text:
What are the products of the hydrolysis of ICl3?
I have learnt that interhalogens on hydrolysis give a halic acid derived from the smaller atom and an oxoacid derived from the larger atom. For example hydrolysis of $\ce{BrF5}$ would give me $\ce{HF}$ and $\ce{HBrO3}$.
Similarly the hydrolysis of $\ce{ICl3}$ should give me $\ce{HCl}$ and $\ce{HIO2}$ but my textbook says that $\ce{HIO2}$ is too unstable to be isolated. What then would be the products of the reaction?

Guess it would disproportionate to $\ce{HIO3}$ and something on the other side.

This book mentions the reaction, however, I am unable to comprehend the paragraph that follows the given reaction.

$$\ce{3ICl3 + 6H2O → 9HCl + HI + 2HIO3}$$

Iodine(III) chloride react with water to produce hydrogen chloride,
  hydrogen iodide and iodic acid. The reaction is called Hydrolysis of
  iodine trichloride.(source)

The reaction is same as @xavier_fakerat noted in the comment section.

Something like this:
$$\ce{2 ICl3 + 3 H2O  -> ICl + HIO3  +  5 HCl}$$

Source: Chemistry of Interhalogen Compounds (Page 16) By P. B. Saxena
  2007

Screenshot (click image for larger variant)

Strangely some suggest that HI is produced too, but I think it depends on reaction conditions note: $${3ICl3 + 6H2O → 9HCl + HI + 2HIO3}$$ but I favor the disproportionation reaction above

@NilayGosh So it depends on reaction conditions right?

Basically yes. Also various iodine species and  other intermediates will form during the reaction course but they are unstable and disproportionates. The type of intermediates forming depens on reaction conditions and concentration of water.


text:
Reducing sugar among the given compounds
The question is to determine which of the following sugars is not reducing?

I have the idea that in case of disaccharide sugars  in which the monosachharide units are bonded via the carbon containing aldehydic group are non reducing.However the present question deals with sugars having monosaccharide units.I have no idea on how to deal with such sugars.Any insights?Thanks.

The English wikipedia page on reducing sugars starts with

A reducing sugar is any sugar that is capable of acting as a reducing agent because it has a free aldehyde group or a free ketone group. All monosaccharides are reducing sugars [...]

Unfortunately, the monosaccharides tend to hide the aldehyde and keto groups by undergoing intramolecular reactions between a carbonyl group and a hydroxy group. This reaction leads to a cyclic hemiacetal.
In order to figure out which of the structures in your question is a monosaccharide in disguise, we need to spot the hemiacetal moiety. With other words:
Which of the structures can open again to show a carbonyl group?
We can definitely rule out (C)! This is a hydroxymethyl-substituted cyclohexane with some hydroxy and a methoxy group.
What about (D)? Close, but no cigar. This is an acetal, which won't open that easy.
I'll leave the last two examples to you ;-)

Thank you.That means that D) which is an acetal doesnot give the open chain form and hence is non reducing.

@aniline Exactly :)

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text:
Are nitrogen oxides always radicals?
Some reactions between nitrogen oxides and ozone can be the following:
$$NO\cdot(g) + O_3(g) \rightarrow NO_2\cdot(g)+O_2(g)$$
$$NO_2\cdot(g) + O\cdot(g) \rightarrow NO\cdot(g)+O_2(g)$$
Where the $NO$ when transformed to $NO_2$ stays a radical. Why is that the case? Wouldn't the unbounded electron left be used to form a double bond between nitrogen and oxygen?

Compounds of nitrogen and oxygen with the general formula $\ce{N}_x \ce{O}_y$ will always be radicals when $x$ is odd. Nitrogen atoms have five valence electrons and oxygen atoms have six valence electrons. $\ce{NO}$ will have 11 valence electrons. One will have to be unpaired. $\ce{NO2}$ has 17 valence electrons. One must be unpaired. However $x$ can be even like in $\ce{N2O}$, which has 20 valence electrons.

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text:
% yield in a Williamson reaction
Calculate % yield in a Williamson reaction using the information provided below. Enter numeric value only (without % sign) rounded to the nearest whole percent.

I've done ether synthesis before but not sure what this is asking...
1.1g/2.1g=52% yield?

I think they're asking for the molar yield (moles of product obtained divided by moles of limiting reagent used). The clue is in the fact that they give you the molecular weights. Should be 44%.
[BTW, what you've calculated is the mass yield, which is sometimes used in process chemistry].

I guess your instructor means that you can calculate the moles of product (1.1 g /172.23 (g/mol) = 0.0064 mol) and divide them by the moles of reagent (2.1 g /144.17 (g/mol) = 0.0146 mol), which in this case are equal to the 'theoretical' moles of product, so 0.0064 mol /0.0146 mol = 44%; or you can calculate the theoretical mass you would get if the reaction had 100% yield (172.23 g/mol *0.0146 mol = 2.51 g) and divide the actual mass you obtained by that (1.1 g /2.51 g = 44%). As you can see, it's mathematically equivalent.


text:
When is a dipole-dipole interaction strongest?
Are there any conditions that'll improve the interaction? Just like, I assume, hydrogen bonds are strongest when the difference in electronegativity is biggest.

To some degree, hydrogen bonding can be thought of as a subset or type of dipole-dipole interaction. This Wikipedia article defines dipole-dipole interactions as:

Dipole-dipole interactions are electrostatic interactions between permanent dipoles in molecules. These interactions tend to align the molecules to increase attraction (reducing potential energy).

The same article states, regarding hydrogen bonding:

The hydrogen bond is often described as a strong electrostatic dipole-dipole interaction. However, it also has some features of covalent bonding.

So, the assumption/example in your question is generally correct: just as a large difference in electronegativity of two atoms within a hydrogen bonding molecule results in stronger hydrogen bonds, the same tends to hold for non-hydrogen bonding dipole-dipole interactions. And while there are other forces affecting dipole-dipole interactions (i.e. solvent effects, molecular geometry/steric effects, etc.) it is the strength of the permanent dipole moment resulting primarily from the bond length and the difference in electronegativity between the atoms at the permanent dipole location that determine how strong the intermolecular dipole-dipole interactions will be.

An electric dipole is formed in a molecule as the vectorial sum of all the partial charges on the atoms multiplied by their separation. Thus if, these charges are large and well separated and not symmetrically situated about the molecule the dipole will be large. For example chloro-benzene has a far larger dipole than toluene. However, 1,4-dichlorbenzene has zero dipole as the charges cancel out. (Molecular point groups can be used to decide if a molecule has a dipole, but not its magnitude. Only molecules in groups $C_1, C_s,C_n,C_{nv}, n>1$ have dipoles). Hydrogen bonds have no special significance with respect to dipoles.
Dipoles interact through space, and the strength of their interaction depends on the distance between the dipoles,  relative orientation and relative permittivity (dielectric constant) of the intervening medium. Each of these have a dramatic effect for any given pair of dipoles.
The interaction energy E for dipole vector $\mu_1, \mu_2$ at separation R and where $\bar R$ is the vector joining the centres of the dipoles is
$$E = \frac{1}{4\pi\epsilon_0\epsilon} \left( \frac{\mu_1\cdot\mu_2}{R^3} - 3\frac{(\mu_1\cdot \bar R)(\mu_2\cdot \bar R)}{R^5}    \right)  $$
The dot product between two vectors e.g. $\mu_1\cdot \bar R = |\mu_1||\bar R|\cos(\theta_1)$ where $\theta_1$ is the angle between the vectors, see figure.
If the vectors are now represented as unit vectors e.g. $\bar \mu_1=\bar u_1 d_1$ where d is the magnitude of the dipole, and $\bar R = \bar u_R R$ then the energy becomes
$$E = \frac{K}{4\pi\epsilon_0\epsilon}\frac{d_1d_2}{R^3}   $$
where
$$K= \bar u_1\cdot \bar u_2 -3(\bar u_1 \cdot \bar u_R)(\bar u_2 \cdot \bar u_R) $$
The figure shows the geometry.

If the dot product $\bar \mu_1 \cdot \bar \mu_2 =0 $ and one  of the two dot products with $\bar u_R $ are $0$ then the energy is zero, and this can occur when dipoles are at 90 degrees and one points to the centre of the other. Physically this makes sense as each end of a dipole sees equal positive and negative charges on the other dipole.


text:
Aryl halides by Hunsdiecker reaction
I wanted to know wether we can prepare aryl halides by Hundiecker reaction or not. if so then what is its mechanism i know the mechanism for alkyl halides. please refer these links and please try to explain with reason.
http://www.organic-chemistry.org/abstracts/literature/918.shtm
http://www.organic-chemistry.org/abstracts/literature/449.shtm

There might be more recent publications, but an old review by Robert G. Johnson and Robert K. Ingham, Chem. Rev., 1956, 56, 219-269, gives some directions.
Based on early attempts with benzoic acid that date back to the time around 1840, is was long believed that the silver salts of aromatic carboxylic acids do not undergo a Hunsdiecker reaction at all. The authors write:

In no case was bromobenzene obtained as a result of the
  action of bromine on silver benzoate. The product of Peligot and Bunge was a monobromobenzoic acid; apparently Kekule’s experiment was completely unsuccessful.

However, it later turned out that the inital verdict wasn't quite true.
Benzoic acids bearing electron-withdrawing substituents, such as $\ce{Cl}$ or $\ce{NO2}$ can be converted to their corresponding bromoarenes in a Hunsdiecker reaction. The authors of the review denote:

Thus, the three isomeric nitrobenzoic acids (as silver salts) were
  converted to the corresponding bromides in excellent yields (ortho, 95 per cent; meta, 89 per cent; para, 79 per cent). The yields from the chlorobenzoic acids were less satisfactory, but sufficiently high to indicate the efficacy of the chloro group.

Sir substituents like NO2 would work in case of carbanionic mechanism but how would this stabilise the intermediate in case of free radical mechanism

There is a recent paper1 which discussed the various ways of decarboxylative halogenation. In the section "2.1. The Hunsdiecker–Borodin Reaction", you will find tons of details including aryl halide synthesis with their yields:

The paper cited two sources for the reaction scheme2,3.
But the problem with these type of reactions is you end up contaminating your end products with chlorinated and mixed halide side products or esters if you mess up the stoichiometry.For e.g. Bromodecarboxylation of Silver Benzoate:

In some case, the aromatic halide end product is volatile e.g. Bromobenzene
Notes and References

Decarboxylative Halogenation of Organic Compounds by Andrii Varenikov, Evgeny Shapiro, and Mark Gandelman, Chemical Reviews 2021 121 (1), 412-484
DOI: 10.1021/acs.chemrev.0c00813

Bockemüller, W.; Hoffmann, F. W. Über Carbonsäure-Hypohalogenite. Liebigs Ann. 1935, 519, 165– 192,  DOI: 10.1002/jlac.19355190115

Bromobenzene is volatile in vacuo at 100°C", but it was isolated contaminated with less than 10% of carbon tetrachloride. The low yield of benzyl bromide may be due to its volatility in vacuo at 100".
Oldham, J. W. H. 22. A Study of the Action of Bromine on the Silver Salts of Organic Acids. J. Chem. Soc. 1950, 100– 108,  DOI: 10.1039/jr9500000100 (PDF)

When the silver of picolinic acid reacts with bromine in hot nitrobenzene, the expected bromopyridine is formed along with 2,2'-dipyridine
Reaction Mechanisms in Organic Chemistry by Metin Balcı, 2021

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text:
Why do different H+ in polyprotic acids have different K?
Why do different $\ce{H}$ atoms in a polyprotic acid have different values of $\pu{K_a}$?
All $\ce{H}$ atoms (which can show acidic nature) have similar properties within the molecule. It does not matter which one reacts. For example, in $\ce{H3PO4}$, all the $\ce{H}$ atoms can be dissociated from the molecule to act as acids. All of them are attached to an $\ce{O}$ atom, which is attached to the central $\ce{P}$ atom. If all the properties are same, then why is $\pu{K_a}$ different?
`

You are correct that all of the $\ce{H}$'s in $\ce{H3PO4}$ are equivalent.
However, once the first H is removed, the remaining two $\ce{H}$'s are no longer equivalent to the $\ce{H}$'s in the original $\ce{H3PO4}$. After the initial deprotination, the molecule gains an overall negative charge making it more difficult to remove the second $\ce{H}$.  Then when the second $\ce{H}$ is removed the molecule gains a second negative charge making it even more difficult to remove the final $\ce{H}$.  Thus, the $K_a$ values for the originally identical $\ce{H}$'s change each time the molecule is deprotinated.

Consider $\ce{H3PO4}$. $K_{a1}$ and $K_{a2}$ for $\ce{H3PO4}$ are different because they concern different chemical species: $K_{a1}$ deals with the $\ce{H3PO4}$/$\ce{H2PO4-}$ equilibrium, whereas $K_{a2}$ deals with the $\ce{H2PO4-}$/$\ce{HPO4^2-}$ equilibrium.


text:
Why is the hydrogen atom adjacent to the carbonyl carbon of aldehyde not acidic?
In $\ce{H^a-CH_2-CO-H^b}$, why is $\ce{H^a}$ more acidic than $\ce{H^b}$ ? From what I know, $\alpha$ hydrogen ($\ce{H^a}$) is highly acidic and is easily abstracted in presence of base $\ce{OH^-}$. But why is $\ce{H^b}$ not abstracted in the similar fashion?
$\ce{C}$ is electronegative because of the $+I$ effect it exerts on $\ce{O}$. Since, $\ce{H^b}$ is more nearer to the $\ce{C}$, shouldn't it be more acidic as the bond pair is more shifted towards the carbonyl carbon?

The problem is that removal of H in the latter case causes resonance stabilization while in the other one it does not.

related http://chemistry.stackexchange.com/questions/12232/acidity-of-aldehydes

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text:
Why aren't extremely dilute acid solutions basic in nature?
Lets say we prepare a solution of $\ce{HCl}$ with a concentration of  $\mathrm{10^{-8}}$. Even on full dissociation (which it will, because it is a strong acid), the $\ce{H+}$ ion concentration due to $\ce{HCl}$ is $\mathrm{10^{-8}}$.
So shouldn't the pH be 8? Is there any factor I am missing? Does the properties of acids change at low concentrations? If so, then how?

@Mithoron I got my answer. How do you get if a question is duplicate? I search for questions and hardly get relevant answers!

Must be my superior searching skills ;) There are some tips (https://chemistry.meta.stackexchange.com/questions/3071/searching-guidelines-for-newbie) but practice is needed to be good at it.

Water is a weak acid, too. At that concentration of your acid, the number of hydronium ions from water will be roughly $10^{-7}$ molar, making the solution roughly neutral.


text:
How does a mixture not retain its chemical identity?
My book says that matter can be classified as an element, compound, or mixture. It further says that a pure substance (or substance) is matter that has distinct properties and a composition that doesn't very from sample to sample. So far, so good. Then it says that

all substances are either elements or compounds

An element is one type of atom (something that "cannot be decomposed into simpler substances") and a compound is composed of two or more elements; these both clearly are substances. Then, it says that

Mixtures are combinations of two or more substances in which each substance retains its chemical identity.

This is where I'm confused. How are mixtures not substances? They have distinct properties, and their composition doesn't vary from sample to sample, if you set a standard for the sample. You can create any arbitrary mixture, and you can create any arbitrary compound, and if you follow the "recipe" for either you will always get the same thing, as far as I know.
Where is my logic flawed?

It's a mixture of substances

@Joel, and a compound (a substance) is a combination of elements (each a substance). I don't see how that's any different.

You can't create any arbitrary compound. You may have $\ce{C2H6}$ or $\ce{C2H4}$, but never $\ce{C2H_{5.5}}$.

@heather , Once the elements have formed a compound it is not considered a mixture of those elements anymore it is a compound and a substance too.

@Joel, right, and I'm saying that why isn't that true for when you combine compounds into a mixture (that it is still a substance).

@IvanNeretin, you still follow a "recipe" though to create a mixture, so it's composition doesn't vary from sample to sample, right?

True, but it may vary if that's your intention.

@heather , that would be a mixture of substances wouldn't it?

Pure ethanol (substance) , ethanol / water (mixture)

You can separate a mixture into its different molecules (substances) , it may not be easy though but can in principle be done. Molecules of each type, e.g. alcohol, water, sugar etc. have their own structure , i.e. all water molecules are the same. A mixture assumes that no chemical reaction  has occurred, however, if a reaction is made to occur in a mixture  then new molecules are made and a different mixture of products results.

@porphyrin, I don't see though how that relates to whether or not it is a substance - the definition of a substance has nothing to do whether or not a chemical reaction has occurred.

Substances, i.e. molecules  are made through chemical reactions. A 'mixture' is a collection of substances, i.e. in chemistry this means molecules. A mixture of fruit, say apples & oranges is no different, the 'substances ' (apples, oranges) retain their identity in the mixture. In chemistry unlike in this fruit example, reactions can also occur and so change the substances (molecules) in the mixture.

As your book uses the terms:

Elements: Single atoms (e.g. O).
Compound: Consists of only one type of molecule, where that molecule consists of more than one element (e.g. H2O).
Mixture: Consists of more than one type of molecule (e.g. Na+ + Cl- + H2O, which is saltwater, or for example the air you breathe, which is mostly N2 with some O2 floating around and a bunch of other stuff).

I'm not entirely sure where your book classifies materials that consist of a single type of molecule where that molecule consists of only one element (e.g. O2) but whatever, you should be able to fit that in somewhere. It probably includes those as "elements".
To use a loose macro analogy (don't deconstruct this too deeply or you will soon run into trouble), let's say "blue paint" and "yellow paint" are elements. When combined they form the compound "green paint". On the other hand say "blue marbles" and "yellow marbles" are elements. When combined, they don't form "green marbles", they just form a mixture of blue and yellow marbles.
Note also that samples of your pile of blue and yellow marbles may vary: If you pick up a handful of them you very well could end up with slightly different proportions each time. On the other hand, every sample you take of the green paint will be identical green paint.

A good example of the difference might be to look at water, $\ce{H2}$, and $\ce{O2}$.
If you have gaseous $\ce{H2}$, and $\ce{O2}$ in a container you have a mixture; you can change it to have $99\%$ oxygen and $1\%$ hydrogen or vice versa or any ratio in between, but it is still a mixture of the two gases. These mixtures will have different properties depending on the ratio of the two gases. Each gas still has its own properties and the mixture exhibits some combination of them depending on the ratio.
If they are allowed to react to form water, now we have a substance. Regardless of how we make water, it will always have the same composition (two hydrogens for each oxygen) and the same properties. Importantly, these properties are completely different from hydrogen, oxygen, or any mixture of the two.

The phrase about retaining the chemical identity refers to the reactivity of the compounds in question. A compound reacts in certain ways that cannot be explained simply by looking at the elements a compound is made up of. For example, sodium bromide $\ce{NaBr}$ will react with chlorine gas to liberate sodium chloride $\ce{NaCl}$ and bromine. However, mixing just the elemental components with chlorine gas will either create a mixture that practically does not react at all (bromine and chlorine) or a mixture that will react strongly in a wildly different way (sodium and chlorine).
Having established that certain compounds react specifically we can look at how mixtures react. For example, you might create a mixture (i.e. solution) of sodium bromide in water. You can then bubble chlorine gas through this solution. The chlorine gas will not react to any noticeable extent with water but it will react with the dissolved sodium bromide — in the exact same way as above, i.e. it will also liberate bromine.
$$\ce{2 NaBr + Cl2 -> 2 NaCl + Br2}$$
Thus, even in the mixture with water sodium bromide has retained its chemical identity, i.e. its reaction with chlorine to give bromine and sodium chloride.
This principle can be extended to any mixture whose components have distinct reactivities. For example, nitrogen monoxide reacts with oxygen to give nitrogen dioxide — and these two will also react in this way (with no other reactions taking place) if, for example, the nitrogen monoxide is mixed with methane and the oxygen with carbon dioxide.


text:
Why are actinides not commonly included in rare earth metals?
According to the German Wikipedia, the rare earth metals include all elements of the third side group except actinium, and all lanthanides.

Zu den Metallen der Seltenen Erden gehören die chemischen Elemente der 3. Nebengruppe des Periodensystems (mit Ausnahme des Actiniums) und die Lanthanoide – insgesamt also 17 Elemente.

https://de.wikipedia.org/wiki/Metalle_der_Seltenen_Erden
Since actinides (like lanthanides) should be part of the third side group, why aren't they considered part of the rare earth metals?

It comes down to the history of their discovery, natural occurrence and applications. It's a fascinating story that you can read about in here:
http://www.vanderkrogt.net/elements/rareearths.php
The rare earths (i.e. lanthanides) were discovered over a few decades in the late 1800s. They were laboriously separated from "rare earths" - the name used back then to describe the rare minerals (silicates, oxides, phosphates, etc) that included these elements. These are all naturally occurring rocks. You can literally pick up from the earth a mineral that has rare earths in it. They are also stable (apart from Pm). Therefore, they are used in industry. There are rare earth mines, rare earth magnets, rare earth lasers. You get the stuff from the earth.
Actinides, were mostly discovered in the early to mid 1900s, after rare earths were a "thing". Other than U and Th, you do not find them in nature. You do not mine them from the earth, and they do not exist in "earths". Their radioactivity precludes their use in industrial applications other than a few niche uses. Although their chemical properties are similar to the lanthanides, there are many differences. They are simply not rare earths. Almost, but not quite.
I'll also add that the definition of "rare earths" is a loose one. The broad chemical definition includes the lanthanides La–Lu, Y and Sc. As far as geologists care, Sc is not a rare earth element, and Y is only "honorary". Industry and mining usually consider the lanthanide oxides as REO (rare earth oxides) and use REY (rare earths + yttrium) when they are talking also about Y, because it's not obvious to them that it is a rare earth.


text:
Why convert benzoate in soda to benzoic acid for spectroscopy experiment?
I did an experiment where I analyzed the amount of caffeine and benzoate in soda via spectroscopy. I've got the absorbances and did all the math for everything, but I'm having a hard time figuring out why we had to add an acid to protonate the benzoate to form benzoic acid before running the spec.
I've been able to find many similar procedures and many with various pre-lab questions, but none seem to address the reason for the conversion from benzoate to benzoic acid. Here is one such procedure: UV Spectroscopic Analysis of Caffeine & Benzoic Acid in Soft Drinks
The only thing I can think of is that it was in an aqueous solution (soda), so perhaps some benzoate will be in the form of benzoic acid and some will still be in the form of benzoate. By adding $\ce{HCl}$, we can force all the benzoate to be in the form of benzoic acid for better quantification.
Is this the correct line of thinking or am I missing something?

To help us help you, it would be great if you could provide relative amounts. Soda is already quite acidic, but if you changing the pH to 1, that's a different story.

Your assumption is correct, the addition of HCl, as already in the earlier publication by McDevitt (J. Chem. Educ., 1998, 75, 625-629 DOI: 10.1021/ed075p625) ensures benzoates are converted into benzoic acids.
The publication by Guo et al., is instructive as it lists several benzoic acids side by side, for example:

source: J. Phys. Chem. A, 2012, 116, 11870–11879 DOI: 10.1021/jp3084293
The shift of the centre of absorption may be more important than a potential change in $\varepsilon$ if you want to use the UV-Vis spectrum recorded in a principal component analysis (PCA), where overlap of absorption bands of your consitiuents in the analyte should be avoided.


text:
Enthalpy change for a van der waal gas

One mole of Van der Waal gas is compressed from $\pu {20 dm^3}$ to $\pu{10 dm^3}$ at $\pu{300 K}$. Given $a = \pu{3.60 dm^6 mol^-2}$ and $b = \pu{0.44 dm^3 mol^-1}$. Find $\Delta H$.

$\Delta H = -\mu C_p \Delta P + C_p \Delta T$
Since for VDW gas $ \displaystyle \mu C_p = {2a\over RT}- b$.
The process took place at $\pu {300 K}$ which means $\Delta T = 0$.
$\therefore  \Delta H = -\mu C_p \Delta P + C_p \Delta T = \left(-{2a \over RT} + b\right)\Delta P = \pu{0.4399 kJ/mol}$.
But the answer is $\pu {-30.5 J/mol}$
I think my mistake is taking $\Delta T = 0 $ but if that were the case what should I take as $\Delta T$, I don't even know the process.
Is the author's answer correct ?

@ChesterMiller I wonder what is wrong with what I am doing ? that will be more helpful.

You can start by telling us where you got our equation for $\Delta H$ from.

The units for a appear to be wrong for a vdw gas, should they not be $\pu{dm^6\cdot bar \cdot mol^{-2}}$ assuming molar volumes.?

@porphyrin Yes, I noticed that too.  Also, that value of a seems very low to me (if the correct units involve Pa).

@Chester Miller, yes, but if we assume numerical value is ok then 3.6 is approx a for $\ce{CO2}$ but then b is almost ten times too big for this gas. In fact it seems to be too large for most any gas in $\pu{dm^3 mol^{-1}}$ .

@ChesterMiller I used $dH = {dH\over dP}_T dP + {dH\over dT}_P dT$.

But you have to integrate, because there is a large volume change and pressure change.

If these constants are supposed to be for CO2, then the correct values are: $ a = 3.6\ bar\ dm^6\ mol^{-2}$ and $b=0.044\ dm^3\ mol^{-1}$

Using these values for the constants, I get a value of about -355 J/mol for $\Delta H$

@ChesterMiller Yes I doubt the credibility of the answer. I also tried your equation and got wrong answer.

and using the values in the question the enthalpy comes to $-301.3 \pu{J/mol}$  so it looks like a double typo in powers of $10$.

The equation for $\Delta H$ for a van der Waals gas should be:
$$\Delta H=C_p^{IG}\Delta T-2a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)+b\left(\frac{RT_2}{V_2-b}-\frac{RT_1}{V_1-b}\right)$$where $C_p^{IG}$ is the heat capacity of the gas in the Ideal Gas (IG) region (low pressures and high specific volumes).  For constant temperature ($T_1=T_2=T$), this equation reduces to:$$\Delta H=-2a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)+bRT\left(\frac{1}{V_2-b}-\frac{1}{V_1-b}\right)$$
Explanation:
In the link https://physics.stackexchange.com/questions/321968/internal-work-done-by-a-gas/322170?noredirect=1#comment725750_322170, I showed that, for a van der Waals gas, the internal energy U is given by:  $$U(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-\frac{a}{V}$$
where $U(T_{ref},\infty)$ refers to the internal energy in the ideal gas reference state $T=T_{ref}$ and $V\rightarrow \infty$ (i.e., a reference state in which the gas exhibits ideal gas behavior).
The enthalpy H of the gas is equal to U + PV:
$$H(T,V)=\left[U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-\frac{a}{V}\right]+\left[-\frac{a}{V^2}+\frac{RT}{V-b}\right]V$$
Combining the terms on the right hand side gives:
$$H(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTV}{V-b}$$
But, $$\frac{V}{V-b}=\left[1+\frac{b}{V-b}\right]$$
So, substituting into the equation for H gives:
$$H(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTb}{V-b}+R(T-T_{ref})+RT_{ref}$$
But, $$U(T_{ref},\infty)+RT_{ref}=H(T_{ref},V)$$and$$\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}+R(T-T_{ref})=\int_{T_{ref}}^{T}{C_p^{IG}(T')dT'}$$where $C_p^{IG}(T)$ is the ideal gas heat capacity at constant pressure.  So, finally, we obtain:$$H(T,V)=H(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_p^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTb}{V-b}$$

How did you get this equation ?

@A---B See my explanation added to the Answer.

Thank you. It might take me some time to digest all this.

In addition to the method detailed by @Chester Miller, I offer an alternative derivation. It is often the case in thermodynamics that there are two or more ways of solving the same problem.
To calculate the change in enthalpy at constant temperature we can start with
$$dH=TdS+VdP$$
and then find the change with volume and integrate in V at constant T.
Thus
$$\left(\frac{\partial H}{\partial V}\right )_T=T\left(\frac{\partial S}{\partial V}\right)_T+ V\left(\frac{\partial P}{\partial V}\right)_T$$
using Maxwell’s equations
$$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$
gives
$$\left(\frac{\partial H}{\partial V}\right )_T=T\left(\frac{\partial P}{\partial T}\right)_V+ V\left(\frac{\partial P}{\partial V}\right)_T$$
which nicely give  derivatives in P and V . Integrating means first working out the derivatives since
$$\int dH=\int \left[ T\left(\frac{\partial P}{\partial T}\right)_V+ V\left(\frac{\partial P}{\partial V}\right)_T\right ] dV $$
For the van der Waals gas
$$P=\frac{RT}{V-b}- \frac{a}{V^2}$$
the derivatives are
$$T\left(\frac{\partial P}{\partial T}\right)_V= \frac{RT}{V-b} $$
 and
$$V\left(\frac{\partial P}{\partial V}\right)_T = -\frac{VRT}{(V-b)^2}+\frac{2a}{V^3} $$
substituting and working out the indefinite integral gives
$$H=-\frac{2a}{V}+\frac{bRT}{V-b} + const$$
Now the constant is the standard state value, chosen, say, to be at infinite volume, but  we  do not need to know what the standard state is as we only need the difference in H.
Now take the difference to get the change in enthalpy
$$\Delta H = -2a \left (\frac{1}{V_2} -\frac{1}{V_1} \right) +RTb\left ( \frac{1}{V_2-b}  -\frac{1}{V_1-b}  \right) $$
( As an aside, as the gas expands the work done can also be calculated using $w=-\int pdV$. This is
$$w = -\int_{V_1}^{V_2} \left (\frac{RT}{V-b}-\frac{a}{V^2}\right )dv = -RT\ln \left(\frac{V_2-b}{V_1-b} \right )-a\left(\frac{1}{V_2}-\frac{1}{V_1} \right)$$
As the work done on an perfect gas is $w_{per}=-RT\ln(V_2/V_2)$ we can see that is the volume is large compared to b the work done by the van der Waals gas is less than that for  the perfect gas as a result of intermolecular interactions.)

Thanks for the answer. I prefer this derivation little more over that one.


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Reaction of concentrated nitric acid react with metallic lead?
Here is my work:
Concentrated $\ce{HNO3}$ should first decompose into $\ce{NO2}$, $\ce{H2O}$ and $\ce{[O]}$. Nascent oxygen should combine with $\ce{Pb}$ to form $\ce{PbO}$. Now $\ce{PbO}$ should neutralize acid to form $\ce{Pb(NO3)2}$.
Therefore, the net products would be lead nitrate, $\ce{NO2}$ and $\ce{H2O}$.
Is this right?

Well, you were going along the right track.... just one correction. The products will be Pb(NO₃)₂, NO₂ and H₂O. Nitric acid is a very powerful oxidizing agent. It oxidized H₂ to H₂O.

Oh! I forgot to add O. I have added it now.

Lead when reacted with nitric acid forms lead(II) nitrate, nitrogen oxide and water.
According to Wikipedia:

Lead(II) nitrate can be obtained by dissolving metallic lead in
  aqueous nitric acid:
$$\ce{Pb + 4 HNO3 → Pb(NO3)2 + 2 NO2 + 2 H2O}$$
Since the solvent is concentrated nitric acid (in which lead(II)
  nitrate has very low solubility) and the resulting solution
  contains nitrate ions, anhydrous crystals of lead(II) nitrate
  spontaneously form as a result of the common ion effect.

According to webelements:

Lead reacts slowly nitric acid, $\ce{HNO3}$. Nitrogen oxides are
  formed together with lead(II) nitrate, $\ce{Pb(NO3)2}$.

The mechanics of the reaction on Pb with concentrated HNO3 likely commences as follows:
$\ce{Pb -> Pb(II) + 2 e-}$
$\ce{HNO3 = H+ + NO3-}$
$\ce{Pb(II) + 2 NO3- = Pb(NO3)2}$
$\ce{H+ + e- = .H}$
$\ce{.H + NO3- = OH- + .NO2}$  (1997 Source)
The literature also contains citations (see, for example, Page 37) of an associated mechanism involving autocatalytic $\ce{.NO2}$:
$\ce{.NO2 +e- = NO2-}$
$\ce{H+ + NO2- = HNO2}$
$\ce{HNO2 +HNO3 = H2O +2 .NO2}$
where the last reaction, I claim, can also proceeds in the reverse direction as follows:
$\ce{.NO2 + .NO2 = N2O4}$
$\ce{N2O4 + H2O = HNO2 + HNO3}$
Another 2019 paper, further cites the radical reactions:
$\ce{.H + .NO2 = .OH + .NO}$
And:
$\ce{.OH + .NO = HONO}$
One could interpret the $\ce{.NO}$ formation reaction above to imply that in dilute solutions, more of an aqueous $\ce{.NO2}$ presence and therefrom $\ce{.NO}$. Alternately, $\ce{.NO}$ could be sourced from the decomposition of any created $\ce{HNO2}$ (including from aqueous $\ce{.NO2}$) per the reaction:
$\ce{HNO2 + H+ + e− ⇌ NO + H2O}$ $\ce{ Eo = +0.98 V}$
Reference Wikipedia


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Finding a stable conjugate base when comparing nitrogen vs. a triple bond
I am trying to understand acid-base reactions and I am trying to figure out which conjugate base is more stable.

The molecule on the top with dashed lines around the $\ce{H}$ is the original molecule and then below it I remove $\ce{H}$ and see what the resultant conjugate base looks like. I don't understand why the bottom left conjugate base is less stable than the one on the right.  Why doesn't nitrogen create a more stable conjugate base than the triple bond?

Maybe it's due to geometry? The triple bonded C has a single lone pair to donate, while the N has two. So that would allow a wider range of acceptable orientations when considering N-/H+ collisions?

Hey, just so you know, the homework tag is being phased out, so please don't tag questions with it. Thanks!

@heather thanks! I won't do that from now on.

@DanBurden could you elaborate some more?

Well I think the crux of this lies in the fact that ammonia has pKa 33 while ethyne has 25, which you can apply here. Why is it so, because I guess a sp carbon is more electronegative than a sp3 nitrogen perhaps.

@Sawarnik I see that pKa is greater for ammonia, but I guess what about sp carbon makes it more electronegative than an sp3?

Well you know that sp carbon is quite a bit more electronegative than a sp3 carbon right? Now why exactly it exceeds that of sp3 nitrogen is something i don't know, or perhaps I may be entirely wrong here.


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Why is mass gained at the cathode?
I'm dealing with a voltaic cell constructed like this:

(From “Electrochemistry”, chapter 19 from the book Principles of General Chemistry (v. 1.0).)
I understand that as oxidation occurs at the $\ce{Zn}$ electrode, the electrode loses mass as $\ce{Zn}$ dissolves into $\ce{Zn^{2+}}$ in the solution and eventually will bond with $\ce{Cl-}$ coming from the salt bridge. But on the $\ce{Cu}$ cathode side, how could it gain mass if initially the $\ce{Cu}$ electrode lost mass after the $\ce{Cu^{2+}}$ ions dissolved in the solution? I get that the $\ce{Cu^{(2+)}}$ ions are re-attracted to the $\ce{Cu}$ electrode once electrons flow from anode to cathode, but wouldn't that mean that the $\ce{Cu}$ will retain the same mass once the cell is complete?

The idea is that the copper ions are already in the solution prior to the electrodes being introduced

Please always attribute your image sources, and always check that you are allowed to use them in the first place. After that, please upload them to the stack exchange server at imgur (via the built in dialogue) so that they remain accessible and don't suffer link rot.

There is an error in the diagram for this post. The oxidation of zinc to Zn2+ is depicted as the process forming Zn2-. The reaction is correct, but the figure shows the wrong charge for the ion produced.

As you have figured out, the zinc electrode is undergoing oxidation since zinc is a stronger reducing agent than copper.
$$\ce{Zn -> Zn^2+ + 2e-} $$
Thus, the zinc electrode becomes thinner as shown in your image. It is oxidizing into soluble $\ce{Zn^2+}$ ions and thus its mass decreases.
In the right solution, $\ce{Cu^2+}$ ions are being reduced to metallic copper which deposits on the copper cathode
$$\ce{Cu^2+ + 2e^- -> Cu} $$
The $\ce{Cu^2+}$ ions are already there to begin with. The copper electrode isn't dissolving to form $\ce{Cu^2+}$ ions to begin with (the copper electrode does not react with the $\ce{Cu(NO3)2}$ solution), so there isn't any loss in mass of the copper electrode initially. Rather, during the operation of the cell, the reduction of copper(II) ions forms metallic copper, which causes the copper electrode to increase in mass and gets thicker, as seen in the image.

Hey. Do you know why the need of NO3- in the glass?

So with the cathode, there were no $\mathrm{Cu}^{2+}$ ions that initially dissolved in the solution. Rather, they were there from the beginning and once the copper metal cathode was added and the cell completed, the copper ions in the solution would then form a solid on the cathode, and thus it would gain mass.

To make a galvanic cell you need two containers, which "initially contains the $\ce{Zn^2+}$ in one container and the other contains $\ce{Cu^2+}$." After that you've dipped those electrodes so that electrons can flow through the cell through the conducting mediums. And therefore, the electrolyte can change to solid electrode and vice versa (as can be seen on $\ce{Cu^2+}$ and $\ce{Zn^2+}$, respectively).


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Find whether a gas cools or heats up on Joule-Thomson expansion

A gas that follows $P(V - nb)= nRT$ is subjected to Joule-Thomson expansion. Tell whether it cools or heats up.

$$\mu = {\partial T \over \partial P} = {\partial P(V - nb)/nR \over \partial P} = \frac1{nR}\left(V - nb + {\partial V \over \partial P}\right) = \frac1{nR}\left(V - nb - {nRT\over P^2}\right)$$
Now how do I determine whether $\mu >0 $ or $\mu < 0$ without knowing anything  about temperature or anything else ?

The equation you gave for $\mu$ is incorrect.  That partial derivative is supposed to be at constant enthalpy H.  Do you know the mathematical relationship between dH, dT, and dP?

Yes if you mean $dH = -\mu C_p dP + C_p dT$.

I set $dH = 0$ then I get $\mu dP = dT$.

Not that part.  The next part.  Do you really think that that expression you wrote  is equal to the partial derivative of T with respect to P at constant H?

There are some details in this answer http://chemistry.stackexchange.com/questions/71543/enthalpy-in-adiabatic-process/71567#71567 . You should find that the coefficient for your gas is $-B/C_p$.

The equation for dH is:  $$dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$


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When is a molecule with fluorine considered more acidic than one with chlorine?
I am learning about acids and bases.  Here is the problem I am stuck on:

According to organic chemistry as a second language the molecule to the right with fluorine is more acidic than the molecule with chlorine.  I would think that since chlorine is bigger it is more stable resulting in a more stable conjugate base?

You are thinking along the right lines, but remember that F is also more electron withdrawing than Cl. So how would that effect the stability of the conjugate base anion?

Chlorine isn't big enough to reduce basicity of conjugate base via steric hindrance.

@airhuff So would F have a greater electron density, and giving a more negative partial charge? I don't know how that impacts stability?

@Mithoron can you explain what that means? what is considered "big enough"? and I will look into what you mean by steric hindrance (haven't covered how reactions occur yet).

The conjugate base has a negative charge on the oxygen. This is what makes the conjugate base basic, and the more electron density you have at that negative charge, the more strongly it will attract a proton to stabilize itself (meaning it's more strongly basic). The fluorine molecules are strongly electron withdrawing, so they can pull some of the electron density away from the negatively charged oxygen, thus partially stabilizing it and reducing the strength with which it attracts protons, making it less basic (more stable).

That means that Cl atoms aren't big enough to block access to oxygen.


text:
Why do heavy metals like mercury and lead have an affinity for sulfur or sulfur compounds?
Many of the heavy metals such as mercury, lead, thallium, cadmium, and arsenic exert toxic effects largely by binding to or otherwise interfering with organic sulfur compounds or groups in the body. What is the chemical reason that many of these heavy metal elements and their compounds seem to bind easily to sulfur and sulfur compounds?

HSAB explains this fairly decently

Strongly related: http://chemistry.stackexchange.com/questions/275/why-do-thiols-have-such-a-propensity-for-bonding-with-mercury/277

Roughly speaking, they both have big squishy electron density, so they smudge together very easily (soft acid-base as orthocresol suggests).


text:
Why is neutralization reaction exothermic in nature?
Why is heat liberated in such reaction. Isn't it just a simple rearrangement of ions? Hydrogen ion joining hydroxyl ion.

You are right in that the reaction is $\mathrm{H}^+ + \mathrm{OH}^- \rightarrow \mathrm{H_2O}$, but recognise that bonds are formed in this reaction. Bond formation releases energy (and similarly bond breakage is endothermic), and thus neutralisation reactions are exothermic.


text:
How would you find the expected molecular orbital energy diagram for compounds such as NO?
How would you find the expected molecular orbital energy diagram for $\ce{NO}$? Would it be the expected one for $\ce{O2}$? Or the one for $\ce{N2}$? Also, would the ionization energy be bigger than that of$\ce{NO+}$?

Welcome to Chemistry StackExchange. Try looking in one of many standard textbooks on physical chemistry and on inorganic chemistry then seek out more specialist books on molecular structure. Wikipedia has a page with some diagrams you could use as a start.


text:
Why decarbonate soda for spectroscopy experiment?
I did an experiment with a nanodrop photometer where I had to find the amount of caffeine and benzoic acid in soda. During the experiment, we were to decarbonate the soda by heating it. I've done a similar experiment before where I had to add salt to decarbonate soda. I've also read similar procedures where soda was decarbonated by shaking.
My question is this: Why should the soda be decarbonated?
My thought is perhaps the air pockets in the soda (due to $\ce{CO2}$) could cause discrepancies in absorbance due to changes in the refractive index.
I also thought that perhaps $\ce{CO2}$ absorbs in the same region as caffeine and/or benzoic acid, but NIST does not appear to have UV/Visible spectrum data for $\ce{CO2}$, whereas it UV/Visible spectrum data for caffeine. Please pardon my naiveté here, but I'm then assuming $\ce{CO2}$ thus does not absorb in UV/Vis?

The primary reason is that when soda warms up, the $\ce{CO2}$ would escape slowly as bubbles. These bubbles would significantly scatter light affecting the absorption spectra. Usually, the effect of such scattering is quite pronounced and the absorption bands of individual molecules will be (might be) buried below the broadened feature in the lower wavelength region.
In colloidal and suspension systems light is scattered by individual particles. These particles are larger in size than the wavelength of light and such scattering phenomena is termed as Tyndal scattering or Tyndall effect.  For such a process, the scattering coefficient has $wavenumber^{4}$ dependence which means that for lower wavelengths the scattering increases significantly. Thus when the sample has particles then absorption spectra is skewed and we see an increasing absorption peak when going to a lower wavelength. (You can yourself try this with milk which is a suspension).
About the absorption spectrum of $\ce{CO2}$:
$\ce{CO2}$ being a simple molecule is expected to have electronic absorption bands below 300nm. It does have electronic absorption between 200 to 270 nm. It is more difficult to find on the internet since usual absorption of $\ce{CO2}$ is referred to the vibrational absorption observed in the in the infrared region. Hence, 'UV absorption of $\ce{CO2}$' is a better search term.
For your reference: T. Joutsenoja, A. D'Anna, A. D'Alessio, and M. I. Nazzaro, "Ultraviolet Absorption Spectra of Carbon Dioxide and Oxygen at Elevated Temperatures," Appl. Spectrosc. 55, 130-135 (2001)

Rayleigh scattering is defined as scattering on particles much smaller than the wavelength of the radiation. I doubt that the $\mathrm{CO_2}$ bubbles are much smaller than 270 nm. Hence, it would be inaccurate to call this Rayleigh scattering and assume a 4th power dependence. It is also misleading to describe a 4th power dependence as exponential.

I agree with the points made by you. I will edit accordingly.


text:
What does the crystal field splitting diagram for trigonal planar complexes look like?
For a trigonal planar complex (e.g. $\ce{[AgCl_3]^2-}$):
According to this image (source):

The $x^2 - y^2$ and $xy$ orbitals are on the same energy level. However, assuming that one of the $\ce{Cl}$ ligands lies on the $y$ axis (or the $x$ axis), wouldn't the $x^2 - y^2$ orbital be higher in energy? Thanks in advance.

Say you put a ligand in the trigonal complex (or an equatorial ligand in a trigonal bipyramidal complex) on the $x$-axis, where $z$ is the "axial" axis.  Then that ligand is directly aligned with an $x^2-y^2$ lobe and falls on a "seam" of the $xy$ orbital.  But the other two ligands in the plane are closer to an $xy$ lobe ($15°$ off) than to an $x^2-y^2$ lobe ($30°$ off).  These effects balance so that both in-plane $d$ orbital see the same total metal-ligand intetaction.

The most basic crystal field argument includes point-symmetric charges approaching the central metal in a way as the ligands would. Then, any orbitals that are symmetry-equivalent will end up at the same energy, and depending on how much these point towards the point-symmetric approaching charges they will be raised or lowered. Therefore, we should be looking at:

which orbitals are symmetry-equivalent, i.e. transform the same way in the point group (here: $D_\mathrm{3h}$. Warning: Link leads to a MathJax-heavy page!)
where the ligands are approaching from

Thankfully, the character tables already include the symmetry of the orbitals: look at the final columns of the table where the orbital designators are written (s orbitals are always in the totally symmetric irreducible representation, i.e. $\mathrm a_1'$ here). We see that $z^2$ is in the row $\mathrm a_1'$, $x^2-y^2$ and $xy$ are both in the row $\mathrm e'$ and $xz$ and $yz$ are both found in $\mathrm e''$. Thus, the orbitals must end up in three different groups of distinct energy:

a single energy level containing only $\mathrm{d}_{z^2}$
a twice-degenerate energy level containing $\mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$
another twice-degenerate energy level consisting of $\mathrm{d}_{xy}$ and $\mathrm{d}_{x^2-y^2}$

The first two should be immediately obvious, the third maybe not. However, it is easily explained if you remember how $\mathrm{d}_{xy}$ and $\mathrm{d}_{x^2-y^2}$ look with respect to the principal axis (the $z$ axis): they are both perpendicular to $z$ and differ only by their alignment, $\mathrm{d}_{x^2-y^2}$ having lobes pointing along the coordinate axes while $\mathrm{d}_{xy}$’s lobes point in-between the coordinate axes. A simple $C_8$ rotation ($45^\circ$) around the principal axis will transform one onto the other.
More importantly, however, one of the main elements of symmetry is $C_3$ — the rotation around the $z$ axis by $120^\circ$. Applying this will transform either of these two orbitals into something that is in-between a pure $\mathrm{d}_{xy}$ and a pure $\mathrm{d}_{x^2-y^2}$ orbital: it can be understood as a linear combination of the two. However, any operation of symmetry must transform everything onto something symmetry-equivalent and thus the two must be symmetry-equivalent and therefore degenerate.
It remains to be shown why your diagram shows the orbitals at the energy levels it does. Simply speaking, there are no ligands in $z$ direction so anything containing $z$ contribution should be stabilised. This obviously applies to $\mathrm{d}_{z^2}$ and also to $\mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$. Why is the former slightly higher in energy than the latter? The crystal field model’s answer is that $\mathrm{d}_{z^2}$ has a central ‘hoop’ that is still pointing towards the ligands in the $xy$ plane while the other two have a nodal plane in the $xy$ plane. Thus, the latter two should have a lower energy.
The highest energy obviously needs to be given to all orbitals in the $xy$ plane as that is the plane in which the ligands are. This applies to both $\mathrm{d}_{xy}$ and $\mathrm{d}_{x^2-y^2}$ as I have shown above that they are symmetry-equivalent. Therefore, these two are found at the highest energy.


text:
crystallinity of seed crystals
Does the crystallinity of seed crystals affect the final crystal structure? and is it different depending on the crystallized material?

Kind of by definition, growth of new material on a seed crystal will assume the crystal structure of the seed. Otherwise the 'seed crystal' is just a nucleation site (which may be called a 'seed' without the 'crystal'). If the 'seed' is not crystalline, then it is not a 'seed crystal'.

In my field, protein crystallography, we often use seeding to improve our crystallization experiments. Typically seeding speeds up crystallization by introducing well defined nucleation cores for crystal formation.
It can often be observed that seeding influences the crystal formation, as crystals grown from seeded crystallization conditions often have better diffraction. This is very important as all our crystals must be formed in their native solvent enviroment (water, lipids/detergents, ions etc).
Sometimes the act of seeding can influence the crystallographic symmetry, it is not uncommon for seeding to cause crystallographic symmetry to turn into non crystallographic symmetry. Personally I have had P12(1)1 crystals generate P1 crystals due to small changes in the asymmetric unit. However my case were also rather extreme as seeding cut my crystallization time from ~3-6 months to 3-6 days.
Also note that crystal seeds, does not necessarily have to be crystalline in nature. Small glass fragments, dust particles and hairs, can also induce crystal growth (I have personally seen examples of all three on a project I have worked on). This probably occurs when the protein being crystallized is highly stable, but cannot form strong inter-molecular interactions in the crystal packing. Then having a foreign surface to interact with may help keep the protein molecules close together and allow weaker indirect interactions (such as hydrogen bonds via water molecules) to form and stabilize the crystal structure.
So in my field, seeding can have an enormous impact on crystallization experiments.


text:
Calculating a theoretical degree of polymerisation of ethylene oxide
This is the problem: Researchers add 88 grams of ethylene oxide. This reacts fully to 92 grams of poly ethylene oxide. Calculate the degree of polymerisation.

I thought: molecular weight of ethylene oxide is 44 grams. So you have 2 moles in 88 grams and when this polymerizes, you'll get 2 moles of monomers and thus n=2. However, the answer states that n=9.
This is the answer included in the test:

Calculating moles ethylene oxide: 88/44, - 1 point
Concluding that 4 g H20 reacted , - 1 point
Amount of moles formed of poly ethylene oxide (= amount of moles H2O reacted) =   4g H2O / 18, - 1 point
average n = amount of moles ethylene oxide that reacted / amount of moles of poly ethylene oxide formed - 1 point
Concluding that n=9, - 1 point

Thanks alot!

Hi orthocresol. U edited my thing but shouldn´t one of the O´s be outside the monomers... like it was before. Now the molecular weight of the monomer is 16 higher.


text:
What is this blue crystal?
I was sorting out some old boxes in one of the rooms in my house, when I found a small container with about 10 grams of this blue crystalline powder. I assume that it contains some form of copper due to the colour, but the label's all ripped and I can't see the name of it.
So what compound is it?

It appears to glow in the image, but this is just the reflection from the camera flash and it doesn't look like this in real life.

Seems like a classic case of $\ce{CuSO4.5H2O}$, you may want to do a Google image search and compare your sample to it (since you have the actual thing you're in a better position to judge whether it looks the same)

You could try warming it. If it's what @orthocresol says it is, it should turn white when heated (i.e. when it loses water and becomes anhydrous). I did this for my CHEM 101 lab last semester :)

@Gallifreyan I tried this and it did in fact turn white

I'm voting to close this question as off-topic because not enough information has been included in the question at the moment to deduce a definitive answer; simply a picture with no other information isn't helpful. I reckon there are a lot of blue-colored crystalline compounds around.

@M.A.R. What other information do you expect me to provide? An answer has already been given, which I managed to test by heating the compound. And whilst there may be other blue crystalline compounds out there, this one is quite clearly the most common.

@George FWIW, this already has "too broad" close votes. I agreed with the close votes but didn't want not to leave a comment, closing a well received question.

Copper(II) nitrate is also a good guess.

@Zhe Would Copper (II) Nitrate experience the same colour change when heated?

I'm having trouble finding a reference to the color of the anhydrous compound. I do see mentions of it subliming, so that might help you determine which copper(II) salt it is.

As orthocresol also mentioned, it's probably copper sulfate. As you said, the blue color essentially gives it away.

(picture taken from Science Madness Wiki)

It's Copper Sulfate, more commonly known as root killer. It's also used in pools to kill algae. In the home, you flush it very slowly down your toilet drain to kill any tree or bush roots growing in the sewage pipes. It's fairly corrosive, so use it sparingly.


text:
Synthesis of a tungsten carbene bromide from [CpW(CO)3]-
I'm currently working through past test and exam papers in order to prepare for my organometallic test at the end of the month, but unfortunately I've gotten stuck on the following question.
Since I don't know how to build molecules on here, I've attached a screenshot of the question.

The biggest problem I have is getting the Br coordinated to the metal while I already have my alkyl $\ce{(CH2CH2CH2Br)}$ bonded to one of the carbonyls. I don't know what reagent to use to get it on there. One idea I have is using $\ce{NaBH4}$ to get a hydride onto Tungsten, then I use R-Br (where R is some alkyl) to essentially swap the H with the Br and form as a by-product, R-H, but I'm not sure if this would work. Below is my complete synthesis.

EDIT: I spoke to my professor and it was unnecessary for me to remove the $\ce{Br-}$ in one of the earlier steps. That $\ce{Br-}$ would've attacked the metal centre later on and the rest of the steps would be elementary.


text:
Function of water and the need of two barrels in a biogas Generator
For my chemistry assignment, I have to build a biogas generator. I found the following diagram,
This is the website that the image is from: http://www.motherearthnews.com/renewable-energy/other-renewables/biogas-generator-zm0z14aszrob
(Manure goes through the feeding tube, then anaerobic bacteria produces methane. The methane then goes to the second tub and is collected)
On the right side of the diagram,

why does the barrel need to be immersed in water? Does it do something to the methane?
why does the methane not immediately go from the first tank through the tube out to the burner but instead goes into the second tank?
Once the methane is inside the second tank, what's stopping it from rising and going out?

Could you provide some additional information. For example, nothing in the diagram is clear about the reaction being done to form the methane. Also, where did this image come from? If you could link to the site, that may help someone else help you interpret it.

Nothing mysterious, just anaerobic digestion https://en.wikipedia.org/wiki/Anaerobic_digestion // The anaerobic digestion creates methane which builds up pressure in first container and then flows into the second container. The rise of the second container is just an indicator that you've actually made some methane.

The reasons are as follows:

The inverted barrel must be full of water. Then, when methane is formed, it will displace the water and the barrel will start floating. Methane does not do anything with water. It is almost insoluble (solubility is around 40 mg/L) and therefore does this setup work.
The second tank is storage. Without storage you have to burn the methane when it is formed. With the second tank you can use it when you need it and you can a higher flow rate.
If to much methane is produced and not consumed at the burner, it will come out of the second tank. But you can control that by stopping the feed in the first tank.


text:
Why do enthalpy changes of combustion and formation not take into account oxygen?
I've seen many enthalpy changes in my studies and I don't understand why the oxygens on either side of reactions are not calculated (especially if the unreacted oxygens don't balance). For example:
$$\ce{2C(s, graphite) + 3H2(g) + 1/2O2(g) ->  C2H5OH(l)}$$

It would be much easier to discuss/answer the question if it contained a specific example.

@airhuff 2C (s, graphite) + 3H2 (g) + 1/2O2 (g) --> C2H5OH (l).

By convention elements in their most stable polymorph at standard conditions have at standard conditions (298.15 K) a standard enthalpies of formation $\Delta{}_fH^\circ$ of zero (one reference).  This is why thermodynamics discerns C(graphite) from C(diamond), and C(fullerene), for example.  Hence on first view it looks like zeros in your example (graphite, Hydrogen, Oxygen) got forgotten; but they are not.


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Thermal decomposition results: repeatable in ordinary temperatures?
In thermal decomposition studies where very high temperatures are used (see below for four reactions and their respective temperatures, coming from one paper), can I directly relate these decomposition processes to the ones that happen in room temperature?
Is it OK to assume those same reactions will take place even if those temperatures are not reached (instead taking more time to happen or with the requirement of other variables to compensate for the absence of the same heat, but still existing the same mechanism)?

117–170° C: $\ce{Cu(CH3COO)2·H2O → Cu(CH3COO)2}$
170–309° C: $\ce{Cu(CH3COO)2 → Cu2O}$
309–479° C: $\ce{Cu2O → CuO}$
946–1016° C: $\ce{CuO → Cu2O}$

Obs.: these reactions are neither complete nor balanced.


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Would an aqueous solution of Zn(2+) develop a positive charge if solid zinc was dipped into it?
Say you have a beaker containing $\pu{1M}$ solution of $\ce{ZnSO4}$. The solution will be electrically neutral because there will be an equal number of cations and anions. If you then put a zinc electrode into the solution, will some of the solid zinc reduce to give the solution a slightly positive net charge(because of the excess zinc ions)? Or will there be an equilibrium between the oxidation and reduction of zinc that would keep the solution electrically neutral?
Will the answer be any different in the case of copper-zinc galvanic cell with no salt-bridge? Will the electrolytic solution in the zinc half-cell of a copper-zinc galvanic cell with no salt-bridge develop a positive charge? The scenario is the same as the one in the previous passage except now the zinc electrode will be connected to a copper electrode.
For some context, I was trying to understand why the electrolytic solution in a galvanic cell needs to be electrically neutral in order for the electrodes to develop a non-zero potential difference.


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Why is NH3 a stronger base than PH3?
I know that the orbitals of nitrogen overlap better with the 1s orbitals of hydrogen, but I don't know how to use this information to say why ammonia is stronger than $\ce{PH3}$.

From the Wikipedia article for phosphine:

The low dipole moment and almost orthogonal bond angles lead to the
  conclusion that in PH3 the P-H bonds are almost entirely pσ(P) – sσ(H)
  and phosphorus 3s orbital contributes little to the bonding between
  phosphorus and hydrogen in this molecule. For this reason, the lone
  pair on phosphorus may be regarded as predominantly formed by the 3s
  orbital of phosphorus. ... This electronic structure leads to a lack
  of nucleophilicity and an ability to form only weak hydrogen bonds.

And, one might assume, likely also explains the lowered basicity of phosphine relative to ammonia.
Ammonia is classically $sp^3$-hybridized, which entails that the lone pair resides in an an $sp^3$-orbital. Based on empirical evidence (some of which is mentioned above), phosphine is less $sp^3$-hybridized, with the lone pair residing in an orbital that is more like an $s$-orbital than an $sp^3$-orbital. Thus phosphine's orbital has more s-character, which means the electron pair will be held closer to the nucleus and be less inclined to bond. This is an argument rationalizing phosphine's lower basicity than ammonia.
Countering this argument, one might argue that phosphine's lone pair orbital is a $3s$, which is inherently bigger than the $2s$-orbital ammonia uses for hybridization. So "after"* hybridization, ammonia's $sp^3$-orbital is certainly bigger than its original $2s$ orbital ($2s$ has more s-character than $sp^3$), however, it may not actually be bigger than phosphine's $3s$. Thus, perhaps ammonia's hybridized $sp^3$-orbital is not so different in size than phosphine's $3s$-orbital, I suppose we would need a quantum computation to know for sure.
At the end of the day, we are only trying to rationalize an observed phenomenon after the fact. In spite all of our human ideas of orbitals, hybridization, etc, nature does as it pleases. Perhaps our models can offer a correct explanation and someone can interpret it better than I did (which shouldn't be too difficult), but also for some phenomena we must realize that our models don't fit perfectly and there are exceptions to every trend.
*Quotes used because ammonia is inherently hybridized, the act of nitrogen's $\{s, p_x, p_y, p_z\}$ atomic orbitals hybridizing when the molecule forms is a human idea that helps explain empirical observation, but may not necessarily take place physically. Indeed, the very $\{s, p_x, p_y, p_z\}$ orbitals themselves are only exact solutions for the hydrogen atom (as treated by quantum mechanics), and are only approximations of the orbitals of all other elements. But let's not get bogged down in all that.


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Why do conjugated dienes undergo Diels-Alder reactions?
If conjugated dienes are so stable, why can they still undergo Diels-Alder reactions ([4+2] cycloaddition)?

Hint: Stable does not mean unreactive.

Their stability is to be balanced against the conditions they are exposed to.  And there are other polyenes even more stable (like benzene) than dienes. In addition, provided enough activation provided (some Diels-Alder reactions already occur at room temperature, for example cyclopentadiene reacting with itself), do not forget to compare thermodynamics, like $\Delta{}_fH$ of starting material(s), on one, and the of the product(s) on the other side of the equation.


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What makes enantiomers so special?
We study enantiomers in detail. But what makes them more important to study than, say, diastereomers? What I'm trying to ask is, enantiomers are simply a type of stereoisomers that can rotate plane polarized light in opposite directions by the same amount. What makes them more important than other stereoisomers?

Enantionmers look the same in many characterization methods, but can have vastly (and sometimes, as in the Contergan-Scandal) devastating differently chemical properties. So its important to study all enantiomers if you have stereo-centers in your molecule.

What makes enantiomers so special? Nothing, except that we are made of them.

@Fl.pf. Better known as thalidomide outside Germany.

What do you mean by "special" or "more important" - both of this is unclear/opinion based, even if many other users seem to not recognise it.

The reason (strict mirror image) enantiomers have more attention paid to them than diasteriomers (at least in Organic Chemistry classes) is because they share the same reactivity toward achiral reagents.
To achiral compounds, both of the enantiomers look the same. There is no way for the achiral substance to determine the difference between the two. If you take a reaction with an enantiomer and flip it through a mirror, there's no chemical difference between the two.
In contrast, there are a number of effects which cause diastereomers to behave differently. (Basically, the chiral centers in one part of the molecule interact with the chiral centers at the other part, and because those pairs are different in two different diastereomers, the reactivity is different.) The upshot is that two diastereomers have as much difference in behavior as compounds which don't share molecular formula. (e.g. the difference in behavior of an ethyl versus methyl variant.)
This has very real implications for practicing organic chemists. If you start with achiral starting material and produce a chiral compound, there is no way (with strictly achiral materials) to produce anything but a racemic mixture of two enantiomers. You can futz with conditions to get specific diasereomers, but not specific enantiomers. In order to get an "enantiomeric excess" of one, you either need to use chiral reactants or a chiral catalyst.
Likewise with separation. If all you have is an achiral chromatography column (e.g. silica) and simple (achiral) solvents, you're not going to be able to separate two enantiomers. They're always going to migrate the same. Again, futzing with conditions can get an achiral chromatography column to resolve diasteromers, but not enantiomers. To resolve enantiomers, you need a chiral column (or less likely, a chiral solvent).
Combined with the importance of particular stereoisomers in biological systems, that's why the emphasis on enantiomers in Organic Chemistry classes. Historically, it was a big sticking point in the development of organic synthesis to a) figure out that enantiomers were a thing, and b) work out how to deal with enantiomers. Even today there's tons of research into chiral control. However, modern practicing organic chemists tend to work on larger and more complex systems, with many stereocenters so the enantiomer/diasteromer divide isn't as important, as often they're working with stereocenters "in isolation".

Enantiomers have different physical arrangement of atoms. This difference plays a crucial role in biology.

But, why do enantiomers have different biological activities?
Recognition of chiral drugs by specific drug receptors is explained by
  a three-point interaction of the drug with the receptor site, as
  proposed by Easson and Stedman. The difference between the interaction
  of the two enantiomers of a chiral drug with its receptor is
  illustrated below.

Source: https://www.khanacademy.org/test-prep/mcat/chemical-processes/stereochemistry/a/chiral-drugs

You've answered why the study of stereoisomers is important. But I want to know why we have we given special names to those stereoisomers that are superimposable on their mirror image? Why is knowing whether a molecule is chiral or not important?

@xasthor Why we give them special names?  Because (R)- and (S)-enantiomer are not identical structures and if they are exposed into an environment that itself is chiral, their interaction with said environment differs.  Say you have gloves, that happen to be chiral, too.  Once in contact with your chiral hands *), their fit differs if you have them on the left, or right hand, right? *) I know, it is a pleonasm.

@Buttonwood But that's not unique to chiral molecules. Any stereoisomer, when exposed to a particular environment will react differently. So why do we give the ones that are non super imposable mirror images of each other special names?

@Xasthor So your question addresses "why some of the stereoisomers are called enantiomers, other diastereomers; why are some called meso-compounds"? If so, then the different classification is highlight what they share, and what they do not share.  Aunt, mother, sister, daughter are all females; and at the same time, orthogonal to this, there is the concept of grandparents, parents, siblings, children.  Yet not every child is female.

The main reason enantiomers are emphasized more comes down to the interesting properties they have, particularly in biological systems.
A pair of enantiomers exhibit identical physical properties (e.g. melting point, dipole moment, etc) except for their optical activity (as you said, they rotate plane-polarized light with the same magnitude, but in the opposite directions). Chemically, however, a pair of enantiomers can behave very differently. This is particularly seen in biological systems, which are often adapted to work properly with only a particular enantiomer of a pair. For example, almost all naturally occurring sugars are the D-enantiomer and almost all naturally occurring amino acids are the L-enantiomer.
This has major implications in medicine; often times only one of the enantiomers is effective and the other can be inert or even harmful. The comments on your question mention thalidomide, a drug that was sold as a sedative and a treatment for morning sickness until it was realized that while one of the enantiomers had the desired effect, the other caused birth defects. Making enantiopure mixtures is necessary to ensure you are truly generating the effect you want (in the case of thalidomide, this still fails because the compound naturally racemizes).
For diastereomers, there is a greater expectation that they would behave differently due to their different physical properties. Hence, a lack of emphasis on diastereomers (which I'm not certain there is a lack of emphasis, at least from the organic chem course I took) may be due to the fact that the relationship between diastereomers is less direct than that between enantiomers and so it is tougher to draw conclusions about what properties they will share.


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Hoffmann's rule in ester pyrolysis
Why is the Hoffmann's product the major product in the pyrolysis of an ester to form a carboxylic acid and alkene?


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If α-glucose is less stable than β-glucose, why is glucose almost always α in its compounds?
Maltose, saccharose, turanose, trehalose, trehalulose, amylopectin, amylose - they are all α. Why? Also, why is cellulose β?

I think it has to do with configurations and conformations of sugars and biological needs.
A sugar with a six-membered ring is known as a pyranose, the simplest compound containing such a ring. Similarly, sugars with five-membered rings are designated furanoses.
The cyclization of a monosaccharide renders the former carbonyl carbon asymmetric. The resulting pair of diastereomers are known as anomers. In the α anomer, the OH substituent to the anomeric carbon is on the opposite side of the sugar ring from the CH2OH group at the chiral center that designates the D or L configuration (C5 in hexoses).

The other anomer is known as the β form. Sugars accurately described
  as cyclic polyhydroxy hemiacetals or hemiketals, because the linear
  form exists in very minute quantities (although β forms comprises
  larger percentages in mutarotation mixtures and under physiological
  conditions)

Hexoses and pentoses may each assume pyranose or furanose forms. The equilibrium composition of a particular monosaccharide depends somewhat on conditions but mostly on the identity of the monosaccharide.
The pyranose ring, like the cyclohexane ring, may assume a boat or a chair conformation.

The relative stabilities of these various conformations depend on the stereochemical interactions between the substituents on the ring. The boat conformer crowds the substituents on its “bow” and “stern” and eclipses those along its sides, so that in cyclohexane it is ~25 kJ /mol less stable than the chair conformer. The ring substituents on the chair conformer fall into two geometrical classes: the rather close-fitting axial groups that extend parallel to the ring’s threefold rotational axis and the staggered, and therefore minimally encumbered, equatorial groups.

Since the axial and equatorial groups on a cyclohexane ring are conformationally interconvertible, a given ring has two alternative chair forms; the one that predominates usually has the lesser crowding among its axial substituents.

The conformational situation of a group directly affects its chemical
  reactivity. For example, equatorial OH groups on pyranoses esterify
  more readily than do axial OH groups. Note that β -D-glucose is the
  only D-aldohexose that can simultaneously have all five non-H
  substituents in the equatorial position. Perhaps this is why glucose
  is the most abundant naturally occurring monosaccharide.

Glycogen and starch ingested in the diet are hydrolysed by α -amylases and glycosidases, enzymes in saliva and the intestine that break (α1n4) glycosidic bonds between glucose units. Most animals cannot utilise cellulose as a fuel source, because they lack an enzyme to hydrolyse the (β1n4) linkages.
In cellulose the glucose residues have the β configuration, whereas in amylose the glucose is in the α configuration. The glucose residues in cellulose are linked by (β1n4) glycosidic bonds, in contrast to the (α1n4) bonds of amylose. This difference gives cellulose and amylose very different structures and physical properties.
The most stable three-dimensional structure for the (α 1n4)-linked chains of starch and glycogen is a tightly coiled helix, stabilized by interchain hydrogen. For cellulose, the most stable conformation is that in which each chair is turned 180° relative to its neighbours, yielding a straight, extended chain. (In contrast, the average plane of each residue along the amylose chain forms a 60° angle with the average plane of the preceding residue, so the helical structure has six residues per turn). All ⎯OH groups are available for hydrogen bonding with neighbouring chains. With several chains lying side by side, a stabilizing network of interchain and intrachain hydrogen bonds produces straight, stable supramolecular fibers of great tensile strength.
Steric factors and hydrogen Bonding influence homopolysaccharide folding
Polysaccharide molecules can be represented as a series of rigid pyranose rings connected by an oxygen atom bridging two carbon atoms (the glycosidic bond).

There is, in principle, free rotation about both C-O bonds linking the
  residues however, rotation about each bond is limited by steric
  hindrance by substituents. The three dimensional structures of these
  molecules can be described in terms of the dihedral angles, ϕ  and Ψ,
  about the glycosidic bond:

The bulkiness of the pyranose ring and its substituents, and electronic effects at the anomeric carbon, place constraints on the angles ϕ and Ψ; thus certain conformations are much more stable than others.
Summarising this, the abundance of the forms of sugars has a lot do with factors affecting conformations such as boat and chair – actually this the reason why , equatorial OH groups on pyranoses readily react compared to axial ones, (sterics, rotation about the C-O bond – this is limited in cellulose as it forms almost linear chains and a strong network of H bonds between adjacent molecules limiting flexibility), and other biological factors such as availability of enzymes to break glycosidic bonds connecting the molecules (as noted animals don’t have the ability to break (β1,4) glycosidic bonds).
Hope this helps
References
Lehninger Principles of Biochemistry
Biochemistry Voet and Voet

An enzyme can use any conformer of monomer it wants as a building block for polymer formation, so there is no violation of a chemical law there.  But from a physiological point of view, why build starch and glycogen with alpha linkages?  The reason is that the polymer forms an open structure, which enzymes such as amylase or phosphorylase can easily attack.  This is exactly the kind of structure you want for a fuel storage polymer.  Also, glycogen and starch granules are more or less water soluble and not damaging to cells.
Formation of beta-linkages creates fibers which are damaging to cells, and if they bind to one another are hard to metabolize.  Cellulose is the perfect example. Fibers are made for structural reasons, not fuel storage.


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Free chemical reactions database
I'm creating a program, which predicts reactions of compound. So, I need a database of chemical reactions. And I need both inorganic and organic reactions. But I can't use Scifinder or Reaxys because they are not free. Is there any free analogs?

Related: http://chemistry.stackexchange.com/q/46725/4945 http://chemistry.stackexchange.com/q/22285/4945 http://chemistry.stackexchange.com/q/55937/4945 http://chemistry.stackexchange.com/q/4040/4945

In the current form the question is addressed, and assuming you took a look at a listing of software already in the field like here, I speculate the sheer number of organic reactions (just some) alone renders such a task as a one-man / one-woman project too large.  There is a lot of work associated to create such a database, or software. Organic Synthesis is one open-access reference for organic reactions, yet the normal output is a html / pdf of the reaction procedure, and not a *.sdf (or other chemically relevant *.xml-like) format.  Inorganic Synthesis is a publication with similar intent as the former, yet not openly accessible.
Nevertheless, if you are interested in this topic, I suggest to consult the recent publications Computer-aided synthesis design: 40 years on (DOI 10.1002/wcms.61) as well as the just about a year old Computer-Assisted Synthetic Planning: The End of the Beginning in Angewandte Chemie (English version DOI: 10.1002/anie.201506101).

Thank you a lot for links: they are very helpful. But as I understand, in Organic Syntesis procedures of reactions are described. Are there any database where only reactions are presented, e.g. "CH4+2O2->CO2+2H2O"?

Beside "the two" (referring Chemical Abstracts by ACS, and Reaxys/Beilstein by Elsevier; both behind a paywall) used day-in day-out, I assume Martin's comment (+1 is not enough) is really a treasure chest worth to explore more in detail.  RSC's Methods in Organic Synthesis (http://pubs.rsc.org/lus/methods-in-organic-synthesis) still allows manual download of single entries as *.cdx which you might convert with openbabel and eventually process further.  As back "old days" Isisbase (MDL) was a program to maintain locally a literature collection allowing to search by chemical structure.

I speculate, facing a new substrate,  in the end such a software "simply" looks for pattern "typical" for reactions already trained / encoded, similar to E. J. Corey's red book "The logic of chemical synthesis".  By virtue of computational power, more rapidly; perhaps even with suggestions no chemist (yet) attempted.

I'm sorry, I don't understand you - English is not my native language.

At what point?  I suggest http://chat.stackexchange.com/rooms/56490/chemical-reactions-database

I'm sorry, I can't answer you in chat: I haven't enough reputation. But I can read.


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Collagen: similarity of human to fish and meat amino acids
I am looking for a collagen supplement for my partner who has joint pain, bone pain, fatigue associated with ligaments, flaccid skin problems, and is just over 40. As she does not eat fish or meat, and also does not eat much at all, hence not enough to absorb the required amino acids required in the synthesis of collagen, I am considering a collagen supplement. The collagen should break down in the body, thus generating the twenty amino acids that make part of the three collagen stands, collagen making up the ligaments, and it should take three months to start generating the extracellular tissue of I understand correctly.
So, I would like to know, how similar are the twenty amino acids that make up the tendons in the human body to the ones found in fish and meat, and what needs to be eaten with these supplements, fish, or meat, for the amino acids contained in collagen to work their way through the body's cells and be reformulated into ligaments outside the cells. I assume supplements such as colvita won't be enough and I want to know what to take with the supplements and food.
Can someone name the names of the twenty amino acids found in human ligaments?
If this post is inappropriate here, then please post to biology stack exchange.
Thanks.

I'm voting to close this question as off-topic because you can easily Google the answer.

I just did a quick search on Wikipedia and found this hope it helps you out .
The 9 essential amino acids that cannot be made by the body and must come from food are phenylalanine . valine . threonine , tryptophan , methionine , leucine , isoleucine , lysine and histadine.
There are 6 other amino acids that are considered conditionally essential in the human diet which means there synthesis may be limited under special pathophysiologoical conditions they are arginine , cysteine , glycine , glutamine , proline , tyrosine.
There are 5 amino acids that are dispensable by humans meaning they can be synthesized by the body, they are alanine , aspartic acid , asparagine , glutamic acid , and serine.
Ref:Wikipedia/9essentialaminoacids

Thanks. So is this the complete lusty of amino acids that are synthesized by the body?

There are 11 non essential amino acids, these are synthesized by the body.


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Suspension or Solution
Tomorrow I have a science test, and in it there are some mixtures, and we have to identify whether they are a solution, a suspension or a colloid. Is there an easy way we can identify this WITHOUT getting the mixtures up and seeing whether they sediment or not?
Note: They are everyday mixtures, like lattes, freshly squeezed orange juice etc.

If the solute particles are very small ($\ce{< 1nm}$), then they form a true solution.
If the solute particles are very large ($\ce{1nm < size < 1000nm}$), then they form a collidal solution.
If the solute particles are very large ($\ce{ > 1000nm}$), then they form a suspension.
The numbers are not exact; they are approximate border ranges. A particle with size $\ce{1100nm}$ can form a colloidal solution. The questions given in your test will give the samples whose size lie in the middle range. Otherwise, it is quite difficult to determine what type of solution they will form; it has to be determined experimentally.

If the analyte happens to be in a transparent container, shine light on it, see if the light is scattered (Tyndall effect) -- if so, you have either suspension, or colloidal matter.  Instead of a laser pointer, a good pocket light (like the ones with halogen bulb halogen or LEDs) may be used as light source, too.

(source)

It's a test. We cannot come to class with bottles of mixtures

If you point an intense light source to the surface of the liquid (either solution, colloïdal mixture) you may see if the light beam passes across the medium "without a trace", or if you observe a milky / foggy scattering that resembles more or less an illuminated rod (like above left hand example) if seen from the side.  The picture above is more visual in comparision, of course, but even for the bottle of $\ce{CoCl2}$ soluted in water you see a deposit.  If the test is tomorrow, familiarize yourself with milk, as one example of colloids. The "fog" is visible even without lifting the jar.


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What can cause the biuret test to produce an orange colored solution?
What can cause the biuret test to produce an orange colored solution?
I did the biuret test to a few number of foods in a school lab assignment. The test with tofu turned orange. I wonder why, since the biuret test is supposed to give violet as positive. I tried looking it up on google with no success.
An explanation I came up with is this: during my search, I found out the composition of the biuret and Fehling reagent is quite similar (Fehling reagent and biuret test). Maybe the orange result has something to do with this similarity, as the Fehling test gives orange as the positive result. But the thing is, the Fehling test I did on the tofu showed a negative result. At first, I suspect I took the test tube with Fehling for the biuret. But this is not the case since my classmates got the same orange result with the biuret.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

I'm also having the same problem with Biuret Test in this case is the Lowry Assay which get the same principe as Biuret Test. My hypothesis comes from my stability testing. By 2 weeks of stability testing, the result giving me orange colours which i think it is either the hyaluronic acid that is contained in the sample interfere the result or the protein is degraded and the negative result shows orange colour


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Can a product be alcohol free while containing isopropyl myristate or isopropyl palmitate?
I always find isopropyl myristate and isopropyl palmitate in cosmetics, and I want to know if they have alcohol in them or not. I read that they are composed of alcohol and acid, but they are considered to be conditioning agents, so does that mean that there is no alcohol left in them? So can a finished product be alcohol free while containing isopropyl myristate or isopropyl palmitate?

If you don't know anything about chemistry, then the first question I think we need to ask is: are you talking about a chemist's definition of "alcohol" or are you taking about ethyl alcohol/ethanol (the one that constitutes alcoholic drinks)?

Isopropyl myristate and palmitate are esters formed from isopropyl alcohol and myristic and palmitic acids. These products will have been purified before being used in the cosmetics to remove any residual isopropyl alcohol. If the labelling does not mention isopropyl alcohol as a component then it will not be present.
Note. I don't know your reason for wanting to know about the presence of isopropyl alcohol but if you are concerned for ethical reasons about the presence of alcohol then you should be aware that what is generally referred to as 'alcohol' is ethanol. Isopropyl alcohol is a different substance.

I've always been curious, is there an equivalent term to ethanol for the non-drinkable alcohols? In other words, do any other groupings of alcohol exist?

@Hashim Currently, alcohol is defined as an organic compound containing the functional group -OH. At one time, ethanol was probably the only known alcohol, so it became synonymous with alcohol.