text:
Does rice protein have a high concentration of arsenic?
According to recent reports rice is high in arsenic. What about rice protein?
I am taking a protein powder that has "yellow pea protein isolate, sprouted whole-grain rice protein concentrate, organic instantized quinoa seed protein, organic Chia Sprout protein".
Would protein concentrate from Rice have high levels of arsenic?
http://www.consumerreports.org/cro/magazine/2015/01/how-much-arsenic-is-in-your-rice/index.htm

sigh Rice doesn't inherently have any arsenic at all. There may be a bit in one sample, a bit more in another. In theory there could be none whatsoever. This question is quite bad for a one which was upvoted 6 times and isn't HNQ.

To the contrary, I'd say arsenic is ubiquitous in soils worldwide and that every plant contains arsenic, it's just a matter of how much. If you don't measure any it's because your instrumentation doesn't have a low enough detection limit, not because the plant doesn't have any arsenic at all. That's true for about any element, but arsenic is a particularly big concern.

The US FDA reported arsenic testing of 12 rice protein powders as follows:

333863609,
  Dietary Supplements,
  Rice Protein Powders,
  Plant-Based Energizer:
  0.1
333511892,
  Dietary Supplements,
  Rice Protein Powders,
  Protein Energizer:
  0.8
333511937,
  Dietary Supplements,
  Rice Protein Powders,
  Protein Energizer:
  0.9
334620956,
  Dietary Supplements,
  Rice Protein Powders,
  Raw Vegan Chocolate:
  0.9
334094485,
  Dietary Supplements,
  Rice Protein Powders
  Plant-Based Drink Mix:
  1.0
333511880,
  Dietary Supplements,
  Rice Protein Powders,
  Raw Protein:
  1.1
333687971,
  Dietary Supplements,
  Rice Protein Powders,
  High Protein Energy Meal:
  1.4
333687931,
  Dietary Supplements,
  Rice Protein Powders,
  High Protein Energy Meal:
  1.6
334093871,
  Dietary Supplements,
  Rice Protein Powders,
  Blend of protein, fruits & veggies:
  1.6
334094067,
  Dietary Supplements,
  Rice Protein Powders,
  Chocolate Plant Protein Shake:
  3.7
334093956,
  Dietary Supplements,
  Rice Protein Powders,
  Blend of protein, fruits & veggies:
  4.4
333688312,
  Dietary Supplements,
  Rice Protein Powders,
  Organic Meal Replacement:
  5.7

Where the units are micrograms of inorganic arsenic per serving.

Thanks, So to cut to the chase, does rice protien have high levels of arsenic? Took a look at the link but looks like it requires a more concentrated sitting to understand. How many units of inorganic arsenic per serving is too high?

@larry909 well the FDA proposed an action level of 100 ppb inorganic arsenic for infants cereal https://www.fda.gov/food/foodborneillnesscontaminants/metals/ucm319870.htm and 1 of the 12 samples above exceeded that level.

Which one? How many micrograms equals 100 pbb?

@larry909 "Blend of protein, fruits & veggies" was 152 ppb of inorganic arsenic (230 ppb total arsenic). It was  4.4 micrograms of inorganic arsenic for a 29 gram serving.  (see page 58 of the pdf in the answer).

So in general, based on your understanding of these numbers, does rice protien powder have a higher level of arsenic per serving than a serving of regular rice? About the same?

@larry909 on average the rice protein powder had about half as much inorganic arsenic per serving as the rice.

To add some more context from ConsumerLab.com: "The cancer-warning threshold of the State of California is 10 mcg of arsenic per daily serving, meaning that if someone consumed more than one serving of some of these products in one day, they could exceed this amount."


text:
Structure of lead(II,IV) oxide
I need help sorting out the molecular structure of tri-lead tetroxide, aka lead(II,IV) oxide $\ce{Pb3O4}$.
This data sheet shows this image:

The Wikipedia's lead (II,IV) oxide's link to a Jmol interactive image looks like this:

Even the Wikipedia article's Talk Page's discussion on the naming and structure makes for interesting reading.
The first drawing shows a $\ce{Pb-O-Pb-O-Pb-O -}$ ring, the second shows a 4-bond $\ce{Pb}$ in the middle with two 2-bond $\ce{Pb}$'s on either side.
Is there a "right way" and a "wrong way" to draw a single molecule of $\ce{Pb3O4}$ that represents how it might actually be configured? Are both, either, or neither of the above representations "right"?

It's not a molecule, it's an extended solid. Drawing this is as wrong as drawing a covalently bound molecule of NaCl. For the structure of Pb3O4 see e.g. Wells Structural Inorganic Chemistry 5ed, p 558

@orthocresol Isn't there some fundamental concept in Chemistry - the solid can be subdivided to individual molecules? At some point can not a single molecule exist? Would it not be stable? If it can not exist, that would be the basis of an answer.

If you have access to scientific papers, this, Fayek, Leciejewi, Zeitsch. f. anorg. u. allg. Chemie, 2004, 336, 104-110 could be worth reading.
Also, what orthocresol said. Its a crystal, painting a molecule is in no case right.

@Fl.pf. I did not ask about a crystal. I asked about a single molecule. Can that simply not exist? Is that the answer? Or perhaps both might? There are molecular beams that can be neutral, there are inorganic molecules on gold or silicon surfaces in ultra-high vacuum just for example.

No, a crystal cannot in general be subdivided into individual molecules. I remember a question on this site about it, I will try to search.

@orthocresol Aren't water and  ice just two examples of distinct water molecules interacting via hydrogen bonds? At least most of them?

Yes, but ice melts to form water molecules. Pb3O4 doesn't melt to form Pb3O4 molecules. In fact it doesn't even melt, it decomposes (Wikipedia). NaCl doesn't melt to form NaCl molecules, it forms Na+ and Cl- ions. Therefore it would not be correct to think of NaCl solid as an aggregation of NaCl molecules.

@orthocresol OK let's not get side-tracked by what "in general" means, and I didn't ask about $\ce{ Pb_3O_4 }$ melting, and I could have just as easily said the water sublimated, right? Here, for this question, I would just like to know about a single molecule of $\ce{ Pb_3O_4 }$, and general guidelines of what happens in general are probably not going to help me right now.

Yes, but even if it sublimates it's still water, with H-O-H covalent bonds. The truth is that Pb-O bonding is not very covalent. As far as I can tell, there is no potential for the formation of discrete Pb3O4 molecules. Maybe it would help if you added some context as to why you're curious about the structure of a single Pb3O4 molecule.

@orthocresol OK that's something I can definitely understand - thanks! As far as the "why", there isn't a particular context as much as that it seemed to be an opportunity to understand chemistry better. Currently I thought of crystals as either of two extremes - just one giant molecule made of unit cells (e.g silicon) or an ordered array of distinct molecules (e.g. ice - no, not quartz!). So this molecule piqued my interest because it didn't seem to fit nicely with  my "two kinds of crystals" thinking.

Come to think of it, the very tags "inorganic-chemistry" and "molecular-structure" are inherently contradictory.

@Fl.pf. ok I've got the paper - neutron diffraction from 1965 - I love the papers when each step in the thinking is explained. Great choice!

@uhoh I'm glad I could be of assistance :)

@IvanNeretin would you say $\ce{N_2}$ is inorganic? And does it have a molecular structure? ;)

Yeah, well, all right, let's say "often contradictory".

@IvanNeretin I've asked a new molecular-structure question.

What you call a molecular structure is in reality an asymmetric unit. These two terms are not related, and in general are not interchangeable. The same way $\ce{Pb3O4}$ is not a molecular formula, but a formula unit.
All pictures you presented are correct in terms that they do represent an asymmetric unit, but actually an IUCr-recommended asymmetric unit (can be generated with Olex2's command standardise) looks like this:
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#575961}{\Large\bullet}~\ce{Pb}$.

... and with $Z = 4$ these are packed in a unit cell with $Pbam$ space group accordingly:

The crystal structure itself can be represented as infinite layers of cross-linked chains made of $\ce{[PbO6]}$ octahedra, which are stacked via $\ce{- O-Pb-O -}$ bridges:

Crystallographic data is taken from ICSD #9753 [1].

Gavarri, J. R.; Weigel, D.; Hewat, A. W. Journal of Solid State Chemistry 1978, 23 (3), 327–339. DOI 10.1016/0022-4596(78)90081-6.

This is very helpful! It's late here and I was just about to log out. I now understand the term asymmetric unit and I'm guessing after tomorrow morning's coffee I will understand how to put them together and still get the stoichiometry right, and why the Pb is sometimes drawn with thee bonds to O, and sometimes with four.

Later than I expected it seems, sorry for the delay and thank you for the clear, concise, and well-sourced answer!

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text:
Why does Fluorine have a high affinity for Calcium?
I keep hearing from several websites and videos that Fluorine seems to like bonding to Calcium specifically; can someone please tell me why? I can't seem to find a clear response on the internet. Fluorine has a high affinity to Calcium, more than other elements. Why is that?

Actually it's not the question of affinity but insolubility of CaF2. As this salt has solubility product of ~4×10−11 which is mostly caused by nice packing and strong ionic interaction (see enter link description here)

The free energy of formation is also quite negative. Combined with the low solubility, it indicates that it will readily form and will not go in to solution, so once formed it is quite happy as is.

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text:
Good chemistry experiments/projects
Im not sure if this is the right place to ask, but what are some good chemistry experiments or projects that could be conducted in grade 12?
Here's what I have found so far:

Preparation of aspirin
Synthesis of dyes
Analysing the percentage of metals present in coins

In my opinion, these projects are not something that grade 12 students should be doing, I'm looking for something more complex, more in touch with research being conducted in universities.
Edit: Please refrain from suggesting topics that involve spectroscopy or crystallography or any other equipment unlikely to be found in a high school lab.

Synthesis of dyes can be as complex as you want it to be.

The school wouldn't let me do it if it involves a lot of heating/high pressures. Furthermore, some student is already doing it and we can't repeat the topics.

Well, it still can be as complex as you want, even without much heating or high pressures. But OK, if this topic is taken, let's forget it. Do anything else. Determine the composition of $\ce{MgCl2}$ solvate(s) with ethanol. Analyze dust in the air. Separate different kinds of chlorophyll by chromatography. After all, the ability to come up with a topic on your own is one of the prerequisites to being a researcher.

@IvanNeretin, Can I have some more information on dyes?

Look at the most complex molecules on this page: https://en.wikipedia.org/wiki/Azo_compound

What about browsing across

the section Laboratory Experiments in the Journal of Chemical Education?  This month (march issue), for example with "Unboiling an Egg: An Introduction to Circular Dichroism and Protein Refolding" by Hoben et al., (DOI:10.1021/acs.jchemed.6b00319), or "Demonstrating the Effect of Surfactant on Water Retention of Waxy Leaf Surfaces" by Chiu et al., (DOI: 10.1021/acs.jchemed.6b00546)
more broad than the journal, visiting educational ressources the American Chemical Society offers
as an example from the other side of the pond, a collection provided by the Royal Society of Chemistry (you may refine your search by age of the audience)
albeit less chemistry is not the barycentre, bring science home in the Scientific American

and then, choosing while taking into account what resources are at your disposition.

How about some biochemistry?

Enzyme characterization (best if combined with biology or biotechnology)
Protein purification (best if combined with either biology or biotechnology)
Protein crystallization (best if combined with physics+math+biology)

All these projects are great, but they require the use of crystallography, which we won't be able to do at school.


text:
why cookware does not melt down?
why cookware does not melt down
On stove supllied with butane gas which has a flame temperature of 1970°C?
http://www.engineeringtoolbox.com/flame-temperatures-gases-d_422.html

Why do you think cookware or even the oven is close to the flame temperature?

You can solder very easily, domestic water pipes for example, but if there is any water in them this is simply impossible, the water conducts the heat away very effectively. Its the same in pans.

This is adiabatic flame temperature - I guess you didn't check the link under table. You can't get such temp. in practice, as it won't burn adiabatically.

Related: What is the meaning when it is said “Temperature of an Oxyacetylene Torch is 3500 degrees Celsius”?

Cookware has a large surface area and radiates heat. Also conducts heat to water or food it is in contact with. You reach a steady state at some temperature that is less than the melting point.
If you used a small piece of metal, you might get some melting. For example, if you put a post-1983 penny on your burner, you can probably metal the zinc inside to form a brass alloy.

i try it to melt a gold foil once but it just turned red :(

Gold melts at 1064 degrees C, while zinc melts at 420 degrees C...

The answer is that cookware transfers the energy from the flame into whatever is being cooked. For instance a pot of boiling water will stay unmelted until all the water has evaporated, after which the pot will start heating up towards the melting point of the metal it is made from.
This is called heat conductance and is used in both cookware and liquid bi-propellant rocket engines (which burn closer to 3000°C). In fact some rocket engines have been made from aluminium with a melting point as low as ~460°C depending on which alloy is used.

so the cookware dissipates heat faster than the heating, let say I have steel or stainless steel pot and I want reach 1000 how doing that?

Well, you have two options. You can keep the heat from dissipating (e.g., something like a furnace or kiln) or you can increase the thermal input (e.g., bigger flames like in a furnace). This is why metals are melted down in furnaces.

@Zhe good advise thank you, does pressure help to ?

Unlikely. You could try a hotter flame though.

Hmmm... I don't know if I am completely satisifed with these answers. There's a company called Kelly Kettle that makes camping kettles, some of them aluminium, and they warn on their web site if you use it without water in them, they can melt, and you can google online and find pictures of melted kettles. They also stopped making their bases from aluminium and switched to steel so that they don't melt. As far as I know the temperature of a wood fire is similar to butane. So I don't know why a frypan is not melting down, I don't think that slice of bacon is helping disappate or absorb much heat


text:
Why is this molecule named 1,1-chloroiodoethane, rather than 1-chloro-1-iodoethane?
$\ce{CH3CHClI}$
If there is more than one substituent on a carbon, shouldn't they be named separately like 1-chloro-1-iodoethane?

I'm pretty sure they should be named separately. Where have you seen that they should be listed together?

On my teacher's online quiz, the program marked my answer as wrong and corrected it to 1,1-chloroiodoethane.

I would ask your teacher about it. Even if the answer they gave is valid (which I'm not certain that it is formally correct), your answer should be valid as well.

Pretty sure your teacher isn't following IUPAC standards.

The correct name is 1-chloro-1-iodoethane since the locants are placed immediately in front of the prefixes that represent the substituent groups.

Furthermore, since the release of the IUPAC recommendations of 1993, generally all locants are placed immediately in front of the part of the name to which they refer (e.g. cyclohex-2-en-1-one). The corresponding rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows.

P-14.3.2 Position of locants
Locants (numerals and/or letters) are placed immediately before that part of the name to which they relate, except in the case of the traditional contracted names when locants are placed at the front of their names.

P-15.1.7.1.4 The longest chain is numbered from one end to the other by arabic numbers, the direction being chosen so as to give the lower locants to the substituent groups (side chains) [see P-14.4(f)]. The lower set of locants is defined as the set that, when compared term by term with other locant sets cited in order of increasing magnitude, has the lower term at the first point of difference (see P-14.3.5). The locants are placed immediately in front of the part of the name to which they refer. Identical simple substituent groups are indicated by multiplicative prefixes, such as ‘di’, ‘tri’, etc. [P-16.3.3(b)]. For compound or complex substituent groups (see P-29.4 and P-29.5), the multiplicative prefixes ‘his’, ‘tris’, ‘tetrakis-‘, etc. (P-14.2.2) are used as described in P-16.3.5(a).


text:
Finding Enthalpy
Given Information
$\ce{CH_4 + 2 O_2 -> CO_2 + 2 H_2O}$
"The Heat of reaction is $-802000\, \frac{J}{mol}$ of $\ce{CH_4}$ at $25\, ^\circ C$.
Rate of Moles of $\ce{CH_4}$ used is $0.625\, \frac{kmol}{h}$.
What is the enthalpy of reaction?"
Someone told me that the answer is $0.625 \cdot (-802000)$
But I dont get the logic behind it.
Thank you.

I am new to stack exchange so I am not sure if this kind of question is an acceptable one

I know that the formula being used is moles*H, but dont we use that formula when there is a phase change ?

for starters what is $kmol$? is that 1000 mol? Than what someone told you can't be right...
always look what unit your solution is and what units the parameters you are given are in. often oyu find the right formula by that

Welcome to Chemistry! This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.


text:
Springer Materials database molecule visualizer's 'crooked' straight-chain alkanes — science or art?
The Springer Materials database includes a molecule visualizer tool:

Enhanced data visualization features display interactive crystal structures, data tables, and phase diagrams with export options for further analysis.

I noticed that some of the straight-chain alkanes $\ce{ C_nH_{2n+2} }$ appear to be clearly curved, other's look more like random displacements, while others nearly straight. I'm wondering if this is:

Just some random noise "as a helpful reminder" to the viewer that these are quantum systems with vibrations
"...helpful reminder" that long chain molecules bend and wiggle depending on temperature
Representation the static shape of the molecules
Something else?

Does octadecane really "look like" a human spine?

above: GIF is made from python-processed screenshots of n-octane through n-octadecane from Springer Materials molecule visualizer. Still are the last three in the series. Most have slight atomic displacements that appear at first to be random, but the last one available $\ce{ C_{18}H_{38} }$ suddenly looks like my spine when I've been slouching over StackExchange all day instead of getting work done. Why?

I guess this is just some random noise, not even quantum in nature.

@IvanNeretin I wonder if the randomness was added as a reminder of either the quantum nature of molecules, or their flexibility. I"m not asking here about the mechanism (which is really interesting but haven't done any research yet), just the intent of the algorithm that springer is using.

Their intent is a mystery, but my bet is on flexibility.

Yes a mystery. If you want a better idea you could (a) do a full MD simulation or (b) more simply simulate what is happening by performing random walks on a tetrahedral lattice and limit close approach to simulate van-der-waals atom sizes.


text:
How to write a reasonable mechanism for the reaction of alcohols to synthase ketone?
In the section of reactions of alcohols in my textbook, there are no mechanisms or explanation for this reaction:

I am interested in Jones Oxidation particularly. I researched a little bit on internet but can't find a satisfying mechanism for it. Thanks in advance!

Going to vary depending on which oxidant you choose

I noticed later the fact you've mentioned and edited the question.

Did you try googling it? Here you go: (http://www.name-reaction.com/jones-oxidation). Note that the key step is an E2 elimination across the C-O bond: H-C-O-LG --> C=O where LG is the chromium leaving group.

but then with Jones oxidation, you form chromic acid isn't it? I could write a simple mechanism if it was basic oxidation. For that try checking the linked posted by @electronpusher

Thanks @electronpusher , in the link, there is what i want. But, i can't get the basic version that xavier_fakerat mentioned.

see http://chemistry.stackexchange.com/questions/43283/mechanism-for-oxidation-of-primary-alcohols-to-carboxylic-acids


text:
Calculating the Activity Coefficient parameters form Henry's Law
Here is the question, I am stuck on part B:

Basically, this is my thought process so far, as $x_a \rightarrow 0$, the Henry's law line becomes tangential to the actual partial pressure curve..
 Henry's law states that:
$$
p_{\alpha} = Hx_{\alpha}
$$
Similarly, using the activity coefficient model:
$$
p_{\alpha} = \gamma_{\alpha}x_{\alpha}p^*_{\alpha}
$$
Where H is Henry's constant and $p^*_{\alpha}$ is the vapour pressure of the pure component. So at small $x_{\alpha}$ values one can assume:
$$
H = \gamma_{\alpha}p^*_{\alpha}
$$
From this I can get values for the $\gamma$ (I've been given values for pure component pressure). Then I can get values for $\ln \gamma_1, \ln \gamma_2$. My next step would be to substitute in values of $x_2$ and $x_1$ into the given correlations. So, for $\gamma_1$ to be correct, $x_1$ must be very small, assume it is zero therefore $x_2 \approx 1$, therefore:
$$
\ln \gamma_1 = \frac{A \times 1^2}{\bigg( (1-1)^2\frac{A}{B} + 1 \bigg)^2} \implies \ln \gamma_1 = A
$$
Clearly, this is wrong because if I run through the same process for $\ln \gamma_2$, I get a different A value. Can someone please advise, thanks.

bump, is there an error in the question??

I think there must be an error, the van Laar activity model uses B as a coefficient on for $\ln \gamma_2$


text:
Determining the product which is not possible in the reaction
This is an objective type question I encountered in a objective test.

NBS (N-bromosuccinimide) is used for the allylic substitution of $\ce{Br}$ in a reaction. In this reaction, since the allylic carbon is the carbon adjacent to the carbon having double bonds, I thought that options B), C) and D) all can be the products but not A). However, the answer key gives the answer as D) and I can't understand why it should be?

This is weird. Unless there is some mechanism that I am overlooking, A1, B2, C2 and D2 should all not be reaction products.

Is there any other information? In particular, does it state that the starting material is racemic?

@jerepirrre no,no further information was provided to us.

D is a product, but only one product. Not two products. Those two are the same molecule flipped on a mirror plane. Were the molecule chiral (having an non-superimposable asymmetric center), then it could be two products. Not as it sits now.
Notice D1 is the same as B1. D2 is also B1, but flipped. Six products.
D2 is an enantiomer of D1, but not identical to it. I cannot puzzle out how D is supposed to be the correct answer.

The structures in D are not the same compound, they are mirror images. They are chiral. You cannot "flip on a mirror plane". You can only rotate (flip) around an axis.

You are correct. The methyl group has shifted from the starting location. Fixing...

@Pete
B1 is same as D1
But, D1 & D2 are not same.
@Aniline
I see some problem in the options while it was framed/typed/published
I have shown the 6 possible stereoisomers, which have have labelled independently as A, B, C, D, E & F
For the sake of clarity have generated their names with some software [not from my own] to be assured that these are actually different or same compounds.
Showing the mechanism, how there is a possibility of getting SIX by a free radical mechanism:

This one a redraw of the options A, B, C & D with their names:

The relationship of D1 & D2 as per question doesn't work to the the same:
In my opinion, there are six stereoisomers possible, but the correct option does not exist, in the question[options given, in fact].
Hope it throws some more light on your query.

The second product in Option D) is not a product of this reaction.
You cannot rotate the substrate molecule in such a way that the Me substituent ends up in the portraied position.

@SirJohnson So that means that the two products in option D are identical.we can rotate the first one 180 degree to get the second one.

To me, it looks like each group of two has an impossible product: the one where the methyl group’s carbon’s stereochemistry is inverted. Can you explain how the methyl group inversion would proceed? Especially since I am getting six products without methyl group inversion …

Absolute stereochemistry is irrelevant if the products in (B) are not wrong.


text:
Helicity in Molecular motors and topological isomers
I'm working on a project to understand the work of the 2016 Nobel laureates and I've come across a problem of not being able to find the right sources to understand basics of helicity (more specifically right-handed
helicity (M) and lefthanded helicity (P)) to understand things like the diagram below (I understand cis, trans, I don't know much about the (P,P) and (M,M)).

I would like some suggestions for books or articles that would help me learn more about these concepts.
Thanks in advance!

I asked a similar question a while ago (http://chemistry.stackexchange.com/questions/58517/m-p-nomenclature-for-axial-chirality), you may find the comment useful

This wiki article pretty much explains the basics https://en.wikipedia.org/wiki/Atropisomer

@Anon I don't think these are atropisomers. There's no free rotation about the double bond.

@Zhe Sure they are not, I linked the page as a reference for the M/P notation, as it includes a clear diagram on how it is assigned.

@Anon Sadly, I would have preferred that they explicitly choose to use either $\ce{A -> A'}$ or $\ce{B' -> A}$ to determine the designation, instead of both.

Any pretty non-old organic stereochemistry book should work.
M/P notation
You present some organic compounds, so let's see definition from the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013:

P-92.1.1.2 The helicity rule: stereodescriptors ‘M’ and ‘P’
Helicity is the chiral sense of a helical, propeller, or screw-shaped molecular entity (see ref. 37). The ‘helicity’ rule is expressed by Prelog and Helmchen (see subsection 5.1, ref. 36) as: “Depending on Whether the identified helix is left- or right-handed it is designated ‘minus’ and marked ‘M’ or ‘plus’ and marked ‘P’ ”.
The application of this system to the description of conformations and configurations considers the torsion angle between two specified (fiducial) groups are attached to the atoms linked by that bond. The sign of the smaller torsion angle between the fiducial groups defines the sense of chirality sense of the helix (see torsion angle, P-94).
P-92.1.1.2.1 The chirality of hexahelicenes is denoted by the stereodescriptors ‘P’ and ‘M’.

  (P)-hexahelicene (PIN)                  (M)-hexahelicene (PIN)
(…)

The description follows further, the rule can be applied also for biphenyls, allenes, or similar structures, I'd say where along the axis some two approx. perpendicular groups are in clockwise or anticlockwise rotated positions.
The arrows show the helicity orientation in the direction from the observer; but may be confusing(opposite), when one applies common hand and thumb methods for determining a screw thread handedness, but results in the same correct helicity orientation/descriptor.

Δ/Λ notation
In the question you're mentioning molecular motors. These can be also propeller like structures. If they were from inorganic chemistry world, they could use different notation Δ/Λ; from IUPAC Red Book 2005:

IR-9.3.4.11 The skew-lines convention
Tris(bidentate) complexes constitute a general family for which a useful, unambiguous convention has been developed based on the orientation of skew lines which define a helix.
  Examples 1 and 2 represent the delta (Δ) and lambda (Λ) forms of a complex such as $\ce{[Co(NH2CH2CH2NH2)3]^{3+}}$.   (…)

          2.
delta (Δ)                      lambda (Λ)

This one is the Δ isomer:

Δ/Λ likely corresponds to organic's P/M respectively (however, in some structures, there might be doubts which parts of the structure should be interpreted as the propeller blades.. see e.g. Q. Molecular knots absolute configuration)

The structures in your ask, 3,3′-dimethyl-2,2′,3,3′-tetrahydro-1H,1′H-4,4′-biphenanthrylidene, do have a combination of a geometrical isomerism at the connector double bond, and two chirality axes (perpendicular to the double bond connector).

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text:
Difference in pH of water and rainwater
In my textbook (Olmsted and Williams 4th ed) it is given that pH of pure water is 7, pH of unpolluted rain water is between 5-6 and pH of acid rain is between 4-5.
It is obvious that pH for acid rain would be low. But how is it acidic, even slightly, for unpolluted rain water? Shouldn't unpolluted mean without any solutes or dust in water? Shouldn't it be similar to pH of pure water (7)?

At first thought, it might seem that rainwater should be clean like distilled water, as it evaporated, condensed, then fell back to earth. It sounds like a condenser or distillation apparatus in a laboratory. But in reality, the process is very different.
To start with, every raindrop must have a particle to condense upon, called a cloud condensation nucleus.  According to this Wikipedia article:

Cloud condensation nuclei or CCNs (also known as cloud seeds) are
  small particles typically 0.2 µm, or 1/100th the size of a cloud
  droplet on which water vapor condenses. Water requires a
  non-gaseous surface to make the transition from a vapor to a liquid;
  this process is called condensation. In the atmosphere, this surface
  presents itself as tiny solid or liquid particles called CCNs.

Furthermore, the composition of CCN is frequently very acidic, having formed from sulfuric acid or sometimes weak organic acids. These types of CCN form in both polluted as well as pristine environments. Both natural and man-made sources of sulfur emissions result in the photochemical oxidation to sulfuric acid, which forms new CCN which are very hygroscopic (water absorbing).
The type of CCN described above are most commonly formed over land. In marine environments, there are natural sources of sulfur gases, but most CCN form from evaporated sea spray, which is of course much more pH neutral.
However, regardless of how the cloud drop was originally formed, they all have another source of acidity from the atmosphere; they absorb carbon dioxide, which is rapidly converted to carbonic acid when dissolved in a cloud droplet.  It is this absorption of naturally occurring, acidic carbon dioxide that is the primary reason that essentially all rain is somewhat acidic.

So the term "unpolluted" is not equivalent to pure

@YAHB , exactly. The natural atmosphere is a pretty complex system, and even in unpolluted environments the composition of a simple raindrop can very greatly, but they will always be somewhat acidic due to CO2 absorption in addition to the other natural compounds discussed above. There is also wash-out of dust particles during the early part of a rainfall event that make natural rain even "dirtier"!

@airhuff - The pH of "unpolluted" rainwater is governed by 2 equilibria constants, that for the Henry's law partitioning of carbon dioxide between air and water, and the other for the dissociation of carbonic acid to proton and bicarbonate. Using p$\ce{CO2}$ of $10^{-3.5}$ and given K${H} = 10^{-1.5}$, and pK${\rm a} = 10^{-6.3}$, we thus get $[\ce{H+}] = 10^{-5.65}$ or pH = 5.65 for rainwater. You might include these details to expand your answer to address the mechanisms controlling the pH more precisely than you have.

@ToddMinehardt , thanks much for your input and work on your calculations. I shied away from further focus on CO2 because though I believe my highlighted statement is very true, the quantitative aspect is much more complex, as even pristine environments can have rain pH much lower than 5 due to natural emissions of sulfur gases, nitric acid formation from lightning, etc.  See here. It would be nice to have some more quantitative calculation, but per my examples above, it is probably too complex for a comprehensive treatment. I think.

The source of the lowering of pH observed in rainwater is frequently presented as simply the dissolution of acid gases. This is displayed, for example, here, which cites CO2, SO2 and NO2. Further, to quote:

The sulfuric and nitric acids formed from gaseous pollutants can easily make their way into the tiny cloud water droplets. These sulfuric acid droplets are one component of the summertime haze in the eastern U.S.

However, this educational source also cites an alternate explanation:

Contrary View:
  The benchmark of "natural rain" is 5.6. Acid precipitation in the range of 4.2-5.0 has been recorded in most of the Eastern United States and Canada. EPRI (Electric Power Research Institute)....
  EPRI also contends that pH 5.6 may or may not be a valid reference point. It should not be considered the "background" or "natural" acidity of precipitation.
Even without man-made influences, there are natural sources of sulfur oxides, nitrogen oxides, and other species important to determining the precipitation acidity at any given time. Hence, trying to quantify man's contribution to the natural condition will never be possible, since the "natural background" condition cannot be known.
In the forest areas of Brazil at the headlands of the Amazon River, an area remote from civilization, the monthly average of 100 rain events in the 1960s ranged from pH 4.3 to pH 5.2, with the median value of pH 4.6 and one reading as low as pH 3.6.

To explain why the rainforest, not a normally cited source of pollutant or volcanic acid gases, is associated with a median pH of 4.6 requires a more complex explanation. Given the effect of sunlight in equatorial regions, I concur with the above comments by AirHuff on photochemical oxidation paths. I suggest the following source 'Participation of the Halogens in Photochemical Reactions in Natural and Treated Waters' by Yi Yang and Joseph J. Pignatello, available online. To quote, in parts:

Abstract: Halide ions are ubiquitous in natural waters and wastewaters. Halogens play an important and complex role in environmental photochemical processes... While inert to solar wavelengths, halides can be converted into radical and non-radical reactive halogen species (RHS) by sensitized photolysis and by reactions with secondary reactive oxygen species (ROS) produced through sunlight-initiated reactions in water and atmospheric aerosols, such as hydroxyl radical, ozone, and nitrate radical. In photochemical advanced oxidation processes for water treatment, RHS can be generated by UV photolysis and by reactions of halides with hydroxyl radicals, sulfate radicals, ozone, and other ROS... Recent studies indicate that halides, or the RHS derived from them, affect the concentrations of photogenerated reactive oxygen species (ROS) and other reactive species; influence the photobleaching of dissolved natural organic matter (DOM); alter the rates and products of pollutant transformations; lead to covalent incorporation of halogen into small natural molecules, DOM, and pollutants; and give rise to certain halogen oxides of concern as water contaminants.

In the current context, aerosols rich in halide salts, transition metal oxides and photo-sensitive organic acids are likely also instrumental in understanding the mechanics of the pH effect.


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Why there is no bichloride or binitrate ion like bicarbonate or bisulphate?
I have been thinking lately that if there is bicarbonate why there is no bichloride or binitrate, please someone  give an easy explanation.

What's the relationship of bicarbonate to carbonate? Bisulfate to sulfate?

Well, I say it's better question then it looks. For example Bifluoride does exist. I wouldn't be so sure that isn't any example of some obscure  bichloride or binitrate compound. It's basicly only a matter of stoichiometry.

Related question (and answer)? http://chemistry.stackexchange.com/questions/31083/why-was-the-prefix-bi-used-in-compounds-such-as-for-bicarb-of-soda
@Mithoron that might explain your Bifluoride. According to the logic in the above question there would be two times as much flouride produced for a given quantity of acid in the reaction.

Okay, you want simple, here's a real simple way to put it.
How many limbs does a dog have? 4 right?
Now if it loses one of its limbs, a prosthetic one can be placed,(We'll get to this: with its whole body, being able to cope with the leg). Now consider the case of a human being, same scenario, the same thing can be done, since we can place the limb in the "slot"(somebody put a better word here), where the leg previously was(although, the limb cant do everything a leg couldve done). Now consider the case of a creature with just one limb, if it loses the sole limb it had, life's gonna be hard, simply because of the amount of discomfort its going to face while adapting to the limb. Similarly, sulphate ($\ce{SO4^2-}$) and carbonate ($\ce{CO3^2-}$) have a valency greater than 1. Hence can accept an H+ ion and still exist as a stable specie, and further accept a metal ion;
In the case of $\ce{HCl}$, it has a valency of just 1, hence upon accepting an $\ce{H+}$ to get the "bi" tag to its name, it has lost its charge, and now is incapable of accepting a metal ion to form a salt(while keeping the $\ce{H+}$ to itself), as it has simply become hydrochloric acid.


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Confusion in comparing melting point
We have to compare melting points of $\ce{LiH}$, $\ce{NaH}$, $\ce{KH}$, $\ce{CsH}$. I know the melting point of an ionic compound is more than that of a covalent compound.
So according to Fajan's rule, ionic character increases in $\ce{LiH}$, $\ce{NaH}$, $\ce{KH}$, $\ce{CsH}$. As ionic character increases, melting point should also increase in $\ce{LiH}$, $\ce{NaH}$, $\ce{KH}$, $\ce{CsH}$.
But actually, the melting point decreases in $\ce{LiH}$, $\ce{NaH}$, $\ce{KH}$, $\ce{CsH}$. What could be the reason behind this?

Why do you think that the ionic character is the factor that matters the most?

@Zhe  because to compare melting point we have to compare bond strength  . And I think ionic bond has strongest bond strength.

So, compare the melting temperature of diamond or tungsten vs these compounds. Note that labeling a bond as ionic, covalent, or metallic says very little about how 'strong' they are.

Okay , I got my misconception . @JonCuster then here what could be the reason .

Consider the sizes of the ions and chemical hardness (according to Pearson). If they are not good matches, a decreased melting point is reasonable.

Melting point is a somewhat tricky concept because it heavily depends on the structure of the bulk solid. Are you sure that the crystal packing in all 4 substances is the same? If not, then you are comparing apples to oranges.

"Are you sure that the crystal packing ... is the same?" Somewhat exceptionally, alkali metal hydrides do all have the rock salt structure where the structure is known.

You can get an estimate of the bond strength by using Coulomb's law and periodic trends for atomic radius. $$F = k{ {q_1q_2} \over r^2}$$The charge of all these alkali metals is the same in these compounds, so the only difference in bond strength comes from the the distance between the charges. The distance between the charges is related to each atoms' atomic radius and that is an increasing trend as you move down the alkali metals.
So, if distance is increasing, the force holding the solid together is decreasing, which would justify the trend $\ce{LiH}>\ce{NaH} >\ce{KH}> \ce{CsH}$ for melting point.


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Kc only effected by temperature?
This makes little sense to me, with a lot of contradictory statements I have learnt.
For example:
$$\ce{N2 + 3H2. <--> 2NH3}$$
($\mathrm{\Delta h = -46.2~kJ~mol^{-1}}$)
I have been told a higher kc = large amount of product being formed in the equilibrium.
So therefore why if I increase the pressure why don't I get a increase in kc as the forward reaction is favoured creating more product?

After asking around I have come to the conclusion that a reaction has many kc values depending on temperature , but then I don't see how the kc equation can work as you can get different concentrations of A B C and have constant temperature by increasing pressure.
And also the statement of higher kc = more product is also wrong.
Obviously I'm wrong but I'm just explaining where my confusion lies in my logic.
Where have I gone wrong? Or is there an explanation that will clear things up?

We want to perform the reaction you mentioned:
$$N_2+3H_2\longrightarrow 2NH_3$$
At the beginning we have certain concentrations of reagents, say $[N_2]_{in}$, $[H_2]_{in}$ and $[NH_3]_{in}$. The reason why I am assuming we have some ammonia is just a way for not dealing with infinite values of energy as it will be clear soon. Since Chemists often work at constant temperature and pressure we will follow the spontaneity of this reaction following the changes in the Gibbs Free energy $G_r(T,P)$ of reaction. A model describing the Gibbs free energy changes is the following:
$$\Delta G_{r}(T,P)=\Delta G_r^0(T,P=1bar) + RT \ln\biggl(\frac{[NH_3]^2}{[N_2][H_2]^3}\biggr)$$
The reason we have chosen not to have a zero concentration of ammonia is because in that case we would get $\Delta G_{in}=-\infty$ and this is not sensible. It is important to note that $\Delta G^0$ only depends on the temperature since the pressure has been fixed at 1 bar. Another aspect to note is that the concentrations in square brackets are the concentration at a certain point of the reaction. Now we are ready to find the answer to the question. What happens at the equilibrium? By definition the equilibrium is reached when the Gibbs free energy is minimum, thus when $\Delta G_r =0$. Taking the above equation and setting this condition, isolating the concentrations from the logarithm, we get that the concentrations of our reagents/product at the equilibrium are
$$\frac{[NH_3]_{eq}^2}{[N_2]_{eq}[H_2]_{eq}^3}=\exp\biggl(-\frac{\Delta G_r^0}{RT}\biggr)=K_{eq}$$
Clearly the equilibrium constant depends only on the temperature and not on the pressure since in $\Delta G_r^0$ pressure has been fixed at 1 bar by definition. Changing the total pressure does not effect the amount of ammonia you get as you can see from a "mathematical" point of view. However in order to increase the product concentration we can increase the concentration of the reagents (in this case the partial pressure since we are in gas phase) since what is really fixed at the equilibrium for a given temperature is $\frac{[NH_3]_{eq}^2}{[N_2]_{eq}[H_2]_{eq}^3}$ and not the concentrations taken singularly.


text:
How would one correct for a blank in a potentiometric titration?
I want to do a potentiometric titration with perchloric acid of a sample consisting of active drug (weak base) and two other components. Based on theory, I already know that these two other components should not interfere with my titration.
However, I still want to do a blank titration with just these two components to make sure. What I would expect with this blank titration is just an exponentially decreasing curve. If this is indeed what I observe, would I have to correct my sample for this blank titration? If so, how would one do this?

Even though the other compounds should theoretically not interfere with your titration, a method/matrix blank like you've proposed is always a good idea.
What I would do is simply subtract the blank potential value from that of the sample for each volume addition, then plot and find the inflection point as you would usually do (i.e. from the derivative plot, etc.).
If you are correct in your assumption that the other compounds won't interfere, then the position of the inflection point should be the same as that calculated without the blank, within experimental error (i.e. within a standard deviation or two or three, etc. of a set of replicate titration values.).  If you are incorrect and there is some interference, then the blank subtraction will correct for it and the blank corrected value is what should be used/reported.


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Stability of Fe(II)
How stable is Fe(II) in a solution? I am doing simple photometry with fresh samples and I feel like it doesn't oxidize that quickly, but have to use a stabilizing agent (dihydroxyfumaric acid) nonetheless. Couldn't find the pKa of the acid, but I suppose the decrease in the pH will stabilize the iron. Is that right? If yes, why?
Also, are the reducing properties of the dihydroxyfumaric acid important for keeping the iron in the +2 state?


text:
Melting points of compounds?

Compare the melting points of:
1) $\ce{BeSO_4}$
2) $\ce{CaCO_3}$
3) $\ce{BaCO_3}$
4) $\ce{SrCO_3}$

Okay, so there are a few statements given by my teacher:
1) Thermal stability of ionic compounds is proportional to the counterbalance effect of electropositive field of cations(magnitude of +ve charge) and electronegative field of an anion (mag. of -ve charge). (I interpret this as the lesser the difference between the magnitude of charge, the more thermally stable the compound.)
2) The real magnitude of charge is directly proportional to the number of electrons lost(basically the number we write next to an element, like $\ce{Ca^{2+}}$, right?) and inversely proportional to the size of the ion.
3) Except the second group elements and halides, all anions can be considered as "large anions" and hence their effective magnitude of charge will be same (roughly).
So since all of them are "large anions", we have to compare the magnitude of charge on the cations. Now, being large, their charge is small. So for least difference, we want the cation with the smallest charge. This means the largest size, which simply means $\ce{Ba}$. So the answer should be 3. To confirm this, I Googled the melting points of these substances.
1) $\pu{110^oC}$
2) $\pu{811^oC}$
3) $\pu{825^oC}$
4) $\pu{1494^oC}$
So according to me, the melting points in decreasing order should be:
3 > 4 > 2 > 1
When it turns out to be:-
4 > 3 > 2 > 1
Where did I go wrong?
I personally think:
1) Since the charge on anions is only roughly same, I think that when we consider the difference between the charges, that would be messed up and so we would end up with wrong answers(possibly).
2) That since this question is supposed to be for high school students, we make a lot of assumptions, which obviously don't hold true in the real world, and hence the answer is wrong? That being said,is my answer correct keeping these assumptions in mind?
3) Since chemistry has a lot of exceptionss , is this another one of those? Am I going to have to memorize this too?

The very odd 2nd group, with queer Mg... down the group : ionic radius drops, resulting weaker metallic bonds. Perhaps you can look into thermal decomposition. I am not an expert to draw an answer that may clarify your doubts.

@bonCodigo I think you are wrong there. Down the group atomic radius increases, right?(The no. of shells increases..)

Blimey! Typo, cations (in this case), ionic radius increases. Frankly your argument is valid, from thermo point of view. @alphonse's answer is pretty comprehensive. And take a look at this link too.

@bonCodigo So yea, at undergrad level, the ans would be 3, right?

If I were you, I would bring those materials and my argument to the teacher. And ask, why are we settling on numbers. So he/she can give you the reasoning behind the question, the answer of their choice to clarify knowledge to you. Yes No? :D

@bonCodigo Actually, I just found out that my teacher gave the ans as 3) too. But, what I am concerned about is if that's the case, then why does google show different values? Is it somehow related to decomposition of the substances at high temp?

The tags seems quite unrelated to the Q or to each others

Point 1 "from the teacher" doesn't make sense in terms of grammar. Moreover stability and melting point have nothing in common. Stability of the ionic structure might be. But again the terminology is confusing.  Check your notes again.  Perhaps the teacher mentioned things before. Without refreshing some special points I myself I want be able to answer according the given order.  Just Be sulphate can be let apart

First, in order to compare mp's you need to know the mp's. You have two choices: 1. run melting point measurements or 2. find literature written by someone who has run the experiments. Perhaps this is a translation problem from your native language? Perhaps you meant predict the ranking of mp's for those 4 compounds.
You state that CaCO3 melts at 811 °C. As far as I know, that is incorrect. First, the thermodynamically stable polymorph is calcite, which according to wikipedia has a mp of 1339 °C but which decomposes at temperatures above ~840 °C (also according to Wikipedia!). I hope you see this is clearly contradictory. My first hand experience is that it does indeed decompose at around 820–840 °C, so I am unable to explain how they were able to measure its melting point 500 °C above that. 500 °C is an enormous gap, but perhaps they measured it under some enormous pressure (far from STP) - I really don't know.
The other major problem I'm having is understanding what you (or your teacher) means by "counterbalance the effect of electropositive field of cations ... and electronegative field of an anion." It seems to me to be meaningless, at least I don't recognize this terminology in reference to the electrostatic forces usually invoked to account for ionic solids. BeSO4 is an obvious problem since Be is known to for its tendency to form covalent compounds. Indeed, according to Wikipedia, anhydrous Be sulfate has Be-S (covalent) bonding. While you could predict that the covalent bonding would reduce the mp, what you'd do with the polymeric nature of the solid is beyond me - I've no idea how you or your teacher can explain that, except by hand waving.
This leaves the Ba and Sr salts. Simple logic of ion size would predict the order Ba < Sr and ignoring the ridiculous assertion on Wikipedia that the Ba salt's melting point is 811 °C while at the same time (you did READ the article, right?) stating that that is the temperature at which it a "polymorphic transformation" occurs. A little further digging provides a safety data sheet which claims a mp of 1400 °C "close to its decomposition temperature", but such data should be suspect.
Your teacher's choice of picking carbonate as the anion was a poor one, in my honest opinion. Because of the tendency of these salts to decompose, rather than melt, you need to be very careful where you obtain your data from. But perhaps that is the lesson your teacher hopes some of you will learn - to not just accept a number because it is readily available (without subjecting it to critical analysis (who, what, where, why, when, how - which in this case means how reliable is the reference? does it reference a primary (academic) source? if not, an authoritative secondary source? if not, then you should seek independent confirmation. Such confirmation is hard to find on the internet, since one lazy site steals from another until the noise drowns out the signal.
I really don't know if density functional theory (DFT) quantum mechanical modeling by supercomputers is able to accurately predict the mp's of these material. My guess is that because of the change in crystal structure(s) that it remains beyond our current capabilities (for some things, calculating the value has become reasonably accurate – as far as I know (and I'm no expert) mp isn't one of the successes. If it was accurate it would be a third way to "compare" the mp's.

So, as a conclusion, we can say that the following salts decompose at a point, and hence their mp/decompostion temp. are all a bit mixed up. And yes, I didn't read the article. I just googled "mp of ____" and got my values from there. So, if this question comes in an objective type q.paper, what would be the correct ans?

No, this isnt a translation from my native language. This is exactly as it appeared( it was already in English. Dot same words/language.) Basically, after giving this question, our teacher told us to arrange the compounds in decreasing order of their mp's.

I think that we should somehow, I dont know, ignore the decomp. stuff and compare the mp's if they didnt decom. at all(maybe??)

@alphonse, please separate the paragraphs in your answer. Plus one, but we can't edit yours without adding 6 chars.

@alphonse I separated the paragraphs (and spelled out DFT to get the additional six characters). It's in the edit queue.

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


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Grignard Reagent R group
Can the R group of a Grignard reagent (RMgX) be an alkene?
Also what groups can't R be?

Yes it can be an alkane or an alkyne too.

related http://chemistry.stackexchange.com/questions/41235/can-grignard-reagent-contain-secondary-amino-group

R cannot have a beta leaving group because elimination occurs

There can't be any acidic groups or carbonyl groups.

No nitro groups.  They oxidize the Grignard reagent.


text:
Why are styrene acrylate copolymers used for toner?
I have been scratching my head lately, after learning that laser printer toners are often made from styrene-acrylate copolymers. As of yet, I have not found information on these polymers in reference texts or papers/research outside of the realm of xerography, electrophotography, and other printing technologies. What properties make these polymers particularly interesting for toners, but completely uninteresting to other fields? At risk of overreaching, I'm also curious how these copolymers came to be used in toner in the first place, given the lack of convenient information?
References:

Reactive Polymers Fundamentals and Applications, chapter 20 "Toners"
Production of carbon black acrylic composite as an electrophotographic toner using emulsion aggregation method:
Investigation the effect of agitation rate, paper by Maryam
Ataeefard
Production of black toner through emulsion aggregation of magnetite, carbon black, and styrene-acrylic co-polymer:
Investigation on the effect of variation in components, paper by
Maryam Ataeefard
Various MSDS pages from toner and laser printer manufacturers

Suitably tuned Tg, so it's a powder but melts to the paper easily. Easiest way to tune is likely to make a copolymer. Low Mw plastisisers are not tolerable, because they would boild out in the fixer. Electrostatic properties, so the transfer to the paper works reliably.


text:
Does this 'Belousov-Zhabotinsky Reaction' kit contain the correct reagents?
I want to run the Belousov–Zhabotinsky reaction as a demonstration. As far as I'm aware, you need sodium bromate, sodium bromide, water to $\pu{100 ml}$, malonic acid, phenanthroline and ferrous complex.  This is the recipe that I know, but I can't get hold of the right ingredients.
I found this kit entitled 'Belousov-Zhabotinsky Reaction' for £15.00. I am not sure it has the right ingredients:

Kit contents: Iodine sodium, malonic acid, Sulfamic Acid, sodium thiosulfate
  Manganese sulfate, starch solution, cups 5 pieces, chemicals, racks etc.

Does this kit have the right ingredients for the Belousov–Zhabotinsky reaction?

No, it doesn't, at least not strictly.
According to Wikipedia (emphasis added):

One of the most common variations on this reaction uses malonic acid $(\ce{CH2(CO2H)2})$ as the acid and potassium bromate $(\ce{KBrO3})$ as the source of bromine. The overall equation is:
$$ \ce{3CH2(CO2H)2 + 4BrO3− → 4Br− + 9CO2 + 6H2O} $$
[ $\ldots$ ]
Many variants of the reaction exist. The only key chemical is the bromate oxidizer.

The kit doesn't contain any bromide or bromate, so by definition it will not allow demonstration of a Belousov–Zhabotinsky reaction.

However: It does appear that the kit may allow demonstration of an iodine clock reaction:

Hydrogen peroxide variation
This reaction starts from a solution of hydrogen peroxide with sulfuric acid. To this is added a solution containing potassium iodide, sodium thiosulfate, and starch.

The ingredients are not a perfect match, and actually appear to be something of a mishmash between the iodine clock reaction and the Belousov–Zhabotinsky reaction, but they're in the right ballpark. Metal ions and their oxides are well known as catalysts for hydrogen peroxide decomposition, and "Sulfamic acid" could be a typo that was supposed to be "sulfuric acid".

https://en.wikipedia.org/wiki/Sulfamic_acid Close enough in terms of properties, I guess.

@orthocresol , as long as it's just acidifying the matrix, it should probably work ok.


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How to select for 1-4 carbon sugar bond in Chemical Glycosylation?
I am interested in Chemical Glycosylation, specifically the production of cellulose by synthetic means. However, it is difficult to ensure that the glycosidic bond is formed between the 1 and 4 carbons.
Are there any techniques that can select for this reaction?
I am looking for good references on this subject.

You have to protect the OH groups you don't want to react with protecting groups.
In this case you could install a benzylidene acetale between the 4 and 6 OH, then protect all the other OH and then do selective reductive opening of the benzylidene so your benzyl group will be on position 6 and you got a free OH.
That said: it's quite expensive, labor intensive and has very bad atom economy. And you might run into problems with deprotection once you got polymer structures, so that's not a feasible option for synthesis of cellulose.
I don't think it's possible to optimize a synthesis without protecting groups, so the best way would be doing it enzymatically, which could be very expensive too.


text:
What is a polyfuranoside?
I came across this molecule on wikipedia, and all that was written about it was two words: Docosanasaccharide and polyfuranoside.

https://en.wikipedia.org/wiki/Chemical_glycosylation#Notable_synthetic_products

The term polyfuranoside is a combination of

the Greek prefix poly, which means many and
the term furanoside, which indicates that the structure is somewhat related to furan, a five-membered (aromatic) heterocycle that contains an oxygen atom. Forget about the aromaticity here, what counts is

five-membered ring
contains an oxygen atom

If you take a look at the large diagonal of the structure, you'll recognize a large number (12) of these structure elements connected by glycosidic bonds, hence the other description, docosanasaccharide.


text:
Why are acid-base reactions so exothermic?
If I mix sulfuric acid with sodium hydroxide, the solution really heats up. Could someone explain why acid-base reactions are exothermic from a molecular perspective?

Sodium hydroxide and Sulphuric acid have $-\Delta_fH^\circ$ around $\pu{1300 kJ/mol}$ and sodium sulfate and water have around $\pu{1500 kJ/mol}$. So we have an exothermic reaction.

When an acid meets a base in a beaker... How easy is it to find a strong acid or base in nature? It isn't - it is hard work to make them. Well, all that work comes out when you mix them back together again. Write out the reactions and see what happens.

Sorry if this is too basic, but imagine you ripped a bunch of water molecules in two, throwing the H+ pieces in 1 jar and the OH- pieces in another. All the energy put into that comes crashing back at you when you mix the solutions together, allowing the oppositely charged pieces of water to come back together again. No matter the path actually taken to get to this state of separated water molecules (H+ in the strong acid, OH- in the strong base), that's the energy that you get back in the form of heat when you mix the two. Is that what you're shooting for when you say "molecular" perspective?

One can think of it as simply because of electrostatic attraction. Basically, an acid produces $\ce{H+}$ ions in water while a base produces $\ce{OH-}$ ions in water. So when you mix an acid and a base (e.g. sulfuric acid and sodium hydroxide) in a beaker, you get a solution containing $\ce{H+}$ and $\ce{OH-}$ ions. So, when these ions are far apart, there exists potential energy between them (because opposite charges attract each and want to come closer together). So when these ions come closer together, the attraction between them becomes strong enough that a bond forms between the ions to form $\ce{H2O}$. During this process, the initial potential energy between the ions decreases and is converted to heat energy. Hence this is why it is exothermic.
So generally speaking, energy is released when a bond is formed, while energy is required (energy is absorbed) to break a bond. So in the case of acid and base reactions, no bond is being broken, bonds are only being formed. So it must be an exothermic reaction then.

For a simple answer, an acid, likes do donate its $\ce{H+}$ ions, and a base likes to give off its $\ce{OH-}$ ions. The bond enthalpy of the A-H and B-OH bonds are low. Now the $\ce{H+ and OH-}$ fuse to form a water molecule, giving rise to new bonds, hence is an exothermic process.

From a purely enthalpic point of view , $H^+$ when combines with $OH^-$ it gives a  water molecule, that is a well known stable molecule. Since the energy must be conserved, when you "becker system" gains energy via the condensation reaction just named, the surrounding must loose energy: this happens since it is heated up by this exothermic reaction. Therefore you can think of the exothermicity as a consequence of the conservation of energy.

This doesn't explain why acid-base reactions are exothermic, except a small hint that water is stable. Therefore a comment and an NAA


text:
Final temperature of a adiabatic process

One molar of a perfect gas of $C_v = \pu{20.18 J/K}$ at $\pu{3.25atm}$ and $\pu{310 K}$ undergoes adiabatic expansion to reach a final state of $\pu{2.50atm}$. Calculate final volume and temperature.

Since $PV^{\gamma} = const$ and $\displaystyle T^{C_v/R}V = const$,
$$V_f = V_i\left(P_i\over P_f \right)^{1\over \gamma}$$
Which I got as $\pu{0.0113m^3}$ and $V_i = \pu{0.00782 m^3}$ .
To get final temperature I used
$$T_f = T_i\left(V_i\over V_f \right)^{R\over C_v} = 310\left(0.00782\over 0.0113 \right)^{8.314\over 20.8} = \pu{267K}$$
But when I use equation of state to get final temperature I get $\pu{344K}$ which is the correct answer.
Why did not equation of state and $\displaystyle T^{C_v/R}V= const$ match in this case ?

How are you obtaining Gamma? Your equation for the volume and temperature also doesn't look right. Where are you getting $\frac {\text{C}_v}{\mathrm{R}}$? https://en.wikipedia.org/wiki/Adiabatic_process

@Tyberius Gamma is $C_p/C_v$ ? and $C_p - C_v = nR $ no ?

@Tyberius Sorry I forgot to include the initial temperature if that is what you asked.

From your equation, I don't get 0.0113 for the final volume.

@ChesterMiller I have linked the calculations. Can you tell me what went wrong ?

You raised it to the $\gamma$ instead of $1/\gamma$

I see now how you are getting (Cv/R). How I notice now that the answer you say is correct can't possibly be correct, as the temperature will always go down in an adiabatic expansion.

@ChesterMiller Still I get $V_f= 0.0094$ and $T_f = 288.016$ which is still very far ?

@Tyberius But I used equation of states to get that answer and it is also given in the book. what have I missed ?

With that new value for $V_f$, the equation of state should also give $T_f=288K$. The textbook has to be wrong, because an increase in temperature cannot happen for an adiabatic expansion.

@Tyberius Thank you, if you don't mind can you write an answer ?

288 is correct.

@ChesterMiller Yes thanks.  Can you write an answer ?

I don't know what you mean.  You already did it.

Your error stems from a miscalculation of $\mathrm{V_f}$ in the first calculation. As mentioned in the comments, it appears that you actually exponentiated by $\gamma$ rather than $\frac 1 \gamma$. When you differentiate by $\frac 1 \gamma$ , you should obtain $\mathrm{V_f}=0.0094 \mathrm{m^3}$.
Plugging this into your expression for $\mathrm{T_f}$, you should obtain:$$T_f = T_i\left(V_i\over V_f \right)^{R\over C_v} = 310\left(0.00782\over 0.0094 \right)^{8.314\over 20.8} = 288\mathrm{K}$$
Similarly, inserting this into the ideal gas equation, you should obtain: $$\mathrm{T_f= {P_fV_f \over R}}={(2.5)(0.0094) \over(8.206\times10^{-5})}=288\mathrm{K}$$ (The ideal equation will be slightly off as written due to rounding).
In regards to the textbook solution, the final temperature could not possibly be higher than the initial. An adiabatic expansion always reduces the temperature. Intuitively, we can understand this by noting that an expansion with no heat exchange should decrease the internal energy and thus decrease the temperature of the system.

Thank you very much. Did you check http://chemistry.stackexchange.com/questions/71629/change-in-internal-not-zero-for-an-isothermal-process ?


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The product of a reaction series involving 1,3-dithiane, 1-chloro-3-iodopropane and butyllithium
I encountered this question while solving an objective paper.

$\ce{BuLi}$ is a strong base and it can be used to abstract a proton from the substrate. Low temperature favours substitution more than elimination. So my idea was that the ring will break up $\ce{Li}$ will be substituted in place of two $\ce{S}$ atoms. The next product should involve formation of a six membered saturated ring but I am unable to proceed further. What is going to happen?

Related: http://chemistry.stackexchange.com/questions/7605

Given the choices, C is the answer. As A --> B is written, the halide is added to the dithiane anion. In the real world this mode of addition, if added slowly, could eventually give dialkylation with the eventual formation of glutaraldehyde. To optimize the formation of cyclobutanone, the dithiane anion should be added slowly to the halide.

Some right ideas, but unfortunately, not correct. The correct answer is (C).
The protons in between the two sulfurs are more acidic because of hyperconjugative stabilization into the sulfur. Addition of 1-iodo-3-chloropropane results in the substitution product as you guessed. Another equivalent of butyllithium does the same thing again, except, this time, we do an intramolecular substitution for ring closure. Mercury has high affinity for sulfur, so that deprotects what is analogous to a ketal to give the ketone.

The key for these kinds of problems is to approach it from both ends. For example, the products are all carbonyl groups. Where would those groups come from? They can only be from the last step because otherwise butyllithium would add into a carbonyl. This gives you some idea for what 'C' would look, and so on.
For the forward direction, it's really important that you not kill all the reactivity by substituting off the heteroatoms. If you do that, you'll end up with an alkane, and there's no good chemistry unless you want $\ce{C-H}$ activation magic. Since butyllithium is not adding into anything, consider acid-base chemistry because it's also a very powerful base.


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Change in internal not zero for an Isothermal process?
(All energy quantities are in $\pu {kJ/mol}$.)

One molar $H_2O$ is condensed isothermally and reversibly at $\pu{373K}$. Given $\Delta_{vap}H^\circ_{100^\circ}[H_2O] = \pu{40.656 kJ/mol}$. Find $q, w, \Delta_r U, \Delta_r H$.

Since the process is isothermal $\Delta U = 0$ and $\Delta H_p = q_p = -w_p = -40.656$.
I got $\Delta H_p = q_p$ part correct but
For other two
author used $\Delta H - \Delta n(g)RT = \Delta U = 37.55$ and then first law to get $w = 3.101$.
Why $\Delta U$ is not zero ?

ΔU is only required to be zero for an isothermal process undergone by an ideal gas. In this case, we are dealing with liquid water, so the ideal gas law will not apply. For a related question on this site:http://chemistry.stackexchange.com/questions/37720/thermodynamics-enthalpy-and-internal-energy-for-isothermal-expansion

@Tyberius Second answer solved my problem. Thank you.


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Ignoring initial H+ Concentration For Addition of Strong Base to a Buffer Solution
When adding a strong base to a buffer solution, with a generic weak acid HA and generic weak conjugate base A-, is the initial amount of H+ ignored because it is so small? Specifically, at least in my text book, we only consider what happens to HA and A- when a strong base, say NaOH, is added.

I guess yes. Then again, is it really ignored?

It really isn't clear what you're asking. Typically for a weak acid you use equal amounts of HA and NaA. If acid is added the NaA protonates to HA, and if base is added the HA deprotonates to A-.

Suppose you have a weak acid and its corresponding base in solution. Suppose the solution has the pH = 5.
HA + H2O = A- + H3O+
Now we add a strong base to the solution. What will happen?
The pH will increase, suppose it will increase to pH = 6.
The weak acid-base system will immediately respond to the pH change of the solution and a new equlibrium will be established defined by the new pH and the equilibrium constant of the weak acid-base system. The equilibrium will be forced to the right, i.e. more A- will be fomed at pH = 6 compared to what we had at pH = 5.
Ok, what has happened to the hydrogen ions present at pH = 5 or those that were formed up to
pH = 6?
They have reacted with the strong base forming water.
Do we think the concentration of the hydroxide ions from the strong base has decreased at pH = 6?
No.
Why?
At pH = 6 the hydrogen ion concentration is lower than at pH = 5. This is equal to say that the hydroxide ion concentration is higher at pH = 6 than at pH = 5. The difference in the amount of hydroxide ions present at pH 6  and 5 will be equal to the amount of hydroxide ions consumed in the transfer from pH 5 to 6. What you put in, you get back in this case. The net effect is that the hydroxide ion concentration of the strong base has not changed at the equilibrium of the process.
I suggest you also look at my answer for another question "Buffer and strong acid".


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Alcohol and acetic acid
I recently did an experiment putting alcohol and acetic acid with conc. Sulfuric acid and heating it up, then i recondense the CH3COOC2H5. After that I put in some water and then NaCl solution I see some semi-transparent and red precipitate, can someone explain what it is

Likely contamination


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The Nomenclature Of Poly substituted benzenes?
I am confused as to the nomenclature of poly substituted benzenes. Would more preference be give to to priority order or the alphabetical order? In this pdf, 4-aminophenol has been cited as more correct rather than 4- hydroxyaniline. However, there are some websites which site the other one as more correct. What would be the correct nomenclature. A reference from one of IUPAc's books itself would be great.
Thanks a lot!!


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Number of electrons transferred & Gibbs free energy
To determine the number of electrons transferred for the following reaction:
$$\ce{CH3OH + \frac{3}{2} O2 -> CO2 + 2 H2O}$$
I found the half reactions to be:
$$\text{Anode: } \ce{CH3OH + H2O -> 6H+ + 6e- + CO2}$$
$$\text{Cathode: } \ce{\frac{3}{2}O2 + 6 H+ + 6 e- -> 3 H2O}$$
The number of electrons transferred is therefore 6. If I was given the following overall reaction:
$$\ce{2 CH3OH + 3 O2 -> 2 CO2 + 4 H2O}$$
Which is just multiplying everything by 2 for the original reaction, would I say 12 electrons have been transferred or still 6? Since the Gibbs free energy change = nFE where n is the number of electrons transferred should surely be the same?

On topic video

You should use $12$ electrons. Another way to think of it is when you calculate the $\Delta G$ of the first reaction, that value will be units of energy per mole of reaction. So, since you doubled the moles of reaction, the $\Delta G$ would double.

It is $12$. Gibbs Free Energy is an extensive thermodynamic property. In electrochemistry, it depends upon the number of electrons taking part in the reaction.
In the second reaction, you are using twice the number of reactants. This implies that twice the number of electrons are used in the redox reaction, so the Gibbs free energy will be doubled.

When you go from $\ce {CH3OH}+1.5\ce {O2}$ to $2\ce {CH3OH}+3\ce {O2}$, you double the amount of reaction you're working with.  Those ciefficients are not just ratios, they're the number of moles of stuff you put in or get out.  So the number of (moles of) electrons and free energy change will scale up accordingly.


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Osmotic pressure and Raoult's law co-relation?
Consider two compartments A and B joined. A contains pure water. B contains brine (dilute). Consider the Vant Hoff factor ($i$).
Obviously pressure over B is less than over A.
We also know that the excess pressure if equal to $iC{\rm R}T$ (C is concentration of brine)
Thus it follows that pressure over B is p - icrt (p is vapour pressure of compartment A, which is pure water in this case).
But from Raoult's law we also know that pressure over compartment B is PX (x is mole fraction of water in brine).
That would mean
$$p - iC{\rm R}T= px$$
Is the equality correct?


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Clothing Fabric Question
How does cotton and polyester come together during the production process to produce clothing items? Also, why is it cheaper to have cotton and polyester in a clothing item versus clothing that is made of just 100% cotton?


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Can water be included in an ICE table if it is not the solvent and is a product?
The question that I have is whether you will count water in a Fischer esterification equilibrium problem where the equation is
carboxylic acid + ethanol --> ethyl acetate + Water
We need to find the concentrations of the different components at equilibrium and are also given the equilibrium constant. I know that if water is a reactant and the solvent you do not count it when finding x for the concentrations, however, in this problem, it is specified that the water is not a solvent and the moles of water are given. I've looked in multiple sources and none have said if there is an exception with water like this.
Also, in the problem, it only refers to the water as "water", so it doesn't state that the water is pure/a pure liquid.
Thanks!

In this sort of a reaction the water counts. // For reactions like $K_a$ and $K_{sp}$ you don't count water because it is assumed that the concentration of water is constant. To be more precise you'd actually use activities of the reactants which does indirectly consider the concentration of water.


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How to remove paraffin wax drop from cam lense glass?
When I put a my cellphone on a table for charging, some drops of paraffin wax were laying there which I couldn't see.  After some time during charging, heat melted the paraffin wax and it stuck on the rear camera lens of my Lenovo k6 Power.
I tried to wipe the lens with a special cloth especially used for cleaning my glasses, but it couldn't remove all the scratches/stains.
Please suggest any solutions.

It'll dissolve in (USP, clear) mineral oil, or (non-pumice-containing) waterless hand cleaner.   The 'special cloth' will just get contaminated, don't want that.

The first thing I would try is to re-heat the phone then try to wipe the paraffin off with your lens cloth. Paraffin wax typically melts between 46 and 68 °C (115 and 154 °F)$^1$. Assuming you don't have a good way to accurately heat the camera to this temperature, I would just repeat what you did to melt the wax in the first place; let the camera discharge it's battery and then recharge it. Once it has heated up, try wiping it off with the lens cloth.
Unfortunately, I have a feeling the above procedure alone may not be sufficient to remove all of the wax. It's worth trying because it's simple and safe for your phone. If that procedure does not remove all of the wax, the next thing I would try is using a solvent in addition to the heating. The wax should dissolve in diethyl ether$^2$, which is the primary component of most engine starting fluid products. You should be able to go to any auto parts store and purchase a can of starting fluid who's primary ingredient is diethyl ether (sometimes just called ether).
The first thing you will want to do with the ether is to verify that it does not attack the material holding/surrounding your lens. If the area around the lens is all metal and glass, you are safe and can skip this step. If the area around the lens is plastic (unlikely) then you will want to test a small area of the plastic to be sure it will not be dissolved or distorted in any way by the ether. Simply soak a corner of the lens cloth with the ether and touch it to the plastic. You have to be fast because ether evaporates very quickly. You might want to use a magnifying glass to look at the area that you whetted with the ether. If there is any change in the color or texture (you should feel the area with your finger), then you have to decide whether you want to proceed and risk ruining your camera. Again, it is unlikely that this will be a problem. I looked at some images of your phone model, and while there are some plastic parts to the construction, the lens assembly appears to be just metal and glass.
If you decide to proceed with cleaning the lens with ether, it should be a pretty simple, straight forward process. I would heat the phone by charging it as discussed above, soak a portion of the lens cloth with the ether-based starter fluid, and gently wipe the remaining wax from the lens. The lens glass and any metal surrounding it are impervious to the ether (and any other ingredients of the starter fluid), but avoid contact with any plastic components.
Given your problem as you've stated it and the make of your phone, I'm pretty confident that your camera lens can be thoroughly freed of the paraffin wax. Don't hesitate to ask for clarifications of anything I've said in the comments, and best of luck!
1) Kaye, George William Clarkson; Laby,Thomas Howell. "Mechanical properties of materials". Kaye and Laby Tables of Physical and Chemical Constants. National Physical Laboratory. Retrieved 25 October 2013.
2) Seager, Spencer L.; Slabaugh, Michael. "Alkane reactions". Chemistry for Today: General, Organic, and Biochemistry. Belmont, CA: Cengage. p. 364. ISBN 978-0-538-73332-8.

Who cares about the lens assembly (typically all metal, like you said)? It is the anti-reflective coating on the lens itself that can be damaged in the first place. Chances are it's ruined already. Then again, I totally support your idea of starting with the diethyl ether. Petroleum ether, maybe. Acetone, definitely no.

@IvanNeretin , excellent point that I really didn't think about. I assume that's some organic polymer and like you said it may be too late as it is. I think if it were my camera, I would probably get it pretty hot and really try to go after it with the lens cloth. If that didn't work I'd figure I have nothing to loose and try ether as my first choice of solvent. I think I better edit that caution into my answer...thanks for the heads up on that oversight.

@IvanNeretin , some quick internet searching seems to indicate that the overwhelmingly most common antireflective camera lens coating is MgF2 because of it's physical properties and optimal refractive index. Wikipedia also said that fluoropolymers are occasionally used, but that's the only mention I've seen of an anti-reflective coating other than MgF2, so the possibility of a solvent soluble coating may be a non-issue.  I'm going to look into this a bit more before making any edits to the question. Any further thoughts you might have would be greatly appreciated.


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Structural determination of carbohydrates(sucrose)
I recently came across a question in which we have to find out the number of moles of $\ce{HIO4}$(Periodic acid) which are required for cleavage of $\ce{C-C}$ bonds in sucrose.
I went to https://en.m.wikibooks.org/wiki/Structural_Biochemistry/Carbohydrates/Carbohydrate_Structure_Elucidation_through_Periodic_Acid_Cleavage . But this article is rather incomplete and doesnot answer the question. In general what are the things to be kept in mind in order to find out the number of moles of periodic acid required for the structural determination of carbohydrates. Any ideas?  Thanks.

You need to be thinking about titrating your periodic acid before and after reaction.

@user1945827 The amount of periodic acid used before and after reaction is to be neglected in the question.

you're contradicting your original question, "we have to find out the number of moles of..." vs " The amount of periodic acid used before and after reaction is to be neglected".  The number of moles is the amount.  ... But I've just realized that this is a homework question.  Work out the stoichiometry for each C(OH)-C(OH) group in glucose ...  Hope that helps.

The correct term of "number of moles" is amount of substance.

First of all, lets be familiar with what per-iodic acid does to organic compounds.
It selects the carbon-carbon bonds that have either an -OH group, a =O(carbonyl) group or if it is the functional group carbon of a carboxylic acid. Both the carbons of the C-C bond must satisfy this condition. Here are some examples of the selection points of per-iodic acid:

Now that per-iodic acid has chosen the center, it cleaves the C-C bond and adds an -OH at each carbon. For every C-C bond, one per-iodic acid molecule is consumed. I am avoiding going throughout the actual mechanism to keep things simple. If any of the carbons has two -OH groups, they will spontaneously dehydrated to give a =O bond instead. For the examples shown above, the final products of the cleavage-oxidation would be:

Now, coming to your main question, the structure of sucrose would be:

By what we have seen, lets mark the C-C bonds that per-iodic acid can cleave:

Since we have 3 centers, so thats why 3 molecules of per-iodic acid are consumed per molecule of sucrose. Now you can even proceed and find out the oxidation products.
I hope this answer helps. Happy carbohydrate breaking!

That is a great explanation, Balagopal
I had just a small doubt...
Acetals [including sucrose] can be hydrolysed in acidic medium.
Is HIO4 sufficiently acidic to bring about hydrolysis of sucrose...
I doubt, if this happens, we need to reconsider the number of moles of HIO4 consumed completely.

@Che mistry, acetals are hydrolysed? As per Wikipedia, it wasn't mentioned that per-iodic acid cleaves acetals, and neither does my textbook.

@Che Mistry, Acetal hydrolysis doesn't involve a $\ce{HIO4}$ molecule. Simply an acidic medium that is. The $\ce{H^+}$ will be returned after hydrolysis is over.

@Pritt , but i have seen that only syn diols reacts with HIO4, as when we see mechanism there is formation of ring with IO4- that is possible only with syn configuration of OH OH, how can anti react ?


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Help with uncertainties
When diluting a mixture of two reactants with lets say a 100ml of H2O, how do you add this to the uncertainties of the concentrations of the reactants? Should the relative uncertainty of the added H2O be added to the relative uncertainty of the volume in both reactants?

When adding values together, you should add together their absolute uncertainties. For example, if you started with $500\pm5\mathrm{mL}$ and add $100\pm1\mathrm{mL}$, your new volume value is $600\pm6\mathrm{mL}$.
You would add the relative uncertainties if you are multiplying values. For example, if you started with $1\pm.06\text{ moles}$ and want to find the molarity in a solution with a volume of $500\pm5\mathrm{mL}$, your molarity will be $2\pm.14\text{M}$. The uncertainty was obtained by finding the sum of the relative uncertainties and multiplying it by the concentration to obtain an absolute uncertainty, as shown below.  $$(\frac{.06}{1}+\frac{5}{500})(2\text{M})=\frac{7}{100}(2\text{M})=.14\text{M}$$
For more details on uncertainty rules, this page from the University of Victoria gives some of the basics.

When adding or subtracting uncertainties they are squared and added then the square root taken. Thus if $z=x+y$ where x and y are two measurements with uncertainties $\sigma_x, \sigma_y$ the combined uncertainty is
$$\sigma_z =  \sqrt{\sigma _x^2 + \sigma_y^2}$$
 and the answer $z= (x+y) \pm \sigma_z$
For more details besides adding values see this answer
Uncertainty Equations Confusion

@L. Wagemaker I would recommend this answer. My method is a simplification, which is easier to compute, but will typically overestimate the uncertainty.


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Why there is no temperature scale tied to normal human body temperature?
Fahrenheit:

The lower defining point, 0 °F, was established as the temperature of
  a solution of brine made from equal parts of ice and salt.

Celsius:

0 °C was defined as the freezing point of water and 100 °C was
  defined as the boiling point of water

Why hasn't anyone used human body normal temperature as 0? It would be much more logical for humans to base temperature scale on our body, not something that is outside.

Please check wikipedia how the Fahrenheit scale was developed. And no, there is no "logic" behind that. Convenience, perhaps.

Interesting point. Water is a pretty important part of human physiology though, so the Celsius scale is  not completely irrelevant to the human body. I suspect water's importance to human physiology was a major reason for using it's properties for a temperature scale.  I just saw DavePhD's answer though and I had no idea there actually was a "physiological scale"! Again, interesting question.

Why on earth should it make sense to have a temperature scale based on human body temperature, which is not very constant to begin with ? And what does this have to do with chemistry? Please close question.

I will not be voting to close the question as it at least loosely fits the history-of-chemistry tag. Virtually every chemical process has some relationship, cause and/or effect, to temperature. Choice of a temperature scale to be used in the sciences has historically not been a simple process. Different temperature regimes are important to different disciplines. Until relatively recently, erroneous concepts of absolute zero abounded. Although the ultimate choice of temperature scale is fundamentally arbitrary, the history of this choice is loaded with scientific invention and discovery.

As @airhuff, I think this question is valid for "history of chemistry" and hence suggest to keep it.  Temperature is one of the fundamental properties of matter; and as science is dynamic, evolving, depending on the field of application some scales are used more frequently, than others.  And in contrast to Karl's second comment I assume, it is thanks to both i) establishing a scale (perhaps "at all"), and ii) learning how to measure accurately "temperature" as comparison with references, that we recognized body temperature varies.

Why nobody used human body normal temperature as 0?

That has been done.
See Medical Thermometry and Human Temperature at page 12:

37° Centigrade scale
29.6° Reaumur
98.6° Fahrenheit
77° Walferdin's tetracentigrade
0° of the physiological scale

Then the physiological scale is further discussed starting on page 265, which includes a drawing of a physiological scale thermometer.

physiological 0 is placed in the centre of the drama, whose acts are health, sickness, and death


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Benzene ring position labeling (ortho vs para)
Why is the ortho position on a benzene ring next to the primary carbon and not in the opposite position, as ortho means "straight", not "next to" as in para?

Not sure, but I'm glad it turned out O-M-P, because it being alphabetical order offers a great mnemonic for new students.

What alphabet are you referring to? In English, M-O-P is alphabetical order, and O-M-P doesn't work.

https://en.wikipedia.org/wiki/Arene_substitution_pattern#Origins

Does this answer your question? Origin of "ortho", "meta", "para"


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Can final volume differ from initial volume in a constant volume process?

A sample of 1.00 mol perfect gas molecules with $C_{p,\mathrm m} = 7R/2$ is put through the following cycle: (a) constant-volume heating to twice its initial volume, ...

(Problem 2.5, Atkins Physical Chemistry 8th ed.)

The problem says that in a constant volume process the final volume is twice the initial volume.
Is this even possible? How can volume become twice if it was constant all the time, it wouldn't have exploded to twice its size suddenly after the process ended right?

The question is obviously wrong. What's your source?

@ToddMinehardt Physical Chemistry by Peter Atkins 8th ed. I took it from my school library.

Very odd. The question makes no sense as written. Constant volume is, well, constant, so you can't get twice the volume if the volume is held constant. Maybe have a look at an earlier or later edition of Atkins, see if this is an error introduced during a revision.

@ToddMinehardt I don't have a previous version. Anyway the question is 2.5 in problems section.

@ToddMinehardt the 8th ed of Atkins indeed has this weird question: https://i.stack.imgur.com/enri8.png. The question has been removed in the 10th ed. It is very obviously wrong, I suspect it was meant to be constant pressure.

Next time, please include the source with it, and quote it verbatim. That allows people to better judge what's going on.

@orthocresol - Thanks for tracking that down. Guessing we can settle/close this question under the "typo" category?

@orthocresol I also thought it meant constant pressure but then in answers they gave work as zero which cannot happen in a constant pressure process.

@orthocresol Yes I do so next time I ask a question.


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Sn1 Stereochemistry and Energy

Sn2 has simultaneous bond breaking and forming because it's concerted.
Sn1....does it occur with a loss of stereochemistry at the reactive center? An online textbook says so.
Sn1 has both retention and inversion of config. So don't know which one fits Sn1. Only one statement is true for SN1.

Which of the energy diagrams shown below represents an exothermic SN1 reaction?

It would be graph A because Sn1 is stepwise two steps. And it's exo because of -E?

Homework questions must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: How do I ask homework questions on Chemistry Stack Exchange?

I'm voting to close this question as off-topic because it's an AMIRITE question, too localized to help future visitors. Please include an answer not in your question, but in the form of an answer with full details of your reasoning.

@Klaus Warzecha Not off-topic because I gave my reasoning. Only statement I couldn't reason is if SN1 "occurs with a loss of stereochemistry at the reactive center" since it's not in my notes.


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Does the size of a hole in a container affect the rate of effusion?
Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of density of the gas.
If there were two containers both with the same gas but one of the containers had a larger hole than the other, would the rate of effusion be quicker for the gas in the container with the larger hole?

A small hole is used. Knudsen conducted experiments similar to  Graham's but approximately 50 years later (in 1909) and used a 0.025 mm diameter hole in a platinum strip of 0.0025 mm thickness. It was observed the size was not critical as long as the mean free path of the gas was ten times larger than the hole, thus the gas was at low pressure.
In these experiments the gas is allowed to stream out of a small hole in the vessel into vacuum. The velocity of the escaping gas is just that of the trapped gas, that would be obtained if the hole was not there, and these velocities would be the molecular velocities. Thus the size of the hole must not disturb the distribution of molecular velocities, achieved by being smaller than mean free path.

Will the shape of hole I.e if it is a square of side a, will that effect Graham's law ?


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Why do you use catalysts if they increase the rate of forwards and backwards reactions equally?
So I wanted to know why catalysts didn't affect concentration; since if they increased rate of reaction, presumably the concentration of the products would increase.
I did read online, however, that catalysts increase the rate of the forwards and backwards reactions equally; and then the concentration not changing makes sense, but there's another question now.
I thought the whole point of a catalyst was to increase the rate of reaction so you can get more of something in less time. If it increases only the rate of reaction and not concentration, the net output of the products would be the same as if you didn't use a catalyst — and in that case, what's the point of even using a catalyst?

You got it absolutely right. The point is not to get more of something in less time (which the catalysts can't do anyway). The point is to get as much of something in less time. Also, welcome to Chem.SE.

Thanks very much for the welcome. :) I wasn't sure what you meant by that though; isn't getting as much of something as you can in less time the same as getting more of something in less time? I mean I intended it as getting as much of something as you can as fast as possible, but maybe my definitions are all messed up...?

In a simplified representation, a catalyst lowers the overall activation energy while going from initial state A to the of the products B along the reaction coordinate. The interference of the catalyst on the reaction alters the reaction mechanism, renders alternative pathways accessible, which then may be  more favourable than for the reaction in absence of the catalyst. Catalysts are deployed regardless if the overall reaction is exothermic, or endothermic. And, because of the relationship described in the Eyring equation, changes in the free activation enthalpy influence rate constants.

For the instance of a heterogeneous catalysis, the picture is simplified as assuming energy chances for adsorption of the starting material(s) to and subsequent desorption of the product(s) from the catalyst were non-significant in the overall picture.
You are right that it may ease both the forward, as well as the backward reaction.  However, in absence of catalyst, overcoming the activation energy at all were just too expensive, even if the reaction $\ce{A -> B}$ drawn here happens to be an exothermic one.
As examples, think about the hydrogenation of benzene; in absence of Nickel there were practically no reaction at all, even at elevated temperatures.  Numerous biochemical reactions were unable to proceed in absence of a catalyst at ambient temperature, too.

I'm nit picking a bit, but the catalyst does not lower the activation energy as such but opens up a new pathway to products and this has a  a lower activation energy. The thermodynamics are always the same as the product and reactant have the same energies as when un-catalysed.

Thanks for replying! I do understand that catalysts provide an alternate pathway with lower activation energy, which leads to a faster rate of reaction; but if more product isnt being produced through that, what is the point? I mean what would be the point of a faster reaction with the same net output as a slower reaction? I don't know, maybe I'm just confusing myself...

@Porphyrin I welcome your comment and consider it as useful nit picking.  The answer given was revised.

@Jasmine Keeping reaction rates aside, thanks to a catalyst's interaction  the conversion along $\ce{A  -> B}$  occurs e.g. already at a lower temperature, so you need less investment in heating the reaction volume.  Or it occurs at a lower pressure, which lowers the cost of the reaction vessel, too.

@Jasmine What is the point of driving someplace, when you can just walk there? Sure, it would be somewhat slower, but the end result is the same, isn't it?

I think OP is of the misunderstanding that all reactions are (appreciably) reversible. If the equilibrium vastly favors the products, such that the reaction is essentially irreversible, then increasing the rate of the backward reaction will cause a negligible loss of product.