text:
Ok to separate steel from aluminum with bronze? (Galvanic action)
I have an aluminum hinge that I want to use with a stainless steel hinge pin. The problem is that I cannot use the steel pin directly because there would be galvanic action between the stainless steel and aluminum which will result in corrosion. Therefore I need to put some kind of bearing liner between them. Note that the hinge will experience strong forces, so using something soft like nylon will not work.
My preference would be for a bronze bushing, but I do not know if there is galvanic action between bronze and stainless steel, or bronze and aluminum.
What material can I use? Is there a simple explanation of how I can predict whether a metallic alloy will react galvanically with a different alloy? I have Mathematica, so if there is a formula I can write in Mathematica, that would be just as good.
-------------------------------- Sacrificial Anode?
One method I have heard of to protect from galvanic corrosion is a "sacrificial anode". If I attach a bar of zinc to the back of the aluminum hinge will that somehow protect it, and allow me to use a stainless steel hinge pin?

Pretty much any two different metals would make a galvanic pair. If anything, bronze bushing in your case is worse than no bushing at all. Then again, all this does not matter unless the hinge is constantly soaked in water. Is that so?

@IvanNeretin No, but in real life, galvanic corrosion will occur without any visible electrolyte. For example, if you screw a stainless steel screw into an aluminum plate, after a few years there will be visible corrosion on the aluminum.

In a sense the hinge will effectively be "soaked" in water at the level of a few to dozens of molecular layers of water, depending upon the relative humidity at the hinge, which is plenty to provide a medium for an electrolyte layer promoting corrosion over months and years.

If you must use an aluminum hinge with a steel pin, anodize the aluminum. Anodizing builds up a coat of very hard, non-conductive $\ce{Al2O3}$, sapphire.
There are commercial metal fishers that will perform this, if you do not want to do so. There is a bit of an art in the process of creating a uniform coat of the required thickness: too thick and it cracks off; too thin and it wears off.

The sacrificial anode is certainly a possible and common approach, but I will propose another approach using organic polymers (not “soft” , not Nylon).
According to the Engineers Edge website:

There are three conditions that must exist for galvanic corrosion to
  occur. First there must be two electrochemically dissimilar metals
  present. Second, there must be an electrically conductive path between
  the two metals. And third, there must be a conductive path for the
  metal ions to move from the more anodic metal to the more cathodic
  metal. If any one of these three conditions does not exist, galvanic
  corrosion will not occur.

Note that if you are not convinced of my proposed organic polymer approach that the link above includes a table of anodic indices for several metals and alloys.
Organic based corrosion inhibitors can strongly chemisorb to and  effectively passivate many metal surfaces, including stainless steel and, I assume, brass.  You only need it on one of the two.  This chemisorbed layer prevents the formation of a water/electrolyte layer at the metal's surfaces.  Additionally, this layer inhibits the transfer of metal ions.  So this mitigates the second two out of the three criteria given above for galvanic corrosion to occur.
Pollyurea-based elastomers are commercially available and should be an excellent candidate in your case.  This would be simple to implement and you have a good choice of commercially available products (here and  here just for example) of varying physical characteristics.  These types of elasomers are used in situations that are probably far more physically rigorous than your hinge would experience: sealing metallic joints in the hulls of maritime cargo vessels for example.
According to this Wikipedia page:

Some polyureas reach strengths of 6000psi (40MPa) tensile and over
  500% elongation making it a tough coating.

Also sited in the same Wikipedia reference is an interesting line of research being taken to even further pollyurea-based elastomers’ already desirable characteristics:

In 2014 a polyurea elastomer-based material was shown to be
  self-healing, melding together after being cut in half, without the
  addition of catalysts or other chemicals. The material also includes
  inexpensive commercially available compounds. The elastomer molecules
  were tweaked, making the bonds between them longer. The resulting
  molecules are easier to pull apart from one another and better able to
  rebond at room temperature with almost the same strength.
The rebonding can be repeated. Stretchy, self-healing paints and other
  coatings recently took a step closer to common use, thanks to research
  being conducted at the University of Illinois. Scientists there have
  used "off-the-shelf" components to create a polymer that melds back
  together after being cut in half, without the addition of catalysts or
  other chemicals.

Literature references are given in the previous link.
The options I’ve discussed or referenced above focus on the choice of metals used for each component of the hinge or as a sacrificial anode (see table in Engineers Edge link) and a polyurea-based coating.  I propose the polymer coating approach for the arguments made above including ready availability, affordability, likely effectiveness and relative simplicity.


text:
How can I choose the right symmetry in CAS calculation using MOLCAS?
With $C_{2\mathrm{v}}$ symmetry and a closed shell configuration, how can I choose the occupation for each irreducible representation in the input file?

Welcome to Chemistry! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using LaTeX syntax. If you receive useful answers, consider accepting one.

Molcas' gv.exe can determine the symmetries that Molcas can handle. Molcas then should autodetect the highest possible symmetry in the input geometry.

Also Molcas can only handle D2h subgroups, afaik.

To perform a CASSCF calculation in MOLCAS, you have to supply a "standard" SCF calculation, specifically the orbitals, in a manner that the CASSCF module of MOLCAS can read them and use them as the one-determinantal basis for the following CASSCF procedure. As your system is closed-shell, a standard Hartree-Fock calculation performed beforehand using MOLCAS will do the trick. So, before you may run the RASSCF module of MOLCAS to perform a CASSCF calculation, you need to run the SCF module of MOLCAS to perform a Hartree-Fock calculation.
If this Hartree-Fock calculation is performed utilizing the $C_{\mathrm{2v}}$ symmetry of the molecule it will yield a set of orbitals, some for each symmetry group, i.e. $\mathrm{a_1, b_1, a_2, b_2}$. As the calculation is closed-shell the occupation will either be zero or 2. If your issue is the detection of symmetry in MOLCAS, please consider the following:
MOLCAS handles symmetry for all calculations via the GATEWAY module, you can specify the generators for the symmetry you deem the molecule to have there. This specification using generators might seem a little peculiar if you are used to programs like Gaussian or Turbomole, however, there are many resources to be found explaining it. See for example here:

Lecture on using symmetry in MOLCAS
Manual page on "Computing high symmetry molecules"

For C2V, if I'm not mistaken, the following keywords for GATEWAY might suit you:
&GATEWAY
coord=yourStructre.xyz
basis=cc-pVDZ
group=xy y

Now, assuming the symmetry was correctly recognized during the run of the GATEWAY submodule, and you have successfully performed a HF calculation utilizing the molecular symmetry, you can the proceed to chose your active space. Picking your active space is actually the tricky part of a CASSCF calculation where experience and chemical intuition come into play. I cannot give a full tutorial on choosing an active space, as this is a huge subject of its own. You can find some information here:

MOLCAS Manual - Tools for selection of the active space
MOLCAS FAQ: How do I choose my active space in a CASSCF/CASPT2
calculation?
Chemistry StackExchange:
How to choose proper active space in method such as CASPT2 and CASSCF?

Once you have chosen the orbitals you want to include in your active space, you need to identify their "number" in the Hartree-Fock calculation in order to be able to specify them. The "number" is of course given due to the energetical ordering of MOs ascending in energy. If your Hartree-Fock calculation yielded for example

10 occupied orbitals of $\mathrm{A_1}$ Symmetry
8 occupied orbitals of $\mathrm{A_2}$ Symmetry
6 occupied orbitals of $\mathrm{B_1}$ Symmetry
4 occupied orbitals of $\mathrm{B_2}$ Symmetry

and you have determined (via chemical intuition, experience and black magic) that the inclusion of the energetically highest two occupied orbitals of each symmetry and the energetically lowest two unoccupied orbitals of every symmetry will yield a fine active space, your input to the RASSCF module of MOLCAS for a ground state calculation could look like this:
&RASSCF
 Symmetry
 3
 Charge
 0
 Spin
 1
 Inactive
 8 6 4 2
 Nactel
 8 0 0
 Ras2
 4 4 4 4
 LumOrb

Please notice the Symmetry keyword for the RASSCF module which is set to "3" in my example above. A $C_{\mathrm{2v}}$ symmetry yields 4 symmetry groups. Different symmetries will yield a different number if symmetry groups, for example $C_2$ symmetry will give you two ($\mathrm{A}$ and $\mathrm{B}$). The number of values used as input for the keywords Inactive and Ras2 correspond to the number of symmetry groups present. You have to specify the desired symmetry of the wavefunction via the Symmetry keyword of the RASSCF module, or else the default, which is $\mathrm{A_1}$, will be calculated.
The total inputfile for a CASSCF calculation in symmetry may look like this:
&GATEWAY
coord=yourStructre.xyz
basis=cc-pVDZ
group=xy y
&SEWARD
&SCF
&RASSCF
 Symmetry
 3
 Charge
 0
 Spin
 1
 Inactive
 8 6 4 2
 Nactel
 8 0 0
 Ras2
 4 4 4 4
 LumOrb
&GRID_IT
 All; Ascii

I hope this helped you.

source: Stack Exchange

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text:
How to convert a linear dicarboxylate species to the dimethyl ester?
I was wondering about the ozonolysis of cyclohexane and further treatment with $\ce{H2O2}$. I understand that the product to this reaction is hexanedioic acid.
How could we substitute the $\ce{-H}$ on each carboxylic group to a methyl, as in the following structure:

I am pretty sure it would be by adding two moles of methanol (Fischer esterification). Is this procedure correct?

I would try $\ce{NaOH}$ and then $\ce{MeI}$. And in one step but more dangerous you can prepare in a solution of $\ce{CH2N2}$.

I think I would prefer sodium hydride to sodium hydroxide. Otherwise, agree with everything else.

Oh, I see. But I thought Williamson synthesis was used for synthesising ethers from alcohols?

You have a couple of options.

Your acid-catalyzed Fisher esterification is one of them. The downside is that water, which is liberated in the reaction, has a higher boiling point than methanol. You can't just destill it off to drive the reaction to completion.
Zhe and  Hexacoordinate-C mentioned diazomethane and methyl iodide which might work well. You might want to make sure that your DNA does not get methylated.
The Steglich esterification, using dicyclohexylcarbodiimide (DCC) and N,N'-dimethylaminopyridine (DMAP) might be an alternative to the Fisher esterifiction.
You might want to try something different, using dimethyl carbonate as a non-cancerogenic alkylation reagent. The procedure is described in Green Chemistry Letters and Reviews, 2013, 61, 89-93 and uses potassium carbonate as a base, dimethyl carbonate as a reagent (and solvent), and tetrabutylammonium chloride as a catalyst.

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text:
Recommendations for electrostatic potential surfaces
Are there standards for density cutoffs, colours and colour mappings for electrostatic potential surfaces?
I observe that people use many different values.
For instance, this article (in Fig. 1a) uses 0.001 au density cutoffs with colours ranging from blue (< 0 kcal/mol) to red (> 20 kcal/mol).
I understand that not every colour/energy range will work every time, but other publications/websites even switch the colour range.
This website, for instance, uses red for negative energies and blue for most positive ones.
Furthermore, the same website suggests a .02 au cutoff.
Are there recommendations on this?
Which values are more commonly used by textbooks?

You have probably found out from playing around that there are many options. For coloring there is no standard, blue and red are common because they are the easiest to see since they contrast well. As far as the density cutoff, there are some papers that recommend values but the main idea is to choose a value that gets close to the Van Der Waals radius for the atoms. This is after all what is typically is of interest in a chemical sense. The electron density at the Van Der Walls surface is what others atoms/molecules "see". Typically I will just plot both the Van Der Waals and electrostatic surface and modify the electrostatic cutoff to make them close. Then you can set the limits for positive and negative colors.

Since it is then a matter of surface accessibility, would you recommend colouring the van der Waals surface instead of an electron density isosurface?

It depends on what you are interested in. If you want to know where there is an excess/depletion of electron density the electron density surface is good for that. If you just want an idea of the bulk and access ability of a part of a molecule the Van der Waals surface is a good idea. In the case of the Van der Walls surface, you might want to consider the Connolly molecular surface or something like that, check Wikipedia for "Accessible surface area" for an idea of what it is.


text:
Calculating the number of neutrons in a sample

Which of the following contains one mole of neutrons?
A. $\pu{1 g}$ of $\ce{^1_1H}$
B. $\pu{1 g}$ of $\ce{^{12}_6C}$
C. $\pu{2 g}$ of $\ce{^{24}_{12}Mg}$
D. $\pu{2 g}$ of $\ce{^{23}_{10}Ne}$

Here's a question which I don't know how to answer. I tried using the equation:
moles=mass/mr
but that did not match the answer given, which is (C). I keep doing 2/24 which doesn't appear to equal 1 mole.

The formula
$$\text{amount of substance} = \frac{\text{mass}}{\text{molar mass}}$$
tells you that you have 1/12 moles of magnesium, since you are taking the mass of magnesium (2 g) and dividing by the molecular mass of magnesium (24 g/mol). However, the question is not asking about the number of moles of magnesium.
Each magnesium atom has 24 - 12 = 12 neutrons. Consequently, 1/12 moles of magnesium has (1/12)(12) = 1 mole of neutrons.

Answer A has no neutrons
Answer B has 1/12 mole of carbon with 6 moles of neutrons per mole.  So there's only one-half mole of neutrons.
Answer C has 2/24 or 1/12 mole of Mg with 12 moles of neutrons (and 12 moles of protons) per mole of Mg.   12/12 = 1, the answer.
Answer D has 2/23 mole Ne with 23-10 moles of Neutrons per mole of Ne.  So is more than 1 mole of neutrons in the answer.


text:
How to know when dissolution is complete?
In a given solubility equilibrium, I know that if $ Q >K_s$, then the salt will precipitate until $ Q =K_s$. If $ Q <K_s$, however, would the precipitate dissolve until $ Q =K_s$ too?

You've described the general tendency for any equilibrium process, including solubility. So, the short answer is yes, you will dissolve more until the reaction quotient is equal to the equilibrium constant.


text:
how do i find how much energy needs to be removed?
How much energy must be removed to convert $8.49\ \mathrm{g}$ of water at $24.5\ ^{\circ}\mathrm{C}$ to an ice cube at $-12.0\ ^{\circ}\mathrm{C}$?  Give your answer as a positive value in Joules since only positive energy can be removed.
$$c_{\mathrm{water}} = 4.184\ \mathrm{J}\,\mathrm{g}^{-1}\,^{\circ}\mathrm{C}^{-1}$$
$$c_{\mathrm{ice}} = 2.03\ \mathrm{J}\,\mathrm{g}^{-1}\,^{\circ}\mathrm{C}^{-1}$$
$$\Delta H_{\mathrm{fusion, water}} = 6.01\ \mathrm{kJ}\,\mathrm{mol}^{-1}$$
O am just learning this material, and I'm struggling on what formula to use. Is it
$$q= n(H_{\mathrm{final}} - H_{\mathrm{initial}})$$

Several points: 1. Generally speaking, figuring out the formula is not the way you want to go because you want to understand the underlying concept. Then, figuring out which formula is free and you can use it correctly. 2. Please define the non-obvious variables in your equation, namely $n$.

On closer inspect, it's not even clean what $H$ is in that final equation.

Firstly let us understand that heat removed would be directly proportional to the decrease in temperature and also the mass. $$\delta Q\,\,\alpha \,\, m\delta T$$  Here the specific heat capacity appears as the constant of proportionality.
Therefore we have, $$\delta Q\,\,=\,\,mc\delta T$$
Now all you have to do is to use this formula to convert 8.49g of water at 24.5°C to water at 0°C.
But after this you need to use the concept of latent heat i.e. heat you need to remove to convert water at 0°C to ice at 0°C.As this value is given as 6.01 kJ/mol you will have to remove that much amount of heat for every mole of water(so you need to convert grams into moles).
Now what you have is 8.49 grams of ice at 0°C. Now just use the formula again to convert ice at 0°C to ice at -12°C (using the value of c of ice this time).
At the end you need to add all those heat energies and you are done !!
The sum of heat energies is the amount of energy you need remove from 8.49g water at 24.5°C to convert it to ice at -12°C.
Hope this helps.


text:
Why is the polar-nonpolar electronegativity difference defined to be 0.4?
In introductory college chemistry, bonds between elements that have an electronegativity difference $\Delta EN > 0.4$ are generally consider to be polar (and thus have a dipole moment), while $\Delta EN < 0.4$ bonds are considered to be nonpolar. I think more advanced chemistry classes generally don't adhere strictly to this threshold, but I'm still curious if this 0.4 threshold has any meaning (or if it's just arbitrary).
For example, if there's an experiment plotting $\Delta EN$ vs. the intermolecular force strength, and there's a big jump on the graph at 0.4, that would be convincing evidence. Does anyone have an explanation along this line of reasoning?
(My question could also apply to the $\Delta EN > 1.7$ threshold for ionic vs. polar covalent. I saw a question on this before here, but I think it's worth further discussion.)

Both the $0.4$ threshold for nonpolar/polar bonds and the $1.7$ threshold for polar/ionic bonds are arbitrary choices. When defining nonpolar, polar and ionic bonds, it is important to allow $\ce{C-H}$ bonds ($\Delta \chi = 0.4$) to be nonpolar — it is the best description of alkanes’ low polarities and it facilitates discussions so much. Likewise, a few compounds commonly known as salts should fall just within the ionic boundary for ease of discussion. Teaching introductory levels is just so much easier if you can draw things black and white.
At higher levels, it should always be remembered that not only the thresholds are arbitrary but even electronegativity as a concept follows quite arbitrary definitions.

Except for one electronegativity scale that links electronegativity to ionisation enthalpy and electron affinity, all definitions and all defined reference points (typically fluorine having $\chi = 4.0$) are arbitrary.
Choosing a definite value for a specific element is also arbitrary since the physical implication of electronegativity depends on:

bonding partners
structure (i.e. hybridisation)


text:
Does naphthalene in moth balls really sublimate or does it evaporate?
The triple point of naphthalene is $\pu{-56 °C}, \pu{1.05 kPa}$ (which is much less than $\pu{1 atm}$). So how can a naphthalene ball sublimate at $\pu{1 atm}$, when considering that its melting point at $\pu{1atm}$ is $\pu{78.2 °C}$ (and the boiling point is $\pu{218 °C}$)?

Naphthalene, as well as any other solid, may sublimate at any temperature and pressure. The triple point parameters are irrelevant. The pressure of other components (that is, air) is irrelevant as well. What is relevant is the vapor pressure of naphthalene itself in the surrounding air, and that is pretty close to $0$. Until it reaches the saturated vapor pressure (or until your mothball is all gone, whatever happens sooner), naphthalene will sublimate.
The solids which we perceive as incapable of sublimation (like steel, glass, or rocks) have very low vapor pressure, so their sublimation would take forever. Other than that, everything above applies to them as well.
Then again, some solids melt upon heating and some do not, but that's another story. We aren't talking about any heating, are we?

By definition, going from solid to gas directly is sublimation.  If moth balls were not solid, but liquid instead, then we'd call the process evaporation.

This doesn't really answer the question.

Actually it does...

In fact, it is my answer that answers the wrong question. But then again, how was I to know what the OP had in mind?

I really don't see how this is an answer.

I got my answer at wikipedia after a little through study, it goes like this - Evaporation that occurs directly from the solid phase below the melting point, as commonly observed with ice at or below freezing or moth crystals (napthalene or paradichlorobenzene), is called sublimation.
So it is actually evaporation but called sublimation
source:https://en.wikipedia.org/wiki/Evaporation


text:
What does "electrochemically active species" mean in the expression for the mass-transfer limited current?
The expression for the mass-transfer limited current $i$ is given by
$i = nFAmC$.
where $C~[\mathrm{mol\, cm^{-3}}$] represents the concentration of the electrochemically active species that gets converted at the electrode; if the entire analyte gets converted, then $C$ is equal to the initial bulk concentration of the analyte.
I am confused as to what electrochemically active species mean? Is that just saying the ions in solution? Or is it the oxidized ion and the reduced ion?

It is the species that can get oxidized or reduced in the system.
It is specifically written out in order to make sure that it doesn't get confused with supporting electrolyte or any other component in the system.
So, for a reduction reaction, it is the concentration of the oxidized species.

wouldn't it be the reduced species for the reduction reaction?

it is the concentration of the reactants. So in $\ce{O + e- -> R}$ which is a reduction, it is the concentration of O, which is the oxidized species.


text:
Does higher [OH] not entail higher pH?

A student has equal volumes of $\pu{1.0 mol dm-3}$ sodium hydroxide and ammonia solutions. Which statement about the solutions is correct?
A. Sodium hydroxide has a lower electrical conductivity than ammonia.
B. Sodium hydroxide has a higher hydrogen ion concentration than ammonia.
C. Sodium hydroxide has a higher pH than ammonia.
D. Sodium hydroxide has a higher hydroxide ion concentration than ammonia.

The answer is C, and I understand why. What I do not understand is why D isn't correct also? Could someone clarify this for me?

Share some reasoning why C is true and why you think D should be true.

At first glance I'm inclined to agree with you. Where is this question from?

D is also correct but as you know sodium hydroxide is strong base and ammonia solution is a weak base. In weak base solution we must know Kb to calculate hydroxide ion concentration. Although a solution of weak base has lower concentration of hydroxide ion compared to a solution of a strong base having same concentration.

See this website. Essentially it's a misprint in the IB Chem testbank.
https://www.thinkib.net/chemistry/blog/8944/multiple-choice-answers
There's a correct rewording for D that makes C the only correct answer, according to the link I just posted.

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text:
How to determine the temperature dependence of equilibrium constants?

Given $K_c=2.8\times10^2$ at $727~^\circ\mathrm{C}$ for
  $$\ce{ 2SO2 + O2 <=> 2SO3},$$
  calculate $K_c$ and $K_p$ when 8 grams of $\ce{SO2}$ result in 0.32 grams
  of $\ce{O2}$ forming at equilibrium (temperature is not $727~^\circ\mathrm{C}$).

Now at first I thought this was straightforward :

determine $K_c$ at this temperature from the equilibrium conditions
apply Van 't Hoff equation to determine the temperature
calculate $K_p$ from $K_c$ and the temperature

However right after that part of the question the next part asks

Is the new temperature higher or lower than $727~^\circ\mathrm{C}$?

This implies that it is possible to calculate $K_p$ without calculating the temperature? With the new $K$ values then provide a qualitative argument for the temperature increasing or decreasing without having to calculate it?

Are you certain there isn't a typo in transcribing the original question?

I agree that the second question seems a bit leading, but personally I would just calculate all quantities of interest as you laid on in your plan. The van't Hoff equation requires the reaction enthalpy change, so I assume using a tabulated value is fair game. If so, you will know whether the forward reaction is endo- or exothermic. If you calculate the equilibrium constant at the unknown temperature as you planned, you can compare it to the equilibrium constant given in the problem (for 727 C). Comparing these two equilibrium constants will tell you which way the reaction proceeded, and with the sign of the enthalpy, you should be able to deduce whether heat was added or lost... i.e. whether the new temperature is higher or lower than the initial state, 727 C (without having to go through the entire calculation).

Thanks, yeah that is exactly what I did. The second question just made me think I was overlooking something.


text:
extraction/isolation of Curcumin from Turmeric using ethanol
I want to extract/isolate Curcumin from Turmeric using ethanol solvent, already know paramaters such as optimum particulate size, optimum temp (140 degrees F) etc, I need to know how to safely set up a basic distillation to recover the ethanol for reuse.  I have basic lab glassware incl and heat plate/stirrer, condenser.
Thank you

First off, welcome.  What kind of scale are you talking about (i.e. mg, g, kg of Temeric?).  Could you list the glassware and other relevant lab equipment you have available (flasks and their volumes, vacuum flasks, a roto-vap, etc.).  Have you done this type of thing before, say in an organic chemistry lab or is this your first go?  Actually my last question is by far the most important.

In their open access article Optimization of the extraction of curcumin from Curcuma longa rhizomes, published in Revista Brasileira de Farmacognosia, 2013, 23, 94-100, the authors systematically examined the effect of various parameters on the extraction yield, namely

drug to solvent weight ratio (1/6 and 1/4 g/g)
agitation speed (30 and 70 rpm)
extraction time (12 and 24 h)
extraction temperature (50 and 80 °C)
ethanolic strength (70 and 90 % v/v)

Based on their analysis, the following optimum conditions were seemingly predicted:

drug to solvent weight ratio: 1/6 g/g
agitation speed: 30 rpm
extraction time: (12 and 24 h
extraction temperature: 80 °C
ethanolic strength: 70% v/v

Please have a look at the full article for further details.

Edit

I need to know how to safely set up a basic distillation to recover the ethanol for reuse.

That is a rather fundamental question which suggests that you don't have a lot of experience in organic benchwork.
You might want to have a look at the English wikipedia page on distillation for the general setup.
Note that you will not be able to fully separate ethanol and water by destillation. Instead, you will obtain an azeotrope, a mixture that has a boiling point lower than that of ethanol and water. In order to check the density of your destillate and to dilute again to an ethanolic strength suitable for extraction, a hydrometer might be a suitable tool.
In terms of lab safety, the following points should be considered:

Do not work alone!
Ethanol vapour is flammable!

Keep open flames and electric sparks away.
Run your still in a well-vented place.
Have a fire extinguisher at hand.
Have a work space with an unblocked emergency exit.

If you use standard lab glassware, a stirrer and an oil bath to heat the still pot, make sure that the tubes for the cooling water that cools your condenser are properly secured by wire or clamps. Remember that the heating bath is hot! The last thing that you want is cooling water splashing into the hot silicone oil.

On a lab scale, I would would not set up a classical distillation apparatus, but remove (and collect the ethanol) at reduced pressure using a rotary evaporator with a water bath!

In any case, use suitable eye protection!

I read the question as "How to safely recover the used ethanol?"


text:
How to dissolve cat hair without effecting fabric?
I am wondering if it is possible to create some sort of a solution that will dissolve cat hair (or dog hair), but, is not harmful to the surrounding fabric? I am picturing a solution you would soak a cat hair covered towel in that will dissolve the hair but leave the towel unharmed. Or, perhaps in a spray bottle that can be sprayed on dog hair covered car cushions, will dissolve the hair but be able to evaporate and not damage the cushions. Is this possible?

I'm afraid chemistry is of no help here. Hair are more durable than most kinds of fabric.

Probably a biological agent, e.g. an enzyme or protein mixture. That has a better chance of degrading or decomposing keratin than even strong acid etc.
But it would have to have a specific, targeted mechanism; I don't know of such a substance

I guess experienced pet owners choose their clothing and upholstery so as to match their cat's color (or the other way around).

No, it most likely isn't. Hair and general "fabric" - which might even be animal hair (wool) share too many properties for this to be possible.
If you specify further, there are possibly certain combinations of cat hair and specific fabrics that might have properties that allows for this - but the number of possible combinations is as you may imagine, high.
PS: The opposite is easier - acetone could remove most synthetic polymer fabrics while leaving the cat hair, but I suspect it would not help you.

To get pet hair off of couches or other upholstery, use a clean, dry kitchen sponge. Just run it slowly across the couch and the fur will cling to it


text:
Why are group 1 elements called alkali metals and group 2 elements are called alkaline earth metals?
I was studying the periodic table. My teacher said that

Group 1 elements are called alkali metals because their oxides are soluble in water and group 2 elements are called alkaline earth metals as their oxides form in the earth and are water soluble.

What did she mean by "oxides"? Aren't alkali metals bases which are soluble in water? Then what is the term "oxides" doing here?
What is the difference between "alkali" and "alkaline"? Why do we call one of them "alkali metals" and the other "alkaline earth metals" when the concept is the same?
Why are the bases of group 1 and group 2 elements soluble in water? What is the reason?

I really don't know where your teacher has this definition from, but the word "alkali" has Arabic origins and means ashes, which is where most of them initially were found. By oxides your teacher probably meant $$\ce{4M + O2 -> 2M2O; \hspace{2em} M2O + H2O -> 2MOH}.$$

alkalis are bases that dissolve in water so my first question is now answered. Thanks!! I just needed to know what she mean by oxides. If she mean pure compounds with oxygen or bases.

I think she probably meant what @Martin-マーチン said she probably meant ;)

The description of the reasoning of the naming is … let’s call it simplified.
First, concerning your first point. Alkali metals are not bases by themselves, they are metals. Metals typically cannot be classified in the original Arrhenius or extended Brønsted-Lowry acid/base classification. The cannot dissociate to liberate either $\ce{H+}$ or $\ce{OH-}$ because they only consist of metal atoms. And they cannot act as a donor or an acceptor for protons — one because of the lack of protons and two because of the lack of favourable lone pairs.
It is only when we oxidise these metals in some way or another that they form a species that can interact with water in the acid-base formalities. One possibility for oxidation is simple burning, i.e. reacting with atmospheric oxygen. Unfortunately, only lithium is nice enough to ‘do what you might expect’, hence why I’m using that as an example in the following equations:
$$\begin{align}\ce{4Li + O2 &-> 2 Li2O}\tag{1}\\[0.4em]
\ce{Li2O + H2O &-> 2 LiOH}\tag{2}\end{align}$$
As we can see, the oxide we generated, lithium oxide, can react with water to form a hydroxide which dissociates into $\ce{OH-}$ ions — an Arrhenius base. We can also say that the oxide anion is capable of accepting protons donated by water — a Brønsted-Lowry base. Therefore, lithium oxide (and also lithium hydroxide) are the bases while lithium metal is not.
The second question is easily answered. Alkali is a noun and alkaline is an adjective. Thus, alkali metals is a compound word formed by two nouns while alkaline earth metals is also a compound word formed by two nouns where the first noun is further specified by an adjective.
The third question is difficult to answer. There is not really an easy or understandable explanation; it all boils down to churning numbers and finding out that the dissolution is favourable, i.e. the corresponding equation represents a decrease in Gibbs free energy: $\Delta G < 0$.
We can break dissolution of solids into two general processes:

breakdown of the ionic lattice, i.e. liberation of individual ions from a solid
$$\ce{NaOH(s) -> Na+(s) + OH- (s)}\tag{3}$$
dissolution of the individual ions
$$\begin{align}\ce{Na+(s) &->[H2O] Na+(aq)}\\[0.3em]
\ce{OH-(s) &->[H2O] OH-(aq)}\end{align}\tag{4}$$

The first of these steps requires energy because you are breaking bonds (this is always unfavourable). It is typically represented by $\Delta_\text{latt} H^0$ — the standard lattice enthalpy.
The second process is exothermic because the individual ions are now again somewhat balanced by the polar water molecules. It is typically represented by $\Delta_\text{solv} H^0$.
Both processes are favourable from an entropic point of view.
Whether the entire reaction $(5)$ happens, depends on the difference between lattice and solvation enthalpies. If the lattice enthalpy is too high to be offset by solvation enthalpies and the entropy effects, an ionic compound will not dissolve. For alkali metal oxides and alkaline earth metal oxides, the lattice enthalpy is typically low and thus they dissolve.
$$\ce{NaOH (s) -> Na+ (aq) + OH- (aq)}\tag{$3+4=5$}$$
The actual reason for their names is in another castle beyond the scope of this answer. At first glance, I would forward you to the corresponding Wikipedia articles.

amazing answer!! But I need to get some things straight. First, check if my understanding of question 3) is correct-

so firstly, ionic compounds can either be broken down when they are in a lattice form by water (which is endothermic as the ions are already being attracted by other ions) and 2nd when water breaks down free floating ionic compounds(which is exothermic as the bond energy in them is released). Now if the breaking of lattice is less endothermic than the breaking of single molecules is exothermic then there would be enough energy for reaction with water and thus the base will be soluble. Otherwise the base wont be soluble as there simply isn't enough energy(heat) there. Am I right?

now, if the understanding is correct, the question about 3) is, how can you even find free floating ions when ions always stay in lattices?

also about 1) I still don't understand..a metal oxide is not a metal base. Then how can we call it an alkali?

@MartianCactus On 1): What is a ‘metal base’ in the first place? The term is not commonly defined. ‘Metal oxide’ is.

@MartianCactus On 3): The dissolution of ions from an ionic lattice into an aquaeous solution is, in fact, a one-step process, but we cannot easily analyse it. This is why we break it into two formal steps, namely breaking the ionic bonds of the lattice (1) and then dissolving the formal free ion in water (2). In reality, these two processes happen simultaneously. When determining whether a transition (here: solid → dissolved) will happen, we need to consider the overall $\Delta G$ change which is influenced by both. The lattice breaking will always be unfavourable (otherwise we wouldn’t …

have a lattice in the first place). The dissolution is typically favourable. But only if the dissolution is more favourable than breaking the lattice, an ionic compound will dissolve.

ok so, we first break the lattice and then break the single free floating units which came out of the lattice. If the total reaction is exothermic then the base will dissolve right?

still dont get it...yes there is no such thing as a metal base, but at the same time you cant call a  "metal oxide" an "alkali" either can you? A metal oxide is not a base in the first place..

@MartianCactus A metal oxide can act as a base, see equation $(2)$. And I think you got the dissolution part.

but it can only act as a base AFTER it has reacted with water, right? Then how can we call it a base before that?

@MartianCactus The acid/base definitions state that something an acid/a base if it reacts in this way without requiring it have already reacted in this way. So it will act as a base upon the reaction with water and thus is classified as a base before.

then can't we call pure Li a base? Becuase it too reacts with oxygen, then wuth water and then becomes base.

@MartianCactus No. We can only call things bases that react like a base with water. In elemental lithium’s case, we first need to react it with something else. (Although lithium can also react with water in a redox reaction liberating hydrogen gas, first, to also give the base lithium hydroxide.)

"liberating hydrogen gas, first, to also give the base lithium hydroxide" meaning?

@MartianCactus $$\ce{2Li (s) + 2H2O (l) -> 2 LiOH (aq) + H2 ^ (g)}\tag{6}$$

so we can still call Li a base right? Because it directly reacts with water to make LiOH

@MartianCactus Unfortunately no. That reaction is a redox reaction, not an acid-base reaction. It would have to be an acid-base one to call lithium a base.

Let us continue this discussion in chat.

Alkali metals when come in contact with water form alkali (base) and form metal oxides when reacts with water. Alkali metal oxides are also bases and form alkaline solution when dissolved in  water. Group 2 metals also form bases during reaction with water and their oxides are  basic and found in earth crust. Be in group 2 is not known as alkaline earth metal because its oxide is amphoteric and are rare.

IUPAC includes Be as an alkaline earth metal. It also includes Sr and Ba, which are certainly "alkaline" but not much "earth" because their oxides/hydroxide are fairly soluble in water.

Group one elements form basic (alkaline) solution when react with water or we can say the oxides of group first elements are basic in nature, as they liberate hydroxide ion.
The second group elements called alkaline earth metals. Earth* because of the reason that their oxides found in earth crust as Be oxides are rare and found in earth crust, and alkaline* because their oxides are basic in nature and form basic solution in water.

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


text:
Acidity of primary and secondary amines
I was reading a book on organic chemistry by David Klein and I came across this question, but I wasn't able to understand why one amine is more acidic than the other.

Have you taken into account that there is a $\ce{-CH3}$ group attached to the top N which has +I effect i.e., electron donating effect?
$\ce{-CH3}$ group will destabilize the negative charge on N formed after the release of proton.
Electron releasing group decreases acidic strength of a compound. So in this question proton of primary amine is more acidic and that of secondary amine is less acidic.

I suppose it is too much of a hint to fully draw out the -CH3 group. I guess the intent of the question is to see who takes the time to visually analyze and digest the question. Or else the instructor has not made it clear that a line is shorthand for -CH3.


text:
How exactly does a battery work (specifically Pb/H2SO4/H2SO4/PbO2 lead acid battery)?
Specifically talking about the lead-acid battery, I understand (or think I understand) this much:
The anode, in the presence of the electrolyte, $\ce{H2SO4}$ will be oxidised:
$$\ce{Pb (s) + SO4^2- (aq) -> PbSO4 (s) + 2e-}$$
At the same time at the $\ce{PbO2}$ electrode, two electrons from the $\ce{PbO2}$ solid are lost to the electrolyte:
$$\ce{PbO2 (s) + 2e- + 4H+ (aq) + SO4^2-(aq) -> PbSO4 (s) + 2H2O}$$
This all occurs without a solid conductive path (say a wire) between the anode and cathode, the cathode will build a positive charge while the anode builds a negative charge and ions will diffuse appropriately during this process.
When a wire is connected between the anode and cathode, electrons that have built up on the surface of the $\ce{Pb}$ will migrate to the $\ce{PbO2}$ which is of lower potential and if a load is placed along the way work can be done.
So batteries will discharge by themselves without even contributing to any work? Is this not true for all batteries?
The part that confuses me is that most people say that batteries store chemical energy.
I understand that the redox reactions release energy which can be used as electrical energy but if batteries discharge without contributing to any meaningful work then why are they still considered to be storing energy? Aren't they converting chemical energy to heat continually unless they are put to use?

Your question could really benefit from some formatting. Yes, the batteries are storing chemical energy. Whether it would be used meaningfully or just wasted into heat is up to you; the energy conservation law does not care about that.

Thanks! I've rarely come a source talking about self-discharging in batteries so I was confused about the how batteries store energy thing.

What's confusing about that? Yes, they store energy. And yes, they do that in a non-ideal manner, much like a leaking fuel tank.

@shA3245699 Batteries do store energy just not forever. Self discharge is usually slow (batteries you buy often have lifetimes of years). But it does represent an efficiency loss. What matters in practice is how big and how fast that loss occurs. If it is slow enough it doesn't stop the battery being useful.

In order for oxidation-reduction chemical reactions to occur, the electrons must somehow go from the anode electrode where the oxidizing (electron donor) takes place to the cathode electrode where the reducing (electron acceptance) occurs.
Ideally, the electrolyte, in this case sulfuric acid, should allow the movement of positively charged ions but prevent the movement of electrons through it.  This forces the electrons to go through a circuit, external to the cell, to go from anode to cathode thus enabling the redox to occur.  Without electron flow, said redox reaction will not occur.
In the real world, there will be some leakage current through the electrolyte and other parts of battery.  We can eliminate the leakage with a more insulating electrolyte, but this would hurt the charging process and make the battery less rechargeable.  So, what we have is an economical compromise.


text:
how do i synthesize (E) 2-methyl-1-phenylpent-1-en-3-one?
i need a way to synthesize the (E) 2-methyl-1-phenylpent-1-en-3-one
it's 2 days and still i can't find a solution.

Welcome to chemistry.SE! Feel free to take a [tour] of the site, and see [ask] for details about asking questions. What are your thoughts on this problem? Did you consider using benzaldehyde and pentan-3-one? What have you found about the E/Z stereoselectivity?

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

Looks like a crossed aldol condensation between benzaldehyde and pentan-3-one.
In a classical approach, you would (under stirring)

Add your base to a dry solvent of your choice
Cool down (ice bath or dry ice/isopropanol)
Add pentan-3-one dropwise to make sure that it's all immediately converted to the corresponding enolate. You do not want self-condensation of the ketone!
Let it come to room temperature (and stir for a while)
Cool down again
Slowly add benzaldehyde. Freshly distilled benzaldehyde is recommended.
Let it come to room temperature again (and stir for a another while)

Check the progress of the reaction by TLC. You should have at least the aldol addition product by now. In order to facilitate elimination of water and thus complete the aldol condensation, it might be necessary to acidify the mixture and heat it.
There are other more modern (e.g. Mukayama aldol reaction, using the silyl enol ether of the ketone) and/or "greener" methods, but they are mostly a variation of the basic approach described above.


text:
Hydrophobicity of Polysaccharides
Are polysaccharides generally considered to be hydrophobic and/or hydrophilic? My reasoning is that although they have polar groups/electronegative atoms, the polymeric nature of polysaccharides excludes water so I'm leaning towards them being hydrophobic.

They are typically full of hydroxy functions and thus hydrophilic. However, that does not mean ‘soluble’ automatically.

What do you mean by 'the polymeric nature of polysaccharides excludes water'?

Also the OP has shown some thought here and it's a reasonably written question so I don't think this is homework.

Are polysaccharides generally considered to be hydrophobic and/or hydrophilic?

They are generally considered to be hydrophilic unless they are highly modified (e.g. to siloxanes).

My reasoning is that although they have polar groups/electronegative atoms, the polymeric nature of polysaccharides excludes water so I'm leaning towards them being hydrophobic.

There is no rule that polymers exclude water. Linear polymers can either stretch out in water (when the interactions with water are favorable) or clump up (if the interactions with itself or another polymer is favorable.
One notable exception is cellulose. The straight chains of cellulose make perfect interactions (involving both hydrogen bonding and dispersion forces), yielding a non-water soluble fiber. Cotton is made of cellulose fibers. The cotton fibers are highly hydrophilic (they are not water repellent) because the surface of the fiber is studded with hydroxyl groups.
Perhaps surprisingly, methyl cellulose (where all hydroxyl group are modified into methyl ether groups) is moderately soluble in cold water. The methyl groups prevent the packing of straight chains to cellulose fibers, and the polymer contains sufficiently hydrophilic groups to be water soluble (similar to polyethylene glycol, which also is an ether-containing polymer).

Maybe worth mentioning that water soluble starch can be dried to the point that the strands hydrogen bond so tightly to each other that water is excluded and it is extremely difficult to rehydrate and dissolve.


text:
Would nodal planes of participating Atomic orbitals be nodal planes of molecular orbital?
As I take sigma 2p antibonding molecular orbital as an example

Would it have 3 or 1 nodal planes. The two supposed nodal planes pass through the centres of the two atoms and are same as in the original atomic p orbitals. By LCAO A and B, I suspect that there will be some electron density due to the wave function of B where A is zero. Is there a fault in my logic?
For pi there is no problem, there are 1 nodal planes for pi bonding mo and 3 for antibonding mo as there is a common nodal plane to both MO.

just so I surely have it down, dyz dyz Pi bond bmo has 1 and abmo has 2 nodal planes? with z axis as internuclear axis.

No, should be more than that. Draw them with blue and red regions for different signs of $\psi$, then you'll see.

http://m.imgur.com/xNFcJKf taking internuclear axis as z

You are right in that node of one AO is not a node of other AO, and hence not a node of the MO (that is, unless we're talking about π bonding between two p-orbitals). In effect, there still would be three nodal planes, but two of them would not quite match those of individual AOs, and might even be not quite planar.
The same applies to any other type of orbitals.
Look at these crude approximations of $\pi$ orbitals formed by two d-type AOs.
Bonding:

Antibonding:

See those vertical nodal "planes" which are not quite planar?
That's why, BTW, I'd rather not ask about nodal planes. By asking that, we get bogged down in a pointless and un-chemical quibbles concerning the definition of a plane. Ask about nodal surfaces, and you'll get a meaningful answer.

Sorry, I accidentally deleted a comment. "Just as cones are not considered nodal planes, these won't either right?"

I started writing a comment, then noticed that your comment is gone, so I just added mine into my answer as the last paragraph.

Just one last example, for delta bond with z as internuclear axis for dxy, dxy is the following correct? (3rd for abmo is z=0 plane) http://m.imgur.com/2wVkIDt

Yeah, looks about right. Then again, you might want to put (+) and (-) all over your drawings to denote the signs of $\psi$.

I forgot to ask, by extension these bendy nodal planes are present jn sigma p too right? so, sigma p bmo has 2 nodal SURFACES and sigma p abmo has 3 with one of them being the central plane.

Yes, should be that way.

AFAIK: The nodal planes are where the wavefunction sign inverses. If you drew the colors according to the sign value then you would see that for antibonding sigma MO there is a virtual nodal plane, since the sign changes between + and - between the two centre orbital lodes. Even if they by some magic grew larger, they can never touch, they interfere destructively. There will always be a plane between them where the wavefunction returns 0. (when the two spheres meet they will form a planar interface. Imagine soap bubbles.)
I found this, it is more descriptive; https://socratic.org/questions/how-can-i-draw-antibonding-orbitals

That is the point I was making, they don't equally get larger, the wave function of B is negative everywhere near A then, Negative part of A grows more than positive side, resulting in nonzero wave function where the nodal plane of A used to be.

But then you are not considering the molecular orbital, you are considering two atomic orbitals. The molecular orbital is combined for both A and B, there is no "B part of orbital". Consider the sum of both A and B, if you will.

I was following the LCAO model. I think Ivan has answered all of the queries.

You were right, but the nodal plane is not the same as that of an participating atom. I forgot that film between the soap bubbles is acutally curved to account for the excess pressure in case of different sized bubbles. :)


text:
Why does the hydrogen atom in HCl (when dissociated in water) bond to the oxygen rather than staying with the chlorine atom?
Why does the hydrogen atom in $\ce{HCl}$ (when dissociated in water) bond to the oxygen rather than staying with the chlorine atom. They ($\ce{Cl}$ and $\ce{O}$) have roughly the same electronegativity (3.5).
I saw this question:
Dissociation of HCl in aqueous solution
but it doesn't answer what I'm asking.

Look at it from the point of view of the chloride ion. In $\ce{HCl}$, the chloride is attracted to only one hydrogen. But when dissolved in water, the chloride is attracted to the partial positive charges on the hydrogens of several water molecules. Each $\ce{H2O-Cl}$ attraction is less than the $\ce{H-Cl}$ attraction but many of them act together to attract the $\ce{Cl}$ ion away from $\ce{HCl}$. Once the $\ce{Cl}$ is attracted away, that leaves a hydrogen ion to be attracted to the partially negative oxygen of a water molecule.
These forces between multiple waters and an $\ce{HCl}$ are stronger than forces between adjacent $\ce{HCl}$ molecules. Otherwise, $\ce{HCl}$ would be a liquid at room temperature instead of a gas.

thanks, your answers have helped me. but i have forgotten to add a second half to the question:    ammonia NH3 is acting with water as base. it draws H atom from the H2O molecule and lives it with OH.    why does the weak (lower electronegativity) nitrogen atom can draw electron from the OXYGEN (higher electronegativity). what's more, that the nitrogen allready have 3 hidrogens and yet "want" to have another one, while the oxygen gives away hydrogen atom that is naturally bonded to him (oxygen have 2 valence electrons that are able to bond to H). please help me digest it  :-)  thanks.

Mr Ytfu Gjuf, please read my answer and stop using electronegativity to answer anything again. My answer also applies to your 2nd question exactly the same way

I still think your education level is important because some details might be lost, but I'm just gonna take a guess and say you are familiar with some concepts but don't have a really solid background. You can do this with some equations, I'm not gonna. So here goes nothing:
First, a mistake in your question: You try to argue with electronegativity.That thing can tell you something about how polar a bond is, not so much about how strong it is. Take for example the Helium dimer (very weak bond) and the Nitrogen molecule (very strong bond). Both have the same electronegativity difference, 0. Your quesiton is of a thermodynamic nature, that means we are interested in energy differences.
Solutions are weird as in we do not really understand them. Especially water. The molecules "know" each other and when something new is introduced, they can tell and react accordingly. So now you introduce an HCl molecule into a solution. The question now is: What is the difference in the Gibbs-Energy (sometimes called free energy) of the solvated HCl molecule compared to the Gibbs-Energy of the solvated proton and the solvated Chlorine ion?
One can say in first approximation that in water, Chlorine anions and protons are stabilized by a solvation shell. It is favorable if the two ions go their separate ways and let water do its thing with them rather than stay together.
Some of the things I wrote are not completely correct. But I hope this is a good trade off. There are a number of possible answers, one can look at it in detail. I think this is enough for now.

if you could expand more. how the free energy takes role here, and why   "the two ions go their separate ways" and how it expalains the ammonia-water reaction.   i am autodidact learning now chemistry in the level of highschool (although much older. have experience in other sciences, not chemistry) thanks

An acid-base reaction is just another chemical reaction. The Gibbs energy of a reaction tells you if it is exergonic. If you are not familiar with those words, they are the topic of another discussion I believe. You can find a good introduction to these concepts in a lot of Chemistry textbooks. A reaction should be exergonic if you want it to happen, that means the Gibbs energy has to be negative. What happens in a water solution is that a solvation shell forms around any molecule introduced, but especially around ions and even more so around protons or hydroxide ions.

Not enough space. You should maybe google what happens when you put a proton for example into water. You sometimes hear things like "H3O+ forms". This is not correct. In fact a giant cluster out of many water molecules forms. You have to take all these things into account and then decide if it's worth dissociating for the two ions or if it isn't. As I implicated, this is a complicated topic that can't be explained by a simple concept like electronegativity.

I’m going to answer to the posted question without reading all of the commentary.
First, the answer should NOT be given in terms of electronegativity. I cannot go into detail here, but the failure of electronegativity theory to predict or explain that HI is a strong acid must not be ignored.
Acidity and basicity are properties of the electron pairs of an atom. We know that HF is tetrahedral so we know a proton is attached to an electron pair in HF. We should infer the electron pairs of a fluoride are also tetrahedral. That means we should judge the basicity of fluoride on the basis of each of its valence electrons and not on the basis of its net charge. No change has occurred to the charge of an electron pair.
If dissolving HCl in water gives hydronium ion and chloride ion, this should tell us the electron pairs of oxygen are more basic than an electron pair of a chloride ion. Again, these are properties of the electron pairs and not a property of the ions per se.
The above are consistent with the long bonds of NaCl and the ease at which the ions separate and easily dissolve in water. A theory of ionic attraction suggests they should form strong bonds. Long bonds belie a strong bond. Na(+) and Cl(-) each have completed octets and do not need to be attracted to the electrons of each other. That is different that the much shorter and stronger bonds of CH3Cl in which an electron pair must be shared. It is also consistent with the properties of sodium metal. Metallic sodium has one additional electron beyond its octet. In solution, it likes to donate this electron. Try adding sodium metal to water and see if this reaction is exothermic and spontaneous (everyone knows the answer).
This should tell us things we already know. If we remove a proton from the nucleus of oxygen, will the valence electrons become more or less basic? This should predict nitrogen and ammonia should be more basic than oxygen and water. We should anticipate HCl will form ammonium chloride with ammonia. The attraction of the ions should be low, each have completed octets. The electron pairs of chloride are not very basic which is why HCl is a strong acid. Water is not as acidic, so a mixture of ammonia in water may form some ammonium hydroxide, most does not. It will be hydrogen bonded, but the protons will mainly stay on the oxygen.
On electronegativity theory, I have talked about it at American Chemical Society meetings. Unfortunately, it is not good science although repeated in virtually every chemistry textbook. Acidity is virtually the gold standard to measure ionization. The failure of electronegativity theory to match acidity should have been a clue as to its validity. Electronegativity theory is a quasi scale of reactivity. Fluorine is very reactive and more reactive than chlorine, etc.
I have not updated this in some time, but search for electronegativity and curvedarrowpress and you can see slides of an old talk there. I am writing a manuscript in which I have a fuller explanation of the data Pauling used to advance electronegativity theory. That will explain why HI is a strong acid and why iodide is a good leaving group. The explanations are much like my discussion above.


text:
Chaotic melting points of n-alkyl carboxylic acids

Is there any trend here at all? This seems very chaotic as a trend.

This paper may help.

It has the common zig-zag practically all properties of alkylic homologues show, and there is a minimum at 5. Looks very regular to me. No.5 cannot decide if it's rather polar or nonpolar, and gets so confused it does not crystallise.

More to the point the odd number carbons have one trend and the even another.

My guess without looking at any literature:
Higher acidity $\implies$ higher hydrogen bonding $\implies$ higher melting point.
Thus, changing from H and Me where there is virtually no +I effect from the chain, to Et, Pr, Bu where the +I effect of the chain decreases the melting point, as the hydrogen bonds get weaker, as the COOH groups are less acidic.
Further increasing the chain length after hexyl does not really lead to any additional +I effects on the COOH group that is now far away. Thus, the melting point increases as there are additional van der Waals interactions between the longer chains.
Then there seems to be a factor that makes crystals of odd chains more favorable.

So there are basicall two opposing trends: longer chains decrease the stability of the coo- anion but then after a certain chain length each methylene group adds additional vdw interactions.

Please, use proper punctuation, spacing, and capitalization, and do not use short forms unless in chemical formulas.

@DHMO Sorry, was kind of in a hurry ; )

I'm afraid only last 2 sentences are OK to have them in answer to this question.

@Mithoron Why exactly?

@Mithoron Actually, that's the common behaviour: H,Me -> No +I, Et,Pr,Bu -> +I gets stronger, Pe,Hex,Hept,... -> +I does not increase significantly, but instead vdw interactions. Obviously, this also affects the melting point.

An obvious rule is that for even $n+1$ the melting point of the acid $\ce{C_{n+1}}$ is higher than the melting point of acid $\ce{C_n}$.

I don't find it obvious. Can you explain why?

The zigzag line in the above image shows that the melting point increases from 1 to 2, from 3 to 4, from 5 to 6, and so on. I'm quite sure that there's no exception from this rule for n-alkyl carboxylic acids.


text:
Which d orbitals can form sigma, pi, delta bonds?
If two $\mathrm d_{xy}$ orbitals approach each other on $x=y,$ $z=0$, would a σ bond be formed? I would think so.
Can $\mathrm d_{z^2}$ form π bond with another $\mathrm d_{z^2}?$ (As all others can on $x$ or $y$ or $z$ as internuclear axis?) I am not sure but overlapping similar to p could be observed.
I would assume all except $\mathrm d_{z^2} - \mathrm d_{z^2}$ form δ bond. (both of same type like $\mathrm d_{xy}$ with $\mathrm d_{xy}.)$
What are the faults in my logic?

Atomic orbitals forming molecular orbitals are a way of thinking about chemical bonding. Neither atomic orbitals nor molecular orbitals really exist so they cannot be observed either. Do you have an example? One can then maybe tell you what "kinds of bonds" the d-Orbitals could form there. The concepts you are referring to have been given up by most people a long time ago and are mainly used today as a tool to teach people wrong things. They can, however, be relevant under very specific circumstances, hence an example would be great.

sigma bond is overlapping on the internuclear axis and pi is perpendicular to it. LCAO, doesn't have mathematical problems to me, I am a novice though. I would just ask what stops from a sigma or pi bond to be formed, if two lobes of same phase overlap, theoretically, some sort of bonding molecular orbital would be formed right and with it a abmo? unless there is +,+ and +,- interference equally, then it would be non bonding.

Sigma, pi and delta denote how many planar nodes are in the bond. Sigma bonds have no node, pi bonds have one and delta bonds have two. You can tell what kind of bond forms by how the orbitals overlap. Two single lobes form a sigma bond, two pairs of lobes form a pi bond and two quartets form a delta bond.

https://goldbook.iupac.org/S05434.html I was wrong

The image makes it seem as if d orbitals were only capable of forming delta bonds which is definitely not the case (they can form pi and sigma bonds too, of course). Please be more explicit.

@Jan Do you know a source which discusses this topic in greater detail? Thanks a lot! And yes I think this is a better representation of different bonding situations: commons.wikimedia.org

@samjoe I am one of the worst people you could ask for sources because all my sources boil down to my lecture notes ^^'

σ bond: $\mathrm s - \mathrm s;$ $\mathrm s - \mathrm p_x;$ $\mathrm s - \mathrm d_{x^2 - y^2};$ $\mathrm s - \mathrm d_{x^2};$ $\mathrm p_x - \mathrm p_x;$ $\mathrm p_x - \mathrm d_{x^2 - y^2};$ $\mathrm p_x - \mathrm d_{z^2};$ $\mathrm d_{x^2 - y^2} - \mathrm d_{x^2 - y^2}.$
π bond: $\mathrm p_y - \mathrm p_y;$ $\mathrm p_z - \mathrm p_z;$ $\mathrm p_y - \mathrm d_{xy};$ $\mathrm p_z - \mathrm d_{xz};$ $\mathrm d_{xy} - \mathrm d_{xy};$ $\mathrm d_{xz} - \mathrm d_{xz}.$
δ bond: $\mathrm d_{yz} - \mathrm d_{yz}.$
$x$ is between atoms line. Other configurations are impossible. So there is only one δ bond configuration (if not count $\mathrm f$ orbitals). There also exists hypothetical φ bond.
P.S. This https://commons.wikimedia.org/wiki/File:Quintuple_bond_orbital_diagram2.png is absolutely incorrect, even for the fact that two different form of orbitals are called the same ($d_{z^2}$) and those π bonds (both) are impossible because they have opposite signs.
https://pubs.acs.org/doi/pdf/10.1021/ja075454b and https://pubs.rsc.org/en/content/articlepdf/2017/ra/c7ra07400j provides some much more complex math behind that, the last also managed to debunk existence of φ bond.

Can you back up your statement that there is only one kind of d orbital overlap that gives rise to delta bonds? Checking actual published literature on higher-order bonding and delta bonds (for example) always shows MO diagrams with two δ-symmetry orbitals, even in systems without available f orbitals.

It seems you went so far as to even edit the Wikimedia Commons page containing the diagram to say it's incorrect, which is uncalled for without adequate evidence.

Yep. I did. Anyway, the picture on Wikimedia is not used anywhere on wikipedia or even on Wikimedia itself. So it does not matter. Anyway, "even for the fact that names of orbitals are not correct" is still valid: two different (!) orbitals are called the same! I think I would have proposed speedy deletion if not for the fact that if z is between atoms line, it looks like there are two delta bond configurations. But if it is x there is only 1.

Interesting. Surely, the $z$-axis is the more conventional choice for the internuclear axis, in which case you have the well-understood MO diagram where $d_{z^2} \to \sigma$, ${d_{xz}, d_{yz}} \to \pi$, and ${d_{xy}, d_{x^2-y^2}} \to \delta$. I don't understand why this should be different if $x$ is instead the internuclear axis? You could easily take linear combinations of the conventional d-orbitals, such that instead of $d_{z^2}, d_{xy}, d_{xz}, d_{yz}, d_{x^2-y^2}$ you had $d_{x^2}, d_{yz}, d_{yx}, d_{zx}, d_{y^2-z^2}$ and then the same mappings would apply.

Unless you mean to say that the number and types of bonds formed depends on which lienar combination of d-orbitals we take -- in which case it would seem like the entire notion of types of bonds is not actually physically meaningful -- is that what you mean? I apologise, but I don't have the time to read the papers properly, not even the one by Hoffmann. It would also be good if you could include a brief explanation of them in the answer. I very much agree that that figure on Wikimedia is incorrect, though. :-/

authors: Валерий Заподовников


text:
How to make Poinsettia pigment into ink?
Like many families, this winter we had bought a red poinsettia plant to decorate our kitchen, and now that it's losing its leaves, I had the genius idea to extract the red pigment. The activity proved a success, and let my 5 yr baby girl amazed (and us too). The pigment was extracted by letting leaves soak in 96% alcohol overnight in a closed container in a hot spot, then filtering the solids and heating the solution until alcohol evaporated for the most part, also adding a few drops of lemon juice to increase acidity and transform the brownish liquid into a blood red tint the that now has the aspect of an ink. Unfortunately, we were unable to use this ink to basically anything, as most mediums like regular paper or white paint are neutral or high pH and turn the poinsettia ink from red to any other color, like blue or green.
As you might have gathered by now, I am not a chemist, and I have no knowledge in the matter beyond high school classes from the old days. I did understand thou after reading this article that changes composition and color with pH.
So what I ask for, if at all possible, is a method described in plain terms and using household accessible chemicals to stabilize this red ink in order to use it, for instance, to paint on paper without losing its intense red color. (at least for a few hours or days)

Soak the paper in acid beforehand, then let it dry.

Welcome to Chemistry.SE! What a lovely idea! In addition to Ivan Neretin's suggestion, you could try to carefully paint over the writing/drawing with (diluted) vinegar. An artistic introduction to anthocyanin inks might give some further directions.

Thank you all for your answers, soaking the paper in lemon juice or vinegar beforehand did cross my mind, also just skipping the reactive mediums altogether and directly draw over a thin plastic like a transparency. Anyhow, I'll look over all the resources and post the method that produced the best results.

I think that would be hard, you could try to use a mordant to convert a dye to a pigment, you can use Alumina $\ce{FeSO4}$  or $\ce{CuSO4}$. There is an article that explains it in detail how to use Poinsettia for colouring silk fabric.

Rawat, Bhawana, et al. "Color fastness properties of silk fabric dyed
  with poinsettia leaves." MAN MADE TEXTILES IN INDIA 48.11 (2005): 418.

You can try to soak a thick paper in a solution of 1-5% of $\ce{FeSO4}$  with your dye heating gently.
Another solution could involve a different medium, like arabic gum mixed with citric acid.

Thank you for the link and the article, dying some wool might came in handy, I already envision granny making some white and red gloves with a poinsettia pattern in it's natural tone.


text:
Is there a substance that is hard (ideally abrasive) in solid form but would dissolve in water?
doing some research for a dishwasher / dish cleaning product and interested in any compounds that could be fashioned into a brush or other abrasive form and then dissolve and wash away with the water cycle?

Would simple granulated salt work? Toxicity sounds like an issue for some stuff.

A usual cleaning product that does not pose any health hazard and that is easily washed away is sodium hydrogenocarbonate (aka baking soda). I use it myself, and it works a charm in the kitchen!


text:
Why does being "bonded" decrease atomic size?
I have a furiously overwhelming question-

Why is the atomic size of noble gases larger than the atomic size of
the previous halogen when atomic sizes decrease across a period?

I did search it on the net and asked my teacher but everyone says that because noble gases are usually not bonded.
But:
Why does being "bonded" decrease atomic size?
Also, we are talking about the atomic sizes of isolated elements?

What data are you using?  According to this Wikipedia article the calculated atomic radius is smaller for a noble gas than it is for the preceding halogen, just as you would expect.

sorry I mean the "previoius halogen", I edited the question. :)

See

As Paul has pointed out, there are many different ways of defining atomic radii.
ron has also pointed out that the trend goes in the opposite direction from what you said.
For instance, according to Ptable,
|Calculated radii |
|  |r(pm)|  |r(pm)|
|H | 53  |He| 31  |
|F | 42  |Ne| 38  |
|Cl| 79  |Ar| 71  |
|Br| 94  |Kr| 88  |

I have added hydrogen here as the natural "halogen" for helium.
Empirical and covalent also follow the same trend as above.
But the van der Waals radii, as shown below, do not,
|    vdW radii    |
|  |r(pm)|  |r(pm)|
|H | 120 |He| 140 |
|F | 147 |Ne| 154 |
|Cl| 175 |Ar| 188 |
|Br| 185 |Kr| 202 |

I am assuming this is the data set you've mentioned.
The reason for the disagreement is that electron clouds may relax or contract depending on their chemical environments and so do atomic radii.
Thus, what this shows is that:

H, F, Cl and Br are larger than He, Ne, Ar and Kr when bonded.
H, F, Cl and Br are smaller than He, Ne, Ar and Kr when not bonded;

The reason for the above is the following:

For covalent radii (source),

Atomic size gradually decreases from left to right across a period of elements.
  This is because, within a period or family of elements, all electrons are added to the same shell.
  However, at the same time, protons are being added to the nucleus, making it more positively charged.
  The effect of increasing proton number is greater than that of the increasing electron number; therefore, there is a greater nuclear attraction.
  This means that the nucleus attracts the electrons more strongly, pulling the atom's shell closer to the nucleus.
  The valence electrons are held closer towards the nucleus of the atom.
  As a result, the atomic radius decreases.

What is being described is also called shielding effect.
For van der Waals radii (source),

[While] van der Waals radius is used to define half of the distance between the closest approach of two non-bonded atoms of a given element.
[...]
[Based on the ideal gas law above], van der Waals equation takes the molecular size and the molecular force into account. As a result, the attractive force (a/V2) is added into the pressure part. Likewise, the volume is subtracted by the molecular volume (b), which is determined by the van der Waals radius.

Here the shielding effect is much less important.
van der Waals radii are actually used to express how much volume each atom occupies on its own.
It is a minimum interatomic contact distance, less than that the atoms are somehow bonded.
And here the most important effect is plain electrostatic repulsion between atoms.
Thus it seems reasonable to me that it is harder to put two Kr atoms close together than two Br atoms, simply because Kr has more electrons and the electrostatic repulsion will be stronger between them.

and what do you mean by "those atoms"? If you mean He, Ne, Ar and K then thats my question. Why are they smaller when those atoms are not bonded?

I edited the original post to clarify it.

Please include definitions of the distances in the tables. For one thing the links may die, for another it is central to the discussion and it is not obvious from what is currently written in the answer.

The first thing you should realize is that there are various definitions of the size of the atom. One distinguishes for example the covalent radius, the Van der Waals radius and the ionic radius. The covalent radius is based on the binding of atoms into molecules and on the resulting bond length. The van der Waals radius is half the minimum distance between two atoms that not form a covalent bond and the ionic radius is derived from crystal structures of ions in salts. The covalent radius cannot be determined for noble gasses as they do not bind under normal conditions. The Van der Waals radius however can be determined and since the noble gasses are not very polarizable, the Van der Waals radius does increase with respect to the preceding halogen. You'll find more in the wikipedia article as mentioned by ron.

I dont really understand what the van der walls force is...why would you take the half of the distance between 2 non bonded elements?

van der Waals forces are actually attractive forces due to dispersion of electron clouds, but they do not play a role in this discussion.
The point here is one of the assumptions of the van der Waals gas equation: atoms occupy space and two of them can't occupy the same volume at the same time; so if you model them as perfect spheres, the closest they can get is actually just twice the radius of that sphere. If both atoms are of the same element, this radius is the van der Waals radius for this element.


text:
Why is sodium cyanide an inorganic compound?
I have been told that sodium cyanide is an inorganic compound during a lab experiment but I don't know the reason for this.
Shouldn't sodium cyanide be an organic compound because it has a carbon atom bonded to nitrogen?
The Wikipedia article  on cyanide also says that it is an inorganic compound but I fail to find why it is so.
How do we differentiate between an organic compound and an inorganic compound?

We don't. It is a historical concept.

Related: http://chemistry.stackexchange.com/questions/22195

Related: http://chemistry.stackexchange.com/questions/6690/what-is-the-definition-of-organic-compounds

So many questions about inorganic/organic division. Nature is impartible, and one shouldn't expect strict borderlines between artificially divided sub fields of chemistry.

Sodium cyanide is classified as an inorganic compound for historic reasons, much like the entire organic-inorganic classification system is historical in reasoning.
The basic initial idea was that forming organic compounds required a vis vitalis, a life force, and could therefore only occur in living systems. On the other side, inorganic compounds were easily synthesised in vitro and have been synthesised even before alchemy was renamed to chemistry. These included most ionic compounds then known.
Naturally, sodium cyanaide is ionic. Most notably, early chemists would have been able to precipitate cyanide from other cyanide sources, redissolve it and add a sodium source (most likely sodium hydroxide) to regain sodium cyanide; thereby classifying it as inorganic.
In 1828, Friedrich Wöhler disproved the vis vitalis assumption and laid the foundation for synthetic organic chemistry. Yet the distinction between inorganic and organic chemistry was still upheld. Rather than defining it by synthesisability, it was said that any molecule containing carbon be organic — or one of many other similar definitions. This could exclude the cyanide salts but does not have to. The only salts and carbon-containing molecules that are always and consistently classified as inorganic are carbon monoxide, carbon dioxide and (hydrogen)carbonates.
By now, the distinction between organic and inorganic chemistry has lost most of its relevance and is only there for historic reasons. In principle, the distinction is that organic chemists are more interested in the carbon-containing compounds while inorganic chemists lean towards other elements more. But then there are numerous in-between fields that don’t really fit anywhere such as organometallic chemistry, bioinorganic chemistry etc.
Therefore, treat the organic/inorganic distinction as a historical residue and don’t read anything into it.


text:
Thermodynamics of DNA double helix formation?
The entropy of a system decreases when 2 single stranded DNA molecules come together and form a double stranded molecule. For this to occur the total entropy of the universe must increase. Thus, heat is released when the double stranded molecule forms. My question is: Where does this heat come from?

Creation of hydrogen bonds.

Hydrogen bonds between the water that is released from the apolar DNA bases?

The entropy of a system decreases when 2 single stranded DNA molecules come together and form a double stranded molecule.

Are you sure?
I would argue that association of two single-stranded DNA strands to form a double-strand is energetically favourable — for entropic reasons. This should be a combination of

the hydrophobic effect
the liberation of a few water molecules per base that are no longer needed in hydrogen bonding.

Point 2 is explained quickly. Upon association of two single strands, the characteristic Watson-Crick base pairs form and hydrogen bonds are established between the DNA bases. These hydrogen-bond donors and acceptors were previously associated with water molecules keeping them close to the DNA single strands and thus rather ordered. These water molecules are now liberated; and since there are at least two per base, this partial process is entropically favourable.
Point 1 may not be as clear as point 2, but it again boils down to the fact that water molecules are liberated from a rather well-defined structure. Large parts of DNA — the area around the bases — are not as hydrophilic as other parts of the molecule or water itself. While they still are somewhat polar, their interaction with water is not ideal. Thus, the ‘surface’ water molecules will form a kind of defined boundary around the molecule. This boundary becomes lesser overall if the surface to volume ratio decreases. And upon association, you have almost double the volume in practically the same surface (i.e. you have halfed your effective surface). Therefore again, a large number of water molecules are liberated.
If DNA association is entropically favoured, there is no need to invoke any heat.

Your explanation is very clear. My teacher told us there is a decrease in entropy in the system (free single stranded DNA versus double stranded DNA). The release of water does increase the entropy of the system, but the decrease of entropy caused by the double strands is greater than this increase. But since the formation of double strands does occur spontaneously there has to be a release of heat. This heat then increases the entropy of the universe more than the decrease in the system, which is why the double strands are formed. This is also what my textbook states.

@guest  You write: "But since the formation of double strands does occur spontaneously there has to be a release of heat."  This is not correct.  Endothermic reactions, which absorb (rather than release) thermal energy, can also occur spontaneously.

@Jan  According to this article (https://academic.oup.com/nar/article/43/17/8577/2414454 ), DNA pairing has an unfavorable (i.e., negative) $\Delta S$, even after accounting for water.  This is in part because "water [is] fixed by the AT base pair in the minor groove and released upon dissociation of this group...."  Having said that, this looks like a tricky problem (that this paper may not resolve), so if you're interested in exploring this further you should also read the articles that cite this paper.


text:
How does a mercury detector work?
I am new to chemistry and saw a "portable mercury vapor monitor." How do these machines work?

There are many ways to detect mercury, and you can find a more detailed analysis of these methods, along with examples of these detectors and their limitations, here.

Gold film sensors

Gold Film Sensors were the first reliable forms of mercury detectors due to gold’s affinity for elemental mercury [...] When a mercury rich air sample passes over a thin gold film, the mercury deposits on the gold and changes the electrical resistance of the foil. This change in resistance is directly proportional to the mass of mercury vapor taken from a known volume of air, which can be calculated in $\mathrm{mg/m^{3}}$.

Cold Vapor Atomic Absorption Spectroscopy (CVAAS)

In mercury CVAAS, a light source of known wavelength and intensity (~254 $\mathrm{nm}$, middle ultraviolet spectrum) is radiated through a sample of air where the light eventually encounters a detector. If mercury is present, electrons from within the mercury atoms will absorb some of this energy from the light source. The difference between the initial energy of the light source and the energy measured by the detector gives you an indirect measurement of how many mercury atoms were initially present.

Atomic Fluorescence Spectroscopy (CVAFS)

Cold Vapor Atomic Fluorescence Spectroscopy (CVAFS) is an improvement upon the traditional CVAAS. When a mercury atom absorbs the energy from the UV wavelength, an electron transitions from a stable ground state to an unstable ‘excited’ state. This excitation event describes atomic absorption as discussed in the [CVAAS] section. However, when the energy source is removed the excited electron returns to its ground state. In doing so, a photon of light is emitted during the loss of potential energy. This fluorescence of light is often unique for various chemical species. Mercury in particular absorbs light at 254 $\mathrm{nm}$ and fluoresces light at the same wavelength. Because the light absorbed and emitted are at the same wavelength, this form of fluorescence is referred to as resonance fluorescence. Other chemicals such as chlorides, sulfides and hydrocarbons absorb light at 254 $\mathrm{nm}$ but either do not fluoresce or fluoresce at a different wavelength.

Most commonly they work by pulling air through a UV absorption cell and simply measuring the absorption at 254 nm.  Often there is a pre-concentration step involved.  This usually involves a gold-plated region prior to the UV cell.  At room temperature mercury adsorbs to the gold, then after some period of time the gold-plated region is flash heated to release all of the mercury at once.


text:
How can absolute binding free energies be calculated?
In the calculation of binding free energies, such as between a protein and a ligand, I learned that absolute values cannot be obtained from simulation (such as taught in this lecture, slide 2). Rather, with the appropriate reference state, only relative free energies can be calculated. However, there are a number of papers that discuss calculations of absolute free energies (ex. recent JACS paper). What does this mean? Are the absolute values--even among different papers--using some kind of general reference state?

These are two very different things - binding energy isn't energy of compound but of reaction.

Free energies are computed by transforming one state into another, e.g. by transforming ligand A into a ligand B (inside a protein in our case). This transformation yields the difference in binding free energies between ligands A and B: $\Delta \Delta G^{bind}_{A \rightarrow B}$. This is a relative binding free energy.
On the other hand, if ligand A is transformed into nothing, it's absolute binding free energy is calculated: $ \Delta G^{bind}_A$.
In both cases ($\Delta G$ and $\Delta \Delta G$), we are calculating differences between two states, and never calculating the absolute free energy $G$ of a single state. This is probably what you meant by: "only relative free energies can be calculated."
PS: since binding occurs in a solvent, it would be necessary to take ligand solvation and protein solvation into account, in order to properly describe the unbound state. For example, in the calculation of $\Delta G^{bind}_A$ described above, ligand A should appear in bulk water instead of disappearing completely.


text:
Redox reaction KMnO4 + FeX
Today I was conducting a redox titration with $\ce{MnO4}$ and unknown amount of $\ce{FeX}$ which through qualitative observation looked like $\ce{FeO(s)}$. I titrated the $\ce{FeX}$(which was dissolved with 10mL $\ce{H2O}$ and 5mL $\ce{H2SO4}$). I had two questions about this experiment:
1) If I am trying to find the molarity of the unknown $\ce{Fe^2+}$ ions (redox reaction), would the increase in volume (mL) of $\ce{MnO4-}$ increase the yield of $\ce{Fe^2+}$ ?
2) When conducting this experiment I did not rely on any indicators yet relied on the $\ce{KMnO4}$ for colour change. I really don't know what the possible sources of error for this experiment are. I need some help trying to come up with 2.

You had a solid FeX and you try to find a concentration, I don't know how this would be possible. Could you add all steps from start to finish describing the experiment you did?


text:
What does the label of stability: 0 mean on an MSDS sheet?
I was reading an MSDS sheet on bleach. There is a section named NFPA. Under that is are the words,

Health Hazard: 3, Flammability:0, Stability:0

Does that mean that Bleach is highly unstable?

Do not confuse GHS (5-1) and NFPA 704 (0-4) placards.

According to the Wikipedia site for the NFPA 704   identification system, a Stability rating of 0 means:

Normally stable, even under fire exposure conditions, and is not
  reactive with water.

The low concentration of the bleach, in other words it's mostly water, is the reason that it fits the "0" category.
Note that the bleaching component of the bleach solution is sodium hypochlorite (see your MSDS for bleach).  Pure solid sodium hypochlorite has a NFPA reactivity rating of "1".


text:
What type of chemical reaction is sulfuric acid with calcium carbonate?
$$\ce{H2SO4 + CaCO3 -> H2CO3 + CaSO4}$$
I thought it would be a double replacement and also a decomposition reaction because $\ce{H2CO3}$ immediately breaks down to $\ce{H2O}$ and $\ce{CO2}$.

The decomposition part is not portrayed in your equation so you can't say that the equation contains decomposition.

While double replacement is correct, you could also classify it as a neutralization reaction (what I think would be the most common answer if one had to pick a single description).  I agree with the comment explaining that you can't consider it a decomposition reaction as written.  You would need to explicitly show that as part of the reaction for that to be an acceptable description.


text:
What is the significance of the kink in the [A] vs T graph for a 2nd order reaction?

The 2nd order graph shows a bit of a kink, whereas the 1st order graph is smooth.

What is the significance of this kink, and why does it appear in only the 2nd order graph.
What is the significance of the numerical value at this kink? The graph below show it occurs at $[A]=1.5-2$.

Can you add an annotation to that upper 2nd-order kinetics plot, to indicate what you're referring to when you say "kink"? I think I know what you mean, but I'd like to be sure.

(Depending on which feature you're asking about, it might just be a plotting glitch.)

The shape is different so it looks like it has a kink, the kink itself isn't important.

@K-Feldspar Ahh, thanks for that edit, that was very helpful. I think there was a small plotting glitch in the prior figure, but it's not in the revised one. ortho's answer is correct to a point, but I think I may be able to provide some further insight, once I get a chance to write up an answer.

Summary
As orthocresol correctly notes, the presence of this kink is simply a feature of the $x$-shifted hyperbola that is the solution to the second-order kinetics problem, and the presence/characteristics of the kink will vary depending on the particular parameters of the problem.
However, this answer aims to demonstrate that:

The kink in the 2nd-order plot occurs due to a rapid change in the slope of the curve from "large" to "small" values.
The kink only occurs in the 2nd-order plot if (a) the initial concentration and rate constant are both "high enough"; and (b) the time scale of interest is "short enough".
A similar kink also occurs with first-order kinetics, again if the rate constant is "large enough" and the time scale is "short enough".

In order to generalize the solutions to the first- and second-order kinetics problems, coming from a chemical engineering background my standard approach is to non-dimensionalize, then solve.
1st-Order Problem
$$
\ce{A ->[k_1] P} \\ ~ \\
{\mathrm dC_\mathrm A\over \mathrm d t} = -k_1 C_\mathrm{A}~;\quad
     C_\mathrm A(0) = C_\mathrm{A}^\mathrm o
$$
2nd-Order Problem
$$
\ce{2A ->[k_2] Q} \\ ~ \\
{\mathrm d C_\mathrm A\over \mathrm d t} = -k_2 C^2_\mathrm{A}~;\quad
    C_\mathrm{A}(0) = C_\mathrm{A}^\mathrm o
$$
(Note that although I'm writing both reactions with $\ce A$ as the reactant, I'm assuming that these reactions do not occur simultaneously.)
To nondimensionalize, you define a new set of variables by dividing each dimensional variable by a scaling factor. Here, it makes the most sense to scale the concentration by the initial value:
$$
\Theta = {C_\mathrm A \over C_\mathrm A^\mathrm o}~,~~\text{or}~~~
     C_\mathrm{A} = C_\mathrm{A}^\mathrm{o}\,\Theta
$$
The scaling factor for the time doesn't have an immediate, natural definition, so I'll use the generic variable $t^*$:
$$
\tau = {t\over t^*}~,~~\text{or}~~~ t = t^*\,\tau
$$
Then, these new variables are substituted into the original equations. The first-order equations become (using subscripts to indicate the order of the reaction under consideration):
$$
{C_\mathrm A^\mathrm o\over t^*}
  {\mathrm d \Theta_1\over \mathrm d \tau} =
    - k_1C_\mathrm A^\mathrm o\, \Theta_1
$$
The scaling constants can then be grouped on the RHS of the differential equation:
$$
{\mathrm d \Theta_1\over \mathrm d \tau} =
    - k_1 t^* \Theta_1
$$
For convenience, define the parameter $\eta = k_1 t^*$, yielding:
$$
{\mathrm d \Theta_1\over \mathrm d \tau} = -\eta\,\Theta_1~;\quad
    \Theta_1(0)=1 \tag{1}
$$
The form of $\eta$ reveals that it is dimensionless, along with all of the other terms in the equation.  This approach allows representation of any 1st-order kinetics problem to be represented by a single equation, once appropriate values for the system parameters $k_1$ and $t^*$, and thus $\eta$, are defined.
Non-dimensionalization of the 2nd-order equation proceeds as follows:
$$
{C_\mathrm A^\mathrm o\over t^*}
  {\mathrm d \Theta_2\over \mathrm d \tau} =
    - k_2\left(C_\mathrm A^\mathrm o\right)^2\, \Theta_2^2 \\
{\mathrm d \Theta_2\over \mathrm d \tau} =
    - k_2 t^* C_\mathrm A^\mathrm o\,\Theta_2^2
$$
$$
{\mathrm d \Theta_2\over \mathrm d \tau} =
     -\kappa\,\Theta_2^2~;\quad\Theta_2(0)=1 \tag{2}
$$
The non-dimensional parameter $\kappa = k_2 t^* C_\mathrm{A}^\mathrm{o}$ is defined analogously to $\eta$, but has a different form due to the different reaction kinetics.
As an aside, note that in Eqs. $(1)$ and $(2)$ all of the information about the scale of each problem has been wrapped into $\eta$ and $\kappa$, respectively.  Taking $\kappa$ as an example, if $C_\mathrm{A}^\mathrm{o}$ is 'large enough' relative to $t^*$ and $k_2$, then $\kappa$ will be large and the reaction will be "fast". Similarly, for a given $C_\mathrm{A}^\mathrm{o}$ and $k_2$, if you extend your time range of interest long enough (viz., increase $t^*$), it will make the reaction term 'seem faster' compared to a shorter time scale.

The solutions to the above differential equations are:
$$
\Theta_1(\tau) = e^{-\eta\,\tau} \tag{3}
$$
$$
\Theta_2(\tau) = {1\over 1 + \kappa\tau} \tag{4}
$$
The derivations of the above are standard enough that I won't go into them in detail.
It will be useful later to have explicit expressions for the rates as functions of $\tau$. Substituting Eq. $(3)$ into Eq. $(1)$:
$$
{\mathrm d \Theta_1 \over \mathrm d \tau} =
   -\eta\, e^{-\eta\, \tau} \tag{5}
$$
And, substituting Eq. $(4)$ into Eq. $(2)$:
$$
{\mathrm d \Theta_2 \over \mathrm d \tau} =
   -{\kappa\over \left(1 + \kappa\tau\right)^2} \tag{6}
$$

Now (finally!) to start tying these equations to the plots in the original question.
First, to demonstrate the utility of the scaled approach, consider the lower plot in the original question, where numerical values for the rate constant and initial concentration are given: $k_2 = 2~{1\over \mathrm{\mu M\, s}}$ (I'm assuming a typo on the units here) and $C_\mathrm{A}^\mathrm{o} = 10~\mathrm{\mu M}$.  Further, the time scale has implicitly been defined by the choice of the domain of the abscissa: $t^* = 2~\mathrm{s}$.  This leads to a value of $\kappa = 40$,  which yields the following plot:

As can be seen, the visual appearance of the profile is essentially identical to that in the OP's plot.
Anticipating subsequent work, I'm actually going to work from here with a slightly higher value, $\kappa=54$:

The highlighted band in the figure marks the range where the (non-dimensional) slope decreases from $4$ to $0.25$, and falls right on the "kink" noted by the OP.  Qualitatively, it marks the transition region from the "high slope" to the "low slope" portion of the curve. It is the narrowness of this transition region $\left(\Delta\tau \ll 1 \right)$ that gives the visual impression of the "kink" in the curve.
There are a couple of ways one could compare this curve to a 1st-order curve.  Naturally, one cannot define a 1st-order curve that lies exactly on the 2nd-order curve, as the two functional forms are different. However, after taking as given that $\Theta_1(0) = \Theta_2(0) = 1$, I see two intuitive ways of selecting a 'comparable' 1st-order curve:

Constrain the initial slopes to be equal [cf. Eqs. $(5)$ and $(6)$]:

$$
\left.{\mathrm d \Theta_1 \over \mathrm d \tau}\right|_{\tau=0} =
   \left.{\mathrm d \Theta_2 \over \mathrm d \tau}\right|_{\tau=0}
   \quad\longrightarrow\quad
   \eta = \kappa
$$

Constrain the 'final values' (here, at $\tau = 1$) to be equal [cf. Eqs. $(3)$ and $(4)$]:

$$
e^{-\eta} = {1\over 1+\kappa} \quad\text{or}\quad \eta =
    \ln{\left(1+\kappa\right)}
$$
For Case #1, I'll plot $\eta = 54$ along with $\kappa = 54$:

As can be seen, in actuality, one can introduce a "kink" into a 1st-order curve, given high enough a value of $\eta$.  In fact, when the initial slopes of the non-dimensional curves are made to be equal, the kink in the 1st-order curve is more dramatic than that in the 2nd-order curve!
On the other hand, Case #2 looks much more like that depicted in the original question, without a visible kink in the 1st-order curve. Plotting for $\kappa=54$ and $\eta = \ln{\left(\kappa+1\right)} \approx 4$ gives:

(This is is why I picked $\kappa = 54$ above, so that the value for $\eta$ here would be approximately an integer value.)
In this case, the slope transition region for the 1st-order curve is broad $\left(\Delta\tau\sim 1\right)$, leading to no perceptible kink.
Further, it’s possible to plot a second-order curve where no kink is present.  Taking $\kappa = 4$, comparable with the above $\eta = 4$ curve, gives the following:

As with the first-order chart with $\eta = 4$, the broader transition region above corresponds to the lack of an observable kink in the curve.

So: To kink, or not to kink?  The scale is the question!

There's no significance to the shape, that's simply how any graph of the form $y = 1/(c+x)$ looks like ($c$ being any constant you like).
Even if the supposed "kink" has some significance, there isn't any significance to the number, especially since the number depends on your system anyway (it's dependent on the value of your rate constant as well as the initial concentration).