text:
Find the initial volume of the weak acid given a titration graph
Question:

Below is a picture of a titration between a weak acid, $\ce{HA}$, and a 0.150 M $\ce{NaOH}$ solution. Find the initial volume of the weak acid in milliliters in this titration.

My work so far:
From the graph, I think that the two most important points are (0,3.0) and (20.00,8.0) - I don't know the reason, but it looks important.
At (0,3.0), there is no $\ce{NaOH}$. This equilibrium takes place:
$$ \ce{HA_{(aq)} + H2O_{(l)} <=> A_{(aq)}- + H3O+_{(aq)} } $$
The constant $K_\text{a}$ can be found from:
$$
 K_\text{a} = \frac{[ \ce{H_3O^+}][\ce{A^-}]}{[\ce{HA}]}
$$
with $[\ce{H_3O^+}] = 10^{-3}$ since $\text{pH}=-\log_{10}[\ce{H_3O^+}]=3$
This is a dead end for me.
At (20.00, 8.0), $[\ce{H_3O^+}] =10^{-8}$, this number has decreased because some $\ce{H_3O+}$ have reacted with the newly dissolved $\ce{OH^-}$, although how much I can't tell. I have reached another dead end
Other facts I do know:

0.150 M of $\ce{NaOH}$ means for every 1 liter of the solution, there will be exactly 0.150 moles of $\ce{NaOH}$.
$\ce{NaOH}$ dissolves completely as a strong base. So, for every 1 mole of $\ce{NaOH}$ there will be 1 mole of $\ce{OH^-}$

Outline
There are three main points of interest in the titration graphic, and two you have found yourself. The three points are

The start at $V_\text{titr}=0$
Halfway to the equivalence point at $V_\text{titr}=10~\mathrm{mL}$
The equivalence point at $V_\text{titr} = 20~\mathrm{mL}$

Why is number 2 important? Because that is where we can read out the $\text{p}K_\mathrm{a}$ value of the acid, because at that point $[\ce{HA}] = [\ce{A-}]$.
Now, armed with this information, here is how to proceed:

Find out total amount of acid
Find out total concentration of acid
Calculate the initial volume from the data gathered in steps 1 and 2

1  Total amount of acid
The total amount of acid can be gleaned from information point 3, the equivalence point.
$$ n_\text{eq} = V_\text{titr,eq} \cdot c_\text{titr} = 20~\mathrm{mL} \cdot 0.150~\mathrm{mol\,L^{-1}} = 0.003~\mathrm{mol} $$
2  Total concentration of acid
The total concentration can be calculated from information points 1 and 2. We know the initial concentrations (denoted by subscript $_0$) because of the pH value at the beginning:
$$ [\ce{A-}]_0 = [\ce{H+}]_0 = 10^{-3}$$
Using the equilibrium constant that we got at point two, we can calculate the initial concentration of the associated (un-dissociated) acid species:
$$ [\ce{HA}]_0 = \frac{[\ce{H+}][\ce{A-}]}{K_\mathrm{a}} = \frac{10^{-6}}{10^{-5}} = 10^{-1} $$
The total initial concentration of all acid is then trivial:
$$ [\ce{A}]_\text{tot,0} = [\ce{HA}]_0 + [\ce{A-}]_0 = 0.101 $$
Note that we here actually calculated the activities and not concentrations. Since we assume an activity coefficient of $\gamma = 1$ because we only consider dissolved species, the concentrations are simply the activity multiplied with the standard concentration:
$$ c(\ce{A})_\text{tot,0} = \gamma \cdot [\ce{A}]_\text{tot,0} \cdot 1~\mathrm{mol\,L^{-1}} = 0.101~\mathrm{mol\,L^{-1}} $$
3  Calculation of the initial volume
The initial volume is now calculated following the simple concentration formula:
$$ c_0 = \frac{n_0}{V_0} \Leftrightarrow V_0 = \frac{n_0}{c_0}$$
Since the amount of acid is constant ($n = n_0\; \mathrm{const}$), we can easily fill in the values:
$$ V_0 = \frac{0.003~\mathrm{mol}}{0.101~\mathrm{mol\, L^{-1}}} \approx 0.0297~\mathrm{L} = 29.7~\mathrm{mL} $$

Extremely comprehensive answer, thanks. One question: what is the activities and how is it different from concentrations ?

Activity is, for the purposes here, just a fancy word for concentration. For more in-depth information, check out this link.

Here is a different approach:
From the titration curve we estimate the following:

The solution of the acid has about the pH = 3 before the titration is initiated.
The half titration point is at about pH = 5, i.e. equal to the pKa of the acid.
At the equivalence point, 20.00 ml of 0.150 M $\ce{NaOH}$ has been added.

We look at the protolysis reaction of the acid:
$$\ce{HA + H2O = A- + H3O+}$$
Before equilibrium:

$\mathrm{[\ce{HA}] = x}$
$\mathrm{[\ce{A-}] = 0}$
$\mathrm{[\ce{H3O+}] = 0}$

After equilibrium:

$\mathrm{[\ce{HA}] = x - 10^{-3}}$
$\mathrm{[\ce{A-}] = 10^{-3}}$
$\mathrm{[\ce{H3O+}] = 10^{-3}}$
$\mathrm{Ka = [\ce{A-}]\times[\ce{H3O+}]/[\ce{HA}] = 10^{-5}}$

Solving the equation for $x$ gives:
$\mathrm{x = \pu{0.101 M} = [\ce{HA}]}$
To convert all $\ce{HA}$ to its corresponding base $\ce{A-}$ required:
$\mathrm{20.00\times0.150~mmol~of~\ce{NaOH}}$.
At the equivalence point $[\ce{A-}] = 20.00\cdot0.150/(20.00 + x)$, where $x$ is the volume of the acid being titrated. At the equivalence point:
$0.101 = 20.00\times0.150/(20.00 + x) \implies x = 9.7 \approx 10 ml$
Answer: The original volume of the acidic solution was 10 ml.


text:
How to recognize Barium sulphite?
Yes it's a white ppt. which is soluble in dil. $\ce{HCl}$, but I'm not sure which of the following happens:
$$\ce{BaSO3 +$dil.$HCl->Ba(HSO)3 +Cl-}\tag1$$
$$\ce{BaSO3 +2H+->Ba^2+ +SO2 ^ +H2O}\tag2$$
Questions:

Which of the above actually happens? I have these two from different sources.
Does it becomes $\ce{BaSO4}$ on standing? Is it afterwards soluble in dil. $\ce{HCl}$, does this change require an oxidising agent, if yes then which one is more suitable - bromine water or $\ce{HNO3}$?

@hey I suppose you don't know that SO3- = sulphite not sulphate!

@hey Thanks for doing some cleanup on other questions.  If you're not sure about something, like with this question, you can leave a comment for the author to get clarification.

There is no compound like $\ce{Ba(HSO)3}$.

Sulphite$(\ce{SO3^2-})$ and Barium Choride$(\ce{ BaCl2})$ or Strontium Chloride$(\ce{ SrCl2})$:
$$\ce{SO3^2- +Ba^2+ ->BaSO3 v}$$
the precipate disoolves in dilute hydrochloric acid, when sulphur dioxide evolves:
$$\ce{BaSO3 v +2H+ ->Ba^2+ +SO2 ^ +H2O}$$
On standing, the precipitate is slowly oxidises to the sulphate and is then insoluble in dilute mineral acids; this change is rapidly effected by warming with bromine water or a little concentrated nitric acid or with hydrogen peroxide:
$$\ce{2BaSO3 v +O2 ->2BaSO4 v }\\
\ce{2BaSO3 v +Br2 + H2O->BaSO4 v +2Br- +2H+}\\
\ce{2BaSO3 v +2HNO3 ->3BaSO4 v +2NO ^ +H2O}\\
\ce{2BaSO3 v +H2O2 ->BaSO4 v +H2O}$$
The solubilites at $\pu{18 ^\circ C}$ of the sulphites of $\ce{Ca,Sr,Ba}$ are respectively $1.25, 0.033, 0.022$ all in $\pu{g/l}$
Source:

Vogel’s Qualitative Inorganic Analysis for the JEE, by Svehla/Sivasankar,  7/e Paperback, 1 January 2013

Please cite a source for the text for this one (there seems to be multiple sources also using this text).

@jonsca done, i think it's not satisfactory for the term source? maybe you'll agree.

I think you've done your due diligence.  It appears as though your instructors are using material without citing it, which isn't something that I can address.  It's always better to put things into your own words, though, which is something that you can address.  Ideally, we should be citing all sources, even those that we draw material from without quoting, but this can be difficult at times.  If you are going to post something from class notes and you have a spare moment, pop it into Google to double check whether it needs a citation.  The one post that is from Wikipedia

should be cited accordingly, even if your instructor did not do so.  Thanks for taking care of these.

@jonsca i mean he might have gathered from various sources and formed for our's(his students') benefit, gave notes for the course. anyways i'll try to find it's source

According to what I've found in russian segment of internet:
$\rm BaSO_3 + {dil.} HCl\rightarrow Ba(HSO)_3 +Cl^-\tag{1}$
$\rm BaSO_3 +2H^+\rightarrow Ba^{2+} +SO_2+H_2O\tag{2}$
$\rm 2BaSO_3+O_2 \xrightarrow{\Delta} 2BaSO_4\tag{3}$
$\rm BaSO_3+H_2O_2 \rightarrow BaSO_4 + H_2O\tag{4}$
I believe that it is stable at room temperature on standing.
Reaction $\pu{(1)}$ and $\pu{(2)}$ are both possible depending on concentration from my point of view.

There is no compound like $\ce{Ba(HSO)3}$. Also, cite your source.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Where can I find a complete and accurate table of CPK colours?
I have been using several chemical drawing softwares including MarvinSketch, Jmol, Accelrys DS Visualizer, Avogadro, etc., all of which I have set to CPK colouring. Despite this supposedly common setting I have noticed that fluorine atoms are coloured rather varyingly from software to software. For example, MarvinSketch uses orange, Accelrys uses cyan as does Avogadro and Jmol uses algae green for fluorine atoms. I have tried Googling for the answer but I have only found incomplete tables, aside from on the Wikipedia website but I am dubious as to using Wikipedia for this information due to its reputation for inaccuracy.

Indeed, this is annoying. Because of this problem, several years ago, one of the Jmol developers and I sat down and worked out a color scheme. It was similar to Accelrys, although we tried to make certain "known colors" match (e.g., rust-like for Fe, golden for Au, etc.)
Certainly, there's been an effort to keep color consistency between Jmol and Avogadro
These colors are open source and available, e.g., through Open Babel
If there are differences between Jmol and Avogadro, I'm sure both the Jmol developers and the Avogadro developers (including myself) would be happy to reconcile them.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Diatomic halogens - bond strength
Why is the $\ce{F-F}$ bond relatively weak? And why does bond strength decrease from $\ce{Cl-Cl}$ to $\ce{I-I}$?

Does this have anything to do with "lone-pair" repulsion on the rather small $\ce{F-F}$ molecule? I haven't asked my professor but I'm suspecting this is going to the first thing on his mind. I'm sure that many of you here will have rebuttals to this argument.

On the other hand does $\ce{F2}$ even have lone pairs? There are no non-bonding orbitals populated on the molecular orbital diagram of fluorine. I wouldn't expect there to be any non-bonding orbitals populated either, as this is a simple diatomic we're dealing with and the atoms involved are identical ... so each atom's orbital should be able to interact with the other atom's orbital effectively.

My initial guess would be that given the high electronegativity of fluorine, the electron density will have an appreciable restriction of motion that will hinder the overlap of electron density in the singly-occupied orbitals used in MO theory. This subdued overlap leads to a weaker bond with respect to the other halogens.

What about the lone pair argument? Yay or nay? @Lordstryker

As for fluorine: antibonding unbond a bit more strong than bonding bond and $\pi$-interaction are the strongest for the second row. As for bromine and iodine: orbitals become increasingly diffuse down a column, so any covalent bonding becomes weaker down a column.

See also http://chemistry.stackexchange.com/questions/16276/what-is-charge-shift-bonding/16278#16278 for a discussion of "charge shift bonding" in $\ce{F2}$ which seems to be a more accurate representation than either VB (lone pairs) or MO in this case.

@Dissenter I would imagine that given the decreased 'diffuseness' of electrons around Fluorine (again F is highly electronegative so it holds on 'tightly' to these electrons more than other elements), and at lower energy orbitals, this would lead to larger electron-electron repulsions (since you are effectively restricting the space electrons can be found in).

I see as I write the answer, that many comments are spot-on.
Whether or not $\ce{F2}$ has lone pairs depends on your view of molecular orbitals and valence bond (or Lewis) descriptions of bonding.
Covalent radii for fluorine compounds are usually around 0.64 Å for F, but yet the F-F bond distance is 1.43 Å.
If we go for the Lewis or valence bond view, we'd argue that the large number of lone pairs on a small atom will be highly repulsive.
If I go for the molecular orbital view, I'd point out that in reality, the anti bonding orbitals $\pi^*$ and $\sigma^*$ are higher in energy than their bonding counterparts, particularly in the 2nd row, so the F-F molecule will be weaker than a simple single bond.
There's a really nice article by Peter Politzer in JACS that I like to use when discussing this "anomaly": "Anomalous properties of fluorine" J. Am. Chem. Soc., 1969, 91 (23), pp 6235–6237

It appears rather that the anomaly involves the fluorine atom, in whatever circumstances, and its interaction with an external electron which enters its outer shell. [...] In any case, it is suggested that because of the exceptionally small size of the fluorine atom (see Table I), the entering electron feels unusually large forces of attraction from the nucleus and repulsion from the electrons already associated with the atom. It appears that the magnitude of the anomaly in the repulsive forces is greater than in the attractive forces.

In other words, regardless of the MO or VB view, the issue is one of electrostatic repulsion.

There's also a nice article I just found on The Antibonding Effect.

Aren't anti bonding orbitals always higher in energy than their corresponding bonding orbitals ?

Yes. But in your figure, the bonding orbitals have the same amount of stabilization as the amount of destabilization in the anti-bonding orbitals. In fact, the $\Delta E$ between the atomic orbital levels and the anti bonding MOs is much greater. I guess I'd say the anti-bonding orbitals are less stable than you'd expect simply from the bonding stabilization.

Yes, but isn't that universally true as well?

Yes. But this depends on the magnitude of the overlap of the orbitals (large in the 2nd row). Since $\ce{F2}$ has a lot of occupied anti bonding orbitals, there's a large amount of destabilization. That is, from an MO picture, the "bond order" might be 1, but it's a weaker single bond than say $\ce{B2}$ because of this destabilization.

Hutchinson so may the lone pairs in the Lewis picture be considered anti-bonding electrons?

Let us continue this discussion in chat.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What are t2g and eg in CFT?
In the crystal field theory (CFT), when the splitting of the d-orbital occurs, it gets divided into two parts.
The upper part with higher energy is the  $\mathrm{e_g}$ and the lower part with lower energy is called the $\mathrm{t_{2g}}$ as in:

So can anyone please explain me what is this $\mathrm{t_2}$ or the $\mathrm{e}$ or the $\mathrm{_g}$ below.

See also http://chemistry.stackexchange.com/questions/8072/splitting-of-d-orbitals-when-ligands-approach-central-metal-ion?rq=1

See also: http://chemistry.stackexchange.com/q/16305

As stated in both the links Geoff and Philipp have kindly commented (1, 2) they are to do with symmetry labels we chemists like to assign to orbitals. Knowing an orbitals symmetry class can lead to a lot of simplifications down the road when you use quantum mechanical calculations and even dictate the reactivity of which orbitals are "allowed" to interact together in reactions.
In this case the $\mathrm{t_2}$ groups three of the metal atom's d-orbitals into a certain class while two of the orbitals belong to the $\mathrm{e}$ class. The $\mathrm{t}$ means triply degenerate while the $\mathrm{e}$ means doubly degenerate (degenerate means have the same energy).
The $\mathrm{g}$ is not about how many energy levels are degenerate rather it is an indication of the answer to a certain operation we can perform on an orbital. It instead relates to how the orbitals behave if we hypothetically were to put a line bisecting the orbitals or "invert" them around a centre point (crudely put "going from one corner to the other". If the phase of the wavefunction (or perhaps more pragmatic) the "colour" of the orbital lobes changes while doing this we label the orbital with a $\mathrm{u}$ if we have no phase change we label it with $\mathrm{g}$. They indicate the orbitals response to an inversion about the centre of symmetry.
We can do lots more of these operations to investigate the symmetry of the orbitals which together lead us to assigning the molecule to a symmetry group. Knowing this is very important for a plethora of reasons.

g and u come from the German words gerade and ungerade, meaning even and uneven, respectively.


text:
Relationship between Equivalence Ratio and Adiabatic Flame Temperature for Fuel Rich Combustion
I'm reading a combustion text and would like help interpreting the following passage (diagram attached). It is regarding the relationship between equivalence ratio and adiabatic flame temperature:

For equivalence ratios between  $\phi$ = 1 and $\phi$ ($T_{max}$), the heat capacity decrease more rapidly with $\phi$ than $\Delta H$, while beyond $\phi$ ($T_{max}$), $\Delta H$ falls more rapidly than does the heat capacity. The decrease in heat capacity is dominated by the decrease in the number of product moles formed per mole of fuel burned, with the decrease in the mean specific heat being less significant.

I’m assuming that we are keeping the total mass of reactants constant (just changing the proportion between them). I get that total heat generated goes down since less fuel is reacting. But I don’t understand why heat capacity (I think this is extensive) decreases. Octane (at least at 300K
) has a higher specific heat than $\ce{CO2}$ and $\ce{H2O}$, so doesn’t having more fuel vapor increase the heat capacity?
For the last sentence, I don’t understand why a decrease in heat capacity is more significant than the decrease in mean specific heat?
The phrase “decrease in the number of product moles formed per mole of fuel burned,” does “mole of fuel burned” include fuel that technically didn’t react, since we have a fuel rich mixture?


text:
Standard list of odors?
Is there such a thing as a list of "standard" odors?  In everyday context, there are various descriptions to odors such as "fruity" and "plastic-like".  But are there more accurate and normalized names of odors?

It's a long time since I worked in that field, but there are standard odours, and perfumery has a vocabulary to describe odours.  Common English is fairly weak in that department.  According to this paper consensus amongst professionals is high, although there are pathological cases (coriander/cilantro).

The problem with odors or taste is that it is a form of Qualia. Whether cyanide smells like bitter almonds or is odorless is subjective. Similarly, whether Phenylthiocarbamide taste extremely bitter or completely tasteless is also subjective (depending on genetics).
Even in everyday life, coffee/wine/spices taste different to different people. Even for fruits, the Durian fruit can taste and smell fragrant to certain people but utterly disgusting to others. This is probably why the perfume industry is making billions - if everyone agree that a certain smell is "nice", everyone would be using the same perfume. (Do Genes determine our perfume preferences?)
So I believe it is difficult to have a standard list of odors. Even in the event that the list is created, it will not be universally agreed upon. There are bound to be disputes. Colours on the other hand, (which is also a form of Qualia), are less subjective (everyone can agree that chlorine is green and bromine is brown).

This argument can be applied to colors as well. How an electromagnetic wave of certain wavelengths is perceived as certain name color is dependent on individual's genetics. But there are still standardized named colors, isn't it? It's a matter of definition and standardization.

Now that I think about this old question, after reading this recent article about "Turing Tests and the Problem of Artificial Olfaction", https://www.technologyreview.com/s/601220/turing-tests-and-the-problem-of-artificial-olfaction/, I think the fact there has not been a list of standard named odors is mostly because odor is harder to measure, as compared to colors. Traditionally, accurate measurement and quantitative description of odors does not exist (?). Giving names and standardizing the definitions can't happen if there is no measurements.


text:
Proof of kinetic energy equation
Is there a way to prove that $\frac{1}{2} mv^2 $$= \frac{3}{2}RT$
The first one is kinetic energy and the second is the average kinetic energy.
Can this be proved by component of velocity.

The proof is here: http://en.wikipedia.org/wiki/Kinetic_theory#Properties

https://physics.stackexchange.com/questions/45838/how-to-deduce-e-3-2kt
http://hyperphysics.phy-astr.gsu.edu/HBASE/kinetic/molke.html#c1
http://en.wikipedia.org/wiki/Kinetic_theory#Properties
These contain proof at various level . Choose according to your mathematical level

Would you be willing to summarize the proof in case the links break? Additionally, none of these links do a good job of explaining the steps.


text:
Is dehydrating copper(II) sulphate pentahydrate a chemical or physical change?
Is removing the waters of hydration from $\ce{CuSO4 * 5H2O}$ a chemical or physical change? Sorry for the trivial question. I get varying answers across the net.

Does the crystal melt, or does it just change color?

I heated it in the lab just until it turned white

The traditional answer is that it's a chemical change when bond-making and bond-breaking is involved.  The material certainly changes appearance, and heating copper sulphate somehow releases bound water.
That said, bonding in metal-aquo complexes runs the gamut from very weak to exceedingly strong - exchange rates span 15 orders of magnitude (Fig. 1 in Jan Reedijk, Metal-Ligand Exchange Kinetics in Platinum and Ruthenium Complexes. Platinum Metals Rev. 2008, 52, 2–11. DOI: 10.1595/147106708X255987), and where do you draw the line between a "physical" electrostatic interaction (in, say, $\ce{Na(OH_2)_6^+}$) and a chemical bond (as seen in $\ce{Cr(OH_2)_6^3+}$?

You heated copper(II) sulfate until it turned white.
This is a quite simple chemical reaction:
\begin{align}
\ce{\underset{(blue)}{CuSO4.5H2O} &->[][100 ^\circ C] CuSO4.H2O + 4 H2O}\\
\ce{CuSO4.H2O &->[][> 200 ^\circ C] \underset{(white)}{CuSO4}}
\end{align}

According to one text,[1] a change in color, release of a gas (and some other changes), are indications of a chemical reaction. Since water vapor is released and the solid changes color, this should be considered as a chemical reaction, not a physical change.
[1] Wikibooks, General Chemistry/Properties of Matter/Changes in Matter

This misses the point of the question. Changes in colour and release of gases are indications that a reaction may have taken place. However, passing an electric current through a grey tungsten filament makes it glow yellow-white but there's no reaction. Heating water to 100C makes it release a gas (steam) but there's no reaction. The asker has observed signs that a reaction may have taken place and is asking if one actually did. What is the chemical change you think the asker observed?

@DavidRicherby There is a difference between the tungsten filament and the compound above in that the above compound has a different color at room temperature after heating. Also, there does not appear to be any water with the starting material, yet it is liberated when heated. As I answered the question as it was written, it is a chemical reaction. The OP did not ask what the reaction was.

A good answer would explain both why there is a reaction and what the reaction is. If you're going to be minimalist, why not just answer "yes"?

By heating copper(II) sulfate pentahydrate until it was white and contained no more water, you undergo a chemical change. The chemical makeup of $\ce{CuSO4.5H2O}$ changed to $\ce{CuSO4}$. The change in color in this situation also indicates a chemical change, but a change in color doesn't always entail a chemical change.

The crystal structure shows that the copper is six coordinated and has four oxygen atoms from water molecules in a plane and the other two octahedral positions are each occupied by one oxygen from an $\ce{SO4}$ group.
The fifth water molecule has its oxygen bridging two $\ce{CuSO4}$ molecules, see the sketch, and is bonded to four oxygens, two from water and two to sulphate oxygens, presumably by hydrogen bonds. The H atoms scatter x ray so weakly that these are not seen.
As these bonds from this oxygen will be weaker than those of other oxygens bonded to the cation they should be broken first on heating so this should be considered to be a chemical reaction. Breaking the other bonds $\ce{Cu-OH2}$ bonds will follow on further heating to form the anhydrous solid.

Structure of $\ce{CuSO4-5H2O}$ determined by x-ray diffraction of crystals.  The sulphate oxygens are shaded, the water oxygens as large open circles and the fifth water oxygen with a bold outline.  The Cu atoms are at the centre of the octahedral structure and drawn with a dot in the centre of the circle.(After Beevers & Lipson, Proc. Roy. Soc. A 1934, v146, 570, image from A. Wells 'Structural Inorganic Chemistry', publ Oxford)

This is not entirely correct to use crystal structure to predict the path for dehydration. According to DTA, copper(II) sulfate looses 4 water molecules first, not 1 as one would think looking at the crystal structure. I touched this issue in one of my answers to related question.

You might well be correct, but energetically to lose four strongly bound waters before losing one other one does not seem to be what one would expect. Only ultra-fast time resolved methods could determine an answer, for example time resolved x-ray diffraction  after fast laser heating of a small crystal.

No, it is not a chemical change! Copper sulfate ($\ce{CuSO4.5H2O}$) is a hygroscopic material. That means it readily co-ordinates with water. Hydrous $\ce{CuSO4}$ is not a bond. Therefore, when you heat copper sulfate crystals, the water evaporates leaving behind anhydrous copper sulfate which is white in colour. There are no bonds breaking or forming (the definition of a chemical change) and so it is a physical change. Moreover, there is no new substance being formed and so it is a physical change.

The definition of a chemical reaction: We call changes that change bonding form of atoms, which results changes in the structure and properties of matter, is called a chemical reaction.
So this reaction isn't a chemical reaction. Copper (II) Sulphite is still Copper (II) Sulphite and water was'nt in bond with Salt it was trapped in crystal system of salt
P.S: SO3 is Sulphite and SO4 is Sulphate

Unfortunately, this is wrong. First, how do you think water is "trapped" within crystal lattice if there were no chemical bonds? Second, crystal structure of copper(II) sulfate pentahydrate consists of [Cu(SO4)2(H2O)4] octahedra with 1 water molecule dwelling in the lattice per copper, and the first dehydration step actually takes four water molecules out, not just one, suggesting that even the first coordination sphere is disturbed.


text:
Conversion atom to another
One child has claimed to have find a solution to all physical problems. On asking for details, he said that all periodic elements has common components, i.e. electrons, protons, neutrons.
The child has suggested a solution: convert atom to another by adding electron. This way one can get substance like $\ce{H2O, Au, He}$ in abundance.
How can it be done?

While what you suggest may sound nice on paper, it has some serious problems.

Getting the elements. Let's say that we're talking about purifying water to remove toxic elements such as Hg or Cd. Extracting the elements out of the water is a feat by itself, for example using reverse osmosis methods. This is a method used for seawater desalination - to turn them from salt water filled with all kinds of elements into drinkable water. This process is very expensive, and a very polluting one as well. Desalination just the amount of water you need for drinking water is complicated, so desalinating an entire reservoir is simply not going to happen.
Let's say you did somehow manage to extract the element in question. Now you need nuclear reactions to transmute one element to another. Not all are possible. For some elements, like Tc or Am this is the only way you can produce them. However, you are going to end up with radioactive nuclear waste.

To sum it up, even if it was possible, you would need so much power and to do it and you will produce some much pollution that it's simply not worth it. Just going and mining the gold will be orders of magnitude cheaper (and probably cleaner) than producing it using nuclear reactions.
If you do manage to somehow extract the the polluting elements, you usually do something else with them (aka recycling) and you do not attempt being an alchemist. Another example is soils contaminated with lead. The solution is to just dig it up, put it somewhere where it is not hazardous to anyone and replace it with clean soil.

I think you are looking for nuclear transmutation which occurs via nuclear reactions or radioactive decay, as explained in the referenced wikipedia page. As you can imagine then, implementation has some important impediments.

Also see synthesis of precious metals.  It would be so nice if nuclear power could bring down the price of ruthenium and rhodium, but there are technical difficulties.

Someone has done what the child suggested. Sorta.
First, some background.
What you're talking about is called transmutation, and for thousands of years it drove nearly all of humanity's research in chemistry (which was called alchemy at the time).
The alchemists' dream of turning "base metals" like lead into gold was handily crushed when modern chemistry was being developed in the victorian era, and it was found that chemical elements were just that: elements. Indivisible.
As knowledge grew and electrons, protons, and neutrons were discovered people still thought that transmutation was simply too difficult to actually pull off. Sure, nuclei could change on their own - people knew about radioactive decay - but it was thought that this process could not be affected from the outside. Intentional transmutation was still largely considered impossible. Then someone did it.
In 1934 Irène Joliot-Curie (Daughter of Pierre and Marie Curie) transmuted one element into another for the first time. Shortly afterwords the Manhattan project got started and transmutation became indispensable.
Transmutation in Your Life
Elements made by transmutation are used in hundreds of ways in modern life, the smoke detector above your head uses 241Am made in a nuclear reactor, the dye used in a PET scan has 18F as an active ingredient, and 239Pu and 233U nuclear fuels are made from 238U and 232Th respectively. Some gunsights even use tritium lighting. These are all examples of bulk materials made by transmutation.
Alright, Here's Where That "Sorta" Comes In
Transmutation has one inherent problem that renders that 'turning harmful stuff into harmless stuff' plan moot.
Transmutation is extremely messy.
These nuclear reactions are less like chess and more like the game risk. Sure, one outcome might be more likely, but not by much - and plenty of other outcomes will also occur. It's impossible, on a fundamental level, to narrow the outcome of one of these reactions. Additionally, you can't make it happen for all of the atoms at once; some atoms will react before the others and if you wait too long the ones that reacted first start doing more reactions in the background.
Remember that 241Am in your smoke detector? Well in order to make it you take 239Pu (which you need a different reaction to make in the first place but let's ignore that for now) and throw neutrons at it. Eventually, it'll catch one and either fission (split) or become 240Pu. If that 240Pu then absorbs another neutron before it happens to decay, you get 241Pu.
All of this happens spread out over some time, with some atoms going from 239Pu all the way to 241Pu in a fraction of a second by just happening to catch two neutrons in a row without fissioning and others will just happen not to have caught any neutrons at all. Some will even go farther, and one of those 241Pu atoms could absorb another neutron and either fission or become 242Pu.
And after all that, you still don't have 241Am! How do you get it? You wait. Over time, 241Pu slowly decays into 241Am. 241Pu has a half-life of 14 years, so after only a few  years a sizable amount of the 241Pu will have turned into 241Am. They're still mixed though, so the final step is to chemically separate them.
Congratulations! Finally, from your starting 239Pu you've made a little 241Am! Also some 240Pu, 241Pu, 242Pu, various decay products, and a mess of fission-fragments!
So while nuclear transmutation is used to make plenty of substances that are indispensable to modern western civilization, it can not be used to change a sample of one element completely into another with no side products.


text:
Is there a substance that binds with the sodium in salt (NaCl) in solution so that the sodium can be removed (precipitated out, for example)?
Is there a substance that binds with the sodium in salt $\ce{NaCl}$ in solution so that the sodium can be removed (precipitated out, for example)?
Could such a substance make salt water potable?

I think your question would benefit from a bit more specificity. In particular, what do you mean by "bind?" Do you want molecules that will chelate the ions while remaining soluble, or do you want something that will precipitate the ions out of solution?

The recent edit makes the question clear. Can something be added to salt water to precipitate out the sodium, leaving a filtrate that could then easily be made drinkable?

$\ce{NaCl}$ itself does not exist as a compound in water. The dissolution of salt in water means that both the sodium cation and the chloride anion are separated from each other and are each surrounded by water molecules. Therefore, no compound can bind to NaCl in water because $\ce{NaCl}$ does not exist in water.
It is, however, possible that a compound "binds" with either sodium or chlorine. For example, by adding copper to the solution and performing a bit of more electrochemical magic (out of the scope of this answer), you can get the solid compound copper chloride ($\ce{CuCl}$).

All compounds of sodium are soluble in water, except $\ce{NaMg(UO2)2(CH3COO)9·6H2O}$. So sodium ions can be recognized in solution by preparing a solution containing magnesium acetate and uranyl acetate. Adding a solution containing some $\ce(Na^+} ions produces a crystalline yellow precipitate. The trouble is that uranium is radioactive, and cannot be used any more in the lab. The best method to determine sodium in a solution is to pulverize the solution in the flame, and analyze the flame spectrum. A yellow line (or double line) shows that the solution contains sodium

Ah, @Maurice, you left out cryolite! ;-)

@Maurice sodium silicates are extremely insoluble in water. NaAlSi2O6, or NaAl3Si3O8 are two examples.

Yes, cryolithe and silicates are insoluble into water. But they cannot be used for detecting sodium in water. Because no aqueous solution exist that can be added to a sodium solution so as to show that this solution contains sodium, and makes a precipitate of silicate or cryolithe.

@Maurice ever heard of TEOS? https://en.wikipedia.org/wiki/Tetraethyl_orthosilicate

@Gimelist. Tetraethyl orthosilicate is decomposed by water, producing some insoluble deposit of silica. So it is useless for analyzing sodium (or any other ion) in water solutions.


text:
What do (R) and (Z) mean in chemical names?
So, I'm confused on the notation my professor uses in a certain question; could someone explain.
He says:
Draw the structure: (R)-(Z)-chloro-4,6,-dimethyl-3-octene
I get everything after '(z)'
I'm just unsure what he means by '(R)-(Z)'...
Any input is appreciated. Thanks.

For R read here and for Z here.

The proper name should, however, be (3 Z , 6 R )-1-chloro-4,6-dimethyloct-3-ene.

@Martin Is it necessary here. I mean it has only one place for Z abd T and only one chiral centre

@DSinghvi I was merely stating the proper IUPAC recommended name. If they make more or less sense is up to everyone else to consider.

The name (R,Z)-chloro-4,6-dimethyloct-3-ene is ambiguous. This could be (R,Z)-1-chloro-4,6-dimethyloct-3-ene, (R,Z)-3-chloro-4,6-dimethyloct-3-ene, (R,Z)-6-chloro-4,6-dimethyloct-3-ene, or (R,Z)-8-chloro-4,6-dimethyloct-3-ene. (2-chloro-4,6-dimethyloct-3-ene, 5-chloro-4,6-dimethyloct-3-ene, and 7-chloro-4,6-dimethyloct-3-ene have one additional stereocenter.) For example:


text:
Proof that Equilibrium Constant is a Ratio of Rate Constants
I have a question about this proof that shows the relationship between the rate constant and the equilibrium constant. In this proof, the assumption is that the reaction is $(a+b)$th order in the forward direction and $(c+d)$th order in the reverse direction.
I had always thought reaction orders are empirically determined (and not necessarily whole numbers); can one simply add the stoichiometric coefficients like this? If not, how does one show that equilibrium constant is a ratio of the rate constants since the equilibrium constant requires the concentrations raised to the order of the species' stoichiometric coefficient?

For a reaction step, the assumption that the rate is proportional to the product of the concentration of reactants raised to their respective powers is valid.
However, generally, reactions are made up of many reaction steps (e.g. SN1 reaction). For such reactions, you cannot use the assumption above. You have to understand the reaction mechanisms in order to determine the rate equation.
(For reactions that involve decimal powers, usually the steady state approximation is applied, and some mathematics is done along with empirical findings in order to determine the rate.)
Addendum: Yes. A reaction step means an elementary reaction. From Wikipedia: "An elementary reaction is a chemical reaction in which one or more of the chemical species react directly to form products in a single reaction step and with a single transition state.". Therefore, there is only one reaction step.
Likewise, for the equilibrium constant,$K$, in your ratio derivation $K_c = k_f/k_r$, the chemical species must also be from that particular reaction step, and not the global reaction step. It makes sense because the equilibrium constant, K, is related to the free energy change, $\Delta{G}$ of a reaction step. In a two step reaction (e.g. SN1), there are two $\Delta{G}$ and therefore two equilibrium constants.
However, in practice, people do use the overall equilibrium constant, $\beta{}$, which is the product of the equilibrium constants at each reaction step. $\beta{} = K_1K_2K_3...$ (E.g.If the first step gives you 70% yield, and the second step gives you 70% yield, the global step would be 49% yield).(http://en.wikipedia.org/wiki/Chemical_equilibrium#Multiple_equilibria)
In short, for your rate constant,$k$ and your equilibrium constant $K$, in $K_c = k_f/k_r$, they must be of that of a single reaction step.

Is a reaction step an elementary reaction? For an elementary reaction aA+bB->cC+dD, Kc = kf/kr = ([C]^c[D]^d)/([A]^a[B]^b), but for a global reaction qQ+pP->tT, Kc = [T]^x/([Q]^y[P]^z) where x,y,z are empirically determined and no longer equal to the stoichiometric coefficients (for H2O formation this would be Kc = [H2O]^x/([H2]^y[O2]^z))? So the fact that Kc = kf/kr is always true, but Kc is not necessarily a ratio of the concentrations raised to their stoichiometric coefficients?

I am confused now. I looked at the derivation for free energy here and it doesn't assume the reaction as elementary or global. This answer and an example in my combustion book for CO2 dissociation (which I do not believe is an elementary reaction) also uses stoichiometric coefficients for Kp. But in another chapter the book states that Rate = k[A]^x[B]^y where x and y are empirically determined.

@Yandle, I've edited my answer to explain further.

So when we derive K from Gibbs free energy (like in the Wikipedia link), the assumption is that the reaction is an elementary reaction? My confusion: For global reaction (A+B->C+D), Rate = kf[A]^x[B]^y where x and y are not the stoichiometric coefficients and hence K=kf/kr would be ratios of the concentrations raised to empirically determined values x,y etc.. However, if I go B=K1K2... I still end up with concentrations raised to the stoichiometric coefficients of the global reaction as if the global reaction was elementary, since intermediate terms cancel.

Yes your assumption that it is an elementary reaction is correct. 2. Yes for global reaction (A+B->C+D), you would not be able to use K = kf/kr. 3. While you can multiply the eqm constants to obtain the global eqm constant, you cannot multiply the rate constants to  obtain the global rate constant (in fact there is no such thing since the overall rate is dependant on many rate constants, for example overall rate = k1/(k2-k1) [A]^3/2 [B]^1/2)). Therefore, you cannot use $\beta{}=kf_{overall}/kr_{overall}$ - it would not be mathematically equivalent. K = kf/kr must be of a reaction step.

If we can simply go β = K1K2 etc., then in the derivation for K from Gibbs free energy, does it even matter whether the reaction is elementary or global, since the intermediate terms would cancel? How come global K (or β) isn't defined as ratio of overall rate constants and is there another relationship between β and kf/kr (global)? I always thought equilibrium constant is found by rearranging kf[A]^w[B]^x=kr[C]^y[D]^z where w,x,y,z may or may not be the stoichiometric coefficients.

Yes you are right, -RTlnβ = -RT(lnK2-lnK1) = ΔG2-ΔG1 = $ΔG_{overall}$. 2. Nope, there is no relationship between β and kf/kr simply because there an overall rate constant doesn not exist.

So only the K=kf/kr derivation assumes elementary reactions? 2. In my combustion text, they do define a global rate constant $k_g$. For F+aOx->bPr it is $d[F]/dt=-k_g[F]^m[Ox]^n$ where the equation is (m+n)th order and m and n are from curve fitting experimental data..

Yes. Remember kf/kr must be of an elementary reaction. 2. The global rate constants is most likely made up of elementary rate constants. usually to avoid writing something complicated like k1k2'/k2-k1, we assign it to $k_g$

So m and n in my original equation would also be something complex for the same reason? How should I interpret the relationship (hand-wavy wise) between global equilibrium coefficient and global rate constant? I used to interpret in the same way as an elementary reaction (kf/kr) which is incorrect.

No, there is no such thing as global rate constant, so you are right in saying that it is incorrect to interpret it the same way as elementary rate constants. $\beta{}≠k_{f,overall}/k_{r,overall}$

And yes, your m and n will probably be fractions - since that particular rate equation is not an elementary equation.

What I meant to ask is, although $β≠k_{f,overall}/k_{r,overall}$ mathematically, can I still interpret $β$ to qualitatively reflect the ratio of the global rate constants (i.e. large $β$ means $k_{f,overall}>k_{r,overall}$), but not using any quantitative definitions between $β$ and global rate constants. In addition, is $β$ also basically the product of kf/kr of each elementary reaction?

Interesting question. I'm not sure whether there is any direct proportionality relationship between $\beta{}$ and $k_g$, (since $k_g$ is different for different situations) but I would be careful before assuming that there is. And yes, $\beta{} = K_1K_2..$ so you are right in saying it is the product of kf/kr of each elementary reaction, although the kr of the first reaction does not necessarily equal to the kf of the second reaction.

So a large $β$ doesn't necessarily mean that the global forward rate constant ($k_{fg}$) is higher than the global reverse ($k_{rg}$), or did I mis-interpret as this implies the equilibrium favours a direction opposite to that implied by the global rate constants? Physically, why does ($k_{fg}/k_{rg}$) deviate from $β$ where for elementary reactions K = rate constant ratio? In other words, what physical effects cause the global rate constant ratio to not simply be a product of all the elementary reaction rate constant ratio.

No, there is no such thing as $k_{fg}$ or $k_{rg}$, take a look at http://www.science.uwaterloo.ca/~cchieh/cact/c123/steadyst.html Example 1

Sorry in my previous question I was using global rate constants $k_{fg}$ and $k_{rg}$ in the context of my combustion book where Foward Rate = $k_fg[A]^m[B]^n$; in the link posted $k_{fg} = k = k_fk_2/(k_b+2k_2)$.

The Kfg in your book will probably comprises of many elementary rate constants such as those posted in the link - they merely simplify it to Kfg so that it looks neater - but it doesn't mean anything since it's a pseudo rate constant.

I think I understand what you are saying. Basically $k_{fg}$ and $k_{rg}$ don't exist because they are dependent on other variables (elementary rate constants)? If I still want to use the two as simplifying representations, can we still qualitatively say, the β we have is very large, hence $k_{fg}$ is likely bigger $k_{rg}$ (not attaching numbers to anything) and forward reaction is favoured?

I'm not sure about that. If the units of $k$ are unknown - you may be taking the ratio of two dimensionally unequal constants(e.g. one may have the units $s^{-1}$, while the other $mol^{2/3}s^{-1/3}$ (just an example)). Qualitatively, I'm also not sure whether there is a definite relationship between $\beta$ and $kfg/krg$. It may be tempting to believe that if $kfg$ is higher than $krg$ then  the reaction may proceed to the right and $\beta$ would be higher, but mathematically I'm not sure how to prove that, without knowing $kfg$ or$ krg$

Okay, so although β value will imply whether reactants/products are favoured, it does not imply anything about the global rate constant itself? In the link you provided, the theoretical derivation showed [N2O5] was raised to the power of 1 in the global rate equation. I know empirical curve fit powers won't be integers, but is it possible for theoretical derivations result in fractional values?

1.Yes. 2. Yes, if you look at the other example in the link I provided, Br has the power of 3/2.

@Yandle Please ask new questions as these issues come up. You can link your subsequent questions to this one if they are follow-ups.  It doesn't help anyone in the future if you discuss them beyond asking for a clarification on what was posted.

@jonsca My apologies I will do this in the future. The only question I had that was not related to my original question was the last one regarding the relationship between between β and global rate constants. Should I post this question?


text:
Empirical formula of Berthollides
Can the empirical formula be defined for berthollides? Is it equal to the molecular formula, or it can't be defined?

"Berthollide" seems to be a word for "non-stoichiometric compound" that has fallen out of use.
You have to look at the structure, as a rule things happen for a reason.  The formula can be derived from the structure.
Consider the series of alums, which is  KAl(SO4)2·12H2O, or more generic:  X+Y3+(SO4)2·12H2O.  The generic structure can accomodate a monovalent cation, provided it's sufficiently large (lithium is out of the question, sodium is an edge case), and a tervalent cation (Al, Cr, Fe, pretty much anything), and the proportion can be freely chosen within the constraints of the structure.

What about compounds with non-integer indexes of elements that build it?

Consider the wuestite structure (Fe(1-x)O).  The ideal structure (FeO) is a sodium chloride structure, but in wuestite phases some octahedral gaps (where you'd ideally find a Fe(2+) ion go unoccupied.

Can an empirical formula be defined in that case?


text:
Resin for making Jewelry
I want to make resin jewelry, but I'm having problems!
I want to create that effect, but I do not know what materials to use

The result is a mix of resin and other such things


text:
How to temper a non-ferrous metal
Can non-ferrous metals/alloys (copper, brass, bronze) be tempered (made harder) by temperature control alone, like ferrous metals? I understand it's simple to anneal copper, brass, etc., just by making them hot enough.
I'm Wondering if it's possible to temper non-ferrous metals without having to either work-harden it or media-blast it (essentially work hardening).

Most non elemental metals can be heat treated, due to the construct of the phase diagram.

Your use of the word "temper" is incorrect for metallurgy in English.  Temper normally means to reheat and partially soften a steel that has been hardened by quenching.  There are a few very unusual steels ( high speed) that may harden during tempering. For nonferrous metals you could be describing "age hardening".

Most non-ferrous alloys are heated in order to form a solution.
Techniques:
Annealing:
Most non-ferrous alloys that are heat-treatable are also annealed to relieve the hardness of cold working. These may be slowly cooled to allow full precipitation of the constituents and produce a refined microstructure.
Different annealing techniques are subjected with non-ferrous alloys like  recrystallization annealing, partial annealing, full annealing, and final annealing.

Quenching
Quenching is a process of cooling a metal at a rapid rate. This is most often done to produce a martensite transformation. In ferrous alloys, this will often produce a harder metal, while non-ferrous alloys will usually become softer than normal.

Reference: Wikipedia

I know you have been doing a lot better (and I appreciate it!), but it's still best to blockquote (using the >  character) quotes that are pulled directly from the text.

@jonsca I think this answer looks better in this way because all the text is from 3 to 4 different paragraph from wikipedia. But yes from next answer i will use blockquote. Even if you think this answer should be in blockquote i can edit that.

Yeah, I can see that most of it is in your own words, but the "Quenching" section is not.  I would do a lead in like "As far as quenching (rapid cooling) goes," (or something like that) and then blockquote the rest.  Anything you use verbatim like that should be quoted, honestly.  I'm not trying to pick on you here (it probably seems like it, but not my intention), but we need to give quotes within sources the proper credit, even if it is Wikipedia.

Thanks, but that does not seem to answer my question: is it possible to harden non-ferrous metals by temperature change alone

You have a misconception that tempering is hardening. It is softening done after hardening to relieve stresses, improve ductility and toughness .
From this source

In nonferrous alloys, eutectoid transformations, which play such a prominent role in steels, are seldom encountered.
On the other hand, the principles associated with chemical homogenization of cast structures are applicable to many alloys in both classes. Examination of the heat treatment used for nonferrous alloys reveals that a wide variety of processes are employed. . Annealing after cold working is a very important heat treatment for nonferrous alloys.

Basically if the non-ferrous alloy has a eutectoid region in its phase diagram , tempering can be used after hardening. But non-ferrous usually are not hardened in this way. Age hardening, Precipitation hardening, Dispersion strengthening are commonly seen in non-ferrous alloys.

No.
Neither ferrous, nor non-ferrous metals can be hardened by tempering.
Tempering generally refers to steel and it does the opposite of hardening. It reduces the hardness and brittleness of the steel (softens it) while improving the toughness and making it more ductile.  This makes the steel better able to handle further cold processing and makes it more impact resistant (which is why e.g. swords need tempering - otherwise they are too hard and brittle to withstand impact).
Annealing is used on non-ferrous metals, generally  for similar reasons (although also used to improve conductivity in some cases).

No.
Neither ferrous, nor non-ferrous metals can be hardened by tempering.
Tempering generally refers to steel and it does the opposite of hardening. It reduces the hardness and brittleness of the steel (softens it) while improving the toughness and making it more ductile.  This makes the steel better able to handle further cold processing and makes it more impact resistant (which is why e.g. swords need tempering - otherwise they are too hard and brittle to withstand impact).
Annealing is used on non-ferrous metals, generally  for similar reasons (although also used to improve conductivity in some cases).
Qualifier: not too sure of this, but I have a vague recollection that some of the popular "american" (as opposed to english) dictionaries actually do define tempering as hardening metal ...


text:
Sodium Bisulfate vs Sodium Bisulfite
I'm wanting to do a follow up to this Q&A on the cooking stack regarding browning avocados. I specifically want to expand my experiment to include the options in Wayfaring Stranger's Answer. Unfortunately, I managed to get the wrong stuff, and shipping rules make exchanging it problematic.
I am the proud owner of $500~\mathrm{g}$ of sodium bisulfate ($\ce{NaHSO4}$), instead of sodium bisulfite ($\ce{NaHSO3}$).
Dammit Jim! I'm a cook not a chemist!
What I have learned is that that both compounds are used as food additives towards the same end, and that my stuff has a slightly higher LD50. That's it.
Can anyone advise as to substituting my compound for the other? I'm not interested in tasty guacamole at this point, I'm really only looking at browning. If I'm trying to more or less duplicate the more successful experiments of the University of Florida, should I start with the same ratios? Or more or less of the additive? I can get sodium metabisulfite (also an anti-oxidant food additive) locally, if that would be a better option.
EDIT: So far, answers have all been consistent in that I have a half-kilo of white powder that is useless to me. I will make a trip to the wine-making shop and acquire some sodium metabisulfite. That still leaves me with the question of how much? The University of Florida got the results I'd like to duplicate with $30~\mathrm{mg}$ of sodium bisulfite per $100~\mathrm{g}$ of avocado. Should I start with the same ratio of sodium metabisulfite?

Sodium bisulphate (NaHSO4): an acid
Sodium bisulphiTe (NaHSO3): an antioxidant
Sodium metabisulphiTe (Na2S2O5): another antioxidant

In short, bisulphate and bisulphite are not interchangeable, but bisulphite and metabisulphite are.
Sodium metabisulphite is prepared from sodium bisulphite via dehydration, as in this equation:
$$\ce{2 NaHSO3 -> Na2S2O5 + H2O }$$
It's reversible in aqueous solution. The assumption that 30 mg of metabisulphite contain the same amount of sulphite as 30 mg of bisulphite is OK, the error is small, about 10 %.  The correct amount would be 27 mg of metabisulphite.
You'd ask yourself if pasteurization (75 deg. C, 30 seconds) would work to deactivate the polyphenoloxidase without affecting flavour too much.
I'd also suggest searching Pubmed (it's a decent alternative to commercial citation databases) for such keywords as "polyphenol oxidase" (that's the enzyme that causes browning), "inhibition" and "sulfite".
If you want hands-on answers I'd suggest going to the library and get (via interlibrary loan) a copy of this paper: Plant Foods Hum Nutr. 1995, 47, 245-56, "The control of polyphenol oxidase activity in fruits and vegetables. A study of the interactions between the chemical compounds used and heat treatment."

From the abstract: ... Interactions between the use of ascorbic acid,
  citric acid, EDTA, sodium metabisulphite and heat treatment (70
  degrees C for 2 min) in the control of PPO activity were studied in
  avocado ..., banana ..., apple ..., pear ..., peach ..., potato ...,
  eggplant ..., mushroom ... and
  hearts-of-palm ... . The results demonstrated that
  PPO of avocado and eggplant was most resistant to inhibition by the
  methods used. ... The results indicated that,
  with the exception of PPO from avocado, the most adequate alternative
  method to substitute for the use of SO2 in the control of PPO was a
  combination of ascorbic acid, citric acid and heat treatment.

This shows that how to get on top of browning while still having a presentable product is a tough problem and an active area of research.  Also, avocado seems a specially hard case.

I accepted this answer mostly because, if I understand the rules correctly, you should have received half the bounty, even though I never did manually award it. Strange. I brought it up in meta.

Generally, sodium hydrogen sulfite and sodium hydrogen sulfate are not interchangeable.
Sodium hydrogen sulfite or “sodium bisulfite” (NaHSO3) is used in food processing as sanitising agent for food containers and fermentation equipment, preservative to reduce or prevent microbial spoilage, selective inhibitor of undesirable microorganisms
in the fermentation industries, and as an antioxidant and inhibitor of enzyme-catalysed
oxidative discoloration and non-enzymic browning. The food additive code (the E number, which is commonly found on food labels) of sodium hydrogen sulfite is E 222.
Sodium hydrogen sulfate or “sodium bisulfate” (NaHSO4) is used as acidity regulator. The food additive code of sodium hydrogen sulfate is E 514.

Sulfate will not provide the antioxidant effect you seek.  Metabisulfite has antioxidant effect.

Sodium bisulfate will also control browning. It works by lowering the pH to inhibit the polyphenol oxidase enzyme. It lowers pH without a sour taste so it does not change the flavor of the food.


text:
How do you explain pKa to non-professional?
Let's say we have substance $\ce{A}$, which is mixed with substance $\ce{B}$ to improve shelf-life because $\mathrm{p}K_\mathrm{a}$ of the substance $\ce{A}$ is $7.9$ and in mix the $\mathrm{pH}$ is $5.2$.
Does this mean that in the solution with $\mathrm{pH}$ near $8$ the substance $\ce{A}$ has multiple molecules in neutral state and not dissociate, thus it forms precipitates?
Does $\mathrm{p}K_\mathrm{a}$ mean that most of the molecules of the substance $\ce{A}$ are in nondissociated state?
 Does lowering $\mathrm{pH}$ in this case causes substance $\ce{A}$ to dissociate?
Are bold parts right? Please explain in plain English without the use of external links.

pKa and pH are related concepts but often confused. pKa is a property of a compound that tells us how acidic it is. The lower the pKa, the stronger the acid. pH is a property of a particular solution that depends on the concentrations and identities of the components.
For this discussion, I'm going to use the terms protonated and deprotonated to mean that a compound is associated or dissociated with a proton.
Based on the relationship between the pKa of a compound and the pH of a solution, we can predict whether a compound will be protonated or deprotonated. If the pH is lower than the pKa, then the compound will be protonated. If the pH is higher than the pKa, then the compound will be deprotonated.
A further consideration is the charge on the compound. Acids are neutral when protonated and negatively charged (ionized) when deprotonated. Bases are neutral when deprotonated and positively charged (ionized) when protonated.
Given the information you provided, if compound A (with pKa 7.9) is in a solution of pH 5.2, compound A will be in the protonated state.
Without knowing anything about the identities of A and B, the following is speculation. Most likely compound A is a base and compound B is an acid. The protonated version of A (its "conjugate acid") is more stable for long term storage than A itself.
Another possibility is that compound A is an acid and B is a stronger acid. If the deprotonated version of A is unstable, then compound B is added to ensure that if the mixture came in contact with a base, something more acidic than A would be present to react with the base.
EDIT BASED ON NEW INFORMATION:
As Mateus B said, lidocaine will be protonated by HCl at the amine nitrogen. This is likely done for two reasons. First, as its hydrochloride, it will be more water soluble. In solution, the compound will dissociate into protonated lidocaine cation and chloride anion. Second, as the nitrogen is protonated, it is less reactive to the environment. Over time, amines can react with oxygen to form the corresponding N-oxide and/or can absorb carbon dioxide.

From your answer I understand more, but If I provide the substances - can the answer be more precise and understandable? A- is lidocaine (local anaesthetic), B -  hydrochloric acid. ?? Thanks! +1

The pKa of a compound also depends on the environment. Most published pKa values are an aqueous pKa, that is they tell you how acidic a compound is in water. However, that compound will likely have a different acidity and pKa, when they are  an isolated molecule, or dissolved in some other solvent, not water, for example: ethanol, dichloromethane, or DMSO. This extra aspect to pKa that isn't necessarily relevant here, but good to keep in the back of your mind when thinking about the protonation state of a compound, you need to think about what it is dissolved in, too.

Let's say we have substance A which was mixed with substance B to improve shelf-life because pKa of the substance A is 7.9 and in mix the pH is 5.2.
  Does this mean that in the solution with pH near 8 the substance A has multiple molecules in neutral state and not dissociate, thus it forms precipitates?

At all pH, the substance A will exist as a mixture of HA and A-. HA may or may not be soluble, so it may or may not precipitate.

Does pKa mean that most of the molecules of the substance A are in nondissociated state?

The higher the pKa, the lower the Ka, the lower the ratio of A- to HA, and thus the higher fraction of molecules of substance A will be in the non dissociated state.

Does lowering pH in this case causes substance A to dissociate?

No. Increasing pH causes substance A to dissociate.

Some additional information:
From the Henderson–Hasselbalch equation,
$\textrm{pH} = \textrm{pK}_{a}+ \log_{10} \left (
> \frac{[\textrm{A}^-]}{[\textrm{HA}]} \right )$
If pH = pKa, the concentration ratio of [A-] to [HA] is 1.
If your pH is higher than pKa, then [A-] must be greater than [HA].
If you lower your pH such that it is lower than pKa, then [HA] must be
  greater than [A-]
Therefore, by increasing the pH, you are dissociating HA into H+ and
  A-. Which makes sense according to the Le Chatelier principle, where
  H+ is released to counteract the increase in pH (or the decrease in
  [H+])

Could you please answer in plain English the question as it was asked?

@Ilan Explained in plain English, although I dare say the previous explanation was as simple as it could be, even to a non professional.

I'll include the original answer for completeness, under additional information.

If the substance A is lidocaine (base?), and B is hydrochloric acid, does this information help to focus the answer?? Thanks.

I think users @jerepierre and @t.c provided excellent answer above, but I will attempt a more simplified answer as requested.
First we need to understand what happens with an acid (with a specified pKa value), when a solution reaches certain pH values.

At a pH above the pKa, more of the acid molecules are deprotonated.
When the pH equals the pKa, the acid molecules are 50% protonated and 50% deprotonated.
At a pH below the pKa, more acid is protonated.

Without knowing more about the specific chemical nature of the molecule, we cannot know more about what will happen. In general, in a water solution more deprotonated or protonated acids will be more soluable (more molecules stays in solution). This is a result of that part of the molecule gaining an electric charge when gaining or losing extra protons. This charge makes the molecule polar, and soluable in polar solutions (such as water).
Solubility of a molecule can be explained using another concept that relates to pKa. This is called the isoelectric point (pI). Sometimes molecules are large, and complex. Several different parts of such a large molecule can be either 'acidic' or 'basic', depending on their chemical nature. They then gain or lose charge differently as the pH of a solution changes. At a certain pH, all the charges even out and the molecule becomes neutrally charged (instead of positive, or negative). It then loses its polarity which makes it soluable in water and it precipitates.

By specifying that A is lidocaine I could say that by adding HCl to it you'd be forming a diferent compound, lidocaine hydrocloride.

Can you elaborate a bit more?  This doesn't really (directly) answer the question as-is.

There isn't much more to elaborate on this. He asked about the influence of pKa on a generic case of two compounds, which has been answered already by the guys above, but then he asks about the specific case of lidocaine and HCl, to which I answered that it doesn't need to be directly related to pKa as the two compounds will react forming the compound shown in the image. In order to elaborate more I could only go toward reaction mechanism, which I don't think will interest a non-professional.

@MateusB. I've read that Lidocaine hydrochloride has longer shelf life than just Lido as is... Could you please answer What are the causes for this extension?


text:
Why doesn't this structure represent a substituted pentane?
In the following ASCI structure (single bond under "$\ce{CH}$" containing $\ce{CH2,CH2, CH3}$)
$$
\begin{align*}
\ce{CH3-CH2-CH2-&CH-CH3}\\
&|\\
\ce{&CH2-CH2-CH3}
\end{align*}
$$
The correct way to name it is "4-methlyheptane".
But when I do it the way I have been taught I get the following "4-propylpentane".
Here are my systematic steps:
Firstly, I count the longest carbon chain. In this case its 5 carbons together so suffix -pentane. Then I look to see if there are any branched chains and number it from left to right. In this case the branched chain is on number 4. The branched chain contains $\ce{CH2-CH2-CH3}$, resp. $\ce{C3H7}$, aka propyl.
Ergo, 4-propylpentane, however, the correct answer as mentioned above is 4-methylheptane.
How is it 'heptane' ? The longest chain has FIVE carbons. I got the '4-' part correct, but how is $\ce{CH2-CH2-CH3}$ Methyl? Is this a mistake in my book?

You need to find the longest continuous carbon chain of:
$$
\begin{align*}
\ce{CH3-CH2-CH2-&CH-CH3}\\
&|\\
\ce{&CH2-CH2-CH3}
\end{align*}
$$
Stretch that out to $\ce{CH3CH2CH2CH(CH3)CH2CH2CH3}$ (with the $\ce{-CH3}$ coming off of the $\ce{CH}$) and you'll see that the book is correct.  The "side chain" of $\ce{-CH2CH2CH3}$ is actually part of the main carbon chain.  As far as the numbering, counting from either side give the 4- position due to the symmetry of the molecule.

If anybody knows how to better jury-rig the diagrams, I'm open to an edit.

Just draw it on paper and take a picture. The built in package is too limited to draw a compound of even this low level of complexity.

@jonsca I found this http://web.chemdoodle.com/demos/sketcher to be of use when no local chemeditor is available.

The structure is misleadingly drawn.  The longest carbon chain is not this:
$$
\begin{align*}
\color{red}{\ce{CH3-CH2-CH2 -}}&\color{red}{\ce{CH-CH3}}\\
&|\\
\ce{&CH2-CH2-CH3}
\end{align*}
$$
but this:
$$
\begin{align*}
\color{green}{\ce{CH3-CH2-CH2 -}}&\color{green}{\ce{CH}}\ce{-CH3}\\
&\color{green}{|}\\
&\color{green}{\ce{CH2-CH2-CH3}}
\end{align*}
$$
That's 1, 2, 3, 4, 5, 6... 7 carbons.  Thus, this is a substituted heptane.  The single remaining carbon forms a methyl group, which is attached to the fourth carbon (from either end) in the heptane chain.  Thus, 4-methylheptane.


text:
Electron interference
Assuming that electrons behave as both waves and particles, and that waves(like in double slit experiment) can interfere with themselves:

Do electrons actually move in any direction, or do they just 'teleport'(similar to how when making a Quantum Leap, they instantaneously appear at a different shell, rather than actually move continuously).
Do electrons 'subtract' with other electrons for the probability function of an electron being present in that location. i.e. if there is a covalent bond, will the electrons interfere with the other probably function aside from repelling it?.
Do electrons only have positive probabilities. Is it more like how gravity never repels, only attracts, or more similar to how waves of light have peaks and troughs. Do electrons ever cancel out with others? If so, how?

Just a short comment to the third question: probability is non-negative by definition. See probability axioms.

And to add to @Wildcat's comments, $\psi^*\psi$ will, by construction, always be positive real. Of course there can be interference - there are places where the probability is zero.

Ok,thanks, but @GeoffHutchison how can they interfere to be zero probability if it is only positive? Wouldn't that cause the probability to add together, not cancel to zero?

@Brayton The wave functions can be negative (or imaginary or complex). The probability is the square of the wavefunction, or more accurately $\psi^*\psi$ so of course the probability cannot be negative.

Traditional understanding of trajectory is of no use for electrons. Granted, on macroscale you may detect something close to movement and trajectory, (electron beams and electron cannons take their names from this) but it still far from what we have in classical mechanics. Ideally you should abandon any attempts to relate quantum and classical mechanics and view quantum mechanics as an abstract mathematical construct.
Incorrect question. If you have a system of two or more particles, you should consider the wavefunction of combined system, not separate wavefunctions for separate particles. Still, for electrons being fermions, there is a strict rule that wavefunction must be zero for positions in phase-space where two electrons of same spin are at same location and since electrons are of same charge, probability of having two electrons of different sping close to each other is usually lower than to have them at some distance.
Yes, only positive probabilities. Though, this does no apply to wavefunction itself, it may be negative at some point (actually, its value can be a complex number).


text:
How to recover 15g gold from 50 gallons of 30% Acid?
Someone mistakenly poured a solution containing $15\mathrm{g}$ of gold into a nearly full waste drum that is around $30$% $\ce{HCl}$ with a little nitric acid in there as well.
Is there a safe way of separating the acid from the gold via distillation?

You can more or less safely use distillation to concentrate the solution. Likely, some amount of chlorine and nitrogen oxides will be produced, but fume hood should be enough to deal with it. Once the solution would be concentrated to amount of several tens of ml, some chemical method to precipitate gold may be used, say, neutralizing the solution followed with reduction by $\ce{Zn}$ followed by washing the precipitate with $\ce{HCl}$

At a rate of USD 1200/oz, it's a lot of work to recover USD 600 worth of gold.

You could distill the waste easily enough, but the distillation gases will be rather nasty. You will also need a condenser to cool everything down again or the environmental penalties (and corrosion damage to the exhaust system) will be a lot more than the gold value.
Once you have the waste boiled down (literally) into a small flask you could use any of the known gold-extraction methods to get it out in elemental form.
At current market prices you have about US$580 of gold in the drum. No idea how much you will spend in fuel and cooling water to recover it but it will be a substantial percentage. Why don't you just tell "Someone" that they can either pay for the gold they dumped or mind the still themselves over the weekend (and pay for the propane). This will also encourage others to pay more attention to what is in the beaker.

Before giving "someone" the choice, consider whether you'd really trust the same person who dumped the gold in the waste drum in the first place to a) choose wisely, or b) carry out the distillation safely and without further material damage.

Thankyou for ideas. I plan to use a glass condenser to avoid corrosion. I'm not sure whether it will be liquid or air cooled. I know the liquid ones are more popular in general, but it would be really nice to use long thin glass tubes that are air cooled. That is another question entirely though.

Also, to answer your question: I have to build the still first.


text:
Why are weak acids weak?
What are weak acids? My chemistry and biology classes skimmed over them briefly but nothing in depth. I understand that an acid is weak if it partially dissociates and breaks down into ionic compounds, but they never stated why.
So my question is why do weak acids break down in water? Do strong acids dissociate at all? What if they were in a different solution (or do they just stay themselves all the time?)

What are weak acids?

Examples of weak acids in water solution would be acetic acid (main component of vinegar) and ammonium ion (conjugate acid of ammonia).

I understand that an acid is weak if it partial dissociates and breaks down into ionic compounds but they never stated why.

An acid is weak if not all of the acid molecules ionize into hydrogen protons and its conjugate base in a particular solvent system. Alternately, if we were to use the broader, Brønsted definition, an acid is weak if it does not completely or nearly completely donate its proton to some base.
Acetic acid is a weak acid in water. Put acetic acid into pure liquid ammonia and now acetic acid will fully dissociate. The proton will be lost from acetic acid to ammonia and this is a favorable reaction. Makes sense, because the only thing that's changed is the solvent that's grabbing the proton - and ammonia is only a few billion times stronger as a base than water.

That and do the strong acids dissociate at all. Like if they were in a different solution or do they just stay themselves all the time.

Strong acids by definition ionize to an extent of 100% or nearly 100%. Strong acids will give up their protons completely to the strongest base in the system.
Note that in the above text I have used the word "ionize" to describe molecular acids because by definition when these acids react, they ionize. Or form ions. On the other hand salts of acids and bases such as sodium hydroxide are said to "dissociate," because ionic compounds are made of ions, and it wouldn't make much sense to say that ions ionize or become ions.

That's right.  The keyword is "levelling effect of the solvent".

There is nothing magic about weak acids.  The dissociation reaction of an acid in water looks like this: $\ce{HA + H2O <=> A- + H3O+}$.  There's an equilibrium constant K that's related to the free enthalpy $\Delta G$ in the usual manner:
$$\Delta G=-RT \cdot \ln K$$
A weak acid is simply an acid where the equilibrium lies far to the left.  A strong acid, on the contrary is an acid where the equilibrium lies far to the right, i.e. it is more acidic than the $\ce{H3O+}$ ion.

I think you meant that weak acids don't break down in water... you see, and acid is usually considered strong if a lot of it ionizes and donate it's protons (H+) to the water in which it's dissolved. While a base is considered strong depending on it's capacity to protonate (take the H+ from the water leaving OH- on the solution).
This sort of behavior is defined by the affinity between the ions (that are in the acid/base) and the H+.
There are more complete definitions for acids and bases that don't necessarily require to consider it in a solution with water, but if your teacher skimmed over more basic definitions, then it's not likely that you will study it unless you get General or Inorganic Chemistry classes in college.
If you're interested in this sort of stuff you can always get a chemistry book on the library. I used "Chemistry The Central Science" by Brown during my General 'n Inorganic Chemistry classes on my first semester on college and I think it is a book that would be easily understood by someone who's still in high school given you put some effort in it.


text:
Who Was the First Chemist?
I know chemistry has been around for a long time, but I was curious who the first chemist in recorded history. I am aware that Antoine Lavoisier was one of the first modern chemistry and is considered to be the "Father of Modern Chemistry", but was he is the first chemist?

I'll write it up later, but Robert Boyle is often considered the first modern chemist.

That depends how you define "chemist."
Today we think of chemists as working at the molecular level. However Lavoisier didn't have any concrete evidence of an atom's existence. He worked with Sulfates (SO4-2), metal oxides, and other molecular groups in an empirical an algebraic fashion.
Throughout much of the 1800s there was enormous debate about the very existence of atoms. Much of the problem arose from confusion over molecular weights and stoichiometry. Physicists were the most stubborn opponents of atoms, which we still can't fully explain even today. Eventually Advodagro's number and accepted stoichiometry combined to form irrefutable evidence of molecules. Many scientists were completely convinced that atoms existed long before they were proven though. Dalton was the first to introduce stoichiometry so he would be a good candidate as well.

The first chemist? The first person who learned that if you mix crushed plant root with the runoff from fermented vegetation (alcohol) you got something you could use to make your robe a bit more colourful? If he experimented with various plants and liquid ratios he would meet the modern definition of "chemist". I recall doing similar things in grade 9 Chemistry class.
That individual is long lost in the mists of time. The main problem with history is we only know what people bothered to write down, or left behind in a durable form. In the days before patents or research for it's own sake all this would have been closely guarded secrets recorded (if at all) in shorthand on small pieces of whatever passed as paper in those days.
The first chemistry research paper was possibly found in the Great Library of Alexandria, which unfortunately burned down on more than one occasion.


text:
Can I make Cobalt Violet with these materials?
I'm given $\ce{CoCl2*6H2O}$ as well as $\ce{Na3PO4*12 H2O}$
The chemical formula for Cobalt Violet is $\ce{Co3(PO4)2}$. I haven't done stoichiometry in a long time but I think that's creatable, right?
Some website says that I need $\ce{NaH2PO4·10H2O}$, instead of the Sodium phosphate dodecahydrate which is what I have.

Welcome aboard. You can format chemical formulas and equations to make them more readable as shown in this guide: http://meta.chemistry.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here/88#88

My best guess that pH control is important - at basic pH (and $\ce{Na3PO4}$ is pretty basic) you will precipitate cobalt hydroxide as well.  To prevent that, the recipe calls for $\ce{NaH2PO4}$ to provides a lower pH.  Cobalt phosphate is so insoluble that it will precipitate even in somewhat acidic medium.

Right, I hadn't considered that. It shouldn't be too difficult to make it work with the other salts and some pH adjustment though.

The dihydrogen phosphate anion is actually a weak acid and will dissociate to make both $\ce{HPO4^{2-}}$ and $\ce{PO4^{3-}}$ when dissolved in solution. If it works with the dihydrogen phosphate salt, it should work with the hydrogen phosphate and phosphate salts, too. (After all, it's the phosphate anion that precipitates out with the cobalt, if your formula is correct.)


text:
Does copper react with hydrochloric acid?
In my chemistry lab, we had 18 different reactions involving many different substances. One of the experiments involved placing a piece of copper metal in hydrochloric acid.
While observing it for a minute, my partner and I did not notice any reaction taking place. However, a minute was all we were allowed, and we could not keep it longer to observe it.
It was the only experiment that looked like it had no reaction, so I felt a little uneasy putting it down. (I didn't do anything wrong, but I guess I was unsure with myself).
I have seen some sources say that no reaction is the current result. However, I have also seen some sources saying that it does react, via $\ce{Cu + 2HCl -> CuCl2 + H2}$, yielding cupric chloride and dihydrogen gas.
Which is the correct result? If no reaction is correct, what would be the correct chemical reaction for this case?

The correct result?  Are you going to argue with observable reality?

@AbelFriedman well, I guess I am concerned that a reaction was occurring at a slow pace that I did not have the time to observe. And the fact that sites say that a reaction should occur.

@erdekhayser this may be helpful http://chemistry.stackexchange.com/questions/16507/copper-doesnt-react-with-hydrochloric-acid

@SherlockHolmes Thanks! That is very informative.


text:
Are there any papers linking the bad taste of orange juice after toothpaste to sulfates?
Wikipedia states:

After using toothpaste, orange juice and other juices have an unpleasant taste. This effect is attributed to products of the chemical reaction between stannous fluoride in toothpaste and the acetic acid in the juices. Sodium lauryl sulfate alters taste perception. It can break down phospholipids that inhibit taste receptors for sweetness, giving food a bitter taste. In contrast, apples are known to taste more pleasant after using toothpaste.

The primary reference for this appears to be a comment on the US Department of Energy website.
There are several references to this.
This seems to be an untested hypothesis. My question is Are there any papers linking the bad taste of orange juice after toothpaste to sulfates?

There may be more, but this is one paper examining the effects of membrane-modifying compounds on taste receptors. It suggests that substances like sodium lauryl sulfate and gymnemic acid affect taste response by altering taste receptors' lipid membranes. Though many anionic surfactants have sulfate functional groups, not all compounds with sulfate groups act as surfactants (nor does the sulfate anion on its own). It sounds like this is something that any compound that can perturb the structure of a taste receptor's membrane can do, not something that necessarily applies to all sulfates.


text:
how to measure dissolved oxygen?
How do you measure dissolved oxygen reliably in a culture of cells? This page has info on one sensor: http://water.me.vccs.edu/concepts/domeasure.html
how would you know that this sensor is working - meaning that only dissolved oxygen triggers the electrical change in the sensor, and not other atoms? and how would you calibrate the sensor to translate changes in electric signal to the change in number of dissolved oxygen atoms?
also, how would one get pure dissolved oxygen in varying amounts as 'positive control' for the sensor?

To start, the figure and explanation in the link are almost comically misleading. Potassium chloride solution does not "attract" oxygen. The oxygen flowing through the wire is pretty silly and

Where the probe joins the wire, oxygen mingles with the electricity.
  Oxygen is not very ionized, meaning that it does not have a negative
  charge as electricity does, so the oxygen dilutes the current at the
  electrode beyond the probe.

is just plain ridiculous. This one is better.
The way this type of probe works is by amperometric reduction of $\ce{O2}$, the first example being the Clark electrode. Essentially, an electrolyte is separated from the measurement solution by a porous film of poly(tetrafluoroethylene) (Teflon). These pores are small enough that only small substances like oxygen can diffuse through. On the electrolyte side, there are two electrodes in one of two possible configurations:

In the original Clark-type electrode, the working electrode is platinum, which is chemically-inert but catalyzes the reduction of oxygen. The counter electrode is a AgCl-coated Ag electrode that acts as a reference for the measurement. (Its potential is supposed to remain constant) To reduce the oxygen, an external potential must be applied across the two electrodes (normally using a potentiostat).
The galvanic cell–type electrode has one electrode made of silver (where the oxygen reduction occurs) and one made of an easily oxidized metal like lead. It works the same way as the Clark electrode, but instead of applying an external potential, the current required to reduce the oxygen is supplied by the oxidation of the lead electrode.

Any oxygen that reaches the working electrode is reduced and produces a current that is measured to produce the analytical signal. The amount of current produced depends on how much oxygen diffuses through the membrane, which depends on the concentration of dissolved oxygen in the test solution.

Now, for your questions:

how would you know that this sensor is working - meaning that only
  dissolved oxygen triggers the electrical change in the sensor, and not
  other atoms?

This one is a bit tricky. You're right that oxygen isn't the only thing that could produce a signal—amperometry isn't a very selective technique on its own. You can control the redox reactions that occur at the working electrode, to a limited degree, by changing the potential you apply, i.e. different substances require different potentials to be reduced and won't be reduced if an insufficient potential is applied. In the case of this specific type of electrode, selectivity comes mainly from the PTFE membrane. Since it limits diffusion to only a few very small molecules, that greatly increases the selectivity of the electrode, though depending on the solution you're trying to measure, there still may be interferences.

and how would you calibrate the sensor to translate changes in
  electric signal to the change in number of dissolved oxygen atoms?

In principle, you could calculate the number of reduction reactions happening at the surface from Faraday's laws, but to turn this into a measure of the exact concentration in solution requires a lot of thought of how the transport of oxygen to the electrode occurs, from diffusion from the electrolyte to the electrode surface and transport from the test solution into the electrolyte. In practice, the assumption that the measured current is directly proportional to the concentration in solution is made (and is valid under many conditions), and a calibration is made using a standard solution.

also, how would one get pure dissolved oxygen in varying amounts as 'positive control' for the sensor?

Calibration solutions seem to be commonly made using tables of oxygen saturation at different temperatures. It's also possible to measure dissolved oxygen through other means, such as titration. Solutions can be made up, titrated, then the value used to calibrate an electrode.

Now, for the measurement specifically in cell cultures, a membrane electrode may not be the best bet, at least not without sample pre-treatment. The PTFE membrane is quite susceptible to adsorption of proteins and the like, which will probably affect the electrode's response. It may be necessary to somehow remove these contaminants (without affecting the dissolved oxygen concentration). Other options that might be viable are colourimetric measurements or titration, though there are problems with these, as well.


text:
What are the product(s) of red phosphorous when burned?
A match box lighter strip typically consists of powdered glass and red phosphorous. When this part is burned, it's residue is like a sticky dust that, when applying friction, will release smoke or vapor. You can do it with you fingers and it looks really cool.
So what's the chemistry behind this?

The major product formed is -  Tetraphosphorus decaoxide ($\ce{P4O10}$)

Infact you could have googled it. There are many interesting videos on youtube.
Check this out -  Burning of Red Phosphorus


text:
Why do acids ionize in solution if they are covalently bonded?
Maybe I am wrong, but I thought acids were covalently bonded since hydrogen doesn't form ionic compounds. For example, wouldn't $\ce{HCl}$ look like:

Since the electrons are shared, why does it break into $\ce{H+}$ and $\ce{Cl-}$ in solution?

Because it is energetically favourable ($\Delta{G}<0$) for hydrogen chloride to react with water to form hydronium( $\ce{ H3O+}$) and chloride ions.
Remember that $\ce{ H+}$ does not exist as $\ce{ H+}$ in water, but rather as $\ce{ H3O+}$.

Why is it energetically favorable for this reaction to occur? Is the enthalpy very low (negative), or is the entropy very high, or both?

Both. for most cases, dissolving a compound to form an acid releases heat (negative enthalpy). Also, entropy is increased due to the addition of chloride (and also the breaking of H-Cl bond).

Typically the curved arrows are used for electron pair transfers.

Your are right. But you have to remember that the bond between hydrogen and chlorine (for example) is covalent polar one. When you introduce this molecule into water which is a polar solvent with a dipole moment of 1.85 D, the polarization of the molecule $\ce{HCl}$ increases. We  have then two solvated ions  $\ce{H_3O^+}$  and and  $\ce{Cl^-}$. And  as water has also a high dielectric constant  $\epsilon_r=80$, the electrostatic force between  $\ce{H_3O^+}$ and $\ce{Cl^-}$ will decrease by a factor of 80. So, the the bond between $\ce{H_3O^+}$ and $\ce{Cl^-}$ will weaken  and we have ionization of the acid.


text:
How to determine peroxy oxygen?

Which of the following compound doesn't contain peroxy oxygen?

$\ce{CrO5}$
$\ce{Cr2O3}$
$\ce{Na2O2}$
$\ce{BaO2}$

I think peroxy oxygen is peroxide, am I right?. I have read in this answer that there is no direct way to recognise peroxide oxygen, and I think that's the reason no one has taught me that in the school.
I am familiar with many peroxides like $\ce{Na2O2}$, $\ce{BaO2}$, etc. But I not able to recognise between 1 and 2.
Is there any way for this type of question to recognise peroxide oxygen?  And if there is no way to determine then really I need to remember all, because we can't use X-ray crystallography in exam or while studying high school chemistry!

Answer given is 2. $\ce{Cr2O3}$

try to draw strutural formula using highest valency for the element. It is pretty straighforward way to get the answer by elimination.

If you first assume that the oxygen has an oxidation state of $-2$ but then the metal ends up with something suspiciously high, you might want to rethink that oxygen. Which is more plausible for CrO5 — Cr(X) or peroxide linkages?

I have become familiar with many peroxide like Na2O2, BaO2, etc. But not able to recognise between 1 and 2. Is there any way for this type of question to recognise peroxide oxygen?

Yes. In the exam, if you do not know whether (1) or (2) contains peroxy, simply look at the charges of Chromium if all the oxygen is  in the -2 oxidation state.
For $\ce{CrO5}$, if all the oxygen have a charge of -2, then Cr would have to have a charge of +10, which is impossible, therefore you know that some of the oxygen must have a charge of -1(peroxy). On the other hand, for $\ce{Cr2O3}$, Cr has a charge of +3, which is possible (Cr(III) salts are plenty)

I'd like to offer an alternative method here, which is generic for all binary compounds (but may also work for ternery, quarternery, etc.).
Look at the highest oxidation state of the metal. Then divide the negative by the number of oxygens. This is how you find the average oxidation state of oxygen. If this is greater than -2, then there has to be a peroxyl oxygen involved. If it is -2, or lower, then there is no peroxyl involved. Additional remark: If the average oxidation state of oxygen is lower than -2, then the metal is not in its highest oxidation state.

"If this is greater than -2, then there has to be a peroxyl oxygen involved." I'm probably missing something, but couldn't the oxygen be a superoxide instead? In a superoxide $\ce{O_2-}$, each $\ce{O}$ atom has an average oxidation state of $-0.5$, whereas in peroxide each oxygen has an oxidation state of $-1$ only...


text:
How is the thermal stability of nitric acid related to its structure?
How is the thermal stability of nitric acid related to its planar structure?
In textbooks it is written that nitric acid is unstable at room temperature and decomposes where $\ce{H2SO4}$ and $\ce{H3PO4}$ are stable and decompose at much higher temperature.$\ce{H2SO4}$ and $\ce{H3PO4}$ have tetrahedral structures.


text:
Can the product be called "vinegar"?
For a product to be called "vinegar", it must contain at least $5.0$% acetic acid by mass. A 10.0 g sample of a "raspberry vinegar" is titrated with $0.125 \mathrm{M}$ barium hydroxide and required $37.50 \mathrm{mL}$ for complete neutralization. Can the product be called "vinegar"?
My professor said that the product is vinegar, but I did not understand how he solved it. Could you please tell me what to do? What formulas should I use to solve this kind of problem?

unless you were given a molecular weight for the raspberry vinegar, I think you'd have to assume that the only species that can react with the barium hydroxide is the acetic acid itself. Therefore you would expect to get barium acetate and water.

0.125M of $\ce {Ba(OH)2}$ means 0.125 moles of $\ce {Ba(OH)2}$ per volume unit (so far I've only seen liters been used as volume unit) .
Assuming that all the $\ce {Ba(OH)2}$ you used was consume on the reaction with the acetic acid, and knowing the ratio of $\ce {Ba(OH)2}$ and acetic acid on the reaction is 1/2 you`d have this reaction:
$$\ce {Ba(OH)2 + 2CH3CO-OH -> Ba(CH3COO)2 + 2H2O}$$
All you need to do is calculate the amount of acetic acid consumed. Knowing the amount of acetic acid consumed based on the amount of moles of $\ce {Ba(OH)2}$ that are present on 37.5mL. Then you need to get the molar mass of acetic acid  and calculate the amount of mass of acetic acid you had. Then divide the total mass of vinegar by the mass of acetic acid  you found, multiply it by 100 and you'll have the mass percentage of acetic acid.


text:
Are Grignards Chiral?
Can Grignards be chiral? I've seen them drawn as ionic structures in class. I've also seen little asterisks on certain positions of Grignards (a common designation of chirality). So this begs the question - are Grignards chiral?
From what I understand Grignards are best described as polar covalent rather than ionic, but that still leaves one more question: can ionic compounds be chiral?

I read the question as asking whether a stereocenter bearing a C-Mg bond will keep its configuration. The answer is yes, but not indefinitely. Unfunctionalized (alkane) Grignards of this type racemize/epimerize with a half life on the order of hours. Specific compounds will have more or less stability. By analogy to organolithium compounds, I would expect having Lewis basic atoms in the proximity to stabilize the configuration. I'm not sure if this is still an active field of research, but it certainly was for a long time. Try searching "configurational stability of Grignard reagents" or similar search terms.
There's an additional complication with these chiral Grignard reagents: they can't be prepared by the usual method of oxidative addition of Mg(0) to an alkyl halide. The mechanism of that reaction goes by a single electron transfer, through a radical intermediate, destroying any stereocenter. Chiral Grignards have to be synthesized using less direct methods.
Ref: Side Reactions in Organic Synthesis: A Guide to Successful Synthesis Design, pp. 200-202.