text:
Finding structure of a compound given its reactions
Question:

An organic compound A (molecular formula $\ce{C6H12O}$) does not change the colour of acidic dichromate solution. Compound A on treatment with $\ce{H2SO4}$ produces alkene which on oxidative ozonolysis gives a molecule $\ce{C6H10O3}$ which gives positive iodoform test. Find out structure of A.

So I know about ozonolysis, but I do not know when acidic dichromate solution changes colour. From the question an alkene is produced so I guess it may be an alcohol, but if someone an explain about it and iodoform test, then I will be able to give another try on this problem.

Welcome on Chemistry SE, your question is nice, but you should be more careful with details (orthography, punctuation etc.) .

I have taken the liberty of identifying your unknown compound.  If you'd like to work through the problem yourself, then carefully scroll down to the part of the answer that deals with the iodoform test.  Good luck!
What we know

Just looking at the molecular formula we see that there are too many hydrogens for the 6 carbons to be in a benzene ring.
does not change the colour of acidic dichromate solution - acidic dichromate readily oxidizes primary and secondary alcohols to aldehydes and ketones respectively; tertiary alcohols are not oxidized. The aldehyde produced from a primary alcohol can be further oxidized to a carboxylic acid, this results in a solution color change from orange to green. But there was no color change, therefore there is no primary $\ce{-OH}$ or aldehyde group in compound A.
Compound A on treatment with H2SO4 produces alkene - this is a common reaction that causes deydration (elimination of water) and converts an alcohol into an olefin.  Therefore compound A is (as you suggested) an alcohol (but not a primary alcohol).
oxidative ozonolysis gives a molecule $\ce{C6H10O3}$ - ozonolysis will cleave the double bond in the compound produced upon dehydration.  The key fact here is that the resulting product still contains 6 carbon atoms.  If the double bond involved carbons outside of a ring we would wind up with an ozonolysis product containing fewer than 6 carbons.

(image source)
Only if the double bond is contained within a ring will we retain all 6 carbon atoms.  Therefore our double bond must be in a ring.

positive iodoform test - This is our final clue, the iodoform test (or more generally the haloform reaction) is a test that detects the presence of methyl ketones and generates iodoform ($\ce{HCI3}$) in the process.

Therefore we know that there is a methyl ketone in our ozonolysis product.
Combining all of this information (non-aromatic, ring, non-primary alcohol, methyl group) leads to the following structure
(mouse over to see the answer)

which is called

 1-methylcyclopentanol.

Note: some methylated-cyclobutanols might also be considered, but when these dehydrate and produce an alkene, the double bond is likely to be placed external to the ring (or [advanced topic] external to a cyclopropane ring formed upon rearrangement) and we would not wind up with 6 carbons in our ozonolysis product.

While permanganate and dichromate are fairly similar with respect to their use as oxidants, it's still a bit confusing to link one under the name of the other.

@ron thank you sir for such effort in the question , as there was four options  given in the question i faced no difficulty in answering it after your kind help . Great explanation!

@KarlRatzsch Thanks for the good catch, I've made the necessary changes.


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How can 32 g of Gatorade powder contain 33 g of carbohydrates?
A screenshot of the nutrition facts of a can of Gatorade powder was posted on my Facebook feed:

The label reads "Per 32 g (about 3 tbsp)", but if you read through the ingredients, the amount of carbohydrates listed exceed that of the serving size (33 g).
My question is, how can this possibly be? Is there a chemical process as some people have suggested that is changing the 32 g of powder into 33 g of carbs once water is added?
Re-asking it here for a chemistry perspective as some users have suggested on skeptics.SE.

Is there a chemical process as some people have suggested that is changing the 32g of powder into 33g of carbs once water is added?

No. Even if there was, it's not relevant in this case: the numbers on the label are an assay of the pure contents of the product before it's mixed with anything, or prepared or altered in any way. Nutrition labels refer to the product as-is.
In the United States, the Food and Drug Administration is in charge of labeling the quantities you're interested in. The relevant document is the Food Labeling Guide, one of many documents which are linked from a more comprehensive list called Labeling and Nutrition Guidance Documents and Regulatory Information. As per the comments (below), I'm noting that it appears as if the product pictured in your post is from the Canadian market and thus subject to nutritional labeling dictated by agencies in Canada. That said, the methodologies used by both the US and Canada are similar.
There are at least 2 regulations detailed in the labeling guidelines that we need to examine here to explain what's going on:
(1) The numbers reported on the container have simply been rounded up or down as per the Food and Drug Administration's Appendix H in the Food Labeling Guide. For instance, they show that for Total carbohydrate and Sugars, anything less than 0.5 g is expressed as 0 g, and quantities greater than or equal to 1 g are to be expressed in the nearest 1-gram increment. This is why of the 33 g of total carbohydrate, only 31 g appears to be accounted for as sugars;
and
(2) Total carbohydrate is calculated in a way that propagates rounding errors introduced as detailed in (1):

Total carbohydrate is calculated by subtracting the weight of crude protein, total fat, moisture, and ash from the total weight of the sample of food.

In summary: The numbers don't add up because quantifying the components of a food to conform to the rules and regulations for reporting them on a Nutrition Facts label allows for the accumulation of rounding and measurement error.

I think you'll find that the controlling agency in this case is the Canadian Food Inspection Agency, not the FDA. The remarks concerning assay and rounding are more-or-less correct, however.

I don't see how this explains the discrepancy. If total carbs is calculated by subtracting a nonnegative quantity from the original total mass, total carbs can't be greater than that original total mass.

@DavidRicherby, 32 g product is rounded once, but 33 g carbs is a sum of several rounded totals.

@CeesTimmerman But the answer says that the 33g total carbs is a difference of rounded terms (see the last quoted section of the answer). A difference of nonnegative values cannot be larger than the largest of those values, whatever rounding scheme is used.

@David, that's for the USA, but the product label is governed by Canada.

So in effect this is just slightly dirty sugar ...

Thanks @CeesTimmerman  - post edited to include your information re: Canada.

Thanks @StanRogers - post edited to include your information re: Canada.

@HagenvonEitzen Yup: https://en.wikipedia.org/wiki/Oral_rehydration_therapy

@ToddMinehardt No, because that explanation also says that total carbohydrates is calculated by subtraction and, however you round it, subtraction of non-negative quantities cannot possibly lead to a larger value than the one you started with.

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What bases are fit for human consumption and can be used in foods?
I know that are several acids that can be found in the kitchen, such as citric acid and vinegar.
However, I never heard of a base that can be used in the kitchen to raise the pH levels of any food. I am well aware that this mix would probably make a salt as a result. Something like hydrogen peroxide doesn't sound safe for food. So, my question is:
What are (if any) the bases that can be used on those acids and still be human-consumable?

I edited your post but didn't correct the part about "a base that can be used in the kitchen to reduce the pH levels of any food" - I'm pretty sure you mean raise the pH levels. You can/should edit your question for clarity on that point.

Thanks @ToddMinehardt, it has been a while since I last took Chemistry. Anyway, I should've checked it before posting, but thanks for the clarification.

Unfortunately Googling for "alkaline food" these days is a bad idea; results are dominated by nonsense sites espousing the "alkaline diet" nonsense / woo.
There aren't many foods that are naturally alkaline.  The two exceptions I know of are:

baking soda
egg white

Sometimes when I'm cooking and accidentally add too much acid (e.g. lemon juice or vinegar) to a broth or sauce, I can partially correct the mistake by adding small amounts of baking soda.  This converts the excess acid to the sodium salt and $\ce{CO2}$ is evolved.  For the example of vinegar (acetic acid):
$\ce{CH3COOH + NaHCO3 -> CO2(g) + CH3COONa + H2O}$
So sodium acetate salt is left behind, which limits the degree to which you can correct overaddition of acid.  (You don't want your food to be too salty.)$%edit$

sodium acetate itself is also basic and sometimes used with food

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Relative strength of desiccants
Is it meaningful to characterize the relative strength of desiccants?  For example, is there a measure of hygroscopy, or an ordering of desiccants, such that higher ones will always dry lower ones?
Let us restrict the question to "practical" desiccants, which are hygroscopic substances that can be "recharged" (i.e., dried out) by heating.  I'm not certain I know all possible mechanisms of water sorption within this class of chemicals.  But is it safe to say that for all such chemicals, at standard temperature and pressure, there is some water vapor pressure around a desiccant that has sorbed any water?
I am trying to imagine a series of tests in which two different desiccants are put in a humid chamber in adequate quantities that either could completely sorb all the water.  At hydrous equilibrium will the "stronger" desiccant contain virtually all of the water?  Or is the equilibrium distribution of water a function of the desiccants' relative "hygroscopy," rates of sorption, or some other factor(s)?

This is ultimately a practical question.  One example: I have a hygroscopic, hydrated salt X.  Can I dry it out by sticking it in a sealed box at STP with a "more powerful" desiccant?  If not what determines the limit of dehydration that would be achieved by exposure to other desiccants?

Probably the easiest way would be to compare equilibrium water vapor pressure over various desiccants.  I have a feeling that I have seen the relevant chart, but I can't remember where.

You also have to consider particle sized. Finely divided desiccants work much better than coarse grains of exactly the same material.

@Michael: Certainly.  That would be part of the practical characterization of the desiccant.  My guess is that the "form" of the desiccant is more indicative of its capacity than of its "strength."  ("Form" could also include such specifications as a desiccant needing to be stirred to maintain its strength.)

Yes, it is meaningful, but often ignored for practical reasons. In the lab, e.g. in a desiccator, you would just use a large excess of the desiccant of choice which would always work. It’s often more meaningful to classify them as to whether they are acidic, basic or neutral.
But let’s assume we wanted to create that scale. For simplicity reasons, let’s check a single desiccant first. We need to consider the following equilibrium:
$$\ce{H2O_{(g)} <=> H2O_{(x)}}$$
With (x) again being the water bound to the desiccant in whichever way. (Do not think of this as chemical bonding; it usually is not). It is important to remember that this is an equilibrium with an equilibrium constant k.
$$k_{\mathrm{x}} = \frac{[\ce{H2O_{(x)}}]}{[\ce{H2O_{(g)}}]}$$
One could rearrange this in numerous ways resulting in a representation that essentially says the partial pressure of water in the surrounding atmosphere is more or less proportional to the concentration of water in the desiccant.
$$p_{(\ce{H2O})} \approx c \cdot [\ce{H2O_{(x)}}]$$
Now let’s add a second desiccant into our consideration. The first equation now rearranges itself to the following:
$$\ce{H2O_{(y)} <=> H2O_{(g)} <=> H2O_{(x)}}$$
For each of the two subsystems we again get an equilibrium constant $k$ which we can rearrange to proportionalise the partial pressure to the concentration of water in x or y. $k_{\mathrm{x}} \neq k_{\mathrm{y}}$. There will be one partial pressure of water where both the left side and the right side of the double equation will be at equilibrium; that partial pressure is the final equilibrium that will be reached. This gives us the following conclusions:

Never will the air be completely dry in the presence of (only) a (or two, or a hundred) desiccant(s).
Neither of the two desiccants will fall completely dry again; rather unlike when heating (in an open atmosphere).
Say you start the experiment with the weaker desiccant and then add a dry sample of the stronger desiccant:

The stronger desiccant will further reduce the vapour pressure;
The reduced vapour pressure will remove water from the weaker desiccant;
That water will partly be absorbed by the stronger desiccant
However neither will the weaker desiccant be completely dried nor will the partial pressure of water in the atmosphere be completely zero.

How dry (or wet) either desiccant will end up/how low the partial pressure of water will end up, depends on the strengths of the desiccants. If you use a very strong and a very weak one, the strong one will absorb much more water than the weak one. If the two are similar in strength, they will absorb similar amounts of water.

I'd say no, because there is also a reactivity dimension to desiccants. Some drying agents react with some solvents and therefore can't be used as a desiccant. This happens for many organic solvent and desiccant pairs. Some desiccants also react as desiccants and this changes their drying power. This is the case for phosphorous pentoxide, for example, which gradually forms a viscous layer of phosphoric acid which stops any drying. So any linear ordering you are attempting is likely to be overwhelmed by the reactivity side effects of the many desiccants.
On the other hand, if you are only interested in some theoretical point in time estimate of desiccant efficiency, this might be possible but would have virtually no practical import. A desiccant has to react over time and the side effects cannot be ignored.

Interesting.  To keep the question tractable let's assume that we're only talking about water in a standard atmosphere; no other volatile compounds present.  The question is still widely applicable with that assumption.

No that's still not true. Phosphorous pentoxide is used as a general very strong desiccant, but still has the viscous phosphorous acid problem. What you are proposing is a very controlled environment which would likely never exist in real conditions (i.e. we always expose new phosphorous pentoxide). This is technically possible, but has almost no practical application. Are you only interested in theoretical but perhaps non-achievable results?

No, I'm interested in the practical application.  All desiccants reach sorption limits.  I found this page which shows capacity and rates for a few common desiccants.  In my proposed test we would put enough phosphorous pentoxide in the test chamber that it could absorb all of the water before its sorption rate slope goes below 1, or hits zero, or something like that.  (Presumably, as $\cf{P4O10}$ is a popular practical desiccant, it is used in quantities or methods in which the $\cf{H3PO4}$ doesn't degrade its desiccation strength.)

At hydrous equilibrium will the "stronger" desiccant contain virtually all of the water? Or is the equilibrium distribution of water a function of the desiccants' relative "hygroscopy," rates of sorption, or some other factor(s)?

"Rates" aren't relevant for questions about equilibrium.

I am trying to imagine a series of tests in which two different desiccants are put in a humid chamber in adequate quantities that either could completely sorb all the water.

For that particular test, consider two dessicants $\ce{X}$ and $\ce{Y}$.  The winner of the test will be determined by the thermodynamics of the two reactions:
$$\ce{H2O(g) <=> H2O_{(X)}}$$
$$\ce{H2O(g) <=> H2O_{(Y)}}$$
which represents the the absorption of water vapor into phase $\ce{X}$ (or $\ce{Y}$) respectively.  The two reactions can be combined to yield
$$\ce{H2O_{(X)} <=> H2O_{(Y)}}$$
and the free energy change $\Delta G_r$ of that reaction will determine which dessicant is "stronger" in the sense of that particular test.
This test doesn't take into account many practical considerations that other answers have noted, such as reactivity.
I'd bet that liquid sulfur trioxide would win most desicating contests.$%edit$

It almost sounds like you have the answer to the question!  If I understand what you're saying, there is a binding energy associated with each desiccant (which may not be constant over hydration level, but which nonetheless can be characterized by a monotonic curve?), and in the limit water will be bound to minimize free energy?  E.g., if X has higher binding strength than Y over all hydration levels attainable in the test then all water will ultimately be bound to X?  If so then what is this "binding strength" desiccant characteristic called in the art?

Note also: Chemicals like $\ce{SO3}$ would not be included in this question as "practical" desiccants because they are not "rechargeable" by simple baking in a standard atmosphere (AFAIK).

It depends on temperature as well. Sure metaphosphoric acid is extremely difficult to dehydrate to phosphorus pentoxide, but with heat and liquid $\ce{SO3}$, it should be rather easy if it's under vacuum, and everything is in separate containers inside a larger one. This way the equilibrium is broken since the $\ce{SO3}$ absorbs the water and can't get back to the resulting pentoxide. So despite $\ce{SO3}$ being a weaker dessicant, it can dehydrate hot metaphosphoric acid under certain conditions.


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Ligand Binding Paradox?
I have a ligand $\ce{L}$ and a receptor $\ce{R}$. As can easily be shown, with dissociation constant $K_\mathrm{D} = \frac{[\ce{L}][\ce{R}]}{[\ce{LR}]}$ the fraction of receptors occupied by ligand is
$$\theta = \frac{[\ce{LR}]}{[\ce{LR}] + [\ce{R}]} = \frac{[\ce{L}]}{[\ce{L}] + K_\mathrm{D}}$$
It's easy to see that the fractional saturation of receptors is independent of receptor concentration. Now here's where I get confused. Let's say we have a constant volume of $\pu{1 L}$ with $\pu{5 mol}$ ligand and a $K_\mathrm{D}$ of 5:
$$L = 5, \quad K_\mathrm{D} = 5 \quad \to \quad \theta = 0.5$$
If we have $\pu{10 mol}$ of receptor total ($[\ce{LR}] + [\ce{R}] = 10$), then $\pu{5 mol}$ will be bound to ligand. Now let's say we increase the number of receptors by $10\times$. Our formula for $\theta$ is independent of $[\ce{R}]$ or $[\ce{LR}]$. We now have $\pu{50 mol}$ receptor bound to ligand.
That being said, we only had $\pu{5 mol}$ ligand to start with, which is itself a paradox. Can someone explain to me where I went wrong?

What do you mean by Kd=5; 5M?

I see where I made my mistake. I set $[\ce{L}] = 5$, but this "$L$" is referring to the equilibrium concentration of free ligand -- not the total ligand concentration. In order to do that I should have constrained the system to $[\ce{L}] + [\ce{LR}] = 5$. Working through what this would do currently.

So in the first case, my assumption actually dictated that there be $\pu{10 mol}$ ligand total instead of the $\pu{5 mol}$ I erroneously stated. I incorrectly equated total ligand and free ligand the first time I did this.

So I see where I messed up, but now I'm curious if the values of $[\ce{L}]$, $[\ce{R}]$, and $[\ce{LR}]$ can be solved as functions of $[\ce{L_{total}}]$, $[\ce{R_{total}}]$ and $K_\mathrm{D}$.

If you still have queries then you should add this as an edit to your question.

Three unknowns (L, R, and LR) per three equations (total L, total R, and the dissociation constant) - should be solvable, don't you think?


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Diluting a sodium chloride solution with specific glassware
I have a $\pu{0.3 M}$ $\ce{NaCl}$ stock solution and need to create a $\pu{0.01 M}$ $\ce{NaCl}$ solution that is $\pu{50 mL}$. The only tools I am given are volumetric pipettes that are $10.0, 5.0, 1.0 \text{ and }\pu{0.5 mL}$ and $\pu{50.0 mL}$ volumetric flasks. I understand the physical process where I would measure out a specific amount and place it in the flask using the pipette(s) and then add the required amount of water.
I need to use the dilution equation $M_1\times V_1 = M_2\times V_2$ which I did and I got $\pu{1.666667 mL}$ of solution that needs to be taken from the $\pu{0.3 M}$ stock solution to create the $\pu{0.01 M}$ solution. The only problem is that I can't do that since the materials I have do not allow me to measure out that amount.
I tried to "convert" the $\pu{0.3 M}$ to various other concentrations that would hopefully get me down to $\pu{0.01 M}$ but have failed so far. I tried to find concentrations that would result in a whole number or a value with $.5$ at the end after dividing by the product from $M_1\times V_1$ but failed every time. In addition, I combined the two concentrations and found the midpoints between them and tried to use those values, but they did not work either. Where is a good place for me to start?

Assuming you can use a beaker (or just one of those 50 ml volumetric flask for mixing), you could first dilute some of the 0.3 M solution to 0.1 M by adding 5 ml of stock to 10 ml of water. Now you have a 0.1 M solution of NaCl which you can calculate how to dilute using the M1*V1=M2*V2 equation. Doing this, I calculate that to make 50 ml of 0.01 M solution, you need 5 ml of the 0.1 M solution you made, which you can do with a pipette.

Generally, the first step is not proper since adding 5 ml of a solution to 10 ml of water does not precisely yield 15 ml.

This question is years old, but I went through the calculations for my own interest.
The stock solution needs to be diluted to 1/30th of its original strength.
Take 5 ml of it and add 5ml water, resulting in 10ml of 0.15M solution.
Add 20 ml water, resulting in 30ml of 0.05M solution.
Take 10ml of that, add 40 ml of water. The result is 50ml of 0.01M solution.
To verify: 10 x 0.05 = 50 x 0.01
That's one way to do it.


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How to obtain nitrogen using only homemade utilities?
I am now constructing a RC plane. I would like to fill the tires of the landing gear with nitrogen to avoid combustion of oxygen during landing. I would therefore like to obtain a small amount of nitrogen using (only) home-made utilities. How can it be done?

You can buy argon or nitrogen in a gas cylinder. It isn't prohibitively costly. However if you only need a gas not supporting oxidation, there is a lot easier way: mix backing soda  with vinegard and pass the product through a filter to stop water drops. The reasulting gas is safe for anything short of active metals.

Crude process, but does the trick: You can use a combustion/oxygen consuming process, like a candle, to burn the excess oxygen. Eventually the candle will be too oxygen starved to sustain a flame, so some residual oxygen would be left (although this would already mean there is no fear of combustion of said tyres). Now to remove the remaining oxygen you'd use something like iron files (since you wanted to use only "household" stuff). Together with some humidity in that air, they will rust rapidly and should drive down the oxygen contents towards negligible. So now we have:

a small trace remainder of oxygen,
the equivalent of $\ce{CO2}$ from the combustion process,
some impurities from the candle wax probably,
trace gases such as helium etc. and ... mostly
nitrogen.

This already is not sustaining any combustion. But if you're a purist or want to use $\ce{N2}$ because it is lighter than $\ce{CO2}$, to remove the $\ce{CO2}$ you could use something like "roasted limestone", i.e. the stuff from you water kettle, which you heat so that you get lime, which then binds with the $\ce{CO2}$ to revert to limestone $\ce{CaCO3}$. Now you have something, I'd assume, like 98% nitrogen and surely to few traces of $\ce{O2}$ to cause any harm to your equipment. Of course, I'd buy bottled $\ce{N2}$, but for the sake of using household implements only, this might do the trick.

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What are the effects of Lithium Dioxide on the body?
My question is about how much science-fiction and how much chemistry is involved in the the 'Randy's Donuts' scene of Iron Man 2:

Agent Romanoff injects Lithium Dioxide into Tony Stark's neck to remove the symptoms of Palladium poisoning.

My Question is: Is it safe to inject Lithium Dioxide into our body?
Does it have any side effects?

A quick google of LiO2 reveals, that the proper name of the compound is Lithium superoxide. Given that it is only stable at about 15-40 K, i.e. colder than -230 deg C, Tony would certainly have a very "stiff" neck after the infusion. It's a movie universe about magic stuff, take it with a wink and a grain of salt.

I'm sorry. I know this question is for scifi.SE but I assumed this site's people are more knowledgeable about Chemical compounds and their effects on organic bodies so if you feel this question is inappropriate for chemistry.SE then please let me know

Maybe rather lithium peroxide but still even touching it wouldn't be good idea.

The better question is "why palladium poisoning" - from the Wikipedia article: Palladium is a metal with low toxicity. ... but there is no clear evidence that the element has any adverse effects on humans.

I was not implying that the question is not appropriate for this site, just saying that it is indeed science fiction and that from a scientific point of view does not make much sense. From this pov, the obvious answer is: If it was possible to do that, it would be very unhealthy.

Is it safe to inject Lithium Dioxide into our body?

No. It will do bad things.

Does it have any side effects?

Yes. It will do bad things.
As lithium is an alkali metal, it does not usually form covalent bonds, so we should treat whatever it makes as ionic. The common form of lithium oxide is $\ce{Li2O}$, which is not that scary. "Lithium dioxide" would be $\ce{LiO2}$. This implies that that lithium is tetravalent, as $\ce{Li^4+}$. This is not possible, as it requires lithium to give away 4 electrons but it does not have 4, only 3. This means that oxygens are in the superoxide form: is $\ce{O2-}$ instead of the more stable $\ce{O^2-}$. That is definitely not stable in isolation at ambient conditions (unlike other alkali superoxides) and I'm not sure where you will get that stuff.
The result of that is that one molecule (or whatever it is) of $\ce{LiO2}$ will really want to convert itself to $\ce{Li2O}$. The superoxide ion is not going to do fun things. For each $\ce{O2-}$ you're going to need 3 more electrons to make it stable, so that is very strong oxidiser.
As a comparison to what this could do to you, I would guess it would be worse than injecting pure hydrogen peroxide (you have about 5% of that in hair bleach), or pure sodium hypochloride (similar 5% in kitchen bleach). This is definitely not going to treat "palladium poisoning" because it will probably kill you beforehand.
That said, using an oxidiser (such as lithium dioxide) to treat noble metal poisoning (such as palladium) seems dubious to me. I'm not a doctor, but I reckon that there are better method to treat that. So why lithium and palladium? I guess is sounds more credible than "using titanium to treat potassium poisoning" but less credible than "using dysprosium to treat beryllium poisoning", both of which make absolutely no sense as "using lithium to treat palladium poisoning".

As an amateur doctor I wonder if there is not some standard chelation therapy that would extract palladium (were that ever, for some reason, necessary).

Also, why wouldn't you assume "lithium dioxide" is a synonym for lithium peroxide ($\ce{Li2O2}$), rather than the implausible lithium superoxide?

@feetwet because Li2O2 would be named dilithium dioxide, or just lithium peroxide. Following the convention for the rest of the periodic table, something dioxide is always XO2.

There is no such thing as lithium dioxide. Lithium oxide however (Li2O) does exist and reacts violently and exothermic with water to produce H2 gas and LiOH, which itself is a strong base. Not something you want in your body which consists mainly of water. There's also lithium peroxide (Li2O2) which you could call lithium dioxide, but it'd be a stretch and not really correct. The peroxide reacts similarly to Li2O with the added bonus of forming hydrogen peroxide.

There is also lithium superoxide, $\ce{LiO2}$ which you could call lithium dioxide, but it is highly unstable and has only been isolated at very low temperatures.


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What Affects Tm and Tr In Chromatography?
In chromatography, a mobile phase with analytes is passed over a stationary phase. A given analyte has different solubility in the two phases and consequently spends different periods of time in them. A particular molecule while in the stationary phase remains immobile, is swept along when present in the mobile phase.
$T_m$, the time spent in the mobile phase, is the same for all analytes. The reason (I'm not so sure) must be since all analytes travel the same distance finally, with the same speed as the mobile phase, so take the same time in so doing. But since some of these spend longer times in the stationary phase than others, the total time to reach the end, that is, $T_m + T_r$ is different ($T_r$ is the time spent in the stationary phase).
But I have a bit of a trouble understanding the rate aspect of the process, and what exactly determines the $T_r$ and $T_m$ - are they the average times spent by an arbitrary molecule in the two phases? If so, are their values dependent on the rate of the process, or the solubilities?

Think of it like a solvent extract. Each solvent extraction is a plate.
So at time 1 I do extraction 1  and move top layer to extraction flask 2.
At time 2 I have two extractions to do. I extract flask 2 and move top layer to flask 3 and then I extract 1 again and move top layer to 2 again.
At time 3 I have three extractions to do. and so one.

Another analogy.
Think of the whole thing as binomial trials. Each molecule has a different binomial probability. So the absolute difference in the peak positions between the molecules gets bigger as the number of trials increases.


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Evaporating water from a combination of dietary minerals: will it work safely or react?
I've combined multiple elements dissolved in water into a preferred ratio, but would like to evaporate some of the water to increase the concentration (let's say at 200 $^{\circ}\mathrm{C}$ in the oven). The elements in question and the amounts are: 2 mg chromium, 5 mg lithium, 2 mg vanadium, 140 mg iron, and 50 mg boron. The volume of water is about 60 ml.
I would like to know if I can safely do this procedure without a) causing a reaction; and b) evaporating any of the elements.

Without info about what salts exactly did you use the may be problem with answering this.

It seems like 200 degrees is overkill. Why not simply boil off the water (maximum temp. 100C), or use vacuum drying at room temperature? I would not expect a reaction from commonly found salts of these elements, plus boron in the form of boric acid.


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Why is the inductive effect for a protonated amino group greater than that of a nitro group?
Why does $\ce{NH3+}$ have a higher $-\ce{I}$ effect ( inductive effect) than $\ce{NO2}$ even though oxygen is more electronegative? Does it have something to do with positive charge on $\ce{N}$?

The short answer is that $\ce{NH3+}$ is a cation, and any cation will always draw electrons towards it stronger than any neutral species.
The electronegativity of the species involved is only a secondary cause. However, one can safely say that $\ce{NH2}$, if nitrogen’s lone pair is (somehow) impeded from excersizing it’s $+\ce{M}$ effect, would have a lesser $-\ce{I}$ effect on the whatever than $\ce{NO2}$ would.

The inductive effect which you are talking about must be from the most common order given for strength of -I.
Here what you are missing is that its not $\ce{NH3+}$ but its - $\ce{NH3+}$ or $\ce{R-NH3+}$.
What I mean to say is some alkyl or aryl group will be attached to it.
Coming to your Question, -I, means that the group attached will pull electrons more towards itself, gaining partial negative charge and hence imparting partial positive charge to the alkyl or aryl group . Though $\ce{O}$ is more EN than $\ce{N}$, $\ce{NO2}$ is less EN comparatively because there is cation on $\ce{NH3+}$, so it will have more tendency to gain or pull an electron.

authors: Михайло Олександрович Іванцов


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Is the axial position more suited for a lone pair or an electronegative element?
In one question, I found that it is given $\ce{XeF2}$ and $\ce{CO2}$ have the same shape.
$\ce{CO2}$ is linear. In $\ce{XeF2}$, if the fluorine is placed in an axial position, then only shape is linear. But in most compounds, lone pairs are placed in an axial position.
Is the lone pair or the electronegative element more appropriate for placement in an axial position?

So in $\ce{XeF2}$ there will be three lone pairs on the xenon, and the xenon will have two covalent bonds to fluorines. The xenon is of course the central atom, on to that we apply VSEPR-theory. We count five electron dense areas (two bonds and three lone pairs) and so VSEPR-theory predicts a trigonal bipyramidal shape for $\ce{XeF2}$

The lone pairs want to stay as far away from each other as possible (in VSEPR). Indeed would there have been just two lone pairs putting them axially would have introduced the maximal possible angle of 180°. Instead we have three lone pairs, so the equatorial positions introduce the best possible angle of 120°. Compare this to the 90° angle introduced if we had put two lone pairs axially and just one equatorial.
What is left now are the axial positions to which we allocate the fluorines, obtaining a linear molecule. In this case the VSEPR prediction has been confirmed experimentally.

I dislike this explanation extremely. A much better explanation would not apply VSEPR but rather say that it is a four-electron three-centre bond with a p-orbital of xenon which means that it must follow a p-orbital’s shape namely linear. VSEPR is a simple concept meant for simple cases.


text:
Which element should be placed in axial position bulky one or the smaller one?
In one question for finding $\ce{P(CH3)2H3}$ structure? Which one should be placed in axial position? I think $\ce{-CH3}$ should be axial position to minimise repulsion.

According to Inorganic chemistry (Meissler and Tarr), electronegativity and symmetry actually play a huge role.  More electronegative groups tend to occupy axial positions, even if they are smaller, and symmetry is always preferred.  For an example of the role of symmetry consider, $\ce{PF3(CF3)2}$ ($\ce{-CF3}$ are axial) vs $\ce{PF2(CF3)3}$ ($\ce{F}$ are axial).  For an example of electronegativity being more important than size look at $\ce{PCl3F2}$ which has the $\ce{F}$ at the axial positions.
In this case, $\ce{-CH3}$ groups are both bigger, more electronegative, and putting them at axial positions will be symmetric (and cause the net dipole to cancel); with all these considerations, the $\ce{-CH3}$ groups will be axial.

No, it's the other way around. Axial positions are less spacious, so you'd want to place your smaller substituents there and the bulky ones to the equatorial.

But axial positions are longer than equitorial ones

Yes, and that's exactly because they are more "narrow", so the substituents occupying them are forced farther away from the center.

I'm afraid Pi5 is right you should cite some source if you think otherwise

Would this count as a source?
https://en.wikipedia.org/wiki/Trigonal_bipyramidal_molecular_geometry


text:
Compare ONO bond angle in nitrogen oxides

Which has maximum $\ce{ONO}$ angle?

$\ce{NO3-}$
$\ce{NO2+}$
$\ce{NO2-}$
$\ce{NO2}$

According to me $\ce{NO2+}$ should have the maximum angle because the structure is linear but in answer it is given $\ce{NO3-}$. We can consider the lone pair in $\ce{NO2+}$ but still also by VSEPR theory I don't think it is possible to decrease angle from 180 to less than 120.

You are right, that's $\mathrm{NO_2^+}$, since it is linear, and whoever says otherwise is wrong.
Also, those (except for NO2) are not compounds, but particles.


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Glycine nucleation simulations
I am trying to run nucleation simulations of glycine in water. To generate a supersaturated solution, a solution that will crystallize, I need to know the solubility. Luckily for me, this data exists. Now at 1 atm, 324 K, glycine's solubility is 0.338 g/ml water.
My question is how to determine the appropriate box size for a given simulation, if I specific the number of glycine molecules. I posted a question similar to this before,in the CCL mailing list, and was told to just assume ml water is ml of simulation box. For instance, if I want 1024 glycine molecules, whose molecular weight is 75.066 g/mol,
  (1024/6.0223*10^(23))*(75.066) = weight = 1.27*10^(-19) g
  weight/0.338 = ml simulation box = 3.78*10^(-19) ml
  box = ml simulation box^(1/3) = 72.28 angstroms

I guess my first question is if this procedure appears to be correct. Specifically, all experimental data for solubility is clearly gathered at 1 atm pressure. However, if I fix the pressure, my understanding is that there is only one equilibrium concentration that the MD will take me to. If I think of the number of DOF's here, temperature is fixed, pressure is fixed, and the mol fraction is fixed, so I don't get to choose the box and hence the solubility. If I work in the NVT ensemble, which I am leaning towards, I don't know if I will actually be supersaturated because the simulation is no longer ran at 1 atm. Is there another definition of solubility that I should be using for MD?
The above scenario is more confusing to me when I think of simulating a "bulk" crystal. If I were to simulate a glycine crystal consisting of 1024 molecules, the above method would tell me a simulation box of 72.28 angstroms and a temperature of 294 K would mean the crystal wouldn't dissolve (assuming  nanoscale stability is similar to bulk stability). However, if I am simulating a bulk crystal, why would the box size matter to the crystal? It should always see a "sea" of solvent, and I thought I should only be concerned with making sure the box is large enough to prevent the cluster from seeing its own periodic images.


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Is so called "Polymerized LiNb (Lithium Niobocene)" a possible compound?
The so called "Polymerized LiNb (Lithium Niobocene)" is a fictional compound.
I heard about this substance being used for an interesting purpose. It reacts to mechanical stress or pressure by creating an electric charge (piezoelectricity) and, conversely, deforms, as if external pressure was applied, when voltage is applied. Apparently it was far more "piezoelectric" than any other similar compounds, i.e. it deformed far more drastically than other piezoelectric materials.
Does such a compound, or similar compounds, exist?

http://pubs.acs.org/doi/pdf/10.1021/om980716r for real niobocene

So... Sorry if I've misunderstood, but I don't think anyone has answered my question. Can Polymerized Lithium Niobocene exist?

similar type of question as here http://chemistry.stackexchange.com/questions/35423/what-are-the-effects-of-lithium-dioxide-on-the-body

Be patient, your question is problematic - the naming is vague and improper as it's fictional material - you care more about polymerization, niobocene, lithium, or properties?

Accepted SCFlint's answer because I think my question was far too lacking in chemical knowledge. Sorry guys, my bad :/ - transfer Op's comment from question

I included the link and made clear that we are talking about a fictional substance. I think the paper @Mithoron linked is very interesting, but only from a structural point of view. It does not have the properties that are assigned to it in the fictional universe.

related https://en.wikipedia.org/wiki/Artificial_muscle

https://chemistry.stackexchange.com/questions/140519/artificial-muscle-hasels-reaction

https://science.sciencemag.org/content/359/6371/61

Materials such as Terfenol-D and Galfenol are magnetostrictive ferrous alloys that expand and contract in the presence of magnetic fields, as well as creating magnetic fields when acted on by a physical force.  Both these materials were developed at Ames Lab (formerly the Naval Ordnance Laboratory, that's why the material names end with -NOL).  Terfenol-D has the highest magnetostriction of any Fe-alloy, but the high rare earth content or Terbium and Dysprosium limits its commercial applications.
The "polymer" moniker for Lithium polymer batteries does no refer to the lithium metal or oxides, it only speak to the polymer electrolyte or a polymer casing.  It just a marketing term, no a chemical designation
So on to your question... Polymer, no.  Monomer, no.  Niobium has an extremely high melting point (nearly 2500C/4500F).  It is a very difficult material to melt.  Lithium boils around 1300C.  From a metallurgical processing point of view, you would need a chamber pressurized to several MPa with an inert gas to prevent the lithium from volatilizing as you attempt to alloy it with niobium.  You would also need some powerful magnetic stirring to attempt to keep the mix homogeneous since the lithium would literally float out of the solution.

Welcome on Chemistry SE! Unfortunately you answered pretty tricky question - OP's compound would be rather piezoelectric organometallic polymer than  magnetostrictive alloy as this LiNb (misleading) symbol suggests, but even then your answer is interesting.

Sorry for my mistake (I'm not well-versed in Chemistry at all), out of curiosity, what should I have written instead of LiNB?

@At0micCyb0rg you wrote as it were and you did OK - one has to work with data he has - it's authors of this info who made it misleading. Also maybe I'll try to answer, or someone else - you should be patient here ;)


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What is an "attack" in Organic Reactions?
I began to study organic reactions and found the term "attack" everywhere, what this "attack" precisely mean?

You might have heard the term "attack" in statements like "the nucleophile attacks the electrophile", and terms like frontside and backside attack, and what this basically means is that the nucleophile approaches the electrophile, and begins to share its electrons with the electrophile to form a bond between them (and possibly break a different bond in the molecule too). Frontside and backside attack mean the nucleophile approaches the front or the back of the molecule and forms a bond.
If you've learnt about arrow pushing, you'll know that an arrow has its tail starting at the nucleophile (the electron rich species) and it's head points to the electrophile (the electron poor species) - you can also think of "attack" as the nucleophile "attacking" the electrophile with electrons to react/form a bond, in the direction the arrow is pointing. I suppose we call it attack because it makes it sound more action-packed :-)


text:
Changes in freezing point of sodium acetate
Background:
I was recently playing around with some reusable heat packs, the kind that operate by phase change of sodium acetate trihydrate and noticed that they didn't reach the temperature that I had expected (58 °C) rather, they stayed at 52 °C.
There were two such packs, roughly 50 mm × 100 mm × 15 mm and placed together with the broad sides in contact. Between them and at the center, there was a calibrated PT-100 RTD (which I certainly trust to be better than 1 °C accurate). The entire mass was then insulated with a few layers of thick wool fabric. The temperature reading became constant at 52 °C within a few minutes of crystallization being initiated in both packs and remained very stable for more than 10 minutes thereafter.
It's worth noting that although the manufacturer labeled the contents as being 'Sodium acetate (food grade)' and water, no proportions are given. Also, the contents of the pack are bright red, indicating the presences of unlisted dyes.
Question:
What would be the most likely causes for the discrepancy? Contaminants, incorrect, proportions something else?
I'm ultimately interested in using this reaction to maintain ~58 °C in an experiment. I'd like to avoid the same result if I mix up a batch of this stuff from pure ingredients.

Its previous environment might be to blame.
But, was this a case of contamination?
Perhaps not!
What conditions could lead to this backward data? Results generate more questions-some which can only be answered by the seller-and, perhaps a few things were used as "replacements", and in combining, the results attached you to this mystery. Keep us updated!

I'd guess that there was something besides sodium acetate in the packs. If there was just sodium acetate then the pack would become a rock when in the solid state. So there is probably extra water with NaCl so that the pack still stays somewhat squishy when the solid acetate forms.

My assumption  is that the addition of water (other than water of hydration) lowers the temperature - consider that the heat of crystallization is shared among the sodium acetate, the added water and even the packaging. Perhaps the manufacturer adds water to lessen the chance of burns.
You'd need to experiment with lab-grade chemicals to be sure, such as this ~99% pure sodium acetate trihydrate.


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Aromatic substitution
Respective aromatic substitution: Why isn't the $\pi$ bond attacked by the catalyst $\ce{FeBr3}$?
$\ce{Fe}$ must have a strong partial positive charge and acts as an electron acceptor to form a $\ce{Fe-Br}$ complex, but why can't it accept the electrons from the $\pi$ bond?


text:
How can I test if a spirit (tequila) has oxidised?
Is there a way to test if tequila has oxidised and if so the level of oxidisation?
It would be best if this could be done without a control sample, but the most important thing is if I can do this with 'household' equipment  - although I do have access to a hydrometer.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. If you receive useful answers, consider accepting one. Can I assume you want to know the percentage of ethanol oxidised to acetic acid? I would assume, that this might be a bit complicated with household items. Tequila itself is a wild mixture of compounds, accurate determination of all compounds is a challenge itself.

I guess, checking the taste isn't good enough for you? ;)

@Martin-マーチン "percentage of ethanol oxidised to acetic acid" sounds like the kind of thing we're looking for.

@Mithoron it's the taste that's made us want to conduct these tests - it was a brand new bottle but seems a little different to the expected taste

To my knowledge tequila and other high percent stuff doesn't have problem with turning alcohol into acid, as acetic fermentation can't proceed  in solutions with high ethanol content


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Covalent bonds other then 2-center, 2-electron bonds
I've heard of single bonds, double bonds, triple bonds, and quadruple bonds between two atoms, but in each of those bonds the two atoms contribute the same number of electrons from each of their valence shells.
My question is, why can't the atoms contribute an uneven number of electrons? For example, why can't there be a situation where a carbon atom contributes two valence electrons while a nitrogen atom only contributes one?
Surely the sigma bonds arising from the overlap of two valence orbitals allow one valence orbital to have two valence electrons while the other only has one. (I don't know about pi-bonds, and I don't know if it is possible for a carbon atom and a nitrogen atom to give two and one valence electrons, respectively, but maybe there is another pair of atoms that would allow this).

Fractional bond order isn't sth unusual,  also there are 2-center-3-electron bonds and other types

You can make a "bond" with one electron. It probably is somewhat more stable than the two single entities. Trouble is, what you then have is still a radical (as one of the two partners was before).

One example to consider is benzene, where each carbon contributes 1 electron to a pi-bonding system of 6 electrons in 3 orbitals. The C-C bond lengths in benzene are intermediate between standard single and double bond lengths. Also take a look at the various boranes such as diborane (B2H6) which contain three-centered (or higher?) bonds involving boron and hydrogen.

the two atoms contribute the same number of electrons from each of their valence shells

While this may not be exactly what you are looking for, bonds in metal complexes such as $\ce{[FeF6]^{3-}}$ are real, although only one of the atoms (an anion, actually) contributed electrons. For other complexes, such as ferrocene or those involving allyl ligands, the accounting can become complicated bordering on hair-splitting.
Other than that, boranes (@iad22agp already mentioned those) and compounds that are often called hypervalent such as $\ce{SF6}$, $\ce{PCl5}$ are examples of three-center bonding.

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Hard-hard and soft-soft acid/base interactions
How does one conclude if a given compound or a complex is a hard or soft acid or base?
What are the factors that contribute to hard-hard, soft-soft, soft-hard, and hard-soft interactions?

This could be helpful: http://chemistry.stackexchange.com/questions/10273/hardness-and-softness-are-terms-used-to-describe-which-aspect-of-acidic-behaviou?rq=1

authors: ضُحــى حمزة حليـم


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Using AC voltage for electroplating
As I know, in electroplating method, the type of voltage used is DC voltage. My question is, does replacing DC current supply by AC one affect the functioning of the cell? And why?

The difference is quite huge. Firstly, the deposition will occur on both electrodes (if possible). Secondly, if the frequency is high enough, there will be no deposition at all as only polarization and depolarization effect take place. The frequency effects mass, transport and electrical double layer. Depending on the electrolyte composition, the dissolution (if possible) will reduce the deposition. Usually pulse electroplating is used.  The problematic is huge, please see some articles for more info:
An Overview of Pulse Plating - Norman M. Osero
and
Pulse and pulse reverse plating—Conceptual, advantages and applications, M.S. Chandrasekar, Malathy Pushpavanam


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What prevents self-condensation in the Perkin reaction?
An acid anhydride is condensed with an aromatic aldehyde in the Perkin reaction. The latter is non-enolisable, and thus the former forms the enolate which attacks as a nucleophile. If the attack is made at the aromatic aldehyde, an aldol-like product is formed which can be hydrolysed to the acid.
However if the acid anhydride enolate attacks another acid enolate, in a nucleophilic addition-elimination step, this results in the leaving of a carboxylate ion and the generation of an active methylene group (similar to Claisen condensation). There may be proton transfer between the two later. Surely this is more favourable than the first mechanism? Why doesn't this occur though?
Similarly, in the Stobbe condensation, instead of attack by the enolate at the ketone or aldehyde, cannot the succinic ester be attacked again?


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Why is the acoustic absorption coefficient of hexane in air less than air?
Why is the acoustic absorption coefficient of hexane in air less than air? A 120 kHz pulse train exhibits a larger amplitude after traveling in a hexane/air mix than in pure air.
My hypothesis is that the hexane molecule has fewer degree of freedom than that of $\ce{N2/O2}$. Is this plausible?

I was taught that linear molecules have 3N-5 DOF while nonlinear ones have 3N-6, where N is the number of atoms. Using that, N2/O2 have 1 DOF each, while hexane has 54 DOF, so it seems that "hexane molecule has fewer degree of freedom" is incorrect. ----- The core question here, however, is quite an interesting one. Do you have a reference for the claim about the 120kHz pulse train?

The only reference is my experimental work, currently unpublished. This observation arose when checking amplitudes. I had set the receiver amplitude to max in air and assumed it would be less in the hexane/air mix. However, the signal saturated the receiver. have not tried it on any other alkanes yet, because I was not looking for the effect. I do know that stuff like dichloromethane really kills the signal.
Maybe 120kHz is some kind of window in its absorption spectrum, for some unknown reason.

Best hypothesis so far is that as gas density increases the transducer couples more efficiently to the medium, giving the impression of a decrease in absorption coefficient due to the increased amplitude at the receiver.


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Thermal stability of charged and uncharged alpha helices (the concept of alpha-helix capping)
I just learned that charged alpha helices are less thermally stable than uncharged ones. The only difference I see between them would be a larger dipole moment in the charged helix, but how would this contribute to stability?
Here a picture for clarification:

I am not sure about the thermal stability but carboxy-terminal amidation can protect from carboxypeptidase digestion. That is just biochemical stability. Can you indicate where exactly you read about what you are asking (add links if you can or paste an excerpt)

It was an exercise sheet: http://goo.gl/x1xiaj - Unfortunately, there are no online solutions, but in class, they told us that charged alpha helices are less thermally stable than uncharged ones (see question).

This might relate to the concept of 'capping' alpha-helices, where the placement of beneficial charges at either end of a helix is shown to increase the overall stability of the protein.  See the article1 published in Nature.
In general, by introducing preferential charges at the ends of alpha-helices you stabilize the inherent dipole moment created by the alignment of amide groups in the protein backbone.

Here is a different citation (found here):

In 1988, Richardson and Richardson observed that certain amino acid
residues are stastically favored as the amino-terminal caps for the
alpha helices in 45 globular proteins. The introduction of a
negatively charged side chain at the N-cap, which can neutralize the
partial dipole created by the unpaired amide protons, has been shown
to increase stability in T4 lysozyme.

I also believe the figure in the question is a bit misleading - as the illustrated charges may be intended on side-chain groups. I therefore assume that this helix is a part of a protein, and that the illustration simply wants to highlight the fact that you place charged side-chain groups on the ends of the helix (i.e. capping) within a protein structure.
Reference

Serrano, L., Fersht, A. Capping and α-helix stability. Nature 342, 296–299 (1989). DOI: 10.1038/342296a0

OP doesn't know what the concept behind it was and that is why they asked, don't add parts that make it seem as if they were asking for confirmation.. (Edit review)

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Why are isotopes of hydrogen the only ones with special names?
Why does no other element but hydrogen get special names for its isotopes?

Because D has different organic chemistry perhaps. http://www.physlink.com/Education/AskExperts/ae685.cfm

The isotopes of hydrogen are much more commonly used in the laboratory and in chemical discussion then the isotopes of other elements.

Dare I suggest the simple answer is that we can't name everything, or we'd run out of names ?

Harold Urey and George Murphy used spectroscopy to identify deuterium late in 1931, announcing it at the 1931 Christmas meeting of the American Physical Society. Picking up out of 'From Nuclear Transmutation to Nuclear Fission, 1932-1939" by Per F. Dahl:

If anything, the naming of the new isotope proved more problematic than its isolation.  At a special session on heavy hydrogen at the general June meeting in 1933 of the APS in Chicago, organized in conjunction with the Century of Progress Exposition, the ensuing discussion on its naming 'threatened to become acrimonious,' according to Francis Aston of the Cavendish Laboratory - the great authority on atomic weight measurements and a guest speaker at the discussion.  The argumentation had to do with whether to retain the name 'hydrogen' for the isotope, as Niels Bohr preferred; after all, it was not a new element and had the atomic number 1.  Both Gilbert Lewis and Ernest Lawrence opted for 'dygen' for the H$^{2}$ isotope and 'dyon' for its nucleus, wheras Rutherford preferred 'diplogen' and 'diplon' instead.  In the end, Urey had the last word, as he was entitled to, settling on 'deuterium' for the isotope and 'deuteron' for its nucleus.

Given the heavyweights in the field wanting a separate name for the isotope, even Neils Bohr could not hold back the tide.  Again, though, you have to remember that this was in the very early days of nuclear physics. While the existence of the nucleus dated back to Rutherford's experiments pre-WWI, they were still very unclear just what the nucleus was constructed out of.
This was all occurring just as Cockroft and Walton were using their new ion accelerator to perform the first human-induced nuclear fission ($^{7}$Li + p -> 2 $^{4}$He), published in the Proceedings of the Royal Society on July 1st, 1932. Alpha, beta, and gamma particles had been identified since the late 1800s, and natural sources were used for nuclear physics experiments. It was readily recognized that a source of energetic protons would be highly desired.
Also, remember that 1932 was the year that Chadwick discovered the neutron as a separate entity. Thus, also from Dahl:

The discovery of the neutron would soon revise Urey's view of his deuteron as consisting of two protons and an electron.  However, as late as October 1932, it was still not obvious among physicists that the deuteron consists of 'one proton and one Chadwick neutron' instead of two protons and one electron.

Against this rapid-fire series of experiments on nuclear physics, it really isn't unusual that deuterium got a separate name - it was still not universally appreciated that it wasn't something weird. Than, once it was accepted, it was too useful to readily distinguish experiments with protons vs deuterons vs tritons to revert their names back to normal isotope nomenclature.

This is obviously the most correct of the answers given since unlike the others it is not a post-factor rationalization of isotope naming but an actual account of how deuterium came to be named.  It should be the accepted answer, IMO.

It is probably the most accurate non-ex-post-facto account of the naming of deuterium, but OP actually asks for the absence of names for all the other isotopes (for which, I believe, there is no "real" reason to pinpoint but only rationalization)

In addition to the reasons ste listed, the isotopes of hydrogen have the greatest differences in mass compared to other elements. Consider that deuterium is twice as heavy as protium, and tritium is three-times as heavy as protium. Isotopes of all elements can be used in kinetic isotope experiments. The dramatic differences in mass among the hydrogen isotopes gives the most pronounced isotope effects. This type of difference in reactivity likely made it easiest to discover the different isotopes, and these differences explain the continued usage of the trivial names even more systematic descriptors are available.

This answer describes interesting chemistry but is unfortunately wrong, given the history provided by Jon Custer.  Deuterium was named because it was discovered first, and was discovered before we knew much about neutrons.  So kinetic isotope effects had little to do with how names were chosen.

@CurtF. I agree that my answer does not explain where the names came from, but I think it does provide some value in explaining why the historical names for the hydrogen isotopes have persisted when more systematic descriptors are available. If this is edging on opinion, I'm happy to remove my answer.

I think there are two reasons. First, it is more convenient to categorize them under the actual element-name to which they belong. If I say "15-Beryllium" everyone knows immediately, what I'm talking about. If we add hundreds of isotope-names, it would be quite a mess. Leading to the second reason: Xenon for example has over known 30 isotopes. There are just too many to name them all.
Hydrogen has, by the way, 7 known isotopes, of which only 3 have names:
1-Hydrogen: Protium (rarely used)
2-Hydrogen: Deuterium (D)
3-hydrogen: Tritium (T)

IUPAC discourages using the abbreviations to avoid confusion with other elements. Although, I see a lot of "D" in the organic literature.
Addendum:
I just saw, that in the novel The Mouse That Roared the isotope 4-Hydrogen has been named quadium (Q). This is neither used nor "official" though.

That immediately raises the question, though, of why hydrogen does get special names. Perhaps not all isotopes should get special names, but why don't certain isotopes from other elements have special names as hydrogen does? (e.g. carbon 14)

Yes, why hydrogen? I think, as many things in trivial nomenclature, there is not a really specific reason behind it but tradition and conventions. For deuterium, I see a lot of use in organic skeletal formulae, because it's probably the most used radioacive marker. Instead of then having a mess of abbreviations, one decided to use a formal nomenclature, as 14C. I don't rule out, that there is more to it, though.

Actually, it seems to go back to the beginnings of nuclear physics, where they (Rutherford et al.) were not completely sure what the particle twice as heavy as a proton actually was ('From Nuclear Transmutation to Nuclear Fission, 1932-1939", Per F. Dahl, is a nice book covering this era in physics history).

I like @JonCusters answer.  This answer is a good post-facto rationalization of current isotope naming practices, but doesn't describe the causal reason for the deuterium isotope name discrepancy (DIND?)   Deuterium was named because it was discovered first, and was discovered before we knew much about neutrons.  So convenience had little to do with how names were chosen.

@R.M. Hydrogen is quite special. It's very easy to ionise it completely (since it only has a single proton and electron), which basically leaves you with a "naked" proton - extremely reactive (that's how most acids work). Unlike other elements, hydrogen isotopes have a  significant effect on chemistry, despite the fact that chemistry mostly cares about outer shell electrons. The chemical and mechanical differences between hydrogen and deuterium are bigger than for any other element. Since other isotopes are interchangeable in chemistry, deuterium and tritium stand out quite a bit.

This is not true. There are more isotopes that had (and have now) its own name.
For example:
210Po - RaF (Radium F); 218Po - RaA (Radium A); 218Pb - RaB; 227Th - RdAc (Radioactinium); and much more
Some of the names are still widely used (f.e. RaF)

Maybe this is country/field dependent, but these don't seem to really be used any more. Most of them were placeholder names that just describe how the isotope was produced. Can you point to a recent example that still uses these names? The most recent I could find was a 1968 paper about Radium A


text:
Which gas among radon and hydrogen chloride has higher ideal behavior?

Which one among the following shows maximum ideal behaviour?

$\ce{NH3}$
$\ce{HCl}$
$\ce{Rn}$
$\ce{H2O}$

Ammonia and water have hydrogen bonding so they're ruled out. That leaves radon and $\ce{HCl}$. Radon is a noble gas which means inertness but it's too big in size. That could mean greater van der Waal forces. And $\ce{HCl}$ on the other hand has covalent bonding.
I want to confirm which one of them is more ideal.

Some molecules are highly polar and thus are subject to electrostatic (dipole-dipole) interactions, which are typically far stronger than the interactions of induced dipoles (commonly referred to as van der Waals forces). See where this gets you.


text:
Methyl Iodide synthesis
I was trying to determine whether I could synthesize methyl iodide by combining methanol, potassium iodide and sulphuric acid together... The way I see it, the sulphuric acid will protonate the OH instead of the I forming a very stable leaving group of water. Then the iodine ions would donate an electron pair to the carbon. You would be left with the salt Potassium bisulfate.... Then to remove my methyl iodide, just distil.
Will this work?

Probably yes, in combination with something like a Dean-Stark apparatus to remove excess water.

I know in theory the reaction of methanol with hydrogen iodide will form methyl iodide and water. This is part of the Monsanto process for acetic acid. But the problem is that back reaction will also occur.
Also if you try to use concentrated sulfuric acid to stop the back reaction (water plus methyl iodide to form methanol and HI) being a problem. But then you will form HI from the KI, the HI tends to decompose into hydrogen and elemental iodine. This will make the reaction very messy.
The Dean-Stark head will be useless for this reaction as methanol (methyl alcohol) and water are totally misicble. A better dehydration reagent would be trimethyl orthoformate. If you were to mix methanol, trimethyl orthoformate, sodium iodide and sulfuric acid I think you would have a better chance of forming methyl iodide. But the purity would not be good as it would have methyl formate mixed with it.
The easy way to make methyl iodide is to react a solution of sodium iodide in water with dimethyl sulfate. If you put the sodium iodide in a pear shaped 50 ml flask, add a stir bar and then the dimethyl sulfate you can then heat it up and the methyl iodide will distill out as it forms. I think that this reaction has much more promise for a methyl iodide synthesis.
Dimethyl sulfate can be formed by the reaction of sulfur trioxide and methanol, so in industry it is a good and cheap methylation reagent. This could then be reacted with sodium or potassium iodide to form the methyl iodide.
I tend to put a still head on top of a 1 inch long fractionation column and then use a liebig condensor to collect the methyl iodide. The only reason a chemist does this is if you want to make I-131 containing methyl iodide. I have found that if you add I-131 as sodium iodide to the synthesis before you add the dimethyl sulfate that you can make radioactive methyl iodide.
When I do this I work in a very good fume hood, while wearing nitrile gloves to handle every object. It is important with this reaction to pay attention to both the radioactive and the non radioactive hazards. The dimethyl sulfate is a very nasty toxic reagent (carcinogenic). I tend at the end of the reaction to allow it to cool. I then add some diethylamine to the still pot and then distill the diethylamine through the glassware (by this time the methyl iodide has been placed inside a glass vial over copper wire). The idea of the diethylamine is to make sure that no dimethyl sulfate remains in the glass equipment.
I then allow it to cool, I then wash out all the glass ware into a plastic waste bottle. I then put the glass equipment inside plastic bags. The glass is allowed to stand in a fume hood for weeks while the radioactivity decays away. The waste is allowed to stand. I have always used I-131 while has a half life of 8 days, but I am sure the work could be done with a longer lived radioiodine.

Instead of Dimethyl sulfate, can I use Trimethyl sulfoxide Bromide? I am quite comfortable working with it and I guess NaI will replace the I with Br and DMSO will become free again and CH3-I will be created? I have a Bromine recovery system at site and I could use the formed DMSO to re use in the reaction. Or maybe I could use HI. I am comfortable with HBr so HI should be easy too

In princible you could combine Me3SO Br with iodide and then heat it to decompose it. Many years ago as a second year undergraduate I did a synthesis project where I reacted methyl iodide with DMSO to make Me3SO I. I found that it did not have a melting point, instead if I heated a little in a melting point tube it eventually decomposed to DMSO and methyl iodide. But if I was you I would not want to do things with methyl iodide outside of a fully equipped chemistry lab. You need a fume hood to use it safely.

I work in an Industrial Lab and I make copious amounts of Me3SO Br daily (like about 500kg daily) so I am pretty comfortable making it. We got a recent order for Methyl Iodide and though they gave us their own proprietary process, I was thinking of recycling my DMSO for this use.

So what I said in the comment still holds: it could work (particularly in combination with a Dean-Stark apparatus to remove excess water), but there is no way to know for sure unless someone tries it out.
The more usual (and early) method is described in Organic Syntheses and utilizes the reaction of methanol with iodine in the presence of phosphorous

I don't know the exact reaction mechanism (if anyone has a clue, please feel free to edit!) but what happens is that $\ce{PI_3}$ is generated in situ from the elemental iodine and phosphorous and reacts with methanol to yield the methyl iodide and phosphorous acid
$\ce{3 CH3OH + PI3 -> 3 CH3I + H3PO3}$
Phosphorous acid does not act as nucleophile and so the problem of a competing backward reaction, which you would have in your proposed synthesis, is largely prevented.
I'd also like to add some caution: the reaction described in Organic Syntheses is not suited for beginners or in amateur setting. The chemicals involved are nasty: highly toxic, highly flammable, alkylating and pyrophoric. Don't try this at home :)

I actually ended up finding something about the proposed reaction. The sulphuric acid ends up being reduced by the Iodide so the methyl iodide is only an intermediate state. I saw results for this describing a 20% yield for methyl iodide (under cold conditions). On the other hand, using phosphoric acid in place of sulphuric acid works with very high yield.

To use a Dean Stark head you need to have water in a solvent which it is not miscible such as benzene or toluene. The idea of the Dean Stark head is that the water will evapourate with the hydrocarbon solvent and then collect in a condensor the water forms a seporate layer which is trapped in the head rather than returning into the flask. If you attempt to use a dean stark head on a mixture of methanol and water then you will never get the two layers. I guess you could use a DS head to remove methanol from hetane however.


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Removing harmful gases from automobile exhaust using electrostatic precipitation
If I attach an electrostatic precipitator to a car's exhaust - and also make the smoke pass through an area containing $\ce{NaOH}$ in order to absorb sulfur dioxide - does that mean that the smoke that emerges will not contain harmful gasses such as $\ce{CO2}$, $\ce{CO}$ and $\ce{SO2}$? If not, what is the reason?


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How can I- be both a nucleophile and a weak base?
I am just confused as to how $\ce{I-}$ cannot attract $\ce{H+}$, but it can attack a carbocation. It is sharing electrons in both cases, so what is the difference?

Nucleophilicity is a kinetic property, basicity is a thermodynamic property. They are friends some of the time, but they don't always hold hands as your example shows.

Also, iodide is a base. Whether it reacts as a base depends on where the proton is coming from.

As Jori pointed out, nucleophilicity is a kinetic property while basicity is a thermodynamic property. Nucleophilicity in part is about the rates of chemical reactions; thermodynamics is in part about the favorability of chemical reactions.
A reaction may be very favorable, i.e. it has a very negative free energy change, but it might take a long time to actually complete. For example, the rusting of iron is favorable from a thermodynamic standpoint. The products, iron oxides, are very stable. However, the process is relatively slow from a kinetic standpoint; putting water on iron isn't going to instantly result in a familiar reddish-brown hue. It might take a day or so.
On the other hand, if you squirt something such as tert-butyllithium into the air, you'll probably catch yourself or something on fire. This is a carboanion salt. Carboanions are strong (Bronsted, and therefore, Lewis) bases. Acid-base reactions also tend to be relatively fast (kinetics) since they involve the transfer of something so small - a hydrogen proton. A graduate student unfortunately exposed some of this carbocation salt to the air and the resulting fireball instantly caught her sweater on fire. The carboanion salt likely grabbed protons from atmospheric moisture - a very favorable process from a thermodynamic standpoint - and kinetically, a very fast process. And so the resulting heat likely ignited the resulting protonated carboanion (alkane in this case) on fire.
In your case, $\ce{I-}$ is a good nucleophile in both protic and non-protic solvents. It's a rather big ion - the biggest of the halide anions. That means its negative charge is relatively dilute. This in turn means it's not as strongly solvated by the solvent molecules; it doesn't have as many tenacious solvent molecules hindering its trek toward a carbocation. So $\ce{I-}$ can reach carbocations relatively quickly. Note that terms dealing with time usually indicate that the writer is speaking about a chemical's kinetic properties.
For the same reason that $\ce{I-}$ is a good nucleophile it's a poor base. It's dilute negative charge (due to its size) means that $\ce{H+}$ has better things to do than be attracted to $\ce{I-}$. $\ce{I-}$ has a dilute amount of negative charge so it cannot effectively stabilize the hydrogen proton as well as many solvents might - especially water, which extensively associates with $\ce{H+}$; $\ce{H3O+}$ only begins to describe how well the proton is associated with water molecules; the proton has been found to be associated with 2, 3, 4, or more water molecules!


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Why is trans-cyclooctene chiral?
How does trans-cyclooctene exhibit chirality if there are no stereocenters?
Related follow-on questions:

Are all higher cycloalkenes chiral?
Do more double bonds cause a bigger number of stereoisomers in cycloalkenes?

related http://chemistry.stackexchange.com/a/31296/9961

Related: http://chemistry.stackexchange.com/questions/24901/do-molecules-with-axial-chirality-have-stereogenic-units/24910#24910

Especially related, since trans-cyclooctene undergoes ring flip very slowly, unlike cyclohexane: http://chemistry.stackexchange.com/questions/34475/determining-chirality-after-considering-conformational-interconversion-in-a-cycl/34480#34480

http://chemistry.stackexchange.com/questions/24901/do-molecules-with-axial-chirality-have-stereogenic-units

I absolutely agree with Brian's comment on the answer, it would be better, if you roll back to the original question and ask a follow up question with the added ones.

Very interesting question! The key word you are looking for is planar chirality. In trans-cyclooctene, the polymethylene bridge can either go "in front of" or go "behind" the plane of the double bond, assuming you fix the double bond and the two hydrogens in place.
As pointed out by @jerepierre, they are considered different molecules due to a high-energy barrier which prevents the interconversion. Cyclooctene is the first cycloalkene to have both stable cis- and trans- isomers. The chain in trans-cyclooctene is not long enough to swing over the double bond. As the chain gets longer, the energy barrier to rotation decreases.
Here are the two mirror images of trans-cyclooctene (image source: own work).

These two molecules are mirror images of each other but are not superimposable. Therefore trans-cyclooctene is chiral despite not having a chiral center.
Source

Edit: this is the first time I learned that a chiral molecule does not have to have a chiral center. After making some searches online, I feel a need to expand the answer to clarify the concept of chirality.
A chiral molecule is one that has a non-superimposable mirror image. Mathematically a molecule is chiral if it is not symmetric under an improper rotation. Chirality arises due to:

point chirality: typically a carbon center with four different substituents;
axial chirality: such as allenes with different substituents on each carbon (see this question);
planar chirality: such as the case of trans-cyclooctene;
inherent chirality: due to the presence of a curvature in a structure that would be devoid of symmetry axes in any bidimensional representation, such as fullerenes.

Hm. Technically, though, it does have a chiral center (at the center of the ring), right?  It just doesn't coincide with a nucleus.

@Brian I just edited the answer for a more comprehensive view of chirality.  I don't think it has a chiral center as the four carbon atoms (the two in the double bond and the two directly bonded to them) are in the same plane and do not form a tetrahedron  around a point.

I'm probably splitting hairs, or up against a nomenclatural definition. Formally speaking, does a 'chiral center' only exist at a nucleus? If so, then the center of mass doesn't qualify as a chiral center, and my initial comment is incorrect.

Thank you. I've expanded my question a little bit, because I still have some doubts.

I'd give +1/2 if I could. An additional key feature of trans-cyclooctene is that it doesn't undergo ring flip (or does so slowly). For a similar question, see: http://chemistry.stackexchange.com/questions/34475/determining-chirality-after-considering-conformational-interconversion-in-a-cycl/34480#34480

@Marko I can't answer your 2nd question right now. Meanwhile you may find this question useful: http://chemistry.stackexchange.com/questions/30940/cis-trans-stability-of-cycloalkenes

@Brian As I was unsure myself, I consulted the goldbook, where it is explicitly stated as "an atom". In the case of planar chirality, the definition also clearly states plane. One rational about this is, that there are multiple atoms that lie in the same plane when considering enantiomers, which is not the necessarily the case for point chirality.

@Martin-マーチン then I have a question: could the plane be one that does not pass through the atoms? The definition linked defines the chiral plane as the plane passing thru the double bond and the linked atoms (at least in cyclohexene), but are the two carbon atoms coplanar with the $\ce{HC=CH}$?

@Marko, your edit would possibly have been better posted as a new question, possibly even two, with crosslinks posted among them.

@user2124232 They will most likely not be perfectly planar because of the strain in the small molecule, but very close to it.

Point four would be a Moebius band.

@KarlRatzsch mind elaborating a bit?

A Moebius band is chiral, you can glue it together leftsided or rightsided. Upon second thought, it might also be a planar chirality. Not sure how to picture it.

I don't understand the point about fullerenes, anyway. Would that be a fullerene with three different substituents?

@KarlRatzsch now I'm imagining how a Möbius band molecule would be like. I haven't figure out fullerene yet because of its complex structure, but other examples given by wiki such as calixarene should be easier to understand.

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

Some of the terminological confusion here come because the terms centre of chirality and chiral centre don't mean exactly the same thing. In normal use a chiral centre is a carbon (or other tetrahedral atom) with 4 different substituents, guaranteeing chirality if it is in a simple molecule (though more complex chiral centres can exist eg in octahedrally coordinated metals with 3 bidentate ligands). But more complex molecules like trans-cyclooctene or hexahelicene there is a centre of chirality that does not correspond to an atom.


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Does precipitated copper carbonate contain impurities
I recently performed a double replacement reaction between copper sulfate and sodium carbonate to yield copper carbonate as a precipitate. I then filtered the precipitate, but I noticed that the precipitate still contained a large amount of water.
Would there be a significant amount of sodium sulfate dissolved in this water, making the copper carbonate impure? If so, how would I separate the remaining sodium sulfate from the copper carbonate?

You rinse the precipitate with water, that's how.

Yes, the water contained in the freshly filtered precipitate will contain dissolved $\ce{Na2SO4}$, as the latter is the soluble by-product of your double displacement reaction. To get rid of it, you need to wash your precipitate thoroughly with distilled water (pH 7). If you really want to make sure that all $\ce{Na2SO4}$ has been removed, you can test samples of the wash water filtrate for $\ce{SO4^2-}$ by adding diluted hydrochloric acid and aqueous $\ce{BaCl2}$. When a white precipitate of $\ce{BaSO4}$ no longer forms, the aqueous sodium sulfate can be considered separated from the $\ce{CuCO3}$ precipitate.


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Which graph shows Chemical Equilibrium?
So my teacher started the explanation of Chemical Equilibrium saying..
The concentration does not change with time.
And, gave us certain graphs asking which ones show chemical equilibrium.
I cannot understand what's written in my copy since I took the lesson two years ago. I need to be explained what I wrote in my copy.
I know that the concentration does not change with time.
If concentration is same or not, does not matter. That only the rate of forward reaction and the rate of backward reaction needs to be same.
These are the graphs: (I am sorry I could not draw better.)
:
Now, I know 1,4,5 and 6 show chemical equilibrium.Reasons:

The concentration of the reactant is lost but same amount is achieved by the product.So here, the rate of forward reaction and rate of backward reaction is same.
For 5 and 6- The concentration of reactant and product may not be the same but it does not change with time.

I need help regarding :
1.Explanation as to how the other graphs do not show chemical equilibrium(if possible in terms of how the answer for the other graphs are explained).
2.Explanation for graph 5 and 6. I cannot understand the reason I had written.
3.An overall explanation and answer of the question.
Thank you very much for your help and understanding!!

The graphs 4, 5, 6 and 7 have the label "Rate" on the y-axis, but in the graph you label the functions with a "c", which I assume is for concentration.

@Martin-マーチン I cannot change the picture now . But I reckon rate could mean rate of reaction and rate of concentration for the other graphs.

Firstly, you need to understand what it means when a system is in equilibrium. Equilibrium is the state of a system where the rate of the forward and reverse reaction are equal.
Note: this doesn't mean that the reaction completely stops, all this means that the concentration of the reactants and products remain constant.
A simple example of this can be shown in an experiment where a cube of sugar is placed in a saturated solution of sugar. If you leave it for a few days and come back, you will find that the shape of the sugar cube has changed. Instead of being a smooth, perfect cube, it is a irregular shape. However when if weigh the solid sugar, you find that it will weight exactly the same as the original sugar cube. Now this can be explained by the fact that the solid sugar is in equilibrium with the dissolved sugar in the saturated solution. Despite the forward and reverse reaction being equal (as no more sugar will dissolve) the reaction is still occurring. At a microscopic level, glucose molecules in the sugar cube are dissolving into the solution while dissolved sugar molecules precipitate out onto the sugar cube. However where they precipitate on the cube is random, hence they don't form a perfect cube. This is why the sugar cube changes shape but has the same original mass.
To do this question, it is important that you understand the concept that the reaction still occurs when a system reaches equilibrium. Now, lets look at each graph separately (remember, when we are looking at these graphs, we are only interested at what is happening at the end, what happens at the start is irrelevant):
1: Here the forward and reverse reaction are equal, so from our definition of equilibrium, this graphs shows a system in equilibrium. You should note that despite the system being at equilibrium, the reaction is still going
2,3: Here in both graphs the forward and reverse reaction are not equal. Therefore the system can't be in equilibrium. This is because in graph 2, the rate of the backward reaction is greater than the rate of the forward reaction. Hence the concentration of the products will be decreasing while the concentration of the reactants will increase. Vice verse for graph 3
4,5,6: Here, since the concentration of the reactants and the products are not changing over time, this means that the forward and reverse reaction is the same. Therefore, the system is in equilibrium
7: In this graph, the concentration of the products is increasing while the concentration of the reactants is decreasing. Hence the forward reaction must be greater than the backward reaction. Therefore, the system isn't in equilibrium.
So to recap the ideas that we heave just learnt, here is the  2 main things to check to work out if a system is in equilibrium or not:

Is the forward and backward reaction equal? if they are equal, this means that the system is in equilibrium. If they aren't equal, the system isn't in equilibrium
Is the concentration of the products and the reactants constant over time? if they are constant, the system is equilibrium. Note, the concentration of the products and reactants don't have to be equal. If concentration of the product and reactants are aren't constant, the system isn't in equilibrium.

Apparantly the chemistry students at the Uni Bochum have a long-term experiment going on as to whether they can create a saturated aquaeous sucrose solution in a coke bottle. I heard that the solution be not saturated yet, although the experiment has been going on for some time.


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Is a supersaturated solution still a solution?
Given that a solution must be a homogenous mixture, and having crystallized solids at the bottom of a solution is most definitely not homogenous, can a supersaturated solution still be technically defined as a solution?
On that note, is it possible for a solution to have more solute than solvent and remain in solution form?

You seem to be getting it wrong. A solution with crystallized solids at the bottom is not the same as supersaturated solution, and has never been called such. It is called "solution with precipitate", "heterogenous mixture of solution and something", "water with mud", or anything in the same vein.
A supersaturated solution is a solution, not just technically, but in all sincerity. It is homogenous, clear, transparent, and you'll never tell it from ordinary solution - I mean, until you touch it, at which point a precipitate would start to form.
As for your second question: yes, there are quite a few compounds that would dissolve in water incredibly well. ZnCl2 comes to mind... Wait, it's simpler than that - common table sugar (sucrose) would do! It dissolves up to 2000 g/L, how'd you like that?


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Variation in atomic sizes in the transition elements
From left to right across a period, effective nuclear charge increases in the transition series, just like in the p block, and so atomic size decreases. Here the electrons are being added to the d orbital, where shielding is relatively poor compared to the p orbital. It should follow that the increase in the effective nuclear charge is more significant for the transition elements than the p block in a given period, leading to greater size variation among successive members - but the size variation is smaller in the transition elements instead. In fact atomic size increases from Ni to Zn. Is there any logical explanation for the fact?

From Ni to Zn there is increase in size because of the completely filled 3d and 4s orbitals. Here the nuclear charge is increased with atomic number but due to more electron -electron repulsion in 3d and 4s orbitals  as compared to force of  attraction between nucleus and outer orbital electrons there is little expansion in size as compared to other elements in 3d series.

From Ni to Zn, Z$_{eff}$ decreases and the shielding effect increases .
Z$_{eff}$ is inversely proportional to the size, so if Z$_{eff}$ decreases, size will increase.
That's why there is some uncertainty in the order of size for 3d series.

In case of d block elements as we move from left to right across the period, atomic number increases. The nuclear charge increases. But the electrons are added to penultimate i.e. (n-1) shell, hence the electron cloud density of inner shells increases which increases the screening effect. Thus nuclear charge increases and screening effect increases. Hence there is decreases in the atomic radius but the extent of variation is very small compared to s block and p block elements. The extent of variation is so small that all of them can be considered to have almost equal atomic radii.


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What causes incomplete combustion?
During the combustion of hydrocarbons, there is a difference between the amounts of carbon or hydrogen that results in incomplete or complete combustion of the material.
My question is, besides from an insufficient amount of oxygen, what can cause incomplete combustion?
What are the physical or chemical properties of the hydrocarbon responsible for incomplete combustion, such as the hydrogen to carbon ratio (mass, volume, number of atoms, etc.) or the saturation.
Please keep the reply suitable for a student studying high school to early university chemistry.

I think insufficient air is only reason for incomplete combustion.

According to the Iowa State University page Carbon Monoxide Poisoning: Checking for Complete Combustion (AEN-175) suggest the following mechanisms that cause incomplete combustion:

Insufficient mixing of air and fuel.
Insufficient air supply to the flame.
Insufficient time to burn.
Cooling of the flame temperature before combustion is complete.

The main notion is the lack of air to cause complete combustion (either through lack of supply, mixing or time for combustion).
However, it should be noted that, practically speaking, according to the ChemWiki page Burning Alkanes, the size of the hydrocarbon has an affect in 'normal' conditions:

Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame.

The reason being (according to ChemWiki) is:

If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and turn to a gas.

So, the larger the hydrocarbon, the less likely it would vaporise sufficiently, thus less chance of the optimal mixing of air and fuel, resulting in incomplete combustion.
However, provide the right conditions with sufficient supply of air and optimal hydrocarbon-air mixing ratio, it is still possible to cause complete combustion, in the general formula (for many hydrocarbons):
$$\ce{C_{x}H_{y} +O2 ->H2O + CO2}$$
A quite comprehensive resource explaining each of the factors is the University of Tulsa chapter Fuels and Combustion.

a particular flame temperature is recommend for each types of sample, if your sample required higher temperature to vaporize and given temperature is not the required one, incomplete combustion may occur

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. Could you add an example for your explanation, I have a hard time understanding it.