text:
Rotational degrees of freedom (3N-5 and 3N-6)
In spectroscopy we described the electric energy with the approximative separability of internal motions as:
\begin{equation}
E=E_e+E_v+E_r+E_{ns}
\end{equation}
(energies: electronic, vibratory, rotatory; nuclear spin (neclected))
With the Born Oppenheimer approximation of nuclei and electrons you get a formula which describes the degrees of freedoms for vibration motions:

3N-5 for linear molecules
3N-6 for non-linear molecules

This formula can be understood by:

the coordinates of 3N atoms.
3 degrees of motions for the three translation of the centre-of-mass motion
2 (for linear) respective 3 (for non-linear) degrees of freedoms reduced for the rotation of the whole molecule

I guess trans buta-1,3-diene is here described as a linear molecule. [added: in the rigid rotor approximation]
First I was thinking why we are so much interested on the number of freedoms for vibrations. We consider only the rotation of the whole molecule but not around bonds.
Why do we not consider rotation about a bond axis, e.g. for pentane a rotation of an ethyle around the propyle group? (rotation about the C(2)-C(3) axis)
May be because it's much another energy range as rotational energy of about 1-100 cm-1 (typical period of 0.1-10 ps).

Rotations around bonds are typically termed "internal rotations", and represent one of the most common problematic cases for the rigid-rotor-harmonic-oscillator (RRHO) model of internal molecular motion.  This is because RRHO assumes that any vibrational amplitudes are "small," and internal rotations are most definitely not small!  Such rotations still involve the internal degrees of freedom of the molecule, though, and thus (as Jan notes) are considered as part of the 'vibration' of the molecule, not its 'rotation.'
Internal rotations typically have some energy cost involved (as in your pentane example), and so cannot be treated as "free" rotations. They are usually termed "hindered rotations," and there is a great deal of literature studying them.  Some citations I know offhand:

Classic Pitzer (J Chem Phys 5 469, 1937), and Pitzer & Gwinn (J Chem Phys 10 428, 1942)
Independent-rotations approximation for treating hindered internal rotation: Pfaendtner (Theor Chem Acc 118 881, 2007)
Other / more-advanced treatments of internal rotation:

Hecht (J Chem Phys 26 31, 1957)
Strekalov (Chem Phys 355 62, 2009)
Vansteenkiste (J Chem Phys 124 044314, 2006)
McClurg (J Chem Phys 106 6675, 1997)

One species of particular interest to me on this topic has been nitromethane, which has an extraordinarily small barrier to rotation around the $\ce{C}-\ce{N}$ bond (see Strekalov, above), such that at ambient temperature it can be considered essentially a free rotation.

trans-buta-1,3-diene is not a linear molecule.  To be linear all atoms must be "in-line", e.g. $\ce{HCCH}$ is linear but $\ce{H2CCH2}$ is not.
The rotation around a bond axis is not free and is therefore described as a vibration. See for example this blog post.  So, for example, for ethane the lowest of the $3N-6$ vibrational modes corresponds to a torsional motion about the $\ce{C-C}$ single bond.  The mental picture here is not 2 methyl groups that rotate freely relative to one another, but rather a rocking back and forth motion in one minimum and then an occasional rotation in to the next minimum.  The vibrationa motion is on the femto second time scale and every few nano-seconds the switch from one minimum to the next will occur. So the "rotation" will happen a thousand times per millisecond and look like free rotation.

I now see that trans-buta-1,3-diene belongs to point group $\ce{C_{2h}}$ and should therefore be an asymmetric top ($\ce{I_{A}<I_{B}<I_{C}}$) and not be considered as linear molecule ($\ce{I_{A}<I_{B}=I_{C}}$) of point groups C_{∞v} or D_{∞h}.

If I'm right the energy difference betweeen anti and staggerend around a C-C single bond is about 5 kcal/mol (converted: 1750 $cm^{-1}$). If there is not as much energy available, I see the "rotation" is like a vibration. I thought a vibrational motion is typically between 300 and 3000 $cm^{-1}$. What is meant with "not free"? Are translational, rotational and vibrational motion considered as "free" motions?

The energy of the motion that describes rotational motion about a CC single bond is included in the vibrational energy ($E_v$).  The frequency of this motion is typically < 500 cm$^-1$, so there is usually several bound levels.  So the expression for $E_v$ for C-C bond rotation looks just that for, say, a C-C stretch.

I'm a little bit surprized that my value of 1750 $\ce{cm^{-1}}$ (eclipsed activation barrier for butane around C(2)-C(3) bond) is different from your value of <500 $\ce{cm^{-1}}$. I don't know if the total "stretching vibration"-like rotation is defined by an average value of the different conformation activation energies or by the maximal value (5kcal/mol). Secondly, is the anharmonic oscillar $\omega_e(v+1/2)$-$\omega_e$ $x_e(v+1/2)^2$+.... still a good approximation to describe this complete (360°) rotation around the single bond C(2)-C(3) of pentane (as described in my initial post) ?

Considering internal and external motion of a molecule, I'd like to ask you something else: If we have an isolated pentane molecule at RT the Boltzmann energy is approx. 0.59 kcal/mol = 207  $\ce{cm^{-1}}$. There is not enough energy for electronic or vibrational excitation, so the molecule is in its electronic and vibratory ground state. However usually 10-20 rotational states are populated at room temperature. Can we attribute mathematically this 207 $\ce{cm^{-1}}$ energy to x % translational energy ($\ce{1/2mv^2}$) and to y % rotational energy?

You should ask this as a separate question.

The two/three questions from Jun 15 as well?

I'll have a look at them tomorrow, gottta run

Considering my newest question, I've started a new question here: http://chemistry.stackexchange.com/questions/33104/how-to-distribute-the-boltzmann-energy-to-internal-and-external-motion

1: the 5 kcal/mol is the barrier height on the potential energy surface (PES). The < 500 cm$^{-1}$ is the spacing between vibrational energy levels.  Two different things.  2: Given enough terms you can describe the complete PES using an anharmonic potential.  In practice only the first few terms are used and this only describes the PES around the minimum and cannot account for complete rotation

If this $\ce{<500 cm^{-1}}$ comes really from the "rotational-like" motion I expect it to be may be turning around +- 60° (avoiding eclipsing interactions), otherwise if the rotation happens around 360° I would expect (as I wrote) also a band at around $\ce{1750 cm^{-1}}$. A small 60° rotation in one rotation really looks like a vibration if there is not as much energy to overgo the transition state of the eclipsing interaction...


text:
CIF fields that are available for some atoms, but not others
I've been working on another CIF file, this time for (-)-threo-methylphenidate based on Pharma Res, 1995 (sadly the relevant crystallography data isn't supplied in the free preview at Springer). I have several questions relating to this file, the most pressing is that so far I have this loop:
loop_
_atom_site_label
_atom_site_chemical_conn_number
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
_atom_site_adp_type
_atom_site_B_iso_or_equiv
_atom_site_type_symbol
N1   1    -0.8852(9) 0.120(1)   -1.0300(8) Biso 3.1(2) N
H1a  2    -0.9650000 0.08500000 -1.0010000             H
H1b  3    -0.8800000 -0.0800000 -1.0330000             H
C2   4    -0.739(1)  0.049(2)   -0.940(1)  Biso 3.9(2) C
H2   5    -0.7460000 -0.0800000 -0.9360000             H
C3   6    -0.614(1)  0.092(1)   -0.988(1)  Biso 5.1(3) C

that is generating this error in enCIFer:
Too many or too few data values in the loop

which I'm going to take a stab in the dark and say that it's because two fields (Biso & $B_{eq}$ value) aren't supplied for each atom in this loop (namely it is missing from the hydrogen atoms). The problem is that the paper I cited didn't specify these for hydrogen atoms. Should I use separate loops for hydrogen & other atoms (i.e., one loop for hydrogens, another for the other atoms)? I ask because I fear that this might make molecular modelling software treat the hydrogens & the other atoms as being part of separate molecules.

Most software I've used gets the atomic positions from the CIF and then interprets molecular identity from connected components and/or the asymmetric unit.

My understanding of the CIF format is exactly what you describe. You must provide everything you declare, so your current loop fails for the reason you suspect.

So can dots (.) be used where the missing data is so that this error won't appear?

It looks like the structure model equally was published in 1998JOC1785 which is entry NODBIR here.  Submitting the .cif to checkcif reveals some alerts, but not a "too few / to many atoms".  Some positional disorder, but ok.

Your loop defines 8 records. That works fine for heavy elements. However, the lines with hydrogen atoms have only 6 items. That's the reason why this error is reported (all you records are not multiplier of 8).
Usually in refinement the hydrogen atom's $U_{iso}$ is 1.2–1.5 bigger than this of atom to which the H is bound (smaller 1.2 for more constrained movements like on aromatic carbons, bigger 1.5 for aliphatic)
Try something like this:
N1   1    -0.8852(9) 0.120(1)   -1.0300(8) Biso 3.1(2) N
H1a  2    -0.9650000 0.08500000 -1.0010000 Biso 4.     H
H1b  3    -0.8800000 -0.0800000 -1.0330000 Biso 4.     H
C2   4    -0.739(1)  0.049(2)   -0.940(1)  Biso 3.9(2) C
H2   5    -0.7460000 -0.0800000 -0.9360000 Biso 4.     H
C3   6    -0.614(1)  0.092(1)   -0.988(1)  Biso 5.1(3) C

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text:
Why does pKa of a acid-base indicator equal to the pH when the equivalence point is reached?
To be honest, I just don't get it.
I had this question while reading a Chemguide article, on this topic:

I think my confusion is caused by the way I look at pKa of acid-base indicator. Should you think of it more as a reaction quotient. Because technically, at equilibrium, the two concentration aren't the same so cannot be cancelled out.

Would you add a link to the Chemguide page so that we can see if the way it is being explained there is part of what is tripping you up?

Just search up acid base indicator Chemguide

This is the best website I've found re explaining this concept re why indicator pH needs to be within +/- one pH unit of equivalence point: https://groups.chem.ubc.ca/courseware/pH/section15/index.html
check it out

Look at the following equation:

$$\mathrm{pH} = \mathrm pK_{\mathrm a} + \log\frac{[\ce{A^-}]}{[\ce{HA}]}$$

At the half equivalence point, say we have 10 moles of $\ce{WA},$ and so there will be 5 moles of $\ce{SB}$ as the name suggests ("half equivalence"). So we know that the 10 moles will be neutralized to 5 moles of $\ce{WA}$ by the $\ce{SB}.$ If we do this then the $\ce{CB}$ is also now equal to the $\ce{WA}$ since it will increase by 5 moles.
$$\ce{[WA]~ =~ [CB]}.$$
Thus,
\begin{align}\mathrm{pH} &= \mathrm pK_{\mathrm a} + \log 1\\  &=\mathrm pK_{\mathrm a}\;.\end{align}
So this is why the the $\mathrm{pH} = \mathrm pK_{\mathrm a}$ at the Half-Equivalence Point.

Vocabulary:

SB : Strong Base. (eg. $\ce{NaOH}$)
CB : Conjugate Base. (eg. $\ce{Ac^{-}}$)
WA : Weak Acid. (eg. $\ce{HAc}$)

NOTE: We can never be 100% accurate here, this is just used to derive that $\mathrm{pH} = \mathrm pK_{\mathrm a}$

No I am talking about the endpoint-which in most case is the equivalence point.

Yeah but the pKa equals the pH at the H-E point

But we are talking about an indicator which itself is a weak acid. I understand what you said is correct, for other situations, but isn't exactly what I am looking for.

OK, I will delete my answer or turn it into a community wiki.

Wait, what I need to know is what causes the two concentration to be equal, when the equivalence point is reached, equal moles of acid and base are added and react.

The equivalence point I am talking about is regarding the acid base in the titration. I think the half equivalence point you are talking about is regarding the indicator itself.

Us adding the the substance reacts with the WA or WB, to make the concentrations equal.

Let us continue this discussion in chat.

No actually I'm talking about the concentrations of the acid and base.

Why does pKa of a acid-base indicator equal to the pH when the equivalence point is reached?

Consider titrating an acid with NaOH. There two pKa values to consider:

The indicator itself has a pKa value
The acid being used has a pKa value

So titrating acetic acid (pKa = 4.76) with NaOH, then phenolphthalein (pka = 10) is a good indicator to use since it will be colored after the neutralization reaction is complete. However bromophenol blue (pKa = 4.75) wouldn't work because it would turn blue before the neutralization reaction is complete.
So the idea is that the indicator should indicate when the reaction is complete, not the equivalence point of the reaction.

Why does pKa of a acid-base indicator equal to the pH when the equivalence point is reached?

It does not. The equivalence point depends on the reaction you are monitoring with the indicator. The $\mathrm{p}K_\mathrm{a}$ of a given indicator is given.
So you would first figure out what the pH at the equivalence point is. Then you would choose an indicator that changes color near that pH. Now, you can ask about the relationship of the pH of color change and the $\mathrm{p}K_\mathrm{a}$ of the indicator.

I think my confusion is caused by the way I look at pKa of acid-base indicator. Should you think of it more as a reaction quotient. Because technically, at equilibrium, the two concentration aren't the same so cannot be cancelled out.

The pH indicator reaction reaches equilibrium after less than a couple of seconds. So if you stop titrating and swirl the reaction mixture a bit, you are at equilibrium and the indicator color reflects the pH. This happens before, at and after the equivalence point (of the titration reaction).
When you are just in the middle of the transition between one and another color of the indicator, you have the same concentration of protonated and deprotonated indicator. Under those circumstances, the two concentrations do cancel out and the pH is equal to the $\mathrm{p}K_\mathrm{a}$.
To summarize, you choose an indicator such that at the equivalence point of your titration reaction, the indicator reaction is roughly halfway between colors. You also make sure the concentration of the indicator is much much lower than that of the analyte so that analyte and titrant determine the pH and the indicator reports on it without skewing the result.

well, weak acids ionize partially n thus we should expect a half ionization . if the acid is xmole at the start, at the end point we should expect the conjugate base to be x/2mole leaving behind x/2mole of the acid. hasselbach equation applies to weak acid n bases

This doesn't really answer the question. It is also very poorly written. Please use proper syntax and avoid non-standard abbreviations like 'n' for 'and'.

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text:
What reactions create a lot of gas really quickly?
I'm looking for a chemical reaction that produces a large amount of a gas in a very short amount of time (like less than a second). What are my options?

What are you trying to blow up?

I'm trying to inflate a small airpocket @ringo

In that case, your problem has been extensively researched both academically and industrially, so that's probably a good place to look for solutions.

try gunpowder ;-)

The reaction used for car airbags involves sodium azide (NaN3). A spark initiates the reaction and it generates gaseous nitrogen really fast.
I don't know much more about it but you can certainly find some good informations on google!

Can you add some source or make it more detailed? I think this answer is considered as law quality.

I agree with Freddy. This reaction is the first that came to me, but your answer should contain more information (for starters, the complete reaction). I think that prior to this a similar reaction was used, but it generated $\ce{H2}$ which was a bit dangerous...

I wish I could give more informations but I looked in Housecroft & Sharpe Inorganic Chemistry which was the reference book for my inorganic chemistry course last semester and there is no mention of this reaction. All I have is a pdf document with class notes produced by my teacher, that's where I found the equation but there isn't much more information.
The only other thing it says is that sodium azide is used in air bags.

Here is the abstract from an article I found on sci. finder:
Source: Diemann, E. Chemistry of airbags, Praxis der Naturwissenschaften, Chemie Volume 42, Issue 8, p.3-4, 1993
A lecture demonstration for school teachers concerning the thermal decompn. of NaN3 into Na and N2 is described.  Information is given on the application of this reaction to the very rapid formation of N2, which can be used to inflate airbags in motor vehicles.  The reaction is initiated by the closing of an elec. switch upon mech. impact, and the dangerous Na, which is also generated, is removed by other means.


text:
Calculation of volume required for titration with dibasic acid
Here is a problem:

Titrate $1 M$ sulfuric acid with $\pu{50 ml}$ of $1 M$ sodium hydroxide solution. What volume of sulfuric acid will be required for neutralization.

This is a simple problem. My reasoning is since both reactants are strong we will have $\pu{0.05 L}$ $\ce{OH-}$ ions and for neutralization we need an equal number of $\ce{H+}$ ions. Since $\ce{H2SO4}$ is a strong acid and ignoring the second dissociation which is weak, we get an equal number of $\ce{H+}$ as $\ce{H2SO4}$ ions. Since concentrations are equal, the volume is simply $\pu{50 mL}$.
Apparently this is wrong. I think the source of confusion is what is meant by "required for neutralization". If you write out the molecular equation for this reaction and do stoichiometry you get a different volume.

"Neutralized" usually means "will not react upon further addition of strong acid/base". What would this mean in your case?

You should get different volume. The only way you should get equal volume is if the ratio between protons and hydroxide are equal. Are they equal in this case?

We'll by definition that's what you would need for neutralization, so yes. Unless neutralization means something different.

Actually, I don't think we need to do stoichiometry at the ionic level, that is to say we don't need to compare ions during stoichiometry, only the compounds.

We first observe the balanced equation. $$\ce {H2SO4 + 2NaOH <=> Na2SO4 + 2H2O}$$ we see the ratio of $\ce {NaOH}$ and $\ce {H2SO4}$ is 2:1. Given $ 0.050~\mathrm{L}$ of $\ce {1M~NaOH}$ we find moles of $\ce {NaOH}$ and use stoichiometry to find how many moles of $\ce {H2SO4}$ is needed. We find the volume to be $25~\mathrm{ml}$.

Yes but then why is my reasoning wrong?

It is wrong because unless you're going to do the full math, you should consider the second dissociation of sulfuric acid strong, and not weak. It might not be complete and as strong as the first, but it's considerable and the error in considering it non-existent is a lot greater than considering it complete.


text:
Lewis structure of chlorite ion
When solving the lewis structure for the $\ce{ClO2-}$ ion, taking into consideration formal charges, the structure is represented as:

A lewis model with 2 double bonds also fits the formal charge and lewis model requirements, however this time the negative formal charge would be on the chlorine atom. Is there a reason this form is not favoured?

When drawing structures and taking into consideration formal charge, negative charges are assigned to the most electronegative atoms. Since the electronegativity of chlorine is less than the electronegativity of oxygen, it follows that the model with 2 double bounds is less favorable.

Perfect! Thanks for the answer, I completely disregarded electronegativity. Would the 2 double bond structure be a part of a resonance form, or is it completely unfavourable?

Yes, although not as much as the more favorable.


text:
Measurement of Water Clarity
Turbidity is essentially the clarity of water. Turbidity can be measured in many ways such as a turbidimeter, secci disk or a turbidity tube (which incorporates a secci disk at the bottom). If you had a small turbid water sample - 20mL, how could you measure the turbidity without an electronic device??

There's a lot of portable turbidity meters with different light sources available. Your choice depends on the the standards that you have to follow: EPA = tungsten filament, ISO = IR leds.

I'm not sure how thats relating to my question?

Let me reiterate the question: Measuring the turbidity of a small sample size (20mL) in the cheapest way.

First of all, please make your question clear. You changed your question here in the comments.

wow, i think a mod changed it sorry

Turbidity is measured by the examining the light passing through a sample.  A sample's turbidity is formally measured using standardized methods (e.g. ISO 7027 or "Standard Methods," No. 2130B. source).  Standardized methods require the instrumentation called for by the standard.
An informal measure of visible differences in turbidity can be obtained by comparing the 20mL sample to a set of calibrated standards.  An example of a set of calibration standards for turbidity are shown in the following link http://www.water.ncsu.edu/watershedss/info/turbid.html.


text:
How to attach a liebig condenser to a flask for reflux heating?
I know it's probably a stupid question, but I can not figure this out. I have a liebig condenser that I usually use for distillation with a Kjeldahl bulb, which makes it easy to attach to a flask because the bulb has a stopper on both ends?
Now I have an experiment that calls for reflux heating, where the condenser needs to be held vertically over the flask. One end of the condenser is a cork-fit opening, and the other end is a spout. The flask also has a cork-fit opening.

Bad idea, see my answer here http://chemistry.stackexchange.com/a/41460/17952

Maybe I don't understand your question well, but often in the lab when we do reflux heating, we don't really attach the liebig condenser.
You fix the flask using a clamp. And gravity holds the liebig condenser into place. You can put a clamp around it to hold it straight but you'd better not tighten it too much because these things are fragile.
I hope it helps you :)
If it's not clear maybe it would be easier to answer if you posted pictures of your glassware!

Those Keck-clips are awesome :)

Thanks for your help, but I don't think I can just let gravity hold the condenser in place because my glassware does not have ground-glass joints. I believe I found the fix, it is a 14mm-18mm male-to-male glass adapter. Apparently they are meant more for the cannabis market, but I think it will serve my purpose.

I think that a PTFE (teflon) sleave is a good idea with ground glass joints as it stops the glass joints getting stuck. Of all things in my lab the most dangerous object is the humble liebig condensor. Take very great care when pushing a glass tube through a rubber bung or a cork. Also take great care when attaching rubber hoses to a glass tube (like the water connections on a liebig). The modern Quickfit liebig with screw on water connections is one of the greatest safety improvements in organic chemistry in my opinion.


text:
Is there any required data missing from my CIF file?
I previously asked a question about a CIF file here (namely CIF fields that are available for some atoms, but not others) I think I have the answer (namely to place dots [.] at places where the two missing terms would normally appear), although I felt I'd wait to see if anyone answers in case I am wrong. Presently I have this CIF file:
#=====================================================================
# X-ray crystallography data from Pharma Res, 1995
#---------------------------------------------------------------------

data_methylphenidate

_cell_length_a                9.401(3)
_cell_length_b                7.3053(7)
_cell_length_c                11.113(3)
_cell_angle_alpha             90
_cell_angle_beta              109.260(2)
_cell_angle_gamma             90
_cell_formula_units_Z         2
_cell_volume                  720.5(3)
_cell_measurement_temperature 295.15
_cell_measurement_reflns_used 1797

_chemical_formula_sum         'C14 H20 Cl N O2'
_chemical_formula_weight      269.77
_chemical_name_systematic     'methyl (2S)-2-phenyl-2-[(2S)-piperidin-2-yl]acetate;hydrochloride'
_chemical_identifier_inchi    'InChI=1S/C14H19NO2.ClH/c1-17-14(16)13(11-7-3-2-4-8-11)12-9-5-6-10-15-12;/h2-4,7-8,12-13,15H,5-6,9-10H2,1H3;1H/t12-,13-;/m0./s1'
_chemical_identifier_inchi_key 'JUMYIBMBTDDLNG-QNTKWALQSA-N'

_refine_ls_R_factor_gt        0.070

_space_group_crystal_system   monoclinic
_space_group_name_H-M_alt     'P 21'
_space_group_IT_number        4

loop_
_symmetry_equiv_pos_as_xyz
'x, y, z'
'-x, y+1/2, -z'

loop_
_atom_site_label
_atom_site_chemical_conn_number
_atom_site_cartn_x
_atom_site_cartn_y
_atom_site_cartn_z
_atom_site_adp_type
_atom_site_B_iso_or_equiv
_atom_site_type_symbol
N1   1  -0.8852(9) 0.120(1)   -1.0300(8) Biso 3.1(2)   N
H1a  2  -0.965     0.0850     -1.001     .      .      H
H1b  3  -0.880     -0.080     -1.033     .      .      H
C2   4  -0.739(1)  0.049(2)   -0.940(1)  Biso 3.9(2)   C
H2   5  -0.746     -0.080     -0.936     .      .      H
C3   6  -0.614(1)  0.092(1)   -0.988(1)  Biso 5.1(3)   C
H3a  7  -0.525     0.0380     -0.933     .      .      H
H3b  8  -0.603     0.2210     -0.990     .      .      H
C4   9  -0.640(1)  0.021(2)   -1.122(1)  Biso 5.0(3)   C
H4a  10 -0.639     -0.110     -1.119     .      .      H
H4b  11 -0.561     0.0640     -1.150     .      .      H
C5   12 -0.792(2)  0.086(1)   -1.213(1)  Biso 5.3(3)   C
H5a  13 -0.786     0.2150     -1.225     .      .      H
H5b  14 -0.812     0.0250     -1.291     .      .      H
C6   15 -0.915(1)  0.049(2)   -1.160(1)  Biso 4.3(2)   C
H6a  16 -1.004     0.1050     -1.214     .      .      H
H6b  17 -0.928     -0.080     -1.159     .      .      H
C7   18 -0.721(1)  0.124(1)   -0.809(1)  Biso 3.3(2)   C
H7   19 -0.737     0.2520     -0.817     .      .      H
C8   20 -0.834(1)  0.051(2)   -0.7527(9) Biso 3.8(2)   C
O9   21 -0.889(1)  0.172(1)   -0.6908(9) Biso 4.9(2)   O
C10  22 -0.998(2)  0.113(2)   -0.634(1)  Biso 5.8(4)   C
H10a 23 -1.028     0.2150     -0.594     .      .      H
H10b 24 -0.956     0.0210     -0.572     .      .      H
H10c 25 -1.085     0.0650     -0.698     .      .      H
C11  26 -0.563(1)  0.093(1)   -0.709(1)  Biso 3.9(3)   C
C12  27 -0.509(2)  -0.091(2)  -0.681(1)  Biso 5.6(3)   C
H12  28 -0.566     -0.192     -0.726     .      .      H
C13  29 -0.376(2)  -0.114(3)  -0.590(2)  Biso 7.2(4)   C
H13  30 -0.343     -0.238     -0.572     .      .      H
C14  31 -0.290(2)  0.009(6)   -0.522(2)  Biso 11.3(8)  C
H14  32 -0.1960    -0.018     -0.460     .      .      H
C15  33 -0.343(2)  0.204(4)   -0.548(2)  Biso 9.0(6)   C
H15  34 -0.284     0.3020     -0.501     .      .      H
C16  35 -0.479(2)  0.242(2)   -0.643(1)  Biso 5.9(3)   C
H16  36 -0.515     0.3640     -0.664     .      .      H
O3   37 -0.871(1)  -0.1124(9) -0.7590(9) Biso 5.7(2)   O
Cl   38 -0.8105(3) 0.5484     -1.0089(3) Biso 4.72(6)  Cl

. The problem is that Avogadro and enCIFer are having trouble rendering it and I don't see what the problem is. e.g., this is what Avogadro is giving me at the moment:

Are _cell_measurement_theta_min and _cell_measurement_theta_max the problem (namely that I haven't specified them)?

I changed some lines and got it right. Can you please check it. https://drive.google.com/file/d/0B_uHH541lFwhS1VJcGpkQ3o4YzA/view?usp=sharing

Sure did. I think I see what you did too, you changed the Beqs to 1, removed the standard deviations & conn number column and added occupancies. Did ya do anything I missed?

(I meant you sure got it right, Avogadro & Mercury are rendering it right.)

Ok, the basic change i did was, to convert cartesian coord to fractional coord system because clearly from your attached figure you can see they are all coalesced in a small area at (000), that means your position should be scaled up.

Mamun's answer enabled me to find the fault in my CIF file. I was using cartn categories for fract data. Here's the correct file:
#=====================================================================
# X-ray crystallography data from Pharma Res, 1995
#---------------------------------------------------------------------

data_methylphenidate

_cell_length_a                9.401(3)
_cell_length_b                7.3053(7)
_cell_length_c                11.113(3)
_cell_angle_alpha             90
_cell_angle_beta              109.260(2)
_cell_angle_gamma             90
_cell_formula_units_Z         2
_cell_volume                  720.5(3)
_cell_measurement_temperature 295.15
_cell_measurement_reflns_used 1797

_chemical_formula_sum         'C14 H20 Cl N O2'
_chemical_formula_weight      269.77
_chemical_name_systematic     'methyl (2S)-2-phenyl-2-[(2S)-piperidin-2-yl]acetate;hydrochloride'
_chemical_identifier_inchi    'InChI=1S/C14H19NO2.ClH/c1-17-14(16)13(11-7-3-2-4-8-11)12-9-5-6-10-15-12;/h2-4,7-8,12-13,15H,5-6,9-10H2,1H3;1H/t12-,13-;/m0./s1'
_chemical_identifier_inchi_key 'JUMYIBMBTDDLNG-QNTKWALQSA-N'

_refine_ls_R_factor_gt        0.070

_space_group_crystal_system   monoclinic
_space_group_name_H-M_alt     'P 21'
_space_group_IT_number        4

loop_
_symmetry_equiv_pos_as_xyz
'x, y, z'
'-x, y+1/2, -z'

loop_
_atom_site_label
_atom_site_occupancy
_atom_site_chemical_conn_number
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
_atom_site_adp_type
_atom_site_B_iso_or_equiv
_atom_site_type_symbol
N1   1.0  1  -0.8852(9) 0.120(1)   -1.0300(8) Biso 3.1(2)   N
H1a  1.0  2  -0.965     0.0850     -1.001     .      .      H
H1b  1.0  3  -0.880     -0.080     -1.033     .      .      H
C2   1.0  4  -0.739(1)  0.049(2)   -0.940(1)  Biso 3.9(2)   C
H2   1.0  5  -0.746     -0.080     -0.936     .      .      H
C3   1.0  6  -0.614(1)  0.092(1)   -0.988(1)  Biso 5.1(3)   C
H3a  1.0  7  -0.525     0.0380     -0.933     .      .      H
H3b  1.0  8  -0.603     0.2210     -0.990     .      .      H
C4   1.0  9  -0.640(1)  0.021(2)   -1.122(1)  Biso 5.0(3)   C
H4a  1.0  10 -0.639     -0.110     -1.119     .      .      H
H4b  1.0  11 -0.561     0.0640     -1.150     .      .      H
C5   1.0  12 -0.792(2)  0.086(1)   -1.213(1)  Biso 5.3(3)   C
H5a  1.0  13 -0.786     0.2150     -1.225     .      .      H
H5b  1.0  14 -0.812     0.0250     -1.291     .      .      H
C6   1.0  15 -0.915(1)  0.049(2)   -1.160(1)  Biso 4.3(2)   C
H6a  1.0  16 -1.004     0.1050     -1.214     .      .      H
H6b  1.0  17 -0.928     -0.080     -1.159     .      .      H
C7   1.0  18 -0.721(1)  0.124(1)   -0.809(1)  Biso 3.3(2)   C
H7   1.0  19 -0.737     0.2520     -0.817     .      .      H
C8   1.0  20 -0.834(1)  0.051(2)   -0.7527(9) Biso 3.8(2)   C
O9   1.0  21 -0.889(1)  0.172(1)   -0.6908(9) Biso 4.9(2)   O
C10  1.0  22 -0.998(2)  0.113(2)   -0.634(1)  Biso 5.8(4)   C
H10a 1.0  23 -1.028     0.2150     -0.594     .      .      H
H10b 1.0  24 -0.956     0.0210     -0.572     .      .      H
H10c 1.0  25 -1.085     0.0650     -0.698     .      .      H
C11  1.0  26 -0.563(1)  0.093(1)   -0.709(1)  Biso 3.9(3)   C
C12  1.0  27 -0.509(2)  -0.091(2)  -0.681(1)  Biso 5.6(3)   C
H12  1.0  28 -0.566     -0.192     -0.726     .      .      H
C13  1.0  29 -0.376(2)  -0.114(3)  -0.590(2)  Biso 7.2(4)   C
H13  1.0  30 -0.343     -0.238     -0.572     .      .      H
C14  1.0  31 -0.290(2)  0.009(6)   -0.522(2)  Biso 11.3(8)  C
H14  1.0  32 -0.1960    -0.018     -0.460     .      .      H
C15  1.0  33 -0.343(2)  0.204(4)   -0.548(2)  Biso 9.0(6)   C
H15  1.0  34 -0.284     0.3020     -0.501     .      .      H
C16  1.0  35 -0.479(2)  0.242(2)   -0.643(1)  Biso 5.9(3)   C
H16  1.0  36 -0.515     0.3640     -0.664     .      .      H
O3   1.0  37 -0.871(1)  -0.1124(9) -0.7590(9) Biso 5.7(2)   O
Cl   1.0  38 -0.8105(3) 0.5484     -1.0089(3) Biso 4.72(6)  Cl

+1 Nicely spotted! On a very minor side note: If a very strict CIF checker complains about line length > 80 characters, as it could for the INCHI key block, you can always use a text block with ; on leading and trailing lines instead of a string in quotation marks on the same line as the _something label. This is what you typically do when submitting CIF files for publication in IUCr journals. Please have a look at DOI for further information.


text:
Are there any (liquid) chemicals that experience rapid, extensive, and controllable volume changes in reaction to heat?
The larger purpose of this question is the engineering of liquid filled sacs that change their volume according to heat. They'd need to have diverse ranges of volume, at least a factor of ten (a 10 ml to 100ml transition).

I don't think you will find any liquids that will fit your requirements. After all, thermometers would tend to use liquids with large thermo expansion coefficients. The liquids in the thermometers don't even double in volume. Mercury has a volume coefficient of $182\times 10^{-6}K^{-1}$, so a temperature change of $10000K$ is needed to double the volume.

http://chemistry.stackexchange.com/questions/29700/what-reactions-create-a-lot-of-gas-really-quickly - you'd need to produce gas


text:
Dimensional Analysis in Chemistry
I was once told on this site that it was incorrect form to use units that specify the chemical being referred to in dimensional analysis. For example:
$$150.~\mathrm{g}~~\ce{KNO3} \cdot \frac{1~\mathrm{mol}~~\ce{KNO3}}{101.103~\mathrm{g}~~\ce{KNO3}} \cdot{} \frac{1~\mathrm{mol}~~\ce{C7H4O}}{6~\mathrm{mol}~~\ce{KNO3}} \cdot \frac{104.106~\mathrm{g}~~\ce{C7H4O}}{1~\mathrm{mol}~~\ce{C7H4O}}=25.7~\mathrm{g}~~\ce{C7H4O}$$
This is the way I learned it, and I feel that this clears up confusion that might otherwise arise when from performing such a calculation.
My overall question is: is it acceptable to do dimensional analysis as I presented it? If not, please explain why it is not correct to do it this way and show me what I should be doing instead.

I don't understand what is contentious.  Everything you did is perfectly fine in my eyes.  I don't know exactly what chemical you mean by C7H4O, but that's irrelevant.  If there is such a chemical, and it reacts with nitrate salts, everything you calculated is fine.

The way I would perform the dimensional analysis you give in your question would be slightly different. That is mainly due to the fact that I don't consider chemical names or formulae as units, but as designators.
As an aside, I think you were starting from a question like the following:

How much $\ce{C7H4O}$ would you have to weigh in for it to fully react
  with $150~\mathrm{g}$ of $\ce{KNO3}$ if the reaction is described by
  the following equation? $$ \ce{C7H4O + 6KNO3 -> Fumes^  + Tar} $$

The result is $m_{\ce{C7H4O}}$, given is $m_{\ce{KNO3}} = 150~\mathrm{g}$.
The amount of $\ce{C7H4O}$ used per $\ce{KNO3}$ is
$$|\nu_{\ce{C7H4O}} / \nu_{\ce{KNO3}}| = \frac{1}{6}$$
As such, the result is composed as follows:
$$\begin{align}
n_{\ce{C7H4O}} &= \frac{1}{6} n_{\ce{KNO3}} \\
\tag{$n=m/M$}\frac{m_{\ce{C7H4O}}}{M_{\ce{C7H4O}}} &= \frac{1}{6} \frac{m_{\ce{KNO3}}}{M_{\ce{KNO3}}} \\ \hline
m_{\ce{C7H4O}} &= \frac{1}{6} \frac{m_{\ce{KNO3}}}{M_{\ce{KNO3}}} M_{\ce{C7H4O}}
\end{align}$$
If we now apply dimension analysis to what we think is the result, we can very quickly see how the two molar masses cancel each others units out and only the mass of $\ce{KNO3}$ remains to give the right hand side of the equation a dimension of $\mathrm{g}$. As the only symbol left on the left hand side is a $m$ we know that it needs a dimension $\mathrm{g}$ and as such we can happily plug in the numbers into our calculator for the result.
Frankly, I am so used to "my" method that I had to look at your example quite closely to see what was actually going on. To me that method just seems unwieldy, as you clutter up equation space with chemical formulae.
That being said; find the method that works the best for you (meaning it is robust enough to also solve your chemical engineering, biology and other problems using dimensional analysis) and stick to it.


text:
Equation for heat loss over distance?
Is there an equation to figure out what the temperature of an the area will be a certain distance from the heat source?
For example, if I know there is a heat source at $0~\mathrm{m}$, and the temperature is $100~\mathrm{^\circ C}$, is there a way to find out what the temperature will be at $50~\mathrm{m}$ away from the heat source? Is there some sort of equation for this? All of this is happening in oxygen.

This is a heat transfer related problem. You can certainly do that. For solid medium heat transfer you need to use Fourier heat transfer equation
\[q=-k\cdot\nabla T\]
Here, k= material's conductivity and $\nabla T$ = 1st derivative of the Temp. along the driving force or heat
For Liquid and gas phase you should use law for thermal convection. And for vacuum medium you should use radiation law. There are lots of ways to do this but from your problem statement i think it's a numerical grid type problem. For your better understanding, i would suggest 'Engineering heat transfer' by william s janna (I may be biased here because i like this book) but you can certainly use any other standard heat transfer book or internet.

Where in the equation does distance from heat source go? And what units do I use?

@Jayden Miller this is a partial differential equation. You will probably have to solve it numerically, which means breaking up the volume you are interested in into small boxes and then solving the heat transfer equation across the boundary of each one, most likely using a computer program. You can use any units you want as long as they are consistent. SI units is probably easiest. If you give more specific details in your problem statement, there may be a way to simplify the problem to one dimension, which would make it a lot easier.


text:
The explanation on atomic size
I always receive two type of explanation for the decreasement of atomic size for elements across the period.
Explanation according to proton number
This explanation is from my teacher.

The proton number of elements increases across the period.
The force of attraction between the proton in the nucleus and the electrons is stronger across the period as the proton will pull the electrons towards the nucleus.
So, the atomic size of the elements will decrease across the period

Explanation according to number of valance electron
This explanation is from my friends

The valance electron of elements increases across the period.
The force of attraction between the valance electrons and the proton in the nucleus is stronger across the period between the valance electrons is pulled towards to the nucleus
So, the atomic size of the elements will decrease across the period.

My question is which explanation is more suitable for the statement

the atomic size of the elements decrease across the period.

iirc, the real explanation is too complicated and needs lots of computing, but the two you explained above are just parts of the real explanation. They're not wrong, but they're not complete either.

Both of those explanations include part of the puzzle, but are not really complete on their own. To get a better understanding, you need to combine them, as well as understand something called Coulomb's law.
The basics of Coulomb's law are this:

The closer two oppositely charged things are, the stronger the attraction
The stronger the charge on either of the things, the stronger the attraction

With those points in mind, let's combine it all. I'll keep my explanation similar to the way you phrased yours:

Each step across a row on the periodic table adds both one nuclear proton and one valence electron. So, that is increasing both the total positive charge of the nucleus and the total negative charge of the valence shell, but not the distance between them (because electrons are added to the same shell).
The force of attraction between two charged objects depends on how strong each of them are (Coulomb's law). Since both the nucleus and the valence shell are stronger than in the previous step, and the distance between them has not increased (the electrons are added to the same shell) there is a stronger force of attraction between them, and the two are pulled closer together.
So, the atomic size of elements decreases across a period.


text:
How does standardising a secondary standard solution make it suitable as a standard solution?
I know that a standard solution in volumetric analysis must have an accurately known solution. Primary standard solutions (e.g. anhydrous sodium carbonate) can be used straight-away as they follow the list of criteria (high level of purity, unaffected by air exposure, etc.)
I also know that secondary standard solutions (hydrochloric acid) cannot be used directly like primary standard solutions as they don't follow the criteria. But I don't understand how standardising hydrochloric acid with a primary solution (e.g. anhydrous sodium carbonate) will make it able to be used as a standard solution.

When talking about standards, there's always the dilemma: How is a standard standardized? Well, you compare it to another standard you already have, and after you consider all the sources of errors, you have a new standard.
In case of the primary standards, you can always promptly trust the label's purity information, as well as your balance's and other measuring equipment (like glassware) accuracy, and at a much lower degree the molar masses, and so on. Having that concentration "certified", anything you calculate using that as a reference will be "certified" as well. However...
If you're going to follow strict analytical rules, than you won't blindly trust those, which is why you cshould calculate the error (or many of the related statistical measures) by considering the accuracy of, like I said, the balance, the glassware, temperature effect on volume (density), etc.
One common way to do this, called the standardized uncertainty, is take the square root of the sum of the squares of each instrument/measure uncertainty divided by it's quantity. You should define a minimal formula which describes the property you want to account the error for, and go for there. Let's take, for example, the concentration of 100 mL of a 0.01 M solution of potassium hydrogen phthalate (KHP), used as a primary standard to standardize sodium hydroxide.
The concentration of the solution is $[KHP]=\frac{n}{vol}$, but we obviously didn't measure molês, so we write
$$[KHP]=\frac{\frac{m}{MM}}{vol}$$
where $m$ is the mass weighed on the balance, and $MM$ the molar mass. Now that we know what influence in your concentration, we have to use some numbers. Let's consider that:

Weighed mass: 0.20422 g, on a balance with uncertainty of ± 0.001 g (1 mg)
Volume measured: 100 mL, in a volumetric flask with uncertainty of ± 0.1 mL

This means that the uncertainty of the prepared solution is
$$\frac{u_{[KHP]}}{0.01} = \sqrt{\left(\frac{0.001}{0.20422}\right)^2 + \left(\frac{0.1}{100}\right)^2} $$
$$u_{[KHP]} \approx 5\times10^{-5}$$
Now note that I didn't take into account the purity of the salt (to make things simpler), and the error on the molar mass (we could assume it's much lower than the others). And what does all this have to do with your question?
My point is, even standards have a variations in them, and when you standardize something, like a HCl or NaOH solution, you could take that into account. The concentration of NaOH, in this case, will have a relative uncertainty higher than the concentration of KHP, because all of the uncertainty of the KHP preparation must be taken into account, as well as the titration variation. The outcome is that a standard based on  a standard of another standard (...) will eventually have a very high uncertainty, which is why we try to sitck to primary and secondary standards.
PS.: The uncertainties of the equipments/glassware are usually available in the manuals or even engraved on them. This values are usually the calibration precision, and should be transformed into uncertainty using some statistical inference, which could be that the variation follows a retangular distribution, resulting in
$$u = \frac{precision}{\sqrt3}$$
Disclaimer: There's a lot more to this than what I summarized above, starting with being able to indentify what should and shouldn't be taken into account, not to mention replicates and a lot of statistics. I don't recommend blindly following this formulae to calculate uncertainties.


text:
Why can't helium be solidified at 'ordinary' pressures?
According to the UC Davis ChemWiki Chemistry of Helium, helium has a comparatively unusual property, specifically:

Helium is the only element that cannot be solidified by lowering the temperature at ordinary pressures.

'Ordinary' referring to standard air pressure (1 atmosphere).  In order to solidify, there needs to be a corresponding pressure increase, with a projected density of:

$0.187 \pm 0.009~\mathrm{g~mL^{-1}}$ at $0~\mathrm{K}$ and $25~\mathrm{bar}$.

So why can't helium be solidified at 'ordinary' pressures?

https://en.wikipedia.org/wiki/Zero-point_energy

The short answer is that even the London dispersion forces between helium atoms (molecules) are extremely weak. So helium boils at a lower temperature than any other element as well.

When the temperature of helium gas is decreased to about 5.2 K, a phase transition to ordinary liquid helium ($\ce{He}$ I) occurs. The behavior of this liquid phase is normal and identical to any other liquid phase.
As temperature is decreased more (at moderate pressures), helium does not solidify. In fact, it undergoes a phase change to a second liquid phase known as $\ce{He}$ II at a temperature of approximately 2.17 K. As shown in the phase diagram below, the $\ce{He}$ II phase persists until absolute zero. Due to quantum effects, $\ce{He}$ II has remarkably unique properties. It is considered a superfluid.

For more details, please click here. Pressure is required to produce solid helium (25 atmospheres or more), because of the small mass and extremely weak forces between the helium atoms.

Great answer, the properties of helium are indeed remarkably unique - are there any other elements/compounds having 2 liquid phases?

@santiago helium-3 is even stranger than helium-4: https://en.wikipedia.org/wiki/Helium-3#Cryogenics

It should be noted that helium is the only substance whose isotopes have radically different phase diagrams.

@santiago Liquid crystalline materials often have even three liquid phases.

There are a numer of liquid sulphur modifications with varying viscosity.


text:
Why does copper precipitate out as a pink solid?
When copper is precipitated out of a solution e.g from CuI, why does it appear pink instead of copper-coloured? Are there other situations where copper appears pink in solid form?

What you describe as pink is the actual colour of copper. The colour of a clean, solid surface of high-purity copper is typically salmon red. In your case, the appearance may be further affected by the particle size of the precipitate.
The familiar reddish-brown colour of copper is caused by copper(I) oxide, which is slowly formed on the metal surface.


text:
Can chemicals change crystal structure by only touching?
Are there any known, or suspected compounds that take the crystal structure of what ever object they are touching, i.e. if it touches $\ce{NaCl}$ it would turn into a face centered cubic crystal structure.

What do you mean by touching? Rubbing to crystals and wow now they have the same structure?

@Mithoron sort of, but like a liquid crystal, more like: Oops spilled this, OMG its now a the structure of that randomly placed salt on the floor

No, there is no "exact mimic" crystal, but the crystal habit can alter the crystal form. Consider that the environment in the uniform inside of a substance is different from that along its boundaries; growing on a different surface can alter the shape of the crystal.
The semiconductor industry, in particular, is often faced with difficulties in growing crystalline semiconductors with layers of varying composition (epitaxy). There is often strain, leading to defects, unless an intermediate layer can smooth the transition.


text:
Solubility of C5H11OH and C5H11Cl
Objectively, what makes the solubility of $\ce{C5H11OH}$ greater than the $\ce{C5H11Cl}$ in water?

http://chemistry.stackexchange.com/questions/10547/the-solubility-of-haloalkanes-in-water

I believe the $\ce{C5H11OH}$ is more soluble because it has dipole dipole, H-bonding and London Dispersion. The $\ce{C5H11Cl}$ only has dipole dipole and London Dispersion. The H-bonding is highly soluble in water

Yes you perfectly right, $\ce{Cl}$ never does H-bonding ;)

In the case of the water solubility what you want to know is how powerful the secondary interaction  of the compound that you are going to dissolve with  water?If the compound that you wish to dissolve have hydrogen bonding then it's definitely dissolve in water .It does not mean others  are  insoluble  in water.
But  the compounds do not have hydrogen bonding are  less soluble in water when compared to    the ones which have hydrogen bonding.Because ones which have hydrogen bonding interact with water molecules easily than others which do not have hydrogen bonding
Now take a look at the image   below,

It shows two water molecules  which posses the hydrogen bonding interaction.
Now give the eye on compounds that you have got,
The  compounds that you going to dissolve  may form  interaction force  with water as above image. If that interaction force is hydrogen bonding ($\ce{C5H11OH}$) then it's solubility is greater than ones which do not have hydrogen bonding($\ce{C5H11Cl}$) (dipole dipole interactions)
So because of that solubility of $\ce{C5H11Cl}$ is less than the solubility of ($\ce{C5H11OH}$) with water.
Then the bottom line is ,the solubility of a compound is vary according the  secondary interaction between the solute and solvent.In the case of water acting as  the solvent solubility is highest in  compounds which having H bonding , than which haaving dipole-dipole forces and  weak dispersion forces.


text:
Is there more than one equation for Delta T(reaction)?
So, I have to find the $\Delta T_{\mathrm{reaction}}$ of a solution for my lab report. Earlier in the lab, it gave the equation as
$\Delta T_{\mathrm{reaction}}=T_{\mathrm{mixture}}+\frac{1}{2}(T_{\mathrm{substance~1}}+T_{\mathrm{substance~2}})$
and all temperatures were at the exact time of mixing found from a linear equation of temperature points on a graph. Now, I have another mixture which is water and a substance and I am supposed to find the $\Delta T_{\mathrm{reaction}}$.
Here's the thing, we don't have the temperature of the substance, only the temp of the water and the solution it made, so how could I go about doing it? I thought about using room temperature since it was a solid substance but it doesn't say anything about that.
The substance is solid $\ce{NaOH}$.

Please avoid Latex in titles due to searching issues.

The solid likely was at room temperature, but it also probably had lower mass and almost certainly had lower heat capacity than the solution to which it was added.  I suspect that the equation you gave is for mixing two dilute solutions of equal volume.  If I'm not mistaken it should be Tmixture - 1/2... rather than + 1/2.

If You are stating you have your initial water temperature and the final solution of your temperature, then your change in temperature is just your final temperature minus your initial temperature. No need for linear equations. The change in temperature for the reaction is the temperature change of the water due to the heat given off or absorbed by the system to the surroundings.


text:
Why is pickle salty? How does salt get into pickle?
I know this may look like some question easily answered with common sense, but please read.
If water is the solvent wouldn't the only thing that moves be water? (Osmosis)
Edit: this is basically what I think. If you imagine the skin of whatever you are pickling as a membrane that prohibit the flow of electrolytes but allows that if water. Then salt, NaCl, cannot move anywhere.

Your concept of a cell is a little vague. Although the cell wall is made from a diglyceride bilayer, it is not impregnable. In fact, there are many proteins that pass through the cell wall where dissolved material can pass. One of the proteins is called the sodium potassium pump. In a living cell, the protein uses ATP to move sodium out of the cell and drag potassium into the cell. Both of these ions are against the gradient. In a non-living cell, such as the pickle, sodium ions can move across the cell membrane and make the cells salty.

So even after being cut off, the cucumber, say for ex, will not absorb any sodium unless it's cell is completely dead?

Also, does Cl also come in like this and get recombined with sodium?

Sodium leaks in even if the cell is alive. There is nothing preventing chloride ions from passing through the cell membrane.


text:
Difference in ozonolysis reagents
What is the difference when following up ozonolysis with DMS (dimethyl sulfide), Zn/water, or just water? Or is there really no substantial difference at all?
In my textbook, the alkenes section uses DMS or Zn/water but then in the alkynes section it uses water with no explanation as to why that happens. My professor also uses them interchangeably, which gets confusing.

Dimethylsulfide (DMS) is commonly added to an ozonolysis reaction after it is complete (e.g. it is the first step in the reaction work-up).  I can see how zinc could serve the same purpose.  Both of these reagents can reduce oxidized materials.  When the ozonolysis is complete the reaction contains the raw ozonide along with numerous undesirable side-products such as peroxides, hydroperoxides and tetroxanes - all of these materials are potentially explosive.  The DMS or zinc will reduce them to carbonyls and alcohols and remove the explosion hazard.  I suppose that tap water is somewhat acidic or basic depending upon where you live, and it might, therefore, also serve to reduce these materials.
It is one thing to say or write "add water during workup" in a lecture or text, it's another thing to actually deal with explosive materials in the lab.  Personally, I wouldn't chance it. I'd use a real reducing agent such as DMS or zinc, instead of a water work-up.


text:
Why electronegativity instead of electropositivity
When I learnt about polarity, I always come to the term electronegativity and always use the electronegativity chart. However, when I studied further, they have the word electroposivity.
So, I've been thinking why we use electronegavity more often than electropositivity? And why we have a electronegativity chart instead of electropositivity chart?

The first thing that should be highlighted here is that electropositivity is simply the opposite of electronegativity, any of the two can be used interchangeably with the necessary modifications to the sentence.
It is true though that electronegativity is more commonly used then electropositivity. For example, the Wikipedia article on the subject is electronegativity, and mentions electropositivity as its opposite. I believe there are two reasons for that, related to each other.
First: Consistency
It's very important in the scientific community, from reasons that go from facilitating students understanding to the writing of papers, that standards are used, and that everyone is talking about the same thing. It's easier if everyone is talking about the same thing, even if they are the complete opposite. It just makes us think faster. Why choose electronegativity then?
Second: How we explain chemistry
It's unecessary to explain why and how the study of the electrons "behaviour" in atoms and molecules are important to chemistry. Usually, when explaining a phenomenon, we talk about where the electrons are "going to". We say a reaction occurs because an electron is taken by an atom, or maybe because it's donated by one. We explain the polarity of the H-O bond by saying that the oxygen will attract the electrons of the covalent bond a lot more than hydrogen will, and not (usually) that the hydrogen doesn't attract the electrons as much as oxygen does.
Again, it's obvious that you can explain everything by means of electropositivity, but, in my experience, we usually explain things by saying which atom/group has the property that makes electrons go to them with higher value, rather than saying which atom/group has the property that makes electrons leave them with higher value.

Electropositivity is not a thing (not matter how many online dictionaries incorrectly define it).  Pauling defined electronegativity as the power of an atom to attract electrons in a bond to itself. Electropositivity, as the opposite of electronegativity, would imply that it is a measure of the power of an atom to repel electrons in a bond from itself. Electrons are always attracted to the atoms involved in the bond, they can just experience a greater force of attraction from one atom over another.  An analogy (flawed as all such analogies are) would be to consider a tug of war.  Both sides pull, one side just pulls harder.  If the tug of war moves from east to west, you would not say that the east side was pushing harder than the west side, you would only ever say that the west side was pulling harder than the east.
Thus, we can refer to atoms in a bond as either more or less electronegative, and not either more electronegative or more electropositive.


text:
Effect of drinking ultra-pure water
What would be the effect if someone were to drink ultra-pure water with an electrical resistivity of $18 \, \mathrm{M} \Omega \! \cdot \! \text{cm}$?
Would they immediately die?  Would they just need to pee more?  Would $\ce{CO2}$ from the air (after the bottle is opened) and whatever's in saliva dissolve into the water making it much less pure before it gets into the important parts of their digestive tract?
What would be the threshold for bad things to happen — a teaspoon, a liter, 12 gallons in 10 minutes (at which point I assume regular water would kill you)?
I'm not sure what the science here would be.

Great question, but I don't get why they would die.

Related questions on other SE sites:  Does drinking distilled water remove needed minerals from your body?; Death because of distilled water consumption?; How safe is drinking distilled water?

What would be the effect if someone were to drink ultra-pure, 18 M-ohm water?

Not much, although if they drank many gallons of the water it could be a problem.

Would they immediately die?

No.

Would they just need to pee more?

Probably.

Would $\ce{CO2}$ from the air (after the bottle is opened) and whatever's in saliva dissolve into the water making it much less pure before it gets into the important parts of their digestive tract?

It would become orders of magnitude less pure as soon as it mixed with their saliva and then their gastric juices.  Same as regular water.

What would be the threshold for bad things to happen -- a teaspoon, a liter, 12 gallons in 10 minutes (at which point I assume regular water would kill you)?

I don't think the effect would be that different from drinking regular water.  All over the internet there is this idea that super pure water is somehow too pure or dangerous, no good evidence or mechanism is ever presented.
Water is toxic primarily because it dilutes the sodium and potassium ions in your body, leading to Hyponatremia or possibly the far more dangerous Hypokalemia.  The normal level of $\ce{Na+}$ in blood plasma is about 135 millimolar (mM), and the normal level of $\ce{K+}$ is about 5 mM.
A very rough approximation to understanding how toxic water can be would then be to compare the concentration difference of sodium (or potassium) between the ingested water and the blood.  A 25 micromolar -- i.e. 25 μM or 0.000025 molar solution of sodium hydroxide has a conductivity of about 6.7 microsiemens per cm, more than 100-fold higher than "pure" 18.2 megohm water (0.055 microsiemens per cm).  According to this rough difference-based model, the ability of the very dilute sodium ion solution to cause problems is proportional to (135 - 0.025) mM, and the ability of the "pure" water to cause problems is (135 - 0) mM.  For nearly biological or medical conditions of interest, 135 mM is not really different from 134.975 mM, and so ultra-pure water is not really more dangerous than everyday drinking water.

Another way to look a this is, how much of your own blood (or other bodily fluid such as cytoplasm) would you need to add to the pure water to make it "safe for consumption" since it's now just as impure as tap water? This mixing can occur internally instead. Considering just Na+, as a simplification, blood has (order of magnitude) 100 times as much as tapwater. The remaining danger is that hypothetically the pure water might on contact cause cells to swell and explode via osmosis before mixing can occur, leading to significant tissue damage. But this turns out not to happen...

Steve, yes, osmotic pressure is another variable that depends on the difference in concentrations, so for osmotic pressure too, ultrapure water isn't significantly worse than regular tap water.

and as far as salts go, most all of these enter the body via our food, so pure water or tap water  in a normal diet will make little difference. Stick to pure beer instead :)

Pure water (i.e. 18 Megohm conductivity) is still just water. If you don't drink so much as to cause water intoxication, it would be no different physiologically from drinking tap water or bottled water.

MegaOhm is unit of resistance and neither conductivity nor conductance nor  resistivity. however forgive if it is a correct in some-other usage that i dont know.

Technically, it is the AC impedance of water measured by meters called "conductivity" meters, since polarization (i.e. bubbles) increase the DC resistance measured. See https://en.wikipedia.org/wiki/Conductivity_(electrolytic)#Measurement

At elementary level, we do learn that salts are important for our body, actually very much for normal day-to-day functioning of human body (as asserted by Curt F.)
So from point of view of elementary level, pure water does causes harm.
But as new research and innovative science suggests, which we do understand from the answer of Curt F.; there might be actually, or in fact (because my textbook also says that Osmotic pressure-Colligative property is one the most important and has fundamental role in earth's biological system.) it is partially wrong to say that salts are important for our body, so we should drink only that water which either contains little or enough salt content in it.
Finally I would be quoting from Curt F. to further extend my point of view:

"I don't think the effect would be that different from drinking regular water. All over the internet there is this idea that super pure water is somehow too pure or dangerous, no good evidence or mechanism is ever presented.
Water is toxic primarily because it dilutes the sodium and potassium ions in your body, leading to Hyponatremia or possibly the far more dangerous Hypokalemia. The normal level of Na+ in blood plasma is about 135 millimolar (mM), and the normal level of K+ is about 5 mM"...
... "According to this rough difference-based model, the ability of the very dilute sodium ion solution to cause problems is proportional to (135 - 0.025) mM, and the ability of the "pure" water to cause problems is (135 - 0) mM. For nearly biological or medical conditions of interest, 135 mM is not really different from 134.975 mM, and so ultra-pure water is not really more dangerous than everyday drinking water."

This means at more advance level, especially research-one; it is easy to conclude that a slight difference wouldn't mind your body. You have saliva in your mouth-the liquid lining on your oesophagus-and digestive juices in your stomach and intestines, which (in my view) has the capacity to reform the content of (practically) any kind of pure water.
Intially I was thinking to write it as a comment in response to Curt F.'s answer but i don't have 50 reputation points. Mind me, for any mistake and therefore correct my even the slightest of slightest mistakes.

Would they immediately die?
No.
I agree, no. But if long time drink pure water, such as 18 M-ohm water, it could cause shortage of u-elements, such as Zn, Se, Fe,..., even all those very,very small quantity, but important for the life, the health...
Normally speaking, drink normal water is the best.
the question is how to get the pure water, 18 M-ohm? Could be expensive.
It can be said using two methods: Physical or Chemical. Steam method is physical. Ion change is chemical method. Both methods changed property of natural/original water. not good. especially, chemical method. More or less, chemical reaction take place, the products changed. pH changed. some thing changed,...    natural/original is the best, need the water, parents, grand parents, grand and grand parents' drinking.

You should expand your answer to specifically address what the OP is asking. As it reads now, your answer is more of a comment.


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What is heat of vaporization? How can it be used at temperature as low as 25 °C?
When I first learned this term. I was introduced to this graph:
The heat in this case is at 100 °C, which I had ever since believed to be the only temperature of vaporization, assuming vaporization = boiling.
Then I saw a table with all the heat of vaporization values for different compounds. Most importantly, all these values apply to reactions at 25  °C.
Different heat of vaporization values apply to different temperature. but water, for ex., cannot boil, with vapor bubbles, unless at 100 °C, thus all the energy used will just be used to simply raise the temperature. wouldn't it?

How can water boil being supplied the heat of vaporization at 25C?

But is it even necessary to make water boil to convert from liquid to gas? I mean since evaporation is constantly happening, isn't this heat of vaporization just making evaporation go faster?

Think of vaporization as the removal of the fastest molecules. The separation of the gaseous form from the more closely-bound liquid therefore removes energy from the liquid. The heat of vaporization can apply at much lower temperatures than the BP of water at atmospheric pressure. Even on a day when the temperature is well blow freezing, laundry will dry on a clothesline because ice evaporates (well, sublimes). Also, water at skin temperature soon feels cold as it loses heat through evaporation.
BTW, the boiling point of water is 100 °C only at 1 atmosphere; see this video of water boiling at room temperature, ~25 °C.

So if I am given a table of heat of vaporization on test, let's say, according what you say, I am then also given the pressure?

No, a liquid can evaporate much below its boiling point.

When water evaporates from a cup is the temperature of the vapor equal to the temperature of liquid water in the cup?

Since the vapor is made of the fastest-moving molecules, which escaped attractive forces in the liquid, which do you think would be warmer?

authors: Анастасия Коновалова


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Formation of radicals from C-H and C-C bonds
Why do radical mechanisms favor the homolytic dissociation of a $\ce{C-H}$ bond over a $\ce{C-C}$ bond?
The $\ce{C-H}$ bond is stronger in magnitude (and it's also slightly polarized), so why would it break more easily as compared to the $\ce{C-C}$ bond?
My instructor explains this by claiming that the products of $\ce{C-H}$ bond dissociation are stabler than$\ce{C-C}$ bond dissociation.
I know that radicals are unstable and highly reactive, so why does it make any difference as to which radical is formed? Wouldn't the radical easiest to form be the one formed the most?

Note that most radical mechanisms do not explicitly feature homolysis of $\ce{C-H}$ bonds or $\ce{C-C}$ bonds. If one were to compare these two processes, say for ethane $\ce{CH3CH3}$, we would find that homlysis of $\ce{C-C}$ is favored over homolysis of $\ce{C-H}$. Obligatory bond dissociation energy references for the organics and here for the halogens.

Homolysis

$\ce{C-C}$
$$\ce{H3C-CH3 -> 2H3C.}\ \ \Delta H^\circ=377\ \text{kJ/mol}$$
$\ce{C-H}$
$$\ce{H3CCH2-H -> H3CCH2. + H.}\ \ \Delta H^\circ=423\ \text{kJ/mol}$$
However, $\ce{C-C}$ and $\ce{C-H}$ homolysis steps are not common steps in radical mechanisms of alkanes. For example, the chlorination of ethane begins with the homolysis of $\ce{Cl2}$, which is much more favorable than homolysis of $\ce{C-C}$ or $\ce{C-H}$.
$$\ce{Cl2 -> 2Cl.} \ \ \Delta H^\circ = 243\ \text{kJ/mol}$$
The chlorine radicals then react with the alkane by abstraction, meaning that bonds are being formed as well as being broken.

Abstraction

$\ce{C-C}$
$$\ce{Cl. + H3C-CH3 -> Cl-CH3 +H3C.}$$
$$\begin{array}{c|c|c|}
 & \text{Broken} & \text{Formed} \\ \hline
\text{bond} & \ce{H3C-CH3}  & \ce{Cl-CH3} \\ \hline
\Delta H^\circ & 377\ \text{kJ/mol} & -350 \ \text{kJ/mol} \\ \hline
\end{array}$$
$$\Delta_r H^\circ = +27 \ \text{kJ/mol}$$
$\ce{C-H}$
$$\ce{Cl. + H3CCH2-H -> Cl-H +H3CCH2.}$$
$$\begin{array}{c|c|c|}
 & \text{Broken} & \text{Formed} \\ \hline
\text{bond} & \ce{H3CCH2-H}  & \ce{Cl-H} \\ \hline
\Delta H^\circ & 423\ \text{kJ/mol} & -432 \ \text{kJ/mol} \\ \hline
\end{array}$$
$$\Delta_r H^\circ = -9 \ \text{kJ/mol}$$
Thus, abstraction of $\ce{H}$ by $\ce{Cl.}$ is more favored because it is coupled to the formation of the $\ce{H-Cl}$ bond.

Okay, that explains the formation in the propagation and termination steps of the reaction. But why does it happen in the initiation of the radical reaction?

It doesn't happen in initiation. Organic radical reactions almost exclusively are initiated by the homolysis (thermally or photochemically) of weaker $\ce{X-X}$ or $\ce{O-O}$ bonds, and not by the direct homolysis of $\ce{C-H}$ or $\ce{C-C}$ bonds. See Wikipedia page on radical initiators.


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Length of C-H bonds in hybridized bonding orbitals
What is the effect of hybridization and double-bond formation on the rest of the atoms/molecules attached to the central atom?
For example, in a benzene molecule, the Carbon-Carbon bonds are intermediate between single and double bonds. But what is the effect of the presence of the double bonds between the carbons on the bonds between the carbon and hydrogen? Do the bond lengths decrease because of the sp2 hybridized bonding orbitals?

Yes, the bond length decreases.  One way to think about it is that a sp3 center with 109.5 degree angles is more crowded than an sp2 center with 120 degree angles.  More crowding = more repulsion = longer bond length.
Another way to think about it is that the e-pair in an s orbital is sitting on the nucleus.  Adding a p-orbital moves the electron pair further away from the nucleus.  More p-orbital character = longer bond lengths.
You can test it for yourself using http://molcalc.org/


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Can water be liquefied or solidified just by adjusting the temperature, regardless of the pressure?

technically the line between liquid and solid phase would go a long way before hitting the y-axis, but the point is it will eventually.
So the question remains, can any liquids, including water (with its unique phase diagram) be liquefied regardless of the pressure?

This is not an accurate Phase diagram of water. Here you can see a better phase diagram. What do you mean by "can any liquids, including water (with its unique phase diagram) be liquefied regardless of the pressure"

@Gowtham my point is since the solid-liquid equilibrium line is sloped inward towards the y axis, theoretically it will hit eventually.

Did you check out the link, Do you ask "can water exist as equilibrium between solid and liquid at some temperature regardless of the pressure "? sorry i really dont get what you are trying to ask.. On the line in the phase diagram, both the phases exists at same time given some T and P in a dynamic equlibrium. If you move away from the line, it will transform to a single phase(if we go to one phase region) or  two-phase(move to another part of a line)

Here is my questions: no matter what the temperature is, just by adjusting temperature, so if you pick a fix y coordinate and change x coordinate, can you ALWAYS liquefy or solidified water? See the picture, above a certain pressure so will be no solid state?

you can add your last comment to your question, so that it is more clear.

see this http://chemistry.stackexchange.com/questions/29729/why-cant-helium-be-solidified-at-ordinary-pressures - in helium solid-liquid equilibrium line does hit y axis

If I actually understood your question correctly, yes there are regions above which there will be only solid state (rather than no solid state). This is limited to the molecular region.

The Roman numerals indicate various ice phases.
From the above phase diagram (which can be found here) the solid-liquid equilibrium line doesn’t touch the y-axis as you suggest. But there will be only solid phase above the $10\ \mathrm{GPa}$ region(approximately).
In the phase diagram you drew, if (a big if) such a PD exists for some element/molecule, then there will be no solid above the point where solid-liquid line meets y-axis.
As pointed out in the comments, when we reach high temperatures and pressures we won’t be seeing the molecular water. It will dissociate into atoms/ions as shown in the figure.

This is correct, though actually even above $1:\mathrm{TPa}$ water doesn't necessarily stay solid no matter what you do. Raise the temperature to solar-interiour values, and the water will stop being water at all...

I assume at some temperature (i.e. in the sun's core there are no molecules) water will decide it wants to freely disassociate into atoms, but I have no idea what that would be (order of magnitude). So presumably there might be a pressure at which "H2O" (what's the disassociation constant in high-temperature ice...) is never not a solid.

http://www.sandia.gov/pulsedpower/prog_cap/pub_papers/Redmer-Icarus2011.pdf has a theoretical phase diagram for water at extremely high pressure and temperature (speculating that these conditions might exist in the interior of Uranus and Neptune). The basic observation appears to be that, at very high pressures, water will go from ice to an exotic "superionic" semi-dissociated semi-fluid phase as you raise the temperature, and from there directly to plasma.

Good discussion.  One minor point in the diagram you cite:  Ice VII is still molecular.  Ice X is when you get to a nonmolecular structure in the fully solid state.

Theoretically, YES!
But why?
In the real world there are a thousand different ways to accomplish a certain task, but only a few will be the most efficient. So in considering your question.

So the question remains, can any liquids, including water (with its unique phase diagram) be liquefied regardless of the pressure?

First of all, I do not like how you put regardless of the pressure, there has to be a certain amount of pressure. Yes you can do this, but there really is an easier way to this. Use a freezer to solidify it, or leave it out in the open to liquify it. :)

I think the OP is confused about phase diagrams , not about practicality of  anything.

He doesn't look confused about it. He just wants to know if this is possible.

I didn't know what is outside the picture students are usually given, the one that shows solid phase above a certain pressure. The one in Gowtham's answer.


text:
Theoretical model for the hydrogen bond
I am curious how, from the perspective of quantum chemistry, the hydrogen bond is modeled. In particular I am interested in using such a model to estimate the amount of energy released per $\ce{H2O}$ molecule, if say two of them approach each other e.g. in case of condensation. Any hint for precise literature would be great!

I am confused by your question.  What do you mean by the amount of energy released per H2O molecule?  There are a lot of models you can use.  One of the simplest would be to consider a water dimer or perhaps an HF dimer.

It's modeled as any bond in quantum chemistry. Plenty of references for that. There are however other models that are slightly more parametrized. Are you interested in those?

Hi! Well I am interested in all sorts of models. If you have references this would be great!

Please go through papers by Prof. Ralf Ludwig. You will certainly gain a lucid understanding of the subject both at experimental end as well as computational end. He advocates use of low frequency vibrational modes such as Far IR along with quantum chemistry for studying hydrogen bonding and allied weak interactions in very simplistic yet the most accurate way.
References:

"Hydrogen Bonding in protic ionic liquids: Reminiscent of Water." K. Fumino, A. Wulf, R. Ludwig., Angew. Chem., 2009, 121, 3230-3233. Angew. Chem. Int. Ed., 2009, 48, 3184-3186.
"Advancing into Water’s “No Man’s Land”: Two Liquid States?" D. Paschek, R. Ludwig, Angew. Chem., 2014, 126 (44), 11888-1890. Angew. Chem. Int. Ed., 2014, 53 (44), 11699–11701.

Thank you for adding references. If you have not already done it, you can take the [tour] to get to know the site. For more specific functions, visit the [help]. You can always [edit] your post to include more contents. I have combined your answers and comments into one. And welcome to Chemistry.se!


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Are molecular optimizations in quantum chemistry really in vacuum?
I know that Gaussian (or other quantum chemical packages) assumes the molecule is alone in vacuum, that is, the molecule doesn't interact with anything, when optimizing the molecule.
But I've found that optimization at HF or DFT levels results are almost the same as experimental data, which were from the solid state by X-ray diffraction. Everyone knows that the solid state is quite different from the vacuum, for molecules interact with each other in the solid state.
Then why do the two generate nearly the same results? Or have I looked at so few samples that I neglect occasions when the two give quite different results?

You raise an excellent question and the short answer is "yes, quantum chemistry calculations are intrinsically in vacuum".
Approximate methods like semiempirical (AM1, PM6, PM7) and hybrid DFT methods (e.g., B3LYP) use experimental data, which often comes from crystal structures.
For many years, single-crystal x-ray diffraction studies were the gold standard of "what is the geometry of a molecule." Performing gas-phase measurements of bond angles and bond lengths are extremely difficult, since they typically require a molecular beam of the intended target and then very high resolution IR,  rotational spectroscopy, or gas electron diffraction.
I once heard a talk by an extremely talented theorist, who asked the question "is an experimental crystal structure accurate enough?" He showed that in many cases, the answer is "no." (Consider that establishing the position of hydrogen atoms is difficult if not impossible with many single crystals.) The method he was proposing (geometry optimization with CCSD(T) and a basis set limit) isn't really appropriate for large molecules.
Until there's some huge database of gas-phase molecular geometry measurements and CCSD(T) calculations, most methods are compared with crystal structures.

I'll point out that for bond lengths and angles of non-hydrogen atoms, the geometry in the solid state often is not that different from the gas phase. Packing effects tend to alter dihedral angles (e.g., flattening the inter-ring biphenyl torsion).

I'm not well enough versed in the method to evaluate it critically, but Neese's DLPNO-CCSD(T) claims excellent accuracy relative to 'canonical' CCSD(T), but with linear scaling with system size. It's implemented for closed-shell species in the current version of ORCA, and might be a first step toward such a practical theoretical database.

In fact, performing a gas-phase measurement is not so difficult, analysis is the tough part: the procedure which spits out geometrical parameters from the experimental data is usually quite cumbersome. With respect to the methods used for gas-phase structural determination @GeoffHutchison definitely forgot to mention GED (Gas Electron Diffraction) technique which is quite frequently used jointly with MW spectroscopy to give the most reliable results.

Oh, yes, yes, yes. Establishing the position of hydrogen atoms in X-ray crystallography always looks like a straight cheating to me. :D Good-ol' "Everybody believes in experimental results, except the one who did it." On the other hand, "Nobody believes in theoretical calculations, except the one who did it." :D

@Wildcat Sorry, I wasn't familiar with GED. Thanks. And yes, the analysis of such measurements is very time-consuming. One of my former colleagues did rotational spectroscopy.

@GeoffHutchison, I did GED in the past. The most interesting part of analysis is that you anyway need quantum chemical calculations (basically, you need a force field to perform the GED results analysis). So usually what is done is a combined analysis of electron diffraction data, rotational constants and ab initio calculations. Anyway, this joint analysis provides quite reliable gas phase geometric parameters. I usually benchmark DFT functionals against such data if available.

Are you sure that DFT parameters are fitted on X-ray geometry?

@Greg this is not as true anymore, but for a very long time functionality would compare against bond lengths and angles from X-Ray.

@GeoffHutchison OK, now I understand, thank you. There are many other inherent biases like this in those parametrizations - frankly, there is a practical advantage of using methods which are accurate for the wrong reasons.

It is possible to do crystal structure optimizations in Gaussian by adding translation vectors (DFT isn't as good with intermolecular reactions, but it will probably be a better approximation than optimizing without PBC, which is akin to a gas phase molecule).
Whether or not solid behavior and gas behavior are similar will depend on the material, and the properties you're measuring, so could you tell us more specifically what you're studying?

Hi Felipe, I'm not talking about a specific molecule. Yesterday I suddenly realised that I used Gaussian to optimise molecules in gas phase but got the same results as by X-ray diffraction. And I remember once in class a teacher talked about a molecule optimised at HF level, he said,"look, the bond length and bond angle are exactly the same as experimental data in this paper, so this level and basis set are suitable for this type of molecule." Since then I've been using solid state data to choose the best method to optimise molecules in gas phase. Now I question this way of studying chemistry.


text:
Water softener and RO filter
I have a water softener for showers and dishwasher and a reverse osmosis filter for drinking water.  For convenience of the install, the RO filter is using softened water (some calcium carbonate has been exchanged with sodium chloride -- this is my basic understanding; correct me if I'm wrong).  I don't fully understand how an RO filter works except that it is essentially just a molecule-sized physical filter.
My question is, Does the RO filter care whether the supply water is salty or hard?  Is it more effective at filtering one or the other?  Is one going to influence the life of the RO filter more than the other.
My apologies if this is not a "true" chemistry question, but I'm looking for a bit more of an thorough explanation than I got at diy.stackexchange.

You are not wrong in your understanding of a water softener, nor are you wrong in your understanding of a reverse osmosis filtration system.

Does the RO filter care whether the supply water is salty or hard?

I would like to make the point that your RO filter is incapable of caring about the hardness of the water. This is more about carefully choosing your words more than anything else. I was also tripped up by this when I first began chemistry, but it was something I learned to do and I challenge you to do the same.

Is it more effective at filtering one or the other?

Yes, but only slightly. Here you can find a list of the typical rejection characteristics of RO filters. The lower bound for the percent amount of multivalent ions ($\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$) filtered is higher than the lower bound for the percent amount of monovalent ions ($\ce{Na+}$ and $\ce{K+}$) filtered, but the upper bounds are very close in value.
Overall you could say that RO systems are more efficient at filtering hard water, but the difference is not very large. Most systems will likely be performing considerably above the lower bounds of the typical rejection characteristics anyway, so this should not be a deciding factor in deciding to feed your RO hard or soft water.

Is one going to influence the life of the RO filter more than the other?

Yes! In an ideal RO filtration system, all of the filtered materials would simply drain away, and pure water would flow to the storage tank, but nothing is ideal. If you feed hard water into your RO system, there will be scaling on the semi-permeable membrane that will eventually reduce its filtered water output. This can cause lack of flow will create a potentially damaging buildup of pressure elsewhere in the system. I would therefore recommend that you keep your setup the way it is!


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Is it possible to predict the lines in the atomic emission spectrum of Na?
The atomic emission spectrum for sodium ($\ce{Na}$) is completely dominated by a line in the range of yellow, about $590~\mathrm{nm}$ (to be more precise, it's a doublet). Here is how it looks like:

This line is due to the transition $3\mathrm{p} \to 3\mathrm{s}$.
I know the Rydberg equation which can be used to predict the transitions in hydrogen. I would like to know if there is a way to predict the transitions for other elements, such as the alkali metal $\ce{Na}$.

When the Schrödinger equation is able to be solved for multi-electron systems, then it should be possible to predict the transition energies.

The answer is no: the spectra are way too complex! Even the neon spectrum is too complex. So we measure the spectra empirically.

The origin of the Rydberg equation is the Bohr model's equation for the energy of a hydrogenic state:
$E_\text{B} = -\dfrac{R_{H}hc}{n^2}$
This is accurate for other hydrogenic atoms, with only one electron too (bearing in mind that $R_{X}$ is proportional to the atomic number squared).
When you get to non-hydrogenic atoms you need to model inter-electron repulsion as well. This can be done by introducing an empirical parameter $\delta$, the quantum defect: $E_\text{B} = -\dfrac{R_{X}hc}{(n-\delta_X)^2}$. For sodium the quantum defect is 1.37, allowing good approximation of the transition energies between n-states in the electronic spectrum.
I don't think there is a similar back-of-the-envelope method to calculate the differences in energies between multi-electron atom l-energy levels (like the 3s-3p). The theory of angular momentum tends to mathematical complexity. There are likely good computationally derived approximations though.


text:
If there are two longest chains possible in an organic compound and both have the same number of substituents, how do we decide the parent chain?
Is the parent chain in the above case determined by atomic mass, alphabetical preference or any other rule?

If two longest chain are possible with same number of side chain in both of them, then we consider sum of locants of the side chain. The longest chain in which the sum of locants of the side chain is least is taken into consideration.

If the sum of the locants is also same, then?

In that case alphabetical order is follwed..

Alphabetical order in what sense?

This is incorrect; we determine the point of first difference of locants between the two possible chains, and prefer the smallest number: https://www.acdlabs.com/iupac/nomenclature/79/r79_933.htm#ID_73zo0o

This answer is wrong. See also IUPAC nomenclature: "Smallest sum of locants"?


text:
Radiation in space and its effects on chemical reactions?
I know that one current area of research is ways to protect astronauts from ionizing radiation when they venture out of the atmosphere of Earth, but would that same ionizing radiation be a cause of concern when performing chemical reactions in space or in the atmosphere of planets with little atmosphere?
I'm aware that radiation like ultraviolet, which is present in sunlight here on Earth, is used in organic chemistry for the homolytic fission of halides, but I'm not sure how much higher energy radiation might affect other reactions.

Nice question and there is also Microwave chemistry and if i am correct space has background microwave radiation

Isn't the clue in the name of the phenomenon,  ionising radiation? The electromagnetic energy is strong enough to remove electrons. This would normally cause chemical changes to occur, much like in a mass spectrometer. If visible light can cleave bromine and UV can cause changes in vitamin D derivatives and stilbene to cyclise, I would anticipate ionising radiation to affect molecules much like it would chemicals in the cells in your body.

It depends....
First let's define the "problem space".
Chemical reactions are typically driven by a few eV (electron volts). "Ionizing radiation" can be keV or MeV which is vastly excessive. In fact that amount of energy is so massive that it would "rip up" the chemical bonding in any material. Only solid metals would be somewhat stable since the 3D lattice locks the atoms into place. $\ce{NaCl}$ for example could decompose to Na metal and $\ce{Cl_2}$ gas.
Any sort of chemical reaction in space has to have a container of some sort which would mitigate the penetration of the ionizing radiation somewhat. But it surely won't stop all of the ionizing radiation. There are nuclear experiments that are conducted a mile underground to try to remove cosmic particles.
The primary source of ionizing radiation in our solar system is the sun. So more distance from the sun is your friend.
"Performing chemical reactions in space" sort of implies that a human is doing the reactions. A biological entity is going to be more effected than any vat of chemicals. So in an environment which humans or plants could safely operate, then performing chemical reactions won't be a problem.


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Intermolecular and Intramolecular Forces (Vapor pressure of n-butylamine and methanol)

What are the different properties that intermolecular forces and Intramolecular forces attribute to? i.e. intermolecular forces contribute to BPs, vapor pressures, etc.

Specific problem: why does n-butylamine have lower vapor pressure at room temperature than methanol? Explain in regards to IMFs.

I added a specific Question regarding vapor pressure.

What are intermolecular molecular forces?
The forces(electrostatic force) between molecules of a compound. These forces are responsible for intermolecular distances and how tightly the molecules are held together. If the intermolecular forces(van der Waals force) are large, it means the molecules are tightly held together in a group and they are fixed in their position.
Now what is boiling point? It's a temperature at which liquid state of a compound changes into its gas state. What's the difference between gaseous and liquid phase? Gaseous phase has very large intermolecular distance when compared to that in liquid phase. If a large force exists between the molecules and they are quite fixed in their position, then large amount energy or heat will be required to move them away from each other. Large heat means greater temperature and hence greater boiling point. So the lesson is that, the stronger the intermolecular forces, the higher the boiling point.
The similar concept in applicable in melting point. the stronger the intermolecular forces, the higher will be the melting point.
You will recall that the vapor pressure is a direct measure of escaping tendency, so we can use these terms more or less interchangeably(In case you may be wondering, how are they related. Click here). If the intermolecular forces are large, then molecule in liquid phase are not able to leave to liquid easily, hence lowering the escaping tendency and therefore lower vapour pressure.
Note: This is a very basic explanation.

@ShiddharthYadav I added a specific vapor pressure problem in my question. Why is n-Butylamine lower? Doesn't it have a lot of LDFs?

@Loong what I think is that was a typo from the answerer. Everywhere else he mentions intermolecular forces. So that seems more like a mistake than deliberate action.

Butylamine does not exhibit H-bonding between each other, where as alcohols do. This added force from the hydrogen bond makes the boiling point of alcohols higher than those of most other ORGANIC molecules. Ionic bonds have the most strength.

I disagree. Nitrogen does participate in hydrogen bonding. But I do agree with your second point, the added force from hydrogen bonding would decrease vapour pressure, thereby increasing boiling point.


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Why is sand/glass polar? How does lattice structure change polarity?
The molecular geometry of sand is linear. Dipole moment is thus cancelled.
But we all know from the meniscus water makes with glass that the dipole-dipole/H-O attraction  between sand and water is stronger than the hydrogen bond between water molecule. maybe I am wrong
My guess is either that electron lone pair makes it polar or the lattice structure does.

I'm not sure where you get the idea that glass is linear. What makes a glass is the lack of long range ordering, though in fused silica, the approximate tetrahedral structure of $\ce{SiO4}$ is maintained (this is a 2D section, the fourth oxygens are not shown):
 Source
In normal glasses, there is no significant net dipole moment across the entire structure, but the reason we observe strong adhesive forces between water and glass is a surface phenomenon—in the bulk of the glass, every silicon atom is connected to four oxygen atoms and each oxygen atom to two silicon atoms, but what happens at a boundary/surface? There are oxygen atoms that only have one bond. When these are exposed to an aqueous solution, some will bear a negative charge, others will pick up a proton from water. The theory behind it is pretty complicated, but you can think of it as a simple acid-base reaction—at low pH more will be protonated:
$$\ce{Si-OH <=> Si-O- + H+}$$
In any case, protonated or not, silanol groups are very happy to associate with water, accounting for the concave meniscus.

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Determining structure factors for crystals from lattice and motif
I understand that the structure factor for a crystal is given by:
$$F_{hkl} = \sum_r f_{r} e^{2{\pi}i(hx_{r}+ky_{r}+lz_{r})} $$
My textbook (''Structures of crystals'', Glazier) suggests that you can calculate structure factors if you know the lattice type and motif of the crystal, simply by finding the product of the lattice and the motif structure factors.
Can this method be used generally, and if so, do you need to define the motif relative to a centre of symmetry in order to get real structure factors for centrosymmetric crystals ?

Q1) A crystal is a convolution of a lattice and motif, so by the convolution theorem, the fourier transform of the crystal (i.e. your structure factor) is the product of the fourier transform of the lattice and the fourier transform of the motif. That is, this is general.
Q2) If the crystal is centrosymmetric* then the structure factor is real - this is a property of the fourier transform (the imaginary sine terms have opposite signs and equal other and are therefore zero). If your lattice and motif generate a centrosymmetric crystal then your fourier transform will be real. The motif doesn't need to be centrosymmetric to generate a centrosymmetric crystal.
*and the atomic form factors are real, but this isn't relevant.