text:
Simple practical, non-toxic, non-flammable reaction for slow release of CO2
Someone I know (not me) has recently had problems with bed bugs. I have advised him his best bet is to (1) Buy a kick-ass steam cleaner and nuke the brutes from orbit, (2) get a professional pest controller in, and (3) Stop buying second hand furniture in future!
But being an inventive sort, he prefers a more subtle approach and is now intent on designing the perfect indoor bed bug trap.
It appears these critters are attracted by a localized $\ce{CO2}$ excess and by body heat, and are repelled by light and (presumably) excess heat. So assuming this trap won't involve a tethered goat or suchlike, he will need a compact, non-flammable, non-toxic, readily and cheaply available substance that can be persuaded to release a steady stream of $\ce{CO2}$ for several hours (without releasing CO or any other noxious or smelly biproduct).
I was thinking perhaps water dripping slowly onto calcium carbide, if there was some way of converting the resulting acetylene to $\ce{CO2}$ without burning it at a readily discernable rate. But a "slow burn", if it could be attained, might provide the heat source (although I reckon a more controllable electric heater would be better for that).
Any ideas? I suppose one obvious answer would be to simply buy a $\ce{CO2}$ cylinder, but could one then buy a sufficiently slow release valve to go with it?

Hi and welcome to chemistry.stackexchange.com. Nice question! Feel free to take the [tour]. I am going to improve your question using MathJax in a few seconds. You can learn more about it in the [help].

It is advisable not to conduct a CO2 producing reaction in a bedroom since it may most probably produce CO as well. And recently I heard a couple died by using a firewood heater in a closed bedroom

@slhulk This is oversimplified. It only applies to reactions where $\ce{CO2}$ is produced by oxidation. However, it is safe to say that no $\ce{CO}$ will be formed if the reaction is an acid-base reaction like the answers suggested below.

Won't excess of CO2 be harmful as well?.

I would recommend not touching or coming near calcium carbide at any rate. The acetylene it creates is explosive in certain concentrations (with a flame to ignite it) and notorious for extremely loud bangs. And it also is rather a diversion for what you want to do.
Instead, consider carbonates. Baking soda is one of the prime household sources of carbonates, and you just need to add an acid to release carbon dioxide; and nicely, the only by-product is water. You should use rather dilute acids so that you can slow down the reaction rate. Vinegar might not be the acid of choice (it has a distinct smell) but either essence of vinegar diluted or citric acid (which you should be able to buy from baking suppliers, too) should work. Or lemon juice.
To make the reaction sufficiently slow, you would need to (as I said) dilute the acid let it be poured into the baking soda solution slowly. Experiment a little! For a heat source, you could likely just put your average heating cushion underneath.
(Personally, though, I would use the steam cleaner because it sounds most kick-arse!)

An easy one is to mix together and then grind up citric acid with sodium bicarbonate (hydrogencarbonate). If you were making bath bombs, it's a similar method but without fragrances or colouring. You will literally need a few drops of water to help bind them and mould the solids, but that's it. When ready simply add water to the pellet/lump. Slow-ish $\ce{CO2}$ release.

Regarding bed-bugs, they are notoriously difficult to remove - I'd highly advise getting a pro involved as soon as possible.
For slow/steady carbon dioxide generation in a general sense, however, I'd go with yeast. East to find bakers yeast and flour in any supermarket. Google a basic bread recipe, and mix accordingly. If you do a slow rise, the dough will slowly release carbon dioxide (and other metabolic by-products) - depending on temperature, pH, and ratios used in the recipe, it could probably keep producing for 24 hours or more. And if you punch the dough down every so often, it may be able to keep generating longer than that. At the end, you can take your dough, throw in into a bread pan or just a sample tray, and make bread!
Pros:

No chemical safety issues
Delicious homemade bread as a by-product

Cons:

More difficult cleanup
Your bread may end up being bed-bug flavored


text:
Why is the total number of electrons in a neutral carbon atom different from the number of dots in a Lewis Dot structure for carbon?
Lewis Structure has 4 but the neutral atom has 6. why?

Because the most involved electrons on bonding are depicted to be the valence electrons.

Only the valence electrons are represented in a Lewis structure. The inner electrons are not important for the bonding, therefore not displayed.

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text:
How to calculate bond angle of simple molecules quantitatively?
Hi I'm new to this forum so spare me if I've made mistakes.
Our high school chemistry teacher explained that bond angles of different molecules varies slightly, e.g. 107 degrees in ammonia and 104.5 degrees in water, which could be predicted by VSEPR theory qualitatively.
I have checked literature either through google or stack-exchange, but none point me to any formula/algorithm/computationally inexpensive methods to  find non-ideal bond angles by calculation.

I was taught that for each unbonded electron pair, the usual bond angle for a certain configuration is reduced by ≈2º.

Your first mistake: This is not a forum! D= No, don't worry, joking ;) Welcome to chemistry.stackexchange.com ^^

@ringo never heard of the approximation - could you point me to the reasoning behind?

This has a simple reason: To the best of my knowledge there is no quantitative way to predict bond angles without resorting to computational chemistry.
This is because the entire subject of predicting bond angles is a lot more tricky than it was displayed to you. Take phosphane and hydrogen sulphide, the heavy analogues to ammonia and water. Their bond angles are nowhere near $109^\circ$ but much closer to $90^\circ$.
If you dig deeper into the matter, you find that ammonia and water are more the exception than the rule, that the VSEPR model fails spectacularly on the long run, and that their geometries are actually finely balanced between an ideal $90^\circ$ angle and the hydrogen-hydrogen repulsion.
In fact, not even with computers can you calculate the bond angle directly with a certain set of axiomic assumptions. Rather, computational chemistry would empirically calculate energies for different angles and deem the angle that has the lowest energy to be ‘correct’ and ‘calculated’.
So no simple quantitative rule exists nor will it exist.

Note: There are, of course, semi-quantitative rules like the one mentioned by Ringo in a comment. But I don’t consider them quantitative, because they still contain approximations.

Thank you for the explanation. But then, does the smaller angle between H2S and PH3 compared to H2O and NH3 related to their hybridization, as S & P are in the same group as O & N?

@MarcoXerox You can kind of say that. $\ce{S}$ and $\ce{P}$ mainly use the p-orbitals for bonding to hydrogen while $\ce{N}$ and $\ce{O}$ use somehow hybridised orbitals. But from a theoretical perspective, it’s rather that $\ce{N}$ and $\ce{O}$ would use only their p-obitals if they weren’s damn small. The hybridisation and their corresponding angles are obtimised by nature.

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Chemical expression and terminology
Is this expression correct "the reaction was not accomplished " when we check the final product finding two or more spots on TLC paper ?

Because "finding two spots" was difficult to understand, my answer to this includes an idealized graphical representation of a TLC.  It consists of two lanes A, and B for two starting materials. There is a reference lane (D) of the already known, confirmed and purified product -- for example obtained by a previous batch of synthesis, or (as met once in lab classes) the TA offers to spot it, using his/her known reference.
If lane C was used to co-spot both the starting material(s) as well the reaction product, yet you do not detect (by UV-Vis, or derivatization) a spot corresponding in $R_f$-value to your reference (in lane D), than your analyte analyzed does not contain significant amounts of your intended product.

Depending on how / if at all a work-up prior to spotting the sample of the unknown was executed, this observation indeed may refer to "the chosen conditions did not yield significant amounts of the intended product".
Yet if you forgot to quench/neutralize (and subsequently extracted) your analyte sample prior to spotting, your product may be still at the ground line -- esp. for salty samples on a silica TLC -- and hence not elute at $R_f = 0.75$. Second potential pitfall I imagine: you quench your reaction with water, and extract at different pH values (separation of strongly acidic / acidic / basic compounds) -- your intended product simply is in the subsequent extract that accidentally was not spot on lane C.

The TLC was idealized that the compounds would nicely separate from each other, without any smearing/fronting/tailing either individually, nor in presence of each other (+ other components of the reaction mixture not detected by the TLC [e.g. solvent(s)]).  Simultaneously, it is assumed there is no decompostion of any of the products while eluting the TLC, that could be checked by 2D-TLC.


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Why shouldn't lithium batteries be discharged below 50%?
I know both from online websites and from experience that discharging a li-ion battery below 50% greatly decreases its life and capacity.
But why is it so?

There are two main effects that cause damage to a Li-ion battery due to "deep cycling":

During the charge/discharge cycle, lithium ions are intercalated within the electrodes. This causes physical stress to the electrode material and spalling of their coatings. Deeper discharge causes increased stress. This effect is made worse with the temperature range from heavy current use as in a laptop.
Lithium metal may accumulate on the anode, taking it out of play. This effect may be due more to over-charging than to deep-discharge, though.

But deep cycling is not below 50% but more below somewhat around 7-10% afaik.

Deep cycling means a large change in charge, not just a specific charge state.

Ok, then go ahead.


text:
Significant figures addition vs multiplication
I was finding the molecular weight of $\ce{Br2}$ the other day and noticed that if I add the weight of $\ce{Br}$ to itself, $79.90+79.90=159.80$, that answer for the weight has $5$ significant figures. But if I just multiply the weight by $2$, $79.90 \times 2 = 159.8$, the weight has $4$ significant figures. If adding something to itself is the same as multiplying it by $2$ mathematically, which accuracy is more justifiable for my calculations in cases like this?

You forgot 2 * 0 = 0 . The multiplied answer contains 5 significant figs too

but you have to round it down to 4 because 79.90 only has 4 sig figs... http://www.sigfigscalculator.com/

According to your argument the added answer should also be rounded off to 4 significant values. It should not be 159.80

no, the rules for addition are that the lowest decimal place is kept, so two decimal places. Again, you can verify this with any online sig. fig. calculator

If this so, your argument "but you have to round it down to 4 because 79.90 only has 4 sig figs..." fails.

That is the rule for multiplication.

you can calculate using this calculator, i have recently used this calculator https://www.sigfigcalculator.net/

Your calculations are in accordance with the rules of the so-called significance arithmetic even though they yield different results.
When multiplying numerical values, according to the rules of significance arithmetic, the result is rounded to the number of significant digits in the factor with the least significant figures. When you multiply the value of the molar mass $M_{\ce{Br}}=79.90\ \mathrm{g/mol}$ by 2, the value of the molar mass $M_{\ce{Br}}$ has four significant digits, whereas the factor 2 is assumed to be exact. Accordingly, the result is rounded to four significant digits:
$$M_{\ce{Br2}}=2M_{\ce{Br}}=2\times79.90\ \mathrm{g/mol}=159.8\ \mathrm{g/mol}\tag1$$
When adding numerical values, according to the rules of significance arithmetic, the result is rounded to the position of the least significant digit in the most uncertain of the numbers being summed. When you add the molar mass $M_{\ce{Br}}=79.90\ \mathrm{g/mol}$ to itself, the value of the molar mass $M_{\ce{Br}}$ is significant to the hundredth place. Accordingly, the result is rounded to the hundredth place:
$$M_{\ce{Br2}}=M_{\ce{Br}}+M_{\ce{Br}}=79.90\ \mathrm{g/mol}+79.90\ \mathrm{g/mol}=159.80\ \mathrm{g/mol}\tag2$$
However, this concept of significant digits is just an oversimplified form of the concept of uncertainty as a quantifiable attribute. In your question, you are testing the limits of the applicability of this simplification, which is certainly a good idea.
When a number is given without any further information, it is generally interpreted so that the last digit is rounded. Thus, the molar mass of $M_{\ce{Br}}=79.90\ \mathrm{g/mol}$ is generally assumed to represent a value between $79.895\ \mathrm{g/mol}$ and $79.905\ \mathrm{g/mol}$, or $M_{\ce{Br}}=\left(79.900\pm0.005\right)\ \mathrm{g/mol}$.
Therefore, the assumed uncertainty of the molar mass $M_{\ce{Br}}$ is $u(M_{\ce{Br}})=0.005\ \mathrm{g/mol}$. (Note that, normally, the uncertainty corresponds to the estimated standard deviation; in this case, however, the uncertainty just indicates an interval corresponding to a high level of confidence.)
When you multiply the value of the molar mass $M_{\ce{Br}}$ by 2, you also have to multiply its uncertainty by 2 (the factor 2 is assumed to be exact):
$$M_{\ce{Br2}}=2M_{\ce{Br}}=2\times79.90\ \mathrm{g/mol}=159.80\ \mathrm{g/mol}\tag{3a}$$
$$u(M_{\ce{Br2}})=u(2M_{\ce{Br}})=2u(M_{\ce{Br}})=2\times0.005\ \mathrm{g/mol}=0.010\ \mathrm{g/mol}\tag{3b}$$
When you add the value of the molar mass $M_{\ce{Br}}$ to itself, you also have to add the uncertainty to itself:
$$M_{\ce{Br2}}=M_{\ce{Br}}+M_{\ce{Br}}=79.90\ \mathrm{g/mol}+79.90\ \mathrm{g/mol}=159.80\ \mathrm{g/mol}\tag{4a}$$
$$\begin{align}u(M_{\ce{Br2}})&=u(M_{\ce{Br}}+M_{\ce{Br}})=u(M_{\ce{Br}})+u(M_{\ce{Br}})=0.005\ \mathrm{g/mol}+0.005\ \mathrm{g/mol}\\&=0.010\ \mathrm{g/mol}\tag{4b}\end{align}$$
(Note that such a simple addition of uncertainties is not correct when the errors of the added values are uncorrelated, for example when calculating $M_{\ce{HBr}}=M_{\ce{H}}+M_{\ce{Br}}$.)
Both calculations yield the same results for the molar mass $M_{\ce{Br2}}$ ($\text{(3a)}$ and $\text{(4a)}$) as well for its uncertainty ($\text{(3b)}$ and $\text{(4b)}$). Thus, the molar mass of $M_{\ce{Br2}}$ is estimated to represent a value between $159.79\ \mathrm{g/mol}$ and $159.81\ \mathrm{g/mol}$, or $$M_{\ce{Br2}}=\left(159.80\pm0.01\right)\ \mathrm{g/mol}$$
By way of comparison, the result of the multiplication according to the rules of significance arithmetic $\text{(1)}$, $M_{\ce{Br2}}=159.8\ \mathrm{g/mol}$, represents a value between $159.75\ \mathrm{g/mol}$ and $159.85\ \mathrm{g/mol}$, or $$M_{\ce{Br2}}=\left(159.80\pm0.05\right)\ \mathrm{g/mol}$$ Thus, this approach overestimates the uncertainty of the result.
The result of the addition according to the rules of significance arithmetic $\text{(2)}$, $M_{\ce{Br2}}=159.80\ \mathrm{g/mol}$, represents a value between $159.795\ \mathrm{g/mol}$ and $159.805\ \mathrm{g/mol}$, or $$M_{\ce{Br2}}=\left(159.800\pm0.005\right)\ \mathrm{g/mol}$$ Thus, this approach underestimates the uncertainty of the result.

If the $0$ in $79.90$ is important, the when you multiply by $2$ also, you get $159.80$ so there is no discrepancy. However, you do realise that suppose the weight was $79.904$ (in fact, it is !), writing 2 decimals only will be $79.90$ and times $2$ will produce $159.80$. However, if you had written $79.904$, you would have got $159.808$, which up to $2$ decimals is $159.81$
In multiplication, the simple rule is, the number of significant digits of the answer is that of the multiplicand with the least number of sinificant digits.
Here is one such guide which you can go through.

When you multiply 2 (exact) times 79.90 (4 sig figs) it equals 159.80, but because you have to round to 4 sig figs, it's 159.8, right?

Try http://www.sigfigscalculator.com/ to see what I mean (enter "2.000000*79.90")


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Does Reverse Osmosis remove elemental ions from water?
From what I understand, RO is able to remove (or at least considerably reduce) the salts dissolved in water.
But thinking about it just doesn't make sense. RO works by filtering the solution using a membrane with pores so fine that only water can go through them. But elemental ions are WAY smaller than molecules of water.
So how does RO prevents salts from going to the other side?

Are you sure those ions are way smaller? Especially if you consider their aqueous layers around them necessary to stabilize their charge?

Elemental ions are definitely not WAY smaller than molecules of water. Besides, they are charged.

The short answer is pressure. What you've described above is basically just osmosis, and without applied pressure to overcome the osmotic pressure of the system, reverse osmosis doesn't happen.
This Wikipedia article gives a good description of the concept, and the difference between filtration, osmosis and reverse osmosis:

Osmosis is a natural process. When two solutions with different
  concentrations of a solute are separated by a semipermeable membrane,
  the solvent has a tendency to move from low to high solute
  concentrations for chemical potential equilibration.
Formally, reverse osmosis is the process of forcing a solvent from a
  region of high solute concentration through a semipermeable membrane
  to a region of low solute concentration by applying a pressure in
  excess of the osmotic pressure. The largest and most important
  application of reverse osmosis is the separation of pure water from
  seawater and brackish waters; seawater or brackish water is
  pressurized against one surface of the membrane, causing transport of
  salt-depleted water across the membrane and emergence of potable
  drinking water from the low-pressure side.
The membranes used for reverse osmosis have a dense layer in the
  polymer matrix—either the skin of an asymmetric membrane or an
  interfacially polymerized layer within a thin-film-composite
  membrane—where the separation occurs. In most cases, the membrane is
  designed to allow only water to pass through this dense layer, while
  preventing the passage of solutes (such as salt ions). This process
  requires that a high pressure be exerted on the high concentration
  side of the membrane, usually 2–17 bar (30–250 psi) for fresh and
  brackish water, and 40–82 bar (600–1200 psi) for seawater, which has
  around 27 bar (390 psi)[6] natural osmotic pressure that must be
  overcome.

I had the same question because yes ions are significantly smaller than a molecule of water.
After some research I concluded that the effective size of ion in a solution is actually dictated by the solvation shell. This arrangement of water molecules around the ion neutralize the charge and make it much larger than an individual water molecule and the pores in the reverse osmosis membrane.

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text:
Why NaBH4 does not reduce carboxylic acid while LiAlH4 does?
My teacher told - "To reduce carboxylic acid  to alcohol, $\ce{LiAlH4}$ is used. But since $\ce{LiAlH4}$ is an expensive reagent, it is commercially reduced by forming an ester and then to alcohol". I asked that we can also use $\ce{NaBH4}$ as it also contains 4 hydrogen bonds. He said that $\ce{LiAlH4}$ is a strong reducing agent that $\ce{NaBH4}$. I was not satisfied by his answer. I want an answer in terms of a reaction mechanism.

Reaction mechanism won't help you much here imo, but thermodynamics could.

The answer is polarity.
Consider the two ions $\ce{AlH4-}$ and $\ce{BH4-}$ (as they exist as ions in a solution).
Now the electro-negativity of $\ce{Al}$ is $1.5$, and that of $\ce{B}$ is $2$, and that of $\ce{H}$ is $2.1$.
So the electro-negativity difference ($\Delta\mathrm{EN}$) between $\ce{Al-H}$ bond is $0.6$ and that of $\ce{B-H}$ bond is $0.1$. It is clearly visible that the $\Delta\mathrm{EN}$ of the $\ce{Al-H}$ bond is greater than that of the $\ce{B-H}$ bond.
Now considering the mechanism, due to the small polarity induced by the carbonyl group of the carboxylic acid, the $\ce{H}$ atom in $\ce{Al-H}$ bond is readily broken due to its high polarity, while the $\ce{H}$ in $\ce{B-H}$ can't do the same because it has a low polarity.
This is the reason why $\ce{NaBH4}$ cannot be used instead of $\ce{LiAlH4}$ for reducing carboxylic acids to alcohols.

Borane $\ce{BH_3}$ can reduce carboxylic acids to alcohols


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Does nitrogen inversion affect the basicity of amines?
If I were to compare the basic strength of 1-azabicyclo[2.2.1]heptane and triethylamine:

Can I say that 1-azabicyclo[2.2.1]heptane is more basic than triethylamine because the lone pair of electrons is less available in the latter due to rapid nitrogen inversion? Nitrogen inversion is not possible in the bicyclic amine.
There is not much difference in the inductive effect. In both cases, the nitrogen atoms are $\mathrm{sp^3}$ hybridized.

Background
Amine basicity correlates with, among other things, the hybridization of the nitrogen orbital that is holding the lone pair of electrons.  The less s-character in this orbital, the more basic the amine. So, as the following figure indicates, a trialkylamine (the lone pair is in an $\ce{sp^3}$ orbital) is more basic (less acidic or higher $pK_{a}$) than the lone pair in pyridine (the lone pair is in an $\ce{sp^2}$ orbital), which is less acidic then the lone pair in a nitrile (the lone pair is in an $\ce{sp}$ orbital).

(image source)
The rate of nitrogen inversion also correlates with, among other things, hybridzation. But in cases where these "other things", such as ring constraints, limit nitrogen inversion, then the rate of nitrogen inversion may not correlate with hybridization. So the rate of nitrogen inversion does not always correlate with hybridization and is therefore not a good indicator of lone pair "availability".
Answer
Looking at your two compounds, a reasonable first guess might be that the nitrogens in these compounds are both roughly $\ce{sp^3}$ hybridized.  Therefore, even though one compound can undergo nitrogen inversion and the other can't, their hybridzations are similar and we would expect similar basicities.
We could refine this prediction by noting that in the bicyclic compound, due to geometric constraints, the carbon-nitrogen bonds around nitrogen are slightly pulled back a bit from the tetrahedral angle.  This would increase the p-character in these bonds (as a bond angle moves from 109° to 90° the p-character increases with 90° being pure p).  As a result, the p-character in the lone pair orbital decreases and the s-character increases.  As mentioned above, more s-character in the lone pair orbital leads to lower basicity.
Therefore we might expect both compounds to have similar basicity with the bicyclic compound perhaps being a bit less basic.
This analysis is corroborated by the $\mathrm pK_\mathrm a$ values of the conjugate acids (a more positive number means that the amine is more basic). According to J. Org. Chem. 1987, 52 (10), 2091–2094, quinuclidine has a $\mathrm pK_\mathrm a$ of 10.90, and 1-azabicyclo[2.2.1]heptane a $\mathrm pK_\mathrm a$ of 10.53. The decrease in basicity can be attributed to the increased strain and greater s-character of the lone pair. Triethylamine itself has a $\mathrm pK_\mathrm a$ of 10.75 (Evans table), making it marginally more basic than 1-azabicyclo[2.2.1]heptane.

Ron, using your explanation, the former should be more basic. Strained systems would show more p character wouldn't they? Take cyclopropane which is actually sp5. If lone pairs are exposed with no N inversion, the compounds do tend to be slightly more basic. Compare piperidine with di or triethylamine

"Strained systems would show more p character"...yes, the strained bonds.  So the bonds to carbon around the nitrogen would have more p-character (right?) leaving more s-character for the lone pair.  More s-character, lower basicity

On the one hand I'm wrong; quinuclidine is not strained as the rings are 6 membered. On the other hand,  quinuclidine is about 0.25 units more basic than triethylamine, using Evan's Pka's table. I think the explanation is almost there

@Mithoron  Does Bredt's rule is followed if we have any other atom(like, here N in this case) ? By Bredt's rule i mean "carbocation or double isn't stable at bridge head C atom"

@safdar Does Bredt's rule is followed if we have any other atom(like, here N in this case) ? By Bredt's rule i mean "carbocation or double isn't stable at bridge head C atom"

There is more steric hindrance to the approach of the N to some acid when you have a non bicyclic amine. With quinuclidine the nitrogen free pair of electrons are more "exposed" and available for bonding to a proton or other acid. With trimethylamine, there is bond rotation about all the single C-C and C-N bonds. The space around the N free electron pair is cluttered(spatially hindred) due to the ethyl groups. Tougher for a proton to get to the N electron pair.

if nitrogen atom on the first compound accepts H+.Since the lone pair is given up, there will be less lone pair bond pair repulsion only bond pair bond pair repulsion making the compound more stable in such strained systems.

The bridge head is the most unreactive, and thus if we see the conjugate acid of the first compound, it will be extremely unstable. Amine flipping does lower the basicity, but still in this context, the basicity of the second amine will be greater than that of the first.

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text:
Reaction of citric acid and calcium chloride
I was planning to extract citric acid from lemon juice by reacting the lemon juice with calcium chloride hoping it would react to form calcium citrate and precipitate according to the reaction
$$\ce{2 C6H8O7(aq) + 3 CaCl2(aq) -> Ca3(C6H5O7)2(s) + 6 HCl(aq)}$$
However I was not sure that the reaction would occur.  I know that the calcium citrate should precipitate out and drive the reaction forward, however it seems that the $\ce{HCl}$ generated reaction would then react with any calcium citrate bringing back to calcium chloride and citric acid.  Which reaction would move to completion or would the two reactions reach an equilibrium.  If this will not work I plan to just use calcium hydroxide instead by reacting calcium chloride with sodium hydroxide.

How do you know that calcium citrate will precipitate out and drive the reaction forward? In fact, it will not. Why in the world would it? Because calcium citrate is poorly soluble (which is true)? Well, that's not enough. Calcium carbonate or calcium phosphate are even more insoluble, yet even they will not precipitate in similar conditions (i.e., in acidic media). That's the problem with the salts of weak acids. (Yes, we may consider phosphoric acid weak, if we are talking of its third proton.) You have a solution with plenty of calcium and citrate (or phosphate, or carbonate), but most of that citrate (or phosphate, or carbonate) is not in the fully deprotonated form, so the solubility limit is not reached, and the precipitate can't appear.
Also, the very word "extract" does not quite apply here anyway.
Also, what was your plan to regenerate citric acid from calcium citrate afterwards?

Well I didn't vote for deletion but I feel that at the moment it reads more like a comment than a proper answer, especially seeing as it throws a lot of questions back at the OP.

I guess it does. Well, let's add some supporting info.

This procedure is done in our laboratory class, but first you have to add sodium hydroxide until the lemon juice is somewhat alkaline. Then add the calcium chloride (use $1~\mathrm{ml}$ for every $2~\mathrm{ml}$ of the lemon juice + $\ce{NaOH}$ solution).
I think the calcium chloride would first react with the hydroxide, producing calcium hydroxide, which would then react with citric acid to form calcium citrate.
$$\ce{2 C6H8O7 + 3 Ca(OH)2 -> Ca3(C6H5O7)2.4H2O + 2H2O}$$
Filter out the salt (precipitate) formed, weigh it, then you would have to add sulfuric acid afterwards to react with the calcium citrate. This would then form citric acid (soluble) and a precipitate which is calcium sulfate.
$$\ce{Ca3(C6H5O7)2.4H2O + 3H2SO4 -> 2C6H8O7 + 3CaSO4.2H2O + 2H2O}$$
You would then have to concentrate the filtrate to obtain the citric acid crystals.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I improved the formatting of your post by using MathJax. You can learn more about it in the [help] or by visiting this meta-post. Please note that there is a difference between uppercase O (the letter) and 0 (the number) and between a subscript 5 (the number) and lowercase s (the letter). I was astounded at $\ce{C6Hs07}$ ;)


text:
Determine the order of the reaction, and the value of the rate constant. Predict the concentration of AB at 25s
I've never posted a question about chemistry before so I'm not really sure how this community works but here is the question.

I do have the answer to this question but I have no idea on going about solving it.
I do know:

that second order is equal to all of the exponents added up
Rate is equal to $k[A]^{p1} [B]^{p2} $
Rate is also equal to $([A_2]-[A_1])/([t_2]-[t_1])$

And I believe $M$ stands for molarity.
I have little to no idea how I am to solve this other than those 4 things.
Here are the answers the book gives me:

Second Order
$k = 2.25\times 10^{-2}$
$[AB]$ at $25s = 0.619 M$

The rate of the reaction depends on the concentration of reactants, not products.

but all they have given me are the products to work with.

By the way, this previous question is the same as yours, just the numbers are changed. The answer should be applicable. And I don't quite get what you're talking about. AB is the reactant and the concentration of AB at various times is given to you in that table.

If n is the order of this reaction, then the rate equation is given by
$$-\frac{d[AB]}{dt}=k[AB]^n$$
If we take the log of both sides of this equation, we obtain:
$$\log\left(-\frac{d[AB]}{dt}\right)=n\log([AB])+\log k$$
This equation is of the mathematical form $y=mx+b$ (i.e., a straight line), where $y=\log\left(-\frac{d[AB]}{dt}\right)$, $x=\log([AB])$, m = n, and $b=\log k$.  So, if we could plot a graph of $y=\log\left(-\frac{d[AB]}{dt}\right)$ vs $x=\log([AB])$, we should get a straight line with a slope of m and and intercept of logk.  So we could determine both the reaction order and the rate constant all in one shot from this single graph.
But how would we reduce the given data to plot such a graph?  Well, we could use finite difference approximations for the average reaction rate and the average concentration over each time interval.  If $[AB]_t$ represents the molar concentration of reactant AB at time t, we would approximate the average reaction rate over each time interval between two successive values of time $t_1$ and $t_2$ in the table as:
$$\left(-\frac{d[AB]}{dt}\right)_{ave}=\frac{[AB]_{t_1}-[AB]_{t_2}}{t_2-t_1}$$
The average concentration over this same time interval would be:
$$[AB]_{ave}=\frac{[AB]_{t_1}+[AB]_{t_2}}{2}$$
So we would plot the log of the average reaction rate as a function of the log of the average reactant concentration for all the time intervals in the table.  The result should be a straight line with a slope equal to the reaction order and an intercept equal to the log of the reaction rate constant.

Whenever you are given concentrations vs. time, think "integrated rate law." There are three integrated rate laws that your instructor will either give you or expect you to know:
(1) for a zero order reaction: $[A]_t = -kt + [A]_0$
(2) for a 1st order reaction: $ln[A]_t = -kt + ln[A]_0$
(3) for a 2nd order reaction: $\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$
An important thing to note is that each of these equations is in the same pattern as the equation for a line, viz., $y=mx+b$.  For example, number 2 above (for the 1st order equation):
$\underbrace{ln[A]_t}_{y} = \underbrace{-kt}_{mx} + \underbrace{ln[A]_0}_{b}$
Why is this important? Because you can plot the data as expected by each of these equations and the one that plots as a straight line tells you the order of the reaction.
So, use your calculator or spreadsheet and enter the data. Then add two columns. One column computes the natural log of the given concentrations. The second column computes the reciprocal of the given concentrations.
Now create three separate plots. Plot "Time" on the horizontal axis of all three plots. Then, on the vertical axis of each separate chart, plot the given concentration in chart1, the natural log of concentration in chart2 and the reciprocal of concentration in chart3. You will find that chart3 is the graph of a straight line. Therefore, this is a second order reaction.
Your spreadsheet or calculator can calculate a linear regression line. The value of $k$ is the slope of that linear regression line. How do we know that? Because $k$ is the slope for each of the integrated rate law equations. In this case, the integrated rate law for this second order reaction is
$\underbrace{\frac{1}{[A]_t}}_{y} = \underbrace{kt}_{mx} + \underbrace{\frac{1}{[A]_0}}_{b}$
as shown above. (The $kt$ term is the $mx$ term in the equation for a line and $m$ is the slope.)
Finally, you can calculate the concentration of AB at 25 seconds using the equation for the regression line, which in this case is $y=0.0225x + 1.06$. Since we calculated the regression line for $\frac{1}{[AB]}$ we need to reverse that, so we want $[AB]_{25s} = \frac{1}{y}$:
$[AB]=\frac{1}{0.0225 \cdot 25 + 1.06}$ (the regression calculated the reciprocal so we need to reverse that)
$[AB]=0.616$


text:
Bond length comparison between two carbon atoms
Why is the bond length of double and triple bonds between two carbon atoms shorter than the single bond length between two carbon atoms?

In the case of a carbon-carbon single bond, 2 electrons are shared in the bond connecting the two carbon atoms. With a carbon-carbon double bond, 4 electrons are shared between the two carbon atoms and 6 electrons are shared in a triple bond.  Having additional electrons between the two atoms 1) improves the bonding overlap (makes the bond stronger) between the two carbon atoms and 2) better screens the two carbon nuclei form each other.
Both of these factors, better bonding overlap and better nuclear screening, will allow the two carbon atoms to approach closer together.  Consequently, the more electrons (or the more bonds) between two carbon atoms, the shorter the distance between them.


text:
Melting and boiling point comparison
Graphite, silica and diamond are covalent compounds and still they have high melting and boiling points. Why?

The very term "covalent compounds" is confusing, so you may just as well stop using it. The elementary substances you specified are covalent crystals (AKA "covalent solids"). A single crystal of any of these is in fact one huge molecule linked by covalent bonds. To melt that crystal, you have to break the covalent bonds, and they are pretty strong.
On the other hand, we have $\ce{N2}$, $\ce{O2}$, and other covalent molecules. They may be quite strong (in that each molecule by itself is pretty hard to tear apart), but the molecules are linked together by the so-called van der Waals forces, and these are quite weak, so these compounds are gases, and it takes some effort to see them turning into a liquid or a solid.


text:
Why do metals form ions in solution?
I'm looking at redox half-reactions, such as the one below:
$$\ce{Zn(s) -> Zn^2+(aq) + 2 e-}$$
Does the zinc react with the water and dissociate into ions? Why doesn't solid zinc have a charge?

How do your questions connect with half-reaction? Zinc can react with water in presence of significant amount of OH- ions, but you need second half reaction

Solids can't have a charge (neither can liquids or gases). The electrostatic forces are immensely great, way greater than you seem to think right now. Just imagine the forces that would be tearing apart a droplet of water (say, 36 microliters) if 0.001 of its molecules (not all of them, God forbid) would suddenly become positive ions. If you are aware of the Coulomb's law, you can help your imagination by actually calculating these forces.
(By now you may wonder how electrical phenomena are possible at all, but that's another story.)
Just in case if you don't like that answer, here is another:
Solid zinc, like any other metal, does have a charge. It consists of $\mathrm{Zn}^{2+}$ ions. These ions are immersed in a sea of electrons, which balances their charge, holds them together, and also is responsible for such properties typical of all metals as electrical conductivity, metallic lustre, etc.

Does the zinc react with the water and dissociate into ions?

Not by itself. You've only shown a "half-cell" reaction in a chemical reaction known as an oxidation-reduction reaction. For the reaction to take place something in the solution would also have to be oxidizing. So if you just stick a strip of zinc in distilled water, the zinc strip won't dissolve.

Why doesn't solid zinc have a charge?

The simple answer is by convention. When we say "solid zinc" we mean a macroscopic size sample of zinc. So we're saying that the whole sample isn't charged. However if you look at an individual zinc atom in the chunk of zinc, then the single atom effectively has a +2 charge and 2 electrons are in "metal conduction band" which means that the electrons are in bonding levels that allow them to conduct electricity.
If you have two metal zinc plates you can make a capacitor from them. You can then use a battery to charge one plate + and one plate -.
Another exception is a thermocouple. You have two different metals pressed together. Think of two wires fused in the middle. At the junction the two metals will develop a charge. The current will be small, but an operational amplifier can easily detect the current.


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Does solid aluminum oxide conduct oxide ions at high temperatures?
Can alumina behave as a solid electrolyte for oxygen ions? Could a cell be constructed with an oxygen cathode and a molten aluminum anode producing more alumina? Would the resulting alumina deposit on the electrolyte? Would it be crystalline?

Alumina is one of the most abundant compounds on Earth. If it would be also a solid electrolyte for oxygen ions, why in the world would anyone go into the trouble of dealing with the whole $\ce{Y2O3/ZrO2}$ thing? No, alumina doesn't and can't behave like that. Its oxygen ions are held in place pretty tightly.
Some forms of alumina may serve as conductors of $\ce{Na+}$ and similar cations, but that's another story.

You might use use yttria-stabilized zirconia because alumina is a poor ionic conductor. Ice is a proton conductor, but nobody uses it as such. Ionic conductivity increases with temperature. The melting point of alumina is over 2000°C. Are you sure there's no ionic conduction?

Why, there might be some, just not enough to make a real thing out of it. I should have said "If it would be also a good solid electrolyte..."

If you read my question carefully, you'll note that I never asked if alumina is a good solid electrolyte, only if it is one. The popularity of zirconia and ceria based electrolytes in SOFCs proves it doesn't make for a practical electrolyte if the intent is to generate electricity.

Hm... that's right. It turns out that in fact I answered another question, albeit somewhat related to yours. Well, let it remain this way.


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Does freezing tights make them more durable?
Various blog posts and articles suggest that freezing a pair of tights will make them last longer.
From a blog named Eclipsed Moon:

Put your tights in the freezer.
I thought that this was a silly old-wives tale, but it actually does
  work! Putting your tights in a plastic bag and storing them in the
  freezer when you aren’t using them helps keep the material firm and
  less prone to ripping.

From an article on PureWow:

Good thing there’s a super-easy trick for extending the life of your
  tights.
You freeze them.
Yes, it sounds crazy, but hear us out: The next time you buy a new
  pair (and before you wear them), run them under water until they’re
  damp. Then place them in a plastic bag and freeze overnight. Take them
  out and give them a day to defrost and thaw out.
You only need to do this once — after that, you’re good to go. The
  chilly temps firm and strengthen the individual fibers of your tights,
  making them less susceptible to runs later on.

I'm skeptical.  My questions are:

Where is the evidence?  Are there any controlled studies confirming this phenomenon?
Does the water make any difference?
Is this advice for storage, one-time treatment, or both?
If it is indeed effective as a one-time treatment, why don't they come pre-treated from the factory?
If it works, why does it work?

Cross-posted here after being closed on Lifehacks, since this seem to be the closest to a Materials Science site.

Did you actually check it? Chemistry is an experimental science, you know.


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Phosphoric Acid Titration
I physics graduate student, but I teach an interdisciplinary lab for the university which incorporates biology, physics, and chemistry.
We recently added a weak-acid titration lab to the course and I got a result which surprised me.

The procedure was to neutralize 100 mL of 0.010 M Phosphoric acid by adding 0.10 M NaOH.
We have 0.001 mole of phosphoric acid in solution, but this acid is triprotic which means that a maximum of 0.003 moles of hydronium can be released into the solution. My understanding of the equillibrium reactions was that as I lowered the ambient hydronium ion concentration more protons would be released by the acid.
So my expectation was that I would need to add 30 mL of the NaOH solution to reach a pH of 7. In practice it only took between 12-15 mL.

I talked this over with a Chemist I work with and she wasn't sure why this discrepancy happened either. One possiblity is that the person who prepared the acid screwed up, but we don't have more stock for me to test that theory.
So my question is do these results sound normal or did I have the right idea? If I didn't have the right idea then what am I missing theoretically speaking?

It's a long story; read about the $\mathrm{pK_a}$ and acid-base equilibrium, otherwise it would not make sense to you. The result is more or less OK.

You are right in calculating that you need 30 mL of NaOH for complete reaction with H3PO4. However, phosphoric acid is a weak acid and NaOH is a strong base. The pH at the end-point will therefore be greater than 7, and the volume of NaOH you need to reach pH 7 will be less than 30 mL.

You might want to read about the pH of a neutralized solution. But in case you don't, one of my answers here on chem.se covers the calculation of titration curves with and without increasing volume. It is quite easy to create a titration curve using an adapted (see exact solution for triprotic acids here) version of equation 8.

$$V(B)=\frac{V(H_3A)~\left(\frac{k_w}{x}-x+C(H_3A)\frac{x^2 k_1+2x k_1 k_2+3 k_1 k_2 k_3}{x^3+x^2 k_1+x k_1 k_2+k_1 k_2 k_3}\right)}{\left(x-\frac{k_w}{x}+\frac{C(B) \cdot k_b}{\frac{k_w}{x}+k_b}\right)}$$

I did that for you, so you can see, why it is not wrong, that you only needed 12-15 ml to reach pH 7 but also that that pH 7 is not what you searched for.

What you probably actually wanted, was to reach one of the equivalence points at 10 or 20 ml. You will also not have a chance to reach the third EP in an aqueous solution, which is why 30 ml is not a good goal to go for.

Thank you for the comprehensive answer. This is exactly what I was looking for.

Why is it not possible to reach the equivalence point in aqueous solution? Is it because the third neutralisation reaction with $\ce {NaOH}$ does not go to completion?

I think it is best to ask this as a separate question.

This is very nice and I am sorry I did not find your answer until a short time ago! I have been using the same equation, with slightly different notation, for many years. It is from an old Schaum’s Outline, as shown in the second Excel spreadsheet screenshot here: https://chemistry.stackexchange.com/a/136203/79678. I will be linking  your answer here every time we get the polyprotic acid problem. I upvoted this answer and several other nice ones you linked.

Phosphoric acid (which is WEAK ACID)  releases it's protons in 3 different stages. It has 3 pKa values for 3 dissociations. So it is impossible to get a 1:1 reaction , reactant condition when titrating.
Here are the pKa values to help you in calculations,
Acidity (pKa)
1 = 2.148
2 = 7.198
3 = 12.319
(at room temperature)


text:
Potassium tert-butoxide reaction with chloroform and picric acid
If potassium tert-butoxide is used in a solution of chloroform and and a strong protic acid like picric acid will the tert-butoxide ion react with picric acid or will it form a dichlorocarbene.
Mechanism for reactions will be much appreciated.
(This is most definitely NOT a homework question!
I need to know this to solve my homework problems)

Well, in both cases, KOtBu acts as a base. If you were a molecule of KOtBu, would you prefer to react with CHCl3 or picric acid?

@orthocresol from what I know, I would react with picric acid because it has a pKa of around .38 ( I think?), much lower than that of chloroform. But I wanted to know if there was some other mechanism that I wasn't aware of which would force it to react with CHCl3 to form a dichlorocarbene.

Depends if you had more equivalents of KOtBu than picric acid.Even if you have some chloroform deprotonation, the protonation from picric acid is so much faster than loss of Cl-.


text:
What are some sources of error in the preparation of a standard solution (of sodium carbonate) and a titration (determining the concentration of HCl)?
When preparing  a standard solution of sodium carbonate, what are some sources of error?
Also, when conducting a titration, what are some errors involved when determining the concentration of hydrochloric acid?
Thank you

This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Most errors of standardization come from weighing the solid ($\ce{Na2CO3}$) or from measuring the liquid added in the dilution of $\ce{HCl}$. In theory you could do the necessary calculations and take the time needed to weigh out the solid/liquid you can create solutions that are incredibly accurate that will be very near the target molarity. As far as standardizations go imprecision of the equipment is going to be the biggest drawback, though ensuring the sodium carbonate is throughly dried could be another source of error.


text:
Why won't ethene and ethyne react with halogens?
With alkanes, a free-radical halogenation (substitution) is observed.
I was told that there will be no reaction in a similar situation with ethene and ethyne — and I am having trouble understanding why.
Is it because the halogen radical will not be able to abstract a hydrogen atom from carbon because it is influenced by the greater electron density in the carbon-carbon double/triple bonds? Or because the $\ce{C-H}$ bonds are shorter in these molecules compared to alkanes? (resonance?)
I've assumed that the problem arises in the propagation step of the most common reaction mechanism seen with alkanes.
Or is it due to sterics? ($\unicode[Times]{x3c0}$-bonds get in the way of the attacking bromine radical or restricted rotation in double/triple bonds get in way of ideal orientation for reaction to proceed to the transition state)
The necessary heat/$h \nu$ is provided for the reaction. Additionally, no other reactants/catalysts are present.
Please keep your answers on a novice college student level.

Both substance classes have an inherently different reactivity. Alkanes are very unreactive — in fact, save radical reactions, there is not much that will adequately attack them. Alkenes and alkynes have $\unicode[Times]{x3c0}$-bonds which are much less energetically stable than the alkanes’ $\unicode[Times]{x3c3}$-bonds.
$\unicode[Times]{x3c0}$-bonds can be attacked by halogens directly, the mechanism is called electrophilic addition and the usual product is an anti-dihalide with the halogen atoms attached to the carbons previously involved in the double bond. An intermediate species of this reaction is the halonium ion, a three-membered ring consisting of the ex-double-bonded carbons and one positively charged halogen. The other halogen can then attack according to $\mathrm{S_N2}$ from behind.
However, if you try hard enough you will also manage to create halogen radicals in the presence of alkenes. The radical will now not attack a $\ce{C-H}$ bond, but rather the energetically less stable $\ce{C=C}$ bond forming a $\ce{C-Hal}$ bond and a new carbon-centred radical. This carbon radical will then most likely polymerise with further double bonds by the same mechanism, creating a carbon polymer rather than a haloalkene.
I’m not entirely sure how alkynes will react. Their triple bond is more stable than a double bond, so selective halogenation of double bonds is possible in the presence of triple bonds. However, in general there is no reason why they couldn’t react according to the same pathways.

@SufyanNaeem Thanks, I keep reconsidering becoming a teacher ;D

It's because the addition of a halogen (at least chlorine, bromine and iodine) with an alkene or alkyne is not a free-radical halogenation, it's (quite rapid) electrophilic addition.  For instance, see this tutorial:  http://www.chemguide.co.uk/mechanisms/eladd/symbr2.html

It would be helpful to provide a summary of the link in the answer, in case the link becomes invalid in the future.

If you generated a halogen radical, wouldn't you end up with the more competitive polymerisation occurring as compared to the benzoyl radical (from benzoyl peroxide)


text:
Change in pressure?
Consider a sample of gas in a fixed volume container . From the arguments of Kinetic Theory of Gases  on quadrupling the temperature what will be the effect on pressure .
Basically I thought that the answer should be that the pressure gets doubled . But the answer is that it gets quadrupled . Can anyone explain how ??

Why do you think it should result in a two fold increase in pressure?

Consider the following diagram of a particle. Now under the assumptions from kinetic theory, the average force on the walls of the container is given by: $$F_{average} = \frac{mNv^2_x}{L}$$ Now assuming that the speed of the particle is random in all directions we get: $$v^2 = v^2_x + v^2_y + v^2_z = 3v^2_x$$ Now pressure is defined as the average force acting on a surface area of $\mathrm{1m^2}$. Therefore the pressure is: $$p = \frac{F_{avg}}{A} = \frac{mNv^2}{3LA} = \frac{mNv^2}{3V} = \frac{2N}{3V}\frac{1}{2}mv^2$$
So we can see that the pressure on the wall is directly proportional to $1/2mv^2$. This relates to the total kinetic energy of the particle. Temperature is defined as the average kinetic energy of a particle, hence $1/2mv^2$ is directly proportional to temperature. Therefore a two fold increase in temperature will result in a two fold increase in kinetic energy which will result in a four two increase in pressure.
Hence $p \propto T$. Therefore a four fold increase in temperature will result in a four fold increase in pressure.

Prove it with Kinetic theory of gases

Sorry I misunderstood your question. I will edit my answer so that it uses Kinetic theory of gases.


text:
Determining pressure of sublimation of iodine

The pressure of sublimation of iodine can theoretically be determined by considering the two hypothetical steps: the sublimation of iodine at $\pu{298 K}$, followed by an isothermal expansion of our system until it reaches the sublimation
  pressure.
a) Determine the enthalpy and entropy change for this process.
  b) Determine the pressure of sublimation.
Given Data

Triple point: $\mathrm{12.07~kPa, 113.0~^\circ C}$
Melting point: $\mathrm{113.7~^\circ C}$
Boiling point: $\mathrm{184.3~^\circ C}$
$C_{\mathrm m,p}{\left(\ce{I2 (s)}\right)} = 55.44 \mathrm{~J~K^{–1}~mol^{–1}}$
$C_{\mathrm m,p}{\left(\ce{I2 (l)}\right)} = 89.38 \mathrm{~J~K^{–1}~mol^{–1}}$
$C_{\mathrm m,p}{\left(\ce{I2 (g)}\right)} = 36.90 \mathrm{~J~K^{–1}~mol^{–1}}$
$\Delta_\text{f}H^\circ_\mathrm m{\left(\ce{I2(g)}\right)} = 62.44 \mathrm{~kJ~mol^{-1}}$
$S^\circ_\mathrm m{\left(\ce{I2(g)}\right)} = 260.6 \mathrm{~J~K^{-1}~mol^{-1}}$
$S^\circ_\mathrm m{\left(\ce{I2(s)}\right)} = 116.1 \mathrm{~J~K^{-1}~mol^{-1}}$

My Attempt
I have calculated the enthalpy and entropy change for the first step by using Hess's Law. They are $\mathrm{62.44~kJ~mol^{-1}}$ and $\mathrm{144.5~J~K^{-1}~mol^{-1}}$ respectively.
Now for the second step which is the isothermal expansion:
$$\Delta H = -nC_{\mathrm m,p}\Delta T = 0$$
For entropy:
$$\Delta S = nR\ln\left(\frac{V_\mathrm f}{V_\mathrm i}\right)$$
If assume that I have 1 mole of iodine I know what $n$ and $V_\mathrm i$ is but how am I suppose to know what $V_\mathrm f$ is?
Also I am struggling to understand the question. It says isothermal expansion of our system until it reaches the sublimation pressure. How are you suppose to know when it reaches sublimation pressure if it has already sublimed?

You know that the change in G in going from the solid to the vapor at the equilibrium sublimation pressure is zero, correct?  You also know that the free energy of formation of the solid at 298 K and 1 atm is zero, correct?  So, all you need to do is determine the free energy of formation of the vapor at the hypothetical state of 298 K and 1 atm, and then determine the pressure change necessary to bring that down to zero.

authors: 北京做假证_北京办文凭毕业证 - spam


text:
Is it possible to disable the splash screen of ChemDraw?
Is it possible to disable the splash screen of ChemDraw? I can't find a corresponding setting in the preferences dialog. Is there maybe a hidden setting / some configuration file that can be modified?

I assume the question seeks out for a way to use ChemDraw more quickly, for example while being confronted with a larger number of chemical structures already generated, and to be processed further.  Then

If it is to browse through a number of structures aiming "to find the right file", to gain an overview, then Thumbfish (Windows only) may be helpful.  Chemdoodle (Win/Mac/Linux) and MarvinSketch offer in their file-open menu a preview as well.
On the other hand, provided the extensive documentation especially of ChemDraw seconds native, the *.cdxml format, the (batchwise) conversion into other formats is easily done with OpenBabel (supported file formats) and Apple Script ChemBot (application here).

Thank you for your suggestions. I am indeed looking for a way to use ChemDraw more quickly. However, I am manually creating/editing many structures..


text:
Hydrogentaion of alkenes in absence of catalysts
I came across this question regarding hydrogenation of alkenes
Reaction A and Reaction B were carried out using the two different substrates as shown
I have depicted how i thought the reaction would occur.This gives me the correct answer for Reaction A but not for Reaction B.
Apparently instead of b,the compound c would be formed,implying simple hydrogenation of the exocyclic bonds without the formation of a transition state as in reaction a .
I cannot figure out the exact cause for this.
Does this have anything to do with the two double bonds being on the opposite sides in coumpound Y ,and therefore not being able to form the same transition state as X?
P.S: No catalysts were mentioned.Only H2(1 mole) and heat were mentioned in the original question.

It has to be catalytic hydrogenation following the standard mechanism. I can't see how they expect you to predict the product of the top one unless it has been taught specifically in class.

Isn't that how hydrogenation occurs?(in absence of a catalyst)
And no, we weren't taught non-catalytic hydrogenation in class, but this isnt a classroom  material.
I came across this question while going through a problem book.

I don't know any examples of non catalytic hydrogenation. For the (catalytic) mechanism see: link

I know the mechanism for catalytic hydrogenation.But i dont see how, if we were using a heterogenous catalyst,the correct products of the above reactions can be predicted

I agree. The bottom one is exocyclic disubst vs endocyclic trisub. But for the top one I can' t see how you can predict the product they ask unless it has been specifically mentioned in theory.

I have experienced a lot of hydrogenation and as depicted, does not match any variations that aren't heterogeneously catalysed. Homogeneous catalysis, metal or non-metal alike do not operate this way either. The diagram suggests a concerted mechanism. Has anyone else seen anything like this? The closest I can suggest is diimide

I am not sure of mechanism at all, it just seemed to answer one of the two questions given. Can you predict a mechanism whereby we may arrive at a from X and c from Y?

The main point of these exercises is that not all C=C bonds are created equal. Their reactivity towards hydrogenation can be influenced by steric effects and specifically applying here, the degree of substitution of the double bond. So for the bottom example, the exocyclic disubstituted double bond will react but not the endocyclic one which is much more stable. Now here there is some uncertainty in this exercise since the actual result is dependent on the exact conditions which are not provided. For example hydrogenations are well known to be very capricious on the selection of metal catalyst and $\ce{Pd}$  can give a different result than using $\ce{Pt}$ or $\ce{Rh}$. And this only an example as there are so many different set of conditions possible with variable results.
The top one is an even worse exercise in the sense that the outcome is very specific on the exact conditions. Now, knowing the exact outcome we can rationalise as follows: Here the electronics also play a role and this diene because of conjugation will be more stable and thus more resistant towards hydrogenation. When one of the exocyclic bonds gets reduced (even if it only partially does so) then in the intermediate formed the other unreacted double bond will isomerise (much much faster than the reduction of the exocyclic conjugated bond in the starting material)  to the more stable product with the  endocyclic double bond i.e. Structure a. This structure will be unreactive towards further reduction and eventually will be the sole product of the reaction.


text:
How to determine age of stones (and other inorganic matter)?
I'd like to ask how can one determine /with relative precision/ the age of stones and stone monuments, like statues, columns, stone walls or stone tablets? I know that one determines the age of organic substances by radiocarbon dating, however, this method is inapplicable for rocks, so what else can we try?
Under age I mean the period since a stone was carved into a historical artifact and I'm interested in periods that are accurate up to a historic age, like Antiquity or the Renaissance - I suppose one cannot ask for more, unless there are visual marks on the artifact, which provide clues?
PS I hesitated much whether my question is Chemistry- or Physics- related and decided to post it here, since it's purely practical. If you decide otherwise, I'll move it to Physics.SE ;)

Are you suggesting that physics is a purely theoretical subject and chemistry a purely practical one? ;)

Definitely no. I really struggle to determine the boundaries between sciences. My logic was that in Physics one usually perform experiments to verify Physical laws, while in Chemistry the experiments usually have practical applications like the one I'm interested in... Thus, I think my question is more about Chemistry rather than Physics /maybe I'm wrong, of course :P/

What timeframes are you interested in? Ancient rocks are usually dated with radioisotope methods, often using uranium. More modern rocks may use other techniques.

I'm interested in narrower time-frames, e.g. centuries or at most millennia. For example, how can we determine whether a statue is from Antiquity or from the Renaissance (assuming there are no erosion and other visual marks on it)?

Are you asking about the age of the rock since it solidified or recrystallized? This is possible by radiochemical means (e.g. potassium–argon dating). Or are you asking about the age of a statue since it was made from that rock? Depending on the circumstances, this may also be possible by different radiochemical means.

I meant the later. I found in Google how to determine the epoch, from which a rock is, but this doesn't help much in determining the age of cultural artifacts...

The question being asked, I believe, is the age of the inscription or statue rather than that of the material from which it is made.
The answer, though, is that it is quite difficult to do on the basis of a single stone object. If it is found in situ, then the age of surrounding items may be a guide.
There are some clues to the age of stone objects, though. For example, if a stone object (e.g. the walls of Siloam's Tunnel) has soot on its surface, such as from torches, then radiocarbon dating shows a minimum age (because the soot could have been deposited some time after the tunnel or object was made).
Weathering of a surface can provide information, providing the object has been exposed to known atmospheric conditions; but a buried statue would erode differently from one on the surface.
For stones, or better, for ceramics, that have been heated, thermoluminescence provides an accurate way to measure age (or at least time since last being heated).
One indication for the age of the Egyptian pyramids is their alignment to a calculated geologic reference that changed with time due to the precession of the Earth.

Radiocarbon dating is one form of isotopic dating, which is applicable to rocks, and is the basis for huge chucks of geology, archaeology, and astrophysics.  How do you think we estimate the age of earth, moon, prehistoric pottery and tools, etc?
There are many ways to do it:
thermoluminescence dating, Cl-36 dating (within 50 years of present due to Nuclear weapons testing), U-Th dating(last 500,000 years), Optical dating (100-200,000 years before present), and I-129 dating (recent).  These just are the ones I can think of off the top of my head, I'm sure there are many more, and any decent archaeology textbook will have a chapter on the subject if you need more specifics.  Any method can be chosen based on what the sample is, how recent it is, and how big it is (ie how much you can destroy to get an answer).

I presume you mean that isotopic dating is applicable to rocks? I did a double-take at the first sentence because I thought you meant that we could use radiocarbon dating to date the moon, which would be quite impressive indeed.

@chipbuster Clarification: Radiocarbon dating is one form of isotopic dating.  Isotopic dating can be applied to many things besides organic matter.


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What is the difference between reactant and reagent?
What is the difference between reactants and reagents? I've seen both used for very similar situations... The dictionary said that reagents are added to a reaction to identify another substance, but wouldn't that be an indicator?

There's no difference imo

SciFinder differentiates between them somehow

see also reactant and reagent in the IUPAC Compendium of Chemical Terminology (Gold Book)


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Why are aromatic compounds so stable?
I've never actually been told deeper reasons for this. In organic chemistry classes (1 and 2) we're usually told that aromatic compounds are very stable whereas anti-aromatic ones are not, and also, we're taught how to identify them. Why are they, though?
(I'm pretty sure this has something to do with the de-localisation of electron pairs, which enhances stability, but I only have a vague idea. Is that the only reason?)

see this earlier answer for a simple explanation as to why delocalization is energy lowering.

Let me start by saying that aromatic compounds are not exceedingly stable, they are just much more stable than antiaromatic compounds. However, they can still be hydrogenated exothermically to the corresponding cycloalkanes.
$$\ce{C6H6 + 3 H2 -> C6H12}$$
But why are they more stable than antiaromatic ones? Let’s take a look at the orbital scheme for cyclopentadienyl $\ce{C5H5^{?}}$ (the scheme is the same for radical, anion or cation — at least theoretically). (Taken from this site).

The pictures of the molecular orbitals correspond to their energies. The e means that the two orbitals next to each other are degenerate, i.e. have the same energy. Let’s fill 6 $\unicode[Times]{x3c0}$-electrons ($\ce{C5H5-}$) into this scheme. We get three completely filled orbitals and two rather antibonding unoccupied orbitals. Looking good.
Now let’s extract a single electron to create the $\ce{C5H5^{.}}$ radical. The electron must come from one of the two HOMOs, so we end up with one degenerate orbital completely occupied, the other one only occupied by a single electron (usually drawn with ↑ or $\unicode[Times]{x3b1}$-spin). Now we’ll abstract another electron to arrive at the $4\unicode[Times]{x3c0}$ $\ce{C5H5+}$ cation. Two of the remaining four electrons can be filled into the lower-lying $\ce{a_1}$ orbital, the other two need to be in $\ce{e1}$ and since the latter is degenerate we arrive at a triplet state: Both electrons have $\unicode[Times]{x3b1}$-spin. This is rather unstable.
The triplet state and hence the instability can be avoided by distorting the molecule. If elements of symmetry ($\ce{C5H5^?}$: most importantly the plane of symmetry) are lost, the equivalence of the orbitals here labelled $\ce{e_1}$ is also lost and there is no reason not to pair the electrons into a single orbital. That’s why cyclobutadiene is not a square but a rectangle.
If I remember correctly, the results of simple approximations are actually mathematically equivalent to taking a circular $\unicode[Times]{x3c0}$-system, putting it on one corner and drawing an orbital energy on every carbon — the cyclic Hückel approximation. And since you will always get pairs of degenerate orbitals by this method using perfect shapes, that gives us the Hückel rule of $4n+2$ — two for the lowest orbital (the corner the circle is resting on) and $4n$ for every pair of orbitals that then have the same energy.

Bond order is directly proportional to stability of a compound. In aromatic compounds bond order is maximum because all the electrons goes in bonding molecular orbitals and all electrons are paired . Thus according to the formula of bond order
bond order = 1/2(number of electrons in bonding molecular orbitals - number of electrons in anti bonding molecular orbitals )
Bond order of aromatic compounds is maximum.


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Why are trifluoromethyl diazirines so stable?
Why are trifluoromethyl diazirines so stable?
Trifluoromethyl diazirines stability is well documented, however I am yet to find an explanation of why...
Can it be homoaromatic or even aromatic?
2 pie e-, in a planar system? Even with the sp3 carbon, couldn't that make it homoaromatic?
Or is it just aided by the trifluoromethyl group withdrawing electrons from the strained ring?
Bonus points of you know the mechanism of Schmitz Diaziridine/diazirine synthesis.
Any resources or insight is greatly appreciated.
Thanks,
James

related http://www.jbc.org/content/255/8/3313.full.pdf

Thanks, yeah I've seen that.
Seems to show the conditions it can survive, but not explanation of why or mechs (unless I missed it).


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Determining the equation of state when given thermodynamic relations

Empirical observations of a thermodynamic system have shown that:
$$\begin{align}C_{\mathrm m,V} &\neq C_{\mathrm m,V}(T) \\[6pt]
\left(\frac{\partial U}{\partial V_\mathrm m}\right)_{T} &= f(V_\mathrm m) \\[6pt]
\left(\frac{\partial U}{\partial p}\right)_{V} &= g(V_\mathrm m) \end{align}$$
where $f$ and $g$ are functions of the molar volume, $V_\mathrm m$. Determine the state equation.

How can I solve this problem?


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Lithium-ion batteries
I understand what occurs on a regular Li-ion battery - the positive Lithium ion moves from one electrode to the other of the battery. But what is exactly the physical mechanism for this intercalation to occur?
Is it that the Lithium in the LiCoO2 electrode lowers the chemical potential between electrodes? I am really curious about the process - any reference would also be helpful!

Yes, lithium ion batteries works with the principle of intercalation and deintercalation of ions from cathode to anode and anode to cathode respectively. To be specific for intercalation, lithium (in cathode) having high oxidation potential oxidize, loosing an electron which goes to the current collector to anode. So lithium ion get positive charge. Since the battery source you use also has positive terminal hence repulsion between ion and positive terminal occurs. Thus positive ion of lithium goes towards anode, which has layered structure thus result in intercalation. Another way of seeing it is because of difference in the concentration the diffusion of ions occur or you can also consider the fact charge neutrality.

Can you be more specific which step you don't understand?
In principle, the main cause for the discharge of batteries is a electrolyte (in the considered case this is some Li salt, usually dissolved in organic solvent), which provides a conductive path for the cations to propagate. A very simple animation and explanation of this process you can see for example here, but I suppose you are already familiar with this.

I understand your point. But specifically speaking, there must be a physical mechanism undergoing on the Li-ion batteries, where the lithium intercalation between the anode and the electrode enable the flows of electrons. What I fail to understand for this specific case is the physical explanation to that.

@EdwardMclaurin There are different electrochemical potentials on the two electrodes. Difference in intensive properties always result in the movement of extensive properties: https://en.wikipedia.org/wiki/Intensive_and_extensive_properties . In the current case the extensive property is charge (ions, electrons) and amount of substance.


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How is the mol highway used in this problem?
How many moles of propane gas, $\ce{C3H8}$, are contained in 11 grams of the gas at standard conditions? Also, can you please explain step-by-step (and include sigfigs?)
I think I might know how to do this, but I'm not sure if this process is correct.
I found online that $\ce{C3H8}$ has a mass of 44.11 grams per mol. So do I just use a proportion to find the number of moles this 11 grams of propane has?

You are on the right track. Every element has a molar mass that is defined under standard conditions, which is in this case the natural abundance of an element taken into account. Pressure, volume and temperature will not change the mass of a compound. Since every element has a specific molar mass, every compound does too. The molar mass is a measurement of how much mass does one mole of a substance have.
The molar mass is usually referred to with $M$, while the mass is referred to as $m$. The amount of substance is $n$. This gives you the following relationship: $$M=\frac{m}{n}$$
Since you have given $m(\ce{C3H8})=11~\mathrm{g}$ and you already looked up $M(\ce{C3H8})=44.1~\mathrm{g\,mol^{-1}}$, you can use this formula to determine $n(\ce{C3H8})$.
In this question it is quite hard to explain the use of significant figures. Those are used to imply a certain inaccuracy. Not enough information is given by the question, as of how accurate the measurement is. It is a mere exercise of converting one property into another. Here you should not worry about it.


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Lemon Battery works on water?
My son and I were using a mini science experiment kit for creating batteries with copper, zinc, and lemonade, similar to the famous lemon battery project.  My son's hypothesis was that the liquid had to taste like lemonade to make a battery, and so he hypothesized (I thought correctly) that lime juice would work, but water and milk would not.
However, when we conducted our tests, the battery successfully ran a digital clock with all three liquids.  I washed the cups and copper and zinc sheets thoroughly between each test, but the non-acidic liquids continued to function as batteries.
Did I do something wrong, or is the liquid not required to be acidic to make a battery?

Milk is slightly acidic with pH of about 6.5

You can use a potato as well - which I don't think is acidic

You might try cooking oil or distilled (not filtered but distilled) water as liquids which should not support the battery.  It takes a remarkably small current to run a digital clock, so even very poor electrolyte solutions can do the job.

For reference, there are a few scholarly sources about lemon batteries. This one is nice and to the point.

Sadly, the lemon battery is a bit of a cheat: it's not the lemon that's providing the energy at all. The very first battery ever was very similar to the lemon cell, but used damp cloth, or cardboard soaked in salty water (might be worth giving it a go, I've never tried).
The energy really comes from the metal of the two electrodes which have had lots of energy piled into them (heat etc) when they were refined. When you wonder where the energy comes from, think of the guys in the refinery and their furnaces, some of that is "in" the metal, and you're getting it back.
The purpose of the solution (or briney felt discs) is to give one part of this reaction (the electrons) a good run for their money, going through the electrical circuit, as they can't travel through the electrolyte, while the other part, the ions, can cross the other way, through the solution. They meet up at the other end, and the reaction is complete and the zinc slowly corrodes away.
So what you're testing is the quality of electrolytes. Salts and acids in solution make good electrolytes: salty water, lemons, potatoes, coke. Milk acts as an electrolyte because it contains some ions from salts (which nutritionists helpfully/confusingly call electrolytes).
Pure water is pretty bad as an electrolyte. There's one of two options. First, it's not absolutely useless as one, and maybe the clock requires so little power that water is good enough. The other option is that tap water is hard, or otherwise has salts (minerals etc) in it which increase the ion concentration greatly. Probably a combination of those two factors. It might be worth trying distilled water (if you have some for the car, for example) which is probably much poorer.
More about the "first battery" is here: it's lots of these metal-electrolyte-other-metal cells end-to-end. The zinc is slowly being corroded by the copper the necessary combination of a. ions (which are big, heavy old things that amble through the electrolyte) and b. poor, zippy, lightweight electrons which can't, and have to go "the long way round" through the electrical circuit. https://en.wikipedia.org/wiki/Voltaic_pile

(Cells and batteries are an active area of research with a lot of money behind them. Parts of this answer are simplified to apply to traditional and educational cells and batteries, and scoot over special cases, which is still a good starting point to learn about more exciting varieties).

What is necessary is a solution with a sufficient conductivity (not necessarily an acid - could be any molecule able in solution, to ionize or dissociate into charged species). I don't know the conductivity of milk but I'd expect it to be very low due to the high content of fat. Anyways, I'd inspect what kind of preservatives or additives they add to the milk that could explain what you observed. :)

Milk usually has not more than ca 3.5% fat - which means that 96.5% are l water. And there are plenty of salts and proteins in milk, which are charged and can conduct electricity. Which are all natural and no additives ;)

In the incredibly unlikely circumstances that this interests someone: the conductivity of milk is a good indicator of mastitis, or related problems, as the low conductivity of milk increases with blood contamination (blood is salty).


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Electrode/solution interface : What's the difference between the electrochemical Peltier heat and the heat caused from the overpotential?
In principle heat effects caused by the passage of an electric current might arise from: electronic peltier effect at the junction of electrode/solution level, Joule effect due to solution resistance, electrochemical Petier effect at the interface level and the thermal effect due overpotential. What is the difference between the last two ones?


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Find the displacement of mercury in a vertical tube
So we have a 1m tube sealed at both ends. It is initially horizontally placed and the middle 0.1m contains mercury and the two ends contain air. Now the tube is turned vertical, we have to find the amount of mercury displaced.
Let the pressure in the upper end be $P_1$ and the pressure in the bottom end be $P_2$. The area of cross-section is $A$ and the lengths in the upper and bottom end are $(0.45-x)$ and $(0.45+x)$,respectively.Let $x$ be the mercury displacement.
By Boyle's Law
$$P_1V_1=P_2V_2$$
$$P_1A(0.45-x)=P_2A(0.45+x)$$
A single equation is insufficient to solve this, what should I do.

You know the value of $P_2-P_1$ as well.

I assume the Hg is a competent barrier between the parcels of air and the parcels both have an initial pressure $P_0$ of 1 bar (760mm of Hg). Each air parcel has an identical initial energy content of $$ P_0V_0 = P_0A(0.45\mathrm{m}) $$
With hydrostatic equilibrium, $\Delta P = P_{\mathrm{bot}} - P_{\mathrm{top}}$ would all be due to the weight of the Hg, IOW, in mm of Hg terms, would be 100mm (0.1m) of Hg. -- This is your second equation. $$\Delta P = P_{\mathrm{bot}} - P_{\mathrm{top}} = 100\, \mathrm{\scriptstyle mmHg} $$From Boyle's Law we know that $$ P_{\mathrm{top}} = P_0 \frac{V_0}{V_{\mathrm{top}}} \quad\quad\text{and}\quad\quad P_{\mathrm{bot}} = P_0 \frac{V_0}{V_{\mathrm{bot}}} $$ Substituting, $$ \Delta P = P_0(\frac{V_0}{V_{\mathrm{bot}}} - \frac{V_0}{V_{\mathrm{top}}}) $$ or $$ (\Delta P) V_{\mathrm{top}} V_{\mathrm{bot}} = P_0V_0(V_{\mathrm{top}} - V_{\mathrm{bot}}) $$ Substituting again, $$ (\Delta P) A^2 ( 0.45\mathrm{m} - x )( 0.45\mathrm{m} + x ) = P_0A^2(0.45\mathrm{m}) ( (0.45\mathrm{m} - x) - (0.45\mathrm{m} + x) ) $$ eliminating $A^2$ and simplifying $$ \Delta P ( (0.45\mathrm{m})^2 - x^2) = -2P_0 x(0.45\mathrm{m}) $$ or in quadratic form $$ x^2 - 2\frac{P_0}{\Delta P}(0.45\mathrm{m})x - (0.45\mathrm{m})^2 = 0 $$ which would be $$ x^2 - (6.84\mathrm{m})x - 0.2025\mathrm{m}^2 = 0 $$ You can take it from there.


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Why iron reacts differently with concentrated and dilute sulfuric acid?
When $\ce{H2SO4}$ is dilute,

$ \ce{Fe +H2SO4 -> FeSO4 + H2} $

and when $\ce{H2SO4}$ is concentrated,

$\ce{3Fe +8H2SO4 -> FeSO4 + Fe2(SO4)3 + 4SO2 +8H2O}$

What made the reaction with dilute acid different from the reaction with concentrated acid?
How did the second equation form, and why there are 2 salts - one with Fe[II] and the other with Fe[III]?

Well, the second equation has the formation of $\ce{SO2}$. How do you think this came about?

related http://chemistry.stackexchange.com/questions/15612/what-will-happen-when-a-metal-reacts-with-an-acid-which-is-neither-dilute-nor-co  http://chemistry.stackexchange.com/questions/19897/why-concentrated-sulfuric-acid-for-dehydration-and-dilute-for-hydration

@orthocresol i don't know

Also http://chemistry.stackexchange.com/questions/28001/why-do-exam-questions-specify-dilute-acid

@bigworld12 You should add your thoughts about that, do you understand what happens in first equation?

@Mithoron i fully understand the first equation as i have been studying it for a long time , the (Fe) is more powerful than the (H) so it takes its place to provide a salt (FeSO4)

@bigworld12 Okay.. think about the oxidation states of S in H2SO4 and SO2. What has happened to sulfur?

@orthocresol it got reduced?

reduced from +7 to +4

@bigworld12 Yeah. Actually it's +6 to +4 but close enough. I'll type up an answer. Hold on.

In the first equation, $\ce{Fe}$ is oxidised to $\ce{Fe^2+}$ by $\ce{H+}$; the $\ce{SO4^2-}$ ions play no role in the reaction. You could replace $\ce{H2SO4}$ with $\ce{HCl}$ and get the same reaction.
In the second equation, the same oxidation of $\ce{Fe}$ to $\ce{Fe^2+}$ by $\ce{H+}$ also occurs. However, on top of that, some of the $\ce{SO4^2-}$ ions are reduced to $\ce{SO2}$, and some of the $\ce{Fe^2+}$ is further oxidised to $\ce{Fe^3+}$. Clearly, in the dilute acid, $\ce{SO4^2-}$ was not strong enough to act as an oxidising agent - but in the concentrated acid it is.

The reason for this can be somewhat explained by Le Chatelier's principle. Consider the redox reaction:
$$\ce{2Fe^2+ + SO4^2- + 4H+ -> 2Fe^3+ + SO2 + 2H2O}$$
In the dilute acid, this equilibrium lies far to the left because the concentrations of $\ce{SO4^2-}$ and $\ce{H+}$ are rather low. However, making the acid more concentrated shifts the equilibrium to the right so that you get some of the products.
As to why you get a mixture of iron(II) and iron(III), the simple answer is that the equilibrium above doesn't lie entirely to the right. $\ce{H2SO4}$ is not that strong an oxidising agent to begin with.

More quantitatively, the redox reaction above can be investigated using standard reduction potentials and the Nernst equation. The following data is taken from Wikipedia (you will get pretty similar values from more "reliable" sources):

$E^\circ(\ce{Fe^3+}/\ce{Fe^2+}) = +0.77 \mathrm{~V}$
$E^\circ(\ce{SO4^2-}/\ce{SO2}) = +0.17 \mathrm{~V}$

Under standard conditions, $E^\circ$ for the full reaction is therefore $0.17-0.77 = -0.60 \mathrm{~V}$; the negative value indicates that the forward reaction is non-spontaneous.
The Nernst equation for this reaction can be written:
$$E = E^\circ - \frac{RT}{nF}\ln Q = E^\circ - \frac{RT}{nF}\ln\left[\frac{(a_{\ce{Fe^3+}})^2(a_{\ce{SO2}})(a_{\ce{H2O}})^2}{(a_{\ce{Fe^2+}})^2(a_{\ce{SO4^2-}})(a_{\ce{H+}})^4}\right]$$
I will make several assumptions here:

Activity coefficients are taken to be unity, meaning that the activities can be replaced by concentrations.
The activities of all the products, as well as $\ce{Fe^2+}$, are unity.
The only source of $\ce{H+}$ is from the dissociation of the $\ce{H2SO4}$, and the dissociation is complete. That means that if $c$ is the formal concentration of $\ce{H2SO4}$, we have $[\ce{H+}] = 2c$ and $[\ce{SO4^2-}] = c$.

Substituting in the values of $E^\circ = -0.60 \mathrm{~V}$, $R = 8.314 \mathrm{~J~K^{-1}~mol^{-1}}$, $T = 298 \mathrm{~K}$, $n = 2$, and $F = 96485 \mathrm{~C~mol^{-1}}$, this allows us to simplify the Nernst equation:
$$E / \mathrm{V} = -0.60 + 0.0128\ln (16c^5)$$
Strictly speaking, I should use $c/c^\circ$ instead of $c$ in order to make the logarithm dimensionless, but I'm fed up of technicalities. Now for the actual calculation. Let's assume that you're using $10\mathrm{~M}$ sulfuric acid, such that $c = 10$ (or $c/c^\circ = 10$, if you want to insist). This gives a value of $E = -0.42 \mathrm{~V}$ - as expected, it is less negative than before, which corresponds to a larger value of $K = \exp(nFE/RT)$. The fact that it is still rather negative just means that there is a difference between the real world and my theoretical treatment. Presumably, those assumptions above are not entirely valid.

Just want to be accurate in concepts:
Concentrated sulfuric acid is at least $18~\mathrm{M}$ and contains only 2% water by mass. In this situation, most sulfuric acid molecules do not ionize due to the lack of water. The oxidizing power of concentrated $\ce{H2SO4(l)}$ does not come from $\ce{H+}$ ions or $\ce{SO4^2-}$ ions, but rather the $\ce{H2SO4}$ molecules.

Oh, so that's what one other question was dedicated to. It'd be great to have a reference confirming that H2SO4 molecules are the key players in concentrated sulfuric acid.

Iron (also steel) is passive in concentrated H2SO4, forming a layer of sulfate on the surface.

And in dilute? The questions is on the difference between them.


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Causing a metal or glass marble to float in an aqueous solution
I am trying to find a way that you could cause metal (Aluminum) or glass to float in a mixture of water and some water-soluble material. I am aware that salt can be used to heighten the density of water, but I doubt it could go to that extent because both aluminum and glass are more dense than salt (possibly flawed logic, I have limited knowledge of the subject). So basically: if there is any way to cause Aluminum or glass to float in water, possibly through adding a water soluble material to the water, please let me know.
Edit: Basically what I'm doing is trying to think of creative ways to sort materials by density, for example a set of marbles. With the example of solutions with salt, you would vary the amount of salt in the mixture to alter what materials would float.

You might be able to find something clear and nontoxic that appears to be water that is dense enough to float glass or aluminum, but an easier option might be trickery, via a hollow metal ball.  Not sure whether this is an option for your application.

Try to make the appropriate surface area of metal or glass . So that they flow rather than this because it is not economic feasible

The best current non-toxic alternative to Clerici Solution appears to be Sodium Polytungstate (SPT).  It also has a neutral (~7) pH, which is important when working with aluminum. By dilution/evaporation, density covers the range $1.1\!\!-\!\!3.1$g/cm$^2$, but above $2.5$g/cm$^2$ it can develop a harmless reversible blue tint when exposed to reduced metals. Aluminum has an oxide patina, so I am not sure if it qualifies as a reduced metal in this setting. SPT is rather expensive though.
Aluminum is $2.7$ g/cm$^2$ and Common glass is $2.52$ g/cm$^2$. So both of those materials, many gemstones and some minerals, will float in SPT.

@JohnDowns cesium tungstate is NOT sold at  www.formatebrine.com. They sell Cesium Formate, a completely different animal. Instead of editing my answer, please create your own independent answer.

Yeah, that was clearly a spam edit and should have been rejected.

@Jan, I would have still tried to incorporate the edit if it had been about cesium [poly-]tungstate, but it wasn't even that.

That would have been way too kind to a spammy edit suggestion ;)

Is there some reason not to use magnesium or beryllium ?

One of the requirements for the liquid base of oilfield drilling fluids is high density. Besides preventing the chips from falling out of suspension, the increased hydrostatic pressure helps prevent the sides of the well from collapsing before a steel casing can be applied.
Oilfield drilling fluids are based on alkali metal formate solutions. Sodium, potassium and even cesium formate solutions are used on a multi-ton scale. Reasons for their success in this application are the high solubility of formates and the low toxicity of the solutions. Sodium and potassium formate are certainly dirt cheap; I am unaware of the price of cesium formate.
According to the relevant document dowloadable at http://www.cabotcorp.com/solutions/products-plus/cesium-formate-brines/formate-technical-manual cesium formate brines of up to 2.50g/cm3 density can be prepared at 15.6C. I would assume solubility increases at higher temperature.
That may not be enough to float regular glass, but should be enough to float borosilicate glass, which according to http://www.udel.edu/chem/GlassShop/PhysicalProperties.htm has a density of 2.23 g/cm3. Similarly, aluminium will not float in this solution but boron will.
Looking at the other answers, I note that Clerici solution uses a combination of formate and malonate, presumably to improve solubility by having a mixture of ions. This strategy should in principle work with cesium. Probably the ultimate non-toxic solution would be a cesium polytungstate solution.
One issue you will find is that materials of similar density tend to have similar refractive index. Transparent glass in a high density transparent solution may be almost invisible. Also, be careful with aluminium in contact with alkaline solutions. If the pH is high enough, they will dissolve the oxide layer, allowing the aluminium underneath to corrode extremely rapidly with evolution of hydrogen.
Edit: Thanks to John Downs for the following info: Cesium Tungstate solution with density of SG 2.80 is available from various sources.  This is interesting though it less than the density of the sodium polytungstate solution mentioned in Aabaakawad's answer.

Well, if a toy boat made out of aluminum or glass is not what you had in mind, let us consider the method of increasing water density with additives. Surely, ordinary table salt will not get you this far, but there are other salts. Clerici solution comes to mind; it is probably not that easy to come by, but certainly dense enough. And yes, it was invented for this very purpose: to separate materials by density.

Make sure you don't drink it - thallium is highly toxic.

Alright, is there maybe some other way of floating metal on water? possibly applying some force through the water, like air, though would the weight be affecting this rather than density? anyway, do you have any ideas for alternatives to the other solution? (forgive the pun)

From a practical point of view: try methylcellulose. It forms a transparent gel with water. From intuition I'd say you can get a saturated enough solution at room temperature to get glass or aluminium marbles to float (warning, I haven't tried it myself or calculated whether it's possible). Alternatively, use another gelling agent with food applications, such as xanthan gum, or even cook a starch pudding (preferably tapioca starch, as other common types such as cornstarch are not as transparent). You can change the concentration to get the density you want.
Unlike lab reagents, food gelling agents are easy to obtain (some more than others) and nontoxic to the point of being safe to ingest. The downside is that the result is visibly viscous. So if you are trying to do construct a magic trick ("look, this glass marble floats in pure water") it won't help.

Is the "floating" due to density effects or viscosity effects? This Dow chemical publication lists the specific gravity of a 10% solution at 1.0245 (table 6), which is well below the expected density of glass/aluminum. While marbles might float "kinetically" on a high-viscosity solution, their equilibrium state might not be - meaning such a solution would be ill suited for separating items of different densities.

@R.M. I really can't say. I'm just speaking from experience using this kind of gels around the home, but I have never floated something as dense as aluminium in them. If somebody here can confirm or disprove my suggestion either by calculation or empirical observation, I'd be very happy to hear it.


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Effectiveness of Grignard reagent
So for my experiment, I am synthesizing grignard's reagent (phenylmagnesium bromide). I am trying to change the dryness of ether and see if there is a change in products. Is there a simple method to quantify the results? I need to figure out the effectiveness of the product. If water is introduced, the reaction will probably form alkanes and magnesium hydroxide rather than phenylmagnesium bromide. It has to be a simple method since I am working in a school lab.

For a grignard reaction to work, the ether must be free from water. This is usually achieved via distillation over sodium/potassium using benzophenone as an indicator. As you are in a school lab, the solvent provided to you is likely already dried/dry enough for purpose.
If the solvent is a little wet, two things will happen:
Firstly, it is more difficult to get the grignard formation to 'initiate' as the water effectively quenches the reagent as it forms. However, with persistence it is usually possible to get them to initiate when making simple grignards such as PhMgBr.
The other issue however, is that as the water is quenching any grignard reagent formed until all the water is used up, you will make less reagent than you think you are (in this sense the reaction is self-drying) meaning if you do think your ether is wet, you should aim to make a greater quantity of reagent than required.
It is possible to titrate a grignard reagent to identify the actual concentration of solution made, although in an undergraduate practical it is likely that you will use the reagent in excess anyway.
In terms of identifying your product, you should carry out the reaction, work it up, purify, and use a technique such as NMR to see if your product has formed. If this is unavailable, simple IR spectroscopy can help you determine if the functional groups in your starting materials have changed (e.g. Carbonyl to alcohol).
Good luck with your reactions


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How do electrolytes work?
I just want to know how to electrons travel from one end to the other.
The concept I have in my mind is that the ions functions as "buses" for the electrons then send them to the other side, but doesn't that make them neutral molecules? (in which case they wouldn't be attracted to the + side.)

In compact matter, electrons "trickle" from a source to one atom, forcing off electrons which move to the next atom, ionizing each in turn. The same holds true in the opposite direction, where electrons are removed and the corresponding "hole", or positive ion, is propagated towards the negative electrode. Individual electrons and ions move slowly at the drift velocity, but electromotive force propagates much faster. This is similar to the motion of the particles in the toy, "Newton's Cradle".
In an ionized gas, electrons and ions may move much faster, actually crossing the full volume of the plasma (though electrons move faster, due to their lesser mass for the same charge).


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Why does equilibrium favour weak acid or weak base?
For example, in the reaction $\ce{NaOH +HCl->NaCl +H_2O}$, the products are favoured by a factor of approximately $10^{24}$.
Is this a general rule? Is it because the products are more stable than the reactants?

Here is a simple intuitive explanation. In your above example, there is actually 2 reactions occurring, the forward and the reverse reaction. Let us consider the forward reaction. Here, both $\ce{NaOH}$ and $\ce{HCl}$ are strong bases and acids. What this means is that HCl has a strong ability to donate a proton and NaOH has a strong ability to accept protons. This means that when NaOH and HCl react together, the reaction will basically go to completion. So you will end up with up with very little NaOH  and HCl and the equilibrium will favour the products.
Now lets look at the reverse reaction which is: $$\ce{Na+ + Cl- + H2O -> NaOH + HCl}$$ Here $\ce{Cl-}$ is the conjugate base of HCl which is  a strong acid. This means that $\ce{Cl-}$ is a really weak base, meaning that it is pretty bad at pulling protons from molecules. Also water is the conjugate acid of NaOH, which is a strong base. This means that water is a really weak acid, meaning that it is also pretty bad at kicking out one of its protons. This means that in this reaction, barely any of the products will react and you will mainly end up with NaCl and water.
As you can see in both reaction the equilibrium will favour the weak acid and base as they aren't able to react that well together. This effect can also be explained by using equilibrium constants but I won't go into that.


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Why does the $V_{\mathrm{max}}$ of an enzyme not change with competitive inhibition? Shouldn't it decrease since there are fewer active sites?
My textbook states that $V_{\mathrm{max}}$ doesn't change for competitive inhibition. This makes no sense to me, since decreasing the amount of active sites available should lower the rate of an enzyme catalyzed reaction.
My attempt at rationalizing this is that $V_{\mathrm{max}}$ is dependent on the rate at which the enzyme/substrate complex dissociates into free enzyme + product by the equation Kcat * [E]. So, since it only takes into account the enzyme/substrate complex, we basically don't consider any enzymes with inhibited active sites.
Is that reasonable?

You can think of $V_{max}$ as describing the nature of the active site - it's deftness at converting substrates to the product, the rest is just probability. Competitive inhibitor does not change properties of the active site - they just hang there for some amount of time until the E-I complex dissociates hence they don't affect the maximum theoretical conversion rate of that enzyme.
Competive inhibitors only decrease the chance of inhibitor binding to the enzyme. Thus you can always raise the concetration of your substrate to the state that probability (now the other way around) of inhibitor binding the enzyme will become negligible with regard to the substrate allowing it to work at his maximum rate.


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IUPAC nomenclature
What is the IUPAC name of the following compound?

The preferred IUPAC name is (5  E)-N-(prop-2-en-1-yl)oct-5-enamide. Please add details to isolate your problem and to narrow the answer set.

(E)-N-allyloct-5-enamide
Start by looking for the longest continuous carbon chain. That's 8 carbons giving you the oct-.
On that chain, theres an alkene at the 5 position, and an amide at the end, giving the 5-enamide. You might also write this as 5-en-1-amide, though it is implied.
The geometry of the alkene as being (E) always goes to the front of the string, as would any R or S declarations. In this case, (E) is unambiguous, but in cases where multiple points of geometry exist, you would define geometry or stereochemistry along with position (5R),(3E) etc.
So, thus far you have E-oct-5-enamide. You now need to describe the other group bound to nitrogen on the amide, this is an allyl group, providing you with the full name (E)-N-allyloct-5-enamide.