even fill a car with petrol.

Got it. Do you mind clarifying your analogy?

@BarryHarrison Sorry, no analogy I was just trying to juxtapose the minuscule amount of monomer potentially present in a plastic syringe versus the concoction of vapors present at an activity most people perform regularly.

Could you comment on propylene glycol leaching/dissolving latex and/or natural rubber? Thanks!

@BarryHarrison To keep this G-rated I will be obscure, but my research shows that it would be about the same situation for latex as for HDPE and PP as propylene glycol is used with a certain latex product for men. Also my google adds are going to be a bit wonky the next few days.

I am sorry to create you any inconveniences! Thank you so much for looking into this.

Plastic syringes are made of polypropylene and the document linked by @A.K. states that polypropylene is not compatible with antifreeze.
When referring to the plastic containers antifreezes are sold in the most common material used I have seen is HDPE and there is no data available on compatibility according to the Geotech data sheet linked.
The thing is that elsewhere in the data sheet it states that polypropylene is compatible with propylene glycol when it can be used as antifreeze as an alternative to ethylene glycol.


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Why oxidation state decreases
oxides of group 1 elements are Li2O , Na2O2 , KO2 , RbO2
Oxygen’s oxidation state in each compound is -2 , -1 , -1/2 ,-1/2 respectively I.e oxide-> peroxide-> superoxide
my question
The oxidation state of the oxygen Is increasing down the group why is that ?
My attempt
the tendency to donate electron increases down the group , by this statement the oxidation state of oxygen in the oxides down the group should decrease as oxygen gets easily reduced.
What am I missing here?

What about the size of the cations?

Firstly, the oxides you have listed are not the only ones that exist. It's that on reacting with excess oxygen, Li forms monoxide, Na forms peroxide, and the elements further down form superoxide. These are not necessarily the strongest oxide form too, infact, according to this link:- K2O2; the most stable oxide of K is peroxide despite being able to form a superoxide. (Also, Li, Na can also make superoxides (though not very stable) and K, Rb can also make monoxides.)
So I think the question changes to why are the elements down the group more able to form oxygen compounds with decreasing oxidation state of oxygen, and that is due to the size of the metal atoms; the larger the metal atom, the higher number of oxygen atom containing species it can (stably) accommodate around it.


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Chemical reaction
Consider the following
Ca(OH)2 + 2NH4Cl → CaCl2 + 2NH3 + 2H2O
(Source: https://chemiday.com/en/reaction/3-1-0-440)
It appears to be double displacement reaction as CaCl2 is formed, but it’s not as NH4OH is not formed instead NH3 is formed
Why NH3 is formed instead of NH4OH ?
Is there intermediate step which transform NH4OH into NH3( I don’t think so but still)

It is postulated that NH4OH as a molecule does not exist in solution. A better way to write might be NH3(aq). As to the evolution of gaseous ammonia, this reaction would only proceed if you heat up (i) a moist paste of NH4Cl and Ca(OH)2 or (ii) strongly heat a solution, otherwise ammonia would remain dissolved in the solution.

$\ce{NH4OH <=>NH3 + H2O}$, that is, they are in equilibrium. Since the solution is weakly ionized, (Kb is ~1.8 * 10-5 at 293 K), there is little $\ce{NH4OH}$ present at a given instant, so ammonia water is only weakly alkaline.
One might say the answer $\ce{NH3 + H2O}$ is 99.998% correct, but $\ce{NH4OH}$ is still 0.002% correct, not entirely wrong.
See also this discussion on what species are present.

As the reaction takes place in water medium ,(More water on RHS) Why can’t we apply le’chatliiers principle and say equilibrium shifts to left hand side. Similar statement is made In the linked answer

I think the debatable and interesting part is the question: Does NH4OH as a molecule exist in solution? It was the same story as the hypothetical H2CO3 but people finally detected it. Not sure about the status of ammonium hydroxide.


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To predict if a substance is deliquescent
FeCl3,FeSO4 and CuSO4 all are deliquescent solids.

Is there any way to predict this without knowing this fact?
  I am looking for a way to predict if a substance is deliquescent for my upcoming exams.

There is no general rule, just partial empirical rules for particular anions, which can be memorized from textbooks.
There is theoretical reasoning why salts  are soluble or not, concerning the energy of the crystal lattice, energy of ion hydration, entropy change etc. But easier way is to go for empirical rules and "chemical sense".


text:
Kb values of N,N-dimethylaniline derivatives
I need the $K_\mathrm{b}$ values for ortho-, meta- and para- positions of the following compounds:

I've looked and couldn't find them anywhere.

I've redrawn the compounds of interest with chemfig to make them more readable. However, I'm not sure I recognized all groups correctly — feel free to proof-read the edit. $\mathrm{\LaTeX}$ source code is available at Pastebin.

chemicalize.com is  a cheminformatics website which is the kind of resource you are looking for.
You can use the search tool to find your compounds. The input is the structure itself, which you can draw in-situ.
Here is the $\mathrm{p}K_\mathrm{a}$ data I retrieved from the aforementioned site. Note that these are calculated / estimated, not measured $\mathrm{p}K_\mathrm{a}$ values.

Table 1. $\mathrm{p}K_\mathrm{a}$ values (of the conjugate acid)
$$
\begin{array}{lccccc}
\hline
& 1 & 2 & 3 & 4 & 5 \\
\hline
\textit{ortho-} & 3.71 & 5.19 & -0.98 & 1.44 & 2.66 \\
\textit{meta-} & 3.89 & 5.15 & 1.73 & 4.77 & 3.45 \\
\textit{para-} & 5.55 & 5.46 & 0.65 & 4.68 & 3.65 \\
\hline
\end{array}
$$
Since the question is asking for $K_\mathrm{b}$ values, I converted them too.
Table 2. $K_\mathrm{b}$ values
$$
\begin{array}{lccccc}
\hline
& 1 & 2 & 3 & 4 & 5 \\
\hline
\textit{ortho-} & \pu{5.13e-11} & \pu{1.55e-9} & \pu{1.05e-15} & \pu{2.75e-13} & \pu{4.57e-12} \\
\textit{meta-} & \pu{7.76e-11} & \pu{1.41e-9} & \pu{5.37e-13} & \pu{5.89e-10} & \pu{2.82e-11} \\
\textit{para-} & \pu{3.54e-9} & \pu{2.88e-9} & \pu{4.47e-14} & \pu{4.79e-10} & \pu{4.47e-11} \\
\hline
\end{array}
$$

Please be kind enough to explain your downvote.


text:
Stability of molecules by resonance
I am pretty confused by some principles of resonance. Well by my study of electrostatics I know that potential energy of a system of  charges of same nature decreases as we increase the separation between them but same doesn't goes with a system of charges with different nature because their potential energy decreases as we decrease their separation as internal conservative forces do positive work.
I am a newcomer in chemistry so I have a doubt that my book says that energy of a molecule is minimum when they are separated by large distance (under the heading of application of resonance) without caring for nature of charges.This fact goes well for charges of same nature but contradicts the fact for system having different nature of charges.
Please tell me  that why this contradiction exists.

Why not to post what exactly the biok is saying? Nature should be sign,  they is presumably charges, and so on. As far the electrostatic potential is at play, stick to what you have already learned.

In chemical resonance, the same principle holds as in electrostatics:
"In general, the boundary structures of a conjugated system with the lowest energy are those
...
- in which the distance between charges of opposite sign is as small as possible and the distance between charges of the same sign is as large as possible
... ."
Wikipedia: Resonance (chemistry) - Major and minor contributors:
"Energetically unfavourable and therefore less favorable structures are minor contributors. With rules listed in rough order of diminishing importance, major contributors are generally structures that
...
3. carry a minimum of formally charged atoms, with the separation for unlike and like charges minimized and maximized, respectively
... ."


text:
How fast is the interconversion triplet → singlet in organic molecules?
When computational studies are performed on transition state geometries, the "thermodynamic" (that is, with no kinetic considerations) energy of the analyzed state is obtained.
I was wondering: if a (multistep) reaction begins in a biradical state, the kinetics of the interconversion from triplet to singlet is so slow that the reaction is likely to proceed in a radical fashion, or if a lower transition state exists for the ionic (singlet) counterpart, then it is likely to proceed in a ionic way, because the rate of interconversion is extremely fast?
I am sorry if the question is dumb, but all I could find was about transition metals, and I don't know if the same applies.

It depends on the organic molecule.
When spin-orbit coupling facilitates intersystem crossing, it can be fast: 100 femtoseconds for nitronapthalene (viz. Chem. Eur. J. 2018, 24, 5379 – 5387).
The Wikipedia article on intersystem crossing cites a timescale for radiative decay (phosphorescence) of triplet to singlet of 10−8 to 10−3 s.

Spin-orbit coupling will cause the spin flip needed. This is always present but becomes larger the heavier the atom are, iodine, xenon for example present either in solvent (such as iodobenzene) or as part of the molecule. Paramagnetic species also work in the same way, e.g. O2, NO as does an external magnetic field because, as the molecule tumbles, the apparent changing magnetic field the molecule experiences causes T-S crossing.
How fast intersystem crossing is will depend largely on the energy gap between singlet and triplet; the larger the gap the slower the process, and  as the strength of spin-orbit coupling increases so does the rate constant. This 'energy gap law' has been confirmed by many experiments on excited states and of aromatic molecules and metal complexes and has the form $k=Ae^{-\beta\Delta E}$ where $A$ depends on the strength of the interaction and $\beta$ on the properties of the molecule, this varies as $\ln(\Delta E)$ so is almost constant as $\Delta E$ changes, typical values are 3 to 5.


text:
Is it safe to boil carbonated water in a kettle?

Isn't it dangerous to boil carbonated water in a kettle?

Whenever I boil water in my kettle, I close its lid. Does this close lid constitute 'under pressure' (and so danger) emboldened beneath?

There was no boiling water explosion, although I more than half expected one.
The only way it could conceivably cause an explosion is if you boiled it under pressure.

Closure discussed in Meta: https://chemistry.meta.stackexchange.com/q/4456/23561

No danger!
If you are worried about flames, carbon dioxide is not flammable.
If you are worried about explosions, gas is not released so fast to cause an explosion.
For comparison, you can free much more energy per time by overheating plain water.
So nope, no danger, even if you place a lid on it (assuming that the lid is free to bump and release CO2/aqueous vapor, but even if it wasn't I doubt that it would be much more dangerous than doing the same with plain water).


text:
About the percentage yield and purity
I was wondering why all of the percentages like mass percent composition, percentage yield and percentage purity are calculated using masses.
Wouldn't it make more sense to use number of moles of the substance since we're counting the number of atoms in the molecules?

It is for the sake of convenience and practical purposes. Analytical balances or any other instrument cannot count the number of atoms/molecules, neither they are calibrated in terms of moles. All we can do is to measure its mass directly. Secondly, "measuring" moles assume that we know the exact molecular mass of that material. What if you make a new unknown product in a synthetic reaction? All you can do is to measure its mass directly and later characterize it.

Oh I see, thank you for your helpful answer!


text:
What provides the energy allowing membrane proteins to facilitate diffusion across the membrane?
We know that facilitated diffusion requires no energy.  But the transport proteins, when bind to a molecule from ECF (extra-cellular face) , the proteins rotate and deliver the molecule into ICF (intra-cellular face). Now my question is that--- Which force provides with the torque to rotate and how it works with out energy?

[...]how it works with out energy!

The transport proteins can change conformation in a stochastic manner. Even if the conformation opening to the extra-cellular side is a bit different in energy than the one opening to the intra-cellular side, there is no net change in energy/entropy/Gibbs energy in a complete cycle. The only thing that changes is that the cargo molecules are on the other side, a process for which the change in Gibbs energy is negative.
This argument is the same for any catalyst. If the catalyst is recovered in the same state as it was at the beginning of the reaction (and it has to be, otherwise it would not be a catalyst), the catalyst does not change the thermodynamic quantities of reactants or products. As a consequence, it speeds up the reaction, but does not influence the equilibrium constant.
If the catalyst has a high activation energy to do what it needs to do, then it is not an efficient catalyst. After all, the role of the catalyst is to lower the activation energy. So the transport proteins must be able to change from one to the other conformation both with cargo and without cargo in a facile manner. Otherwise, they could not go through the cycle of 1) picking up cargo 2) switching conformation 3) releasing cargo 4) switching conformation again.

Which force provides with the torque to rotate [...]

Random forces associated with thermal energy are responsible for the conformational change. The binding constant for cargo has to be moderate so the the cargo dissociates easily. One the other hand, the binding constant for non-cargo has to be so low that nothing other than cargo binds.

As another answer correctly points out, the redistribution of energy is enough to keep a protein in constant random motion, including motions that result in transport of a solute from the outside to the inside of the cell by attaching temporarily to the protein. A passive diffusion process does not require bond breaking/formation as is usual in a directed process.
Passive diffusion depends on a concentration gradient. So unless the concentration (or, more accurately, activity) of the diffusing solute is lower inside the cell, there will not be net transport into the cell. The point is not that facilitated diffusion does not require energy — it does — just not directed motion (work) based on active chemical modification of the protein or its environment. Random thermal motion (energy) will do.
This is really a rewording of what another answer explains, namely that for net passive diffusion into the cell to occur, the chemical potential (or free energy) of the solute inside the cell must be lower than that outside. The chemical potential gradient results from some other process that does not involve the protein (it was created perhaps by removing or exhausting the solute inside of the cell, or by increasing the concentration of solute outside the cell). The origin of the potential energy gradient is not a result of chemical modifications (work) at or around the protein itself. Instead it is only a result of the solute concentration difference, and it is therefore we refer to the process as "passive" diffusion.


text:
Friedel Crafts Acylation: Reaction of acyl halide with catalyst
To form an electrophile for this reaction to occur, the acyl halide interacts with the catalyst. However, on reviewing the conditions for the reaction, I have encountered that the 2 equivalence of the acyl halide is required.
In the case of the acyl chloride, there are 2 ways the acyl chloride can react with the catalyst.

(Sorry for the sloppy diagram)
In the case of the first reaction, Cl is good leaving group, but what is the reasoning of the second reaction to occur?

Strength of the O-Al bond is a significant factor

@Waylander, could it be that Al has vacant p-orbital where the 1 lone pair from oxygen atom is donated, to create a bond forming the adduct (similar as in BF3, since both are in the same group)?

That is exactly what occurs


text:
Why is molybdenum that important in inorganic synthesis as well as biosynthesis of ammonia?
We know that Mo (Molybdenum ) is used as  a catalyst for Haber's process, during synthesis of Ammonia from Nitrogen. Again, in biological systems, Nitrogen fixation (ALSO forming Ammonia from Nitrogen) uses Mo as activator of Nitrogenase enzyme. My question is "why Mo is so special for Nitrogen?"

Helpful: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2999624/

Two processes running at very different temperatures and pressures. I'd wager this is coincidence.

Why Molybdenum is Used in the Manufacture of Ammonia?
Molybdenum tubing has particularly good resistance to corrosion by
mineral acids, provided oxidizing agents are not present. It is also
resistant to many liquid metals and to most molten glasses. Molybdenum
is relatively inert in hydrogen, ammonia, and nitrogen up to about
1100 C, but a superficial nitride case may be formed in ammonia or
nitrogen.(Ref. 1)

Also, molybdenum has a special capability to form stable complexes with nitrogen atoms and hence molybdenum goes very well in biological reactions as well(See the link @user79161 has posted in the comment section). Recently, a group of scientists has developed a method for synthesis of ammonia using a dinitrogen-bridged dimolybdenum complex as catalyst and under very mild conditions of room temperature and at atmospheric pressure. However, the reaction mechanism of this new method is still unclear.

Reference

https://samaterials.podbean.com/e/why-molybdenum-is-used-in-the-manufacture-of-ammonia/
https://www.u-tokyo.ac.jp/focus/en/articles/a_00259.html (Picture)


text:
Does the main washing effect of soap come from foam?
I am curious if the main cleaning effect of soap comes from its foam (or, maybe, it is just a by-product).
Here are some  points, that might clarify what I mean:

Does foam sort of "carry" dirt and germs away from the surface?
If it is not the "main motor", which makes the soap function: is it possible to make a soap which will still be as effective without it producing any foam?
What percent of the cleaning work, if any, does foam undertakes (i.e. how effective would soap be against germs and dirt without its foaming ability)?

Foam is a side effect of using a tensio-active agents (though some tensio-active agents are engineered to produce as little foam as possible, e.g. in washing machines). In fact, you could consider that, if soap has nothing better to do, it will form foam. This answers some of your questions: foam does not (really) carry away dirt and its percentage of efficiency in the whole process is low.
Molecules of soap are made of 2 parts: a hydrophobic tail (which loves fat, grease, lipids, whatever you call them) and a hydrophilic head which loves water, and hate lipids.
When the hydrophobic tails are busy linking to a greasy stain or grease on your hands, they no longer participate into forming bubbles/foam.
The easiest way to verify that is to pour a little bit of oil in your hands and rub them. Wash your hands with soap a first time: there will be almost no foam. Now rinse your hands and wash them again: there should be much more foam!
Note that it does not mean that your hands are cleaner, regarding bacteria, but just that they are less greasy!
If you want to remove bacteria, rub your hands with soap for at least 30 seconds. So-called "anti-bacterial" soaps do not achieve better and promote bacterial resistance.

Anti-bacterial soaps are better at killing bacteria. It's just that normal soap is already amazing at killing bacteria, and humans are resistant enough that the difference is essentially non-existent for a normal healthy human. It's certainly very noticeable where people aren't healthy.

@Luaan Normal soap doesn't "kill" bacteria at all. Rather, the surfactants in soap allow the bacteria to be washed away. This is why hand-washing technique is also so important, and why water alone (no soap) is still surprisingly effective when no other options are available. If you want a biocidal effect, you need an anti-bacterial soap. But, as pointed out here, the efficacy levels are not significantly higher (although they do increase the longer you rub the soap around, unlike the surfactant effect, which peaks after a certain minimal amount of time), and they promote resistance.

The foaming action of soap was put there as a sort of "indicator" that something is happening. It's a marketing strategy because people are ignorant about how soap works. The bubbles are just air pockets and have no effect on cleanliness.

This is not true for "normal" soap, but is true for some "detergents".

Actually, it is. You can make some soap yourself and it'll only foam slightly, if you don't add foaming agent like producers do.

Made soap yesterday and have been making soap for personal use for several years. I use the standard oil/lard process. This soap foams quite well, in my experience. The bubbles seem finer, and the froth seems "creamier".


text:
How were the Hoffman's and the Zaitsev's rule formulated even before the discovery of the electron?
I recently studied Zaitsev's and Hoffman's rules for deciding which product is formed via elimination and noticed that the rules for formulated much before the discovery of electrons and any conceivable atomic theory. So how did they work out such rules without even knowing how these compounds bond and behave at microscopic levels.

As Wikipedia notes, Zaitsev's rule is 'empirical' and 'subject to many exceptions'. They did a lot of chemistry and noticed trends. It didn't matter exactly how it worked, they pieced together a phenomenological theory that worked some of the time.

Yeah I read that but how did they even talk about there being double bonds after elimination if there wasn't any theory describing their existence. Also can you please elaborate on what does empirically exactly mean in this context and can I find somewhere what trends did they notice or how they concluded?

The term "double bond" stems from their time, afaik. Since the ~1930ies we know that the two bonds are not a pair of equals. They knew about the molecular geometry (or rather connectivity), and about addition and elimination.

History of chemistry can be very confusing. So hard to keep track who knew what, and when, when the old journals are hard to get, and textbooks even harder, and the nomenclature wasn't as uniform yet. Good opportunity to practice your school French and German, though. ;-)

Thanks a lot. This really clears things up. Also do you know of any standard textbook or source which has a comprehensive timeline of development of chemistry?

This might be a starting point for the history of chemistry: http://web.lemoyne.edu/~giunta/papers2.html#organic

This excerpt is from an article in J Chem Ed (J. Chem. Educ., 1961, 38 (6), p 297
DOI: 10.1021/ed038p297) describing the contents of Markovnikov's 1870 paper in Liebig’s Annalen (translated into English, I guess):

So this was before the discovery of the electron, but as you can see from the diagrams, the concept of chemical structure, atoms and valence was already in place.

Right what I wanted to add. Chemistry is first of all an experimental discipline, I always like to mention that structures like phtalocyanines or something like that were deduced solely by building/dismantling them on the bench! And just using melting points and the likes!


text:
Why do Hydrogen Halides(with the exception of HF) form mostly strong acids?
Why do Hydrogen Halides(with the exception of HF) form mostly strong acids?
My guess is that the negatively charged Halide part of the molecule is attracted by the partially positive part of the water molecule. Thus it is separated more easily. Although because it has one more electron it is more stable once it has dissociated

This question has already been answered:https://chemistry.stackexchange.com/questions/34818/why-is-hf-the-least-acidic-hydrogen-halide

I would like to correct a few points firstly that the halide part is not negatively charged. Also the electronegativity of the halide is the dominating effect in this case.
Hydrofluoric Acid as we know is not a very strong acid as compared to other hydrogen halide acids, had it been the case of electronegativity this would not have been the case.
The dominating factor here is the sizes of the atoms forming the molecule. As the size of halide increases, the bond strength of the H-X bond decreases, making it easier to break the Hydrogen halide bond and act as a stronger acid.
So the acidic strength comes out to be:
              HF < HCl < HBr < HI

Agree mostly but why "the halide part is not negatively charged"? What about $F^-$, $Cl^-$, $Br^-$ ,$I^-$ ...?

The question stated that the negative charged fluoride existed before dissociation, which wont be the case

Ah, I interpreted that to mean the halogens are electronegative, but ok.


text:
"Estrontium" on poster
I was looking at a poster for an international chemistry olympiad. Its background contains some random elements as periodic table tiles.
One of them said "Estrontium." A google search returned nothing but clip art.
Is this an accepted spelling internationally (similar to the way Spanish speakers prefix "e" on words starting with "s" followed by a consonant) or a careless typo on their part? My chemistry teacher was equally confused.

Sounds like Spanish to me. Olympiads may be international, but element names definitely aren't.

That's pretty weird, since in spanish it would be "estroncio". The name strontium comes from that of a scotish village. What was the language of the poster?

Probably multiculturalism at its best. When I studied in a public high school, there was a periodic table in the classroom where element thallium had a handwritten prefix "geni-". "eStrontium" isn't that bad in comparison though:)

The entire poster was in English. The olympiad itself took place in Paris.

Like all good Americans, the only language I know is English. So if it isn't English, it is Greek to me.

Is it just strontium?

Did the tile in question contain more than just the element name (atomic number or weight, for example)? Any chance you took a picture?

@andselisk I'm puzzled how you get "multiculturalism" from a typographical error or a high-school student with a middle-school sense of humor?

@jeffB Episode with Tl is an unrelated story, just a memory to spice things up. As to why I call the precedent multicultural: an incorrectly formed word using Spanish prefix, Scottish name and Latin suffix has somehow managed to pass quality control and show up on a poster of an international olympiad. I thought it was obvious:)

Seems like a typo, as all the spanish periodic tables I could find give the spanish translation of strontium as estroncio. None of the elements seem to use the "-ium" suffix when translated into spanish, so it looks like they just forgot to fully translate the word.

"Estrontium" is not used as an element name in any language. It appears that the error traces to a single user by the name of Alejo Miranda (listed as from Ecuador) who has posted a large collection of periodic tables and individual element clip art to shuttersock, iphoto, etc. in which "estrontium" is incorrectly used as the English name for strontium.  The Spanish periodic table correctly uses estroncio.  The link to the gllery at shutterstock is
https://www.shutterstock.com/g/Alejo+Miranda?searchterm=estrontium&search_source=base_gallery&language=en&sort=popular&measurement=px&safe=true


text:
Corrosion of iron: no electron pushing mechanism?
Why does the oxidation-reduction reaction between iron, water, and $\ce{O2}$ gas to form rusts lack an electron pushing mechanism and a specific sequence of steps?
$$\ce{4Fe^0(s) + 3 O2(g) + 2n H2O(l) -> 2 Fe2O3·nH2O(s)}$$
I've noticed that most of the reactions described in undergraduate organic chemistry textbooks have detailed and widely accepted electron pushing mechanisms whereas most inorganic reactions (such as the corrosion of iron) lack electron pushing mechanisms. Why?
Also, which type of $\ce{O2}$ is involved in the corrosion of iron? The singlet excited state or the triplet ground state or doesn't matter?
$$
\underset{\text{triplet oxygen}~\ce{^3O2}}{
\ce{
·\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}-\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}·
}
}
\qquad
\underset{\text{singlet oxygen}~\ce{^1O2}}{
\ce{
\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}-\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}
}
}
$$

My tip to the hat (i.e., a compliment) for typing this nicely the two Lewis formulae with plain LaTeX' \underset instead of using a mol sketcher.

Arrow pushing mechanism is just a formal way of thinking how the reaction might be occurring. Unfortunately, organic chemistry textbooks present it as if all of these reactions were occurring in real time, step by step, and we are able to watch it like a movie. The reality must be far more complex. If you were truly investigating a reaction mechanism, we have to design very clever experiments (there are books on this topic). It is experimental data that determines the true outcome of an organic reaction not the arrow pushing formalism. Usually, when chemists find out the experimental products, then they postulate arrow pushing mechanisms.
Coming to the rust question. Rust has puzzled chemists and engineers for decades because it causes millions of dollars of losses every year.  Your equation is a very simplistic way of thinking that rust forms in a single step.
$$\ce{4Fe^0(s) + 3 O2(g) + 2n H2O(l) -> 2 Fe2O3·nH2O(s)}$$
It does not proceed in a single step. Our atmospheric chemistry is very complex, you have carbon dioxide, you have sulfur dioxide, nitrogen oxides, ozone, water vapors, sunlight, plenty of free radicals and plenty of minor trace components. Then there are different phases of rust too. You can have a quick look at Google Scholar and one such representative abstract [1].
In short, people still do PhD in this field and one can only imagine how complex corrosion science is.
References

Misawa, T.; Hashimoto, K.; Shimodaira, S. On the Mechanism of Atmospheric Rusting of Iron and Protective Rust Layer on Low Alloy Steels. Corrosion Engineering 1974, 23 (1), 17–27. https://doi.org/10.3323/jcorr1974.23.1_17.


text:
Electrolytes are ionic compounds that conduct electricity
Electrolytes are ionic compounds that conduct electricity, or the flow of charged particles, when dissolved in water. This occurs via the separated ions acting as those charged particles, or charge carriers, when a voltage is applied. Which compound would not create an electrolyte solution when dissolved in water?
A) ammonium acetate, $\ce{NH4C2H3O2}$
B) sucrose, $\ce{C12H22O11}$
c) sodium hydroxide, $\ce{NaOH}$
D) potassium nitrate, $\ce{KNO3}$

You have to improve your questions because it looks like homework.Most likely your future questions will get flagged.Write the problem you are having when you are asking a question.

Welcome to Chem.SE. For these questions, can you show us and edit in your way of approaching the question so we can help you better? Because what you said sems like you just copy from a textbook and not understanding what it means

Sucrose is an covalent compound and it does not ionize.
$\ce{KNO3}$, $\ce{NaOH}$ are good ionic compounds hence the can form electolytic solutions.
Ammonium acetate also can ionize.
Hence the answer is 2

Hey! It's not a hw question, its asking about Which compound would not create an electrolyte solution when dissolved in water

We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you.
Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework.
Don't worry, they're not banned. But, we require a minimal effort.
Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.
See homework


text:
The products of a chain reaction beginning with acetone and sodium amide
I encountered the following question in an exam:

$\ce{CH3COCH3 + NaNH2}$  giving out $\ce{A}$.
$\ce{A}$ then reacts with $\ce{C2H2}$ giving out $\ce{B}$.
$\ce{B}$ then reacts with $\ce{H+}$ to give $\ce{C}$.
$\ce{C}$ then reacts with 1 equivalent of $\ce{H2}$ in presence of $\ce{Pd}$ to give $\ce{D}$.
$\ce{D}$ then is heated to $\pu{400 ^{\circ}C}$ in presence of $\ce{Al2O3}$ to give $\ce{E}$.

I reasoned it as the first reaction could be an acid-base reaction, but that didn't help much. Then I tried a nucleophilic substitution, but there aren't any good leaving groups and still we get stuck after the first step.
One of the teachers reasoned it as the first reaction being reversible and it goes back to $\ce{CH3COCH3}$ which then reacts with all other and it ends up at isoprene which is the correct answer according to the answer key, but this explanation seems very flawed to me as then what was the purpose of the first step? Can someone please check and write the steps and products in each reaction?

Acetone and sodamide give acetone sodium enolate A, nothing further happens until acetylene is introduced. The pKas of acetone and acetylene are both around 25 so an exchange occurs reforming acetone and creating sodium acetylide. This seems an unusual way of making sodium acetylide, but I don't think this is a real world example (unless someone can find a reference).
Sodium acetylide adds to acetone giving the sodium alkoxide of tertiary alcohol B
Work up with dilute acid gives the tertiary alcohol C.
Hydrogenation over Pd catalyst reduces the triple bond to a double bond giving D.
Passing D over alumina at 400C dehydrates the tertiary alcohol to give isoprene.

Nice answer to a poorly formulated reaction sequence.  Given the pKa's of acetone (~20) and acetylene (~25), a practicing chemist would prepare sodium acetylide first and then add acetone. One equivalent of hydrogen in the presence of "Pd" will not be selective, also giving over reduction of the initially formed alkene. That's why we have Lindlar's catalyst. Alas!

Is it possible for an aldol self-condensation to happen in the first step? I went along a different line of thought. The product then undergoes DA reaction with acetylene.

An aldol, as the alkoxide, is likely reversible. Aldol, itself,  is difficult to prepare in the presence of base IN the solution of acetone. A Soxhlet technique is used. You are presuming that mesityl oxide is formed. Less likely. Unsure what you had in mind for a Diels-Alder between mesityl oxide and acetylene.