text:
Oxidation state of the sulfur atoms in the thiosulfate Ion
What are the oxidation states of the two inequivalent sulfur atoms of the $\ce{S2O3^{2-}}$ molecule? For all I know is that Lewis Structure does matter in most of the cases.

According to Oxidation state of sulfur in thiosulfate and implications for anaerobic energy metabolism

according to the currently held view, the two sulfur atoms of thiosulfate exist in the oxidation state of sulfate (+6) and sulfide (−2) and do not change their respective oxidation states upon disproportionation. Our results based on XANES spectroscopy indicate that the two different sulfur atoms in thiosulfate have charge densities corresponding to +5 and −1 oxidation states

and in Using X-ray absorption to probe sulfur oxidation states in complex molecules

the oxidation states were established for the outer and inner sulfur atoms of thiosulfate as −1 and +5, respectively. The revised values give a far better understanding of the role of thiosulfate as an energy substrate for bacteria because intra-molecular electron transfer, which generates metabolic energy with the production of sulfide and sulfate, cannot be envisaged with the previous view that the oxidation states were −2 and +6 for the outer and inner sulfur atoms of thiosulfate

Although both of the above articles are by the same author.
On the other hand, Determination of Changes in Sulfur Oxidation States
in Prostate Cancer Cells

there is a difference between formal and apparent
  oxidation state received from XANES measurements, for
  example the apparent oxidation states of sulfur in thiosulfates,
  obtained from XANES data, are -1 and +5,
  while the formal ones are -2 and +6. Such a difference
  could occur due to the calibration curve which has been done
  on the basis of chemical compounds in which there are
  ionic bonds which isn't true in the case of thiosulfates.

However, it must be kept in mind that according to the current IUPAC definition, the oxidation state is not an actual property of a substance but instead a charge "an atom might be imagined to have", specifically:

[Oxidation State] is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules: (l) the oxidation state of a free element (uncombined element) is zero; (2) for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion; (3) hydrogen has an oxidation state of 1 and oxygen has an oxidation state of -2 when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of -1 in hydrides of active metals, e.g. LiH, and oxygen has an oxidation state of -1 in peroxides, e.g. H2O2); (4) the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion.

The current IUPAC definition of oxidation state doesn't specify how to apportion electrons of bonds between two atoms of the same element.  This causes various oxidation states to meet the IUPAC definition depending upon how one chooses to apportion the electrons of the S-S bond between the two S atoms.  Also, experimentally, the S-S is somewhere between a single bond and a double bond based on its length, so the number of electrons to be apportioned is also in question.  One could say the oxidation states are of the S atoms are -2, +6; -1, +5 or 0, +4 without contradicting the IUPAC definition.
In "Expanded Definition of the Oxidation State" there is a proposed addition to the definition:

Bonds between like atoms (having the same formal charge) are cleaved homolytically.

If an S=S double bond of thiosulfate is cleaved homolytically, 0 and +4 oxidation states result.
If an S-S single bond of thiosulfate is cleaved homolytically, -1 and +5 oxidation states result.
However, the article concludes 0 and +4 oxidation states even with the single bond, because when there is a single S-S bond the two S-S atoms do not have the same formal charge. They give both electrons from the S-S single bond to the central S. Below is Fig. 4 from the article:

The most recent (June 2014) information (and probably most authoritative) I can find is Toward a comprehensive definition of oxidation state (IUPAC Technical Report) Pure and Applied Chemistry. Volume 86, Pages 1017–1081
In this article, another definition of Oxidation State is proposed requiring

Bonds between two atoms of the same element are divided equally.

with no exceptions for formal charges.  This clearly rules out the possibility of -2, +6 oxidation states for the S atoms in thiosulfate, but leaves ambiguity between-1, +5
 if the S-S bond is single and 0, +4 if the S=S bond is double.
Specifically, concerning thiosulfate the article states:

Structural properties of sodium thiosulfate suggest that all its terminal atoms carry some of the anion
  charge. In other words, all form ionocovalent bonds to a cation. The calculated Mulliken charges (−1.4 for
  O and −1.1 for terminal S) confirm this but, being different, they suggest unequal SS and SO bond orders. The
  SS bond distance of 2.025 Å (202.5 pm) is shorter than the single bond of 2.055 Å (205.5 pm) in crystalline
  $\ce{S8}$ or 2.056 Å (205.6 pm) in the $\ce{H_2S_2}$ gas, but substantially longer than the double bond of 1.883 Å
  (188.3 pm) in $\ce{S2O}$ or 1.889 Å (188.9 pm) in $\ce{S2}$. Given this uncertain situation, the approach chosen
  here is to draw in Fig. 24 two limiting Lewis formulas.

Then they proceed to draw two Lewis formulas, similar to above, one on the left with S-S single bond and -1 and +5 oxidation states for the sulfurs and one on the right with S=S double bond and 0 and +4 oxidation states for the sulfurs, and go on to say:

The result in Fig. 24 means that the individual OS of the two sulfur atoms in thiosulfate are ambiguous,
  and no formula in-between these two limits yields reasonable values. The formula on left provides an
  unusual yet not entirely illogical OS of −1 for the terminal sulfur, reminiscent of the oxygen OS in peroxides.
  The formula on right, while having an SS bond order that is too high, suggests OS values that at times
  are used...

The article also comments that:

A nominal OS is one that has been nominally adjusted against the generic OS definition in order to suit a specific purpose. That purpose may be...polarizing
  a homonuclear bond in thiosulfate to OS [of S] = +6 and −2 in order to emphasize the uniqueness of the central atom and similarity of all four ligands.

@Del Pate wrote: "All I want to know now is , Since oxidation state is a theoretical concept , should we rely on the actual charge densities (as mentioned in most research papers) of the atoms (which is a practical approach) or stick to the theoretical concept of Electronegativity differences ? The latter gives the answer as - -2 for terminal. +6 for central sulfur ."

@Del Pate I wouldn't say it is a theoretical concept, more like and ficticious or imaginary concept.  The current IUPAC definite of oxidation state includes the phrase "the charge an atom might be imagined to have" http://goldbook.iupac.org/O04365.html So if you need a value for actual experimental or theoretical research you should use an actual value not an imaginary value.  (continues below)

(continued from above) Also, I don't understand why you say electronegativity consideration necessarily yields -2 and 6.  "Towards a comprehensive..." says "EN scales have been developed, and it is clear that none of them will provide the absolute OS result in each and every chemical compound" and that oxidation state is ambiguous for thiosulfate because the bond order is uncertain.

@DavePhD The original goldbook definition was based on electronegativity which has now been changed to a bit random definition by IUPAC . So for a theoretical approach we use the EN difference of the two atoms which form the bond .(MARamezani's answer is based on EN)

@DelPate But looking a Fig. 24 of the "Toward a comprehensive definition..." article, one doesn't necessarily arrive at -2 and +6 by that method because, in the left example, the SO bond order is 5/3 and SS bond order is 1, while on the right the SO bond order is 4/3 and the SS bond order is 2.

The last statement of that paragraph on thiosulphate suggests that textbooks go for +6,-2 approach . Its also mentioned in the report that EN is the most used and accepted approach for its simplicity . Also refer the 1st cautionary note on pg1060 . Practical electron densities are not matched with the concept of oxidation states .

@DelPate I have trouble understanding the meaning of the sentence that mentions textbooks.  I thought it meant current textbooks use -2 and +6, rather than recommending anything.

I think it means standard texts provide -2/+6 answer . I think this textual answer is based EN method and maybe considered as the actual answer while your answer on electron density might be just for practical purposes(read the cautionary note of that report) @DavePhD

@DelPate S and S have the same electronegativity so how is giving zero electrons to the central S and 8 electrons to the outer S following the EN rule. If the S-S bond is single, the central S should get 1 electron and be +5 oxidation state and if the S=S bond is double the central S should get 2 electrons and be +4 oxidation state.

@DavePhD The external S is sp2 hybridized while the central one is sp3 hybridized . Hence the external one is more electronegative and will get both the electrons of the double bond and become -2 .

@DavePhD I dont why but your last comment got deleted . You said that "the current IUPAC definition doesnt tell how the electrons are distrubuted for similar atoms " and your link to the IUPAC report seems to claim "homolytic distribution for like atoms"

@DavePhD My reply: Martins answer over here ( http://chemistry.stackexchange.com/questions/10944/electronegativity-considerations-in-assigning-oxidation-states/10946#10946 )suggests that Iupac definition isnt up to the mark .Thiosulfate cant be resolved with their set of rules as the rules dont have anything about similar atoms . The IUPAC technical report that you quoted says "homolytic distribution for like atoms" .But the two sulfur are not alike (structurally and also on basis of hybridization and EN).This (http://pubs.acs.org/doi/pdf/10.1021/ed1005213) suggests "heterolytic distrubution".

@DelPate, yes I deleted to edit.  The current IUPAC definition doesn't specify how to apportion the electrons of a bond when the atoms are the same.  The article you cite says "Expanded Definition of the Oxidation State" says to cleave homolytically and gives two sp2 + sp3 examples (acetic acid and acetalydehyde) where they cleave homolytically.

@DelPate I edited the answer, what do you think about it now?

Its perfect and I hope that IUPAC redefine the term so that we get exact answers . +1 for the help.

This answer started off so well but then uses expanded octets for the central sulphur. Sorry, I’ll have to give this $-1$.

oxidation state between two same atom is not necessarily to be zero and how could we know that the elements of same atoms does not share equal electrons

The IUPAC has redefined oxidation state in 2016, I finally got around to update (some/most) of my answers, but the most comprehensive one is here: Electronegativity Considerations in Assigning Oxidation States. Thiosulfate is mentioned in the sources specifically, but it appears to be the same as the article you cited already. cc @DelPate

I'm not unsympathetic to MAR's answer. It is logical to consider sulfur's "usual" oxidation state to be the same as oxygen: -2. However, the fact that the two sulfurs are bonded to each other suggests that the oxidation state should be considered differently. Below is a systematic method that gives an answer (+5/-1) that matches the experimental data nicely summarized by DavePhD.
Let's say we start with disulfide dianion. Overall, the two sulfurs carry two negative charges. Similar to peroxide (as described in the IUPAC Gold Book), each sulfur is considered oxidation state -1.
Now, we add an oxygen to one of those sulfurs. Oxygen is (almost) always -2. To keep the overall charge -2, that means that the newly bonded sulfur is oxidation state +1.
Oxidizing that sulfur again gives sulfurs of +3 and -1. Finally, in thiosulfate the oxidations states are +5 and -1.

+1 for the effort . Thanks . I was just wondering whether the oxidation states of the sulfur atoms would remain the same in thiosulfuric acid ? I think it should but just to confirm.

@DelPate yes, thiosulfuric acid would have the same oxidation states as thiosulfate. This is always true for acid/conjugate base pairs.

But if two of the oxygens are bonded to a hydrogen and we have a double bond between the two sulfurs , I think +6/-2 should come . There is no negative charge on the oxygen now for the delocalisation .

It's exactly as I have outlined above except one of the oxygens is a hydroxyl instead. Adding a proton has no effect in the oxidation states.

Finally, the correct answer to the question using correct structural formulae.

"The S–S bond length is 201.3 pm, which indicates essentially single-bond character, while the mean S–O bond length is 146.8 pm, which indicates considerable $\pi$ bonding between the S and O atoms" https://books.google.com/books?id=wz8TDAAAQBAJ&pg=PA640&lpg=PA640&dq=thiosulfate+bond+lengths&source=bl&ots=wnfD05BXHI&sig=E-P0baLqI8XaeRuXHB137WrBWik&hl=en&sa=X&ved=0ahUKEwjQxuaSs6XPAhVE4D4KHdPZD48Q6AEIMDAC#v=onepage&q=thiosulfate%20bond%20lengths&f=false  The S-O bonds are not just single bonds.

Yes, lewis structure does matter.
Draw the lewis structure of thiosulfate.

  Source

How is oxidation state measured? It's a crude way of attributing some charge to the atoms.
O is more electronegative than S. You have S as the central atom, and another S "around" the tetrahedron structure. What would that do to each S?

 The central S is expanding octet, while at the same time having a positive oxidation state due to the bonds it makes. The other S is having a negative oxidation state.

What oxidation state does an O atom usually have? You can apply it to this ion as well. And we have a rule that says the sum of the oxidation numbers is equal to the charge of the ion. The "noncentral" S will get the same oxidation number as O atoms and then you can measure the central S atom's oxidation state by calculating all of the oxidation states to be -2.

I think the OP's point is that there are multiple Lewis structures that could be drawn, like the two in Fig. 24 of "Toward a Comprehensive Definition of Oxidation State" in addition to what you have drawn.  I don't think this is a homework question.

@DavePhD I don't think the OP actually wanted to dig that deep. If the article you mention was his\her problem, s\he would definitely have noted it. If the problem was that deep, the OP would mention the reason for what caused the confusion. Your answer is awesome; but it's correct, not high-school. In other words, I guess that was a bit random.

"What oxidation state does an O atom usually have? You can apply it to this ion as well. " @MARamezani How can the external Sulfur get -2 just by comparison ?

@DelPate In chemistry at a high school level (we're past the thing that you're a high school wanderer though.) Sulfur is known to "behave" like another of surrounding O atoms. It's an incomplete (but basic) notion. Don't let it bother you.

@MARamezani I have 2 question about your answer. 1. Why do you say "the central S is an expanding octet", but draw S with only an actual octet?, and 2. Why do you give both electrons from the S-S bond to the outer S and none to the central S in assigning oxidation state? Compare what you are saying to the Wikipedia article http://en.wikipedia.org/wiki/Oxidation_state#Inequivalent_atoms_of_an_element and decide if you agree or disagree with the article.

@DavePhD - 1. Edited and added a more credible image (the former Lewis structure is apparently unacceptable, why? No idea. Maybe you know) 2. Wikipedia is a bit dizzy when getting to the oxidation numbers: http://en.wikipedia.org/wiki/Thiosulfate#Nomenclature . +5 and -1 are correct in my idea too, but as a high schooler, a poor student has to close his\her eyes and memorise: Oxidation number: The noncentral S: -2 and the central one: +6. They wouldn't know about calibration curves or whatever. I wanted to add something too, but I just forgot what I was about to say!

@MARamezani A high school student should know the rule that two atoms of the same element share the electrons equally.

@MARamezani your wiki link says the direct result while Dave's suggests "The central sulfur is assigned only one valence electron from the S-S bond and no valence electrons from the S-O bonds" which I feel is incorrect since the terminal S atom is more electronegative and should get both the electrons of the sulphur-sulphur double bond . That would yield your answer .

@DelPate well of course. I didn't post this answer blindfoldedly! :)

@jerepierre unfortunately in the case of thiosulfate things aren't that simple. Read Dave's answer for why they're believed to have those even oxidation numbers.

Guys I love this debate!

@MARamezani Yes,its quiet interesting . So there is some mistake in Wikipedia's subtopic of Inequivalent Structures . Here is a link which supports your theory of EN as basis of oxidation states -     http://pubs.acs.org/doi/pdf/10.1021/ed1005213 (and I dont think the high school information and the actual information would be different as +6 / -2 answer looks most convincing theoretically )

The article says to cleave bonds between like atoms homolytically.  Experimentally, the S-S bond is somewhere between a single bond and a double bond (at least based upon its length).  So the central atom should be attributed at least one, maybe two electrons.

@DavePhD I feel that the two sulfur atoms cannot be considered like atoms as they possess different properties . If experimental data is used does the topic shift towards the actual charge denisties(which is away from OS) ?

@DavePhD anything to add?

@DelPate I don't really understand your last two comments.  If you mean you are personally advocating for a definition of oxidation state whereby a bond is heterolytically cleaved unless it is between 2 atoms of the same element and the molecule has a plane or axis of symmetry going perpendicularly through the center of the bond, that's interesting but I'm not aware of anyone else proposing such.  The "Extended Definition" article says to cleavebonds between atoms of the same element homolytically unless the atoms have different formal charges (continues below)

(continues from above) So for thiosulfate they conclude 0, +4 for the 2 sulfurs regardless of whether the S-S bond is single or double.  It think some experimental data is always need to determine what atoms are bonded and sometimes bond order (single, double,etc.), to determine oxidation state.

@DavePhD Ya , I meant that but was not proposing a new definition ( that would actually require some serious research and going against the IUPAC,{who have still stuck to a strange definition} ) but wanted to know what exactly they meant by "similar atoms" . Since they have concluded using bond lengths which comes under a practical/experimental approach I felt a theoretical approach should have been proposed . Thanks , anyway.

$-1$. Sulphur does not ‘expand its octet’.

How do we get to know that the two atoms of same element is not sharing equal electrons

@koolman draw a Lewis structure, and see for yourself what each atom of the same element is bonded two. That should give you some ideas.

Why -2 for non central S atom?

@M.A.R. There's some great info here.  Would you mind integrating anything pertinent into your answer so we can clean these up a bit?


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What is 'Retention period' in chromatography and what is its importance?
I heard people quite often speak about retention time/retention period in Chromatography and Mass Spectrometry. I see somewhere that it is the time duration between the injection time and the target compound peak. I have the following questions:
1)What is the purpose of defining that time?
2)If the compound peak is absent, how one can speak about retention time (I mean if the compound completely fragments)?
3)Is it has importance in mass spectrometry as well?

To answer your three questions in series:
1) The time is a measure of how strong the interaction of a compound is with the column used. The stronger the interaction the longer the compound will stay on the column and this is therefore a way to separate two compounds. For example one molecule with a weak interaction spending a minute on the column, the other spending 5. Collecting fractions of solvent from a liquid chromatography column will yield pure compounds. Note that the time is also a measure for the amount of 'solute' flushed over the column. Many chromatograms also report retention in column volumes of solute. I will not go into detail about that here. It might be worth a separate question.
2) If no peak is found, no retention time can be reported. In chromatography it never (actually almost always) good when a compound fragments/degrades. That means the technique is not suited for purification or analysis. In unique cases, the degradation can be used to analyse whether or not a compound will degrade when interacting with certain materials.
3) In mass spectrometry a retention time in the machine is, as far as I am aware a 'Time of Flight'. Compounds are quickly brought into a gaseous and charged state and are pulled through a magnetic field. Depending on the weight of a compound, the time it flies, is shorter or longer. That is a way of calculating the weight of a molecule. In some cases, for example small proteins, the molecules are purposefully fragmented. The fragments can then be used to reconstruct of what amino acids the protein is made of.
Hope this answers your question! If any clarification is needed, please, do ask :)

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Why does the phosphorus trifluoride molecule have dipole moment?
My book says that from drawing the Lewis structure for the molecule below, you can conclude that it has dipole moment. Can someone please explain why?

I understand that the dipole moment expresses the polarity of the bond. And that the dipole moment of a molecule is the sum of the dipole moments for each bond in the molecule.

Fluorine is more electronegative than phosphorus.
The phosphorus is at the apex of a pyramid, the base of the pyramid being an equilateral triangle with a fluorine atom at each vertex.
The F-P-F angles are 96 degrees.
Each F-P bond contributes to the net dipole moment, as a vector from the P to the F.  The net dipole moment is the vector sum of the vectors along the three P-F bonds.
If all four atoms were in a plane (a trigonal planar geometry), there would be no net dipole moment.  But because the geometry is pyramidal, the vector sum is not zero and there is a net dipole moment.
In some molecules lone pairs also contribute to the net dipole moment, in this particular case the lone pair of phosphorus is in a mostly s-like orbital, so in first approximation it can be neglected.

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Lifetime measurement: double peak in instrument response function
I am measuring flourescence lifetimes with a picosecond pulse laser, using the LifeSpec II with an MCP detector. Instead of a single peak rapidly decaying, as would be expected, my IRF contains two peaks. I am suspecting some sort of reflection causing this. Does anyone get what I mean here? Could add a graph soon if that might help.
EDIT: I have realized this might not be an error, but I would still like to know why.

Welcome to chemistry.SE! Don't hesitate to add your graph. Though I'm no expert in spectroscopy, I'm definite it will help.

What is the ratio of peak 1 to peak 2?

The ratio can now be seen in the graph above. Note the logscale.

When you say "instrument response" in the question, does that mean the data is for some kind of a blank sample?  What can you tell us about the sample chamber?  Is it a capillary, a cuvette, or...?  Is it aqueous, full of air, or a solvent, or...?  How are you measuring the IRF?

I use a piece of semitransparent frosted glass to reflect light from the incoming beam into the detector, to get a crude emulation of scattering. The chamber containing the sample holder is full of air.

I am suspecting some sort of reflection causing this. Does anyone get what I mean here?

The time between the two peaks looks like 0.4 nanoseconds.
This time (0.4 ns) should be compared to the dimensions of the vessel (chamber, cuvette, I don't know the right word) and the speed of light in the medium being observed (water, air, diamond).
So for example in vacuum or gas the light would travel 0.4ns x (0.3m/ns) = 0.12m.
In water where the index of refraction is 1.33 more like 0.09m.

When I take the IRF I use a piece of frosted glass reflecting part of the incoming light onto the detector. On the side of the chamber opposite to the detector there is a black plastic wall that possibly could be hit by light from the sample and reflect it back through the sample into the detector. It doesn't look very reflective though. Also the light would have to travel further than 12 cm extra if there was a reflection from there.

It is quite common to observe secondary peaks or "bumps" in the instrumental response function for a laser excitation. It is not always easy to pinpoint the origin of the second peak, but it is typically not associated with a reflection in the instrument. If any reflection were to be involved, it would be one within the cavity of the laser itself. It is however more often the case that these secondary peaks are just an adventitious product of the laser production process (no laser is perfect). In fact, manufacturers sometimes will suggest keeping an eye on the relative ratio of these peaks in the IRF to gauge the health status of your laser.
For instance, observe the secondary peak in the red line below from a PicoQuant FluoTime instrument with a picosecond laser:

(original source)
You can also see an example of a much more bumpy IRF in a leaflet published by Edinburgh, the makers of your own instrument LifeSpec II spectrometer.

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Determine most electronegative element based on successive ionization energy data

X, Y and Z are three unknown elements whose first 5 ionization energies are given below. Which of the 3 is the most electronegative?$$
\begin{array}{|c|c|c|c|c|c|}\hline
&\text{IE}_1&\text{IE}_2&\text{IE}_3&\text{IE}_4&\text{IE}_5\\\hline
\ce{X}&577&1815&2740&11600&15000\\\hline
\ce{Y}&800&2427&3660&25026&32827\\\hline
\ce{Z}&738&1451&7733&10543&13630\\\hline
\end{array}
$$

Because of the IE ratios, I concluded that X and Y are in the third column and Z is in the second column. I know that EN increases from left to right for elements in the same row. Is there anything else I should know in order to solve this?

Let's look at what amounts of energy we get from IEs.
$$\ce{X}: 577 \rightarrow 1815 \rightarrow 2740 \rightarrow 11600 \rightarrow 15000$$
After taking out three electrons, we see a first jump in IEs. Thus, there are three electrons in the valence shell.
$$\ce{Y}: 800 \rightarrow 2427 \rightarrow 3660 \rightarrow 25026 \rightarrow 32827$$
Again, after three electrons out, we have the first jump. The result is the same as X: Three electrons in the valence shell.
$$\ce{Z}: 738 \rightarrow 1451 \rightarrow 7733 \rightarrow 10543 \rightarrow 13630$$
You see the first bump in IEs in between IE2 and IE3. So, the third electron comes out of a lower shell and thus, two electrons in the valence shell.
The more the electrons, the more shells they will fill and the farther they'll be from the nucleus (among other lesser important factors). That means X has more electrons than Y because of relatively lower IEs, and therefore Y is more electronegative as it's closer to the fluorine.
You're correct that from left to right we get more electronegativity for the atoms, but note that
though with exceptions, such as Al and Ga, from top to bottom you get lesser electronegativities.
Even in the case of Al and Ga, the one that's more electronegative will get relatively more IE to lose electrons. So that's not what you should worry about. So let's conclude:

If you see two elements that are in the same group, the one with a generally higher IE is more electronegative.

As a side note, a really nice way to compare electronegativities is through atomic radii.


text:
How to know it when I see a covalent network?
This is a well-known (better said: well-discussed) question in the internet. When you look for answers for popular questions, you usually see them with a variable degree of reliability and complexity. Unfortunately, for this one, I only observed very very crude and general rules of thumb. So let's get a real answer:

A network solid or covalent network solid is a chemical compound (or element) in which the atoms are bonded by covalent bonds in a continuous network extending throughout the material. In a network solid there are no individual molecules, and the entire crystal may be considered a macromolecule. Formulas for network solids, like those for ionic compounds, are simple ratios of the component atoms represented by a formula unit. Covalent network, wikipedia

Diamond and SiO$_2$ are really great examples of covalent networks-lattices. So enough with stories:
If you face a new chemical formula, how would you assume it's a covalent network? (In case it is) Is it somehow done by drawing the Lewis structure? Is there a rule for this? Or is it only possible to know such thing with experimental data?

Fun fact: I'm preparing for an internal chemistry olympiad. The examiners usually take all the exceptions into account. So general rules such as "if it contains Si, it's a covalent network" aren't really reliable for me. Better phrased, I'm looking for a rule with the least exceptions.

A good source of information on this topic is the website Tetrahedra of Structure, Bonding & Material Type, from which the above image is taken.
An extreme example of the metal corner should be cesium, the ionic corner CsF, and the molecular corner $\ce{F2}$.  At the top is diamond carbon.
The network covalent apex is truncated on the basis that a network covalent material requires an element with the electronegativity of carbon or less (less than or equal to 2.55).
Then the challenge is to distinguish a boundary between network covalent and the other three types
network covalent vs. metal
network covalent vs. ionic
network covalent vs. molecular
Examples of network covalent compounds other than diamond and silicon dioxide include silicon carbide, silicon nitride, boron nitride, aluminum phosphide, gallium arsenide, aluminum oxide.  However, anytime there is an electronegativity difference there is some ionic component.
Tin can exist in metallic and network covalent allotropes.
Selenium can exist in a network covalent allotrope or as molecular $\ce{Se8}$
Going from network covalent to molecular covalent, think of how there is diamond, graphite, nanotubes, fullerenes and even small molecules like $\ce{C4}$


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Clarifications about acid-base theory
Considering the Arrhenius theory of acids and bases:

Is every existing compound either an acid or a base?
Why is a compound such as $ \ce {NH_3}$ considered a base, if it doesn't even have an oxygen atom and a base is supposed to release $ \ce {OH-}$ in water?
If $X$ is a base according to Arrhenius theory, will $X$ still be a base according to the other two theories (and the same if $X$ were an acid)?

According to the Brønsted-Lowry theory:

There are no absolute acid and bases, but only acid-base reaction in which one compound acts as a base and the other as an acid. So, is every reaction of two compounds an acid-base one?
Except for water, are there any other substances which in some reactions act as bases while in others as acids?

Ammonia is the very reason chemists had to rethink the Arrhenius theory.

A couple of thoughts:
There are mainly 3 different definitions of acids and bases:

The Arrhenius formulation: Acid = proton donor in water,  Based = hydroxide ion donor in water
The Brønsted-Lowry formulation: Acid = potential proton donor, Base = potential proton acceptor
The Lewis formulation: Acid = Electron pair acceptor, Base = Electron pair donor

The Lewis definition is the most general, followed by the BL and finally by the Arrhenius definition.
You are right, for example HCl is considered a strong acid in water, but it is very weakly acidic in glacial acetic acid. If you follow the Lewis definition a large number of reactions will be considered acid base reactions. An exception are free-radical based reactions
There are many compounds that are amphoteric, but it really also depends on the definition you use.


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How to get a "rechargable" chemical reaction for a glow stick
I want to create my own glow stick that lasts a while while ligting up the surroundings. How would I go about this? I know there are youtube videos and such. But the way they do it doesn't allow for any way of "recharging"
How can i make a "rechargable" glow stick?

What qualifies as recharging?  Adding more of a compound?  Heating/cooling?  Waiting a while?  Toxicity requirements?  If you're looking for a cheap, nontoxic glowstick that will relight after you leave it in the freezer for an hour, that would be a seriously valuable commodity.

Chemoluminescent processes based on the reaction of diaryloxalates with hydrogen peroxide in the presence of a fluorescent dye are not reversible.
While the dye remains unchanged, both the oxalates and the hydrogen peroxide are consumed. "Recharging" a device (glow stick) isn't possible. When performing the reaction in a flask, only could probably prolong the light emission by subsequently adding portions of the starting materials over time.
Photoluminescent glow-in-the-dark devices based on the emission of long-lived excited triplet states can indeed be "recharged" by holding the device in the light again. But this has nothing to do with the reactions happening in glow sticks.


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How does an ice bath change the rate of an reaction?
If I had hydrogen peroxide decompose with the help of manganese dioxide and measured the time taken for the reaction to complete and the volume of oxygen collected. I will get a graph that looks like this:

Now what will happen to the shape of the graph if I placed the reaction vessel in ice?
I thought the following. The graph will rise but at a less rate and less steep than in the graph above. Then it will get to the point where the reaction ceases and doesn't go to completion because there are no more collisions.

Placing the reaction vessel in ice will lower the temperature of the reaction and thus slow down the reaction. Temperature is a measure of the average kinetic energy of the molecules, so lowering the temperature lowers their kinetic energy. In order for reactants to form products, they must collide with a certain amount of energy. Thus, at a lower temperature it will take longer for the same amount of product to be formed.
You also need to keep in mind that at lower temperatures the equilibrium constant will change. Since this is an exothermic reaction, at lower temperatures there will be more product formed (Le Chatelier's principle). There will be more moles of oxygen present in the iced reaction. However, you're measuring the volume of oxygen, and a colder gas will have a lower volume (for the same number of moles of gas). So you've got two opposing trends: the ice will cause more oxygen to be produced (which would normally correlate with a larger volume) but the colder temperature will also cause the gas to shrink. Without more information, such as the temperature of the two reactions, I don't think you can determine whether you will wind up with a larger or smaller volume, so it would be hard to determine the shape of the graph for the colder reaction.


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Dialkylation reactions
In the synthesis of ethyl 2-benzyl-2-methyl-3-oxobutanoate, is the nucleophilic substituion of a benzyl (chloride) group an Sn1 reaction?
The experiment is conducted in ethanol (protic) and is under reflux, but most textbooks have it as an Sn2?
Will it then be a mixture of the two?

Do you mean ethyl 2-benzyl-2-methyl-3-oxobutanoate? Do you mean addition of the benzyl group (not benzene group)?

Yes, sorry the nucleophilic substitution occurring in the presence of benzyl chloride.

what are you reacting with the benzyl chloride to give the desired product

benzyl chloride plus sodium ethoxide and ethyl 3-oxobutanoate

Both mechanisms will probably be present but $S_N2$ will dominate because the enolate anion is a strong nucleophile. $E2$ will not occur because there are no $\beta$ hydrogens available on the benzyl chloride. Despite the relative stability of the benzyl cation, especially in protic solvents, the strong nature of the enolate nucleophile and the lack of an $E2$ pathway means that $S_N2$ will probably dominate.
For a comparison of alkyl halide reactivities with different nucleophiles, see this link.

if you were to perform a second alkylation, the yield would be quite low given the steric effects of having another benzyl group added. However, why is it necessary to have a stronger base for the second alkylation? The statement that alkyl group groups are electron-donating is so vague..

sorry i'm slightly confused. do you mean adding the methyl group as a second alkylation. if so how are you adding it?

Ok, so if I started with ethyl 2-benzyl-3-oxobutanoate. If I added methyl iodide for instance, would I need a stronger base than sodium ethoxide? It's really just a question of whether that benzyl group is electron-donating?

a strong base is used to generate significant quantities of the enolate. in this case the main factor is steric hindrance due and so using a small alkylating agent is a more important consideration. methyl iodide is probably the best option as you have suggested. i don't think there's any reason why you couldnt use sodium ethoxide apart from the possible side reaction of it substituting the methyl iodide to form methyl ethyl ether

Yes, that's a moot point (Sn1). If water was to be added to ethyl 2-benzyl-3-oxobutanoate, would it act in a Sn1 reaction with benzyl chloride -seems unlikely? Or would it act as a nucleophile, although this again seems unlikely?


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Why is it that weak bases usually contain nitrogen?
Why do weak bases usually contain nitrogen? I know there are two electrons on top for a bond of $\ce{H}$ but why is it mostly nitrogen? I know there are other elemental bases too but why do I keep seeing $\ce{N}$'s mostly?
Is there something special about nitrogen in this respect?

This is actually not true, water is a weaker base than ammonia, and the hydroxide ion is a weaker base than the amide ion.

The short answer is that yes, the fact that many weak bases contain nitrogen does have to do with nitrogen itself. However, it's not really accurate to say that weak bases "usually" contain nitrogen.
For simplicity, I'm limiting my explanation to aqueous solutions, the Brønsted-Lowry definition, and main-group compounds. A base must do one of two things: either remove a proton ($\ce{H+}$) from water or dissociate to produce something else (typically $\ce{OH-}$) that can remove a proton from water. Most compounds that produce $\ce{OH-}$ directly are ionic; even if they don't dissolve much (like alkaline earth hydroxides), these ionic compounds dissociate completely and are therefore strong bases.
A base that removes protons from water must have a free pair of electrons for the proton to bond to. In period 2, boron tends to form compounds that tie up all three of its electrons. Carbon compounds, likewise, tend to have no free electron pairs. Oxygen and fluorine compounds (e.g., water and hydrogen fluoride) have free electron pairs, but because of their high electronegativity these elements keep their electrons to themselves and are unlikely to grab protons from water if they already have a full octet. Nitrogen is the only element in this period that balances two factors: it has a free electron pair in many of its compounds (e.g., ammonia and organic amines) and has a low enough electronegativity that it can share those electrons with a proton to form a cation (e.g., ammonium, $\text{NH}_3$).
Covalent compounds of elements in the third period and below tend to be very weak bases, if at all. I'm not sure why, but I suspect that it's because bonds involving these elements tend to form much weaker dipoles than bonds involving nitrogen. The weaker dipole would limit their ability to forcibly pull protons off of water molecules.
All that said, there are many ionic compounds that form basic solutions due to containing the cation of a weak acid. In these cases the base is a cation, not a molecule. Certainly some of these cations contain nitrogen, but many do not. Examples without nitrogen include any soluble carbonates and bicarbonates and any salts of carboxylic acids.

I think the higher periods (3+) forming weaker bases is due to the lone pair having a much larger s-character than in the second period with nitrogen.


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What is buffer capacity and how can a buffer "run out"?
I'm a little confused on something:
A buffer has a certain capacity and then runs out.
But, if the buffer is given by: $$\ce{A- + H2O <=> HA + OH-},$$ when the $\ce{A-}$ reacts with the water, the reaction shifts to the left. So how can it "run out"? If $Q$ just keeps becoming $0$, shouldn't the reaction keep shifting to maintain $K$? Does the right side, perhaps, approach $0$ as it keeps decreasing in concentration to maintain $K$, but it can never become $0$ molar, because that wouldn't maintain $K$?
I guess if $\ce{H2O}$ was "aqueous"/had a concentration and was therefore used in the $K$ expression, this the $K$ would always be maintained.
But how does the buffer "run out" of $\ce{A-}$ to react when $\ce{H2O}$ is liquid (not aqueous)? It doesn't make sense as wouldn't the reaction always adjust to make more $\ce{A-}$ to maintain $K$?

If strong acid is added to a buffer system, conjugate base removes it.  Eventually, the concentration of conjugate base falls to such a low level that its effect on pH can be neglected.  At this point, the solution behaves more like a weak acid solution.  If even more acid is added, the system behaves like a strong acid system in terms of calculating pH.
If strong base is added to a buffer system, conjugate acid removes it.  Eventually the concentration of conjugate acid falls to such a low level that its effect on pH can be neglected.  At this point. the solution behaves more like a weak base solution.  If even more base is added, the system behaves like a strong base system in terms of calculating pH.
In the equation you wrote, $$\ce{A- + H2O <=> HA + OH-},$$ think about the effect of adding acid.  Acid reacts with the $\ce{OH-}$, producing water and removing $\ce{A-}$ from solution.  The $\ce{OH-}$ and $\ce{A-}$ cannot be removed from solution but their concentrations can be driven to negligible levels by the addition of sufficient quantities of acid.


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Why are enantiomeric excess % and specific rotation considered?
I was wondering, when working in a lab in an area related to organic compounds, when and why would a chemist consider the enantiomeric excess % and how would a chemist used that information?? Same Thing goes with specific rotation, why, how, and when would a chemist use this? I was just curious because I saw this in my organic chemistry book and they relate the problems to lab work and I want to get a good grasp of tis context and it made me wondered.

Enantiomeric excess is very important concept in organic chemistry because it clearly states the purity of one isomer over the other. For example, have a look at the structure of Thalidomide which is racemic (50:50 —R:S isomers) in nature. Between the two form, one is having very important medical property where as another form is sedative. In these cases, synthesizing enantiomerically pure (>99%) form is mandatory and in January 1996, the FDA announced it would consider further incentives for developing single isomer drugs, owing to their better pharmacokinetics prosperity, safety, and tolerability.
http://en.wikipedia.org/wiki/Thalidomide
These are the reason which makes chemists to target enantimerically pure form using latest available technologies like, Chiral GC or HPLC.
Specific optical rotation is very old concept to judge the enatiopurity of the compounds but its not accurate because it depends on various factors like concentration of the sample, solvent which we use, temperature, etc. Even small impurity in the sample gives drastic change in the rotation value. Thats why chemists adapt latest technique like HPLC of GC, nevertheless, chemists still carry out specific optical rotation to make sure that the enantiomerically excess compound has some rotation.

While one enantiomer of thalidomide is a nasty teratogen (causing horrible deformities in foetuses) and the other is a useful drug, it is important to note that the two versions interconvert in the body. So, in this case there is little reason to make the pure enantiomer. But the general idea has influenced the FDA in its thinking for other compounds.

But anyways, welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

To be specific, the enantiomeric excess is useful because it gives a limit on yield based on recrystallization given certain assumptions. If you assume that the enantiomers themselves crystallize better than the racemate, the excess is the theoretical maximum yield because every bit of the opposite enantiomer reduces the amount of the one you can get back.
This is historically why e.e. is used over other measurements of purity like e.r. (enantiomeric ratio).
Specific rotation is not so useful these days because we have other analytical tools. With appropriate reference, chiral chromatography, like HPLC, will tell you what you have. Specific rotation was originally used to help identify which of the two enantiomers you had. This doesn't tell you about the internal structure, since there's no easy way to figure out the rotation based on the structure (unless you use computational chemistry, but that wasn't available back then either).

Measurement of how much one enantiomer is present in excess of the racemic mixture is known as enantiomeric excess(optical purity). It is often denoted by ee.
ee=% of one enantiomer -% of another enantiomer.
It is important in the production of chiral drugs and discovery of asymmetric catalysts. It can be used to measure the purity of a compound. Also, enantiomeric excess is important when determining the optical activity of a mixture of two enantiomers.


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Thermodynamic properties software
I am looking for a software (can be freeware or purchased) to look up water and steam thermodynamic properties (e.g. density, enthalpy, entropy) preferably a macro in excel. What do you use for this application?

I have access to several programs or similar solutions for the work with so-called steam tables. I usually prefer to use the NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9. It is a program that can calculate the thermodynamic and transport properties of water and steam as well as some other important fluids and their mixtures. It was developed by the National Institute of Standards and Technology (NIST) and is commercially available here.
Although I usually don’t do this myself, REFPROP can be used with Microsoft Excel. This is described in the corresponding Help file as follows:

Excel Spreadsheets
The REFPROP software includes dynamic link library ('.dll') support that allows a broad variety of applications to take advantage of REFPROP's capabilities.  REFPROP is distributed with front-end software that links Microsoft Excel applications to fluid property functions.
The REFPROP Excel front-end is supplied as a Microsoft Visual Basic module.  Excel automatically incorporates Visual Basic as a macrofunction execution, scripting and interface utility.  When the REFPROP front-end module is imported into an Excel spreadsheet, REFPROP functions become accessible to the user.  The proceedure for doing this is described below.
The information given below demonstrates the front end usage; however, the user is not bound to this format; rather, the preferred method would be to write one's own visual basic code that calls the DLL directly.  A dedicated module will increase the speed of the application and give the ability to use the REFPROP functions to their fullest.
The following supplementary instructions are intended for users who wish to create and modify their own spreadsheets that utilize the REFPROP functions based on the front-end module.  The Excel application included with the software is called REFPROP.xls and is located in the directory where the REFPROP program was installed.  Sample FORTRAN and Visual Basic files are also included.
Properties are calculated by calling various functions.  For example, the following line would be placed in a cell to calculate the density of argon at 300 K and 10 MPa:
=Density("argon","TP","SI",300,10)
Other properties can be calculated by using any of the following functions:
(…)

I downloaded a macro for this many years ago and have since lost it.  It might have been this one: http://www.mycheme.com/steam-tables-in-excel/ .  That link is worth a look either way.

You can also use Engineering Equation Solver or EES (http://www.fchart.com/ees/).
You get a free Student Version if you buy Cengel's books (http://novellaqalive.mhhe.com/sites/0072383321/student_view0/ees_software.html#)
You can even solve equations, manipulate units, create tables and plot different parameters.
With this software you can simply write rho = density('Substance',T=Temperature,P=Pressure).
Take a look at this screenshot:

And the solutions:

The power of it stands in the ability to use the units. You can also change the references and the standards of tables, without the need to write extra codes.

You should check out CoolProp library. It is a C++ library that implements:

Pure and pseudo-pure fluid equations of state and transport properties for 122 components
Mixture properties using high-accuracy Helmholtz energy formulations
Correlations of properties of incompressible fluids and brines
Computationally efficient tabular interpolation
Highest accuracy psychrometric routines
User-friendly interface around the full capabilities of NIST REFPROP
Fast IAPWS-IF97 (Industrial Formulation) for Water/Steam
Cubic equations of state (SRK, PR)

Other free options based on CoolProp

SmoWeb for online calculation
ThermoState for offline calculation

Note: I am the developer of ThermoState


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Can we force a reversible reaction in equilibrium to achieve 100% completion?

Can we force a reversible reaction in equilibrium to become a "complete" one? If so, will the equilibrium constant change?

It's said about reversible reactions that they never go to completion & I know that's because in reversible reactions, the mixture obtained in equilibrium is more stable than the extremes (i.e. either reactant or product) But if we force the reaction to proceed in a certain direction, say in the direction of products, will it ever go to 100% completion or near 100%?
My second question is regarding the equilibrium constant. I have read about equilibrium constant that its value doesn't change if we alter the initial concentration of reactants, which I understand why, but consider this situation:
 if we force equilibrium to proceed in the direction of products by altering $p,T,V$ (only and not adding or removing any substance), the equilibrium will get disturbed and obtain a new equilibrium. Wouldn't the equilibrium constant for this new equilibrium be different from the earlier one?

My bad , i should have been clearer, the first one is also a Q, my Q is that can rev che rexns be made to go to completion somehow by altering the parameters , or will the eqb keep on acquiring new eqb states & not go to completion.

The two questions are sort of related, so i'll answer to the best of my ability.
A reaction in dynamic equilibrium can get close to completion depending on its equilibrium constant, which is a ratio between products and reactants at a certain temperature. No, the equilibrium constant won't change if you add more reactants in an attempt to push the reaction to completion. We refer to Le Chatelier's principle which states that if one were to add another parameter, say a reactant, the reaction will act to stabilize the stress that has been placed on it.
Take this reaction $\ce{PCl3 + Cl2 <=> PCl5}$. If I were to add more $\ce{PCl3}$, the change we would witness is a shift in equilibrium, that is to say, the system will react to consume more reactants and produce more products. What we see essentially is the system reacting to stabilize the changes we made.
This principle isn't just limited to adding reactants or products, however. We see the system tries to stabilize changes made to pressure, temperature, and volume. If we increase pressure, how will the system act to stabilize it? Pressure is a result of the gas particles hitting the walls of the container. Thus we would see the system reduced the number of gas particles in order to reach equilibrium again. Essentially, it will shift to the side with the least amount of gas particles. Volume and pressure are obviously related in this principle. An increase in volume is a decrease in pressure and a decrease in volume is an increase in pressure.
Take this reaction for example $\ce{CaCO3(s)<=>CaO(s) + CO2(g)}$. If we increase the volume (decrease the pressure), we would see the system shift to the right because it has the most moles of gas and we need more gas particles to fill the volume and stabilize the pressure.
Lastly, when changing heat we view it as a product or a reactant for exothermic and endothermic reactions respectively. An exothermic reaction has heat written as a product $\ce{A <=> B} + \text{heat}$ and an endothermic reaction has heat written as reactant. Thus, when we change the heat of an endothermic or exothermic reaction, we are changing its reactants or products and we treat it in the same way we treat a change in concentration.
The point is that the equilibrium constant remains constant, otherwise it would be called the equilibrium variable. We do, however, see a change in equilibrium position.

I downvoted this answer because it does not answer the question "But if we force the reaction to proceed in a certain direction, say in the direction of products, will it ever go to 100% completion or near 100%?". This answer currently just repeats what is commonly given in introductory texts about the Le Chatelier's Principle.

To make it short and simple no you cannot because that is how the reaction works, however you can change the equilibrium constant by increasing or decreasing the temperature. One thing to point out is that adding or decreasing pressure, concentration can effect the equilibrium and make it shift to a certain direction. K will change if you make a change that the reaction cannot fix like temperature.
But one thing to note is that if we do get the reaction to go one direction then there is no concept of equilibrium and reversible reactions. There won't be a K because it will always go a certain way. You can't change the K when there isn't one.

You can force it.
No equilibrium constant if it is a "complete" reaction.

This is just my insight into this with what I've learned in college chemistry.


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Why does Ammonium sulfate cause proteins to precipitate?
I'm trying to understand an observed reaction from a chem practical:
You have a solution of ammonium sulfate, which you add to a bovine serum albumin solution. The proteins precipitate, and the solution becomes cloudy. w/v % of $\ce{(NH4)2SO4}$ at this point is 57%.
When you add water (in a ratio of 2 parts water: 1 part serum solution)to the mix, the process reverses, and the solution becomes clear again.
This can be explained by the ammonium sulfate solution 'stealing' water from the proteins, which causes them to precipitate, and then the process reversing once more water is added. (Right?)
Question: Does the ammonium sulfate 'steal' the water because it wants to reach equilibrium, or because it's supersaturated, or what?

https://en.wikipedia.org/wiki/Ammonium_sulfate_precipitation https://en.wikipedia.org/wiki/Salting_out and https://en.wikipedia.org/wiki/Protein_precipitation have info about that.

Well, I wouldn't know how to sum it up better than chemwiki.ucdavis:

To achieve this conformation the molecule folds in such a way that all of the hydrophobic parts of a molecule are aggregated together and the hydrophillic groups are left to interact with the water. In the case of proteins it is the charged amino acids that allow selective salting out to occur. Charged and polar amino acids such as glutamate, lysine, and tyrosine require water molecules to surround them to remain dissolved. In an aqueous environment with a high ionic strength, the water molecules surround the charges of the ions and proteins. At a certain ionic strength, the water molecules are no longer able to support the charges of both the ions and the proteins. The result is the precipitation of the least soluble solute, such as proteins and large organic molecules.

If you ask what the aminoacid content of bovine serum albumin is, I'll cite:

Glycine,
  illanine,
  Valine,
  Leucine,
  Isoleucineh,
  Proline,
  Phenylalanine,
  Tyrosine,
  Tryptophan,
  Serine,
  Threonine,
  Half-cystinee,
  Methionine,
  Arginine,
  Histidine,
  Lysine,
  Aspartic acid and
  Glutamic acid. Source


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Rate constant and Michaelis Menten
Please consider the following figure of a concentration over time (msec) diagram with variables for an enzyme, a substrate bound enzyme and a product.

Recall that $\ce{E + S \leftrightharpoons ES \rightharpoonup E + P}$ and also recall that the velocity constant of the respective reactions are noted $k_1 , k_2, k_3$.
What I would like to know is:
On the plot provided, is the slope of $[ES]$ equal to $k_2$ and is the slope of $[E]$ equal to $k_1 + k_3$ thereby giving some insight/explanation for $k_m$ which is used to determine the affinity of an enzyme?
Any more details on Michaelis Menten is greatly appreciated but please note that my level of understanding is limited on the topic.

The data in the graph are not really corresponding to the chemical equation.  Assuming the data for [E] and [ES] are correct, [P] vs t should have increasing slope (be a concave-up curve, nearly-parabolic because [ES] vs t is nearly linear) until [ES] reaches a constant value around 2msek, beyond which point [P] vs t should be linear with a substantially positive slope.

Given $\ce{E + S <=>ES -> E + P}$
$\frac{d[\ce{ES}]}{dt}=-\frac{d[\ce{E}]}{dt}=  k_1[\ce{E}][\ce{S}]-k_2[\ce{ES}] -k_3[\ce{ES}]  $

is the slope of $[ES]$ equal to $k_2$ ?

no

is the slope of $[E]$ equal to $k_1 + k_3$

no

On the plot provided, is the slope of [ES] equal to k2 and is the
  slope of [E] equal to k1+k3 thereby giving some insight/explanation
  for km which is used to determine the affinity of an enzyme?

As DavePhD said, no to both counts.  First note that in your system, $\ce{E_0=E +ES}$, where $\ce{E_0}$ is the total amount of enzyme added, and $\ce{E_0}$ is conserved -- enzymes, as catalysts, are created or destroyed in the reaction.  This means $\frac{d\ce{E_0}}{dt}=0$, which implies that $\frac{d\ce{E}}{dt}=-\frac{d\ce{ES}}{dt}$.  So the slope of [ES] is negative the slope of [E].
Second, as you study more about Michaelis Menten kinetics, you will come to appreciate that this data is (mostly) for the unsteady-state regime, also often called the presteady-state regime.


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Wavelength of iPhone 4S camera light for visible spectroscopy calibration
I ordered a DIY spectroscopy kit from public-lab, however don't have access to a halogen lamp to calibrate the wavelengths.
I'm wondering if it would somehow be possible to calibrate the wavelength using the iPhone camera light?

According to the site, wavelength calibration is meant to be done with a compact fluorescent lamp (for the mercury lines), but at the bottom there's a tool for calibrating using any two known wavelengths. If the LED on the phone has two distinct lines that you can calibrate against, it may work, but something like a pair of monochromatic LEDs or laser pointers (say one blue and one red) or something will probably work better.

Just out of curiosity, if I have a band of red/blue/green, would these wavelengths be fixed for the standard wavelengths of red/blue/green, regardless of the incoming light source? Could I then calibrate with two of these colours?

Only if you know exactly what the wavelength is. The nice thing about LEDs is that you can look at the datasheet and it'll tell you the maximum and the tolerance.

I don't know about the exact calibration setup of the spectroscopy kit, which sounds interesting, but i doubt that the cell phone lamp will be useful here.
Both lamps seem to emit while light, but both the ways how light is generated, as well as the emission spectra of cell phone camera light and a halogen lamp are rather different.
Cell phone camera lights are typically blue light emitting diodes partly covered with one one more phosphors (not, it isn't phosphorous) which absorb the blue light and emit yellow (or orange) light. The emission spectrum of the devices consists of some distict bands.
Halogen lamps are tuned heavily incandescent lamps with a tungsten filament. Here, the spectrum is more similar to a black body radiator.
Consequently, I suggest to get a halogen lamp. A cheap 12V lamp with a socket (often GU4) and a power supply will probably work fine.
"Naked lamps" do emit in the UV too; Lamps with a reflector to provide a collimated beam have an additional glas cover that shield the UV part. Check the manual of the spectrometer kit for some directions.

My dad told me about building a fluorescent detector with a halogen headlamp bulb from a car when he was in grad school 30 years ago. I assume a headlight bulb would be an ok source? I know you can get a replacement socket with wires you can strip and solder, and a 9V battery with 2 AA batteries in series would provide 12V to light it up, but I don't know how long it would run.

I'm not a spectroscopist but I wonder if wavelength calibration is all that is required (not intensity calibration), isn't the fact that cell phone camera lights emit in "some distinct" bands a good thing?  If you can figure out precisely what wavelength those bands are, doesn't that permit calibration?

@CurtF. Intensity calibration would probably be included in the process of blanking the spectrophotometer and creating a calibration curve for your current compound of interest.

Interesting. If I bought a "naked lamp" for the UV, would a small cheap usb webcam be able to pick this up, or would I need a special one?


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Entropy - Negligible?
In electrochemistry, the Nernst equation tells us the cell potential for non-standard conditions. I read that the factor $\frac{-RT}{nF}\ln{Q}$ accounts for difference in cell potential due to differing Gibbs Free Energy due to changing temperature and concentration.
This difference in Gibbs Free Energy is only due to entropy right? Because enthalpy is a constant? So obviously, entropy significantly interferes in the process of electrolysis.
And yet, it is ignored in virtually all spheres of chemistry. I'm beginning to doubt my chemical intuition because of entropy effects, and faith in it can only be restored if entropy really was negligible. So it is negligible or not? If not, why is it always ignored?

This difference in Gibbs Free Energy is only due to entropy right?

This is mostly true.  (I think it is 100% true when concentrations can be used in the reaction quotient $Q$, but am not sure if it is strictly true when solutions are not ideal and activities have to be used instead.)

Because enthalpy is a constant?

Enthalpy is a "constant" only because of the way electrochemical standard states are defined.

So obviously, entropy significantly
  interferes in the process of electrolysis.

Have no idea what this means.  How does it "interfere"?

And yet, it is ignored in
  virtually all spheres of chemistry.

False.

I'm beginning to doubt my chemical
  intuition because of entropy effects, and faith in it can only be
  restored if entropy really was negligible. So it is negligible or not?
  If not, why is it always ignored?

It isn't negligible and is not ignored.


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Why is Ag+ a soft acid?
A soft species is typically large, polarizable and has a low charge (magnitude). The silver cation is not highly charged (this supports softness) however, it is anomalously small due to high nuclear charge and poor shielding by the d-orbitals. Consequently, shouldn't it also be less polarizable?
It does have a tendency to form bonds that are significantly covalent in nature which backs up the assertion that $\ce{Ag^+}$ is soft; however, I don't understand why.

Well, it's still big enough...

According to HSAB theory, Soft Acids have:
large atomic/ionic radius;
low or zero oxidation state bonding;
high polarizability;
low electronegativity.
The ion $\ce{Ag+}$ satisfies all the aforementioned properties.
I don't agree with you when you say it is anomalously small due to high nuclear charge and poor shielding by the d-orbitals. The effective nuclear charge has nothing to do here.
According to http://www.webelements.com/silver/atom_sizes.html: The radius of this ion is $114\ \mathrm{pm}$ when it's in 4-coordinate, tetrahedral geometry and it's $116\ \mathrm{pm}$ when  it's 4-coordinate, square-planar geometry. Please just compare these radii with the radius of  $\ce{Al^3+}$ ion : $53.5\ \mathrm{pm}$ which is considered as a hard acid.
I hope it's clear now!

Hi, sorry for reopening this topic, but the K+ ion is considered hard and its radius according to your source is 152pm, which is larger than the silver+ ion, so what causes K+ to be hard?

Polarizability enters as well.  The $4d$ electrons in silver ion are relatively polarizable and so soften this ion beyond what you might expect from ionic radius alone.  Potassium ion does not have that.

probably it is related to its filled $\mathrm{d}$ orbitals. Filled or nearly filled orbital elements counted as soft acids, and $\ce{Ag+}$ has totally filled $\mathrm{d}$ orbitals, and it can donate to ligands, making ligand-$\pi$ bonding, which causes stronger interaction.

Neither zinc(II) nor copper(II) would qualify as soft in any HSAB definition of the word.

Filled or nearly d orbitals is one of the factors should be considered to label an acid soft (http://www.rbmcollege.ac.in/sites/default/files/files/reading%20material/hard-and-soft-acids-and-bases.pdf) but not the sole condition. Several properties should be considered when deciding an acid is soft or hard.  Zinc(II) and copper(II) are smaller in ionic radius (both around 70 pm, while Ag(I) around 110 pm) and also they are highly charged.


text:
pH Curves for Titrations
Why does the pH rapidly increase near the equivalence point during--why does the slope of the graph of a pH curve sharply increase around this area?

I'll try to answer your question firstly qualitatively, then quantitatively (using mathematical equation):
Let's consider the case of  the titration of strong acid (volume $V_a$ and concentration$C_a$: unknown) with a strong base (volume $V_b$ and concentration$C_b$), and we'll denote $V_b(eq)$ the volume of base at the equivalence:

The pH increases slowly at first because the pH scale is
logarithmic, which means that a pH of 1 will have 10 times the
hydronium ion concentration than a pH of 2. Thus, as the
hydronium ion is initally removed, it takes a lot of base to change
its concentration by a factor of 10, but as more and more
hydronium ion is removed, less base is required to change its
concentration by a factor of 10. Near the equivalence point, a
change of a factor of 10 occurs very quickly, which is why the
graph is extremely steep at this point. As the hydronium ion
concentration becomes very low, it will again take a lot of base to
increase the hydroxide ion concentration by 10 fold to change the
pH significantly.
Let's find the equation of the titration curve in the the region $0<V_b<V_b(eq)$:

The number of moles of $\ce{H3O+}$ before the titration is $C_aV_a$
The number of moles of $\ce{H3O+}$ after adding the volume   $V_b$ of the base is $C_aV_a-C_bV_b$
Then, the concentration of $\ce{H3O+}$ is $$[\ce{H3O+}]=\frac{C_aV_a-C_bV_b}{V_a+V_b}$$
So, the titration curve in the region $0<V_b<V_b(eq)$  $$p\mathrm{H}=-\log\frac{C_aV_a-C_bV_b}{V_a+V_b}$$
If you trace this mathematical function $p\mathrm{H}=f(V_b)$, you will have a smooth slope at the beginning and as $V_b \rightarrow V_b(eq)$, the slope of the graph sharply increases as the function becomes not defined: $C_aV_a=C_bV_b(eq)$.

A while back I made a more formal derivation in a similar question, at the expense of making it significantly harder to read. Perhaps the asker can also browse the answers there.

You're right, the answer given there is really nice. Thank you


text:
Chemical potential of mixtures
I feel that there are very few textbooks that treat the chemical potential of mixtures in an understandable clear way, which is why I wanted to ask here about certain things. Although I do not have a single question, but 5 of them, I think that it is merely impossible to separate them thematically, which is why I hope to meet an expert here.
If we have an ideal gas then if we know the chemical potential for a particular $(p,T)$ we have:
$$\mu(p_2,T) \approx \mu(p_1,T) + \left(\frac{\partial \mu}{\partial p} \right)_T(p_1,T) (p_2-p_1) = \mu(p_1,T) + \frac{V}{N}(p_2-p_1).$$
Similarly for changes with respect to temperature we have
$$\mu(p,T_2) \approx \mu(p,T_1) + \left(\frac{\partial \mu}{\partial T} \right)_p(p,T_1) (T_2-T_1) = \mu(p,T_1) - \frac{S}{N}(T_2-T_1).$$
Now, if we have either a gas or a liquid (Question 1: is this true, does this law hold for both fluids?) $\ce{A}$ and put another material $\ce{B}$ into it, then we have by Raoult's law:
$$\mu_{\mathrm{A,~with~mixed~B}}(p,T) = \mu_A(p,T) + RT\ln\left( \frac{N_A}{N_A + N_B} \right)$$
Question 2: Now, is the chemical potential of the total fluid given by $A$ and $B$ then the sum $\mu_{\mathrm{A+B}} = \mu_{\mathrm{A,~with~mixed~B}}(p,T) + \mu_{\mathrm{B,~with~mixed~A}}(p,T)$?
Let's imagine that $A$ and $B$ would be gases, would this mean that
Question 3 : $$\mu_{\mathrm{A,~with~mixed~B}}(p,T) = \mu_\mathrm{A}(p,T) + RT\ln\left( \frac{p_\mathrm{A}}{p_\mathrm{A} + p_\mathrm{B}} \right) ?$$
Question 4 : And what if $\ce{A}$ and $\ce{B}$ would be two two miscible liquids? could we still use the vapour pressures to calculate the chemical potential of the liquids mixture?
$$\mu_{\mathrm{A,~with~mixed~B}}(p,T) = \mu_A(p,T) + RT\ln\left( \frac{p_\mathrm{A}}{p_\mathrm{A} + p_\mathrm{B}} \right) ?$$
Now, imagine that I have a liquid $\ce{A}$ and salt $\ce{B}$ in it. Also, I have a gas layer $\ce{C}$ of $\ce{A}$ above the liquid. What exactly is now in equilibrium: $\mu_{\mathrm{A+B}} = \mu_\mathrm{C}$, or $\mu_{\mathrm{A,~with~mixed~B}} = \mu_\mathrm{C}$?
Question 5: My last question would be: Imagine that you have a liquid (let's say water) and some gaseous water phase above (according to the vapour pressure). Now, you add oxygen to this. What would this mean for the chemical potentials of the liquid/gas water?

It's much simpler than all that.
Pure gas
$$\mu_g(T,P)=\mu^0(T)+RT\ln P$$
where $\mu^0(T)$ is the molar free energy of the pure gas at temperature T and pressure 1 Bar, and P is the pressure expressed in bars.
Pure liquid
If $P_{sat}(T)$ is the equilibrium vapor pressure at temperature T, then, at pressure $P_{sat}(T)$, we know that the free energy of the liquid is equal to the free energy of the vapor.  So, $$\mu_l(T,P_{sat})=\mu^0(T)+RT\ln P_{sat}(T)$$If we neglect the effect of pressure on the chemical potential of the liquid (because the specific volume is very small), then, for a pure liquid at temperature T and pressure P:$$\mu_l(T,P)=\mu^0(T)+RT\ln P_{sat}(T)$$
Ideal gas mixture
For species A in an ideal gas mixture, the chemical potential of species A is the same as the chemical potential of the pure species, except evaluated at the partial pressure of A in the gas mixture:$$\mu_{Ag}(T,P,y_A)=\mu^0_A(T)+RT\ln p_A=\mu^0_A(T)+RT\ln (Py)_A=\mu_A(T,P)+RT\ln y_A$$
where $p_A$ is the partial pressure of A (bars), $\mu^0_A(T)$ is the free energy of pure A at temperature T and pressure of 1 bar, P is the total pressure of the mixture (bars), $\mu_A(T,P)$ is the free energy of pure A at the temperature and  pressure of the mixture, and $y_A$ is the mole fraction of A in the mixture.  So the chemical potential of A in the mixture is equal to the chemical potential of pure A at the same temperature and pressure as the mixture plus the correction term for mixing $RT\ln y_A$.
Ideal liquid mixture
By analogy to an ideal gas mixture, the chemical potential of species A in an ideal liquid solution is given by:
$$\mu_{Al}(T,P,x_A)=\mu_{Al}(T,P)+RT\ln x_A=\mu^0_A(T)+RT\ln P_{A,sat}(T)+RT\ln x_A$$
where $\mu^0_A(T)$ is the free energy of pure A at temperature T and pressure of 1 bar, $P_{A,sat}(T)$ is the saturation vapor pressure of species A at temperature T (bars), and $x_A$ is the mole fraction of species A in the liquid.

In the last equaiton does "pure A" means pure liquid or gas?

The superscript 0 means pure gas.  I guess there should also be a Poynting correction for the pressure.


text:
What is the density of compressed natural gas (CNG) used in vehicles?
I would like to know the density of the compressed natural gas used for natural gas vehicles in order to complete the formula in this Mathematics Stack Exchange post.

The density of CNG at the pressure of 200 bars (which is optimized for use in vehicles) is 435 $\frac{kg}{m^3}$. However, it's (not usually) 175 $\frac{kg}{m^3}$ in some vehicles in some countries. Note that it differs according to your home country. Source 1 and source 2.

If I plug that number (175) into the formula in the other post. I get 74,399.95 miles per gallon. That does not make sense with the data I have been given. This does not mean that your answer is wrong though.


text:
Chemical bonding and hybridization of organic compounds
What happens to the lobe of the p-orbital during sp² hybridization?
During the formation of $\ce{C2H4}$ molecule:
We know that both the carbon atoms will have a total of 3 p-orbitals and they will be involved in hybridization.
Let us number the carbon atoms as $\ce{C}1$ and $\ce{C}2$.
Now the p-orbital of $\ce{C}1$ along $x$-axis will form a $\unicode[Times]{x3C3}$ bond with the p-orbital of $\ce{C}2$ and a $\unicode[Times]{x3C3}$ bond with the s-orbital of hydrogen.
And similarly the p-orbital of $\ce{C}2$ along x-axis will form a $\unicode[Times]{x3C3}$ bond with hydrogen.
So both the lobes of the p-orbital of $\ce{C}1$ and $\ce{C}2$ along $x$-axis are involved in sigma bond formation.
Lets say the p-orbital of $\ce{C}1$ and $\ce{C}2$ along the $y$-axis forms a $\unicode[Times]{x3C0}$ bond.
So here also both the lobes of p-orbital of $\ce{C}1$ and $\ce{C}2$ are involved in hybridization.
Let say that the p-orbital of $\ce{C}1$ which is along the $z$-axis forms a $\unicode[Times]{x3C3}$ bond with the s-orbital of hydrogen.
And similarly the p-orbital of $\ce{C}2$ along $z$-axis forms a $\unicode[Times]{x3C3}$ bond with the s-orbital of the other hydrogen.
Now my confusion is that here two p-orbitals along the $x$ and $y$-axes have four lobes and all of them are involved in bond formation.
But the p-orbital which is along $z$-axis has only one of its lobes involved in bonding. What happens to the other lobe of this p-orbital?

You talk about hybridisation but you haven't actually looked at any hybrid orbitals in your description. Since the geometry around the carbons is trigonal planar, each carbon hybridises its $\ce{2s}$, $\ce{2p}_x$ and $\ce{2p}_y$ orbitals to form three $\ce{sp^2}$ orbitals which lie in the $xy$ plane at $120^\circ$ angles to each other. One $\ce{sp^2}$ orbital on each carbon points towards the other carbon and so these orbitals overlap to form a $\ce{C-C}$ $\sigma$ bond. The other two $\ce{sp^2}$ orbitals on each carbon overlap with the $\ce{1s}$ orbitals on the hydrogens forming $\ce{C-H}$ $\unicode[Times]{x3C3}$ bonds. The remaining unhybridised $\ce{2p}_z$ orbitals on each carbon overlap above and below the plane of the $\unicode[Times]{x3C3}$ bonds to form a $\ce{C-C}$ $\unicode[Times]{x3C0}$ bond.
While $\ce{sp^2}$ orbitals still have two lobes, one lobe is substantially larger than the other and only that takes part in bonding. Note that this is the usual case: even in molecules that solely use p-orbitals, such as chlorine, only one lobe of a p-orbital will form a $\unicode[Times]{x3C3}$ bond; the other of the same orbital will not.
This link provides a nice visualisation of the different orbitals involved in ethene bonding.


text:
What kind of interacion is there between a nanoparticle and its functionalizing coating?
The functionalization of nanoparticles (with coatings such as polymer ligands; see image) are very common (for example, DNA-functionalizied gold nanoparticles). What's the type of interaction between nanoparticles and their coatings? And why is it more favorable than forming aggregations of the the nanoparticles?

In small nanoclusters/nanocrystals/quantum clusters (my goodness, they have so many names nowadays) in the size range around 13, 55 and 144 metal atoms,1 the ligands enter a real orbital interaction (dative bond2). They are ligated to the metal core (Puls et. al., 2014).
In DNA-ligated gold clusters, the phosphate groups are able to bond to the metal core and substitute the phosphine ligands it had before (Yunping et. al., 2003).
In larger nanoparticles,3 the tendency of the particles to agglomerate is much smaller. Some of them can be measured with TEM (Transmission Electron Microscopy) without full agglomeration. This is not given for the small 13 metal atom clusters. The big particles enter interactions with their surfactants for which the main contribution is given either by van der Waals forces, or by electrostatic interactions.

1 Those are some of the most common numbers in gold cluster chemistry, like $\text{Au}_{13}(\text{PR}_3)_{12}$ for example. 55 and 144 can be found in thiolated gold clusters.
2 It has covalent portions as well as ionic. It is, however, better understood if you think of coordination complexes. Only in this case, there is not only a single metal atom in the centre
3 i.e. 5 nm, 12 nm. I stated this because I am not aware of any case in which a 13-metal-atom cluster was analysed by TEM (Transmission Electron Microscopy). The conditions are simply too harsh, the ligand shell is ripped off and the little clusters experience a very strong tendency to agglomerate. You need specialised TEM equipment to acquire meaningful data. Nanoparticles of a few nanometers are less prone to show such a fast agglomeration.

where do the numbers 13,55,144 come from?

Those are some of the most common numbers in gold cluster chemistry, like $\ce{Au13(PR3)12}$ for example. 55 and 144 can be found in thiolated gold clusters.

what exactly do you mean by "enter orbital interaction"? covalent bond?

In this case I mean a dative bond. http://www.chemguide.co.uk/atoms/bonding/dative.html It has covalent portions as well as ionic. It is, however, better understood if you think of coordination complexes. Only in this case, there is not only a single metal atom in the centre.

you said larger nanoparticles agglomerate less. by "larger" you mean larger than 13 metal atoms?

In this case I meant 5 nm, 12 nm. I stated this because I am not aware of any case in which a 13-metal-atom cluster was analysed by TEM (Transmission Electron Microscopy). The conditions are simply too harsh, the ligand shell is ripped off and the little clusters experience a very strong tendency to agglomerate. You need specialised TEM equipment to acquire meaningful data. Nanoparticles of a few nanometers are less prone to show such a fast agglomeration.

Sorry, I just removed the "superatom model" part. I confused that with a different consideration. Therein, the AuL units are viewed as ligands to the central atom in $\ce{[M(AuPMe3)11(AuCl)][GaCl4]3}$ with (M = Ni, Pd, Pt). The twelve AuL "ligands" then form molecular orbitals which are reminiscent of atomic orbitals and thus can be described with the superatom model. (http://dx.doi.org/10.1002/anie.201310436)


text:
In a monomolecular reaction, how can a reactant experience collision to react?
Monomolecularity means no molecules which experience collision with each other taking part in reaction. So how it is possible that in monomolecular reaction only one reactant experiences collision?

One reactant, e.g. O3, a pure chemical species, has many molecules that are in continual motion and can collide with each other (as well as with the walls of a container or with a catalyst), so you'll get some O2 generated. Or the converse, if you have UV photons colliding with pure O2 gas.


text:
"Editing" atom nucleus
First of all. I am really bad at chemistry and all that. So this might or more likely will sound at least primitive... But this question has been bothering me for far too long.
So, I was reading about an atom. How all different kinds of it differ by the number of protons in its nucleus. And that gave me this utopian thought: maybe in far (like, really far) future we will have some sort of super "printers" which could take anything as their "ink" and somehow turn that anything (a brick for example) into manageable atoms.
Then it probably should identify each atom, by counting its protons and edit it by "deleting", "adding" or separating needed number of the protons and neutrons.
So my question is... Do you think this is theoretically possible?
Are there any known ways of changing the number of protons?
*Edit: as it turns out, there is a talk going on about this kind of utopia: https://www.youtube.com/watch?v=RzgVWpa4fzU
Big thanks for your answers! I wasn't expecting to receive any answers :)*

Theoretically possible perhaps, but I would be amazed if such a thing ever became practical. In a very direct sense, you start off by inverting a nuclear weapon; that is, you'd be taking the equivalent of the explosive yield of a nuclear device to pulverize a couple of kilograms of matter into ink. After that's done, then by setting the ink, you're redetonating the nuclear device that was just inverted. Managing that amount of energy for so little product is absurd, not to mention that even an outstanding energy efficiency of 99% would imply a colossal power input to keep it going.

Your talking about sth what is called nanoassemblers, but generally they aren't thought to change atoms they are given as input - if you'd want assembler to make sth from iron, you should give it iron, not for ex. titanium and expect it to turn into iron via nuclear reactions.

Transforming a nucleus isn't really chemistry but nuclear physics. Also to transform a nucleus to a different atomic number takes a lot of energy. To look at it another way, the atomic bomb dropped on Hiroshima converted less than the mass of a US penny into energy.

Theoretically it's possible to create atoms from raw particles and to assemble them; science fiction posits "replicators" in Star Trek, http://en.wikipedia.org/wiki/Replicator_%28Star_Trek%29, and elsewhere.
It is done, a few billion or trillion atoms at a time, when matter is bombarded by protons in a cyclotron or linac. Americium, 241Am, for example, was first produced that way, and you probably have some in smoke detectors in your house, though now it is extracted from the random products of nuclear fission reactions. See http://en.wikipedia.org/wiki/Americium.

In part, your vision already exists and uses electrons as "ink". In a process kown as electron capture, a proton "swallows" an electron of sufficient energy and is thus converted to a neutron.
$$\ce{^{A}_{Z}X + e- -> ^{A}_{Z-1}Y + \nu_{e}}$$
As a consequence, the atomic mass number A remains the same, but the atomic number Z is decreased by one: Element X is converted to element Y.
A typical example would be the conversion of aluminium to magnesium:
$$\ce{^{26}_{13}Al + e- -> ^{26}_{12}Mg + \nu_{e}}$$


text:
Is it possible to derive integrated rate law algebraically?
Here is the calculus way:

I have no prior knowledge of calculus. But my teacher asks me to derive the integrated rate law for any order between one to ten.
It would be fine for me to memorize the derivations for zero to third orders. But I can't do all ten for sure.

What is your question exactly ... ? And what have you tried ?

My question is how to derive integrated rate law algebraically. I found what is image.  I can googlr all the formulas out. But It all seemed to me that the only way is by calculus.

For a function $f(x) = x^n$, the primitive integral is $F(x) = \vert\ln(x)\vert$ if $n=-1$, otherwise $F(x) = \frac{1}{n+1}x^{n+1}$.

We are fortunate to have a very nice $\LaTeX$ capability on the site.  Please consider writing out the equations rather than pasting in a picture.

Calculus is pretty much just a shorthand for algebra (plus the whole limits concept), so the answer is technically yes.
Your teacher might be intending for you to use the pattern @Klaus Warzecha mentioned:

For a function $f(x) = x^n$, the primitive integral is $F(x) = \vert\ln(x)\vert$ if $n=−1$, otherwise $F(x) = \frac{1}{n+1}x^{n+1}$

So in that sense, you can use algebra if you let
$$\int_{A_0}^{A_\tau} \! \frac{1}{[A]^n} \, \mathrm{d}[A] = \frac{1}{n - 1} \left[ \frac{1}{[A]^{n-1}_{\tau}} - \frac{1}{[A]^{n-1}_{0}} \right]$$
for $n \neq 1$
and
$$\int_{A_0}^{A_\tau} \! \frac{1}{[A]} \, \mathrm{d}[A] = \ln \frac{[A]_\tau}{[A]_0}$$
for $n = 1$
Note that this only works for single-species rate equations - if you have multiple species the integration result changes. Note that this also isn't really using algebra - it's memorizing and applying one of many general integration results. Which is what you do in integral calculus.
In any case, it doesn't seem fair to me to expect you to know how to derive integrated rate laws if integral calculus isn't a prerequisite for your course. If it is a prereq, then now you are seeing why. If it isn't, then what your teacher is asking you to do is, in my opinion, unrealistic.


text:
Grignard formation and vicinal dihalides
So let's say I have a symmetric 1,2-dihalide (the halides are identical) and I wanted to make a Grignard and I only add 1 eq. of magnesium metal ...
Will I get 1 eq. of asymmetric Grignards (i.e. only one side of the product has the C-MgBr bond) as my product or will:
a) Some of the product be "double" Grignards - i.e. both sides of what used to be the 1,2-dihalide now have C-MgBr bonds?
b) I get some E2 elimination product from Grignards robbing protons from unreacted 1,2-dihalides?
c) A combination of the above?
d) Something else entirely?

Let's say the above reaction works the way I want it to and we get the 1 eq. of asymmetric Grignard reagent.
We then react it with a carbonyl containing compound such as acetone ... do we get an epoxide? Are there even any routes to an epoxide? This suggested as a possible product ...
But I don't see the epoxide.

Definitely can't be an epoxide... I'm pretty sure you'd just get ethylene from this reaction. The anion (corresponding to the Grignard) will just kick out the other bromide.

If a ring closure would occur after having added $\ce{-CH2-CH2-Br}$ the product wouldn't be an epoxide but an oxetane.

Taken from the Virtual Textbook of Organic Chemistry, which is maintained by William Reusch.

If two halogen atoms are present in a given compound, reactions with
  reducing metals may take different paths depending on how close the
  carbon-halogen bonds are to each other. If they are separated by four
  or more carbons, as in the first example below, a bis-organometallic
  compound may be formed. However, if the halogens are bonded to
  adjacent (vicinal) carbons, an elimination takes place with formation
  of a double bond.

He does however not provide any explanation for this behavior. It could be that this result is largely empirically based, about some aspects of the Gringard reaction mechanism there is not much known about (or consensus among chemists). I looked at Clayden et al. and Vollhardt & Shore and neither give a satisfying answer. If I find any in an other perhaps more advanced textbook, I'll edit this answer.

1,2-Dibromoethane is often used to activate Mg in ether prior to formation of a Grignard reagent. Yes, ethylene is formed.


text:
How does VSEPR theory explain the formation of different bond angles in PCl₅?
We know, that in the exited state, electrons of Phosphorus could go to higher states and form $\ce{dsp^3}$ hybridization. Therefore it will create five equal energy states of electrons. But my question is that, why are the angles of the bonds in $\ce{PCl5}$  90 and 120 degree instead of equally distributed? How does VSEPR theory explain the difference in the angles?
Why do the  five equal bonding pairs of electrons have a trigonal
 bipyramidal arrangement, with two different internal bond angles -
 120° in the plane and 90° between the plane and the axial atoms?

There are only five Platonic solids:

tetrahedron,
octahedron,
cube,
dodecahedron, and
isocahedron.

Therefore five atoms cannot be arranged in five equivalent positions, unless all five lie in a plane.
Trigonal bipyramidal is the minimum energy arrangement of five charges on a sphere. See Kevin Brown's MathPages Min-Energy Configurations of Electrons On A Sphere (Stable link via The Internet Archive).
Also there are not five equal energy states as demonstrated in the lecture notes by Marcel Schlaf and Kathryn Preuss from the University of Guelph:
Chapter 5.5 (continued 2) MO Theory ICl4, PCl5. (pdf) (Stable pdf link via The Internet Archive)

Here's a proof that you can't place 5 ligands equidistant from each other and from the central atom in a nonplanar molecule:

Place two points anywhere you like on the surface of a sphere around the central atom.
They will define a great circle (an "equator") that cuts the sphere in half.
You have 3 points left. You can't place them all off the "equator" without putting two in one hemisphere and 1 in the other. You cannot make all the points equidistant. The only way you can balance the number of points in each hemisphere is to put one in one hemisphere, one in the other, and one on the "equator" with the first two you placed. But the distance between the equatorial points will not be the same as all of the distances to points in the hemisphere.

Why is it trigonal pyramidal, though? The best we can do is make subsets of the ligands equidistant. Place three equidistant points on the equator, and two at the "poles". Imagine a linear shape (the axial positions) passing at right angles through a trigonal planar shape (the equatorial positions). You have a trigonal bipyramidal shape.
The dsp3 hybridization is a futile attempt to explain 5 orbitals around the central atom by mixing the central atom's orbitals alone. In fact back in the early 1990's it was shown that this is a poor explanation (source: E. Magnusson, Hypercoordinate molecules of second-row elements: d functions or d orbitals? J. Am. Chem. Soc. 1990, 112 (22), 7940–7951. DOI: 10.1021/ja00178a014.)


text:
Clarification for metallic bond
Recently, I learn about bonding and my teacher only focus on ionic and covelant bond, not metallic bond. So, I go on google search but I have no idea of what is it? How electrons can suddenly appear to form the bond? How can the metal highest energy level overlap if it need to obey the octet rule

Metallic bonds depend on the Pauli exclusion principle to cause a single energy level to be separated into a sea of very closely spaced levels.
Consider metallic Ca, with its s2 filled with two electrons of opposing spin state. You cannot have another s2 electron in the same atom. Now, add a second Ca atom that can freely interchange electrons with the first (e.g. metallic conductivity). Quantum mechanically, the electrons are indistinguishable, and an electron of the first atom might be found on the second one, so the combination acts as a "super-atom" with two nuclei but a single electron cloud. To allow this without violating the Pauli exclusion principle, one s2 level "moves up" and the other "moves down", splitting the level into two bands.
As more atoms are added, the Fermi "sea" becomes populated to the extent that the energy levels merge into a continuum.
See http://fog.ccsf.cc.ca.us/~wkaufmyn/ENGN45/Course%20Handouts/15_ElectricalProps/02_ConductivityConductor.html and http://www.chemexplore.net/sodium.htm for further explanation.

But how does this sea of electrons cause the atoms to stick together? Is it like with molecular orbital theory where there are a number of bonding orbitals and a number of antibonding orbitals and their ratio determines whether the atoms are repelled or attracted? So there'd be bonding bands and antibonding bands?

good question. My intuitive answer is that the well shielded s orbitals decoy the nuclei to form bonds with the non valence electrons in the closepacked nuclei. This attraction is greater than the shielded repulsion of the nuclei.  The QM of this evades my ken.